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8,088,367,722,740,815,000 | \documentclass[12pt]{article} \newcommand{\ds}{\displaystyle} \newcommand{\pl}{\partial} \newcommand{\dy}{\frac{dy}{dx}} \parindent=0pt \begin{document} {\bf Question} Obtain the general solution of the following differential equations: \begin{enumerate} \item $\ds x\dy +y = 2e^{-x}\qquad (*)$ \item $\ds (y-x^3)+(x+y^3)\dy=0$ \item $\ds x\dy = 3y + x^4$ \item $\ds \dy={{x^2-xy+y^2} \over {xy}}$ \item $\ds \dy+y\cot x=2x\hbox{ cosec} x\qquad (*)$ \item $\ds (3x^2y^2+\sin x)\dy +(2xy^3+y\cos x)=0$ \end{enumerate} \vspace{0.25in} {\bf Answer} \begin{itemize} \item[a)] $\ds\frac{dy}{dx}+\frac{1}{x}y=2\frac{1}{x}e^{-x} \Rightarrow I(x)=\exp(\int\frac{1}{x}dx)=\exp(\ln x)=x$ $\ds\frac{d}{dx}(xy)=2e^{-x} \Rightarrow xy=-2e^{-x}+c \Rightarrow y=\frac{c-2e^{-x}}{x}$ \item[b)] $\ds p=y-x^3, \,\,\,\, q=x+y^3, \,\,\,\, \frac{\pl p}{\pl y}=1, \,\, \frac{\pl q}{\pl x} \Rightarrow$ exact. $\ds \frac{\pl F}{\pl x}=y-x^3 \Rightarrow F=xy-\frac{1}{4}x^4+f(y) \Rightarrow \frac{\pl F}{\pl y}=x+\frac{df}{dy}$ $\ds \frac{\pl F}{\pl y}=x+y^3 \,\,\,$ hence need $\ds\frac{\pl f}{\pl y}=y^3 \,\,\,$ so $\ds f=\frac{1}{4}y^4+c$. There is a solution if $\ds F(x,y)=xy-\frac{1}{4}x^4+\frac{1}{4}y^4=A$ \item[c)] $\ds\frac{dy}{dx}-\frac{3}{x}y=x^3 \Rightarrow I(x)=\exp(\int-\frac{3}{x}dx)=\exp(-3\ln x)=x^{-3}$ $\ds\frac{d}{dx}\left(\frac{1}{x^3}y\right)=1 \Rightarrow \frac{1}{x^3}y=x+c \Rightarrow y=x^4+cx^3$ \item[d)] $\ds\frac{dy}{dx}=\frac{1-\frac{y}{x}+\left(\frac{y}{x}\right)^2}{\frac{y}{x}}$, put $\ds y=xv \Rightarrow x\frac{dv}{dx}+v=\frac{1-v+v^2}{v}$ $\ds x\frac{dv}{dx}=\frac{1-v+v^2}{v}-v=\frac{1-v}{v} \Rightarrow \int\frac{v}{1-v}dv=\int\frac{1}{x}dx$ $\ds\int-1+\frac{1}{1-v}dv=\ln|x|+c \Rightarrow -v-\ln|1-v|=\ln|x|+c$ $\ds e^{-v}\frac{1}{1-v}=Ax \Rightarrow e^{-\frac{y}{x}}\frac{1}{x-y}=A$ \item[e)] $\ds\frac{dy}{dx}+\frac{\cos x}{\sin x}y=2x\frac{1}{\sin x} \Rightarrow I(x)=exp\left(\int\frac{\cos x}{\sin x}dx\right)$ $\ds I(x)=\exp(\ln(\sin x))=\sin x \Rightarrow \frac{d}{dx}((\sin x)y)=2x$ $\ds\Rightarrow (\sin x)y=x^2+c \Rightarrow y=\frac{x^2+c}{\sin x}$ \item[f)] $\ds p=2xy^3+y\cos x, \,\,\,\, q=3x^2y^2+\sin x$ $\ds\frac{\pl P}{\pl y}=6xy^2+\cos x, \,\,\,\, \frac{\pl q}{\pl x}=6xy^2+\cos x \Rightarrow$ exact. $\ds\frac{\pl F}{\pl x}=2xy^3+y\cos x \Rightarrow F=x^2y^3+y\sin x+f(y)$ $\ds \Rightarrow \frac{\pl F}{\pl y}=3x^2y^2+\sin x+\frac{df}{dy}$ $\ds\frac{\pl F}{\pl y}=3x^2y^2+\sin x$ hence need $\ds\frac{df}{dy}=0 \Rightarrow f=c$ solution is $\ds F(x,y)=x^3y^3+y\sin x=A$ \end{itemize} \end{document} | {
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9,056,932,489,119,640,000 | Written by Joachim Lambek
Written by Joachim Lambek
foundations of mathematics
Article Free Pass
Written by Joachim Lambek
Gödel
Implicit in Hilbert’s program had been the hope that the syntactic notion of provability would capture the semantic notion of truth. Gödel came up with the surprising discovery that this was not the case for type theory and related languages adequate for arithmetic, as long as the following assumptions are insisted upon:
1. The set of theorems (provable statements) is effectively enumerable, by virtue of the notion of proof being decidable.
2. The set of true statements of mathematics is ω-complete in the following sense: given any formula ϕ(x), containing a free variable x of type N, the universal statement ∀xNϕ(x) will be true if ϕ(n) is true for each numeral n—that is, for n = 0, n = S0, n = SS0, and so on.
3. The language is consistent.
Actually, Gödel also made a somewhat stronger assumption, which, as the American mathematician J. Barkley Rosser later showed, could be replaced by assuming consistency. Gödel’s ingenious argument was based on the observation that syntactical statements about the language of mathematics can be translated into statements of arithmetic, hence into the language of mathematics. It was partly inspired by an argument that supposedly goes back to the ancient Greeks and which went something like this: Epimenides says that all Cretans are liars; Epimenides is a Cretan; hence Epimenides is a liar. Under the assumptions 1 and 2, Gödel constructed a mathematical statement g that is true but not provable. If it is assumed that all theorems are true, it follows that neither g nor ¬g is a theorem.
No mathematician doubts assumption 1; by looking at a purported proof of a theorem, suitably formalized, it is possible for a mathematician, or even a computer, to tell whether it is a proof. By listing all proofs in, say, alphabetic order, an effective enumeration of all theorems is obtained. Classical mathematicians also accept assumption 2 and therefore reluctantly agree with Gödel that, contrary to Hilbert’s expectation, there are true mathematical statements which are not provable.
However, moderate intuitionists could draw a different conclusion, because they are not committed to assumption 2. To them, the truth of the universal statement ∀xNϕ(x) can be known only if the truth of ϕ(n) is known, for each natural number n, in a uniform way. This would not be the case, for example, if the proof of ϕ(n) increases in difficulty, hence in length, with n. Moderate intuitionists might therefore identify truth with provability and not be bothered by the fact that neither g nor ¬g is true, as they would not believe in the principle of the excluded third in the first place.
Intuitionists have always believed that, for a statement to be true, its truth must be knowable. Moreover, moderate intuitionists might concede to formalists that to say that a statement is known to be true is to say that it has been proved. Still, some intuitionists do not accept the above argument. Claiming that mathematics is language-independent, intuitionists would state that in Gödel’s metamathematical proof of his incompleteness theorem, citing ω-completeness to establish the truth of a universal statement yields a uniform proof of the latter after all.
Gödel considered himself to be a Platonist, inasmuch as he believed in a notion of absolute truth. He took it for granted, as do many mathematicians, that the set of true statements is ω-complete. Other logicians are more skeptical and want to replace the notion of truth by that of truth in a model. In fact, Gödel himself, in his completeness theorem, had shown that for a mathematical statement to be provable it is necessary and sufficient that it be true in every model. His incompleteness theorem now showed that truth in every ω-complete model is not sufficient for provability. This point will be returned to later, as the notion of model for type theory is most easily formulated with the help of category theory, although this is not the way Gödel himself proceeded. See below Gödel and category theory.
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-5,716,262,472,832,958,000 | Updated date:
Math: How to Find the Roots of a Quadratic Function
Author:
I studied applied mathematics, in which I did both a bachelor's and a master's degree.
Quadratic Function
Quadratic Function
Quadratic Functions
A quadratic function is a polynomial of degree two. That means it is of the form ax^2 + bx +c. Here, a, b and c can be any number. When you draw a quadratic function, you get a parabola as you can see in the picture above. When a is negative, this parabola will be upside down.
What Are Roots?
The roots of a function are the points on which the value of the function is equal to zero. These correspond to the points where the graph crosses the x-axis. So when you want to find the roots of a function you have to set the function equal to zero. For a simple linear function, this is very easy. For example:
f(x) = x +3
Then the root is x = -3, since -3 + 3 = 0. Linear functions only have one root. Quadratic functions may have zero, one or two roots. An easy example is the following:
f(x) = x^2 - 1
When setting x^2-1 = 0, we see that x^2 = 1. This is the case for both x = 1 and x = -1.
An example of a quadratic function with only one root is the function x^2. This is only equal to zero when x is equal to zero. It might also happen that here are no roots. This is, for example, the case for the function x^2+3. Then, to find the root we have to have an x for which x^2 = -3. This is not possible, unless you use complex numbers. In most practical situations, the use of complex numbers does make sense, so we say there is no solution.
Strictly speaking, any quadratic function has two roots, but you might need to use complex numbers to find them all. In this article we will not focus on complex numbers, since for most practical purposes they are not useful. There are however some field where they come in very handy. If you want to know more about complex numbers you should read my article about them.
Ways to Find the Roots of a Quadratic Function
Factorization
The most common way people learn how to determine the the roots of a quadratic function is by factorizing. For a lot of quadratic functions this is the easiest way, but it also might be very difficult to see what to do. We have a quadratic function ax^2 + bx + c, but since we are going to set it equal to zero, we can divide all terms by a if a is not equal to zero. Then we have an equation of the form:
x^2 + px + q = 0.
Now we try to find factors s and t such that:
(x-s)(x-t) = x^2 + px + q
If we succeed we know that x^2 + px + q = 0 is true if and only if (x-s)(x-t) = 0 is true. (x-s)(x-t) = 0 means that either (x-s) = 0 or (x-t)=0. This means that x = s and x = t are both solutions, and hence they are the roots.
If (x-s)(x-t) = x^2 + px + q, then it holds that s*t = q and - s - t = p.
Numerical Example
x^2 + 8x + 15
Then we have to find s and t such that s*t = 15 and - s - t = 8. So if we choose s = -3 and t = -5 we get:
x^2 + 8x + 15 = (x+3)(x+5) = 0.
Hence, x = -3 or x = -5. Let's check these values: (-3)^2 +8*-3 +15 = 9 - 24 + 15 = 0 and (-5)^2 + 8*-5 +15 = 25 - 40 + 15 = 0. So indeed these are the roots.
It might however be very difficult to find such a factorization. For example:
x^2 -6x + 7
Then the roots are 3 - sqrt 2 and 3 + sqrt 2. These are not so easy to find.
The ABC Formula
Another way to find the roots of a quadratic function. This is an easy method that anyone can use. It is just a formula you can fill in that gives you roots. The formula is as follows for a quadratic function ax^2 + bx + c:
(-b + sqrt(b^2 -4ac))/2a and (-b - sqrt(b^2 -4ac))/2a
This formulas give both roots. When only one root exists both formulas will give the same answer. If no roots exist, then b^2 -4ac will be smaller than zero. Therefore the square root does not exist and there is no answer to the formula. The number b^2 -4ac is called the discriminant.
Numeric example
Let's try the formula on the same function we used for the example on factorizing:
x^2 + 8x + 15
Then a = 1, b = 8 and c = 15. Therefore:
(-b + sqrt(b^2 -4ac))/2a = (-8+sqrt(64-4*1*15))/2*1 = (-8+sqrt(4))/2 = -6/2 = -3
(-b - sqrt(b^2 -4ac))/2a = (-8-sqrt(64-4*1*15))/2*1 = (-8-sqrt(4))/2 = -10/2 = -5
So indeed, the formula gives the same roots.
Quadratic Function
Quadratic Function
Completing the Square
The ABC Formula is made by using the completing the square method. The idea of completing the square is as follows. We have ax^2 + bx + c. We assume a = 1. If this would not be the case, we could divide by a and we get new values for b and c. The other side of the equation is zero, so if we divide that by a, it stays zero. Then we do the following:
x^2 + bx + c = (x+b/2)^2 -(b^2/4) + c = 0.
Then (x+b/2)^2 = (b^2/4) - c.
Therefore x+b/2 = sqrt((b^2/4) - c) or x+b/2 = - sqrt((b^2/4) - c).
This implies x = b/2+sqrt((b^2/4) - c) or x = b/2 - sqrt((b^2/4) - c).
This is equal to the ABC-Formula for a = 1. However, this is easier to calculate.
Numerical Example
We take again x^2 + 8x + 15. Then:
x^2 + 8x + 15 = (x+4)^2 -16+15 = (x+4)^2 -1 = 0.
Then x = -4 + sqrt 1 = -3 or x = -4 - sqrt 1 = -5.
So indeed, this gives the same solution as the other methods.
Summary
We have seen three different methods to find the roots of a quadratic function of the form ax^2 + bx + c. The first was factorizing where we try to write the function as (x-s)(x-t). Then we know the solutions are s and t. The second method we saw was the ABC Formula. Here you just have to fill in a, b and c to get the solutions. Lastly, we had the completing the squares method where we try to write the function as (x-p)^2 + q.
Quadratic Inequalities
Finding the roots of a quadratic function can come up in a lot of situations. One example is solving quadratic inequalities. Here you must find the roots of a quadratic function to determine the boundaries of the solution space. If you want to find out exactly how to solve quadratic inequalities I suggest reading my article on that topic.
Higher Degree Functions
Determining the roots of a function of a degree higher than two is a more difficult task. For third-degree functions—functions of the form ax^3+bx^2+cx+d—there is a formula, just like the ABC Formula. This formula is pretty long and not so easy to use. For functions of degree four and higher, there is a proof that such a formula doesn't exist.
This means that finding the roots of a function of degree three is doable, but not easy by hand. For functions of degree four and higher, it becomes very difficult and therefore it can better be done by a computer.
Comments
Santosh Sahu from Bangalore on April 25, 2020:
A nice article.
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4,349,972,220,464,549,400 | LARA
Compositional VCG
Let $P, Q$ be formulas and $c$ a command of our language using variables $V = \{x_1,\ldots,x_n\}$. To check $\{P\}c_1\{Q\}$ we compute formula $F$ containing $x_1,\ldots,x_n,x'_1,\ldots,x'_n$ specifying the semantics of c_1, that is, formula $F$ such that
\begin{equation*}
r_c(c_1) = \{ (\{("x_1",x_1),\ldots,("x_n",x_n)\},\{("x_1",x'_1),\ldots,("x_n",x'_n)\}) \mid F_1 \}
\end{equation*}
Denote the truth value of formula in two states by $P_F(F_1)(s,s')$ (predicate for the formula) using an auxiliary definition of $f_T(s,s')$ (function for the term):
\begin{equation*}
\begin{array}{l}
P_F(F_1 \land F_2)(s,s') = P_F(F_1)(s,s') \land P_F(F_2)(s,s') \\
P_F(F_1 \lor F_2)(s,s') = P_F(F_1)(s,s') \lor P_F(F_2)(s,s') \\
P_F(\lnot F_1)(s,s') = \lnot P_F(F_1)(s,s') \lor P_F(F_2)(s,s') \\
P_F(t_1 = t_2)(s,s') = (f_T(t_1)(s,s') = f_T(t_2)(s,s')) \\
P_F(t_1 < t_2)(s,s') = (f_T(t_1)(s,s') < f_T(t_2)(s,s')) \\
f_T(t_1 + t_2)(s,s') = f_T(t_1)(s,s') + f_T(t_2)(s,s') \\
f_T(t_1 - t_2)(s,s') = f_T(t_1)(s,s') - f_T(t_2)(s,s') \\
f_T(K * t_1)(s,s') = K \cdot f_T(t_1)(s,s') \\
f_T(x)(s,s') = s(x) \\
f_T(x')(s,s') = s'(x)
\end{array}
\end{equation*}
Then we can construct formula $F_1$ such that $r_c(c_1) = \{ (s,s') \mid P_F(F_1)(s,s') \}$. Denote such formula $F_1$ by $F_c(c_1)$. To can then represent the validity of a Hoare triple
\begin{equation*}
\forall s,s'.\ (f_T(P)(s) \land (s,s') \in r_c(r) \rightarrow f_T(Q))
\end{equation*}
by formula
\begin{equation*}
\forall x_1,\ldots,x_n,x'_1,\ldots,x'_n.\ (P \land F_c(c_1) \rightarrow Q)
\end{equation*}
Rules for Computing Formulas for Commands
Therefore, to prove Hoare triples, we just need to compute for each command $c_1$ the formula $F_c(c_1)$. We next show how to do it. These rules all follow from the semantics of our language.
Assignment
\begin{equation*}
F_c(x=e) = (x'=e \land \bigwedge_{v \in V \setminus \{x\}} v=v')
\end{equation*}
Assume
\begin{equation*}
F_c({\it assume}(e)) = (e \land \bigwedge_{v \in V} v=v')
\end{equation*}
Havoc
\begin{equation*}
F_c({\it havoc}(x)) = (\bigwedge_{v \in V \setminus\{x\}} v=v')
\end{equation*}
For simplicity of notation, in the sequel we work with state that has only one variable, x.
Union
Note
\begin{equation*}
\{(\vec x,\vec x') \mid F_1 \} \cup \{(\vec x,\vec x') \mid F_2 \} = \{(\vec x,\vec x') \mid F_1 \lor F_2 \}
\end{equation*}
Therefore,
\begin{equation*}
F_c(c_1 [] c_2) = F_c(c_1) \lor F_c(c_2)
\end{equation*}
Sequential composition
Note that
\begin{equation*}
\{(\vec x,\vec x') \mid F_1 \} \circ \{(\vec x,\vec x') \mid F_2 \} = \{ (\vec x,\vec z) \mid F_1[\vec x':=\vec z] \} \circ \{ (\vec z,x') \mid F_2[\vec x:=\vec z] \} = \{ (\vec x,\vec z) \mid \exists \vec z. F_1[\vec x':=\vec z] \land F_2[\vec x:=\vec z] \}
\end{equation*}
Therefore,
\begin{equation*}
F_c(c_1\ ;\ c_2)\ =\ (\exists \vec z. F_c(c_1)[\vec x':=\vec z] \land F_c(c_2)[\vec x:=\vec z])
\end{equation*}
Above, $F_c(c_1)[\vec x':=\vec z]$ denotes taking formula $F_c(c_1)$ and replacing in it occurrences of variables $\vec x'$ by variables $\vec z$.
To avoid re-using variables, introduce always a fresh variable as $\vec z$ and denote it $\vec z_i$.
Using Computed Formulas
Note: the result will be disjunctions of such existential quantifications. We can always move them to top level.
Resulting formula:
\begin{equation*}
\forall x,x'.\ (P \land (\exists z_1,\ldots,z_n. F_c(c_1))\ \rightarrow Q)
\end{equation*}
which is equivalent to
\begin{equation*}
\forall x,x',z_1,\ldots,z_n. (P \land F_c(c_1) \rightarrow Q)
\end{equation*}
Conclusions: we can just generate fresh variables for intermediate points and prove the validity of the resulting quantifier free formula $(P \land F_c(c_1) \rightarrow Q)$.
Optimizations: assignments and assume statements generate equalities, many of which can be eliminated by one-point rule
\begin{equation*}
(\exists x. x=t \land F(x)) \leftrightarrow F(t)
\end{equation*}
Example
Take the program in the example below:
(if (x < 0) x=x+1 else x=x);
(if (y < 0) y=y+x else y=y);
It translate into the following relation:
\begin{equation*}
\begin{array}{l}
(assume(x<0) \circ (x = x+1)\ \cup\ assume (\lnot (x<0)) \circ skip) \circ \\
(assume(y<0) \circ (y = y+x)\ \cup\ assume(\lnot(y<0)) \circ skip)
\end{array}
\end{equation*}
By distribution of composition over union the relation above becomes:
\begin{equation*}
\begin{array}{l}
(assume(x<0) \circ (x = x+1) \circ assume(y<0) \circ (y = y+x)) \cup \\
(assume (\lnot (x<0)) \circ assume(y<0) \circ (y = y+x)) \cup \\
(assume(x<0) \circ (x = x+1) \circ assume(\lnot(y<0))) \cup \\
(assume (\lnot (x<0)) \circ assume(\lnot(y<0)))
\end{array}
\end{equation*}
It can be easily proven that $\{P\}\ \cup_i p_i\ \{Q\} \Leftrightarrow \wedge_i \{P\}\ p_i\ \{Q\}$:
\begin{equation*}
\begin{array}{1}
\{P\}\ \cup_i p_i\ \{Q\} \Leftrightarrow \\
sp(P, \cup_i p_i) \subseteq Q \Leftrightarrow \\
\cup_i sp(P, p_i) \subseteq Q \Leftrightarrow \\
\wedge_i sp(P,p_i) \subseteq Q \Leftrightarrow \\
\wedge_i \{P\}\ p_i\ \{Q\}
\end{array}
\end{equation*}
We can take all the terms of the union one by one as in the following small example:
\begin{equation*}
\begin{array}{1}
(assume(x<0) \circ (x = x+1) \circ assume(y<0) \circ (y = y+x)) \cup \\
assume(\lnot (x<0)) \Leftrightarrow \\
\exists x_1,y_1.(x<0 \wedge y_1 = y \wedge x_1 = x) \wedge (x'=x_1 + 1 \wedge y'=y_1)) \vee \\
(\lnot (x<0) \wedge x'=x \wedge y'=y)
\end{array}
\end{equation*}
Using valid formulas we can move the existential quantifiers outside the formula.
However, we can avoid expanding all paths and instead compute relations by following program structure.
Size of Generated Formulas
The compositional approach generates a formula polynomial in the size of the program. Indeed, fix the set of variables $V$. Then:
• the size of formula for each basic command is constant
• non-deterministic choice is disjunction
• sequential composition is conjunction (along with renaming that does not affect size–or affects at most $\log n$)
Moreover, formula generated in such a way looks very much like the program itself, converted to static single assignment form, see
Further reading | {
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4,428,130,694,711,949,000 | Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » Add fractions
Add 5/2 and 9/8
1st number: 2 1/2, 2nd number: 1 1/8
5/2 + 9/8 is 29/8.
Steps for adding fractions
1. Find the least common denominator or LCM of the two denominators:
LCM of 2 and 8 is 8
Next, find the equivalent fraction of both fractional numbers with denominator 8
2. For the 1st fraction, since 2 × 4 = 8,
5/2 = 5 × 4/2 × 4 = 20/8
3. Likewise, for the 2nd fraction, since 8 × 1 = 8,
9/8 = 9 × 1/8 × 1 = 9/8
4. Add the two like fractions:
20/8 + 9/8 = 20 + 9/8 = 29/8
5. So, 5/2 + 9/8 = 29/8
In mixed form: 35/8
MathStep (Works offline)
Download our mobile app and learn to work with fractions in your own time:
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Related:
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-9,177,297,022,883,210,000 | Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
This question is from Dummit and Foote's Abstract Algebra, page 638, question 20. It gives a nice paragraph of hints that basically guides one through the problem, but I'm very stuck at a crucial junction. Any useful hint is much appreciated. I have detailed what I know and what I do not know, but if you just want the tl;dr, just read the question, which is the following sentence.
"Let $p$ be a prime. Show that any solvable subgroup of $S_p$ of order divisible by $p$ is contained in the normalizer of a Sylow $p$-subgroup of $S_p$. [...]
Hint: Let $G \leq S_p$ be a solvable subgroup of order divisible by $p$. Then $G$ contains a $p$-cycle, hence is transitive on $\{1, \ldots, p\}$. Let $H < G$ be the stabilizer in $G$ of the element $1$, so $H$ has index $p$ in $G$. Show that $H$ contains no nontrivial normal subgroups of $G$ (note the conjugates of $H$ are the stabilizers of the other points). Let $G^{(n-1)}$ be the last nontrivial subgroup in the derived series for $G$. Show that $H \cap G^{(n-1)} = 1$ and conclude that $\lvert G^{(n-1)}\rvert = p$, so that the Sylow $p$-subgroup of $G$ (which is also a Sylow $p$-subgroup of $S_p$) is normal in $G$."
Here are the things I do know:
1. $H$ has an order that divides $(p-1)!$ since it has index $p$ in $G$, and $G$ has order $pu$ for some $u$ not divisible by $p$.
2. Everything up to and excluding the part where I am asked to prove that $H \cap G^{(n-1)} = 1$.
3. I know how to prove the next part where I'm asked to prove that $|G^{(n-1)}| = p$ provided I know how to do that previous part!
4. I know that $\lvert S_p \rvert = p!$, so any Sylow $p$-subgroup of $S_p$ has size $p^1 = p$, since no other factors of $p!$ can contain $p$ as a prime factor.
Now here are the things I do not know:
1. I am terribly stuck at the step where I have to show $H \cap G^{(n-1)} = 1$. I tried showing that this is normal, so I can use the result immediately preceding to conclude that it is trivial. But I'm having major problems. I may just be missing something extremely obvious.
2. Even if I can do that part, the next part asks us to conclude that this Sylow $p$-subgroup is normal in $G$, which I can't immediately see how to derive. I'm assuming ``this Sylow $p$-subgroup'' is referring to the size $p$ subgroup $G^{(n-1)}$---it has the right size to be a Sylow $p$-subgroup.
share|improve this question
Just to define some terms in case the definition you are used to is different from mine. A finite group $G$ is solvable if the derived series $G^{(0)} := G, G^{(k)} := [G^{(k-1)}, G^{(k-1)}]$ for $k \geq 1$ eventually becomes the trivial subgroup $\{e\}$. Here, $[G^{(k)}, G^{(k)}]$ is the subgroup generated by all the "commutators" of the form $g^{-1}h^{-1}gh$, where $g, h \in G^{(k)}$. This condition is equivalent to saying that the finite group $G$ has a composition series whose factors are Abelian. The derived series is not to be confused with the lower central series. – vwxf Mar 25 '11 at 7:02
Note that $G^{(n-1)}$ is an abelian group, and $H$ is maximal; Thus $H\cap G^{(n-1)}$ is a normal subgroup of $G$. From what came before, this is the trivial subgroup; It then follows that $G=HG^{(n-1)}$ and so $|G| = |H|\cdot |G^{(n-1)}|$. – user641 Mar 25 '11 at 8:04
Thanks for your reply! I didn't realize that $G^{(n-1)}$ was Abelian until you mentioned it. But I can't see why an intersection of a maximal subgroup and an Abelian one means it is normal. – vwxf Mar 25 '11 at 8:55
Yeah, I don't see why Abelian intersect maximal implies normal. Any suggestions? – vwxf Mar 25 '11 at 12:01
Thanks for everyone's help! In case this is of any use to anyone else in the future, I will summarize various hints on how to tackle (1) and (2) in "things I do not know": (1) Try looking ahead. In order to prove $|G^{(n-1)}| = p$, what new subgroup are you going to construct? This construction, along with the maximality of $H$, may give you a hint as to how to prove $H \cap G^{(n-1)}$ is normal in $G$. (2) Is $G^{(n-1)}$ normal in $G$? What is its order (size)? What should the size of a Sylow $p$-subgroup be? – vwxf Mar 25 '11 at 23:38
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It is false in general that "abelian intersect maximal implies normal": $A_5$ is maximal in $S_5$, the subgroup generated by $(1,2,3,4)$ is abelian, but the intersection of the two is nontrivial (contains $(1,3)(2,4)$) and not normal in $S_5$.
However, it is true that the intersection of a maximal subgroup and an abelian normal subgroup is normal.
Proposition. Let $G$ be a group, and $H$ a maximal subgroup of $G$. If $N$ is an abelian normal subgroup of $G$, then $N\cap H$ is normal in $G$.
Proof. If $N\subseteq H$, then $N\cap H = N\triangleleft G$ and we are done.
If $N$ is not contained in $H$, then maximality of $H$ and normality of $N$ imply that $HN=G$ (since $HN$ is a subgroup). Let $x\in H\cap N$ and $g\in G$. Then we can write $g = hn$ with $h\in H$ and $n\in N$. Then $$gxg^{-1} = (hn)x(hn)^{-1} = h(nxn^{-1})h^{-1} = hxh^{-1}$$ with the last equality since $N$ is abelian. Now, $x,h\in H$, so $hxh^{-1}\in H$. And $x\in N$, so $hxh^{-1}\in N$. Thus, $hxh^{-1}\in H\cap N$, proving that $H\cap N$ is normal in $G$. QED
Added. More generally: note that $H\cap N$ is certainly normal in $H$. If $HN=G$, then you only need to show that $N$ normalizes $H\cap N$: then $(hn)x(hn)^{-1} = h(nxn^{-1})h^{-1}$, which will lie in $H\cap N$ if $nxn^{-1}\in H\cap N$. Therefore:
Proposition. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. Then $H\cap N\triangleleft HN$ if and only if $N\subseteq N_{HN}(N\cap H)$, where $N_{NH}(N\cap H)$ is the normalizer of $N\cap H$ in $NH$.
Proof. If $N\subseteq N_{NH}(N\cap H)$, then the argument proceeds as above. Conversely, if $N\cap H\triangleleft NH$, then $N\subseteq NH=N_{NH}(N\cap H)$. $\Box$
In particular, if $N$ is abelian then $N\subseteq C_{NH}(N\cap H)\subseteq N_{NH}(N\cap H)$; and if $H$ is maximal, then this gives the proposition above in the nontrivial case.
Now apply this to $H$ and the abelian normal subgroup $G^{(n-1)}$ to conclude that $H\cap G^{(n-1)}\triangleleft G$.
Now, you know that $G^{(n-1)}$ is normal in $G$ (because the derived terms are always normal in $G$), and has order $p$. Since a $p$-Sylow subgroup of $G$ has order $p$, then $G^{(n-1)}$ is a $p$-Sylow subgroup of $G$. Since it is normal in $G$, the $p$-Sylow subgroup of $G$ is normal in $G$. So $G$ is contained in the normalizer of the $p$-Sylow subgroup $G^{(n-1)}$ of $S_p$.
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Thank you very much for your help! This step was in fact like a fibre bundle lol (locally trivial). I guess if I worked on it longer without thinking too hard about it, I guess I might be able to stumble across this solution. – vwxf Mar 25 '11 at 23:10
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Arturo's solution follows the hint and is correct. But I did not find the suggestion to show that $N \cap H = 1$ particularly helpful.
You could reason alternatively as follows. Use the same argument as Arturo to show that $NH = G$. Since $|NH| = |N||H|/|N \cap H|$, and $p$ does not divide $|H|$, it follows that $p$ divides $|N|$. Since $N$ is abelian, it has a unique Sylow $p$-subgroup of order $p$, which must be normal in $G$.
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Thanks for this additional speed-up. This would circumvent that step indeed. It does require the extra theorem about Abelian groups though, but I guess that is fair. Once again, I learn something new everyday. :) – vwxf Mar 25 '11 at 23:12
@vwxf: It's not really a theorem, but an observation: any two Sylow subgroups are conjugate, but in an abelian group, all subgroups are normal; if you had $S_1$ and $S_2$ both Sylow $p$-subgroups, there is a $g$ with $S_2 = gS_1g^{-1}$, but $gS_1g^{-1}=S_1$. – Arturo Magidin Mar 26 '11 at 21:46
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3,876,177,818,885,065,700 | top
河內塔程式執行步驟
各位前輩，
小弟了解河內塔的運作方式，
將n-1盤子 ⇒ 由來源木棒(A)移到輔助木棒(B)
將最大盤子 ⇒ 由來源木棒(A)移到目的木棒(C)
將n-1盤子 ⇒ 由新的來源木棒(B)透過新的輔助木棒(A)移到目的木棒(C)
程式碼如下:
private static void hanoi(int n, char sour, char aux, char dest){
if(n == 1){
System.out.printf("Move disk %d from %c to %c\n", n, sour, dest);
System.out.println("n = 1");
}
else{
//將A上n-1個盤子借助C移到B
hanoi(n-1, sour, dest, aux);
System.out.printf("Move disk %d from %c to %c\n", n, sour, dest);
System.out.println("print");
//將B上n-1個盤子借助A移到C
hanoi(n-1, aux, sour, dest);
}
}
想很久(想破頭了)還是想不通他的執行順序和結果為什麼是這樣?
Move disk 1 from A to C
Move disk 2 from A to B
Move disk 1 from C to B
Move disk 3 from A to C
Move disk 1 from B to A
Move disk 2 from B to C
Move disk 1 from A to C
問題一: 遞迴不是"先進後出"嗎? 請教一下哪裡想錯了?
第一次hanoi呼叫
n=3，hanoi(2, A, C, B)，結果Move disk 2 from A to B
n=2，hanoi(1, A, C, B)，結果Move disk 1 from A to B
n=1，結果Move disk 1 from A to C
輸出把順序顛倒
n=1，結果Move disk 1 from A to C
n=2，hanoi(1, A, C, B)，結果Move disk 1 from A to B
n=3，hanoi(2, A, C, B)，結果Move disk 2 from A to B
中間輸出Move disk 3 from A to C
第二次hanoi呼叫
n=3，hanoi(2, B, A, C)，結果Move disk 2 from B to C
n=2，hanoi(1, B, A, C)，結果Move disk 1 from B to C
n=1，結果Move disk 1 from A to C
輸出把順序顛倒
n=1，結果Move disk 1 from A to C
n=2，hanoi(2, B, A, C)，結果Move disk 1 from B to C
n=3，hanoi(1, B, A, C)，結果Move disk 2 from B to C
問題二: 每呼叫一次方法，都會執行到print，那方法裡面不是有兩個hanoi，hanoi也會被呼叫兩次?
一、因為它將輸出直接寫在方法裏了，就變成先遇到先輸出了。
二、是的。
TOP
本帖最後由 allenbrian 於 2015-6-30 10:33 編輯
謝謝版主回覆，
問題一還是不了解他的執行跟輸出的順序間的關係，
煩請再說明詳細一點，感謝!
問題二它將輸出直接寫在方法裏了，就變成先遇到先輸出了。
可以說明一下它遇到的過程嗎? 感謝!
TOP
建議你動手在紙上一個步驟一個步驟畫畫看，或者是開Debugger來逐步查看各變數狀況，這會是最清楚的。
TOP
回復 4# codedata
我實際有用紙筆算過了，只是不了解程式執行的步驟，
謝謝版大提醒要用debug，第一次用debug，超強大的!
終於了解程式怎麼執行的了，謝謝版大!
TOP | {
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-528,096,162,651,516,160 | Fonction polynomiale
En mathématiques, une fonction polynomiale (parfois appelée fonction polynôme) est une fonction obtenue en évaluant un polynôme.
Par abus de langage, on appelle parfois une fonction polynomiale un polynôme, confondant ainsi la notion de fonction polynomiale avec celle de polynôme formel. Cette confusion est sans gravité dans le cadre des polynômes à coefficients réels ou complexes (ou plus généralement à coefficients dans un corps infini) mais peut conduire à des contresens en général (par exemple pour les polynômes à coefficients dans un corps fini).
Fonctions polynomiales réelles ou complexes modifier
Un exemple de fonction polynomiale réelle de degré 5
Un cas courant est celui où le polynôme est à coefficients réels ou complexes. Plus précisément, on considère le polynôme P de la forme
où les ak sont des nombres réels ou des nombres complexes. La fonction polynomiale f associée est alors définie par
où la variable x peut être elle-même réel ou complexe.
Voici les exemples les plus courants[1] :
Dans le cadre des fonctions polynomiales réelles ou complexes, on définit le degré d'une fonction polynomiale comme le degré du polynôme auquel elle est associée (avec la convention que le degré vaut −∞ si la fonction est nulle).
Puisqu'un polynôme réel ou complexe non constant de degré n a au plus n racines d'après le théorème de d'Alembert-Gauss, on en déduit qu'une fonction polynomiale réelle ou complexe non constante de degré n a au plus n zéros. Autrement dit, deux fonctions polynomiales réelles ou complexes de degrés inférieurs ou égaux à n et coïncidant sur plus de n points sont nécessairement identiques (c'est-à-dire qu'elles ont même degré et mêmes coefficients).
La fonction polynomiale f réelle ou complexe est infiniment dérivable (elle est même analytique) et la k-ième dérivée de f est exactement la fonction polynomiale associée à la k-ième dérivée formelle de P. En particulier, les dérivées d'ordre k > n de fonctions polynomiales de degré n sont identiquement nulles.
Cas où
Cas où
Cas où
Par exemple, la dérivée formelle de P est donnée par
et on a
De même, les primitives de f sont exactement les fonctions polynomiales associées aux primitives formelles de P, c'est-à-dire de la forme
C est une constante réelle ou complexe arbitraire.
Fonction polynomiale sur un corps quelconque modifier
Plus généralement, il est possible de considérer un polynôme P à coefficients dans un corps K (commutatif) quelconque:
où les aj sont des éléments de K. La fonction polynomiale f associée est alors la fonction de K dans lui-même définie par
Dans le cas où le corps est infini (par exemple dans le cas du corps des nombres réels ou des nombres complexes traité plus haut), on peut encore identifier polynômes et fonctions polynomiales. Autrement dit, l'application qui à un polynôme à coefficients dans K associe la fonction polynomiale correspondante est une injection de l'ensemble des polynômes à coefficients dans K dans l'ensemble des applications de K dans lui-même. Par conséquent, on peut définir comme à la section précédente la notion de degré d'une fonction polynomiale comme le degré du polynôme correspondant et deux fonctions polynomiales sont identiques si et seulement si leurs polynômes associés sont les mêmes.
Dans le cas où K est un corps fini, ce qui précède n'est plus vrai. Par exemple, dans le corps à deux éléments, le polynôme X(X-1) n'est pas le polynôme nul mais la fonction polynomiale associée est identiquement nulle. Il n'est alors pas possible de définir la notion de degré d'une fonction polynomiale et deux fonctions polynomiales peuvent être identiques sans que leurs polynômes associés soient égaux. Cela montre qu'il est nécessaire dans ce cadre de distinguer la notion de fonction polynomiale de celle de polynôme formel.
Morphisme d'évaluation vis-à-vis d'une algèbre associative modifier
On peut également considérer un polynôme P à coefficients dans un anneau A quelconque:
où les aj sont des éléments de A. On peut alors comme ci-dessus définir la fonction polynomiale associée.
De manière plus générale, il est possible de considérer une algèbre associative E (unitaire) sur A et l'application qui à un élément e de E associe l'élément P(e) de E défini par
Cette application est un morphisme d'anneaux appelé morphisme d'évaluation.
Un cas d'usage très courant est celui où A est un corps K (commutatif) et où E est l'ensemble des endomorphismes d'un espace vectoriel sur K. Ainsi, si u est un tel endomorphisme, on a
où les puissances correspondent à la composition de fonctions et id est l'application identité de l'espace vectoriel. Ainsi, pour tout polynôme P et tout endomorphisme u, P(u) est un endomorphisme. La notion de polynôme d'endomorphisme joue un rôle central pour la réduction d'endomorphisme.
De manière complètement analogue, il est possible de définir la notion de polynôme de matrice.
Voir aussi modifier
Notes et références modifier
1. On considère dans chacun des cas que le coefficient dominant, c'est-à-dire an est différent de 0. | {
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} | 0c090d63199a0a01e3b08e4a255778a0 |
-1,552,361,216,097,588,700 | Exercism v3 launches on Sept 1st 2021. Learn more! 🚀🚀🚀
Avatar of PatrickMcSweeny
PatrickMcSweeny's solution
to Roman Numerals in the JavaScript Track
Published at Aug 26 2019 · 0 comments
Instructions
Test suite
Solution
Note:
This exercise has changed since this solution was written.
Write a function to convert from normal numbers to Roman Numerals.
The Romans were a clever bunch. They conquered most of Europe and ruled it for hundreds of years. They invented concrete and straight roads and even bikinis. One thing they never discovered though was the number zero. This made writing and dating extensive histories of their exploits slightly more challenging, but the system of numbers they came up with is still in use today. For example the BBC uses Roman numerals to date their programmes.
The Romans wrote numbers using letters - I, V, X, L, C, D, M. (notice these letters have lots of straight lines and are hence easy to hack into stone tablets).
1 => I
10 => X
7 => VII
There is no need to be able to convert numbers larger than about 3000. (The Romans themselves didn't tend to go any higher)
Wikipedia says: Modern Roman numerals ... are written by expressing each digit separately starting with the left most digit and skipping any digit with a value of zero.
To see this in practice, consider the example of 1990.
In Roman numerals 1990 is MCMXC:
1000=M 900=CM 90=XC
2008 is written as MMVIII:
2000=MM 8=VIII
See also: http://www.novaroma.org/via_romana/numbers.html
Setup
Go through the setup instructions for Javascript to install the necessary dependencies:
https://exercism.io/tracks/javascript/installation
Requirements
Install assignment dependencies:
$ npm install
Making the test suite pass
Execute the tests with:
$ npm test
In the test suites all tests but the first have been skipped.
Once you get a test passing, you can enable the next one by changing xtest to test.
Source
The Roman Numeral Kata http://codingdojo.org/cgi-bin/index.pl?KataRomanNumerals
Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
roman-numerals.spec.js
import { toRoman } from './roman-numerals';
describe('toRoman()', () => {
test('converts 1', () => expect(toRoman(1)).toEqual('I'));
xtest('converts 2', () => expect(toRoman(2)).toEqual('II'));
xtest('converts 3', () => expect(toRoman(3)).toEqual('III'));
xtest('converts 4', () => expect(toRoman(4)).toEqual('IV'));
xtest('converts 5', () => expect(toRoman(5)).toEqual('V'));
xtest('converts 6', () => expect(toRoman(6)).toEqual('VI'));
xtest('converts 9', () => expect(toRoman(9)).toEqual('IX'));
xtest('converts 27', () => expect(toRoman(27)).toEqual('XXVII'));
xtest('converts 48', () => expect(toRoman(48)).toEqual('XLVIII'));
xtest('converts 59', () => expect(toRoman(59)).toEqual('LIX'));
xtest('converts 93', () => expect(toRoman(93)).toEqual('XCIII'));
xtest('converts 141', () => expect(toRoman(141)).toEqual('CXLI'));
xtest('converts 163', () => expect(toRoman(163)).toEqual('CLXIII'));
xtest('converts 402', () => expect(toRoman(402)).toEqual('CDII'));
xtest('converts 575', () => expect(toRoman(575)).toEqual('DLXXV'));
xtest('converts 911', () => expect(toRoman(911)).toEqual('CMXI'));
xtest('converts 1024', () => expect(toRoman(1024)).toEqual('MXXIV'));
xtest('converts 3000', () => expect(toRoman(3000)).toEqual('MMM'));
});
const NUMERALS = {
1: "I",
4: "IV",
5: "V",
9: "IX",
10: "X",
40: "XL",
50: "L",
90: "XC",
100: "C",
400: "CD",
500: "D",
900: "CM",
1000: "M"
};
export const toRoman = input => {
let number = input;
return Object.keys(NUMERALS)
.reverse()
.map(key => {
let numeral = "";
if (key <= number) {
let times = Math.floor(number / key);
number -= key * times;
for (let i = 1; i <= times; i++) {
numeral += NUMERALS[key];
}
}
return numeral;
})
.join("");
};
Community comments
Find this solution interesting? Ask the author a question to learn more.
What can you learn from this solution?
A huge amount can be learned from reading other people’s code. This is why we wanted to give exercism users the option of making their solutions public.
Here are some questions to help you reflect on this solution and learn the most from it.
• What compromises have been made?
• Are there new concepts here that you could read more about to improve your understanding? | {
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-6,480,860,834,386,887,000 | Very Basic Model Theory
by So8res15 min read31st Oct 201315 comments
35
Logic & Mathematics
Personal Blog
In this post I'll discuss some basic results of model theory. It may be helpful to read through my previous post if you haven't yet. Model Theory is an implicit context for the Heavily Advanced Epistemology sequence and for a few of the recent MIRI papers, so casual readers may find this brief introduction useful. And who knows, maybe it will pique your interest:
A tale of two logics
propositional logic is the "easy logic", built from basic symbols and the connectives "and" and "not". Remember that all other connectives can be built from these two: With Enough NAND Gates You Can Rule The World and all that. Propositional logic is sometimes called the "sentential logic", because it's not like any other logics are "of or relating to sentences" (/sarcasm).
first order logic is the "nice logic". It has quantifiers ("there exists", "for all") and an internal notion of equality. Its sentences contain constants, functions, and relations. This lets you say lots of cool stuff that you can't say in propositional logic. First order logic turns out to be quite friendly (as we'll see below). However, it's not strong enough to talk about certain crazy/contrived ideas that humans cook up (such as "the numbers").
There are many other logics available (second order logic AKA "the heavy guns", ω-logic AKA "please just please can I talk about numbers", and many more). In this post we'll focus on propositional and first-order logics.
Training Wheels
As a warm-up, let's define "models" for propositional logic.
Remember, logics together with languages produce sentences. Languages of propositional logic are sets of basic symbols. From these symbols and the two connectives of propositional logic (∧ and ¬) we construct our sentences.
For example, from the language {x, y} we can build the sentences x, ¬x, x∧y, and many more. We cannot build sentences like hello (which doesn't use the right symbols) or ¬xy (which breaks the rules of the logic).
We want to study interpretations for such sentences, where an "interpretation" is some object which assigns a truth value to each sentence. This is denoted by the ⊧ operator: A⊧B expresses some form of "A models B as true". This operator is used frequently and in many contexts throughout model theory.
What we're looking for is some object (set) such that there is a relation (⊧) between the object and all sentences generated by the language under consideration. There are many object/relation pairs that assign truth values in a stupid way. For example, an interpretation that assigns "true" to every sentence is quite useless.
Many other interpretations are somewhat useful but still "wrong". For example, there are interpretations of sentences which treat like implication instead of conjunction. Somebody may have use for such things, but we certainly don't.
We want to narrow our consideration to interpretations where means "and" and ¬ means "not", so we explicitly require object/relation pairs such that whenever the object models both φ and ψ, it also models φ∧ψ. (And X⊧φ iff it is not the case that X⊧¬φ, where X is the object under consideration.)
If we can find any objects that work like this, we'll be justified in calling them "models". However, we haven't defined any such objects yet — we've only constrained their potential behavior.
Any interpretation of the sentences generated from a language L of propositional logic must assign truth values to all basic sentence symbols in L. It turns out, once we select which sentence symbols are "true", we're done. The rest of the behavior is completely specified by the rules about and ¬.
In other words, any object/relation which assigns truth values to sentences according to the above rules is isomorphic to a set S of sentence symbols with the operator defined in the obvious way (starting with S⊧x iff x∈S).
Exploring this idea gives you a feel for how to define models of a logic.
More power
Propositional logic is pretty boring. If you want to say anything interesting, you've got to be able to say more than just "and" and "not". Let's move on to a stronger logic.
First-order logic uses the symbols ( ) ∀ ∃ ∧ ¬ ν ' ≡ or equivalent (where ν ν' ν'', etc. are variables) with the familiar syntactic rules. A language of first-order logic has three different types of symbols: relation symbols, function symbols, and constant symbols. The rules of logic are designed such that symbols act as you expect, given their names.
An interpretation of the sentences generated by some language in first-order logic is (as before) an object that assigns each sentence a truth value (via a relation). We narrow these objects down to the ones that treat all the symbols in ways that justify their names.
More specifically, we consider interpretations that have of some "universe" (set) of items. Constant symbols are interpreted as specific items in the universe. Relation/function symbols are interpreted as relations/functions on the universe.
We further restrict consideration to interpretations where the logical symbols act in the intended fashion. For example, we require that the interpretation hold c≡d true (where c and d are constant symbols) if and only if the interpretation of c in the universe is equal to the interpretation of d in the universe.
All this is pretty mechanical. The resulting objects are "models". Some of the early results in model theory show that these mathematical objects have indeed earned the name.
Completeness
Completeness is the poster-child of model theory, and it warrants quite a bit of exploration. It's actually saying a few different things. I'll break it down:
1. Theorems of first-order logic are true in every model of first-order logic.
A theorem of first-order logic is something we can prove from the logic alone, such as (∀ν)(ν≡ν). Contrast this with a sentence like (∃ν)(F(ν)≡c), where F is a function symbol and c is a constant symbol — the truth of this sentence depends entirely upon interpretation.
So what (1) says is that there aren't any models that deny tautological truths of first-order logic. This result should not be surprising: this claim merely states that we picked a good definition for "models". If instead we found that there are models which deny theorems, this would not be some big grand proof about the behavior of first-order logic. Rather, it would mean that we put the "model" label on the wrong group of thingies, and that we should go back and try again.
2. Any sentence true in every model of first-order logic is a theorem of first-order logic.
This is the converse to (1), and it's quite a bit more powerful. This states that the theorems of a language are only the things true in every model of that language. There's no mystery sentence that is true in every model but not provable from the syntax of the logic.
From one point of view, this says that there are no "missing" models that "would have" said the mystery sentence was false (hence "completeness"). From another, this says that all sentences are "well behaved": no sentence "outwits" all models.
It's worth noting that this is where completeness fails for models of second order logic. From one perspective, there must be some "missing models" in second-order logic. From the other perspective, there must be certain sentences which can "outwit" their models (presumably Gödel sentences). I haven't spent any time formally examining these intuitive claims yet; I have much to learn about second-order logic.
I should also note that I've been implicitly assuming completeness for the last two posts by ignoring the syntactic rules of the logics under consideration. The completeness theorem lets me do this, because it states that the results interpreted by first-order models are exactly the same as the results derived from the syntactic rules of first-order logic. Because the completeness theorem holds, I'm allowed to treat the logical sentences as dead symbols and consider the binding M⊧φ∧ψ iff M⊧φ and M⊧ψ to be the means by which obtains its meaning.
When the completeness theorem fails (as it does in stronger logics) then there is a gap between what the rules of the logic allow and what the models of the logic can say. In that case I must separately consider what the rules say and what the models model. This is the edge of my comfort zone; more exploration with second-order logic is necessary.
Fortunately, everything is well-behaved in first order logic. In fact, (1) and (2) above can be made stronger:
3. A set Σ of sentences is consistent if it has a model.
This is another sanity check that we've put the "model" label on the right thing. A set of sentences is inconsistent if it can derive both φ and ¬φ for some sentence φ. (More specifically, Σ is inconsistent if it can derive everything. If there is any sentence that Σ cannot derive, Σ is consistent. Remember that from a contradiction, anything follows.)
Our models are defined such that whenever they model φ, they do not model ¬φ (and vice versa). Thus, if a set of sentences has a model (Σ "has a model" when there is some model that holds true every sentence σ in Σ) then there must be some sentences which Σ cannot deduce (¬σ for each σ in Σ, for instance). Thus, Σ is consistent.
4. A set Σ of sentences has a model if it is consistent.
This is where things get interesting again. This is a stronger version of (2) above: every consistent theory has a model.
This makes a lot of sense after you understand (2) and (3), but don't underestimate its importance. This tells us that for every consistent theory there is some interpretation following the rules which we laid out. Again, this says something positive about our models (they are strong enough to handle any consistent theory) and something negative about the logic (it's not strong enough to permit a consistent theory stating "I cannot be modeled").
Note: I'm not sure what it would mean for a theory to state "I cannot be modeled", nor am I convinced that the idea is meaningful. Again, we're nearing the edge of my comfort zone.
The point is, the completeness theorem for first order logic says "if you hand me a consistent theory, I can build a model of it". This is very useful. We don't have to waste any time worrying whether or not there's an interpretation available that satisfies all our stringent rules about how actually means "equals" and so on: if the theory is consistent, a model exists.
Witnesses
There's been a lot of hand-waving going on for the past four points. Let's get back to the models.
In model theory, you're manipulating interpretations for theories. Due to the generality of such work, there are few tools available to manipulate any abstract model. One of the most useful tools in model theory is the ability to extend a model.
Ideally, when you extend a model, you want to make it more manageable without changing its behavior. Our first example of such a technique involves extending a model to add "witnesses".
A "witness" is a constant symbol that witnesses the truth of an existentially quantified sentence. For example, in the language {S, +, ✕, 0} of arithmetic, the constant symbol 0 is a witness to the sentence (∃ν)(Sν≡S0) (because 0 makes (Sν≡S0) true). However, there is no witness to the sentence (∃ν)(ν≡S0), because 1 is not a constant symbol of the language.
However, we can extend a language to add new constant symbols. Specifically, given any model, we can extend the language to contain one constant symbol ā for each element a in the universe. Then we can extend the model to interpret each ā by a. Such extension does not change the behavior of the model, but it does give the model a number of nice properties.
A model is said to "have witnesses" if for every existentially quantified sentence shaped like (∃ν)ψ that it models, there is some constant c such that the model models ψ(ν\c) (ψ with ν replaced by c).
Before we discuss them, note a few things:
1. We can also reduce a model & language extended in this way by eliminating the added constants.
2. We can talk about sets of sentences with witnesses if, whenever a sentence shaped like (∃ν)ψ is in the set of sentences Σ, there is some constant c such that ψ(ν\c) is also in Σ.
3. Just as we can extend a model & language so that the model has witnesses, we can extend a set of sentences and language so that the set of sentences has witnesses (by adding a new constant symbol for each existentially quantified sentence). Such extension does not threaten the consistency of a set of sentences.
4. It's easy to construct a model from a consistent set of sentences that has witnesses. The universe is the set of all constant symbols (technically, one element for each equivalent set of constant symbols), and the behavior of function/relation symbols is forced by their behavior on the constant symbols.
Formalize these four points and put them together, and you've just proved the completeness theorem. (Given any consistent set Σ of sentences, add witnesses then make a model then reduce it to the original language, you now have a model of Σ).
Compactness
The compactness theorem states that A set Σ of sentences has a model iff every finite subset of Σ has a model.
This leads to a pretty surprising result. Let's break it down.
1. If Σ has a model then every finite subset of Σ has a model.
This direction is obvious: any model of Σ is also a model of all subsets of Σ: if a model holds true all sentences σ in Σ then it obviously holds true all sentences σ in any subset of Σ.
2. If every finite subset of Σ has a model then Σ has a model.
This is where things get interesting. Note that we measure the size of a model by measuring the size of its universe. So when we say "a countably infinite model" we mean a model with a countably infinite universe.
The proof of the above is actually quite easy: in first order logic, all proofs are finite. Therefore, any proof of contradiction (and thus inconsistency) must be finite. Since every finite subset of Σ has a model, no finite subset of Σ is inconsistent. Because all inconsistencies are finite, Σ is not inconsistent. Then, by completeness, because Σ is consistent it has a model.
Or, in other words, the compactness theorem says
If Σ wants to be inconsistent it can do it in a finite subset.
When we know that no finite subset of Σ is inconsistent, we know that Σ has a model.
The surprising result that this leads to is as follows:
If a theory T allows arbitrarily large finite models then it has an infinite model.
If you hand me a theory that allows arbitrarily large finite models, I can extend the language by adding countably many constants to the language. Then I can consider the set Σ of sentences built from T joined with sentences of the form "there are at least n distinct constants" for all finite n. Clearly, every finite subset of Σ is satisfied by one of the finite models of T, which come in arbitrarily large sizes — just pick one that fits. Then, by compactness, Σ has a model. The model of Σ has infinitely many constants.
I have just given a fully general recipe for building an infinite model given a theory T that admits arbitrarily large finite models. What does this mean? It means that no theory in first order logic is capable of saying "I admit only finite models". Sentences can say "models must have no more than 10 elements" or "models are allowed to be infinite", but sentences cannot say "My model is finite". This follows directly fro the fact that all proofs of contradiction are finite. You either have a specific limit, or you don't get a limit at all.
MORE POWER
If you hand me a theory that allows arbitrarily sized finite models, I can hand you back an infinite model.
It gets better.
If you hand me an infinite theory, I can expand it.
Using a method similar to the one above, I can expand T with sentences of the form "there are at least β distinct constants", for all β less than my chosen cardinal α. Following the same argument as above, all finite subsets of this theory have a model so this theory has a model, which obviously is of power α.
In other words, a theory in first-order logic can either say "I am at most this big", where this is a specific finite number, or "I am very big", where very is can be any infinite cardinal that you like.
Example: The theory of arithmetic doesn't have a finite cutoff point. There's no maximum number. Countably infinite models are allowed.
Therefore, arbitrarily large infinite models are allowed.
There are countable models of arithmetic (at least one of which you'll find familiar), and there are also models of arithmetic where there are as many "numbers" as there are reals. Then there are bigger models of arithmetic that are saturated, no, dripping with numbers.
Now you understand why first order logic cannot discuss the "standard" model of number theory. No first-order theory can specifically pinpoint "countable" models! A first order theory must either specify a finite maximum size explicitly, or allow models of unfathomable size.
Even more
I was hoping to get farther than this, but this is a good stopping point. Hopefully you've learned something about model theory. There are many more interesting results yet.
The book Model Theory by Chang and Keisler primarily focuses on introducing new ways to extend arbitrary models (giving them nice properties in the process) and then discusses results that can be gleaned from models extended thus. Focus is also given to special cases where models are well behaved, such as atomic models (which are a sort of minimal model for a given theory) and saturated models (which are a sort of maximal model for a given theory, within a given cardinality. There Are Always Bigger Models).
There's also quite a bit of exploration into manipulating languages (Eliminate Quantifiers To Win Big Prizes!!!) and even manipulating logics (Q. How far beyond first order logic we can go before sacrificing things like completeness and compactness? A. Not very.)
If you're interested in learning more, then… well, I'm probably not the person to talk to. I'm not even halfway through this textbook. But if you're feeling gutsy, consider picking up Model Theory and getting started. Some of the harder concepts would be much easier to work through with other people rather than alone.
In fact, I have a number of notes about trying to learn something difficult on my own (what worked, what didn't) that I plan to share. But that's a story for another day.
Finally, please note that my entire exposure to model theory is half a textbook and some internetting. I guarantee I've misunderstood some things. If you see errors in my explanations, don't hesitate to let me know! My feedback loop isn't very strong right now.
35
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Something that kind of interests me; to the Pythagoreans mathematics was magic, involving mystical insight into the ultimate nature of reality. Similar ideas continued in Plato and those he influenced, with echoes down to the early modern rationalists. But Pythagorean mathematics was exceedingly primitive and limited. Modern logic and mathematics are vastly more powerful and sophisticated, and yet the Pythagorean mysticism has almost completely disappeared. Why were people more impressed with mathematics when it was more limited? I imagine novelty was a factor, and they say familiarity breeds contempt, but I have a couple of other hypotheses. First, there seem to be obvious places here and there in modern logic and mathematics where we seem to face choices, where it looks like a matter of "doing this is convenient for this purpose" rather than "this is the only way things could be." This tends to weaken the idea that the fundamental nature of reality is being revealed. Second, there are parts of modern mathematics that are deeply weird (e.g. many things about how infinite cardinals work). I imagine there are people who find it hard to accept that those parts of mathematics at least could be describing any real world.
I'm certainly not arguing for returning to Pythagorean mysticism, but of the two reasons I propose why people might be ruling that out, I'm inclined to think the first reason looks fairly good while the second strikes me as highly suspect. I'm quite curious as to which has been most influential (or whether it's some other factor or factors, perhaps familiarity really is the driving force, or perhaps external influences, say from hostility to mysticism in other fields, are important).
Tegmark level IV is the modern version of "Pythagorean mysticism" (it states that "the ultimate nature of reality" is math). So, there is no need to return to anything, we are already there. Well, some of us are, even on this forum. As for "the fundamental nature of reality", there is no indication for it to be a real thing, the deeper we dig, the more we find. But yes, in the map/territory model, math helps build better maps of the territory, whether or not it is the "ultimate" territory. The "weird" parts may or may not be useful in building better maps, it's hard to tell in advance. After all, the (freshly revealed) territory ls also weird.
Cantor who first did the first work on infinite cardinals and ordinals seemed to have a somewhat mystic point of view some times. He thought his ideas about transfinite numbers were communicated to him from god, whom he also identified with the absolute infinite (the cardinality of the cardinals which is too big to itself be a cardinal). This was during the 19th century so quite recently.
I'd say that much mysticism about foundational issues like what numbers really are, or what these possible infinities actually mean, have been abandoned by mathematicians in favour of actually doing real mathematics. We also have quite good formal foundations in terms of ZF and formal logic nowadays, so discussions like that do not help in the process of doing mathematics (unlike, say, discussions about the nature of real numbers before we had them formalised in terms of Cauchy sequences or Dedekind cuts).
I don't think those attitudes are quite as gone as you seem to think although that might be the mind projection fallacy. It's better understood, to the mystery and superstition are gone, but the senses of transcendent awe and sacred value are still there. Certainly, with Tegmarkian cosmology (that I tend to hold as axiomatic, in the literal "assumption++" sense), it's very much the "one ultimate nature of reality".
Those are officially my names for those logics now. Any other suggestions?
I don't think I can do funny ones on demand, unfortunately :(
Too much pressure!
I don't see how 3 and 4 are stronger than 1 and 2. They are just the special cases of 1 and 2 where the sentence is a contradiction.
Arbitrarily large finite models are certainly not allowed in the theory of arithmetic.
3 and 4 are generalizations to sets of sentences. But you're right, the generalization is pretty simple.
Arbitrarily large models are allowed in the first-order theory of arithmetic, and no first-order theory of arithmetic can restrict models to only the integers. This is one of the surprising results of compactness.
You said arbitrarily large finite models, however. First-order arithmetic has no finite models. : )
Oh, yeah, that's a typo. Fixed, thanks.
From what I know, Chang & Keisler is a bit dated and can create a wrong perspective on what model theorists are researching nowadays. Maybe you should also look at a modern textbook, like the ones from Hodges, Marker or Poizat.
Which of the 3 would you recommend? Does someone know why MIRI recommends Chang and Keisler if it is somewhat outdated?
Marker is the closest to the state of the art. Hodges is a bit verbose and for beginners. Poizat is a little idiosyncratic (just look at the Introduction!).
I am also interested in the basis of MIRI's recommendation. Perhaps they are not too connected to actual mathematicians studying it, as model theory is pretty much a fringe topic.
Poizat is a little idiosyncratic (just look at the Introduction!)
Here is the last paragraph, which you can read via Google Books:
Indeed, I wrote the outline of this book while wandering across India, so that, in my mind, Henkin's method is inextricably linked to the droves of wild elephants that I met while crawling among the swamp plants of the preserves of Kerala; the elimination of imaginaries, to the gliding of the vultures above the high Himalayan peaks; and the theorem of the bound, to the naked bodies of the Mauryan women that the traveler saw on the bends of a jungle trail, before they had time to cover themselves. I dare hope only that this book will evoke similarly pleasant images in my reader; I wish only that it will be a pleasant companion for you, as it was for me.
The first paragraph of the preface to the English translation (which I found via Amazon's "look inside" feature) says this:
It was written in a dialect of Latin that is spoken as a native language in some parts of Europe, Canada, the U.S.A., the West Indies, and is used as a language of communication between several countries in Africa.
which sounds extremely weird until it transpires that he means it was written in French. (It appears that the book was less well received than the author hoped, at least partly on account of its having been written in French, and he is still cross about it, which I think is why he expresses himself in such a peculiar way: he's both parodying and criticizing the idea that there's something obscure about French.)
So, yeah, "idiosyncratic" is right. As for the mathematical approach, here's another quotation from the Introduction:
[...] I have deliberately chosen an unusual approach to the foundations of logic: The idea that I take as primitive is that of the back-and-forth construction in the style of Fraissé, rather than that of satisfaction of a formula.
(Gosh!)
Why not link to the books or give their ISBNs or something?
There are at least two books on model theory by Hodges: ISBN:9780521587136 and ISBN:9780511551574 | {
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Home / Math / Introduction to variance: Definition, Types, and Calculations
Introduction to variance: Definition, Types, and Calculations
introduction to variance definition types and calculations with blue background
Variance is a very important term in statistics, and in probability theory, it also plays an important role. Variance generally refers to the measurement of the scattered set of numbers or values that are spread out from their average value.
The variance was first introduced in 1918, and Mr. Ronald fisher gave the concept of this term. He wrote the concept of variance in his scientific paper which was published by the royal society of Edinburgh. In this article, we will study the basic definition of variance and its types, there is also an example section.
Definition of variance:
“In probability theory and statistics, variance is the expectation of the squared deviation of a random variable from its population mean or sample mean.” The data values could be sample or population.
Sample:
Sample refers to a random collection of data from a large data (population) i.e. a factory produces 1000 bulbs on daily basis and we want to check the quality of all bulbs then we will take 10 bulbs out of that 1000 (1 from every 100). The chances of error in the sample are more than in the population.
Population:
The population data is the total number of relevant observations. It is typically larger than a sample, and the ratio of errors is very less as compared to the sample data i.e. in the above example, the total number of bulbs (1000) is the population. The population is always greater than the sample.
Types of variance:
There are two basic types of variance:
1. Sample variance
2. Population variance
1. Sample variance:
The variance of sample data is represented by S2 and is given by:
Sample variance calculation formula
• S2 is the variance of the sample data.
• xi is the number of terms.
• is the mean of the sample data.
• n is the total terms.
2. Population variance:
The population variance is represented by σ2 and is given by:
population variance calculation formula
• σ2 is the variance of the population data.
• xi is the number of terms.
• µ is the population mean.
• N is the total terms.
How to calculate the variance of sample data?
There are a few steps to calculate the variance, the steps are as follows.
1. First of all, find the mean of the given sample data i.e.
2. Find the difference between the values and the mean.
3. Find the square of the difference.
4. Sum up the squared values.
5. The last step is to divide the calculated value by “n – 1” (for sample variance) to get the final answer.
How to calculate the variance of population data?
There are a few steps to calculate the variance of population data.
1. First of all, find the mean of the given population data i.e.µ
2. Find the difference between the values and the mean (deviation)
3. Find the square of the difference.
4. Sum up the squared values. (Sum of squares)
5. Divide the calculated value by “N” (for population variance) to get the final answer.
Examples of variance:
Example 1: For Sample Variance.
Calculate the variance of the sample data 12, 45, 23, 86, 43, 97
Solution:
Let x = 12, 45, 23, 86, 43, 97
Total terms = n = 6
Formula:
Step 1: Find the mean of the sample data
Mean = = (x) / n
Mean = = (12 + 45 + 23 + 86 + 43 + 97) / 6
Mean = = 306 / 6
Mean = = 51
Step 2: Find the difference between the values and the mean.
12 – 51 = – 39
45 – 51 = – 6
23 – 51 = – 28
86 – 51 = 35
43 – 51 = – 8
97 – 51 = 46
Step 3: Find the square of the difference.
(– 39)2 = 1521
(– 6)2 = 36
(– 28)2 = 784
(35)2 = 1225
(– 8)2 = 64
(46)2 = 2116
Step 4: Sum up the squared values.
Sum of squares = 1521 + 36 + 784 + 1225 + 64 + 2116
Sum of squares = 5746
Step 5: Divide the calculated value by “n – 1” because we are finding the variance of the sample data
Variance = S2 = (5746) / (6 – 1)
Variance = S2 = (5746) / 5
Variance = S2 = 1149.2
A variance calculator by Allmath is an online resource to get rid of the above lengthy calculations.
example of calculation result
Example 2: For Population Variance.
Calculate the variance of the population data 92, 54, 71, 20, 67, 34, 12
Solution:
Let X = 92, 54, 71, 20, 67, 34, 12
Total terms = N = 7
Formula:
Step 1: Find the mean of the population data µ
Mean =µ = (X) / N
= (92 + 54 + 71 + 20 + 67 + 34 + 12) / 7
= 350 / 7
= 50
Step 2: Find the difference between the values and the mean (deviation)
92 – 50 = 42
54 – 50 = 4
71 – 50 = 21
20 – 50 = – 30
67 – 50 = 17
34 – 50 = – 16
12 – 50 = – 38
Step 3: Find the square of the difference.
(42)2 = 1764
(4)2 = 16
(21)2 = 441
(– 30)2 = 900
(17)2 = 289
(– 16)2 = 256
(– 38)2 = 1444
Step 4: Sum up the squared values. (Sum of squares)
Sum of squares = 1764 + 16 + 441 + 900 + 289 + 256 + 1444
Sum of squares = 5110
Step 5: Divide the calculated value by “N” because we are finding the variance of the population data
Variance = S2 = (5110) / (7)
Variance = S2 = 730
Summary
In this article, we have studied the basic definition of variance, and the general history of this term. We have also read about the types of variance and the difference between the sample and the population. We have also gone through the method of finding the variance.
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1,026,270,914,439,970,200 | Instructions
Study the given information and answer the given question.
Eight people—K, L, M, N, W, X, Y, Z are sitting around a circular table facing the centre, with equal distances between each other but not necessarily in the same order. K is not an immediate neighbour of W and Z. M sits third to the right of W. L is an immediate neighbour of W. Only two people sit between L and X. Y sits to the immediate right of N and only one person sits between K and Y. W is not an immediate neighbour of Z.
Question 170
Who amongst the following sits second to the right of M?
Solution
M sits third to the right of W and L is an immediate neighbour of W, => let L sits to the immediate right of W.
Only two people sit between L and X, => X sits to the immediate right of M.
K is not an immediate neighbour of W and Z, => K sits to the immediate right of X
Y sits to the immediate right of N, => Y sits to the immediate left of W.
K sits 2nd to the right of M.
=> Ans - (B)
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Download App | {
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-8,139,122,223,188,912,000 | Call Us 04 421 6939 or LOCATE CENTER
skyline-mathnasium-uae
Are you a collector? What do you like to collect? Coins? Stuffed animals? Action figures?
In this week’s Word Problem Wednesday Eli is collecting seashells and dividing them up to share with friends. Exercise your math-muscle with Eli in this week’s challenge, and don’t forget to check back tomorrow for the solution!
Eli picks up 288 seashells at the beach. He divides them evenly among 12 boxes. Then he gives 7 boxes of seashells to Lincoln. How many seashells does Lincoln get?
Here’s our solution: When Eli divides the 288 seashells into 12 boxes, he puts 288 ÷ 12 = 24 seashells in each box. Since Lincoln gets 7 boxes, he gets 24 × 7 = 168 seashells.
How does your solution match up with ours?
168 is a whole lot of seashells! What would you do with that many seashells? Do you think any of those shells were big enough for Eli or Lincoln to put to their ears and hear the ocean?
(Photo by en:user:sannse from Wikimedia Commons)
First published By Mathnasium on Apr 5, 2017 | {
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Dilation in Science and Math
Dilation Dilation has been used for millions of years. Even in the ancient times and still we use it until this day. An example of dilation used in ancient times is when ancient Egyptians built the pyramids. The pyramids were built in different sizes, but proportional. Now in this day and time we use dilation in many aspects. Dilation is used in both science and math. In science the microscope shows dilation, without microscopes many of the scientific discoveries wouldn't be possible! In math dilation mainly is used in Geometry to draw figure of different sizes in proportional sizes. In art dilation is used widely for, example architecture, paintings, and statues. In our everyday life we have many examples of dilation like, binoculars, toy cars, little ornaments that represent larger ones in a smaller version. This involves the use of dilations, that is, transformations of the plane that are either contractions or expansions about a point (the center of the dilation), by a constant (positive) ratio. A dilation can either be an expansion (if the ratio is larger than one) or a contraction (if it is smaller than one). Look at the figure below.Construct a point C in the plane, and mark it as the center of dilation. Now draw any polygonal figure, and dilate it about the center C by a fixed ratio (1/2, or 3, or whatever). Drag around this polygon, and observe how the image changes. In particular look at the vertices, their images and the center. Can you see any relation among them? To find the scale factor we have to add one side of both corresponding sides and divide them by the corresponding side of the preimage. For example, side A for the preimage is equal to 5 and side A for the image is equal to 10. Thus, 5+10/5 so the scale factor will equal 3!!!! Preimage Image When I first learned about dilation, I thought that is wasn't important and that there was no use for it at all, but doin this project made my perspective of dilat
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-2,566,402,443,616,304,600 | 0.10096 en tant que fraction
Résultats décimal à fraction pour: 0.10096
Partie entière du nombre entier: empty
Partie décimale à fraction: .10096 = 10096/100000
Fraction simple: 10096/100000
= 5048/50000
= 2524/25000
= 1262/12500
= 631/6250
Faites défiler vers le bas pour personnaliser l'activation du point de précision 0.10096 à décomposer en un nombre spécifique de chiffres. La page comprend également des représentations graphiques 2-3D de 0.10096 en tant que fraction, les différents types de fractions et quel type de fraction 0.10096 est une fois converti.
Numérateur et dénominateur pour 0.10096 en tant que fraction
0.10096 = 0 10096/100000
numerator/denominator = 10096/100000
Niveau de précision pour 0.10096
Le niveau de précision correspond au nombre de chiffres à arrondir. Sélectionnez un point de précision inférieur ci-dessous pour casser la décimale 0.10096 vers le bas plus loin sous forme de fraction. Le point de précision par défaut est 5.
Si le dernier chiffre de fin est "5", vous pouvez utiliser les options "arrondir la moitié vers le haut" et "arrondir la moitié vers le bas" pour arrondir ce chiffre vers le haut ou vers le bas lorsque vous modifiez le point de précision.
Par exemple 0,875 avec un point de précision de 2 arrondi à moitié vers le haut = 88/100, arrondi à moitié vers le bas = 87/100.
select a precision point:
10096/100000
= 5048/50000
= 2524/25000
= 1262/12500
= 631/6250
Représentation graphique de 0.10096 en tant que fraction
Représentation en camembert de la partie fractionnaire de 0.10096
Est 10096/100000 un nombre mixte, entier ou une fraction appropriée?
Un nombre mixte est composé d'un nombre entier (les nombres entiers n'ont pas de partie fractionnaire ou décimale) et d'une partie fractionnaire appropriée (une fraction où le numérateur (le nombre supérieur) est inférieur au dénominateur (le nombre inférieur). Dans ce cas la valeur du nombre entier est empty et la valeur de fraction appropriée est 10096/100000.
Toutes les décimales peuvent-elles être converties en une fraction?
Toutes les décimales ne peuvent pas être converties en fraction. Il existe 3 types de base qui comprennent:
Terminer les décimales ont un nombre limité de chiffres après la virgule décimale.
Exemple: 2587.46 = 2587 46/100
Récurrentes les décimales ont un ou plusieurs nombres répétés après la virgule décimale qui continuent indéfiniment.
Exemple: 3976.3333 = 3976 3333/10000 = 333/1000 = 33/100 = 1/3 (rounded)
Irrationnelles les décimales durent indéfiniment et ne forment jamais un motif répétitif. Ce type de décimal ne peut pas être exprimé sous forme de fraction.
Exemple: 0.393001415.....
Fraction en décimal
Vous pouvez également voir la conversion inverse, c'est-à-dire. comment la fraction 10096/100000 est converti en décimal.
Retour d'information
Convertisseur décimal en fraction
Entrez une valeur décimale:
Conversions décimales courantes en fractions
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8,634,932,801,985,868,000 | Sign up ×
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If $p<1$ and $X$ is a random variable distributed according to the geometric distribution $P(X = k) = p (1-p)^{k-1}$ for all $k\in \mathbb{N}$, then it is easy to show that $E(X) = \frac 1p$, $\mathop{Var}(X)=\frac{1-p}{p^2}$ and $E(X^2) = \frac{2-p}{p^2}$.
Now consider a "conditional" geometric distribution, defined as follows (if there is standard terminology for this, let me know and I'll call it that):
1. Fix a set $J\subset \mathbb{N}$ and a number $\mu>0$ (this will eventually be large).
2. Let $P(X=k) = C \gamma^k$ if $k\in J$ and $P(X=k)=0$ otherwise, where $C>0$ and $\gamma<1$ are chosen so that probabilities sum to $1$ and $E(X) = \mu$.
I'm trying to understand how $E(X^2)$ (or equivalently, $\mathop{Var}(X)$) depends on $J$ and $\mu$. In the case where $J=\mathbb{N}$ the standard results show that $p=\frac 1\mu$ and so $E(X^2) = \mu^2(2-\frac 1\mu)$. I'm interested in the case where $\mu$ becomes very large and would like to obtain a similar estimate $E(X^2) \approx A\mu^2$, for some constant $A>1$, in a more general setting.
The example I'm working with at the moment is $J= \{2^n \mid n\in \mathbb{N}\}$, but ideally I'd like some conditions on the set $J$ that would guarantee an estimate of the above form.
Is there a standard name for these distributions, or a reference where I can read more about them? Are estimates of this form known?
Edit: As Brendan McKay pointed out below, this boils down to understanding the behaviour of the function $g(\gamma) = \sum_{j\in J} \gamma^j$, and in fact the issue that motivated the question I posed can be stated more directly in terms of this function.
The condition $E(X) = \mu$ is equivalent to the equation $\mu = \gamma g'(\gamma) / g(\gamma)$, which determines $\gamma$ implicitly as a function of $\mu$. We would like to understand how $g(\gamma)$ grows as $\mu\to\infty$, and hence $\gamma\to 1$. (In particular, this means we're really interested in the case where $J$ is infinite.)
In the case $J=\mathbb{N}$, one has $g(\gamma) = \frac\gamma{1-\gamma} = 1 - \frac 1{1-\gamma}$, and so $\mu = \gamma (\frac{\gamma}{(1-\gamma)^2}) (\frac{1-\gamma}\gamma) = \frac{\gamma}{1-\gamma}$, so that in fact $g(\gamma(\mu)) = \mu$ and the two quantities go to infinity together.
In the more general case a reasonably simple argument shows that $\lim_{\mu\to\infty} g(\gamma(\mu)) = \infty$ provided $J$ is infinite, but it's not at all clear to me how the rate at which $g$ grows (in terms of $\mu$) depends on $J$ for more general sets.
That's the original motivation -- after some messing around we decided that we could figure out the growth rate if we knew something about $E(X^2)$ as suggested above, and since it was phrased in terms of what seemed to be a reasonably natural probability distribution, we decided to ask it in that form. But now you have the whole story...
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3 Answers 3
up vote 3 down vote accepted
I'm not sure what you really want but here is a couple of simple minded inequalities that can serve as a baseline.
Below $g=\sum_{k\in J}\gamma^k$, $M=\sum_{k\in J}k\gamma^k$, so $\mu=\frac Mg$. We'll need the counting function $F(n)=\#\{k\in G: k\le n\}$ of the set $J$. I will assume that $F$ is extended as a continuous increasing function to the set $[1,+\infty)$ and that $g\ge 1$.
1) For every $N$, we have the trivial estimate $g\le F(N)+\frac MN$. Taking $N=2\mu$, we get $g\le F(2\mu)+\frac g2$, i.e.,
$$ g\le 2F(2\mu) $$
2) Let $\nu$ satisfy $F(\nu)=3g$. Since $g\ge F(\nu)\gamma^\nu$, we conclude that $\gamma^\nu\le \frac 13$ so $1-\gamma>\frac 1\nu$. Now, for every $N$, we have $$ M\le Ng+(N+\frac 1{1-\gamma})\frac 1{1-\gamma}\gamma^N\le Ng+(N+\nu)\nu e^{-N/\nu}\. $$ Since we clearly have $\nu\ge F(\nu)=3g$, we can choose $N=\nu\log\frac\nu g\ge \nu$. For this choice, the second term on the right is at most $2Ng$, so, dividing by $g$ we get $\mu\le 3N$, i.e., $$ \mu\le 3F^{-1}(3g)\log\frac{F^{-1}(3g)}{g} $$
Examples of what these inequalities yield:
1) Dense set ($F(n)\approx n$). Then $g\approx\mu$
2) Power lacunarity ($F(n)\approx n^p$, $0<p<1$). Then $g$ is between $\mu^p(\log\mu)^{-p}$ and $\mu^p$ up to a constant factor.
3) Geometric lacunarity ($F(n)\approx\log n$). Then $g\approx \log\mu$.
As you see, one can lose a logarithm sometimes but the advantage is that I do not make any regularity assumptions here. Of course, if $F$ is regular enough, you can, probably, do a bit better.
share|cite|improve this answer
I like these estimates -- this is the sort of thing I was looking for. The truth is that I didn't have a particularly clear idea of exactly what I wanted when I asked the question, which is why it never really came out as clearly as I'd have liked. It came up in some work a colleague and I are doing, where we started by maximising the entropy of a probability distribution on $\mathbb{N}$ with a fixed mean -- which led to the geometric distribution -- and then wanted to consider the case where the support of the distribution was forced to lie in $J$. (ctd...) – Vaughn Climenhaga Nov 28 '11 at 5:45
(ctd...) In order to get the sorts of estimates we wanted for the application we had in mind, we thought we'd need some more detailed information about the relationship between $g$ and $\mu$ in terms of the structure of $J$, and so I asked this question in a rather vague and open-ended attempt to see what might be true. In the end we found another way to deal with the issue we were faced with, that doesn't require dealing with conditional geometric distributions, but I still find this question interesting for its own sake. – Vaughn Climenhaga Nov 28 '11 at 5:48
Define $g(\gamma) = \sum_{j\in J} \gamma^j$. The condition $E(X^2)\sim A\mu^2$ as $\mu\to\infty$ seems to be equivalent to $$ \frac{g(\gamma) g''(\gamma)}{(g'(\gamma))^2} \to A $$ as $\gamma\to 1$ from below. Alternatively define $h(x)=\sum_{j\in J} ~e^{-jx}$ and then you want $$ \frac{h(x)h''(x)}{(h'(x))^2} \to A$$ as $x\to 0$ from above.
Of course these are translations of the problem rather than solutions, but I mention them as someone will probably see what to do next.
share|cite|improve this answer
If $J$ is finite then this is obviously a closed-form solution. Otherwise the answer depend heavily on what form $J$ is in. e.g. if $J$ is not a decidable set then very few digits of $A$ should be computable. For any set whose generating function has a nice closed form, there will be a nice formula for $A$. It seems like the only fully general question left is whether $A$ is always defined. – Will Sawin Nov 18 '11 at 7:41
@Brendan: I seem to recall seeing one or two equations like this as we derived the question that I posed from the question that originally motivated it... I'll edit the original question to include the motivation as well. Certainly we'd be very happy to understand the limits you point out, and that would suffice... – Vaughn Climenhaga Nov 18 '11 at 18:05
@Will: We're mostly interested in what happens when $J$ is infinite, and in particular in quantifying the behaviour under some conditions on (say) the growth rate of the gaps in $J$, or something like that. (If $J$ has bounded gap size then this should be more or less comparable to the case $J=\mathbb{N}$.) – Vaughn Climenhaga Nov 18 '11 at 18:07
As a consequence, in the case of a $J$ obtained repeating a finite set $F\subset [0,n)$ with periodicity $n$, i.e. $J=F+n\mathbb{N}$, we have $g(x)=(1-x^n)^{-1}P(x)$ with $P(x):=\sum_ {k\in F} x^k$; so for $x\to1$, $g'(x)=nx^{n-1}(1-x^n)^{-2}P(x)+O((1-x)^{-1})$ and $g''(x)=n^2x^{2n-2}(1-x^n)^{-3}P(x)+O((1-x)^{-2})$, whence by Brendan's formula $g''g/(g')^2\to 2$ as $x\to 1$. So every periodic $J$ has $A=2$. – Pietro Majer Nov 19 '11 at 21:16
Can you estimate $C$ and $\mu$ etc...using the first term say $j=\min J$?
It seems to me for instance that $1/C=\gamma^j+\gamma^{j_2}+\cdots\leq \sum_{k=j}^\infty \gamma^k=\gamma^j/(1-\gamma)$.
share|cite|improve this answer
The issue is that we're really interested in an estimate from the other direction, of the form $1/C \geq$ some function of $\gamma$; in other words, we want a lower bound on how $E(X^2)$ grows in terms of $E(X)$, which is going to depend much more on the behaviour of the tail of $J$ then on the initial terms. – Vaughn Climenhaga Nov 18 '11 at 18:02
Thanks. There's a typo in our edit. It's $g(\gamma)=\frac{1}{1-\gamma}-1$ – Pietro Poggi-Corradini Nov 19 '11 at 1:41
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-8,861,259,540,262,751,000 | Ladybugs/Ladybirds make for Math/Maths Book Magic
Aren’t ladybugs magical? Did you know that these adorable polka-dotted crimson creatures are only called ladybugs in the U.S.. In the U.K., and other English speaking countries, they’re called ladybirds. Ben Orlin, wrote and sketched some math-related differences between U.K. and U.S. here on his awesome blog, Math with Bad Drawings. Here’s an example of a U.S. and U.K difference from Orlin’s post.
20150514160636_00010
Drawing by Ben Orlin @ Mathwithbaddrawings.com
What the heck does math/s have to do with ladybugs? Well, this week’s magical math/maths book has ladybirds/ladybugs galore.
The Book
Written and illustrated by Alison Limentani, this picture book was published in 2016.
51svg0qvu-L._SX496_BO1,204,203,200_
The simple text calls for some powerful mathematical thinking. The book begins with the statement:
“10 ants weigh the same as 1 ladybug.”
“9 lady bugs weigh the same as 1 grasshopper.”
The book continues in a similar way comparing the weights of the previous animal with the next, counting down from 10 ants, 9 ladybugs, 8 grasshoppers, all the way to 1 swan.
At the end of the book, the ladybugs/birds make a bountiful return when compared with the weight of one swan. Limentani’s vibrant and endearing illustrations on a bright blue backdrop make sure the multiplicative relationships in this book take center stage.
The Math
Since the number of animals varies from 10 to 1, there are opportunities for young ones to count. However the main mathematical idea in this book is multiplicative reasoning, and more generally proportional reasoning.
The essence of proportional reasoning is the consideration of a number in relative terms, rather than absolute terms. For example, 1 ladybug weighs the same as 10 ants. 9 ladybugs weighs the same as 1 grasshopper. And the related implicit question, how many ants does it take to equal the weight of one grasshopper?
Here is a nice summary document outlining the concept of Proportional Reasoning from the Ontario Ministry of Education (2012). Below is a web of interconnected Proportional Reasoning Concepts from the document. Note the Scaling concept from last week’s post. A key component of proportional reasoning is to reason multiplicatively instead of additively.
In addition to comparing weights, the end pages provide an opportunity to discuss quotative division with decimals. Here is a post by Chris Hunter about using Limentani’s book as a jumping off point for quotative division.
img_8753-e1511808452736.jpg
Finally, the last page shown above gives the weights of each animal which connects to systems of measurements. Note the use of ounces with decimals. 3.2 oz? Ugh! [ Why don’t we use metric in the U.S. again? Perhaps it has something to do with Pirates?]
The Magic
The book afforded skip-counting opportunities for my daughter. She used a counting by ten strategy that she is working on (she is in kindergarten) and was able to get 90 ladybugs weigh the same as 1 grasshopper.
Towards the beginning of the book, my son surprised me with his multiplicative reasoning and creative mental math skills. As the number got larger, he listened more and added aloud less. At the end, I challenged him to figure out how there came to be 362, 880 ladybugs. I told him: “This is a big problem. So you’ll need to take your time.”
Less than 5 minutes later he came back in the kitchen asking: “Can use a calculator?”
I paused, “Where are you?” He explained how he got 72 ladybugs in 1 stickleback fish. Using addition similar to this: 9+9=18 AND 18+18=36 AND 36+18=54 AND 54+18=72.
After seeing his work, I thought again about the calculator. It will make it more fun for him. Realizing that, I agreed. And that is where it took a magical turn.
Here is his work.
IMG_8735
He described adding on the calculator, until here (pointing to 15,120): “That’s when I started multiplying. ”
Surprised I asked: How did you know to multiply?
He replied: “It is faster. It was likes groups of.” [Using a repeated addition connection to multiplication.]
Then, as if that wasn’t enough, he started analyzing that last page on his own.
img_8753-e1511808452736.jpg
“That’s interesting.” He remarked, noticing a pattern on righthand page. We discussed and discovered much to chat about regarding decimal numbers, patterns and relationships (also that lb means pounds). For us, the math/maths magic in How much does a ladybug weigh? was immeasurable. Making it is #21 on our list list of magical math books.
Have a magical math book you’d like share? Please go to the Shared booklist to find out how. If you’d like to receive these magical math book posts each Monday, be sure to follow this blog in the side bar of this page.
Thanks and see you next Monday! #mathbookmagic
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-5,896,437,585,948,402,000 | Graph of Cubic Functions/Cubic Equations for zeros and roots (-11,-20,-2)
Let us consider the cubic function f(x) = (x+11)(x+20)(x+2) = x3 + 33x2 + 282x + 440.
We will inspect the graph, the zeroes, the turning and inflection points in the cubic curve curve y = f(x).
Cubic Polynomials and Equations
A cubic polynomial is a polynomial of degree 3.
where a is nonzero.
An equation involving a cubic polynomial is called a cubic equation and is of the form f(x) = 0. There is also a closed-form solution known as the cubic formula which exists for the solutions of an arbitrary cubic equation. A cubic polynomial is represented by a function of the form. And f(x) = 0 is a cubic equation. The points at which this curve cuts the X-axis are the roots of the equation.
Graph of y = f(x) = (x+11)(x+20)(x+2) = x3 + 33x2 + 282x + 440
Cubic Polynomial Curve Plot on Graph
A few computed points on the curve, apart from the zero(s) which are known:
(-21,-190), (-18,224), (-15,260), (-12,80), (-9,-154), (-6,-280), (-3,-136), (0,440), (3,1610)
Characteristics of the Graph Plot, Curve Sketching of Cubic Curves
If the given cubic function is: f(x) = ax3 + bx2 + cx + d
The derivative of this function is: f'(x) = 3ax2 + 2bx + c
The function given to us us f(x) = (x+11)(x+20)(x+2) = x3 + 33x2 + 282x + 440
And the derivative for this is f'(x) = 3x2 + 66x + 1
Consider the cubic equation f(x) = (x+11)(x+20)(x+2) = x3 + 33x2 + 282x + 440 = 0
The roots of this cubic equation are at:
(x - (-11)) = 0 => x = -11,
OR (x - (-20)) = 0 => x = -20,
OR (x - (-2)) = 0 => x =-2
This cubic equation has real and unique roots at -11, -20, -2.
The plotted curve cuts the x-axis at these values of x: i.e, these are the zeroes of the given cubic polynomial. Note, how there is a turning point between each consecutive pair of roots. This is consistent with what one would expect from the Rolle's Theorem which states that if a function f (x) is derivable in an interval (a, b) and continous in the interval [a,b] and also f (a) = f (b), then there exists atleast one value c of x lying within (a, b) such that f'(c) = 0: that is, there exists a point where the first derivative (the slope of the tangent line to the graph of the function) is zero.
The turning or stationary points is where f'(x) = 0 => 3x2 + 66x + 1 = 0 => x = -16.2, x = -5.8
These are also called the "critical" points where the derivative is zero.
Coming to other geometrical features of this curve: What we see here is the graph of a nonlinear function. The y-intercept of this curve is at y=440.
And the second derivative of this curve becomes zero at x = -11.0. At this point the curve changes concavity. A cubic curve has point symmetry around the point of inflection or inflexion. These are just some of the important features and aspects to keep in mind while trying to visualize and analyze a plot of an algebraic function. A graphical and visual inspection helps in several ways.
The zeroes of a polynomial, if they are known, and the coefficients of that polynomial are two different sets of numbers that have interesting relations. If we know the zeroes, then we can write down algebraic expressions for the coefficients. Going the other way is much harder and cannot be done in general.
A cubic function has a bit more variety in its shape than the quadratic polynomials which are always parabolas. We can get a lot of information from the factorization of a cubic function. We get a fairly generic cubic shape when we have three distinct linear factors
Compute the Area between the given curve and the X-Axis
Here are some examples of computing the area under curves or between a given curve and the X-Axis
We'll use integral calculus and definite integrals here. Skip this part if you haven't yet started integral calculus.
F(x) = Integral of f(x) = ∫ ( x3 + 33x2 + 282x + 440) dx = x4/3 + 33x3/2 + 282x2 + 440x + C
Where C is the constant of integration.
Values of F at each of the roots -20,-11,-2:
F(-20) = (-20)4/3 + (33 * (-20)3)/2 + 282*(-20)2 + 440*(-20) + C = 25333.33 + C
Similarly, we can compute: F(-11) = 12200.83 + C and F(-2) = 121.33 + C
Where C is the constant of integration.
Area between the curve and the X-axis = |Area enclosed for x in interval (-20,-11)| + |Area enclosed for x in interval (-11,-2)|
= |Area enclosed for x in interval (-20,-11)| + |Area enclosed for x in interval (-11,-2)|
= |Definite integral of f(x) between limits -20 and -11| + |Definite integral of f(x) between limits -11 and -2|
= |F(-11) - F(-20)| + |F(-2) - F(-11)|
= |12200.83 + C - (25333.33 + C)| + |121.33 + C - (12200.83 + C)|
Terms involving the constant of integration cancel out.
Area = |-13132.5| + |-12079.5|=> Area = -1053.0 Square Units
Check the plot of another cubic curve here with roots at -8, -20, -2
Here's another cubic curve here with roots at -11, -10, -2
Here's another cubic curve here with roots at -11, -20, 4
Many of these concepts are a part of the Grade 9,10,11,12 (High School) Mathematics syllabus of the UK GCSE/GCE curriculum, Common Core Standards in the US, ICSE/CBSE/SSC/NTSE syllabus in Indian high schools. You may check out our free and printable worksheets for Common Core and GCSE.
ą
Prashant Bhattacharji,
Mar 21, 2017, 1:33 AM
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-4,337,422,030,861,731,000 | Waste less time on Facebook — follow Brilliant.
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Derivative of The Sine Function
In my last note I proved that \( \quad\displaystyle \lim_{\theta\to 0}\frac { \theta}{sin\theta} = 1\)
If you have not read that I would recommend reading that first as this limit is used in proving the derivative of the sine function. Here is the link.
\( \qquad \qquad\qquad\qquad\qquad\qquad \displaystyle \lim_{\theta\to 0}\frac { \theta}{sin\theta} = ?\)
To prove the derivative of the sine function we will be using the first principles of derivatives.
\(\qquad\qquad\qquad\qquad\qquad\displaystyle \lim_{\Delta\theta\to 0}\frac{\sin(\theta + \Delta\theta) - \sin\theta}{\Delta\theta}\)
Use the addition formula for \(\sin(\theta + \Delta\theta)\).
\(=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta \times \cos \Delta\theta + \sin\Delta\theta \times \cos\theta - \sin\theta}{\Delta\theta}\)
rearrange the equation and factor out \(\sin\theta\)
\(=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta \times \cos \Delta\theta - \sin\theta + \sin\Delta\theta \times \cos\theta }{\Delta\theta}\)
\(=\qquad\qquad\qquad\quad\displaystyle\lim_{\Delta\theta\to 0} \frac{\sin\theta( \cos \Delta\theta - 1) + \sin\Delta\theta \times \cos\theta }{\Delta\theta}\)
break apart the fraction
\(=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\frac{\sin\theta( \cos \Delta\theta - 1)}{\Delta\theta} + \frac{\sin\Delta\theta \times \cos\theta }{\Delta\theta}\)
\(=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta} + \frac{sin\Delta\theta}{\Delta\theta} \times cos\theta\)
We wIll now evaluate the two limits seperately
\(\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} \frac{sin\Delta\theta}{\Delta\theta} \times cos\theta\)
In my last note I proved that \(\displaystyle\lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta}=1\)
\(=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} 1 \times cos\theta\)
\(=\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0} cos\theta\)
That solves that limit now the other limit.
\(\qquad\qquad\qquad\quad\displaystyle \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta} \)
divide out the \(\sin\theta\)
\(=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta} \)
multiply by the conjugate
\(=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos \Delta\theta - 1)}{\Delta\theta} \times \frac{\cos\Delta\theta+1}{\cos\Delta\theta+1}\)
\(=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{( \cos^{2} \Delta\theta - 1)}{\Delta\theta(\cos\Delta\theta+1)} \)
Rewrite the the equation using the Pythagorean identity. \( \cos^{2}\theta + \sin^{2}\theta=1 \)
\(=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} \frac{-\sin^{2} \Delta\theta }{\Delta\theta(\cos\Delta\theta+1)} \)
break apart the fraction
\(=\qquad\qquad\qquad\quad\displaystyle \sin\theta \lim_{\Delta\theta\to 0} (-sin\Delta\theta) \times \frac{\sin\Delta\theta}{\Delta\theta} \times \frac{1}{\cos\Delta\theta + 1}\)
evaluate the limits
\(=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times (-sin0) \times 1 \times \frac{1}{\cos0 + 1}\)
\(=\qquad\qquad\qquad\quad\displaystyle \sin\theta \times 0 \times 1 \times \frac{1}{2}\)
\(=\qquad\qquad\qquad\quad\displaystyle 0\)
Now add both limts.
\( \displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad \lim_{\Delta\theta\to 0}\sin\theta \times \frac{( \cos \Delta\theta - 1)}{\Delta\theta} + \lim_{\Delta\theta\to 0}\frac{sin\Delta\theta}{\Delta\theta} \times cos\theta\)
\( \displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad0 + \cos\theta\)
\( \displaystyle \frac{d}{d\theta}=\qquad\qquad\qquad\quad \cos\theta\)
Note by Brody Acquilano
2 years, 4 months ago
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$\begingroup$
I've just had to do a homework on binomial expansion for approximation:
$1.07^9$
so:
$(1+0.07)^9$
To do binomial expansion you need a calculator for the combinations button (nCr), so why would use a more complicated method, which only gives an approximation be used over just typing 1.07^9 into a calculator?
(or is this never done in real life, and it's just a homework?)
$\endgroup$
4
• 2
$\begingroup$ We might need the full problem to determine the context, but I believe the point is that you get a good approximation by only taking the first few terms of the binomial expansion. This works because higher powers of $.07$ are small. E.g., adding the first $3$ of $10$ terms gives 1.8064, compared to the correct result of about 1.83845921. You don't need a calculator to find 9Cr, especially when r=0,1,2. $\endgroup$ Jan 31, 2011 at 21:36
• $\begingroup$ That is about as much context as there is. in nCr, if r = 0 then nCr = 1, and if r = 1, then nCr = n, but when you get 2 it's impossible to learn off by heart, for all the possible values of n. eg 9C2 = 36. So you've used a calculator to find 9C2, and your about 0.03 out, whereas you could have just done 1.07^9? $\endgroup$
– Jonathan.
Jan 31, 2011 at 22:03
• $\begingroup$ nC2 = n(n-1)/2 is probably worth knowing. If you don't want to memorize it (I'm with you there), keep in mind that nCr tells you how many subsets of size r there are in a set of size n. For nC2, there are n choices for the first element, (n-1) choices for the second element, and you divide by 2 because you've just counted each set twice. So for example 9C2=9*8/2 = 9*4=36. Since either n or n-1 is even, it can usually be computed quickly by hand. On the other hand, 1.07^9 would be quite tedious by hand. $\endgroup$ Jan 31, 2011 at 23:12
• $\begingroup$ Another way to find $nCr$ by hand (if $r$ is not too large) is to write down Pascal's triangle row by row just by adding (each number is the sum of the two above it). en.wikipedia.org/wiki/Pascal%27s_triangle $\endgroup$ Feb 1, 2011 at 4:40
1 Answer 1
6
$\begingroup$
Expanding the whole thing using Binomial Theorem gives you an exact value. Not an approximation.
To get an approximation you can consider a few terms from the expansion.
For instance, for "small" $x$, $1+nx$ is a "reasonable" approximation for $(1+x)^n$.
Notice that this corresponds to picking the first two terms from the binomial theorem expansion $(1+x)^n = 1 + \binom{n}{1} \ x + \binom{n}{2}\ x^2 + \dots + x^n$.
For example
$1.0007^9 \approx 1 + 9\times 0.0007 = 1.0063$ which agrees with $1.0007^9 = 1.0063176688422737867054812736724$ upto $4$ decimal places.
Depending on how accurate you want it, you could consider more terms from the binomial expansion.
This is based on the fact that for small $x$, as the power $r$ of $x$ gets larger, the term $x^r$ becomes small quite fast.
$\endgroup$
2
• $\begingroup$ To expand the whole thing using Binomial Theorum, you need a calculator. You might as well use the power button and get the answer a lot quicker. If you didn't have a calculator, then it would probably be quicker to go and buy one. $\endgroup$
– Jonathan.
Jan 31, 2011 at 22:05
• 1
$\begingroup$ @jonathan: I think you missed the point. We are talking about an approximation. The expand the whole thing gives you an exact value. To get an approximation you don't have to expand the whole thing! Notice that $1+nx$ are the first two terms in the expansion of $(1+x)^n$ $\endgroup$
– Aryabhata
Jan 31, 2011 at 22:28
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Ëèíåéíîå óïîðÿäî÷åíèå
Ðàññóæäåíèÿ – ýòî, â ñóùíîñòè, ïðàâèëüíî âûñòðîåííûå ôàêòû.
Á Ô Àíäåðñîí (Anderson  F, 1980, ð. 62)
Äæîýëü ñèëüíåå Áèëëà, íî ñëàáåå Ðè÷àðäà. Ðè÷àðä ñèëüíåå Äæîýëÿ, íî ñëàáåå Äîíàëüäà. Êòî èç íèõ ñàìûé ñèëüíûé, à êòî – íà âòîðîì ìåñòå ïî ñèëå?
Õîòÿ ÿ óâåðåíà â òîì, ÷òî âû íèêîãäà â æèçíè íå âñòðå÷àëèñü ñ Äæîýëåì, Äîíàëüäîì, Ðè÷àðäîì è Áèëëîì, ÿ óáåæäåíà, ÷òî âû ñìîæåòå îòâåòèòü íà ìîé âîïðîñ Ïîñûëêè èëè óòâåðæäåíèÿ â ýòîé çàäà÷å ñîäåðæàò èíôîðìàöèþ îá óïîðÿäî÷åííûõ ñâÿçÿõ ìåæäó òåðìèíàìè, ïîýòîìó òàêîé òèï çàäà÷ íàçûâàþò ëèíåéíûì óïîðÿäî÷åíèåì, èëè ëèíåéíûì ñèëëîãèçìîì. Êàê è âî âñåõ çàäà÷àõ íà äåäóêòèâíûå ðàññóæäåíèÿ, ïîñûëêè ñëóæàò îñíîâîé äëÿ âûâîäà âàëèäíîãî çàêëþ÷åíèÿ – çàêëþ÷åíèÿ, èñòèííîãî ïðè óñëîâèè âåðíîñòè ïîñûëîê.  çàäà÷àõ ñ ëèíåéíîé ñòðóêòóðîé ìû ñòàëêèâàåìñÿ ñ óïîðÿäî÷åííûìè ñâÿçÿìè, â êîòîðûõ îòíîøåíèÿ ìåæäó òåðìèíàìè ìîæíî ïðåäñòàâèòü â âèäå ïðîñòðàíñòâåííîãî ðÿäà.
Ëèíåéíûå ñõåìû
Êàê âû ðåøàëè çàäà÷ó ïðî Äæîýëÿ, Äîíàëüäà, Ðè÷àðäà è Áèëëà? Áîëüøèíñòâî ëþäåé ðåøàåò òàêèå çàäà÷è ïîýòàïíî, ðàññòàâëÿÿ ëþäåé ñîãëàñíî óñëîâèÿì:
Óñëîâèå «Äæîýëü ñèëüíåå Áèëëà, íî ñëàáåå Ðè÷àðäà» ïðåîáðàçóåòñÿ â ñëåäóþùóþ ñõåìó:
Óñëîâèå «Ðè÷àðä ñèëüíåå Äæîýëÿ, íî ñëàáåå Äîíàëüäà» óêàçûâàåò íà òî, ÷òî â ñàìóþ âåðõíþþ ñòðîêó ñõåìû íàäî ïîìåñòèòü Äîíàëüäà:
Òàêèì îáðàçîì, ëåãêî «óâèäåòü», ÷òî Äîíàëüä – ñàìûé ñèëüíûé, à Ðè÷àðä íà âòîðîì ìåñòå. Èçó÷åíèå ëèíåéíûõ ñèëëîãèçìîâ ïîêàçàëî, ÷òî ïðè îòâåòå íà âîïðîñ ëþäè, ïî êðàéíåé ìåðå ÷àñòè÷íî, ïîëàãàþòñÿ íà ïðîñòðàíñòâåííîå âîîáðàæåíèå èëè êàêîãî-ëèáî ðîäà ïðîñòðàíñòâåííîå ïðåäñòàâëåíèå çàäà÷è.
Ïîðàáîòàéòå íàä ïðèâåäåííûìè íèæå ïàðàìè ëèíåéíûõ ñèëëîãèçìîâ. Ïîïðîáóéòå îïðåäåëèòü, êàêîé èç ñèëëîãèçìîâ â êàæäîé ïàðå ðåøèòü ëåã÷å.
1. à) Äæóëèî óìíåå, ÷åì Äèàíà.
Äèàíà óìíåå, ÷åì Ýëëåí.
Êòî èç íèõ ñàìûé óìíûé? Äæóëèî, Äèàíà, Ýëëåí èëè ýòî íåèçâåñòíî?
ÈËÈ
á) Äæîàíí âûøå ðîñòîì, ÷åì Ñüþçåí. Ðåáåêêà âûøå ðîñòîì, ÷åì Äæîàíí. Êòî íèæå âñåõ ðîñòîì? Äæîàíí, Ñüþçåí, Ðåáåêêà èëè ýòî íåèçâåñòíî?
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-3,586,883,206,359,015,000 | Try NerdPal! Our new app on iOS and Android
Find the derivative of $3\sin\left(x-2\right)sinh\left(x\right)$
Step-by-step Solution
Go!
Go!
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Final Answer
$3\left(\cos\left(x-2\right)sinh\left(x\right)+\sin\left(x-2\right)cosh\left(x\right)\right)$
Got another answer? Verify it here!
Step-by-step Solution
Problem to solve:
$\frac{d}{dx}\left(3\sin\left(x-2\right)\cdot sinh\left(x\right)\right)$
Specify the solving method
1
The derivative of a function multiplied by a constant ($3$) is equal to the constant times the derivative of the function
$3\frac{d}{dx}\left(\sin\left(x-2\right)sinh\left(x\right)\right)$
Learn how to solve differential calculus problems step by step online.
$3\frac{d}{dx}\left(\sin\left(x-2\right)sinh\left(x\right)\right)$
Unlock the first 3 steps of this solution!
Learn how to solve differential calculus problems step by step online. Find the derivative of 3sin(x-2)sinh(x). The derivative of a function multiplied by a constant (3) is equal to the constant times the derivative of the function. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=\sin\left(x-2\right) and g=sinh\left(x\right). The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if {f(x) = \sin(x)}, then {f'(x) = \cos(x)\cdot D_x(x)}. The derivative of a sum of two or more functions is the sum of the derivatives of each function.
Final Answer
$3\left(\cos\left(x-2\right)sinh\left(x\right)+\sin\left(x-2\right)cosh\left(x\right)\right)$
SnapXam A2
Answer Assistant
beta
Got another answer? Verify it!
Go!
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acsch
Useful tips on how to improve your answer:
$\frac{d}{dx}\left(3\sin\left(x-2\right)\cdot sinh\left(x\right)\right)$
Main topic:
Differential Calculus
Used formulas:
6. See formulas
Time to solve it:
~ 0.07 s | {
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A certain characteristic in a large population has a [#permalink] New post 04 May 2006, 09:56
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A certain characteristic in a large population has a distribution that is symmetric about the mean M. If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distribution is less than M+D?
a. 16%
b. 32%
c. 48%
d. 84%
e. 92%
Please show your work.
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Re: PS: percent [#permalink] New post 04 May 2006, 10:00
kuristar wrote:
A certain characteristic in a large population has a distribution that is symmetric about the mean M. If 68 percent of the distribution lies within one standard deviation D of the mean, what percent of the distribution is less than M+D?
a. 16%
b. 32%
c. 48%
d. 84%
e. 92%
Please show your work.
using standard normal distribution that would be 84%
as it is expected that 68% of the population is within 1 standard deviation, and if is symetric 34% would be above the mean that is 0.5
so add 34%+50% = 84%
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[#permalink] New post 04 May 2006, 10:07
should be 84 %
since it is symmetric about mean ... m 50 %
between m-d and m+d it has 68 % so d =34 %
so 50 +34 =84 % below m + d
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[#permalink] New post 04 May 2006, 15:10
This one was discussed recently, answer is 84%
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[#permalink] 04 May 2006, 15:10
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} | 0c090d63199a0a01e3b08e4a255778a0 |
3,885,945,968,266,102,300 | Fourier analysis, also known as spectral analysis, encompasses all sorts of Fourier expansions, including Fourier series, Fourier transform and the discrete Fourier transform (and relatives). The non-commutative analog is (representation-theory).
learn more… | top users | synonyms
1
vote
0answers
28 views
Proof by construction of Wiener's tauberian theorem on $\mathbb{R}^n$
Would anyone have a reference to a proof by construction of Wiener's tauberian theorem in $\mathbb{R}^n$? As a reminder the theorem goes as follows: Theorem (Rudin, Functional Analysis Th. 9.5): ...
1
vote
0answers
37 views
Fourier transformation of the symmetric group $S_3$
I am trying to compute the Fourier transformation of the symmetric group $S_3$ following the section 4 of Quantum Computing and the Hunt for Hidden Symmetry. The multiplication table of $S_3$ is as ...
1
vote
0answers
35 views
an application of the Poisson summation formula
It is written in a paper that I was reading that "by an application of the Poisson summation formula" we have $\sum_{n \ne 0} |n|^{-1} e^{inx} = C \ln |x| + \phi(x)$ with some smooth function ...
1
vote
0answers
23 views
Fourier transform of $e^{-\delta (x)} \cos\left(\frac{1}{x}\right)$
I have a function $$ f(\omega) = \exp\left(-\frac{\gamma}{\gamma^2+\omega^2}\right)\cos\left(\frac{\omega}{\gamma^2+\omega^2}\right), $$ and I'm trying to calculate its Fourier transform at the ...
1
vote
0answers
22 views
Calculate Norm Operator
I'm trying to solve this exercice: Let $\omega(y)=y^{-4}$ and $L^{1}(\mathbb{R},\omega)$ the space of measurable functions $g:\mathbb{R}\rightarrow\mathbb{R}$ so that $g\omega$ is Lebesgue ...
1
vote
0answers
15 views
Does inverse of all Fouriers transforms have a corresponding function in time domain?
I am trying to cancel out the following transfer function of a system: $$\frac{( 1 - e^{(i*k*T)} ) }{ (i*k)}$$ I thought it would work if I find the inverse Fourier transform of $$\frac{ (i*k)}{( 1 ...
1
vote
0answers
23 views
Kernel of Helmholtz Equation on a plane
On the $z=0$ plane I have the boundary conditions $V=\delta(x)\delta(y)$ I want to solve for $z>0$. Helmholtz equation is $\nabla ^2 V +k^2 V=0$ I though that spherical harmonics would be useful. ...
1
vote
0answers
36 views
What is the Fourier transform of $1/|x|$?
I looked it up in several tables and calculated it in Mathematica and Matlab. Some tables say that the answer is simply $$\frac{1}{|\omega|}$$ and in other table it is ...
1
vote
0answers
47 views
Fourier transform of $1/x$, $x>0$
I'm new to Fourier transform. From Does the Fourier Transform exist for f(t) = 1/t? ,I saw that the Fourier transform of $1/x$ is $\text{sgn}(x)$. Define $$f(x)=\begin{cases} \frac{1}{x} ...
1
vote
0answers
37 views
Why does inputting complex exponentials into a system give its frequency response?
Let's say I have an FIR filter with the equation: $$ y[n] = \sum_{i=0}^{N-1} h[i] x[n-i] $$ I know that to find the frequency response of this filter, I need to input a complex exponential in place ...
1
vote
0answers
34 views
Can someone explain negative frequencies when doing the Fourier transform?
I apologize if this question has been asked before. I have looked and have not found a clear explanation. When doing the discrete Fourier transform (e.g. fft in MATLAB) for a vector of discrete time ...
1
vote
0answers
31 views
Fourier sums convergence
I am given the Sin series expansion: (1)$$x=2\sum_{1}^{\infty}\frac{(-1)^{n+1}}{n}Sin(nx)$$ But how do I deal with a solution given by: ...
1
vote
0answers
27 views
Littlewood-Paley theorem at endpoints
Littlewood-Paley theorem says that the $L^p$ norm of the square function associated with $f$ is equivalent to the $L^p$ of $f$ when $p\in (1,\infty)$. I'm interested in the endpoint case. Why it ...
1
vote
0answers
22 views
prove that if $f(x)$ is real, then $\left | F(s) \right |^2$ is an even function
I understand that the fourier transform of f(x) can be broken up into odd and even parts such that the transform of $f(x)$ can be represented by $$F(s) = 2\int_{0}^{\infty}E(x)cos(2\pi x s)dx - ...
1
vote
0answers
44 views
Numerical integration with FFT
Suppose I am faced with the following integral: \begin{equation} F(\bf{x}) = \int \frac{d^{2}\bf{k}}{(2\pi)^{2}} exp(-i\bf{k}\cdot \bf{x})f(\bf{k}) \end{equation} where $f$ is some (known) function ...
1
vote
0answers
33 views
Computing Fourier transform of $e^{-ax^2 - b|x|}$
recently I've been working on some estimation problems in $L_2$ space and came across the problem of computing the following Fourier transform for constants $a,b>0$: $$ F(w) := ...
1
vote
0answers
49 views
Expanding Fourier Series of $f(x)=x^2$ where $0<x<1$ (even and odd)
I tried to solve Fourier series (which appeared on title) and ended up to below solution : on even state : $ \phi(x)= \begin{cases} x^2 & 0<x<1 \\ x^2 & -1<x<0 \end{cases} $ ...
1
vote
0answers
18 views
What is $ \mathcal S ( \mathbb R_+ )$?
Does Schartz space of functions defined on positive halfline exists in a meaningful way? How does such functions look like and do you know any applications of such space?
1
vote
0answers
43 views
Fourier transform step-by-step example: $f(x) = 1/2$ where $x\in[0,1]$ otherwise $f(x) = 0$
I'm trying to understand the general procedure for finding the fourier transform of a function f(x). I've seen the general theory, but feel It would help with a concrete example to see how it is ...
1
vote
0answers
18 views
Fourier Transform of Distribution Equal to the Distribution Itself
We define $T(t)=a\delta^{(n)}+bt^n$ for $a, b$ nonzero complex constants and $n$ a nonnegative integer. I want to find the combinations of $a, b, n$ such that the fourier transform of $T$ is equal to ...
1
vote
0answers
35 views
CT fourier Transform Proof
I am confused trying to understand the Proof of Fourier Transform from Oppenheim book Signals and Systems. I am pasting the equations directly from the book: ...
1
vote
0answers
25 views
Criterium forSubspace of tempered distributions
I have a question concerning the subspace $\mathcal{S}'_h$ of tempered distributions defined by $u\in\mathcal{S}'_h\Leftrightarrow\lim_{\lambda\rightarrow\infty}\Vert\theta(\lambda ...
1
vote
0answers
36 views
Convergence of the sum of fourier coefficients
I have a function $f$ defined on $[-\pi, \pi]$, $f(\pi) = f(-\pi)$, and has continuous first derivative. How can I prove that the sum of the absolute value of the Fourier coefficients converges? The ...
1
vote
0answers
38 views
Finding a Fourier pair satisying some conditions
I want to find a Fourier pair $(r,\hat{r})$ satisfying some conditions listed below and making $\hat{r}(0)$ as small as possible. The requirements for $(r,\hat{r})$: $r,\hat{r}\in L^1(\mathbb{R})$, ...
1
vote
0answers
35 views
Eigenfunctions of a 4th Order PDE on 2D domain
What technique would you use to find the eigenfunctions of this problem? $$\nabla^2(\nabla^2 u) = 10$$ $0 < y < G$ $0 < x < L$ With Pure Homogeneous Dirichlet boundary conditions and ...
1
vote
0answers
29 views
Piecewise linear approximation of a (low order) trigonometric polynomial, with quantization
The Problem Given a trigonometric polynomial of order $K$: $$y(t)=\sum_{k=-K}^K c_k \ e ^ {j k \bar \omega t} \ , \qquad c_{-k}=\bar c_k$$ we want to find the best approximation to it using a ...
1
vote
0answers
123 views
wave front set - directions of singularities
I am learning about the wave front set of a distribution but am having difficulty understanding some details, which to me seem counter intuitive. We know the fourier transform of a smooth function ...
1
vote
0answers
46 views
Fourier transforms of distributions
I am reading a proof claiming that every partial differential operator $P(D)$ has a fundamental solution $E$. It says that "if we have a distribution $u$ on $R^n$ with $u(P(D)\phi)=\phi(0)$ for ...
1
vote
0answers
71 views
A continuous function whose Fourier Series diverges
I have read several times that there exist many continuous functions whose Fourier series diverge at some points (sometimes even on a dense subset of the domain!). I tried to find an explicit example ...
1
vote
0answers
30 views
I have derived a short proof that answers 'DTFT assumes the sequence to be periodic and hence the expression for the same'. Does my proof make sense?
Define: Function '$f$' defined at all (-oo, oo). $U(to)$ is a step function that equals 1 when $t \geq to$ $\sum_{n \epsilon Z}\delta(t - nT)$ is an impulse train that is non-zero for all $t = nT|{n ...
1
vote
0answers
35 views
Question about Fourier Transform?
Could someone please explain how they got from the first step to the next? I have no idea how the second step follows...
1
vote
0answers
61 views
The derivative of the trigonometric polynomial
For $n\in\mathbb{N}$, let $\varphi:\mathbb{R}\to\mathbb{R}$ be a $4n-$ periodic function s.t. $$\varphi(t)=n-|n-t|,\quad -n\leq t\leq 3n$$ For each $j\in\mathbb{Z}$ define $$c_j=\int_0^1\varphi(4n ...
1
vote
0answers
33 views
Derivation of First order derivation of Fourier transform
Let $R$ be $(-\infty,+\infty)$. Starting with: $$x(t)=\int_{R}X(f)e^{2i\pi ft}df$$We have its Fourier transform: $$F\{x(t)\}(f)=X(f)=\int_{R}x(t)e^{-2i\pi ft}dt$$ Its derivation is derived as ...
1
vote
0answers
35 views
Fourier Transform of a CFT two point function
In the study of Conformal Field Theory in physics, one encounters the following function $$ \left( \frac{1}{\sinh(x+i\epsilon)} \right)^{2\Delta}. $$ It appears as the correlation function of certain ...
1
vote
0answers
37 views
Find the Fourier Sine series of $x-\frac{1}{2}$ on the interval $0<x<1$
Two Parts to this question are: (a) Find the Fourier Sine series of $x-\frac{1}{2}$ on the interval $0<x<1$ (b) Let $a \notin Z$ and find the Fourier Cosine Series of $\cos ax$ on the ...
1
vote
0answers
62 views
Fourier transform on Laplace equations
We can solve some Laplace equations on the lectangular area by using Fourier transform. However many textbooks only introduce the cases when the area is given as a half plane or a strip. ; $y>0, ...
1
vote
0answers
49 views
Generalization of a FFT with powers of 3
I'm not satisfied with the current answers asked about this on MathSE. So I'm going to ask: Describe the generalization of the FFT algorithm to the case in which n is a power of 3. What's the ...
1
vote
0answers
45 views
problem 9.26 from Folland's real analysis Fourier Transform of $ G(x,t) = (4t\pi)^{(-n/2)} e^{{-|x|^2}/{4t}} \chi_{(0,\infty)}(t)$
I was just given this question from Folland's real analysis second edition dealing with tempered distributions and their Fourier transforms Exercise 26 on page 300 : On $ R^n \times R $ let $ ...
1
vote
0answers
26 views
Extension of the Fourier transform, proof-read
I'm writing a Bachelor-thesis in mathematics which is to be submitted in a couple of days, and would be thankfull if the following arguments concerning the extension of the Fourier transform from the ...
1
vote
0answers
64 views
Techniques in analytic number theory
I'm fairly new to the subject and trying to figure things out. Would be nice to hear some ideas and trickery for what follows. Suppose we wish to show there exists an integer $x$ in some finite set ...
1
vote
0answers
14 views
Autocorrelation of a signal known only on an interval of finite size
Let's consider we have a continuous random signal ${ t \in ] - \infty \,;\, + \infty [ \mapsto b (t)}$. We assume this signal to be stationary, so that when ensemble-averaged, one may introduce the ...
1
vote
0answers
36 views
Zero padding property of FFT
I wonder if the following basic property I thought up is a real property of FFT (or more specifically the discrete version of Fourier series), and if so what is it officially called? The property ...
1
vote
0answers
55 views
Inverse Fourier Transform involving inverse square root
I'm currently working on this paper: http://web.calstatela.edu/faculty/rcooper2/article.pdf and I want to proof Lemma 3.0 in the case of $n=2$, on page 441. It seems that the Ph.D. thesis the author ...
1
vote
0answers
15 views
Fourier transform of $1 - \cos(xe^{-x^2})$
Is there a closed form expression or maybe an infinite series? If not is there a "good" approximation to it? Even a "good" approximation of the fourier transform close to zero frequency would do. Can ...
1
vote
0answers
30 views
Are the Spherical harmonics the S^2 equivalent of the exp(i \pi n) function series?
As I understand it, the Spherical harmonics and the "Fourier functions" $\exp(i\pi n)$ with $n\in\mathbb{N}$ have much in common: Both are eigenfunctions of the angle part of the Laplace operator. ...
1
vote
0answers
139 views
Fourier series of half of $\sin(\pi x)$
So my question is: Find the Fourier series (using integrals) for the half wave rectified sine function: $$f(x)= \begin{cases}0&-1<x<0\\ \sin(\pi x)& 0<x<1\end{cases}$$ ...
1
vote
0answers
110 views
A discussion on fourier and laplace transforms and differential equations …?
i have read many of the answers and explanations about the similarities and differences between laplace and fourier transform. Laplace can be used to analyze unstable systems. Fourier is a subset of ...
1
vote
0answers
1k views
Fourier transform of the Cosine function with Phase Shift?
How can i calculate the Fourier transform of a delayed cosine? I haven't found anywhere how to do that. This is my attempt in hoping for a way to find it without using the definition: $$ x(t) = ...
1
vote
0answers
35 views
What are the statistics of the discrete Fourier transform of a Bernoilli process?
The problem I would like to understand the statistics of the discrete Fourier transform of a sequence of uncorrelated events $\{x_n\}$ each of which takes the value $\pm1$ with probability $1/2$. In ...
1
vote
0answers
54 views
Diagnalization of block matrix with circulat blocks
I have the following Matrix $A = \begin{pmatrix} X \\ Y \end{pmatrix}$ Where X, and Y are circulant Matrices. I want to diaganlize $AA^T$. I tried the following: $AA^T = \begin{pmatrix} ... | {
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-4,356,644,236,237,217,300 | Everything about 5153
Discover a lot of information on the number 5153: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
Mathematical properties of 5153
Questions and answers
Is 5153 a prime number? Yes
Is 5153 a perfect number? No
Number of divisors 2
List of dividers 1, 5153
Sum of divisors 5154
How to write / spell 5153 in letters?
In letters, the number 5153 is written as: Five thousand hundred and fifty-three. And in other languages? how does it spell?
5153 in other languages
Write 5153 in english Five thousand hundred and fifty-three
Write 5153 in french Cinq mille cent cinquante-trois
Write 5153 in spanish Cinco mil ciento cincuenta y tres
Write 5153 in portuguese Cinco mil cento cinqüenta e três
Decomposition of the number 5153
The number 5153 is composed of:
2 iterations of the number 5 : The number 5 (five) is the symbol of freedom. It represents change, evolution, mobility.... Find out more about the number 5
1 iteration of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
1 iteration of the number 3 : The number 3 (three) is the symbol of the trinity. He also represents the union.... Find out more about the number 3
Mathematical representations and links
Other ways to write 5153
In letter Five thousand hundred and fifty-three
In roman numeral MMMMMCLIII
In binary 1010000100001
In octal 12041
In hexadecimal 1421
In US dollars USD 5,153.00 ($)
In euros 5 153,00 EUR (€)
Some related numbers
Previous number 5152
Next number 5154
Next prime number 5167
Mathematical operations
Operations and solutions
5153*2 = 10306 The double of 5153 is 10306
5153*3 = 15459 The triple of 5153 is 15459
5153/2 = 2576.5 The half of 5153 is 2576.500000
5153/3 = 1717.6666666667 The third of 5153 is 1717.666667
51532 = 26553409 The square of 5153 is 26553409.000000
51533 = 136829716577 The cube of 5153 is 136829716577.000000
√5153 = 71.784399419372 The square root of 5153 is 71.784399
log(5153) = 8.5473343483282 The natural (Neperian) logarithm of 5153 is 8.547334
log10(5153) = 3.7120601424611 The decimal logarithm (base 10) of 5153 is 3.712060
sin(5153) = 0.70897809341048 The sine of 5153 is 0.708978
cos(5153) = 0.705230503498 The cosine of 5153 is 0.705231
tan(5153) = 1.0053139929341 The tangent of 5153 is 1.005314
Some random numbers
93
17
16
16
22
57
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73
34
53
22
44
43
39
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6,594,061,815,472,492,000 | Common Factors
Common factors of two or more numbers are a number which divides each of the given numbers exactly.
For examples
1. Find the common factor of 6 and 8.
Factor of 6 = 1, 2, 3 and 6.
Factor of 8 = 1, 2, 4 and 8.
Therefore, common factors of 6 and 8 = 1 and 2.
[It can easily be shown that each common factor (1 and 2) divides the given numbers 6 and 8 exactly].
2. Find the common factor of 8 and 12.
Factor of 8 = 1, 2, 4 and 8.
Factor of 12 = 1, 2, 3, 4, 6 and 12.
Therefore, common factors of 8 and 12 = 1, 2 and 4.
`
3. Find the common factor of 10 and 15.
Factor of 10 = 1, 2, 5 and 10.
Factor of 15 = 1, 3, 5 and 15.
Therefore, common factors of 10 and 15 = 1 and 5.
4. Find the common factor of 14 and 21.
Factor of 14 = 1, 2, 7 and 14.
Factor of 21 = 1, 3, 7 and 21.
Therefore, common factors of 14 and 21 = 1 and 7.
● Factors.
Common Factors.
Prime Factor.
● Repeated Prime Factors.
● Highest Common Factor (H.C.F).
● Examples on Highest Common Factor (H.C.F).
Greatest Common Factor (G.C.F).
Examples of Greatest Common Factor (G.C.F).
Prime Factorisation.
To find Highest Common Factor by using Prime Factorization Method.
Examples to find Highest Common Factor by using Prime Factorization Method.
To find Highest Common Factor by using Division Method.
Examples to find Highest Common Factor of two numbers by using Division Method.
To find the Highest Common Factor of three numbers by using Division Method.
`
5th Grade Numbers Page
5th Grade Math Problems
From Common Factors to HOME PAGE
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. | {
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3,874,305,579,062,674,000 | Excel Probability
In this article, we have used PROB, fractional method, NORM.DIST and NORMDIST function to find out Excel probability.
Unraveling uncertainty in Excel becomes effortless with its array of powerful probability functions. From the straightforward PROB and fractional techniques to the more advanced NORM.DIST and NORMDIST functions, Excel equips you to unveil the odds and predict outcomes. In this guide, we’ll explore how Excel’s user-friendly tools allow you to effortlessly compute probabilities, offering insights into a world of data-driven decision-making. Whether you’re a beginner or an experienced user, Excel’s probability features pave the way for confident analysis and informed choices.
Overview of Excel probability
Download Practice Workbook
Excel PROB Function
The PROB function calculates the probability based on a range comparing limits.
Syntax:
PROB(x_range, prob_range, lower_limit, [upper_limit])
5 Examples of Excel Probability
In this section, we are going to demonstrate 5 different examples to find out the probability in a dataset. We have used PROB function, fractional method, Z Score, and NORM.DIST function to find out probability.
1. Calculate Probability Using the PROB function
In this section, we will use the PROB function to calculate probability with lower and upper limits.
Here, we have a dataset of car sales with their probability value. Now, we will calculate the total probability based on lower and upper limits set on cells C12 and C13 respectively.
• Write the formula below in cell C14.
=PROB(C5:C10,D5:D10,$C$12,$C$13)
Probability using Prob function
The formula calculates the probability of a random variable falling within a specified range, given a set of data. The range of values for the random variable is provided in cells C5 to C10, while the corresponding probabilities for those values are in cells D5 to D10. The 3rd and 4th arguments $C$12 and $C$13 are used to specify the lower and upper limits of the desired range.
2. Calculating Probability Without Upper Limit
Again, we will use the PROB function with the same dataset but the upper limit is missing here.
• We will use the formula below in cell C13 to get the sales probability.
=PROB(C5:C10,D5:D10,$C$12)
Calculating probability without upper limit
In this context, the values representing the random variable are located in cells C5 to C10, and the corresponding probabilities for those values are found in cells D5 to D10. The 3rd argument $C$12 serves as the lower limit for the desired range. By utilizing these inputs, the formula computes the cumulative probability of the random variable falling within the specified range, taking into account the provided data and probability values, while considering the designated lower limit.
3. Getting Individual Probability with Mathematical Fractional Method
Here, we will calculate the individual probability based on total production and sold amount. You can consider this as a ratio, not as a probability based on the total sales. We will use simple mathematical fractions for this method.
• Here we have divided Sold Quantity / Production Quantity to find individual probability using the formula below.
=D5/C5
Using the Mathematical fractional method
4. Getting Dice Probability with Mathematical Fractional Method
Here we have counted the probability of summing two dice. Here we gave thrown 2 dice and given the summation 2 to 12. Then find out the probability of the sum.
• Use the formula below to find the chances of the number of rolls.
=COUNTIF($C$5:$H$10,B13)
Finding chances of Rolls
The formula counts how many times the value in cell B13 appears within the range C5:H10. The value of cell B13 is the sum value of two dice.
• Now find the probability by the formula below to get the individual chances.
=C13/36
• Again, drag the Fill Handle.
Dice Probability
5. Calculating Probability in Excel with Mean and Standard Deviation
In this section, we will calculate the probability using the Mean and Standard Deviation.
5.1 Using Z Score
We will use the AVERAGE and STDEV.P functions to calculate mean and standard deviation respectively.
• First, we have to find the Mean and Standard Deviation on cells C12 and C13 using the following formulas respectively.
For Mean:
=AVERAGE(C5:C10)
For Standard Deviation:
=STDEV.P(C5:C10)
Finding Standard deviation and average
• Now we have to find the Z score using the formula below on cell C5 and drag the fill option from C5 to C10.
=(C5-$C$12)/$C$13
Finding Z score
=NORMSDIST(D5)
• After that, drag the Fill Handle.
Probability using Z score
5.2 Using NORM.DIST
We already calculated the Mean and Standard Deviation in the previous section. Now, we will use them to calculate probability with NORM.DIST function.
• First, we have to find Normal Distribution of Production Quantity by the DIST function on cell D5.
=NORM.DIST(C5,$C$12,$C$13,FALSE)
Finding normal distribution
The formula calculates the cumulative normal distribution value for a specific input value of cell C5, given the mean ($C$12), standard deviation ($C$13), and uses a cumulative distribution. This function is commonly used in statistics to determine the probability that a random variable from a normal distribution falls below a certain value.
• Again use the NORM.DIST function to get the probability.
=NORM.DIST(G4,C12,C13,TRUE)
Probability with NORM.DIST
The formula calculates the cumulative probability that a sales value (represented by G4) is less than 65, assuming a normal distribution with a given mean (C12) and standard deviation (C13). The TRUE parameter indicates that the function computes the cumulative distribution, providing the probability that a randomly selected value from the specified normal distribution is less than the value in G4 (65 in this case), based on the provided mean and standard deviation. This can be interpreted as the probability that sales fall below 65 in a distribution with the given characteristics.
Common Errors When Using PROB Function
1. Misaligned Data: Ensure that the data ranges for values and probabilities are properly aligned. A mistake in aligning these ranges can lead to incorrect calculations.
2. Incorrect Probability Format: Probability values should be valid and within the range of 0 to 1. Mistakenly using percentages or decimal values outside this range can result in erroneous results.
3. Inconsistent Ranges: Ensure that the data ranges for values and probabilities have the same number of elements. Mismatched ranges will lead to incorrect probabilities.
4. Order of Arguments: Make sure you’re using the correct order of arguments in the function. For example, ensure the lower and upper limits you’re providing in the right sequence.
5. Misinterpretation of Limits: Be clear about whether you’re using inclusive or exclusive limits for the lower and upper bounds. Mistakes in understanding this can lead to wrong probability calculations.
6. Incorrect Parameters: Ensure that you’re using the correct parameters for the specific distribution or function you’re working with. Different distributions may have variations in input parameters.
7. Incorrect Distribution Assumption: Using the wrong probability distribution assumption can lead to inaccurate results. Ensure that the distribution you’re using accurately reflects the nature of your data.
8. Using the Wrong Function: Make sure that the function you’re using is appropriate for the problem you’re trying to solve. Different functions have different purposes, such as calculating cumulative probabilities or probability density functions.
9. Not Considering Assumptions: Understand the assumptions of the probability model you’re using. Ignoring or misunderstanding these assumptions can lead to misleading results.
10. Rounding Errors: Be cautious with rounding numbers during calculations, as cumulative probabilities can be sensitive to even small rounding errors.
Why Isn’t My PROB Function Working?
If your PROB function isn’t producing the expected results, consider these steps:
1. Sum of Probabilities: Make sure the total sum of probabilities in the range_prob is exactly 1. If it’s not, adjust the probabilities to ensure they add up to 1.
2. Matching Cell Counts: Ensure that the number of cells in range_x (values) and range_prob (probabilities) are the same. They should align properly for the function to work correctly.
3. Numeric Values: Confirm that all values within range_x and range_prob are numbers. The function requires numeric inputs to calculate probabilities accurately.
4. Syntax Check: Double-check the formula for any syntax errors. Ensure you’ve provided all the required arguments correctly, following the formula’s structure.
Probability and Inverse Probability
Here is an explanation of probability and inverse probability, along with some examples.
Probability: Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. It is expressed as a value between 0 to 1, where 0 represents an impossible event, 1 represents a certain event, and values in between represent varying degrees of likelihood. Probability can be used to model uncertain outcomes in various scenarios, such as games of chance, weather forecasting, and financial risk assessment.
Example of Probability: Consider rolling a fair six-sided die. The probability of rolling a 4 is 1/6, as there is one favorable outcome (rolling a 4) out of six possible outcomes (rolling numbers 1 through 6).
Inverse Probability: Inverse probability, also known as the quantile function or percent-point function, is the opposite of the probability concept. It helps you determine the value for which a certain probability is reached or exceeded in a probability distribution. In other words, given a probability, the inverse probability function helps you find the value that corresponds to that probability.
Example of Inverse Probability: Let’s say you have a standard normal distribution (mean = 0, standard deviation = 1). Using the inverse probability function, you can find the value at which the cumulative probability is, for instance, 0.95. This helps you identify the threshold beyond which only 5% of the data lies.
Things to Remember
• Stay updated with new features and changes in Excel. Newer versions may introduce improvements or enhancements to probability functions.
• Be prepared to troubleshoot and resolve errors that may arise during probability calculations. Understanding common error messages will help you identify and fix issues.
Frequently Asked Question
1. Can Excel probability analysis be used in finance and statistics?
Absolutely, Excel probability analysis has wide-ranging applications in finance, statistics, and many other fields. It can assist in financial modeling, risk assessment, investment analysis, and statistical research by quantifying uncertainty and aiding in decision-making processes.
2. Are there any limitations to using Excel for probability analysis?
While Excel is a powerful tool, it may have limitations when dealing with extremely complex or specialized probability calculations. In such cases, dedicated statistical software might be more suitable. Additionally, ensuring accurate data input and understanding the assumptions of the probability models is crucial.
3. What is the difference between NORM.DIST and NORMINV functions?
NORM.DIST calculates the cumulative probability of a value, while NORMINV calculates the inverse probability – it finds the value that corresponds to a given cumulative probability.
Conclusion
In summary, Excel’s probability functions provide a versatile and essential toolkit for quantifying uncertainty, predicting outcomes, and making informed decisions in various fields, from finance and logistics to research and analysis, underscoring the increasing significance of data-driven approaches in today’s complex world.
Excel Probability: Knowledge Hub
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Joyanta Mitra
Joyanta Mitra
Joyanta Mitra, a BSc graduate in Electrical and Electronic Engineering from Bangladesh University of Engineering and Technology, has dedicated over a year to the ExcelDemy project. Specializing in programming, he has authored and modified 60 articles, predominantly focusing on Power Query and VBA (Visual Basic for Applications). His expertise in VBA programming is evident through the substantial body of work he has contributed, showcasing a deep understanding of Excel automation, and enhancing the ExcelDemy project's resources with valuable... Read Full Bio
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6,557,369,989,487,341,000 | Brain freezer a.k.a. MATH
posted by Crystal
Please Help!
Are all squares similar?
I think that they aren't but i'm not sure. I think this because rectangles can be squares and rectangles can take different forms and shapes. Please help
What do you mean by square? A square will always have a 4 sides and angles equal. Opposite sides are parallel. Is this what you needed? Please be more clear.
I mean if you compare squares will they always be similar such as same shape diferrent size. For example all rectangles are NOT similar because some are thin others can be wider. But I think i already found an answer THANKS FOR HELPING ME ANYWAY!!! :)
All squares are similar.
are all squares similar???????
How are they all similar?
Respond to this Question
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8,263,308,603,563,221,000 | 'Statistics are figures, but all figures are not statistics'. Justify the statement
Statistics is often regarded as being a means by which observations are expressed numerically in order to investigate casual relationship between the variables. Any fact, to be called statistics must be numerically expressed (so that it can be counted, divided or be subject to mathematical analysis) and should be placed in relation to each other. But qualitative data cannot be included in statistics unless they are quantified by assigning some figures for assessment. However, not all numbers are comparable and measurable.
For example : The fact that height of a student is five feet tells nothing, unless it is comparable. Thus for figures to be included in statistics, they must be aggregate of facts and not individual figures. Thats why it is rightly said that statistics are figures but all figures are not statistics.
sir / madam , is this answer applicable for this question too ?
_**“All statistics must be expressed in terms of numbers but all numbers are not statistics.” Explain with**_
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2,357,358,624,958,942,700 | Python Numbers
Here's an overview of the various numeric types in Python, and how to work with them.
Python has three distinct numeric types: integers, floating point numbers, and complex numbers. These are usually referred to as int, float, and complex types. Also, the boolean type is a subtype of the integer type.
int
Refers to an integer. An integer is a whole number (i.e. not a fraction). Integers can be a positive number, a negative one, or zero. Examples of integers: -3, -2, -1, 0, 1, 2, 3
float
Refers to a floating point number. Floating point numbers represent real numbers and are written with a decimal point dividing the integer and the fractional parts. Floating point numbers can also be in scientific notation, with E or e indicating the power of 10 (eg, +1e3 is equivalent to 1000.0). Examples of floats: 1.0, 12.45, 10.4567, -10.0, -20.76789, 64.2e18, -64.2e18.
complex
A complex number takes the form a + bj where a is a real number and b is an imaginary number. Each argument an be any numeric type (including complex). The first argument can also be a string (but the second argument can't). Examples: 1.4j, -1.4j, 2+18j, -2.18j, 5.14-7j, 5.14e+45j, -5.14e+45j.
In Python, all numeric types are immutable. If you want to change any part of a number you need to reassign the number itself.
Here's an example of creating some number objects, then printing each number along with its type:
Result
<class 'int'> 1
<class 'int'> -1
<class 'float'> 1.0
<class 'float'> -1.0
<class 'float'> 2000.0
<class 'float'> -2000.0
<class 'complex'> 3.14j
<class 'complex'> (-0-3.14j)
Random Numbers
Python provides various functions that allow you to generate random numbers. These are made possible by the random module. The random module must be imported using the import keyword before you can use any of the random number functions. Here are a few:
Result
0.8296271520534051
34
90
[60, 70, 20]
[90, 50, 80, 20, 70]
Of course, this is only an example of the results that could be returned. Seeing as the results are randomly generated, they will be different each time it's run.
Minimum and Maximum Numbers
You can use the min() and max() functions to return the smallest or largest number within a group of numbers. You can supply the numbers as multiple parameters or as a list. Like this:
Result
3
1
77
11
Type Conversion
Python has an inbuilt ability to convert numbers to a single type when performing calculations. For example, say you want to do this:
These are two different number types. The 100 is an integer while the 2.5 is a float. However, Python can handle this. It will make the resulting number a float.
Here's a demo:
Result
100 <class 'int'>
2.5 <class 'float'>
102.5 <class 'float'>
Conversion Functions
You can also use functions such as int(), float(), and complex() to make an explicit conversion between one number type and another.
Here are some examples:
Result
<class 'int'>
<class 'float'>
<class 'complex'>
<class 'bool'>
Numbering Systems
The decimal numbering system is the most widely used system in the modern world. Also called base-ten, the decimal system has 10 as its base, and uses the digits 0 to 9.
There are other numbering systems though, that don't use 10 as its base. The binary system is base-two (uses the digits 1 and 0), the octal system is base-eight (uses digits 0 to 7), and the hexadecimal system is base-sixteen (uses digits 0 to 9, and letters A to F). These numbering systems tend to be more popular in mathematics and computing.
In Python, you can specify the numbering system a number uses by using a two-digit prefix as follows:
Numbering System Prefix
Binary 0b or 0B
Octal 0o or 0O
Hexadecimal 0x or 0X
The prefix can be uppercase or lowercase.
Here's an example of printing out various numbers from different numbering systems. Although we provide the numbers in binary, octal, and hexadecimal, the output is in decimal.
Result
3
13
14
The following example prints out a range of numbers from each system. You can see how each value in the code maps to a base-ten number in the output:
Result
... 1 2 3 4 5 ...
... 5 6 7 8 9 10 11 12 13 ...
... 7 8 9 10 11 12 13 14 15 16 18 19 ...
You can also use functions such as hex() and oct() to return an integer as a hexadecimal or octal number.
Result
0o144
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Fonctions de plusieurs variables
November 1, 2004
1
1.1
Di erentiabilit e
Motivation
Pour une fonction dune variable f , d enie au voisinage de 0, etre d erivable en 0, cest admettre un d eveloppement limit ea ` lordre 1, f (x) = b + ax + x (x). Alors b = f (0) et a = f (0). Interpr etation g eom etrique. La courbe repr esentative de f poss` ede en (0, a) une tangente, la droite d equation y = b + ax. On veut faire pareil pour une fonction de deux variables. La courbe repr esentative est remplac ee par une surface repr esentative d equation z = f (x, y ), la droite tangente par un plan tangent d equation z = c + ax + by . La tangence sexprime en disant que la distance entre le point (x, y, f (x, y )) de la surface et le point (x, y, c + ax + by ) du plan est petite devant la distance de (x, y ) ` a lorigine. Exemple 1.1 f (x, y ) = x2 + y 2 .
1.2
Di erentiabilit e dune fonction de deux variables
D enition 1.2 Soit f une fonction de deux variables, d enie au voisinage de (0, 0). On dit que f est di erentiable en (0, 0) si elle admet un d eveloppement limit e` a lordre 1, i.e. si on peut ecrire f (x, y ) = c + ax + by + x2 + y 2 (x, y ),
o` u (x, y ) tend vers 0 lorsque x et y tendent vers 0. Dans ce cas, f admet des d eriv ees partielles en (0, 0), et c = f (0, 0), a= f (0, 0), x f (0, 0). y
La di erentiabilit e de f en un point quelconque (x0 , y0 ) se traduit par le d eveloppement limit e f (x0 + u, y0 + v ) = f (x0 , y0 ) + f f (x0 , y0 )u + (x0 , y0 )v + x y u2 + v 2 (u, v ),
o` u (u, v ) tend vers 0 lorsque u et v tendent vers 0. Exemple 1.3 f (x, y ) = x(2 x + y ) + y (1 x y ) est di erentiable ` a lorigine. En eet, f (x, y ) = = 2x + y x2 y 2 2x + y + 1 x2 + y 2 (x, y ),
o` u (x, y ) = x2 + y 2 tend vers 0 quand x et y tendent vers 0. Th eor` eme 1 Soit f une fonction de deux variables d enie au voisinage de (0, 0). Si les d eriv ees f partielles f et sont d e nies au voisinage de (0 , 0) et continues en (0 , 0) , alors f est di e rentiable x y en (0, 0), et son d eveloppement limit e` a lordre 1 s ecrit f (x, y ) = f (0, 0) + f f (0, 0)x + (0, 0)y + x y x2 + y 2 (x, y ).
Exemple 1.4 f (x, y ) = x(2 x + y ) + y (1 x y ) est di erentiable en tout point. En eet, on na qua utiliser le th eor` eme 1. On peut aussi calculer directement f (x0 + u, y0 + v )
2 2 2 = 2x0 + 2u + y0 + v x2 0 2x0 u u y0 2y0 v v 2 2 2 = 2x0 + y0 x2 0 y0 + (2 2x0 )u + (1 2y0 )v u v
2 2x0 + y0 x2 0 y0 + (2 2x0 )u + (1 2y0 )v +
u2 + v 2 (u, v ).
1.3
Gradient
D enition 1.5 Soit f une fonction de deux variables, di erentiable tout point dun domaine D. Son gradient est le champ de vecteurs d eni sur D par f : (x, y )
f x (x, y ) f y (x, y )
Exemple 1.6 Le gradient de la fonction d enie sur R2 par f (x, y ) = x2 est le champ de vecteurs 2x horizontal (x,y) f = . 0
1.4
Interpr etation du d eveloppement limit e
Proposition 1.7 Si f est di erentiable en P , alors pour toute droite t P + tv passant par P , la fonction t f (P + tv ) est d erivable, et d f (P + tv )|t=0 = P f v. dt On verra plus loin (th eor` eme 2) que cette formule est vraie pour toute courbe, et non seulement les droites, sous la forme d f (c(t)) = c(t) f c (t). dt
1.5
Lignes de niveau
D enition 1.8 On appelle lignes de niveau de f les ensembles de la forme Lw = {(x, y ) ; f (x, y ) = w}. Exemple 1.9 Les lignes de niveau de la fonction f (x, y ) = x2 + y 2 sont des cercles concentriques. Celles de la fonction f (x, y ) = xy sont des hyperboles, ` a lexception de la ligne de niveau 0, qui est la r eunion de deux droites.
Proposition 1.10 Le gradient dune fonction est un vecteur perpendiculaire aux lignes de niveau, pointant dans la direction dans laquelle la fonction augmente. Sa longueur est dautant plus grande que la fonction varie rapidement, i.e. que les lignes de niveau sont rapproch ees. Le gradient indique la direction de plus grande pente. Preuve. Soit t c(t) une ligne de niveau. Alors t f (c(t)) est constante, donc 0= d f (c(t)) = c(t) f c (t), dt
ce qui montre que le gradient est orthogonal ` a la tangente ` a la ligne de niveau. Lorsque lon se d eplace dans la direction du gradient, par exemple, par t c(t) = P + tP f , d f (c(t))|t=0 = P f c (0) = P f dt donc f augmente, dautant plus vite que Soit v un vecteur unitaire. Alors P f est grand.
2
> 0,
d f (P + tv )|t=0 = P f v dt est maximum lorsque v est colin eaire et de m eme sens que P f , donc P f indique la direction de plus grande pente.
1.6
G en eralisation
De la m eme fa con, on peut parler de d eveloppement limit e et de di erentiabilit e pour une fonction 2 ), puis pour une application Rn Rp . de n variables (remplacer x2 + y 2 par x2 + + x n 1 Dans ce cas, les coecients du d eveloppement limit e sont des vecteurs de Rp . Exemple 1.11 Soit I un intervalle de R et c : I R2 une courbe. Calculer un d eveloppement limit e de c en 0, cest calculer des d eveloppements limit es des fonctions coordonn ees x(t) = a0 + a1 t + t (t), y (t) = b0 + b1 t + t (t), et former le d eveloppement limit e vectoriel c(t) = a0 b0 +t a1 b1 + t (t).
Proposition 1.12 Une application F = (f1 , . . . , fp ) : Rn Rp est di erentiable si et seulement si chacune de ses composantes lest.
1.7
La di erentielle
D enition 1.13 Soit F := (f1 , . . . , fp ) : Rn Rp une application di erentiable en P . Sa di erentielle en P est lapplication lin eaire de Rn dans Rp qui appara t comme le terme non constant du d eveloppement limit e` a lordre 1 en P . Sa matrice, appel ee matrice jacobienne, a pour coecients les d eriv ees partielles, f1 f1 . . . x x1 n . . . Jf (P ) = . . . . fp fp . . . xn x1 Exemple 1.14 Si A est une matrice, alors lapplication lin eaire fA : Rn Rp quelle d enit est di erentiable, et sa matrice jacobienne est A en nimporte quel point. Exemple 1.15 Soit f (x, y ) = 2x + y x2 y 2 . Sa matrice jacobienne est 2 2x 1 2y . 3
Autrement dit, la matrice jacobienne dune fonction, cest son gradient vu comme un vecteur ligne. Exemple 1.16 Soit F (t) = cos(t) . Sa matrice jacobienne est sin(t) sin(t) . cos(t)
Autrement dit, la matrice jacobienne dune courbe, cest sa d eriv ee vue comme un vecteur colonne. Exemple 1.17 Soit F (r, ) = (r cos(), r sin()). Sa matrice jacobienne est cos() r sin() . sin() r cos()
1.8
Matrice jacobienne dune fonction compos ee
Il sagit de g en eraliser la formule (g f ) = (g f )f . Th eor` eme 2 Soient f : Rn Rp et g : Rp Rq des applications. On suppose f di erentiable en P et g di erentiable en f (P ). Alors g f est di erentiable en P , et Jgf (P ) = Jg (f (P ))Jf (P ). Preuve. Si v Rn , f (P + v ) = f (P ) + Jf (P )v + On pose w = f (P + v ) f (v ). Alors g (f (P ) + w) = g (f (P )) + Jg (f (P ))w+ Autrement dit, g f (P + v ) = g f (P ) + Jg (f (P ))(Jf (P )v + v = g f (P ) + Jg (f (P ))Jf (P )v + v (v ))+ (v ), w (w) w (w). v (v ).
car
est born e.
Corollaire 1.18 Soit I un intervalle de R, soit c : I R2 une courbe dans le plan. Soit f : R2 R une fonction sur le plan. Alors (f c) (t) = Jg c (t) = c(t) f c (t) = f f (c(t))x (t) + (c(t))y (t). x y
Corollaire 1.19 Soit f : R2 R une fonction sur le plan. Soit g : R R une fonction dune variable. Alors Jgf = g (f )Jf , i.e. P g f = g (f (P ))P f.
Corollaire 1.20 Soit F : R2 R2 , F (r, ) = (r cos(), r sin()), le changement de coordonn ees polaires. Soit c : R R2 une courbe param etr ee, vue en coordonn ees cart esiennes (x(t), y (t)) ou polaires (r(t), (t)). Alors la vitesse en coordonn ees cart esiennes sobtient en appliquant la matrice jacobienne de F ` a la d eriv ee des coordonn ees polaires, x y = cos() r sin() sin() r cos() r = r er + re .
1.9
Condition dextremum
Proposition 1.21 Soit f une fonction ` a valeurs r eelles d enie au voisinage dun point P de Rn . Si P est un minimum local (resp. maximum local) de f , alors le gradient de f sannule en P . Preuve. Cas n = 2. Soit P = (x0 , y0 ). A fortiori, x0 est un minimum local (resp. maximum local) de la fonction x f (x, y0 ), donc sa d eriv ee en x0 est nulle. Or celle-ci vaut f x (P ). De f m eme, x (P ) = 0, donc P f = 0. Remarque 1.22 En g en eral, la r eciproque est fausse. On peut donner des conditions suivantes plus fortes, faisant intervenir les d eriv ees secondes. Cest lobjet du paragraphe suivant.
2
2.1
D eveloppement limit e` a lordre 2
Motivation
On sint eresse au mouvement dans un champ de forces d erivant dun potentiel V . Les positions d equilibre correspondent aux points o` u les d eriv ees partielles de V sannulent. Pour quune position d equilibre P soit stable, il vaut mieux que V poss` ede un minimum local strict en P , i.e., que pour v = 0 assez petit, V (P + v ) > V (P ). Soit f une fonction dune variable. Supposons que f admet un minimum en 0. Alors sa d eriv ee f (0) sannule. La r eciproque nest pas vraie : la fonction d enie sur R par f (x) = x3 a une d eriv ee nulle en 0 mais nadmet pas de minimum local. Une condition susante fait intervenir la d eriv ee seconde. Proposition 2.1 Soit f une fonction dune variable. Supposons que f (0) = 0 et f (0) > 0. Alors f poss` ede un minimum local strict en 0 : pour x = 0 susamment petit, f (x) > f (0). Preuve. Le d eveloppement limit e de Taylor-Young donne 1 f (x) = f (0) + f (0)x2 + x2 (x). 2 Alors f (x) f (0) 1 = f (0) + (x) > 0 x2 2 pour x assez petit. On peut aussi parler de d eveloppement limit e` a lordre 2 pour une fonction de plusieurs variables. Cest li e aux d eriv ees partielles secondes, cela donne un condition susante pour un minimum local strict.
2.2
D enition
Proposition 2.2 Soit m(x, y ) = axr y s un polyn ome de degr e r + s. Alors on peut ecrire m(x, y ) = u (x, y ) tend vers 0 quand x et y tendent vers 0 ( x2 + y 2 )r+s1 (x, y ) o` Autrement dit, d` es que r + s 2, un mon ome axr y s peut etre mis dans le reste dun d eveloppement limit e` a lordre 1. Il ne reste donc dans le d eveloppement limit e` a lordre 1 dune fonction f que des termes de degr e 0 (le terme constant f (0, 0)) et 1 (la di erentielle de f en (0, 0)). On va voir que les mon omes axr y s tels que r + s 3, peuvent etre mis dans les restes des d eveloppements limit es ` a lordre 2. Ceux-ci ne comportent donc que des termes de degr es 0, 1 et 2. Les termes de degr e 2 sont de la forme px2 + rxy + sy 2 , o` u p, q et r sont des constantes. Cela motive la d enition suivante. 5
D enition 2.3 Soit f une fonction de deux variables d enie au voisinage de 0. On dit que f admet un d eveloppement limit e` a lordre 2 en (0, 0) si on peut ecrire f (x, y ) = c + ax + by + px2 + qxy + ry 2 + (x2 + y 2 ) (x, y ), o` u (x, y ) tend vers 0 lorsque x et y tendent vers 0. Plus g en eralement, on dit que f admet un d eveloppement limit e ` a lordre 2 en (x0 , y0 ) si on peut ecrire f (x0 + u, y0 + v ) = c + au + bv + pu2 + quv + rv 2 + (u2 + v 2 ) (u, v ), o` u (u, v ) tend vers 0 lorsque u et v tendent vers 0. Th eor` eme 3 (D eveloppement limit e de Taylor-Young). Soit f une fonction de deux variables 2 2f f d enie au voisinage de 0. On suppose que f admet des d eriv ees partielles secondes x2 , xy et et que celles-ci sont continues au voisinage de 0. Alors f admet un d eveloppement limit e` a lordre 2, f (x, y ) = f (0, 0) + f f 1 2f 2f 2f (0, 0)x + (0, 0)y + ( 2 (0, 0)x2 + 2 (0, 0)xy + 2 (0, 0)y 2 ) x y 2 x xy y +(x2 + y 2 ) (x, y ).
2f y 2 ,
Autrement dit, la plupart des fonctions quon rencontrera admetteront un d eveloppement limit e. Exemple 2.4 f (x, y ) = cos(x) cos(y ) admet en (0, 0) le d eveloppement limit e f (x, y ) 1 1 = (1 x2 + x2 (x))(1 y 2 + y 2 (y )) 2 2 1 1 = 1 + x2 + y 2 + (x2 + y 2 ) (x, y )). 2 2
En ( eveloppement limit e 2 , 2 ), elle admet le d
f(
+ u, + v ) 2 2
= sin(u) sin(v ) = (u + u2 (u))(v + v 2 (v )) = uv + (u2 + v 2 ) (u, v ).
Dans les deux cas, on reconna t les d eriv ees partielles secondes dans les coecients de u2 , uv et v 2 .
2.3
Signe
Pour une fonction dune variable de la forme px2 , le signe ne d epend que du signe de p. Pour une fonction de deux variables de la forme px2 + qxy + ry 2 , l etude du signe se ram` ene ` a celui du trin ome du second degr e Z pZ 2 + qZ + r. En eet, si on pose Z = x/y , px2 + qxy + ry 2 = x2 (pZ 2 + qZ + r). Par cons equent, Proposition 2.5 Si q 2 4pr < 0 et p > 0, alors pour tout (x, y ) = (0, 0), px2 + rxy + sy 2 > 0.
Si q 2 4pr = 0, p 0 et r 0, alors pour tout (x, y ), px2 + qxy + ry 2 0. Si q 2 4pr > 0, la fonction px2 + qxy + ry 2 prend les deux signes au voisinage de 0. Th eor` eme 4 Soit f une fonction de deux variables d enie au voisinage de 0. On suppose que f admet un d eveloppement limit e` a lordre 2 au voisinage de (0, 0), de la forme f (x, y ) = c + ax + by + px2 + qxy + ry 2 + (x2 + y 2 ) (x, y ). 6
Si (0, 0) est un minimum local pour f , alors a = b = 0, q 2 4pr 0, p 0 et s 0. De m eme, si (0, 0) est un maximum local pour f , alors a = b = 0, q 2 4pr 0, p 0 et s 0. R eciproquement, si a = b = 0, q 2 4pr < 0 et p > 0, alors (0, 0) est un minimum local pour f. De m eme, si a = b = 0, q 2 4pr < 0 et p < 0, alors (0, 0) est un maximum local pour f . Exemple 2.6 La fonction f (x, y ) = cos(x) cos(y ) de lexemple 2.4 admet en (0, 0) un minimum ete f comme local strict. En revanche, en ( 2 , 2 ), il ne sagit pas dun minimum local. Si on interpr` le relief dune table bossel ee, une bille qui roule sur la table sarr etera dans un creux (par exemple, en (0, 0)), mais pas dans un col comme ( 2 , 2 ). Exemple 2.7 On sint eresse aux bo tes en forme de parall epip` ede. On cherche, parmi les bo tes de contenance donn ee 1, ` a minimiser laire. Montrer que laire atteint un minimum local pour la bo te cubique. Notons x et y les longueurs de deux des c ot es. Si la contenance vaut 1, alors la hauteur vaut 1 . Laire de la bo te, somme des aires des 6 faces, vaut z = xy f (x, y ) = 2xy + 2yz + 2zx = 2xy + 2 2 + . x y
La bo te cubique correspond ` a x = y = 1. On applique le th eor` eme 3 ou on d eveloppe f (1 + u, 1 + v ) 2 2 + 1+u 1+v = 2 + 2u + 2v + 2uv + 2 2u + 2u2 + 2 2v + 2v 2 + u2 (u) + v 2 (v ) = 6 + 2u2 + 2uv + 2v 2 + (u2 + v 2 ) (u, v ). = 2(1 + u)(1 + v ) +
Le discriminant q 2 4pr = 12 < 0, donc le crit` ere 4 sapplique, et la bo te cubique est bien un minimum local de laire. | {
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-8,288,712,069,822,790,000 | Sunday
January 22, 2017
Homework Help: geometry
Posted by alisa on Tuesday, November 15, 2011 at 8:04pm.
Two similar solids have a scale factor of 3:5. If the height of solid 1 is 3cm, what is the height of solid 2?
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-1,004,769,158,580,979,700 | Mathematics Questions and Answers – Rational Numbers Between Two Rational Numbers
«
»
This set of Mathematics Quiz for Class 8 focuses on “Rational Numbers Between Two Rational Numbers”.
1. Which number lies between the numbers \(\frac{1}{4}\) and \(\frac{1}{6}\)?
a) \(\frac{1}{5}\)
b) \(\frac{1}{6}\)
c) \(\frac{1}{3}\)
d) \(\frac{1}{2}\)
View Answer
Answer: a
Explanation: The given fractional numbers represents \(\frac{1}{4}\) = 0.25 and \(\frac{1}{6}\) = 0.17
Our options are \(\frac{1}{5}\) = 0.20
\(\frac{1}{6}\) = 0.17
\(\frac{1}{3}\) = 0.33
\(\frac{1}{2}\) = 0.50
The only option between 0.17 and 0.25 is \(\frac{1}{5}\) i.e. 0.20.
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2. The rational number which is not lying between 6 and 7 is ________
a) \(\frac{36}{7}\)
b) \(\frac{45}{7}\)
c) \(\frac{46}{7}\)
d) \(\frac{13}{2}\)
View Answer
Answer: a
Explanation: The only number which is not between 6 and 7 is \(\frac{36}{7}\)
\(\frac{36}{7}\) = 5.14
\(\frac{45}{7}\) = 6.42
\(\frac{46}{7}\) = 6.57
\(\frac{13}{2}\) = 6.5.
3. Between any consecutive two integers, there are always infinite integers.
a) True
b) False
View Answer
Answer: b
Explanation: The set of integers is […., -1, 0, 1, ….] there cannot be any integers between two integers.
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4. The maximum number of integers between two consecutive natural numbers is ________
a) zero
b) 2
c) 3
d) infinite
View Answer
Answer: d
Explanation: There can be an infinite number of rational numbers between any two consecutive natural numbers. But there cannot be any integers between two consecutive natural numbers.
5. Which on the following number lies between \(\frac{3}{10}\) and \(\frac{3}{10000}\).
a) \(\frac{3}{1000}\)
b) \(\frac{3}{10000}\)
c) \(\frac{3}{100000}\)
d) \(\frac{3}{1000000}\)
View Answer
Answer: a
Explanation: The given fractional numbers represents \(\frac{3}{10}\) = 0.3 and \(\frac{3}{10000}\) = 0.00003
Our options are in fractions, converting it into decimal form we get,
\(\frac{3}{1000}\) = 0.0003
\(\frac{3}{10000}\) = 0.00003
\(\frac{3}{100000}\) = 0.000003
\(\frac{3}{1000000}\) = 0.0000003.
The only option between 0.3 and 0.00003 is \(\frac{3}{1000}\) i.e. 0.0003.
6. \(\frac{32}{9}\) lies between which pair?
a) \(\frac{6}{5}\) and \(\frac{7}{2}\)
b) \(\frac{3}{2}\) and \(\frac{5}{2}\)
c) \(\frac{7}{2}\) and \(\frac{10}{2}\)
d) \(\frac{7}{2}\) and \(\frac{6}{5}\)
View Answer
Answer: c
Explanation: When we place the 4 given pairs and the number on the number line we observe that the number \(\frac{32}{9}\) lies between the pair \(\frac{7}{2}\) and \(\frac{10}{2}\), while other options fail to capture the given number between them. Hence the correct pair is \(\frac{7}{2}\) and \(\frac{10}{2}\).
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7. Which of the following is the greatest number among the given rational numbers?
a) \(\frac{139}{26}\)
b) \(\frac{72}{21}\)
c) \(\frac{99}{36}\)
d) \(\frac{101}{20}\)
View Answer
Answer: a
Explanation: The numbers when placed on the number line, the number which is placed farthest from 0 is the greatest. Here the numbers are, \(\frac{139}{26}\) = 5.34
\(\frac{72}{21}\) = 3.42
\(\frac{99}{36}\) = 2.75
\(\frac{101}{20}\) = 5.05
So, we conclude that the greatest number is \(\frac{139}{26}\) and hence would be placed farthest from 0.
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8. [….., -1, 0, 1, …….] the given set shows which type of numbers?
a) Rational numbers
b) Integers
c) Natural numbers
d) Whole numbers
View Answer
Answer: b
Explanation: The given set is of integers.
The sets of other numbers are,
Whole number -> [0, 1, 2, 3, 4, ……]
Natural number -> [1, 2, 3, 4, …….]
Rational number -> [……, 0, ……..] This set consists of all numbers on the number line.
9. Pick the smallest number of the following.
a) –\(\frac{139}{26}\)
b) –\(\frac{72}{21}\)
c) –\(\frac{99}{36}\)
d) –\(\frac{101}{20}\)
View Answer
Answer: a
Explanation: The numbers when placed on the number line, the number which is placed nearest from 0 is the smallest. Here the numbers are, –\(\frac{139}{26}\) = -5.34
–\(\frac{72}{21}\) = -3.42
–\(\frac{99}{36}\) = -2.75
–\(\frac{101}{20}\) = -5.05
So, we conclude that the smallest number is –\(\frac{139}{26}\) and hence would be placed smallest from 0. Thing to remember: When the numbers are placed on the left side of the number line, the number farthest from the center is the smallest number. This is due to the negative sign.
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10. Pick the option which has the ordered pair of BIG and SMALL number.
a) \(\frac{6}{3}\) and \(\frac{7}{3}\)
b) \(\frac{12}{11}\) and \(\frac{11}{12}\)
c) \(\frac{13}{22}\) and \(\frac{33}{11}\)
d) \(\frac{12}{4}\) and \(\frac{13}{5}\)
View Answer
Answer: d
Explanation: Here we have to find the ordered pair of numbers and the order should be big following by small. This condition is only satisfied by one pair of numbers which is \(\frac{12}{4}\) and \(\frac{13}{5}\).
Sanfoundry Global Education & Learning Series – Mathematics – Class 8.
To practice Mathematics Quiz for Class 8, here is complete set of 1000+ Multiple Choice Questions and Answers.
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Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses. | {
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-5,863,664,236,860,968,000 | What is 590 divided by 886 using long division?
Confused by long division? By the end of this article you'll be able to divide 590 by 886 using long division and be able to apply the same technique to any other long division problem you have! Let's take a look.
Want to quickly learn or show students how to solve 590 divided by 886 using long division? Play this very quick and fun video now!
Okay so the first thing we need to do is clarify the terms so that you know what each part of the division is:
• The first number, 590, is called the dividend.
• The second number, 886 is called the divisor.
What we'll do here is break down each step of the long division process for 590 divided by 886 and explain each of them so you understand exactly what is going on.
590 divided by 886 step-by-step guide
Step 1
The first step is to set up our division problem with the divisor on the left side and the dividend on the right side, like we have it below:
886590
Step 2
We can work out that the divisor (886) goes into the first digit of the dividend (5), 0 time(s). Now we know that, we can put 0 at the top:
0
886590
Step 3
If we multiply the divisor by the result in the previous step (886 x 0 = 0), we can now add that answer below the dividend:
0
886590
0
Step 4
Next, we will subtract the result from the previous step from the second digit of the dividend (5 - 0 = 5) and write that answer below:
0
886590
-0
5
Step 5
Move the second digit of the dividend (9) down like so:
0
886590
-0
59
Step 6
The divisor (886) goes into the bottom number (59), 0 time(s), so we can put 0 on top:
00
886590
-0
59
Step 7
If we multiply the divisor by the result in the previous step (886 x 0 = 0), we can now add that answer below the dividend:
00
886590
-0
59
0
Step 8
Next, we will subtract the result from the previous step from the third digit of the dividend (59 - 0 = 59) and write that answer below:
00
886590
-0
59
-0
59
Step 9
Move the third digit of the dividend (0) down like so:
00
886590
-0
59
-0
590
Step 10
The divisor (886) goes into the bottom number (590), 0 time(s), so we can put 0 on top:
000
886590
-0
59
-0
590
Step 11
If we multiply the divisor by the result in the previous step (886 x 0 = 0), we can now add that answer below the dividend:
000
886590
-0
59
-0
590
0
Step 12
Next, we will subtract the result from the previous step from the fourth digit of the dividend (590 - 0 = 590) and write that answer below:
000
886590
-0
59
-0
590
-0
590
So, what is the answer to 590 divided by 886?
If you made it this far into the tutorial, well done! There are no more digits to move down from the dividend, which means we have completed the long division problem.
Your answer is the top number, and any remainder will be the bottom number. So, for 590 divided by 886, the final solution is:
0
Remainder 590
Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 590 Divided by 886 Using Long Division?". VisualFractions.com. Accessed on August 16, 2022. http://visualfractions.com/calculator/long-division/what-is-590-divided-by-886-using-long-division/.
• "What is 590 Divided by 886 Using Long Division?". VisualFractions.com, http://visualfractions.com/calculator/long-division/what-is-590-divided-by-886-using-long-division/. Accessed 16 August, 2022.
• What is 590 Divided by 886 Using Long Division?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/long-division/what-is-590-divided-by-886-using-long-division/.
Extra calculations for you
Now you've learned the long division approach to 590 divided by 886, here are a few other ways you might do the calculation:
• Using a calculator, if you typed in 590 divided by 886, you'd get 0.6659.
• You could also express 590/886 as a mixed fraction: 0 590/886
• If you look at the mixed fraction 0 590/886, you'll see that the numerator is the same as the remainder (590), the denominator is our original divisor (886), and the whole number is our final answer (0).
Long Division Calculator
Enter another long division problem to solve
/
Next Long Division Problem
Eager for more long division but can't be bothered to type two numbers into the calculator above? No worries. Here's the next problem for you to solve:
What is 590 divided by 887 using long division?
Random Long Division Problems
If you made it this far down the page then you must REALLY love long division problems, huh? Below are a bunch of randomly generated calculations for your long dividing pleasure: | {
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Math Help - propositional functions and set operators
1. #1
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Question propositional functions and set operators
Hi guys,
I'm new on forum. With searching I got some answers of my questions. TY for that. But I couldn't find exact answer for this Q's.
1) Are all functions are propositional functions or not?
2) How can we relate union, intersection, complement with logical connectives (AND, OR, NOT)?
Thanks
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2. #2
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Re: propositional functions and set operators
Quote Originally Posted by eaksoy View Post
1) Are all functions are propositional functions or not?
No, functions from real numbers to real numbers are not propositional just because their domain and codomain (range) are not the set {T, F} of truth values.
Quote Originally Posted by eaksoy View Post
2) How can we relate union, intersection, complement with logical connectives (AND, OR, NOT)?
The connection is made using characteristic functions. Given a subset S of some universal set U, the characteristic function of S, \chi_S:U\to\{T,F\}, is defined as follows.
\chi_S(x)=\begin{cases}T & x\in S\\F & x\notin S\end{cases}
(Usually 1 and 0 are used instead of T and F.) Then for any sets A, B, we have \chi_{A\cap B}(x)=\chi_A(x)\text{ AND }\chi_B(x), and similarly for other set operations.
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首页 > 脚本专栏 > python > Python numpy.transpose
Python numpy.transpose使用详解
作者：November丶Chopin
本文主要介绍了Python numpy.transpose使用详解，文中通过示例代码介绍的非常详细，对大家的学习或者工作具有一定的参考学习价值，需要的朋友们下面随着小编来一起学习学习吧
前言
看Python代码时，碰见 numpy.transpose 用于高维数组时挺让人费解，通过一番画图分析和代码验证，发现 transpose 用法还是很简单的。
注：评论中说的三维坐标图中的 0 1 2 3 标反了，已经修正，感谢大家提醒（2019.02）。
正文
Numpy 文档 numpy.transpose 中做了些解释，transpose 作用是改变序列，下面是一些文档Examples：
代码1：
x = np.arange(4).reshape((2,2))
输出1：
#x 为：
array([[0, 1],
[2, 3]])
代码2：
import numpy as np
x.transpose()
输出2：
array([[0, 2],
[1, 3]])
对于二维 ndarray，transpose在不指定参数是默认是矩阵转置。如果指定参数，有如下相应结果：
代码3：
x.transpose((0,1))
输出3：
# x 没有变化
array([[0, 1],
[2, 3]])
代码4：
x.transpose((1,0))
输出4：
# x 转置了
array([[0, 2],
[1, 3]])
这个很好理解：
对于x，因为：
代码5：
x[0][0] == 0
x[0][1] == 1
x[1][0] == 2
x[1][1] == 3
我们不妨设第一个方括号“[]”为 0轴 ，第二个方括号为 1轴 ，则x可在 0-1坐标系 下表示如下：
这里写图片描述
代码6：
因为 x.transpose((0,1)) 表示按照原坐标轴改变序列，也就是保持不变
而 x.transpose((1,0)) 表示交换 ‘0轴’ 和 ‘1轴’，所以就得到如下图所示结果：
这里写图片描述
注意，任何时候你都要保持清醒，告诉自己第一个方括号“[]”为 0轴 ，第二个方括号为 1轴
此时，transpose转换关系就清晰了。
我们来看一个三维的：
代码7：
import numpy as np
# A是array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15])
A = np.arange(16)
# 将A变换为三维矩阵
A = A.reshape(2,2,4)
print(A)
输出7：
A = array([[[ 0, 1, 2, 3],
[ 4, 5, 6, 7]],
[[ 8, 9, 10, 11],
[12, 13, 14, 15]]])
我们对上述的A表示成如下三维坐标的形式：
在这里插入图片描述
所以对于如下的变换都很好理解啦：
代码8：
A.transpose((0,1,2)) #保持A不变
A.transpose((1,0,2)) #将 0轴 和 1轴 交换
将 0轴 和 1轴 交换：
在这里插入图片描述
此时，输出
代码9：
A.transpose((1,0,2)) [0][1][2] #根据上图这个结果应该是10
后面不同的参数以此类推。
到此这篇关于Python numpy.transpose使用详解的文章就介绍到这了,更多相关Python numpy.transpose内容请搜索脚本之家以前的文章或继续浏览下面的相关文章希望大家以后多多支持脚本之家！
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-13,801,553,250,401,110 | MixedRadix
MixedRadix[{b1,,bn}]
represents the list of bases of a numerical system in which different digits have different bases.
Details
• A numeric system with bases {b1,,bn} can express numbers from 0 to b1×b2××bn-1. Larger numbers are represented using an extended list of bases, effectively prepended with base Infinity.
Examples
open allclose all
Basic Examples (2)
Express an integer as a list of digits in a mixed radix numerical system:
Reconstruct that number:
Make explicit the bases used:
Scope (5)
Use IntegerDigits with a mixed radix:
Use FromDigits with a mixed radix:
Use BaseForm with a mixed radix:
Use IntegerLength with a mixed radix:
Use IntegerReverse with a mixed radix:
Applications (3)
A primorial number system uses a mixed radix of primes:
A factorial number system uses a range of integers as mixed radix. The last digit is always 0:
Construct a list from the names of tactical units in a Roman army:
A legion was made of 10 cohorts, a cohort of 6 centuries, a century of 10 contuberniae, and a contubernia of 8 soldiers:
Decompose a number of Roman soldiers in these tactical units:
Conversely, this is the number of soldiers in a legion:
Properties & Relations (6)
IntegerDigits with a single base is equivalent to a MixedRadix list repeating that base:
The inverse operation of IntegerDigits with a mixed radix is performed by FromDigits with the same mixed radix:
The digit at a given position can be between 0 and the corresponding base minus one:
The next number will need one more digit:
That result is equivalent to using a list of bases prepended with Infinity:
Any positive integer is then representable:
The use of the Infinity base is made explicit by BaseForm:
An empty list of bases is effectively equivalent to the list {Infinity}:
IntegerDigits with a MixedRadix specification performs a NumberDecompose operation:
FromDigits with a MixedRadix specification performs a NumberCompose operation:
Wolfram Research (2015), MixedRadix, Wolfram Language function, https://reference.wolfram.com/language/ref/MixedRadix.html.
Text
Wolfram Research (2015), MixedRadix, Wolfram Language function, https://reference.wolfram.com/language/ref/MixedRadix.html.
CMS
Wolfram Language. 2015. "MixedRadix." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/MixedRadix.html.
APA
Wolfram Language. (2015). MixedRadix. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/MixedRadix.html
BibTeX
@misc{reference.wolfram_2022_mixedradix, author="Wolfram Research", title="{MixedRadix}", year="2015", howpublished="\url{https://reference.wolfram.com/language/ref/MixedRadix.html}", note=[Accessed: 04-June-2023 ]}
BibLaTeX
@online{reference.wolfram_2022_mixedradix, organization={Wolfram Research}, title={MixedRadix}, year={2015}, url={https://reference.wolfram.com/language/ref/MixedRadix.html}, note=[Accessed: 04-June-2023 ]} | {
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} | 0c090d63199a0a01e3b08e4a255778a0 |
-7,603,534,345,765,457,000 | Lines r and s are parallel
advertisement
Lines r and s are parallel.
Think about it:
You already know that when any two lines intersect, vertical angles are always congruent and adjacent angles are always supplementary. You also know that when two parallel lines are intersected by a transversal, corresponding angles are congruent.
Answer the following questions.
1.
1 is congruent to 3 because they are _________________ angles.
2.
1 is supplementary to 4 because they are ________________ angles.
3.
1 is congruent to 5 because they are ___________________ angles.
4.
So 5 would also be ___________________ to 4.
Try it:
1.
List all angles that are congruent to 1._____________________________
2.
List all angles that are supplementary to 1.__________________________
3.
Are 2 and 6 congruent or supplementary?
4.
Are 2 and 5 congruent or supplementary?
5.
Suppose the measure of 2 is 140 o
, what is the measure of 8?
6.
Suppose the measure of 1 is (x + 15) o
and the measure of 8 is 130 , then what is the value of x?
7.
Suppose the measure of 1 is (3x + 25) o
and the measure of 5 is (x + 75) o
, then what is the value of x?
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-2,326,288,392,116,558,000 | What is the compound interest on $82815 at 20% over 28 years?
If you want to invest $82,815 over 28 years, and you expect it will earn 20.00% in annual interest, your investment will have grown to become $13,651,610.71.
$
$
If you're on this page, you probably already know what compound interest is and how a sum of money can grow at a faster rate each year, as the interest is added to the original principal amount and recalculated for each period.
The actual rate that $82,815 compounds at is dependent on the frequency of the compounding periods. In this article, to keep things simple, we are using an annual compounding period of 28 years, but it could be monthly, weekly, daily, or even continuously compounding.
The formula for calculating compound interest is:
$$A = P(1 + \dfrac{r}{n})^{nt}$$
• A is the amount of money after the compounding periods
• P is the principal amount
• r is the annual interest rate
• n is the number of compounding periods per year
• t is the number of years
We can now input the variables for the formula to confirm that it does work as expected and calculates the correct amount of compound interest.
For this formula, we need to convert the rate, 20.00% into a decimal, which would be 0.2.
$$A = 82815(1 + \dfrac{ 0.2 }{1})^{ 28}$$
As you can see, we are ignoring the n when calculating this to the power of 28 because our example is for annual compounding, or one period per year, so 28 × 1 = 28.
How the compound interest on $82,815 grows over time
The interest from previous periods is added to the principal amount, and this grows the sum a rate that always accelerating.
The table below shows how the amount increases over the 28 years it is compounding:
Start Balance Interest End Balance
1 $82,815.00 $16,563.00 $99,378.00
2 $99,378.00 $19,875.60 $119,253.60
3 $119,253.60 $23,850.72 $143,104.32
4 $143,104.32 $28,620.86 $171,725.18
5 $171,725.18 $34,345.04 $206,070.22
6 $206,070.22 $41,214.04 $247,284.26
7 $247,284.26 $49,456.85 $296,741.12
8 $296,741.12 $59,348.22 $356,089.34
9 $356,089.34 $71,217.87 $427,307.21
10 $427,307.21 $85,461.44 $512,768.65
11 $512,768.65 $102,553.73 $615,322.38
12 $615,322.38 $123,064.48 $738,386.86
13 $738,386.86 $147,677.37 $886,064.23
14 $886,064.23 $177,212.85 $1,063,277.08
15 $1,063,277.08 $212,655.42 $1,275,932.49
16 $1,275,932.49 $255,186.50 $1,531,118.99
17 $1,531,118.99 $306,223.80 $1,837,342.79
18 $1,837,342.79 $367,468.56 $2,204,811.35
19 $2,204,811.35 $440,962.27 $2,645,773.61
20 $2,645,773.61 $529,154.72 $3,174,928.34
21 $3,174,928.34 $634,985.67 $3,809,914.01
22 $3,809,914.01 $761,982.80 $4,571,896.81
23 $4,571,896.81 $914,379.36 $5,486,276.17
24 $5,486,276.17 $1,097,255.23 $6,583,531.40
25 $6,583,531.40 $1,316,706.28 $7,900,237.68
26 $7,900,237.68 $1,580,047.54 $9,480,285.22
27 $9,480,285.22 $1,896,057.04 $11,376,342.26
28 $11,376,342.26 $2,275,268.45 $13,651,610.71
We can also display this data on a chart to show you how the compounding increases with each compounding period.
As you can see if you view the compounding chart for $82,815 at 20.00% over a long enough period of time, the rate at which it grows increases over time as the interest is added to the balance and new interest calculated from that figure.
How long would it take to double $82,815 at 20% interest?
Another commonly asked question about compounding interest would be to calculate how long it would take to double your investment of $82,815 assuming an interest rate of 20.00%.
We can calculate this very approximately using the Rule of 72.
The formula for this is very simple:
$$Years = \dfrac{72}{Interest\: Rate}$$
By dividing 72 by the interest rate given, we can calculate the rough number of years it would take to double the money. Let's add our rate to the formula and calculate this:
$$Years = \dfrac{72}{ 20 } = 3.6 $$
Using this, we know that any amount we invest at 20.00% would double itself in approximately 3.6 years. So $82,815 would be worth $165,630 in ~3.6 years.
We can also calculate the exact length of time it will take to double an amount at 20.00% using a slightly more complex formula:
$$Years = \dfrac{log(2)}{log(1 + 0.2)} = 3.8\; years$$
Here, we use the decimal format of the interest rate, and use the logarithm math function to calculate the exact value.
As you can see, the exact calculation is very close to the Rule of 72 calculation, which is much easier to remember.
Hopefully, this article has helped you to understand the compound interest you might achieve from investing $82,815 at 20.00% over a 28 year investment period. | {
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6,529,067,404,109,244,000 | Kyoya and Train
题目描述
Kyoya Ootori wants to take the train to get to school. There are $n$ train stations and $m$ one-way train lines going between various stations. Kyoya is currently at train station $1$, and the school is at station $n$. To take a train, he must pay for a ticket, and the train also takes a certain amount of time. However, the trains are not perfect and take random amounts of time to arrive at their destination. If Kyoya arrives at school strictly after $t$ time units, he will have to pay a fine of $x$.
Each train line is described by a ticket price, and a probability distribution on the time the train takes. More formally, train line $i$ has ticket cost $c_i$, and a probability distribution $p_{i, k}$ which denotes the probability that this train will take $k$ time units for all $1 \le k \le t$. Amounts of time that each of the trains used by Kyouya takes are mutually independent random values (moreover, if Kyoya travels along the same train more than once, it is possible for the train to take different amounts of time and those amounts are also independent one from another).
Kyoya wants to get to school by spending the least amount of money in expectation (for the ticket price plus possible fine for being late). Of course, Kyoya has an optimal plan for how to get to school, and every time he arrives at a train station, he may recalculate his plan based on how much time he has remaining. What is the expected cost that Kyoya will pay to get to school if he moves optimally?
题意概述
有$n$座城市和$m$条铁路，第$i$条铁路从$a_i$号城市出发，到达$b_i$号城市，票价为$c_i$。某人要在$t$个单位时间内从$1$号城市到$n$号城市，如果超过时间就会被罚款$x$。已知每次搭乘第$i$条铁路有$p_{i, k}$的概率需要$k \; (1 \le k \le t)$个单位时间。求在采取最优策略的情况下总花费的期望。
数据范围：$2 \le n \le 50, \; 1 \le m \le 100, \; 1 \le t \le 20000, \; 0 \le x \le 10^6$。
算法分析
首先考虑最暴力的DP。令$f_{i, j}$表示当前时刻为$j$，在最优策略下从$i$号城市到$n$号城市的总花费的期望。那么
$$
f_{i, j}=
\begin{cases}
\min(c_e+\sum_{k=1}^t p_{e, k} \cdot f_{b_e, j+k} \mid e \in [1, m] \land a_e=i), & i \lt n \land j \le t \\
d_{i, n}+x , & i \lt n \land j \gt t \\
0, & i=n \land j \le t \\
x, & i=n \land j \gt t
\end{cases}
$$
第二部分中的$d_{i, n}$表示在只考虑铁路票价的情况下从$i$号城市到$n$号城市的最小花费（因为已经超时了，所以不如走总票价最少的路）。
这样做的复杂度是$O(mt^2)$的，需要对它进行优化。令
$$S_{e, j}=\sum_{k=1}^t p_{e, k} \cdot f_{b_e, j+k}$$
转移方程的第一部分变为
$$f_{i, j}=\min(c_e+S_{e, j} \mid e \in [1, m] \land a_e=i)$$
对于$S$，它类似于卷积，可以将其中一部分翻转后用FFT求值；对于$f$，可以用CDQ分治，统计较大的$j$对较小的$j$的贡献。
具体来说，假设我们要计算$l \le j \le r$的$f_{i, j}$，令$mid=\lfloor {l+r \over 2} \rfloor$，那么先计算$mid \lt j \le r$的$f_{i, j}$，用这些来更新$l \le j \le mid$的$S_{e, j}$（$m$条边分别用FFT更新，复杂度$O(m(r-l)\log (r-l))$），最后计算$l \le j \le mid$的$f_{i, j}$。当$l=r$时$f$可以直接由$S$得到。总复杂度降至$O(mt\log^2t)$。
实现时细节很多，要注意DP边界条件和FFT下标变换。
代码
/*
* Your nature demands love and your happiness depends on it.
*/
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
static int const N = 55;
static int const M = 105;
static int const T = 100000;
static double const PI = acos(-1);
int rev[T];
class Point {
public:
double x, y;
Point(double _x = 0, double _y = 0) : x(_x), y(_y) {}
std::pair<double, double> pair() {
return std::make_pair(x, y);
}
friend Point operator+(Point const &a, Point const &b) {
return Point(a.x + b.x, a.y + b.y);
}
friend Point operator-(Point const &a, Point const &b) {
return Point(a.x - b.x, a.y - b.y);
}
friend Point operator*(Point const &a, Point const &b) {
return Point(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}
Point operator/(double const &n) {
return Point(x / n, y / n);
}
} wn[T], A[T], B[T];
int init(int n) {
int m = n, l = 0;
for (n = 1; n <= m; n <<= 1, ++l)
;
for (int i = 1; i < n; ++i)
rev[i] = rev[i >> 1] >> 1 | (i & 1) << l - 1;
for (int i = 0; i < n >> 1; ++i)
wn[i] = Point(cos(2 * PI / n * i), sin(2 * PI / n * i));
return n;
}
void fft(Point *a, int n, int inv = 0) {
for (int i = 0; i < n; ++i)
if (i < rev[i])
std::swap(a[i], a[rev[i]]);
for (int i = 1; i < n; i <<= 1)
for (int j = 0; j < n; j += i << 1)
for (int k = 0; k < i; ++k) {
Point x = a[j + k], y = wn[n / (i << 1) * k] * a[j + k + i];
a[j + k] = x + y, a[j + k + i] = x - y;
}
if (inv) {
std::reverse(a + 1, a + n);
for (int i = 0; i < n; ++i)
a[i] = a[i] / n;
}
}
int n, m, t, x, mp[N][N];
double p[M][T], s[M][T], f[N][T];
struct Line {
int a, b, c;
} li[M];
void update(int l, int r) {
int mid = l + r >> 1, len = init(r - l + r - mid - 2);
for (int i = 1; i <= m; ++i) {
for (int j = 0; j < len; ++j)
A[j] = B[j] = 0;
for (int j = mid + 1; j <= r; ++j)
A[j - mid - 1] = f[li[i].b][r - j + mid + 1];
for (int j = 1; j <= r - l; ++j)
B[j - 1] = p[i][j];
fft(A, len), fft(B, len);
for (int j = 0; j < len; ++j)
A[j] = A[j] * B[j];
fft(A, len, 1);
for (int j = l; j <= mid; ++j)
s[i][j] += A[r - j - 1].x;
}
}
void solve(int l, int r) {
if (l == r) {
for (int i = 1; i <= m; ++i)
f[li[i].a][l] = std::min(f[li[i].a][l], s[i][l] + li[i].c);
return;
}
int mid = l + r >> 1;
solve(mid + 1, r), update(l, r), solve(l, mid);
}
int main() {
scanf("%d%d%d%d", &n, &m, &t, &x);
memset(mp, 0x3f, sizeof mp);
for (int i = 1; i <= m; ++i) {
scanf("%d%d%d", &li[i].a, &li[i].b, &li[i].c);
mp[li[i].a][li[i].b] = std::min(mp[li[i].a][li[i].b], li[i].c);
for (int j = 1; j <= t; ++j)
scanf("%lf", &p[i][j]), p[i][j] /= 100000;
}
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
for (int k = 1; k <= n; ++k)
mp[j][k] = std::min(mp[j][k], mp[j][i] + mp[i][k]);
for (int i = 1; i <= n; ++i)
for (int j = 0; j <= t << 1; ++j)
if (i == n)
if (j <= t)
f[i][j] = 0;
else
f[i][j] = x;
else
if (j <= t)
f[i][j] = 1e9;
else
f[i][j] = x + mp[i][n];
update(0, t << 1), solve(0, t), printf("%.8lf\n", f[1][0]);
return 0;
}
The Child and Binary Tree
题目描述
Our child likes computer science very much, especially he likes binary trees.
Consider the sequence of $n$ distinct positive integers: $c_1, c_2, \ldots, c_n$. The child calls a vertex-weighted rooted binary tree good if and only if for every vertex $v$, the weight of $v$ is in the set ${c_1, c_2, \ldots, c_n}$. Also our child thinks that the weight of a vertex-weighted tree is the sum of all vertices’ weights.
Given an integer $m$, can you for all $s \; (1 \le s \le m)$ calculate the number of good vertex-weighted rooted binary trees with weight $s$? Please, check the samples for better understanding what trees are considered different.
We only want to know the answer modulo $998244353$ ($7 \times 17 \times 2^{23}+1$, a prime number).
题意概述
有$n$种点，每种点有无限个，第$i$种点的权值为$c_i$。定义一棵二叉树的权值等于它所有点的权值之和。求对于所有$s \in [1, m]$，权值为$s$的二叉树有几棵。两棵二叉树不同当且仅当它们左子树或右子树不同，或者根节点权值不同。
数据范围：$1 \le n, m, c_i \le 10^5$。
算法分析
令$f(x)$表示权值为$x$的二叉树个数，$F(x)$为其生成函数（$F(x)=\sum_{i \ge 0} f(i)x^i$）。
令$C(x)$为给定$c$的集合的生成函数（$C(x)=\sum_{i=1}^n x^{c_i}$）。
根据DP转移方程，易知
$$
f(x)=\sum_{w \in {c_1, c_2, \ldots, c_n}} \sum_{i=0}^{x-w} f(i)f(x-w-i)
$$
$$
F(x)=C(x)F(x)^2+1
$$
解得
$$
F(x)={1 \pm \sqrt{1-4C(x)} \over 2C(x)}={2 \over 1 \pm \sqrt{1-4C(x)}}
$$
显然，若取减号，则当$x$趋近$0$时分母为$0$，因此只能取加号。接着就是多项式开根和多项式求逆了。
• 多项式求逆：
求$GF \equiv 1 \pmod {x^n}$。
假设已知$G_0F \equiv 1 \pmod {x^{\lceil n/2 \rceil}}$
$G-G_0 \equiv 0 \pmod {x^{\lceil n/2 \rceil}}$
$G^2-2GG_0+G_0^2 \equiv 0 \pmod {x^n}$
$G-2G_0+G_0^2F \equiv 0 \pmod {x^n}$
$G \equiv 2G_0-G_0^2F \pmod {x^n}$
• 多项式开根：
求$G^2 \equiv F \pmod {x^n}$。
假设已知$G_0^2 \equiv F \pmod {x^{\lceil n/2 \rceil}}$
$(G_0^2-F)^2 \equiv 0 \pmod {x^n}$
$(G_0^2+F)^2 \equiv 4G_0^2F \pmod {x^n}$
$\left({G_0^2+F \over 2G_0}\right)^2 \equiv F \pmod {x^n}$
$G \equiv {G_0+G_0^{-1}F \over 2} \pmod {x^n}$
代码
#include <algorithm>
#include <cstdio>
#include <cstring>
static const int N = 500000;
static const int MOD = 998244353;
static const int G = 3;
static const int INV2 = 499122177;
int n, m, c[N], C[N], rev[N], wn[N], tmp[N], tmp2[N], tmp3[N];
int power(int a, int b) {
int ret = 1;
for (a %= MOD, b %= MOD - 1; b; b >>= 1)
b & 1 && (ret = 1ll * ret * a % MOD), a = 1ll * a * a % MOD;
return ret;
}
void init(int &n) {
int m = n << 1, l = 0;
for (n = 1; n < m; n <<= 1, ++l)
;
for (int i = 1; i < n; ++i)
rev[i] = rev[i >> 1] >> 1 | (i & 1) << (l - 1);
}
void ntt(int *a, int n, bool inv) {
for (int i = 0; i < n; ++i)
if (i < rev[i])
std::swap(a[i], a[rev[i]]);
wn[0] = 1, wn[1] = power(G, (MOD - 1) / n);
for (int i = 2; i < n >> 1; ++i)
wn[i] = 1ll * wn[i - 1] * wn[1] % MOD;
for (int i = 1; i < n; i <<= 1)
for (int j = 0; j < n; j += i << 1)
for (int k = 0; k < i; ++k) {
int x = a[j + k], y = 1ll * wn[n / (i << 1) * k] * a[j + k + i] % MOD;
a[j + k] = (x + y) % MOD, a[j + k + i] = (MOD + x - y) % MOD;
}
if (inv) {
for (int i = 1; i < n >> 1; ++i)
std::swap(a[i], a[n - i]);
int rec = power(n, MOD - 2);
for (int i = 0; i < n; ++i)
a[i] = 1ll * a[i] * rec % MOD;
}
}
void get_inv(int *f, int *g, int n) {
if (n == 1)
return void(g[0] = power(f[0], MOD - 2));
int rec = n;
get_inv(f, g, (n + 1) >> 1), init(n);
for (int i = (rec + 1) >> 1; i < n; ++i)
g[i] = 0;
for (int i = 0; i < rec; ++i)
tmp[i] = f[i];
for (int i = rec; i < n; ++i)
tmp[i] = 0;
ntt(g, n, 0), ntt(tmp, n, 0);
for (int i = 0; i < n; ++i)
g[i] = 1ll * g[i] * (MOD + 2 - 1ll * g[i] * tmp[i] % MOD) % MOD;
ntt(g, n, 1);
for (int i = rec; i < n; ++i)
g[i] = 0;
}
void get_sqrt(int *f, int *g, int n) {
if (n == 1)
return void(g[0] = 1);
int rec = n;
get_sqrt(f, g, (n + 1) >> 1);
for (int i = 0; i<(n + 1)>> 1; ++i)
tmp2[i] = g[i];
for (int i = (n + 1) >> 1; i < n; ++i)
tmp2[i] = 0;
get_inv(tmp2, tmp3, n), init(n);
for (int i = (rec + 1) >> 1; i < n; ++i)
g[i] = 0;
for (int i = 0; i < rec; ++i)
tmp2[i] = f[i];
for (int i = rec; i < n; ++i)
tmp2[i] = 0;
ntt(tmp2, n, 0), ntt(tmp3, n, 0);
for (int i = 0; i < n; ++i)
tmp3[i] = 1ll * tmp3[i] * tmp2[i] % MOD;
ntt(tmp3, n, 1);
for (int i = 0; i < rec; ++i)
g[i] = 1ll * (g[i] + tmp3[i]) * INV2 % MOD;
for (int i = rec; i < n; ++i)
g[i] = 0;
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i < n; ++i)
scanf("%d", &c[i]);
for (int i = 0; i < n; ++i)
if (c[i] <= m)
++C[c[i]];
for (int i = 1; i <= m; ++i)
C[i] = (MOD - (C[i] << 2)) % MOD;
C[0] = 1;
get_sqrt(C, c, m + 1), ++c[0], get_inv(c, C, m + 1);
for (int i = 1; i <= m; ++i)
printf("%d\n", (C[i] << 1) % MOD);
return 0;
}
序列求和 V4
题意概述
给定$n$和$k$，求$\sum_{i=1}^n i^k \bmod 10^9+7$。有$T$组数据。
数据范围：$1 \le T \le 500, \; 1 \le n \le 10^{18}, \; 1 \le k \le 50000$。
算法分析1
设答案为$f(n)$。这是一个$(k+1)$次多项式，只要确定$(k+2)$个点就可以确定$f$。
我们可以令$x=1\ldots k+2$，在$O(k\log k)$时间内计算出它们对应的$y$。
利用拉格朗日插值法。假设有$n$个点$(x_i, y_i)$，那么
$$f(x)=\sum_{i=1}^n {\prod_{j \in [1, n] \land j \neq i} (x-x_j) \over \prod_{j \in [1, n] \land j \neq i} (x_i-x_j)}y_i$$
考虑分子部分。在给定$x$的情况下，这一部分可以$O(k)$预处理$O(1)$求值。
考虑分母部分。它形如$\prod_{j \in [i-k-2, i-1] \land j \neq 0} j$，可以在插值时$O(1)$维护。
总时间复杂度为$O(Tk\log k)$。
代码1
#include <cstdio>
#define int long long
void read(int &n) {
char c; while ((c = getchar()) < '0' || c > '9') ; n = c - '0';
while ((c = getchar()) >= '0' && c <= '9') (n *= 10) += c - '0';
}
static const int K = 50005;
static const int MOD = 1000000007;
int n, k, a[K], p[K], q[K], inv[K], base[K];
int power(int a, int b) {
int ret = 1; a %= MOD, b %= MOD - 1;
while (b) b & 1 && ((ret *= a) %= MOD), (a *= a) %= MOD, b >>= 1;
return ret;
}
int calc(int n) {
if (n <= k + 2) return a[n]; n %= MOD;
int w = power(base[k + 2], MOD - 2), ans = 0;
p[0] = q[k + 3] = 1;
for (int i = 1; i <= k + 2; ++ i) p[i] = p[i - 1] * (n - i) % MOD;
for (int i = k + 2; i; -- i) q[i] = q[i + 1] * (n - i) % MOD;
for (int i = 1; i <= k + 2; ++ i) (ans += a[i] * w % MOD * p[i - 1] % MOD * q[i + 1]) %= MOD, w = w * (i - k - 2) % MOD * inv[i] % MOD;
return (ans + MOD) % MOD;
}
signed main() {
int T; read(T), base[1] = 1;
for (int i = 1; i < K; ++ i) inv[i] = power(i, MOD - 2);
for (int i = 2; i < K; ++ i) base[i] = base[i - 1] * (MOD + 1 - i) % MOD;
while (T --) {
read(n), read(k);
for (int i = 1; i < k + 3; ++ i) a[i] = (a[i - 1] + power(i, k)) % MOD;
printf("%lld\n", calc(n));
}
return 0;
}
算法分析2
此题也可以用伯努利数解决，因为
$$\sum_{i=1}^n i^k={1 \over k+1} \sum_{i=0}^k (-1)^i {k+1 \choose i} B_in^{k+1-i}$$
其中伯努利数由其指数型生成函数${x \over e^x-1}$定义。易知
$$\sum_{i=0}^{\infty} B_i{x^i \over i!}={x \over e^x-1}={x \over \sum_{i=1}^{\infty} {x^i \over i!}}={1 \over \sum_{i=0}^{\infty} {x^i \over (i+1)!}}$$
只要对分母求多项式逆元，即可得到伯努利数，时间复杂度$O(k\log k)$。其它部分均可$O(k)$预处理，答案也可以$O(k)$计算。因此总时间复杂度是$O(k\log k+Tk)$。
代码2
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#define int long long
static int const N = 200000;
static int const MOD = 1000000007;
static int const M = 31622;
static double const PI = acos(-1);
int b[N], rev[N], fac[N], inv[N];
class complex {
private:
double x, y;
public:
complex(double _x = 0, double _y = 0) : x(_x), y(_y) {}
double real() { return x; }
friend complex operator+(complex const &a, complex const &b) {
return complex(a.x + b.x, a.y + b.y);
}
friend complex operator-(complex const &a, complex const &b) {
return complex(a.x - b.x, a.y - b.y);
}
friend complex operator*(complex const &a, complex const &b) {
return complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}
} wn[N], A[N], B[N], C[N], D[N], E[N], F[N], G[N];
int power(int a, int b) {
int ret = 1;
for (; b; b >>= 1)
b & 1 && ((ret *= a) %= MOD), (a *= a) %= MOD;
return ret;
}
int init(int m) {
int n = 1, l = 0;
for (; n <= m; n <<= 1, ++l)
;
for (int i = 1; i < n; ++i)
rev[i] = rev[i >> 1] >> 1 | (i & 1) << l - 1;
for (int i = 0; i < n >> 1; ++i)
wn[i] = complex(cos(2 * PI / n * i), sin(2 * PI / n * i));
return n;
}
void fft(complex *a, int n, bool inv = 0) {
for (int i = 0; i < n; ++i)
if (i < rev[i])
std::swap(a[i], a[rev[i]]);
for (int i = 1; i < n; i <<= 1)
for (int j = 0; j < n; j += i << 1)
for (int k = 0; k < i; ++k) {
complex x = a[j + k], y = wn[n / (i << 1) * k] * a[j + k + i];
a[j + k] = x + y, a[j + k + i] = x - y;
}
if (inv) {
std::reverse(a + 1, a + n);
for (int i = 0; i < n; ++i)
a[i] = a[i].real() / n;
}
}
int get_mod(double t) {
return (int)(fmod(t, MOD) + MOD + 0.5) % MOD;
}
void get_inv(int *f, int *g, int n) {
if (n == 1)
return void(g[0] = power(f[0], MOD - 2));
get_inv(f, g, n + 1 >> 1);
int len = init(n << 1);
for (int i = 0; i < n + 1 >> 1; ++i)
A[i] = g[i] / M, B[i] = g[i] % M;
for (int i = n + 1 >> 1; i < len; ++i)
A[i] = B[i] = 0;
for (int i = 0; i < n; ++i)
C[i] = f[i] / M, D[i] = f[i] % M;
for (int i = n; i < len; ++i)
C[i] = D[i] = 0;
fft(A, len), fft(B, len), fft(C, len), fft(D, len);
for (int i = 0; i < len; ++i) {
E[i] = 0 - A[i] * C[i];
F[i] = 0 - A[i] * D[i] - B[i] * C[i];
G[i] = 2 - B[i] * D[i];
}
fft(E, len, 1), fft(F, len, 1), fft(G, len, 1);
for (int i = 0; i < n; ++i) {
int x = get_mod(E[i].real()) * M % MOD * M % MOD;
int y = get_mod(F[i].real()) * M % MOD;
int z = get_mod(G[i].real());
int w = (x + y + z) % MOD;
C[i] = w / M, D[i] = w % M;
}
for (int i = n; i < len; ++i)
C[i] = D[i] = 0;
fft(C, len), fft(D, len);
for (int i = 0; i < len; ++i) {
E[i] = A[i] * C[i];
F[i] = A[i] * D[i] + B[i] * C[i];
G[i] = B[i] * D[i];
}
fft(E, len, 1), fft(F, len, 1), fft(G, len, 1);
for (int i = 0; i < n; ++i) {
int x = get_mod(E[i].real()) * M % MOD * M % MOD;
int y = get_mod(F[i].real()) * M % MOD;
int z = get_mod(G[i].real());
g[i] = (x + y + z) % MOD;
}
}
int get_c(int n, int m) {
return fac[n] * inv[m] % MOD * inv[n - m] % MOD;
}
signed main() {
int T;
fac[0] = 1;
for (int i = 1; i < N; ++i)
fac[i] = fac[i - 1] * i % MOD;
inv[N - 1] = power(fac[N - 1], MOD - 2);
for (int i = N - 1; i; --i)
inv[i - 1] = inv[i] * i % MOD;
get_inv(inv + 1, b, 50001);
for (int i = 0; i < 50001; ++i)
(b[i] *= fac[i]) %= MOD;
for (scanf("%lld", &T); T--;) {
int n, k, ans = 0;
scanf("%lld%lld", &n, &k), n %= MOD;
for (int i = k, f = k & 1 ? -1 : 1, p = n; ~i; --i, f = -f, (p *= n) %= MOD)
ans += (MOD + f) * get_c(k + 1, i) % MOD * b[i] % MOD * p % MOD;
printf("%lld\n", ans % MOD * power(k + 1, MOD - 2) % MOD);
}
return 0;
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-6,082,868,271,612,487,000 | 2
$\begingroup$
I am trying to evaluate the following integral with Mathematica:
\begin{align} I = \int_{0}^{\infty} da \, \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \mbox{sinc}\left(\tfrac{w}{2} a \right) \delta' \left( \frac{D^2}{a}- a \right), \end{align} where the prime on the delta function denotes differentiation with respect to the argument of the Delta function. When I evaluate this integral with Mathematica as:
Integrate[Exp[-a^2/(4 s^2)]/a^2 Sinc[w a / 2] Derivative[1][DiracDelta][D^2/a - a],{a,0,Infinity}, Assumptions -> s > 0 && w > 0 && D > 0]
I get the result: \begin{align} I_{Mathematica} = \frac{e^{-\frac{D^2}{4 s ^2}} }{4 D^4 s ^2 w } \left[\left(D^2+6 s ^2\right) \sin \left(\frac{D w }{2}\right)-D s ^2 w \cos \left(\frac{D w }{2}\right)\right]. \end{align}
However, if I evaluate this integral analytically, using the fact that \begin{align} \frac{d}{da} \delta\left( \frac{D^2}{a}- a \right) = - \delta' \left( \frac{D^2}{a}- a \right) \left(\frac{D^2}{a^2}+1\right) \implies \delta' \left( \frac{D^2}{a}- a \right) = - \left[\frac{d}{da} \delta\left( \frac{D^2}{a}- a \right) \right] \left(\frac{D^2}{a^2}+1\right)^{-1}, \end{align} I get the following result: \begin{align} I_{analytic} &= \int_{0}^{\infty} da \, \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \delta' \left( \frac{D^2}{a}- a \right) \\ &=- \int_{0}^{\infty} da \, \left[\frac{d}{da} \delta\left( \frac{D^2}{a}- a \right) \right] \left(\frac{D^2}{a^2}+1\right)^{-1} \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \\ &= \int_{0}^{\infty} da \, \delta\left( \frac{D^2}{a}- a \right) \left[\frac{d}{da} \left(\frac{D^2}{a^2}+1\right)^{-1} \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \right] \\ &= \int_{0}^{\infty} da \, \frac{\delta\left( D - a \right)}{2} \left[\frac{d}{da} \left(\frac{D^2}{a^2}+1\right)^{-1} \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \right] \\ &= - \frac{e^{-\frac{ D^2 }{ 4s^{2}}}}{4 D^{4} s^{2} w} \left[ \left(D^{2}+4 s^{2}\right)\sin\left( \frac{ Dw}{2} \right) - D s^{2} w \cos \left( \frac{ Dw}{2} \right) \right], \end{align} which differs from $I_{Mathematica}$ by an overall negative sign and the prefactor in front of $s^2$ in the first term.
I'm not sure if the issue is with the way Mathematica handles the derivative of the delta function or if I've made a mistake in my analytic calculation. Any help would be much appreciated, I've been staring at this for days!
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17
• $\begingroup$ Perhaps your analytical formula is wrong? As far as I know DiracDelta'[x]==-DiracDelta[x]/x , which is different from your "fact". $\endgroup$ Mar 31 '20 at 18:53
• 1
$\begingroup$ The formula holds for arbitrary argument! $\endgroup$ Mar 31 '20 at 19:44
• 1
$\begingroup$ Thanks for the discussion! Note that the bounds are differnt so that Integrate[ x Derivative[1][DiracDelta][x], {x, 0, Infinity}] == - 1 + HeavisideTheta[0]. $\endgroup$
– e4alex
Mar 31 '20 at 20:25
• 1
$\begingroup$ Also, ``Integrate[ x Derivative[1][DiracDelta][x], {x, 0, Infinity}]'' is not equal to Integrate[ x^2 Derivative[1][DiracDelta][(x - 1)^2], {x, 0, Infinity}]. However, presumably when you say `The formula holds for any argument', then also the integration measure would change. I believe this is equivalent to the "fact" I stated above. Note I use quotations around "fact" to leave open the possibility that there may be a mistake there, but I do not see one. $\endgroup$
– e4alex
Mar 31 '20 at 20:27
• 1
$\begingroup$ @e4alex s/he says that every time the Dirac delta-function is mentioned, don't worry. $\endgroup$
– Roman
Apr 1 '20 at 8:16
4
$\begingroup$
Let's talk about the Dirac $\delta$-"function". Strictly speaking, it's a linear functional $$\delta:C^\infty(\mathbb R)\to\mathbb R\qquad\qquad\delta(f)=f(0).$$ However, we usually use the notation $$\int_{-\infty}^\infty\delta(x)f(x)dx$$ to denote the evaluation $\delta(f)$. The derivative of the $\delta$-"function" is computed via formal integration by parts: $$\delta'(f)=\int_{-\infty}^\infty\delta'(x)f(x)dx=-\int_{-\infty}^\infty\delta(x)f'(x)dx=-f'(0).$$ Your integral has the additional complications that there is a function inside the argument of $\delta'(x)$, and that the integral is not taken over all of $\mathbb R$. Composing distributions with functions is, in general, not possible, but in this case we can appeal to a theorem of Hormander:
Theorem: Suppose $f:M\to N$ is a smooth function whose differential is everywhere surjective. Then there is a linear map $f^*:\mathscr D(N)\to\mathscr D(M)$ such that $f^*u=u\circ f$ for all $u\in C(N)$.
For our purposes, this means $\int_{-\infty}^\infty\delta'(f(x))g(x)dx$ makes sense provided $f(x)$ is smooth and $f'(x)$ never vanishes. Similarly, reducing the domain of integration is, in general, not possible, but we have:
Theorem Suppose $E_1$ and $E_2$ are disjoint closed sets, and let $\mathscr D_{E_i}$ denote the set of distributions which coincide with a smooth function on $E_i^c$ for $i=1,2$. Then there is a bilinear map $$m:\mathscr D_{E_1}\times\mathscr D_{E_2}\to\mathscr D(\mathbb R^n)$$ such that $m(u,v)=uv$ when $u$ and $v$ are continuous.
In our case, we would like to compute the integral $$\int_0^\infty\delta'\left(\frac{D^2}{x}-x\right)g(x)dx=\int_{-\infty}^\infty\chi_{(0,\infty)}(x)\delta'\left(\frac{D^2}{a}-a\right)g(x)dx,$$ where $\chi_{(0,\infty)}$ is the characteristic function of the half-line $(0,\infty)$. The theorem says that the product $$\chi_{(0,\infty)}(x)\delta'\left(\frac{D^2}{x}-x\right)$$ makes sense whenever the singular support of $\chi_{(0,\infty)}$, namely $\{0\}$, does not intersect the singular support of $\delta'\left(\frac{D^2}{x}-x\right)$, namely $\{D,-D\}$. Thus when $D\neq 0$, our integral makes sense and $$\int_0^\infty\delta'\left(\frac{D^2}{x}-x\right)g(x)dx=\begin{cases}g'(D),&D>0\\g(-D),&D<0\end{cases}.$$ To compute your integral, just plug in your particular function $g(x)$. When you're working with distributions (like $\delta$) you need to be very careful about what you do with them. I don't know how Mathematica conceptualizes the $\delta$-distribution, but I wouldn't be inclined trust that it would go through the necessary analytical reasoning and get the right answer.
TL;DR: Do your distributional calculus by hand.
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10
• $\begingroup$ Thank you so much for those theorems that rigorously justify the composition and multiplication of distribution, placing the integral we seek to evaluate on solid ground! The detailed reply is much appreciated. $\endgroup$
– e4alex
Apr 2 '20 at 23:47
• $\begingroup$ @ AestheticAnalyst However, I'm not sure I agree with the evaluation of the integral in your last equation, or perhaps it was misunderstood what was meant by $\delta'(f(x))$. Consider $\frac{d}{dx} ( \frac{D^2}{x} - x )= - \delta'( \frac{D^2}{x} - x ) \left(\frac{D^2}{x^2}+1 \right)$. $\endgroup$
– e4alex
Apr 3 '20 at 0:36
• $\begingroup$ @ AnethesticAnalyst It follows that: \begin{align} \int_0^\infty \delta'\left( \frac{D^2}{x} - x \right) g(x) \, dx &= - \int_0^\infty \left(\frac{D^2}{x^2}+1 \right)^{-1} g(x) \frac{d}{dx} \delta \left( \frac{D^2}{x} - x \right) \, dx \\ &= \int_0^\infty \frac{d}{dx} \left[ \left(\frac{D^2}{x^2}+1 \right)^{-1} g(x) \right] \delta\left( \frac{D^2}{x} - x \right) \\ &= \frac{1}{2} \frac{d}{dx} \left[ \left(\frac{D^2}{x^2}+1 \right)^{-1} g(x) \right]_{x = D} \end{align} Would you agree? $\endgroup$
– e4alex
Apr 3 '20 at 0:36
• $\begingroup$ @AestheticAnalyst Thank you for this interesting contribution. One remark: Assuming ( Wikipedia ) Derivative[1][DiracDelta][x]==-DiracDelta[x]/x I would get something like Integrate[g[x],x] Derivative[1][DiracDelta][x],{x,0,Infinity}]=-g[a]/(d^2 -a) /.a->d without distinguishing cases of d! $\endgroup$ Apr 3 '20 at 6:39
• 1
$\begingroup$ @yarchik Regarding that prior post, feel free to take my comments as: "The math does not support the given result. Unevaluated would be better. A message that the singular integral is undefined would also be nice." $\endgroup$ Apr 3 '20 at 15:02
1
$\begingroup$
Here my attempt to solve the integral Integrate[f[a] Derivative[1][DiracDelta][d^2/a - a],{a,0,Infinity}]:
f[a_] := Exp[-a^2/(4 s^2)]/a^2 Sinc[w a/2]
Substitution u[a]=d^2/a-a (integrationlimits change to u[0]=Infinity],u[Infinity]=-Infinity)
u[a_] := d^2/a - a
sola = Solve[u == d^2/a - a, a][[2]] (*solution a>0*)
Now Mathematica is able to solve the integral
int=Integrate[f[a/.sola] Derivative[1][DiracDelta][u]/u'[a]/.sola ,{u, Infinity,-Infinity}]
(*(E^(-(d^2/(4 s^2))) (d s^2 w Cos[(d w)/2] - (d^2 + 4 s^2) Sin[(d w)/2]))/(4 d^3 Sqrt[d^2] s^2 w)*)
Hope it helps solving your problem!
$\endgroup$
15
• $\begingroup$ Thanks a lot for your reply! I agree with your approach and it agrees with the result I get from $I_{analytic}$. This is the result I trust, however, when I evaluate Integrate[Exp[-a^2/(4 s^2)] Sinc[w a / 2] Derivative[1][DiracDelta][D^2/a - a],{a,0,Infinity}, Assumptions -> s > 0 && w > 0 && D > 0], which appears to be the same integral, I get $I_{Mathematica} \neq I_{analytic}$. This is the source of my confusion. Do you see why these two integrals give differnt results? Is it something wrong with how code $I_{Mathematica}$? $\endgroup$
– e4alex
Apr 1 '20 at 16:38
• $\begingroup$ @e4alex I agree and think that Mathematica result, which only differs in sign, is wrong. You get the Mathematica result if you take the first branch of sola[…][[1]]! One additional remark: Substitution d^2->d2 helps mathematica a lot... $\endgroup$ Apr 2 '20 at 5:54
• $\begingroup$ @UlrichNeumann I do not think MMA result is wrong. The fact that Dirac's delta only has meaning within an integral does not mean that you can change the limits of integration at will. The limits of integration are an integral part 😁 of the definition. $\endgroup$ Apr 2 '20 at 14:38
• $\begingroup$ @SolutionExists Thanks for your comment, I didn't change the integration limits at will. My point is the transformation a->u , which must cover the integration range 0<a<Infinity . That's why I think sola[[2]] is the right branch (and Mathematica perhaps took the wrong) $\endgroup$ Apr 2 '20 at 14:42
• $\begingroup$ @UlrichNeumann I didn't mean you personally, I meant in general. The integral in the OP goes from zero to ∞, but the definition of Dirac's delta must use an integral from -∞ to ∞. Changing the limits of the integral to [0,∞) is equivalent to multiply to test function by a Heaviside theta, so the OP integral is the integral of a distribution times a distribution. That is doable but not trivial. $\endgroup$ Apr 2 '20 at 14:49
0
$\begingroup$
My previous answer and comments were wrong. I didn't notice the argument of the δ function was not linear in the integration variable (and I wasn't even drunk).
In the Wikipedia page, there is this paragraph
In the integral form the generalized scaling property may be written as $∫_{-∞}^∞ f ( x ) δ ( g ( x ) ) d x = ∑_i f ( x_i ) / | g ′ ( x_i ) | $.
The Jacobian of the transformation is 1/g'(x). Please note the absolute value in the denominator.
Basically, find the zeroes of the argument of the δ, and integrate around them (by parts if necessary). Also,
The distributional derivative of the Dirac delta distribution is the distribution δ′ defined on compactly supported smooth test functions φ by $δ ′ [ φ ] = − δ [ φ ′ ] = − φ ′ ( 0 )$ .
(1) Finding the zeroes:
Solve[-a + Δ^2/a == 0, a]
(2) Finding the Jacobian:
jac = Solve[Dt[-a + Δ^2/a == u[a]], u'[a]]
/. Dt[Δ] → 0 /. a → Δ // FullSimplify
(3) Evaluating the integral by parts (don't forget the minus sign in front):
v1 = -D[(E^(-(a^2/(4 s^2))) Sinc[(a w)/2])/a^2, a] / Abs[jac]
/. a → Δ // FullSimplify
(4) Divide the integral by the Jacobian (the previous division was because of the integration by parts, this one because of the scaling):
v1 / Abs[ jac ]
The answer is the same as $I_{MMA}$. By the way, MMA is simply using
Integrate[ f[a] DiracDelta'[2 a], {a, -∞, ∞}]
(*-(1/4) f'[0]*)
Prove that analytically and you will find the error in your analytic calculation.
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15
• $\begingroup$ Many thanks for replies in answering this question. I am able to prove the last expression you suggest by the methods I use above, so I still don't see the error in my analytic calculation. Observe $\frac{d}{da} \delta(2 a) = 2 \delta'(2a)$. It then follows that: \begin{align} \int da \, f(a) \delta'(2a) &= \int da \, f(a) \frac{1}{2} \frac{d}{da} \delta(2 a) \\ &= -\frac{1}{2} \int da \, \left( \frac{d}{da} f(a) \right) \delta(2 a) \\ &= -\frac{1}{4} \int da \, \left( \frac{d}{da} f(a) \right) \delta(a) \\ &= -\frac{1}{4} f'(0) \end{align} $\endgroup$
– e4alex
Apr 3 '20 at 15:41
• $\begingroup$ @e4alex ok, now we agree that the Jacobian is in the denominator twice. So, the only discrepancy should be the sign. Remember that Dirac's delta is an even function, so only the minus sign from integration by parts is necessary. $\endgroup$ Apr 3 '20 at 16:18
• $\begingroup$ For example, the following minus sign in your post should not be there \begin{align} I_{analytic} &= \int_{0}^{\infty} da \, \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \delta' \left( \frac{D^2}{a}- a \right) \\ &=- \int_{0}^{\infty} da \, \left[\frac{d}{da} \delta\left( \frac{D^2}{a}- a \right) \right] \left(\frac{D^2}{a^2}+1\right)^{-1} \frac{e^{-\frac{a ^2}{4s^2}} }{a^2} \frac{\sin \left(\tfrac{w}{2} a \right) }{\tfrac{w}{2} a} \end{align} Also, it seems that there are algebraic errors (you have a 4 where I got a 6). $\endgroup$ Apr 3 '20 at 16:18
• $\begingroup$ I'm not sure I agree that the Jacobian appears twice. The first factor of two comes from the chain rule, while the second comes from the scaling property of the delta function. So the negative sign you believe should not be there follows from the chain rule which I state in the original post. Is it that you do not agree with the way I apply the chain rule? $\endgroup$
– e4alex
Apr 3 '20 at 16:36
• $\begingroup$ It appears you get the answer that agrees with Mathematica. Note that I believe $I_{Mathematica}$, as stated in the original post, agrees with your results (note the appearance of a 6 instead of a 4). $\endgroup$
– e4alex
Apr 3 '20 at 16:37
0
$\begingroup$
speculation: Mathematica cannot handle Derivative[1][DiracDelta][1/x-x]in a right way?
Here I'll give a simplified example which perhaps shows that Mathematica gives a wrong result, when applied to Derivative[1][DiracDelta][1/x-x]!
Let's consider the integral
Integrate[Derivative[1][DiracDelta][1/x - x], {x, 0, Infinity} ]
(*0*)
which MMA (v12) evaluates to zero!
Alternatively integration with substitution u=1/x-x, x=-u/2+Sqrt[1+(u/2)^2] (see my first answer)
us=D[1/x-x,x]/. x->-u/2+Sqrt[1+(u/2)^2];
Integrate[ Derivative[1][DiracDelta][u]/us, {u, Infinity,-Infinity} ]
(*1/4*)
To "proof" the last result I'll consider the deltadistribution as a well known limit
dirac = Function[x, Exp[-(x^2/(2 eps))]/Sqrt[2 Pi eps]] (* eps->0 *)
int=Integrate[dirac'[1/x - x], {x, 0, Infinity} ]
(*(E^(1/eps) (-BesselK[0, 1/eps] + BesselK[1, 1/eps]))/(eps^(3/2) Sqrt[2 \[Pi]])*)
eps->0
Simplify[ Normal[Series[int, {eps, 0, 0}]], eps > 0]
(*1/4*)
Why can't Mathematica find this result? What's wrong here?
$\endgroup$
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4,853,410,596,845,671,000 | The angles of the triangle ABC are alpha = 35°, beta = 48°. Perimeter Of Triangle is the sum of all the sides of the triangle. Triangle Calculators. Also explore many more calculators covering geometry, math and other topics. Thanks. 1 Other formulas that calculate the same Output, Semiperimeter Of Triangle =Perimeter Of Triangle/2. Semiperimeter of a triangle Solve. Calculator shows law of cosines equations and work. $('#content .addFormula').click(function(evt) { Semiperimeter Of Triangle and is denoted by s symbol. In geometry, the semiperimeter of a polygon is half its perimeter. Given theorem values calculate angles A, B, C, sides a, b, c, area K, perimeter P, semi-perimeter s, radius of inscribed circle r, and radius of circumscribed circle R. Goal: Theory: Part 1. What is the measure of the semi-perimeter and what is the area of the triangle? Perimeter semicircle = (circumference circle / 2) + 2 * radius. Assuming you know all three lengths, a, b, and c. The semi-perimeter (1/2 of the perimeter) of the triangle is s. Knowing that, you may determine the area based on these calculations: s = (a + b + c) / 2 or 1/2 of the perimeter of the triangle Area of Triangle when semiperimeter is given, Side a of a triangle given side b, angles A and B, The Semiperimeter Of the Triangle is half of the measurement of the perimeter of the triangle. The triangle perimeter is the sum of the lengths of its three sides 2. You can input only integer numbers or fractions in this online calculator. 3. The formula is based on all three sides of the triangle. Semiperimeter Of Triangle calculator uses Semiperimeter Of Triangle =Perimeter Of Triangle/2 to calculate the Semiperimeter Of Triangle, The Semiperimeter Of the Triangle is half of the measurement of the perimeter of the triangle.. Semiperimeter Of Triangle and is denoted by s symbol. Geometry calculator for solving the semiperimeter of a scalene triangle given the length of sides a, b and c. By finding the base and height of the triangle. 2. Assuming you know all three lengths, a, b, and c. The semi-perimeter (1/2 of the perimeter) of the triangle is s. Knowing that, you may determine the area based on these calculations: s = (a + b + c) / 2 or 1/2 of the perimeter of the triangle Charismatic, though you might b Heron's Formula is used to calculate the area of a triangle with the three sides of the triangle. Side Angle Side is a theorem used to find the area of the triangle. Calculate angles or sides of triangles with the Law of Cosines. Geometry; Recently viewed formulas. You must activate Javascript to use this site. Semiperimeter of the triangle The semiperimeter of the triangle is half its perimeter. This is the semi-perimeter (s) of the solved triangle. N.B. Calculates triangle perimeter, semi-perimeter, area, radius of inscribed circle, and radius of circumscribed circle around triangle. Free Triangle Area & Perimeter Calculator - Calculate area, perimeter of a triangle step-by-step This website uses cookies to ensure you get the best experience. How to calculate Semiperimeter Of Triangle? Free Triangle Perimeter Calculator - Find perimeter of triangles step-by-step This website uses cookies to ensure you get the best experience. More in-depth information read at these rules. where, s = (a + b + c) / 2. It is not possible for a triangle to have more than one vertex with internal angle greater than or equal to 90°, or it would no longer be a triangle. Here is how the Semiperimeter Of Triangle calculation can be explained with given input values -> 7 = 14/2. // event tracking Area of a triangle base on the base and height; Area of a triangle using two sides and the interior angle. The triangle circumcenter calculator calculates the circumcenter of triangle with steps. If you have any difficulties with units conversion, you can use the length converter. The abbreviations denote our starting measurements. Perimeter semicircle = π * r + 2 * r = r * (π + 2) or. Semiperimeter Of Triangle calculator uses Semiperimeter Of Triangle =Perimeter Of Triangle/2 to calculate the Semiperimeter Of Triangle , The Semiperimeter Of the Triangle is half of the measurement of the perimeter of the triangle. Description. nDEF > by SAS 4. Menu. Determine the magnitudes of all angles of triangle A'B'C '. Usually by the length of three sides (SSS) or side-angle-side or angle-side-angle. 1. The formula is based on all three sides of the triangle. window.jQuery || document.write(' | {
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Thread: Simple trigonometric equation can you help?
1. #1
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Simple trigonometric equation can you help?
tan theta = - root 3 is the question, obviously I'm trying to find theta here.
I used inverse tan root 3 to get a value of 60, and then drew a trigonometric graph with active tan and got 150 and 300 for values of theta. Can anyone confirm this is right?
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2. #2
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\arctan(-\sqrt{3}) = -60^{\circ}<br />
-60^{\circ} + 360^{\circ} = 300^{\circ}
other angle is back in quad II, 180 degrees out
300^{\circ} - 180^{\circ} = 120^{\circ}
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The question is from CAT Percentages. 2IIMs CAT question bank provides you with CAT questions that can help you gear for CAT Exam CAT 2023. This question is from the topic percentages using integers. In this question, Two workers are planting trees. Later another worker is added and overall efficiency is increased. So, we need to find the no of trees planted by the second worker as a percentage of the number of trees planted by first worker.
Question 32 : In a field, two workers are planting trees. After sometime, a third worker is added and the number of trees planted becomes half as large. How many trees can the second worker plant as a percentage of the number of trees planted by first worker if it is given that efficiency of second worker is 1/3 of 1st and 3rd worker combined.
1. 65%
2. 60%
3. 70%
4. 75%
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Explanatory Answer
Method of solving this CAT Question from Percentages: Going by options might help at times.
Half as large means an increase of 50%.
=> 3rd worker = \\frac{1st Worker + 2nd worker}{2}\\) --------Equation (1)
Also given, number of trees planted by 1st + 3rd = 3 * 2nd --------Equation (2)
This problem can be solved easily using options:
a) If trees planted by 2nd worker = 65 1st worker = 100 3rd worker = 82.5 (from Equation(1))
Putting in Equation (2) => 182.5 = 3 * 65 = 185 (Not satisfied)
b) If trees planted by 2nd worker = 60
1st worker = 100
3rd worker = 80 (from Equation(1))
Putting in Equation(2) => 180 = 3 * 60 = 180 (Satisfied)
The question is "How many trees can the second worker plant as a percentage of the number of trees planted by first worker if it is given that efficiency of second worker is 1/3 of 1st and 3rd worker combined. "
The second worker can plant 60% of the number of trees planted by first worker
Hence, the answer is 60%
Choice B is the correct answer.
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Trying to solve a rather difficult differential equation
1. Apr 9, 2015 #1
1. The problem statement, all variables and given/known data
Consider a system composed of two species X and Y with fractional populations x and y, respectively, where x+y=1. The two species interact in such a way that the differential equation for x is:
\begin{equation}
\frac{dx}{dt}=xyA_{0}e^{-\alpha t}
\end{equation}
where $A_{0}$ and $\alpha$ are non-negative constants. Solve the equation by separation of variables and hence show that the solution for x(0) = $x_{0}$ is:
Photo attached- too long to write out!
2. The attempt at a solution
Again... attached. The problem that I am having is that I can't make x the subject of the equation because I end up with x/(x-1) on the left hand side.
Attached Files:
Last edited: Apr 9, 2015
2. jcsd
3. Apr 9, 2015 #2
Are you saying that you don't know how to solve that final algebraic equation for x?
4. Apr 9, 2015 #3
SammyS
User Avatar
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
Those sideways images are very difficult to read.
I did use the 'Windows' snipping tool to show the solution you are to verify, then pasted it into a word processor app. & rotated it. Capture4.PNG
Chet's got the rest.
5. Apr 10, 2015 #4
Don't worry, I was a little tired last night doing a 5 hour practice paper. I've got it now... silly me.
6. Apr 10, 2015 #5
SammyS
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Staff Emeritus
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Good.
To make x the subject either of the following forms might have helped.
##\displaystyle\ \frac{x}{x-1}=\frac{1}{1-1/x}=1+\frac{1}{x-1}\ ##
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The probability P of some event E, denoted P(E), is defined with respect to a "universe", or sample space \Omega, of all possible elementary events in such a way that P must satisfy the Kolmogorov axioms.
Alternatively, a probability can be interpreted as a measure on a σ-algebra of subsets of the sample space, those subsets being the events, such that the measure of the whole set equals 1. This property is important, since it gives rise to the natural concept of conditional probability. Every set A with non-zero probability (that is, P(A)> 0 ) defines another probability
P(B \vert A) = {P(B \cap A) \over P(A)}
on the space. This is usually read as "probability of B given A". If the conditional probability of B given A is the same as the probability of B, then A and B are said to be independent.
In the case that the sample space is finite or countably infinite, a probability function can also be defined by its values on the elementary events \{e_1\}, \{e_2\}, ... where \Omega = \{\,e_1, e_2, \dots\,\}.\,
Kolmogorov axioms Edit
The following three axioms are known as the Kolmogorov axioms, after Andrey Kolmogorov who developed them. We have an underlying set Ω, a sigma-algebra F of subsets of Ω, and a function P assigning real numbers to members of F. The members of F are those subsets of Ω that are called "events".
First axiom Edit
For any set E\in F, that is, for any event E, we have P(E)\geq 0.
That is, the probability of an event is a non-negative real number.
Second axiom Edit
P(\Omega) = 1.\,
That is, the probability that some elementary event in the entire sample set will occur is 1. More specifically, there are no elementary events outside the sample set.
This is often overlooked in some mistaken probability calculations; if you cannot precisely define the whole sample set, then the probability of any subset cannot be defined either.
Third axiom Edit
Any countable sequence of pairwise disjoint events E_1, E_2, ... satisfies P(E_1 \cup E_2 \cup \cdots) = \sum P(E_i).
That is, the probability of an event set which is the union of other disjoint subsets is the sum of the probabilities of those subsets. This is called σ-additivity. If there is any overlap among the subsets this relation does not hold. Some authors consider merely finitely-additive probability spaces, in which case one just needs an algebra of sets, rather than a σ-algebra.
For an algebraic alternative to Kolmogorov's approach, see algebra of random variables.
Lemmas in probability Edit
From the Kolmogorov axioms one can deduce other useful rules for calculating probabilities:
P(A \cup B) = P(A) + P(B) - P(A \cap B)
This is called the addition law of probability, or the sum rule. That is, the probability that A or B will happen is the sum of the probabilities that A will happen and that B will happen, minus the probability that A and B will happen. This can be extended to the inclusion-exclusion principle.
P(\Omega\setminus E) = 1 - P(E)
That is, the probability that any event will not happen is 1 minus the probability that it will.
Using conditional probability as defined above, it also follows immediately that
P(A \cap B) = P(A) \cdot P(B \vert A)
That is, the probability that A and B will happen is the probability that A will happen, times the probability that B will happen given that A happened; this relationship gives Bayes' theorem. It then follows that
A and B are independent if and only if P(A \cap B) = P(A) \cdot P(B).
See also Edit
External linksEdit
• The Legacy of Andrei Nikolaevich Kolmogorov Curriculum Vitae and Biography. Kolmogorov School. Ph.D. students and descendants of A.N. Kolmogorov. A.N. Kolmogorov works, books, papers, articles. Photographs and Portraits of A.N. Kolmogorov.
de:Kolmogorow-Axiom
es:Axiomas de probabilidad fr:Axiomes des probabilités gl:Axiomas de probabilidade he:אקסיומות ההסתברות nl:Axioma's van de kansrekeningro:Axiomele probabilităţii ru:Аксиоматика Колмогорова th:สัจพจน์ของความน่าจะเป็น zh:概率公理
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Related Articles
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Java Program to Find the Largest of three Numbers
• Difficulty Level : Medium
• Last Updated : 23 Aug, 2021
Problem Statement: Given three numbers x, y, and z of which aim is to get the largest among these three numbers.
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Example:
Input: x = 7, y = 20, z = 56
Output: 56 // value stored in variable z
Flowchart For Largest of 3 numbers:
Algorithm to find the largest of three numbers:
1. Start
2. Read the three numbers to be compared, as A, B and C
3. Check if A is greater than B.
3.1 If true, then check if A is greater than C
If true, print 'A' as the greatest number
If false, print 'C' as the greatest number
3.2 If false, then check if B is greater than C
If true, print 'B' as the greatest number
If false, print 'C' as the greatest number
4. End
Approaches:
• Using Ternary operator
• Using if-else
Approach 1: Using Ternary operator
The syntax for the conditional operator:
ans = (conditional expression) ? execute if true : execute if false
• If the condition is true then execute the statement before the colon
• If the condition is false then execute a statement after colon so
largest = z > (x>y ? x:y) ? z:((x>y) ? x:y);
Illustration:
x = 5, y= 10, z = 3
largest = 3>(5>10 ? 5:10) ? 3: ((5>10) ? 5:10);
largest = 3>10 ? 3 : 10
largest = 10
Java
// Java Program to Find the Biggest of 3 Numbers
// Importing generic Classes/Files
import java.io.*;
class GFG {
// Function to find the biggest of three numbers
static int biggestOfThree(int x, int y, int z)
{
return z > (x > y ? x : y) ? z : ((x > y) ? x : y);
}
// Main driver function
public static void main(String[] args)
{
// Declaring variables for 3 numbers
int a, b, c;
// Variable holding the largest number
int largest;
a = 5;
b = 10;
c = 3;
// Calling the above function in main
largest = biggestOfThree(a, b, c);
// Printing the largest number
System.out.println(largest
+ " is the largest number.");
}
}
Output
10 is the largest number.
Approach 2: Using the if-else statements
In this method, if-else statements will compare and check for the largest number by comparing numbers. ‘If’ will check whether ‘x’ is greater than ‘y’ and ‘z’ or not. ‘else if’ will check whether ‘y’ is greater than ‘x’ and ‘z’ or not. And if both the conditions are false then ‘z’ will be the largest number.
Java
// Java Program to Find the Biggest of 3 Numbers
// Importing generic Classes/Files
import java.io.*;
class GFG {
// Function to find the biggest of three numbers
static int biggestOfThree(int x, int y, int z)
{
// Comparing all 3 numbers
if (x >= y && x >= z)
// Returning 1st number if largest
return x;
// Comparing 2nd no with 1st and 3rd no
else if (y >= x && y >= z)
// Return z if the above conditions are false
return y;
else
// Returning 3rd no, Its sure it is greatest
return z;
}
// Main driver function
public static void main(String[] args)
{
int a, b, c, largest;
// Considering random integers three numbers
a = 5;
b = 10;
c = 3;
// Calling the function in main() body
largest = biggestOfThree(a, b, c);
// Printing the largest number
System.out.println(largest
+ " is the largest number.");
}
}
Output
10 is the largest number.
My Personal Notes arrow_drop_up
Recommended Articles
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123,239,032,761,110,450 |
Results 1 to 11 of 11
Thread: exponential graphs
1. #1
Member i_zz_y_ill's Avatar
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Question exponential graphs
part one: given y=0.5e^-t -0.2e^-2t +0.1sint -0.3cos t
and another curve
y=Ae^-t + Be^-2t + 0.1e^t(sint -cost) A B constants
How do you know that for large values of t that the first one is bounded and the second one is unbounded? And what does bounded mean?
part two: Given x = Ae^-0.04t +Be^-0.02t +1875
and y = -Ae^-0.04t + Be^-0.02t +625
How do you know that they grow exponentially like a normal exponential. I would have thought that they would be reflected in the y or x axis(corrosponding to each) one. And the y curve has a dip in it, how are you supposed to know that.
Tanswers to these questions would be greatly aprecciated I have an exam soon and this is quite hard. Thanks!!!!
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2. #2
MHF Contributor Mathstud28's Avatar
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Quote Originally Posted by i_zz_y_ill View Post
part one: given y=0.5e^-t -0.2e^-2t +0.1sint -0.3cos t
and another curve
y=Ae^-t + Be^-2t + 0.1e^t(sint -cost) A B constants
How do you know that for large values of t that the first one is bounded and the second one is unbounded? And what does bounded mean?
part two: Given x = Ae^-0.04t +Be^-0.02t +1875
and y = -Ae^-0.04t + Be^-0.02t +625
How do you know that they grow exponentially like a normal exponential. I would have thought that they would be reflected in the y or x axis(corrosponding to each) one. And the y curve has a dip in it, how are you supposed to know that.
Tanswers to these questions would be greatly aprecciated I have an exam soon and this is quite hard. Thanks!!!!
Bounded means that for sufficiently large numbers a function is always smaller or larger than another function
For example
\sin(x) is bounded above by 2 and below by -2\
Does that make senes?
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3. #3
Member i_zz_y_ill's Avatar
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Not really. I don't think it means in relation to each function. The mark scheme says that the first one(x) is bounded oscillations and the second one (y) is unbounded oscillations. I can't figure out what it means by large t. I know this would make the first part of the functios zero but then for the sin and cos parts it doesn' really make sense. Is it referring to the fact that for th y function the line connecting its msximum points as t tends to large is a decsyng eponential in which case it is 'not bounded' and for the second one in theory it would be a straight line making it bounded? From an anayltical standpoint however putting a large value of t into the sin and cos values doesnt really get you anywhere,,,,,,is this correct what i'm thinking thnx
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4. #4
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Quote Originally Posted by i_zz_y_ill View Post
given y= 0.5 e^{-t } -0.2e^{-2t} +0.1\sin t -0.3\cos t
and
y=Ae^{-t} + Be^{-2t} + 0.1e^{t}(\sin t -\cos t)
How do you know that for large values of t that the first one is bounded and the second one is unbounded? And what does bounded mean?
For large values of t you may neglect the value of e^{-at} (where a is a positive constant) as it becomes very close to zero.
So for large values of t the first function approaches y= 0.1\sin t -0.3\cos t to simplify this you need to use an "R" formula. are you familiar with how to write 0.1\sin t -0.3\cos t as R \sin ( t - \theta ) ? but it should be clear that is the a simple sine curve and cannot exceed a particular value.
for the second one you apply a similar argument so the function approaches y= 0.1e^{t}(\sin t -\cos t) then simplify \sin t -\cos t into the form of R \sin ( t - \alpha) so your function becomes y = 0.1R e^{t}\sin( t - \alpha).
as -1 \leq \sin ( t - \alpha) \leq 1 then by multiplying by 0.1R e^{t} you get -0.1R e^{t} \leq 0.1R e^{t}\sin( t - \alpha) \leq 0.1R e^{t} . What can you deduce form this inequality ?
Bobak
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5. #5
Member i_zz_y_ill's Avatar
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Oh ri thanks. So for the first one the max value is R. I still don't see how that makes it bounded erm and the second one if its limits are between |0.1Re^t| then how is it unbounded. I would have thought in theory this would decay and have limits(max and min) according to its exponential curve. Sill a bi confused sorry!
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6. #6
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Quote Originally Posted by i_zz_y_ill View Post
Oh ri thanks. So for the first one the max value is R. I still don't see how that makes it bounded
It is bounded because the curves lines between the lines y = R and y = -R
I attached a graph to illustrate this.
the second one if its limits are between |0.1Re^t| then how is it unbounded.
because the limits the seconds ones lies between increase without limits for large t. have you tired sketching the curve e^{x} \sin x ? I have attach a image of that to help you get the general idea.
Attached Thumbnails Attached Thumbnails exponential graphs-picture-16.png
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7. #7
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oh right yeah!!! I guess this means it is unbounded by having an exponential boundary. Ok so ill use the form Rsin(t+/- alpha) since this is easier to analyse for larger t I suppose. K thanks alot I think I understand now if im correct in presuming this???
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8. #8
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Quote Originally Posted by i_zz_y_ill View Post
oh right yeah!!! I guess this means it is unbounded by having an exponential boundary. Ok so ill use the form Rsin(t+/- alpha) since this is easier to analyse for larger t I suppose. K thanks alot I think I understand now if im correct in presuming this???
Yes if the bounding function is exponentially increasing then it is unbounded. also your not using Rsin(t+/- alpha) because it is easier to analyse for large t, you use it because the values of R makes your maximum and minimum values apparent.
Bobak
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9. #9
Member i_zz_y_ill's Avatar
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Oh right yeah!!!! I understand, i appreciate ur time/effort thanks alot!
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10. #10
Member i_zz_y_ill's Avatar
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Oh and you wouldn't happen to know the second part to my original question would you? part 2: I can't draw those graphs
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11. #11
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Quote Originally Posted by i_zz_y_ill View Post
Oh and you wouldn't happen to know the second part to my original question would you? part 2: I can't draw those graphs
I am unsure about what your question is for that part. but it look easier than the first question. Just remember the e^{-at} terms vanish for large values of t. so you should be able to easily determine what values both x and y are approaching for large values of t.
Bobak
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494,376,012,588,165,600 | Sum of Absolute Values
Problem
$\begin{align} f(x)=|x|&+2|x-1|+|x-2|+|x-4|\\ &+|x-6|+2|x-10|. \end{align}$
Find the minimum value of $f(x),$ for $x\in\mathbb{R}.$
Brute force solution
$f(x)$ is the sum of functions with $V$-shaped graphs. The minimum is bound to occur at one of the "break" points. There are only six of them: $x=0,1,2,4,6,10.$ Let's check:
$\begin{align} f(0)&=0+2+2+4+6+20=34\\ f(1)&=1+0+1+3+5+18=28\\ f(2)&=2+2+0+2+4+16=26\\ f(4)&=4+6+2+0+2+12=26\\ f(6)&=6+10+4+2+0+8=30\\ f(10)&=10+18+8+6+4+0=46 \end{align}$
The minimum is achieved at two points, $x=2$ and $x=4$ and is equal to $26.$
Looking back
Function $f$ being the sum of functions with $V$-shaped graph is piecewise linear. It is linear between any two successive breakpoints and outside the smallest interval that contains all of them, $[0,10]$ in this problem. So the fact that the minimum is achieved at two points, means that on the interval $[2,4]$ the function is constant. Indeed, here is its graph:
graph of f(x)=|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|
Now, is there anything about this problem that suggests a more insightful approach? There is indeed. A similar problem has been popularized by Paul J. Nahin:
Suppose we have a real line before us (labeled as the \(x\)-axis), stretching from \(-\infty\) to \(+\infty\). On this line there are marked \(n\) points labeled in increasing value as \(x_1 \lt x_2 \lt \ldots \lt x_n\). Let's assume that all the \(x_i\) are finite (in particular \(x_1\) and \(x_n\)), and so the interval of the \(x\)-axis that contains all \(n\) points is finite in length. Now, somewhere (anywhere) on the finite \(x\)-axis we mark one more point (let's call it \(x\)). We wish to pick \(x\) so that the sum of the distances between \(x\) and and the original points is minimized. That is, we wish to pick \(x\) so that
\(S = |x - x_{1}| +|x - x_{2}| + \ldots + |x - x_{n}|\)
is minimized.
The crucial insight is already apparent in the simplest graph of that sort, say, that of $g(x)=|x-1|+|x-2|:$
graph of f(x)=|x-1|+|x-2|
Any function $g(x)=|x-a|+|x-b|$ is constant in the interval $[a,b]$ (assuming $a\lt b)$ where it takes the value of $b-a.$ This is how it help to solve our problem: the given function is the sum of the simple ones (each of only two terms):
|x|+2|x-1|+|x-2|+|x-4|+|x-6|+2|x-10|
$\begin{align} f(x)&=|x|+|x-10|\\ &+|x-1|+|x-10|\\ &+|x-1|+|x-6|\\ &+|x-2|+|x-4|. \end{align}$
The first of these is constant on the interval $[0,10],$ the others, successively, on $[1,10],$ $[1,6],$ $[2,4],$ the last of which is common to all the pairs. We may even generalize.
Generalization
Given an increasing sequence of real numbers $a_{1}\lt a_{2}\lt a_{3}\lt\ldots\lt a_{n}$ and a sequence of as many positive real coefficients $\{\alpha _{i}\},$ $i=1,2,\ldots ,n.$ Function $\displaystyle f(x)=\sum_{i=1}^{n}\alpha_{i}|x-a_{i}|$ achieves its minimum at one of the intervals $[a_{k},a_{k+1}]$ if and only if the coefficients $\{\alpha _{i}\}$ can be split into two groups with equal sums. Otherwise, the minimum is achieved at one of the given points $\{a_{i}\}.$
Proof
The proof is by induction.
The claim is obvious when there are just one or two terms.
For a larger $n,$ assume without loss of generality that $\alpha _{1}\le\alpha _{2}.$ Then,
$\begin{align} f(x)&=\bigg[\alpha _{1}|x-a_{1}|+|x-a_{n}|\bigg] \\ &\space\space +\bigg[\alpha _{2}|x-a_{2}|+\ldots +\alpha _{n-1}|x-a_{n-1}|+(\alpha _{n}-\alpha_{1})|x-a_{n}|\bigg], \end{align}$
where the function in brackets has fewer than $n$ break points. Also, the coefficients $\alpha_{2},\ldots,\alpha_{n-1},\alpha_{n}-\alpha_{1}$ could be split into two groups of equal sums only if that is true of the original $n$ coefficients. To complete the induction, it suffices to observe that the interval $[a_{1},a_{n}]$ where the first term is constant, contains all the break points of the second term.
Modification
Given an increasing sequence of real numbers $a_{1}\lt a_{2}\lt a_{3}\lt\ldots\lt a_{n}.$ Find the minimum of function $\displaystyle f(x)=\sum_{i=1}^{n}(x-a_{i})^{2}.$
Solution
Notice that
$\begin{align} f(x)-f(y)&=\sum_{i=1}^{n}\bigg[(x-a_{i})^{2}-(y-a_{i})^{2}\bigg]\\ &=\sum_{i=1}^{n}\bigg[(x^{2}-y^{2})-2(x-y)a_{i}\bigg]\\ &=n(x^{2}-y^{2})-2(x-y)\sum_{i=1}^{n}a_{i}. \end{align}$
Letting $\displaystyle y_{0}=\frac{a_{1}+\ldots+a_{n}}{n}$ gives
$\begin{align} f(x)-f(y_{0})&=n(x^{2}-y_{0}^{2})-2(x-y_{0})\sum_{i=1}^{n}a_{i}\\ &=n(x^{2}-y_{0}^{2})-2nxy_{0}+2ny_{0}^{2}\\ &=n(x^{2}-2xy_{0}+y_{0}^{2})=n(x-y_{0})^{2}\ge 0, \end{align}$
implying that the minimum is achieved at $\displaystyle y=\frac{a_{1}+\ldots+a_{n}}{n}.$
Is there a further modification/generalization?
Acknowledgment
The example came from an excellent book by Xu Jiagu, Lecture Notes on Mathematical Olympiad Courses, v 8, (For senior section, v 1), World Scientific, 2012, p.54
Absolute Value
|Contact| |Front page| |Contents| |Algebra|
Copyright © 1996-2018 Alexander Bogomolny
63706026
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-2,628,581,593,770,184,000 | Deriver [Functional Terms, Identity, First Order Theories, Set Theory—Gentzen Syntax]
Logical System:
12/26/13
Welcome!
The tutorials presented here look at some topics in logic to the level of intermediate to advanced Predicate Calculus. They build on the Tutorials of Easy Deriver which provided the introductory material to Propositional and Predicate Calculus.
The program, widgets, or Notes, should be accompanied by a suitable textbook, such as:
M.Bergmann, J.Moor, J.Nelson, The Logic Book
A.Hausman, H.Kahane, P.Tidman, Logic and Philosophy
W.Hodges, Logic
C.Howson, Logic with Trees
R.C.Jeffrey, Formal Logic: Its Scope and Limits
H.Leblanc and W.Wisdom, Deductive Logic
B.Mates, Elementary Logic
M.D.Resnick, Elementary Logic
Unfortunately these textbooks use slightly different choices of rules and symbols one from another. To adjust to this the Notes are in different major sections, with the sections tailored to particular texts.
You are invited to review
Notation
Not all logicians, and logical texts, use the same symbols for the so-called 'logical connectives'. Nor do they use the same sequences of symbols for 'well formed formulas'.
Here are typical possibilities for symbols
'not' : ∼ (the 'tilde'), ¬ (looks like the top right corner of a box)
'and': ∧, & (the ampersand), . (just a period)
'or': ∨ (usually just this, vel)
'implication': ⊃ , →
'equivalence': ≡, ↔
'existential quantifier': ∃, ∑
'universal quantifier':∀, ∏
So, in a logic book, you might see (A&B)→C and that is just the same as (A∧B)⊃C.
And you might see (∀x)(Fx ⊃ Gxy) and that might be just the same as ∀x(F(x)→G(x,y)).
The software running here can easily manage or render any of these. But we should explain what we favor, and help you find what you prefer.
The 'default' system
We like the use of notation like R(a,b,c) for the application of a predicate R to the arguments or terms a, b, c, and the use of f(a,b,c) for the application of a functor or function f to the arguments a, b, c. In the simple form, this employs the upper case letters A-Z, perhaps followed by subscripts, to be predicates, so, for example, R, S₁, T₁₁ are all predicates. And it employs lower case letters ie [a--v], perhaps followed by subscripts, to be constant terms or functions. But an extension is useful.
Often, when working informally, authors will write Red(x) to mean that the predicate Red is applied to the variable x. The default system will accept this also (with these new style of predicates also optionally followed by subscripts). So Red, Soft₁, Tiger₁₁ are all predicates. The rule or convention here is that the first letter of a predicate is upper case, then any sequence of upper or lower case letters can follow, and the predicate can be completed with subscripts. So 'AVeryLongPredicate' is a predicate. A similar approach is applied to constant terms or functions, only this time, the term must start with a lower case letter ie [a--v]. So aFunctionAppliedToATerm(aTerm) is a perfectly good term of the form f(a). Variables, though, consist of lower case [w-z] only, optionally followed by subscripts [ie there can be no sequence of upper or lower case letters in between].
So the argument. Socrates is a man, All men are mortal, therefore, Socrates is mortal can be symbolized
M(s), ∀x(M(x)→M₂(x)) ∴ M₂(s)
But, the software will also accept
Man(socrates), ∀x(Man(x)→Mortal(x)) ∴ Mortal(socrates)
Notice here that there are no parantheses around the quantifiers (after all, the parantheses not needed). [Of course, the reason brackets are needed for predicates, terms etc. is help determine what, say, Redbox, is supposed to mean, when predicates and terms need not have constant length.]
The default system also uses ~, &, v, →, ≡, so a typical formula is ∀x(F(x)&~H(x) → G(x,y))
The 'gentzen' system
This uses Rabc for the application of a predicate R to the arguments or terms a, b, c, and the use of f(abc) for the application of a functor or function f to the arguments a, b, c. i.e the predicates and atomic terms are of length 1. It uses the upper case letters A-Z to be predicates, so, for example, R, S, T are all predicates. And it employs lower case letters ie [a--v], perhaps followed by subscripts, to be constant terms or functions. Variables, consist of lower case [w-z]. There are parantheses around the quantifiers. The gentzen system also uses ~, ∧, v, ⊃, ≡, so a typical formula is (∀x)(Fx~Hx Gxy).
The 'howson' system
The Colin Howson book uses a notation like R(a,b,c) for the application of a predicate R to the arguments or terms a, b, c. It employs the upper case letters A-Z, perhaps followed by subscripts, to be predicates, so, for example, R, S₁, T₁ are all predicates. Variables consist of lower case [w-z] only, optionally followed by subscripts. There are no parantheses around the quantifiers. The howson system also uses ~, &, v, →, ≡, so a typical formula is ∀x(F(x)&~H(x) → G(x,y)). | {
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1,673,480,671,827,541,200 | VideoPress Test
Patience, persistence, and perspiration make an unbeatable combination for success. – Napoleon Hill
Lecture - 9 Chapter 13 Limits and Derivatives
All rules for direct differentiation Detailed Explanation and Tricks for direct derivative in case of, Addition and Subtraction, Multiplication (Product Rule)
NCERT EXERCISE 13.2
Question 11. Find the derivative of the following functions:
(i). \sin x \cos x
(ii). \sec x
(iii). 5 \sec x + 4 \cos x
(iv). \cosec x
(v). 3 \cot x + 5 \cosec x
(vi). 5 \sin x – 6 \cos x + 7
(vii). 2 \tan x – 7 \sec x
Question 9. Find the derivative of
(i). 2x – \frac{3}{4}
(ii). (5x^3+3x-1)(x-1)
(iii). x^{-3}(5+3x)
(iv). x^5 (3-6x^{-9})
(v). x^{-4}(3-4x^{-5})
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8,201,104,315,743,225,000 | 8.2first Fundamental Theorem Of Calculusap Calculus
1. The First Fundamental Theorem of Calculus. Let be a continuous function on the real numbers and consider From our previous work we know that is increasing when is positive and is decreasing when is negative. Moreover, with careful observation, we can even see that is concave up when is positive and that is concave down when is negative.
2. $ begingroup$ no, but '(the two) fundamental theorem of derivative and integrals' would probably be a better name than 'fundamental theorem of calculus' $ endgroup$ – reuns Apr 2 '16 at 6:37 add a comment.
1. 8.2first Fundamental Theorem Of Calculus Ap Calculus Free
2. 8.2first Fundamental Theorem Of Calculus Ap Calculus Answers
It's called the fundamental theorem of calculus. And we'll be abbreviating it FTC and occasionally I'll put in a 1 here, because there will be two versions of it. But this is the one that you'll be using the most in this class. The fundamental theorem of calculus says the following. It says that if F' = f, so F'(x) = f(x), there's a capital F.
The Fundamental Theorem of Calculus (FTC) shows that differentiation and integration are inverse processes.
Part (1) (FTC1)
If (f) is a continuous function on (left[ {a,b} right],) then the function (g) defined by
[{{gleft( x right)} = intlimits_a^x {fleft( {t} right)dt},;;}kern0pt{{a le x le b}}]
is an antiderivative of (f), that is
[{g^primeleft( x right) = fleft( x right);;text{or};;}kern0pt{frac{d}{{dx}}left( {intlimits_a^x {fleft( t right)dt} } right) }={ fleft( x right).}]
If (f) happens to be a positive function, then (gleft( x right)) can be interpreted as the area under the graph of (f) from (a) to (x.)
The first part of the theorem says that if we first integrate (f) and then differentiate the result, we get back to the original function (f.)
Part (2) (FTC2)
The second part of the fundamental theorem tells us how we can calculate a definite integral.
If (f) is a continuous function on (left[ {a,b} right]) and (F) is an antiderivative of (f,) that is (F^prime = f,) then
[{intlimits_a^b {fleft( x right)dx} }= {Fleft( b right) – Fleft( a right);;}kern0pt{text{or};;{intlimits_a^b {{F^primeleft( x right)}dx} }= {Fleft( b right) – Fleft( a right)}.}]
To evaluate the definite integral of a function (f) from (a) to (b,) we just need to find its antiderivative (F) and compute the difference between the values of the antiderivative at (b) and (a.)
So the second part of the fundamental theorem says that if we take a function (F,) first differentiate it, and then integrate the result, we arrive back at the original function, but in the form (Fleft( b right) – Fleft( a right).)
Thus, the two parts of the fundamental theorem of calculus say that differentiation and integration are inverse processes.
The Area under a Curve and between Two Curves
The area under the graph of the function (fleft( x right)) between the vertical lines (x = a,) (x = b) (Figure (2)) is given by the formula
[S = intlimits_a^b {fleft( x right)dx} = {Fleft( b right) – Fleft( a right).}]
Let (Fleft( x right)) and (Gleft( x right)) be antiderivatives of functions (fleft( x right)) and (gleft( x right),) respectively.
If (fleft( x right) ge gleft( x right)) on the closed interval (left[ {a,b} right],) then the area between the curves (y = fleft( x right),) (y = gleft( x right)) and the lines (x = a,) (x = b) (Figure (3)) is given by
[
{S = intlimits_a^b {left[ {fleft( x right) – gleft( x right)} right]dx} }
= {Fleft( b right) – Gleft( b right) }-{ Fleft( a right) + Gleft( a right).}
]
The Method of Substitution for Definite Integrals
The definite integral (intlimits_a^b {fleft( x right)dx} ) of the variable (x) can be changed into an integral with respect to (t) by making the substitution (x = gleft( t right):)
[{intlimits_a^b {fleft( x right)dx} }={ intlimits_c^d {fleft( {gleft( t right)} right)g’left( t right)dt} .}]
The new limits of integration for the variable (t) are given by the formulas
[{c = {g^{ – 1}}left( a right),;;}kern-0.3pt{d = {g^{ – 1}}left( b right),}]
where ({g^{ – 1}}) is the inverse function to (g,) that is (t = {g^{ – 1}}left( x right).)
Integration by Parts for Definite Integrals
In this case the formula for integration by parts looks as follows:
[{intlimits_a^b {udv} }={ left. {uv} right _a^b – intlimits_a^b {vdu} ,}]
where (left. {uv} right _a^b) means the difference between the product of functions (uv) at (x = b) and (x = a.)
Solved Problems
Click or tap a problem to see the solution.
Example 1
Calculate the derivative of the function (gleft( x right) = intlimits_1^x {sqrt {{t^3} + 4t} dt} ) at (x = 2.)
Example 2
Calculate the derivative of the function (gleft( x right) = intlimits_{ – large{frac{pi }{2}}normalsize}^x {sqrt {{{sin }^2}t + 2} dt} ) at (x = large{large{frac{pi }{6}}normalsize}.)
Example 3
Find the derivative of the function (gleft( x right) = intlimits_3^{{x^2}} {large{frac{{dt}}{t}}normalsize}.)
Example 4
Find the derivative of the function (gleft( x right) = intlimits_0^{{x^2}} {sqrt {1 + {t^2}} dt}.)
Example 5
Find the derivative of the function (gleft( x right) = intlimits_1^{{x^3}} {{t^2}dt}.)
Example 6
Find the derivative of the function (gleft( x right) = intlimits_{{x^2}}^{{x^3}} {tdt}.)
Example 7
Calculate the derivative of the function (gleft( x right) = intlimits_{sqrt x }^x {left( {{t^2} – t} right)dt} ) at (x = 1.)
Example 8
Evaluate the integral (intlimits_0^2 {left( {{x^3} – {x^2}} right)dx}.)
Example 9
Evaluate the integral (intlimits_{ – 1}^1 {left( {{t^2} + {t^{21}}} right)dt}.)
Example 10
Calculate the integral (intlimits_0^1 {left( {sqrt[large 3normalsize]{t} – sqrt t } right)dt}.)
Example 11
Evaluate the integral (intlimits_0^1 {{largefrac{x}{{{{left( {3{x^2} – 1} right)}^4}}}normalsize} dx}.)
Example 12
Evaluate the integral (intlimits_1^e {left( {t + large{frac{1}{t}}normalsize} right)dt}.)
Example 13
Evaluate the integral (intlimits_0^{ln 2} {x{e^{ – x}}dx}.)
Example 14
Evaluate the integral (intlimits_{ – 1}^1 {left {x – large{frac{1}{2}}normalsize} right dx}.)
Example 15
Evaluate the integral (intlimits_{ – 2}^1 {left {{x^2} – 1} right dx}.)
Example 16
Find the area bounded by the curves (y = {x^2}) and (y = sqrt x.)
Example 17
Find the area bounded by the curves (y = 2x – {x^2}) and (x + y = 0.)
Example 18
Find the area of the triangle with vertices at (left( {0,0} right),) (left( {2,6} right)) and (left( {7,1} right).)
Example 19
Find the area inside the ellipse ({largefrac{{{x^2}}}{{{a^2}}}normalsize} + {largefrac{{{y^2}}}{{{b^2}}}normalsize} = 1.)
Example 1.
Calculate the derivative of the function (gleft( x right) = intlimits_1^x {sqrt {{t^3} + 4t} dt} ) at (x = 2.)
Solution.
We apply the Fundamental Theorem of Calculus, Part (1:)
[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_a^x {fleft( t right)dt} } right) }={ fleft( x right).}]
Hence
[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_1^x {sqrt {{t^3} + 4t} dt} } right) }={ sqrt {{x^3} + 4x} .}]
Substituting (x = 2) yields
[{g^primeleft( 2 right) }={ sqrt {{2^3} + 4 cdot 2} }={ sqrt {16} }={ 4.}]
Example 2.
Calculate the derivative of the function (gleft( x right) = intlimits_{ – large{frac{pi }{2}}normalsize}^x {sqrt {{{sin }^2}t + 2} dt} ) at (x = large{large{frac{pi }{6}}normalsize}.)
Solution.
We use the Fundamental Theorem of Calculus, Part (1:)
[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_a^x {fleft( t right)dt} } right) }={ fleft( x right).}]
Then
[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_{ – frac{pi }{2}}^x {sqrt {{{sin }^2}t + 2} dt} } right) }={ sqrt {{{sin }^2}x + 2} .}]
Note that the lower limit of integration ({ – large{frac{pi }{2}}normalsize}) does not affect the answer.
Now we compute the value of the derivative for (x = large{frac{pi }{6}}normalsize :)
[{g^primeleft( {frac{pi }{6}} right) }={ sqrt {{{sin }^2}frac{pi }{6} + 2} }={ sqrt {{{left( {frac{1}{2}} right)}^2} + 2} }={ sqrt {frac{9}{4}} }={ frac{3}{2}.}]
Example 3.
Find the derivative of the function (gleft( x right) = intlimits_3^{{x^2}} {large{frac{{dt}}{t}}normalsize}.)
Solution.
We introduce the new function
[{hleft( u right) = intlimits_3^u {frac{{dt}}{t}}.}]
Using the FTC1, we have
[{h^primeleft( u right) }={ frac{1}{u}.}]
As (gleft( x right) = hleft( {{x^2}} right),) then by the chain rule
[{g^primeleft( x right) = left[ {hleft( {{x^2}} right)} right]^prime }={ h^primeleft( {{x^2}} right) cdot left( {{x^2}} right)^prime }={ h^primeleft( {{x^2}} right) cdot 2x }={ frac{1}{{{x^2}}} cdot 2x }={ frac{2}{x}}]
Example 4.
Find the derivative of the function (gleft( x right) = intlimits_0^{{x^2}} {sqrt {1 + {t^2}} dt}.)
Solution.
Since the upper limit of integration is not (x,) we apply the chain rule. Let (u = {x^2},) then (u^prime = 2x.)
Consider the new function
[hleft( u right) = intlimits_0^u {sqrt {1 + {t^2}} dt} .]
By the FTC1, we can write
[h^primeleft( u right) = sqrt {1 + {u^2}} .]
As (gleft( x right) = hleft( {{x^2}} right),) we have
[{g^primeleft( x right) = left[ {hleft( {{x^2}} right)} right]^prime }={ h^primeleft( {{x^2}} right) cdot left( {{x^2}} right)^prime }={ sqrt {1 + {{left( {{x^2}} right)}^2}} cdot 2x }={ 2xsqrt {1 + {x^4}} .}]
Example 5.
Find the derivative of the function (gleft( x right) = intlimits_1^{{x^3}} {{t^2}dt}.)
Solution.
8.2first Fundamental Theorem Of Calculus Ap Calculus Free
Calculus
Let (u = {x^3},) then (u^prime = 3{x^2}.)
We introduce the new function
[hleft( u right) = intlimits_0^u {{t^2}dt} .]
Using the FTC1, we obtain
[h^primeleft( u right) = {u^2}.]
Since (gleft( x right) = hleft( {{x^3}} right),) we have
[{g^primeleft( x right) = left[ {hleft( {{x^3}} right)} right]^prime }={ h^primeleft( {{x^3}} right) cdot left( {{x^3}} right)^prime }={ {left( {{x^3}} right)^2} cdot 3{x^2} }={ {x^6} cdot 3{x^2} }={ 3{x^8}.}]
Example 6.
Find the derivative of the function (gleft( x right) = intlimits_{{x^2}}^{{x^3}} {tdt}.)
Solution.
We split the interval of integration (left[ {{x^2},{x^3}} right]) using an intermediate point (c,) so that (c in left[ {{x^2},{x^3}} right].) Hence the derivative of (gleft( x right)) is written in the form
[{g^primeleft( x right) }={ frac{d}{{dx}}left( {intlimits_{{x^2}}^{{x^3}} {tdt} } right) }={ frac{d}{{dx}}left( {intlimits_{{x^2}}^c {tdt} + intlimits_c^{{x^3}} {tdt} } right) }={ frac{d}{{dx}}left( {intlimits_c^{{x^3}} {tdt} – intlimits_c^{{x^2}} {tdt} } right) }={ frac{d}{{dx}}left( {intlimits_c^{{x^3}} {tdt} } right) – frac{d}{{dx}}left( {intlimits_c^{{x^2}} {tdt} } right).}]
We calculate both terms using the FTC1 and the chain rule:
[{frac{d}{{dx}}left( {intlimits_c^{{x^3}} {tdt} } right) }={ {x^3} cdot left( {{x^3}} right)^prime }={ {x^3} cdot 3{x^2} }={ 3{x^5};}]
[{frac{d}{{dx}}left( {intlimits_c^{{x^2}} {tdt} } right) }={ {x^2} cdot left( {{x^2}} right)^prime }={ {x^2} cdot 2x }={ 2{x^3}.}]
Then
[g^primeleft( x right) = 3{x^5} – 2{x^3}.]
Example 7.
Calculate the derivative of the function (gleft( x right) = intlimits_{sqrt x }^x {left( {{t^2} – t} right)dt} ) at (x = 1.)
Solution.
We split the integral function into two terms:
[{gleft( x right) }={ intlimits_{sqrt x }^x {left( {{t^2} – t} right)dt} }={ intlimits_{sqrt x }^c {left( {{t^2} – t} right)dt} + intlimits_c^x {left( {{t^2} – t} right)dt} }={ intlimits_c^x {left( {{t^2} – t} right)dt} – intlimits_c^{sqrt x } {left( {{t^2} – t} right)dt},}]
where (c in left[ {{x^2},{x^3}} right].)
Find the derivative of (gleft( x right)) using the FTC1 and the chain rule (for the second term):
[{frac{d}{{dx}}intlimits_c^x {left( {{t^2} – t} right)dt} }={ {x^2} – x;}]
[{frac{d}{{dx}}intlimits_c^{sqrt x } {left( {{t^2} – t} right)dt} }={ left( {{{left( {sqrt x } right)}^2} – sqrt x } right) cdot left( {sqrt x } right)^prime }={ left( {x – sqrt x } right) cdot frac{1}{{2sqrt x }} }={ frac{{sqrt x }}{2} – frac{1}{2}.}]
Then
[{g^primeleft( x right) }={ left( {{x^2} – x} right) }-{ left( {frac{{sqrt x }}{2} – frac{1}{2}} right) }={ {x^2} – x – frac{{sqrt x }}{2} + frac{1}{2}.}]
At the point (x = 1,) the derivative is equal to
[{g^primeleft( 1 right) }={ {1^2} – 1 }-{ frac{{sqrt 1 }}{2} + frac{1}{2} }={ 0.}]
Example 8.
Evaluate the integral (intlimits_0^2 {left( {{x^3} – {x^2}} right)dx}.)
Solution.
Using the Fundamental Theorem of Calculus, Part (2,) we have
[
{intlimits_0^2 {left( {{x^3} – {x^2}} right)dx} }
= {left. {left( {frac{{{x^4}}}{4} – frac{{{x^3}}}{3}} right)} right _0^2 }
= {left( {frac{{16}}{4} – frac{8}{3}} right) – 0 }={ frac{4}{3}.}
]
Example 9.
Evaluate the integral (intlimits_{ – 1}^1 {left( {{t^2} + {t^{21}}} right)dt}.)
Solution.
An antiderivative of the function ({{t^2} + {t^{21}}}) is (large{frac{{{t^3}}}{3} + frac{{{t^{22}}}}{{22}}}normalsize.) Then using the Fundamental Theorem of Calculus, Part (2,) we have
[require{cancel}{intlimits_{ – 1}^1 {left( {{t^2} + {t^{21}}} right)dt} }={ left. {left[ {frac{{{t^3}}}{3} + frac{{{t^{22}}}}{{22}}} right]} right _{ – 1}^1 }={ left( {frac{{{1^3}}}{3} + frac{{{1^{22}}}}{{22}}} right) }-{ left( {frac{{{{left( { – 1} right)}^3}}}{3} + frac{{{{left( { – 1} right)}^{22}}}}{{22}}} right) }={ frac{1}{3} + cancel{frac{1}{{22}}} + frac{1}{3} – cancel{frac{1}{{22}}} }={ frac{2}{3}.}]
Example 10.
Calculate the integral (intlimits_0^1 {left( {sqrt[large 3normalsize]{t} – sqrt t } right)dt}.)
Solution.
[
{intlimits_0^1 {left( {sqrt[large 3normalsize]{t} – sqrt t } right)dt} }
= {intlimits_0^1 {left( {{t^{largefrac{1}{3}normalsize}} – {t^{largefrac{1}{2}normalsize}}} right)dt} }
= {left. {left( {frac{{{t^{largefrac{1}{3}normalsize + 1}}}}{{frac{1}{3} + 1}} – frac{{{t^{largefrac{1}{2}normalsize + 1}}}}{{frac{1}{2} + 1}}} right)} right _0^1 }
= {left. {left( {frac{{3{t^{largefrac{4}{3}normalsize}}}}{4} – frac{{2{t^{largefrac{3}{2}normalsize}}}}{3}} right)} right _0^1 }
= {left( {frac{3}{4} – frac{2}{3}} right) – 0 }={ frac{1}{{12}}.}
]
Example 11.
Evaluate the integral (intlimits_0^1 {{largefrac{x}{{{{left( {3{x^2} – 1} right)}^4}}}normalsize} dx}.)
Solution.
First we make the substitution:
[
{t = 3{x^2} – 1,;;}Rightarrow
{dt = 6xdx,;;}Rightarrow
{xdx = frac{{dt}}{6}.}
]
Determine the new limits of integration. When (x = 0,) then (t = -1.) When (x = 1,) then we have (t = 2.) So, the integral with the new variable (t) can be easily calculated:
[
{intlimits_0^1 {frac{x}{{{{left( {3{x^2} – 1} right)}^4}}}dx} }
= {intlimits_{ – 1}^2 {frac{{frac{{dt}}{6}}}{{{t^4}}}} }
= {frac{1}{6}int {{t^{ – 4}}dt} }
= {frac{1}{6}left. {left( {frac{{{t^{ – 3}}}}{{ – 3}}} right)} right _{ – 1}^2 }
= { – frac{1}{{18}}left( {frac{1}{8} – 1} right) }
= {frac{7}{{144}}.}
]
Example 12.
Evaluate the integral (intlimits_1^e {left( {t + large{frac{1}{t}}normalsize} right)dt}.)
Solution.
An antiderivative of the function ({t + large{frac{1}{t}}normalsize}) has the form (large{frac{{{t^2}}}{2}}normalsize + ln t.) Hence, by the Fundamental Theorem, Part (2,) we have
[{intlimits_1^e {left( {t + frac{1}{t}} right)dt} }={ left. {left[ {frac{{{t^2}}}{2} + ln t} right]} right _1^e }={ left( {frac{{{e^2}}}{2} + ln e} right) }-{ left( {frac{{{1^2}}}{2} + ln 1} right) }={ frac{{{e^2}}}{2} + 1 – frac{1}{2} – 0 }={ frac{{{e^2}}}{2} + frac{1}{2}.}]
Example 13.
Evaluate the integral (intlimits_0^{ln 2} {x{e^{ – x}}dx}.)
Solution.
We can write
[
{I = intlimits_0^{ln 2} {x{e^{ – x}}dx} }
= { – intlimits_0^{ln 2} {xdleft( {{e^{ – x}}} right)} .}
]
Apply integration by parts: ({largeintnormalsize} {udv} ) (= uv – {largeintnormalsize} {vdu} .) In this case, let
8.2first fundamental theorem of calculus ap calculus free
[
{u = x,;;}kern-0.3pt{dv = dleft( {{e^{ – x}}} right),;;}Rightarrow
{du = 1,;;}kern-0.3pt{v = {e^{ – x}}.}
]
Hence, the integral is
[
{I = – intlimits_0^{ln 2} {xdleft( {{e^{ – x}}} right)} }
= { – left[ {left. {left( {x{e^{ – x}}} right)} right _0^{ln 2} – intlimits_0^{ln 2} {{e^{ – x}}dx} } right] }
= {{ – left. {left( {x{e^{ – x}}} right)} right _0^{ln 2} }+{ intlimits_0^{ln 2} {{e^{ – x}}dx} }}
= {{ – left. {left( {x{e^{ – x}}} right)} right _0^{ln 2} }-{ left. {left( {{e^{ – x}}} right)} right _0^{ln 2} }}
= { – left. {left[ {{e^{ – x}}left( {x + 1} right)} right]} right _0^{ln 2} }
= {{ – {e^{ – ln 2}}left( {ln 2 + 1} right) }+{ {e^0} cdot 1 }}
= { – frac{{ln 2}}{2} – frac{{ln e}}{2} + ln e }
= {frac{{ln e}}{2} – frac{{ln 2}}{2} }
= {frac{1}{2}left( {ln e – ln 2} right) }
= {frac{1}{2}ln frac{e}{2}.}
]
Example 14.
Evaluate the integral (intlimits_{ – 1}^1 {left {x – large{frac{1}{2}}normalsize} right dx}.)
Solution.
We rewrite the absolute value expression in the form
[left {x – frac{1}{2}} right = begin{cases}– x + frac{1}{2}, & text{if }x lt frac{1}{2}x – frac{1}{2}, & text{if }x ge frac{1}{2}end{cases}]
and split the interval of integration into two intervals such that
[{intlimits_{ – 1}^1 {left {x – frac{1}{2}} right dx} }={ intlimits_{ – 1}^{frac{1}{2}} {left {x – frac{1}{2}} right dx} }+{ intlimits_{frac{1}{2}}^1 {left {x – frac{1}{2}} right dx} }={ intlimits_{ – 1}^{frac{1}{2}} {left( { – x + frac{1}{2}} right)dx} }+{ intlimits_{frac{1}{2}}^1 {left( {x – frac{1}{2}} right)dx} .}]
Now we can apply the Fundamental Theorem of Calculus, Part (2,) to each of the integrals:
[{intlimits_{ – 1}^1 {left {x – frac{1}{2}} right dx} }={ intlimits_{ – 1}^{frac{1}{2}} {left( { – x + frac{1}{2}} right)dx} }+{ intlimits_{frac{1}{2}}^1 {left( {x – frac{1}{2}} right)dx} }={ left[ { – frac{{{x^2}}}{2} + frac{x}{2}} right]_{ – 1}^{frac{1}{2}} }+{ left[ {frac{{{x^2}}}{2} – frac{x}{2}} right]_{frac{1}{2}}^1 }={ left[ {left( { – frac{1}{8} + frac{1}{4}} right) – left( { – frac{1}{2} – frac{1}{2}} right)} right] }+{ left[ {left( {cancel{frac{1}{2}} – cancel{frac{1}{2}}} right) – left( {frac{1}{8} – frac{1}{4}} right)} right] }={ frac{9}{8} + frac{1}{8} }={ frac{{10}}{8} }={ frac{5}{4}.}]
Example 15.
Evaluate the integral (intlimits_{ – 2}^1 {left {{x^2} – 1} right dx}.)
Solution.
We represent the absolute value expression as follows:
8.2first fundamental theorem of calculus ap calculus pdf
[{left {{x^2} – 1} right text{ = }}kern0pt{begin{cases}{{x^2} – 1}, & text{if }x in left( { – infty , – 1} right] cup left[ {1,infty } right)1 – {x^2}, & text{if }x in left( { – 1,1} right)end{cases}}]
So we can split the initial integral into two integrals:
[{intlimits_{ – 2}^1 {left {{x^2} – 1} right dx} }={ intlimits_{ – 2}^{ – 1} {left {{x^2} – 1} right dx} }+{ intlimits_{ – 1}^1 {left {{x^2} – 1} right dx} }={ intlimits_{ – 2}^{ – 1} {left( {{x^2} – 1} right)dx} }+{ intlimits_{ – 1}^1 {left( {1 – {x^2}} right)dx} .}]
Using the Fundamental Theorem of Calculus, Part (2,) we obtain:
[{intlimits_{ – 2}^1 {left {{x^2} – 1} right dx} }={ intlimits_{ – 2}^{ – 1} {left( {{x^2} – 1} right)dx} }+{ intlimits_{ – 1}^1 {left( {1 – {x^2}} right)dx} }={ left[ {frac{{{x^3}}}{3} – x} right]_{ – 2}^{ – 1} }+{ left[ {x – frac{{{x^3}}}{3}} right]_{ – 1}^1 }={ left[ {left( { – frac{1}{3} – left( { – 1} right)} right) }right.}-{left.{ left( { – frac{8}{3} – left( { – 2} right)} right)} right] }+{ left[ {left( {1 – frac{1}{3}} right) – left( { – 1 – left( { – frac{1}{3}} right)} right)} right] }={ frac{2}{3} + frac{2}{3} + frac{2}{3} + frac{2}{3} }={ frac{8}{3}.}]
Example 16.
Find the area bounded by the curves (y = {x^2}) and (y = sqrt x.)
Solution.
First we find the points of intersection (see Figure (6)):
[
{{x^2} = sqrt x ,;;}Rightarrow
{{x^2} – sqrt x = 0,;;}Rightarrow
{sqrt x left( {{x^{largefrac{3}{2}normalsize}} – 1} right) = 0,;;}Rightarrow
{{x_1} = 0,;{x_2} = 1.}
]
As you can see, the curves intercept at the points (left( {0,0} right)) and (left( {1,1}right).) Hence, the area is given by
[
{S = intlimits_0^1 {left( {sqrt x – {x^2}} right)dx} }
= {left. {left( {frac{{{x^{largefrac{1}{2}normalsize + 1}}}}{{frac{1}{2} + 1}} – frac{{{x^3}}}{3}} right)} right _0^1 }
= {frac{1}{3}left. {left( {2sqrt {{x^3}} – {x^3}} right)} right _0^1 }={ frac{1}{3}.}
]
Example 17.
Find the area bounded by the curves (y = 2x – {x^2}) and (x + y = 0.)
Solution.
First we find the points of intersection of the curves (see Figure (7)):
[
{2x – {x^2} = – x,;;}Rightarrow
{{x^2} – 3x = 0,;;}Rightarrow
{xleft( {x – 3} right) = 0,;;}Rightarrow
{{x_1} = 0,;{x_2} = 3.}
]
The upper boundary of the region is the parabola (y = 2x – {x^2},) and the lower boundary is the straight line (y = -x.)
The area is given by
[
{S = intlimits_0^3 {left[ {2x – {x^2} – left( { – x} right)} right]dx} }
= {intlimits_0^3 {left( {2x – {x^2} + x} right)dx} }
= {left. {left( {{x^2} – frac{{{x^3}}}{3} + frac{{{x^2}}}{2}} right)} right _0^3 }
= {left. {left( {frac{{3{x^2}}}{2} – frac{{{x^3}}}{3}} right)} right _0^3 }
= {frac{{27}}{2} – frac{{27}}{3} }={ frac{9}{2}.}
]
Example 18.
Find the area of the triangle with vertices at (left( {0,0} right),) (left( {2,6} right)) and (left( {7,1} right).)
Solution.
First we find an equation of the side (OA) (Figure (8)):
[
{frac{{x – {x_O}}}{{{x_A} – {x_O}}} = frac{{y – {y_O}}}{{{y_A} – {y_O}}},;;}Rightarrow
{frac{{x – 0}}{{2 – 0}} = frac{{y – 0}}{{6 – 0}},;;}Rightarrow
{frac{x}{2} = frac{y}{6},;;}Rightarrow
{y = 3x.}
]
Similarly, we find an equation of the side (OB:)
[
{frac{{x – {x_O}}}{{{x_B} – {x_O}}} = frac{{y – {y_O}}}{{{y_B} – {y_O}}},;;}Rightarrow
{frac{{x – 0}}{{7 – 0}} = frac{{y – 0}}{{1 – 0}},;;}Rightarrow
{frac{x}{7} = frac{y}{1},;;}Rightarrow
{y = frac{x}{7}.}
]
Next, find an equation of the side (AB:)
[
{frac{{x – {x_B}}}{{{x_A} – {x_B}}} = frac{{y – {y_B}}}{{{y_A} – {y_B}}},;;}Rightarrow
{frac{{x – 2}}{{7 – 2}} = frac{{y – 6}}{{1 – 6}},;;}Rightarrow
{frac{{x – 2}}{5} = frac{{y – 6}}{{ – 5}},;;}Rightarrow
{y = 8 – x.}
]
As you can see from Figure (8,) the area of the this triangle can be calculated as the sum of two integrals:
[
{S = {I_1} + {I_2} }
= {intlimits_0^2 {left( {3x – frac{x}{7}} right)dx} }+{ intlimits_2^7 {left( {8 – x – frac{x}{7}} right)dx} }
= {left. {left( {frac{{10{x^2}}}{7}} right)} right _0^2 }+{ left. {left( {8x – frac{{4{x^2}}}{7}} right)} right _2^7 }
= {frac{{10 cdot 4}}{7} + left( {56 – frac{{4 cdot 49}}{7}} right) }-{ left( {16 – frac{{4 cdot 4}}{7}} right) }={ 20.}
]
Example 19.
Find the area inside the ellipse ({largefrac{{{x^2}}}{{{a^2}}}normalsize} + {largefrac{{{y^2}}}{{{b^2}}}normalsize} = 1.)
Solution.
By symmetry (see Figure (9)), the area of the ellipse is twice the area above the (x)-axis.
The latter is given by
[
{{S_{frac{1}{2}}} }={ intlimits_{ – a}^a {sqrt {{b^2}left( {1 – frac{{{x^2}}}{{{a^2}}}} right)} dx} }
= {frac{b}{a}intlimits_{ – a}^a {sqrt {{a^2} – {x^2}} dx} .}
]
To calculate the last integral, we use the trigonometric substitution (x = asin t,) (dx = acos tdt.)
Refine the limits of integration. When (x = -a,) then (sin t = -1) and (t = – {largefrac{pi }{2}normalsize}.) When (x = a,) then (sin t = 1) and (t = {largefrac{pi }{2}normalsize}.) Thus we get
[
{{S_{frac{1}{2}}} }={ frac{b}{a}intlimits_{ – a}^a {sqrt {{a^2} – {x^2}} dx} }
= {frac{b}{a}intlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {sqrt {{a^2} – {a^2}{{sin }^2}t}, }}kern0pt{{ acos tdt} }
= {abintlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {{{cos }^2}tdt} }
= {abintlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {frac{{1 + cos 2t}}{2}dt} }
= {frac{{ab}}{2}intlimits_{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} {left( {1 + cos 2t} right)dt} }
= {frac{{ab}}{2}left. {left( {t + frac{{sin 2t}}{2}} right)} right _{ – largefrac{pi }{2}normalsize}^{largefrac{pi }{2}normalsize} }
= {frac{{ab}}{2}left[ {frac{pi }{2} + frac{{sin pi }}{2} }right.}-{left.{ left( { – frac{pi }{2}} right) – frac{{sin left( { – pi } right)}}{2}} right] }
= {frac{{pi ab}}{2}.}
]
Hence, the total area of the ellipse is (pi ab.)
8.2first Fundamental Theorem Of Calculus Ap Calculus Answers
Many people believe that mathematics is about number-crunching, but much more importantly, math is about reasoning. For example, when you solve a word problem, you are using your reasoning skills to put together the given information in just the right way.
In a way, AP Calculus is all about reasoning. You have to interpret each problem and correctly apply the appropriate methods (limits, derivatives, integrals, etc.) to solve it. However sometimes we have to take it one step further and reason with theorems and definitions as well, gluing our thoughts together with mathematical logic.
Why is this important? Well using nothing more than a handful of assumptions and plenty of definitions, theorems, and logic, Euclid developed the entire subject of Geometry from the ground up! If that’s not a reason to respect the power of definitions and theorems, then nothing else is.
What are Definitions and Theorems?
In mathematics, every term must be defined in some way. A definition of a mathematical object is formal description of the essential properties that make that object what it is. For instance,
• Definition: A triangle is a three-sided polygon.
It’s very important to understand the definitions of our mathematical terms so that we can employ just the right tool in each specific case. So if you see a three-sided polygon in a problem, then you know that it’s a triangle by definition. Then you may use a property or formula related to triangles as part of your reasoning steps.
We also rely on general statements of truth called theorems in order to reason about a specific situation. Speaking of triangles, perhaps one of the most famous (and useful) theorems of all time is the Pythagorean Theorem. (By the way, this theorem shows up in Book 1 of Euclid’s Elements, over 2000 years ago!)
Diagram for Pythagoras theorem by Drini (Pedro Sanchez)
Reasoning with Definitions
On the AP Calculus exams, you must know and be able to apply the definitions of calculus. Let’s see what that means in an example problem.
Example 1
If , which of the following is true?
(A) f(x) is continuous and differentiable at x = 3
(B) f(x) is continuous but not differentiable at x = 3
(C) f(x) is neither continuous nor differentiable at x = 3
(D) f(x) is differentiable but not continuous at x = 3
In order to properly address this question, we must know the definitions of continuous and differentiable.
• A function f is continuous at a point x = a if
• A function f is differentiable at a point x = a if the derivative f ‘(a) exists.
Example Solution — Continuity
First let’s determine if the function is continuous at x = 3. Because f is defined piece-wise, we must compute both the left and right hand limits.
Now because the left and right hand limits agree, we know that the two-sided limit as x → 3 exists and equals 0. Next, check the function value at x = 3.
Therefore, since the limiting value equals the function value (both are 0), the function f is continuous at x = 3 by definition. (For more about this topic, check out AP Calculus Exam Review: Limits and Continuity.)
Example Solution — Differentiability
Moving on to differentiability, now we must check whether f ‘(3) exists. Again, because f is defined piece-wise, we must be careful at the point where the function changes behavior. First find the derivative of each piece.
Note, there is no typo here — the derivative of the first piece can only be found when x < 3. In fact it takes more analysis to figure out what happens atx = 3.
Because the derivative itself is actually a certain kind of limit (by definition!), we’ll have to see what the limiting values for f ‘ are as x → 3. As before, examine each piece separately.
This time there is a mismatch. Because the left and right derivatives do not agree (18 ≠ -9), the derivative does not exist at x = 3. Thus by definition, f is not differentiable at x = 3.
In summary, f is continuous, but not differentiable at x = 3. Choice (B) is correct.
Reasoning with Theorems
Remember, a theorem is a true mathematical statement. Typically theorems are general facts that can apply to lots of different situations. Here is a small list of important theorems in calculus.
• Intermediate Value Theorem
• Extreme Value Theorem
• Mean Value Theorem for Derivatives
• Rolle’s Theorem
• Fundamental Theorem of Calculus (two parts)
• Mean Value Theorem for Integrals
A Theorem by any other Name…
There are many other results and formulas in calculus that may not have the title of “Theorem” but are nevertheless important theorems. Every one of your derivative and antidifferentiation rules is actually a theorem. Here is a partial list of other theorems that may not be explicitly identified as theorems in your textbook.
• Differentiability implies continuity
• The first derivative rule for increase and decrease
• The second derivative rule for concavity
• First and second derivative rules for relative extrema
• Product Rule, Quotient Rule, Chain Rule, etc.
• Additivity and linearity of the definite integral
• Techniques of antidifferentiation such as substitution, integration by parts, etc.
• Various tests for convergence of series
Now let’s see if we can use the right theorems to crack the next example.
Example 2
Find .
Notice that this is a derivative of an integral. That means we may be able to apply the Fundamental Theorem of Calculus. There are two parts to the theorem, but the one we need is:
However, before we can apply this theorem, we must change the form of the integral. The theorem requires that the lower limit of integration must be a constant. By using the rule for switching the order of integration (another theorem!), we may write:
Next, because the upper limit of integration is not a simple variable, x, we must use yet another theorem: the Chain Rule. Here, the “inside function” is u = x3.
It’s interesting to note in this case that no other method could have led to the solution. It is impossible to write down an antiderivative for the function, sin t2. Fortunately the Fundamental Theorem of Calculus in the form we used it avoids the antidifferentiation step altogether.
Conclusion
Definitions and theorems form the backbone of mathematical reasoning. Knowing your definitions means knowing which tools can apply in each situation. And by understanding the theorems, you can avoid doing a lot of unnecessary or difficult work.
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About Shaun Ault
Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!
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-1,653,459,266,811,357,700 | 読者です 読者をやめる 読者になる 読者になる
あのKTOKの数学blog
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Riemann積分のすっごーいところとLebesgue積分のすっごーいところ
数学 解析 Riemann積分 Lebesgue積分
面積を求めたり、函数を調べたり、不変量を求めたりと、いろいろなところで活躍している積分。ところで積分といえば、Lebesgue積分やRiemann積分があります。今回はLebesgue積分、Riemann積分のそれぞれの優れているところを取り上げます。
前提知識: 微積分(Riemann積分)
今回はLebesgue積分を学んだことありましたら、より楽しめるのではないかと思われます。
(わからない部分は流し読みすることをお勧めします。)
§1 Lebesgue積分は極限との相性が良い。
積分論の歴史を振り返ります。積分論が発展した理由の一つとして、B. Riemannが初めて厳密に積分を定義することができました。これにより微分積分学の基本定理など様々な積分の命題が証明されたが、この積分方法では極限との相性が悪いです。この問題の解決の糸口を開いたのがA. Lebesgueでした。特に、単調収束定理、Lebesgueの優収束定理によって、積分と極限の相性がとても良いものになりました。
さて、Lebesgue積分が極限と相性の良いことを見ていきましょう。簡単のため、 [ a, b ]閉区間(ただし a bは実数もしくは - \infty \infty )で考えます。Lebesgueの単調収束定理とは次に述べる定理です。
定理1. \{f_n \} はLebesgue積分可能な函数列で次の条件を満たしているとしています。
1. f_n(x) \leq f_{n+1} (x)
すると、ほとんどいたるところで f_n (x) \displaystyle \lim_{n \to \infty} f_n(x)に収束する。
このとき、
\displaystyle \lim_{n \to \infty} \int_a^b f_n (x) dx = \int_a^b \lim_{n \to \infty} f_n (x) dx
が成立する。
『ほとんどいたるところ』がわからなければ、各点と思って読めばよいでしょう。また、Lebesgue積分は高校でやっている積分と計算結果は変わりませんLebesgue積分可能がわからなければ、そういう函数の集合があると思えばよいでしょう。
では、この命題を狭義(resp.広義)Riemann積分の範疇に変えると次のような定理になります。
定理2. \{f_n \} は狭義(resp.広義)Riemann積分可能な函数列で次の条件を満たしているとしています。
1. f_n(x) \leq f_{n+1} (x)
すると、ほとんどいたるところで f_n (x) \displaystyle \lim_{n \to \infty} f_n(x)に収束する。さらに、
2. \displaystyle \lim_{n \to \infty} f_n (x) は狭義(resp. 広義)Riemann積分可能である。
このとき、
\displaystyle \lim_{n \to \infty} \int_a^b f_n (x) dx = \int_a^b \lim_{n \to \infty} f_n (x) dx
が成立する。
Lebesgue積分より条件が多いです。問題なのは2.です。Lebesgue積分では、収束先である fはLebesgue可積分になります。しかし、Riemann積分だと、 fは必ずしもRiemann可積分になるとは限りません。なので、余分な条件が必要になります。
ではLebesgue積分の場合の使いやすさを実感しましょう。具体的に、ゼータ函数と熱核の関係式について見てみましょう。
\displaystyle n^{-s} = \frac{1}{\Gamma (s)} \int_0^{\infty} e^{-n t } t^{s-1} dt
であるから、
\displaystyle \zeta (s) = \sum_{n=1}^{\infty} n^{-s}
\displaystyle = \sum_{n=1}^{\infty } \frac{1}{\Gamma (s)} \int_0^{\infty} e^{-n t } t^{s-1} dt
\displaystyle = \frac{1}{\Gamma (s)} \int_0^{\infty} \sum_{n=1}^{\infty} e^{-n t } t^{s-1} dt
となります。最後の式変形についてです。Lebesgueの場合、 \sum_{n=1}^{m} e^{-n t} t^{s-1} m が増えるにつれて単調増加するので、そのまま単調収束定理を用いることにより最後の式変形が従います。一方、Riemann積分の場合、 \sum_{n=1}^{\infty} e^{-n t } t^{s-1}がRiemann積分可能かいちいち確認しなくてはいけません。Lebesgueのほうが議論が楽であることが実感できるでしょう。
Lebesgue積分のある種の完備性、つまり極限との相性がよいので、複素解析函数解析など、解析のあらゆるところに貢献しています。こうみると、Riemann積分は劣った積分に見えます。ではRiemann積分ではできて、Lebesgue積分できないことはあるのでしょうか？次の節で見ていきましょう。
§2 Riemann積分はWeylの一様分布定理と相性が良い。
Riemann積分の良さを話す前に、Weylの一様分布定理について述べておきましょう。次の問題を考えます。
問題3. \alpha 無理数とする。 a_n = n \alpha - \lfloor n \alpha \rfloor とする。つまり、 a_n n \alpha の小数部分である。このとき、 \{ a_n \} [ 0 , 1 ]で稠密か。
この問題3を解決するために、これより強い問題を解きます。
定義4. [ 0, 1 ] に属する数列 \{ \xi_n \} 一様分布列であるとは任意の [0 , 1 ]区間に含まれる区間 [a,b]に対して
\displaystyle b-a = \lim_{N \to \infty} \frac{\# \{ \xi_n \in [ a , b ] | n= 1 , ... ,N \} }{N}
が成り立つことをいう。
問題5. 数列 \{ a_n \} は一様分布列か？
問題5は問題3より強い主張です。問題5の答えは次の通りになります。
定理6. 数列 \{ a_n \} は一様分布列である。
この定理はWeylの一様分布定理と呼ばれています。証明の概略を述べておきしょう。 f 函数とします。次の条件を考えます。
\displaystyle \lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N f( a_n ) = \int_0^1 f(x) dx
上の条件を条件Ａと呼ぶことにします。
Step1. kを任意の整数として f(x) = e^{i \pi k x}は条件Ａを満たす
これはそのまま代入して級数の収束について考えます。
Step2. 函数 f g条件Ａを満たすとき、 af + bgも条件Ａを満たす。
Step3. Weiesrtrass三角函数近似定理*1により、 f が連続函数なら条件Ａを満たす
Step4. f(x) =\begin{cases}1 \,\, (x \in [ a , b ] ) \\ 0 \,\, (x \not \in [a , b ] )\end{cases} は条件Ａを満たす。つまり、Weylの一様分布定理が成り立つ。
これは fが上からも下からも連続函数で抑えることができることから従います。詳しくは参考文献「2.」をご覧ください。
さて、条件Ａはどのような函数だったら成り立つのでしょうか。実は次の主張が成り立ちます。
命題7. f がRiemann積分可能なとき、条件Ａは成立する。
証明の概略についてです。 f がRiemann積分可能なので、てきとうな区間を考えて単函数で上からと下からで抑えることができます。単函数というのはStep4.で挙げた函数の線形和であらわしたものです。このことから従います。
ではLebesgue積分可能な函数ならどうでしょう。これは反例があります。
反例8. X=\{ x | x = \alpha q - \lfloor \alpha q \rfloor , q \in \mathbb{Q} \} とする。
このとき f(x) =\begin{cases}1 \,\, (x \in X ) \\ 0 \,\, (x \not \in X )\end{cases} は条件Ａを満たさない。
実はこの函数はLebesgue積分可能だが、Riemann積分不可能である函数です。よって、条件ＡはRiemann積分可能では成り立つがLebesgue積分可能では成り立つとは限らないことになります。
ここで話は終わりません。数学の恒例行事である、逆を考えてみましょう。つまり、条件Ａが成り立つならRiemann積分可能かを考えてみます。これは反例があります。
反例9. f(x) =\begin{cases}1 \,\, (x \in \mathbb{Q} ) \\ 0 \,\, (x \not \in \mathbb{Q} )\end{cases} は条件Ａを満たす。
条件Ａでは緩いことがわかります。 f( a_n ) =0となってしまいます。 a_n \in \mathbb{Q} が原因でしょう。この反省を踏まえるとRiemann積分可能と同値な命題を得ることができます。
命題10. f有界でLebesgue可測な函数(つまりLebesgue積分可能)とする。このとき、次の主張は同値である。
• fがRiemann積分可能
• 任意の一様分布列 \{ \xi_n \} に対して次が成り立つ。
\displaystyle \lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N f( \xi_n ) = \int_0^1 f(x) dx
参考文献(ABC順)
1. 小平 邦彦, "解析入門II," 岩波書店, 1976年.(復刊版: 小平邦彦, "軽装版 解析入門＜1＞," 岩波書店, 2003年.)]
2. Elias M. Stein, Rami Shakarchi 著, 新井 仁之, 杉本 充, 高木 啓行, 千原 浩之 訳, "プリンストン解析学講義, フーリエ解析入門," 日本評論社, 2007年.
今回の題材は主に「2.」です。Weylの一様分布定理の章を題材に今回の話を組み立てています。また、「1.」にArzelàの定理と呼ばれる、Lebesgueの優収束定理の元となった定理が掲載されています。Lebesgueの優収束定理はArzelàの定理に比べて条件が緩くなっていることがわかります
次回は3/19までに作りたいと思います。
*1:連続函数が三角函数の線形和で一様に近似できるという定理 | {
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2,276,157,615,690,385,700 | Kalkulus Contoh
Langkah 1
Konversikan pertidaksamaan ke persamaan.
Langkah 2
Kurangkan dari kedua sisi persamaan tersebut.
Langkah 3
Faktorkan menggunakan metode AC.
Ketuk untuk lebih banyak langkah...
Langkah 3.1
Mempertimbangkan bentuk . Tentukan pasangan bilangan bulat yang hasil kalinya (Variabel1) dan jumlahnya . Dalam hal ini, hasil kalinya dan jumlahnya .
Langkah 3.2
Tulis bentuk yang difaktorkan menggunakan bilangan bulat ini.
Langkah 4
Jika faktor individu di sisi kiri persamaan sama dengan , seluruh pernyataan akan menjadi sama dengan .
Langkah 5
Atur agar sama dengan dan selesaikan .
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Langkah 5.1
Atur sama dengan .
Langkah 5.2
Tambahkan ke kedua sisi persamaan.
Langkah 6
Atur agar sama dengan dan selesaikan .
Ketuk untuk lebih banyak langkah...
Langkah 6.1
Atur sama dengan .
Langkah 6.2
Kurangkan dari kedua sisi persamaan tersebut.
Langkah 7
Penyelesaian akhirnya adalah semua nilai yang membuat benar.
Langkah 8
Gunakan masing-masing akar untuk membuat interval pengujian.
Langkah 9
Pilih nilai uji dari masing-masing interval dan masukkan nilai ini ke dalam pertidaksamaan asal untuk menentukan interval mana yang memenuhi pertidaksamaan.
Ketuk untuk lebih banyak langkah...
Langkah 9.1
Uji nilai pada interval untuk melihat apakah nilai ini membuat pertidaksamaan bernilai benar.
Ketuk untuk lebih banyak langkah...
Langkah 9.1.1
Pilih nilai pada interval dan lihat apakah nilai ini membuat pertidaksamaan asal bernilai benar.
Langkah 9.1.2
Ganti dengan pada pertidaksamaan asal.
Langkah 9.1.3
Sisi kiri tidak lebih kecil dari sisi kanan , yang berarti pernyataan yang diberikan salah.
Salah
Salah
Langkah 9.2
Uji nilai pada interval untuk melihat apakah nilai ini membuat pertidaksamaan bernilai benar.
Ketuk untuk lebih banyak langkah...
Langkah 9.2.1
Pilih nilai pada interval dan lihat apakah nilai ini membuat pertidaksamaan asal bernilai benar.
Langkah 9.2.2
Ganti dengan pada pertidaksamaan asal.
Langkah 9.2.3
Sisi kiri lebih kecil dari sisi kanan , yang berarti pernyataan yang diberikan selalu benar.
Benar
Benar
Langkah 9.3
Uji nilai pada interval untuk melihat apakah nilai ini membuat pertidaksamaan bernilai benar.
Ketuk untuk lebih banyak langkah...
Langkah 9.3.1
Pilih nilai pada interval dan lihat apakah nilai ini membuat pertidaksamaan asal bernilai benar.
Langkah 9.3.2
Ganti dengan pada pertidaksamaan asal.
Langkah 9.3.3
Sisi kiri tidak lebih kecil dari sisi kanan , yang berarti pernyataan yang diberikan salah.
Salah
Salah
Langkah 9.4
Bandingkan interval untuk menentukan mana yang memenuhi pertidaksamaan asal.
Salah
Benar
Salah
Salah
Benar
Salah
Langkah 10
Penyelesaian tersebut terdiri dari semua interval hakiki.
Langkah 11
Hasilnya dapat ditampilkan dalam beberapa bentuk.
Bentuk Ketidaksamaan:
Notasi Interval:
Langkah 12
Masukkan Soal
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-4,945,966,196,158,124,000 | What is a ratio variable?
Ratio-variable meaning Filters. (statistics) A variable with the features of interval variable and, additionally, whose any two values have meaningful ratio, making the operations of multiplication and division meaningful.
What is an example of a ratio scale variable?
The common example of a ratio scale is length, duration, mass, money age, etc. For the purpose of marketing research, a ratio scale can be useful to evaluate sales, price, share, and a number of customers.
What is ratio data example?
An excellent example of ratio data is the measurement of heights. Height could be measured in centimeters, meters, inches, or feet. In ratio data, the difference between 1 and 2 is the same as the difference between 3 and 4, but also here 4 is twice as much as 2.
What is an example of ratio?
In mathematics, a ratio indicates how many times one number contains another. For example, if there are eight oranges and six lemons in a bowl of fruit, then the ratio of oranges to lemons is eight to six (that is, 8∶6, which is equivalent to the ratio 4∶3). Equal quotients correspond to equal ratios.
Is age a ratio variable?
Age, money, and weight are common ratio scale variables. For example, if you are 50 years old and your child is 25 years old, you can accurately claim you are twice their age.
What is a ratio scale in geography?
Map scale refers to the relationship (or ratio) between distance on a map and the corresponding distance on the ground. For example, on a 1:100000 scale map, 1cm on the map equals 1km on the ground. For example, a 1:100000 scale map is considered a larger scale than a 1:250000 scale map.
What do you mean by ratio scale give an example?
A ratio scale is a quantitative scale where there is a true zero and equal intervals between neighboring points. Unlike on an interval scale, a zero on a ratio scale means there is a total absence of the variable you are measuring. Length, area, and population are examples of ratio scales.
Which of the following is an example of ratio measurement?
A ratio scale is the most informative scale as it tends to tell about the order and number of the object between the values of the scale. The most common examples of this scale are height, money, age, weight etc.
Which is the best example of ratio data?
Examples of Ratio Data
• Kelvin Scale: One most noted example of Ratio Data is the temperature on a Kelvin scale.
• Height: Height or length is measured in meters, inches, or feet.
• Speed: Speed can also be an example of a ratio scale.
• The ratio between 44.738 mph to 22.369 mph is 2.
Is age an example of ratio data?
Age is frequently collected as ratio data, but can also be collected as ordinal data.
What is a real life example of a ratio?
Examples of ratios in life: The car was traveling 60 miles per hour, or 60 miles in 1 hour. You have a 1 in 28,000,000 chance of winning the lottery. Out of every possible scenario, only 1 out of 28,000,000 of them has you winning the lottery.
What is a ratio easy definition?
1a : the indicated quotient of two mathematical expressions. b : the relationship in quantity, amount, or size between two or more things : proportion. 2 : the expression of the relative values of gold and silver as determined by a country’s currency laws.
What is an example of fixed ratio schedule?
Fixed refers to the delivery of rewards on a consistent schedule. Ratio refers to the number of responses that are required in order to receive reinforcement. For example, a fixed-ratio schedule might be delivery a reward for every fifth response.
What is an example of interval variable?
Intervals variables are very usefull to compare the size of social inequalities. Another examples of interval variables would be: gross domestic product, number of siblings or the number of years studied.
What is fixed interval ratio?
Variable-Interval (VI) schedule. A fixed-ratio schedule of reinforcement means that reinforcement should be delivered after a constant or “fixed” number of correct responses. For example, a fixed ratio schedule of 2 means reinforcement is delivered after every 2 correct responses.
What is fixed ratio schedule?
In operant conditioning, a fixed-ratio schedule is a schedule of reinforcement where a response is reinforced only after a specified number of responses. Essentially, the subject provides a set number of responses and then the trainer offers a reward. | {
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Square Numbers Worksheets
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Square numbers are used to express numbers multiplied by themselves in a simple way. This helps to make them more accessible when your child is first introduced to them when learning the maths curriculum. Our square numbers worksheets help children understand why we use square numbers, how to express them using the correct notation and help them start to learn some square numbers. Square numbers are denoted using a small 2 to the right of a number - for example, 3² is equivalent to 3 x 3 which equals 9. Our square number worksheets are a valuable resource that helps your child with other maths worksheets, including our multiplication worksheets, area worksheets and much more.
Learn about square numbers worksheets
Our worksheets on square numbers have been created with children in mind; designed by our team of UK teachers, we’ve made sure there’s plenty of fun square number activities, questions and answers and practice tests to keep learning interesting! Our square numbers worksheet content is aligned with the National Curriculum, helping your child to succeed and build their confidence as they progress throughout their academic journey. They’re perfect as an additional learning aid or resource for a homeschooling programme. Keep reading to find out how you can use our square numbers worksheets with your child.
Key stage 2 square number worksheets are where your child comes across simple square numbers. Starting to understand square numbers in year 5, our worksheets start with identifying square numbers, as well as starting to add them together. Your child will also be encouraged to start working backwards, helping them to understand the simple principles of square numbers and their roots.
In our year 5 square numbers worksheets and year 6 square number worksheets, your child will be able to identify whether a number is square by understanding its factor pairs. They will also begin to memorise and consolidate their knowledge of square numbers and be able to fill in the gaps of a sequence of squared numbers.
By the time your child reaches our key stage 3 square number worksheets, they should have a good understanding of what square numbers are and be able to identify them. In our year 7 square number worksheets, your child will learn how to add and subtract square numbers, helping them improve their confidence in answering any square number questions they come across. They will also learn how to square numbers with decimal points.
Our year 8 square numbers worksheet activities and year 9 square numbers worksheets will apply your child’s previous knowledge to square units of measurement, as well as encourage your child to start being able to answer square number mental maths questions. Giving your child all the tools they’ll need in square number assessments and practice tests, our square number worksheet learning resources are here to help your child achieve academic success and beyond!
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2,013,203,755,121,519,900 | Mathematics
Question
which of the following equations correctly represents the law of sines?
A.) a sin B
----------
sin B
B.) a sin A
----------
sin B
C.) b sin B
----------
sin A
D.) b sin A
----------
sin B
2 Answer
• It seems as if D is the answer
The law of sines is usually written as
[sin (A) / a] = [sin (B) / b] which can be manipulated algebraically as
[b * sin (A) / sin (B)] = a
• The equations represent the law of sines correctly would be option D that is [tex][b \times\dfrac{ sin (A) }{ sin (B)}] = a[/tex].
What is the law of sines?
For any triangle ABC, with side measures |BC| = a. |AC| = b. |AB| = c,
we have, by law of sines,
[tex]\dfrac{sin\angle A}{a} = \dfrac{sin\angle B}{b} = \dfrac{sin\angle C}{c}[/tex]
Remember that we took
[tex]\dfrac{\sin(angle)}{\text{length of the side opposite to that angle}}[/tex]
The law of sines is can be written as
[tex]\dfrac{sin\angle A}{a} = \dfrac{sin\angle B}{b} = \dfrac{sin\angle C}{c}[/tex]
which can be manipulated algebraically as
[tex][b \times\dfrac{ sin (A) }{ sin (B)}] = a[/tex]
Learn more about the law of sines here:
https://brainly.com/question/17289163
#SPJ2
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-8,569,619,473,154,604,000 | What is universal set? - Definition from WhatIs.com
Part of the Mathematics glossary:
A universal set is the collection of all objects in a particular context or theory. All other sets in that framework constitute subsets of the universal set, which is denoted as an uppercase italic letter U. The objects themselves are known as elements or members of U.
The precise definition of U depends the context or theory under consideration. For example, we might define U as the set of all living things on planet earth. In that case, the set of all dogs is a subset of U, the set of all fish is another subset of U, and the set of all trees is yet another subset of U. If we define U as the set of all animals on planet earth, then the set of all dogs is a subset of U, the set of all fish is another subset of U, but the set of all trees is not a subset of U.
Some philosophers have attempted to define U as the set of all objects (including all sets, because sets are objects). This notion of U leads to a contradiction, because U, which contains everything, must therefore contain the set of all sets that are not members of themselves. In 1901, the philosopher and logician Bertrand Russell proved that this state of affairs leads to a paradox. Today, mathematicians and philosophers refer to this result as Russell's Paradox.
This was last updated in July 2012
Contributor(s): Stan Gibilisco
Posted by: Margaret Rouse
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-7,529,145,925,312,109,000 | Recurring Decimal
Use your calculator to find which whole number divided by another whole number gives the answer:
1.36363636...
A Mathematics Lesson Starter Of The Day
Share
Topics: Starter | Arithmetic | Calculator | Fractions
• N. Cox, Woodbridge Suffolk
•
• It would have been good if this was one of those starters where I could get a new question by using the refresh button.
[Transum: Thanks for your suggestion, there is now a button lower down this page to 'Change Numbers'. Also if you click on the link lower down the page (Transum.org/go/?to=recurring) for the related student activity you will find more recurring decimals.]
• The Best Maths Class Ever (7cd/M2), King Alfred's College
•
• Many of us found it too hard at the beginning but when we realised what we were doing, we managed to think of lots of answers.
• Miss Groves, Edinburgh
•
• Class 1B3 from Forrester High School Edinburgh were enjoying this very much. We found the answer and then we multiplied up and divided to get other results.
• J. Miley, Kingsbury School, Birmingham
•
• I agree, thank you, though ... it got them thinking!
• Jill, Knowling
•
• A very good starter although once completed you should be able to refresh it.
• Transum,
•
• Thanks for your comments. There is now a button below allowing you to generate a different recurring decimal. There is also a link to a self marking quiz related to this starter below. Did you know that a fraction in lowest terms with a prime denominator other than 2 or 5 always produces a repeating decimal?
• Mr Kennelly, 3rd Class Mayo
•
• My class found it hard to do but then we figured it out. The best one yet I think and the class loved it to :).
• Ms Polius-Curran, Basildon Upper Academy
•
• This starter has to be one of the best differentiated starters I have come across, my Foundation group enjoyed using the calculator to find the two whole numbers whilst my Higher group solved this without the calculator by using the method to convert recurring decimals to fractions.
• Mr Okungbowa, Northbury Junior School
•
• This is very good. I would like more questions like this one because they are really mind boggling and it takes a lot of working out to find correct answers.
• Sapphire Class, Gloucester
•
• We tried this in Y5/6 - the closest we could get is 1000/748 = 1.3368983
We also noticed that 52/22 = 2.363636.
How did you use this starter? Can you suggest how teachers could present or develop this resource? Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for Maths teachers anywhere in the world.
Click here to enter your comments.
If you don't have the time to provide feedback we'd really appreciate it if you could give this page a score! We are constantly improving and adding to these starters so it would be really helpful to know which ones are most useful. Simply click on a button below:
Excellent, I would like to see more like this
Good, achieved the results I required
Satisfactory
Didn't really capture the interest of the students
Not for me! I wouldn't use this type of activity.
This starter has scored a mean of 3.0 out of 5 based on 130 votes.
Previous Day | This starter is for 6 July | Next Day
Answer
Let the number be represented by x
x = 1.36363636... [1]
100x = 136.363636... [2]
Subtract [1] from [2]
99x = 135
x = 135/99
x = 15/11
The two whole numbers could be 15 and 11
What else could they be?
Note to teacher: Doing this activity once with a class helps students develop strategies. It is only when they do this activity a second time that they will have the opportunity to practise those strategies. That is when the learning is consolidated. Click the button above to regenerate another version of this starter from random numbers.
A spreadsheet could be used to quickly find a solution to this kind of problem.
Spreadsheet method
Christmas Present Ideas
It is often very difficult choosing Christmas presents for family and friends but so here are some seasonal, mathematics-related gifts chosen and recommended by Transum Mathematics.
Equate board game
Here's a great board game that will give any family with school-aged kids hours of worthwhile fun. Christmas is a time for board games but this one will still be useful at any time of year. Games can be adapted to suit many levels of Mathematical ability.
For Maths tutors working with just one or small groups of pupils this game has proved to be an excellent activity for a tutorial. Deciding on the best moves can spark pertinent discussions about mathematical concepts.
Equate looks a bit like Scrabble--for aspiring mathematicians, that is. Designed by a real mathematician, it works like this: You put down tiles on a board and make points by correctly completing simple equations. Your nine tiles include both numbers and mathematical symbols; you can add on to previous plays both vertically and horizontally. more...
How Not To Be Wrong
The maths we learn in school can seem like an abstract set of rules, laid down by the ancients and not to be questioned. In fact, Jordan Ellenberg shows us, maths touches on everything we do, and a little mathematical knowledge reveals the hidden structures that lie beneath the world's messy and chaotic surface. In How Not to be Wrong, Ellenberg explores the mathematician's method of analyzing life, from the everyday to the cosmic, showing us which numbers to defend, which ones to ignore, and when to change the equation entirely. Along the way, he explains calculus in a single page, describes Gödel's theorem using only one-syllable words, and reveals how early you actually need to get to the airport.
What more could the inquisitive adult want for Christmas? This book makes a cosy, interesting read in front of the fire on those cold winter evenings. more...
Graphic Display Calculator
This handheld device and companion software are designed to generate opportunities for classroom exploration and to promote greater understanding of core concepts in the mathematics and science classroom. TI-Nspire technology has been developed through sound classroom research which shows that "linked multiple representation are crucial in development of conceptual understanding and it is feasible only through use of a technology such as TI-Nspire, which provides simultaneous, dynamically linked representations of graphs, equations, data, and verbal explanations, such that a change in one representation is immediately reflected in the others.
For the young people in your life it is a great investment. Bought as a Christmas present but useful for many years to come as the young person turns into an A-level candidate then works their way through university. more...
iPad Air
The analytics show that more and more people are accessing Transum Mathematics via an iPad as it is so portable and responsive. The iPad has so many other uses in addition to solving Transum's puzzles and challenges and it would make an excellent Christmas gift for anyone.
You have to hold iPad Air to believe it. It’s just 7.5 millimeters thin and weighs just one pound. The stunning Retina display sits inside thinner bezels, so all you see is your content. And an incredible amount of power lies inside the sleek enclosure. So you can do so much more. With so much less. more...
Before giving an iPad as a Christmas gift you could add a link to iPad Maths to the home screen.
Aristotle's Number Puzzle
It’s a bit of a tradition to give puzzles as Christmas Gifts to nieces and nephews. This puzzle is ideal for the keen puzzle solver who would like a challenge that will continue over the festive period (at least!).
This number puzzle involves nineteen numbers arranged into a hexagon. The goal of the puzzle is to rearrange the numbers so each of the fifteen rows add up to 38. It comes in a wooden style with an antique, aged look.
Keep the Maths in Christmaths with this reasonably priced stocking filler. more...
The Story Of Maths [DVD]
The films in this ambitious series offer clear, accessible explanations of important mathematical ideas but are also packed with engaging anecdotes, fascinating biographical details, and pivotal episodes in the lives of the great mathematicians. Engaging, enlightening and entertaining, the series gives viewers new and often surprising insights into the central importance of mathematics, establishing this discipline to be one of humanity s greatest cultural achievements. This DVD contains all four programmes from the BBC series.
Marcus du Sautoy's wonderful programmes make a perfect Christmas gift more...
Click the images above to see all the details of these gift ideas and to buy them online.
Online Maths Shop
Laptops In Lessons
Teacher, do your students have access to computers?
Do they have iPads or Laptops in Lessons?
Whether your students each have a TabletPC, a Surface or a Mac, this activity lends itself to eLearning (Engaged Learning).
Laptops In Lessons
Here a concise URL for a version of this page without the comments.
Transum.org/go/?Start=July6
Use a spreadsheet to help investigate which pairs of numbers, when divided, give a recurring decimal answer.
Here is the URL which will take them to a student version of this activity.
Transum.org/go/?to=recurring
Apple
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-1,151,229,563,346,761,200 | Integral test examples pdf
The double integrals in the above examples are the easiest types to evaluate because they are examples in which all four limits of integration are constants. This happens when the region of integration is rectangular in shape. In non-rectangular regions of integration the limits are not all constant so we have to get used to dealing with
Exercises and Problems in Calculus John M. Erdman Portland State University Version August 1, 2013 INTEGRATION OF FUNCTIONS OF A SINGLE VARIABLE 87 Chapter 13. THE RIEMANN INTEGRAL89 13.1. Background89 problem15in Chapter29, for example, where the background is either minimal or largely irrelevant to the solution of the problem. ix.
1.1.2. Evaluating Integrals. We will soon study simple and ef-ficient methods to evaluate integrals, but here we will look at how to evaluate integrals directly from the definition. Example: Find the value of the definite integral R1 0 x2 dx from its definition in terms of Riemann sums.
Calculus - Integral Test (examples, solutions, videos) If is convergent then is convergent. If is divergent then is divergent. Example: Test the series for convergence or divergence. Solution: The function is continuous, positive, decreasing function on [1,∞) so we use the Integral Test: Since is a convergent integral and so, by the Integral test, the series is convergent. Integral test (practice) | Khan Academy Use the integral test to determine whether a given series is convergent or divergent. If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the … Integral Test
Calculus II - Integral Test - Lamar University May 31, 2018 · The original test statement was for a series that started at a general n = k and while the proof can be done for that it will be easier if we assume that the series starts at n = 1. Another way of dealing with the n = k is we could do an index shift and start the series at n = 1 and then do the Integral Test. Integral Calculus - Exercises INTEGRAL CALCULUS - EXERCISES 45 6.2 Integration by Substitution In problems 1 through 8, find the indicated integral. 1. R (2x+6)5dx Solution. Substituting u =2x+6and 1 2 du = dx,youget Z (2x+6)5dx = 1 2 Z u5du = 1 12 u6 +C = 1 12 (2x+6)6 +C. 2. R [(x−1)5 +3(x−1)2 +5]dx Solution. Substituting u = x−1 and du = dx,youget Z £ (x−1)5 +3(x−1) 2+5 ¤ dx = Z (u5 +3u +5)du = = 1 6 The Integral Test - Math24
infinite series, part ii: divergence and integral tests 12. EXAMPLE 14.2.8. Determine whether the series. •. Â k=1 cos 1 n converges. Solution. Notice that lim n!•. The Integral test. Theorem. A series ∑an composed of nonnegative terms converges if and only if the sequence of partial sums is bounded above. these types will help you decide which tests or strategies will be most useful in as in the example to the left. If you integral, the integral test may prove useful:. Example 2.1. Does the harmonic series ∑. ∞ n=1. 1 n converge? Let us use the integral test. Note that. ∫ ∞. 1. 1 x dx is a divergent integral from the p-test for commonly used convergence tests for positive series (the Integral Test, diverges: if the terms grow very large and positive, for example, it is natural to want to N[EulerGamma,20] 0.57721566490153286061. Example 2.2.5 Euler- Mascheroni Constant. Consider a sequence of partial sums
May 31, 2018 · The original test statement was for a series that started at a general n = k and while the proof can be done for that it will be easier if we assume that the series starts at n = 1. Another way of dealing with the n = k is we could do an index shift and start the series at n = 1 and then do the Integral Test.
As a general rule, the integral test tends to be quite useful for series in the vicinity of this barrier. EXERCISES 1–4 çEvaluate the following improper integrals. 1. 2.( (" È ! _ _ B" "B B.B .B / 3. 4.( (! ! È _ _ #" "" B.B .B " B 5. Find a general formula for , assuming .(" _: " B.B : " 6–9 ç Use Integral test to determine whether the given series Lecture 25 : Integral Test Integral TestIntegral Test ExampleIntegral Test Examplep-seriesComparison TestExample 1Example 2Example 3Example 4Example 5Example 6Limit Comparison TestExampleExampleExampleExampleExampleExample Example 4 Section 11.3: The Integral Test because if we tried to integrate from 0 to ∞, the integral will have diverged. Of course, the key point is that the first few terms will not affect divergence or convergence - it is the ultimate behavior which counts and this is measured by the integral. We illustrate the power of the integral test with a few examples. Example 2.2. 11.3: The Integral Test The Integral Test 1. For a positive decreasing (or eventually decreasing) sequence a n and corresponding function f, the series P 1 n=1 a n converges if and only if R 1 f(x)dxconverges. 2. R n 1 f(x)dx P i=1 a n a 1 + R n 1 f(x)dx. 3. If s= P a n and s n is the nth partial sum, then Z 1 n+1 f(x)dx R n = s s n Z n f(x)dx. Example: Since a n = 1
13.3 The Integral Test
Calculus II - Integral Test - Lamar University
Note appearance of original integral on right side of equation. Move to left side and solve for integral as follows: 2∫ex cosx dx = ex cosx + ex sin x + C ∫ex x dx = (ex cosx + ex sin x) + C 2 1 cos Answer Note: After each application of integration by parts, watch for the appearance of a constant multiple of the original integral. | {
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Lamp and Shadow
A lamp is placed on the ground 100 feet away from a wall. A man six feet tall is walking at a speed of 10 ft/sec from the lamp to the nearest point on the wall. When he is midway between the lamp and the wall, the rate of change in the length of his shadow is (in ft/ sec)?
Discussion:
773Let the length of the shadow at any point of time me y and distance of the man from Lamp be x.
By similarity of triangles we can say \(\frac {y}{100} = \frac {6}{x} \).
Differentiating both sides with respect to time t we get \(\frac {1}{100} \frac {dy}{dt} = 6 \cdot \frac {-1}{x^2} \frac {dx}{dt} \). Replace dx/dt by 10, x by 50 we get dy/dt = -2.4 ft/sec where the negative sign implies reduction in the length of the shadow.
October 27, 2013
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2,834,223,074,599,555,000 | Thursday
February 11, 2016
Homework Help: statistics
Posted by Cameron on Monday, June 3, 2013 at 8:00am.
A specific study found that the average number of doctor visits per year for people over 55 is 8 with a standard deviation of 2. Assume that the variable is normally distributed.
1. Identify the population mean.
The population mean is 8.
2. Identify the population standard deviation.
The population standard deviation is 2.
3. Suppose a random sample of 15 people over 55 is selected. What is the probability that the sample mean is above 9?
4. Suppose a random sample of 100 people over 55 is selected. What is the probability that the sample mean will be below 7?
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-6,594,906,165,851,983,000 | Riddle #669
Question: Five pirates are parting ways after finding a treasure of 100 pieces of gold. The pirates decide to split it based on a vote. Each pirate, from oldest to youngest, gets to propose a plan on how to split the gold.
If at least 50 percent of the other remaining pirates agree on the plan, that is how they will split the gold. If less than 50 percent of the pirates agree, the pirate who came up with the plan will be thrown overboard. Each pirate is smart, greedy, and wants to throw as many others overboard as possible without reducing the amount of gold they get.
What plan can the first (oldest) pirate propose to live and get as much gold as possible?
Riddle Discussion
By: mikage on 29/6/16
I made an error in my answer. Pirate 5 does not need three votes to win, he only needs two. He can give one gold to pirate 3, and two to either 4 or 5, leaving him 97.
By: mikage on 29/6/16
As explained in previous comments, there are some flaws in the answer.\n\nThe starting point is correct: If there are 2 pirates, the youngest one will deny the oldest pirate's plan and get all the gold. What you forgot to mention is that the oldest pirate will get killed as a result.\n\nIf there are 3 pirates, the oldest pirate is certain his plan will pass, WHATEVER that plan is, because the second pirate will vote for it. If he did not vote for it, he would become the oldest pirate left and be in the situation described above (where he dies). So in a 3 pirate scenario, the oldest pirate takes all the gold.\n\nIf there are 4 pirates, the second pirate will never vote for the plan because he is assured to have all the gold when it is his turn. Even if the first pirate offers to give the second pirate all the gold, the second pirate would prefer killing the oldest AND taking all the gold, so he will refuse the plan. The only way for the oldest pirate to win is to get the votes of pirates 3 and 4. Since they get nothing in the 3-pirate case, he only has to offer 1 gold to them. He therefore gets 98 gold (and pirate 2 gets nothing).\n\nWith 5 pirates, the oldest pirate knows that if his plan is rejected, the second pirate will propose the plan described above, giving the two youngest pirates 1 gold each, and nothing to pirate 3. To win, he needs to offer two gold to pirates 4 and 5, and one gold to pirate 3. He will get 95 gold.
By: prisonmatch on 30/9/15
I think for the solution to be true the statement 'If at least 50 percent of the other REMAINING pirates agree on the plan' should be 'If at least 50 percent of the pirates agree on the plan' else the 4 pirate condition wont work.\nIn the current condition \nIf there are 4 pirates the best plan by oldest is to keep 97 and give 0 to 2nd, 2 to 3rd and 1 coin to 4th. Because in the 3 pirates situation youngest will get nothing and 2nd pirate will get just 1 instead of 2 coins( in 4 pirates condition).\n\nIn the solution given 5th pirate(youngest) will definitely refuse the offer as he is greedy and hope for a better plan by 2nd oldest pirate because in that case too he is at least getting 1 coin and gets to throw 1 overboard. as said by funkychicken. so answer is 97(oldest) 1(3rd pirate) 2 (5th pirate) ,
By: prisonmatch on 30/9/15
I think for the solution to be true the statement 'If at least 50 percent of the other REMAINING pirates agree on the plan' should be 'If at least 50 percent of the pirates agree on the plan' else the 4 pirate condition wont work.\nIn the current condition \nIf there are 4 pirates the best plan by oldest is to keep 97 and give 0 to 2nd, 2 to 3rd and 1 coin to 4th. Because in the 3 pirates situation youngest will get nothing and 2nd pirate will get just 1 instead of 2 coins( in 4 pirates condition).\n\nIn the solution given 5th pirate(youngest) will definitely refuse the offer as he is greedy and hope for a better plan by 2nd oldest pirate because in that case too he is at least getting 1 coin and gets to throw 1 overboard. as said by funkychicken. so answer is 97(oldest) 1(3rd pirate) 2 (5th pirate) ,
By: prisonmatch on 30/9/15
I think for the solution to be true the statement 'If at least 50 percent of the other REMAINING pirates agree on the plan' should be 'If at least 50 percent of the pirates agree on the plan' else the 4 pirate condition wont work.\nIn the current condition \nIf there are 4 pirates the best plan by oldest is to keep 97 and give 0 to 2nd, 2 to 3rd and 1 coin to 4th. Because in the 3 pirates situation youngest will get nothing and 2nd pirate will get just 1 instead of 2 coins( in 4 pirates condition).\n\nIn the solution given 5th pirate(youngest) will definitely refuse the offer as he is greedy and hope for a better plan by 2nd oldest pirate because in that case too he is at least getting 1 coin and gets to throw 1 overboard. as said by funkychicken. so answer is 97(oldest) 1(3rd pirate) 2 (5th pirate) ,
By: MikeHolmes on 18/9/14
4 pirate have to vote, 2 vote is 50percent .\nSo da plan would be..\n''Count to 4 and Eliminate the 4th pirate (count starts from himself) and repeat the process ''.\n THE OLDEST PIRATE Gets all the gold.\nHere, the 2nd,3rd and 5th pirate will vote (because they won't be eliminated ,not yet that is). \nSo what do you think of my answer?
By: McJagger on 26/7/14
One pirate kills everyone. Viola probably solved
By: Twhit on 25/7/14
the three oldest pirates split it.
By: Funkychicken on 8/7/14
There's actually an inconsistency in your logic here. the highest amount that he can get for himself is 97 pieces of gold. The flaw comes in your proposed plan for 4 pirates. Because it has to be voted on by the "remaining" pirates that plan would only be approved of by 1/3rd of the pirates, instead he has to offer 2 pieces of gold to the second youngest pirate to outbid the previous offer. this means that for the oldest pirate to get the most gold he has to offer the 3rd pirate 1 gold piece, and the youngest pirate 2 gold pieces to outbid the offer that would come otherwise.
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5,203,540,654,165,921,000 | How to find the middle element of a square array
조회 수: 221(최근 30일)
Kevin Carty
Kevin Carty 2020년 3월 12일
답변: Anthony Olivares 2022년 1월 21일
function middleElement = FindMiddle(squareArray)
% FindMiddle: Return the element in the center of squareArray
% Inputs: squareArray - n x n input array, where n is odd
%
% Outputs: selectedData - center element of squareArray
% Assign elementIndex with location of middle row/col
% Hint: Use the size() function to deterimine the dimension of squareArray
elementIndex = size(squareArray)
% Assign middleElement with the center element of squareArray
middleElement = squareArray(elementIndex)
end
How do I start this? I figure I would have to set the elementIndex to an integer but I don't know how to do that. Any guidance would be appreciated.
답변(4개)
Bhaskar R
Bhaskar R 2020년 3월 12일
편집: Bhaskar R 2020년 3월 12일
function middleElement = FindMiddle(squareArray)
% FindMiddle: Return the element in the center of squareArray
% Inputs: squareArray - n x n input array, where n is odd
%
% Outputs: selectedData - center element of squareArray
% Assign elementIndex with location of middle row/col
% Hint: Use the size() function to deterimine the dimension of squareArray
center_ind = (size(squareArray)+1)/2;
% Assign middleElement with the center element of squareArray
middleElement = squareArray(center_ind(1), center_ind(2))
end
댓글 수: 2
Walter Roberson
Walter Roberson 2020년 3월 12일
N is odd. If you divide it by 2 you will get a leftover 1/2, but you cannot use 1/2 as an index.
Example: n=7. 7/2=3.5 but you cannot use 3.5 as an index. What is the proper index?
1234567
!
The ! has the same number of elements before and after so the index of the center is 4.
Now take (7+1)/2 and you get 4 which is the correct index. This can also be thought of as 7/2 + 1/2 = 3.5+.5 = 4: just group the operation differently, 7/2+1/2 times 2 is 7 + 1, so divide by the 2 is (7+1)/2
댓글을 달려면 로그인하십시오.
Jada Roth
Jada Roth 2020년 9월 10일
function middleElement = FindMiddle(squareArray)
% FindMiddle: Return the element in the center of squareArray
% Inputs: squareArray - n x n input array, where n is odd
%
% Outputs: selectedData - center element of squareArray
% Assign elementIndex with location of middle row/col
% Hint: Use the size() function to deterimine the dimension of squareArray
elementIndex = ceil(size(squareArray)/2);
% Assign middleElement with the center element of squareArray
middleElement = squareArray(elementIndex(1),elementIndex(2));
end
Piyush Lakhani
Piyush Lakhani 2020년 3월 12일
Hi Kevin,
The 'size' function returns the two element array 'row and column'. As what i can understand you want a single value that determines the variable 'n'.
so may be 'length' function gives the output you wants. Try in following way.
function middleElement = FindMiddle(squareArray)
% FindMiddle: Return the element in the center of squareArray
% Inputs: squareArray - n x n input array, where n is odd
%
% Outputs: selectedData - center element of squareArray
% Assign elementIndex with location of middle row/col
% Hint: Use the size() function to deterimine the dimension of squareArray
elementIndex = length(squareArray)
% Assign middleElement with the center element of squareArray
middleElement = squareArray(elementIndex)
end
Anthony Olivares
Anthony Olivares 2022년 1월 21일
Here is what I did:
function middleElement = FindMiddle(squareArray)
% FindMiddle: Return the element in the center of squareArray
% Inputs: squareArray - n x n input array, where n is odd
%
% Outputs: selectedData - center element of squareArray
% Assign elementIndex with location of middle row/col
% Hint: Use the size() function to deterimine the dimension of squareArray
% Here I used the size() function to obtain a row vector with the
% dimensions of squareArray. Therefore, this would return [3, 3] for a
% 3x3 array.
elementIndex = (size(squareArray) + 1) / 2;
% Assign middleElement with the center element of squareArray
% Here, we simply plug in the values of the row vector obtained with
% the size() function to get the middle element in an odd 2D array.
middleElement = squareArray(elementIndex(1), elementIndex(2));
end
If we were to run FindMiddle([1, 2, 3; 4, 5, 6; 7, 8, 9]) :
• elementIndex would return [3, 3], since it is a 3x3 array.
• Therefore, squareArray(elementIndex(1), elementIndex(2)); would return a 5, since squareArray(3, 3) is 5.
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A certain office supply store stocks 2 sizes of self-stick
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A certain office supply store stocks 2 sizes of self-stick [#permalink]
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New post 04 Apr 2006, 02:43
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: blue, green, yellow, or pink. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color or 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible?
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New post 04 Apr 2006, 02:54
The Total types wud be
Small Size: - each carrying the same color (blue, green, yellow, or pink) = 4
Large Size: - Each carrying the same color (blue, green, yellow, or pink) = 4
small Size different colors = 4C3 (as order is not important) = 4
Large Size different colors = 4C3 (as order is not important) = 4
Total = 4 + 4 + 4 +4 =16
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Re: Combination [#permalink]
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New post 04 Apr 2006, 03:14
jodeci wrote:
A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: blue, green, yellow, or pink. The store packs the notepads in packages that contain either 3 notepads of the same size and the same color or 3 notepads of the same size and of 3 different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible?
Thanks
1. Notepad1
same colour: 4
different colours: we can choose 3 out of 4 in 4c3=4 ways.
2. Same with notepad2
So, 4*4=16
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New post 04 Apr 2006, 06:47
assume the two sizes are 1 and 2
So we have: B1,G1,Y1,P1 and B2,G2,Y2,P2
Same COlor, Same Size:
8 possibilites --> B1B1B1, B2B2B2, G1G1G1, G2G2G2.... etc
Same size, different color:
4C3 * 2 = 8 possibilites.
Total = 16
[#permalink] 04 Apr 2006, 06:47
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The number of zeros that come after 1 for 10 crores is
A. 6
B. 7
C. 8
D. 9
seo-qna
Last updated date: 20th Jun 2024
Total views: 413.4k
Views today: 12.13k
Answer
VerifiedVerified
413.4k+ views
Hint: We can simply write the amount in figures. We can put zeros by counting from the unit place, tens place, hundredth place, thousandth place, ten-thousandth place, lakh place, ten lakh place, crore place, and in the 10 crores’ place, we must write 1. Then we can count the number of zeros that come after 1.
Complete step by step Answer:
We are asked to find the number of zeros in 10 crores. We can put zeros by counting from the unit place, tens place, hundredth place, thousandth place, ten-thousandth place, lakh place, ten lakh place, crore place and we must write 1 in the 10 crores’ place. We can write 10 crores in figures as follows,
${\text{10,00,00,000}}$
From the above digits, we can easily count the number of zeros after 1. By counting, we will get the number of zeros to be 8.
So, there are 8 zeros that come after 1 in 10 crores.
Therefore, the correct answer is option C.
Note: We must use commas to separate the digits while writing large figures. This helps us for writing the digits and also to count the digits. Another method for doing this problem is to convert 10 crores into exponential form to the base 10 and the power of 10 will be the number of zeros. We can also count the number of zeros that must be added to 1 to get 10 crores. The answer will be the same method we do. | {
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Homework Help: Input/output signal
1. Sep 10, 2005 #1
Not even sure how to title this thread. I am saying honestly: we did not cover this in class but he gave us this in homework!!!
Two identical linear systems are connected in cascade. When the input signal to the first stage is x(t) the output signal from the first stage is y(t), the output of the second stage is z(t). If an input signal, tu(t) is applied to a single stage, the output signal of that stage is : [tex] (e^{-2t}-1)u(t)[/tex] find z(t) if x(t) = u(t).
u(t) is a step function, u(t) = 1 if t >=0, and 0 otherwise;
what i was thinking of doing is converting the e...-expression to frequency domain and also the tu(t) and then seeing how input is modified to get the output and apply the same thing to u(t) which is the input into cascaded system, but TA said that there's multiplication involved in s-domain. I am sorta clueless...
If someone could explain or direct to a source, that would be very much appreciated, 'coz my book does not these examples and i don't even know what to seach for on Google! :cry:
maybe i am not making a connection to something...the latest we covered was circuit analysis in frequency domain doing nodal analysis.
Last edited: Sep 10, 2005
2. jcsd
3. Sep 10, 2005 #2
Astronuc
User Avatar
Staff Emeritus
Science Advisor
4. Sep 10, 2005 #3
sorry, should have given more detail: i understand laplace transform.
What i don't understand is how to make use of the given info that input is one function and output signal is a different function and what to do in s-domain (laplace "world") to apply that to u(t) which is the input signal to the 2-stage linear system.
5. Sep 11, 2005 #4
EDIT: deleted my previous garbage...
EDIT: hold on...have another idea...would this work?:
[tex]H(s) = \frac{L(output)}{L(input)}[/tex]
then when i find this factor i will have to multiply my given L(input) by it in s-domain and take L-1 to get the time-domain signal...
would that work?
Last edited: Sep 11, 2005
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-1,215,885,021,574,988,800 | Tuesday
March 31, 2015
Homework Help: Calculus
Posted by Ally on Monday, October 22, 2012 at 1:01pm.
Im finding local Min and max of x^2/x-2
I know I take the derivative of the first one but how do i derive this one. And the second one especially don't know how to do.
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-8,582,462,272,312,351,000 | "ПРОГНОЗ" и "ПРОГНОЗ". Функции ЛИНЕЙНАЯ
В этой статье описаны синтаксис формулы и использование прогноза. ФункцииLINEAR и FORECAST в Microsoft Excel.
Примечание: В Excel 2016 функция ПРОГНОЗ была заменена функцией ПРОГНОЗ. LINEAR в составе новых функций прогнозирования. Синтаксис и использование этих двух функций будут одинаковыми, но старая функция ПРЕДСПРОС в конечном итоге будет отознана. Он по-прежнему доступен для обратной совместимости, но мы думайте об использовании новой функции ПРОГНОЗ. Вместо этого функция ЛИНЕЙЛ.
Описание
Вычислять и прогнозировать будущее значение с помощью существующих значений. Будущая стоимость — это значение y для заданного значения x. Существующие значения — это известные значения x и y, а будущее значение предсказывается с помощью линейной регрессии. Эти функции можно использовать для прогнозирования будущих продаж, требований к запасам или потребительских тенденций.
Синтаксис
ПРЕДСКАЗ.ЛИНЕЙН(x;известные_значения_y;известные_значения_x)
- или -
ПРЕДСКАЗ(известные_значения_y;известные_значения_x)
ПРОГНОЗ или ПРОГНОЗ. Аргументы функции ЛИННЕЯ следующую:
Аргумент
Обязательный
"Ссылки"
X
Да
Точка данных, для которой предсказывается значение.
Известные_значения_y.
Да
Зависимый массив или интервал данных.
Известные_значения_x.
Да
Независимый массив или интервал данных.
Замечания
• Если x не является числом, FORECAST и FORECAST. ЛИННА возвращает #VALUE! значение ошибки #ЗНАЧ!.
• Если known_y или known_x пустые или одна из них имеет больше точек данных, чем другие, ТО ЕСТЬ ПРОГНОЗ и ПРОГНОЗ. ЛИНЛИН возвращает значение #N/Д.
• Если дисперсия для known_x равна нулю, то ПРОГНОЗ и ПРОГНОЗ. ЛИННА возвращает #DIV/0! значение ошибки #ЗНАЧ!.
• Уравнение для ПРОГНОЗ и ПРОГНОЗ. ЛиНЕЙНАЯ — это a+bx, где:
Уравнение
и
Уравнение
где x и y — средние значения выборок СРЗНАЧ(известные_значения_x) и СРЗНАЧ(известные_значения_y).
Пример
Скопируйте образец данных из следующей таблицы и вставьте их в ячейку A1 нового листа Excel. Чтобы отобразить результаты формул, выделите их и нажмите клавишу F2, а затем — клавишу ВВОД. При необходимости измените ширину столбцов, чтобы видеть все данные.
Известные значения y
Известные значения x
6
20
7
28
9
31
15
38
21
40
Формула
Описание
Результат
=ПРОГНОЗ. ЛИНЛИН(30;A2:A6;B2:B6)
Предсказывает значение y, соответствующее заданному значению x = 30
10,607253
См. также
Функции прогнозирования (справочник)
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cieracabrera Group Title
-4x-6y=12 6x+12y=-18
• 11 months ago
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solve for x ???
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6(x - 2y + 3z)
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\[(-4x-6y)+6y=2+6y\]
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d|dw:1386135666024:dw||dw:1386135721010:dw|
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then plug x in for y
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\[-4x-6y=6y=6y=2 so x=-\frac{ 3y|1 }{ }\]
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\[x=- \frac{ 3y|1 }{ 2 }\]
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thanks i guess
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-8,707,282,199,857,002,000 | login
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A123711 Indices n such that A123709(n) = 8 = number of nonzero terms in row n of triangle A123706. 6
12, 18, 20, 24, 28, 36, 40, 44, 45, 48, 50, 52, 54, 56, 63, 68, 72, 75, 76, 80, 88, 92, 96, 98, 99, 100, 104, 108, 112, 116, 117, 124, 135, 136, 144, 147, 148, 152, 153, 160, 162, 164, 171, 172, 175, 176, 184, 188, 189, 192, 196, 200, 207, 208, 212, 216, 224, 225 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Triangle A123706 is the matrix inverse of triangle A010766, where A010766(n,k) = [n/k].
It appears that this equals A200511, numbers of the form p^k q^m with k,m >= 1, k+m > 2 and p, q prime. - M. F. Hasler, Feb 12 2012
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
MATHEMATICA
Moebius[i_, j_] := If[Divisible[i, j], MoebiusMu[i/j], 0]; A123709[n_] :=
Length[Select[Table[Moebius[n, j] - Moebius[n, j + 1], {j, 1, n}], # != 0 &]]; Select[Range[500], A123709[#] == 8 &] (* G. C. Greubel, Apr 22 2017 *)
PROG
(PARI)
CROSSREFS
Cf. A123706, A123709, A123710, A123712; A010766.
Sequence in context: A036456 A102467 A126706 * A200511 A317711 A323055
Adjacent sequences: A123708 A123709 A123710 * A123712 A123713 A123714
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 09 2006
STATUS
approved
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2,802,385,916,156,476,000 | RSA加密(二) 加密过程
上篇文章我们介绍了密码学的基础，相信你已经对密码学及相关的知识有所了解，如果还没有阅读过，可以点击这里查看👉 RSA加密(一) 密码学基础
这篇文章主要给大家讲解下RSA加密的过程，但在讲加密过程前我们需要了解几个数学公式。都很简单，相信大家看了就会明白，文中遇到有链接的地方，可先跳过不影响阅读。下面我们先来看看RSA加密的介绍。
RSA加密简介
本文开篇的这张图就是RSA加密的三位作者，RSA算法是由Rivest、Shamir和Adleman在1977年共同提出，他们当时正好都在麻省理工学院工作，RSA就是他们三人姓氏开头字母拼在一起组成的。
当年的照片，留心下他们的发际线对比😅
但其实在RSA算法提出的4年前的1973年，在英国政府通讯总部工作的数学家克利福德·柯克斯（Clifford Cocks）在一个内部文件中便提出了一个与之等效的算法，但该算法当时被列入机密，直到1997年才得到公开（稍微有一些些可惜了~）。
从上篇文章我们了解到，RSA加密算法是通过分解质因数的困难性来实现的。换言之，对一个极大整数做因数分解愈困难，RSA算法愈可靠。假如有人找到一种快速因数分解的算法的话，那么用RSA加密的信息的可靠性就会极度下降。但找到这样的算法的可能性是非常小的。今天只有短的RSA钥匙才可能被强力方式破解。到当前为止，世界上还没有任何可靠的攻击RSA算法的方式。只要其钥匙的长度足够长，用RSA加密的信息实际上是不能被破解的，正因为RSA加密的安全性，所以RSA加密算法在当今互联网信息传递中被广泛的使用。
可以毫不夸张的说：只要有计算机网络的地方就有RSA加密算法。
数学公式
前面说到要理解RSA加密需要先了解一些数论相关的知识，这里我总结了几个必要的公式，方便后面能够更好的理解加密过程。
1.互质关系
1. 质数定义：除了1和它本身，不能被其它数整除的数。例如：2、3、5、7、11、13、17、19、23、29…
2. 任意两个质数构成互质关系。例如5和7，除1之外没有其他数能够整除5和7。
3. 质数A与不是它的倍数的数构成互质。例如5能和1、2、3、4、6、7、8、9、11构成互质关系。
2.欧拉函数
对于任意正整数n，欧拉函数就是计算比n小的数中与n互质的数有多少个，用φ(n)表示，例：
计算整数8的欧拉函数：与8形成互质关系的是1、3、5、7共四个数，所以 φ(8) = 4。
从上面互质关系中的第三条我们可以知道，当n是一个质数时，比n小的所有数都与n形成互质关系，所以有：
n是质数：φ(n) = n-1
根据中国剩余定理(又称孙子定理)我们可以得到：两个互质整数p、q乘积的欧拉函数为：
φ(pxq) = φ(p)φ(q) （即可分开计算再相乘）
所以当p、q都是质数的时候，根据上面的公式 φ(n)=n-1 我们可以得到：
φ(pxq)=(p-1)(q-1)
我们知道上面这种情况的欧拉公式就够了，关于更多欧拉函数的相关知识我们可以查看👉这里
3.欧拉定理
介绍欧拉定理之前，先说下取余操作。大家都知道 5除4余1，在数学中表示为 5=1(mod4) ，因为在计算机中计算余数的符号为百分号%，所以下面公式统一用 5%4=1 这种既简洁又好理解的形式表示。
欧拉定理：若正整数a和n互质则有
$a^{φ(n)}$ % n=1
正如上面的式子，伟大的定理往往都相当简洁！关于欧拉定理如何证明，感兴趣的同学可以查看👉这里
另外与之相关的一个概念，若正整数a和n互质则存在整数b使得：
ab % n=1
b就叫做a的模反元素，那b就一定存在吗？欧拉定理我们可以写成:
a x $a^{φ(n)-1}$ % n=1
所以b=$a^{φ(n)-1}$时上式成立，故b必然存在。
对于后面的加解密过程，我们主要用到如下两个公式：
RSA加密、解密过程
生成公钥、私钥
1. 随机找出两个不相等的质数 p、q
这里方便起见我们取 p=5，q=11（实际应用中p和q越大越难破解）
2. 计算p、q的乘积 n = p x q
n = 5 x 11 = 55（n的二进制长度称为RSA加密的密钥长度，这里55表示成二进制是110111，长度只有6位，实际应用中为1024位或更高的2048位）
3. 计算n的欧拉函数φ(n)，根据上面我们所推导的欧拉函数，得到：φ(n)=(p-1)(q-1)
φ(n) = 4 x 10 = 40
4. 随机选择一个整数e，条件是1< e < φ(n)，且e与φ(n) 互质
e和40要互质，为了方便这里我们取 e = 3.（ 实际应用中这里一般取65537 ）
5. 计算e对于φ(n)的模反元素d
即 ed % φ(n)=1可以写成：3d = k40+1，当k=2时，我们得到d=27.
6. 将e和n封装作为公钥(e,n)，d和n封装作为私钥(d,n)，即：
公钥：（e=3，n=55）
私钥：（d=27，n=55）
加密
上面我们已经计算出了公钥和私钥，根据上篇文章中的非对称加密流程。发送发收到公钥后开始进行加密操作。
假设现在我们要发送一段消息m=4，公钥为 (e=3,n=55)
m必须是一个整数（可将发送的消息通过ASCII码转换成十进制），且m<n，这就导致RSA加密的内容长度受到了限制。
计算密文：c = $m^e$ % n
即密文c = $ 4^3$ % 55 = 64%55 = 9
发送方便将密文9，通过网络等途径发送给接收方。
解密
接收方收到密文c=9后，使用自己创建的私钥 (d=27,n=55) 进行解密操作。
计算原文：m = $c^d$ % n
即原文m = $9^{27}$ % 55 = 4 （简单推导见👉这里，也可通过编程验证）
于是接收方便得到了发送方想要发送的消息，整个加密和解密过程就结束了。
总结
可以看到RSA加密和解密的过程并不复杂，用到的公式也只有仅仅两个，但这里面还有一些问题等待着我们去探索，例如：
• m<n，这就导致RSA加密的内容长度受到了限制，那如何加密发送一段很长的文本呢？
• RSA在公钥和密文传输的过程中就真的很安全吗？
• 为什么解密的时候原文：m 就等于 $c^d$ % n呢？
欢迎在留言区说出你的想法，我们将在下一篇文章中给大家带来解答。 | {
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660,024,425,888,909,800 | « Common Core Math--Ramping up Discourse with Sketchpad | Main | Photos + Phones = Math HW Shifting to Ubiquitous Use »
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Debi
What is the grade level and CCSS mathematics domain and standard/s for this lesson? Thank you for posting it!
Marsha
I use this with my 8th graders.
CCSS.Math.Content.HSG-GPE.B.5 Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems (e.g., find the equation of a line parallel or perpendicular to a given line that passes through a given point).
It's a starting step towards being able to do this.
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4,341,114,256,022,284,000 | Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » Multiply fractions
Multiply 45/3 with 45/54
1st number: 15 0/3, 2nd number: 45/54
This multiplication involving fractions can also be rephrased as "What is 45/3 of 45/54?"
45/3 × 45/54 is 25/2.
Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 45/3 × 45/54 = 45 × 45/3 × 54 = 2025/162
3. After reducing the fraction, the answer is 25/2
4. In mixed form: 121/2
MathStep (Works offline)
Download our mobile app and learn to work with fractions in your own time:
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Related:
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-7,849,094,951,971,586,000 | Sommatie
Uit Wikipedia, de vrije encyclopedie
Ga naar: navigatie, zoeken
Sommatie is het optellen van een groep getallen, het resultaat hiervan is de som of het totaal. Een oneindige som wordt vaak gezien als een reeks.
Notatie[bewerken]
In de wiskunde is er een compacte en effectieve manier om een sommatie aan te geven. Dit wordt in de wiskunde gedaan met Σ, de hoofdletter sigma uit het Griekse alfabet, vaak in groter formaat, zoals \sum.
Syntaxis[bewerken]
Een sommatie ziet er zo uit:
\sum^n_{i=m} x_i = x_m + x_{m+1} + x_{m+2} + \dots + x_{n-2} + x_{n-1} + x_n
Het getal i - een zogeheten dummyvariabele - duidt de index aan. Het getal m is de ondergrens van de sommatie, en het getal n de bovengrens van de sommatie. De vermelding van i=m geeft aan dat de sommatie begint met de term met als index de waarde m. Elke volgende term heeft een index 1 hoger dan de vorige, en de sommatie eindigt met de term met index n.
Zo is met a_i=i^2:
\sum^4_{i=2} a_i =\sum^4_{i=2} i^2 = 2^2 + 3^2 + 4^2 = 29
Dat de index een dummyvariabele is, houdt in dat voor de index ook een andere letter, bijvoorbeeld de letter k, gebruikt kan worden:
\sum^n_{i=m} x_i =\sum^n_{k=m} x_k
Met k als index wordt de bovengenoemde sommatie:
\sum^4_{k=2} a_k =\sum^4_{k=2} k^2 = 2^2 + 3^2 + 4^2 = 29
Of met de letter n als index en als algemene term a_n=2^n:
\sum^{12}_{n=6} 2^n = 2^6 + 2^7 + 2^8 + 2^9 + 2^{10} + 2^{11} + 2^{12} = 8128
Niet iedere term hoeft de index te bevatten. Bijvoorbeeld geldt dat:
\sum^{8}_{n=4} 1 = 1 + 1 + 1 + 1 + 1 = 5
Einstein-sommatieconventie[bewerken]
Volgens de einstein-sommatieconventie, veel gebruikt in theoretische fysica en met name voor algemene relativiteitstheorie worden sommatietekens weggelaten en wordt er automatisch over een index gesommeerd als deze in een uitdrukking zowel covariant als contravariant voorkomt.
In programmeertalen[bewerken]
De sommatie kan ook worden gebruikt in programmeertalen, een paar voorbeelden:
In Python:
sum(x[m:n+1])
En dit in Fortran (of Matlab):
sum(x(m:n))
In C/C++/C#/Java kan men deze code gebruiken, mits n, m en x zijn int-types. Dat x een array is. En dat m <= n:
int i;
int som = 0;
for (i = m; i <= n; i++)
som += x[i];
Eindige sommen[bewerken]
Voor diverse eindige sommen is de uitkomst samen te vatten in een eenvoudige formule. Bijvoorbeeld
\sum_{ i \mathop =1}^ni = \frac{n(n+1)}{2}
En
\sum^n_{i=1} i^2 = \frac n6 (2n+1)(n+1)
En
\sum^n_{i=1} i^3 = \frac {n^2(n+1)^2}4
Een vaak gebruikte formule is
\sum^n_{k=1} a^k = \frac {a^{(n+1)}-a}{a-1} voor a\neq 1
Oneindige sommaties[bewerken]
Een oneindige sommatie wordt ook wel een reeks genoemd. Het eventuele resultaat is de limiet van de partiële sommen.
Uit de bovenstaande formule voor \sum^n_{k=1} a^kvolgt als we geen n maar oneindig veel termen sommeren de formule voor de meetkundige reeks
\sum^\infty_{k=1} a^k = \frac {a}{1-a} voor |a| < 1
Voorbeeld van een reeks met π in de uitkomst[bewerken]
Een formule van Leonhard Euler luidt:
\frac{\pi^2}{6} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots = \sum^\infty_{n=1} \frac{1}{n^2}
Andere voorbeelden staan bij het lemma reeks. | {
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458,389,513,460,443,140 | Take the 2-minute tour ×
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What is a good way to test the first row of a Matrix to check if one of the values equal a value. For instance, In this example, I want to get vector in the second row that has the bigger value in the first row, 10.
I can quickly do it with a lop, but don't think it's the best way...
enter image description here
share|improve this question
Perhaps MemberQ[m[[1]],n] ? – David Carraher Oct 6 '12 at 20:56
What does bigger value mean in I want to get vector in the second row that has the bigger value ? – David Carraher Oct 6 '12 at 20:59
I meant that in the example, I want to know that the bigger value is 10, and therefor the vector I'm interested in is {1,1,1}. In another example of a 3x3, the values at the first row may be 1, 4, 3 and I need to detect that the bigger value is now in position 2 (number 4), and will use that to get the vector in the second row, below the number 4. – whynot Oct 6 '12 at 21:28
5 Answers 5
up vote 2 down vote accepted
Data
m = {{10, -2, 1}, {{1, 2, 3}, {4, 5, 6, 7}, {8, 9, 10}}}
Code
This gives the vector in row 2 that is opposite (i.e. under) the largest value (or first of multiple largest values) in row 1:
m[[2, Position[m[[1]], Max[m[[1]]]][[1, 1]]]]
{1, 2, 3}
Analysis
Max[m[[1]]] gives the largest value in row 1: 10
Position[m[[1]], Max[m[[1]]]][[1, 1]] gives the column of interest: 1
share|improve this answer
So for your general question you will want to use Position/Extract, let say for the example you give to get the vector that corresponds to 1 (which is neither the largest, nor the smallest ...) you could do:
Extract[mat[[2, All]], Position[mat[[1, All]], 1]]
Now if you want to generally get results that deal with numeric ordering, you can use b.gatessucks answer, or the one I prefer is:
mat[[2, Ordering[mat[[1, All]], -1]]]
Where -1 gives the largest value, 1 gives the smallest, and any other integer gives the relative largest/smallest (if negative)
share|improve this answer
If you define your input as
matrix = {{10, -2, 1}, {a, b, c}};
then you can do
Last@Last@SortBy[Transpose[matrix], #[[1]] &]
share|improve this answer
A variant is to use Ordering: Last@(Transpose@mat)[[Last@Ordering@First@mat]] – murray Oct 6 '12 at 21:34
The question of taking (or applying a transformation to) some part of a rectangular array based on the ordering of another part has come up in many variations. It is related to the general concept of Concomitants of order statistics (see, for example, Yang).
Using the built-in function Ordering (as in Gabriel's answer), one can define a function to extract the concomitants in a dataset as follows:
concomitantRows[list_, cols_, ordcol_, ranks_, orderingF_: Less] :=
Part[list[[All, cols]], Ordering[list[[All, ordcol]], ranks, orderingF]]
which takes a list, and extracts cols associated with the ranks in the ordcol using the ordering function orderingF.
testdata = {RandomSample[Range[6]], RandomChoice[Range[5], 3] & /@ Range[6],
RandomChoice[CharacterRange["A", "F"], 2] & /@ Range[6], RandomSample[Range[6]]};
Transpose@testdata // Grid
enter image description here
Take the rows in the sub-array formed by columns {2,3,4} associated with the top 2 elements in col 1 based on default ordering Less:
concomitantRows[Transpose@testdata, {2, 3, 4}, 1, 2]
(* Out[] = {{{3, 3, 4}, {"B", "D"}, 5}, {{4, 4, 1}, {"C", "C"}, 4}} *)
Take the ones associated with the third element:
concomitantRows[Transpose@testdata, {2, 3, 4}, 1, {3}]
(* Out[] = {{{1, 1, 5}, {"A", "B"}, 3}}*)
share|improve this answer
One can use Pick[] for the purpose:
m = {{10, -2, 1}, {{1, 1, 1}, {-3, 1, 3}, {1, -2, 1}}};
First@Pick[Last[#], First[#], Max[First[#]]] &@m
{1, 1, 1}
share|improve this answer
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Mad Ade's Garden
Logic puzzles require you to think. You will have to be logical in your reasoning.
Puzzle ID:#14619
Fun:*** (2.26)
Difficulty:*** (3.09)
Category:Logic
Submitted By:mad-adeAhu****
Mad Ade, Who had suddenly and unexpectedly after years of selfless dedication to Kebabs, decided to grow some vegetable for himself and subsequently planted his garden late in May. He planted five different vegetables: Tomatoes, Peppers, Radishes, Lettuce, and Broccoli. For each he planted a different number of rows from one to five. The hard work involved with each planting was accomplished on a separate day, from May 25th to May 30th (On Sunday May 26th he didn't work in the garden). Each vegetable was planted in a different part of the garden, the north end, the south side, the east side, the west side or the northwest corner. Can you tell which vegetable was planted in each part of the garden, the day planted and the number of rows?
1. Mad Ade turned the soil for the peppers three days after he dug hand-size rocks out of the radish patch. Mad Ade's bright rows of radishes went deeper than the broccoli but were not as deep as the tomatoes. Mad Ade rolled his trusty wheelbarrow into the west end of the garden one day before he planted the broccoli. Mad Ade believes the morning sun is bad for tomatoes and so kept those plants out of the eastern side of his garden.
2. Mad Ade labored hard in the south side of the garden and was mighty proud of the three perfect rows there. On one week's last day, Mad Ade planted the east side of the garden. Mad Ade didn't put the peppers or the tomatoes in the northwest corner. Mad Ade planted an even number of rows in the Northwest corner of his garden.
3. Mad Ade planted three rows before he planted two rows, but planted them after he planted four rows. Mad Ade figured the lettuce wouldn't last long so he planted fewer than three rows. Mad Ade planted the south garden one day before he planted the lettuce. Mad Ade planted more rows of peppers than he did broccoli (which was not planted in the west side of the garden and was not planted the 29th).
Which vegetable was planted in how many rows, on which day of May and in which side of Mad Ade's garden?
Answer
Please Note that 1.1 refers to clue 1, sentence 1, and thus 2.3 would be clue 2 sentence 3.
Explanation:
1.1 Mad Ade turned the soil for the peppers three days after he dug hand-size rocks out of the radish patch.
Radishes are 25th or 27th.
Peppers are 28th or 30th.
1.3 Mad Ade rolled his trusty wheelbarrow into the west end of the garden one day before he planted the broccoli
Broccoli is 28, 29 30 (and West is 27, 28, 29)
3.3 Mad Ade planted the south garden one day before he planted the lettuce.
Lettuce 28, 29 30 (and South 27, 28, 29 )
Tomatoes can only be 25th or 27th.
2.2 On one week's last day, Mad Ade planted the east side of the garden.
Saturday the 25th is East.
1.4 Mad Ade believes the morning sun is bad for tomatoes and so kept those plants out of eastern side of his garden
So East, 25th is not Tomatoes.
So Radishes 25th and Tomatoes 27th.
And 3 days later 28th = peppers.
3.4 Mad Ade planted more rows of peppers than he did broccoli (which was not planted in the west side of the garden and was not planted the 29th).
So Broccoli 30th and Lettuce 29th.
1.3 Mad Ade rolled his trusty wheelbarrow into the west end of the garden one day before he planted the broccoli.
West is 29th.
3.2 Mad Ade figured the lettuce wouldn't last long so he planted fewer than three rows.
So Lettuce is one or two rows.
2.4 Mad Ade planted an even number of rows in the Northwest corner of his garden.
Northwest is 2 or 4 rows.
1.2 Mad Ade's bright rows of radishes went deeper than the broccoli but were not as deep as the tomatoes.
Radishes are not one or five.
Broccoli is not five or four.
Tomatoes is not one or two.
2.3 Mad Ade didn't put the peppers nor the tomatoes in the northwest corner.
Northwest is broccoli.
Lettuce has one row.
2.1 Mad Ade labored hard in the south side of the garden and was mighty proud of the three perfect rows there.
So South is three rows.
Radishes are four.
3.3 Mad Ade planted the south garden one day before he planted the lettuce.
So planted 3 in south on 28.
Five rows of tomatoes were planted in the North Garden.
Summary:
Tomatoes, Five, 27th, North
Peppers, Three, 28th, South
Radishes, Four, 25th, East
Lettuce, one, 29th, West
Broccoli, two, 30th, Northwest
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Comments
Smithy*en*
Sep 11, 2003
Next time mad put some line breaks in.... it's hard to read!
thetrickster
Sep 22, 2003
Mad Ade, you little plagurist you (sp?) lol. Good teaser all the same
DiNamite
Oct 05, 2003
This got kind of long and boring-sry, but, I got too bored and left.
tenzin_gyatsu
Dec 18, 2003
I feel the length and level of detail involved in solving the problem make it truly harder than something one can solve in 15 or 30 minutes, (I think I spent about 2 hours [in multiple intervals] solving this).
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1,111,374,470,499,863,300 | Abacus and Slide Rule
Ye Olden Tools of Yore
I’ve been meaning to write an Abacus post for years. I used one in my first job, back in high school, and they’ve appealed to me ever since. Many years ago I learned there were people who had no idea how an abacus worked. Until then I hadn’t internalized that it wasn’t common knowledge (maybe a consequence of learning something at an early age).
Recently, browsing through old Scientific American issues before recycling them, I read about slide rules, another calculating tool I’ve used, although, in this case, mainly for fun. My dad gave me his old slide rule from when he considered, and briefly pursued, being an architect.
So killing two birds with one stone…
It was back in 1998 or 1999, I think. We threw a large Halloween party. (The whole nine yards with decorations: dry ice fog, orange and black colored food.)
But the party was too big. We’d invited people from three separate social groups, and they didn’t blend well. (Not foreseeing that possibility, we didn’t plan any mixer activities.)
Therefore the party fragmented — one might even say coagulated — and no one mingled. (Minnesotans can be shy.)
The little groups enjoyed themselves okay, but the party had no flow (lesson learned). A smaller clot was a couple, friends of my wife that I’d gotten to know.
I noticed them fiddling with my abacus. They seemed engrossed, but puzzled. It turned out they had no idea how it worked, so I gave them a demo.
Their reaction seemed on the, “Whoa! Cool!” side, so with that in mind, here’s how an abacus works:
§ §
At heart, an abacus is a scratchpad (although “memory register” is a more exact analogy).
An abacus in the starting (aka reset, clear, “all zeros”) state.
Each column of an abacus represents a digit. The one shown above (and in all examples here) has 13 columns, so it can represent a 13-digit number (or multiple numbers with fewer digits).
Each column consists of a set of seven “stones” that slide up and down on a rail. The lower set of five stones each have a value of one. The upper set of two stones each have a value of five.
The condition shown above is the “all zeros” state. When the lower stones are down, they are not counted. When the upper stones are up, they are not counted.
What is counted is the stones touching the middle crosspiece.
The configuration here is designed to show off the full range of normal digits as well as some transitory states.
For example, notice how there are two ways to represent 5 and two ways to represent 10.
In both cases, the left-hand one is the “normal” way. When five stones accumulate in the lower part (as in the right-hand examples), you usually transition that to left-hand version.
As the left-most column shows, digits in an abacus can represent values up to 15. Values over 9 are usually carried to the column to the left.
Essentially, an abacus works almost exactly the same way as you do math by hand with paper and pen. Both deal with digits, add and subtract, carry and borrow. The abacus just acts as a memory register so you’re not scribbling all over the paper.
§
Here’s a very simple step-by-step example that illustrates adding along with a little subtracting.
We’re going to start with 27 (two right columns), and we’re going to add 48 (two left columns).
Normally, we wouldn’t bother showing the 48, but doing so here provides the example of a little subtraction.
The first step is to add 8 to the 7 (note how we also subtract 8 from the left columns)…
In abacus terms, 8 is 5+3, so we’ll first move a 5-stone down, which results in a total of 12. That has to be carried to the next column to the left. (Note we also removed 5 from the left columns.)
We carry the 12 by resetting the two 5-stones and adding a 1-stone in the next column:
Now we add three 1-stones to complete the addition of 8 (and subtract three from the left side):
There are now five stones in the lower half, so we transition that by resetting those five and sliding down a 5-stone:
(As a self-check, 27+8=35, and 48-8=40, so we’re good so far.)
Now all we have to do is add the 4 to the 2 (which is now a 3 from the carry).
But there’s a problem: there are only two stones remaining, and we need four. So we’ll two-step by taking first taking two stones:
Which gives us five that we transition to the normal form:
Then we can do the other two:
And we’re done. (27+48=75)
§
Larger numbers are just more of the same. A typical use involves summing a list of numbers and only the running total is maintained on the abacus.
Subtraction is a similar process, except sometimes we have to borrow from the left rather than carry.
It’s fairly easy to multiply any number by a small number, but multiplying two large numbers gets a bit tricky. Even multiplying by, say, 25 is better done as two operations of multiplying by 5.
I may post a Sidebar getting into more examples, including multiplication, especially if there’s an interest, but for now, that’s the basics of how an abacus works.
Note that the 5+2 design isn’t the only design. A fairly common alternate is 5+1, with only a single 5-stone.
§ §
While an abacus is discrete (it uses digits), in contrast, a slide rule is analog — it uses a logarithmic scale.
Slide rules have multiple scales along their length. How many, and what type, varies. Mine (above) has the very standard C/D scales as well as the CI (Inverted) and CF/DF (Folded) scales.
These are all variations of the C/D scales. Note that the C and D scales are identical, and so are the CF and DF scales.
The C/D scale is a logarithmic scale that runs from 1.00 to 10.0. If you recall your high school math, logarithms transform multiplication into addition:
If we take two ordinary foot-long rulers, and place them end-to-end, we have a two-foot long ruler. We’ve added them.
If we place them next to each other, scales touching, and slide one 6 inches along the length, we are again adding. For example, the 2 of the ruler we slid now lies opposite the 8 of the other ruler: 6+2=8
However, when we do this with logarithmic scales, we end up multiplying.
Two logarithmic scales that run from 1.00 to 10.0, placed end-to-end, create a logarithmic scale that runs from 1.00 to 10.0 and 10.0 to 100, which comes from multiplying 10×10.
§
One thing about slide rules: They have an accuracy of about three digits, at best (two digits in some cases).
In the days before calculators they provided a “good enough” handy multiplication device (“handy” in all senses of the word).
And, the truth is, for a lot of work, two or three digits of precision are quite good enough, and there are ways to refine an answer if needed.
Another property of slide rules is that, when set for a specific answer, they actually provide a continuum of answers along their length.
§
Here’s an example:
The slide rule above is set to 1/12 (on the lower scales labeled C and D at the far left).
The 1.00 of the C scale is directly above the 1.20 of the D scale, but we can call it 12.0, if we like. This brings up another aspect of slide rules: we have to keep track of our decimal points.
The reticle (sliding piece with vertical hairline) is set to 1.50 over 1.80, but since we’re moving the decimal point (1.20 = 12.0), we see it as being 1.50 over 18.0 (which is the same as 1/12).
Over on the far right, you can see 2.00 over 24.0, which is also 1/12. That fraction is expressed along the entire length all the way to the end:
On the left of the image above, 5.00 over 60.0 (keeping track of our decimal point), is the same as 1/12, and so is 6.00 over 72.0 (or 7.50 over 90.0).
Everywhere you look, 1/12.
§
The CI scale inverts the C scale, and provides a quick inverse for any number on the C scale.
For example, in the image above, on the left side, the red 2 is directly above the 5. The inverse of 5.00 is 0.20, and the inverse of 2.00 is 0.50.
The CF/DF scales just fold the C/D scales, which make it easier to deal with numbers that are on the edges of the C/D scale. Rather than trying to read a value of the ends of the ruler, you can be comfortably in the middle.
§ §
An abacus might seem like a cumbersome tool, but it’s surprising how a bit of practice makes it fairly easy and fast.
That first high school job was a retail position, and at the end of the night we (two of us) had to “count our banks” (add up the cash in our cash register drawers and reconcile it with our sales).
The boss, a Chinese man, had a mechanical desk calculator — the kind with 10 buttons for each digit and a level you pulled to cycle the mechanism. (This was 1971 or so.) Our boss also had an abacus.
So I’d take the abacus, and my co-worker would take the adding machine, and we’d race to see who finished first.
Believe it or not, sometimes I won. When I lost, it wasn’t by much.
Stay calculating, my friends!
About Wyrd Smythe
The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe
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1,741,861,281,820,394,200 | Mostre que o triângulo de vertices (2,4) (5,1) e (6,5) é isosceles e calcule seu perimetro.
Peça mais detalhes ao usuário Thatysousa
Respostas Sair
Vamos dar letras aos pontos:
(2,4) => A
(5,1) => B
(6,5) => C
Vamos calcular a distância entre A e B
d = √(5-2)² + (1-4)²
d = √(3)² + (-3)²
d = √9 + 9
d = √18
Vamos calcular a distância entre B e C
d = √(6-5)² + (5-1)²
d = √(1)² + (4)²
d = √1 + 16
d = √17
Agora a distância de C até A
d = √(6-2)² + (5-4)²
d = √(4)² + (1)²
d = √16 + 1
d = √17
Triângulo isósceles: dois de seus lados possuem a mesma medida
Pelos cálculos está provado que este triângulo tem dois lados de mesma medida, que valem √17, e um diferente, que vale √18
Fatorando o √18 = 3√2
Portanto, o perimetro é: √17 + 3√2 + 3√2
√17 + 6√2
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2,952,170,318,666,971,000 | What is 61 689/796 as a decimal?
It's very common when learning about fractions to want to know how convert a mixed fraction like 61 689/796 into a decimal. In this step-by-step guide, we'll show you how to turn any fraction into a decimal really easily. Let's take a look!
Before we get started in the fraction to decimal conversion, let's go over some very quick fraction basics. Remember that a numerator is the number above the fraction line, and the denominator is the number below the fraction line. We'll use this later in the tutorial.
When we are using mixed fractions, we have a whole number (in this case 61) and a fractional part (689/796). So what we can do here to convert the mixed fraction to a decimal, is first convert it to an improper fraction (where the numerator is greater than the denominator) and then from there convert the improper fraction into a decimal/
Step 1: Multiply the whole number by the denominator
61 x 796 = 48556
Step 2: Add the result of step 1 to the numerator
48556 + 689 = 49245
Step 3: Divide the result of step 2 by the denominator
49245 ÷ 796 = 61.865577889447
So the answer is that 61 689/796 as a decimal is 61.865577889447.
And that is is all there is to converting 61 689/796 to a decimal. We convert it to an improper fraction which, in this case, is 49245/796 and then we divide the new numerator (49245) by the denominator to get our answer.
If you want to practice, grab yourself a pen, a pad, and a calculator and try to convert a few mixed fractions to a decimal yourself.
Hopefully this tutorial has helped you to understand how to convert a fraction to a decimal and made you realize just how simple it actually is. You can now go forth and convert mixed fractions to decimal as much as your little heart desires!
Cite, Link, or Reference This Page
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 61 689/796 as a decimal?". VisualFractions.com. Accessed on August 11, 2022. http://visualfractions.com/calculator/mixed-to-decimal/what-is-61-689-796-as-a-decimal/.
• "What is 61 689/796 as a decimal?". VisualFractions.com, http://visualfractions.com/calculator/mixed-to-decimal/what-is-61-689-796-as-a-decimal/. Accessed 11 August, 2022.
• What is 61 689/796 as a decimal?. VisualFractions.com. Retrieved from http://visualfractions.com/calculator/mixed-to-decimal/what-is-61-689-796-as-a-decimal/.
Mixed Fraction to Decimal Calculator
Mixed Fraction as Decimal
Enter a whole number, numerator and denominator
Random Mixed Fraction to Decimal Calculation | {
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4,071,260,211,699,705,300 | На этой странице вы сможете посмотреть несколько примеров для нахождения экстремумов функции, в каждом из них есть своя уникальность, поэтому рекомендую посмотреть все.
Здесь часто используется нахождение производной, что бы лучше понимать, как её надо находить, то сначала посмотрите мои таблицу производных.
1. Имеем функцию:
экстремум функции
Найдём её производную:
Найдём производную фукции
Прировняем производную к нулю и найдём значение переменной.
Прировняем производную к нулю
Наносим x=0 на координатную прямую и смотрим, где производная будет отрицательной, а где положительной. То есть до нашей точки (для этого берём любое значение до ноля ну, например, -1 и подставляем его в формулу с производной, видим что выйдем -2, то есть знак минус) и после неё (всё точно также берём любое число по праву сторону от ноля, например, 1 результат будет 2 – значит знак плюс).
Наносим x=0 на координатную прямую
Видим, что при прохождении через точку x=0, производная меняет знак с плюса на минус, то значит, что это будет точка минимума.
2. Всё аналогично делаем и в следующем примере.
5
Наносим точку x=0 на координатную прямую, и вычисляем соответствующие значения.
точки «подозрительные» на экстремум
Видим, что здесь знак производной не меняется, то есть данная точка не будет экстремумом.
3. Приступим к следующему примеру:
знак функции
Как всегда найдём производную и прировняем её к нулю. Поскольку в нас дробь, то к нолю надо приравнивать, только числитель.
производная дроби
Ещё надо учитывать точки разрыва, при которых знаменатель будет равен нулю.
точки разрыва
Наносим все эти данные на координатную прямую и находим знак производной на каждом из промежутков.
промежутки на координатной прямой
Видим, что при прохождении через точки -1 и 1 производная не меняет знака, эти точки не будут экстремумами, а при прохождении через 0 меняет с плюса на минус, поэтому точка x=0 будет максимумом.
4. Ну и рассмотрим ещё один небольшой пример:
92
Опять находим производную и приравниваем её к нолю:
максимум и минимум функции
Полученные значения переменных наносим на координатную прямую и высчитываем знак производной на каждом из промежутков. Ну например, для первого возьмём -2, тогда производная будет равна -0,24, для второго возьмём 0, тогда производная будет 2 и для третьего возьмём 2, тогда производная будет -0,24. И проставим соответствующие знаки.
максимум и минимум графика фукции - экстремумы
Видим, что при прохождении через точку -1 производная меняет знак с минуса на плюс, то есть это будет минимум, а при прохождении через 1 – меняет знак и плюса на минус, соответственно это будет максимум.
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7,029,019,168,551,612,000 | 1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
1st Grade Thanksgiving Printables
Grade Levels
Common Core Standards
Product Rating
4.0
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File Type
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12 MB|20 pages
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Product Description
This fun packet includes 20 printables that are great for the Thanksgiving season. Please note: All the activities are common core aligned. The title and objective for each printable is listed below.
MATH PRINTABLES:
Adding and Subtracting by 10’s: The students will practice adding and subtracting by 10’s. The students can refer to the turkey number line on the bottom of their worksheet for assistance. CCSS.MATH.CONTENT.1.NBT.C.4
Identifying Simple Fractions: The students will practice identifying fractions that are ½, 1/3, and ¼. The students will be asked to draw lines on a pie to create these different fraction amounts. CCSS.MATH.CONTENT.1.G.A.3
Place Value: The students will count ten rods and ones cubes to determine a given number. CCSS.MATH.CONTENT.1.NBT.B.2
Find the Missing Pie: The students will look at a sequence of numbers and determine the missing number. CCSS.MATH.CONTENT.1.NBT.A.1
Time to Gobble Gobble: The students will be asked to draw the hands on clocks to display times on the hour and half hour. CCSS.MATH.CONTENT.1.MD.B.3
Comparing Numbers: The students will compare two different numbers and record the correct greater than, less than, or equal to symbol. CCSS.MATH.CONTENT.1.NBT.B.3
Truthful Turkeys: The students will look at a simple addition or subtraction problem and determine if the equation is true or false. CCSS.MATH.CONTENT.1.OA.D.7
If, Then.. : The students will practice the basic principle for the commutative property while matching if and then statements. CCSS.MATH.CONTENT.1.OA.B.3
Skip Counting: The students will count by 2's and fill in the missing numbers.
Part Part Whole: The students will determine the unknown number in addition problems and part part whole relationships. CCSS.MATH.CONTENT.1.OA.B.4
LITERACY PRINTABLES:
Punctuation Pies: The students will read a sentence and add the correct punctuation mark to the end of the sentence. CCSS.ELA-LITERACY.L.1.2.B
Match Up: the students will look at a root word and match it to the correct inflectional forms. CCSS.ELA-LITERACY.L.1.4.C
Capitalize It: The students will read different sentences and underline the words that need to be capitalized. CCSS.ELA-LITERACY.L.1.2
Pronouns Verb Sort: The students will look at different words and determine if each word is a pronoun or a verb.CCSS.ELA-LITERACY.L.1.1.D
Dot the Digraph: The students will look at a picture and mark the digraph for each illustration. CCSS.ELA-LITERACY.RF.1.3.A
CVC Hats: The students will color in the letters to spell CVC words.
CCSS.ELA-LITERACY.RF.1.2.C
Noun Verb Sort: The students will look at different words and determine if each word is a noun or pronoun. CCSS.ELA-LITERACY.L.1.1.D
Recipe for a Good Book: This is a fun way to have the students identify the author, illustrator, characters, and setting of a book or story. CCSS.ELA-LITERACY.RL.1.3
Let’s Eat: This is a journal prompt writing activity, wherein the students will draw a picture of their favorite Thanksgiving meal. Then, the students will be asked to write a few sentences describing their meal.
Past or Present: The students will cut, sort, and glue pictures that represent the past and present.
If you like this packet, you'll also enjoy
1st Grade Christmas Printables
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20 pages
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N/A
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Overview of Lecture Factorial Designs Experimental Design Names
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Presentation on theme: "Overview of Lecture Factorial Designs Experimental Design Names"— Presentation transcript:
1 Overview of Lecture Factorial Designs Experimental Design Names Partitioning the Variablility The Two-Way Between Groups ANOVA Evaluating the Null Hypotheses Main effects Interactions Analytical Comparisons
2 Factorial Design Much experimental psychology asks the question: What effect does a single independent variable have on a single dependent variable? It is quite reasonable to ask the following question as well. What effects do multiple independent variables have on a single dependent variable? Designs which include multiple independent variables are known as factorial designs.
3 An example factorial design
If we were looking at GENDER and TIME OF EXAM, these would be two independent factors GENDER would only have two levels: male or female TIME OF EXAM might have multiple levels, e.g. morning, noon or night This is a factorial design
4 Experimental Design Names
The name of an experimental design depends on three pieces of information The number of independent variables The number of levels of each independent variable The kind of independent variable Between Groups Within Subjects (or Repeated Measures)
5 Experimental design names
If there is only one independent variable then The design is a one-way design (e.g. does coffee drinking influence exam scores) If there are two independent variables The design is a two-way design (e.g. does time of day or coffee drinking influence exam scores). If there are three independent variables The design is a three-way design (e.g. does time of day, coffee drinking or age influence exam scores).
6 Experimental Design Names
If there are 2 levels of the first IV and 3 levels of the second IV It is a 2x3 design E.G.: coffee drinking x time of day Factor coffee has two levels: cup of coffee or cup of water Factor time of day has three levels: morning, noon and night If there are 3 levels of the first IV, 2 levels of the second IV and 4 levels of the third IV It is a 3x2x4 design E.G.: coffee drinking x time of day x exam duration Factor coffee has three levels: 1 cup, 2 cup 3 cups Factor time of day has two levels: morning or night Factor exam duration has 4 levels: 30min, 60min, 90min, 120min
7 Experimental Design Names
If all the IVs are between groups then It is a Between Groups design If all the IVs are repeated measures It is a Repeated Measures design If at least one IV is between groups and at least one IV is repeated measures It is a Mixed or Split-Plot design
8 Experimental design names
Three IVs IV 1 is between groups and has two levels (e.g. a.m., p.m.) IV 2 is between groups and has two levels (e.g. coffee, water). IV 3 is repeated measures and has 3 levels (e.g. 1st year, 2nd year and 3rd year). The design is: A three-way (2x2x3) mixed design.
9 Experimental design names
The effect of a single variable is known as a main effect The effect of two variables considered together is known as an interaction For the two-way between groups design, an F-ratio is calculated for each of the following: The main effect of the first variable The main effect of the second variable The interaction between the first and second variables
10 Analysis of a 2-way between groups design using ANOVA
To analyse the two-way between groups design we have to follow the same steps as the one-way between groups design State the Null Hypotheses Partition the Variability Calculate the Mean Squares Calculate the F-Ratios
11 Null Hypotheses There are 3 null hypotheses for the two-way (between groups design. The means of the different levels of the first IV will be the same, e.g. The means of the different levels of the second IV will be the same, e.g. The differences between the means of the different levels of the interaction are not the same, e.g.
12 An example null hypothesis for an interaction
The differences betweens the levels of factor A are not the same.
13 Partitioning the variability
If we consider the different levels of a one-way ANOVA then we can look at the deviations due to the between groups variability and the within groups variability. If we substitute AB into the above equation we get This provides the deviations associated with between and within groups variability for the two-way between groups design.
14 Partitioning the variability
The between groups deviation can be thought of as a deviation that is comprised of three effects. In other words the between groups variability is due to the effect of the first independent variable A, the effect of the second variable B, and the interaction between the two variables AxB.
15 Partitioning the variability
The effect of A is given by Similarly the effect of B is given by The effect of the interaction AxB equals which is known as a residual
16 The sum of squares The sums of squares associated with the two-way between groups design follows the same form as the one-way We need to calculate a sum of squares associated with the main effect of A, a sum of squares associated with the main effect of B, a sum of squares associated with the effect of the interaction. From these we can estimate the variability due to the two variables and the interaction and an independent estimate of the variability due to the error.
17 The mean squares In order to calculate F-Ratios we must calculate an Mean Square associated with The Main Effect of the first IV The Main Effect of the second IV The Interaction. The Error Term
18 The mean squares The main effect mean squares are given by: The interaction mean squares is given by: The error mean square is given by:
19 The F-ratios The F-ratio for the first main effect is: The F-ratio for the second main effect is: The F-ratio for the interaction is:
20 An example 2x2 between groups ANOVA
Factor A - Lectures (2 levels: yes, no) Factor B - Worksheets (2 levels: yes, no) Dependent Variable - Exam performance (0…30) Mean Std Error LECTURES WORKSHEETS yes 19.200 2.04 no 25.000 1.23 16.000 1.70 9.600 0.81
21 Results of ANOVA When an analysis of variance is conducted on the data (using Experstat) the following results are obtained Source Sum of Squares df Mean Squares F p A (Lectures) 1 37.604 0.000 B (Worksheets) 0.450 0.039 0.846 AB 16.178 0.001 Error 16 11.500
22 What does it mean? - Main effects
A significant main effect of Factor A (lectures) “There was a significant main effect of lectures (F1,16=37.604, MSe=11.500, p<0.001). The students who attended lectures on average scored higher (mean=22.100) than those who did not (mean=12.800). No significant main effect of Factor B (worksheets) “The main effect of worksheets was not significant (F1,16=0.039, MSe=11.500, p=0.846)”
23 What does it mean? - Interaction
A significant interaction effect “There was a significant interaction between the lecture and worksheet factors (F1,16=16.178, MSe=11.500, p=0.001)” However, we cannot at this point say anything specific about the differences between the means unless we look at the null hypothesis Many researches prefer to continue to make more specific observations. Mean Std Error LECTURES WORKSHEETS yes 19.200 2.04 no 25.000 1.23 16.000 1.70 9.600 0.81
24 Simple main effects analysis
We can think of a two-way between groups analysis of variance as a combination of smaller one-way anovas. The analysis of simple main effects partitions the overall experiment in this way Worksheets Yes No Lectures (Yes) Worksheets Worksheets Yes No Yes No Lectures (No) Yes Lectures No Worksheets (yes) Worksheets (no) Yes Yes Lectures Lectures No No
25 Results of a simple main effects analysis
Using ExperStat it possible to conduct a simple main effects analysis relatively easily Source of Sum of df Mean F p Variation Squares Squares Lectures at worksheets(yes) worksheets(no) Error Term Worksheets at lectures(yes) lectures(no)
26 What does it mean? - Simple main effects of Lectures
No significant simple main effect of lectures at worksheets (yes) “There was no significant difference between those students who did attend lectures (mean=19.20) or did not attend lectures (mean=16.00) when they completed worksheets (F1,16=2.226, MSe=11.500, p=0.155).” Significant simple main effect of lectures at worksheets (no) “There was a significant difference between those students who did attend lectures (mean=25.00) or did not attend lectures (mean=9.60) when they did not complete worksheets (F1,16=51.557, MSe=11.500, p<0.001). When students who attended lectures did not complete worksheets they scored higher on the exam than those students who neither attended lectures nor completed worksheets.”
27 What does it mean? - Simple main effects of worksheets
Significant simple main effect of worksheets at lectures (yes) “There was a significant difference between those students who did complete worksheets (mean=19.20) or did not complete worksheets (mean=25.00) when they attended lectures (F1,16=7.313, MSe=11.500, p=0.016). Students who attended the lectures and completed worksheets did less well than those students who attended lectures but did not complete the worksheets.” Significant simple main effect of worksheets at lectures (no) “There was a significant difference between those students who did complete worksheets (mean=16.00) or did not complete worksheets (mean=9.60) when they did not attend lectures (F1,16=51.557, MSe=11.500, p<0.001). When students who attended lectures did not complete worksheets they scored higher on the exam than those students who neither attended lectures nor completed worksheets.”
28 Analytic comparisons in general
If there are more than two levels of a Factor And, if there is a significant effect (either main effect or simple main effect) Analytical comparisons are required. Post hoc comparisons include tukey tests, Scheffé test or t-tests (bonferroni corrected).
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-1,329,727,507,271,405,300 | Is there parallel/simultaneous substitution in Mathematica?
I know this is simple but I’m not proficient at Mathematica and cannot seem to find it online or in the help. Also, Replace[] says it does the assignments in order, and instead of a complicated ternary-thing, I was hoping for something simple, to avoid a mistake.
I would like to make parallel, or simultaneous, substitutions. E.g. in Maxima:
(%i1) psubst ([a^2=b, b=a], sin(a^2) + sin(b));
(%o1) sin(b) + sin(a)
(%i2) subst ([a^2=b, b=a], sin(a^2) + sin(b));
(%o2) 2 sin(a)
Thank you all for answering my simple question!
=================
1
Have you tried doing this in Mathematica though? ReplaceAll aka /. works just like this. Yes it does the assignments in order, but once a rule has been applied to a specific part, it won’t apply any other rules to that part.
– Marius Ladegård Meyer
May 31 at 21:19
Yes, I did this: Sin[a^2] + Sin[b] /. a^2 -> b /. b -> a but it gave me back 2 Sin[a]
– nate
May 31 at 21:27
=================
1 Answer
1
=================
You can achieve either result depending on the order of evaluation. As Marius pointed out, once a rule out of a set of rules has been applied to a specific part of an expression, no further rules will be applied to that expression.
Notice the difference between applying both rules at the same time, and applying them one after the other:
(* At the same time, only the first one applies *)
Sin[a^2] + Sin[b] /. {a^2 -> b, b -> a}
(* Out: Sin[a] + Sin[b] *)
(* One after the other *)
Sin[a^2] + Sin[b] /. a^2 -> b /. b -> a
(* Out: 2 Sin[a] *)
If you want a group of rules to be applied to an expression repeatedly until no further change is possible, then ReplaceRepeated (//.) will achieve that:
Sin[a^2] + Sin[b] //. {a^2 -> b, b -> a}
(* Out: 2 Sin[a] *)
Curly brackets! Thanks! I did note that if I change the order I get it right, but because my expression is extremely large I can’t visually check it. But this is much better than something like: Sin[a^2] + Sin[b] /. b -> A /. a^2 -> b /. A -> a
– nate
May 31 at 21:38
@nate Yes, you can reproduce any of those behaviors you mentioned at will, it’s just a question of choosing the right setup 🙂
– MarcoB
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7,147,659,537,667,744,000 | Quantitative Aptitude Questions for IBPS PO and Clerk Mains 2017
Dear Students,
Quantitative-Aptitude-Questions-for-IBPS-PO-Exam-2017
Quantitative Aptitude Questions for IBPS Exam
Quantitative Aptitude is a very important section you must prepare if you are aiming for a job in Bank or Insurance sector. These two weeks are very important as IBPS PO and Clerk Mains are lined up. So, these 15 questions can help you practice three very important topics of Quant Section.
Directions (Q1-5): What should come in place of question mark (?) in the following questions?
Q1. 18.6 × 3 + 7.2 – 16.5 = ? + 21.7
(a) 35.7
(b) 21.6
(c) 24.8
(d) 27.6
(e) None of these
Directions (6-10): Study the following pie chart and table carefully to answer the following questions that follow:
Percentage break up of foreigners who toured of various states of India and the ratio of men to women in them
Total Number of foreigners = 1800
Percentage Break up of foreigners
Q6. What is the number of male foreigners who toured of Goa and MP together?
(a) 632
(b) 674
(c) 692
(d) 649
(e) None of these
Q7. The number of female foreigners who toured of MP forms approximately what percent of the total number of foreigners who toured of all states together?
(a) 7
(b) 5
(c) 19
(d) 15
(e) 10
Q8. What is the respective ratio of the number of female foreigners who toured of Uttarakhand and Jammu & Kashmir together and the total number of foreigners toured in these states together?
(a) 353 : 433
(b) 234 : 551
(c) 427 : 558
(d) 3 : 7
(e) None of these
Q9. What is the approximate average no. of male foreigners who toured all the states together?
(a) 212
(b) 210
(c) 202
(d) 208
(e) 206
Q10. The number of male foreigners who toured Goa forms what percent of the total number of foreigners who toured in Goa? (rounded off to two digits after decimal)
(a) 89.76
(b) 91.67
(c) 88.56
(d) 94.29
(e) None of these
Q11. A train crosses a platform and a tunnel in 18 and 32 seconds respectively. The speed of the train and length of the train are 45 kmph and 140 metres respectively. Find the length of the platform is approximately what percent less than the length of the tunnel?
(a) 72%
(b) 67%
(c) 82%
(d) 61%
(e) None of these
Q12. A dealer sold a refrigerator at a loss of 12%. Had he sold it for Rs.2205 more, he would have gained 24%. For what value should he sell it in order to gain 20%?
(a) Rs. 8120
(b) Rs. 7750
(c) Rs. 6825
(d) Rs. 7350
(e) None of these
Q13. A man has whisky worth Rs. 32 a litre and another lot worth Rs 28 a litre. Equal quantities of these are mixed with water to obtain a mixture of 75 litres worth Rs 24 a litre. Find how much water the mixture contains?
(a) 5 litres
(b) 10 litres
(c) 15 litres
(d) 20 litres
(e) None of these
Q14. Shreekant’s monthly income is 25% more than that of Vimal. Vimal’s monthly income is 15% less than that of Vishnu. If the difference between the monthly income of Shreekant and Vishnu is Rs 2750. What is the monthly income of Vimal?
(a) Rs 11000
(b) Rs 10000
(c) Rs 12000
(d) Data Inadequate
(e) None of these
Q15. If the difference between SI and CI a certain amount at the rates 13% (for SI) and 15%(for CI) for 2 years is Rs.790.Find the principal amount.
(a) Rs.12,640
(b) Rs.10,500
(c) Rs.9,500
(d) Rs.8,500
(e) None of these
You may also like to Read:
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-5,424,083,330,885,669,000 | Системы счисления
Упражнения
В системе счисления с основанием больше 10, цифры записываются так: 0, 1, 2, ..., 9, A, B, C, ...
В тексте программ на языке Python можно использовать целочисленные константы, записанные в двоичной (префикс 0b), восьмеричной (префикс 0o) и шестнадцатеричной (префикс 0x) системах счисления. После указанного префикса идут цифры, которые в двоичной системе счисления могут быть только 0 или 1, в восьмеричной — от 0 до 7, в шестнадцатеричной — от 0 до 9, а также буквы латинского алфавита от A до F (могут быть как строчными, так и заглавными). Например, десятичной число 179 можно задать несколькими способами.
A = 179
A = 0b10110011
A = 0o263
A = 0xB3
Если вы знаете стандартные функции языка Python для перевода представления чисел между различными системами счисления, то этими функциями пользоваться нельзя. Также нельзя использовать функции типа eval, exec и т.д.
Если программа выводит результат в системе счисления с основанием больше 10, то цифры записываются так: 0, 1, 2, ..., 9, A, B, C, ...
A: Шестнадцатеричная цифра - 1
Дана шестнадцатеричная цифра, которую необходимо считать в величину типа str. Выведите ее десятичное значение.
Программа должна содержать функцию перевода hex2int(c). Аргумент функции имеет тип str, результат — int.
Ввод Вывод
9
9
F
15
B: Шестнадцатеричная цифра - 2
Решите задачу, обратную предыдущей.
Ввод Вывод
9
9
15
F
C: Из двоичной в int
Дано число, записанное в двоичной системе счисления. Переведите его в тип int и выведите на экран в десятичном виде. Исходное число необходимо считать в переменную типа string, для перевода реализовать функцию bin2int(s). Аргумент функции имеет тип str, результат — int.
Ввод Вывод
10110011
179
D: Из шестнадцатеричной в int
Решите предыдущую задачу в случае, когда входное число задано в шестнадцатеричном виде. Соответствующая функция должна называться hex2int(s). Аргумент функции имеет тип str, результат — int.
Ввод Вывод
B3
179
E: Из int в двоичную
Переведите число из десятичной системы в двоичную. Соответствующая функция должна называться int2bin(n). Аргумент функции — число типа int, результат имеет тип str.
Ввод Вывод
179
10110011
F: Из int в шестнадцатеричную
Переведите число из десятичной системы в шестнадцатеричную. Соответствующая функция должна называться int2hex(n).
Ввод Вывод
179
B3
G: Из любой в любую
Напишите программу, переводящую запись числа между двумя произвольными системами счисления.
На вход программа получает три величины: n, A, k, где n и k – натуральные числа от 2 до 36: основания системы счисления, A – число, записанное в в системе счисления с основанием n, A<231.
Необходимо вывести значение A в системе счисления с основанием k без лидирующих нулей.
Решение должно содержать две функции перевода — из числа в произвольной системе счисления, записанного в переменной типа str в переменную типа int и обратно.
Ввод Вывод
2
101111
16
2F
10
35
36
Z
H: Из шестнадцатеричной в двоичную
Переведите число из шестнадцатеричной системы счисления в двоичную. Исходное число может быть очень большим (до \(2\times10^5\) символов). Необходимо вывести результат без лидирующих нулей.
Ввод Вывод
2F
101111
I: Из двоичной в шестнадцатеричную
Переведите число из двоичной системы счисления в шестнадцатеричную. Исходное число может быть очень большим (до \(1{,}2\times10^6\) символов).
Ввод Вывод
101111
2F
J: Из уравновешенной троичной в int
В уравновешенной троичной системе счисления используется основание 3 и три цифры: 0, 1 и -1. Цифру -1 будем обозначать знаком $. Достоинство уравновешенной троичной системы счисления: простота хранения отрицательных чисел и удобство нахождения числа, противоположному данному.
Вот как записываются небольшие числа в уравновешенной троичной системе счисления:
Десятичная -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9
Уравнов. троичная $00 $01 $1$ $10 $11 $$ $0 $1 $ 0 1 1$ 10 11 1$$ 1$0 1$1 10$ 100
Подробней о уравновешенной троичной системе счисления можно прочитать в Википедии (статья Троичная система счисления, там используется термин "троичная симметричная система счисления").
Дана запись числа в уравновешенной троичной системе счисления. \(10^5\)
Ввод Вывод
$01
-8
K: Из фибоначчиевой в int
Рассмотрим последовательность Фибоначчи: F1=1, F2=2, Fn=Fn-1+Fn-2 при n>2. В частности, F3=2, F4=3, F5=5, F6=8 и т.д.
Любое натуральное число можно представить в виде суммы нескольких членов последовательности Фибоначчи. Такое представление будет неоднозначным, но если наложить дополнительное условие, что в представлении нет двух соседних членов последовательности Фибоначчи, то представление становится единственным.
Будем говорить, что число A представимо в фибоначчиевой системе счисления в виде akak-1...a2, где ai∈{0;1}, если A=akFk+...+a2F2 и в записи akak-1...a2 нет двух единиц подряд.
Вот как записываются небольшие числа в фибоначчиевой системе счисления:
Десятичная 0 1 2 3 4 5 6 7 8 9 10 11 12 13
Фибоначчиева 0 1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000
Подробней о фибоначчиевой системе счисления можно прочитать в Википедии (статья Фибоначчиева система счисления).
Дана запись числа в фибоначчиевой системе счисления. Запишите его в десятичной системе счисления.
Программа получает на вход строку из символов 0 и 1 и должна вывести одно целое число. Гарантируется, что результат может принимать значения от 0 до 2·109.
Ввод Вывод
10101
12
L: Из int в уравновешенную троичную
Дано целое число oт -2·109 до 2·109. Выведите его представление в уравновешенной троичной системе счисления без лидирующих нулей.
Ввод Вывод
-8
$01
M: Из int в фибоначчиеву
Дано целое число oт 0 до 2·109. Выведите его представление в фибоначчиевой системе счисления без лидирующих нулей.
Ввод Вывод
12
10101
N: Инкремент
Первая строка входных данных содержит последовательность символов '0', ..., '9', 'A', ..., 'Z', являющейся записью некоторого неотрицательного числа в системе счисления с основанием base. Длина числа не превосходит 100000 символов. Вторая строка входных данных содержит основание системы счисления base, не превосходящее 36.
Увеличьте это число на 1 и выведите результат в той же системе счисления.
Ввод Вывод
19A
11
1A0
O: Декремент
Первая строка входных данных содержит последовательность символов '0', ..., '9', 'A', ..., 'Z', являющейся записью некоторого положительного числа в системе счисления с основанием base. Длина числа не превосходит 100000 символов. Вторая строка входных данных содержит основание системы счисления base, не превосходящее 36.
Уменьшите это число на 1 и выведите результат в той же системе счисления.
Ввод Вывод
1A0
11
19A
P: Инкремент в уравновешенной троичной системе счисления
Дана запись некоторого числа в уравновешенной троичной системе счисления, длина записи не превосходит 100000 символов.
Увеличьте это число на 1 и выведите его значение в той же системе.
Ввод Вывод
$01
$1$
Q: Декремент в уравновешенной троичной системе счисления
Уменьшите на 1 длинное число, записанное в уравновешенной троичной системе счисления.
Ввод Вывод
$1$
$01
R: Фибоначчиев инкремент
Дано целое неотрицательное число n, записанное в фибоначчиевой системе счисления, длина числа не превосходят 100000 символов. Выведите значение числа n+1 в фибоначчиевой системе счисления.
Ввод Вывод
100101
101000
S: Фибоначчиев декремент
Дано целое положительное число n, записанное в фибоначчиевой системе счисления, длина числа не превосходят 100000 символов. Выведите значение числа n-1 в фибоначчиевой системе счисления.
Ввод Вывод
101000
100101
T: Шестнадцатеричное сложение
Дано два шестнадцатеричных числа, длиной до 100000 символов каждый. Вычислите их сумму и выведите результат в шестнадцатеричной системе счисления.
Ввод Вывод
F1
F
100
U: Уравновешенное троичное сложение
Дано два числа, записанных в уравновешенной троичной системе счисления. Выведите их сумму без лидирующих нулей. Длины входных чисел не превосходят 100.000 символов.
Ввод Вывод
1$$$
$0$
11
Пример соответствует выражению 14+(-10)=4.
V: Фибоначчиево сложение
Даны два числа, записанные в фибоначчиевой системе счисления, длины чисел не превосходят 100.000 символов. Выведите значение их суммы в фибоначчиевой системе счисления.
Ввод Вывод
10010
10101
1000001
W: Марсианские факториалы
В 3141 году очередная экспедиция на Марс обнаружила в одной из пещер таинственные знаки. Они однозначно доказывали существование на Марсе разумных существ. Однако смысл этих таинственных знаков долгое время оставался неизвестным. Недавно один из ученых, профессор Очень-Умный, заметил один интересный факт: всего в надписях, составленных из этих знаков, встречается ровно \(K\) различных символов. Более того, все надписи заканчиваются на длинную последовательность одних и тех же символов.
Вывод, который сделал из своих наблюдений профессор, потряс всех ученых Земли. Он предположил, что эти надписи являются записями факториалов различных натуральных чисел в системе счисления с основанием \(K\). А символы в конце — это конечно же нули — ведь, как известно, факториалы больших чисел заканчиваются большим количеством нулей. Например, в нашей десятичной системе счисления факториалы заканчиваются на нули, начиная с \(5!=1\times2\times3\times4\times5\). А у числа \(100!\) в конце следует \(24\) нуля в десятичной системе счисления и \(48\) нулей в системе счисления с основанием 6 — так что у предположения профессора есть разумные основания!
Теперь ученым срочно нужна программа, которая по заданным числам \(N\) и \(K\) найдет количество нулей в конце записи в системе счисления с основанием \(K\) числа \(N!\), чтобы они могли проверить свою гипотезу. Вам придется написать им такую программу!
В первой строке входных данных содержатся числа \(N\) и \(K\), разделенные пробелом, (\(1\le N \le 10^9\), \(2 \le K \le 1000\)). Выведите число \(X\) — количество нулей в конце записи числа \(N!\) в системе счисления с основанием \(K\).
Ввод Вывод
5 10
1
100 10
24
100 6
48
3 10
0
X: Счастливые цифры
Школьнику Васе нравятся числа, которые заканчиваются счастливыми для него цифрами \(k\). Поэтому каждый раз, когда он видит какое-нибудь натуральное число n, он сразу пытается подобрать такое \(d\) (\(d \ge 2\)), что число \(n\) в системе счисления с основанием \(d\) заканчивается как можно большим количеством цифр \(k\).
Требуется написать программу, которая по заданным числам \(n\) и \(k\) найдет такое \(d\), чтобы число \(n\) в системе счисления с основанием \(d\) заканчивалось как можно большим количеством цифр \(k\).
Вводятся два целых десятичных числа \(n\) и \(k\) (\(1 \le n \le 10^{11}\), \( 0 \le k \le 9\)).
Выведите два числа: \(d\) — искомое основание системы счисления и \(l\) — количество цифр \(k\), которым заканчивается запись числа \(n\) в этой системе счисления. Если искомых \(d\) несколько, выведите любое из них, не превосходящее \(10^{12}\) (такое всегда существует).
Ввод Вывод
49 1
3 2
7 5
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-3,907,511,852,266,034,000 | Saturday, April 27, 2013
Java Brew Coffee House and Other Saturday Adventures
I hope everyone is doing well today! I am coming to you form Java House Coffee Shop in Montrose, CA.
This morning I went to the Glendale College Library. Guests have to park on the top lot. To get to the campus required a combination stair climbing and taking elevators. I love the view from the top.
Among subjects studied today are centriods, inverse problems, and surface integrals. Three hours went by and I left 20 minutes before the library closed, which only proves the saying "So much to do, and so little time."
Here is some of what studied today:
------
Centriods:
Points (x,y) each with mass m:
X-bar: Σ(x*m)/Σm, Y-bar: Σ(y*m)/Σm
Continuous functions, uniform thickness and density, f(x) ≥ g(x), limits a ≤ x ≤ b,
M_y = ∫(x * (f(x) - g(x)) dx, a, b)
M_x = 1/2 * ∫( (f(x))^2 - (g(x))^2, x, a, b)
M = ∫( f(x) - g(x) dx, a, b)
X-bar: M_y/M, Y-bar: M_x/M
Jon Guilleberg. "Mathematics: From the Birth of Numbers." W.M. Norton & Co. 1997
Normal Vectors
If a vector is perpendicular to another vector, the former vector is known as a normal vector. We can find a normal vector to any two vectors U and V by the cross product.
N = U × V
And the Unit Normal Vector:
n = (U × V)/|U × V| = vector/(norm of a vector)
Applying the normal unit vector to a said function z = f(x,y):
n = [-df/dx, -df/dy, 1] / √( 1 + (df/dx)^2 + (df/dy)^2 )
A unit vector has length 1.
Pretend that "d" is the partial derivative operator.
Surface area of z = f(x,y) over region R is:
∫ ∫ (√( 1 + (df/dx)^2 + (df/dy)^2 ) dx dy
H.M. Schey "div grad curl and all that" W.M. Norton & Co. 1992
Yes the title was in all lowercase. I get weirdly fascinated by all-lowercase titles.
Going to finish my lunch, and do some programming. Love to all of you!
Eddie
This blog is property of Edward Shore. 2013
Tuesday, April 23, 2013
HP 39gii Programming Tutorial - Part 8: Functions in Programs
HP 39gii Programming Tutorial - Part 8: Functions as Arguments
Apologies for the delay.
In part 8, I will show you several was you can use functions as arguments in programs. First, some pointers to help make the programming experience less daunting:
1. The HP 39gii is not a CAS machine. Everything must be evaluated to a number.
2. Functions are entered either as strings (expressions surrounded by quotation marks) or by the use of the QUOTE command. The QUOTE command can be found by pressing [Math], [ 1 ], [ 2 ]. It may help to store the required equation/function in E#/F# first before running the program, which is the case with PARTDER.
3. When using HP mathematical commands, you may have to store pass-through variables into global variables first. I am not sure why this is true for some commands but not for others.
4. The use of expressions and functions don't always work in programming. Use caution. I have particular trouble with SOLVE and FNROOT in this situation.
--------
Note: Recall there are two ways to store objects, one the store arrow ( →) and the assign operator ( := ). I usually code the latter, but for most applications, either way is acceptable. I use both methods in today's programming examples. And yes, we can use both methods in one program.
Hint: I often test commands and their usage at the Home screen to get a better understanding on how mathematical commands work. I also note their syntax by both using the command and accessing the help menu ([SHIFT] [Views]).
------
The Program PARTDER
PARTDER takes the partial derivatives of the function f(X,Y) at the point (x, y) and returns a list of the values {df/dx, df/dy}.
PARTDER takes two pass-through arguments:
x = value of x
y = value of y
Define f(X,Y) as a string into E0 outside the program.
EXPORT PARTDER(a,b)
BEGIN
a → A; b →Y; ∂(E0, X=A) → C;
b → B; a → X; ∂(E0, Y=B) → D;
RETURN {C, D};
END;
Examples:
"SIN(X*Y)+Y^2"→E0
PARTDER(π, 1) returns { -1, -1.14159265359 }
QUOTE(X^(2*Y+1))→E0
PARTDER(4,1) returns {48, 177.445678223}
Eddie
This blog is property of Edward Shore. 2013
Sunday, April 21, 2013
My Slide Rule Collection
I have a small collection of slide rules. Slide rules, which first made their appearance in the 1620s, was the calculator before electronic calculators became prominent in the 1970s.
Having been born three years after the slide rule disappeared from the mainstream I first came across slide rules for the first time when I found my grandfather's book "The Slide Rule and How To Use It". It took me a while to get an understanding how they work, but I think I have a beginner's idea how slide rules work, or getting used to the idea of logarithmic scales.
The scales I know about are:
C and D: Used for multiplication and division
CI and DI: Reciprocal of the C and D scales
A: the square of scale C and D
K: the cube of the scale C and D
L (and LL): logarithmic scales
S: sine
T: tangent
A → D: √
D → A: x^2
K → D: cube root
D → K: cube
S → D: (sin x°)/10
D → S: asin(x/10)... I think
D → L: log x
L → D: 10^x
I think the LL scale is a natural logarithm/exponential scale.
Thank you William Oughtred for inventing the slide rule.
Here is my collection:
Lawrence Deluxe Slide Rule with the book it came with: "The Slide Rule and How to Use It".
This what I found in the garage. Now this sits in my bookshelf.
Scales: A, B, C, CL, D, K
Keuffel & Essel Co. Model 4098A (I think I am correct on the model number)
I picked up this slide rule last year at a swap meet in Azusa for two bucks. What is interesting about this slide rule is that it is six inches long, has a five inch/13 centimeter ruler in the back. The middle part of the scale is removable to reveal the S, L, and T scales in the back.
Scales: A, D. Middle part: B, C. (S, L, T in the back)
Ditizgen Polymath Multiplex
I bought this slide rule in Pomona several months ago for $7. Unfortunately, the case that it came in crumbled. It now has a towel as a case.
Scales:
Front: LL01, K, A, CF, CIF, L, CI, C, D, DI, LL1
Back: LL02, LL03, DF, B, T<45°, T>45°, ST, S, D, LL3, LL2
Keuffel & Esser Co. 4181-3
My dad and I went to the swap meet at Glendale College this morning. Pretty good swap meet despite it being on the small side. A lot of high-end stuff. I got this slide rule for $7.
Slides:
Front: LL02, LL03, DF, CF, CIF, CI, C, LL3, LL2
Back: LL01, L, K, A, B, T, ST, S, D, DI, LL1
I like to work with slide rules every once in a while - fun!
Eddie
This blog is property of Edward Shore. 2013
Thursday, April 18, 2013
HP 39gii Programming Tutorial Update
There will be a Part 8 of the HP 39gii calculator tutorial series. I plan to have it posted by next week.
Part 8 will involve programs that use functions and equations as parameters.
Hope everything is well. It is a crazy world out there!
Eddie
This blog is property of Edward Shore. 2013
Saturday, April 13, 2013
The Weekend's Here: A preview of what's coming next.
Hi everyone! I am at a local Starbucks. Love the coffee shop atmosphere - and coffee.
Way I doing today and over the next few weeks is taking some programs from the classic Hewlett Packard manuals, particularly the 1970s (25, 45, 67, 33E, etc) and programming them on today's calculators (35S, 39gii, 15c LE, 50g). I expect that I will end up using resources from Texas Instruments and Casio as well. I aim to post some programs on my blog starting in May.
Have a great and fun weekend! Thank you for your continual support and comments!
Eddie
This blog is property of Edward Shore. 2013
Tuesday, April 9, 2013
TI-84 Plus C Silver Edition Review
TI-84 Plus Color Silver Edition Review
The changes are mostly positive as TI takes another step forward while maintaining the classic system TI-84+ users are accustomed to.
The good:
No steep learning curve from older 82/83/84 models - just a few more additional commands, mainly revolving around color. You can pretty much pick the TI-84+ CSE up and operate it like any other version of the 82/83/84.
15 colors to work with in graphing and drawing.
Rechargeable batteries - and they are much better than TI nSpire CX's. A charge can last a week or better pending on regular use, a day on heavy use (like plotting every point on the graph screen), charges to full quickly.
Backlit screen
Current mode settings are now on top of the screen at all times.
Up to 10 pictures can are stored in archive memory and still be recalled without having to take up RAM.
The not so good:
Decrease in RAM for 24,000 bytes to 21,000 bytes. This really does not affect you unless you do a lot of programming (like me). We still have a TON of archive ROM to store programs away if you don't plan on using them for a while.
Native trig functions still do not support complex numbers.
Graph screen does not take the whole screen - there is a border.
Graphing equations is slightly slower than its predecessors; however it is not a deal breaker.
Price: around $130, depending on vendor. I got mine for $129. Will the TI-84+ CSE replace the monochrome TI-84+ Silver Edition when the color edition hits the mainstream? I do not know.
Worth buying, especially if
(1) you have not purchased an 84+ before,
(2) am a TI and/or calculator enthusiast, or
(3) want a color calculator with a classic operating structure.
Overall, I enjoy using this calculator.
I leave you with this tip and some screen shots.
Tip: Run programs in Classic mode, they run faster than programs ran in MathPrint mode. This is for any TI-84+ models.
Have a great day! Eddie
This blog is property of Edward Shore. 2013
Monday, April 8, 2013
Using Color Calculators to Plot Patterns (TI-84+CSE, Casio Prizm)
We are in the age of color graphing calculators (and mobile apps). While the Casio Prizm and TI nSpire CX (with and without CAS) got the age started, it could be the next crop of color calculators, such as the TI-84+ Color Silver Edition that puts color into the mainstream.
These patterns are generated by using four colors and a function f(x,y) where
f(x,y) = remainder( int(Φ(x,y)) / 4)
with Φ(x,y) is a positive function. The value of f(x,y) determines on of four colors. I used the DRAWPATTERN (page 269 of the HP 39gii manual, 1st edition) as a template.
With the TI-84+ Color Silver Edition it takes about 40-45 minutes to complete. The real surprise (to me at least), was I did one with the Casio Prizm, which it took about six hours to complete; I was surprised there was any battery left. This is where the advantage of rechargeable battery of the TI-84+ CSE comes in. I am happy then to report that the battery of the TI-84+ CSE held up well.
Without any further ado, here are the graphics created! Nice way to spend a Sunday afternoon, I think.
With each picture I will list Φ(x,y). The window, except noted, is Xmin=-10, Xmax=10, Ymin=-10, Ymax=10.
Recall: f(x,y) = remainder( int(Φ(x,y)) / 4)
Until next time,
Eddie
This blog is property of Edward Shore. 2013
Friday, April 5, 2013
HP 39gii Programming Part 7: Using App Specific Commands
I hope you are all enjoying the day/night so far. The subject of Part 7 of the HP 39gii will be App-Specific commands.
We will work with app and global variables, as well as switching applications and doing App specific commands to our advantage. The two primary commands we will work with are STARTAPP and STARTVIEW.
STARTAPP and STARTVIEW
STARTAPP does what it advertises: starts the specific application called. The required application is stated as a string. For example, STARTAPP("Function") starts the Function graphing app. Running this command is key if you want the user to access specific screens, such as the Symbolic, Plot, and Numeric screens.
Keyboard sequence for STARTAPP is: [Shift] [Math] [ F1 ] [ 1 ] [ 2 ].
Applications include:
"Function" - Function Graphing
"Parametric" - Parametric Graphing
"Polar" - Polar Graphing
"Sequence" - Recurring Sequence Graphing
"Solve" - Solver Interface
"Triangle Solver" - Application for solving sides and angles of triangles
"Finance" - Time Value of Money Application
"Statistics 1Var" - 1 Variable Statistics
"Statistics 2Var" - 2 Variable Statistics
"Inference" - Hypothetical Testing and Confidence Intervals
"Linear Explorer" - Teaching application regarding linear equations
"Quadratic Explorer" - Teaching application regarding quadratic equations
"Trig Explorer" - Teaching application regarding trigonometric functions
"Data Streamer" - Application that works with Hewlett Packard's StreamSmart devices (data collection from various apparatus)
STARTVIEW leaves the user at a specific view, such as the Home screen, Plot screen, etc. Each screen has a specific view code, with syntax STARTVIEW(code). Some view codes are:
0: Symbolic Screen
1: Plot Screen
2: Numeric Screen
3: Symbolic Setup Screen
4: Plot Setup
5: Numeric Setup
6: App Information (accessed by [Shift] [Apps])
7: Views Key (same as if you pressed the [Views])
-1: Home Screen
-2: Mode Setup Screen
Keyboard sequence for STARTVIEW is: [Shift] [Math] [ F1 ] [ 1 ] [ 3 ].
App Variables
App variables are global. Some are specific to one app while others are applicable to several apps. I usually type of the variables, but you can find the app variables by pressing [Vars] [ F2 ]. The order of the apps shift, with the current app on top. I will present some app variables. This is not the extensive list by any means, so please check out the HP 39gii manual for a complete list.
App Variables Specific to One App
Function Symbolic - There are ten functions that are named F#, with # representing a digit 0 through 9. F1 is at top of the Symbolic view, F0 is at the bottom.
Parametric Symbolic - Similar to Function; names are X1 - X9, X0 and Y1 - Y9, Y0.
Polar Symbolic - Similar to Function; names are R1 - R9, R0
Sequential Symbolic - Similar to Function; names are U1 - U9, U0
Solve Symbolic - you can store ten equations in the solve variables E1 - E9, E0
You can store graphing expressions and equations to the appropriate variable by using a string around your expression/equation. You can use both methods of storing (store arrow and the equals-colon assignment syntax) Examples:
"SIN(X^2)" → F1 (store to Function F1)
E3 := "A^2 + B^2 = C^2" (store to Equation E3)
Statistics - 1 Variable:
Data: D1 - D9, D0 (lists)
Analysis: H1 - H5
Set the sample by the command SetSample(H#, D#);
Statistics - 2 Variable:
Data: C1 - C9, C0 (lists)
Analysis: S1 - S5
Set the data to be analyzed by the commands:
SetIndep(S#, C#);
SetDepend(S#, C#);
App Variables used by Several Apps
The Plot window: Xmin, Xmax, Ymin, Ymax, Xscl, Yscl, Tmin, Tmax, Tstep, θmin, θmax, θstep
Three ways to access θ:
1. Be in Polar mode and press [X,T,θ,N].
2. Echo θ from the Characters Menu: [Shift] [Vars] [ F2 ] [ 7 ] (Greek and Coptic) [ F4 ] (Page Down) [ left ] [ left ] [ up ] [ up ] [ F6 ] (OK)
3. char(952) returns "θ" while expr(char(952)) returns θ.
CHECK and UNCHECK
CHECK turns a certain equation/expression on. UNCHECK, as you probably guessed, turns named function off.
Syntax:
CHECK(n)
UNCHECK(n)
Just the number is accepted: n refers to the equation/expression listed in the current app. Despite what the manual says, CHECK (F1) produces an error, n must be numerical.
With that long intro done, let's get to the programming!
The Program SSAT (Simple Statistics)
SSAT takes a list, turns on the Statistics 1-Variable app, and returns:
* The number of elements in the list
* The mean
* The standard deviation
After the results are displayed, you are returned to the home screen.
Note: The commands SetSample and Do1VStats are found in then Command-App-Statistics 1 Variable menu. ([Shift] [Math] [ F2 ])
Program:
EXPORT SSTAT(D)
BEGIN
D1:=D;
STARTAPP("Statistics 1Var");
SetSample(H1,D1);
Do1VStats(H1);
PRINT();
PRINT("n="+Nbltem);
PRINT("Mean="+MeanX);
PRINT("SDev="+sX);
STARTVIEW(-1);
END;
Example:
SSTAT({3,4.5,4.4,4.6,4,3.9,3.8,3,3.2}) returns
n=9
Mean=3.822222222
SDev=.630035272381
The Program PLOTF
The program PLOTF takes three arguments.
PLOTF(function of X as a string, left limit, right limit)
The function F0 is turned on while the other functions F1 - F9 are turned off. The Function app is tuned on the program terminates at the plot screen.
Program:
EXPORT PLOTF(f,a,b)
BEGIN
LOCAL c,d,K;
STARTAPP("Function");
Xmin:=a;
Xmax:=b;
c:=0;
d:=0;
f → F0(X);
CHECK(0);
FOR K FROM 1 TO 9 DO
UNCHECK(K);
END;
FOR K FROM a TO b STEP (b-a)/256 DO
IF F0(K) < c THEN
c:=F0(K); END;
IF F0(K) > d THEN
d:=F0(K); END;
END;
Ymin:=c-1;
Ymax:=d+1;
Xtick:=10^(ROUND(LOG(b-a),0));
Ytick:=10^(ROUND(LOG(d-c),0));
STARTVIEW(1);
END;
Example: Plot y = 2x^3 - x + 0.5 from -5 ≤ x ≤ 5
PLOTF("2X^3-X+0.5",-5,5) gets this:
The Program PROJECTILE
PROJECTILE does the following:
1. Switch the 39gii to the Parametric app and Degrees mode
2. Determines the projectile's range and height. No air resistance is assumed. You determined whether you are working in: SI (meters) or US (feet).
3. The path of the projectile is graphed.
Input: PROJECTILE(initial height, initial velocity, initial angle in degrees)
The program will ask you to select the value of the gravitational constant:
g = 9.80665 m/s^2
Or
g = 32.17405 ft/s^2
Program:
EXPORT PROJECTILE(y0, v0, a0)
BEGIN
LOCAL K;
STARTAPP("Parametric");
CHOOSE(K,"g=","9.80665 m/s²","32.17405 ft/s²");
IF K==0 THEN KILL; END;
IF K==1 THEN G:=9.80665; END;
IF K==2 THEN G:=32.17405; END;
HAngle:=1;
I:=y0;
V:=v0;
A:=a0;
"V*COS(A)*T" → X0(T);
"I+V*SIN(A)*T-0.5*G*T²"→Y0(T);
CHECK(0);
FOR K FROM 1 TO 9 DO
UNCHECK(K); END;
W:=FNROOT(Y0(T)=0,T,10);
R:=X0(W);
S:=FNROOT(X0(T)=R,T,10);
H:=(V²*(SIN(A))²)/(2G);
Xmin:=0;
Xmax:=R+2;
Ymin:=-1;
Ymax:=H+2+I;
Tmin:=0;
Tmax:=S;
Tstep:=S/240;
MSGBOX("Range: "+R);
MSGBOX("Height: "+H);
STARTVIEW(1);
END;
Example:
Initial height=2.1 ft
Initial velocity=56.5 ft/s
Initial angle=46.76°
PROJECTILE(2.1,56.5,46.7), select g=32.17405 ft/s^2
Results: (in feet)
Range: 100.984422673
Height: 26.2756069802
This concludes Part 7 of the HP 39gii programming tutorial. Until next time,
Eddie
Have a great day! Much happiness!
This blog is property of Edward Shore. 2013
Monday, April 1, 2013
HP 39gii Programming Tutorial Part 6: Making programs interactive with GETKEY
Welcome to Part 6 of the HP 39gii programming tutorial. In today's session, it is all about GETKEY. The GETKEY command allows programs to be interactive during execution.
GETKEY and Key Codes
Keystroke sequence for GETKEY:
[SHIFT] [Math] [ F1 ] [ 5 ] [ 5 ]
GETKEY returns the key number of the last key pressed. If there has not been a key pressed, GETKEY returns -1.
GETKEY assigns a key code to each of the keys, starting with 1 on the upper left corner of the keyboard, the [ F1 ], going right then down, to the lower right corner of the keyboard, [ENTER], which has code 50.
The key codes for the arrow keys are as follows:
[ up ] : 9
[ right ] : 10
[ left ] : 14
[ right ] : 15
Below is the map of key codes. (Source: HP 39gii User's Guide, Edition 1)
How to use GETKEY
In today's tutorial, we will demonstrate two uses of GETKEY:
1. Execute commands until any key is pressed.
General structure:
WHILE GETKEY==-1 DO
(Commands)
END;
2. Assign certain actions to certain keys.
Let E be the code for the exit key. Press the key that has the corresponding code E, and the loop ends.
REPEAT
...
GETKEY:=K;
....
IF K==(number) (and/or other conditions) THEN
(do this); END;
...
UNTIL K==E;
Let's demonstrate this in a couple of programs.
The Program SPIN1
Spinning Game #1: Use GETKEY to repeat an action until a key (any key is pressed)
Three spinners spin between 0 and 4. Press any key at any time and try to get the highest score possible.
EXPORT SPIN1()
BEGIN
RANDSEED();
RECT();
TEXTOUT_P("Ready!",1,1);
WAIT(0.5);
WHILE GETKEY==-1 DO
RECT();
INT(RANDOM(0,5))→A;
INT(RANDOM(0,5))→B;
INT(RANDOM(0,5))→C;
TEXTOUT_P(string(A),60,53);
TEXTOUT_P(string(B),145,53);
TEXTOUT_P(string(C),230,53);
WAIT(0.25);
END;
A+B+C→S;
TEXTOUT_P("Your score is "+string(S)+".",1,73);
WAIT(0);
END;
The MOVEM Program
The program MOVEM uses GETKEY to make an interactive program. Use the arrow keys to move the "M" character around. Exit the program with by pressing the [ F6 ] key.
The characters of the large font has a size of 10 x 12 pixels.
To erase a character at a position, use
TEXTOUT_P(character, x, y, font size, 3)
The last 3 tells the 39gii to type the character in white. This is necessary to give the "M" movement in MOVEM.
EXPORT MOVEM()
BEGIN
RECT();
A:=100;
B:=60;
REPEAT
TEXTOUT_P("M",A,B,2);
K:=GETKEY;
IF K≠1 THEN
TEXTOUT_P("M",A,B,2,3);
END;
IF K==14 AND A>0 THEN
A:=A-10; END;
IF K==10 AND A<240 THEN
A:=A+10; END;
IF K==15 AND B<108 THEN
B:=B+12; END;
IF K==9 AND B>0 THEN
B:=B-12; END;
UNTIL K==5;
END;
The screen shot below shows MOVEM in action. The can be a base for a simple game.
Next time, in Part 7, we'll show you how to use App commands in a program.
This blog is property of Edward Shore. 2013
TI-84+ Color Silver Edition vs Casio Prizm Color Command Comparison
Hello everyone! Is it April already? I could swear 2013 just started.
This is just a comparison between two graphing calculators: TI-84+ Color Silver Edition (TI-84+ CSE) vs Casio Prizm (fx-CG 10/20) Color Command Comparison. I own one of both models (no joke!) and enjoy using them. But it's fun to compare the two - even if it is out of pure curiosity.
Year of first release:
Casio Prizm: 2011
TI-84+ CSE: 2013
Colors Available:
Casio Prizm, 8 colors: Black, Blue, Red, Magenta, Green, Cyan, Yellow, White
TI-84+ CSE, 15 colors: Blue, Red, Black, Magenta, Green, Orange, Brown, Navy, Light Blue, Yellow, White, Light Gray, Medium Gray, Gray, Dark Gray
How to access the colors:
Casio Prizm: [SHIFT] [ 5 ] (FORMAT)
TI-84+ CSE: Either the last sub-menu of the VARS menu or if you are in program editing mode, the third sub-menu of the PRGM menu
Here are some of the programming commands that use color:
Output Text - Terminal/Home Screen:
Casio Prizm:
color "Text" (◢ if desired)
color Locate row, col, "text"
TI-84+CSE:
N/A
Text on the Graph Screen:
Casio Prizm: 187 × 379
color Text row, column, "text"/expression
TI-84+ CSE: 265 × 165
TextColor(color)
Text(row, column, "text"/expression)
Drawing Lines: (Points)
Casio Prizm:
Plot/Line-Color color
FLine x1, y1, x2, y2
TI-84+ CSE
Line(x1,y1,x2,y2,color)
Drawing Circles: point (x,y) with radius r
Casio Prizm:
color Circle x,y,r
TI-84+ CSE:
Circle(x,y,r,color)
Plotting Functions
Casio Prizm:
"f(X)"→Y# (Y from the VARS GRAPH menu)
Set-G Color color, #
DrawGraph
TI-84+ CSE:
"f(X)" → Y# (Y from the VARS Y-VARS menu)
GraphColor(#, color)
DispGraph
Happy programming and graphing!
Eddie
This blog is property of Edward Shore. 2013
Casio fx-5800P and TI-84 Plus CE: Simplify Radicals and August TI-95 ProCalc Month
Casio fx-5800P and TI-84 Plus CE: Simplify Radicals and August TI-95 ProCalc Month Introduction The program SIMPRAD simplifies ra... | {
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1,859,565,821,587,728,600 |
Results 1 to 3 of 3
Thread: another trig id problem
1. #1
Newbie
Joined
Oct 2011
Posts
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another trig id problem
How do I do this?
(1+sinx - sin^2x)/(cosx) = cosx + tanx
This makes no sense....
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2. #2
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Re: another trig id problem
Quote Originally Posted by dazedandmathfused View Post
How do I do this?
(1+sinx - sin^2x)/(cosx) = cosx + tanx
\frac{1+\sin{x}-\sin^2{x}}{\cos{x}} =
\frac{1+\sin{x}-(1-\cos^2{x})}{\cos{x}} =
\frac{\sin{x}+\cos^2{x}}{\cos{x}} =
\frac{\sin{x}}{\cos{x}} + \frac{\cos^2{x}}{\cos{x}} = ?
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3. #3
Newbie
Joined
Oct 2011
Posts
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Re: another trig id problem
Quote Originally Posted by skeeter View Post
\frac{1+\sin{x}-\sin^2{x}}{\cos{x}} =
\frac{1+\sin{x}-(1-\cos^2{x})}{\cos{x}} =
\frac{\sin{x}+\cos^2{x}}{\cos{x}} =
\frac{\sin{x}}{\cos{x}} + \frac{\cos^2{x}}{\cos{x}} = ?
Thank you this identity trig is hard.
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-1,579,692,104,544,999,700 | Rounding, Addition, and Subtraction
Students estimate quantities using rounding and develop fluency with the standard algorithm of addition and subtraction. Students focus on the precision of their calculations, and use them to solve real-world problems.
Math
Unit 1
3rd Grade
Unit Summary
In the first unit of 3rd grade, students will build on their understanding of the structure of the place value system from 2nd grade (MP.7) to estimate values by rounding them (3.NBT.1) and develop fluency with the standard algorithm of addition and subtraction (3.NBT.2). Throughout the unit, students attend to the precision of their calculations (MP.6) and use them to solve real-world problems (MP.4).
In 2nd grade, students developed an understanding of the structure of the base-ten system as based in repeated bundling in groups of 10. With this deepened understanding of the place value system, students "add and subtract within 1000, with composing and decomposing, and they understand and explain the reasoning of the processes they use" (NBT Progressions, p. 8). These processes and strategies include concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction (2.NBT.7). As such, at the end of 2nd grade, students are able to add and subtract within 1,000 using a variety of strategies including algorithms, but are not yet fluent with them.
Because all students are not yet fluent with place value strategies, Unit 1 starts off by reinforcing some of the place value understanding that students learned in 2nd grade. Students use this sense of magnitude and the idea of benchmark numbers to first place numbers on number lines of various endpoints and intervals, and next use those number lines as a model to help them round two-digit numbers to the tens place as well as three-digit numbers to the hundreds and tens place (3.NBT.1). Next, students focus on developing their fluency with the addition and subtraction algorithms up to 1,000, making connections to the place value understandings and other models they learned in 2nd grade (3.NBT.2). Last, the unit culminates in a synthesis of all learning thus far in the unit, in which students solve one- and two-step word problems involving addition and subtraction and use rounding to assess the reasonableness of their answer (3.OA.8), connecting the NBT and OA domains. These skills are developed further and built upon in subsequent units in which students estimate and solve two-step word problems that also involve multiplication and division.
This builds toward an even deeper understanding of the place value system that students learn in 4th Grade Math. In 4th grade, students learn about multiplicative comparison, i.e., a value being x times as many as another value. Thus, students’ understanding of the place value system is more precisely refined as "a digit in one place represents ten times what it represents in the place to its right" (4.NBT.1, emphasis ours). Further, students learn to round any multi-digit number to any place. They also use the standard algorithm to solve addition and subtraction problems to the new place values they encounter at this grade level, namely, to one million. While the majority of the content learned in this unit is additional cluster content, they are deeply important skills necessary to be proficient with the major work of the grade with 3.OA.8, as well as a foundation for rounding and the standard algorithms used to any place value learned in 4th grade (4.NBT.1—4) and depended on for many grade levels after that.
Pacing: 16 instructional days (14 lessons, 1 flex day, 1 assessment day)
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Assessment
The following assessments accompany Unit 1.
Pre-Unit
Have students complete the Pre-Unit Assessment and Pre-Unit Student Self-Assessment before starting the unit. Use the Pre-Unit Assessment Analysis Guide to identify gaps in foundational understanding and map out a plan for learning acceleration throughout the unit.
Mid-Unit
Have students complete the Mid-Unit Assessment after lesson 7.
Post-Unit
Use the resources below to assess student understanding of the unit content and action plan for future units.
Expanded Assessment Package
Use student data to drive instruction with an expanded suite of assessments. Unlock Pre-Unit and Mid-Unit Assessments, and detailed Assessment Analysis Guides to help assess foundational skills, progress with unit content, and help inform your planning.
Unit Prep
Intellectual Prep
Unit Launch
Before you teach this unit, unpack the standards, big ideas, and connections to prior and future content through our guided intellectual preparation process. Each Unit Launch includes a series of short videos, targeted readings, and opportunities for action planning to ensure you're prepared to support every student.
Intellectual Prep for All Units
• Read and annotate "Unit Summary" and "Essential Understandings" portion of the unit plan.
• Do all the Target Tasks and annotate them with the "Unit Summary" and "Essential Understandings" in mind.
• Take the Post-Unit Assessment.
Unit-Specific Intellectual Prep
• Read the following table that includes models used in this unit.
pictorial or concrete base ten blocks
Example: Represent 342 with base ten blocks.
place value chart
Ten thousands Thousands Hundreds Tens Ones
4 8 0 5 0
number line
standard algorithm for addition
standard algorithm for subtraction
tape diagram
Example: A grocery store sells 172 red apples and 86 green apples. How many apples did the grocery store sell?
Essential Understandings
• When rounding a number, the goal is to approximate the number by the closest number with no units of smaller value (e.g., so 456 rounded to the nearest ten is 460 and to the nearest hundred is 500). When a number that is being rounded has a 5 in the place being considered and 0 in all smaller places, it is equidistant from the two benchmarks. Therefore, it is simply a convention that the number is rounded to the greater benchmark.
• Fluently adding and subtracting depends on a deep knowledge of the place value system and flexibility in solving. The numbers themselves should dictate which strategy is used, as sometimes that may be the standard algorithm, and other times it may be a mental strategy (such as converting 199 + 456 to 200 + 455).
• The standard algorithm for addition and subtraction is based on the idea of needing to add like-units together and the idea that one can regroup 1 of any unit to be 10 of the next smallest unit (and vice versa).
• Rounding values before computing with them can result in an estimate that is either too high or too low, depending on how the numbers were originally rounded, what computation is being performed, and where in the equation or expression it’s located.
• Rounding numbers can help one to determine whether an answer is reasonable, based on whether the estimate is close to the computed answer or not.
• Making sense of problems and persevering in solving them is an important practice when solving word problems. Key words do not always indicate the correct operation.
Vocabulary
algorithm
approximate/approximation
approximately equal sign, $${\approx}$$
digit
estimate/estimation
place
reasonable
round
value
To see all the vocabulary for Unit 1, view our 3rd Grade Vocabulary Glossary.
Materials
• Tic-Tac-Toe Boards (1 per pair of students)
• Optional: Markers or crayons (2 of different colors per pair of students) — Students could use a pen and a pencil instead.
• Optional: Paper base ten blocks (Total of 9 per student or small group) — Students might not need these depending on their reliance on concrete materials.
• Optional: Base ten blocks (Maximum of about 5 hundreds, 20 tens, 20 ones per student or small group) — Students might not need these depending on their reliance on concrete materials. Or, students can use Paper base ten blocks cut into units instead.
• Thousands Place Value Chart (Total of 6 per student) — Students might need more or less depending on their reliance on this tool.
Unit Practice
Word Problems and Fluency Activities
Help students strengthen their application and fluency skills with daily word problem practice and content-aligned fluency activities.
Lesson Map
Topic A: Foundations of Place Value
Topic B: Rounding to the Nearest Ten and Hundred
Topic C: Addition and Subtraction Within 1,000
Common Core Standards
Key
Major Cluster
Supporting Cluster
Additional Cluster
Core Standards
Number and Operations in Base Ten
• 3.NBT.A.1 — Use place value understanding to round whole numbers to the nearest 10 or 100.
• 3.NBT.A.2 — Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction.
Operations and Algebraic Thinking
• 3.OA.D.8 — Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. This standard is limited to problems posed with whole numbers and having whole-number answers; students should know how to perform operations in the conventional order when there are no parentheses to specify a particular order (Order of Operations).
Foundational Standards
Measurement and Data
• 2.MD.B.6
Number and Operations in Base Ten
• 2.NBT.A.1
• 2.NBT.A.2
• 2.NBT.A.3
• 2.NBT.B.5
• 2.NBT.B.7
• 2.NBT.B.8
• 2.NBT.B.9
Operations and Algebraic Thinking
• 2.OA.A.1
Future Standards
Number and Operations in Base Ten
• 4.NBT.A.3
• 4.NBT.B.4
Standards for Mathematical Practice
• CCSS.MATH.PRACTICE.MP1 — Make sense of problems and persevere in solving them.
• CCSS.MATH.PRACTICE.MP2 — Reason abstractly and quantitatively.
• CCSS.MATH.PRACTICE.MP3 — Construct viable arguments and critique the reasoning of others.
• CCSS.MATH.PRACTICE.MP4 — Model with mathematics.
• CCSS.MATH.PRACTICE.MP5 — Use appropriate tools strategically.
• CCSS.MATH.PRACTICE.MP6 — Attend to precision.
• CCSS.MATH.PRACTICE.MP7 — Look for and make use of structure.
• CCSS.MATH.PRACTICE.MP8 — Look for and express regularity in repeated reasoning.
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Unit 2
Multiplication and Division, Part 1
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-3,529,202,359,628,893,700 | Answers
Solutions by everydaycalculation.com
Answers.everydaycalculation.com » Multiply fractions
Multiply 35/9 with 35/4
1st number: 3 8/9, 2nd number: 8 3/4
This multiplication involving fractions can also be rephrased as "What is 35/9 of 8 3/4?"
35/9 × 35/4 is 1225/36.
Steps for multiplying fractions
1. Simply multiply the numerators and denominators separately:
2. 35/9 × 35/4 = 35 × 35/9 × 4 = 1225/36
3. In mixed form: 341/36
MathStep (Works offline)
Download our mobile app and learn to work with fractions in your own time:
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-5,588,542,338,421,319,000 | Register to reply
Vector Calculus: Area and Mass of a Region
by Kushwoho44
Tags: calculus, mass, region, vector
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Kushwoho44
#1
Dec1-12, 02:40 AM
P: 10
Hi y'all.
Here is exactly what is stated on the theory page of my book:
Example: Area of a Region
The area of a region R in the xy-plane corresponds to the case where f(x,y)=1.
Area of R= ∫∫dR
Example: Mass of a Region
The mass of a region R in the xy-plane with mass density per unit area ρ(x,y) is given by:
Mass of R= ∫∫ρdR
I'm not at all understanding this first part of the theory, why is it that the area of the region R in the xy-plane is the case f(x,y)=1 and how did they obtain that express for the area R?
All help is immensely appreciated as I'm tearing my hair out over this.
Thanks!
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HallsofIvy
#2
Dec1-12, 08:44 AM
Math
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P: 39,542
What do you understand about integrals? Most people are first introduced to integrals in the for [itex]\int_a^b f(x)dx[/itex] where it is defined as "the area of the region bounded by the graphs of y= f(x), y= 0, x= a, and x= b".
Once you start dealing with double integrals, since [itex]f(x)= \int_0^{f(x)} dy[/itex], it is easy to see that we can write that integral, and so that area, as [itex]\int_a^b\int_0^{f(x)} dy dx[/itex]. From that we can see that "dxdy" acts as a "differential of area". That is, we find the area of any region by integrating [itex]\int\int dx dy[/itex] over that region. Similarly, in three dimensions, we can find the volume of a region by integrating [itex]\int\int\int dxdydz[/itex] over that region.
Another way of reaching the same idea is to divide the region into small rectangles with sides parallel to the x and y axes and identifying the lengths of the sides as "[itex]\Delta x[/itex]" and "[itex]\Delta y[/itex]". Of course, the area of each such rectangle is [itex]\Delta x\Delta y[/itex] and the area of the whole region can be approximated by [itex]\sum \Delta x\Delta y[/itex]. "Approximate" because some of the region, near the bounds, will not fit neatly into those rectangles. But we can make it exact by taking the limit as the size of [itex]\Delta x[/itex] and [itex]\Delta y[/itex] go to 0. Of course, I have no idea what method your texts or courses used to introduce the double integral so I cannot be more precise.
Kushwoho44
#3
Dec3-12, 09:01 PM
P: 10
Thanks for that, the stupid part I wasn't understanding was that f(x,y) maps onto z.
Makes perfect sense now.
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Thread: Integration of 2/(3sin(2x)+4cos(2x))
1. #1
Member ssadi's Avatar
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Integration of 2/(3sin(2x)+4cos(2x))
Hello:
I am stuck at this sum:
\int\frac{2}{3 sin 2x+4 cos 2x}
Help please
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2. #2
Air
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Quote Originally Posted by ssadi View Post
Hello:
I am stuck at this sum:
\int\frac{2}{3 sin 2x+4 cos 2x}
Help please
\int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x
Let u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}
\therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u
You have to use T-Substitution and solve: t=\tan \left(\frac{u}{2}\right)
Note: \cos u = \frac{1-t^2}{1+t^2} and \sin u = \frac{2t}{1+t^2}.
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3. #3
Member ssadi's Avatar
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Quote Originally Posted by Air View Post
\int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x
Let u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}
\therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u
You have to use T-Substitution and solve: t=\tan \left(\frac{u}{2}\right)
Note: \cos u = \frac{1-t^2}{1+t^2} and \sin u = \frac{2t}{1+t^2}.
The t substitution, it props up everywhere and yet no mention of its uses are in the book. Thanks a lot.
t=tan x, can you give me any site address where it and the places where it can apply is discussed, I have yet to get used to this substitution.
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4. #4
Member ssadi's Avatar
Joined
Oct 2008
Posts
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Quote Originally Posted by Air View Post
\int\frac{2}{3 \sin 2x+4 \cos 2x}\ \mathrm{d}x
Let u=2x \implies \mathrm{d}x= \frac{\mathrm{d}u}{2}
\therefore \int \frac{1}{3 \sin u+4 \cos u}\ \mathrm{d}u
You have to use T-Substitution and solve: t=\tan \left(\frac{u}{2}\right)
Note: \cos u = \frac{1-t^2}{1+t^2} and \sin u = \frac{2t}{1+t^2}.
I noticed I gave you your first thanks
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5. #5
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Thanks
15
Actually, those 2x work well, since
\begin{aligned}<br />
4\cos 2x+3\sin 2x&=2\left( 2{{\cos }^{2}}x-2{{\sin }^{2}}x+3\sin x\cos x \right) \\ <br />
& =2\Big(\left( 2{{\cos }^{2}}x-\sin x\cos x \right)+\left( 4\sin x\cos x-2{{\sin }^{2}}x \right)\Big) \\ <br />
& =2(2\cos x-\sin x)(2\sin x+\cos x).<br />
\end{aligned}
Then the integral becomes,
\int{\frac{dx}{(2\cos x-\sin x)(2\sin x+\cos x)}}=\int{\frac{{{(2\cos x-\sin x)}^{2}}+{{(2\sin x+\cos x)}^{2}}}{5(2\cos x-\sin x)(2\sin x+\cos x)}\,dx}, which equals \frac{1}{5}\left( \int{\frac{2\cos x-\sin x}{2\sin x+\cos x}\,dx}+\int{\frac{2\sin x+\cos x}{2\cos x-\sin x}\,dx} \right),
and these are natural logs.
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5,848,670,646,355,449,000 | topic badge
Middle Years
4.03 Percentage of a quantity
Worksheet
Percentage of a quantity without a calculator
1
Evaluate the following to the nearest 5 cents:
a
10\% of \$16.20
b
10\% of \$696.60
c
10\% of \$955.97
d
5\% of \$41.70
e
5\% of \$319.80
f
5\% of \$466.11
g
1\% of \$66.20
h
1\% of \$14.37
i
1\% of \$888.60
j
15\% of \$86
2
Evaluate the following:
a
10\% of 5607
b
25\% of 2900
c
50\% of 1800
d
28\% of 5000
e
25\% of 7900 grams
f
30\% of 7 metres
3
Consider the following quantities:
• 55\% of 10 metres
• 55\% of 1 metre
• 5\% of 100 metres
• 50\% of 10 metres
State the largest quantity.
4
Answer the following questions in order to find 45\% of 5 hours without a calculator.
a
How many minutes are there in 5 hours?
b
Find 10\% of 300 minutes.
c
Find 5\% of 300 minutes.
d
Hence, find 45\% of 300 minutes.
5
What percentage of the squares are shaded?
6
Use rounding to approximate 7.8\% of \$89\,099.99.
Percentage of a quantity with a calculator
7
Evaluate the following to the nearest 5 cents:
a
10\% of \$321.06
b
26\% of \$126.60
8
Evaluate the following to two decimal places, if necessary:
a
51.3\% of 240
b
59\% of 440 kilometres
c
5.2\% of 80 kilograms
d
900\% of 90 metres
e
14\% of 100 grams
f
74\% of 4600 kilometres
g
16\% of 4590 \text{ kg}
h
2\dfrac{1}{4}\% of 8500
9
Evaluate the following in metres:
a
25\% of 2 kilometres
b
12.5\% of 2 kilometres
c
37.5\% of 2 kilometres
10
Evaluate the following quantities as improper fractions:
a
24\% of 272
b
51\% of 169
11
Evaluate 61\% of 496, as a mixed number.
Applications
12
A salesperson earns a 13\% commission on their total sales each week. In one week, their sales amounted to \$640.
a
Find 10\% of their total sales.
b
Find 3\% of their total sales, correct to two decimal places.
c
Hence, find the total commission they made that week, correct to two decimal places.
13
A telephone marketer earns a 13\% commission on their total sales each week. In one week, their sales amounted to \$970.
a
Find 10\% of their total sales.
b
Find 3\% of their total sales, correct to two decimal places.
c
Hence, find the total commission they made this week, correct to two decimal places.
14
A car manufacturer estimated that 17\% of 2700 cars produced are faulty. The steps to find 17\% of 2700 cars are given in random order.
When put in the correct order, which step should come second?
A
\dfrac{45\,900}{100}
B
17\% \times 2700
C
459
D
\dfrac{17}{100} \times 2700
15
Lisa scored 70\% on her Maths exam, which was marked out of 140. What was Lisa's actual mark out of 140?
16
Jay scored 92\% of a possible 600 marks. He had aimed to score 96\% of the marks. How many more marks were required to reach his goal?
17
Valentina has completed 25\% of the necessary 70 hours of pilot training.
a
Find 25\% of 70 hours, correct to one decimal place.
b
Find 2.5\% of 70 hours, correct to two decimal places.
c
Quentin has completed 27.5\% hours of the necessary training. Find how many hours he has completed, correct to two decimal places.
18
Liam works 0.8 days for every day of the five day work week.
a
What percentage of the work week does Liam work?
b
What percentage of the five day week does Liam not work?
c
What percentage of the whole 7 day week does he work? Express your answer as a percentage correct to one decimal place.
19
Ben is going to purchase some sports gear on layby. This involves paying some money as a deposit, and paying the remainder later. The price of the gear is \$65.
a
If he needs to pay 25\% deposit, how much is this?
b
Calculate the remaining balance on the layby purchase.
20
When tickets to a football match went on sale, 29\% of the tickets were purchased in the first hour.
a
If the stadium seats 58\,000 people, find the number of seats still available after the first hour.
b
If the stadium seats 60\,000 people, find the number of seats still available after the first hour.
21
A theatre has a capacity of 1000 people. On opening night the theatre was 95\% full. On the second night it was 70\% full. Find the total number of people that attended the theatre on the first two nights.
22
In a recent survey of planes landing at a major airport, it was observed that 19\% of the flights were delayed less than an hour and 3\% of the flights were delayed an hour or more but less than two hours.
a
Find the percentage of flights that are delayed less than two hours.
b
If 9000 flights arrive at the airport in a day, how many flights are delayed less than two hours?
23
A frozen apple pie weighing 331 grams is advertised as being 90\% fat free. How many grams of fat are there in the pie? Round your answer to one decimal place.
24
Valentina heated 500 \text{ mL} of solution. After a few minutes, 25\% of the solution had evaporated. Find the quantity of the solution that remained.
25
Ellie bought a 454 \text{ mL} drink that claimed to be orange juice. In the ingredients list it said that orange juice made up 17\% of the drink. To estimate the amount of orange juice in the drink, which of the following quantities would give the closest answer:
A
10\% \times 454
B
20\% \times 454
C
10\% \times 400
26
Inflation between the years 2009 and 2010 was 1.7\%, such that a purchase made in 2009 for \$100 would be worth \$101.70 in 2010. A product was valued at \$3200 in 2009.
a
Find 1\% of the product's value.
b
Find 7\% of the product's value.
c
Find 0.7\% of \$3200.
d
What is 1.7\% worth?
e
Find the value of the product in 2010.
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-3,288,199,409,118,780,400 | Park City Mathematics Institute
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Zome: Truncated 120-cell
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The giant Zome model pictured here is a "truncated 120-cell."
click on the image to view a larger image
A 120-cell is a four dimensional polytope related to the dodecahedron. In the same way a polyhedron has polygon faces, a 4d polytope has polyhedra "hyperfaces" -- here, 120 dodecahedra. In the first week we made a projection ("shadow") of a 120-cell in just white & warm colors.
The new object is what happens when each of the 330 nodes is replaced by a tetrahedron. Shaving down a polyhedron node gives a new polygon face; truncating a polytope node gives a new polyhedron.
There are beautiful symmetries to be seen from various points of view: 2, 3, 4, 6-fold symmetries, 'tunnels,' and much more.
Construction notes: 1260 nodes, 780 B1 struts, 800 Y1, 480 R1, and 600 RO (projects like this prompted Zome to make these shorter reds). It helps to have a 120 cell as a guide (see the book Zome Geometry). Study the nodes from the center outward, which partitions them as 330 = 20 + 20 + 30 + 60 + 60 + 60 + 20 + 60. Each node will be replaced by one of eight types of tetrahedra, detailed in step #3 of Zome: Model of the Month - April 2007 (for a different project). Make the tetrahedra first -- this will use all the nodes, so that everything else is connecting these with struts. Conveniently, the tetrahedra types are numbered 1 through 8 corresponding to their order from center to boundary. Build from the inside out. We succeeded with only a few missteps; the wonderfully engineered Zome system forces many of the decisions.
all the type 1 tetrahedra
all type 1 & 2 tetrahedra
all type 1, 2, & 3 tetrahedra
all type 1, 2, 3, and 4 tetrahedra
the completed object
Unless otherwise indicated all of these materials are licensed under a
Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
_____________________________________
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© 2001 - 2018 Park City Mathematics Institute
IAS/Park City Mathematics Institute is an outreach program of the Institute for Advanced Study, 1 Einstein Drive, Princeton, NJ 08540.
Send questions or comments to: Suzanne Alejandre and Jim King
With major funding from Math for America
With generous support from Robert and Lynn Johnston | {
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-5,140,406,082,927,048,000 | What is the Greatest Common Factor (GCF) of 2 and 1214?
Are you on the hunt for the GCF of 2 and 1214? Since you're on this page I'd guess so! In this quick guide, we'll walk you through how to calculate the greatest common factor for any numbers you need to check. Let's jump in!
First off, if you're in a rush, here's the answer to the question "what is the GCF of 2 and 1214?":
GCF of 2 and 1214 = 2
What is the Greatest Common Factor?
Put simply, the GCF of a set of whole numbers is the largest positive integer (i.e whole number and not a decimal) that divides evenly into all of the numbers in the set. It's also commonly known as:
• Greatest Common Denominator (GCD)
• Highest Common Factor (HCF)
• Greatest Common Divisor (GCD)
There are a number of different ways to calculate the GCF of a set of numbers depending how many numbers you have and how large they are.
For most school problems or uses, you can look at the factors of the numbers and find the greatest common factor that way. For 2 and 1214 those factors look like this:
• Factors for 2: 1 and 2
• Factors for 1214: 1, 2, 607, and 1214
As you can see when you list out the factors of each number, 2 is the greatest number that 2 and 1214 divides into.
Prime Factors
As the numbers get larger, or you want to compare multiple numbers at the same time to find the GCF, you can see how listing out all of the factors would become too much. To fix this, you can use prime factors.
List out all of the prime factors for each number:
• Prime Factors for 2: 2
• Prime Factors for 1214: 2 and 607
Now that we have the list of prime factors, we need to find any which are common for each number.
In this case, there is only one common prime factor, 2. Since there are no others, the greatest common factor is this prime factor:
GCF = 2
Find the GCF Using Euclid's Algorithm
The final method for calculating the GCF of 2 and 1214 is to use Euclid's algorithm. This is a more complicated way of calculating the greatest common factor and is really only used by GCD calculators.
If you want to learn more about the algorithm and perhaps try it yourself, take a look at the Wikipedia page.
Hopefully you've learned a little math today and understand how to calculate the GCD of numbers. Grab a pencil and paper and give it a try for yourself. (or just use our GCD calculator - we won't tell anyone!) | {
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991,809,900,394,457,600 | 10
2
A comment in A007018 a(n) = a(n-1)^2 + a(n-1), a(0)=1 claims
Subsequence of squarefree numbers (A005117). - Reinhard Zumkeller, Nov 15 2004
Is it really so?
As far as I know, it is open problem if a polynomial $f \in \mathbb{Z[x]}$ of degree $\ge 5$ can be squarefree infinitely often (some source require $f$ to be irreducible).
If the OEIS comment is correct, the sequence will give infinite family of (irreducible) polynomials which are squarefree infinitely often.
Let $a_n$ is OEIS A007018. Set $a_n = x$ and $$f(x)=a_{n+4}=x \cdot (x + 1) \cdot (x^{2} + x + 1) \cdot (x^{4} + 2 x^{3} + 2 x^{2} + x + 1) \\ \cdot (x^{8} + 4 x^{7} + 8 x^{6} + 10 x^{5} + 9 x^{4} + 6 x^{3} + 3 x^{2} + x + 1)$$
$f(a_n)=a_{n+4}$ will be squarefree infinitely often (including the irreducible degree 8 factor) and iterating $x \mapsto x^2+x$ will produce infinite family of polynomials with this property.
Added For reference of squarefree values of polynomials the search terms are square free values of polynomials. E.g. here p.1 and here 11. Squarefree values of polynomials.
flag
Do you have a reference for that question, about some polynomials of high degree being squarefree infinitely often? – Per Alexandersson Jan 4 at 13:00
6
As noticed here: mathworld.wolfram.com/SylvestersSequence.html, it is not known whether all of them are square-free. – Ilya Bogdanov Jan 4 at 13:15
4
Sylvester’s sequence is your $a(n)+1$. Since $a(n)=a(n-1)(a(n-1)+1)$, your sequence consists of squarefree numbers if and only if Sylvester’s sequence does. – Emil Jeřábek Jan 4 at 13:35
1
For the first 100 primes, the sequence becomes periodic modulo $p^2$ without passing through $0$. – David Speyer Jan 4 at 13:39
2
Hm, Mathworld only says that the sequence is squarefree for the first 10e15... – Per Alexandersson Jan 4 at 14:09
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anonymous
• one year ago
2+3x=4y+5z ______ 6 for Z... All answers are some variation of z=12+18x+24y _____________ 5 I'm doing the line system because I was afraid it could be confused with just 5z/6 instead of 4y+5z all over 6.
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1. jim_thompson5910
• one year ago
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you can type it like this 2+3x=(4y+5z)/6
2. anonymous
• one year ago
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Okay, I just wasn't sure if putting those in ()'s changed the equation at all.
3. jim_thompson5910
• one year ago
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anyways, first you multiply both sides by 6 2+3x=(4y+5z)/6 6*(2+3x)=6*(4y+5z)/6 12+18x=4y+5z
4. jim_thompson5910
• one year ago
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what's next?
5. anonymous
• one year ago
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12+18x=24y+30z?
6. anonymous
• one year ago
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Then 12+18x-24y=30z, right?
7. anonymous
• one year ago
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Should I not have multiplied 5 and 6? Or what am I supposed to do now?
8. jim_thompson5910
• one year ago
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when you multiply (4y+5z)/6 by 6, the two '6' terms divide to 1 1 times anything is the same number, so the '6's go away more or less
9. jim_thompson5910
• one year ago
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(4y+5z)/6 will turn into 4y+5z after you multiply the right side by 6
10. anonymous
• one year ago
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Okay. So z=(12+18x-4y)/5.
11. jim_thompson5910
• one year ago
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correct
12. anonymous
• one year ago
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Thank you!
13. jim_thompson5910
• one year ago
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sure thing
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45,636,924,240,807,080 | X2 < 10 what are the inequalities?
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X2 < 10 what are the inequalities?. Are You guys own this kind of concern?, If do then please check the good tips below this line:\r\n
Hello Pal,
Solving linear inequalities is almost exactly like solving linear equations.
Solve x + 3 < 0.
If they`d given me "x + 3 = 0", I`d have known how to solve: I would have subtracted 3 from both sides. I can do the same thing here:
Then the solution is:
x < –3
"Set notation" writes the solution as a set of points. The above solution would be written in set notation as "x ", which is pronounced as "the set of all x-values, such that x is a real number, and x is less than minus three". The simpler form of this notation would be something like " x
< –3", which is pronounced as "all x such that x is less than minus three".
"Interval notation" writes the solution as an interval (that is, as a section or length along the number line). The above solution, "x < –3", would be written as "", which is pronounced as "the interval from negative infinity to minus three", or just "minus infinity to minus three". Interval notation is easier to write than to pronounce, because of the ambiguity regarding whether or not the endpoints are included in the interval. (To denote, for instance, "x < –3", the interval would be written "", which would be pronounced as "minus infinity through (not just "to") minus three" or "minus infinity to minus three, inclusive", meaning that –3 would be included. The right-parenthesis in the "x < –3" case indicated that the –3 was not included; the right-bracket in the "x < –3" case indicates that it is.)
The last "notation" is more of an illustration. You may be directed to "graph" the solution. This means that you would draw the number line, and then highlight the portion that is included in the solution. First, you would mark off the edge of the solution interval, in this case being –3. Since –3 is not included in the solution (this is a "less than", remember, not a "less than or equal to"), you would mark this point with an open dot or with an open parenthesis pointing in the direction of the rest of the solution interval:
Inequalities crop up frequently in math and science. You may be familiar with plotting these on a number line using open and closed circles to represent strict inequality and "or-equal-to" inequalities, or you may be new to plotting inequalities on a number line altogether. No matter the convention used (circle or brackets and parentheses), it displays a set of values a variable can take on while obeying an inequality.
1.Mark all strict inequalities with parenthesis, using the appropriate parenthesis. All "greater than" inequalities should be marked with a left parenthesis, and all "less than" inequalities with a right parenthesis. For example, mark "x > 3" with "(" at 3, and mark "x < 5" with ")" at 5.
2.Mark all "or-equal-to" inequalities with square brackets, using the appropriate bracket. All "greater-than-or-equal-to" inequalities should be marked with a left bracket and all "less-than-or-equal-to" inequalities with a right bracket. For example, mark "x >= 4" with "[" at 4, and mark "x <= 1" with "]" at 1.
3.Embolden the region of the number line between the highest left bracket or parenthesis and the lowest right bracket or parenthesis. Many inequalities describe a variable as being both greater than one number and less than another as in "4 < x <= 8," and emboldening the region between the "(" at 4 and the "]" at 8 indicates that x may be anywhere in that region. If only one bound is provided, like "x > 4," then embolden until the end of the number line.
Thanks !
Have a Great Day.
Professor.A | {
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