text
stringlengths 83
79.5k
|
|---|
H: Probability inequality for sum of two random variables
Does anyone know if the following inequalities are right? If yes, what is the reference for them?
For random variable $x$ and $y$:
\begin{equation}
\mathbb{P}(x+y \ge x_0 + y_0) \le \mathbb{P}(\{x\ge x_0\} \text{or} \{y\ge y_0\}).
\end{equation}
AI: It should be clear that
$$\mathbb{P}(\{x < x_0\} \text{ and } \{y < y_0\} ) \leq \mathbb{P}(\{x+y < x_0 + y_0 \})$$ holds.
Using
$$ \mathbb{P}(\{x \geq x_0\} \text{ or } \{y \geq y_0\} ) = 1- \mathbb{P}(\{x < x_0\} \text{ and } \{y < y_0\} )$$
and
$$ \mathbb{P}(\{x+y \geq x_0 + y_0 \}) = 1 - \mathbb{P}(\{x+y < x_0 + y_0 \})$$
yields your first statement. Repeating this should give the more general statement.
|
H: Find closed form for $1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, \ldots$
Is there any closed form for the following?
$$1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, \ldots$$
I tried to find one, but I failed.
I saw solution on Wolfram Alpha, but I didn't understand it:
Generating function: $$\mathcal G_n(a_n)(z)=\dfrac{z+1}{(z-1)^2(z^2+z+1)}$$
What does that function mean, and how does it give me the solution to my question?
AI: Although Did gave you a very nice closed form, I'll explain a bit about what Wolfram|alpha gave you. A generating function $f$ for a sequence $\{a_k\}_{k = 0}^{\infty}$ is the formal power series defined by
$$
f(z) = \sum_{k = 0}^{\infty}a_k z^k.
$$
So, a generating function for $\{a_k\}$ does not give you the $n$-th term of the sequence when you plug in $n$ - that's what the closed form does. Rather, the generating function encodes information about your sequence by using the terms as coefficients. Generating functions can be used to solve counting problems and to come up with closed forms for sequences; I don't know of any really good references for generating functions, but I'm sure someone can guide you in the right direction if that's what you're interested in.
It's often desirable to have a closed form for the function $f$ rather than just the series, and it appears your sequence has a generating function with a relatively nice closed form, so that's what Wolfram|alpha gave you. If you write out the series explicitly and perform some careful manipulations, you'll find that
$$
1 + 2z + 2z^2 + 3z^3 + 4z^4 + 4z^5 + 5z^6 + \dots = \frac{z + 1}{(z - 1)^2(z^2 + z + 1)}.
$$
(For some region around the origin and provided Wolfram didn't mess up!)
Edit: Here's the way I would derive the generating function for your sequence. The following calculation is formal, and without regard for convergence - as it usually is with generating functions.
\begin{align*}
G(z) &= 1 + 2z + 2z^2 + 3z^3 + 4z^4 + 4z^5 + 5z^6 + \dots\\
&= 1 + z + z^2 + \ldots + z(1 + z + 2z^2 + 3z^3 + 3z^4 + 4z^5 + \dots)\\
&= \frac{1}{1 - z} + z(1 + z + z^2 + z^3 + \dots) + z(z^2 + 2z^3 + 2z^4 + 3z^5 + \dots)\\
&= \frac{1}{1 - z} + \frac{z}{1 - z} + z^3(1 + 2z + 2z^2 + 3z^3 + \dots)\\
&= \frac{1}{1 - z} + \frac{z}{1 - z} + z^3 G(z)
\end{align*}
So,
\begin{align*}
G(z)(1 - z^3) &= \frac{1}{1 - z} + \frac{z}{1 - z}\\
G(z)(1 - z^3) &= \frac{z + 1}{1 - z}\\
G(z) &= \frac{z + 1}{(1 - z)(1 - z^3)}\\
&= \frac{z + 1}{(1 - z)((1-z) (1+z+z^2))}\\
&= \frac{z + 1}{(1 - z)^2(z^2 + z + 1)}.
\end{align*}
|
H: Discrete dynamic models
We have the equation
$$x_{n+1} = ax_n(1-x_n) - v_n$$
Why are there only fixed points for $(a-1)^2 - 4av_0 \geq 0$?
Show that if $ 1<a<4$, there are 2 fixed points with $0<p_1 < p_2 <1$
For the first problem, I was able to calculate that the fixed points would be given by:
$$ p_{1,2} = \dfrac{a-1}{2a} \pm \sqrt{(\dfrac{a-1}{2a})^2 - \dfrac{v_o}{a}}$$
But after that I'm stuck. The second problem I have no idea how to prove but I think that is because I don't understand the first problem.
AI: Taking from where you stopped, consider the expression under the root and factor out $(2a)^2 = 4a^2$:
$$
\begin{split}
p_\pm = \frac{a-1}{2a} \pm \sqrt{\left( \frac{a-1}{2a}\right)^2 - \frac{v_0}{a}}
= \frac{a-1}{2a} \pm \frac{1}{2a} \sqrt{(a-1)^2 - 4av_0},
\end{split}
$$
which is a real number provided $(a-1)^2 \geq 4av_0$, other wise the $\sqrt{\cdot}$ is complex.
2nd Problem
For the last problem it suffices to prove that
$$
0 \leq \frac{a-1}{2a} \pm \frac{1}{2a} \sqrt{(a-1)^2 - 4av_0} \leq 1,
$$
or equivalently that
$$
0 \leq a-1 \pm \sqrt{(a-1)^2 - 4av_0} \leq 2a.
$$
Right inequality:
(assuming $1\leq a \leq 4$), certainly $a - 1 \leq a$ and also
$$
\sqrt{(a-1)^2 - 4av_0} \leq \sqrt{(a-1)^2} = (a-1) < a,
$$
so the sum (and surely the difference) should be no more than $a+a = 2a$.
Left inequality
Since
$$
\sqrt{(a-1)^2 - 4av_0} \leq \sqrt{(a-1)^2} = a-1
$$
and so subtracting both sides from $a-1$ yields the result.
|
H: calculate riemann sum of sin to proof limit proposition
$$\lim_{n \to \infty}\frac1n\sum_{k=1}^n\sin(\frac{k\pi}{n})$$
I'm having trouble expressing $\sin(x)$ differently here in order to calculate the riemann sum.
I want to show that this converges to $\frac{2}{\pi}$ so it equals to $\int_0^1 \sin(x\pi)$.
Is there any easy way to express $\sin(x)$ different here?
Added:
$$\frac{1}{2i}(\frac{\cos(\frac{(n+1)\pi}{n})+i\sin(\frac{(n+1)\pi}{n})-\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})}{\cos(\frac{\pi}{n})+i\sin(\frac{\pi}{n})-1}-\\\frac{\cos(-\frac{(n+1)\pi}{n})+i\sin(-\frac{(n+1)\pi}{n})-\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})}{\cos(-\frac{\pi}{n})+i\sin(-\frac{\pi}{n})-1})$$
AI: The sum can have the closed form
$$\sum_{k=1}^n\sin\left(\frac{k\pi}{n}\right)= \frac{\sin \left( {\frac {\pi }{n}} \right)}{ \left( 1-\cos \left( {\frac {
\pi }{n}} \right) \right)} .$$
Added: To prove the above identity, you need the two facts
1) $$ \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}, $$
2) $$ \sum_{k=1}^{n}x^m={\frac {{x}^{n+1}-x}{-1+x}}. $$
|
H: What does $\sum_{n\in\mathbb N}\frac{n^x}{x^n}$ converge to when $x\in\mathbb R^+$ and $x>1$?
What does $\sum_{n\in\mathbb N}\frac{n^x}{x^n}$ converge to when $x\in\mathbb R^+$ and $x>1$?
I'm looking for a hint of how to tackle this problem.
AI: From the definition of the polylogarithm this is :
$$\sum_{n\in\mathbb N^{*}}\frac{n^x}{x^n}=\sum_{n\in\mathbb N^{*}}\frac{\left(\frac 1x\right)^n}{n^{-x}}=\operatorname{Li}_{-x}\left(\frac 1x\right)$$
This simplifies for non negative integer values of $x$ (and $-1$) as seen here.
|
H: Non symmetric urn and balls problem.
Consider an urn containing $N$ balls, of which m are white and $N - m$ are black. Balls are randomly selected from the urn according to the following rule:
If a black balls is selected, it is "observed" and the it is returned to the urn.
If a white balls is selected, it is "observed" and then it is replaced with a black ball.
Let $X$ be a random variable that counts the number of white balls in a sample of size $n$.
Determine the expected value of $X$.
To clariy the problem consider the case $N=8$, $m=3$ and $n=2$, then:
$P\{X=0\} = \frac{5}{8}\times\frac{5}{8}=\frac{25}{64}$
$P\{X=1\} = \frac{5}{8}\times\frac{3}{8} + \frac{3}{8}\times\frac{6}{8}=\frac{33}{64}$
$P\{X=2\} = \frac{3}{8}\times\frac{2}{8} = \frac{6}{64}$
It is easy to see that the total number of balls in the urn remains fixed, and that "equivalent" permutations of balls do not occur with the same probability, so the usual procedure of counting permutations with the desired property does not work.
Any help would be great.
Thanks in advance.
Edit: Fixed a typo.
AI: Let the white balls be numbered $1$ to $m$. Don't bother to number the black balls. For $i=1$ to $m$, let $X_i=1$ if ball $i$ is selected at least once, and let $X_i=0$ otherwise. Then the number of white balls recorded is $X_1+\cdots+X_m$.
The probability that ball $i$ is not selected $n$ times in a row is $\left(\frac{N-1}{N}\right)^n$. Thus
$$\Pr(X_i=1)=1-\left(\frac{N-1}{N}\right)^n.$$
This is $E(X_i)$. By the linearity of expectation, the expected number of white balls recorded is
$$m\left(1-\left(\frac{N-1}{N}\right)^n\right).$$
Remark: Trying to get at the distribution of the number of white balls recorded is likely to be a painful process. However, as in many other problems, the linearity of expectation makes the expectation quite accessible.
|
H: Closed rectangle contained in set that does not have measure zero.
I was doing some analysis when it occurred to me that if I knew the answer to the following question many of the exercises would be simpler. However, I haven't been able to make much headway. Any help will be appreciated.
Question:
If I have a set $S \subset \mathbb{R}^n$ such that $S$ is not of measure 0, is it always true that there exists some closed rectangle $C$ such that $C \subseteq S$?
AI: No. Think of $\mathbb R \backslash \mathbb Q$.
|
H: (1)Questions about differentiable functions
1)The functions $f$ and $g$: $\mathbb{R} \rightarrow \mathbb{R} $ shall be 3-times differentiable.
Calculate $(f \cdot g)^{(3)}$.
1) $(f \cdot g)'=(f'g+fg')$
$(f'g+fg')'= (f''g+f'g')+(f'g'+fg'')= f''g+2f'g'+fg''$
$(f''g+2f'g'+fg'')'=(f'''g+f''g')+2(f''g'+f'g'')+(f'g''+fg''')$
$=f'''g+3(f''g'+f'g'')+fg'''=(f \cdot g)^{(3)}$
2)Find a function f:$\mathbb{R} \rightarrow \mathbb{R} $, which is 2-times differentiable on $\mathbb{R}$
2)$f(x)=x^2$
$f'(x)=2x$ and $f''(x)=2$
Are my solutions correct or did I sth. wrong?
AI: Yes, indeed, your solutions are correct. Nice work.
Did you note the pattern that terms and the coefficients of $(fg), (fg)', (fg)'', (fg)'''$ follow?
Does Pascal's triangle look familiar?:
See Binomial Theorem and note the correlation.
|
H: Subset of Cantor set that isn't compact
How to prove that the Cantor set has a subset that is not compact? Actually, I want to prove that every infinite set $X\subset\mathbb{R}^n$ has a subset $Y$ that is not compact. If $X$ isn't bounded, then $X$ has a unbounded subset $Y$ that is not compact. If $X$ is bounded and includes some ball, then $X$ includes a open ball $Y$ that is not compact. But if $X$ is bounded and includes no ball, like the Cantor set, I don't know. I think can be easier start by Cantor set, but I'm not sure. Can you help me?
Thanks.
AI: HINT: If no point of $X$ is a limit point of $X$, then you can take $Y=X$. If there is a point $p\in X$ that is a limit point of $X$, show that there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X\setminus\{p\}$ that converges to $p$, and let $Y=\{x_n:n\in\Bbb N\}$.
This works with any infinite $X\subseteq\Bbb R^n$.
|
H: Showing that $\frac{2}{3\pi}\leq\int_{2\pi}^{3\pi}\frac{\sin x}{x} \, dx \leq \frac{1}{\pi}$
Please help showing that,
$$\frac{2}{3\pi}\leq\int_{2\pi}^{3\pi}\frac{\sin x}{x} \, dx \leq \frac{1}{\pi}$$
AI: Make use of the fact that $\dfrac1x$ is a decreasing function, to conclude what you want, i.e., for $0<a<b$ we have
$$\dfrac1b\int_a^b \sin(x) dx < \int_a^b \dfrac{\sin(x)}x dx < \dfrac1a\int_a^b \sin(x) dx$$
Also, note that
$$\int_{n \pi}^{(n+1)\pi} \sin(x)dx = (-1)^n \cdot 2$$
|
H: Proving LHS and RHS
I just came across this problem from proving an equality for combinatorics. In general if I ask you to prove
Prove that
Expression X = Expression Y
Normally wouldn't one read from left to right and start showing the left is equal to the right?
Then I thought of Euclid's elements and the definition of the equality and I thought wouldn't it be the same if i prove Expression Y = Expression X? Or would that be, by human (or western, because I know some cultures read from right to left) convention, considered "circular logic" because I am assuming the truth of the statement and only showing the result?
AI: No: I would do whatever seems most convenient and works. If it’s a matter of algebraic manipulation, this might be manipulating the lefthand side into the righthand side, but it might also be manipulating the righthand side into the lefthand side or manipulating both sides until they meet in the middle. In some cases I might even assume that the two expressions are not equal and somehow derive a contradiction.
|
H: Can I derive $i^2 \neq 1$ from a presentation $\langle i, j \mid i^4 = j^4 = 1, ij = j^3 i\rangle$ of Quaternion group $Q$?
(This question is related to the previous post I've posted few hours ago: (Dummit's AA, 1.5, P3) Are these presentations of the Quarternion group equivalent?)
I was trying to prove that the presentation
$$\langle i, j \mid i^4 = j^4 = 1, ij = j^3 i\rangle
$$
generates Quaternion group. The only thing that I couldn't derive was that $i^2 \neq 1$. I tried to derive a contradiction from the supposition $i^2 =1$ but this didn't give me a contradiction (at least not yet) but rather it brought me to an Abelian group that consists of 4 elements with every element other than 1 having order 2.
I'm not sure where I'm wrong.
Does indeed the above presentation generate an Abelian group with its order 4? (I think this can't happen according to the previous post!!)
Or $i^2 = 1$ is a indeed contradictory?
If the latter is the case, would you please show me the derivation?
AI: You're confused, I think, about the definition of a presentation of a group. That's fair enough, since the idea is slightly subtle.
Formally, the idea of a presentation is as follows: you define some generators like $i,j$ and then consider all expressions in the free group on those symbols, which are strings like $iij^{-1}iji^{-1}jjij$ and so on. Then you impose any rules which the relations, like $i^4=1$, tell you. So e.g. $iiiijjij\equiv jjij$. You also assume $ii^{-1}=i^{-1}i=1$ and so on.
You're not allowed to assume that any other rules for manipulating these things apply. Then you ask how the resulting set of strings works under concatenation. This defines unambiguously the group multiplication.
So for example $\left<i|i^n=1\right>$ is a presentation of the cyclic group of order $n$. The trivial group, even though it satisfies without contradiction the relations, is not defined by this presentation, because you cannot use the relations given to deduce $i^1=1$.
Edit: As you say, the idea is that $i^2=1$ in the group defined by the above cyclic relation ($n>2$) holds $\iff$ it is possible to derive this from the relation. The important part of the $\iff$ part is the forwards implication, the contrapositive of which is "if it cannot be shown, then it is not true". Unprovability is essentially what you need to prove! Unsurprisingly this is hard, hence the Novikov–Boone theorem.
The way to deal with these problems is to show that these relations hold $\iff$ (proving both ways) another canonical set do. Then the groups must be the same.
(As pointed out in your other question, you can deduce unprovability by explicitly constructing a model which violates the statement, though this in general is hard. For simple problems it should work.)
|
H: Weakly closed implies sequentially closed
Another problem involving the weak topology:
Let $X$ be a normed space and $A \subset X$ weakly closed. Then $A$ is sequentially closed, that is: If $(x_n) \subset A$ and $x_n \xrightarrow{w}x$, then $x \in A$.
I know this characterisation is also used as definition of weakly closed. So I guess it should be easy to prove. Yet I have trouble doing so. ;(
I tried proving it directly and towards a contradiction without success. I'm afraid my problem is a lack of understanding of weak closedness. I know how weakly open sets are generated, but this doesn't give me a concrete representation them, or of a weakly closed subset.
However, I know that $A$ is also closed with respect to the norm of $X$. Weak convergence of $(x_n)$ to $x$ means $f(x_n) \rightarrow f(x)$ for every $f \in X^*$. But it doesn't give me any statement related to the norm convergence of $(x_n)$, at least not that I know of. So the norm-closedness of $A$ doesn't really help. I could also find out that norm closedness is not sufficient for closedness with respect to weak convergence - I think convexivity has to be added to make the implication valid, is that correct?
So of course I have to fail if I weak convergence with norm convergence alone.
I've tried to work with balls around $f(x)$, too (for a contradiction). But again, the continuousness of $f$ only gives me control over $|f(x_n)-f(x)|$ if I have some bound for $||x_n-x||$. And I want it the other way around. I have a feeling that this way is wrong because it would need some implication between weak and norm convergence that I know isn't there...
It makes me mad that the proof should be rather simple, yet I'm not able to do it. Some hint please!? :(
AI: HINT: If you understand the product topology on an arbitrary product of topological spaces, you can use that to get a better handle on the weak topology on $X$: like the weak topology in $X$, the product topology is an example of an initial topology.
The weak topology on $X$ is the coarsest topology on $X$ that makes all $f\in X'$ continuous. For each $f\in X'$ let $R_f$ be a copy of $\Bbb R$ with the usual topology. Then the map
$$\varphi:X\to\prod_{f\in X'}R_f:x\mapsto\langle f(x):f\in X'\rangle$$
is an embedding. Thus, we can use the usual base for the product topology to define a base for the weak topology on $X$:
Let $\tau$ be the topology on $\Bbb R$. For each finite $F\subseteq X'$ and $U:F\to\tau$ let $$B(F,U)=\{x\in X:f(x)\in U(f)\text{ for each }f\in F\}\;.$$ The family of all such sets $B(F,U)$ is a base for the weak topology on $X$.
Now fix $x\in X$. For each finite $F\subseteq X'$ and $\epsilon>0$ let $$B(F,\epsilon)=\{y\in X:|f(x)-f(y)|<\epsilon\text{ for each }f\in F\}\;;$$ the family of all such sets is a local nbhd base at $x$ in the weak topology. In particular, if $A\subseteq X$ is weakly closed, and $x\notin A$, then there are a finite $F\subseteq X'$ and an $\epsilon>0$ such that $B(F,\epsilon)\cap A=\varnothing$, i.e., such that
$$\text{for each }y\in A\text{ there is an }f\in F\text{ such that }|f(x)-f(y)|\ge\epsilon\;.\tag{1}$$
Now suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $A$ that converges weakly to $x$. Use $(1)$ and the finiteness of $F$ to $x$ to get a contradiction.
|
H: Union of connected subsets is connected if intersection is nonempty
Let $\mathscr{F}$ be a collection of connected subsets of a metric space $M$ such that $\bigcap\mathscr{F}\ne\emptyset$. Prove that $\bigcup\mathscr{F}$ is connected.
If $\bigcup\mathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty subsets $A,B$. Let $x$ be a point in $\bigcap\mathscr{F}$. Then either $x\in A$ or $x\in B$. I don't know where to go from here.
AI: HINT: You’re actually about halfway there, though you omitted an important qualification of $A$ and $B$: if $\bigcup\mathscr{F}$ is not connected, then it can be partitioned into two disjoint, non-empty, relatively open subsets $A$ and $B$. Now fix $x\in\bigcap\mathscr{F}$, and without loss of generality assume that $x\in A$. $B\ne\varnothing$, so pick any $y\in B$. Then there is some $F\in\mathscr{F}$ such that $y\in F$, and of course $x\in F$. Thus, $x\in A\cap F$, and $y\in B\cap F$, so $A\cap F\ne\varnothing\ne B\cap F$. Why is this a contradiction?
|
H: problem with partial fraction decomposition
I want to do partial fraction decomposition on the following rational function:
$$\frac{1}{x^2(1+x^2)^3}$$
So I proceed as follows:
$$\begin{align}
\frac{1}{x^2(1+x^2)^3} &= \frac{A}{x} + \frac{B}{x^2} + \frac{Cx + D}{1 + x^2} + \frac{Ex + F}{(1 + x^2)^2} + \frac{Gx + H}{(1 + x^2)^3} \\
1 = &Ax(1+x^2)^3 + B(1+x^2)^3 + (Cx + D)x^2(1+x^2)^2 \\
& + (Ex + F)x^2(1+x^2) + (Gx + H)(x^2) \\
1 = &(A+C)x^7 + (B+D)x^6 + (3A+2C+E)x^5 + (3B+2D)x^4 \\
& + (3A+C+E+G)x^3 + (3B+D+H)x^2 +(A)x + (B)\\
\end{align}$$
But from this, $B=1$, $B=-D$, and $3B+2D=0$, which is an inconsistent system. What am I doing wrong?
AI: Your final expression is almost correct; you're just forgetting the $F$. The coefficient of $x^4$ should be:
$$
3B + 2D + F
$$
and the coefficient of $x^2$ should be:
$$
3B + D + F + H
$$
|
H: A subspace of $\mathcal{B}(X;Y)$ isometric to $Y$
Let $X$ and $Y$ be normed spaces. If $\mathcal{B}(X,Y)$ is the space of all continuous linear functions of $^{X}Y$, I'm trying to prove that $\mathcal{B}(X,Y)$ has a closed subspace isometric to $Y$. Can someone give me only a clue.
AI: Hint: On a basis $e_n$ of $X$, $f\in \mathcal B (X,Y)$ is basically (roughly) the same as a specification of $f_n\equiv f(e_n)\in Y$ for all $n$. That is, there is one value in $Y$ for each $n$. You want only one value in $Y$. What set of functions could you consider?
|
H: Prove that $\{x_n\}_n$ is convergent.
Here is an exercise:
Let $x_n=1+2+\dots+\frac1n-\ln n$. Prove that $\{x_n\}_n$ is convergent.
(I believe that this can be found in the site, however I cannot find immediately, so I post it here.)
The hints are much appreciated. I don't want complete proof.
Thanks for your help.
AI: Let's present a different proof than any I've seen on this site, via a picture and a hint.
Look carefully at the following and remember that $\displaystyle \int_1^x \frac 1t dt = \ln x$:
Justify that the area in blue is the limit you're looking for, and for that matter the first $n$ blue bits corresponds to the $n$th term of the sequence. Show the limit exists and is finite (which can be done entirely with the picture too).
|
H: Show that any continuous $f:[0,1] \rightarrow [0,1]$ has a fixed point $\zeta$
Be a continuous function $f:[0,1] \rightarrow [0,1]$.
Show that there is a $\zeta \in [0,1]$ with $f(\zeta)=\zeta$ ($\zeta$ is called fixed point).
Consider the function $g:[0,1] \rightarrow [-1,1]$, $g(x):= f(x)-x$.
$g$ is continuous.
Because of $f(0),f(1) \in [0,1]$ is $g(0)\geq0$ and $g(1)\leq 0$.
Because $f(0)$ has a value between $0$ and $1$, $f(0)\geq 0$.
$g(0) = f(0)-0= f(0) \geq 0 - 0 = 0$.
Because $f(1)$ has a value between $0$ and $1$, $ f(1) \leq 1 $.
$g(1)=f(1)-1 \leq 1 - 1 = 0$ $\Leftrightarrow g(0)\geq 0$ and $g(1)\leq 0 $
After the IVT: $\exists \zeta \in [0,1]:g(\zeta)=0 \Leftrightarrow f(\zeta) = \zeta $
$\zeta$ is a fixed point of $f$. $\Box$
My questions are:
Is this proof done in the correct way or have I missed something?
Is there something I can improve?
AI: Yes, your proof is pretty much correct. I might change it slightly so that it's presented better.
Let $f:[0,1] \rightarrow [0,1]$ be a continuous function. Show that there is a $\zeta \in [0,1]$ with $f(\zeta)=\zeta$ (that is, show that there exists some fixed point $\zeta$).
Consider the function $g:[0,1] \rightarrow [-1,1]$ defined by $g(x):= f(x)-x$. Note that since $f$ is continuous and polynomials are continuous, $g$ is also continuous.
Now observe that since $f(0)\in [0,1] \implies f(0) \ge 0$, we have:
$$
g(0) = f(0)-0= f(0) \geq 0
$$
Likewise, since $f(1) \in [0,1] \implies f(1) \leq 1$, we have:
$$
g(1)=f(1)-1 \leq 1 - 1 = 0
$$
Thus, since $g(1) \leq 0 \leq g(0)$, we know by IVT that $\exists \zeta \in [0,1]$ such that $g(\zeta)=0 \iff f(\zeta) = \zeta $. Hence, $\zeta$ is a fixed point of $f$, as desired. $\Box$
|
H: Jordan Canonical Form of a matrix
Determine the Jordan canonical form of the matrix: $\quad\begin{bmatrix} -7 & 9 \\ -4 & 5 \end{bmatrix}$
AI: Hints, to find the Jordan Normal Form, we can use the following approach (and others are possible too):
(1) Find the eigenvalues (we have a multiplicity $2$ eigenvalue $\lambda_{1,2} = -1$)
(2) Find the eigenvectors (we find one eigenvector and a second generalized eigenvector)
(3) Write $A = P \cdot J \cdot P^{-1}$, where $J$ is the Jordan block (eigenvalues) and P is the corresponding eigenvectors.
Updates
To find the eigenvalues/eigenvectors, we set up and solve the characteristic polynomial $|A - \lambda I| = 0$, so we get ($v_i's$ are the eigenvectors):
$\lambda_1 = -1, ~v_1 = (3, 2)$
$\lambda_2 = -1, ~v_2 = (-\dfrac{1}{2}, 0)$
Jordan Form
The JF is given by:
$$A = PJP^{-1} = \begin{bmatrix}3 & -1/2\\2 & 0\end{bmatrix} \cdot \begin{bmatrix}-1 & 1\\0 & -1\end{bmatrix} \cdot \begin{bmatrix}0 & 1/2\\-2 & 3\end{bmatrix}$$
Note what makes up $J$ and the columns of $P$.
|
H: Probabilty of A and D but not B and not C
I am trying to work a probability question and I am stumped.
Imagine there are 4 signs on the side of the road and each sign has a "chance of being seen" equal to the following
sign A: .75
sign B: .82
sign C: .87
sign D: .9
What are the chances of A AND D being seen, but NOT B and NOT C. (I saw the first and fourth signs, but I did not see the 2nd or the third).
AI: Supposing independent events, you can compute $$0.75\times 0.9\times (1-0.82)\times (1-0.87)?$$
The idea is: you want $A \wedge D\wedge \neg B \wedge \neg C$.
|
H: How should I handle the limit with the floor function?
I wanted to show that
$$ \lim_{n\to\infty} \bigg( \frac{n+\lfloor x \sqrt{n} \rfloor}{n-\lfloor x \sqrt{n} \rfloor}\bigg)^{\lfloor x \sqrt{n} \rfloor} =e^{2x^2} $$
After applying $x \sqrt{n} -1 \leq \lfloor x \sqrt{n} \rfloor \leq x \sqrt{n}$ and replacing $\sqrt{n}$ with $k$, I tried to squeeze it. It worked out quite nicely with l'Hospital's rule.
Yet soon I realized that given $n\in \mathbb N$ I can neither replace nor use l'Hospital's rule. I recalled Stolz–Cesàro theorem yet it seems to be a nightmare to apply it here.
What should I do in this particular case? Is there a way to justify the replacement and the use of the l'Hospital's rule?
AI: After thinking for a while I am ready to answer my own question:
Both the replacement and the use of l'Hospital's rule are justified here since if $\lim_{x\to\infty}f(x)$ exists then $\lim_{n\to\infty}f(n)$ exists as well (since it's merely a subsequence) and is equal the former. Yet the opposite it's true since $\lim_{n\to\infty}\cos(2\pi n)$ exists but $\lim_{x\to\infty}\cos(2\pi x)$ doesn't.
|
H: Solving limits of trig functions.
a) $$\lim_{x\to\pi/2^+} \frac{\ln(x-\pi/2)}{\tan(x)}$$
b) $$\lim_{x\to 0^+} \frac{\ln(x)}{\arctan(x)}$$
How does one approach this question?
AI: Second Question: As $x\to 0^+$, $\arctan x$ approaches $0$ through positive values. In the meantime, $\ln x$ becomes very large negative, so for positive $x$ near $0$, the ratio is very large negative.
Depending on the conventions you use, you can say that the limit is $-\infty$ or that it doesn't exist.
First Question: This was asked and answered a few hours ago.
|
H: Verification and hint for my answers to a basic statistics table
Sorry if this question seems very basic. I tried to find the answer by goolging. But because I do not know the correct keyword cannot find anything. This is a basic statistics problem. I attempted questions but I need them to get verified. I also do not know the answer of question 5.
The following table was drawn up to show the results of a quiz:
$$
\begin{array}{c|l}
\text{x} & \text{f} & \text{fx} & \text{c.f.} \\
\hline
5 & 8 & & \\
6 & 12 & & \\
7 & 15 & & \\
8 & 17 & & \\
9 & 9 & & \\
\end{array}
$$
Complete the $f \times x$ column.
Complete the cumulative frequency column.
Calculate the mean correct to one decimal place.
What is the mode?
What is the Median?
My answers:
$$
\begin{array}{c|l}
\text{x} & \text{f} & \text{fx} & \text{c.f.} \\
\hline
5 & 8 & 40 & 40 \\
6 & 12 & 72 & 112 \\
7 & 15 & 105 & 217 \\
8 & 17 & 136 & 353 \\
9 & 9 & 81 & 434 \\
\end{array}
$$
Answer of Q3. $(40 + 72 + 105 + 136 + 81) \div 5 = 86.8$
Answer of Q4. Mode is 8 because it has frequency of 17.
AI: Answer to Q3: You should divide by the sum of the frequencies and not by $5$.
Answer to Q5: Identify the $x$ value where the cumulative percentage crosses 50%. In other words, at what $x$ value do the cumulative number of observations equal 50% of total observations?
|
H: Can the semidirect product of two groups be abelian group?
while I was working through the examples of semidirect products of Dummit and Foote, I thought that it's possible to show that any semdirect product of two groups can't be abelian if the this semidirect product is not the direct product.
Here is my simple idea:
Suppose $H,K$ are two groups and $H\rtimes K$ be the semidirect product of $H$ and $K$. Let $f:K \rightarrow Aut(H)$ be our homomorphism from $K$ into $Aut(H)$, now we know that $H \unlhd H\rtimes K$ but not necessarily $K$.
If $H\rtimes K$ is abelian, then every subgroup of it is normal, so $K$ must be normal but if $K$ is normal then $f$ is the trivial homomorphism and so the semidirect product turns into the direct product, so the semdidirect product in this case is the direct product, which is a contradiction since we supposed that this semidirect product of $H,K$ is not their direct product.
Is this true? Or have I made a mistake and there exists a counterexample?
AI: You are correct. A semi-direct product of $H$ and $K$ is abelian iff $H$ is abelian, $K$ is abelian, and the semi-direct product is explicitly direct (the action is trivial).
When $H$ and $K$ are not abelian you can have bizarre things like $H \rtimes K$ is (isomorphic to) a direct product of $H$ and $K$, even though the action is not trivial. Take $H=K$ for example (to be non-abelian of order 6 to make it nice and clear). I learned this for $A_5 \rtimes A_5$, and was recently reminded it was true for all non-abelian groups $H=K$.
|
H: Let $\alpha$ be an ordinal then $\alpha=\cup\alpha$ or $\alpha=s(\cup\alpha)$.
Let $\alpha$ be an ordinal then $\alpha=\cup\alpha$ or $\alpha=s(\cup\alpha)$.
Attempt:
Since $\alpha$ is transitive $\alpha=\cup\alpha$ (here we are done) or $\alpha\supsetneq\cup\alpha$.
If $\alpha\supsetneq\cup\alpha$, since $\cup\alpha$ is an ordinal we have $\cup\alpha\in\alpha$. Regularity axiom implies $s(\cup\alpha)=\alpha$ (here we are done) or $s(\cup\alpha)\in\alpha$.
But I can't see any contradiction assuming $(\cup\alpha\in\alpha) \land (s(\cup\alpha)\in\alpha)$.
AI: To get a contradiction from $s(\bigcup\alpha)\in\alpha$, note that, by definition of $s$, we also have $\bigcup\alpha\in s(\bigcup\alpha)$. So $\bigcup\alpha$ is a member of a member of $\alpha$. By definition of $\bigcup$, we have $\bigcup\alpha\in\bigcup\alpha$, contradicting the axiom of regularity.
|
H: Normalizing a continuous distribution
I work at a help/tutoring center at my university. Today a kid came in with this problem. I've only studied math and haven't drifted into physics, but he had this problem:
Let $P(x)=Ne^{-\frac{|x|}{a}}$. Then:
(a) Find $N$ such that $P(x)$ is properly normalized.
(b) Find $\langle x \rangle$ and $\langle x^2 \rangle $
Attempt at solution:
$1=\displaystyle\int_{-\infty}^{\infty}Ne^{-\frac{|x|}{a}}=2\int_{0}^{\infty}Ne^{-\frac{x}{a}}=2\int_{0}^{\infty}Ne^{-u}a\,du=2Na\int_{0}^\infty e^{-u}du=2Na.$
So $N=\frac{1}{2a}$
I'm not sure how to do part (b). The text, Robinett's Quantum Mechanics, was a bit beyond me. Could I get some guidance on how to solve this problem...it seems really interesting.
AI: For part a) you should check that the symmetry of $P(x)$ with respect to $x$ gives
$$\int_{-\infty}^{\infty}Ne^{-|x|/a} dx=2\int_{0}^{\infty}Ne^{-x/a}dx$$
and then do a change of variable $\displaystyle u=\frac{x}{a}$
For part b) note that by definition $\displaystyle\langle f(x) \rangle=\int_{-\infty}^{\infty}f(x)Ne^{-|x|/a}dx$ and that both $x$, $x^2$ and $P(x)$ have defined symmetry with respect to $x$ and you can exploit that symmetry to know the value of some of them (in particular $\langle x \rangle$ without actually computing the integrals
|
H: Triple Scalar Product $=0\implies $Spanning a Plane? (and Checking for Colinearity)
If I have three vectors, $\vec{a},\vec{b},\vec{c}$, and their scalar triple product equals zero, that is $\vec{a}\times \vec{b} \cdot \vec{c}=0$, then I understand that it means the vectors are coplanar, and so they can not span $\mathbb{R}^3$.
My first question is can I say they span a plane? My book really doesn't talk about it much.
My second question is there a way to check for two vectors being colinear in a similar fashion?
AI: First question: No. They could span a line and still have a zero triple product.
Second question: Yes. If two 3D vectors are co-linear their cross product will be zero. Non-zero cross product means that they span a plane.
|
H: Given governing equations, determine the condition for a solution and if they exist, solve the system.
Given the governing equations:
1a + 2b = 2c
2a + 4b = d
2a + 5b = e
3a + 9b = f
What conditions are required for a solution x = [a,b]^T to exist?
If these conditions exist, solve the system for a and b.
I set up Ax=b, did some Gaussian Elimination and came up with:
1a + 2b = 2c
0a + 1b = e - 4c
0a + 0b = d - 4c
0a + 0b = f - 3e + 6c
I'm not sure how to answer the questions.... I understand that the last two equations should be my constraint equations. ?? Any help would be greatly appreciated! :)
AI: First, Gaussian eliminate a bit further. You can get rid of the $2b$ in the first equation by subtracting twice the second equation.
Now you're left with $a$ as a function of $e$ and $c$, $b$ as a function of $e$ and $c$, $d$ as a function of $c$, and $f$ as a function of $e$ and $c$. Can you turn this into a two dimensional subspace of $\mathbb{R}^6$ in which solutions exist?
(REPLY TO COMMENT)
So $d$ and $f$ are arbitrary here - they can be anything we want, and so long as $a,b,c,$ and $e$ obey the relations you've obtained there, the tuple $$\left(\begin{array}{c}a\\b\\c\\d\\e\\f\end{array}\right)=\left(\begin{array}{c}(7/2)d+(2/3)f\\(-1/2)d+(1/3)f\\(1/4)d\\d\\(-1/2)d+(1/3)f\\f\end{array}\right)$$ will be a solution to the governing equation. Note that we can separate this: $$\left(\begin{array}{c}(7/2)d+(2/3)f\\(-1/2)d+(1/3)f\\(1/4)d\\d\\(-1/2)d+(1/3)f\\f\end{array}\right)=\left(\begin{array}{c}7/2\\-1/2\\1/4\\1\\-1/2\\0\end{array}\right)d+\left(\begin{array}{c}2/3\\1/3\\0\\0\\1/3\\1\end{array}\right)f.$$
Now, let me define something for you:
Definition. Given vectors $v_1,\ldots, v_r$, the span of $v_1,\ldots, v_r$ is the set $\{\sum_{i=1}^rav_i|a_i\in\mathbb{R}\}$.
In other words, the span of vectors $v_1,\ldots, v_r$ is all vectors which are linear combinations of $v_1,\ldots, v_r$.
You'll notice that this definition relates to the above problem: my claim is that the solutions to the governing equations is the span of the vectors $$v=\left(\frac{7}{2},-\frac{1}{2},\frac{1}{4},1,-\frac{1}{2},0\right)^\intercal\text{ and }w=\left(\frac{2}{3},\frac{1}{3},0,0,\frac{1}{3},1\right)^\intercal.$$ (The relationship between "span" and the terminology "$2$-dimensional subspace" is a bit more involved, but you can read about it on wikipedia. Loosely, the span of vectors is always a subspace, and it is two dimensional because there are two vectors here which are not multiples of one another.) So there are infinitely many solutions to your governing equations, and the solutions are exactly those vectors contained in the span of $v$ and $w$.
|
H: Tensor Product is distributive.
Tensor Product is distributive.
I get stuck at the proof:
If $T$ is a $p$-tensor and $S, U$ a $q$ tensor. Then I need to show that
$$ T \otimes (S \cdot U) = (T \otimes U) \cdot (S \otimes U).$$
Denote $S(u_1, \ldots, u_q) \cdot U(w_1, \ldots, w_q) = S \cdot U (v_1, \ldots, v_q)$. Then
\begin{eqnarray*}
&&T \otimes (S \cdot U)(v_1, \ldots, v_p, v_{p+1}, \ldots, v_{p+q})\\
& =& T (v_1, \ldots, v_p) \cdot (S \cdot U)(v_{p+1}, \ldots, v_{p+q})\\
& =& T (v_1, \ldots, v_p) \cdot (S (u_{p+1}, \ldots, u_{p+q}) \cdot U(w_{p+1}, \ldots, w_{p+q}))\\
&? =& (T \otimes U) \cdot (S \otimes U)
\end{eqnarray*}
AI: Let $V$ be a vector space and $V^*$ the dual space. Suppose $\alpha,\beta \in V^*$. Let $\alpha \otimes \beta :V \times V \rightarrow \mathbb{R}$ be defined by
$$ \alpha \otimes \beta (x,y) = \alpha(x)\beta(y) $$
for all $x,y \in V$. The proof that $\alpha \otimes (\beta + \gamma) = \alpha \otimes \beta + \alpha \otimes \gamma$ is as follows: let $x,y \in V$,
$$ \begin{align} \alpha \otimes (\beta + \gamma)(x,y)
&= \alpha(x)(\beta + \gamma)(y) \\
&= \alpha(x)[\beta(y) + \gamma(y)] \\
&= \alpha(x)\beta(y) + \alpha(x)\gamma(y) \\
&= \alpha \otimes \beta(x,y) + \alpha \otimes\gamma(x,y) \\
&= (\alpha \otimes \beta + \alpha \otimes\gamma)(x,y). \\
\end{align} $$
As $x,y$ were arbitrary the identity $\alpha \otimes (\beta + \gamma) = \alpha \otimes \beta + \alpha \otimes \gamma$ follows. The steps above are justified by the point-wise defined addition of multilinear mappings and the vector space structure of $V$.
A similar proof can be given for the tensor product of other tensors. I merely give the proof for dual-vectors, or one-forms if you prefer. Hope this helps. I took the liberty of replacing your $\cdot$ with $+$.
|
H: Prove that $\exp(x)=3x$ has at least one solution for $x\in [0,1]$
Prove that $\exp(x)=3x$ has at least one solution $x \in [0,1]$.
$$e^x=3x$$
$$\Leftrightarrow e^x-3x=0$$
Let $$f(x) = e^x - 3x$$
$$f(0)=e^0 - 3 \cdot 0 = 1 > 0$$
$$f(1)= e^1-3 \cdot 1 = e - 3 < 0$$
Thus, since $f(1) < 0 < f(0) $, by the IVT:
$$\exists \zeta \in [0,1]\text{ such that }f(\zeta)=0 \Leftrightarrow e^\zeta = 3\zeta$$
$\Box$
Is that correct? Is there something I can improve?
AI: Well done. Don't be afraid, though, to use words: E.g. Why not write out "...Then there exists a $\zeta$..."
Final sentence: say (write) $\quad e^\zeta = 3\zeta,\;$ and not $\;f(\zeta) = \zeta$.
And...write out "Intermediate Value Theorem", instead of IVT. Certainly, it is fine to abbreviate when you are simply proving something in your own notes. But if you are attempting to communicate to another (e.g., homework, presentation, etc), unless you have previously made explicit what IVT means, "write it out."
Acronyms are handy for saving time while taking notes or writing on a blackboard, but if you are using them to present solutions, write out the theorem's "name" and if you must, follow with a parenthetical acronym if you need to refer to it often: E.g. "By the Intermediate Value Theorem (IVT), we know ... "
|
H: Prove "casting out nines" of an integer is equivalent to that integer modulo 9
Let $s(x)$ be an abstraction for casting out nines of integer $x$. For all integers $x$, prove $s(x) \equiv x$ mod $9$
I'm not asking for an answer more of a way to attack this problem. Can't think of where to start
AI: Hint: Because $10\equiv 1\pmod{9}$, we have $10^k\equiv 1\pmod 9$ for any non-negative integer $k$.
And note for example that $2013=2\times 10^3+0\times 10^2+1\times 10^1+3\times 10^0$.
Added: Consider for example the $4$-digit number "$N=a_3a_2a_1a_0$" where the $a_i$ are digits. For example, with the $4$-digit number $7145$ we have $a_0=5$, $a_1=4$, $a_2=1$ and $a_3=7$.
We want to show that $N$ and the sum $a_3+a_2+a_1+a_0$ have the same remainder on division by $9$, that is, are congruent to each other modulo $9$. Note that
$$N=a_3\times 10^3+a_2\times 10^2+a_1\times 10^1+a_0\times 10^0.$$ Because $10^k\equiv 1\pmod 9$, we have
$$N\equiv a_3\times 1+a_2\times 1+a_1\times 1+a_0\times 1=a_3+a_2+a_1+a_0\pmod{9},$$
which is what we wanted to show.
We can also do it without congruences. For $$(1000a_3+100a_2+10a_1+a_0)-(a_3+a_2+a_1+a_0)=a_3(999)+a_2(99)+a_1(9),$$
and the right-hand side is clearly divisible by $9$.
|
H: Why the two results are different?
I'm trying to evaluate $\int_C\dfrac{z+1}{z^2+9}$ where $C$ is a positively oriented simple closed contour containning $z=\pm3i.$
Method 1: Let $f(z)=\dfrac{z+1}{z^2+9}.$ Then the given integral is equal to $2\pi i\text{ Res}_{z=0}\dfrac{1}{z^2}f\left(\dfrac{1}{z}\right)$$=2\pi i\text{ Res}_{z=0}\dfrac{z(z+1)}{9z^2+1}=0$ since $\dfrac{z(z+1)}{9z^2+1}$ being analytic at $0$ it has a taylor series representation about $0$ so that the coefficient of $\dfrac{1}{z}$ is $0.$
Method 2: $f(z)=\dfrac{\phi_1(z)}{z-3i}$ where $\phi_1(z)=\dfrac{z+1}{z+3i}$ is analytic and non-zero at $z=3i.$ So $f$ has a simple pole at $3i$ and $\text{Res}_{z=3i}f(z)=\phi_1(3i)=(3i+1)/6i=(3-i)/6$
Similarly $\text{Res}_{z=3i}f(z)=(3+i)/6$
So the given integral equal $2\pi i\times$sum of residues $=2\pi i$
Why the two results are different?
AI: In your first method, you forgot to divide by ${z^2}$ after finding $f\left(\frac{1}{z}\right)$.
The correct function is $\frac{z+1}{z(9z+1)}$, and multiplying by $z$ then setting $z=0$ shows that the residue is $1$, as expected.
|
H: Simple yet tricky trigonometry
This might seem silly to ask, but how can I solve a trigonometry problem for the unknown $h$ in the form of:
$x + 45 = h/\tan 30$ and
$x = h/\tan 50$
AI: Since $h$ is the variable we are solving for, express both equations as functions of $h$:
$$x + 45 = \dfrac h{\tan (30^\circ)} \iff x = \dfrac h{\tan(30^\circ)} - 45\tag{1}$$
$$x = \dfrac h{\tan (50^\circ)}\tag{2}$$
Now, since the right-hand and left-hand sides of the equation are each equal to $x$, they are equal to each other. So we have:
$$\dfrac h{\tan(30^\circ)} - 45 = \dfrac h{\tan (50^\circ)} $$
Now we have an equation in one variable, and can solve for $h$. Once you solve for $h$, you can evaluate the tangent of the angles, perhaps using an identity or two?
(Recall that $\tan(45)$ and $\tan(50)$ are constants (have a constant value) - their values do not depend here on $h$ or $x$).
|
H: Proof of bipartite graph formula
I've come across a question that has got be stuck for hours.
I need to proof that:
Let $G$ be a graph $=(V,E)$, a bipartite graph with $n$ vertices and $e$ edges. Show that $$e\leqslant \left\lceil \frac{n}{2} \right\rceil \cdot \left\lfloor \frac{n}{2} \right\rfloor\;.$$
I can't understand that even though this graph
qualifies as bipartite, it doesn't follow the equation showed above.
Any helps would be appreciated.
p.s the red dots indicate a vertex. Sorry if it isn't clear, my paint skills arent top of the art.
AI: That graph is $K_4$, the complete graph on $4$ vertices; it’s not bipartite. Remember, two vertices in the same part of a bipartite graph are not connected by an edge. In $K_4$ every pair of vertices is connected by an edge, so no two vertices could be in the same part.
HINT: Suppose that $G=\langle V,E\rangle$ is a bipartite graph on $n$ vertice, and let $A$ and $B$ be the two parts of $V$, so that $E$ is the set of edges $v_0v_1$ with $v_0\in A$ and $v_1\in B$. Let $a=|A|$ and $b=|B|$; then $e=ab$. (Why?) To prove the result, you need to show that $a$ and $b$ are positive integers whose sum is $n$, then $$ab\le\left\lceil\frac{n}2\right\rceil\left\lfloor\frac{n}2\right\rfloor\;.\tag{1}$$
If $n$ is even, say $n=2m$, then $(1)$ is just $ab\le m^2$, and since $a+b=n$, you can even simplify it to $a(2m-a)\le m^2$. That inequality can in turn be written $m^2-2am+a^2\ge 0$. Can you simplify the lefthand side in a way that shows that this inequality is always true? Once you’ve done that, you’ll have shown that $(1)$ is true when $n$ is even.
To finish the proof, suppose that $n$ is odd, say $n=2m+1$. Now express $\left\lceil\frac{n}2\right\rceil$ and $\left\lfloor\frac{n}2\right\rfloor$ in terms of $m$, and try to make a similar argument.
|
H: Map values 0:90 to 2:0
I have values that range from 0 and 90, and I need to map their values to between 2 and 0. Examples would be 90 = 2, 0 = 0, 45 = 1. How can I go about getting these values converted down to the proper range?
Edit: I messed up on my examples.
Examples should be:
90 = 0, 0 = 2, 45 - 1
AI: Use a straight line with equation:
$$y=2-\frac{1}{45}x $$
A general recipe is as follows:
Suppose that you have two points $(x_1,y_1)$ and $(x_2, y_2)$ and you need to identify a straight line that goes through those points. The equation for a straight line can be written as: $y = mx+c$ where $m$ is the slope and $c$ is the $y$-axis intercept when $x=0$.
Thus, we have:
$y_1 = mx_1+c$
and
$y_2 = mx_2 +c$
Solving the above system of equations gives us:
$m = \frac{y_2-y_1}{x_2-x_1}$
and
$c = y_1 - \frac{y_2-y_1}{x_2-x_1}x_1$
|
H: Handshake problem. discrete math.
at a party, 25 guests mingle and shake hands. prove that at least one guest must have shaken hands with an even number of guests.
AI: HINT: Number the guests $1$ through $25$. Let $a_k$ be the number of guests with whom guest $k$ shook hands. Suppose that the numbers $a_1,a_2,\dots,a_{25}$ are all odd.
Is $\sum_{k=1}^{25}a_k$ odd, or is it even?
Each handshake involves two people, so each handshake is counted twice in $\sum_{k=1}^{25}a_k$, once for each of the two people involved. What does this imply about whether $\sum_{k=1}^{25}a_k$ is odd or even?
|
H: Show that $(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z)$.
If $x>0,y>0,z>0$ and $x+y+z=1$, prove that $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z).$$
Trial: Here $$(1+x)(1+y)(1+z)\ge 8(1-x)(1-y)(1-z) \\ \implies (1+x)(1+y)(1+z)\ge 8(y+z)(x+z)(x+y)$$ I am unable to solve the problem. Please help.
AI: $$(1+x)(1+y)(1+z) \ge 8(1-x)(1-y)(1-z) \Leftrightarrow $$
$$(2x+y+z)(x+2y+z)(x+y+2z) \ge 8(y+z)(x+z)(x+y)$$
Let $a=x+y, b=x+z, c=y+z$. Then the inequality to prove is
$$(a+b)(a+c)(b+c) \ge 8abc \,,$$
Which follows immediately from AM-GM:
$$a+b \ge 2 \sqrt{ab}$$
$$a+c \ge 2 \sqrt{ac}$$
$$b+c \ge 2 \sqrt{bc}$$
Simplification The solution above can be simplified the following way:
By AM-GM
$$2\sqrt{(1-y)(1-z)}\le 1-y+1-z=1+x \,.$$
Similarly
$$2\sqrt{(1-x)(1-z)}\le 1+y \,.$$
$$2\sqrt{(1-x)(1-y)}\le 1+z \,.$$
|
H: how do you determine the derivative is true or false?
$$\frac{d}{dx} \cos(y^2)=-2y \sin(y^2)$$
I need to find out if this statement is true or false. My answer is false, but I'm not sure.It is false because the variables disagree.
AI: More formally, use the Chain Rule:
$$
\frac{d\cos(y^2)}{dx} = -\sin(y^2) \frac{d y^2}{dx} = -2y \sin(y^2) \frac{dy}{dx}.
$$
If you are not told anything about the relationship of $y$ to $x$, they are assumed independent, so $y$ is constant with respect to $x$ and $dy/dx = 0$....
|
H: Proving a lemma about the union of sets, linear independence, spans
OK, here’s another lemma I’m being asked to prove and I am trying to see if I am in the right ballpark.
Let $L = \{y, y_1, y_2, y_3, \ldots, y_m\}$ be a linearly independent subset of $V$, a vector space over field $K$.
Let $S = \{x_1, x_2, \ldots, x_r\}$ be a subset of $V$ such that $\{y_1, y_2, \ldots, y_m, x_1, x_2, x_3, \ldots, x_r\} = \big(L \backslash \{y\}\big) \cup\ S$ spans $V$.
Show that $\exists$ an element $x_i\in\ S$ s.t. $L \cup (S \backslash \{x_i\}) = \{y, y_1, y_2, \ldots, y_m, x_1, x_2, x_3, \ldots, x_{i-1}, x_i, x_{i+1}, x_n\}$ spans $V$.
I approached it like this:
Given that $\big(L \backslash \{y\}\big) \cup S$ spans $V$, that means that $y$ is linearly dependent, and thus a linear combination of $L \cup\ S$, so it can be expressed as:
$$y = \lambda_1y_1 + \lambda_2y_2 +\ldots + \lambda_m y_m + \mu_1x_1 + \mu_2x_2+ \ldots + \mu_rx_r.$$
By construction, $L$ is a linearly independent set, so $\exists$ scalars $c$ such that $cy + c_1y_1 + c_2y_2 + \ldots + c_my_m = 0 $.
But since we have $L$ as linearly independent, that means the coefficients have to be zero. That’s true for both $L$ as a linearly independent set with scalar coefficient $c$ and $\lambda_i$.
That in turn reduces $y = \lambda_1y_1 + \lambda_2y_2 + \ldots + \lambda_m y_m + \mu_1x_1 + \mu_2x_2+ \ldots + \mu_rx_r$ to $y = \mu_1x_1 + \mu_2x_2+ \ldots + \mu_rx_r$.
But $y$ isn’t linearly independent. That means at least one of the coefficients $\mu_i \neq 0$, therefore there is some element $x_i \neq 0$.
The span of $V$ is the set of all vectors in $V$. And we’re trying to show that the union of $L$ and $S$ with that nonzero $x_i$ subtracted is the span of $V$. But that’s where I get a little stuck; I know I’ve almost got the last step.
(Also, any critique on how to write this proof “properly” is much appreciated.)
AI: You don't know that $\{ y, x_1, \dots, x_r \}$ is a linearly dependent subset! You are misunderstanding the definitions. The problem is that when you write
$$
y = \lambda_1 y_1 + \dots + \lambda_m y_m + \mu_1 x_1 + \dots + \mu_r x_r,
$$
you cannot assume that all coefficients $\lambda$ will be zero. Think of $V = \mathbb R^2$ with basis $\{e_1,e_2\}$, $L = \{ e_1, e_2 \}$ and $S = \{e_1 + e_2\}$ and see what happens with your equations.
The right proof would go like this : since $\{ y_1,\dots,y_m , x_1, \dots, x_r\}$ spans $V$, we have
$$
y = \lambda_1 y_1 + \dots + \lambda_m y_m + \mu_1 x_1 + \dots + \mu_r x_r,
$$
and we cannot have all $\mu_i$'s equal to zero, because this would give a linear dependence on the set $\{y,y_1,\dots, y_m\}$. Therefore one of the $\mu$'s is not zero. Say $\mu_i \neq 0$. Then
$$
x_i = \frac {-1}{\mu_i}\left( \lambda_1 y_1 + \dots + \lambda_m y_m + \mu_1 x_1 + \dots + 0 x_i + \dots + \mu_r x_r - y \right),
$$
which means even though the set $\{y,y_1, \dots, y_m, x_1, \dots, x_r \}$ spans $V$, the element $x_i$ is not necessary to span $V$, so we can remove it from this set and still have a generating set for $V$.
Hope that helps,
|
H: Closed and bounded subsets of a complete metric space
Let $\{X_n\}$ be a sequence of closed and bounded subsets of a complete metric space such that $X_n\supset X_{n+1}$ for every positive integer $n$ and $\lim_{n\rightarrow\infty}(\text{diam }X_n)=0$. Prove that $\bigcap_{n=1}^{\infty}X_n$ contains exactly one point.
Every closed subset of a complete metric is complete, so $X_1,X_2,\ldots$ are complete. Also, the intersection of closed subsets is closed, so $\bigcap_{n=1}^{\infty}X_n$ is closed. Suppose there are two points $a,b$ in $\bigcap_{n=1}^{\infty}X_n$. Then $\lim_{n\rightarrow\infty}(\text{diam }X_n)\geq d(a,b)$, a contradiction. Now suppose there is no point in $\bigcap_{n=1}^{\infty}X_n$. How is this a contradiction?
AI: HINT: For each $n\in\Bbb Z^+$ let $x_n\in X_n$. Show that $\langle x_n:n\in\Bbb Z^+\rangle$ is a Cauchy sequence. Where must its limit be?
|
H: Aquaintance problem in discrete math. induction proof.
I'm supposed to prove this by induction. I already proved it by contradiction, but I am lost on how to set it up for induction.
Prove that if at least two people are at a party, at least two of them know the same number of people. We assume that if person 1 knows person 2, then person 2 knows person 1. We also assume that a person doesn't count themselves in how many people they know.
Any ideas?
AI: Since the statement is only true for $n\geq2$ people, the case $n=2$ must be the base case. As usual the base case is easily proved: the two people either both know $0$ or both know $1$ other person.
Now suppose $n>2$ and that it has been shown that among any group of $n-1$ people there are always two that know the same number of other people in that group. We distinguish two cases.
At least one of the $n$ people knows none of the other people. As the relation is symmetric, the remaining $n-1$ people don't know that person either, and considering just them as a group does no change the number of people known by any of them; the induction hypothesis shows that two among them know the same number of other people.
Everyone knows at least one of the other people. Now we have $n$ people, each of which know a number $i\in\{1,2,\ldots,n-1\}$ other people. By the pigeonhole principle, at least one value of $i$ is shared by two or more people.
Added. In reply to the comment by Brian M. Scott, I note that when wrinting this proof I originally had three bullets: the two above, and a second inductive case between them for when at least one person knows all the others (removing that person decreases everyone else's number of acquaintences by$~1$, and hence does not change their relation of having the same number of them). However I realised that this reduced the number of remaining values for $i$ to $n-2$, giving me more than I actually needed to apply the pigeonhole principle, so I then dropped the second bullet.
Taking this up differently, this also allows to do things without using induction at all (though that would invalidate the answer): one clearly cannot both have a person that knowns nobody, and a person that knows everybody (would they know each other?), so in any case only $n-1$ of the $n$ values $0,1,\ldots,n-1$ are left to be taken, and the pigeonhole principle can be applied directly.
|
H: changing summation limits
Let us say we have the following summation
$$\tag 0 \sum_{k=0}^{\infty }\sum_{l=0}^{\infty }g_{k}h_{l}\delta (t-(l+k)T)$$
Now, we let $n = l + k$. Then $l = n - k$ which becomes
$$\tag 1 \sum_{n=0}^{\infty }(\sum_{k=0}^{n }g_{k}h_{n-k})\delta (t-nT)$$
Note: (1) is the correct answer according here: http://puu.sh/3jQxR.png
(1) = solution manual answer
(2) = my solved answer
Now, if I try to solve on my own starting from (0), if we replace every "l" with "n-k", we can end up with (my answer when I solve on my own)
$$\tag 2 \sum_{k=0}^{\infty }\sum_{n-k=0}^{\infty }g_{k}h_{n-k}\delta (t-(n)T)$$
The question is, how can we make my answer (2) similar to the answer in solution manual (1)?
Thanks
AI: Let me first explain how one arrives at $(1)$. The summation in $(0)$ is over all pairs $\langle k,\ell\rangle$ of non-negative integers. For non-negative integers $k$ and $\ell$ let
$$a(k,\ell)=g_kh_\ell\delta\big(t-(k+\ell)T\big)\;;$$
then we can abbreviate $(0)$ as
$$\sum_{k=0}^\infty\sum_{\ell=0}^\infty a(k,\ell)\;.\tag{3}$$
Think of the numbers $a(k,\ell)$ as being laid out in an infinite array with rows and columns each numbered $0,1,2,\dots$; $a(k,\ell)$ is the number in row $k$, column $\ell$. $(3)$ calculates the sum of the entries in this array by first summing each row: the sum of the entries in row $k$ is $\sum_{\ell=0}^\infty a(k,\ell)$. It then adds up the row sums to get the grand total. (I’m assuming that in context there is information to guarantee convergence of these infinite sums.)
But assuming convergence we can sum the entries in other orders. In particular, for each non-negative integer $n$ we can sum along the anti-diagonal $k+\ell=n$, and we can then add up these anti-diagonal sums. (The diagonal of this infinite array consists of the entries $a(k,k)$; the set of entries $a(k,\ell)$ with $k+\ell=n$ for some fixed $n$ is perpendicular to the diagonal, i.e., an anti-diagonal.) When $n=0$, $\langle k,\ell\rangle$ can only be $\langle 0,0\rangle$; when $n=1$, $\langle k,\ell\rangle$ can be $\langle 0,1\rangle$ or $\langle 1,0\rangle$; when $n=2$, $\langle k,\ell\rangle$ can be $\langle 0,2\rangle$, $\langle 1,1\rangle$, or $\langle 2,0\rangle$; and so on. As you can see, for each value of $n\ge 0$, the possible values of $k$ are $0,1,\dots,n$, and of course in each case $\ell=n-k$.
If $n=k+\ell$, we have $a(k,\ell)=a(k,n-k)$, and the sum of the entries along the $n$-anti-diagonal is $\sum_{k=0}^n a(k,n-k)$. The grand total is the sum of these anti-diagonal sums, which is
$$\sum_{n=0}^\infty\sum_{k=0}^n a(k,n-k)\;.\tag{4}$$
Now
$$a(k,\ell)=a(k,n-k)=g_kh_{n-k}\delta(t-nT)\;,$$
so when we expand $(4)$ we get
$$\sum_{n=0}^\infty\sum_{k=0}^ng_kh_{n-k}\delta(t-nT)\;.\tag{5}$$
The nice thing about this arrangement is that the expression $\delta(t-nT)$ depends only on $n$, not on $k$, so we can factor it out of the inner summation to get
$$\sum_{n=0}^\infty\left(\sum_{k=0}^ng_kh_{n-k}\right)\delta(t-nT)\;,\tag{1}$$
which we could just as well write
$$\sum_{n=0}^\infty\delta(t-nT)\sum_{k=0}^ng_kh_{n-k}\;.$$
Your $(2)$ is not wrong, but it will be easier to work with if you observe that $n-k\ge 0$ if and only if $n\ge k$, so that it can be rewritten
$$\sum_{k=0}^\infty\sum_{n=k}^\infty g_kh_{n-k}\delta(t-nT)\;.$$
Getting the clutter out of the way again, we can write this as
$$\sum_{k=0}^\infty\sum_{n=k}^\infty a(k,n-k)\;.\tag{6}$$
Turning $(6)$ into $(4)$ is a matter of reversing the order of summation. What are the possible values of $n$ in $(6)$? They depend on $k$, of course, since we always have $n\ge k$, but since $k$ can be as small as $0$, so can $n$. Thus, $n$ ranges over the non-negative integers, and when we reverse the order of summation, the outer summation must be $\sum_{n=0}^\infty$. For each value of $n$, what are the possible values of $k$? Again, we know that $k\le n$, and $k$ can be as small as $0$, so for each value of $n$ we must let $k$ run from $0$ through $n$. In other words, when we reverse the order of summation and revise the limits appropriately, $(6)$ becomes
$$\sum_{n=0}^\infty\sum_{k=0}^n a(k,n-k)\;,$$
which is exactly $(4)$, as we wished.
|
H: Estimating a series $\sum_{k=1}^\infty e^{-k^2t}\leq \frac{1}{2}\sqrt{\frac{\pi}{t}}?$
Can we prove such an estimate $$\sum_{k=1}^\infty e^{-k^2t}\leq \frac{1}{2}\sqrt{\frac{\pi}{t}}?$$ I need it in my research....
AI: HINT
Note that for a monotonically decreasing function, we have that
$$\sum_{k=1}^{\infty} f(k) \leq \int_0^{\infty} f(x) dx$$
Move your mouse over the gray area for the solution.
In your case, $f(k) = e^{-k^2t}$. Hence, we have$$\sum_{k=1}^{\infty} e^{-k^2t} \leq \int_0^{\infty} e^{-x^2t} dx = \dfrac12 \sqrt{\dfrac{\pi}t}$$ from the following link: Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \dfrac{\sqrt \pi}{2}$
|
H: Calculate the probability of two teams have been drawns
If we know that team A had a $39\%$ chance of winning and team B $43\%$ chance of winning, how we can calculate the probability of the teams drawn?
My textbook mention the answer but I cannot understand the logic behind it. The answer is $18\%$. As working is not shown I guess that this is how the find $18\%$ probability of two teams withdrawn:
$$ (100\% - 39\%) - 43\% = 18\%$$
But I cannot understand the logic behind it. I appreciate if someone can explain it to me.
AI: The sum of all events' probabilities is equal to 1. In this case, there are three disjoint events: team A winning, team B winning or a draw. Since we know the sum of these probabilities is 1, we can get the probability of a draw as follows:
$$
Pr(\text{Draw})=1-Pr(\text{Team A wins})-Pr(\text{Team B wins})=1-0.39-0.43=0.18
$$
|
H: Simple question about splitting fields
Let f(x) be a polynomial in F[x]. Let K be the splitting field of f over F.
Let a be an element of F. Is K also the splitting field of (x-a)f(x)? I think it should be.
I just want to make sure I'm not being stupid!
AI: Yes, you're right.
The splitting field of $f\in F[x]$ is the field obtained by adjoining all the roots of $f$ to $F$, the roots of $(x-a)f$ are just the roots of $f$ together with $a$ (if it wasn't one already), and adjoining $a$ does nothing because it's already in $F$. Thus, the splitting field of $(x-a)f$ is the same as the splitting field of $f$.
|
H: Integration of complex function?
I have to find $$\int_C \frac{dz}{(z^2+9)^2} \,dz$$ if
C is the circle with the radius 3 and with the center at the point 2i.
Now,I know how to find the above integral. $$\int_C \frac{dz}{(z-a)^n} \,dz$$ But I have no clue how to solve and find this one.HELP :/
AI: Just consider $C$ to be the circle centered at $a$ with radius $r$, $z-a=re^{i\theta},$ then
$$\int_C \frac{dz}{(z-a)^n} \,dz= \int_{0}^{2\pi} \frac{ire^{i\theta}d\theta}{re^{in\theta}}. $$
Now, I let you to finish it. The integral should be $0$ for every $n$ except for $n=1$ where it is $2\pi i$.
Added: Note that, $n=1$ is a special case, now, for $n\neq 1$, we have
$$ \int_{0}^{2\pi} \frac{ire^{i\theta}d\theta}{re^{in\theta}}=i\int_{0}^{2\pi} {e^{i\theta(1-n)}d\theta} = \frac{e^{i\theta(1-n)}}{1-n}\Big|_{0}^{2\pi}= \frac{1}{1-n}( e^{i2\pi(1-n)}-1 )=0 ,$$
since $e^{i2m\pi} = 1,\, \forall m\in \mathbb{Z}. $
I think you can do it for $n=1$. Just subs $n=1$ in the integral and the result follows.
|
H: Commutativity of reals using Cauchy seq.
I am trying to prove that real numbers are commutative using the definition of real numbers as the equivalence class of Cauchy sequences of rational numbers.
The following is what I know and what I plan.
RTP If $x,y \in \Bbb R$, then $x+y = y+x$ and $xy = yx$.
My understanding is that if $x \in \Bbb R$ it means that $x$ is the equivalence class of the rational sequence $\{x_j\}$.
Since rationals are commutative, it's clear that $\{x_j+y_j\} \equiv \{y_j +x_j\}$ so I am planning to use this idea with multiplication as well.
Now this is my question.
Do I have to show not only the sequence which the terms are switched, but also all the equivalent sequences, as in
RTP If $\{x_j\} \equiv \{x'_j\}$ and $\{y_j\} \equiv \{y'_j\}$ then $\{x_j+y_j\} \equiv \{y'_j+x'_j\}$ and something analogous to this with multiplication ?
I know how to prove it, I just need advice if this part is actually necessary or if I am missing more.
AI: You don't need the argument in the second box if you've already shown that addition and multiplication of Cauchy sequences is well-defined on equivalence classes. Once you've shown that, it suffices to work with any arbitrary choice of representatives. (Because the operations being well-defined on equivalences classes means precisely that you can pick any pair representatives, add them, and all the resulting sequences will be in the same equivalence class.)
I assume you showed addition and multiplication were well-defined when they were first introduced, so you probably don't need the second argument.
|
H: Is a normed topological space metrizable?
As stated in the title:
If there is a norm on a topological space, then we get a metric induced by the norm.
Is this true?
AI: Yes: if $\|\cdot\|$ is a norm on $X$, then $d(x,y)=\|x-y\|$ is a metric on $X$. See this Wikipedia article, especially this section to get started.
|
H: Normal subgroups of $\langle(123),(456),(23)(56)\rangle$
Let $G$ be a subgroup of the symmetric group $S_6$ given by
$G=\langle(123),(456),(23)(56)\rangle$. Show that $G$ has four normal
subgroups of order 3.
I may be missing something, but I can only find two of them: $\langle(123)\rangle$ and $\langle (456)\rangle$. Suppose there is another one. It must be singly-generated, either by an element of the form $(abc)$ or $(abc)(def)$. The subgroup must be closed under conjugation with $(123),(456)$, etc. I always end up with more than three elements.
AI: Hint: $(123)$ and $(456)$ commute, so $(123)(456)$ has order $3$.
|
H: Are the subsets of homeomorphic spaces also homeomorphic?
Let $W$ be a subset of an $n$-dimensional complex topology vector space $Y$ such that $0\notin W$.We have known that $Y$ is homeomorphic to $C^n$ and let $S$ be the unit sphere of $C^n$.Can anyone show me that $W$ is homeomorphic to $S$?
AI: This is certainly not true. For a counterexample, let $W$ be the set $\{|z|=1\}\cup\{|z|=2\}$, where $Y=\mathbb C$. The unit sphere $S$ is connected, but $W$ is not, so they cannot be homeomorphic.
|
H: Does ${\rm Res}_{z=z_0}\frac{p(z)}{q(z)}=\frac{p(z_0)}{q^{(m)}(z_0)}$ for all $m\ge 1?$
I know the following result involving pole:
If $p(z),q(z)$ be two functions analytic at $z=z_0$ and $p(z_0)\ne0$ and $q(z)$ has a zero of order $m$ at $z_0$ (i.e. $q(z)=(z-z_0)^mf(z)$ where $f(z)$ is analytic and nonzero at $z_0$) then $\dfrac{p(z)}{q(z)}$ has a pole of order $m$ at $z_0.$
Now if $m=1$ then ${\rm Res}_{z=z_0}\dfrac{p(z)}{q(z)}=\dfrac{p(z_0)}{q'(z_0)}.$
Can this result be generalized for $m>1$ i.e. whether ${\rm Res}_{z=z_0}\dfrac{p(z)}{q(z)}=\dfrac{p(z_0)}{q^{(m)}(z_0)}$ for all $m\ge 1?$
AI: You assertion is false. By Cauchy's formula, for any closed curve $\gamma$ around $z_0$,
$\frac{1}{2\pi i}\int_\gamma\frac{p(z)}{q(z)}dz=\frac{1}{2\pi i}\int_\gamma\frac{p(z)/f(z)}{(z-z_0)^m}dz=\left. \frac{d^{m-1}p(z)/f(z)}{dz}\right|_{z=z_0}$.
In particular if $m=1$, you get $\frac{p(z_0)}{f(z_0)}$ which also happens to be the right answer when calculating the Residue via limit for a simple pole: $\lim_{z\rightarrow z_0}(z-z_0)\left[\frac{p(z)}{(z-z_0)f(z)}\right]=\frac{p(z_0)}{f(z_0)}$. Notice that $q'(z)=f(z)+(z-z_0)f'(z)$ , so $q'(z_0)=f(z_0)$ but this falls apart for $m>1$ since you have to evaluate $m$ derivatives of $p(z)/f(z)$ which is more complicated.
|
H: Convergence of the sum of the subsequence $\sum_{k=1}^{\infty } 2^{k}x_{n_{k}}$
Let $\{x_{n}\}$ be a sequence such that $\lim_{n \to ∞}x_{n}=0$. Prove that there exists a sub sequence $\{x_{n_{k}}\}$ of $\{x_{n}\}$ such that $\sum_{k=1}^{\infty } 2^{k}x_{n_{k}}$
converges and $|\sum_{k=1}^{\infty } 2^{k}x_{n_{k}}|$≤1.
AI: Since the $x_n$ go to zero, you can pick arbitrarily small elements of the sequence if you make $n$ large enough.
Let $y_1$ be equal to the first $x_n$ such that $|2x_n|<\frac{1}{2}$.
Let $y_2$ be equal to the next $x_n$ such that $|2^2x_n|<\frac{1}{2^2}$.
Let $y_3$ be equal to the next $x_n$ such that $|2^3 x_n|<\frac{1}{2^3}$.
Continue in this fashion, defining the general $y_k$ term similarly. Then the $y_k$ form a subsequence of the $x_n$, and $\sum_1^\infty 2^k y_k$ converges absolutely to a number $a$ with $|a|\le 1$.
|
H: A first order theory whose finite models are exactly the $\Bbb F_p$
Is there a first order theory $T$ in the language of rings such that its finite models are exactly the fields $\Bbb F_p$ with $p$ prime (but no $\Bbb F_q$ with $q$ a proper prime power is a model of $T$)?
EDIT: Since this question turned out to be trivial, I asked if it is also possible with a finite theory. See here
AI: Let $T$ be the theory of fields with the additional schema:
If $p\cdot1=0$ then for every $x$ it holds $x=0$ or $x=1$ or $x=1+1$ or ... $x=(p-1)\cdot 1$.
Now if $F$ is a finite model of $T$ then $F$ is a field, and it has exactly the same number of elements as its characteristics.
Note that $F$ is an infinite model of $T$ if and only if it is a field of characteristics zero.
|
H: Evaluating the derivative of a Cantor-Vitali function
Let $\varphi \colon [0,1] \to \mathbb R$ a "Cantor-Vitali function", viz. take $x \in [0,1)$ and write it as
$$
x = \sum_{j=1}^{\infty} \frac{a_j}{3^j}, \quad a_j \in \{0,1,2\}
$$
with $a_j$ non definitely equal to 2. Then
$$
\varphi (x) := \begin{cases}
\sum_{j=1}^{k-1}\frac{a_j}{2^{j+1}} & a_k=1 \text{ and } a_j \ne 1 \quad \forall j \in \{1,2, \ldots , k-1\} \\
\sum_{j=1}^{\infty}\frac{a_j}{2^{j+1}} & a_j \ne 1\quad \forall j \in \mathbb N
\end{cases}
$$
Moreover, put $\varphi(1):=1$.
Question. Evaluate $\varphi^{\prime}(x)$ for every $x \in [0,1]$.
Well, I think it is not difficult to prove the existence (at least a.e.) of $\varphi^{\prime}$, since the function should be monotone (increasing?), hence differentiable a.e. by Lebesgue's theorem.
The problem is that I cannot understand how to calculate $\varphi^{\prime}(x)$. Is it zero a.e.? How to prove this? How can I do? Thanks in advance.
AI: Hint: $\phi$ is constant on each of the open intervals making up the complement of the Cantor set.
|
H: Fourier analysis on finite abelian groups
Can someone help me show if $f$ is a character of a finite abelian group then for all $a\in G$,
$$\sum_{[f]}f(a)\stackrel{}{=}
\begin{cases}
|G| & \text{if $ a$ is the identity} \\
0 & \text{otherwise}
\end{cases}$$
Where the sum runs over all characters of $G$,
I was thinking of using the fact that the characters themselves form an abelian group under multiplication, so that I have$$f_k(a)\sum_{[f]}f(a)=\sum_{[f]}f(a)f_k(a)=\sum_{[f]}f(a)$$
So that, $$(1-f_k(a))\sum_{[f]}f(a)=0$$
But I would need $\exists k|f_k(a)\ne 1, \forall a \in G, a\ne e$
In order to divided both sides by $(1-f_k(a))$, which I can't seem to get either.
AI: The proof (which is very standard) can for instance be found in $\S 4.3$ of Appendix B of these notes.
If you take a look, you'll see that you're on the right track, but perhaps you've been stymied by a quantifier error. You want to show your identity for each fixed nonidentity element $a \in G$. So you don't need to find a character which is nontrivial on all nonidentity elements (which is not always possible; consider e.g. $G = Z/2 \times Z/2)$. You just need that each nonidentity element is not killed by at least one character on $G$. There is still something to show here, and the notes explain this point carefully. (It follows from something called the Character Extension Lemma...)
|
H: A generalization of abelian categories including Grp
The category of groups shares various properties with abelian categories. For example, the Five lemma and Nine lemma hold in Grp. Is there a weakened notion of abelian category which also includes Grp such that the 5- and 9- lemmas are still provable by arrow chasing?
AI: Semi-abelian categories are good contexts for doing homological algebra, and $\mathbf{Grp}$ is semi-abelian. One of the axioms is the split short five lemma, and the five lemma, nine lemma, and snake lemma are theorems. A category is abelian if and only if it is semi-abelian and its opposite is semi-abelian.
|
H: A sequence such that every rational is written infinitely
Since $\Bbb Q $ is a denumerable set $ \Bbb Q =\{ r_n: r \in \Bbb Q \}$ where $r_m \neq r_n $ if $ n \neq m $
Find a sequence $\langle x_n\rangle$ of real numbers such that for each $ n \in \Bbb N $, $\{m\in \Bbb N : x_m=r_n\}$ is infinite.
I found this difficult and didn't have an idea to construct the sequence. I think what you need is a sequence such that every rational number is written infinitely, any help would be appreciated
AI: $0/1,1/1,0/2,1/2,2/2,3/2,4/2,0/3,1/3,2/3,3/3,4/3,5/3,6/3,7/3,8/3,9/3,\dots$ gets you every positive rational infinitely often, you can work out a way to get the negative ones in there.
|
H: Inf of the measure of the neighbourhood of a compact set
Let $X$ bea compact metric space, and $\mu$ a Borel probability measure.
For $A\subset X$ de define $b_{\epsilon}(A):=\left\{ x\mid d(x,A)\leq
\epsilon\right\} .$
Suppose $A$ is compact and $\mu(A)=0,$ does this mean $\inf\mu(b_{\epsilon
}(A))=0?$
If this is not true is it true if you add extra hypothesis?
AI: Since $\mu$ is a probability measure we have $$\mu(A) = \mu\biggl(\bigcap_{n=1}^\infty b_{1/n}(A)\biggr) = \lim_{n\to\infty} \mu(b_{1/n}(A))$$
and it doesn't matter whether $A$ is a null set or not. We only use that $A$ is closed and compactness of $X$ is not used either.
Further details:
Suppose $A$ is closed. Then $A = \bigcap_{n=1}^\infty b_{1/n}(A)$:
Clearly $A \subseteq b_{1/n}(A)$. On the other hand, if $x \in \bigcap_{n=1}^\infty b_{1/n}(A)$ then for each $n$ there is $a_n \in A$ such that $d(x,a_n) \leq 2/n$ by definition of $b_{1/n}(A)$. Then $a_n \to x$, hence $x \in A$ since $A$ is closed.
If $B_1 \subseteq B_2 \subseteq \cdots$ are measurable then $\mu\Big(\bigcup_{n=1}^\infty B_n\Big) = \lim_{n\to\infty} \mu(B_n)$ (easy consequence of $\sigma$-additivity of $\mu$, special case of monotone convergence).
Set $B_n = X \setminus b_{1/n}(A)$ then
$$\bigcup_{n=1}^\infty B_n = X \setminus \bigcap_{n=1}^\infty b_{1/n}(A) = X \setminus A$$
Since $\mu(X) \lt \infty$ we have $\mu(B_n) = \mu(X) - \mu(b_{1/n}(A))$, so $$\mu(X) - \mu(A) = \mu(X \setminus A) = \lim_{n\to\infty} \mu(B_n) = \lim_{n\to\infty}\left( \mu(X) - \mu(b_{1/n}(A))\right) = \mu(X) - \lim_{n\to\infty}\mu(b_{1/n}(A))$$
and we can rearrange this to
$$
\mu(A) = \lim_{n\to\infty}b_{1/n}(A)
$$
as claimed above.
|
H: Unprovability of $i^2 =1$ from $\langle i \mid i^4 =1\rangle$ and similar problems
This question is related to Can I derive $i^2 \neq 1$ from a presentation $\langle i, j \mid i^4 = j^4 = 1, ij = j^3 i\rangle$ of Quaternion group $Q$?
I know I'm going too far but let me just ask...
1) Is indeed $\langle i \mid i^4 =1\rangle$ a presentation for the cyclic group of order 4?
2) If so, then do we just assume that $i^2 =1$ or should we prove that $i^2 =1$ cannot be derived from $i^4 =1$, i.e., the unprovability of $i^2 =1$ from $i^4=1$?
For the below questions I need answers only if the latter is the case in the above question 2.
3) Then how do we rigorously prove the unprovability?
4) As for
$$\langle i, j \mid i^4 = j^4 = 1, ij = j^3 i\rangle = Q
$$
(Quaternion group), how do I prove unprovability of $i^2 =1$ from the given presentation? Or how do I even guess that $i^2 \neq 1$?
AI: 1) yes.
2) can't be derived because there is a model for $i^4=1$ in which $i^2\ne1$, namely, $\{{1,\sqrt{-1},-1,-\sqrt{-1}\}}$.
3) See 2).
4) again, by constructing a model in which $i^2\ne1$. One way is by giving the group table for the quaternions.
|
H: Consistency Trapezoidal rule
I want to prove that the consistency order of the trapezoidal rule is actually second order, that means that the error of the actual solution $x(t_{k+1})$ where we can restrict ourselves to the equation $x'(t)=\lambda x(t)$ and the approximation in each step $x_{k+1}=x(t_k)+\frac{\lambda \tau}{2}(x(t_k)+x_{k+1})$ is $O(\tau^3)$. I guess one has to use taylor approximation somewhere, but this does not work.
AI: This is a sketch which shows that solving an ODE with the trapezoidal rule has global error $O(h^2)$.
Consider the differential equation $\dfrac{dx}{dt}=\lambda x$. Integrating this from $t_n$ to $t_{n+1}=t_n+h$ gives
$$
\int_{t_n}^{t_{n+1}}\dfrac{dx}{dt}dt=\int_{t_n}^{t_{n+1}}\lambda x\,dt \Longrightarrow x(t_{n+1})-x(t_n)=\lambda\int_{t_n}^{t_{n+1}} x\,dt.\tag{1}
$$
Evaluating the integral on the RHS:
$$
\int_{t_n}^{t_{n+1}} x\,dt=\frac{h}{2}(x(t_{n+1})+x(t_n))-\frac{h^3}{12} x''(\xi),
$$
where $t_n\leq\xi\leq t_{n+1}$, provided that $x(t)\in \mathcal{C}^2$.
This gives the update formula $(1)$ as
$$
x(t_{n+1})=x(t_n)+\dfrac{h\lambda}{2}\left(x(t_{n+1})+x(t_n)\right)-\dfrac{h^3}{12} x''(\xi).
$$
Solving for $x(t_{n+1})$ gives
$$
x(t_{n+1})=\dfrac{2+h\lambda}{2-h\lambda}x(t_n)-\dfrac{h^3}{6(2-h\lambda)}x''(\xi)=\dfrac{2+h\lambda}{2-h\lambda}x(t_n)+O(h^3).
$$
Thus, the local error is $O(h^3)$.
Considering the global error, integrating the DE from $0$ to $t_n$:
$$
\int_{t_0}^{t_{n}}\dfrac{dx}{dt}dt=\int_{t_0}^{t_{n}}\lambda x\,dt \Longrightarrow x(t_{n})-x(t_0)=\lambda\int_{t_0}^{t_{n}} x\,dt.
$$
The integral can be computed using the trapezoidal rule
$$
\int_{t_0}^{t_n}x\,dt=\frac{h}{2}\left[x(t_0)+2\sum\limits_{i=1}^{n-1}x(t_i)+x(t_n)\right]-\frac{(t_n-t_0)h^2}{12}x''(\xi),
$$
where $t_0\leq\xi\leq t_n$. Following similarly as above, the global error can be shown to be $O(h^2)$.
This is one example of a Runge-Kutta method, this one from the second order family.
|
H: Taylor series of $ f = e^{x^2 + y^2}$ near $(0,0)$
I have to compute the second order Taylor series of the function
$ f = e^{x^2 + y^2}$ near $(0,0)$.
The Jacobian is:
$$ Df(x,y) = (2\ x\ e^{x^2 + y^2}, 2\ y\ e^{x^2 + y^2}) $$
and the Hessian:
$$ D^2f(x,y) = \left( \begin{array}{cc} 4\ x^2\ e^{x^2 + y^2} & 4\ x\ y\ e^{x^2 + y^2} \\ 4\ x\ y\ e^{x^2 + y^2} & 4\ y^2\ e^{x^2 + y^2} \end{array}\right) $$
both of which are $0$ at $(0,0)$.
So the second order taylor series degenerates to
$$ T_2(x,y) = f(0,0) = 1$$
... which sounds like a rather poor approximation for any $x, y$, e.g. $(0.2,0.2)$.
Also it's a bit confusing for me that both Jacobian and Hessian are $0$ at that point.
Do I miss something here, or is the function itself simply not favorable for Taylor approximation?
AI: I think you've miscomputed the Hessian. But anyway you can use $$e^{x^2+y^2}=1+(x^2+y^2)+(1/2)(x^2+y^2)^2+\cdots$$ to see what the series out to second degree looks like.
|
H: Why does the limit of $ f(x) = \int_{0}^{x} e^{-t^2/2} dt$ exist?
For a function
$$ f(x) = \int_{0}^{x} e^{-t^2/2} dt$$
for $x \ge 0$ one has to argue whether $ \lim_{x \to \infty} f(x) $ exists or not.
I thought about the following:
$$ f'(x) = e ^ {-x^2/2} $$
$$ f''(x) = -x\ e ^ {-x^2/2} $$
Given $x \ge 0$, it's obvious that $f(x)$ is strictly increasing and concave. Also, it is clear that $ e^{-x^2/2} \to 0$ as $x \to \infty$, so $f$ has to converge.
Is this enough of a 'proof', or is there more to it?
AI: If you want to compute the limit as $x$ goes to infinity, the standard argument is as follows.
We have
$$
f(x)^2 = \left( \int_0^x e^{-t^2/2} \, dt \right) \left( \int_0^x e^{-s^2/2} \, ds \right) = \int_0^x \int_0^x e^{-t^2/2} e^{-s^2/2} dt ds = \iint_{[0,x] \times [0,x]} e^{-(t^2+s^2)/2} \, dt\, ds.
$$
We have $[0,x] \times [0,x] \subseteq \{ (u,v) \, | \, u,v \ge 0, \, \sqrt{u^2 + v^2} \le \sqrt 2 \, x \} \subseteq [0,\sqrt 2 \, x] \times [0, \sqrt 2 \, x]$. These inclusions imply that if we call the left-most region $A_x$, the middle one $B_x$ and the right-most one $C_x$,
$$
\iint_{A_x} e^{-(t^2+s^2)/2} \, dt\, ds \le \iint_{B_x} e^{-(t^2+s^2)/2} \, dt\, ds \le \iint_{C_x} e^{-(t^2+s^2)/2} \, dt\, ds
$$
and since all double integrals are increasing functions of $x$, the limit is defined (i.e. it is either positive real or infinite), and together with $C_x = A_{\sqrt 2 \, x}$, we have the inequalities
$$
\lim_{x \to \infty} f(x)^2 = \lim_{x \to \infty} \iint_{A_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
\le \lim_{x \to \infty} \iint_{B_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
\le \lim_{x \to \infty} \iint_{C_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
= \lim_{x \to \infty} \iint_{A_x} e^{-(t^2+s^2)/2} \, dt\, ds \\
= \lim_{x \to \infty} f(x)^2.
$$
Thus, it suffices to compute
$$
\lim_{x \to \infty} \iint_{B_x} e^{-(t^2 + s^2)/2} \, dt \, ds.
$$
Using a polar change of coordinates : $t = r \cos \theta$, $s = r \sin \theta$, $dt \, ds = r dr d\theta$, we get
$$
\iint_{B_x} e^{-(t^2 + s^2)/2} \, dt \, ds = \int_0^{\pi/2} \int_0^x r e^{-r^2/2} \, dr d\theta = \frac{\pi}2 \left( 1 - e^{-x^2/2} \right).
$$
by the change of variables $u = r^2/2$. Letting $x \to \infty$ leads to
$$
\lim_{x \to \infty} f(x)^2 = \frac{\pi}2,
$$
hence by continuity of the square root function, $\lim_{x \to \infty} f(x) = \sqrt{\frac{\pi}2}$.
Hope that helps,
P.S. : This has probably been already computed on this website.
|
H: How many collections of four numbers, from 1 to 25, have a sum multiple of 5?
How many collections of four numbers, from 1 to 25, added together are a multiple of 5?
I.E the collection $${18, 23, 24, 25}$$ has a sum of 90, which is the maximum sum multiple of 5 that can be constructed with unique numbers from 1 to 25.
I wrote a list of combinations that I'd find with some brute-forcing if you could call it that way, in a attempt to decipher a formula out of them but i don't seem to be getting anywhere.. Obviously the solution should be expressed in a more general form but I clearly don't know what I'm doing or where to begin.
If there is any soul that can give me some hint on this, would greatly appreciate it :)
AI: If $\{a,b,c,d\}$ has a sum congruent to $x\ (\mathrm{mod}\ 5)$, then
$$\{a+1\ (\mathrm{mod}\ 25), b+1\ (\mathrm{mod}\ 25), c+1\ (\mathrm{mod}\ 25), d+1\ (\mathrm{mod}\ 25)\}$$
has a sum congruent to $x+4\ (\mathrm{mod}\ 5)$. Given any $x$, this clearly provides a bijection between the collection of 4-sets whose sums are congruent to $x$ and the collection of 4-sets whose sums are congruent to $x+4$. Since this holds for any $x$, and since $\mathrm{gcd}(4,5) = 1$, it follows that all 5 collections of 4-sets have the same size. In particular, the collection of 4-sets whose sum is a multiple of 5 is exactly one fifth of the collection of all 4-sets. That whole collection has size ${25}\choose{4}$, and one fifth of that is:
$$2530$$
|
H: The difference between inside and outside
If I parameterise $\mathbb{R}^n$ with generalised polar coordinates $(r, \Theta)$ it is possible to partition $\mathbb{R}^n$ into three parts
$$A = \{x \in \mathbb{R}^n \mid r < 1\}$$
$$B = \{x \in \mathbb{R}^n \mid r = 1\}$$
$$C = \{x \in \mathbb{R}^n \mid r > 1\}$$
$A$ is the inside and $C$ is the outside, of a sphere given by $B$. I can then define a function the maps most of the outside to the inside, for example:
$$f: C\to A$$
$$f(r,\Theta) = (1/r, \Theta)$$
This function works almost everywhere except for where $x\in A$ is the origin. It would seem impossible to define a bijective function that completely maps $A$ to $C$ and that you'll always have a single point which is problematic. Basically, $C$ has a hole and $A$ does not. My question is, is the presence of this single point generally considered the basis on which inside and outside are different?
Similarly, if we add a point at infinity to $\mathbb{R}^n$, what does this do to the notion of inside and outside, is there still a difference?
AI: Yes. Often when you look at the complement of an embedded sphere, not just the standard embedding $r=1$ in your question, the two components are referred to as "the bounded component" and "the unbounded component." The unbounded component can be distinguished topologically by the fact that it is not compact.
When you add a point at $\infty$, then the two components are homeomorphic, so there is no distinction. Indeed, for the $r=1$ sphere, the two components correspond to the upper and lower hemisphere of $S^3$ under stereographic projection.
|
H: Find unitary matrix $U$ such that $U^*AU$ is diagonal, where $A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}$
Given:
$$A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}$$
Find a unitary matrix $U$ such that $U^{*} A U = D$ where $D$ is diagonal.
Now the eigenvalues of $A$ are $0$ with eigenvector $(i,1)$ and $2$ with eigenvector $(-i,1)$ and they are not orthogonal to one another. So we need to apply gram-schmidt process to orthogonalize this basis.
But once I started, the norm of the vectors is $0$ so I can not normalize them! What do I do in such a case?
AI: The norm of the vectors can't be zero as they both are non-zero vectors. Assuming you're working with the usual euclidean inner (scalar) product, we have:
$$||(i,1)||^2=\langle (i,1)\,,\,(i,1)\rangle=i\cdot\overline i+1\cdot\overline 1=i(-i)+1\cdot 1=2\\||(-i,1)||^2=\langle (-i,1)\,,\,(-i,1)\rangle=(-i)\cdot\overline{(-i)}+1\cdot\overline 1=(-i)(i)+1\cdot 1=2$$
so both vectors' norm is $\;\sqrt2\;$ ...
|
H: Prove sequence is not convergent in a complete metric space
I'm looking over some previous analysis exams and I've come across this question:
Consider the set $\mathbb{R}$ of real numbers and the metric function
defined as:
$$d(x, y) = \begin{cases} 1 ~ \text{if} ~ x \ne y \\ 0 ~ \text{if} ~ x
= y \end{cases}$$
Is the sequence $(x_n)_{n = 1}^\infty$ defined by $x_n = \frac{1}{n}$ convergent in this metric space?
Is this metric space complete?
For part 1:
Assume that $(x_n)_{n = 1}^\infty$ is convergent to a real $a$. Then, for every $\epsilon > 0$, there exists a $N \in \mathbb{N}$ such that:
$$d(x_n, a) < \epsilon ~ ~ \text{for all} ~ ~ n \geq N$$
Let $\epsilon = \frac{1}{2}$. Then $d(x_n, a) < \frac{1}{2} ~ ~ \implies ~ ~ x_n = a ~ ~ \implies ~ ~ a = \frac{1}{n} ~ ~ \text{for all} ~ ~ n \geq N$. This is a contradiction, so $(x_n)_{n = 1}^\infty$ does not converge in this metric space.
For part 2:
A metric space is complete iff every Cauchy sequence is convergent. A Cauchy sequence is a sequence $(x_n)_{n = 1}^\infty$ such that for every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that:
$$d(x_n, x_m) < \epsilon ~ ~ \text{for all} ~ ~ n, m \geq N$$
Reusing part 1's approach, we let $\epsilon = \frac{1}{2}$. Then if $(x_n)_{n = 1}^\infty$ is a Cauchy sequence, it must follow that $x_n = x_m$ for all $n, m \geq N$. So the sequence is constant for $n \geq N$, and as such trivially converges.
We have proved that if $(x_n)_{n = 1}^\infty$ is a Cauchy sequence in this metric space, then it must be convergent. Therefore this metric space is complete.
Are these proofs valid? In particular, is the use of contradiction sound in part 1 and does the reasoning follow in part 2? I've learned to be careful with analysis arguments, so could someone check over my work?
I am confident about part 1 but there's something about part 2 that confuses me. The completeness property only applies to Cauchy sequences, so if my results are correct then the sequence in part 1 is not Cauchy. So did I make a mistake or was $d$ carefully chosen to defy intuition?
AI: You are all right, the metric $d$ makes your metric space being discrete topological space. In the latter, convergence of $(x_n)$ means that there exists $a$ such that $x_n = a$ for all $n$ big enough. You perfectly showed this using $\epsilon = \frac12<1$. As a result, the sequence $\frac1n$ does not converge.
Any Cauchy sequence thus has to have only constant terms after some $n$ because of the Cauchy condition $|x_n -x_m|<\epsilon$ for all $m,n$ big enough. As a result, you obtain a desired sequence with only constants in the tail, so it clearly converges thanks to the first argument. And surely, the sequence $\frac1n$ is not Cauchy w.r.t. the metric $d$.
|
H: Find a number $A$ so that $\lfloor A^{3^n} \rfloor$ are always odd
Find a number $A$ so that
(1) $\lfloor A^{3^n} \rfloor$ is always odd for $n\geq 1$;($\lfloor x \rfloor$ is the largest integer not greater than $x.$)
(2) $A>1$ and $A^{3^n}$ is never an integer for $n\geq 1$;(This is the place I have edited.)
(3) There is a closed-form formula for $A.$(Series and integral can also be seen as a closed form.)
Mills' constant satisfies (1) and (2) but not (3) (so far).
We can construct a number satisfies (1) and (2) more easily than Mills' constant:
Let $a_1=1,a_{n+1}=a_n^3+2~(n\geq1).$ Denote $$A=\lim_{n\to \infty}a_n^{\frac{1}{3^n}}.$$
Then we can prove that $\lfloor A^{3^n} \rfloor=a_n,$ hence $A>1$ and $a_n$ is odd. The proof is very like to this. But I cannot find a closed-form formula for $A$ again.
Thanks in advance !
AI: For any integer $m \ge 2$, let $A = m + \sqrt{m^2-1}$ and define a sequence $(T_n)_{n\in\mathbb{N}}$ by:
$$T_n = A^n + A^{-n}$$
It is easy to see $T_0 = 2$, $T_1 = 2m$ and $T_{n+2} = 2mT_{n+1} - T_{n}$ for any $n \ge 0$. From this, we can conclude $T_n$ is an even integer for any $n \ge 0$.
It is also clear $A > 1$ and hence for all $n \ge 1$, we have:
$$\begin{align}0 < A^{-(3^n)} < 1 \implies & T_{3^n} - 1 < A^{3^n} = T_{3^n} - A^{-(3^n)} < T_{3^n}\\ \implies & \lfloor A^{3^n} \rfloor = T_{3^n} - 1 \text{ is an odd integer.}
\end{align}$$
|
H: What is the difference between a population and a data set?
A data set is defined as a collection of data, while a population is defined as a collection of items under consideration.
What is the main difference between a data set and a population?
I have been seeing these two terms being used interchangeablely.
AI: A population in statistics is a set of individuals, companies, etc, on which we'd like to draw inference. For example, if we want to make inference about the proportion of HIV positive American citizens, then American citizens is our population. For obvious reasons, we cannot sample the entire population. Thus, we use a subset of our population and use some sort of sample. When the sampling procedure is done, the data is usually collected in a data set, which we use to make analyses in for instance R, Excel, EViews, and so on.
|
H: Regularity of a solution of Laplace equation
Assume $\Omega$ is some open, bounded domain with smooth boundary - say $\Omega = B(0,1) \subset \mathbb{R}^3$. Let $v$ be a solution of the Laplace equation \begin{equation} \begin{cases} \Delta v =0 & \mbox{on } \Omega \\ v=f|_{\partial\Omega} & \mbox{on } \partial \Omega \end{cases} \end{equation} and assume furthermore $f \in L^1_{loc} (\overline{\Omega}) \cap L^4(\overline{\Omega})$ and $Df \in L^2(\overline{\Omega})$.
Can one prove that $v \in L^4$?
What I found so far: this works for $f \in C(\overline{\Omega})$ by the Poisson Integral Formula, but my $f$ is not as smooth (as I can't assert that $f \in H^4$).
Any help is much appreciated!
AI: First note that $f\in L^4(\Omega)$ implies $f\in L^1(\Omega)$, hence the hypothesis $f\in L^4(\Omega)\cap L^1_{loc}(\Omega)$ is unnecessary. Now, consider the problem
$$\tag{P}
\left\{ \begin{array}{cc}
\Delta v=0 &\mbox{ in $\Omega$} \\
v=f &\mbox{ in $\partial\Omega$}
\end{array} \right.
$$
It is well know that if $f\in H^{1/2}(\partial\Omega)$, then problem $(P)$ has a unique solution $v\in H^1(\Omega)$ satisfying $(P)$ in the weak sense. But $H^1(\Omega)$ is contained in $L^{2^\star}$ (Sobolev Embedding), where in our case $$\tag{1}2^\star=\frac{2N}{N-2}=6$$
From $(1)$ we conclude that $v\in L^4(\Omega)$.
Remark 1: $f\in L^4(\Omega)$ with $Df\in L^2(\Omega)$ implies that $f\in H^1(\Omega)$, which implies that $\operatorname{trace}(f)\in H^{1/2}(\partial\Omega)$
Remark 2: To solve problem $(P)$ we procced as follows:
Claim: The solution $v\in H^1$ is characterized by $$\tag{2}\int_\Omega |\nabla v|^2=\min\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$$
Denote $\mathcal{K}=\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$
First note that $\mathcal{K}$ non empty, because $\operatorname{trace}:H^1(\Omega)\to H^{1/2}(\Omega)$ is surjective, hence, we can take a minimizing sequence in $\mathcal{K}$, i.e. a sequence $u_n\in\mathcal{K}$ satisfying $$\int_\Omega |\nabla u_n|^2\to \inf\mathcal{K}$$
Try to prove that $\|\nabla u_n-\nabla u_m\|_2\to 0$ as $n,m\to\infty$. Note that $u_m-u_n\in H_0^1(\Omega)$, hence, by Poincare inequality we can conclude that $$\|u_n-u_m\|_{1,2}\to 0$$
This implies the existence of some $v\in H^1(\Omega)$ such that $u_n\to v$ in $H^1$. Now you can conclude.
Remark 3: Let me propose you another way to solve this problem.
Let $K=\{\int_\Omega |\nabla u|^2: u\in H^1(\Omega)\ \mbox{and}\ \ \operatorname{trace}u=f\}$ and define $F:K\to\mathbb{R}$ by $$\tag{3}F(u)=\int_\Omega|\nabla u|^2$$
First note that $K$ is closed and convex. Now, try and show:
I - $F$ is coercive, i.e. if $\|u\|_{1,2}\to\infty$ in $K$, then $F(u)\to\infty$,
II - $F$ is weakly lower semicontinuous, i.e. if $u_n\in K$ weakly converges to $u\in K$, then $F(u)\leq\liminf F(u_n)$,
III - $F$ is convex.
I, II and III implies that $F$ is minimized by some $v\in K$ which implies that $F'(v)=0$.
|
H: For a fixed and small $\epsilon$, finding the number of real roots of $x^{2}+e^{-\epsilon x}-2+\sin(\epsilon x)$
I saw the following question in an introduction to applied mathematics
exam (this is only the first part of the question):
Assume $0<\epsilon\ll1$ . Denote $$ f(x,\epsilon):=x^{2}+e^{-\epsilon
x}-2+\sin(\epsilon x) $$
How many real roots does $f(x,\epsilon)$ have ?
The answer stated that:
$$ f(0,\epsilon)=1-2=-1 $$ $$ f_{x}(0,\epsilon)=0 $$
hence $0$ is an extremum. Since $$ f_{xx}(0,\epsilon)>2 $$
we have it that $0$ is a minimum.
Now, since this function is increasing for a positive $x$, for a small
enough $\epsilon$, and $f(x,\epsilon)\to\infty$ as $x\to\infty$ than
there is exactly one positive root.
A similar argument show that there is also exactly one negative root.
My main problem with this answer is that I don't understand why for
a positive $x$ the function is increasing : its easy to see that
$x^{2}-2$ is increasing, but the other two elements $e^{-\epsilon x},\sin(\epsilon x)$
are not.
Can someone please explain the claim that $f(x,\epsilon)$ is increasing
for a positive $x$, and/or suggest an alternative method for solving
this question ?
AI: Hints:
For very small $\;\epsilon >0\;$ we have that
$$x^2+e^{-\epsilon x}-2+\sin\epsilon x \cong x^2+1-2+0=x^2-1$$
and the last expression on the right indeed has two zeros, one positive and one negative...and no need to go into two variable functions, partial derivatives and stuff, but:
$$f'(x)=2x-\epsilon e^{-\epsilon x}+\epsilon\cos\epsilon x\cong 2x-\epsilon+\epsilon = 2x>0\;,\;\;\text{for very small}\;\epsilon>0\ldots$$
and this is the reason they say the function's increasing for $\,x>0\,$
|
H: A small calculation .
How
$\sum_{k=0}^n (-1)^n\times(-1)^{n-k}=\sum_{k=0}^n(-1)^k$
i got it
$\sum_{k=0}^n(-1)^n\times(-1)^{n-k}=\sum_{k=0}^n(-1)^{2n-k}$
And
is that $\mathbb E[\mathbb E(X)]=\mathbb E(X)$ ?
AI: Ad 1.: $(-1)^{2n-k} = (-1)^{2(n-k)+k} = ((-1)^2)^{n-k}(-1)^k = 1^{n-k}(-1)^k = (-1)^k$. So:
$$\sum_{k=0}^n(-1)^k = \sum_{k=0}^n(-1)^{2n-k}$$
More generally, $(-1)^p$ depends only on the parity of $p$. Hence if $p$ and $p'$ have the same parity, then $(-1)^p = (-1)^{p'}$. In particular, $(-1)^p=(-1)^{-p}$ and $(-1)^p = (-1)^{p+2q}$.
Ad 2.: $\Bbb E[X]$ is a number, a constant, say $N$; hence $\Bbb E[\Bbb E[X]] = \Bbb E[N] = N = \Bbb E[X]$.
|
H: How to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$?
I don't know how to compute $\sum_{n=0}^{\infty}\frac{(2n+1)}{n!}x^{2n+1}$,appreciate any help!
Is there any general rule for solving such problems?
AI: $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
$$xe^{x^2}=\sum_{n=0}^\infty\frac{x^{2n+1}}{n!}$$
$$\frac{d}{dx}xe^{x^2}=\sum_{n=0}^\infty\frac{(2n+1)x^{2n}}{n!}$$
$$xe^{x^2}(2x^2+1)=\sum_{n=0}^\infty\frac{(2n+1)x^{2n+1}}{n!}$$
|
H: Riesz-Representation theorem for a special class of functions
The original riesz representation theorem states
Let $X$ be locally compact hausdorff space. Then for any nonnegative functional $\Lambda$ on $C_c(X)$, there is a unique regular borel measure $\mu$ on $X$ such that $$\Lambda(f)=\int f\mu(dx)$$
for all $f\in C_c(X)$.
Suppose there is a measured space $(\Omega,\mathcal{A},P)$. I define the space of measurable functions $L^0(P)$ is the equivalence class of measurable functions, which are $P$-a.s. equal. This is a vector space and often used in mathematical finance. If we suppose that $\Omega$ is finite, then we identify $L^0$ with $\mathbb{R}^n$ (for $Y\in L^0$, $Y(\omega_i):=y_i)$.
I was able to prove that a given functional $\Gamma$ on $L^0$ is nonnegative. Can I use the Riesz-Representation theorem to conclude that $$\Gamma(Y)=\int Y R(dw)$$
for a measure $R$?
The problem is the hyptothesis about $C_c(X)$. My notes say there is such a measure, but I'm not sure, why I can apply Riesz.
AI: If $\Omega$ is finite then $\Gamma$ is essentially a row vector with positive entries and "integration against the measure $R$" reduces to a dot product against this vector.
|
H: This set of ordinals is $\in$-transitive?
I was reading this problema about ordinals. I'm following the Z.F theory.
Let $A$ be a nonempty set of ordinals. Prove that if $\cup A \notin A$ then $\cup A$ it's a limit ordinal, i.e $\cup A $ is nontempty but that is directly, and also is not a sucesor of any ordinal.
I proved that $ A\subset \cup A$. In the following way:
if $\alpha \in A$ , since $\cup A \notin A$ then $ \alpha \ne \cup A$ and since $\cup A$ is an ordinal, then $\alpha \in \cup A$ (since if $\cup A \in \alpha$ $\wedge$ $\alpha \in A$ $\Rightarrow$ $\cup A \in \cup A$ a contradiction if we consider the regularity axiom).Thus we have that $ A \subset \cup A$
Remark: I'm using the fact that $\cup A$ is an ordinal, and also that given two ordinals $\alpha$, $\beta$ are always comparable, in the sense that some of this occurs $\alpha \in \beta$, $\beta \in \alpha$ or $ \alpha=\beta$. And this is a theorem.
I don't know what can I do with the other inclusión $ A \subset \cup A$, it's easy to prove that a set A is $\in$-transitive ( $x\in A$ $\wedge y\in x$ $\Rightarrow y\in A$) if and only if $\cup A \subset A$ but it is always true that a set of ordinals is always $\in$-transitive? or this happens only in this case? Please I want a counterexample or the proof of that part )= , because I'm a Little confused
AI: Note that transitive sets of ordinals are ordinals. So if $A$ is a transitive set then it is in fact an ordinal.
However what you can prove is that $\bigcup A=\sup A$, so whenever $\alpha\in A$ we have that $\alpha\leq\sup A=\bigcup A$. If $\bigcup A\notin A$ then $A$ does not have a maximal element, and therefore $\sup A$ must be a limit ordinal.
|
H: Finding a formula to a given $\sum$ using generating functions
Find a close formula to the sum $\sum_{k=0}^{n}k\cdot 5^k$
I tried using generating functions using the differences sequences with no luck.
AI: $$\sum_{k=0}^nx^k=\frac{x^{n+1}-1}{x-1}$$
$$\sum_{k=0}^n kx^{k-1}=\frac{d}{dx}\frac{x^{n+1}-1}{x-1}$$
$$\sum_{k=0}^n kx^{k}=x\frac{d}{dx}\frac{x^{n+1}-1}{x-1}$$
$$\sum_{k=0}^n kx^{k}=\frac{x^{n+1}((x-1)n-1)+x}{(x-1)^2}$$
$$\sum_{k=0}^n k5^{k}=\frac{5^{n+1}(4n-1)+5}{16}$$
|
H: Silverman Adv. Topics example
I would like to refer you to Silverman's Advanced Topics in the Arithmetic of Elliptic Curves example 10.6:
Let $D$ be a nonzero integer, $E:y^2=x^3+D$ with complex multiplication by $\mathcal{O}_K$ where $K=\mathbb{Q}(\sqrt{-3})$. Let $\mathfrak{p}$ be a prime of $\mathcal{O}_K$ with $\mathfrak{p}\nmid 6D$. Since $\mathcal{O}_K$ is a PID, write $\mathfrak{p}=(\pi)$ and one can check that there is a unique $\pi$ generating $\mathfrak{p}$ which satisfies $\pi\equiv 2 (\text{mod }3)$.
My question is now this, if we let $\mathfrak{p}=(2+\delta)$, where $\delta=\frac{1+\sqrt{-3}}{2}$, then because it has norm 7, it is a prime. So, taking $\pi=2+\delta$ how to we get the result that $\pi\equiv 2 (\text{mod }3)$? I've tried to multiply $\pi$ by units in $\mathcal{O}_K$, but don't seem to get anything useful. Think may be missing something obvious here.
Edit: Are you able to prove in general that there is a unique $\pi$ generating $\mathfrak{p}$ which satisfies $\pi\equiv 2 (\text{mod }3)$?
Thanks for the help!
AI: The short answer is...it isn't true for $\pi$.
It is the ideal that possesses such a generator. This doesn't mean all generators satisfy the congruence.
In fact $\pi \equiv -\bar{\delta} \bmod 3$ so you should find that $\pi\delta \equiv -\bar{\delta}\delta \equiv -1 \equiv 2 \bmod 3$.
Since $\delta$ is a unit in $\mathfrak{O}_K$ we find that $\mathfrak{p}=(\pi) = (\pi\delta)$ and so you have found the right generator.
|
H: Fourier Integral problem
Let $f(x)= \begin{cases} 3, & x \in [0,\pi]\\ 2x, & x \in(\pi,2\pi]
\\ 0, & x > 2 \pi \end{cases} $
Express $f$ as a Fourier integral.
I don't know which type of integral I'm supposed to use it. Normally I'd think they mean:
$$f(x) = \int_{0}^{\infty} a(u) \cos ux + b(u) \sin ux \ du$$
Where $$a(u) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(x) \cos ux \ dx$$
$$b(u) = \frac{1}{\pi} \int_{-\infty}^{\infty} f(x) \sin ux \ dx$$
But since $f$ is undefined for negative $x$, how would I calculate these integrals? Could I just 'forget' the negative bits and calculate that in the bounds between $(0,\infty)$? Or should I perhaps use the even (or odd) extension of $f$ and then express it as:
$$f(x) = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty} F_c (u) \cos ux \ du$$
Where $$F_c (u) = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty} f (x) \cos ux \ dx \ ? $$
I would be inclined to do it the second way because I was under the impression that the first way only makes sense for $f$ defined on whole $\mathbb{R}$ ?
AI: It looks to me the way the problem is stated (and assuming $f(x) = 0 \, \forall x < 0$), that you are asked to find
$$\int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x} = 3 \int_0^{\pi} dx \, e^{i k x} + 2 \int_{\pi}^{2 \pi} dx \, x \, e^{i k x}$$
You should note that, had you been asked to find a Fourier series, the function $f$ would be periodic and not $0$ when $x > 2 \pi$.
|
H: Where does the minimal polynomial lie?
Let $\theta = a + b \sqrt{D_1} + c \sqrt{D_2} + d \sqrt{D_1 D_2}$, where $a,b,c,d,D_1,D_2$ are integers. Is there any reason to believe the minimal polynomial for $\theta$ over $\mathbb{Q}$ should be an element of $\mathbb{Z}[x]$, without resorting to the notion of Galois conjugates and algebraically bashing out the particular polynomial? What about if we further assume that $\mathbb{Q}(\sqrt{D_1},\sqrt{D_2})$ is biquadratic over $\mathbb{Q}$?
AI: $\theta$ is an algebraic integer because it is a linear combination of the algebraic integers $1$, $\sqrt{D_1}$, $\sqrt{D_2}$, $\sqrt{D_1 D_2}$ with integer coefficients. Hence its minimal polynomial has integer coefficients.
|
H: What is the maximum possible value of determinant of a matrix whose entries either 0 or 1?
My question is simply the title:
What is the maximum possible value of determinant of a matrix whose entries either 0 or 1 ?
AI: Quoting my question in another thread:
In fact, I don't even know how large the determinant of a 0-1 matrix can be. The Hadamard's bound for the absolute determinant of an $n\times n$ 0-1 matrix is $\frac{(n+1)^{(n+1)/2}}{2^n}$ (online ref. 1 and ref. 2), and the bound is sharp if and only if there exists a Hadamard matrix of order $n+1$. Yet, to my knowledge, there is no known sharp upper bound for the absolute determinant of a general $n\times n$ 0-1 matrix.
|
H: Finding a function with properties
I am looking for a function $f(x)$ with the following properties:
Positive for $x\in(-\infty, 0)$ but tangent to the x-axis at $x=-1$
A root at $x=0$ and negative for $x\in(0, 2)$
A root at $x=2$ and positive for $x\in(2, \infty)$
I thought $f(x)=x(x-2)(x+1)^2$ would do the trick but it does not. The graph I have drawn on paper has a "w" shape with a local minimum at $x=-1$, a local maximum between $x=-1$ and $x=0$ and a local minimum between $x=0$ and $x=2$.
All help is greatly appreciated!
AI: I believe that your function works:
|
H: Embedding torsion-free abelian groups into $\mathbb Q^n$?
Glass' Partially Ordered Groups states without proof:
Every torsion-free abelian group can be embedded into a rational vector space (as a group).
Can someone link me to a proof of this? It seems to me like it's probably false: $\mathbb Q^n$ is countable, so how can an uncountable group be embedded (unless the space has uncountable dimension)?
Does "embedded as a group" mean "embedded in a way that doesn't preserve order"? I'm not sure what the "as a group" modifier indicates.
AI: Let $G$ be an abelian torsion-free group. Then $G$ is a torsion-free $\mathbb Z$-module and consider $S=\mathbb Z-\{0\}$. Then we have a canonical homomorphism $\varphi$ from $G$ to the module of fractions $S^{-1}G$ given by $\varphi(x)=x/1$. Since $G$ is torsion-free we have that $\varphi$ is injective, so $G$ is embedded in $S^{-1}G$. But $S^{-1}G$ is an $S^{-1}\mathbb Z=\mathbb Q$-module, that is, a $\mathbb Q$-vector space.
|
H: Clearing up a property of orthonormal basis
I'm learning the concept of principal component analysis and in one explanation I found the following statement:
Let's take an arbitrary $n$-dimensional vector $\textbf{v} = (v_1, ..., v_n)$ and orthonormal basis $\textbf{e}_1, \textbf{e}_2, ..., \textbf{e}_n$. We can write vector $\textbf{v}$ as:
$\textbf{v} = \sum_{i=1}^{n} \textbf{e}_i * \langle\textbf{e}_i, \textbf{v}\rangle $
and
$\sum_{i=1}^{n}\langle\textbf{e}_i, \textbf{v}\rangle^2 = 1 $ .
Why is the latter part true? Could someone maybe prove this? :) The $*$-sign denotes ordinary multiplication and $\langle \rangle$ denotes dot product :)
Here is the post i'm using as reference
https://stats.stackexchange.com/questions/2691/making-sense-of-principal-component-analysis-eigenvectors-eigenvalues/33654#33654
Thank your for any help!
AI: The second equation is equivalent to state that $\|v\|^2=1$. As easy example, consider the vector $v=v_1e_1+v_2e_2$ in $\mathbb R^2$, with $e_1=(1,0)$ and $e_2=(0,1)$. Then $v_1=\langle v,e_1 \rangle$ and similarly for $v_2$ as $\langle e_i,e_j \rangle=\delta_{ij}$. So you can write $v=\langle v,e_1 \rangle e_1+\langle v,e_2 \rangle e_2$. In the end
$$\|v\|^2=\langle v,e_1 \rangle^2+\langle v,e_2 \rangle^2 $$
|
H: What is a "foo" in category theory?
While browsing through several pages of nlab(mainly on n-Categories), I encountered the notion "foo" several times. However, there seems to be article on nlab about this notion. Is this some kind of category theorist slang? Please explain to me what this term means.
AI: It's slang, which I've mostly seen used in the context of computing rather than category theory; foo is just a placeholder for something else, as is bar. A logician I know likes talking about widgets and wombats $-$ it all serves the same purpose.
For example, you might say "an irreducible foo is a foo with no proper sub-foos".
|
H: Find all generators of $ (\mathbb{Z}_{27})^{\times} $
"Find all generators of $ (\mathbb{Z}_{27})^{\times} $"
My attempt is below.
Since $ (\mathbb{Z}_{n})^{\times} $ is a cyclic if and only if $ n = 1, 2, 4, p^n, 2p^n $, $ (\mathbb{Z}_{27})^{\times} $ is cyclic.
And the order of $ (\mathbb{Z}_{27})^{\times} $ is $ 3^3 - 3^2 = 18 $ by the Euler's phi function.
Thus $ (\mathbb{Z}_{27})^{\times} \cong \mathbb{Z}_{18} \cong \mathbb{Z}_2\times\mathbb{Z}_9 $.
But I don't know how to get the 'all' generators.
('2' seems to be the 'one' generator, since 2 is of order 9 in $\mathbb{Z}_{18}$.)
AI: Hint: If you are viewing the group as $\Bbb Z_2\times\Bbb Z_9$, you are just looking for elements (pairs!) which have additive order 18. The orders of elements of the form $(a,0)$ and $(0,b)$ are all clear to you... but do you realize what the orders of elements of the form $(a,b)$ are?
If you really intend to chase these elements back to $(\Bbb Z_{27})^\times
$, you will have to explicitly write the map you have between this and $\Bbb Z_2\times\Bbb Z_9$.
With lhf's hint, this is the way to go.
|
H: Define parameter in linear system to get different number of solutions
For what values of the parameter $t$ does the following system of
equations have
no solution
more than one solution
exactly one solution
$$ (I):x+y+t\cdot z=-1$$ $$(II):3x+(t+1)y+(t-1)z=-1$$
$$(III):tx+2y+z=0$$
I'm not quite sure how to handle this. First of all I wrote that as an augmented coefficient matrix:
$$A=
\begin{pmatrix}
1 & 1 & t &|&-1 \\
3 & (t+1) & (t-1)&|&-1 \\
t & 2 & 1&|&0 \\
\end{pmatrix}
$$
I suppose there'd be no solution if one of the lines showed inequality and exactly one if the rank of A is equal to the number of unknowns, but I don't know how to apply this here.
Furthermore: what's the requirement for this system to have more than one solution?
Thanks in advance
AI: If the $\det A_0 \ne 0$ ($\operatorname{rank} A_0=3$), then the solution exists and is unique.
This gives you a cubic equation on $t$, whose solutions have to be investigated individually.
For each such $t$, the solution exists (and is not unique) if $\operatorname{rank} A = \operatorname{rank} A_0$ and there are no solutions otherwise.
|
H: Proving minima, maxima, and saddle points don't exist.
I've got the next function: $f(x,y)=x^3y^3$ where $x,y\in \mathbb R$
I need to determine whether there is minima, maxima or saddle point.
Easily enough, after doing the partial derivatives
$f_x'(x,y)=3x^2y^3$
$f_y'(x,y)=3y^2x^3$
I get the point $(0,0)$.
Now, how can i "officially" prove that that point isn't a minima, maxima or saddle point? Is it enough to show that hessian matrix gives 0? Do i need to show that the function can get higher [/lower] values using other points on the function?
AI: Use the definition of maxima/minima. Try to find an open neighborhood $U$ containing $(0,0)$ s.t. for all $(x,y)\in U$ you have either
$$f(x,y)-f(0,0)>0 $$
or
$$f(x,y)-f(0,0)<0 $$
I supposed that the domain of $f$ is the whole $\mathbb R^2$.
The answer to your question is then equivalent to study either
$$x^3y^3>0 $$
or
$$x^3y^3<0 $$
With an $(x,y)$ plot and focusing on what happens around $(0,0)$ you can arrive quite easily to the answer (=negative one, i.e. there exists no such neighborhood $U$, so $(0,0)$ cannot be either a max or a min).
Edit: I add the plot of $f$
Wolfram plot and contour plot of $f$ around $(0,0)$
|
H: Derivative of inverse function, why do I get this contradiction?
Consider two functions, $f(x)=\sin x$ $\;$ and $g(x)=\arcsin x$. Then, $f'(x)=\cos x$ $\;$ and $g'(x)=\frac{1}{\sqrt{1-x^2}}.$ We know that $g[f(x)]=x$, so $f'(x)=\frac {1}{g'[f(x)]}$ $\;$. $\:$ Substituting in, we have $\cos x=\frac{1}{\sqrt {1-\sin^2 x}}$ $\;$, which yields $\cos x=\frac{1}{\cos x}$ . Why am I getting this result? Thanks!.
AI: You substituted $1/\sqrt{1-u^2}$ for $1/g'(u)$, when you meant to substitute $1/\sqrt{1-u^2}$ for $g'(u)$.
|
H: Point on an algebraic surface closest to another one
Given an algebraic surface $F(x,y,z)=0$, $F\in\mathbb{R}[x,y,z]$, and a point $P_0=(x_0,y_0,z_0)\in\mathbb{R}^3$, is there a possibility to (algorithmically) determine a point on $F(x,y,z)=0$ that is closest to $P_0$ with respect to the Euclidean distance?
AI: Yes. You can combine algebraic techniques with Lagrange multipliers.
More precisely, you want to minimize the function $$D(x,y,z)=(x-x_0)^2+(y-y_0)^2+(z-z_0)^2$$ subject to $F(x,y,z)=0$. To do this, introduce an additional variable $\lambda$, and solve the two equations $\nabla(F)=\lambda \nabla D$ and $F(x,y,z)=0$. So you have four equations in four unknowns, all algebraic equations.
To find solutions, you could use Gröbner basis techniques, which are implemented in for example Macaulay2.
|
H: Checking if a function is injective
Let $\mathbb{C}$ be a small category, whose objects are thought of as "admissible worlds" and whose arrows as "temporal admissible developments". Let $X:\mathbb{C}^{\operatorname{op}}\rightarrow \operatorname{Sets}$ be a presheaf of sets defined on $\mathbb{C}$. Denote by $\sigma X$ the image of the world $\sigma$ via the functor $X$. An element $s\in\sigma X$ is called "a person" of $X$, living in the world $\sigma$.
For every admissible temporal development $\pi:\rho\rightarrow\sigma$, denote by $\pi X:\rho X\gets\sigma X$ the image of $\pi$ via $X$, which is a function that associates to each person $s\in\sigma X$, living in $\sigma$, a person $\pi s\in\rho X$, living in $\rho$ ($\pi s$ is short notation for $(\pi X)(s)$).
We'll think of $\pi s$ as the ancestor of $s$, with respect to the past temporal development $\pi:\rho\rightarrow \sigma$.
To each person $s\in\sigma X$ we associate the whole family of ancestors of $s$, denoted by
$$ (-)\cdot s=\{\pi s\}_{\rho;\pi:\rho\rightarrow\sigma}$$
indexed by the set of arrows of $\mathbb{C}$, having codomain $\sigma$, the world where $s$ lives.
I wish that this function were injective, meaning that
$$(-)\cdot s=(-)\cdot s'\Rightarrow s=s'$$
but i can't prove it. Could someone help me?
AI: I hope you excuse me if I will use more standard notation.
Let $\sigma\in\mathbb{C}$, $s,s'\in X(\sigma)$, we have to prove, that if $s\ne s'$, then $(-)\cdot s\ne(-)\cdot s'$. Actually, $(-)\cdot s$ is a mapping $Ob(\mathbb{C}\downarrow\sigma)\to\bigcup_{c\in\mathbb{C}}X(c)$.
Let $s\ne s'$. Then, we have to prove, that there exists such $p\in\mathbb{C}$ and $\pi\colon p\to\sigma$, such that $((-)\cdot s)(\pi)\ne((-)\cdot s')(\pi)$. Put $p=\sigma$ and $\pi=id(\sigma)$. Then we have $((-)\cdot s)(id(\sigma))=id(X(\sigma))(s)=s$ and $((-)\cdot s')(id(\sigma))=id(X(\sigma))(s')=s'$, but they are not equal, QED.
|
H: Unique uniformly continuous function into complete space
Let $M_1,M_2$ be metric spaces such that $M_2$ is complete. Let $f$ be a uniformly continuous function from a subset $X$ of $M_1$ into $M_2$. Suppose that $\overline{X}=M_1$. Prove that $f$ has a unique uniformly continuous extension from $M_1$ into $M_2$ (that is, prove that there exists a unique uniformly continuous function $g$ from $M_1$ into $M_2$ such that $g|X=f$.)
I'm not sure where to start on this one... how can I extend a uniformly continuous function from $X$ to $M_1$?
AI: Because $X$ is dense in $M_1$, any $x \in M_1$ is the limit of a sequence $(x_n)$ in $X$.
Show that $(f(x_n))$ is a Cauchy sequence. In particulier, $(f(x_n))$ converges.
Show that if $(y_n)$ is another sequence converging to $x$, then $(f(x_n))$ and $(f(y_n))$ have the same limit.
Thus, you can extend $f$ by $f(x)=\lim\limits_{n \to + \infty} f(x_n)$.
To conclude, show that if $g$ is another extension of $f$, then $f=g$ (the key property is that $X$ is dense in $M_1$).
|
H: Vorticity equation in index notation (curl of Navier-Stokes equation)
I am trying to derive the vorticity equation and I got stuck when trying to prove the following relation using index notation:
$$
{\rm curl}((\textbf{u}\cdot\nabla)\mathbf{u}) = (\mathbf{u}\cdot\nabla)\pmb\omega - ( \pmb\omega \cdot\nabla)\mathbf{u}
$$
considering that the fluid is incompressible $\nabla\cdot\mathbf{u} = 0 $, $\pmb \omega = {\rm curl}(\mathbf{u})$ and that $\nabla \cdot \pmb \omega = 0.$
Here follows what I've done so far:
$$
(\textbf{u}\cdot\nabla) \mathbf{u} = u_m\frac{\partial u_i}{\partial x_m} \mathbf{e}_i = a_i \mathbf{e}_i \\
{\rm curl}(\mathbf{a}) = \epsilon_{ijk} \frac{\partial a_k}{\partial x_j} \mathbf{e}_i = \epsilon_{ijk} \frac{\partial}{\partial x_j}\left( u_m\frac{\partial u_k}{\partial x_m} \right) \mathbf{e}_i = \\
= \epsilon_{ijk}\frac{\partial u_m}{\partial x_j}\frac{\partial u_k}{\partial x_m} \mathbf{e}_i + \epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i \\
$$
the second term $\epsilon_{ijk}u_m \frac{\partial^2u_k}{\partial x_j \partial x_m} \mathbf{e}_i$ seems to be the first term "$(\mathbf{u}\cdot\nabla)\pmb\omega$" from the forementioned identity. Does anyone have an idea how to get the second term?
AI: The trick is the following:
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_m}{\partial x_k} = 0 $$
by antisymmetry.
So you can rewrite
$$ \epsilon_{ijk} \frac{\partial u_m}{\partial x_j} \frac{\partial u_k}{\partial x_m} = \epsilon_{ijk} \frac{\partial u_m}{\partial x_j}\left( \frac{\partial u_k}{\partial x_m} - \frac{\partial u_m}{\partial x_k} \right) $$
Note that the term in the parentheses is something like $\pm\epsilon_{kml} \omega_l$
Lastly use the product property for Levi-Civita symbols
$$ \epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl} $$
|
H: I haven't understood how does the line, $ H(Y)=-\int_D p(x)\log [\frac{1}{h}p(x)]dx$ appear?
Let $X$ has a pdf $p(x)$.
let $$H(X)=-\int_D p(x)\log p(x)dx$$
where $D=[x:p(x)>0]$
And,
$Y=a+hX$ ; $-\infty<a<\infty$ , $h>0$
so $Y$ has the pdf $g(x)=\frac{1}{h}p(\frac{x-a}{h})$
let $$H(Y)=-\int_{D_1} g(x)\log g(x)dx$$
where $D_1=[x:g(x)>0]$
$H(Y)=-\int_{D_1} g(x)\log g(x)dx$
$\Rightarrow H(Y)=-\int_{D_1} \frac{1}{h}p(\frac{x-a}{h})\log [\frac{1}{h}p(\frac{x-a}{h})]dx$
But i haven't understood how does the next line appear?
$\Rightarrow H(Y)=-\int_D p(x)\log [\frac{1}{h}p(x)]dx$ ?
AI: The transformation is
$$u=\frac{x-a}{h}$$
The range of $u$ is $D$ if the range of $x$ is $D_1$. They just slightly abused notation by reusing $x$ instead of $u$.
|
H: The product of two natural numbers with their sum cannot be the third power of a natural number.
I wanted to know, how can i prove that the product of two natural numbers with their sum cannot be the third power of a natural number.
Any help appreciated.
Thanks.
AI: You're talking about the solvability of the Diophantine equation $xy(x+y)=z^3$.
If a prime $p$ divides $x$ and $y$, then it must divide $x+y$ and $z$ and we can cancel $p^3$ in the equation. So we can assume that $x$ and $y$ are coprime. This implies that $x$, $y$, $x+y$ are coprime. Therefore, if their product is a cube, then each factor must be a cube: $x=u^3$, $y=v^3$, $x+y=w^3$. But then $u^3+v^3=w^3$, which has no integer solutions, as proved by Euler.
|
H: Can you prove this three-way linear map composition?
OK, this was an example that my prof gave when talking about surjective, injective and bijective functions. I also am curious if I am approaching this the right way. (Everyone here has been a really big help in pointing out to me where I am messing up).
It comes after one remark he made: the composition of two bijective (linear) maps are bijective.
So the statement: let $f$ be a function from $A \rightarrow \ B$ and $g$ be a function from $A \rightarrow\ C$.
Assume $f$ is surjective.
$\exists$ a function $h:B\rightarrow C$ such that $h\circ f = g \iff \forall x,y \in A, \ f(x) = f(y)$ implies $g(x) = g(y)$
OK. I approached this as follows: given that if $f$ and $g$ are both surjective, $g \circ f$ will be surjective also. It is also true that if $g \circ f$ is surjective, $g$ is surjective.
Now, the only way $f(x) = f(y)$ implies $g(x) = g(y)$ is if g is surjective as well. Is this the case? I am trying to be sure I understand this.
Meanwhile the statement says that there has to be a function h that when applied to f maps to g. But only if the surjectivity of f implies the surjectivity of g.
But that is about as far as I got. And I don't even know if I understood the premise correctly.
AI: It's really quite simple. If $f(x) = f(y)$, then $(h(f(x)) = (h(f(y))$, so $g(x) = g(y)$.
On the other hand, suppose $f(x) = f(y)$ implies $g(x) = g(y)$. For any $x$,
you want $h(f(x)) = g(x)$. So just define $h$ on the range of $f$
by $h(t) = g(x)$ where $t = f(x)$. This is unambiguous because if $t = f(x) = f(y)$, $g(x) = g(y)$. And on the complement of the range of $f$, define $h$ as you wish.
There is no need for any surjectivity here. And $f(x) = f(y)$ implying $g(x) = g(y)$ has nothing to do with whether $f$ or $g$ is surjective.
|
H: Number of possible outcomes in a license plate
If a license plate consists of 3 letters followed by 3 digits and having at least one digit or letter repeated .. How many outcomes are there?
26 * 26 * 10* 10 * 10 .. Is that right?
AI: That is incorrect.
First, let's find the number of license tags, total, that can be made with $3$ letters and $3$ digits. This is:
$$26\cdot26\cdot26\cdot10\cdot10\cdot10 = 26^310^3$$
Second, look at how many tags can be made without repeating any character (letter or number):
$$26\cdot25\cdot24\cdot10\cdot9\cdot8$$
Now, if we must repeat at least one character, then number of tags that satisfy are:
$$\text{Total} - \text{ThoseThatDon'tRepeat} = 26^310^3-26\cdot25\cdot24\cdot10\cdot9\cdot8$$
|
H: Convergence or divergence of $\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2}$
$$\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2}$$
The series is convergent or divergent?
Would you like to test without the full ...
I've thought of using the comparison test limit, but none worked, tried searching a number smaller or larger compared to use, but I got no ... Could anyone help me?
Please write correctly, because I am Brazilian and use translator ....
AI: First proof" Cauchy's Condensation Test (why is it possible to use it?):
$$2^na_{2^n}=\frac{2^n\log(2^n)}{2^{2n}}=\frac{n}{2^n}\log 2$$
And now it's easy to check the rightmost term's series convergence, say by quotient rule:
$$\frac{n+1}{2^{n+1}}\frac{2^n}n=\frac12\frac{n+1}n\xrightarrow[n\to\infty]{}\frac12<1$$
Second proof: It's easy to check (for example, using l'Hospital with the corresponding function) that
$$\lim_{n\to\infty}\frac{\log n}{\sqrt n}=0\implies \log n\le\sqrt n\;\;\text{for almost every}\;\;n\implies$$
$$\frac{\log n}{n^2}\le\frac{\sqrt n}{n^2}=\frac1{n^{3/2}}$$
and the rightmost element's series converges ($\,p-$series with $\,p>1\,$) , so the comparison test gives us that our series also converges.
|
H: Boundedness and total boundedness
We say that a metric space $M$ is totally bounded if for every $\epsilon>0$, there exist $x_1,\ldots,x_n\in M$ such that $M=B_\epsilon(x_1)\cup\ldots\cup B_\epsilon(x_n)$.
Prove that if $M$ is a totally bounded metric space, then $M$ is bounded. Given an example to show that the converse is false.
The "prove" part is routine. Given points $a,b$, suppose $a$ is in the ball of $x_i$ and $y$ is in the ball of $x_j$. Then $d(a,b)\le d(a,x_i)+d(x_i,x_j)+d(x_j,b)<\epsilon+d(x_i,x_j)+\epsilon$. Since there are only finitely many values of $d(x_i,x_j)$, we are done.
For the "example" part, the space must be bounded, but somehow there exists $\epsilon$ such that the space cannot be covered with finitely many balls of radius $\epsilon$. I can't think of what that space should look like... things like $[a,b]$, open ball, etc. are totally bounded.
AI: Hint: Consider an infinite set in the discrete metric. (Every point is at distance $1$ to every other point.)
|
H: How to determine the MU in economics?
I currently have a table like so:
Hours spent on Activity X | Total Utility
120
220
300
360
396
412
I know that Marginal Utility is calculated use slope formulate (delta Y / delta X).
I am asked to calculate the MU for 5 hours:
So in this case, $delta X = 5 - 1 = 4$ and $deltaY = 396 - 120 = 276$
Then, MU = $deltaY/deltaX = 276/4 = 69$
Next I am supposed to calculate the MU per dollar, which is determined by:
$deltaY/deltaX * 1/Price$ . The question states the price of activity X is $5/hour.
Then, I would have $MU per dollar = 69/5 = 13.8$ This gives me 13.8 dollars as the MU per dollar, however, the answer to this question is 7.2 dollars. Can anyone please explain where I've gone wrong?
AI: Actually you are calculating MU wrong
$$\frac{\Delta Y}{\Delta X}=\frac{Y_5-Y_4}{X_5-X_4}=\frac{396-360}{5-4}=36$$
$$\Rightarrow \text{MU}_\text{per dollar}=36/5=7.2$$
|
H: What is an example of monotonically decelerating/accelerating function?
I came across the term monotonically decelerating function. How can it be written as a mathematical function?
AI: Consider the simple case where an object moves along a $1$-dimensional axis.
Usually, the acceleration of an object whose position is represented by $f(t)$ is identified with $f''(t)$. (Recall $f'(t)$ is the speed.) So in this case, $f''(t)$ would have to be a monotonically decreasing function.
|
H: Solve the following equation algebraically for n. Show your working using factorial notation.?
a) $\dfrac{n!}{(n-2)!} = n(n-1)$
so the expression equivalent to $$\dfrac{n(n-1)}{6} = 12 \implies n(n-1)=72$$
I think its obvious the required solution is $8 \cdot 9 =72$, so $n=9$
b) $8Pn = \dfrac{8!}{(8-n)!} = 6720$
$$(8-n)! = \frac{8!}{6720} = 6 = 3!$$
so $8-n =3 \iff n=5$
AI: Looks good. Your answers are correct! Nice work.
You could also algebraically solve for $n$ in $(a)$, though I agree the solution seems "obvious": $$n(n-1) = 72 \iff n^2 - n - 72 = 0 $$ $$\iff (n-9)(n+8) = 0 \iff n = 9, n=-8$$
But since we are talking about non negative integers $n$ in these equations (since we are evaluating expressions with factorials), we can "throw out" $\;n = -8\,$ as an option, and we are left with $\;n = 9.$
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.