text
stringlengths 83
79.5k
|
|---|
H: Limit of $\lim_{x \to 0}\left (x\cdot \sin\left(\frac{1}{x}\right)\right)$ is $0$ or $1$?
WolframAlpha says $\lim_{x \to 0} x\sin\left(\dfrac{1}{x}\right)=0$ but I've found it $1$ as below:
$$
\lim_{x \to 0} \left(x\sin\left(\dfrac{1}{x}\right)\right) = \lim_{x \to 0} \left(\dfrac{1}{x}x\dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\right)\\
= \lim_{x \to 0} \dfrac{x}{x} \lim_{x \to 0} \dfrac{\sin\left(\dfrac{1}{x}\right)}{\dfrac{1}{x}}\\
= \lim_{x \to 0} 1 \\
= 1?
$$
I wonder where I'm wrong...
AI: $$\lim_{h\to0}\frac{\sin h}h=1$$ but $$\lim_{h\to0}\frac{\sin \frac1h}{\frac1h}=\lim_{h\to0}h\cdot \sin \frac1h$$
Now, $-1\le \sin \frac1h\le 1$
$$\implies \left|h\cdot \sin \frac1h\right|\le \left|h\right| $$
Now, $\lim_{h\to0}h=0$
$$\implies \lim_{h\to0}h\cdot \sin \frac1h =0 $$ |
H: Some Results in $\mathbb{Z} [\sqrt{10}]$
This is a question from an old Oxford undergrad paper on calculations in $\mathbb{Z} [\sqrt{10}]$. We equip this ring with the Eucliden function $d(a+b\sqrt{10})=|a^2-10b^2|$. I want to prove the following results:
If $d(x)=1$, then $\frac{1}{x} \in \mathbb{Z} [\sqrt{10}]$
Any non-zero element of $\mathbb{Z} [\sqrt{10}]$ which is not a unit can be expressed as a product of finitely many irreducibles in $\mathbb{Z} [\sqrt{10}]$
The ideal generated by $2$ and $\sqrt{10}$ is not principal in $\mathbb{Z} [\sqrt{10}]$
Thought so far
Suppose $x=a+b\sqrt{10}$. Clearly if $x$ is a unit then $d(x)=1$, though I'm not sure if this helps. Are we OK simply to note that $\frac{1}{x}=\frac{a-b\sqrt{10}}{a^2-10b^2}$ and since $d(x)=1$ then the deonminator is either $1$ or $-1$.
I know this is true in general in a principal ideal domain and every Euclidean ring is a principal ideal domain, but this proof is lengthy. Is there any calculation one can perform in $\mathbb{Z} [\sqrt{10}]$ to demonstrate this property more quickly.
Any help would be appreciated; I'm not actually too sure what this ideal looks set. Could someone put it in a set notation for me?
Many thanks.
AI: It seems to me that (1) and (3) are dealt with adequately in the comments (but I would be happy to incorporate that here if you need). For (2), the key point is that the ring is Noetherian, which is effectively a consequence of the Hilbert basis theorem, implying that a polynomial ring $\mathbb{Z}[x]$ is Noetherian, and the fact that a quotient ring of a Noetherian ring is Noetherian. Given this, every element can be written as a product of irreducibles, since otherwise you obtain an infinite ascending chain of ideals.
Here are some more details for (3): First, observe that $d(xy)=d(x)d(y)$ for all $x,y$ and that $d(2)=4$ and $d(\sqrt{10})=10$. Thus any common divisor $x$ has $d(x)|2$. If $x$ is a non-unit this implies $d(x)=2$. Thus assuming a non-unit common divisor $x$ exists, there are integers $a,b$ with
$$\pm 2=a^2-10b^2.$$ Consider this equation modulo $10$. It implies that either $2$ or $8$ is a square mod $10$, but the squares modulo $10$ are just $0,1,4,9,6,5$. Hence there is no non-unit common divisor $x$. On the other hand, the ideal is proper since its elements are all of the form $a+b\sqrt{10}$ with $a$ even. Therefore it is not principal. |
H: Find $y$-Lipschitz constant
$$f(x,y)=x^3e^{-xy^2}, 0\leq x\leq a, y\in \mathbb R, a>0$$
I need to find $K>0$ such that $$|f(x,y_1)-f(x, y_2)|\leq K|y_1-y_2|$$ for all $0\leq x\leq a$ and $y_1,y_2\in \mathbb R$
I did this $$|x^3e^{-xy_1^2}-x^3e^{-xy_2^2}|=x^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|e^{-xy_1^2}-e^{-xy_2^2}|$$
now I don't know what to do.
AI: Try to find a Lipschitz constant of $x \mapsto \mathrm{e}^{-x}$, $x \ge 0$. Hint: Fundamental Theorem of Calculus.
Then, you can use this in your Lipschitz estimate. |
H: Subrings of $\mathbb{Q}$
Let $p$ be prime. Suppose $R$ is the set of all rational numbers of the form $\frac{m}{n}$ where $m,n$ are integers and $p$ does not divide $n$.
Clearly then $R$ is a subring of $\mathbb{Q}$.
I now want to show that if $\frac{m}{n}$ belongs to any proper ideal of $R$ then $p|m$.
Can someone point me in the right direction.
Thanks
AI: HINT: If $p\nmid m$, then $\frac{n}m\in R$. |
H: solve $y(x)=\cos \left(y'(x)\right) + y'(x)\sin (y'(x)), y(0)=1$
solve $$y(x)=\cos (y'(x)) + y'(x)\sin (y'(x)), y(0)=1$$
with wolfram alpha I got that a solution is $y(x)=x\arcsin x+\cos (\arcsin x)$
but I have no idea how to find it.
I tried transforming into an exact equation by letting $u=y'$ then I get $y=\frac{u^2}{2}$
I just realised this is wrong (because $\frac{d}{dx}\frac{u^2}{2}=u'u\neq y'$) so there's no point posting the rest of my work.
any hint?
AI: Hint: Apply the chain rule of calculus. |
H: Smooth maps on a manifold lie group
$$
\operatorname{GL}_n(\mathbb R) = \{ A \in M_{n\times n} | \det A \ne 0 \} \\
\begin{align}
&n = 1, \operatorname{GL}_n(\mathbb R) = \mathbb R - \{0\} \\
&n = 2, \operatorname{GL}_n(\mathbb R) = \left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}\Bigg| ad-bc \ne 0\right\}
\end{align}
$$
$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a group.
$AB$ is invertible if $A$ and $B$ are invertible.
$A(BC)=(AB)C$
$I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$A^{-1}$ is invertible if $A$ is invertible.
$$(\operatorname{GL}_n(\mathbb R) := \det{}^{-1}(\{0\}))$$
$\det{}^{-1}(\{0\})$ is open in $M_{n\times n}(\mathbb R)$.
$\det : M_{n\times n}(\mathbb R) \to \mathbb R$ is continuous, why?
$\dim \operatorname{GL}_n(\mathbb R) = n^2 - 1$, why?
$(\operatorname{GL}_n(\mathbb R),\cdot)$ is a Lie group if:
$$
\mu : G\times G \to G \\
\mu(A,B) = A\cdot B \:\text{ is smooth} \\
I(A) = A^{-1} \:\text{ is smooth}
$$
How can I show this?
I want to show that general special linear group is Lie group. I could not show the my last step. How can I show that $AB$ and $A^{-1}$ are smooth? please help me I want to learn this. Thanks
Here the my handwritten notes https://i.stack.imgur.com/tkoMy.jpg
AI: Well first you have to decide what exactly the "topology" on matrices is. Suppose we considered matrices just as vectors in $\mathbb{R}^{n^2}$, with the usual metric topology. Matrix multiplication by say $A$ take a matrix $B$ to another $\mathbb{R}^{n^2}$ vector where the entries are polynomials in the components of $B$.
The other bit is similar, if you know what crammer's rule is. |
H: Closed sets in a given topology
I came across a topology on $ \mathbb Z \times \mathbb Z $ whose basis is defined as follows:
$ B(m,n) = \lbrace (m,n) \rbrace $ if both m and n are odd
$ B(m,n) = \lbrace (m+a, n) | a = -1, 0 ,1 \rbrace $ when m is even and n is odd
$ B(m,n) = \lbrace (m, n+a) | a = -1, 0 ,1 \rbrace $ when m is odd and n is even
$ B(m,n) = \lbrace (m+a, n+b) | a,b = -1, 0 ,1 \rbrace $ when both m and n are even
I was trying to determine the smallest possible closed sets given a point, as the problem requires me to. I showed that when both the coordinates are even, its just the singleton set which is closed.
But in the remaining cases, I think it has to be the entire set itself. However I am lacking a rigorous proof to establish my claim. Can anyone guide me??
AI: HINT: Consider the point $\langle 0,1\rangle$, say. A point $p\in\Bbb Z\times\Bbb Z$ is in the closure of $\{\langle0,1\rangle\}$ if and only if every open nbhd of $p$ contains $\langle 0,1\rangle$. This is the case if $p$ is $\langle 0,2\rangle$ or $\langle 0,0\rangle$, but that’s it. To see this, let $p=\langle m,n\rangle$. If $m<-1$ or $m>1$, $p\notin B(m,n)$, and similarly if $n<0$ or $n>2$. Thus, we need only consider the nine points in the set $\{-1,0,1\}\times\{0,1,2\}$, and in all cases except the two already noted, $p\notin B(m,n)$.
This argument easily generalizes to show that the closure of $\{\langle 2k,2\ell+1\rangle\}$ is the three-point set $\{\langle 2k,2\ell\rangle,\langle 2k,2\ell+1\rangle,\langle 2k,2\ell+2\rangle\}$ for any integers $k$ and $\ell$. That is, if $p$ is an even-odd point, so to speak, the only points that can’t be separated from it are the two adjacent to it vertically.
Now try to analyze the remaining cases in the same general fashion. |
H: How to determine a function of 2 variables from its derivative?
Please even the slightest advice would help!
If I have a function $V$ made of 2 variables $x_1$ and $x_2$,
and its derivative $$\frac{dV}{dt} = \frac{dV}{dx_1}\frac{dx_1}{dt} + \frac{dV}{dx_2}\frac{dx_2}{dt},$$
how do I find $V$? I don't know how to integrate $dV/dt$. Both $dx_1/dt$ and $dx_2/dt$ are given.
Here is an example:
$$dx_1/dt = -10x_2$$
$$dx2/dt = 2x_1 - 20x_2 $$
$$dV/dt = -2x_1^2 - 2x_1x_2 - x_2^2$$
What is $V$?
AI: To start you can make the following substitutions to find $x_1$ and $x_2$ :
from second equation we got : $x_1=1/2(dx_2/dt)+10x_2$ now differentiate $x_1$ with respect to
$t$ : $dx_1/dt=1/2(d^2x_2/dt^2) + 10dx_2/dt$ now putting it in the first equation you have a
differential equation in one variable. $10x_2=-1/2(d^2x_2/dt)-10dx_2/dt$ this is a simple diff. equation that you can solve with the approach: $a\exp(\lambda t)$. From here on the problem gets a bit simpler... |
H: How to show that there does not exist any integer $b$ with $f(b)=14.$
Let $f(x)$ be a polynomial with integer coefficients. Suppose that there exist distinct integers $a_1,a_2,a_3,a_4,$ such that $f(a_1)=f(a_2)=f(a_3)=f(a_4)=3.$ Then show that there does not exist any integer $b$ with $f(b)=14.$
AI: We have $f(x)=3+(x-a_1)(x-a_2)(x-a_3)(x-a_4)g(x)$ for some polynomial $g$ with integer coefficients. But then $(b-a_1)(b-a_2)(b-a_3)(b-a_4)g(b)=11$. Use the primality of $11$ to get a contradiction. |
H: A calculation of the norm of an ideal
Let $L$ be a number field of degree $n$ over $\mathbb{Q}$ and $\mathfrak{a}$ a non-zero ideal of the ring of integers $\mathcal{O}_L$. Suppose that $X=\{x_1,...,x_n\}$ is a $\mathbb{Z}$-basis of $\mathcal{O}_L$, and $Y=\{y_1,...y_n\}$ is a $\mathbb{Z}$-basis of $\mathfrak{a}$.
How can one show that the norm of $\mathfrak{a}$ is equal to the absolute value of the determinant of the change of basis matrix $M=(m_{ij})$ from $Y$ to $X$?
I can show this in the special case of a simultaneous basis, but I'm having trouble with the general case. The problem is trying to show that $$\vert\frac{\oplus_{i=1}^{n}\mathbb{Z}x_{i}}{\oplus_{i=1}^n\mathbb{Z}y_i}\vert=det(M)$$
where $y_i=\sum_{j}m_{ji}x_j$.
Many thanks for your answers.
AI: So you're trying to show this?
Theorem. Let $G$ be a free abelian group of rank $r$, $H\le G$. Then $G/H$ is finite if and only if $G$ and $H$ are of the same rank. In this case, given two $\mathbb{Z}$-bases $X$ and $Y$ for $G$ and $H$ respectively, we can write $Y=SX$ for some invertible matrix $S$, and $|G/H|=|\text{det}(S)|$.
Let $G$ and $H$ have rank $r$ and $s$ respectively. Then we can choose $\mathbb{Z}$-bases $u_1,\dots,u_r$ of $G$ and $v_1,\dots,v_s$ of $H$ with $v_i=\alpha_iu_i$. Clearly $G/H$ is the direct product of finite cyclic groups of orders $\alpha_1,\dots,\alpha_s$, and $r-s$ infinite cyclic groups. Hence $G/H$ is finite if and only if $r=s$, and in that case $|G/H|=\alpha_1\cdots\alpha_r$.
Now
$$
\begin{align*}
u_i&=\sum b_{ij}x_j, & v_i&=\sum c_{ij}u_j, & y_i&=\sum d_{ij}v_j
\end{align*}
$$
and the matrices $B=(b_{ij})$ and $D=(d_{ij})$ are unimodular. Now $C=(c_{ij})$ is the diagonal matrix with the $\alpha_i$ down the diagonal. So $S=BCD$, and taking determinants the result follows. |
H: Why is boundary information so significant? -- Stokes's theorem
Why is it that there are so many instances in analysis, both real and complex, in which the values of a function on the interior of some domain are completely determined by the values which it takes on the boundary?
I know that this has something to do with the general version of Stokes's theorem, but I'm not advanced enough to understand this yet -- does anyone have a (semi) intuitive explanation for this kind of phenomenon?
AI: When a function has zero derivative within a volume, its values are determined by sources outside the volume, which are measured only by values of the field on the boundary.
Got that? Here's the mathematical explanation.
There are many functions $f$ that obey the basic differential equation $\nabla F = 0$. We could be talking about a scalar field, or if we extend the idea of the "gradient" to act on vector fields, then this condition is that the field is both divergenceless and curlless. Let's not fret on how the gradient can be extended in this fashion--that's a lot of tensor stuff that we don't really need to worry about. Let's just keep in mind that this is the condition being imposed.
We often say that equations of the type $\nabla F = j$ represent a field $F$ that permeates space with a source $j$. Let's understand first how that source $j$ determines the field.
To do that, we first need the Green's function for $\nabla$. We'll call it $G$, and it obeys $\nabla G = \delta$, the Dirac delta function.
This is where the generalized Stokes theorem comes in. It tells us that
$$\oint_{\partial M} G(r-r') \, dS' \, F(r') =- \int_M \delta(r-r') \, dV' \, F(r') + (-1)^{n-1} \int_M G(r-r') \, dV' \, j(r')$$
Or, equivalently,
$$F(r) = (-1)^{n-1} \oint_{\partial M} G(r-r') \, \hat n |dS'| \, F(r') + \int_M G(r-r') j(r') \, |dV'|$$
It tells that, at any point $r$, the function $F(r)$ is determined by the values on a boundary plus the values of its derivative throughout the volume.
Moreover, only when that derivative is zero is the function determined entirely by its values on the boundary.
In physics, this is common in electromagnetism problems: there may be no currents in a given volume, but there is some source outside the volume instead. These produce values of the field on the boundary, which in turn affect values within the volume.
Edit: In short, if you know a field's derivative everywhere, you should be able to reconstruct the field itself. Stokes' theorem allows us to divide that information into two categories: sources outside a given volume, and sources inside. Information from sources outside is entirely captured by the surface integral; information from sources inside has to be computed through the volume integral.
Functions entirely determined by their surface values are those that, in particular, have no sources within the volume. |
H: The relationship between plane curves and the derivative of the Wronskian
I have found a theorem but I did not understand the proof. I'm looking for a clarification of the proof or a different proof.
Let $f_1, f_2, f_3$ be the three components of a curve in $R^3$ parameterized by $t$. Let $W$ be the Wronskian matrix of $f$. whose $(i,j)$ entry is $f_j^{(i-1)}$.
Theorem: If $|W'| = 0$ then $f$ is a planar curve i.e. it satisfies $a_1 f_1 + a_2 f_2 + a_3 f_3 + a_0 = 0$ for some constant $a_i$s.
The proof goes something like this:
We can find $b_i$ which are ordinary functions of $t$ such that $W' b = 0$, that is, $\sum b_i f_i' = 0, \sum b_i f_i'' = 0, \sum b_i f_i''' = 0$.
Although the proof does not explain why such functions $b_i$ exist, I imagine I could watch the null space of $W'$ evolve smoothly with time, and select an evolving vector $b$ living inside that nullspace.
By takng the derivative of the equations, we can arrive at $\sum b_i'f_i' = 0, \sum b_i f_i'' = 0$. Then the proof says something I don't understand: "The $b$'s are proportional respectively to the cofactors of the elements of the last row in the determinant of $W'$. The same is true of the $b'$'s"
Then we get $\frac{b_1'}{b_1} = \frac{b_2'}{b_2} =\frac{b_3'}{b_3} = \phi(t)$ and then $b_i = a_i e^\phi$ so $\sum a_i e^\phi f_i' = 0$ so $e^\phi\sum a_i f_i' = 0$ and then we remove the exponential and integrate to get the final result.
AI: $W'=W(f'), f=(f_1,f_2,f_3)$
If a set of analytic functions has a wronskian of $0$ then the functions are linearly dependent, since the $0^{th}$ to $n-1^{th}$ derivatives of each function can be considered as components of a vector, and checking if the determinant is $0$ is a standard test of linear dependence. If they are linearly dependent, there will be one coefficient for each function, $n$ in total, which is roughly why the derivatives up to $n-1$ contain enough information to determine linear dependence/independence.
$W'=0$ implies that there exist $a_1,a_2,a_3$ such that $a_1f_1'+a_2f_2'+a_3f_3'=0$. Integrating gives the required result. |
H: permutation/combination problem
There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door and leave from another door?
I have done like this: They way of entering is 3 and exiting is also
3, therefore the total way will be 3*3=9.
AI: That’s the total number of ways in which he can enter and leave the room. The number of ways in which he can enter the room by one door and leave by another door, however, is only $3\cdot 2=6$: he has only $2$ choices of doors through which he may exit, since he must not leave by the door through which he entered. |
H: Limit: $\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$
How can I find the following limit?
$$\lim_{x \to 0}{[1+\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)]}^{1/x}.$$
It's a limit of type $\displaystyle 1^{\infty}$ and if I note with $\displaystyle f(x)=\ln{(1+x)}+\ln(1+2x)+\ldots+\ln(1+nx)$ then I must do:
$$\displaystyle \lim_{x \to 0}{(1+f(x))}^{1/x}=\lim_{x \to 0}[(1+f(x))^{\frac{1}{f(x)}}]^{\frac{f(x)}{x}}. $$
Is ok? Thanks :)
AI: Write the expression above as
$$\lim_{x \to 0}\left [ 1 + \sum_{k=1}^n \log{(1+k x)}\right ]^{1/x}$$
Note that $n$ remains finite, so we can Taylor expand the logs to get
$$\lim_{x \to 0}\left [ 1 +\sum_{k=1}^n k x\right]^{1/x} = \lim_{x \to 0}\left(1+\frac12 n (n+1) x\right)^{1/x} = e^{n(n+1)/2}$$
Note that the error term in the Taylor expansion is $O(x^2)$, which in this limit produces an additional term as
$$(1+n(n+1)/2 x + c x^2)^{1/x} \sim \left(1+\frac12 n (n+1) x\right)^{1/x} \left ( 1+\frac{c x^2}{1+n (n+1) x/2}\right )^{1/x} \sim\left(1+\frac12 n (n+1) x\right)^{1/x} e^{c x}$$
The error is then
$$(1+n(n+1)/2 x + c x^2)^{1/x} - (1+n(n+1)/2 x )^{1/x} = \left(1+\frac12 n (n+1) x\right)^{1/x} (e^{c x}-1)$$
As $x \to 0$, the error $\sim c x$. |
H: Characterization for compact sets in $\mathbb{R} $ with the topology generated by rays of the form $\left(-\infty,a\right) $
I'm trying to find a sufficient and necessary condition for a subset to be compact in $\mathbb{R} $ when the topology is generated by the basis $\left\{ \left(-\infty,a\right)\,|\, a\in\mathbb{R}\right\} $. I thought it might be something similar to the standard characterization for compact sets in $\mathbb{R}$ with the standard topology like perhaps "closed and upper bounded" but that hasn't really led anywhere.
Help would be appreciated.
AI: HINT: Let $A\subseteq\Bbb R$. Suppose that $A$ has a largest element, say $a$; what happens when you’ve covered $a$ with an open set? Now suppose that $A$ does not have a largest element, and consider the cover $\{(\leftarrow,a):a\in A\}$. |
H: How many friends for birthday party every weekend?
A friend claims that he is invited to a birthday party every weekend.
I know he needs to have at least 52 friends for that, but what is the "realistic" amount of friends that chances are 100% for beeing invited every weekend
AI: There is on $100\%$ guarantee, but you can pick any lower chance and find a number of friends. Say we want a $50\%$ chance that he gets invited every weekend of the year. If you have a favorite number, you can follow this through. We need the probability $p$ of getting invited any given weekend to satisfy $p^{52}=\frac 12$, which gives $p = \left(\frac 12 \right)^{\frac 1{52}}\approx 0.98676$. Now with $n$ friends, the chance of getting an invitation on a given weekend is $1-(1-\frac 1{52})^n$, so we want $$1-\left(1-\frac 1{52}\right)^n= \left(\frac 12 \right)^{\frac 1{52}}\\1-\left(\frac 12 \right)^{\frac 1{52}}=\left(1-\frac 1{52}\right)^n\\ \frac{\log\left(1-\left(\frac 12 \right)^{\frac 1{52}}\right)}{\log\frac {51}{52}}=n\\ n \approx 223 $$ This seems reasonable. On average he will get about $4.288$ invitations per weekend. Using a Poisson model, that says the chance of a miss on a given weekend is one in $\boxed{73}$ |
H: possible combinations of 3-digit
How many possible combinations can a 3-digit safe code have?
Because there are 10 digits and we have to choice 3 digits from this,
then we may get $10^P3$ but A author used the formula $n^r$, why is that. What the problem in my calculations?
AI: The error is exactly the one that I suspected in my comment. $_{10}P_3=\frac{10!}{3!}=10\cdot9\cdot8$. This is the number of ways of choosing a $3$-digit combination in which the three digits are all distinct: there are $10$ ways to choose the first digit, but then only $9$ choices remain for the second digit, and once those have been chosen only $8$ choices remain for the last digit.
The problem, however, does not require the digits to be distinct. Thus, there are $10$ choices for the first digit, and since you’re allowed to repeat digits, there are still $10$ choices available for the second digit, and again $10$ for the third. These choices can be made in $10\cdot10\cdot10=10^3$ different ways, so that’s the number of $3$-digit combinations.
As an independent check, notice that the $3$-digit combinations are just the integers from $0$ through $999$, padded on the left with zeros to bring them up to $3$ digits. There are $999$ integers from $1$ through $999$, and $000$ makes the thousandth, so there are $1000=10^3$ such integers. |
H: Drawing balls with replacement, until I have one of each.
A urn has (n+1) types of balls, n of unique colors and the rest black. When picking a ball randomly from the urn, a colored (non black) ball has a probability of p of being picked. Each ball of color has equal probability of being picked, ie each has a (p/n) chance of being picked. The urn has infinite balls/we are picking with replacement.
Let X be the number of picks until we get at least 1 of each n colored balls.
What is the Expected value of X ? What is the probability distribution of X ?
AI: Hint: you have to pick $\frac 1{p}$ balls to get a colored one, so multiply that by the solution to the Coupon collector's problem |
H: Solve the recursion, $a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$
Bring the following recursion relation to an explicit expression:
$$a_n = 3a_{n-1}-3a_{n-2}+a_{n-3}+8$$
$a_{0} = 0$, $a_1 = 1$, $a_2 = 2$
All the examples I have seen were with maximum 2 steps back ($a_{n-2}$) and I thought I know how to solve those but I'm having a hard time both with the Generating function and the Characteristic polynomial methods.
The Generating function should start from $n = 3$ but what would happen to the defined values?
For the Characteristic polynomial, Does it mean I'll have to find the roots of a polynomial from a 3rd degree?
AI: If you use the characteristic polynomial, it will indeed be a cubic, but it’s a cubic that factors very easily: it’s
$$x^3-3x_2+3x-1=(x-1)^3\;.$$
If you use generating functions, note that you can write the recurrence as
$$a_n=3a_{n-1}-3a_{n-2}+a_{n-3}+8-8[n=0]-7[n=1]-9[n=2]\;,\tag{1}$$
valid for all $n\in\Bbb Z$ if you make the blanket assumption that $a_n=0$ for all $n<0$. The last three terms are Iverson brackets and are there to make the recurrence yield the correct initial values.
Now multiply $(1)$ by $x^n$ and sum over $n\ge 0$:
$$\sum_{n\ge 0}a_nx^n=3\sum_{n\ge 0}a_{n-1}x^n-3\sum_{n\ge 0}a_{n-2}x^n+\sum_{n\ge 0}a_{n-3}x^n+8\sum_{n\ge 0}x^n-8-7x-9x^2\;.\tag{2}$$
If the generating function is $A(x)$, $(2)$ can be rewritten as
$$A(x)=3xA(x)-3x^2A(x)+x^3A(x)+\frac8{1-x}-8-7x-9x^2\;,$$
which you can then solve for $A(x)$ and resolve into partial fractions. |
H: Square-Trangular Numbers Checking Answer
Problem: The first 2 numbers that are both squares and triangles are 1 and 36. Find the next one and if possible, the one after that.
Answer: 1225, 41616
Problem: Can you figure out an efficient way to find triangular-square numbers?
Answer: $s^2 = t(t+1)/2$, where $s,t \epsilon{} \textbf{Z}$.
Problem: Do you think that there are infinitely many?
Answer: Yes. (Can someone give me an justification for my answer?)
AI: Your answer is correct: the third and fourth both-triangular-and-square numbers are 1225 and 41616. There are infinitely many such numbers, all of them being solutions to the equation $s^2 = \frac{1}{2}t(t+1)$ in integers. What you didn't provide is an efficient method for finding further solutions to this equation.
The equation can be rewritten as $(2t+1)^2 - 8s^2 = 1$, which is just a slightly disguised Pell's equation. There are a few general methods for solving such equations. Applying one of them to your case yields a simple recurrence for values of $s$: $s_0=0$, $s_1=1$, $s_{k+2} = 6s_{k+1} - s_k$. |
H: Integrate ${\sec 4x}$
How do I go about doing this? I try doing it by parts, but it seems to work out wrong:
$\eqalign{
& \int {\sec 4xdx} \cr
& u = \sec 4x \cr
& {{du} \over {dx}} = 4\sec 4x\tan 4x \cr
& {{dv} \over {dx}} = 1 \cr
& v = x \cr
& \int {\sec 4xdx} = x\sec 4x - \int {4x\sec 4x\tan 4xdx} \cr
& \int {4x\sec 4x\tan 4xdx} : \cr
& u = 4x \cr
& {{du} \over {dx}} = 4 \cr
& {{dv} \over {dx}} = \sec 4x\tan 4x \cr
& v = {1 \over 4}\sec x \cr
& \int {4x\sec 4x\tan 4xdx} = x\sec 4x - \int {\sec 4x} dx \cr
& \int {\sec 4xdx} = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr} $
I don't know where to go from here, everything looks like it equals 0, where have I went wrong?
Thank you!
EDIT: Is there an easier way to do this?
AI: Let $4x = t$, i.e., $4dx = dt$.
$$I=\int \sec(4x) dx = \dfrac14\int \sec(t) dt = \dfrac14\int \dfrac{dt}{\cos(t)} = \dfrac14 \int \dfrac{\cos(t) dt}{\cos^2(t)} = \dfrac14 \int \dfrac{\cos(t) dt}{1-\sin^2(t)}$$
Now let $\sin(t) = y$, to get $\cos(t) dt = dy$. Hence, we get that
$$I = \dfrac14 \int \dfrac{dy}{1-y^2} = \dfrac18\left(\int\dfrac{dy}{1+y} + \int\dfrac{dy}{1-y}\right) = \dfrac18\left(\log(\vert 1+ y \vert) - \log(\vert 1 - y \vert)\right)+\text{const}$$
This gives us
$$I = \dfrac18 \log \left(\left\vert \dfrac{1+y}{1-y}\right\vert \right) + \text{const} = \dfrac18 \log \left(\left \vert \dfrac{1+\sin(t)}{1-\sin(t)}\right \vert\right)+\text{const} = \color{red}{\dfrac18 \log \left(\left \vert \dfrac{1+\sin(4x)}{1-\sin(4x)}\right \vert\right)+\text{const}}$$
EDIT
You can simplify it further as much as you like.
\begin{align}
\dfrac{1+\sin(4x)}{1-\sin(4x)} & = \dfrac{1+\sin(4x)}{1-\sin(4x)} \times \dfrac{1+\sin(4x)}{1+\sin(4x)} = \dfrac{(1+\sin(4x))^2}{1-\sin^2(4x)}\\
& = \dfrac{(1+\sin(4x))^2}{\cos^2(4x)} = (\sec(4x) + \tan(4x))^2
\end{align}
Hence,
$$\boxed{\color{blue}{I = \dfrac{\log(\vert \sec(4x) + \tan(4x)\vert)}4 + \text{const}}}$$ |
H: Elliptic Surfaces: a naive question
Ground field $\Bbb{C}$. Algebraic category. Smooth surfaces.
Let $S$ be a minimal elliptic surface $p:S\rightarrow C$ the elliptic fibration (general fiber = elliptic curve). Suppose the $m$-canonical system is non-empty and let $D\in \lvert m K \rvert$.
Why can we say that $D.F=0$, where $F$ is a fiber of $p$ ?
It appears to be very obvious, but not for me. Can't it be positive?
Enlighten me, s'il vous plait !
AI: Recall that by adjunction, $2g-2=F^2+K.F$ where $g$ is the genus of $F.$ But in this case, $g=1,$ and $F^2=0,$ so we end up with $F.K=0.$ This implies in particular that any $D\sim mK$ also satisfies $D.F=0.$ |
H: Can someone clarify this implication
I'm reading a finance book, and I saw this implication that I don't understand. I mean where this g function come from? If someone can clarify this I would appreciate. Thanks.
If a have a function like $f(x,t)$ and the following equation
$ \frac{d f}{f}=\sigma dx \Rightarrow lnf - lng(t)=\sigma x$
AI: What you actually are looking at is the equation
$$\frac{df(x,t)}{dx}=\sigma f(x,t)$$
One common trick of solving this, is abusing the notation and writing this as
$$\frac{df}{f}=\sigma dx$$
and then integrating both sides. This is what was done in the book.
Note however that we are differentiating with respect to $x$, although the function depends on $x$ and $t$. One way of understanding this, is that you solve the equation for every possible $t$, one at a time. So imagine a fixed $t$ and solve the differential equation. You will get a constant from the integration. But for any 'other' given $t$ the constant you get, might be a different one. So if you look at the global picture, the constants depend on $t$. In other words, they form a function $g(t)$.
If you want to do a reality check and just convince yourself that the solution makes sense, rearrange the solution for $f$ and differentiate. You will see that the function $g(t)$ dies during the process. |
H: rounding up to nearest square
Say I have x and want to round it up to the nearest square. How might I do that in a constant time manner?
ie.
$2^2$ is 4 and $3^2$ is 9. So I want a formula whereby f(x) = 9 when x is 5, 6, 7 or 8. What it does when x = 4 or 9 doesn't really matter.
I could write a function to do this easily enough. ie.
while (sqrt(x) is not int) {
x++;
}
But I'd rather not do a brute force approach. Any ideas?
Thanks!
AI: You can take the ceiling of $\sqrt{x}$ and square it. |
H: Can someone clarify Example I.I.2 from Hardy's Course of Pure Mathematics?
"If $\lambda, m,$ and $n$ are positive rational numbers, and $m > n$, then $\lambda(m^2 − n^2), 2\lambda mn$, and $\lambda(m^2 + n^2)$ are positive rational numbers. Hence show how to determine any number of right-angled triangles the lengths of all of whose sides are rational."
What does he mean by "how to determine any number of right angled triangles"?
AI: We know that $3$ lengths can form a right angled triangle iff those lengths satisfy $a^2+b^2=c^2$.
But no matter what we choose for $m, \; n$ and $ \lambda$.
$$(2\lambda mn)^2+(\lambda(m^2-n^2))^2=4\lambda^2m^2n^2+\lambda^2m^4-2 \lambda^2m^2n^2+\lambda^2n^4$$
$$=\lambda^2m^4+2\lambda^2m^2n^2+\lambda^2n^4=(\lambda(m^2+n^2))^2$$ as we wanted.
So we can choose any number of rationals $m, \; n$ and $\lambda$ to produce any number of right-angled triangles.
A simple example of an infinite solution set is if you keep $m$ and $n$ constant, but vary $\lambda$, you will get infinite solutions as there are an infinite amount of rational numbers. |
H: equality of integrals without trigonometry
can someone show the equality of these two integrals without using any trigonometry;
$$ \int_{0}^{1} \frac{dt}{\sqrt{1-t^2}} = 2 \int_{0}^{1} \sqrt{1-t^2} \, dt $$
i'm working through a derivation of relationship between a circle's circumference and area and this equality isn't obvious to me. that is I can't see how to do it without trigonometric substitutions which I don't want to make since $\pi$ hasn't been defined yet...
AI: Integrate by parts. First, show that
$$\int_0^1 dt \, \sqrt{1-t^2} = \int_0^1 dt \frac{t^2}{\sqrt{1-t^2}}$$
You do this by noting that
$$\int_0^1 dt \, \sqrt{1-t^2} = [t \sqrt{1-t^2}]_0^1 - \int_0^1 dt \, t \frac{d}{dt} \sqrt{1-t^2}$$
Then note that
$$\int_0^1 \frac{dt}{\sqrt{1-t^2}} - \int_0^1 dt \frac{t^2}{\sqrt{1-t^2}} = \int_0^1 dt \, \sqrt{1-t^2}$$
Combining these two equations, you get the stated relation. |
H: Convergence of sequence
Does the following:
$$
\begin{align}
x_0 & = a \\
x_1 & = x_0 + \frac{1}{1}(x_0 + x_0(c - 1)) \\
x_2 & = x_1 + \frac{1}{2}(b + x_1(c - 1)) \\
x_3 & = x_2 + \frac{1}{3}(b + x_2(c - 1)) \\
& {}\,\vdots \\
x_n & = x_{n-1} + \frac{1}{n}(b + x_{n-1}(c - 1))
\end{align}
$$
where $x_{0}$ is switched to b in the $x_{2}$ term onwards, converge to:
$$\frac{b}{1 - c}\quad\text{?}$$
AI: First, the constant $c$ is slightly misleading. Why not take $d := 1-c$ so that: $$x_{n+1} = \left(1-\frac{d}{n}\right) x_{n} + \frac{b}{n}$$
Let $y_n := x_n - t$ for some $t$ that we'll establish in a second. Then:
$$ y_{n+1} + t = \left(1-\frac{d}{n}\right) y_{n} + \left(1-\frac{d}{n}\right)t + \frac{b}{n}$$
which is more naturally written as:
$$ y_{n+1} = \left(1-\frac{d}{n}\right) y_{n} + \frac{b-td}{n}$$
Thus, let $t := \frac{b}{d}$, so that we have:
$$ y_{n+1} = \left(1-\frac{d}{n}\right) y_{n} $$
This allows you to compute $y_n$ almost explicitly:
$$ y_{n+1} = y_k\prod_{m=k}^n \left(1-\frac{d}{m}\right) $$
where $k$ is reasonably small (say, $k=3$). It is a classical theorem in analysis that because $\sum_m \frac{d}{m}$ diverges to $+\infty$, the product $\prod_{m=k}^n \left(1-\frac{d}{m}\right) $ converges to $0$ as $n \to \infty$. Thus, we have $y_n \to 0$ as $n \to \infty$.
We may now recover $x_n$ and find its limit: $x_n = y_n + t \to t = \frac{b}{1-c}$. Hence, the conjectured convergence is true. And it doesn't really matter is anything changes at the initial steps in the induction. |
H: Integral of $\cot^2 x$?
How do you find $\int \cot^2 x \, dx$? Please keep this at a calc AB level. Thanks!
AI: We use the identity: $$\cot x = \pm \sqrt{\csc^2 x - 1} \implies \cot^2 x = \csc^2 x - 1$$
So we can rewrite the integral as follows:
$$\int \cot^2 x \,dx = \int \left(\csc^2x - 1\right)\, dx$$
$$ \int \left(\csc^2x - 1\right) dx \; = \; \int \csc^2 x \, dx\; -\; \int \,dx \;\;= \;\;-\cot x - x + \text{constant}$$
Recall, $$\dfrac{d}{dx}\left(\cot x\right) = - \csc^2 x$$ hence $$\int \csc^2 x \,dx = -\cot x + C$$ |
H: If $\mathbb f$ is analytic and bounded on the unit disc with zeros $a_n$ then $\sum_{n=1}^\infty \left(1-\lvert a_n\rvert\right) \lt \infty$
I'm going over old exam problems and I got stuck on this one.
Suppose that $\mathbb{f}\colon \mathbb{D} \to \mathbb{C}$ is analytic and bounded. Let $\{a_n\}_{n=1}^\infty$ be
the non-zero zeros of $\mathbb{f}$ in $\mathbb{D}$ counted according to multiplicity. Show that
$$ \sum_{n=1}^\infty \left( 1 - \left|a_n \right|\right)\lt\infty $$
I can understand that $\left|a_n \right|$ goes to $1$ since zeros are isolated, but it doesn't help showing the series is convergent. Any help will be appreciated!
AI: This is known as Blaschke's condition and is in fact also true for functions in the so called Nevanlinna class. The simplest way to prove this is using Jensen's formula.
Assume $f \in H^\infty$. You may as well assume that $f(0) \neq 0$ and that $f$ has infinitely many zeros. Let $n(r)$ be the number of zeros in the disc $D_r$. Fix any integer $k$ and choose $r < 1$ so large that $n(r) > k$. By Jensen's formula,
$$
|f(0)| \prod_{n=1}^{n(r)} \frac{r}{|a_n|} = \exp\left( \frac{1}{2\pi} \int_0^{2\pi} \log|f(re^{i\theta})|\,d\theta \right).
$$
Hence (if $|f(z)| < M$ on $D$):
$$
|f(0)| \prod_{n=1}^{k} \frac{r}{|a_n|} \le |M|.
$$
In other words
$$
\prod_{n=1}^{k} |a_n| \ge \frac{|f(0)|}{|M|} r^k
$$
for every $k$. Letting $r \to 1$ and $k\to\infty$ it follows that
$$
\prod_{n=1}^{\infty} |a_n| \ge \frac{|f(0)|}{|M|} > 0,
$$
which implies that $\sum_{n=1}^\infty (1-|a_n|) < \infty$. |
H: How to integrate $\cos\left(\sqrt{x^2 + y^2}\right)$
Could you help me solve this?
$$\iint_{M}\!\cos\left(\sqrt{x^2+y^2}\right)\,dxdy;$$
$M: \frac{\pi^2}{4}\leq x^2+y^2\leq 4\pi^2$
I know that the region would look like this and I need to solve it as difference of two regions. Lets say $C=A\cup B$, then $A = C - B$
I was able to get the integrals, but don't know how to solve them.
$$\int_{0}^{2\pi}\!\int_{0}^{\sqrt{4\pi^2-x^2}}\!\cos\left(\sqrt{x^2+y^2}\right)\,dydx - \int_{0}^{\pi/2}\!\int_{0}^{\sqrt{\frac{\pi^2}{4}-x^2}}\!\cos\left(\sqrt{x^2+y^2}\right)\,dydx$$
AI: Change to polar coordinates. Thus $\sqrt{x^2+y^2}$ gets replaced by $r$. and $dx\,dy$ gets replaced by $r\,dr\,d\theta$. So we end up needing to integrate $r\cos r\, dr\,d\theta$ over our region.
Note that the limits of integration for $r$ are $\frac{\pi}{2}$ and $2\pi$. If the region you are interested in is just the first quadrant region, as in the picture, then $\theta$ will run from $0$ to $\frac{\pi}{2}$.
The integration with respect to $r$ is done using integration by parts. Let $u=r$ and $dv=\cos r\,dr$. Then $du=dr$ and we can take $v=\sin r$. So
$$\int r\cos r\,dr=r\sin r-\int \sin r\, dr=r\sin r +\cos r+C.$$
Added: So for the first quadrant region, our integral is
$$\int_{\theta=0}^{\pi/2} \int_{r=\pi/2}^{2\pi} r\cos r\,dr\,d\theta.$$
The integration with respect to $\theta$ will be easy, since there is no dependence on $\theta$. |
H: $\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_2\cdot\aleph_1^{\aleph_0}$
I've seen this statement in multiple posts (e.g. here and here), but I can't seem to understand it. I can see why
$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0},$$
by noting that every $\alpha<\omega_2$, so $|\alpha|\leq\aleph_1$, meaning
$$\sum_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\max_{\alpha<\omega_2}|\alpha|^{\aleph_0}=\aleph_1^{\aleph_0}.$$
So how does $\aleph_2$ enter the equation?
AI: The sum has $\aleph_2$ terms, each of which is at most $\aleph_1^{\aleph_0}$, so it’s bounded above by $\aleph_2\cdot\aleph_1^{\aleph_0}$. On the other hand, it’s clearly at least $\aleph_2$ and at least $\aleph_1^{\aleph_0}$, so its bounded below by their maximum, which is simply their product. |
H: $X$ topological space. $A$ open $A \cap Y = \emptyset \ \ \Longrightarrow A \cap \bar{Y} = \emptyset$?
I know this is an easy question, but I cannot demonstrate it properly.
Suppose by contradiction that $A \cap \bar{Y} \neq \emptyset$. Then $\exists \ x \in A \cap \bar{Y}$.
I need help formalizing this reasoning (or correcting if it is wrong)
By the definition of $A$ open, $\exists \ U \subset A$ open neighborhood of $x$. Then my idea is to prove that $U \cap Y \neq \emptyset$ because $U$ in somewhat sense "touches" $Y$, because it's a neighborhood of $x$. But i can't write down a formal demonstration.
Any help is appreciated.
Thanks^^
AI: Suppose that $A$ is open, and $A\cap Y=\varnothing$. Let $F=X\setminus A$; then $F$ is closed, and $Y\subseteq F$. Now take closures: $\operatorname{cl}Y\subseteq\operatorname{cl}F=F$, since $F$ is closed. But this immediately implies that $\operatorname{cl}Y\cap A=\varnothing$. |
H: Simplify with fractional exponents and negative exponents
I am trying to simplify
$$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$
It seems pretty simple at first. I know that a negative exponent means you flip a fraction. So I flip it.
$$ \left(\frac{x^2 y^{-1/2}}{3x^{3/2}y^3}\right)^2$$
Now I need to square it, which is tricky because there are a lot of weird rules with squaring. This is probably where I went wrong.
$$ \left(\frac{x^2 y^{-1/2}}{9x^{3/2}y^3}\right)$$
Now I need to try and simplify things. I know that I can get rid of the $x$ on top since there is a larger one on the bottom.
$4/2 - 3/2 = 1/2$
$$ \left(\frac{x^{1/2} y^{-1/2}}{9y^3}\right)$$
Now I need to get rid of the $y$ exponent.
I am not sure how that is possible.
AI: So we have $$ \left(\frac{3x ^{3/2}y^3}{x^2 y^{-1/2}}\right)^{-2} $$
Yes you are right you can "flip" the fraction to remove the negative exponent.
$$ \left(\frac{x^2 y^{-1/2}}{3x^{3/2}y^3}\right)^2$$
Here you multiply every exponent by 2 to square the expression inside the brackets. This is where you went wrong here.
$$ \left(\frac{x^4 y^{-1}}{9x^{3}y^6}\right)$$
But $y^{-1}=1/y$
so the above is equal to
$$ \left(\frac{x^4}{9x^{3}y^7}\right)$$
Now cancel the $x^3$
$$ \left(\frac{x}{9y^7}\right)$$
And we're done. |
H: Exponential integral question
How would I solve the following problem?
$$f(x)=\int\!\frac{4}{\sqrt{e^x}}\,dx$$
Using $u$ substitution I have set $u=e^x$ andd $du=e^x dx$
so would I have $$4\int\!\frac{du}{\sqrt{u}}$$
What would I do from here?
AI: Notice: You may use your method, but in doing so, recall that $\color{blue}{u=e^x}$ andd $du=e^x dx \implies dx = \dfrac{du}{\color{blue}{\bf e^x}} = \dfrac{du}{\color{blue}{\bf u}}$
This gives us
$$4\int \frac{1}{\sqrt{u}}\frac{du}{u}=4\int u^{-3/2}\,du$$ |
H: Express $4+\sqrt{-2}$ as a product of irreducibles
This is part of an old Oxford Part A exam paper. (1992 A1)
Suppose we equip $R=\mathbb{Z}[\sqrt{-2}]$ with the Euclidean function $d$ defined by $$d(m+n\sqrt{-2})=|m+n\sqrt{-2}|^2$$
I want to determine the units of $R$ and express $4+\sqrt{-2}$ as a product of irreducibles and to use this to determine how many ideals of $R$ contain $4+\sqrt{-2}$.
Progress
I think I have shown that the only units are $1,-1$, but cannot see how one can show this element is a product of irreducibles. In general, we have that in a Euclidean Domain every element is a finite product of irreducibles, but I cannot see how to calculate them.
Thanks
AI: Hint: The key is using the Euclidean function. Like many subrings of the complex plane, we have Euclidean function you give, $\phi$, such that $\phi(a+bi) = a^2+b^2$. We know that $\phi$ is multiplicative on $\mathbb{C}$ and we can show that in the case of $\mathbb{Z}[\sqrt{-2}]$, the division algorithm works.
To determine the units of $R$, if $z$ is a unit then $zz^{-1}$=1 so $\phi(z)\phi(z^{-1})=1.$ Thus either $\phi(z)= 1$ or, without loss of generality, $\phi(z)<1$. Could we achieve this in the given ring?
To express $4+\sqrt{-2}$ as a product of irreducibles, if you can't see how to factorise directly, suppose $ab=4+\sqrt{-2}$ (where neither are units). It follows that $\phi(a)\phi(b)=18$. Using that $\mathbb{Z}$ is a UFD, and using our knowledge gained previously about what values of $\phi$ non-unit elements can take, we can determine some possible values of $\phi$ for the factors. Playing around with possible elements of a given "size" should allow you to factorise $4+\sqrt{-2}$ into smaller parts. To show that these each factor is irreducible, suppose that it is expressible as a product of non-unit elements, determine possible the values of $\phi$ the divisors can take and then show that this is not possible. |
H: Generalising the Chinese Remainder Theorem
We have that for $I,J$ ideals of some ring $R$ with $R=I+J$, $$\frac{R}{I\cap J} \cong \frac{R}{I} \times \frac{R}{J}$$
My question is whether the analogous expression for three ideals $I,J,K$ where $R=I+J+K$ is true?
I think I have found a counterexample with $R=\mathbb{Z}$, $I=2\mathbb{Z}$, $J=2\mathbb{Z}$, $K=3\mathbb{Z}$.
Here, $R=I+J+K$ and $\frac{R}{I\cap J\cap K}=\frac{R}{I \cap K}=\frac{\mathbb{Z}}{6\mathbb{Z}} \cong \frac{R}{I} \times \frac{R}{K}$
We have that $\frac{R}{I} \times \frac{R}{K}$ is not isomorphic to $\frac{R}{I} \times \frac{R}{J}\times \frac{R}{K}$ and so this is our counterexample.
Can someone please check this counterexample is sound? Thanks
AI: Your counterexample is correct. The correct generalization of the Chinese Remainder Theorem, obtained by inducting on the statement you gave is:
Suppose $I_1,\ldots,I_n$ are ideals of a ring $R$ which are pairwise comaximal, i.e. for any $1\leq i\ne j\leq n$ we have $I_i+I_j=R$. Then
$$\frac{R}{I_1\cap\cdots\cap I_n}\cong \frac{R}{I_1}\times\cdots\times\frac{R}{I_n}$$ |
H: Dice probability when a number is disallowed in the first round
In the game Settlers of Catan a player starts each turn rolling 2 six sided dice. There's a variation of the game where if a 7 is rolled in the first round of a game (a 'round' is when each player has taken a turn) it doesn't count an needs to be re-rolled. This must lower the overall probability of a 7 being rolled for the rest of the game (and I'd imagine increase the odds for all the other numbers). How would this probability be calculated?
AI: There are exactly 6 ways to roll a seven with two dice:
\begin{array}{|c|c|}
\hline
\text{First die} & \text{Second die} \\
\hline
1 & 6 \\ \hline
2 & 5 \\ \hline
3 & 4 \\ \hline
4 & 3 \\ \hline
5 & 2 \\ \hline
6 & 1 \\ \hline
\end{array}
The total number of ways of rolling two dice is $6 \times 6 = 36$, since there are $6$ ways of rolling each die individually. Therefore, if you don't reroll sevens, you will get a seven with probability $\frac{6}{36} = \frac16$.
If you do reroll sevens, during the time you reroll them
you will have $0$ probability of getting a seven. You will keep rerolling until you get some other case, and each other case is still equally likely, so now there are only $36 - 7 = 29$ cases left, instead of the usual $36$.
The effect of the first round of the game has no bearing on the probability of a seven being rolled for the rest of the game. This is the nature of probability; the two parts of the game are entirely independent. However, one other thing we can compute that maybe you wanted to know is how much rerolling sevens increases the probability of rolling other things. Well, each individual case (such as 1, 1 or 3, 1) now has probability $\frac1{29}$ of being rolled instead of the original $\frac{1}{36}$, so the probability has increased by
$$
\frac{1/29}{1/36} = \frac{39}{29} \approx 1.24137
$$
when sevens are rerolled.
So each number (2,3,4,5,6,8,9,10,11,12) appears about $24$ percent more often than normal during the first round. |
H: Calculating in quotient ring of $\mathbb{R}[X]$
Part of an old Oxford exam (1992 A1)
We want to find which elements of the quotient ring $\mathbb{R}[X]/(x^3-x^2+x-1)$ are equal to their own square.
Now, we note first that $x^3-x^2+x-1=(x-1)(x^2+1)$
Let $f(x)=(x-1)(x^2+1)$. Clearly we have $[1+(f(x))]^2=[1+(f(x))]$ but I cannot see how to find others. Any guidance would be appreciated.
Thanks.
AI: Your previous question concerns generalized CRT. So you know that
$$\frac{{\bf R}[x]}{(x-1)(x^2+1)}\cong\frac{{\bf R}[x]}{(x-1)}\times \frac{{\bf R}[x]}{(x^2+1)}\cong {\bf R}\times{\bf C}$$
Find the idempotents in $\bf R$ and $\bf C$ to find the idempotents in ${\bf R}\times{\bf C}$, then pull them back through the isomorphisms implicit above.
Indeed the idempotents of ${\bf R}\times{\bf C}$ are $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$. The first and last correspond to $0$ and $1$ in $S={\bf R}[x]/(x-1)(x^2+1)$. To find what the others correspond to, solve e.g.
$$f(x)\equiv\begin{cases}0 & \mod x-1, \\ 1 & \mod x^2+1. \end{cases} $$
without loss of generality with $\deg f(x)\le 2$. Thus we can write $f(x)=(ax+b)(x-1)$ and solve for $a,b$ by distributing out this product, reducing mod $x^2+1$ and setting equal to zero. The other case proceeds in exactly the same way. |
H: How can I evaluate $\sum_{n=1}^{\infty} \frac{1+\sin^2 n}{n+n^{1.5}}$?
How can I evaluate this sum? My teacher asked but I can't get it.
$$\sum_{n=1}^{\infty} \frac{1+\sin^2 n}{n+n^{1.5}}$$
AI: You cannot expect a nice closed form answer to this. However, it is easy to show that it is bounded and thereby it converges. We have $\sin^2(n) \in (0,1)$. Hence, $1+\sin^2(n) \in (1,2)$.
We also have $n+n^{1.5} \in (n^{1.5}, 2n^{1.5})$. Hence, we have
$$\dfrac{1+\sin^2(n)}{n+n^{1.5}} \in \left(\dfrac1{2n^{1.5}}, \dfrac2{n^{1.5}} \right)$$
We hence have
$$\sum_{n=1}^{\infty}\dfrac{1+\sin^2(n)}{n+n^{1.5}} \in \left(\sum_{n=1}^{\infty}\dfrac1{2n^{1.5}}, \sum_{n=1}^{\infty}\dfrac2{n^{1.5}} \right) = \left(\dfrac{\zeta(3/2)}2, 2 \zeta(3/2)\right)$$ |
H: Counting Problem - N unique balls in K unique buckets w/o duplication $\mid$ at least one bucket remains empty and all balls are used
I am trying to figure out how many ways one can distribute $N$ unique balls in $K$ unique buckets without duplication such that all of the balls are used and at least one bucket remains empty in each distribution?
Easy, I thought. I'll just hold a bucket in reserve, distribute the balls, and place the empty bucket. I get:
$ K\cdot N! / (N-K-1)! $
Even were I sure this handles the no duplicates condition, what if $K \geq N$?
Then I get a negative factorial in the denominator. Is the solution correct and/or is there a more general solution?
Thanks!
AI: If the number of buckets is greater than the number of balls, then all distributions qualify, so there are $K^N$ ways to do the job.
If $K\le N$, we can use Inclusion/Exclusion. There are $(K-1)^N$ ways to distribute the balls so that bucket $i$ is empty. So our first estimate is $K (K-1)^N$. But this double counts, for each $i$ and $j$, the $(K-2)^N$ distributions that have $i$ and $j$ empty, So from $K(K-1)^N$ we must subtract $\binom{K}{2}(K-2)^N$. But we have subtracted once too many times the $(K-3)^N$ distributions in which $i$, $j$, and $k$ remain empty. And so on. We end up with
$$K(K-1)^N -\binom{K}{2}(K-2)^N+\binom{K}{3}(K-3)^N -c\dots.$$ |
H: Topological extension property
Let $X$ and $Y$ be topological spaces. We say that the extension property holds if, whenver $S$ is a closed subset of $X$ and $f:S\rightarrow Y$ is continuous, $f$ can be extended to a continuous map on $X$.
A question asks whether the extension property holds for $X$ being any topological space and $Y=[-1,1]$.
I have no idea, but I am thinking whether there is some $X$ on which continuous functions are either constant or unbounded.
Can anyone offer any ideas? Thanks!
AI: Let $X$ be any space that is not completely regular, and let $p\in X$ and $F\subseteq X$ be a point and a closed set such that $p\notin F$, and there is no continuous function $f:X\to[0,1]$ such that $f(p)=0$ and $f(x)=1$ for each $x\in F$. Then it’s easy to check that there is no continuous function $f:X\to[0,1]$ such that $f(p)=-1$ and $f(x)=1$ for each $x\in F$.
Let $S=\{p\}\cup F$, and let
$$f:S\to[-1,1]:x\mapsto\begin{cases}-1,&\text{if }x=p\\1,&\text{if }x\in F\;;\end{cases}$$
then $f$ is continuous, but it cannot be extended continuously to $X$. And as long as $X$ is $T_1$, $S$ is closed in $X$.
This answer has a nice example, due to John Thomas, of a $T_3$-space $X$ with distinct points $p$ and $q$ such that every continuous $f:X\to[0,1]$ satisfies $f(p)=f(q)$, which is enough: you don’t actually need each continuous $f:X\to[0,1]$ to be constant, though Eric van Douwen showed some years ago that any space like the Thomas example can be used to build a $T_3$-space on which every continuous real-valued function is constant. |
H: Does $n n^{1/n} =O(n)$?
I was asked does $n n^{1/n} =O(n)$ ?
I can see that the left hand side is always bigger than $n$ but how would you prove the equality is false?
AI: The equality is true. Recall that $\mathcal{O}(n)$ means the function is eventually $\leq C n$, where $C$ is a positive constant.
In your case, show that $n^{1/n} < e^{1/e}$ by considering the function $x^{1/x}$. We hence have for all $n \in \mathbb{N}$, $$n \cdot n^{1/n} < e^{1/e} \cdot n$$ Since, we have $n \cdot n^{1/n} < \underbrace{e^{1/e}}_{\text{const}} \cdot n$, $\forall n \in \mathbb{N}$, we have that $n \cdot n^{1/n} \in \mathcal{O}(n)$. |
H: $\pi$ is just a number, or also the circumference of a sub-unit circle?
A unit circle defined in the Cartesian plane has a radius of $1$ and a diameter of $2$. So making a full round is $2 \pi$. Now, $\pi$ is the ratio of the circumference over the diameter, so if I have a circle with diameter $1$ (radius $0\mathord{,}5$), the circumference would be $\pi$ because its $c/d = c/1$ must equal $\pi$. Hence $c= \pi$. So $\pi$ is actually the circumference of a "subunit" circle? or I am missing something here?
AI: I think you're confusing yourself by talking about two different circles, but you are correct that $\pi$ is the circumference of a circle of radius $0.5$.
For a circle of radius $1$, we have
$$r\text{ (radius)}=1\qquad d\text{ (diameter)}=2\qquad c\text{ (circumference)}=2\pi$$
The ratio of the circumference to the diameter is
$$\frac{c}{d}=\frac{2\pi}{2}=\pi.$$
For a circle of radius $0.5$, we have
$$r\text{ (radius)}=0.5\qquad d\text{ (diameter)}=1\qquad c\text{ (circumference)}=\pi$$
The ratio of the circumference to the diameter is
$$\frac{c}{d}=\frac{\pi}{1}=\pi.$$ |
H: Please help finishing the calculation to find the Entropy of Pareto distribution.
Let $X$ follow Pareto distribution with parameters $\alpha, a, h$. That is, $X\sim Pa(\alpha,a,h)$, where $\alpha>0$ is the shape parameter, $-\infty < a < \infty$ is the location parameter, and $h>0$ is the scale parameter.
Then the pdf of $X$ is $p(x) = \left(\alpha\cdot h^{\alpha}\right)/(x-a)^{\alpha+1}$; $x\geq a+h$.
Entropy, $H(X) = -\int_{a + h}^{\infty}p(x)\log p(x)\,dx$
$$H(X) = -\int_{a + h}^{\infty}\frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \log\left(\frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \right)\,dx$$
$$= -\int_{a +h}^{\infty} \frac{\alpha h^{\alpha}}{(x-a)^{\alpha+1}} \left[\log(\alpha h^{\alpha}) -(\alpha+1)\log(x-a) \right]\,dx$$
$$= -\alpha h^{\alpha}\int_{a + h}^{\infty}\frac{1}{(x-a)^{\alpha+1}} \left[\log(\alpha) + \alpha \log(h) -(\alpha+1) \log(x-a) \right]\,dx$$
$$= -\alpha h^{\alpha}\int_{a + h}^{\infty} \frac{ \log(\alpha) + \alpha \log(h)}{(x-a)^{\alpha+1}} dx + \alpha h^{\alpha}\int_{a + h}^{\infty} \frac{(\alpha+1) \log(x-a)}{(x-a)^{\alpha+1}}dx$$
$$= -\alpha h^{\alpha}(\log(\alpha) + \alpha \log(h))\int_{a + h}^{\infty} \frac{ 1}{(x-a)^{\alpha+1}} dx + \alpha h^{\alpha}(\alpha+1)\int_{a + h}^{\infty} \frac{ \log(x-a)}{(x-a)^{\alpha+1}}dx $$
Then how can I end up with the result
$$H(X) = -(1+2\alpha)\log(h) – \log(\alpha) + 1 + (1/\alpha)$$
AI: By using the following (which can be computed using standard integrals, integration by parts and this limit)
$$\int^\infty_{a+h}\frac{1}{(x-a)^{\alpha+1}}dx=\left.-\frac{1}{\alpha(x-a)^\alpha}\right|^\infty_{a+h}=\frac{1}{\alpha h^\alpha},$$
$$\int^\infty_{a+h}\frac{\log(x-a)}{(x-a)^{\alpha+1}}dx=\left.-\frac{\alpha\log(x-a)+1}{\alpha^2(x-a)^\alpha}\right|^\infty_{a+h}=\frac{\alpha\log(h)+1}{\alpha^2h^\alpha},$$
I get that $H(x)=-\log(\alpha)+\log(h)+1+\frac{1}{\alpha}$ which seems to match up with the entropy listed on wikipedia - are you sure that the end result you list above is correct? I could have mucked a minus sign somewhere - but I can't seem to find it. |
H: How can I evaluate this sum?
How can I evaluate this sum ?
$$\sum_{n=1}^{m}n^p$$
I know when $p\in \Bbb N$ but when $p\in\Bbb R$ what do I do ?
please help and thanks for all
AI: If $p \in \mathbb{N}$, you have the Faulhaber's formula, i.e.,
$$\sum_{n=1}^m n^p = \dfrac1{p+1} \sum_{j=0}^p (-1)^j \dbinom{p+1}j B_j m^{p+1-j}$$
where $B_j$ are the Bernoulli numbers, with $B_1 = - \dfrac12$. If $p \not\in \mathbb{N}$, we can still obtain the asymptotic behavior as
$$\sum_{n=1}^m n^p \sim \begin{cases}\dfrac{m^{p+1}}{p+1} & \text{ if } p \neq -1\\ \log(m) + \gamma & \text{ if } p =-1\end{cases}$$ |
H: Why do these two volume integrals express different values?
This is a question from a textbook I am reading:
Find the volume when the region R is revolved about the y-axis
when R is defined as the region bounded by $x = \sqrt{16-y^2}$ and
$x = 0$.
Because the function is symmetrical about the y-axis and being revolved about the y-axis, why don't these two integrals evaluate to the same value?
Integral one: $$\int_0^4\!\pi\left(\sqrt{16-y^2}\right)^2\,dy = \frac{128\pi}3$$
Integral two: $$\int_{-4}^4\!\pi\left(\sqrt{16-y^2}\right)^2\,dy = \frac{256\pi}3$$
It seems to me that they represent the same volume--like creating a semicircular volume by revolving a fourth of a circle about its base, or by revolving a semicircle along its base.
I think that the second equation overestimates the volume because it counts the same volume twice. (The textbook says otherwise!)
AI: The function is not symmetrical about the $y$-axis, indeed the curve "lives" in the first and fourth quadrants.
There is symmetry about the $x$-axis, which is why one integral is twice the other.
Remark: Since you are much more familiar with integration with respect to $x$, you might interchange the roles of $x$ and $y$ (geometrically: reflect in the line $y=x$). Then you will see more clearly what is going on. The first integral gives the volume of a half-sphere. The second gives the volume of a full sphere. |
H: How to deal with polynomial quotient rings
The question is quite general and looks to explore the properties of quotient rings of the form $$\mathbb{Z}_{m}[X] / (f(x)) \quad \text{and} \quad \mathbb{R}[X]/(f(x))$$
where $m \in \mathbb{N}$
Classic examples of how one can treat such rings is to find relationships like $$\mathbb{Z}[x]/(1-x,p) \cong \mathbb{Z_{p}}$$ for prime $p$.
However, I often struggle to intuitively understand what the elements of such rings are, and they to compute using them.
For example, what do the elements of the set $\mathbb{Z}_{2}[X] / (x^4+1)$ actually look like. Can this set be described in a clearer way to help understand the way the rings work.
Is there some general way of describing elements of such rings so that isomorphisms and computations become easier to handle?
Apologies if this question is not sufficiently clear.
AI: To treat your special case
$$\mathbb{Z}_{2}[X] / (x^4+1),$$
its elements are the (classes) of the remainders of Euclidean divisions by $x^4+1$, so
$$\mathbb{Z}_{2}[X] / (x^4+1) = \{ [a_0 + a_1 x + a_2 x^2 + a_3 x^3] : a_i \in \mathbb{Z}_{2} \}.$$
(Here I am using brackets to denote residue classes.)
You sum the classes normally, and as to the product, you first take the product, and then take the remainder of Euclidean division by $x^4+1$.
Alternatively, you take the product, and then use the relation $x^4 \equiv -1$ (which is the same as $x^4 \equiv 1$ in this case, as $1 = -1$ in $\mathbb{Z}_{2}$) to reduce the result. So it's not really different from computing in $\mathbb{Z}_{m}$, where you first take an ordinary sum or product, and then take the remainder modulo $m$.
So for instance
$$
[1 + x^2] \cdot [1 + x^3] = [1 + x^2 + x^3 + x^5] = [1 + x + x^2 + x^3],
$$
using one of the two methods above.
The general case of $F[x] / (f(x))$, with $F$ a field, is similar, you get the classes of the remainders of Euclidean division by $f(x)$. |
H: Identity with Bernoulli numbers: $\sum\limits_{k=1}^{n}k^p=\frac{1}{p+1}\sum\limits_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j}$
How I can prove that
$$\sum_{k=1}^{n}k^p=\frac{1}{p+1}\sum_{j=0}^{p}\binom{p+1}{j}B_j n^{p+1-j},$$
where $B_j$ is the $j$th Bernoulli number?
I hope to find the answer. Thanks for help.
AI: This is a consequence of the use of the Bernoulli polynomials. We can define them by $$B_0(x)=1$$ $$B'_{n}(x)=nB_{n-1}(x)$$and $$\int_0^1 B(x)dx=0$$
They have the particular property that $$B_{p+1}(x+1)-B_{p+1}(x)=(p+1)x^{p}$$ which is why we use them to evaluate such a sum, also $B_{n}(0)=B_{n}$, the $n$-th Bernoulli number.
You can find any $B_n(x)$ with the above, and thus evaluate any sum $$\sum_{k=1}^nk^p$$
Give it a try! Set $x=k$ and sum away, to obtain that $$\sum\limits_{k = 1}^n {{k^p} = } \frac{{{B_{p + 1}}\left( {n + 1} \right) - {B_{p + 1}}\left( 0 \right)}}{{p + 1}}$$
In fact, to get your sum to look like in your question you will have to show that $$B_n(x)=\sum_{j=0}^n \binom{n}{j}B_jx^{n-j}$$ where $B_j:=B_j(0)$. Induction should be enough.
ADD This is Faulhaber's Formula. |
H: Quadratic Equation with "0" coefficients
Let's say I have two objects $x$ and $y$ whose position at time $t$ is given by:
$$
x = a_xt^2+b_xt+c_x \\
y= a_yt^2+b_yt+c_y
$$
And I want to find which (if any) values of $t$ cause $x$ to equal $y$. That is
$$
(a_x-a_y)t^2+(b_x-b_y)t+(c_x-c_y) = 0
$$
This can easily be solved with quadratic equation. But what about the case when $a_x = a_y$? Now obviously this makes the problem much simpler to solve by hand! But if I were writing, lets say a simulation, where $a_x$ didn't necessarily equal $a_y$, but it could, how would I go about solving the equation? Preferably I'm looking for some sort of algorithm/equation that I can use to avoid having to write separate logic for the cases where $a_x = a_y$ and $a_x \neq a_y$.
Quick background: I have knowledge of math up to Linear Algebra (I feel a very simple answer lies here, but I can't quite work it out) and ~1/2 a course in Differential Equations (though I think that doesn't really apply here), but I'm more than willing to learn something new if it provides an easy way of solving my problem. Thanks!
AI: You can write a simulation to handle both cases if you resort to numerical solutions as opposed to analytical.
1). You try using Newton-Raphson method, $x_{n+1} = x_{n}-f(x_{n})/f'(x_{n}) $. This method does not care about what the order of f is, and will solve an approximate solution regardless.
2). you do a brute force way of putting values for t, until you get zero, or as close to zero as you can at a reasonable resolution and time. (not recommended) |
H: find recursive solution $T(n)=2T(n/2)+n-1$
I want to solve this:
$$T(n) = 2 T\left(\frac{n}{2}\right) + n - 1 $$
I try :
\begin{align*}
n &= 2^m \\
T(2^m) &= 2T(2^{m-1}) + 2^m -1 \\
2 ^ m &= B \\
T(B) &= 2T(B-1) + B -1 && (1) \\
r - 2 &= 0 \implies r = 2 \implies T (B) = C*2^B && (h)
\end{align*}
particular :
$$
T(B) = AB + D \quad \text{(A and D are const)}
$$
1:
\begin{align*}
AB + DB &= 2(AB + DB) + B -1 \\
A &= -1 \text{ and } D = -1 \\
T(B) &= c \cdot 2^B -1 -B
\end{align*}
and I can't continue solving currectly
please help me
AI: As I said in my comment, you have an error fairly early on, so I am not going to address what all went wrong in your proof.
Here is how I would do it: Let $ k = \log_2 n $ and $ g(k) := 2^k $. Hence $$ T(g(k)) = 2T(g(k - 1)) + 2^k - 1 $$ Let $ f:= T \circ g $. Substituting this in: $$ f(k) = 2f(k - 1) + 2^k - 1 $$ Next, let, $ h(x) := 2^{-x}f(x) $. Hence, $$ h(k) = h(k-1) + 1 - 2^{-k} $$ From here, it is easy to determine a function for the case where the domain of $ h $ is the natural numbers: $$ h(k) = h(0) + \sum_{n = 1}^k 1 - 2^{-n} = h(0) + k + 1 - 2^{-k}$$ Because $ h(x) = 2^{-x} \left(T \circ g\right)(x) $, $ T(x) = 2^x h\left(g^{-1}(x)\right) = 2^{\log_2 x} \left(h(0) + \log_2 x + 1 - \frac{1}{x}\right)$, which reduces to $$ T(x) = Cx + x + x \cdot \log_2 x - 1 $$ where $ C = T(1) $.
The hard part is proving that this is the unique definition across the real numbers. While it is easy to verify that it works on such a domain, that doesn't necessarily imply uniquness. |
H: Linear Transformation Orthogonality
True or False:
If $T$ is a linear transformation from $R^n$ to $R^n$
such that
$$T\left(\vec{e_1}\right), T\left(\vec{e_2}\right), \ldots, T\left(\vec{e_n}\right) $$
are all unit vectors
then $T$ must be an orthogonal transformation
The answer is ?
I know a linear transformation is orthogonal if it preserves the length of vectors.
And I understand a linear transformation is orthogonal if $$T\left(\vec{e_1}\right), T\left(\vec{e_2}\right), \ldots, T\left(\vec{e_n}\right) $$ form an orthonomal basis.
But how do I prove that withough knowing what the $T$, transformaton, does to the elementary column vectors...?
Could anybody help me prove this?
AI: False. Let $T$ be the linear transformation represented by the matrix whose first row is all 1's, and the other rows are all 0's. $T(e_i)=e_1$ for all $i$. |
H: Evaluating the integral: $\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx$
I am interested in evaluating the following integral:
$$
\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx
$$
Using Matlab, Numerically it seems that the integral is convergent,
but I'm not sure about it. How can we prove that the integral is
convergent or not?
Many Thanks in advance.
AI: The integral converges. There is potential trouble at $0$ and "at" infinity. So we split the integral into (i) the part from $0$ to $1$, and (ii) the part from $1$ to $\infty$.
(i) Look at the power series expansion of $2-2\cos x-x\sin x$. The first few terms are $2-2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\right) -x\left(x-\frac{x^3}{31}\right)$. There is cancellation, and the first non-zero term is the $x^4$ term. So our function behaves well as $x$ approaches $0$ from the right: It approaches $\frac{1}{12}$. If we define your integrand to be $\frac{1}{12}$, the resulting function is continuous on $[0,1]$, and hence integrable.
(ii) For large $x$, the integrand is $\lt \frac{K}{x^3}$ for some constant $K$, and we know that $\int_1^\infty \frac{dx}{x^3}$ converges. |
H: Linear Algebra determinant and rank relation
True or False?
If the determinant of a $4 \times 4$ matrix $A$ is $4$
then its rank must be $4$.
Is it false or true?
My guess is true, because the matrix $A$ is invertible.
But there is any counter-example?
Please help me.
AI: You're absolutely correct. The point of mathematical proof is that you don't need to go looking for counterexamples once you've found the proof. Beforehand that's very reasonable, but once you're done you're done.
Determinant 4 is nonzero $\implies$ invertible $\implies$ full rank.
Each of these is a standard proposition in linear algebra. |
H: Deducing a coefficient from a cubic polynomial?
I fully answered the question, and got that $k=-3$, but the answer says it's positive. Can anyone show me my mistake?
"Given that $x-2$ is a factor of the polynomial $x^3 - kx^2 - 24x + 28$, find $k$ and the roots of this polynomial."
Using factor theorem, I realised that $P(2)$ is equivalent to $0$, therefore $2^3 - 2^2k - 24(2) + 28 = 0$
I solved it algebraically and got that $k=-3$, but the answers say it was $k=3$. Did I make a simple error?
Any help would be appreciated.
AI: No error on your part:
If you've copied the problem correctly, then your solution is correct: $k = -3$.
I was very careful in calculating, as I'm sure you were, in double checking, so if $(x - 2)$ is a factor for your given polynomial, then $k$ must be $-3$.
Typo/misprint I suspect, in your text: a typo in the solution, or a misprint of the desired polynomial.
E.g. If the polynomial had been
$$x^3 \color{blue}{\large \bf +} 4k^2 - 24x + 28\quad \text{and}\;\; (x-2) \;\text{is a factor}$$
then $k = 3$. |
H: On Landau notations
How common it is to write e.g. $1-o(1)$ for a function that eventually approaches $1$ from below (or eventually equals $1$)? Would a better notation be $1-|o(1)|$ or what is meant is already obvious from $1-o(1)$ ? Obviously, precisely defining everything will work eventually, but I was wondering about the general convention. Similarly, are notations like $|\omega(1)|$ common?
Edit: In other words, would you say "What the hell is this?" is you saw both $1+o(1)$ and $1-o(1)$ in the same sentence, or would you say "Hmm, ok I got it, these are all assumed to be positive"?
AI: The definition of little-Oh I recall is:
$$
f=o(g) \Leftrightarrow \exists\varepsilon, \varepsilon\to 0, f=\varepsilon g
$$
In particular, it doesn't really make sense to write "$-o(1)$"; according to this definition, if $f$ is $o(g)$, $(-f)$ is also $o(g)$. Another convention to circumvent that is to write $f\to 1^-$ or $f\to 1^+$ (if the limit is reached resp. from below or above). Or $\lim f= 1^-$ ($\lim f= 1^+$).
Rk: I've also seen teachers write $f\nearrow 1$ and $f\searrow 1$, but I'm not sure it is standard. |
H: Matrix involving values of polynomials
I've been doing this problem but im stuck.
Be $f_1 f_2 f_3 \in \mathbb{R}_2$[$x $]. Proove that {$f_1$,$ f_2$,$ f_3$} form a base of $\mathbb{R}_2$[$x $] as $\mathbb{R}$ vector space, if and only if in $\mathbb{R}^{3x3}$ this matrix is invertible:
$ \begin{bmatrix} f_1(1) & f_1(2) & f_1(3) \\ f_2(1) & f_2(2) & f_2(3) \\ f_3(1) & f_3(2) & f_3(3)\end{bmatrix}$
my attempt: $\leftarrow$) i want to proove that $ \lambda_1 f_1 +\lambda_2 f_2 +\lambda_3 f_3 = 0 $ then $ \lambda_i = 0 $ $ $ $ \forall i$ so they're linearly independant and then form a base. first i transpose the matrix, because i know that if a square matrix is invertible its transpose is too. Now i know that if i multiply the matrix by a vector ($ \lambda_1 \lambda_2 \lambda_3 $ ) and ask for the homogeneous solution i get that this ocurr if and only if $\lambda_i = 0 $ for $i= 1,2,3$ Then they're linearly independant and form a base.$ \\ $
$\\ \rightarrow$)...
AI: $\Rightarrow$ By contrapositive: suppose the matrix is not invertible — in this case, its transpose is not either, and there exists $\begin{pmatrix} a \\b \\c\end{pmatrix}\neq \begin{pmatrix} 0 \\0 \\0\end{pmatrix}$ such that
$$
\begin{pmatrix} f_1(1) & f_2(1) & f_3(1) \\ f_1(2) & f_2(2) & f_3(2) \\ f_1(3) & f_2(3) & f_3(3)\end{pmatrix}\begin{pmatrix} a \\b \\c\end{pmatrix}=\begin{pmatrix} 0 \\0 \\0\end{pmatrix}
$$
i.e.
$$
\begin{align*}
a f_1(1) + b f_2(1) + c f_3(1) &= 0 \\
a f_1(2) + b f_2(2) + c f_3(2) &= 0 \\
a f_1(3) + b f_2(3) + c f_3(3) &= 0
\end{align*}
$$
En posant $P\stackrel{\rm{}def}{=}af_1+bf_2+cf_3\in\mathbb{R}_2[X]$, we have
$$
P(1)=P(2)=P(3)=0
$$
and therefore $P=0$. But since by assumption $(a,b,c)\neq(0,0,0)$, it means that $(f_1,f_2,f_3 )$ is linearly dependent, thus not a basis of $\mathbb{R}_2[X]$. |
H: A question on linear operators
This is a problem I’ve been working on as part of my studies for an upcoming comprehensive exam:
Let $F$ be a field, let $V\in F$-$\mathrm{Mod}$ be a finite-dimensional left $F$-vector space, and let $T\in\mathrm{End}_{F}\left(V\right)$ be an $F$-linear operator.
(a) Suppose that $\mathrm{rk}(T)=\mathrm{rk}\left(T^2\right)$. Prove that $\mathrm{ker}(T)\cap T[V]=\left\{0_V\right\}$, and deduce from this that $V=\mathrm{ker}(T)\oplus T[V]$.
(b) Prove that $V=\mathrm{ker}\left(T^k\right)\oplus T^{k}[V]$ for some $k\in\mathbb{N}$.
I’ve solved part (a); if $v\in\mathrm{ker}(T)\cap T[V]$, then $v=T(w)$ for some $w\in V$ and $T(v)=0_V$, so $w\in\mathrm{ker}\left(T^2\right)$. By assumption, $\mathrm{rk}(T)=\mathrm{rk}\left(T^2\right)$. By the Dimension Theorem, $\dim(V)=\mathrm{rk}(T)+\dim\left(\mathrm{ker}(T)\right)=\mathrm{rk}\left(T^2\right)+\dim\left(\mathrm{ker}\left(T^2\right)\right)$, so $\dim\left(\mathrm{ker}(T)\right)=\dim\left(\mathrm{ker}\left(T^2\right)\right)$. Since, clearly, $\mathrm{ker}\left(T^2\right)\supseteq\mathrm{ker}(T)$ (as $T$ is $F$-linear), we have that $\mathrm{ker}(T)=\mathrm{ker}\left(T^2\right)$ (as $V$ is finite-dimensional). Hence $v=0_V$. The Dimension Theorem then allows us to conclude that $V=\mathrm{ker}(T)\oplus T[V]$.
A solution to part (b) has as yet eluded me. Any hints would be appreciated.
AI: Hint: Notice that the ranks of $T^k$ form a decreasing integer sequence bounded below by 0. Hence eventually they become constant, and thus the ranks of $T^n$ and $T^{?}$ for all $?>n$ must be the same. What choice of $?$ allows you to use a result you already know? |
H: Is it possible that a vector space can be a finite union of proper subspaces?
Let $V$ be a vector space, and let $V_i$ for $i=1,\ldots, n$ be non-zero subspaces of $V$. Is it possible that $V=\cup_{i=1}^n V_i?$
The underlying field is assumed to be infinite.
AI: Let $V$ be a vector space over an infinite field $k$. Assume $V=\cup_{i=1}^nV_i$, where $V_i$ are all proper subspaces. Let $u\in V_1$, and take any $v\in V\setminus V_1$. There are infinitely many vectors of the form $u+cv$ for $c\in k^*$, none of which are in $V_1$. Since there are infinitely many of them, and only finitely many subspaces, there must be some $V_j$ containing at least two of them. It follows that both $v$ and $u$ are in $V_j$ so that $V_1\subset \cup_{i=2}^nV_i$.
Continuing this process, we can show that $V=V_n$, contradicting the assumption that the subspaces are proper.
I guess the assumption that $k$ is infinite is a bit overkill. We only need that there are at least as many elements in the field as subspaces in the union to derive a contradiction. This shows that even over finite fields, we need at least as many subspaces as elements in the field to obtain such a decomposition. |
H: Set Notation (Axiom of Replacement)
This question is related to the one I asked yesterday here in that it's related to another one of the Zermelo-Fraenkel Axioms. After looking over the notation used to describe the axiom, that is:
$$ \forall \space x \space \forall \space y \space \forall \space z \space [\varphi (x,y,p) \wedge \varphi(x,z,p) \Rightarrow y = z] \Rightarrow \forall \space X \space \exists \space Y \space \forall \space y \space [y \in Y \equiv (\exists \space x \in X) \varphi(x, y, p) ] $$
I believe I understand most of it, but I'm unsure of why we need to involve the variable z, so I thought I'd just write how I'm interpreting this and have someone correct me where it starts to get fuzzy.
Current Interpretation: For all the elements of the three sets $X, Y, Z,$ if the property $\varphi$ holds under some parameter $p$ for $x, y $ and $x, z$ conjointly implies that $y$ equals $z$ then for any set X there exists a set Y such that for any element of $Y$ there exists an element of X such that property $\varphi$ holds under both the element of $Y$ and the chosen element of $X$ for that property $p$.
What I'm confused about is the purpose of the extra parameter $p$ and the set $Z$ why couldn't you just say something like this:
$$ \forall \space x \space \forall \space y \space \varphi (x,y) \Rightarrow \forall \space X \space \exists \space Y \space \forall \space y \space [y \in Y \equiv (\exists \space x \in X) \varphi(x, y, p) ] $$
What am I missing here? Also if someone could clear up my interpretation that would be awesome.
AI: Use $x=x$ for $\varphi(x,y)$. Then your version of the axiom gives us a universal set $Y$: every $y$ is in $Y$. So it leads to an inconsistent theory.
The part that says that $\varphi(x,y,p) \land \varphi(x,z,p)$ implies $y=z$ says that the relation $\varphi$ is "function-like." It is a very strong set construction principle, but does not (one hopes) lead to inconsistency. |
H: Find the factorization of the polynomial as a product of irreducible on rings R[x] and C[x]
Find the factorization of the polynomial $x^5-x^4+8x^3-8x^2+16x-16$ as a product of irreducible on rings $\Bbb R[x]$ and $\Bbb C[x]$
Testing with the simplest possible root in this case, $P(1) = 0$
Applying the schema of Ruffini
$
\begin{array}{c|lcr}
& 1 & -1 & 8 & -8 & 16 & -16 \\
1 & & 1 & 0 & 8 & 0 & 16 \\
\hline
& 1 & 0 & 8 &0 &16 & 0 \\
\end{array}
$
$P(x)=(x-1).(x^4+8x^2+16)$
So, I need to make it irreducible in $R[x]$, there's no more roots, but I don't know how to reduce it, I tried factoring:
$x^4+8.(x^2+2).(x-1)$, but I can not keep working it.
What I can do to solve it?
AI: Remember $a^2+2ab+b^2=(a+b)^2$. Thus $x^4+8x^2+16=\,?$ Any polynomial $x^2+a$ with the number $a>0$ is irreducible over $\bf R$, but it should be clear how to reduce them over $\bf C$. |
H: Probability of senior citizens in a one million residence
In a city of over $1000000$ residents, $14\%$ of the residents are senior citizens. In a simple random sample of $1200$ residents, there is about a $95\%$ chance that the percent of senior citizens is in the interval [pick the best option; even if you can provide a sharper answer than you see in the choices, please just pick the best among the $5\%$ interval options ]
$N=1000000$ residents;
$p=14\%=0.14$ Senior Citizens
$n=1200$ residents simple random sample
$p=95\%=0.95$ chance that the $\%$ Senior Citizens is in the which interval?
$\quad\big[(9\%-19\%)\,$ or $\,(10\%-18\%)\,$ or $\,(11\%-17\%)\,$ or $\,(12\%-16\%)\,$ or $\,(13\%-15\%)\big]\,?$
AI: Let $X$ be the number of seniors in the sample. We are presumably sampling without replacement, but $1000000$ is very large compared to the sample size, so we can assume that the distribution of $X$ is binomial, $n=1200$, $p=0.14$. So $X$ has mean $(1200)(0.14)$ and standard deviation $\sqrt{1200(0.14)(0.86)}$.
Let $Y=\frac{X}{1200}$, the sample proportion of seniors.
Then $Y$ has mean $0.14$, and standard deviation $\frac{\sqrt{(0.14)(0.86)}}{\sqrt{1200}}\approx 0.0100$.
Note that the distribution of $Y$ is very close to normal. So with probability $0.95$, $Y$ is within $(1.96)(0.0100)$ of the mean $0.14$. The nearest in the given list is $12\%$ to $16\%$. |
H: Dimensions of vector subspaces in a direct sum are additive
$V = U_1\oplus U_2~\oplus~...~ \oplus~ U_n~(\dim V < ∞)$ $\implies \dim V = \dim U_1 + \dim U_2 + ... + \dim U_n.$ [Using the result if $B_i$ is a basis of $U_i$ then $\cup_{i=1}^n B_i$ is a basis of $V$]
Then it suffices to show $U_i\cap U_j-\{0\}=\emptyset$ for $i\ne j.$ If not, let $v\in U_i\cap U_j-\{0\}.$ Then
\begin{align*}
v=&0\,(\in U_1)+0\,(\in U_2)\,+\ldots+0\,(\in U_{i-1})+v\,(\in U_{i})+0\,(\in U_{i+1})+\ldots\\
& +\,0\,(\in U_j)+\ldots+0\,(\in U_{n})\\
=&0\,(\in U_1)+0\,(\in U_2)+\ldots+0\,(\in U_i)+\ldots+0\,(\in U_{j-1})+\,v(\in U_{j})\\
& +\,0\,(\in U_{j+1})+\ldots+0\,(\in U_{n}).
\end{align*}
Hence $v$ fails to have a unique linear sum of elements of $U_i's.$ Hence etc ...
Am I right?
AI: Yes, you're correct.
Were you second guessing yourself? If so, no need to:
You're argument is "spot on".
If you'd like to save yourself a little space, and work, you can write your sum as:
$$ \dim V = \sum_{i = 1}^n \dim U_i$$
"...If not, let $v\in U_i\cap U_j-\{0\}.$ Then
$$v= v(\in U_i) + \sum_{\large 1\leq j\leq n; \,j\neq i} 0(\in U_j)$$ |
H: Another Information Theory Riddle
The following nice riddle is a quote from the excellent, free-to-download book: Information Theory, Inference, and Learning Algorithms, written by David J.C. MacKay.
In a magic trick, there are three participants: the magician, an assistant, and a volunteer.
The assistant, who claims to have paranormal abilities, is in a soundproof room. The magician gives the volunteer six blank cards, five white and one blue. The volunteer writes a different integer from 1 to 100 on each card, as the magician is watching. The volunteer keeps the blue card. The magician arranges the five white cards in some order and passes them to the assistant.
The assistant then announces the number on the blue card.
How does the trick work?
AI: The five cards are uniquely identified by their numbers (low to high). There are $5!=120$ possible orderings for the five cards, which is more than enough to encode the number on the sixth card. In fact, by orienting the cards carefully, there are 4 ways to orient each card and still make a pile (180 degree rotation, turn upside down), there could be $4^55!=122880$ possibilities. I'd only try this variation with a lot of practice, though. |
H: Cubing a simple thing
I am trying to expand $\quad (x + 2)^3 $
I am actually not to sure what to do from here, the rules are confusing. To square something is simple, you just foil it. It is easy to memorize and execute. Here though I am not sure if I need to do it like multiplication where I take one $(x + 2)$ term and multiply by another or if I need to multiply all $(x + 2)$ terms by it.
I want to treat it like how I would square it so I just square it and then I am left with the result and the $(x + 2)$ term. This is wrong and I do not know why. I get this
$$(x^2 + 4x + 4)(x + 2)$$
This is wrong and I am not sure why. So not I try the other way, multiplying everything by everything. This leaves me with
$$(x^2 + 4x + 4)(x^2 + 4x + 4)$$
Which is again wrong. I have exhausted all my options and nothing results in a correct answer and I am not sure why.
AI: The following is correct, though not fully expanded:
$$(x^2 + 4x + 4)(x + 2) = (x+2)^2(x+2) = (x+2)^3$$
Consider writing this as $(x + 2)(x^2 + 4x + 4)$, and then distribute (multiply) each term in the first factor, with each term of the second factor.
$$
\begin{align} (\color{blue}{\bf x + 2})(x^2 + 4x + 4)
& = \color{blue}{\bf x}(x^2 + 4x + 4) + \color{blue}{\bf 2}(x^2 + 4x + 4) \\ \\
& = (x^3 + {\bf 4x^2} + \color{blue}{\bf 4x}) + ({\bf 2x^2 }+ \color{blue}{\bf 8x} + 8) \\ \\
& = x^3 + {\bf 6x^2} + \color{blue}{\bf 12x} + 8\\ \\
\end{align}
$$ |
H: Integration $\int \left(x-\frac{1}{2x} \right)^2\,dx $
Evaluate $$\int\!\left(x-\frac{1}{2x} \right)^2\,dx. $$
Using integrating by substitution, I got $u=x-\frac{1}{2x},\quad \dfrac{du}{dx} =1+ \frac{1}{2x^2}$ , and $dx= 1+2x^2 du$. In the end, I came up with the answer to the integral as :
$$\left(\frac{1}{3}+\frac{2x^2}{3}\right)\left(x-\frac{1}{2x}\right)^3.$$
Any mistake ?
AI: Just expand
$$
\left(x-\frac{1}{2x}\right)^2 = x^2 - 1 +\frac{1}{4x^2}
$$
and integrate term by term. |
H: Integrate $\int {{{\left( {\cot x - \tan x} \right)}^2}dx} $
$\eqalign{
& \int {{{\left( {\cot x - \tan x} \right)}^2}dx} \cr
& = {\int {\left( {{{\cos x} \over {\sin x}} - {{\sin x} \over {\cos x}}} \right)} ^2}dx \cr
& = {\int {\left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {\sin x\cos x}}} \right)} ^2}dx \cr
& = \int {{{\left( {{{\cos 2x} \over {{1 \over 2}\sin 2x}}} \right)}^2}dx} \cr
& = \int {{{\left( {2\cot 2x} \right)}^2}} \cr
& = \int {4{{\cot }^2}2xdx} \cr
& = \int {4\left( {{{\csc }^2}2x - 1} \right)dx} \cr
& = \int {\left(4{{\csc }^2}2x - 4\right)dx} \cr
& = 4 \times {{ - 1} \over 2}\cot 2x - 4x + C \cr
& = - 2\cot 2x - 4x + C \cr} $
Where have I gone wrong? I've tried to spot an error so many times yet I can't find it, I need another pair of eyes.. Thanks.
AI: Why do you think that you've gone wrong? Your answer is correct.
Edit. To show why the solution is the same as the one in your answers, we just need to show that $\tan x-\cot x=-2\cot 2x$.
$$
\begin{align*}
&\tan x-\cot x+2\cot 2x \\
&=\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}+2\frac{\cos 2x}{\sin 2x} \\
&=\frac{2\cos x\cos 2x\sin x-\cos^2 x\sin 2x+\sin^2 x\sin 2x}{\cos x\sin x\sin 2x} \\
&=\frac{2\cos x(\cos^2 x-\sin^2 x)\sin x-\cos^2 x(2\sin x\cos x)+\sin^2 x(2\sin x\cos x)}{\cos x\sin x\sin 2x} \\
&=\frac{2\cos^3 x\sin x-2\cos x\sin^3 x-2\sin x\cos^3 x+2\cos x\sin^3 x}{\cos x\sin x\sin 2x} \\
&=0
\end{align*}
$$ |
H: Find area of triangle ABC
BD Perpendicular AC , AB =BC=a
Find the area of triangle ABC
I have tried Googling , I used formula 1/2 (base X Height) . Used Pythagorean theorem. Anyone can suggest me solution.
AI: Let $y$ represent the distance $CD$. Then $y^2+x^2=a^2$, or $y=\sqrt{a^2-x^2}$. Now you have enough to find the area of triangle $BCD$ using your area formula. Double this to find the area of the whole triangle. |
H: A question about the degree of an element over a field extension.
Say $K$ is a field extension of field $F$. If element $b$ is algebraic with degree $n$ over $F$, we know that $[F(b):F]=n$.
Why is it that $[K(b):K]\leq n$?
AI: Recall the following facts: if $M/L$ is a field extension, then for any $a\in M$,
$$[L(a):L]=\deg(h)$$
where $h\in L[x]$ is the minimal polynomial of $a$ over $L$. The minimal polynomial of $a$ is the monic polynomial of smallest degree having $a$ as a root. Consequently, if $p\in L[x]$ is any other polynomial having $a$ as a root, we must have $h\mid p$.
Let $f\in F[x]$ and $g\in K[x]$ be the minimal polynomials of $b$ over the fields $F$ and $K$, respectively. Because $F\subseteq K$, we have that $f\in K[x]$. Because $f(b)=0$, we must have $g\mid f$, hence
$$[K(b):K]=\deg(g)\leq\deg(f)=[K(b):K].$$ |
H: Why is $[\mathbb{Q}(e^{2\pi i/p}):\mathbb{Q}] = p-1 $?
I know that $e^{2\pi i/p}$ is a root of $x^{p}-1$ and we can write:
$ x^{p}-1=\left(x-1\right)\left(\sum_{i=0}^{p-1}x^{i}\right) $
So $e^{2\pi i/p}$ is a root of $\sum_{i=0}^{p-1}x^{i}$, which means $[\mathbb{Q}(e^{2\pi i/p}):\mathbb{Q}] \leq p-1 $.
Can someone show the other direction?
AI: You can get both directions from the fact that $e^{\frac{2 \pi i}{p}}$ is a root of $P(X)=X^{p-1}+X^{p-1}+..+X+1$, and that $P(X)$ is irreducible over $\mathbb Q[X]$.
You can prove the irreducibility by observing that $P(X)=\frac{X^p-1}{X-1}$ for $x\neq 1$, thus, for $X \neq 0$ we have
$$P(X+1)=\frac{(X+1)^p-1}{X}=\sum_{k=1}^p \binom{p}{k}X^{k-1}$$
use Eisenstein Criteria. |
H: Factoring the third degree polynomial $x^3 - 3x^2 - 4x + 12$ using long division
I am sure there is a better strategy that someone smarter than me would use, but I am not that person.
I am trying to factor$$x^3 - 3x^2 - 4x + 12 .$$
I do not know how, so I attempt to guess with long division. I cheat and look at the answer so find out one of the factors to save myself time, so I try $x - 2$.
I am not sure how to type out long division, but I get$$ (x - 2) \mid (x^3 - 3x^2 - 4x + 12) .$$
So I know I can have an $x^2$ for how many times $x$ goes into the leading term.
Subtracting it all, I am left with$$ (x - 2 ) \mid (-x^2 - 4x + 12) .$$
I know that the leading term goes into the inner leading term $-x$ times or however you say that.$$ (x - 2 ) \mid (- 2x + 12) .$$
Now $-2$:$$ (x - 2 ) \mid 8 .$$
Now I do not know what to do, how did this go so wrong? I have $x^2 - x -2$ on top, and I have a remainder of $8$. This cannot be right; I cheated, so I know that this should be a factor.
AI: Your first term, upon division: $x^2$ is correct $\large\checkmark$ (in the quotient), leaving
$$ x - 2 | -x^2 - 4x + 12\quad\large\checkmark$$
I know that the leading term goes into the inner leading term $-x$ times or however you say that.
$$ x - 2 | - 2x + 12 \quad \longleftarrow \text{error}$$
Here you subtracted incorrectly: We should have $x^2 - x$ in the quotient, that's correct, but multiplying $-x(x - 2) = -x^2 + 2x$
So when we subtract, we subtract, from $(-x^2 - 4x + 12) - (-x^2 + 2x) = -6x + 12$.
Now, we have $$ x -2 \mid -6x + 12$$ and so our ongoing quotient becomes $x^2 - x {\bf - 6}$
which, leaves a zero remainder since $-6(x-2) = -6x + 12$, as desired.
So...we have that $$\frac{x^3 - 3x^2 -4x + 12}{x - 2} = x^2 - x - 6$$
Or, that is, $$x^3 - 3x^2 - 4x + 12 = (x -2)(x^2 - x - 6) = (x-2)(x +2)(x - 3)$$ |
H: How to minimize studying using mathematics?
A friend asked me this question earlier today, and it made me wonder how to come up with a general solution (where each variable is an integer):
I have a vocabulary test tomorrow at school. On it, there will be $W$
terms listed, and I will have to define $X$ of them. However, before
the test, I am given a list of $Y$ possible terms that could be on the
list given on the test. How many terms must I be able to define in order to
assure that I get at least a mark of $Z$% correct on the test?
I've tinkered with some numbers a tad, and came up with some possibilities; i.e. if $W = 10$, $X = 4$, $Y = 19$, and $Z = 100$, then the individual needs to know $13$ terms. I'm having a bit of trouble, however, coming up with a generalized solution.
Any help is appreciated.
AI: To guarantee a particular minimum grade, simply assume that every term you don't study will be on the test.
For the example given, if every term not studied is on the test, to get 100%, you need to find 4 terms that you've studied, no matter which 10 terms appear. Hence you can only afford to skip 6 terms, and must therefore study $19-6=13$ terms.
More details, as per request:
To achieve grade of $Z$, you must get at least $\frac{Z}{100}X$. Hence you can only afford to skip $W-\lfloor \frac{Z}{100} X\rfloor$ terms. Consequently, you must study at least
$$Y-\left(W-\bigg\lfloor \frac{Z}{100} X\bigg\rfloor\right)=Y-W+\bigg\lfloor \frac{Z}{100} X\bigg\rfloor$$
terms. Note that $\lfloor \cdot \rfloor$ denotes the floor function. |
H: On convergence of problematic series.
Determine if the following series is converges or not
$$\sum_{n=2}^{\infty}\frac{1}{\ln^{\ln(\ln n)} (n)}$$
AI: Note that $$\ln\left(\frac{1}{\ln^{\ln(\ln n)}(n)}\right)=\ln1-\ln(\ln^{\ln(\ln n)}(n))=-\ln(\ln n)\ln(\ln n)=-\left[\ln(\ln n)\right]^2$$
Therefore
$$\sum_{k=2}^\infty\frac{1}{\ln^{\ln(\ln n)}(n)}=\sum_{k=2}^\infty e^{\ln\left(\frac{1}{\ln^{\ln(\ln n)}(n)}\right)}=\sum_{k=2}^\infty e^{-\left[\ln(\ln n)\right]^2}$$
It's not too hard to show that $\left[\ln(\ln n)\right]^2<\ln n$ for sufficiently large $n$. Therefore, for sufficiently large $n$, we have $e^{-\left[\ln(\ln n)\right]^2}>e^{-\ln n}=\frac{1}{n}$, so the series diverges (by comparison with the harmonic series). |
H: Closed form for $\sum^{\infty}_{{i=n}}ix^{i-1}$
How can I find a closed form for:
$$\sum^{\infty}_{{i=n}}ix^{i-1}$$
It looks like that's something to do with the derivative
AI: $$\sum^\infty_{r=n}rx^{r-1}=\dfrac {d\left(\sum^\infty_{r=n}x^r\right)}{dx}$$
From here, $\sum^\infty_{r=n}x^r=\dfrac{x^n}{1-x}$ if $|x|<1$
Alternatively,
let $S_n=\sum^m_{r=n}rx^{r-1}=n\cdot x^{n-1}+(n+1)x^n+(n+2)x^{n+1}+\cdots+(m-1)x^{m-2}+mx^{m-1}$
So, multiplying by $x,$ $x\cdot S_n=\sum^\infty_{r=n}rx^r=n x^n+(n+1)x^{n+1}+(n+2)x^{n+2}+\cdots+(m-1)x^{m-1}+mx^m$
On subtraction, $$(1-x)S_n=nx^{n-1}+(x^n+x^{n+1}+\cdots+x^{m-1})-mx^m$$
$$=nx^{n-1}+x^n\frac{(1-x^{m-n})}{1-x}-mx^m=nx^{n-1}+\frac{(x^n-x^m)}{1-x}-mx^m$$
Now, if $m\to\infty$ and $|x|<1,x^m\to0$
From this, $\lim_{m\to\infty}mx^m=0$ if $|x|<1$ |
H: Power series of $f(x)=\sqrt{\frac{1+x}{1-x}}$
How do I find the power series form of $\,f(x)\,$:
$$\displaystyle f(x)=\sqrt{\frac{1+x}{1-x}}$$
I tried to multiply the fraction by $\,\dfrac{1+x}{1+x}\,$ but it didn't help...
AI: Try writing it as $ f(x) = \sqrt{ 1- x^2 } \times 1/(1-x) $ . Then write $ 1/(1-x) $ as a geometric series and the other term can be expanded using binomial theorem. Hope that helps. |
H: Algebra / Equation with 2 or more vaiables
A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds
respectively. What is the speed of the train?
I'm really confused if there are 2 or more variables. Can someone explain this in step by step? How x is removed until the end. Thanks.
Or a link on any tutorial or article regarding this. I don't know what to search, I only find simple examples with 1 variable. Thanks.
AI: Let the length of the train is $x\,\, m$ and speed of the train is $y \,\,m/s$
So you have $x=8y$ and $x+264=20y$.Hence your result is true.
If you have two variable then you must have two equation and from that you can solve the problem. |
H: solving for one variable in terms of others
A question from Steward's Precalculus textbook 5th, Pg 55,
the original formula is $$h=\frac{1}{2}gt^2+V_0t$$
the question asks to write the formula in terms of $t$, the answer is $$t=\dfrac{-V_0\pm\sqrt{v_0^2+2gh}}{g}$$
I don't know the steps on how to get there
AI: $$h=\frac{1}{2}gt^2+V_0t$$
$$2h=gt^2+2V_0t$$
$$gt^2+2V_0t-2h=0$$
now use this formula for Quadratic equation:$ax^2+bx+c=0$;
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
so: $$t=\dfrac{-2V_0\pm\sqrt{4V_0^2+8gh}}{2g} $$
$$t=\dfrac{-V_0\pm\sqrt{V_0^2+2gh}}{g} $$ |
H: Finding the Taylor series of $f(z)=\frac{1}{z}$ around $z=z_{0}$
I was asked the following (homework) question:
For each (different but constant) $z_{0}\in G:=\{z\in\mathbb{C}:\,
z\neq0$} find a power series $\sum_{n=0}^{\infty}a_{n}(z-z_{0})^{n}$
whose sum is equal to $f(z)$ on some subset of $G$. Please specify
exactly on which subset this last claim holds.
Suggestion: Instead of calculating derivatives of f, try
using geometric series in some suitable way.
What I did:
Denote $f(z)=\frac{1}{z}$ and note $G\subseteq\mathbb{C}$ is open.
For any $z_{0}\in G$ the maximal $R>0$ s.t $f\in H(D(z_{0},R))$
is clearly $R=|z_{0}|$.
By Taylor theorem we have it that $f$ have a power series in $E:=D(z_{0},R)$
(and this can not be expended beyond this point, as this would imply
that $f$ is holomorphic at $z=0$.
I am able to find $f^{(n)}(z_{0})$ and to solve the exercise this
way: I got that $$f^{(n)}(z)=\frac{(-1)^{n}n!}{z^{n+1}}$$ hence $$f(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{z_{0}^{n+1}}(z-z_{0})^{n};\, z\in E$$
but I am not able to follow the suggestion, I tried to do a small
manipulation $$f(z)=\frac{1}{z-z_{0}+z_{0}}$$ and tried to work a bit
with that, but I wasn't able to bring it to the form $$\frac{1}{1-\text{expression}}$$
or something similar that uses a known power series .
Can someone please help me out with the solution that involves geometric
series ?
AI: I think you were on the right track:
$$\frac1z=\frac1{z_0+(z-z_0)}=\frac1{z_0}\cdot\frac1{1+\frac{z-z_0}{z_0}}=\frac1z_0\left(1-\frac{z-z_0}{z_0}+\frac{(z-z_0)^2}{z_{0}^2}-\ldots\right)=$$
$$=\frac1z_0-\frac{z-z_0}{z_0^2}+\frac{(z-z_0)^2}{z_0^3}+\ldots+\frac{(-1)^n(z-z_0)^n}{z_0^{n+1}}+\ldots$$
As you can see, this is just what you got but only using the expansion of a geometric series, without the derivatives explicitly kicking in...The above is true whenever
$$\frac{|z-z_0|}{|z_0|}<1\iff |z-z_0|<|z_0|$$ |
H: How to show that these integrals converge?
What test do I use to show that the following integral converges?
If you could provide me with the process that leads to the answer that would really help.
$\displaystyle \int_{0}^{1}\frac{x^n}{\sqrt{1-x^4}}\,dx$
$\displaystyle \int_{0}^{\pi /2}\frac{\ln(\sin(x))}{\sqrt{x}}\,dx$
Thanks
AI: I like to use the comparison tests for these kind of problems.
For the first integral, use the fact that $x^n \leq x$ on $x \in [0,1]$ $\forall n \in \mathbb{N}$, and that $x^4<x^2$ on $[0,1]$, giving
$$\frac{x^n}{\sqrt{1-x^4}} \leq \frac{x}{\sqrt{1-x^2}}$$
and use the fact that $f\leq g \implies \int f \leq \int g$ on any bounded interval.
Hence,
$$\begin{align}
\int_0^1 \frac{x^n}{\sqrt{1-x^4}} \mathrm{d}x &\leq \int_0^1 \frac{x}{\sqrt{1-x^2}} \mathrm{d}x
\\
&=\left. -\sqrt{1-x^2} +c \right|_0^1
\\
&= 1
\end{align}$$
For the second use a similar idea, except here note that $\ln x < x$ $\forall x>0$ and that $\sin x< x$ for all $x$.
We do all this to get the integral of $g$ on the right side to be one which we can compute in closed form, which gives a bound on the integral of $f$ which you want to show converges, which is of course the standard comparison test. |
H: Infinite products of a (finite) group
So I'm having a little trouble understanding the concept of infinite (cartesian) products of a group -- specifically, my notes (and, of course, homework questions) have concepts of, say $S_3^\mathbb{Z}$ and $S_3^\mathbb{R}$ (where $S_3$ is the permutation group of 3 elements, of course), but I'm having a hard time conceptualising the difference between these.
If I assume that we're taking what seems to be the canonical approach: $G^I = \Pi_{i \in I} G$, how do I begin to understand the notion of, say, 3.141519 copies of $G$ (or in my case, $S_3$)? Is there a distinction between $\mathbb{Z}$ and $\mathbb{N}$ copies of a group? Or $\mathbb{Z}$ and $\mathbb{R}$?
It feels like there's something obvious I'm missing, but, of course, I don't know it!
AI: The direct product of copies of a single group $G$ over an indexing set $I$ can be thought of as the space of functions $f:I\to G$ with componentwise multiplication (i.e. $(f_1f_2)(i)=f_1(i)f_2(i)$ for all $i\in I$). This makes sense; any coordinate vector $(a_1,a_2,\cdots)$ can be thought of as a function from $\bf N$ into whatever, as $i\mapsto a_i$, and conversely any function $f:{\bf N}\to{\rm blah}$ can be thought of as a coordinate vector via $(f(1),f(2),f(3),\cdots)$. The utility of the functional approach is that it's kind of hard to write a "vector" with uncountably many coordinates (there are too many coordinates to enumerate), but functions are perfectly acceptable to talk about.
If $I$ and $J$ have the same size, then $\prod_IG\cong\prod_JG$. This can be accomplished by any bijection given by $\varphi:I\xrightarrow{\sim}J$; given $x\in\prod_IG$, define $\bar{x}\in\prod_JG$ to be the element whose $\varphi(i)$th coordinate is equal to the $i$th coordinate of $x$. So for example, let $\varphi:1,2,3,\cdots\mapsto0,1,-1,2,-2,\cdots$. Then
$$(a_1,a_2,a_3,\cdots)\mapsto (\cdots,a_5,a_3,a_1,a_2,a_4,\cdots)$$
is an isomorphism $\prod_{\bf N}G\cong\prod_{\bf Z}G$ for any group $G$. In the language of functions this means that the map $f\mapsto f\circ \varphi^{-1}$ is an isomorphism $\prod_IG\to\prod_JG$.
If $I,J$ are different sizes, then $G^I\not\cong G^J$ since in the first place they do not have the same sizes as sets. If we drop the condition that $G$ is finite though then I'm not so sure what can be said.
It does not make sense to take the direct product of a non-integer number of copies of $G$, any more than it makes sense to have a set of non-integer size. (Makes even less sense; you can define "fuzzy" sets, but I wouldn't know how to begin defining a direct product indexed by a fuzzy set.)
It's noteworthy that when working with an infinite number of nontrivial groups, direct sums and direct products are two different things. The direct product $\prod_I G_i$ can be formalized as the space of functions $f:I\to\bigcup_I G_i$ such that $f(i)\in G_i$ for all $i\in I$, with pointwise multiplication. The direct sum $\bigoplus_IG_i$ is a subgroup of the direct product consisting of those functions with "finite support," i.e. $f(i)$ is not an identity element of $G_i$ for only finitely many $i\in I$. |
H: what is the sum of this?$\frac12+ \frac13+\frac14+\frac15+\frac16 +\dots\frac{1}{2012}+\frac{1}{2013} $
What is the sum of $$\frac12+ \frac13+\frac14+\frac15+\frac16 +\dots\frac{1}{2012}+\frac{1}{2013} $$
AI: Not really an interesting question, I fear ...
The exact answer is $\frac{A}{B}$, where $A$ is the 873-digit number
$$\begin{align}
&1457823020375738882085285521838680376351317016179578495942597370\\
&0378320415510098781465535340716404121275266248327296358362232678\\
&6278734102622894043296263865840984434831046940807020935913518007\\
&6610170430184975025171489501039855136340044946766172114345187347\\
&2525258101947297211093973306792764671036527339494431750637074683\\
&4230708772438305435131071598225161542024364751281758482910493532\\
&5272679486529134147964088721658417270988342669017294670746222437\\
&5396289282709392060556292768642859506675716675390362111468228302\\
&0782368060066857220428567636667351228472222194509213532517098318\\
&4215312313922564589890501941681764939250636021442764267189354097\\
&6967312781103861891868999835897977944251765509632397992463764707\\
&2757539640418390812471213879592790834646976169124993489059261606\\
&2602876187541263449739774491663332258026609790766133246441424484\\
&757471204138450579265132408964711222835023\\
\end{align}$$
and $B$ is the 872-digit number
$$\begin{align}
&2029024882178588859070537995606394892063138617868032497494729352\\
&4603117588071109954985856100075037278586690682862226370908784076\\
&1373758182460385685570815580907584983469652245087586518224049939\\
&8538189200052467764823885503628456029563567766917663813400872806\\
&6581413837126629349013539786812278235204246553266267204666411453\\
&3233291465507116087957721742791662582248630820924043188884944781\\
&6407535614283438262641159101052916955096435138537168813836363531\\
&9456214006726533268868764071437503549723903396960519957683391782\\
&6358829094665518541761643350701837468138883360957730534259985433\\
&0461170299850901340867381633962516193737332035551187788058506648\\
&9862570396408121482924662574281092579439686439440266363727224229\\
&3000091507088457965475315993866302318289137980710843830365091268\\
&4178987345171765323904440193241153257507552225072663118382054707\\
&65887615406383472808692605376427780480000\\
\end{align}$$
Also, the first digits are 7.184845455 ... |
H: How to form a cubic equation with the substitution method?
I had this question:
"Find the cubic equation whose roots are twice the roots of the equation $3x^3 - 2x^2 + 1 = 0$"
In my first attempt, I solved it through the use of simultaneous equations, where I let the cubic equation be: $x^3 + bx^2 + cx + d$. However, I've been told that there is a more efficient method for higher degree polynomials - and it has been labelled as "The Substitution method".
It involved something like "Let $y = x^2$" then form a new equation using that.
Could someone solve this question and explain how to do it for me?
Any help would be greatly appreciated, thanks!
The solution with the substitution method worked like this:
Let $y=2x$,
$x=y/2$
then it was substituted back into the given equation.
Can someone explain this?
AI: Suppose you have a root $x$ of $$3x^3-2x^2+1=0$$
If $y=2x$, then $x=\cfrac y 2$ and $$3(\frac y 2)^3-2(\frac y 2)^2+1=0$$
Clear the fractions and you obtain an equation with integer coefficients satisfied by $y$.
The key to the substitution method is to find an expression for $x$ (a root of the original equation) and substitute it into the original equation. It depends on doing the same thing to each of the roots. |
H: How to prove two polynomials have no zeroes in common?
The question asked:
Divide the polynomial $P(x) = x^3 + 5x^2 - 22x - 6$ by $G(x) = x^2 - 3x + 2$. I did, and got the answer: $(x+8)(x^2-3x+2)-22$.
However, it now asks to: "Show that $P(x)$ and $G(x)$ have no zeros in common."
How do I prove this?
Thanks.
AI: You have that $P(x) = (x+8)G(x) - 22$. What happens when $G(x)$ is zero? |
H: Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$
I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.
Any help is appreciated! Thanks.
Edit: solution I require is $\frac{1}{2} (3^{n-1}+1-2^n)$
Edit2: Solutions to the homogeneous equation would be of the form $h_n = A(\alpha)^n + B(\beta)^n$, and $p_n$ will exist such that $s_n = h_n + p_n$
AI: Let $ t_n := \frac{s_n}{3^n} $. Hence, $ t_n = t_{n-1} + 3^{-n}\left(2^{n-2} - 1\right) $. Also, $ t_1 = 0 $. Hence, $$ t_n = \sum_{k = 2}^n \frac{1}{4}\left(\frac{2}{3}\right)^n - 3^{-n} $$ This is a geometric series and easily evaluated to arrive at $$ t_n = \frac{1}{2}\cdot 3^{-n} \cdot \left(1 - 2^{n}\right) + \frac{1}{6} \implies s_n = \frac{1 + 3^{n-1}- 2^{n}}{2} $$ |
H: Conversion: dec to bin
We have:
$$\begin{align}
A&=\frac{19}{32} \\[0.3em]
B&=\frac{21}{32}\\[0.3em]
C&=\frac{19}{64}\\[0.3em]
D&=\frac{21}{64}.\\[0.3em]
\end{align}$$
In binary it is:
$$\begin{align}
A&=0,10011\strut \\
B&=0,10101\strut\\
C&=0,010011\strut\\
D&=0,010101\strut\\
\end{align}$$
Is it correct?
Does it mean that the denominator says how many positions I have after comma?
AI: The values given are indeed correct. 64 is 1,000000
1/64 1/32
19 * 0.000 001 = 0. 010 011 0. 10011
21 * 0.000 001 = 0. 010 101 0. 10101
Dividing by 1000000 in binary is the same as in decimal: shift the comma six places to the left (with zero-filling as needed). So to make 1 000 00 into one, you shift the comma five places to the right, and to divide 10101 by the same amount, shift the comma five to the left, ie 0,10101.
If you need further places, zero-fill to the left. Divide 10011 by 1000000, you need an extra place before the comma, so eg 000000 10011, becomes 00000,0 10011, and delete unnecassary zeros -> 0,010011. |
H: Develop the next function:$f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0)$
Develop the next function:$\displaystyle f(x)=\frac{4x+53}{x^2-x-30}$ into power series, Find the radius on convergence and find $f^{(20)}(0).$
For the first part:
$\displaystyle\frac {4x+53}{(x-6)(x-5)}\to\frac {4x+53}{(x-6)(x+5)}=\frac{A}{(x-6)}+\frac{B}{(x+5)}=\frac{7}{(x-6)}+\frac{-3}{(x+5)}=\frac{7}{-6(1-\frac x6)}+\frac{-3}{5\left(1-(-\frac x5)\right)}$
$\displaystyle=-\frac76\sum_{n=0}^\infty{\left(\frac x6\right)^n}-\frac35\sum_{n=0}^\infty(-1)^n\left(\frac x5\right)^n$
Now i'm having trouble with finding the radius of convergence and $f^{(20)}(0)$.
What does $f^{(20)}(0)$ means? 20 derivatives and $x=0$? That means i should find a some-kind of pattern and then place $x=0$?
AI: Hints:
1) For the radius of convergence: do you know what is the radius of convergence of $\sum_{n=0}^\infty t^n$? if so, find the radius of convergence for each series separately. Then, the radius of convergence for the sum will be the minimal (do you see why?)
2) You just found the taylor series around $x=0$. The general form of Taylor series is $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. So, what is $f^{(20)}(0)$? |
H: $V_\omega$ is countable
Is there an easy way to prove this? I found a book that suggests the injection $h:V_\omega\to\omega$ defined by
$$h(\{x_1,x_2,\dots,x_n\})=2^{h(x_1)}+2^{h(x_2)}+\cdots+2^{h(x_n)},$$
but I hit some snags while proving that the function is well-defined. Are there any other clever ways to prove this fact? (Note: I am making a formal proof, so I need to be able to justify my steps.)
Edit: Let me see if I can make my difficulty more clear. Before I can start the proof by induction, I need a precise statement of the induction hypothesis, and this is what I am trying to hammer out. The hypothesis:
For each $n\in\omega$, there exists a unique bijection $h_n:V_n\to|V_n|$ such that $$h_n(x)=\sum_{m<|V_{n-1}|}\operatorname{if}(h_{n-1}^{-1}(m)\in x,2^m,0).$$
Moreover, the $h_n$ as defined satisfies $h_n\restriction |V_{n-1}|=h_{n-1}$.
Does this hypothesis contain enough to prove the claim? The reason for the odd way of stating the sum is that it avoids the issue in the linked question (we need to show that addition is a well-defined operation over sets), but as a consequence I have to prove that
the $h_n$ is a bijection and not just an injection, and I think I also need the restriction part in order to prove that the union $h=\bigcup_{n\in\omega}h_n$ is the required bijection.
AI: To prove that $h$ is well-defined and injective, it is most natural to proceed by induction on the rank of sets. The rank $\mathsf{rk}(x)$ of a set $x$ is an ordinal, defined recursively as follows (see also MathWorld for a definition for all sets):
$$\mathsf{rk}(x):= \sup\{\mathsf{rk}(y)+1: y \in x\}$$
Lemma: The rank of a set is well-defined.
This lemma can be proved using the Axiom of Foundation, see also this question. Since all suprema involved in computing $\mathsf{rk}$ for $x \in V_\omega$ are finite, the results are also finite ordinals.
The only set of rank zero is of course $\varnothing$; we define/obtain $h(\varnothing) = 0$.
Now given a set $X = \{x_1, \ldots, x_m\}$ of rank $n+1$, we use the Axiom of Replacement for $f(x) = 2^{h(x)}$, well-defined and injective by induction hypothesis.
We now have the set $Y = \{2^{h(x_1)},\ldots, 2^{h(x_m)}\}$; using your earlier question this gives rise to a well-defined number $h(X) = \sum Y$, so $h$ is well-defined on $X$ as well.
To show that $h$ is injective, let $X'$ be such that $\mathsf{rk}(X') \le \mathsf{rk}(X)$. Then we can obtain the sets $\{h(x_1),\ldots, h(x_m)\}$ and $\{h(x'_1),\ldots, h(x'_{m'})\}$ by the Basis Representation Theorem. The Axiom of Replacement can be applied to these sets since $h^{-1}$ is a functional relation, and we recover $X$ and $X'$; injectivity of both the Basis Representation and $h^{-1}$ ensures that $h(X) = h(X')$ implies $X = X'$. Thus $h$ is injective and well-defined on sets of rank at most $n+1$.
In conclusion, $h$ is well-defined on $V_\omega$; its injectivity on $V_\omega$ is trivially verified. |
H: Is the ascending union of contractible spaces contractible
Let $\{Y_i\}_{i \in \mathbb{N}}$ be a collection of subspaces of $X$ such that each $Y_i$ is contractible and $Y_{i} \subseteq Y_{i+1}$. Is $\bigcup_{i\in \mathbb{N}}Y_i \subseteq X$ also contractible?
AI: No. Consider the half-open arcs
$$Y_n=\{e^{i\theta}\mid \theta\in[0,2\pi-\tfrac{1}{n})\}$$
on the circle $X$. |
H: Finding the limit of a function?
The limit is actually easy: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t}$
One can use hopitals rule and say that ultimately the upper function will be reduced to a constant while the lower function will remain the same. Hence, the limit is zero. I was wondering if there is another way or a more mathematical representation of the answer.
AI: Alternatively, we can proceed by induction on $k\in \mathbb{N}$.
Base Case: For $k=0$, we need only apply L'Hôpital's rule once: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t}{e^t} = \lim \limits_{t\to\infty}\dfrac{1}{e^t} = 0$.
Inductive Hypothesis: Assume that $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t} =0$ holds for $k=r$.
It remains to prove that the equation holds for $k=r+1$. Then by one more application of L'Hôpital's rule and the induction hypothesis, observe that:
$$\lim_{t\to\infty}{\frac{t^{r+2}}{e^t}}=\lim_{t\to\infty}{\frac{(r+2)t^{r+1}}{e^t}}=(r+2)\lim_{t\to\infty}{\frac{t^{r+1}}{e^t}}=(r+2)(0)=0$$
as desired. |
H: how to show that $f(x)$ can be expressed uniquely as follows: $f(x)=\prod_{i=1}^k[f_i(x)]^{n_i}$
Let $f(x)\in F[x],~F$ being a field, be monic. Then how to show that $f(x)$ can be expressed uniquely as follows:
$$f(x)=\prod_{i=1}^k[f_i(x)]^{n_i}$$
for some monic irreducible polynomial $f_i(x)\in F[x]$ and $k\in \mathbb N~(n_i$'s are positive integers)
AI: If $F$ is a field, $F[X]$ is a Principal Ideal Domain (take an ideal $I\neq 0$ of $F[x]$ and a polynomial of minimal degree in it, then conclude by Euclide's algorithm). Since any PID is also a Unique Factorization Domain (a proof of this fact can be found in most of abstract algebra's book, such as in Jacobson's Basic Algebra I), you're done. |
H: Does there exist $g$ s.t $g'=f$?
I have the following homework question:
Let G be the bounded open set shown in gray in this picture, whose
boundary consists of eight line segments. The endpoints of those
segments are, as shown, the points $-2,-1,-1+4i,1+4i,1,2,2+5i,-2+5i$.
Let $f:\, G\to\mathbb{C}$ be an arbitrary function which is
holomorphic in $G$. Does there a function $g:\, G\to\mathbb{C}$ which
satisfies $g'(z)=f(z)$ for all $z\in G$ ? one tool that might be
useful here is the Identity Theorem.
What I did:
I believe that I can construct such a function, but I am unsure if
my construction is correct:
I would take some sequence of points $\{z_{i}\}_{i=1}^{\infty}$,
s.t there exist $\{r_{i}\}_{i=1}^{\infty}$ s.t $D(z_{i},r_{i})\subseteq G$
and s.t $$\cup_{i=1}^{\infty}D(z_{i},r_{i})=G$$
I believe that such a sequence of points can be obtained by choosing
all the points $x+iy\in G$ s.t $x,y\in\mathbb{Q}$.
Moreover, I think that the points can be arranged s.t $$D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$
We first consider $D(z_{0},r_{0})$ - there is some $g_{0}:\, D(z_{0},r_{0})\to\mathbb{\mathbb{C}}$ s.t
$g_{0}'=f$ in $D(z_{0},r_{0})$, since $f$ is holomorphic.
We continue with $z_{1}$and get $\widetilde{g_{1}}$, since $g_{0}'=\widetilde{g_{1}}'$
for all points in $D(z_{0},r_{0})\cap D(z_{1},r_{1})$ then $g_{0}-\widetilde{g_{1}}$
is a constant $c$, and we can pick $g_{1}=\widetilde{g_{1}}+c$.
We take $g(z)$ will be eventually defined for every $z\in G$ as
one of the $g_{i}$'s.
This construction looks a bit fishy to me, I am not sure about the
"moreover" part and even if so, I am not totally convinced I can
arrange the constants to fit to get a holomorphic function $g$.
ADDED:
Instead of $$D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$
we can relax the condition to be $$\cup_{i=1}^{r}D(z_{i},r_{i})\cap D(z_{i+1},r_{i+1})\neq\emptyset$$
Is my construction correct ?
I would also appreciate to see another approach (maybe one that uses
the Identity Theorem as suggested)
AI: You don't need so many disks. The region $G$ is the union of three rectangles $R_i$ intersecting in two squares $Q_1=R_1\cap R_2$ and $Q_2=R_2\cap R_3$. In each of the $R_i$ the given function $f$ has a primitive $F_i$, by the most elementary version of Cauchy's theorem. In $Q_1$ the function $F_2-F_1$ has derivative zero, so it is a constant $c_1$, and similarly in $Q_2$ the function $F_3-F_2$ is a constant $c_2$. It follows that the function
$$F(z):=\cases{F_1(z)+c_1\quad&$(z\in R_1)$ \cr
F_2(z)\quad&$(z\in R_2)$ \cr
F_3(z)-c_2\quad&$(z\in R_3)$ \cr}$$
is a well defined primitive of $f$ in $G$.
Of course observing that $G$ is simply connected makes all of the above superfluous. It is a standard theorem of complex analysis that a function $f$ holomorphic in a simply connected (i.e., without holes) domain $G$ has a primitive on $G$. |
H: $\lim_{y \to \infty}\int_{R}f(x-t)\frac{t}{t^2 +y^2}dt=0?$ for $f\in L^{p}$, $p \in [1,\infty)$
For $f\in L^{p}$, $p \in [1,\infty)$
we want to prove:
$$\lim_{y \to \infty}\int_{R}f(x-t)\frac{t}{t^2 +y^2}dt=0$$
I'm not sure whether we can exchange the limit and the integral, cuz I cannot find the integral function $g(t)$ such that $|f(x-t)\frac{t}{t^2 +y^2}| \leq |g(t)|$.
How could I argue this? Appreciate with any hints!
AI: If $p>1$, use Hölder's inequality. Then write
$$\int_{\Bbb R}\left|\frac{t}{t^2+y^2}\right|^qdt\leqslant \int_{\{|t|\geqslant R\}}\frac 1{|t|^q}dt+(2R)^{Q+1}y^{-2q}.$$
If $p=1$, use dominated convergence theorem, where $|f(x-t)|\frac{|t|}{t^2+1}$ is a dominating function ($y>1$). |
H: If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge?
Let $ \{a_n\}_{n=1}^\infty $ be a sequence such that $\displaystyle \sum a_n $ that is divergent to $+\infty$.
What can be said about the convergence of $\displaystyle \sum \frac{a_n}{1 + a_n} $?
Any hints, thoughts or leads would be greatly appreciated.
Thanks!
AI: It seems that one assumes that $a_n\geqslant0$ for every $n$. Let $b_n=\frac{a_n}{1+a_n}$. If $a_n\geqslant1$ infinitely often, $b_n\geqslant\frac12$ infinitely often hence $\sum\limits_nb_n$ diverges. Otherwise, $a_n\leqslant1$ for every $n$ large enough, say for every $n\geqslant N$, hence $b_n\geqslant\frac12a_n$ for every $n\geqslant N$ and $\sum\limits_{n\geqslant N}b_n\geqslant\frac12\sum\limits_{n\geqslant N}a_n$, which implies that $\sum\limits_nb_n$ diverges.
Finally, if $a_n\geqslant0$ for every $n$ and if $\sum\limits_na_n$ diverges, then $\sum\limits_n\frac{a_n}{1+a_n}$ diverges.
A classical counterexample when the nonnegativity condition fails is $a_n=\frac{(-1)^{n+1}}{\sqrt{n+1}+(-1)^n}$ for every $n\geqslant1$. Then $\sum\limits_na_n$ diverges because $a_n=\frac{(-1)^{n+1}}{\sqrt{n+1}}+\frac1{n+1}+O\left(\frac1{n^{3/2}}\right)$ but $b_n=\frac{(-1)^{n+1}}{\sqrt{n+1}}$ hence $\sum\limits_nb_n$ converges. |
H: If $x^p P(|X|>x|)=o(1)$, then $E(|X|^{p-\epsilon})<\infty$ for $0<\epsilon
If $p>0$ and $x^p P(|X|>x|)=o(1)$ as $x\to\infty$, then $E(|X|^{p-\epsilon})<\infty$ for $0<\epsilon<p$.
It feels like the assumptions should lead to something like $\sum_n^\infty ((n+1)^{p-\epsilon}-n^{p-\epsilon})P(|X|>n)<\infty$ but I don't see what to do.
AI: A consequence of Fubini-Tonelli's theorem is that for any non-negative random variable $Y$ and $q>0$,
$$E(Y^q)=q\int_0^{+\infty}t^{q-1}P(|Y|>t)dt.$$
Using this with $q:=p-\varepsilon$ gives the result. |
H: How do I work out what percent of my customers will be girls and what percent will be boys?
I know that 33.3333% of all girls questions would buy my product and that 80% of all boys questioned would buy it.
What i don't know is how to work out is statistically what percentage of our customers will be boys and what percentage will girls.
AI: Hint: Let $G$ = the person is a girl and $C$ = the person is a customer. Assume that a customer is defined to be anyone who buys your product. Then we are given that:
$P(C|G)=0.333...=1/3$
$P\left(C|\overline{G}\right) = 0.8 = 4/5$
We are asked to find: $P(G|C)$ and $P\left(\overline{G}|C\right)$
The following identity will be useful:
$$P(A|B) = \frac{P(A \text{ and } B)}{P(B)} = \frac{P(A)P(B|A)}{P(A)P(B|A) + P\left(\overline{A}\right)P\left(B|\overline{A}\right)}$$ |
H: $\sum_{k=1}^{n} \binom{n}{k}k^{r}$
Find:$$\sum_{k=1}^{n} \binom{n}{k}k^{r}$$
For r=0 the sum is obviously $2^{n}$.
For r=1 the sum is $n2^{n-1}$.
For r=2 the sum is $n(n+1)2^{n-2}$.
Here's what I've tried:
$$\frac{d(1+x)^{n}}{dx}=n(1+x)^{n-1}=\binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3}x^{2}+\cdots+n\binom{n}{n}x^{n-1}$$
$$\frac{d[x\cdot n(1+x)^{n-1}]}{dx}=\binom{n}{1}+2^{2}\binom{n}{2}x+3^{2}\binom{n}{3}x^{3}+\cdots+n^{2}\binom{n}{n}x^{n-1}$$
So if we continue like this and put x=1 we will get our result.However I cannot generalize this.
I think the answer has something to do with the Stirling Numbers of the Second kind.
Another way to present this question:
Find $A_{1},A_{2},\cdots ,A_{r}$ such that $$k^{r}=A_{1}k+A_{2}k(k-1)+A_{3}k(k-1)(k-2)+\cdots$$
If someone can answer the second question I can solve the first one.
AI: You're right that it has to do with Stirling numbers. Wikipedia's article on Stirling numbers of the second kind answers your second question:
If we let $$(x)_n=x(x-1)(x-2)\cdots(x-n+1) \,$$ (in particular, $(x)_0 = 1$ because it is an empty product) be the falling factorial, we can characterize the Stirling numbers of the second kind by $$ \sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}(x)_k=x^n$$ |
H: How to write in $2^x=5$ in logarithmic form?
How do I write:
$$2^x = 5$$
In a logarithmic form?
I've looked for a solution for some time now, so I decided to try here.
AI: $2^x = 5$
Now 'take log' on both sides:
$\log_2 2^x = \log_2 5$.
We can now use the property $\log a^b = b\times \log a$.
Thus,
$x \times \log_2 2 = \log_2 5$.
But $\log_2 2 = 1$, therefore,
$x = \log_2 5$. |
H: Solve a polynomial involving geometric progression?
I have had trouble with this question:
"Solve the equation $8x^3 - 38x^2 + 57x -27 = 0$" if the roots are in geometric progression.
Any help would be appreciated.
AI: Let the roots be $a, a\cdot r, a\cdot r^2$
So using Vieta's formula $a+ a\cdot r+ a\cdot r^2=\frac{38}8\implies a(1+r+r^2)=\frac{19}4$
and $a( a\cdot r)+ a\cdot r(a\cdot r^2)+a\cdot r^2(a)=\frac{57}8\implies a^2r(1+r^2+r)=\frac{57}8$
Divide to get $ar=\frac{\frac{57}8}{\frac{19}4}=\frac32$ as $a(1+r+r^2)\ne0$
Put $a=\frac3{2r}$ in the first equation |
H: Which is true $A$ is subset of $B$ or $B$ is subset of $A$.
Consider the sets dened by the real solutions of the inequalities $$A=\{(x,y):x^2+y^4\le 1\}$$ and $$B=\{(x,y):x^4+y^6\le 1\}$$Then which is true $A$ is subset of $B$ or $B$ is subset of $A$.
This is Question No. $10.$ of this paper. Please help me.Thanks in advance.
AI: Hints:
$$(x,y)\in A\implies x^2+y^4\le 1\implies |x|,|y|\le 1$$
so
$$ x^4+y^6\le x^2+y^4\implies \ldots\ldots$$ |
H: How the inverse of this matrix be found?
How can the inverse of matrix
$A = \left( \begin{smallmatrix} 6&5\\5&4 \end{smallmatrix} \right)$ be $A^{-1} = \left( \begin{smallmatrix} -4&5\\ 5&-6 \end{smallmatrix} \right)$ where $\frac{1}{ad-bc} = \frac{1}{24-25} = \frac{1}{-1}$?
I thought that an inverse to this matrix was not possible, but my textbox and Wolfram Alpha says otherwise. Can someone tell me how this is possible, or if I have misunderstood the formula?
AI: You can see that this matrix is invertible in couple of ways.
For example:
$|A|=6\cdot4-5\cdot5=-1\neq0$
The rows of $A$ are linearly independent .
The columns of $A$ are linearly independent .
$0$ is not an eigenvalue of $A$ (you can calculate the eigenvalues
by finding the characteristic polynomial of $A$)
You have found a matrix $B$ s.t $AB=BA=I$ |
H: When is the quotient algebra of a unital C* algebra helpful?
Let $\mathcal A$ be a unital C* algebra.
Which properties does $\mathcal B \subset \mathcal A$ has to have for it to make sense to form the quotient algebra $\mathcal A / \mathcal B$?
In cases where this construction makes sense, does $\mathcal A / \mathcal B$ have any special structure/properties that are helpful?
Put differently, for what kind of standard questions does it help to consider $\mathcal A / \mathcal B$ because it has desired properties?
AI: If $\mathcal{B}$ is a closed $*$-ideal of $\mathcal{A}$, then $\mathcal{A}/\mathcal{B}$ is again a unital C*-algebra. This can be found in every introduction to C*-algebras. And of course this basic construction is used everywhere, so that it doesn't make sense to start a list of applications here. A well-known example is the Calkin algebra. In the commutative case, we have $C(X)/I(A) \cong C(A)$ for closed subspaces $A \subseteq X$ with vanishing ideal $I(A)=\{f : f|_A=0\}$. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.