问:
On group such that the group of inner automorphisms of it is isomorphic to $S_{3}$.
Let $\frac{G}{Z(G)}\cong S_{3}$, such that $S_{3}$ is permutation group on 3 letters and $Z(G)$ is non trivial central subgroup of $G$. Then does there exist an automorphism $\alpha$ of $G$ such that $\alpha(g)\neq g$ for $g\in G-Z(G)$? Thanks.</p>

答:
Let $G=S_3\times \mathbb{Z}_5,$ then $Z(G)=Z(S_3)\times Z(\mathbb{Z}_5)=\{\omega\}\times\mathbb{Z}_5,$ where $\omega$ is the trivial permutation of $S_3.$ Also $$G/Z(G)\cong S_3,$$ and since $\operatorname{gcd}(5,6)=1$ we have $$\operatorname{Aut}(G)\cong S_3\times\mathbb{Z}_4$$ (In fact one can compute both of these groups explicitly, if needed).

Let's choose a particular isomorphisms $\psi : S_3\to S_3$ such that $\psi(\rho)=\rho^2, \psi(\sigma)=\rho\sigma,$ where $\rho=(1\,2\,3), \sigma=(1\,2),$ and $\varphi:\mathbb{Z}_5\to\mathbb{Z}_5$ be such that $\varphi(1)=2.$ Note that $\psi(\omega)=\omega, \psi(\rho^2\sigma)=\rho^2\sigma$ and $\varphi(0)=0$ are the only fixed points. Moreover, $\alpha=\psi\times\varphi$ is an automorphism of $G$ satisfying
$$\alpha(g)\neq g$$ for any $g\in G\setminus\{(\omega, 0), (\rho^2\sigma, 0)\}.$ However since any automorphism of $S_3$ has at least two fixed points this construction cannot implement further. All the central extensions of $S_3$ are of the form $G\cong S_3\times A,$ and in general, $${\rm Aut}(S_3)\times{\rm Aut}(A)\subseteq{\rm Aut}(S_3\times A).$$ So, I guess one can find a perfect counter example by picking up $A$ such that above is a proper inclusion. But I cannot think of anything off the top of your head now.


问:
Finding an error in trigonometric relations in triangles
I am trying to derive an approximation formula in my book and I have a mistake somewhere I can't find. As you can see, I already wrote the most important relations into the sketch:

The only relation, which is not based on geometry but on physics, is the law of refraction:
$$\sin\beta=\frac{n_1}{n_2}\,\sin\alpha \tag1$$
In general, the equation should read like this:
$$\theta_2 = \delta - \beta = \delta - \arcsin\left [\frac{n_1}{n_2}\,\sin{(\theta_1+\delta)}\right], \hspace{20pt} \delta = \arcsin\frac{h}{R} \tag2$$

Now I look for small angles. This makes $\sin x \approx x$. The new relations follow: $\delta = \dfrac{h}{R}$, $\beta = \dfrac{n_1}{n_2}\alpha$,
\begin{align*}
\theta_2 = \delta - \beta &\approx \frac{h}{R} - \frac{n_1}{n_2}\alpha \\ &\approx \frac{h}{R} - \frac{n_1}{n_2}\left(\frac{h}{R} + \theta_1\right) \\
&\approx -\frac{n_1}{n_2}\theta_1 + \frac{n_2-n_1}{n_2}\frac{h}{R} \tag3
\end{align*}
The problem is that the solution in the book is as follows:
$$\theta_2\; \approx \;\frac{n_1}{n_2}\theta_1 - \frac{n_2-n_1}{n_2}\frac{h}{R} \tag4$$

So exactly multiplied by the factor -1. But I do not see where the minus could be missing.

Can you help me?

答:
Everything seems fine in absolute value, so that means that you somewhere are inconsistent with the way you measure angles and/or distances. So, if we call $P$ the point where the rays intersect the curved surface, the angle from $PP_2$ to $OP_2$ is positive. But if you measure the angle from $OP_2$ to $PP_2$, then the angle is negative. And similar with distances.

Let's see what's happening if $n_2\gg n_1$. Physically, it means that the angle $\beta$ goes to $0$, so the point $P_2$ goes to $C$, and $\theta_2\to\delta$. In your formula, it means $\delta\approx\frac hR$, while in the solution $\delta=-\frac hR$.

ContentLicense: CC BY-SA 4.0


问:
How to compute $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}/ \langle(2,1,2)\rangle$?
I am trying to compute the factor group  $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}/ \langle(2,1,2)\rangle$. Here is what i did:

First i expand  $\langle(2,1,2)\rangle$, it is equal to {(2,1,2),(0,0,4),(2,1,6),(0,0,0)}. Then, since order of $\mathbb{Z}_{4} \times\mathbb{Z}_{2} \times \mathbb{Z}_{8}$ is 64, we have 64/4=16. So the order of factor group is 16 and so it must be isomorphic to one of the following: $\mathbb{Z}_{16}$,  $\mathbb{Z}_{2} \times\mathbb{Z}_{8}$,  $\mathbb{Z}_{4} \times\mathbb{Z}_{4}$,  $\mathbb{Z}_{2} \times\mathbb{Z}_{2}  \times\mathbb{Z}_{4}$ or $\mathbb{Z}_{2} \times\mathbb{Z}_{2}  \times\mathbb{Z}_{2} \times\mathbb{Z}_{2}$. But from here on i do not know how to continue. Can anyone help? How can i eliminate  $\mathbb{Z}_{16}$ etc.?

Edit: Actually, i have a solution, saying that consider 4(a,b,c)=(4a,4b,4c)=(0,0,4c). 4c is either 0 or 4, in both cases both (0,0,0) and (0,0,4) are in $\langle(2,1,2)\rangle$ so any coset in the factor group has order at most 4. I do not understand why this is the case. Can anyone explain it?

Thanks

答:
An alternate approach is to convert the problem to an integer problem: $\mathbb{Z}/4 \times \mathbb{Z}/2 \times \mathbb{Z}/8$ is isomorphic to the quotient of $\mathbb{Z}^3$ by the (integer) row space of the matrix

$$ \left( \begin{matrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \end{matrix} \right) $$

The quotient group you are interested in, then, is obtained as a quotient of $\mathbb{Z}^3$ by the rowspace of

$$ \left( \begin{matrix} 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 8 \\ 2 & 1 & 2 \end{matrix} \right) $$

We can use elementary integer row operations to perform row reduction (in the spirit of the Euclidean algorithm!) to simplify this matrix without changing its rowspace. However, we can do more: elementary integer column operations have the effect of a change of basis on $\mathbb{Z}^3$, so we can do those too in our quest to find the quotient group.

Doing both column and row operations to simplify the matrix will result in a diagonal matrix, which makes it easy to see the structure of the quotient group.

ContentLicense: CC BY-SA 3.0


问:
Distributing n distinct objects to r different people - Why is my method wrong?
I saw this problem:
How do I distribute 5 different candies to 3 different students?

I decided to try and make a modified Stars and Bars formula to handle the different candies.

My Method:

Suppose I had 5 different candies:
A B C D E

and 2 bars for 3 different students: ||

combined: A B C D E | |

To find all the different combinations of candies and students, I need to do

$${\frac{7!}{2!}}$$

However, my answer was wrong. I'm not sure what the correct answer is, because the quiz I took did not show it.

Why is my method wrong?

答:
You just have to affect one student to each candy. For each candy, you have $3$ different possibilities, so this gives
$$3^5=243$$
possibilities.


问:
GCD of elementary symmetric functions
It is easy to show that p/ $\sigma_j$ ( where $\sigma_j$ is the sum of all products of j distinct members of the set {1,2,...,p-1}) for all $1 \leq j \leq p-2$, but how would you go about showing that the greatest common divisor of $\sigma_1,\sigma_2,...,\sigma_{p-2}$ is in fact p itself?

My first thought was to consider $
\sigma_1 = 1+2+3+\cdots p-1=\frac{p(p-1)}{2}$.  Since we know that p divides each $\sigma_j$, we consider $\frac{\sigma_j}{p}=\frac{p-1}{2}$.  Now if I can show that no prime dividor of $\frac{p-1}{2}$ divides any other $
\sigma_j$, then I'm done.  However, I've run into problems with this and I am not entirely sure this is the proper way to approach the problem.  Any suggestions would be nice.

答:
Hint: We want to show that if $q$ is a prime factor of $p-1$, then $q \nmid \sigma_i$ for some $i$. Consider

$$(x-1)(x-2) \ldots (x-(p-1)) \equiv [x(x-1) \ldots (x-(q-1))]^{\frac{p-1}{q}} \pmod{q}$$

Hint 2: We have $$(x-1)(x-2) \ldots (x-(p-1))=x^{p-1}+\sum_{i=1}^{p-2}{\sigma_i(-1)^ix^{p-1-i}}+(-1)^{p-1}(p-1)!$$

If $q$ were to divide $\sigma_i$ for all $i$, then the above will be congruent to $x^{p-1} \pmod{q}$. Can $[x(x-1) \ldots (x-(q-1))]^{\frac{p-1}{q}}$ be congruent to $x^{p-1} \pmod{q}$?

Title: GCD of elementary symmetric functions


问:
Lindelof's Theorem And Second Countable Topological Space
Let $X$ be a second countable topological space and let a non empty open set $G$ can be written as a union of a class of $G_i$ of open sets, then $G$ can be written as a countable union of $G_i$'s.

It is given that
$$G = \bigcup G_i.$$
Now for any $x \in G$ we have $x \in G_{i_0}$. Now as $X$ is a second countable topological space and hence it has a countable base say
$$B_1, B_2, .........$$
Now as $x \in G_{i_0}$, so there is some $B_{n_0}$ such that
$$x \in B_{n_0} \subset G_{i_0}.$$
And as $x$ is an arbitrary element of $G$ so we have
$$G \subset \bigcup_{k=1}^{\infty} B_{n_k}.$$
Also for each $B_{n_k}$ there exists a $G_{i_k}$ such that
$$B_{n_k} \subset G_{i_k}$$
with the above property and hence we have
$$G \subset \bigcup_{k=1}^{\infty}B_{n_k} \subset \bigcup^{\infty}_{k=1}G_{i_k} \subset \bigcup G_i = G.$$
And hence we get
$$G = \bigcup^{\infty}_{k=1}G_{i_k}.$$
And hence $G$ can be written as a countable union of $G_i$'s.

Is My Proof Correct??

答:
Your understanding of the result is correct, but both the result and your proof are written badly enough that they are unnecessarily hard to follow. The result that you want to prove is this:

Let $X$ be a second countable space and $G=\bigcup_{i\in I}G_i$ a non-empty open set, where each $G_i$ is also an open set. Then there is a countable $I_0\subseteq I$ such that $G=\bigcup_{i\in I_0}G_i$.

(Note that it is incorrect to write $G=\bigcup G_i$: $\bigcup G_i$ actually means $\bigcup_{A\in G_i}A$, the set of all elements of elements of $G_i$. Of course here the elements of $G_i$ are just points of the space $X$, so this doesn’t actually make much sense, and most readers will realize that you actually mean $\bigcup_{i\in I}G_i$, but it’s better to be accurate and clear.)

The best place to start your proof is by identifying the countable base that you’re going to use.

$X$ is second countable, so let $\{B_n:n\in\Bbb N\}$ be a countable base for $X$.

Now for each $x\in G$ you want to choose a member of the base that contains $x$ and is a subset of some $G_i$. The clearest way to do this is to name that member in a way that shows its association with the point $x$.

For each $x\in G$ there are an $n(x)\in\Bbb N$ and an $i(x)\in I$ such that $x\in B_{n(x)}\subseteq G_{i(x)}$. Clearly $$G\subseteq\bigcup_{x\in G}B_{n(x)}\subseteq\bigcup_{x\in G}G_{i(x)}\subseteq G\,,$$ so $$G=\bigcup_{x\in G}B_{n(x)}=\bigcup_{x\in G}G_{i(x)}\,.$$

Now comes the one bit that’s slightly tricky to do right. Our base is countable, so there are only countably many different sets $B_{n(x)}$, but there might be uncountably many different sets $G_{i(x)}$: there might be distinct points $x,y\in G$ such that $n(x)=n(y)$, but $i(x)\ne i(y)$. Here’s one way to fix that.

Let $M=\{n(x):x\in G\}$, for each $m\in M$ let $x_m\in G$ be such that $n(x_m)=m$, and let $i_m=i(x_m)$. Then $I_0=\{i_m:m\in M\}$ is a countable subset of $I$, and $$G=\bigcup_{x\in G}B_{n(x)}=\bigcup_{m\in M}B_m\subseteq\bigcup_{i\in I_0}G_i\subseteq G\,,$$ so $G=\bigcup_{i\in I_0}G_i$, as desired.


问:
Show that for all $\epsilon>0$ exists a continous function $g$ such that $\|g-f\|_2<\epsilon$
Suppose that $f$ is a continous function at $[a,b]$ except for a jump discontinuity at $c\in(a,b)$. Show that for all $\epsilon>0$ exists a continous function $g$ such that $\|g-f\|_2<\epsilon$

My try:

We just need to show that $\|g-f\|_2<\epsilon_0$ for one jump discontinuity and then add the error of the finite number of them in orther to get $\|g-f\|_2<\sum_{i=0}^{n}\epsilon_i=\epsilon$.

Let $S_n(f)$ be the Fourier series of $f$

Now let $N>m$ such that $\|S_N(f)-f\|_2<\epsilon_0$.

Then the sum of all the jump dicontinuities must be $$\|S_N(f)-f\|_2 = \|g-f\|_2<\sum_{i=0}^{n}\epsilon_i=\epsilon$$

Any suggestions would be great!

答:
First, reduce our consideration to a step function as follows. Let $c_1,\dots,c_n$ denote the values of $x \in (a,b)$ at which $f$ has a jump discontinuity.
Define $h_j(x)$ by
$$
h_j(x) = \begin{cases}
0 & x < c_j\\
f(c_j) - \lim_{x \to c_j^-}f(x) & x = c_j\\
\lim_{x \to c_j^+}f(x) - \lim_{x \to c_j^-} f(x) & x > c_j
\end{cases}
\qquad j = 1,\dots,n.
$$
Define $\phi(x) = f(x) - (h_1(x) + \cdots + h_n(x))$. We see that $\phi(x)$ is continuous. Now, suppose that $\gamma_1,\dots,\gamma_n$ are continuous functions with $\|\gamma_j - h_j\|_2 < \epsilon/n$ for each $j = 1,\dots,n$. If we define
$$
g(x) = \phi(x) + \gamma_1(x) + \cdots + \gamma_n(x),
$$
then we find that
$$
\begin{align}
\|g - f\|_2 &=
[\|\phi + \gamma_1 + \cdots + \gamma_n] - [\phi + h_1 + \cdots + h_n]\|_2
\\ & = \|(\gamma_1 - h_1) + \cdots + (\gamma_n - h_n)\|_2
\\ & \leq \|\gamma_1 - h_1\|_2 + \cdots + \|\gamma_n - h_n\|_2 < n \cdot (\epsilon/n) = \epsilon.
\end{align}
$$
That is, it suffices to consider the problem in which we have
$$
f(x) = \begin{cases}
0 & x< c\\
k_1 & x = c\\
k_2 & x > c.
\end{cases}
$$
One choice of $g$ that works well is a piecewise-linear function that satisfies $g(x) = f(x)$ for all $x$ with $|x - c| > \delta$, for some $\delta > 0$ (that depends on $k_2$).


问:
Frobenius norm of product of matrix and vector
I need to find the condition for $\|Ax\|_F^2$=$\|A\|_F^2$$\|x\|_F^2$, where A is a matrix and x is a vector.
Can anyone help me? Thank you.

答:
The Frobenius norm is unitarily invariant, i.e. $\|U A V\|_F = \|A\|_F$, for any $U$ and $V$ unitary. So we may assume that $A = \text{diag}(d_1, d_2, \cdots, d_n)$ is diagonal. (The left hand side can be rewritten as $\|U D V^\dagger (V y)\|_F$, where $UDV^\dagger$ is the singular value decomposition of $A$ and $y = V^\dagger x$). Letting $y = (y_1, y_2, \cdots, y_n)^T$, we have
$$\|UDV (Vy)\|_F^2 = \|D y\|_F^2 = \sum_{i = 1}^n |d_i y_i|^2$$
We want this value to equal $\|A\|_F^2 \|x\|_F^2 = \|D\|_F^2 \|y\|_F^2$, namely we want
$$\sum_{i = 1}^n |d_i y_i|^2 = \left(\sum_{i = 1}^n |d_i|^2 \right) \left(\sum_{i = 1}^n |y_i|^2 \right)$$
For equality to hold we must have
$$\sum_{i \neq j} |d_i|^2 |y_j|^2 = 0$$
As all terms in the summation are non-negative, we must have each term be zero. So ultimately, we can either have $D = 0$ (implying $A = 0$) or $y = 0$, or have exactly one index $i$ such that $d_i, y_i \neq 0$. So in summary, the equality holds if and only if one of the following is true:

$A = 0$

$y = 0$

$A$ is a rank one matrix (i.e. its singular values are all zero except one, say $d_1$) and $y_1$ is a multiple of the right-singular vector corresponding to the nonzero singular value. An example is
$$A = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0\end{bmatrix}, ~~ x = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0\end{bmatrix}$$

We can also conclude from our steps that in all other cases, we have $\|A x\|_F^2 < \|A \|_F^2 \|x\|_F^2$ (the Frobenius norm is sub-multiplicative).

Addendum: Why must $y$ be a right-singular vector?

Answer: If we order the singular values (all of which are zero except one) of $A$ in descending order, i.e. $D = \text{diag}(d_1, \cdots, d_n)$ with $d_1 > 0$ and $d_2 = d_3 = \cdots = d_n = 0$, then we want $V^\dagger y = x = \hat{e}_1$ (here $\hat{e}_1$ is the first standard basis vector). Thus $y = V \hat{e}_1$ (which picks out the first column of $V$). This is the definition of the right-singular vector corresponding to the nonzero singular value (the columns of $V$ are the right-singular vectors of $A$).


问:
Calculating limit of sequence using only $1/n\to0$
First of all, thank you for visiting this thread, and thanks for whom will try helping.
So, the teacher explained to us how to calculate the limit of the sequence by only using one fact: $1/n$ goes to $0$.
However, I can't find the solution for this question, using only this fact.
In red, I wrote in Hebrew "Goes to zero".
In addition, there are about 4 or 5 more students who shared their way to solve this question, and tried the same way as I wrote. But I think we're missing some rule here.

Is it even possible to calculate the sequence's limit using only "$1/n$ goes to $0$"?
Again, thanks in advance. :)

答:
$$-(4\sqrt{n+3}-\sqrt{n-1}-5\sqrt{n+7}+2\sqrt{n-3})
\\=-(4\sqrt{n+3}-4\sqrt{n+7}+\sqrt{n-3}-\sqrt{n-1}+\sqrt{n-3}-\sqrt{n+7})
\\=4\frac{4}{\sqrt{n+3}+\sqrt{n+7}}+\frac{2}{\sqrt{n-3}+\sqrt{n-1}}+\frac{10}{\sqrt{n-3}+\sqrt{n+7}}
\\\le\frac{28}{2\sqrt{n-3}}.$$

Now if we take $n-3=\dfrac{m^2}{196}$, where $m$ goes to $\infty$ with $n$, this expression is positive and bounded above by

$$\frac1m$$ so it tends to zero.


问:
On using Big-O-notation to derive a limit
Let's consider the following limit:
$\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n^2}\frac{1}{e^n}=\lim_{n \to \infty}e^{n^2\cdot\ln\left(1+\frac{1}{n}\right)-n}=\lim_{n \to \infty}e^{n^2\cdot\left(\frac{1}{n}-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)\right)-n}=\lim_{n \to \infty}e^{-\frac{1}{2}+O\left(\frac{1}{n}\right)}=e^{-\frac{1}{2}}$

Intuitively this makes sense. However, I am not sure how to write this out rigorously by using the standard rules of limits.

My attempt:

Firstly, I will try to find an upper bound of $n^2\cdot\ln\left(1+\frac{1}{n}\right)-n$ by using the Taylor expansion of $\ln(x)$ at point $1$:

For all $m>2$ there exists a constant $C$ such that for a large enough $n$ it holds:
$$\sum\limits_{k=3}^{m}\frac{(-1)^{k+1}}{kn^{k-2}}\leq \Big|\sum\limits_{k=3}^{m}\frac{(-1)^{k+1}}{kn^{k-2}}\Big|\leq \sum\limits_{k=3}^{m}\Big|\frac{(-1)^{k+1}}{kn^{k-2}}\Big|\leq C\Big|\frac{1}{n}\Big|\implies \sum\limits_{k=3}^{m}\frac{(-1)^{k+1}}{kn^{k-2}} \in\mathcal{O}(\frac{1}{n}).$$
Hence, $n^2\cdot\ln\left(1+\frac{1}{n}\right)-n\leq  C\Big|\frac{1}{n}\Big|$.
By using the monotonicity of $e^x$ we write:
$$e^{\left(-\frac{1}{2}+\sum\limits_{k=3}^{m}\frac{(-1)^{k+1}}{kn^{k-2}}\right)}\leq e^{\left(-\frac{1}{2}+C\frac{1}{n}\right)}=e^{-\frac{1}{2}}e^{\left(C\frac{1}{n}\right)}.$$
So for $n$ which are large enough we have:

$\left(1+\frac{1}{n}\right)^{n^2}\frac{1}{e^n}=e^{n^2\cdot\ln\left(1+\frac{1}{n}\right)-n}\leq e^{-\frac{1}{2}+\left(C\frac{1}{n}\right)}$ and therfore $\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n^2}\frac{1}{e^n}\leq e^{-\frac{1}{2}}$.

This doesn't yield $e^{-\frac{1}{2}}$ as limit but only as an upper bound.

So how do I show that $e^{-\frac{1}{2}}$ is indeed the limit? Using Big-O-notation seems handwaving in this context.

EDIT

To clarify things, the expression:
$$\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^{n^2}\frac{1}{e^n}=\cdots=\lim_{n \to \infty}e^{-\frac{1}{2}+O\left(\frac{1}{n}\right)}=e^{-\frac{1}{2}}$$

doesn't mean that the limit of $\left(1+\frac{1}{n}\right)^{n^2}\frac{1}{e^n}$ attains the value of $e^{-\frac{1}{2}}$ but only that the limit of $\left(1+\frac{1}{n}\right)^{n^2}\frac{1}{e^n}$, which still is unknown, has the same asymptotic behavior as $e^{-\frac{1}{2}}$. The limit of $\left(1+\frac{1}{n}\right)^{n^2}\frac{1}{e^n}$ might be $5e^{-\frac{1}{2}}$ or $6e^{-\frac{1}{2}}$ or $Ce^{-\frac{1}{2}}$ where $C$ can be any constant. To derive the exact limit one must use the standard rules of limits, as MarkViola did.

Is this correctly summarized?

答:
Using the inequalities $\frac1n-\frac1{2n^2}\le \log\left(1+\frac1n\right)\le\frac1n-\frac1{2n^2}+\frac1{3n^3}$, we assert that

$$-\frac12\le n^2\log\left(1+\frac1n\right)-n\le -\frac12+\frac1{3n}$$

Therefore, we have

$$e^{-1/2}\le\left(1+\frac1n\right)^{n^2}e^{-n}\le e^{1/2}e^{1/3n}$$

whence applying the squeeze theorem yields the coveted limit.

EDIT:  TO ANSWER THE EDIT IN THE OP

If $f(n)=O\left(\frac1n\right)$, then there exists a number $C>0$ such that for sufficiently large $n$, $|f(n)|\le C\frac1n$.

We will show that $\lim_{n\to\infty}e^{f(n)}=1$.  To do so, we will use the fact that $e^x$ monotonically increases along with the inequality $e^x<\frac1{1-x}$ for $x<1$.   Equipped accordingly, we now proceed.

Let $\varepsilon>0$ be given and let $n>C$.  Then, we can write

$$\begin{align}
|e^{f(n)}-1|&\le  e^{C/n}-1\\\\
&<\frac{C}{n-C}\\\\
&<\varepsilon
\end{align}$$

whenever $n>C\left(1+\frac1\varepsilon\right)$.  We conclude that $\lim_{n\to\infty}e^{f(n)}=1$ when $f(n)=O\left(\frac1n\right)$.  And we are done.


问:
Eigenvalue/eigenvector matrix A^(-1)B - Super hard question linear algebra
My question is the following: Given that L is an eigenvalue of both A and B with eigenvectors u_1 and u_2, respectively. Is L an eigenvalue of A^(−1)*B? Is this the case? If yes, can the eigenvector of A^(−1)*B be written in terms of u_1 and u_2.

We know that:

A*(u_1) = L*(u_1)

B*(u_2) = L*(u_2)

We are looking for:

A^(-1)B*(u) = L*(u)

And we want to figure out whether L is an eigenvector of this matrix and what the eigenvector would be.

I have tried to manipulate the two equations that we know to the one we are looking for, but I fail to do so. Could it be the case that there is no eigenvalue for that matrix? How would one show that?

答:
The answer is no. For example, consider $L=1$ and the matrices
$$
A = \pmatrix{2&0\\0&1}, \quad B = \pmatrix{1&0\\0&3}.
$$
In general, if there is a question about whether something always holds, it's good to begin by checking whether it holds for simple examples.


问:
Contour Integral
I have this question:

I'm aware that $e^{iz^2}$ is analytic, and hence $I_R = 0$ by Cauchy's Integral theorem. I'm not really sure what to do from there. Thanks!

答:
The closed contour is split into 3 pieces:

The original integral:

$$\int_0^R dx \, e^{i x^2}$$

The integral over the angled line of the contour; parametrize with $z=e^{i \pi/4} t$:

$$e^{i \pi/4} \int_R^0 dt \, e^{- t^2}$$

(because $i e^{i \pi/2} = -1$).  The third piece is the integral over the circular arc subtending an angle of $\pi/4$ with respect to the origin.  Parametrize using $z=R e^{i \theta}$, $\theta \in [0,\pi/4]$:

$$i R \int_0^{\pi/4} d\theta \, e^{i \theta} \, e^{-R^2 \sin{2 \theta}} e^{i R^2 \cos{2\theta}}$$

which, in absolute value, is

$$\le R \int_0^{\pi/4} d\theta \, e^{-R^2 \sin{2 \theta}} \le R \int_0^{\pi/4} d\theta \, e^{-4 R^2 \theta/\pi} \le \frac{\pi}{4 R}$$

as $R \to \infty$.  Note that I used the relation $\sin{2 \theta} \ge \frac{4 \theta}{\pi}$ when $\theta \in [0,\pi/4]$.  So when we set $R \to \infty$, this integral vanishes.  Thus, in this limit, we are left with

$$\int_0^{\infty} dx \, e^{i x^2} - e^{i \pi/4} \int_0^{\infty} dt \, e^{- t^2} = 0$$

Thus, you may now evaluate the integral on the left in terms of the integral on the right, which you know converges.

Title: Contour Integral


问:
Quasicompact scheme vs quasicompact morphism
We say $X$ is a quasicompact scheme (qc) if every open cover has a finite subcover. We also say that a morphism $f:X\rightarrow Y$ is quasicompact if $f^{-1}(U)$ is qc for every affine open set $U$ of $Y$.

Why is $X$ being qc equivalent to the morphism $X\rightarrow\operatorname{Spec}\mathbb{Z}$ being qc? Clearly, if $f$ is quasicompact, then $f^{-1}(\operatorname{Spec}\mathbb{Z})=X$ is quasicompact. But the opposite direction is what gives me trouble.

I know that $\operatorname{Spec}\mathbb{Z}$ is the final object in the category of schemes, but I'm not sure where to go from there.

答:
As @KReiser pointed out in the comment if $f:X\rightarrow \operatorname{Spec}\mathbb Z$ is quasi-compact, then $f^{-1}(\operatorname {Spec}\mathbb Z)=X$ is a quasi-compact topological space.
 Conversely assume $X$ is a quasi-compact scheme. Any open affine subset of $\operatorname{Spec}\mathbb Z$  is a finite union of distinguished open affines $D(n)$ for $n\in \mathbb Z$. So if $X=\bigcup_{i\in \mathscr F}X_i$ is a finite union of open affines, then $f^{-1}(D(n))=\bigcup_{i\in \mathscr F}{X_i}_{f^\#(n)}$ which is a finite union of open affines and hence quasi-compact. (${X_i}_{f^\#(n)}$ is a distinguished open set subset of the affine scheme $X_i$ and hence affine) Thus $f^{-1}(U)$ is quasi-compact for every open affine $U\subset \operatorname{Spec}\mathbb Z$. So $f$ is quasi-compact.


问:
Can anyone help with solving these questions on vectors?
Given the vectors:
a =(−7,3,−10),b =(2,1,5), and c =(18,4,m), with m being a real number.

e) Use the determinant det (a,b,c) to determine m so that the three vectors are linear
dependent.

f) Calculate for the m value from the previous part the coefficients α and β in the linear
combination: c = αa + βb.

Thanks

答:
e) The vectors are linearly dependent if the determinant ($510-13m$) is zero.  Solve that.

f) Plug in your $m$ and solve for your $\alpha$ and $\beta$.


问:
Differentiation wrt matrix involvoing Khatri-rao product
I got a following minimization problem

$$\min_{\mathbf{X}^{(1)}, \, \mathbf{X}^{(2)}} \;\left\| \mathbf{B} - \mathbf{A} (\mathbf{X}^{(1)} \odot \mathbf{X}^{(2)}) \right\|^{2}_{F},$$

where the matrices $\mathbf{B}\in \mathbb{R}^{100 \times 3}$, $\mathbf{A}\in \mathbb{R}^{100\times 36}$, $\mathbf{X}^{(1)}\in \mathbb{R}^{9 \times 3}$ and $\mathbf{X}^{(2)}\in \mathbb{R}^{4 \times 3}$. The operation $\odot$ refers to the Khatri-rao product.

Given matrices $\mathbf{A}$ and $\mathbf{B}$, my problem is to find out matrices $\mathbf{X}^{(1)}$ and $\mathbf{X}^{(2)}$  such that

$$\mathbb{f} = \left\| \mathbf{B} - \mathbf{A} (\mathbf{X}^{(1)} \odot \mathbf{X}^{(2)}) \right\|^{2}_{F}$$

is minimized.

My idea is to compute the gradient of $\mathbb{f}$ with respect to $\mathbf{X}^{(1)}$ and $\mathbf{X}^{(2)}$ respectively.

My question is, how to do differentiation with respect to a matrix? I have consulted a reference but the situation seems different because it involves a Khatri-rao product in $\mathbb{f}$. Thanks in advance.

$\dfrac{\partial \mathbf{f}}{\partial \mathbf{X}^{(1)}}$ and $\dfrac{\partial \mathbf{f}}{\partial \mathbf{X}^{(2)}} $?

答:
$\def\p{\partial} \def\bb{\mathbb}$
Given two matrices with the same number of columns, e.g.
$$A\in{\bb R}^{a\times n} \qquad B\in{\bb R}^{b\times n}$$
their Khatri-Rao $(\boxtimes)$ product can be written in terms of the Kronecker $(\otimes)$ and Hadamard $(\odot)$ products and all-ones vectors ${\tt1}_p\in{\bb R}^{p}\;$
$$\eqalign{
A\boxtimes B &= (A\otimes {\tt1}_b)\odot({\tt1}_a\otimes B) \\
}$$
We'll also need one more product, the trace/Frobenius product, i.e.
$$\eqalign{
A:C &= {\rm Tr}(A^TC)\qquad &\big({\rm for}\;\, C\in{\bb R}^{a\times n}\big) \\
A:C &= C:A \\
A:A &= \big\|A\big\|^2_F &\big({\rm useful\,identity}\big) \\
A:(C\odot M) &= (A\odot C):M &\big({\rm also\,useful}\big) \\
}$$

Since this derivation is going to be complicated enough without the distraction of superscripts, let's define the following matrices for ease of typing
$$\eqalign{
X &= X^{(1)}\in{\bb R}^{9\times 3}\qquad &Y = X^{(2)}\in{\bb R}^{4\times 3} \\
Z &= X\boxtimes Y\in{\bb R}^{36\times 3}\qquad &W = A^T(AZ-B)\in{\bb R}^{36\times 3} \\
w &= {\rm vec}(W)\qquad &{\cal W} = {\rm Diag}(w)\in{\bb R}^{108\times 108} \\
}$$
Write the objective function in terms of these variables
then calculate the differential.
$$\eqalign{
 f &= \tfrac 12(AZ-B):(AZ-B) \\
df &= (AZ-B):(A\,dZ) \\
 &= W:dZ \\
 &= W:(dX\boxtimes Y) + W:(X\boxtimes dY) \\
}$$
Now calculate the gradient with respect to $X$.
$$\eqalign{
df &= W:(dX\boxtimes Y) \\
 &= W:(dX\otimes{\tt1}_4)\odot({\tt1}_9\otimes Y) \\
 &= W\odot({\tt1}_9\otimes Y):(dX\otimes{\tt1}_4) \\
 &= w\odot{\rm vec}({\tt1}_9\otimes Y):{\rm vec}(dX\otimes{\tt1}_4) \\
 &= {\cal W}\cdot{\rm vec}({\tt1}_9\otimes Y):{\rm vec}(dX\otimes{\tt1}_4) \\
 &= {\cal W}\cdot{\rm vec}({\tt1}_9\otimes Y)
  :(I_{27}\otimes{\tt1}_4)\cdot{\rm vec}(dX) \\
 &= (I_{27}\otimes{\tt1}_4^T)\cdot{\cal W}\cdot{\rm vec}({\tt1}_9\otimes Y):dx \\
\frac{\p f}{\p x}
 &= (I_{27}\otimes{\tt1}_4^T)\cdot{\cal W}\cdot{\rm vec}({\tt1}_9\otimes Y) \\
}$$
And with respect to $Y$.
$$\eqalign{
df
 &= {\cal W}\cdot{\rm vec}(X\otimes{\tt1}_4)
  :{\rm vec}({\tt1}_9\otimes dY) \\
 &= {\cal W}\cdot{\rm vec}(X\otimes{\tt1}_4)
  :({\tt1}_9\otimes I_{12})\cdot{\rm vec}(dY) \\
 &= ({\tt1}_9^T\otimes I_{12})\cdot{\cal W}\cdot{\rm vec}(X\otimes{\tt1}_4):dy \\
\frac{\p f}{\p y}
 &= ({\tt1}_9^T\otimes I_{12})\cdot{\cal W}\cdot{\rm vec}(X\otimes{\tt1}_4) \\
}$$
The gradients can be converted between vector and matrix forms, e.g.
$$\eqalign{
\frac{\p f}{\p y} &= {\rm vec}\left(\frac{\p f}{\p Y}\right)
\qquad\iff\qquad
\frac{\p f}{\p Y} &= {\rm unvec}\left(\frac{\p f}{\p y}\right) \\\\
}$$
These results are in a different form than Florian's,
but they should be equivalent.


问:
Find the limit of $2+\left(-\frac{2}{e}\right)^n$, as $n\to\infty$, if it exists
I'm absolutely unsure about how to approach this. I've considered changing it to $-2=\left(-\frac{2}{e}\right)^n$ and then using the properties of lograrithms, but $\ln(-2)$ is undefined, as is $\ln(-\frac{2}{e})$.

I can't use L'Hopital's rule either because the limit is of the form $1^{\infty}$, and I have no clue how to manipulate it so I can apply it. I'm tempted to say that it approaches infinity, but I can't be confident that there isn't another way to solve it.

Any tips? Thanks.

答: What is the absolute value of $2/e$? Is it less than $1$ -- and if so, what does this imply for the second term of the sum?</p>


问:
How to simplify $\left(x+i\pi\right)^{1+x}+\left(x-i\pi\right)^{1+x}$ for $x>0$
I know according to Wolfram Alpha that, for all $x>0$, the following expression is real : $\left(x+i\pi\right)^{1+x}+\left(x-i\pi\right)^{1+x}$

Can anyone help me find the simplified form ?
This would net me a closed form for the integral I asked in the following post : Closed form of $I = \int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$

答:
That the expression $$f(x) = (x+i\pi)^{1+x} + (x-i\pi)^{1+x}, \quad x > 0$$ has zero imaginary component is immediately appreciable by noting that the arguments of $x+i\pi$ and $x-i\pi$ are equal in magnitude and opposite in sign, thus by De Moivre's theorem, the arguments of $(x+i\pi)^{1+x}$ and $(x-i\pi)^{1+x}$ are also equal in magnitude and opposite in sign.  Their sum therefore has argument $0$.

To find a closed form, we can let $\theta = \tan^{-1} \frac{\pi}{x}$ and $r = \sqrt{x^2+\pi^2}$, hence $$x \pm i \pi = re^{\pm i\theta},$$ and $$(x\pm i \pi)^{1+x} = r^{1+x} e^{\pm (1+x) i \theta},$$ and $$f(x) = r^{1+x} (e^{(1+x)i \theta} + e^{-(1+x)i\theta}) = 2r^{1+x} \cos\left( (1+x) \theta \right) \\ = 2(x^2 + \pi^2)^{(1+x)/2} \cos \left( (1+x) \tan^{-1} \frac{\pi}{x} \right).$$


问:
Find the Maclaurin series for $f(x) = (1+ x)^{\frac{1}{3}}$. Find the interval of convergence. (Please use Maclaurin Series)
I have found the terms $a_0$, $a_1$, $a_2$ basically the terms for the series. I can't get the general term $a_n$ to use for the ratio test.

答:
Here is a possibility to find it, notice that $f(x)-3(1+x)f'(x)=0$.

Now from $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$, find the induction relation $a_{n+1}=\dfrac{(\frac 13-n)}{n+1}a_n$ by formally derivating the series, re-indexing and identifying the coefficient of $x^n$ with zero.

Find back the "known" binomial coefficients $a_n=\dfrac {\frac 13(\frac 13-1)(\frac 13-2)\cdots(\frac 13-n+1)}{n!}$

For the ratio test, well, you have it already from the induction relation...

Look at the compendium for all common expansions.


问:
$A$ is a $n \times n$ matrix. Prove $\bar{A}A$ is similar to $A\bar{A}$ (or disprove)?(Solved, it is true.)
$A$ is a $n \times n$ complex matrix. Prove $\bar{A}A$ is similar to $A\bar{A}$? I assume $\bar{A}$ here means just the complex conjugate of $A$. It is not the conjugate transpose as I know in that case they are similar.

Any ideas that could help? Thanks! If anyone could disprove it that is also good.

Update This problem has been solved. See the comments below. Once figured out that $AB$ and $BA$ has the same nonsingular Jordan block, the singular part could be matched by considering that the solutions of $(\bar{A}A)^kx=0$ and $(A\bar{A})^kx=0$ are conjugate to each other (thus one-to-one).

答: It's true. Two complex square matrices $A$ and $B$ are said to be consimilar if $A=\overline{S}BS^{-1}$ for some invertible complex matrix $S$. It is straightforward to verify that if $A$ and $B$ are consimilar, then $A\overline{A}$ and $B\overline{B}$ are similar. Now, every complex square matrix $A$ is consimilar to $\overline{A}$ (see e.g. corollary 4.11 of Hong and Horn, A canonical form for matrices under consimilarity, LAA 102: 143-168 (1988)). Therefore $A\overline{A}$ is similar to $\overline{A}A$.


问:
Is it a general truth that when two equations can be described with the same sentence, only varying intonation they are equivalent?
My question is the effect of something that I have noticed during some of the courses I am taking as a first year maths bachelor, that in tutorials we will be asked to proof that two statements are equal but that when I try to describe the left and right hand side I end up with one and the same sentence, I will give three examples:
\begin{equation}
    (A^m)^{-1} = (A^{-1})^m
\end{equation}
Both the left and right hand side are described by the sentence "the inverse of matrix A to the power of m"
\begin{equation}
f^{-1}(A\cap B) = f^{-1}(A) \cap f^{-1}(B)
\end{equation}
With both A and B being supsets of the real numbers, both the left and right hand side are described by the sentence "the values that when the function f is applied to them end up inside set A and also inside set B", something similar can be done replacing the intersect with a union and the 'and' with an 'or'.
\begin{equation}
    (AB)C = A(BC)
\end{equation}
This example is discussed in three blue one browns series on linear algebra, both the left and right hand side are described as "first apply transformation C, then transformation B, and finally transformation A"

Of course, the fact that two sides are described the same is not a rigorous proof since it could be that sneaky business is applied in words with double meanings - although often that should be easy enough to spot - but I do think this is a rather interesting patern and I would absolutely love to know if anyone can come up with an example where both the left and right hand side are described the same and still not equal.

答:
The sentences you provide lack punctuation. Hence they are (potentially) ambiguous. Whereas "the inverse of matrix A to the power of m" uses correct syntax, semantically it is ambiguous. It does not adequately describe either $(A^m)^{-1}$ or $(A^{-1})^m$.

Intonation is a form of verbal punctuation which is applied by someone who interprets a sentence. Verbally there are homophones, homonyms and homographs - so it depends on whether the sentence is spoken or written. Context can also be necessary as a clue to the intended meaning. Furthermore, the structure of sentences (syntax) can differ between languages.

So I think there is no general answer to your question, and the question itself is not mathematically interesting. It merely highlights the inadequacies of spoken languages for mathematics.

In mathematics 'punctuation' is provided by brackets (and other conventions such as BODMAS) rather than by intonation. Sometimes also the context is crucial in interpreting a mathematical statement. In mathematics it is essential to avoid all ambiguity. Mathematics is a written language rather than a spoken language. Verbal intonation has no role.


问:
Is the large Sigma notation (summation notation) used for coproducts?
Binary products in a category are denoted $A_1 \times A_2$, while arbitrary products are denoted $\prod_{i \in I} A_i$. Binary coproducts are (sometimes) denoted $A_1 + A_2$,* so I expected that arbitrary coproducts would (sometimes) be denoted $\sum_{i \in I} A_i$.**

As an explicit example (in $\textbf{Set}$), the binary disjoint union of sets is sometimes denoted by $A_1 + A_2$, and the arbitrary disjoint union of sets by $\sum_{i \in I} A_i$, but I wasn't able to find a reference where $\sum$ is used for coproducts in a general category (I checked Mac Lane's and Awodey's books on category theory, as well as doing a fairly comprehensive Google search).

So my question is: does anybody know of a reference where $\sum$ is used for coproducts? (I would be interested to know if anyone reading this uses this notation personally, even if they do not know of any reference.)

Thanks

*although the notations $A_1 \sqcup A_2$ and $A_1 \coprod A_2$ are perhaps more common (and  according to wikipedia $A_1 \oplus A_2$ is also used).

**the notations $\bigsqcup_{i \in I} A_i$ and $\coprod_{i \in I} A_i$ are common (and  according to wikipedia $\bigoplus_{i \in I} A_i$ is also used).

答:
In the interest of providing an answer, I'll restate my comments.

Yes, $\sum A_i$ is used to denote a coproduct by many people (myself included). I've seen this used in plenty of talks, and there was never any confusion.

For an explicit instance of coproducts being introduced this way in print, see the section on "Sums" in Leinster's Basic Category Theory. In the linked edition, it's page 127.

I hope this helps ^_^


问:
Basis and dimension with finite vectors
I am trying to solve the following problems:

Considering the linear independet vectors $\vec T_1, ..., \vec T_s \in \mathbb{R}^s$ and the numbers $n_1, ... n_s \in \mathbb{R} - \{0\}$,

How can I check the vectors

$n_s \vec T_s, n_s \vec T_s + n_{s-1} \vec T_{s-1}, ...,  n_s \vec
   T_s + n_{s-1} \vec T_{s-1} + ... + n_1 \vec T_1$

compose a $\mathbb{R}^s$ basis?

Considering $1 \le p \le s$ and the subspace

$\begin{equation} W = \langle n_s    \vec T_s, n_s \vec T_s + n_{s-1}
   \vec T_{s-1}, ...,  n_s \vec       T_s + n_{s-1} \vec T_{s-1}    +
   ... +    n_p \vec T_p \rangle    \end{equation}$,

how can I calculate $W$ dimension?

答:
The first one is easy. Just check
$$a_sn_s \vec T_s+a_{s-1}( n_s \vec T_s + n_{s-1} \vec T_{s-1})+ ...+ a_1(  n_s \vec
   T_s + n_{s-1} \vec T_{s-1} + ... + n_1 \vec T_1)=0$$
Rearrange it and obtain
$$(a_s+a_{s-1}+\cdots+a_1)n_s\vec T_s+\cdots+(a_2+a_1)n_2\vec T_2+a_1n_1\vec T_1=0$$
Since $(\vec T_s, \cdots, \vec T_1)$ is linearly independent, $(a_s+a_{s-1}+\cdots+a_1)n_s=0, \cdots, (a_2+a_1)n_2=0$, and $a_1n_1=0$. Now since $n_i\ne 0$, all $a_i=0$. Hence, $(n_s \vec T_s, n_s \vec T_s + n_{s-1} \vec T_{s-1}, ...,  n_s \vec
   T_s + n_{s-1} \vec T_{s-1} + ... + n_1 \vec T_1)$ is linearly independent. Since the length of the list is $s$, it is a basis of $\mathbb{R}^s$.

I am learning LA right now. Haven't got that far. Don't know the meaning of the second part.

Title: Basis and dimension with finite vectors


问:
monomorphism from $\mathrm{GL}(n,q^2)$ to $\mathrm{GL}(2n,q)$
Construct an explicit monomorphism from $GL(n,q^2)$ to $GL(2n,q)$. Will $GL(n,q^m)$ always have a subgroup isomorphic with $GL(mn,q^2)$ for every $m,n\in \mathbb{N}$, and prime power $q$?</p>

答:
$\mathbb{F}_{q^m}$ is an $m/n$ dimensional vector space over $\mathbb{F}_{q^n}$ for $n|m$.  multiplication by an element $\alpha\in\mathbb{F}_{q^m}$ gives an $\mathbb{F}_{q^n}$-linear map $\alpha:\mathbb{F}_{q^m}\to\mathbb{F}_{q^m}$, which can be represented by an $m/n\times m/n$ matrix (given a choice of basis for $\mathbb{F}_{q^m}$ over $\mathbb{F}_{q^n}$).  so in terms of matrices, replace each entry $a_{ij}$ in a matrix $A\in{\rm GL}_k(\mathbb{F}_{q^m})$ with an $m/n\times m/n$ matrix from ${\rm GL}_{m/n}(\mathbb{F}_{q^n})$ representing multiplication by $a_{ij}$.  this gives an injection from ${\rm GL}_k(\mathbb{F}_{q^m})$ to ${\rm GL}_{km/n}(\mathbb{F}_{q^n})$.  (replace general linear groups with matrix groups etc if youre nitpicky).

as per another answer consider the $2\times2$ complex matrix
$$
\left(
\begin{array}{cc}
1+i&2+3i\\
-i&3\\
\end{array}
\right)
$$
this becomes a $4\times4$ real matrix
$$
\left(
\begin{array}{cccc}
1&-1&2&-3\\
1&1&3&2\\
0&1&3&0\\
-1&0&0&3\\
\end{array}
\right)
$$
under the identification of multiplication by $a+bi$ wrt the basis $1,i$ of $\mathbb{C}$ over $\mathbb{R}$ with the matrix
$$
\left(
\begin{array}{cc}
a&-b\\
b&a\\
\end{array}
\right)
$$


问:
Trees and leaf: Graph Theory
Let T be a tree with at least two vertices and let $v \in V(T)$. If T-v is a tree, then v is a leaf.

Attempt: Let T be a tree and let v be a leaf of T. Then T-v is a tree.

Let $a,b \in V(T-v)$. We must show that there is an (a,b) path in T-v. We know that since T is connected, there is (a,b)-path P in T. I claim that O does not include the vertex v. Since v is neither the first nor the last vertex on this path, it has two distinct neighbors on the path, contradicting the fact that $d(v)=1.$ Therefore P is an (a,b)-path in t-v and T-v is connected and a tree.

I just proved the converse of my original question. But a little stuck proving original question.

答: Suppose $v$ is not a leaf, then we will show that $T-v$ is disconnected. Use induction; one can check that this is true if $|V(T)|=3$. Now suppose $|V(T)|=n$, then there exists a smaller subtree $T_1$ such that $|V(T_1)|<n$ and $T_1$ has $v$ as a non-leaf node. By induction hypothesis $T_1-v$ is disconnected and so is $T$. </p>

Title: Trees and leaf: Graph Theory


问:
Why is the curl of the gradient of the polar angle phi zero, counter to my intuition?
I was trying to test my intuition about curl and divergence by constructing vector functions whose curl and divergence I think I have intuitive expectations about. An idea I came up with was to think of a scalar function with some shape whose gradient I would intuitively expect to have the desired property. I thought, for example, the function $\phi=\arctan(y/x)$ can be geometrically thought of as a staircase of linearly increasing angle winding around the origin. Therefore I thought that $\vec{\nabla}\phi$ would be a vector field whirlpooling around the z-axis (i.e. in the x-y plane), and therefore would have a nonzero curl in the z direction at (0,0). However in this case $\vec{\nabla}\times\vec{\nabla}\phi = \vec{0}$ for all (x,y), contrary to my expectation. What is wrong about my intuition here?

答:
The curl of the gradient is zero away from the origin, but the singular contribution at the origin integrates with the help of Stokes' theorem to
$$\int\int_S \hat{z}\cdot(\nabla\times\nabla\phi)\,dxdy=\oint_{\delta S}\nabla\phi\cdot dl=2\pi.$$
Here $\hat{z}$ is a unit vector in the $z$-direction, $\phi$ is the polar angle in the $x$-$y$ plane, $S$ is any surface in the $x$-$y$ plane containing the origin, $\delta S$ is its perimeter and the final integral equals the amount the angle $\phi$ increases as we go once around the perimeter.

This integral is expressed by the identification
$$\nabla\times\nabla\phi=2\pi\delta(x)\delta(y)\hat{z}.$$


问:
Do there exist measurable transformations which are recurrent over spaces of infinite measure?
I am trying to find a measurable space $(X, \mathcal{B})$, with measure $\mu$ such that $\mu(X)=\infty$, and with a corresponding measure preserving transformation $T:X\rightarrow X$ such that $T$ is recurrent over $X$.  That is, $T$ is such that for all $B\in \mathcal{B}$, there exists infinitely many $n>0$ such that $T^{n}(B) \cap B \not= \emptyset$.

Does such a dynamical system exist?  If so, can you provide example or reference? If no such system exists, is there a particular reason (ie. it is not possible, or one has not been found)?

答: Your question is related to, what is called, infinite ergodic theory. There are many interesting examples. An excellent reference to learn about this is the book of Jon Aaronson "An introduction to infinite ergodic theory"</p>


问:
Baer's criterion for injective modules
In Dummit and Foote section 10.5, proposition 36 says
Let $R$ be a PID. $Q$ be a $R$ module. $Q$ is injective if for all $r\in R$, $rQ=Q.$

$\cfrac{\mathbb{Z}}{4\mathbb Z}$ is injective as $\cfrac{\mathbb{Z}}{4\mathbb Z}$ module but $\bar{2}\cfrac{\mathbb{Z}}{4\mathbb Z}\neq  \cfrac{\mathbb{Z}}{4\mathbb Z}$ .

Please tell me what I'm missing.

答:
As your counterexample shows, the proof of this result requires $R$ to be a principal ideal domain, not just a principal ideal ring. Dummit and Foote do not explicitly state this hypothesis where it is used, but it is required for the converse direction of the proof, which implicitly relies on the fact that $R\cong rR$ (as $R$-modules) for any $r\in R\setminus\{0\}$ if $R$ is a pid.

In fact, this latter fact holds in any integral domain, not just in pids. This follows from the "cancellation" rule of integral domains: if $R$ is an integral domain and $r\neq 0\in R$, and $sr=tr$ for some $s,t\in R$, then $s=t$. (To see this, note that $sr=tr\implies (s-t)r=0\implies s-t=0$, since $r$ is assumed nonzero). Because of this, the canonical map $\phi_r:R\rightarrow Rr$ given by $s\mapsto sr$ is injective. $\phi_r$ is also an $R$-module homomorphism (since $\phi_r(st)=str=s(tr)=s\phi_r(tr)$ for any $s,t\in R$), and is surjective by definition, and hence gives an isomorphism from $R$ to $Rr$.

Now, let's return to the case where $R$ is a pid. Dummit and Foote's argument for the implication that $Q\text{ injective}\implies rQ=Q$ for any $r\in R\setminus\{0\}$ runs as follows: if $Q$ is injective, and $q\in Q$, $r\in R\setminus\{0\}$ are arbitrary, then we may define a map $f:Rr\rightarrow Q$ by $f(r)=q$. Since $Q$ is injective, by Baer's criterion this may be extended to a map $F:R\rightarrow Q$, and then $F(1)$ will witness divisibility of $q$ by $r$. (Since $rF(1)=F(r1)=f(r)=q$.)

As you say, the problem is in defining the map $f:Rr\rightarrow Q$. Dummit and Foote are implicitly assuming that, for any $q\in Q$, we may define a map $f:Rr\rightarrow Q$ by extending $r\mapsto q$. This is true if $R$ is a pid. Indeed, in this case we may define a map $\tilde{F}: R\rightarrow Q$ by extending $\tilde{F}(1)=q$ (this is the universal property of free $R$-modules), and then compose with $\phi_r^{-1}$ to get a map $f:=\tilde{F}\circ\phi_r^{-1}:Rr\rightarrow Q$ such that $f(sr)=\tilde{F}(s)=sq$ for any $s\in R$. In particular, $f(r)=q$, as desired.

However, if $R$ is not a pid, then this map $\phi_r$ will not necessarily be invertible, so we can't form this composition and hence the result fails. Indeed, your counterexample shows this, for given (for instance) $q:=\bar{1}\in \mathbb{Z}/4\mathbb{Z}=:Q$, and $r:=\bar{2}\in\mathbb{Z}/4\mathbb{Z}=:R$, there is no way to define a map $f:\bar{2}\mathbb{Z}/4\mathbb{Z}\rightarrow\mathbb{Z}/4\mathbb{Z}$ such that $f(\bar{2})=\bar{1}$.

There is a way to salvage the converse direction of this proposition for more general principal ideal rings, which is with the notion of "divisibility". An $R$-module $Q$ is said to be divisible if $rQ=Q$ for any non-zero divisor $r\in R$. Note that, if $R$ is an integral domain, this coincides with the condition that $rQ=Q$ for any $r\in R\setminus \{0\}$. Now, if $r$ is a non-zero divisor then we do have that $R\cong Rr$, even if $R$ is not an integral domain. (Try going through the proof of this fact yourself; it's essentially identical to the proof above.) Then we can apply the technique above to show that, if $R$ is a principal ideal ring (not necessarily a domain) and $Q$ is an injective $R$-module, then $Q$ is a divisible $R$-module. (Again, try going through these details yourself!)

Title: Baer's criterion for injective modules


问:
How to calculate the period of this pendulum
I have a pendulum with the following properties:

The diameter of the sphere is 1 meter.

The sphere of the pendulum is made of pure titanium.

The pendulum is connected to the roof with a rigid string 12 meters long. It does not have mass. The union between the string and the sphere is rigid, so it cannot rotate.

The oscillation angle is 60 degrees from the vertical.

We suppose there's no air, so there's no friction.

With this data, I need to know the period of this pendulum. I have not worked with pendulums so I do not know how to start. Any hint/guide?

I know that with 1. and 2. we can obtain the mass of the pendulum, using the density of the Titanium, $d_{titanium}=4506\;kg/m^3$. The volume of the sphere is $V=\frac{4}{3}\pi$, so the mass is $m=d\cdot V=4506\cdot\frac{4}{3}\pi=6803\pi$.

答:
Let's assume that the gravitational acceleration $g$ is known. You did not specify where the string is connected to the sphere (center or a point on the surface). I will use the center. If it's at the surface, all you need to do is change the distance to the center by adding half the diameter. The moment of inertia of the sphere with respect to the center of the mass is $$I_{CM}=\frac 25 MR^2$$
With respect to the point on the roof around which it oscillates it is $$I=I_{CM}+ML^2=M\left(\frac25 R^2+L^2\right)$$
If you want, you can say that $R^2\ll L^2$. In that case the problem will not depend on the radius of the sphere.

Then using the zero of potential energy at $\theta=0$, where $\theta$ is measured from the vertical axis, you have $$MgL(1-\cos\theta)+\frac12I\dot\theta^2=MgL(1-\cos\theta)+\frac12M\left(\frac25 R^2+L^2\right)\dot\theta^2=MgL(1-\cos\theta_0)$$
$\theta_0$ is the initial angle of $60^\circ$.
Calculating $\dot\theta$ from here, you get
$$\dot\theta=\sqrt{\frac{2gL}{\frac25 R^2+L^2}(\cos\theta-\cos\theta_0)}$$
Using $\dot\theta=\frac{d\theta}{dt}$, you get $dt=\frac{d\theta}{\dot\theta}$. To get the period$$T=2\int_{-\theta_0}^{\theta_0}\frac{d\theta}{\dot\theta}=2\sqrt{\frac{L^2+\frac25R^2}{2Lg}}\int_{-\theta_0}^{\theta_0}\frac{d\theta}{\sqrt{\cos\theta-\cos\theta_0}}$$


问:
General Solution of $y''+4y=\frac{3}{\sin(2t)}$
I have taken a course in ODEs before, but have forgotten how to solve for certain particular solutions. I came across this differential equation recently and would like to solve for the general solution.

The equation is $y''+4y=\frac{3}{\sin(2t)}=3\csc(2t).$

I know how to solve for the homogenous solution as I should since I just take $r^2-4=0\implies r=\pm 2i$, so we have $y_h=c_1\cos(2t)+c_2\sin(2t)$. I am really unsure of hot to solve the particular part of this ODE. I know that if this $\sin(2t)$ was on the numerator rather than denominator, then we simply plug in something like $y=At\cos(2t)+Bt\sin(2t)$ and solve for $A$ and $B$, however I have no idea how to solve this particular part with a cosecant term and would greatly appreciate any help in solving this.

Also, does anyone has any good resources that I can use to review what tricks to use to solve for the particular solutions of certain uncommon ODEs? Thanks in advance.

答:
$$y''+4y=\frac{3}{\sin(2t)}$$
$$\sin (2t)y''\color{red}{+2\cos (2t)y'-2\cos(2t) y'}+4\sin (2t)y=3$$
$$(\sin (2t)y')'-2(\cos(2t)y)'=3$$
Integrate.
$$\sin (2t)y'-2\cos(2t)y=3t+C_1$$
$$\left ( \dfrac y {\sin (2t)}\right )'=\dfrac {3t+C_1}{\sin ^2(2t)}$$
$$ y(t)= {\sin (2t)} \ \int \dfrac {3t+C_1}{\sin ^2(2t)}\ dt$$

Note that:
$$ \int \dfrac {C_1}{\sin ^2(2t)}\ dt=-\dfrac {C_1}2\cot (2t) +C_2$$
For the second integral, perform an integration by part.


问:
on the commutator subgroup of a special group
Let $G'$ be the commutator subgroup of a group $G$ and $G^*=\langle g^{-1}\alpha(g)\mid g\in G, \alpha\in Aut(G)\rangle$.

We know that always $G'\leq G^*$.

It is clear that if $Inn(G)=Aut(G)$, then $G'=G^*$.

Also if $G$ is a non abelian simple group or perfect group, then  $G'=G^*=G$.

Now do there exist a group such that  $Inn(G)\neq Aut(G)$ and $G'=G^*\neq G$?

Thank you

答: Take $G = S_6$, the symmetric group of degree $6$. It is known that $G' = A_6$ is the subgroup consisting of all even permutations. It is also not difficult to see that when $a \in \mathrm{Aut}(G)$ and $g \in G$, permutations $g$ and $a(g)$ must have the same sign (because $\mathrm{Aut}(G)$ acts trivially on $G/G' \simeq \mathbb{Z} / 2 \mathbb{Z}$), which means that $g^{-1}a(g) \in G'$. So, $G^* = G' \neq G$. But it is known that $\mathrm{Inn}(G) \subsetneq \mathrm{Aut}(G)$, so we have the desired example.</p>


问:
Show that $A+A^{-1}\geq 2I$.
I was looking for questions about linear algebra and I found the following statement:

If $A$ is a real positive definite symmetric matrix, then $A+A^{-1}-2I$ is positive definite matrix. Can we assume that $A+A^{-1}-2I>0$ entrywise? Or $\geq 0$? What if $A$ is not symmetric?.

答:
The property of being positive definite/semi-definite does not change if we switch from a matrix $A$ to a similar matrix $PAP^{-1}$ with an invertible matrix $P$. In particular, we can diagonalise $A$ using an orthogonal matrix $P$, i.e. $D=P^{-1}AP$ for some orthogonal matrix $P$ and diagonal matrix $D$. All the entries on the diagonal of $D$ are positive as $A$ is a positive-definite matrix.

Now, if $A$ is similar to $D$, then $A+A^{-1}-2I$ is similar to $D+D^{-1}-2I$ because:

$$P^{-1}(A+A^{-1}-2I)P=D+D^{-1}-2I$$

However, $D+D^{-1}-2I$ is also diagonal, and if the entries on the diagonal of $D$ are $d_i, i=1,\ldots,n$, then the entries on the diagonal of $D+D^{-1}-2I$ are $d_i+\frac{1}{d_i}-2=\frac{1}{d_i}(d_i-1)^2\ge 0$. Thus, the matrix $D+D^{-1}-2I$ is positive semi-definite, and so the matrix $A+A^{-1}-2I$ is positive semi-definite.

Therefore, the matrix $A+A^{-1}-2I$ is always positive semi-definite. It may not be positive definite, as the example $A=I$ shows.

Title: Show that $A+A^{-1}\geq 2I$.


答:
Since $A$ is real and symmetric, it has a decomposition $A=P\Lambda P^\top$ where $\Lambda$ is the diagonal matrix of eigenvalues and $P$ is orthogonal. Hence

$A+A^{-1} -2I = P(\Lambda + \Lambda^{-1} -2)P^\top$

Using the fact that $\lambda+\frac{1}{\lambda} \geq 2$, we have $A+A^{-1} -2I$ is positive semi-definite.

Title: Show that $A+A^{-1}\geq 2I$.

Body:
I was looking for questions about linear algebra and I found the following statement:

If $A$ is a real positive definite symmetric matrix, then $A+A^{-1}-2I$ is positive definite matrix. Can we assume that $A+A^{-1}-2I>0$ entrywise? Or $\geq 0$? What if $A$ is not symmetric?.


问:
$P \mapsto f^{-1}(P) : \operatorname{Spec} B \to \operatorname{Spec} A$ is a continuous map between Zariski topologies.
Let $A, B$ be commutative rings with $1$.

Let $f: A \to B$ be a ring homomomorphism.  For $P \in \operatorname{Spec} B$, $f^{-1}(P) \in \operatorname{Spec} A$, so that $g : \operatorname{Spec} B \to \operatorname{Spec} A$ defined by $g(P) = f^{-1}(P)$ is a function.  The goal is to show that for any $V(I) \subset \operatorname{Spec} A$ a closed set, that $g^{-1}(V(I)) = V(f(I)B)$.  Since then $g^{-1}$ takes closed sets to closed sets and so is continuous.

So far I have:

$$
P \in g^{-1}(V(I)) \\ \iff g(P) = f^{-1}(P) \in V(I) \\ \iff ?
$$

I know that if $P \supset f(I)B$ then $P \supset f(I)$ since $P$ has to be a proper ideal.  Not sure where that plugs into the proof though.

答:
Thanks @Thorgott for teaching me.

The step I was missing was if $I \subset f^{-1}(\mathfrak{p})$ then $f(I) \subset ff^{-1}(\mathfrak{p}) \subset \mathfrak{p}$.

Thus, we have:

$$
g^{-1}(V(I)) = \{ P \in \operatorname{Spec} B: g(P) \in V(I) \} \\
= \{ P \in \operatorname{Spec} B: f^{-1}(P) \in V(I) \} \\
= \{ P \in \operatorname{Spec} B: I \subset f^{-1}(P) \} \\
= \{ P \in \operatorname{Spec} B: f(I) \subset ff^{-1}(P) \subset P \} \\
= V(f(I)B) = V(IB)
$$

The last equality is just by notational definition of $IB \equiv f(I)B$ as used in Matsumura's text.  That's the extension of an ideal along $f$.  And $A \cap J \equiv f^{-1}(J)$ is the contraction of $J$ along $f$.  All just notational.


问:
Closed form of $I = \int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$
I need help with this integral:

$$I = \int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$$

Where $κ,λ>0$.

Neither Mathematica nor Maple could find a closed form for this integral.

Let $G$ follow a $\Gamma\left(\kappa+1,\lambda\right)$ distribution, i.e. its density can be written as $f_G\left(x\right)=\mathbb{I}_{\mathbb{R}_+^\ast}\left(x\right)\ x^\kappa\ e^{-\ \frac{x}{\lambda}}\frac{1}{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}$.

For fixed values of κ and λ, I can use Monte-Carlo to simulate : $$I=\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}\int_{0}^{+\infty}{f_G\left(t\right)\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}=\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}\ \mathbb{E}\left(\sin^2{\left(\frac{G\pi}{2\kappa\lambda}\right)}\right)$$

But I would actually rather have a closed form. Any help or insight will be very much appreciated.

PS : although fluent in english, I mostly study math in french...

Edit : following another user's suggestion, I know have the following :

$$I=\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\sin^2{\left(\frac{\pi t}{2\kappa\lambda}\right)}dt}$$

$$I=\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\frac{1-\cos{\left(\frac{\pi t}{\kappa\lambda}\right)}}{2}dt}$$

$$I=\frac{1}{2}\left(\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}dt}-\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\cos{\left(\frac{\pi t}{\kappa\lambda}\right)}dt}\right)$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}}\left(e^{i\frac{\pi t}{\kappa\lambda}}+e^{-i\frac{\pi t}{\kappa\lambda}}\right)dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}+i\frac{\pi t}{\kappa\lambda}}\ dt}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{t}{\lambda}\ -\ i\frac{\pi t}{\kappa\lambda}}\ dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{\left(\kappa-i\pi\right)t}{\kappa\lambda}\ }dt}-\frac{1}{4}\int_{0}^{+\infty}{t^\kappa e^{-\ \frac{\left(\kappa+i\pi\right)t}{\kappa\lambda}\ }dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\int_{0}^{+\infty}{\left(\frac{\kappa\lambda}{\kappa-i\pi}u\right)^\kappa e^{-u\ }\frac{\kappa\lambda}{\kappa-i\pi}du}-\frac{1}{4}\int_{0}^{+\infty}{\left(\frac{\kappa\lambda}{\kappa+i\pi}u\right)^\kappa e^{-u\ }\frac{\kappa\lambda}{\kappa-i\pi}du}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa-i\pi}\right)^{\kappa+1}\int_{0}^{+\infty}{u^\kappa e^{-u\ }du}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa+i\pi}\right)^{\kappa+1}\int_{0}^{+\infty}{u^\kappa e^{-u\ }dt}$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa+i\pi}\right)^{1+\kappa}\Gamma\left(1+\kappa\right)-\frac{1}{4}\left(\frac{\kappa\lambda}{\kappa-i\pi}\right)^{1+\kappa}\Gamma\left(1+\kappa\right)$$

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{4}\left(2-\left(\frac{\kappa}{\kappa+i\pi}\right)^{1+\kappa}-\left(\frac{\kappa}{\kappa-i\pi}\right)^{1+\kappa}\right)$$

Not sure what to do about the complex numbers I get in the end though...

答:
I can answer myself following the helpful answers from this post : How to simplify $\left(x+i\pi\right)^{1+x}+\left(x-i\pi\right)^{1+x}$ for $x>0$.

$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{4}\left(2-\left(\frac{\kappa}{\kappa-i\pi}\right)^{1+\kappa}-\left(\frac{\kappa}{\kappa+i\pi}\right)^{1+\kappa}\right)$$
$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{4}\left(2-\kappa^{1+\kappa}\left(\frac{1}{\left(\kappa-i\pi\right)^{1+\kappa}}+\frac{1}{\left(\kappa+i\pi\right)^{1+\kappa}}\right)\right)$$
$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{4}\left(2-\left(\frac{\kappa}{\kappa^2+\pi^2}\right)^{\kappa+1}\left(\left(\kappa+i\pi\right)^{1+\kappa}+\left(\kappa-i\pi\right)^{1+\kappa}\right)\right)$$
$$I=\frac{\Gamma\left(\kappa+1\right)\ \lambda^{\kappa+1}}{2}\left(1-\left(\frac{\kappa}{\sqrt{\kappa^2+\pi^2}} \right)^{\kappa+1}\cos{\left(\left(1+\kappa\right)\arctan{\frac{\pi}{\kappa}}\right)}\right)$$


问:
Is this a valid proof for showing two continuous functions that coincide on the rationals, are in fact the same function?
I am trying to prove the following, but I am unsure about a particular step in my proof. I am aware of other proofs for this statement, but I am trying to prove it in a specific way.

Statement:

" Suppose $f,g:\mathbb{R}\to\mathbb{R}$ are two continuous functions such that $f(x)=g(x)$ for all $x\in{}\mathbb{Q}$. Then $f=g$ everywhere."

My proof:

Since $f$ and $g$ are continuous, for each $\epsilon>0$ choose respective $\delta_1,\delta_2>0$ that satisfy the continuity definition. Now define the following: $$\delta=\min(\{\delta_1,\delta_2\}).$$

Let $c\in\mathbb{R}\setminus\mathbb{Q}$ with $|x-c|<{}\delta{}.$ Consider the following:

$$|f(c)-g(c)|$$
$$=|f(c)-f(x)+f(x)-g(x)+g(x)-g(c)|$$
$$\le|f(c)-f(x)|+|f(x)-g(x)|+|g(x)-g(c)|$$
$$<|f(x)-g(x)|+2\epsilon{}$$

This holds for all $x\in(c-\delta,c+\delta),$ so must hold for any $x\in(c-\delta,c+\delta)\cap\mathbb{Q}\neq\emptyset,$ which would lower the upper bound to simply $2\epsilon$ (since $|f(x)-g(x)|=0$ on $(c-\delta,c+\delta)\cap\mathbb{Q}$). Since that is true for every $\epsilon>0$ we have that $|f(c)-g(c)|$ is forced to be $0$, hence $f(c)=g(c)$ for all irrational $c$. Hence $f=g$ everywhere.

my issue is with making the $|f(x)-g(x)|=0$ where $x$ is rational and in the aforementioned open interval. Since it's true for all $x$ in that range, surely we can lower the upper bound?

答:
That is not correct. Note that the $(c-\delta,c+\delta)$ interval is not a fixed interval, since $\delta$ depends upon $c$. So, when $\varepsilon$ gets smaller, then so does $\delta$ and therefore saying that $f=g$ on the interval $(c-\delta,c+\delta)$, as if it was a fixed interval, makes no sense.

If $h=f-g$, then $h$ is continuous and it is $0$ on the rationals. If $h(x_0)\ne0$ for some $x_0$, then there is a $\delta>0$ such that if $x\in(x_0-\delta,x_0+\delta)$, then $\bigl|f(x)-f(x_0)\bigr|<\bigl|f(x_0)\bigr|$, which implies that $f(x)\ne0$. Can you take it from here?


问:
Find the limit of $2+\left(-\frac{2}{e}\right)^n$, as $n\to\infty$, if it exists
I'm absolutely unsure about how to approach this. I've considered changing it to $-2=\left(-\frac{2}{e}\right)^n$ and then using the properties of lograrithms, but $\ln(-2)$ is undefined, as is $\ln(-\frac{2}{e})$.

I can't use L'Hopital's rule either because the limit is of the form $1^{\infty}$, and I have no clue how to manipulate it so I can apply it. I'm tempted to say that it approaches infinity, but I can't be confident that there isn't another way to solve it.

Any tips? Thanks.

Body:
$e>2$ so $\frac{2}{e}<1$ so

$$\lim_{n \to \infty}{\left(-\frac{2}{e}\right)^n}=0$$ so

$$\lim_{n \to \infty}{2+\left(-\frac{2}{e}\right)^n}=2$$


问:
Given an equation, I'm having trouble finding a solution to another equation. Help proving whether or not it can be solved? (Differential Equations)
This is a problem I've come up with, not a homework problem. Ideally I'd like to find a solution, but if I can prove that it can't be solved, that's okay too. The problem is this:

Given $n'' = hn$, find $\frac{n'''n'}{(n'')^2}$ in terms of $h$ and $z$ if necessary. This is using the standard notation of each $'$ being a derivative. h and n are both functions of the variable z, that is h -> h(z) and n -> n(z). All derivatives are with respect to z.

I've done so many different things to solve this and just can't figure it out. If you have any ideas on how to approach this, I'd be really grateful.

Edit: For context, I'm very familiar with the rules of calculus. I haven't posted any attempts because I've made so many and they all just become crazy and insane. I've been working on this problem for ~ 2.5 years, and only just thought now I'd ask for help because I'm beginning to think that it isn't possible.

答:
hint

You have

$$n''(z)=h(z)n(z)$$

this gives after a differentiation with respect to $ z $

$$n'''(z)=h'(z)n(z)+h(z)n'(z)$$
or
$$n'''=h'n+hn'$$


问:
How to prove recursive sequence converges? - $a_{n+1}=1/(a_n+1)$
I have a recursive sequence defined by $$a_{n+1}=\frac{1}{a_n+1}$$, where $a_0=7$.

The sequence is bounded, but it's not monotonic. So how does one show, that the sequence converges?
I already found it's limit under the assumption, that it is converging. Which is $a=\frac{\sqrt5-1}{2}$.

Note, that I'm not allowed to use epsilon proofs or proof by induction.

答: By the recurrence, $$(a_n+1)(a_{n-1}+1)=(\frac{1}{a_{n-1}+1}+1)(a_{n-1}+1)\\=2+a_{n-1}>2$$ since with $a_0=7$ all the terms are obviously positive. Therefore, $$|a_{n+1}-a_n|=\frac{|a_n-a_{n-1}|}{(a_n+1)(a_{n-1}+1)}<\frac12\,|a_n-a_{n-1}|.$$ The rest is a special case of the Banach fixed point theorem: when the distances between consecutive points decrease by a factor less than $1$ the sequence converges (in a complete metric space). As applied to this case, we have by induction: $$|a_{n+1}-a_n|<\frac1{2^n}\,|a_1-a_0|,$$ so the differences decrease at least geometrically. Hence $$|a_{n+m}-a_n|\\\leq (|a_{n+m}-a_{n+m-1}|+\dots+|a_{n+1}-a_n|)\,|a_1-a_0|\\<\left(\frac1{2^{n+m-1}}+\dots+\frac1{2^n}\right)|a_1-a_0|\to0$$ when $n,m\to\infty$ because the geometric series converges. Therefore, $a_n$ is a Cauchy sequence and must converge.


问:
Prove $\lim\limits_{n \to \infty} na_n=0.$

Suppose $\sum_{n=1}^{\infty} a_n$ be a convergent positive series, and
satisfy $\lim\limits_{n \to \infty}\frac{a_{n+1}}{a_n}=1$. Prove
$\lim\limits_{n \to \infty} na_n=0.$

First, we may consider applying Cauchy's condensation test. Since $\sum_{n=1}^{\infty}a_n$ is convergent, $\sum_{n=1}^{\infty} 2^n a_{2^n}$ is convergent as well, which implies $\lim\limits_{n \to \infty}2^n a_{2^n}=0$. Hence, there already exists a subsequence of $\{na_n\}$ convergent to zero. But how to go on ?

答:
This result is wrong.

Counter Example

For all $k \ge 0; 0 \le m \le 2^{k}-1$, define $a_{2^k+m}$ as follows:

$$ a_{2^k+m}= 2^{-k}.\max \left( 2^{-m\frac{k^2}{2^k} } ,2^{-1-(2^k-m)\frac{k^2}{2^k} }\right) $$
This defintion is nothing extraordinary if you look at the line graph of the log sequence $\left( \log(a_n), n \ge 1\right)$ . It is just a piecewise linear curves whose maximas are attained at points of form $2^k$.

Also, the slope of that graph becomes more flatter and eventually becomes a horizontal lines.( Because $\frac{k^2}{2^k} \rightarrow 0$ ).

And we can even check it by some simple algebraic arguments that:
$$ \lim \frac{a_n}{a_{n+1}}= 1$$
While we can check that :
$$ \sum_{ m \ge 0}^{2^k-1} a_{2^k+m} \le \left(2^{-k}+2^{-k-1} \right)\left(
 \frac{ 1-2^{-k^2}}{1-2^{-k^2/2^k}} \right)$$
( replacing the max by the sum of each components)

$$ LHS \le 2.2^{-k}. \underbrace{ \frac{4.2^k}{ k^2 } }_{1-2^{-k^2/2^k} \ge \frac{k^2}{4.2^k}}.(1-2^{-k^2}) \le \frac{8}{k^2} $$
Thus $\sum_{n} a_n < \infty$

However , by definition
$$\limsup na_n=1$$

Discussion:

There must be some kind of additional condition in order for that to be true.

As shown in many previous topics, the condition of monotonicity is a pertinent choice. To me, that's conditional is somehow pretty tight...

As I have shown in my counterexample above: a condition on the convergence rate of $ \limsup \frac{a_{n+1}}{a_n} =1$ is  also crucial.


问:
What is the difference between max/min and sup/inf.
Say there exists a series {A}, which has an upper bound, meaning it has a supremum, call is "a" ( sup(A) = a ).
My question is, is {A}≤a or {A}<a?
My confusion arises from the fact that we know that if "a" is included in {A} then it is also its maximum, meaning {A}≤a, does that still stand for the supremum?
I know that if there is a x∈A, st x≤a than a is the maximum of {A}, so consequently if sup(A) would have the same definition, meaning {A}≤a, wouldn't it be a maximum too?
I would really appreciate some help to clear up this confusion, thank you.</p>

答:
If a set $A$ has an upper bound, then $A$ has a least upper bound (in $\mathbb{R}$). Then for every $a \in A$, the supremum $\sup A$ is an upper bound of $A$ so $a \leq \sup A$. So, to abuse notation a bit, $A \leq \sup A$.

The supremum is meant to generalize the idea of a maximum. You can show that if the set contains an upper bound, then supremum and maximum coincide. In particular, on finite sets the maximum and the supremum are precisely the same.


答:
the first few values are $$  7, \frac{1}{8},  \frac{8}{9}, \frac{9}{17} $$
and the last two are both below $1$ and a bit larger than $1/2.$  Then the Mean Value Theorem says, defining $L = \frac{\sqrt 5 - 1}{2},$  that
$$ \frac{f(x) - L}{x-L}  = f'(\xi)  $$
with some $\xi$ between $x$ and $L.$  The derivative is $-1/(1+x)^2 $ and has absolute value below $4/9$ once $ \xi > 1/2.$

Once $1/2 < x < 1 $ we always have $1/2 < x < 1$
So
so $$ \left|\frac{f(x) - L}{x-L} \right| <  \frac{4}{9}  $$
Repeating $f$ says the distance for $L$ is below a power of $4/9$ times the beginning distance.

Title: How to prove recursive sequence converges? - $a_{n+1}=1/(a_n+1)$

Body:
I have a recursive sequence defined by $$a_{n+1}=\frac{1}{a_n+1}$$, where $a_0=7$.

The sequence is bounded, but it's not monotonic. So how does one show, that the sequence converges?
I already found it's limit under the assumption, that it is converging. Which is $a=\frac{\sqrt5-1}{2}$.

Note, that I'm not allowed to use epsilon proofs or proof by induction.


答:
One way to solve this problem is to use Variation of Parameters

This is an algorithmic approach. Please see the examples (specifically Example $1$)

Here $y_1 = \cos 2t$ and $y_2 = \sin 2t$

Check that the Wronskian is $2$

The PI is given by

$$- \cos 2t \int \frac{\sin 2t}{2} \frac{3}{\sin 2t} dt + \sin 2t \int \frac{\cos 2t}{2} \frac{3}{\sin 2t} dt$$

This simplifies to $$-\frac{3t \cos 2t}{2} + \frac{3}{4} \sin 2t \ln|\sin 2t|$$
$$$$

Title: General Solution of $y''+4y=\frac{3}{\sin(2t)}$

Body:
I have taken a course in ODEs before, but have forgotten how to solve for certain particular solutions. I came across this differential equation recently and would like to solve for the general solution.

The equation is $y''+4y=\frac{3}{\sin(2t)}=3\csc(2t).$

I know how to solve for the homogenous solution as I should since I just take $r^2-4=0\implies r=\pm 2i$, so we have $y_h=c_1\cos(2t)+c_2\sin(2t)$. I am really unsure of hot to solve the particular part of this ODE. I know that if this $\sin(2t)$ was on the numerator rather than denominator, then we simply plug in something like $y=At\cos(2t)+Bt\sin(2t)$ and solve for $A$ and $B$, however I have no idea how to solve this particular part with a cosecant term and would greatly appreciate any help in solving this.

Also, does anyone has any good resources that I can use to review what tricks to use to solve for the particular solutions of certain uncommon ODEs? Thanks in advance.


问:
is possible to calculate Net Present Value with different discount rates per year?
The formular for NPV is: Net Present value formula
$$\sum_{t=1}^n= \frac{Rt}{(1+i)^t}$$

where:
Rt =Net cash inflow-outflows during a single period t

i=Discount rate or return that could be earned in alternative investments

t=Number of timer periods

Which means taking projected cash flow for each year and dividing it by (1 + discount rate). For example this is the formula for a 5 year project and 3% discount rate:
$$\sum_{t=1}^n= \frac{Rt}{(1+0.03)^5}$$

This formula has one serious problem: considers that the discount rate does not change throughout the project. Which means for example that the risk free rate dont change in 5 years ( premise totally out of reality ). My question is very simple: is possible to consider differents rates per year for Net Present Value calculation? for example for the first year 3% for the next year 3.4% etc. If i calculate the npv for the first year with 3% of discount rate and sum by the npv for the second year with 3.4% discount rate??

To solve this doubt, I created a cashflow, separate it into two cash flows, calculated the NPV for each separately and added the same discount rate.

```<code># R CODE
library(FinCal)
total_cashflow=c(10,10,10,20,60)
split1_cashflow=c(10,10)
split2_cashflow=c(10,20,60)

#Net present Value for the complete cashflow 3% discout rate
npv(0.03,total_cashflow)
> 100.7468
#Sum of Net present Value for the two split cashflows 3% discout rate
npv(0.03,split1_cashflow)+npv(0.03,split2_cashflow)
> 105.682
</code>```

Why are the results different?

答:
If we assume that the first payment is made immediately then your first result is right. It starts at $t=0$ and ends at $t=4$

$$PV_1=\frac{10}{1.03^0}+\frac{10}{1.03^1}+\frac{10}{1.03^2}+\frac{20}{1.03^3}+\frac{60}{1.03^4}=100.7468 \ \ \color{\limegreen}{\checkmark}$$

The split approach begins for the first two payments at $t=0$ and ends at $t=1$. For the next three payments it begins once again at $t=0$ and ends at $t=2$.

$$PV_2=\underbrace{\frac{10}{1.03^0}+\frac{10}{1.03^1}}_{\textrm{split 1}}+\underbrace{\frac{10}{1.03^0}+\frac{20}{1.03^1}+\frac{60}{1.03^2}}_{\textrm{split 2}}=105.682 \ \ \color{red}{\times}$$

This is not the intention of the first approach.


问:
Do we have $u^2 \leq u$ when $\|u \| \leq 1$ in a $C^*$-algebra?
Let $u$ be a positive element in a $C^*$-algebra $A$ with $\|u \|\le 1$. Then is it true that $u^2 \leq u$?

Attempt: This is true in the commutative $C^*$-subalgebra $C^*(u)$, since by the Gelfand-Naimark theorem, $C^*(u) \cong C_0(X)$ where $X$ is some locally compact Hausdorff space. Thus, this also holds in $A$ since positivity does not depend on the ambenient $C^*$-subalgebra.

Is this correct?

答: Without loss of generality, assume $A$ is unital.  Then $C^*(u,1)\cong C(\sigma(u)$, the continuous functions on the spectrum of $u$, and this isomorphism takes $u$ to the inclusion function, $\sigma(u)\ni\lambda\mapsto\lambda\in\mathbb C$. The conditions that $u$ is positive and $\|u\|\leq1$ then imply that $0\leq \lambda\leq 1$ for all $\lambda\in\sigma$, and for such $\lambda$, $\lambda^2\leq\lambda$.  Using the above isomorphism, this yields that $u^2\leq u$.


问:
How to calculate conditional expectation $\mathbb{E}[g(x)\ | \ x\geq Q]$ for $x\sim \exp(1)$
Given that $X$ is exponentially distributed continuous random variable $X\sim \exp(1)$ and $g(x)$ is as below. How can I find the Expectectaion of $g(x)$ for the condition that $x\geq Q$, i.e. $\mathbb{E}[g(x)\ | \   x\geq Q]$.

$$g(x) = \frac{A}{\exp(-bQ+c)}\Big(\frac{1 + \exp(-bQ+c)}{1 + \exp(-dx+c)}-1\Big)$$

I suppose I should start by considering an event $\Phi$ such that $\Phi = \mathbb{P}[x \geq Q]$. However, I am stuck about how to go around this condition.

All constants are positive real values.

EDIT: Is the following solution correct for the above question? If it is OK, then I have found the solution. But I will really appreciate if someone can let me know if the following method is correct.

$$\mathbb{E}[g[x]] = \int_Q^\infty g(x). f(x) dx$$

$$\mathbb{E}[g[x]]= \int_Q^\infty\frac{A}{\exp(-bQ+c)}\Big(\frac{1 + \exp(-bQ+c)}{1 + \exp(-dx+c)}-1\Big) e^{-x} dx$$

答:
I think the solution you gave is off by a normalization constant.

To see this, consider $g(x) = 1$. Then your formula gives $\mathbb{E}[g(x)\ |\ x\geq Q] = \int_Q^\infty f(x) dx \leq 1$, while the correct answer should be identically one: the expectation of a constant is always just that constant.

The correct expression can be found by augmenting the "problem space" $X$ to $(X,U)$, where $X \in \mathbb{R}^+$ is the original random variable of your question and $U \in \{0,1\}$ is a discrete binary variable which is $1$ if $X \geq Q$ and $0$ otherwise. Then we can write the problem in a more manageable form:
$$
\begin{align}
\mathbb{E}[g(x)\ |\ x\geq Q] &\equiv \int dx \ g(x) \ p(x \ | \ x\geq Q) \\
&= \int dx \ g(x) \ p(X = x \ | \ U = 1).
\end{align} \tag{1} \label{1}
$$
Now we can use the product and sum rules:
$$
p(X = x \ | \ U = 1) = \frac{p(U = 1 \ | \ X = x) \ p(X = x)}{\int dy \ p(U = 1 \ | \ X = y) \ p(X = y)}. \tag{2} \label{2}
$$
With the conditional pmf
$$p(U = 1 \ | \ X = x) = \begin{cases}
0 & x < Q \\
1 & x \geq Q
\end{cases} \tag{3} \label{3}
$$
and setting $p(X = x) = f(x)$, we get after substituting \eqref{3} in \eqref{2} and then \eqref{2} into \eqref{1}:
$$
\mathbb{E}[g(x)\ |\ x\geq Q] = \frac{\int_Q^\infty dx \ g(x) \ f(x)}{\int_Q^\infty dx \ f(x)}.
$$


问:
Can the Stolz-Cesàro theorem be used to prove L'Hospital's rule, or viceversa?
Stolz-Cesàro theorem is usually stated to be the discrete version of L'Hospital's rule. I was merely wondering whether one of these theorems could be used to prove the other (I couldn't find any proof that does this online).</p>

答:
I found a proof of L 'Hopital's rule that makes use of Stolz-Cesaro for the case $g(x)\rightarrow \infty$.

I've noticed recently that in the hypothesis of L'Hopital's rule it is often assumed that $f$ and $g$ need be defined and differentiable at the point $p$ which is being approached in the limit (this is done in the proof I mentioned and in baby Rudin). Yet the theorem still holds even when $f$ and $g$ are not defined nor differentiable at $p$, so I will rewrite the proof with care so as to prove the theorem in the more general case.

Theorem: Let $f$ and $g$ be real functions differentiable in the open interval $I$ except possibly at the point $p\in I$. Assume that $g'(x)\neq 0$ for every $x\in I-\{p\}$ and that $$\lim_{x\rightarrow p}\frac{f'(x)}{g'(x)}$$ exists and equals $L$. If either

i) $\lim_{x\rightarrow p}g(x)=\pm \infty$, or

ii) $\lim{x\rightarrow p}f(x)=g(x)=0$,

then $\lim_{x\rightarrow p}f(x)/g(x)$ exists and equals $L$.

Proof:

i) Suppose $\lim_{x\rightarrow p}g(x)=\infty$. Consider all points $x\in I$ less than $p$, since $g'(x)\neq 0$ we have that $g'$ is either positive or negative, yet $\lim_{x\rightarrow p}g(x)=\infty$ implies it can only be positive. We may also assume $g(x)\neq 0$ by making $I$ smaller if necessary.

It suffices to show that for any monotone sequence $x_n\rightarrow p$  we get $f(x_n)/g(x_n)\rightarrow L$, so let $x_n$ be such a sequence. Regardless of whether it is approaching $p$ from the right or the left, we have that $g(x_{n+1})>g(x_n)$. Stolz-Cesaro gives

$$\liminf \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}\leq \liminf \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f(x_n)}{g(x_n)}\leq \limsup \frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}$$

By Cauchy's Mean Value Theorem there exists $y_n$ between $x_n$ and $x_{n+1}$ such that $$\frac{f(x_{n+1})-f(x_n)}{g(x_{n+1})-g(x_n)}=\frac{f'(y_n)}{g'(y_n)}$$

noting $\lim_{n\rightarrow \infty}f'(y_n)/g'(y_n)=L$ gives the desired result.

ii) Let $\epsilon >0$. Then there exists a neighborhood $U=(u,p)$ of $p$ such that $x\in U$ implies $$\Big| \frac{f'(x)}{g'(x)} -L\Big| <\epsilon$$

Let $y_n\rightarrow p$ be a monotone sequence in $U$. By Cauchy's Mean Value Theorem there is a sequence $z_n\in (x,y_n)$, so that $z_n\rightarrow p$, such that $$\frac{f(x)-f(y_n)}{g(x)-g(y_n)}=\frac{f'(z_n)}{g'(z_n)}\in (L-\epsilon,L+\epsilon)$$ therefore $\lim_{n\rightarrow \infty}\frac{f(x)-f(y_n)}{g(x)-g(y_n)}=\frac{f(x)}{g(x)}\in [L-\epsilon,L+\epsilon]$ for any $x\in U$, so $\lim_{x\rightarrow p^-}f(x)/g(x)=L$. The other direction is similar.


问:
Can any one prove this claim?
Let $S^n\subset \mathbb{R}^{n+1}$ be the unit sphere. For any given point
$$x=(x_1, \cdots, x_i, \cdots, x_j, \cdots, x_{n+1}) \in S^n,$$
exchange any two coordinate $x_i$ and $x_j$  of $x$ to get a new point
$$x'= (x_1, \cdots, x_j, \cdots, x_i, \cdots, x_{n+1})$$
(may or may not be the same as $x$).
For a given $x\in S^n$, we denote the set of points (including $x$ itself) obtained by this way as $\Gamma_x$.

My claim is: if $x\in S^n$ satisfies $\sum_1^{n+1} x_i = 0$, $\Gamma_x$ can linearly span to the hyperplane $\sum_1^{n+1} x_i =0$.

How to prove this claim?

答:
The condition that $x$ lies on the sphere can be replaced with the condition that $x$ is nonzero. For if $x$ is nonzero, one can divide $x$ by its length. The condition that $x$ is nonzero and that the $x_i$ sum to $0$ implies that the $x_i$ take at least two distinct values.

For $i\ne j$ define $y_{ij}$ to be the vector with $x_i$ and $x_j$ swapped. Denote the standard basis vector for $1\le i\le n+1$ by $e_i$ and write
$$
z_{ij}=y_{ij}-x=(x_j-x_i)e_i+(x_i-x_j)e_j.
$$
If $x_i$ is distinct from all the $x_j$, $j\ne i$, then the set of $n$ vectors $z_{ij}$, $j\ne i$, spans the hyperplane $\sum_{i=1}^{n+1}x_i=0$.

If an $x_i$ distinct from $x_j$ for all $j\ne i$ cannot be found, then let $I$ be the set of indices $i$ for which $x_i=x_1$ and let $J$ be the set of indices $j$ for which $x_j\ne x_1$. Let $j_1$ be an element of $J$. Then the set of $n$ vectors
$$
\{z_{1j}\mid j\in J\}\cup\{z_{ij_1}\mid i\in I\setminus\{1\}\}
$$
spans the hyperplane $\sum_{i=1}^{n+1}x_i=0$.

For example, if $x=\begin{bmatrix}1 & 1 & 1 & -1 & -1 & -1\end{bmatrix}$, a set of spanning vectors is $\{z_{14},z_{15},z_{16},z_{24},z_{34}\}$ with
\begin{align}
z_{14}&=\begin{bmatrix}-2 & 0 & 0 & 2 & 0 & 0\end{bmatrix}\\
z_{15}&=\begin{bmatrix}-2 & 0 & 0 & 0 & 2 & 0\end{bmatrix}\\
z_{16}&=\begin{bmatrix}-2 & 0 & 0 & 0 & 0 & 2\end{bmatrix}\\
z_{24}&=\begin{bmatrix}0 & -2 & 0 & 2 & 0 & 0\end{bmatrix}\\
z_{34}&=\begin{bmatrix}0 & 0 & -2 & 2 & 0 & 0\end{bmatrix}.
\end{align}

Title: Can any one prove this claim?


问:
What is the difference between Emil Artin's Geometric Algebra and David Hestenes Geometric Algebra?
Emil Artin wrote a book on Geometric Algebra. Is this the same as the geometric algebra of David Hestenes?Please correct me if I'm wrong on this, I think David Hestenes geometric algebra is just clifford algebra. Did David change anything from Clifford' original geometric algebra or is he just trying to make it more well known; it seems to be very useful. Anyway, I am just curious if the textbook Geometric Algebra by Artin is also just clifford algebra or geometric algebra? I don't really know the difference between the two. Thank you.</p>

答:
No, Artin's Geometric Algebra is not about what David Hesetenes is talking about.

Artin's book is a much more general course on the connection between algebra and geometry via groups.  It does have a section on Clifford algebras, of which "Hestenes geometric algebras" are a special case.

So far, I've gotten a lot more use out of my copy of Artin's Geometric Algebra than out of any of the "Hestenes geometric algebra" material I've read.

I think Hestenes' principal effect has been to raise the profile of using Clifford algebras practically. I don't think I've ever seen anything about truly new results coming out of the program, but indeed since he raised the profile I think people have been taking a closer look at the Hestenes school perspective.

At one of the spectrum, I have seen some things to the effect that phrasing things with geometric algebra can make some things clearer or more memorable.

On the other end of the spectrum, I have seen people say that geometric algebra and the associated geometric calculus is just "for people uncomfortable with basic manifold theory." The whole situation reminds me a little bit of the quaternion/vector-algebra war.

Just remember not to get caught up in hype, because there is a bit of that. Personally, I someday plan to sit down and weigh these things on their virtues, but for now I'm still brushing up my manifold theory :)


问:
General Solution of $y''+4y=\frac{3}{\sin(2t)}$
I have taken a course in ODEs before, but have forgotten how to solve for certain particular solutions. I came across this differential equation recently and would like to solve for the general solution.

The equation is $y''+4y=\frac{3}{\sin(2t)}=3\csc(2t).$

I know how to solve for the homogenous solution as I should since I just take $r^2-4=0\implies r=\pm 2i$, so we have $y_h=c_1\cos(2t)+c_2\sin(2t)$. I am really unsure of hot to solve the particular part of this ODE. I know that if this $\sin(2t)$ was on the numerator rather than denominator, then we simply plug in something like $y=At\cos(2t)+Bt\sin(2t)$ and solve for $A$ and $B$, however I have no idea how to solve this particular part with a cosecant term and would greatly appreciate any help in solving this.

Also, does anyone has any good resources that I can use to review what tricks to use to solve for the particular solutions of certain uncommon ODEs? Thanks in advance.

答:
The equation can be solved through variation of parameters. As it seems that OP is not familiar to (or has forgotten) this technique, let me demonstrate the idea. First, rewrite the ODE as

$$ \frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix} y \\ y' \end{pmatrix}
= \begin{pmatrix} y' \\ y'' \end{pmatrix}
= \underbrace{ \begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} }_{=\mathbf{A}} \begin{pmatrix} y \\ y' \end{pmatrix} + \underbrace{\begin{pmatrix} 0 \\ 3\csc(2t) \end{pmatrix}}_{=\mathbf{g}(t)}. $$

We know that the solution of the homogeneous equation $y_\text{h}'' + 4y_\text{h} = 0$, or equivalently

$$ \frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix} y_\text{h} \\ y_\text{h}' \end{pmatrix} = \mathbf{A} \begin{pmatrix} y_\text{h} \\ y_\text{h}' \end{pmatrix}, $$

is exactly $ y_{\text{h}} = c_1 \cos(2t) + c_2 \sin(2t) $, or equivalently,

$$ \begin{pmatrix} y_\text{h} \\ y_\text{h}' \end{pmatrix}
= \underbrace{\begin{pmatrix} \cos(2t) & \sin(2t) \\ -2\sin(2t) & 2\cos(2t) \end{pmatrix}}_{=\mathbf{X}(t)} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix}. $$

It is easy to check that $\mathbf{X}' = \mathbf{A}\mathbf{X}$, i.e., each column of $\mathbf{X}$ solves the homogeneous equation. Then we make an ansatz that the solution of the original, non-homogeneous equation takes the form

$$ \begin{pmatrix} y \\ y' \end{pmatrix} = \mathbf{X} \mathbf{u} $$

for some vector-valued function $\mathbf{u}$. Plugging this ansatz to the original equation,

$$ \require{cancel} \cancel{\mathbf{X}'\mathbf{u}} + \mathbf{X}\mathbf{u}' = \cancel{\mathbf{A}\mathbf{X}\mathbf{u}} + \mathbf{g}, $$

and so, the equation reduces to

$$ \mathbf{u}' = \mathbf{X}^{-1}\mathbf{g} =  -\frac{3}{2} \begin{pmatrix} 1 \\ \cot(2t) \end{pmatrix} $$

Integrating both sides,

$$ \mathbf{u} = -\frac{3}{4} \begin{pmatrix} 2t \\ \log\left| \sin (2t) \right| \end{pmatrix} + \mathbf{c} $$

for some constant vector $\mathbf{c}$. Therefore

$$ \begin{pmatrix} y \\ y' \end{pmatrix}
= \mathbf{X} \mathbf{u}
= -\frac{3}{4} \begin{pmatrix} \cos(2t) & \sin(2t) \\ -2\sin(2t) & 2\cos(2t) \end{pmatrix} \begin{pmatrix} 2t \\ \log\left| \sin (2t) \right| \end{pmatrix} + \mathbf{X}\mathbf{c}. $$

Note that $\mathbf{X}\mathbf{c}$ describes the general-solution term. Then reading out the first row of the product in the right-hand side yields

$$ y = c_1\cos(2t) + c_2\sin(2t) -\frac{3}{2} t\cos(2t) + \frac{3}{4} \sin(2t) \log \left| \sin (2t) \right|. $$


问:
How to show that $f(x) = o{(x^2)}$ as $x \to \infty$. when $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as $x\ \to \infty$
Suppose $f$ is differentiable on $(0, \infty)$ and $f'(x) = o(x)$ as
$x \to \infty$.

How to show  that $f(x) = o{(x^2)}$ as $x  \to \infty$?

答:
Cauchy's Mean Value Theorem says that
$$
\frac{f(b)-f(a)}{b^2-a^2}=\frac{f'(c)}{2c}\text{ for some $c$ between $a$ and $b$}
$$
Since $f'(x)=o(x)$, for any $\epsilon>0$ there is an $a$ so that if $c\ge a$ then $\dfrac{f'(c)}{c}\le2\epsilon$. Cauchy's Mean Value Theorem then says that $\dfrac{f(b)-f(a)}{b^2-a^2}\le\epsilon$. This says that $f(x)=o(x^2)$ as $x\to\infty$.</p>


问:
Which of the two methods is correct and why?

Problem

A deck of card is shuffled then divided into two halves of 26 cards
each. A card is drawn from one of the halves, it turns out to be an
ace. The ace is then placed in the second half-deck. The half is then
shuffled, and a card is drawn from it. Compute the probability that
this drawn card is an ace.

my attempt:-

let A = number of aces in the second deck before an ace added to it

let B = number of aces in the first deck before an ace is removed from it

let D = event that an ace is drawn from the second deck after an ace is added to it

Now, what we are asked to find is: $$\sum_{i=0}^3P(A=i|B>0)\cdot P(D) = \sum_{i=0}^3P(A=i|B>0)\cdot \frac{i+1}{27} $$

now, there are two ways of calculating $P(A=i|B>0)$

(Protocol A)restrict the sample space way:-
since B has at least one ace, we restrict the sample space to 51 cards out of which 3 are aces. $$P(A=i|B>0) = \frac{{{3}\choose{i}}{{48}\choose{26-i}}}{{{51}\choose{26}}}$$

(Protocol B)$P(A|B) = \frac{P(A\cap B)}{P(B)}$:-
$$P(A=i|B>0) = \frac{P(A=i\cap B>0)}{P(B>0)} = \frac{{{4}\choose{i}}{{48}\choose{26-i}}\bigg{/}{{52}\choose{26}}}{\bigg{[}{{52}\choose{26}}-{{4}\choose{0}}{{48}\choose{26}}\bigg{]}\bigg{/}{{52}\choose{26}}}$$

My question is : (out of Protocols A and B) which method is the correct one? and Why?

Protocol A has already been used here to answer this question correctly. So, protocol A is obviously correct. So, I guess my question becomes: Why is protocol B incorrect?

答:
I am having difficulty reading either your "protocol B" or quasi's answer and understanding where you are getting your numbers or what events you think you are using.  To address this, I will go through a similar derivation below explaining my thoughts to organize myself and share it with you after it is done.  Hopefully then we can see what went wrong.

We are drawing a card from the first half, seeing it is an ace, and then placing it in the second half.  We then shuffle the second half and draw again and ask what the probability is that the next card drawn is an ace.

Let us define some events.  $X$ is the event we drew an ace from the first deck.  $Y$ is the event we drew an ace from the second deck.  $A_i$ is the event that we began with $i$ aces in the first deck (thus making it $4-i+1$ aces in the second deck after the swap).

$\Pr(Y\mid X) = \dfrac{\Pr(Y\cap X)}{\Pr(X)}=\dfrac{\Pr(Y\cap X \cap (A_1\cup A_2\cup A_3\cup A_4))}{\Pr(X)}$

$=\dfrac{\Pr(Y\cap X\cap A_1) + \Pr(Y\cap X\cap A_2)+\dots + \Pr(Y\cap X\cap A_4)}{\Pr(X)}$

$=\dfrac{\Pr(Y\mid X\cap A_1)\Pr(X\mid A_1)\Pr(A_1) + \dots + \Pr(Y\mid X\cap A_4)\Pr(X\mid A_4)\Pr(A_4)}{\Pr(X)}$

$\dfrac{\frac{4}{27}\cdot \frac{1}{26}\cdot \binom{4}{1}\binom{48}{25}/\binom{52}{26} + \dots + \frac{1}{27}\cdot \frac{4}{26}\cdot \binom{4}{4}\binom{48}{22}/\binom{52}{26}}{\frac{4}{52}}$

$=\frac{43}{459}\approx 0.09368\dots$

Looking at portions of this calculation in greater detail, here we use that $A_0,A_1,\dots,A_4$ partitions the sample space into disjoint events and that $A_0\cap X=\emptyset$ so may be ignored.  We then use the law of total probability to split the numerator up into separate probabilities.  We then use the product rule to split each of these up further.

Now, as for $\Pr(Y\mid X\cap A_k)$ we have $4-k+1$ aces out of $27$ cards in the second deck at the time of drawing from the second deck, so the probability here is $\frac{4-k+1}{27}$.  $\Pr(X\mid A_k)$ is us drawing one of $k$ aces from the first half deck with $26$ cards in it, and $\Pr(A_k)$ is a straightforward hypergeometric probability $\binom{4}{k}\binom{48}{26-k}/\binom{52}{26}$.  In related posts there are claims about using $\binom{52}{26} - \binom{48}{26}$ as the denominator, but here we are looking purely at $\Pr(A_k)$ and are not conditioning on there being an ace available to draw at this point.  Alternatively phrased, we could also have had $\Pr(X\mid A_0)\Pr(A_0)$ appearing in the calculations but this would have equated to zero.

As to what went wrong in your calculation?  Looking it over, I see now that you had been conditioning not on having successfully selected an ace in the first deck, but rather you were conditioning on it being possible to select an ace in the first deck whether or not one of the aces available to select happened to be the top card of the deck.


问:
If $f(nx)=f(x)^n$ then $f(x+y)=f(x)f(y)$
If $f:\mathbb{R}\to\mathbb{R}_{+}$ is injective monotone function and if, $f(nx)=f(x)^n$ for all $n\in\mathbb{Z}$ and for all $x\in\mathbb{R}$, then,
$$f(x+y)=f(x)f(y),\text{ for all }x,y\in\mathbb{R}$$

It is easy to check the equality if $y$ is a multiple of $x$.

I thought about proving the inequalities, $f(x+y)\le f(x)f(y)$ and $f(x+y)\ge f(x)f(y)$, but I couldn't get to them.

Assuming that $f$ is increasing and $x<y$, then $y=kx+r$, for some $k\in\mathbb{Z}$ and $0< r\le \lvert x \rvert$, so

$$f(x+y)=f(x+kx+r)=f(x(k+1)+r)\ge f(x(k+1))=f(x)^{k+1}=f(x)f(x)^k=f(x)f(kx)$$

That is, $f(x+y)\ge f(x)f(kx)$, but the problem is that $f(kx)<f(y)$ so I can't continue the demonstration.

Body:
Let $$f(\frac{p}{q}\cdot x) =f(p\cdot\frac{x}{q})=f(\frac{x}{q})^p.$$
But
$$f(x)=f(q\cdot\frac{x}{q})=f(\frac{x}{q})^q.$$
Hence
$$f(x)^{\frac{1}{q}}= f(\frac{x}{q}).$$
Thus
$$f(\frac{p}{q}\cdot x)= f(x)^{\frac{p}{q}}.$$


问:
How would one go about finding the inverse of a matrix containg only variables?
Lets say I have the following matrix :
$$
   \left[\begin{array}{cc}
   z & y & x & w   \\
   s-a & t-b & u-c & v-d   \\
   a & b & c & d   \\
   3s & 3t & 3u & 3v   \\
   \end{array} \right]
$$
How would I go about demonstrating whether or not this matrix can be inverted?
I've been ready about the (many) criteria for inverting a matrix and yet I just can't figure this out. All I know is that it's det=-30 from a previous question.

答:
Expanding the determinant as with numbers, you will obtain a polynomial. The matrix will be invertible for all values of the variables, except where the polynomial vanishes (roots).

But in your particular case, the polynomial is... particular.


问:
proof verification on the distance of a metric space
let $A\subseteq  \mathbb{R}$ and  $\delta (A)=\sup\left \{  {| d(x,y)| :x,y\epsilon A} \right \}$ and $f:\mathbb{R}\rightarrow \mathbb{R}$ cont. if $\delta (A)<+\infty $ then $\delta (f(A))<+\infty$

my solution,

we know that $\delta (A)=\delta (\bar{A})\Rightarrow \delta (\bar{A})<+\infty $ ,since $\bar{A}$ is closed and bounded we have that is compact ,now $f$ is cont. and we know that $f(\bar{A})$ is also bounded so $\delta (f(\bar{A}))<+\infty$ but

$\delta (f(A)) \subseteq  \delta(f(\bar{A}))$(is = true??)

we have that  $\delta (f(A))<+\infty$

答:
It is true that $f(A) \subseteq f(\bar A)$; you should be able to convince yourself that in general, if $P \subseteq Q$, then $f(P) \subseteq f(Q)$.

From this, it follows that $\delta(f(A)) \leq \delta(f(\bar A))$.


答:
You may know how operations on the rows (or columns) of a matrix affect its determinant:

Adding a multiple of a row/column to another row/column leaves the determinant unchanged

Multiplying a row/column with a constant multiplies the determinant with the same constant

Swapping two rows/columns multiplies the determinant by $-1$.

Now, in your matrix, do the following:

Add the third row to the second.

Add the fourth row multiplied by $-1/3$ to the second.

Verify that you will obtain a matrix in which the second row will be all-zeros. Thus its determinant is zero. Thus, the determinant of the original matrix is also zero.

Title: How would one go about finding the inverse of a matrix containg only variables?

Body:
Lets say I have the following matrix :
$$
   \left[\begin{array}{cc}
   z & y & x & w   \\
   s-a & t-b & u-c & v-d   \\
   a & b & c & d   \\
   3s & 3t & 3u & 3v   \\
   \end{array} \right]
$$
How would I go about demonstrating whether or not this matrix can be inverted?
I've been ready about the (many) criteria for inverting a matrix and yet I just can't figure this out. All I know is that it's det=-30 from a previous question.


问:
What is relation between topological entropy and sensitivity?
It is known that every expansive homeomorphism has finite topological entropy, also expansivity implies sensitivity.

I would like to know that relation between topological entropy and sensitivity. Please help me to know it or introduce a reference for it.

Thanks a lot.

答:
There is none (maybe in some specific setting you could come up with something, but in this generality no). Sensitivity is a global property, but topological entropy is not. For example, you could have a dynamical system with an invariant set $\Lambda$ such that the topological entropy of $f|_{\Lambda}$ is positive (which implies that the entropy of $f$ is positive), but at the same time there is an open set $U$ in your space such that $f|_U$ is the identity (which prevents sensitivity).

In the result you mentioned there is a stronger global property present, expansivity.


问:
An inequality for nonnegative symmetric matrix
Consider $H$ to be a $n \times n$ symmetric matrix with nonnegative entries, that has only one positive eigenvalue and other $n-1$ eigenvalues are negative.
I want to prove that for every nonzero and nonnegative real vector $u$ and every real $v$ nonparallel to $u$, it holds:

$(u^T H v)^2 > (u^T H u)(v^T H v)$.

I can't prove the strict inequality, the best I can prove is this:

$(u^T H v)^2 >= (u^T H u)(v^T H v)$.

Is there any hint how to prove the strict version?

So many thanks in advance!

答:
This isn't true. Consider e.g.
$$
H=\pmatrix{0&1&1\\ 1&0&1\\ 1&1&1},\ u=\pmatrix{1\\ 0\\ 0},\ v=\pmatrix{0\\ 1\\ -1}.
$$
$H$ has one positive eigenvalue $1+\sqrt{2}$ and two negative eigenvalues $-1,1-\sqrt{2}$. Since $v^THu=0=u^THu$, the strict inequality in question doesn't hold.


问:
Is this a valid proof for showing two continuous functions that coincide on the rationals, are in fact the same function?
I am trying to prove the following, but I am unsure about a particular step in my proof. I am aware of other proofs for this statement, but I am trying to prove it in a specific way.

Statement:

" Suppose $f,g:\mathbb{R}\to\mathbb{R}$ are two continuous functions such that $f(x)=g(x)$ for all $x\in{}\mathbb{Q}$. Then $f=g$ everywhere."

My proof:

Since $f$ and $g$ are continuous, for each $\epsilon>0$ choose respective $\delta_1,\delta_2>0$ that satisfy the continuity definition. Now define the following: $$\delta=\min(\{\delta_1,\delta_2\}).$$

Let $c\in\mathbb{R}\setminus\mathbb{Q}$ with $|x-c|<{}\delta{}.$ Consider the following:

$$|f(c)-g(c)|$$
$$=|f(c)-f(x)+f(x)-g(x)+g(x)-g(c)|$$
$$\le|f(c)-f(x)|+|f(x)-g(x)|+|g(x)-g(c)|$$
$$<|f(x)-g(x)|+2\epsilon{}$$

This holds for all $x\in(c-\delta,c+\delta),$ so must hold for any $x\in(c-\delta,c+\delta)\cap\mathbb{Q}\neq\emptyset,$ which would lower the upper bound to simply $2\epsilon$ (since $|f(x)-g(x)|=0$ on $(c-\delta,c+\delta)\cap\mathbb{Q}$). Since that is true for every $\epsilon>0$ we have that $|f(c)-g(c)|$ is forced to be $0$, hence $f(c)=g(c)$ for all irrational $c$. Hence $f=g$ everywhere.

my issue is with making the $|f(x)-g(x)|=0$ where $x$ is rational and in the aforementioned open interval. Since it's true for all $x$ in that range, surely we can lower the upper bound?

答:
The issue is that you have that continuity at points, so the $\varepsilon - \delta$ statements must be about the point under consideration. Here is how I would do it instead if I am not required to use $\varepsilon - \delta$ statements:

We know that $f,g$ are everywhere continuous, and $f|_\mathbb{Q} = g|_\mathbb{Q}$. Let $c$ be an irrational number, and let $a_n \subset \mathbb{Q}$ be a sequence that converges to $c$. Then we have the following:

\begin{equation}
f(c) - g(c) = \left(f(\lim_\limits{n} a_n)\right) - \left(g(\lim_\limits{n} a_n)\right) = \left(\lim_\limits{n} f(a_n)\right) - \left(\lim_\limits{n} g(a_n)\right) = \lim_\limits{n} \left(f(a_n) - g(a_n) \right) = \lim_\limits{n} 0 = 0.
\end{equation}

Since you are specifically trying to use the $\varepsilon - \delta$ approach, here is an alternative:

Let $c$ be an irrational number. Since $f,g$ are continuous at $c$, for every $\varepsilon > 0$ there exists $\delta > 0$ such that the typical statements hold. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, there exists $p \in \mathbb{Q} \cap (c - \delta, c+ \delta)$. So then:

\begin{equation}
|f(c) - g(c)| \leq |f(c) - f(p)| + |f(p) - g(p)| + |g(p) - g(c)| < 2 \varepsilon
\end{equation}
since $|f(p) - g(p)| = 0$.