问:
Show that : $\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$
I found it with help of Wolfram alpha .

$$\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$$

This integral admits an antiderivative (I can add it if so) so I can conclude that this result is true .I would like to see a proof with the use of complex integration or a 'real' proof . For that I think we can put $y=\frac{2}{x^2+1}$ and use integration by parts (perhaps?).For the proof with complex integration(if it exists I'm new for it )  can you detail the solution ?

Thanks!

答:
To get things started, the integral definition of the arctangent, switching the order of integration, and a change of variables gives us

$$\begin{align}
\int_0^\infty\arctan\left(2\over x^2+1\right)\,dx
&=\int_0^\infty\int_0^{2/(x^2+1)}{1\over1+y^2}\,dy\,dx\\
&=\int_0^2\int_0^\sqrt{(2/y)-1}{1\over1+y^2}dx\,dy\\
&=\int_0^2{\sqrt{(2/y)-1}\over1+y^2}\,dy\\
&=\int_\infty^0{\sqrt{z^2}\over1+\left(2\over1+z^2 \right)^2}\cdot{-4z\,dz\over(1+z^2)^2}\\
&=4\int_0^\infty{z^2\over(1+z^2)^2+4}\,dz\\
&=2\int_{-\infty}^\infty{z^2\over z^4+2z^2+5}\,dz
\end{align}$$

The integral is now ripe for residues. The singularities are at $z=\pm\sqrt{-1\pm\sqrt{1-5}}=\pm\sqrt{-1\pm2i}$. Some straightforward algebra shows that the ones in the upper half plane are at

$$z=\pm{1\over\sqrt\phi}+i\sqrt\phi$$

Calling these $c_+$ and $c_-$ and noting that $c_\pm^2+1=\pm2i$, the residue theorem tells us

$$\begin{align}
2\int_{-\infty}^\infty{z^2\over z^4+2z^2+5}\,dz
&=4\pi i\left({c_+^2\over4c_+^3+4c_+}+{c_-^2\over4c_-^3+4c_-} \right)\\
&=4\pi i\left( {c_+\over8i}+{c_-\over-8i}\right)\\
&={\pi\over2}(c_+-c_-)\\
&={\pi\over\sqrt\phi}
\end{align}$$


问:
prove that $E(X^{n}) = 0 $
Prove that for any odd $n \in \mathbb{N}$ and $X \sim \mathcal{N}(0,\sigma^{2})$, $E (X^n) = 0$.

I started with the fact that:
$$E(X^{n}) = {{1}\over{\sqrt{2\pi} \sigma}} {\int_{-\infty}^{\infty} x^{n} e^{{-x^2}/{2\sigma^2}}} dx = {{2}\over{\sqrt{2\pi} \sigma}} {\int_{0}^{\infty} x^{n} e^{{-x^2}/{2\sigma^2}}} dx.$$
Then I used the substitution $$t={{x}\over{\sigma}} \to {{2\sigma^n}\over{\sqrt{2\pi}}} {\int_{0}^{\infty} t^{n} e^{{-t^2}/{2}}} dt.$$
But I have a problem how to solve it. Does anyone know how this equality can be proved? The task tip says to use the Gamma function in the calculation, but I don't see it..

答:
You can use the moment generating function, express it in tayor series and take the derivative.

$$M_X(t)=e^{\frac{\sigma^2t^2}{2}}$$

and

$$\mathbb{E}[X^n]=\frac{d^n}{dt^n}e^{\frac{\sigma^2t^2}{2}}\Bigg|_{t=0}$$

Expanding MGF in taylor series you get

$$e^{\frac{\sigma^2t^2}{2}}=\sum_{n=0}^{\infty}\frac{\Big(\frac{\sigma^2t^2}{2}\Big)^n}{n!}=\sum_{n=0}^{\infty}\frac{\sigma^{2n}\cdot t^{2n}}{2^n\cdot n!}$$

Thus it is evident that, for every integer $n\geq 0$

$$\mathbb{E}[X^{2n}]=\frac{\sigma^{2n}(2n)!}{2^n\cdot n!}$$

and

$$\mathbb{E}[X^{2n+1}]=0$$

ContentLicense: CC BY-SA 4.0

Title: prove that $E(X^{n}) = 0 $


问:
I need a concrete example to show this equality is false
If $(X,\tau_{1})$ and $(X,\tau_{2})$ are
$T_0$-space then  $(X,\tau_{4})$ is not
necessarily a $T_0$-space . Give a concrete
example.

If $\tau_{4}. $is defined as $\tau_{4}$=$\tau_{1} \cap \tau_{2}$ then $\tau_{4}$ is a topology on $X$.

( That I did)

答: Take $X=\{0,1\}$, $\tau_1=\bigl\{\emptyset,\{0\},X\bigr\}$, and $\tau_2=\bigl\{\emptyset,\{1\},X\bigr\}$, in which case $\tau_1\cap\tau_2=\{\emptyset,X\}$.


问:
Sylow's second theorem explanation
I am reading the book Contemporary Abstract Algebra by Gallian and I can't understand some specific part of the proof of Sylow's second theorem that has been given here.

Theorem : If H is a subgroup of a finite group $G$ and $|H|$ is a power of a prime $p$ then $H$ is contained in some Sylow p-subgroup of $G$.

Proof : Let $K$ be a Sylow p-subgroup of $G$ and let $C = \{K_1, K_2, ... , K_n\}$ with $K=K_i$ be the set of all conjugates of $K$ in $G$. Since conjugation is an automorphism, each element of $C$ is a Sylow p-subgroup of $G$. Let $S_C$ denote the group of all permutations of $C$. For each $g \in G$, define $\phi_g : C \to C$ by $\phi_g(K_i) = gK_ig^{-1}$. It is easy to show that each $\phi_g \in S_C$.

Now define $T:G \to S_C$ by $T(g) = \phi_g$. Since $\phi_{gh}(K_i) = (gh)K_i(gh)^{-1} = g(hK_ih^{-1})g^{-1} = g\phi_h(K_i)g^{-1} = \phi_g(\phi_h(K_i)) = (\phi_g\phi_h)(K_i)$, we have $\phi_{gh} = \phi_g\phi_h$ , and therefore $T$ is a homomorphism from $G$ to $S_C$.

Next consider $T(H)$, the image of $H$ under $T$. Since $|H|$ is a power of $p$, so is $|T(H)|$. Thus, by the Orbit-Stabilizer Theorem, for each $i$, $|orb_{T(H)}(K_i)|$ divides $|T(H)|$, so that $|orb_{T(H)}(K_i)|$ is a power of $p$. Now we ask: Under what condition does $|orb_{T(H)}(K_i)| = 1$? Well, $|orb_{T(H)}(K_i)| = 1$ means that $\phi_g(K_i) = gK_ig^{-1} = K_i$ for all $g \in H$; that is, $|orb_{T(H)}(K_i)| = 1$ iff $H\le N(K_i)$. But the only elements of N(K_i) that have orders that are powers of $p$ are those of $K_i$. Thus, $|orb_{T(H)}(K_i)| = 1$ iff $H \le K_i$.

So, to complete the proof, all we need to do is to show that for each $i, |orb_{T(H)}(K_i)| = 1$. Analogous to Theorem 24.1, we have $|C| = |G:N(K)|$. $ \color {orange}{\text{And  since  $|G:K| = |G:N(K)| |N(K):K|$} }$   is not divisible by $p$, neither is $|C|$. $\color {brown}{\text{Because the orbits partition $C$ , $|C|$ is the sum of powers of p. If no orbit has size 1, then p divides each summand}}$ and, therefore, $p$ divides $|C|$, which is a contradiction. Thus, there is an orbit of size $1$, and the proof is complete.

I can't understand the two coloured lines. Especially about the second coloured line I don't get that why orbits partition C, previously nowhere it's written. And I only know about orbits as much as it's given in this book which is just the definition and some examples. And the result that order of the orbit divides the order of the group.

答:
Let G be a group of permutation of a set S. for each i in S, let
$${orb_G(i)={\{\phi(i)| \phi \in G\}}}$$

The set ${orb_G(i)}$ is a subset of S called the orbit of i under G.

so, orbit of i is everything that element is mapped to under the permutation. That is why it is subset of S. hence, it partitions the set S.

Now, intuitively if you think of a rubik's cube, the element at the top left corner of red color , it can move(permute) only to certain point in the rubik's cube, that is the orbit of that  element (3x3x3)  rubic's cube.

Also, you mentioned, no orbit has size 1 which means, any center color of rubik's cube can't move(permute) to any other position. so it has not orbit.

Title: Sylow's second theorem explanation


问:
Find non trivial homomorphism $\mathbb{Z}/q\mathbb{Z} \rightarrow \text{Aut}(\mathbb{Z}/p\mathbb{Z})$
Let $H= \mathbb{Z}/p\mathbb{Z}$ and $F=\mathbb{Z}/q\mathbb{Z}$ with $p,q$ primes. As indication, I got that $${\rm Aut}(H) \cong (\mathbb{Z}/p\mathbb{Z})^\times \cong \mathbb{Z}/(p-1)\mathbb{Z}.$$

I have to show that if $q$ divides $p-1$ there exist a non-trivial homomorphism $\psi$: $F \rightarrow{\rm Aut}(H)$.
I understand if $q$ divides $p-1$ then there is an element of order $q$ in $\mathbb{Z}/(p-1)\mathbb{Z}$. But, the problem I have, is that I don't really understand how to construct homomorphisms so I'd like to have a step-by-step explanation if possible, please. For me, this non-trivial homomorphism is simply of form $\psi([x]_q): y \rightarrow y \cdot g^x$, where $g$ is generator of $(\mathbb{Z}/p\mathbb{Z})^\times$. But I was told that it is wrong.

Thanks in advance and have a nice day.

答:
The map in your problem could represent the morphism induced by a semidirect product of a $p$-Sylow and a $q$-Sylow $\mathbb Z_p\rtimes_{\psi}\mathbb Z_q$.

Consider the homomorhism $\psi: \mathbb Z_q\to \operatorname{Aut}(\mathbb Z_p)$ you have that $|\psi(\mathbb Z_q)|$ divides $|\mathbb Z_q|$ and $|\operatorname{Aut}(\mathbb Z_p)|$, where $|\operatorname{Aut}(\mathbb Z_p)|=\varphi(p)=p-1$ (for example the two groups could be $\mathbb Z_2$ and $\mathbb Z_3$).

Note: Since $q$ divides $|\operatorname{Aut}(\mathbb Z_p)|$, which is a cyclic group, you could take the homomorphism that maps a generator of $\mathbb Z_q$ to a generator of the (single) $q$-group in $\operatorname{Aut}(\mathbb Z_p)$ (in a $p$-cyclic group $P$ every element of $P$ has the same order of the group and generate it).

If $q|p-1$, we can assume $|\psi(\mathbb Z_q)|=q$ and for the first theorem of isomorphism, we know that $$|\psi(\mathbb Z_q)|=q=|F:Ker(\psi)|\implies|Ker(\psi)|=\dfrac{|\mathbb Z_q|}{q}=1 \implies \psi\ne id_{\mathbb Z_q}$$
Since the order of the Kernel of the morphism is not equal to the order of the group, this shows that $$\exists\psi\in \operatorname{Hom}(\mathbb Z_q,\operatorname{Aut}\mathbb Z_q)\text{, where }\psi\text{ is non-trivial and it is also injective}.$$


问:
Range of the function $f(x) = [x] + (x - [x])^2$
If $f(x) = [x] + (x - [x])^2$, $x \in \mathbb{R}$.

Then what is $Im(f)$?

$ \text{My attempt :}$

It's easy for $x \in \mathbb{Z}$. Because then $f(x) = x$.

But the problem I'm having is in the case when $x \notin \mathbb{Z}$.

I've thought this case in two subcases.

$\text{Subcase I : }$ when $x \in \mathbb{{Q}-{Z}}$ then as rational numbers are countable, I can think of the images and the pre-images are in bijection.

$\text{Subcase II : }$ when $x \notin \mathbb{Q}$, then I can sense about the same bijection again
but can't prove it mathematically.

And from these I would've said that as $f$ is bijection on $\mathbb{R}$ thus $\text{Im(f)} = \mathbb{R}$.

Can someone help me building a proof of these two subcases?

$[x] = \mathbb{N} \ni k: k \leq x \wedge k + 1 > x$ - greatest integer less than or equal to x.

答: We can notice, that when we set $k \leq x < k + 1$, then $[x] = k$ and $(x - [x])^2$ takes all the values in range $[0, 1)$, because $\{x\} = x - [x]$ does. $\mathrm{Im}(\{x\}) = [0, 1) \implies \mathrm{Im}((\{x\})^2) = [0, 1) $.


问:
Extending measure on Field to $\sigma$-Field
Let $\mu_1$ and $\mu_2$ be two finite measures defined on $\sigma(\mathcal{F})$ such that, $\forall A \in \mathcal{F}$, $\mu_1(A)=\mu_2(A)$. Show that they must agree on $\sigma(\mathcal{F})$.

I attempted proof of this as follows:

let $\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) | \mu_1(A)=\mu_2(A) \rbrace$. If we show $\mathcal{M}$ is a monotone class, we are done by the Monotone Class Theorem.

To that end, let $\lbrace A_n \rbrace$ be a sequence of non-decreasing sets from $\mathcal{M}$. We "disjointify" the sets :
Let $B_k=A_k- \bigcup_{j=1}^{k-1}A_j$. Note that $\lbrace B_i \rbrace$ are pairwise disjoint and $\bigcup_{j=1}^{\infty}A_j=\bigcup_{j=1}^{\infty}B_j$. Now, since $\mu_1$ and $\mu_2$ are countably additive, $$\mu_1(\bigcup\limits_{j=1}^{\infty}B_j)= \sum\limits_{j=1}^{\infty}\mu_1(B_j),$$ $$\mu_2(\bigcup\limits_{j=1}^{\infty}B_j)= \sum\limits_{j=1}^{\infty}\mu_2(B_j).$$ Because $B_i$ is in $\mathcal{F}$, we have $\mu_1(B_i)=\mu_2(B_i), \forall i \in \mathbb{N}$, and hence $$\mu_1(\bigcup\limits_{j=1}^{\infty}B_j)=\mu_2(\bigcup\limits_{j=1}^{\infty}B_j)$$ We conclude that $\mathcal{M}$ is closed under increasing union. Now, to show $\mathcal{M}$ is closed under decreasing limits of sets,
Let $\lbrace C_n \rbrace$ be non-increasing sets from $\mathcal{M}.$ We want to show $\lim C_n = \bigcap_{n=1}^{\infty}C_n$ is in $\mathcal{M}$. For this, define $S_k=\mu_1(C_k)$ and $T_k=\mu_2(C_k)$ as sequences of real numbers. Then, $\lim_{k \rightarrow \infty}S_k=\lim_{k \rightarrow \infty}\mu_1(S_k)=\lim_{k \rightarrow \infty}\mu_1(T_k)=\mu_1(\bigcap_{n=1}^{\infty}C_n)=\mu_2(\bigcap_{n=1}^{\infty}C_n)$, using continuity from above of a finite measure. Thus, $\mathcal{M}$ is a monotone class of $\mathcal{F}$ and by monotone class theorem, $\mathcal{M}(\mathcal{F})=\sigma(\mathcal{F})$.

Does this look alright?

答:
Your proof is almost correct, but it is not OK.
You defined
$\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) | \mu_1(A)=\mu_2(A) \rbrace$.
Then you take $\lbrace A_n \rbrace$ be a sequence of non-decreasing sets from $\mathcal{M}$. But when you "disjointify" the sets, taking

$B_k=A_k- \bigcup_{j=1}^{k-1}A_j$, all you can say is that $B_i$ is in $\sigma(\mathcal{F})$. You can not say $B_i$ is in $\mathcal{F}$ nor $B_i$ is in $\mathcal{M}$.

The way to correct your proof is simple. Here it is:

Let $\mu_1$ and $\mu_2$ be two finite measures defined on $\sigma(\mathcal{F})$ such that, $\forall A \in \mathcal{F}$, $\mu_1(A)=\mu_2(A)$. Show that they must agree on $\sigma(\mathcal{F})$.

Let $\mathcal{M}= \lbrace A \in \sigma(\mathcal{F}) \,| \, \mu_1(A)=\mu_2(A) \rbrace$. Let us show that $\mathcal{M}$ is a monotone class.

First, let $\lbrace A_n \rbrace$ be a monotone non-decreasing sequence of sets from $\mathcal{M}$.

So, for all $n$, $\mu_1(A_n) =\mu_2(A_n)$.

Since $\lbrace A_n \rbrace$ is be a sequence of non-decreasing sets in $ \sigma(\mathcal{F})$, for any measure $\nu$ defined on $ \sigma(\mathcal{F})$, we have
$$\nu\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\nu(A_n)$$
So applying this to $\mu_1$ and $\mu_2$, we get
$$\mu_1\left (\bigcup\limits_{n=1}^{\infty}A_n \right) = \lim_{n \to \infty}\mu_1(A_n)=\lim_{n \to \infty}\mu_2(A_n) =\mu_2\left (\bigcup\limits_{n=1}^{\infty}A_n \right) $$
So $\bigcup\limits_{n=1}^{\infty}A_n \in \mathcal{M}$, and we conclude that $\mathcal{M}$ is closed under monotone non-decreasing union.

Now, to show $\mathcal{M}$ is closed under decreasing limits of sets.

Let $\lbrace C_n \rbrace$ be a monotone non-increasing sequence of sets from $\mathcal{M}.$

So, for all $n$, $\mu_1(C_n) =\mu_2(C_n)$.

Using continuity from above of a finite measure, we have
$$\mu_1\left (\bigcap\limits_{n=1}^{\infty}C_n \right) = \lim_{n \to \infty}\mu_1(C_n)=\lim_{n \to \infty}\mu_2(C_n) =\mu_2\left (\bigcap\limits_{n=1}^{\infty}C_n \right) $$
So $\bigcap\limits_{n=1}^{\infty}C_n \in \mathcal{M}$, and we conclude that $\mathcal{M}$ is closed under monotone non-increasing intersection.

Thus, $\mathcal{M}$ is a monotone class containing $\mathcal{F}$ and by monotone class theorem, $\mathcal{M} \supseteq \sigma(\mathcal{F})$. So, for all $A\in\sigma(\mathcal{F})$, we have $\mu_1(A)=\mu_2(A)$.

Remark: As defined, $\mathcal{M}$ may be bigger than $\sigma(\mathcal{F})$, but all we need to prove the result is $\mathcal{M} \supseteq \sigma(\mathcal{F})$.


问:
m stations are connected to a single router. The first station transmits p packets per unit time and the i-th station is twice as fast as i-1 station
Question: $m$ stations are connected to a single router. The first station transmits $p$ packets per unit time and the $i-th$ station is twice as fast as the $(i-1)st$ station. How many packets does the router receive per unit time?

Approach:  Total stations = $2^m$ I would assume the router per unit would receive: $p*(2^m -1)$

Does this calculation make sense to approach this problem?

答:
First station sends $p$ packets.

$k$-th station sends $2^{k-1}p$ packets.

Total packets sent:

$$\sum_{i=0}^{m-1}2^ip = p\sum_{i=0}^{m-1}2^i = p(2^m-1)$$

So, yes, your calculation is correct, but writing it out will make that more clear.


问:
Show that if $fd'=f'd $ and the pairs $f, d $ and $f',d' $ are coprime, then $f=f' $ and $d=d' $.
Show that if $fd'=f'd $ and the pair $f, d $ are coprime, that is gcd($d,f $)=1, as is the pair $f',d' $, then $f=f' $ and $d=d' $.

This should be simple, but I couldn't verify it.
Thanks in advance!

答:
Another solution, using
Corollary 31.4 from CLRS:

For all integers a and b and any nonnegative integer n,
$ (an, bn) = n*(a, b); $

$ (f, d) = 1 => (ff', df') = f' $ (multiplying by f')

$ (f', d') = 1 => (ff', d'f) = f $ (multiplying by f)

Now it's easy:

$ (ff', df') = (ff', d'f) => f' = f $ (because df' = d'f)

$ df' = df = d'f => d' = d $

ContentLicense: CC BY-SA 3.0


问:
Finding an area of the portion of a plane?
I need help with a problem I got in class today any help would be appreciated!

Find the area of the portion of the portion of the plane $6x+4y+3z=12$ that passes through the first octant where $x, y$, and $z$ are all positive..

I graphed this plane and got all the vertices but I am not sure how my teacher wants us to approach this problem..

Do I calculate the line integral of each side of the triangle separately and add them together? because we are on the section of line integrals, flux, Green's theorem, etc..

答:
The $x$, $y$ and $z$ intercepts are $(2,0,0)$, $(0,3,0)$ and $(0,0,4)$ respectively. Fix one of the points, say $(2,0,0)$, and create a vector $\vec u$ from $(2,0,0)$ to $(0,3,0)$ and $\vec v$ from $(2,0,0)$ to $(0,0,4)$. Then one half of the magnitude of the cross product will give us the area.

$\vec u=(0,3,0)-(2,0,0)=(-2,3,0)=-2\hat i+3\hat j$ and $\vec v=(0,0,4)-(2,0,0)=(-2,0,4)=-2\hat i+4\hat k$.

The area = $\frac{1}{2}|\vec u\times \vec v|$ = $\frac{1}{2}|12\hat i+8\hat j+6\hat k|=\frac{1}{2}\sqrt{12^2+8^2+6^2}=\frac{1}{2}\sqrt{244}= \sqrt{61}$.


答:
I may certainly not be the most qualified person to answer this question but yes you're right. The most expressive part of your problem is the phrase ' i-th station is twice as fast as the (i-1)th station'. This one clearly models a geometric progression since the ratio between two consecutive terms in a geometric distribution is constant, in this case equal to 2.

So you can consider the number of terms as m, ratio q= 2 and first term p. Now you can apply the sum formula for the series:

$ S_m = p * \frac{q^m - 1}{q -1 }$

Plugging in the values we come up with the formula:

$ S_m = p * (2 ^m -1)$

$S_m$ where m is finite is the number of packets the router gets per unit of time from m sources.
Also this question has nothing to do with probability and $2^m$ as far as I'm aware doesn't have any physical meaning in this case.

Title: m stations are connected to a single router. The first station transmits p packets per unit time and the i-th station is twice as fast as i-1 station

Body:
Question: $m$ stations are connected to a single router. The first station transmits $p$ packets per unit time and the $i-th$ station is twice as fast as the $(i-1)st$ station. How many packets does the router receive per unit time?

Approach:  Total stations = $2^m$ I would assume the router per unit would receive: $p*(2^m -1)$

Does this calculation make sense to approach this problem?


问:
Does the product of standard deviations of two random variables give their covariance?
Does the product of standard deviations of two random variables always give the covariance of these random variables?

If not, when does it specifically hold?

答:
Not true in general. Covariance of independent variables is $0$ but the product of their deviations is not.

The ratio of the covariance of random variables and the product of their standard deviations is called the correlation coefficient between the random variables. $$r_{XY}=\frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y}\in[-1,1]$$$r_{XY}=1$ corresponds to a positive linear relation between $X,Y$ i.e. $Y=aX+b,a>0$.


问:
Examples of noncommutative ring where Fermats Last Theorem fails for n=4
Does anyone have an example where this is the case?</p>

答:
The ring of $2\times 2$ real matrices, for instance. Where we have
$$
X^4+X^4=(\sqrt[4]2 X)^4
$$
for any element $X$.


问:
How to find unitary matrix from multiple eigenvector
enter image description here

Find the unitary matrix $U$ and the upper triangular matrix $T$ such that $U^{-1}AU = T$ where $A = \pmatrix{3&2\\-2&-1}$ has the eigenvalue $\lambda = 1$ (twice) corresponding to the eigenvector $\vec x = \pmatrix{1\\-1}$.

Hi guys.

I am stuck in finding unitary matrix.

As I know, the unitary matrix must be invertible but this matrix has eigenveector of multiplicity.

How can I find unitary matrix here?

答: Hint: The columns of $U$ must form an orthonormal basis of $\Bbb C^2$, and its first column must be a multiple of the eigenvector $\vec x = \pmatrix{1\\-1}$.


问:
Will learning category theory lead to a better and clearer understanding of mathematics?
I read the first chapter on a book about category theory Conceptual Mathematics:A first introduction to categories.In the preface the authors say:
It has been the good fortune
of the authors to live in these interesting times, and to see how the fundamental
insight of categories has led to clearer understanding, thereby better organizing, and
sometimes directing, the growth of mathematical knowledge and its applications.

The introduction was about how the flight of a bird is a function from time to space.It was a very neat explanation.Is category theory about stuff similar to this ?

So my question is: Will learning category theory lead to a better and clearer understanding to mathematics and are there any prerequisites to learning category theory ?

答:
I believe it will. Category theory barely existed before Stefan Banach passed in 1945, but this quote makes me think of category theory:

"...the ultimate mathematician is one who can see analogies between analogies."

The reason I think Banach may have considered category theory relevant to his quotation is that the mathematicians who invented it (e.g., Sanders Mac Lane) are the type who see analogies between analogies. It gives you a wider perspective.


问:
Matrix for which $A^\ast A = 1$, but $A A^\ast \ne 1$ (or reversed)
Title says it all: unitary matrices have the property that $AA^\ast = A^\ast A=1$. I want to find an example as in the title. For finite matrices, this doesn't seem to work out. How can I find an infinite-dimensional matrix satisfying the expression in the title?

Thanks.

答:
For finite matrices this does still work. You just can't use square matrices. For example consider the isometry that embeds $\mathbb{C}^2 \rightarrow \mathbb{C}^3$ defined as
$$
V = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}.
$$
Then $V^* V = I_{\mathbb{C}^2}$ but $VV^* \neq I_{\mathbb{C}^3}$.


问:
Is the set of all step functions countable or uncountable
I've got $f:\mathbb{N}\to \mathbb{N}$ a step function if $f (1)= 1$ and for every $n\geq 2$ we have either
$f(n)=f(n−1)$ or $f(n)=f(n−1)+1.$

I need to show the set of all step functions is countable or
uncountable?

I know that if its countable i need to form a surjection between the set and the set of natural numbers but dont quite know how to show this.

答: One thing to note is that showing asset is countable means there exists an injective function into $\mathbb{N}$, not a surjective function. To show that a set is uncountable, you must show that there cannot exist an injective function


问:
Blowing up of the curve $x^4+y^4-xyz^2$ in $\mathbb{P}^2$
[Problem] (Fulton's Algebraic curves Problems 7.9)

Let $C=V(x^4+y^4-xyz^2)$. Write down equations for a nonsingular curve $X$ in some $\mathbb{P}^N$ that is birationally equivalent to $C$. (Use the Segre imbedding)

[My attempt]

First, I found the multiple points (singular points) by using gradient.
The only singular point is $P=[0:0:1]$.

Next, I tried to blow up this curve by considering $[x:y:z] \times [u:t] \in \mathbb{P}^2 \times \mathbb{P}^1$ $$ \begin{cases} x^4+y^4-xyz^2=0 \\xt-yu=0  \end{cases}$$

Then by multiplying $u^4$ to the first equation, we obtain $x^4u^4+y^4u^4-xyz^2u^4= x^4u^4+(xt)^4-xz^2u^3(xt)=x^2(x^2u^4+x^2t^4-z^2u^3t) =0$

Therefore this $x^2u^4+x^2t^4-z^2u^3t=0$ is the first result of blowing up of the curve.

However, at $[x:y:z] \times [u:t]=[0:0:1] \times [0:1]$ , by observing affine case ($z=1, t=1$), $x^2u^4+x^2+u^3=0$ have singularity.

I'm stuck here. I don't know how can I proceed the blowing up this curve one more time and how to use the Segre imbedding..

Thank you for your attention.

答:
I find a tiny but nice trick in this situation.

The curve $ x^2u^4+x^2t^4-z^2u^3t=0$ have an open subset $x^2u^4+x^2t^4-z^2u^3t=0 , u \neq 0$.

These two varieties are obviously birationally equivalent since it is just an open dense subset.

Now,
$$ x^2u^4+x^2t^4-z^2u^3t=0 , u \neq 0 $$
$$ \Rightarrow x^2u^2+\frac{x^2t^4}{u^2}-z^2ut=0  \Leftrightarrow x^2u^2+y^2t^2-z^2ut=0$$
(in the last part, use $xt=yu$)

Therefore, we find the blowing up $x^2u^2+y^2t^2-z^2ut=0$ which has $(2,2)$ bi-degree.

Finally, we can use Segre embedding and the answer is $T_{1,1}^2+T_{2,2}^2-T_{3,1}T_{3,2}=0$ in $\mathbb{P^5}$.


问:
prove that $\log_2(n+1)-\log_3(2n+1) \le c$ eventually
How can I prove that there is $c>0$ such that starting from a specific $n$ the following is true:
$$
\log_2(n+1) - \log_3(2n+1) \le c
$$

答:
HINT
$$
\log_2(n+1) -\log_3(2n+1) \le \log_2(n+1)
$$
so suffices to prove that $\log_2(n+1) \le cn$ for some $c \in \mathbb{R}^+$, which is equivalent to proving
$$
n+1 = 2^{\log_2(n+1)} \le 2^{cn} = \left(2^c\right)^n = k^n
$$
for $k=2^c \in \mathbb{R}^+$.

Can you finish?

UPDATE

Now that you updated the question, it became a more interesting problem. One way is to consider the function
$$
f(x)
 = \log_2(x+1) - \log_3(2x+1)
 = \frac{\ln(x+1)}{\ln 2} - \frac{\ln(2x+1)}{\ln 3}
$$
over $x \ge 0$ and prove that the function is bounded by showing that it achieves a maximum or reaches a horizontal asymptote.

Achieving maximum (if true) can be shown by solving $f'(x)=0$ and computing $f''(x)$ to establish concavity with the 2nd derivative test.

Horizontal asymptotes are also part of standard calculus class.

Can you finish?

UPDATE 2

It appears this claim is not true. Note that as $x \to \infty$, we have
$$
\begin{split}
f(x)
 &\approx \log_2(x) - \log_3(2x) \\
 &= \log_2(x) - \log_3(x) - \log_3 2 \\
 &= \frac{\ln x}{\ln 2} - \frac{\ln x}{\ln 3} - \log_3 2 \\
 &= \ln x \left(\frac{1}{\ln 2} - \frac{1}{\ln 3}\right) - \log_3 2 \\
 &= a\ln(x) - b,
\end{split}
$$
where $a = \frac{1}{\ln 2} - \frac{1}{\ln 3} > 0$, so $f(x) \to \infty$ as $x \to \infty$. That means the difference between your logarithmic factors diverges with increasing $n$...


答:
For a given function $f$ define $ e_n \in \{0,1\}$ by $f(n)=f(n-1)+e_n$. So a given function is characterised by the vector $(e_1,e_2,\cdots)$. Which looks pretty uncountable ($2^{ \aleph_0}$).

Indeed a standard diagonal arguement will show it is uncountable.

Title: Is the set of all step functions countable or uncountable

Body:
I've got $f:\mathbb{N}\to \mathbb{N}$ a step function if $f (1)= 1$ and for every $n\geq 2$ we have either
$f(n)=f(n−1)$ or $f(n)=f(n−1)+1.$

I need to show the set of all step functions is countable or
uncountable?

I know that if its countable i need to form a surjection between the set and the set of natural numbers but dont quite know how to show this.


答:
Let $T$ be the shift operator on the Hilbert space $H=l^\infty$ with basis $e_0$, $e_1$, $e_2$, ... which is determined by
$$
Te_k=e_{k+1}\ .
$$
Then its adjoint $T^*$ is determined by $T^*e_k=e_{k-1}$, with the convention that $e_{-1}=0$. Or explicitly,
$$
T^*e_k=
\begin{cases}
0 &\text{ for } k=0\ ,\\
e_{k-1}&\text{ for } k>0\ .
\end{cases}
$$
To see this, it is enough to check $(T^*e_j,e_k)=\delta_{j-1,k}=\delta_{j,k+1}=(e_j,Te_k)$. Now compute $T^*T$ and $TT^*$ (especially on $e_0$).

Title: Matrix for which $A^\ast A = 1$, but $A A^\ast \ne 1$ (or reversed)

Body:
Title says it all: unitary matrices have the property that $AA^\ast = A^\ast A=1$. I want to find an example as in the title. For finite matrices, this doesn't seem to work out. How can I find an infinite-dimensional matrix satisfying the expression in the title?

Thanks.


问:
How to calculate the pseudoinverse of a symmetric matrix given its eigenvalue decomposition?
Given a singular symmetric matrix $A$ and the system $Ax = b$, knowing that there exists a solution, how could one find a pseudoinverse given the eigenvalue decomposition s.t. $x = A^*b$ ?

答:
Consider $A$ to be a symmetric matrix. By SVD, $A=P^{T}DP$ where $P$ is orthonormal and $D=\text{Diag}(\lambda_{i})$. Define $A^{-1}=P^{T}D^{-1}P$, where $D^{-1}=\text{Diag}(\lambda_{i}^{*})$ where :
$$
\lambda_{i}^{*}=
\begin{cases}
\displaystyle\frac{1}{\lambda_{i}}& \text{if $\lambda_{i}\neq0$} \\
0 &\text{if $\lambda_{i}=0$}
\end{cases}
$$
You should know that $A=AA^{-1}A$. Thus, we shall show that $A^{-1}=A^{-1}AA^{-1}$ where $A^{-1}$ is also symmetric. Let $A^{-1}=P^{T}D^{-1}P$, we have that :
$$
(A^{-1})^{T}=(P^{T}D^{-1}P)^{T}=P^{T}D^{-1}P^{T}=A^{-1}
$$
Thus, $A^{-1}$ is symmetric. Now we have that :
$$
A^{-1}AA^{-1}=P^{T}D^{-1}PP^{T}DPP^{T}D^{-1}P=P^{T}D^{-1}P=A^{-1}
$$


问:
Recommendation for good online videos for stochastic differential equations
I have a background in statistics and was interested in learning more about modeling with Stochastic Differential Equations. I understand the basics of stochastic processes and measure theory, etc., as well as differential equations and numerical methods. I started to read Evan's short guide to SDEs which is nice. But I find myself getting a bit lost in a lot of the formalisms for defining martingales and then brownian motion, etc. I tend to find it hard to simply read equations instead of seeing someone explain the derivations--especially for a new topic.

I was hoping someone could recommend some online videos or online course on Stochastic Differential Equations. I found a lot of videos on Youtube and google, but was not sure which of them are good. Many of those videos are very incomplete, etc. Any help would be appreciated.

Thanks.

答:
I did find this set of videos very helpful. They are by MIT with a financial focus, but the lectures by Choongbum Lee were at the right level of formalism versus intuition for me. He does a good job of explaining

https://www.youtube.com/playlist?list=PLUl4u3cNGP63ctJIEC1UnZ0btsphnnoHR

In lecture 5, Lee goes over the basics of stochastic processes. Then the real materials starts in lecture 17 where he goes over continuous time processes and derives Ito's Lemma. Lecture 18 continues with Ito's lemma, integration of the Ito integral, and finally a discussion on Girsanov's theorem.

After watching these videos I feel a lot more comfortable reading Evan's book on SDEs or Kloeden, etc.


答: Each subset of $\mathbb{N}$ uniquely defines a step function (it corresponds to the locations of the $+1$ jumps in value), and the set of subsets of $\mathbb{N}$ is uncountable.


问:
prove that $E(X^{n}) = 0 $
Prove that for any odd $n \in \mathbb{N}$ and $X \sim \mathcal{N}(0,\sigma^{2})$, $E (X^n) = 0$.

I started with the fact that:
$$E(X^{n}) = {{1}\over{\sqrt{2\pi} \sigma}} {\int_{-\infty}^{\infty} x^{n} e^{{-x^2}/{2\sigma^2}}} dx = {{2}\over{\sqrt{2\pi} \sigma}} {\int_{0}^{\infty} x^{n} e^{{-x^2}/{2\sigma^2}}} dx.$$
Then I used the substitution $$t={{x}\over{\sigma}} \to {{2\sigma^n}\over{\sqrt{2\pi}}} {\int_{0}^{\infty} t^{n} e^{{-t^2}/{2}}} dt.$$
But I have a problem how to solve it. Does anyone know how this equality can be proved? The task tip says to use the Gamma function in the calculation, but I don't see it..

答:
In fact, for any symmetric distribution $X$, $E[X^{2n+1}] = 0$ for all $n \in \mathbb{N}$.  Simply split into positive and negative parts and observe that the probabilistic density function $f_X$ is symmetric on $\mathbb{R}$.
$$\begin{aligned}
E\left[X^{2n+1}\right] &= \int_{\mathbb{R}} x^{2n+1}f_X(x) \,\mathrm{d}x \\
&= \int_{-\infty}^0 x^{2n+1} f_X(x)  \,\mathrm{d}x + \int_0^{+\infty} x^{2n+1} f_X(x)  \,\mathrm{d}x \\
&= 0
\end{aligned}$$

In the first integral, make the substitution $y = -x$, and you'll see that it cancels the second integral.

$$\int_{-\infty}^0 x^{2n+1} f_X(x) \,\mathrm{d}x = \int_{0}^{+\infty} -y^{2n+1} f_X(y) \,\mathrm{d}y$$

Title: prove that $E(X^{n}) = 0 $


问:
Forms of integral equality
I am trying to solve this integral equality however I believe there is no closed form for the value of a that satisfies it. However i don't know how to create a sum that converges relatively fast. Here is how far i got:
$$\int_{0}^{1}\left(\int_{0}^{1}\frac{\left(1-t^{x}\right)}{1-t}dt\right)dx=\int_{0}^{1}\frac{\left(1-t^{a}\right)}{1-t}dt$$
$$\int_{0}^{1}\frac{1-t+\ln\left(t\right)}{\left(1-t\right)\ln\left(t\right)}dt-\int_{0}^{1}\frac{\left(1-t^{a}\right)}{1-t}dt=0$$
$$\int_{0}^{1}\left(\frac{1-t+\ln\left(t\right)}{\left(1-t\right)\ln\left(t\right)}-\frac{\left(1-t^{a}\right)}{1-t}\right)dt=0$$

答:
$$\int_0^1 \frac{1-t^x}{1-t} \, dt=H(x)$$
$H(x)$ is the harmonic number

$$H(n)=\sum_{n=0}^{\infty}\frac{1}{n}=\int _{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx$$

It can be proved that $H(x)=\psi ^{(0)}(x+1)+\gamma$

where $\psi ^{(0)}$ is the polygamma function
$$\int_0^1 \int_0^1 \frac{1-t^x}{1-t} \, dt= \int_0^1H_x \, dx=\gamma$$
RHS is
$$\int_0^1 \frac{1-t^a}{1-t} \, dt=H(a)$$
therefore equality is true if
$$H(a)=\gamma$$
$$\psi ^{(0)}(a+1)+\gamma =\gamma$$
$$\psi ^{(0)}(a+1)=0$$
There is no closed form for the inverse of the polygamma function, so I found an approximate value for positive $a$
$$a\approx 0.461632$$

Edit

$\psi (1)=-\gamma;\;\psi (2)=1-\gamma$

therefore there exists $a\in (0,1)$ such that $\psi^{(0)}(a+1)=0$

With Newton method, starting from $x_0=0.5$

$$
\begin{array}{l|r|r}
 n & x_n & error\\
\hline
 0 & 0.5 & 0.039035 \\
 1 & 0.460965 & 0.000666909 \\
 2 & 0.4616319 & 2\times 10^{-7} \\
 3 & 0.4616321449683 & 1.9\times 10^{-14} \\
\end{array}
$$

Title: Forms of integral equality


问:
When can we ignore small-oh inside of a limit and when we cannot?
Here's an example.
$$\lim_{x \to 0} \frac {\ln\left(1 + x + \sqrt{1 + x^2} - 1\right) - x}{x^3}.$$
$\ln(1 + x + \sqrt{1 + x^2} - 1) \sim  x + \sqrt{1 + x^2} - 1$ (the limit of their division is equal to 1). And so the left side we can replace with $x + \sqrt{1 + x^2} - 1 + o(x + \sqrt{1 + x^2} - 1)$. Now if we write it inside of the limit we'll have
$$
\lim_{x\to0} \frac {x + \sqrt{1 + x^2} - 1 + o\left(x + \sqrt{1 + x^2} - 1\right) - x}{x^3}
$$
Here we can cancel the two x-es. Now what I'm wondering about is why can't I ignore the small-oh in this limit? In fact the limit of the expression is not the same with and without the small-oh (with small-oh I mean the $o(x + \sqrt{1 + x^2} - 1)$).
And more generally where can we ignore and when we cannot?
For example
If we have
$\lim_{x \to 0} \dfrac {\ln(1+x)} {x}$ here
$\ln(1+x) = x + o(x)$ and so if we ignore $o(x)$ then we'll have this limit $\lim_{x \to 0} x/x = 1$ which's correct. So what's the difference and how can I detect it in problems to avoid errors?

答: Well, it may sound a bit redundant, but you can ignore the small-oh when the order of convergence of $o$ is smaller than that of the denominator, what I'm saying is, if you have $f(x)=g(x)+o(x)$ where $\lim_{x\to a}o(x)=0$ then $\lim_{x\to a} \frac{f(x)}{h(x)}=\lim_{x\to a}\frac{g(x)}{h(x)}$ if $\lim_{x\to a}\frac{o(x)}{h(x)}=0$, the better $g$ aproximates $f$ then the statement $\lim_{x\to a} \frac{f(x)}{h(x)}=\lim_{x\to a}\frac{g(x)}{h(x)}$ will be true for more $h$, but there is no universal rule because the only perfect aproximation of $f$ is $f$.


问:
If all points of a coordinate nbhood are parabolic, then the unique asymptotic curve is a straight line

Do Carmo's exercise: Let $p$ be a point of an oriented surface $S$ and assume that there is a neighborhood of $p$ in $S$ all points of which are parabolic. Prove that the (unique) asymptotic curve through $p$ is an open segment of a straight line.

At the end of the book the following hint is given:

Parametrize a neighborhood of $p\in S$ in such a way that the lines of curvature are the coordinate curves and that $v= const. $ are the asymptotic curves. It follows $e_v=0$, and from the Mainard-Codazzi equations, we conclude that $E_v= 0$. This implies that the geodesic curvature of $v=const.$ is zero.

Since all the points are not umbilical, I know that we may parametrize a neighborhood of $p$ in such a way that the coordinate curves are lines of curvature. I can't see though why we can make $v=const.$ be also an asymptotic curve. Assuming that we may do this, I also can't see why it implies $e_v=0$. The remaining affirmations I agree, after all the coordinate curves being lines of curvature implies $f=F=0$ and hence from the Mainard equations we have $0 = e_v=E_v/2(e/E+g/G)$ which implies $E_v=0.$ Since the geodesic curvature of the $u$-coordinate curve is $-E_v/(2E\sqrt G),$ it follows that the coordinate curve is geodesic and asymptotic simultaneously. Therefore, by another exercise in this same section, we conclude that it is a segment of a line.

Any help in clarifying these claims is much appreciated

答:
Correct me if I am wrong, the following notations agree with those in Do Carmo's textbook.

Since $u=const.$ is line of curvature and $v=const.$ is asymptotic curve, it follows $X_u$ is asymptotic direction, and $X_v$ is principle direction, therefore $N_v=\lambda X_v$ for some real valued function $\lambda$, also $II(X_u)=\langle -dG(X_u),X_u\rangle=\langle -N_u,X_u\rangle$=0. By definition, $e=\langle N,X_{uu}\rangle$, thus
\begin{align*}
e_v&=\langle N_v,X_{uu}\rangle+\langle N,X_{uuv}\rangle=\lambda\langle X_v,X_{uu}\rangle-\langle N_u,X_{uv}\rangle\\
&=\lambda\langle X_v,X_{uu}\rangle-\langle N_u,X_u\rangle_v+\langle N_{uv},X_u\rangle\\
&=\lambda\langle X_v,X_{uu}\rangle-(0)_v+\langle (\lambda X_v)_u,X_u\rangle\\
&=\lambda\langle X_v,X_{uu}\rangle+\lambda\langle X_{uv},X_u\rangle+\lambda_u\langle X_v,X_u\rangle\\
&=\lambda(\langle X_v,X_{uu}\rangle+\langle X_{uv},X_u\rangle)+\lambda_uF=\lambda F_u+\lambda_u F=0
\end{align*}
Then the result follows by the op's argument.


问:
Solving a cubic equation
I have a math problem in front of me, and after struggling three days with it, I cannot find the solution. I need to find the roots of this polynomial equation: $x^3+x^2+1=0$.

Anyone have a solution to this?

Any help is appreciated..

Title: Solving a cubic equation

Body:
Solved the equation using Tartaglia's method
$$ x^3 + x^2 + 1$$
Make then substitution $ x = t + h $
$$ t^3 + 3t^2h+3th^2 + h^3 + t^2 +2th+h^2 + 1 = 0$$
$$ t^3 + t^2(3h+1)+t(3h^2+2h) + (h^3 + h^2 + 1) = 0$$

In order to eliminate the second degree term we add a condition $ h  = -\frac{1}{3}$

$$ t^3 -\frac{1}{3}t + \frac{29}{27} $$

We make $t = u+v$

$$ u^3 + 3u^2 v + 3uv^2 + v^3 -\frac{1}{3}u - \frac{1}{3}v + \frac{29}{27}=0$$
$$ u^3 + v^3 + \frac{29}{27} + 3u^2 v -\frac{1}{3}u + 3uv^2  - \frac{1}{3}v =0$$
$$ u^3 + v^3 + \frac{29}{27} + u\left(3u v -\frac{1}{3}\right) + v\left(3uv  - \frac{1}{3}\right) =0$$
$$ u^3 + v^3 + \frac{29}{27} + (u+v)\left(3u v -\frac{1}{3}\right) =0$$

We make a system of equation
$$u^3 + v^3 + \frac{29}{27} = 0; (u+v)\left(3u v -\frac{1}{3}\right) =0$$

For (2) we have:
$ u \ne v $ because does not satisfy (1), then
$$u = \frac{1}{9v} \rightarrow u^3 = \frac{1}{729v^3}$$

For (1) we have:
$$u^3 + v^3 + \frac{29}{27} = 0$$
$$\frac{1}{729v^3} + v^3 + \frac{29}{27} = 0$$
$$\left(v^3\right)^2 + \frac{29v^3}{27} + \frac{1}{729} = 0$$

Using  general quadratic formula:

$$ v^3 = \frac{-\frac{29}{27} \pm \sqrt{\left(\frac{29}{27} \right)^2 - 4\frac{1}{729} }}{2}$$
$$ v^3 = -\frac{29}{54} \pm \frac{1}{2}\sqrt\frac{31}{27} $$

We hope $u^3$ to be the conjugate of $v^3$ because they are interchangeable. (Try solving $v^3$ to prove it).

So, $$u+v= \sqrt[3]{-\frac{29}{54} + \frac{1}{2}\sqrt\frac{31}{27}} + \sqrt[3]{-\frac{29}{54} - \frac{1}{2}\sqrt\frac{31}{27}}$$

So, we can now have the value of $x$:
$$ x = t + h = u + v + h = \sqrt[3]{-\frac{29}{54} + \frac{1}{2}\sqrt\frac{31}{27}} + \sqrt[3]{-\frac{29}{54} - \frac{1}{2}\sqrt\frac{31}{27}} - \frac{1}{3}$$

$$ x \approx -1.46557 $$

Good luck finding the exact complex solutions ._.


问:
Summation of n(2^n)
I was doing a question in which i had to find the summation of the expression $n(2^n)$ from n=1 to n=9. </p>

<p>I used wolfram alpha to calculate thid sum, but i was wondering if there is an easier way to calculate it? </p>

答:
Let's assume n is finite. We will now derive an explicit formula for calculating the desired sum for any $x$, you can plug in $2$ into final formula.
 The sum: $S_1=\sum_{k=0}^{n} kx^{k}$ looks a lot like: $S_2=\sum_{k=0}^{n} x^{k}$. $S_2$ is of course $\mathbb{geometric}$ series: $S_2 = \frac{1-x^{n+1}}{1-x}$.
Since n is finite, we can safely take the derivative of $S_2$: $S_3 = \frac{d}{dx}(\sum_{k=0}^{n} x^{k})=\sum_{k=0}^{n} kx^{k-1}=\frac{d}{dx}(\frac{1-x^{n+1}}{1-x})=...=\frac{-nx^{n}+nx^{n+1}-x^{n}+1}{(1-x)^2}$.

We wanted a sum of $kx^k, x = 2$ and we derived explicit formula for sum of $kx^{k-1}$. We now just need multiply $S_3$ by $x$: $xS_3 = x\sum_{k=0}^{n} kx^{k-1} = \sum_{k=0}^{n} kx^{k} = S_1 = x\frac{-nx^{n}+nx^{n+1}-x^{n}+1}{(1-x)^2} = \frac{-nx^{n+1}+nx^{n+2}-x^{n+1}+x}{(1-x)^2}$.

For $x=2$ we get: $\sum_{k=0}^{n} k2^{k} = -n2^{n+1}+n2^{n+2}-2^{n+1}+2 = 2^{n+1}(n-1)+2$

Title: Summation of n(2^n)


问:
Are limits of functions different in the complex plane
I recently learnt about complex numbers and this got me thinking about the limits of functions for example when we say that $\lim\limits_{x \to 4} x^2=16$ we assume that $x^2$ is a function defined for all real numbers and we consider the limit as $x$ approaches $4$ from any direction within the domain of $x^2$ .however why don't we consider the limit as $x$ approaches $4$ from the complex numbers is it because $x^2$ is not defined for complex numbers?

答: You can think of convergence in $\mathbb C$ as convergence in $\mathbb R^2$ (they have the same topology). So if you take $\lim_{z\to a}f(z)$ where $f(z)$ is a complex function and $a$ is real and this limit exists, then it doesent matter how you aproach $a$ it will have the same limit so if you take the restriction of $f$ to the real line, namely $g=f|_{\mathbb R}$ then $\lim_{z\to a}f(z)=\lim_{x\to a}g(x)$ where the second limit is on the real line.


问:
Range space of $A$ and Null space of $A^*$
Let $A$ be $p \times q$ matrix with complex entries. Then we define $R(A) = \{Ax : x \in \mathbb{C}^q\, \}$ and $Ker(A) = \{ x \in \mathbb{C}^q : Ax = 0\}$. Let $A^*$ denote conjugate transpose of $A$ i.e. Hermitian operator.

I have to prove that $R(A) \cap Ker(A^*) = \{0\}$

To do this $y$ be in the intersection. Then there is a $x$ in $\mathbb{C}^q$ such that $y = Ax$ and $A^*y=0$. Then $A^*(Ax)=0$. How can I proceed further?

答:
Hint: consider $\langle y, y \rangle$ making use of the facts:

$y = Ax$

$A^*y = 0$

$\langle Au, v \rangle = \langle u, A^*v \rangle$ for any $u, v$.


问:
Modular arithmetic $7^{60} \pmod{77}$

How do you solve $7^{60} \pmod{77}$?

For $5^{60} \pmod{77}$, it's equal to $1$ because I think the $\gcd(5,77) = 1$ and using the FLT we can find that $5^{(7-1)(11-1)} = 1 \pmod{7*11}$. But how do you get $56$ from $7^{60} \pmod{77}$?

答:
First, $7^{60}\equiv 0\pmod{7}$. Then:

$$7^{60}\equiv \big(7^{10}\big)^{6}\equiv \big(7^{\phi(11)}\big)^{6}\equiv 1^{6}\equiv 1\pmod{11}$$

Then, $7^{60}\equiv 0\pmod{7}$ and $7^{60}\equiv 1\pmod{11}$, so $7^{60}$ is equivalent to a unique value $\pmod{77}$ by the Chinese Remainder Theorem. And since $56\equiv 0\pmod{7}, 1\pmod{11}$, we must have $\boxed{7^{60}\equiv 56\pmod{77}.}$

Title: Modular arithmetic $7^{60} \pmod{77}$


问:
Sign change of a polynomial between roots
Let $p(x)$ be a polynomial with $n$ distinct real roots. Label the roots of $p(x)$ as $x_1, x_2, ...,x_n,$ such that $x_1 < x_2 < ... < x_n$. Suppose that $p(x)$ has a root that is also a stationary point, i.e. $\exists \ i \in \{1,2, ... ,n\}$ such that $p(x_i) = p'(x_i)=0$. Is it then true that the sign of $p(x)$ over the interval $\left(x_{i-1},x_i\right)$ is the same as the sign of $p(x)$ over $\left(x_{i},x_{i+1}\right)?$ Intuitively this seems true, but does a rigorous proof for this exist?

Edit: As commented in one of the answers, this claim is not always true. In particular, if one has the polynomial $p(x)=x^3(x-1)(x+1)$, where the root $x=0$ is a point of inflection, then there is a sign change. Thus a root which is a stationary point does not imply that there is a sign change. However, would the converse be true then? If there is no sign change at a root, does that mean that the root is a stationary point?

答:
Well, in case of a double root at $x_i$, the sign does not change. This happen for each root which is two-fold, four-fold and so on.
The condition $p(x_i)=p'(x_i)=0$ and $p''(x_i)\ne 0$ means that the root $x_i$ is two-fold.


问:
Finding the $p$ norm of a row matrix
Let $u \in \mathbb{C}^n$. It is well-known that $||u^*||_2 = ||u||_2$, where the norm on the LHS is the matrix norm and the norm on the RHS is the vector norm. I was trying to see if we can extend this to general $p$ norms. By Holder's inequality we have, for $x \in \mathbb{C}^n$ with $||x||_p = 1$,
$$|u^*x| \le ||u||_{\frac {p}{p-1}}||x||_p = ||u||_{\frac {p}{p-1}}$$

So $||u^*|| \le ||u||_{\frac {p}{p-1}}$. But I have not been able to show equality.

答:
Let $q = \frac p{p-1}$. For $p \notin \{1,\infty\}$, take $x = (x_1,\dots,x_n)$ with
$$
x_j = \begin{cases}
\|u\|_q^{1-q} \cdot |u_j|^q/u_j & u_j \neq 0\\
0 & u_j = 0.
\end{cases}
$$
Verify that we have $\|x\|_p = 1$ and $u^*x = \|u\|_q$.

Title: Finding the $p$ norm of a row matrix


问:
Raising the floor function to a power
I've made a few plots and noticed that $\lfloor

Is it true that for positive $x > 1$ and $n \in \mathbb N,\quad n>=2$ the following holds:

$$(\lfloor x \rfloor + 1)^n >= \lfloor x^n \rfloor $$

If it is, how can it be proven?
If it is not, will that at least hold when $n=2$?
I am interested in the latter case, actually.

Thank you!

答: Hint: $\lfloor x \rfloor + 1 \ge x$</p>

Title: Raising the floor function to a power

Body: Hint: $\lfloor x \rfloor + 1 \ge x$


问:
Fourier series for the solution of $\frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0$
Consider the following problem:

Let $\Omega = (0, \pi) \times (0, 1)$ and let $u$ be a solution of
$$
\begin{cases}
\frac{\partial^2 u}{\partial x^2} + y^2 \frac{\partial^2 u}{\partial y^2} = 0 \quad \text{ in } \Omega \\
u(0, y) = u(\pi, y) = 0 \quad \text{ for } 0 \leq y \leq 1
\end{cases}
$$
Expanding $u$ in a Fourier series,
$$
u(x, y) = \sum_1^\infty c_k(y) \sin(kx)
$$
Determine the coefficients $c_k(y)$ and show that every bounded solution $u$ satisfies
$$
u(x, 0) = 0 \quad \text{ for } 0 \leq x \leq \pi.
$$

I don't know much about Fourier series, so I tried to follow Strauss' PDE book. I began by separating variables:
$$
u(x, y) = f(x)g(y),
$$
which gives us the differential equations
$$
f''(x) + \lambda f(x) = 0, \quad y^2 g''(y) - \lambda g(y) = 0.
$$
From the first equation and the given boundary conditions, it is easy to see where the factors $\sin(kx)$ come from. Now, Wolfram Alpha gives
$$
g(y) = c_1 y^{\frac 12 - \frac 12 \sqrt{4k^2 + 1}} + c_2 y^{\frac 12 + \frac 12 \sqrt{4k^2 + 1}}
$$
for the second equation (already using that $\lambda = k^2$). The problem is that I am out of ideas about how to compute the constants $c_1$ and $c_2$ (which is a necessary step for the second part of the problem, I believe), since there are no other boundary conditions given.

Please, what am I missing? Any hints will be appreciated.

Thanks in advance.

答:
Denoting

$$\displaystyle f_k(x)=a_k\sin kx,\quad g_k(x)=b_{1,k}y^{\frac{1+\sqrt{1+4k²}}{2}}+b_{2,k}y^{\frac{1-\sqrt{1+4k²}}{2}}, $$

and

$$u_k(x,y)=f_k(x)g_k(y), $$

by the superposition principle, the solution is

$$u(x,y)=\sum_{k=1}^{\infty}u_k(x,y)=\sum_{k=1}^{\infty}a_k\sin kx\cdot g_k(y)=\sum_{k=1}^{\infty}c_k(y)\sin kx, $$

where $$c_k(y)=a_kg_k(y) $$

So,

$$u(x,0)=\sum_{k=1}^{\infty}c_k(0)\sin kx.. $$

Since $$c_k(0)=a_k g_k(0)=0 $$

we can conclude that

$$u(x,0)=0 $$

-----------EDIT-------------

What I wrote about $c_k(0)$ is wrong, because you would be dividing by zero.

First, you need bounded solutions. Since the sine terms are bounded, we actually need to look $c_k(y).$

For a expression like $y^r$ be bounded in the the interval $(0,1)$ we need $r\geq 0$, so I will apply this idea to the two exponents terms in $c_k(y)$

The first one:

$$ \frac{1+\sqrt{1+4k^2}}{2}\geq 0 \iff \sqrt{1+4k^2}\geq -1$$

This is satisfied for all $k$, so, these terms are bounded.

Now, I will look at the second term. We need

$$\frac{1-\sqrt{1+4k²}}{2}\geq 0
\implies \textrm{some manipulation} \implies k^2\leq 0 \implies k=0
 $$

But this is a contradiction, because $k$ is a positive integer. So none of these therms are bounded, therefore we need $b_{2,k}=0$

Conclusion:

$$c_k(y)=a_k b_{1,k}y^{\frac{1+\sqrt{1+4k^2}}{2}}\implies c_k(0)=0.$$

Now we can conclude $$u(x,0)=0 $$


问:
Does a bounded from below linear operator with target set equal to the domain always have closed image?
I'm using the definition of a bounded from below linear operator as a linear operator $X\overset{T}{{\to}}Y$, where $X$ and $Y$ are Banach spaces and the operator satisfies: for all $x \in X$, $\lVert T(x) \rVert \geq C \lVert x \rVert$, for some constant $C \gt 0$.

In this question i will refer only to such an operator in a Banach space $B$, $ T \colon B{{\to}}B$.

Note that I don't assume it to be continuous. With only these conditions, can we guarantee that the image of $T$ is closed in $B$? (Which is equivalent to asking if these assumptions imply continuity of the operator, by the open mapping theorem applied to the inverse of the operator, which exists since $\lVert T(x) \rVert = 0 \Rightarrow \lVert x \rVert = 0$).

I have tried to prove that it is closed with only these conditions and checked out other posts that have similar questions, but they don't seem to answer this one, at least in an immediate way. At this point, I believe that it is not generally true, but I also haven't been able to write a counter-example.

I appreciate any help and thank you in advance!

答:
Reardless of my comment above, the result is false, and it is in fact not so difficult to give a counter example.

Let $H$ be a separable, infinite dimensional Hilbert space with basis
$$
  E=\{e_n\}_{n\in \mathbb N}.
  $$

Consider  a Hamel basis for $H$ containing the linearly independent set $E$,  which we will write as the disjoint union
$$
  E\cup F.
  $$

Defining
$$
  N = \{(e_n, ne_n):n\in \mathbb N\},
  $$
we leave for the reader to verify  that the subset $C$ of $H\times H$ defined below is  linearly independent:
$$
   C = (0\times  E) \cup  (0\times  F) \cup  N.
  $$

Next consider a Hamel basis $ D$  of $H\times H$ containing $ C$, which we write as the disjoint union
$$
   D =\underbrace{(0\times  E) \cup  (0\times  F) \cup  N}_{ C} \ \cup  \ M,
  $$
and define
$$
  K= \text{span} (N\cup M).
  $$
where we insist that the span refers to the linear span,  no closure.
It is then easy to see that
$$
  H\times H = (0\times H) \oplus  K,
  $$
where "$\oplus $" is to be interpreted as algebraic direct sum, as opposed to orthogonal
direct sum.

For that reason, the canonical projection $\pi _1: H\times H\to H\times 0$,  once restricted to $K$ becomes a bijection,  and
hence we may speak of its inverse
$$
  \pi _1^{-1}: H\times 0 \to  K\subseteq H\times H.
  $$

Being the inverse of a bounded bijective map,  it is clear that   $\pi _1^{-1}$ is bounded below,  but it is certainly
discontinuous since
$$
  \|\pi _1^{-1}(e_n)\| = \|(e_n,  ne_n)\| = \sqrt{1+n^2}.
  $$

As noted by the OP, the range of   $\pi _1^{-1}$ cannot be closed.


问:
Book about semimartingales and stochastic integral w.r.t semimartingales
Is there an English version of Dellacherie C and Meyer P ,,probabilites et potentiel''? Or what books on semimartingals and stochastic calculus do you recommend?</p>

答: There is an English version, it can be found here .</p>


问:
Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$

For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$

This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.

Is my reasoning fully acceptable throughout my solution?

答:
If $A(z_A),B(z_B),M(z_M)$ are distinct points in Argand plane, then $\arg \left(\frac{z_M-z_B}{z_M-z_A}\right)$ is an oriented angle between vectors $(\vec{AM},\vec{BM}).$

In the present case $z_M=z,z_A=2,z_B=8.$

$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ says that the vectors are orthogonal, and $M$ lies on an open semi-circle with diameter $AB.$

(It is the upper half of the circle with diameter $AB,$ without $A,B.$)


问:
Is my proof correct? 'let a,b∈ Z. We write A | B if A divides B. Is the relation |, symmetric, transitive and/or reflexive?'
The relationship is not symmetrical. When a relationship is symmetrical: if xRy implies yRx for all x, y ∈ A (where A is a non-empty set, and R is a relation in A)

If a, b ∈ Z, and as a | b means that there is an integer k such that b = a k ∙. When we take a = 1 and b = 2 applies: b = a ∙ k → k = 2. So for a = 1 and b = 2 holds a | b, for k ∈ Z.
If a, b ∈ Z, and if b | a is that there is an integer k such that a = b ∙ k. When we take again a = 1 and b = 2, a = b ∙ k → k = 1/2. So for a = 1 and b = 2 is not that b | a ┤ because k ∉ Z.
This implies that f is not symmetrical.

The relationship is transitive. When a, b, c ∈ Z, and as a | b and b | a, is that there is an integer k such that b = a ∙ k and l is an integer that there exists such that c = b ∙ l. This means that
c = a b ∙ ∙ l. Because a, b, c ∈ Z means that the b ∙ l ∈ Z (an integer multiplied by an integer indicates an integer). This means that there exists an integer m (where m = a ∙ b) such that c = m ∙ A. It follows that a | c, so the relationship is the transition.

The relationship is reflexive when (∀a ∈ Z)a|a. because every a ∈ Z can be divided by itself, we say that it is reflexive.

After Looking it through again I now have my proof like this:

(a) Since 0 does not divide 0,  "|" is not reﬂexive.
(b) 2 divides 4 so 2 | 4. But 4 does not divide 2, so 4 does not divide 2. Thus, "|" is not symmetric.
(c) To see that  is transitive, let a, b, c be integers. Suppose that a| b and b |c. Thus,
a divides b and b divides c so there exist integers k and l such that b = ak and c = bl. This
gives c = bl = (ak)l = a(kl). Therefore, a divides c so a | c.

Is this prove correct?

答:
Answer: Not reflexive, not symmetric, not anti-symmetric, transitive

Reason:

Reflexive: for all x ∈ Z, R(x,x) is reflexive, but here R(0,0) is a violation (0/0 is undefined), as 0 belongs to the set of integers but does not satisfy this relation.

Symmetric: for all x ∈ Z, R(x,y)and R(y,x) is symmetric and clearly this relation cannot be symmetric. Considering the example {1,2}, for this to be symmetric (1,2)(2,1) both ordered pair should be present but it is also a violation (2 can be divided by 1 but 1 cannot be divided by 2).

Anti-symmetric: for all x ∈ Z, R(x,y) and R(y,x) then x=y is anti-symmetric. But here, we can easily make a violation as R(-2,2) and R(2,-2) are not anti-symmetric (-2/2 or 2/-2 is -2).

Transitivity: for all x ∈ Z , R(x,y), R(y,z) then R(x,z). This is always true for the above divides relation.

So the given relation "|" is not reflexive, not symmetric, not anti-symmetric but transitive only.


问:
Powers mod prime

I want to find the powers of $2$, $2^{20}$ mod a prime $53$

Is there a way to simplify this problem? I know I could take $(2^{5})^4$ but this is already taking $32^4$ mod $53$ which isn't easy without a calculator.

答:
Take the complement, factorize, and also keep the number small should do the trick.

\begin{align}
2^{20} &\equiv 32^4 \pmod{53} \\
&  \equiv (-21)^4 \pmod{53}\\
&\equiv 7^4 \cdot 3^4 \pmod{53} \\
&\equiv 49^2\cdot (81 - 53) \pmod{53}\\
&\equiv (-4)^2 \cdot(28) \pmod{53}\\
&\equiv 7 \cdot 4^3 \pmod{53}\\
&\equiv 7 \cdot (64-53) \pmod{53}\\
&\equiv (77-53) \pmod{53}\\
&\equiv 24 \pmod{53}
\end{align}

Title: Powers mod prime


答:
Fleshing out a comment, it's easy to see that $2^8=256=265-9=5\cdot53-9\equiv-9$ mod $53$. Thus $2^{16}\equiv81\equiv28$ mod $53$, so $2^{17}\equiv56\equiv3$ mod $53$, so finally $2^{20}\equiv8\cdot3=24$ mod $53$.

The trick here is looking for "easy" steps that give small remainders mod $53$. The search for easy steps can take as much time as simply taking a brute-force (or, somewhat more slickly, a repeated squaring) approach, but can be fun (and instructive) in its own right.

Added later: Another way to get from $2^8$ to $2^{20}$ using $265\equiv0$ mod $53$ can be summarized as

$$2^8=256\equiv-9\implies2^{10}\equiv-36\equiv17\implies2^{20}\equiv289\equiv24$$

(This is similar to Bill Dubuque's derivation, but differs in the way the congruence $2^{10}\equiv17$ is obtained. It also assumes $17^2=289$ is an "easy" square.)

Title: Powers mod prime

Body:
I want to find the powers of $2$, $2^{20}$ mod a prime $53$

Is there a way to simplify this problem? I know I could take $(2^{5})^4$ but this is already taking $32^4$ mod $53$ which isn't easy without a calculator.


答:
To do this we can make use of the identity: (a*b)%m = [(a%b)^2]%b

If we say that a=32^2=1024 and b=53, we get [(1024%53)^2]%53

53 * 19 = 1007 which means that 1024%53=17
Then we do 17^2=289and 289%53 = 24
Therefore your answer is 24.

Title: Powers mod prime


问:
How can you use a (fair) coin to draw straws among 3 people? (Information Theory)
The following nice riddle is a quote from the excellent, free-to-download book: Information Theory, Inference, and Learning Algorithms, written by David J.C. MacKay.

How can you use a (fair) coin to draw straws among 3 people?

答: To expand on Tobias' comment, designate that Player 1 wins on TT (Tails followed by Tails), Player 2 wins on TH, and Player 3 wins on HT. On HH, simply discard the result and flip again. Since all outcomes are equally likely and each player has a single winning outcome, the winner will be chosen uniformly at random.</p>


答: I happen to know that $2^{20}=1048576$, so I used long division to compute the remainder of $1048576\div53$ and got remainder $24$.  It really wasn't hard to do with pen and paper!

Title: Powers mod prime


问:
Locally Lipschitz with respect to a variable and uniformly respect the other
Let:

$\mathbf{f}:D\subseteq \mathbb{R}^{n+1} \to  \mathbb{R}^{n} \ \  \ D \ \text{open}$

$\  \ \ \ \ \ (t,\mathbf{y}) \mapsto \mathbf{f}(t,\mathbf{y})$

Where $t\in \mathbb{R}$ and $\mathbf{y} \in \mathbb{R}^{n}$.

$\mathbf{f}$ is locally Lipschitz in $D$ with respect to $\mathbf{y}$ and uniformly respect $t$ if and only if:
$$\forall (t,\mathbf{y})\in D   \ \exists B_r(t,\mathbf{y}) : \exists L\in\mathbb{R}_+ : \ \forall (t,\mathbf{z}) \in B_r(t,\mathbf{y}) \\ ||\mathbf{f}(t,\mathbf{y})-\mathbf{f}(t,\mathbf{y})||\leq L||\mathbf{y}-\mathbf{z}||$$
Where $L$ doesn't depend upon $t$.

The following proof is left as exercise to the reader

Let  $\mathbf{f}(t,\mathbf{y})=(f_1(t,\mathbf{y}),...,f_n(t,\mathbf{y}))$ and $\mathbf{y}=(y_1,...,y_n)$ then:
$$\mathbf{f}\text{ and }\frac{\partial f_j}{\partial y_s } \ \text{continous in } D \ \  \forall j,s\in\{1,...,n\} \implies \mathbf{f} \text{  is locally Lipschitz in }D\text{ with respect to }\mathbf{y}\text{ and uniformly respect }t $$

I'm trying to prove it. Here is my attempt:

Lemma

Since mean value theorem is not true for vectorial functions, my idea was to use this weaker inequality:

$\mathbf{g}:A\subseteq \mathbb{R}^n \to \mathbb{R}^n \ \ \ \mathbf{g}\in C^1(A)$

$\text{The segment } [\mathbf{y},\mathbf{z}]\subset A$

$\implies ||\mathbf{g}(\mathbf{y})-\mathbf{g}(\mathbf{z})||\leq \sqrt{n}\alpha||\mathbf{y}-\mathbf{z}||$

Where $$\alpha=\max \left\{\max_{[\mathbf{y},\mathbf{z}]} (||\nabla g_1(\mathbf{x})||),\max_{[\mathbf{y},\mathbf{z}]} (||\nabla g_2(\mathbf{x})||),...,\max_{[\mathbf{y},\mathbf{z}]} (||\nabla g_n(\mathbf{x})||)\right\}$$
The existences of the maximum of the $||\nabla g_j||$ is granted by Weierstrass Theorem, since $[\mathbf{y},\mathbf{z}]$ is compact.

Attempt of proof

I can define a family of functions $\mathbf{g}_t: \mathbf{y} \mapsto \mathbf{f}(t,\mathbf{y})$(I'll indicate $g_{t,j}$ its $j$-th component). One can easily verify the following:

$$ \nabla f_j(t_0,\mathbf{y}_0)=(D_t f_j(t_0,\mathbf{y}_0),\nabla g_{t_0,j}(\mathbf{y}_0)) \ \ \forall(t_0,\mathbf{y}_0)\in D $$

In fact:

$$D_{y_s} f_j(t_0,\mathbf{y}_0)=\lim_{h \to 0} \frac{f_j(t_0,\mathbf{y}_0+h\mathbf{e}_s)-f_j(t_0,\mathbf{y}_0)}{h}=\lim_{h \to 0} \frac{g_{t_0,j}(\mathbf{y}_0+h\mathbf{e}_s)-g_{t_0,j}(\mathbf{y}_0)}{h}=D_{y_s} g_{t_0,j}(\mathbf{y}_0)$$

This means that the following implication holds:
$$D_{y_s} f_j(t,\mathbf{y})\leq N \ \ \ \forall(t,\mathbf{y})\in K\subset D \implies ||\nabla g_{t,j}(\mathbf{y})||\leq M  \ \ \ \forall(t,\mathbf{y})\in K $$
In particular the LHS proposition is true for every $j$(and this obviously makes true also RHS for every $j$) by Weierstrass theorem if I choose $K$ to be a closed ball of generic center $(\overline{t},\mathbf{\overline{y}})$(this ball  exists for every choice of  $(\overline{t},\mathbf{\overline{y}})$, since $D$ is open).

After this premises, since $D_{y_s} \mathbf{g}_t:t \mapsto D_{y_s} \mathbf{f}(t,\mathbf{y})  $ and thanks to our hyphothesis it's easy to see that:

$\mathbf{g}_t:A_t\subseteq \mathbb{R}^n \to \mathbb{R}^n \ \ \ A_t=\{\mathbf{y}:(t,\mathbf{y})\in D \} $

$\mathbf{g}_t \in C^1(A_t)$

Moreover since a ball is connected $\forall (t,\mathbf{z})\in K \ \ $ the segment  $[ (\overline{t},\overline{\mathbf{y}}),(t,\mathbf{z})]\subseteq K $ and this implies that $[\mathbf{\overline{\mathbf{y}}},\mathbf{z}]\subseteq A_t$(this can be easily proved showing that the pythagorean distance of $(\overline{t},\overline{\mathbf{y}})$ from $(t,q\overline{\mathbf{y}}+(1-q)\mathbf{z})$ with $q\in [0,1]$ is $\leq$ than the distance $d((\overline{t},\overline{\mathbf{y}}),(t,\mathbf{z}))$) , so by our lemma:
$$\forall (t,\mathbf{z})\in K \ \ ||\mathbf{g}_t(\mathbf{\overline{y}})-\mathbf{g}_t(\mathbf{z})||\leq \sqrt{n}\alpha_t||\mathbf{\mathbf{\overline{y}}}-\mathbf{z}||$$

If we prove that $\alpha_t$ is bounded we are done because we substitute it with its upper bound, so that the inequality doesn't depend upon $t$.Notice that:

$$\alpha_t=\max \left\{\max_{[\mathbf{\overline{y}},\mathbf{z}]} (||\nabla g_{t,1}(\mathbf{y})||),\max_{[\mathbf{\overline{y}},\mathbf{z}]} (||\nabla g_{t,2}(\mathbf{y})||),...,\max_{[\mathbf{\overline{y}},\mathbf{z}]} (||\nabla g_{t,n}(\mathbf{y})||)\right\}$$

Since we know that for every $j \ \ $, $||\nabla g_{t,j}(\mathbf{y})||\leq M \ \ \forall (t,\mathbf{y})\in  K$ ,we have just to prove that $\forall \mathbf{y} \in [\mathbf{\overline{y}},\mathbf{z}], (t,\mathbf{y})\in K$ , this can be done particularly easily showing that $ d((t,\mathbf{y}),(\overline{t},\mathbf{\overline{y}}))\leq d((t,\mathbf{z}),(\overline{t},\mathbf{\overline{y}}))$ (it's straight forward after you use the fact that $\mathbf{y}$ is convex combination of $\mathbf{\overline{y}}$ and $\mathbf{z}$). In conclusion:
$$\alpha_t \leq M$$
And this implies that:
$$\forall (t,\mathbf{z})\in K \ \ ||\mathbf{g}_t(\mathbf{\overline{y}})-\mathbf{g}_t(\mathbf{z})||\leq \sqrt{n}M||\mathbf{\mathbf{\overline{y}}}-\mathbf{z}||$$
By the definition of $\mathbf{g}_t$ and defining $L=M \sqrt{n}$:
$$\forall (t,\mathbf{z})\in K \ \ ||\mathbf{f}(t,\mathbf{\overline{y}})-\mathbf{f}(t,\mathbf{z})||\leq L||\mathbf{\mathbf{\overline{y}}}-\mathbf{z}||$$
Since $(\overline{t},\overline{\mathbf{y}})$ was a generic point of $D$, this concludes the proof.

Is it correct?

答:
It is correct, but a bit messy to read. In exchange for the first three formulas you write, you could simply state that, by definition, $$ \nabla f_j(t,\cdot) = \nabla g_{t,j}. $$
You also never define $N,M$ or $K$ which is bad form. You also should mention explicitly where you use the assumptions that are necessary for this to hold, such as the continuity of the partial derivatives, which I can't really find here. Here's how I would write up the proof (using your notation):

Fix $(t_0,\mathbf{y}) \in D$. Choose $r>0$ such that $B_r (t_0,\mathbf{y}) \subseteq D$, which is possible as $D$ is open. Defining $\mathbf{g}_t$ as you did, we have
$$ \nabla f_j(t,\cdot) = \nabla g_{t,j}. $$ As $B_r(\mathbf{y})$ is compact, and the partial derivatives $\frac{\partial{f}_j}{\partial y_s}$ are continuous on $B_r(t_0,\mathbf y)$, it follows that there is a constant $\alpha$ such that
$$ \sup_{(t,\mathbf x) \in B_r(t_0,\mathbf y)} \left|\frac{\partial{f}_j}{\partial y_s}(t,\mathbf x)\right| < \alpha, \quad j=1,\dots,n.$$
In particular, this means that for $(t,\mathbf x),(t,\mathbf z) \in B_r(t_0,\mathbf y)$,
$$ \alpha_t=\max \left\{\max_{[\mathbf{x},\mathbf{z}]} (||\nabla g_{t,1}(\mathbf{y})||),\max_{[\mathbf{x},\mathbf{z}]} (||\nabla g_{t,2}(\mathbf{y})||),...,\max_{[\mathbf{x},\mathbf{z}]} (||\nabla g_{t,n}(\mathbf{y})||)\right\} < \alpha. $$
where we use that if $(t,\mathbf{x}), (t,\mathbf z) \in B_r(t_0,\mathbf y)$, any segment $\{t\} \times [\mathbf{x}, \mathbf{z}]$ lies in $B_r(t_0,\mathbf y)$ Applying the Lemma as you propose, we then have
$$ \forall (t,z),(t,\mathbf{z})\in B_r(t_0,\mathbf{y}): \ \ ||\mathbf{g}_t(\mathbf{x})-\mathbf{g}_t(\mathbf{z})||\leq \sqrt{n}\alpha_t||\mathbf{x}-\mathbf{z}|| \leq \sqrt{n}\alpha||\mathbf{x}-\mathbf{z}||. $$
Choosing $L := \sqrt{n}\alpha$ then yields the claim.


Title: Powers mod prime

Body:
$2^{21}\equiv 128^3 \equiv 22^3 \equiv 48 \pmod{53}$

Therefore $2^{20} \equiv 24 \pmod{53}$


问:
Dimensions of vector subspaces in a direct sum are additive
$V = U_1\oplus U_2~\oplus~...~ \oplus~ U_n~(\dim V < ∞)$ $\implies \dim V = \dim U_1 + \dim U_2 + ... + \dim U_n.$ [Using the result if $B_i$ is a basis of $U_i$ then $\cup_{i=1}^n B_i$ is a basis of $V$]

Then it suffices to show $U_i\cap U_j-\{0\}=\emptyset$ for $i\ne j.$ If not, let $v\in U_i\cap U_j-\{0\}.$ Then
\begin{align*}
v=&0\,(\in U_1)+0\,(\in U_2)\,+\ldots+0\,(\in U_{i-1})+v\,(\in U_{i})+0\,(\in U_{i+1})+\ldots\\
& +\,0\,(\in U_j)+\ldots+0\,(\in U_{n})\\
=&0\,(\in U_1)+0\,(\in U_2)+\ldots+0\,(\in U_i)+\ldots+0\,(\in U_{j-1})+\,v(\in U_{j})\\
& +\,0\,(\in U_{j+1})+\ldots+0\,(\in U_{n}).
\end{align*}
Hence $v$ fails to have a unique linear sum of elements of $U_i's.$ Hence etc ...

Am I right?

答:
Yes, you're correct.

Were you second guessing yourself? If so, no need to:

You're argument is "spot on".

If you'd like to save yourself a little space, and work, you can write your sum as:

$$ \dim V = \sum_{i = 1}^n \dim U_i$$

"...If not, let $v\in U_i\cap U_j-\{0\}.$ Then

$$v= v(\in U_i) + \sum_{\large 1\leq j\leq n; \,j\neq i}  0(\in U_j)$$


问:
Consequence of a Power Series Convergence at a Point
My Elementry Differential Equations book by Boyce, (10th edition) claims the following about power series convergence:

"If the power series $\sum_{n=0}^{\infty} a_{n} (x-x_{0})^n$ converges at $x= x_{1}$, then it converges absoultely for $\left | x-x_{0} \right | < \left | x_{1}-x_{0} \right |$"

This statement makes sense if the series converges absolutely at $x = x_{1}$. But is the claim also true if the series converges conditionally at $x = x_{1}$?

答:
Let $ R $ be the radius of convergence.

this means that if $ \sum a_n(x_1-x_0)^n $ converges, then

$$|x_1-x_0|\le R$$

So,
$$|x-x_0|<|x_1-x_0| \implies$$
$$|x-x_0|<R \implies $$
$$\sum a_n (x-x_0)^n \; is \; absolutely\; convergent$$


问:
Simultaneous diagonalisation of matrices by diagonalising a random linear combination of these matrices
According to a reply on Mathematica SE numerically finding a matrix that simultaneously diagonalizes a set of pairwise commuting diagonalizable finite complex matrices $\{M_1,\ldots,M_n\}$ can be done by diagonalizing a 'random' linear combination $\sum_{i=1}^nc_iM_i$ of these matrices. Of course not all linear combinations work (take, e.g., all $c_i = 0$) but 'most' of them should work.

I have asked for a reference to a more exact statement (with a proof) but all the references given either proved similar statements for very special matrices or just assumed this was a trivial statement.

Does anyone know of a theorem of the form: a matrix that diagonalizes a linear combination of simultaneous diagonalizable matrices also diagonalizes the individual matrices?
(I've chosen to make this statement somewhat vague since I don't know what possible extra conditions or generalizations there might be)

答:
Since they are all simultaneously diagonalizable, there exists a basis in which they are all diagonal. We fix this as our "standard" basis. Suppose in this basis $M_i$ has $d_{j; i}$ on as the $j$th diagonal entry. Then a linear combination $\sum c_i M_i$ is also diagonal and has  $\sum c_i d_{j;i}$ as the $j$th diagonal entry.

Now what we  want will follow from the two lemmas:

Lemma 1: A diagonal matrix stays diagonal in precisely those bases for which the change of basis matrix is block diagonal, with blocks corresponding to the same values on the diagonal.

Lemma 2: For "generic" $(c_1, \ldots, c_n)$ and any fixed pair $j_1, j_2$ we have $\sum c_i d_{j_1;i}=\sum c_i d_{j_2;i}$ if and only if for all individual $i$ we have $d_{j_1;i}=d_{j_2;i}$.

Indeed, the lemmas (which we prove later, making the notion of "generic" clear in the process) imply that for a generic $(c_1, \ldots, c_n)$ the only bases in which the sum is diagonal are those differing from the standard one by block-diagonal change-of-basis matrices which also keep all individual $M_i$ diagonal as well. This is exactly saying that diagonalizing the sum will diagonalize all of the $M_i$s.

Proof of lemma 1: A matrix is diagonal precisely in those bases that consist of eigenvalues. Any two such bases differ by changing bases for each eigenspace separately, meaning the change of bases matrix is block diagonal, with blocks corresponding to distinct eigenvalues. QED.

Proof of lemma 2: One way implication is clear: if individual entries are equal then the linear combinations are equal, for all $\vec{c}=(c_1, \ldots, c_n)$.

Suppose now for a given pair $j_1$ $j_2$ there exists $k$ such that $d_{j_1,k}\neq d_{j_2,k}$. Then the equation $\sum c_i d_{j_1;i}=\sum c_i d_{j_2;i}$ is of the form $c_k(d_{j_1;k}-d_{j_2;k})=\sum_{i\neq k} c_i(d_{j_1;i}-d_{j_2;i})$ and is a non-trivial linear equation. This means it is satisfied by a codimension one set of $\vec{c}$s.

Thus for all $\vec{c}$ outside of a finite collection of hyperplanes an equation $\sum c_i d_{j_1;i}=\sum c_i d_{j_2;i}$  implies $d_{j_1;i}=d_{j_2;i}$ for all $i$. Those are our "generic" $\vec{c}$s, and for them the lemma holds. QED.


问:
Asymptotic mean in an one-dimensional chaotic system
I have the following one-dimensional dynamical system
$$x_{t+1}=\begin{cases}
ax_t,\quad\mbox{if }x_t\leq 1,\\
ax_t^{c},\quad\mbox{if }x_t\geq 1,\\
\end{cases}$$
where $x_0>0$ is an arbitrary initial value, $a>1$ and $c<-1$.

I know that the fixed point is $x^*=a^\frac{1}{1-c}$ and it is unstable. My understanding is that the system is chaotic. However, I was wondering whether I can analytically compute the asymptotic mean of $x_t$, i.e., $\lim_{t\rightarrow\infty}\mathbb{E}[x_t]$? Do I need to derive first the invariant distribution of $x$? If so, what is the procedure?

Edit:
Moreover, I have a variable $y_t$ related to $x_t$ in the following way
$$
y_t=\begin{cases}1,\quad\mbox{ if }x_t\leq 1,\\
x_t,\quad\mbox{if }x_t\geq 1.\\\end{cases}
$$
I simulated the trajectory of $x_t$ for arbitrary $x_0$ and found that
$$
\lim_{t\rightarrow\infty}\left(\prod_{s=0}^ty_s\right)^\frac{1}{t+1}=x^*.
$$
Is it possible to show it analytically?

答:
I've spotted that taking the logarithm of the difference equation might make things simpler. We get the new difference equation

$$g_{t+1}=\begin{cases}
a' + g_t,\quad\mbox{if }g_t\leq 0,\\
a' + c g_t,\quad\mbox{if }g_t\geq 0,\\
\end{cases}$$

where $g_t=\log x_t$ and $a' = \log a$. Further, we can rewrite this equation into a more comfortable form by using the heaviside theta function

$$g_{t+1} = a' + g_t + (c-1)g_t\Theta(g_t)$$

To get some non-trivial dynamics I have chosen $a'=1$, $c=-3.5$. The velocity plot is as follows

The point where the velocity 0 zero is an attractor, because the the velocity points towards it from both sides. However, the step size is very big, so the system never really lands onto the attractor, just bounces around it. The equilibrium value obtained from setting velocity to zero is $g^* = \frac{a'}{1-c}$

Just to check, I have attempted to simulate $g_t$. Your $y_t$ we will replace with $h_t = \Theta(g_t) g_t$. The logarithm of the desired geometric mean will convert to the arithmetic mean.

$$\frac{1}{T}\sum_t h_t =^? g^*$$

It indeed appears that for reasonable values of $a$ and $c$ that $g_t$ oscillates in some complicated way around its equilibrium value $g^*$, and that $h_t$ converges to that equilibrium value over time.

Edit: Ok, here's a trick that works for the general problem, but it is indeed quite lucky that it does. Notice that in the original equation, the last term is simply $h_t$

$$g_{t+1} = a' + g_t + (c-1)h_t$$

then we can take the sum of both sides

$$\sum_{\tau=0}^{t} g_{\tau+1} = \sum_{\tau=0}^{t} a' + g_\tau + (c-1)h_\tau$$

Let $s^g_t = \sum_{\tau=0}^{t} g_\tau$ and $s^h_t = \sum_{\tau=0}^{t} h_\tau$

$$s^g_{t+1} - g_0 = a't + s^g_t + (c-1)s^h_t$$

Let $\mu^g_t = \frac{s^g_t}{t}$ and $\mu^h_t = \frac{s^h_t}{t}$

$$\mu^g_{t+1} - \frac{g_0}{t+1} = \frac{t}{t+1}(a' + \mu^g_t + (c-1)\mu^h_t)$$

In the limit of $t \rightarrow \infty$

$$\mu^g_{\infty} = a' + \mu^g_{\infty} + (c-1)\mu^h_\infty$$

This is where we get even more lucky, and $\mu^g$ simply cancels out from both sides, and

$$\mu_h^{\infty} = \frac{a'}{1-c} = g^*$$


问:
Hint for Spivak's Calculus Problem 2-4-a
(note: this is very similar to a related question but as I'm trying to solve it without looking at the answer yet, I hope the gods may humor me anyways)

I'm self-learning math, and an answer to an /r/math post about self-learning was finally enough to motivate me to try getting feedback on this site :) . After reading throughout the internet, I've decided to start with Spivak's Calculus. I'm loving the book thankfully, but I'm stuck on this problem and I don't want to look at the answer quite yet.

4 . (a) Prove that
$$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}$$.
Hint: Apply the binomal theorem to $(1+x)^n(1+x)^m$ .

I've done all the prior problems, including Problem 3 (proving the Binomial Theorem) which is obviously closely tied to this, but even with the hint it feels like there's too much I don't know. I applied the hint and found:

$$\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k$$

And, setting $x=1$ got something even more interesting:

$$\left(\sum_{i=0}^n \binom{n}{i}\right)\left(\sum_{j=0}^m \binom{m}{j}\right) = \sum_{k=0}^{n+m} \binom{n+m}{k}$$

However, I don't know where to go after that...is there some property of multiplying sums that I need to prove first, to relate the multiplication of two sums of binomials with the sum of the multiplication of two binomials?

Thank you!

答:
The key to using the below identity is to look at the coefficient of $x^l$ on both sides. Since the polynomials in $x$ are equal, their coefficients must be as well.
$$\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k$$
The coefficient of $x^l$ of the summation on the right hand side is obviously $\binom{n+m}{l}$. The question is, what is the coefficient of the $x^l$ in the product of summations on the left?

It might help to look at a smaller example: say $n=2,m=3,l=3$. The left hand side of the above equation becomes
$$
\left(\binom{2}0x^0+\binom21x^1+\binom22x^2\right)
\left(\binom{3}0x^0+\binom31x^1+\binom32x^2+\binom33x^3\right)
$$
Now, expand out this product, and collect terms with equal powers of $x$. The the $x^3$ term of the result is
$$
\left(\binom22\binom31+\binom21\binom32+\binom20\binom33\right)x^3
$$
Therefore, this summation must be equal to the coefficient of $x^3$ in $\sum_{k=0}^{2+3}\binom{2+3}{k}x^k$, which is $\binom{2+3}3$.


问:
Better method to show $-x^3 + 1 = (-x + 1)(x^2 + x + 1)$
Is there a rule, algorithm, or theorem used for this equality below:

$$-x^3 + 1 = (-x + 1)(x^2 + x + 1)$$

I know the RHS can be established using trial and error distributing the first term through the second, or alternatively same with FOIL.  I'm curious if there is a better method?

Image below from a CAS.

答:
$$1-x^3=1^3-x^3$$

In general, we have$$a^n-b^n=(a-b)\sum_{k=0}^{n-1}a^{n-1-k}b^{k}$$

$a-b$ is clearly a factor of $a^n-b^n$ as when $a=b$, $a^n-b^n$ vanishes.

Another way is to view $\sum_{k=0}^{n-1}a^{n-1-k}b^{k}$ as a geometric sum.


问:
Prove that {$v,T(v)$} are L.I.Describe the matrix of $T$ w.r.t. basis.
The dimension of $V$ is $2$.So I have to prove that $v,T(v)$ are Linearly Independent where $T$ is not a scalar operator.

Since $T$ is not a scalar operator so there exists a vector $v \in V$ such that $T(v) \ne \lambda v$.

Proving Linear Independence

$c_1(v)+c_2T(v)=0$.

We see that $v=ae_1+be_2$ and $T(v)=b_1e_1+b_2e_2$ where {$e_1,e_2$} are standard basis of $V$.

Now $(c_1a_1+c_2b_1)e_1+(c_1a_2+c_2b_2)e_2=0$.

Then $a_1=(-\frac{c_2}{c_1}b_1)$ which is possible only when $c_2=0$ otherwise $b_1$ and $b_2$ will both become a scalar multiple of $a_1,a_2$ contrary to our assumption. From here we can conclude that $T(v),v$ are linearly independent.

Now since they are independent so they form the basis of $V$ and $T(v)= 0.v+1.T(v)$
$T(T(v))=av+bT(v)$.

This has been my attempt .can someone point out my mistake instead of suggesting another method and how I can rectify it.Also I dont have any idea how to create a matrix in stack exchange .Sorry for the inconvenience.

答:
Since $T$ is not a scalar operator so there exists a vector $v∈V$ such that $T(v)≠λv.$

let $v$ and $T(v)$ be linearly dependent then there exists $c_1,c_2$ such that

$c_1v+c_2T(v)=0$ $\Rightarrow$ $c_2T(v)=c_1v$

$c_2T(v)=-c_1v$ $\Rightarrow$ $T(v)=\frac{-c_1}{c_2}v=\lambda v$ for $\lambda=\frac{-c_1}{c_2}$

that is a contradiction so $v$ and $T(v)$ must be linearly independent

$dimV=|\{v,T(v)\}|=2$ $\Rightarrow$ $\{v,T(v)\}$ is a basis


问:
How is the resultant $R(z) =\text{res}_θ(1 - \frac{z}{x}, θ) = 1 - \frac{z}{x}$ calculated?
The resultant of two polynomials is defined as the determinant of the Sylvester matrix. If the polynomials are of degree $n$ and $m$, than the Sylvester matrix will be of dimension

$(m+n)\times (m+n)$.

In the book Algorithms for Computer Algebra on page 534 (example 12.6) the following resultant is calculated:

$R(z) =\text{res}_θ(1 - \frac{z}{x}, θ) = 1 - \frac{z}{x}$

How is this resultant calculated?

答: I guess the point is to consider $\theta$ as the variable. So the resultant in this case is a $1\times 1$-matrix (since the degree of $1-\frac{z}{x}$ as a polynomial in the indeterminate $\theta$ is $0$ and the degree of $\theta$ is $1$), precisely $[1-\frac{z}{x}]$.</p>


问:
prove that $\log_2(n+1)-\log_3(2n+1) \le c$ eventually
How can I prove that there is $c>0$ such that starting from a specific $n$ the following is true:
$$
\log_2(n+1) - \log_3(2n+1) \le c
$$

答:
$\log_2(n+1)-\log_3(2n+1)=\dfrac{\log_3(n+1)}{\log_32}-\log_3(2n+1)=$

$=\log_23\cdot\log_3(n+1)-\log_3(2n+1)=$

$=\left[\log_2\left(\dfrac32\right)+1\right]\cdot\log_3(n+1)-\log_3(2n+1)=$

$=\log_2\left(\dfrac32\right)\cdot\log_3(n+1)+\log_3(n+1)-\log_3(2n+1)=$

$=\log_2\left(\dfrac32\right)\cdot\log_3(n+1)+\log_3\left(\dfrac{n+1}{2n+1}\right)>$

$>\log_2\left(\dfrac32\right)\cdot\log_3(n+1)+\log_3\left(\dfrac12\right)\to+\infty$

as $\;n\to+\infty\;.$

Hence, there does not exist any $\;c>0\;$ such that

$\log_2(n+1)-\log_3(2n+1)\le c\;$ eventually.


答:
Let $\zeta$ be a primitive cube root of unity, then $\zeta^3=1,\zeta^2+\zeta+1=0$ and hence $1-x^3=-(x^3-1)=-(x-1)(x-\zeta)(x-\zeta^2)=(1-x)(x^2-(\zeta+\zeta^2)x+\zeta^3)=(1-x)(x^2+x+1)$

Note: FOIL, which is the distributive property of multiplication and addition, is unavoidable in pretty much any proof. To try to avoid using FOIL in a proof would be like trying to avoid using the fact that $1+1=2\ne 0$ in a field of characteristic $\gt 2$ or $0$.

Title: Better method to show $-x^3 + 1 = (-x + 1)(x^2 + x + 1)$

Body:
Is there a rule, algorithm, or theorem used for this equality below:

$$-x^3 + 1 = (-x + 1)(x^2 + x + 1)$$

I know the RHS can be established using trial and error distributing the first term through the second, or alternatively same with FOIL.  I'm curious if there is a better method?

Image below from a CAS.


问:
Abelianization of subgroups of $GL(n,\mathbb{C})$ and $SL(n,\mathbb{C})$
Consider the following subgroups of $GL(n,\mathbb{C})$ and $SL(n,\mathbb{C})$

$$H_G = \{A\in GL(n,\mathbb{C})\: | \: AA^t = \lambda_AI, \text{for some } \lambda_A\in\mathbb{C}^{*}\}\subset GL(n,\mathbb{C})$$

$$H_S = \{A\in SL(n,\mathbb{C})\: | \: AA^t = \lambda_AI, \text{for some } \lambda_A\in\mathbb{C}^{*}\}\subset SL(n,\mathbb{C})$$

Note that $H_S = H_G\cap SL(n,\mathbb{C})$ and that for $H_S$ the complex numbers $\lambda_A$ are $n$-th roots of unity.

So we have a morphism $H_S\rightarrow \mathbb{Z}/n\mathbb{Z}$ mapping $A\mapsto \lambda_A$. The abelianization of $H_S$ is $\mathbb{Z}/n\mathbb{Z}$.

On the other hand, the determinant gives a morphism $H_G\rightarrow \mathbb{C}^{*}$. So on this side it seems that the abelianization of $H_G$ is $\mathbb{C}^{*}$.

Note that $GL(n,\mathbb{C})/H_G\cong SL(n,\mathbb{C})/H_S$. In particualr, the character groups of $H_G$ and $H_S$ should coincide. But this is not the case for what I said before. One of them is $\mathbb{Z}$ and the other one $\mathbb{Z}/n\mathbb{Z}$.

I wanted to ask you if someone can see what I am doing wrong here.

Thanks a lot.

答:
There are two homomorphisms $H_G\to\mathbf{C}^*$, namely $\delta=\det|_{H_G}$ and $\lambda$, and $H_S=\operatorname{Ker}\delta$. First consider $f=\delta\times\lambda: H_G\to\mathbf{C}^*\times\mathbf{C}^*$.

By definition, the kernel of $f$ (which is also the kernel of $\lambda|_{H_S}$) is $\mathrm{SO}(n,\mathbf{C})$. This is known to be a perfect group for every $n$. It follows that $f:H_G\to f(H_G)$ and $\lambda:H_S\to\lambda(H_S)$ are the abelianization homomorphism.

It remains to determine these images. For $n=0$ all groups are trivial, so assume $n\ge 1$.

Then in restriction to $\mathbf{C}^*I_n$, both $\delta$ and $\lambda$ are surjective, given by $\delta(tI_n)=t^n$ and $\lambda(tI_n)=t^2$. Hence, if $(u,v)$ belongs to $f(H_G)$, and $v=w^2$, then $(uw^{-n},1)$ belongs to the image as well, i.e., is the determinant of some orthogonal matrix, which forces $uw^{-n}=\pm 1$, i.e., $u^2=v^n$.

Conversely, if $u^2=v^n$, consider the scalar matrix $wI_n$. Then $f(wI_n)=(w^n,w^2)$. If $w^n=u$, then this equals $(u,v)$ and we are done. Otherwise, $w^n=-u$. Then pick a matrix $M$ in $\mathrm{O}(n)\smallsetminus\mathrm{SO}(n)$: then $f(wM)=(u,v)$.

Hence the image of $f$ is $\{(u,v)\in\mathbf{C}^*\times\mathbf{C}^*:u^2=v^n\}$. If $n=2m$ is even, this is the same as $\{(u,v)\in\mathbf{C}^*\times\mathbf{C}^*:u=\pm v^m\}$ isomorphic to $\mathbf{C}^*\times C_2$ ($C_2$ cyclic of order $2$ generated by $(-1,1)$ and the identity component being $\{u=v^m\}$. If $n=2m+1$ is odd, this abelianization is isomorphic to $\mathbf{C}^*$, parameterized as $\{(w^n,w^2):w\in\mathbf{C}^*\}$ (where $w=uv^{-m}$).

If we restrict to determinant 1, the condition $u^2=v^n$ becomes $u=1$, $v^n=1$. So the image of the abelianization $\lambda:H_S\to\mathbf{C}^*$ is just the group $\mu_n$ of $n$-roots of unity.


Title: Powers mod prime

Body: $\!\bmod \color{#c00}{53}\!:\ 2^{10}\! = 1024 =\! \overbrace{\color{#c00}{50}}^{\large\color{#c00}{-3}}(20)\!+\!24 \equiv 17\, $ $\Rightarrow 2^{20}\equiv 17^2\equiv 24$


问:
Evaluate the double integral $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$
$\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$

I considered changing the order of integration but that I do not think it does anything as it just leads me to evaluating dy still.

Another possibility I considered is converting the integral to polar coordinates but again the square root bogs me in solving this integral. Any help or hint is appreciated. Thank you.

Body:
$\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x=\int_{0}^{2} \int_{0}^{y^4} \sqrt{2 y^{5}+1} d x d y=\int_{0}^2y^4\sqrt{2 y^{5}+1}dy$

$u=2y^5+1 \Rightarrow du=10y^4dy$

$\int_{0}^2y^4\sqrt{2 y^{5}+1}dy=\frac{1}{10}\int_{1}^{65}\sqrt{u}du=\frac{1}{15} \left(65 \sqrt{65}-1\right)$


问:
Does probability theory suffer from Gödel's incompleteness theorem?
Let us consider the following two theorems by Gödel:

Any consistent formal system $F$ within which a certain amount of elementary arithmetic can be carried out is incomplete; i.e., there are statements of the language of $F$ which can neither be proved nor disproved in $F$;

For any consistent system F within which a certain amount of elementary arithmetic can be carried out, the consistency of F cannot be proved in $F$ itself.

Like others, probability theory is also an axiom-based system with 3 axioms and supports minimum arithmetic.

From the above information, it looks like probability theory also suffers from Gödel's incompleteness theorem. Can't it? If not, then where I am going wrong?

Thus is there a possibility that there may exist two statements in probability theory that contradict each other and the existence of unprovable facts?

答:
This can't really be answered without specifying exactly what "probability theory" is. You say that it's an axiom-based system, but what are those axioms?

You mention that there are specifically three. Based on this I suspect you're referring to the Kolmogorov axioms, which roughly say:

The expected value function $\mathsf{E}$ is non-negative.

The maximum that $\mathsf{E}$ attains is exactly $1$, that is, the probability of the "trivial event" is $1$.

$\mathsf{E}$ is countably additive.

Even ignoring some fundamental issues (see below), these axioms are obviously incomplete: they do not decide for example whether there is an event with probability strictly between $0$ and $1$, and this is definitely something we should be able to express. However, being incomplete is not the same as being susceptible to Godel's theorem: a theory can be incomplete for "non-Godelian reasons." What exactly "Godelian" means isn't clear, but certainly any Godelian theory should have the following property:

A computably axiomatizable theory is essentially undecidable iff every computably axiomatizable extension of that theory is incomplete.

For example, the theory of algebraically closed fields is incomplete (it doesn't pin down the characteristic of the field) but not essentially incomplete (for each $p$, the theory of algebraically closed fields of characteristic $p$ is computably axiomatizable and complete). Since there are very simple probability systems out there (e.g. consider a theory which admits only the always-happens and never-happens events), any reasonable implementation of the Kolmogorov axioms is not essentially incomplete. So in a precise sense, even though those axioms (modulo their issues - again, see below) are incomplete, they are not susceptible to Godel's theorem. The culprit, unsurprisingly, is expressive power.

This raises the question:

Is there a natural theory extending Kolmogorov's which is essentially undecidable?

At this point we do have to address the above-mentioned issues with the Kolmogorov axioms. There are two:

We need to say more about the codomain: presumably we at least want to say that $\mathsf{E}$ takes values in an ordered monoid, and we probably want to say more than that.

Axiom $3$ is not first-order - we either have to $(i)$ whip up a first-order approximation to it, $(ii)$ shift the "naive" context of probability theory (= event space + $\mathbb{R}$) to a richer one (= something with a set-theoretic apparatus allowing us to talk about arbitrary countable sets), or $(iii)$ work in a richer logic than first-order logic, such as $\mathcal{L}_{\omega_1,\omega}$.

I'll take these in reverse order.

Option $(iii)$ doesn't seem appropriate here, although I personally like it. Perhaps surprisingly, Godelian ideas don't go out the window altogether even when we leave the realm of first-order logic, so $(iii)$ actually is a priori something that could yield interesting results. However, it quickly becomes rather technical and I suspect it's not what's wanted here, so let's ignore it here.

Option $(ii)$ is arguably the best one here, but the set-theoretic apparatus we would need to bring into play will on its own be susceptible to Godel's incompleteness theorem (see e.g. here). So at this point we do get Godelian, but not for reasons related to the actual topic at hand.

This leaves $(ii)$ as the only real faithful option. So what we need to do now is $(a)$ identify whatever additional axioms we need to "make Kolmogorov work right," and in particular $(b)$ pin down a "first-order approximation" to Axiom $3$. The good news is that there's a natural approach to $(b)$, namely to work with a scheme of axioms addressing each first-order definable instance of the principle at hand separately. This is the approach taken for example in (first-order) $\mathsf{PA}$ and $\mathsf{ZFC}$ (consider induction and separation/replacement respectively), and is extremely flexible.

So this suggests the following framework for first-order probability theory:

We have two sorts, an event sort and a probability sort.

On the probability sort we have binary operations $+$ and $\times$, a binary relation $<$, and constants $0$ and $1$; and we have axioms asserting that these turn that sort into an ordered field.

On the event sort we have two binary operations $\cap$ and $\cup$, a binary relation $\subseteq$, and two constants $\emptyset$ and $\Omega$, and axioms asserting that this turns that sort into a Boolean algebra with bottom $\emptyset$ and top $\Omega$.

Connecting both sorts, we have a function $\mathsf{E}$: its domain is the event sort and its codomain is the probability sort. We add axioms saying that $ran(\mathsf{E})\subseteq [0,1]$, $ran(\Omega)=1$, and $ran(\emptyset)=0$; and axioms saying that $\mathsf{E}$ is $2$-additive (and hence finitely additive).

We have the countable completeness scheme asserting that there is no definable failure of countable completeness.

And finally, we leave the door open for additional structure and axioms. For example, maybe we want to say that for every event $X$ there is some $Y$ such that $\mathsf{E}(Y)={\mathsf{E}(X)\over 2}$; or maybe we want to view events as subsets of a vector space. Note that I'm even allowing additional sorts (e.g. if we want events to be subsets of a vector space, we'll also want a sort for vectors and a sort for scalars).

That last bulletpoint lets us frame the question more correctly in my opinion:

What are some natural additional structure/axioms which make the resulting probability theory essentially incomplete?

The point is that, since as mentioned above there are very simple $\sigma$-algebras, if we don't add any such additions then we certainly won't get essential incompleteness. (Crucially, the real numbers are less complicated than the naturals from the point of Godel, and indeed much of logic: the theory of real closed fields is decidable!)

The most natural candidates I can think of are additions which describe contexts where events are arithmetical in some sense. There are of course no interesting uniform probability measures when $\Omega=\mathbb{N}$, but there are very natural non-uniform ones (e.g. the binomial distribution). If we want to look at such a measure in the context of number theory, then we presumably want to put enough structure on the event space('s underlying set) to do number theory - and this immediately brings in Godel.

However, again we're seeing incompleteness for reasons unrelated to probability. So really the right question to ask is:

What are some natural additional structure/axioms which make the resulting probability theory essentially incomplete, but which do not automatically yield essential incompleteness on their own?

I know of no natural candidates for such a thing, but since probability isn't my area I'm not sure how much that's worth.


问:
Addition vs multiplication as binary operators
It is known that arithmetic operations of addition and multiplication are binary operations that take in two inputs and give out a single output. However, consider the following scenario:

Suppose I have a jar containing some number of balls, and I give you some balls to add to the jar. If I asked you to add three balls into the jar, then that's easy, you'd just toss em in there. However, let's say I asked you to double the number of balls in the jar, you wouldn't be able to do that unless I showed you many balls were in the jar.

This is sort of strange: We require to know both inputs for multiplication but we don't require to know both to do addition. Why is this so when they are both known to be binary operators?

答:
It has nothing to do with the operators being binary, but rather how those operations deals with the operands.

Let's assume the jar has $n$ balls, and let that be the left operand. Then, adding $2$ balls to the jar would correspond to the operation $n+2$, while doubling the balls would correspond to the operation $n \times 2$

Now, addition can be seen as a repeated process of adding 1 ball to the jar some number of times. How many times?  Well, that's what the right operand says. OK, but how many each time?  Exactly $1$. So, we know everything to do what needs to be done from the right operand alone.

For example, adding 2 balls means adding 1 ball exactly 2 times. No need to know the left operand (i.e. what's already in the jar).

On the other hand, multiplication can be seen as a repeated process of adding n balls to the jar some number of times. How many times? Again, that's what the right operand says (well, you need to subtract $1$ if you start out with the $n$ balls already in the jar).  OK, but how many each time? Well, that's the number $n$, i.e. th number of balls already in the jar i.e. the left operand.

For example, multiplying the number of balls by $2$ (i.e. doubling the number of balls) means adding n balls to the already existing $n$ balls. So this time, yes, we need to know the left operand.


答:
I think changing the order of the integration is best:

$$\sqrt[4]{x}\leq y \leq 2$$
$$0 \leq x \leq 16$$

is the same as

$$0\leq x \leq y^4$$
$$0 \leq y \leq 2$$

from which the integral becomes $\int_0^2\int_0^{y^4}\sqrt{2y^5 +1}dxdy=\int_0^2 {y^4}\cdot \sqrt{2y^5 +1}dy$

which is easy to solve given the substitution mentioned in the comments.

let $u=2y^5 +1$, then $du= 10y^4 dy$ and $dy=\frac{du}{10y^4}$

$$\int_0^2 {y^4}\cdot \sqrt{2y^5 +1}dy = \int_1^{65} {y^4}\cdot \sqrt{u}\frac{du}{10y^4}= \frac{1}{10}\int_1^{65} \sqrt{u} du = \frac{1}{10}\cdot \frac{2\cdot u^{\frac{3}{2}}}{3}\Bigg|_1^{65}$$ $$= \frac{1}{15} \cdot \left(65^{\frac{3}{2}}-1\right)$$

Title: Evaluate the double integral $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$

Body:
$\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$

I considered changing the order of integration but that I do not think it does anything as it just leads me to evaluating dy still.

Another possibility I considered is converting the integral to polar coordinates but again the square root bogs me in solving this integral. Any help or hint is appreciated. Thank you.


答:
Let

$$S_n:=1+x+x^2+\cdots x^n.$$

Then mentally, by cancellation of the terms,

$$S_n-xS_n=1-x^{n+1},$$

and of course,

$$S_n-xS_n=(1-x)S_n.$$


答: If two $n$th degree polynomials agree on $n+1$ points, then the polynomials are equal.  You have $3$rd-degree polynomials, so choose $4$ nice values for $x$ and plug them in on both sides.  How about $x=-1, 0, 1,$ and $2$.   If you get the same value on both sides all four times, the polynomials are equal.


问:
Evaluating the integral: $\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx$
I am interested in evaluating the following integral:
$$
\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx
$$
Using Matlab, Numerically it seems that the integral is convergent,
but I'm not sure about it. How can we prove that the integral is
convergent or not?

Many Thanks in advance.

答:
let $I=\int_{0}^{\infty}\dfrac{2-2\cos{x}-x\sin{x}}{x^4}dx$

then using integration by parts we have
$$I=-\dfrac{1}{3}\dfrac{2-2\cos{x}-x\sin{x}}{x^3}|_{0}^{\infty}+\dfrac{1}{3}\int_{0}^{\infty}\dfrac{\sin{x}-x\cos{x}}{x^3}dx$$

then $$I=\dfrac{1}{3}\int_{0}^{\infty}\dfrac{\sin{x}-x\cos{x}}{x^3}dx$$
and application of integration by parts yields
$$I=-\dfrac{1}{6}\lim_{x\to 0^{+}}\dfrac{\sin{x}-x\cos{x}}{x^2}+\dfrac{1}{6}\int_{0}^{\infty}\dfrac{x\sin{x}}{x^2}dx=\dfrac{\pi}{12}$$


问:
How I am suppose to deal with indeterminations in a given relation?
Suppose a relation $S = \{(x, y)\in\mathbb{Z}\times\mathbb{Z}: x \text{ divides } y \}$

I'm curious about if $S$ is reflexive, symmetric, transitive, or antisymmetric.

The relation is not symmetric because considering $(\forall x\in \mathbb{Z})(\forall y\in \mathbb{Z})(xSy \to ySr)$ for $x=2$ and $y=4$, $\dfrac{1}{2}$ is not an integer.

The problem I see in the reflexive, transitive, and antisymmetric concerns the indetermination $\dfrac{0}{0}$. So the relation is not reflexive, symmetric, transitive, neither antisymmetric? If $(x, y)\in\mathbb{Z}\times\mathbb{Z}/\{0\}$ so the Relation S would be reflexive, transitive and antisymmetric?

答:
By definition, x divides y if $\exists a\in\Bbb Z$ such that $y=ax$. Any integer $a$ satisfies $0=0a$, so $0$ divides $0$. So the relation is actually reflexive.

If $x\mid y,y\mid z$, write $y=ax,z=by$ whence $a,b\in \Bbb Z\implies z=b(ax)=(ba)x$, so z divides x, and thus $S$ is transitive in fact.


问:
Clarification on differentiability definitions
How do I prove that $h_x(0,1)$ and $h_y(0,1)$ exist?

Suppose I were to prove that $h_x(0,1)$ and $h_y(0,1)$ exist, given $h(x,y)$

Do I use

$h_x(0,1)=\lim_{x\to 0} \frac{h(x,1)-h(0,1)}{x-0}$ and
$h_y(0,1)=\lim_{y\to 1} \frac{h(0,y)-h(0,1)}{y-0}$

or is the numerator/denominator incorrect?

答:
As you correctly noted, this is a matter of definition. Such questions are actually open ended, because we can chose the definition to be whatever we want. This said, you want
$$
    h_x(0,1)=\lim_{x\to 0} \frac{h(x,1)-h(0,1)}{x-0}
$$
and
$$
    h_y(0,1)=\lim_{y\to 1} \frac{h(0,y)-h(0,1)}{y-1}.
$$
So, the second denominator was incorrect, easy to fix!

The general pattern is
$$
    h_x(a,b)=\lim_{x\to a} \frac{h(x,b)-h(a,b)}{x-a}
$$
and
$$
    h_y(a,b)=\lim_{y\to b} \frac{h(a,y)-h(a,b)}{y-b}.
$$