问:
A positive semidefinite matrix multiplied by any matrix and its transpose is positive semidefinite?
Let $P$ be a positive semidefinite matrix and $A$ any matrix. Is $X=APA^\intercal$ positive semidefinite, if so give a proof, if not a counterexample. (not sure if the context of the problem means that A should be square)

I doubt X is positive semidefinite, but need help with a counterexample in a 2x2 case.

Example that does not work:
$\left(\begin{matrix}-1&1\\0&0\end{matrix}\right)\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}-1&0\\1&0\end{matrix}\right)=\left(\begin{matrix}-1&1\\0&0\end{matrix}\right)\left(\begin{matrix}-1&0\\1&0\end{matrix}\right)=\left(\begin{matrix}1&0\\0&0\end{matrix}\right)$

Maybe if you got a positive semidefinite matrix without zeros in the off diagonals and tried something it could work. This problem is only worth a few points on my homework but I feel bothered by it.

答:
For real matrices, this is true, but for complex matrices, this is false!

Consider first the (counter)example for $2\times 2$ complex matrices: Let $P=\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}$ and $A=\begin{pmatrix}
i & 0 \\
0 & i
\end{pmatrix}$. Then, $APA^T=\begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}$.

Now for real matrices, for any vector $v$, $\langle APA^T v, v \rangle=\langle P(A^T v), (A^T v) \rangle \geq 0$, as $P$ is positive semidefinite.

For the result to remain true for complex matrices as well, one must replace the transpose by the adjoint.


问:
Gamma function integral inequality
In appendix 1 of the following paper, I found
\begin{align}
\int_s^\infty \exp(-\frac{1}{2}z^\beta)dz&<\frac{1}{\beta}2^{1/\beta}\Gamma\left(\frac{1}{\beta}\right)
\end{align}
however I can't figure out how they got that inequality. I tried writing out the gamma function as an integral and taking logs of both sides, but that got me nowhere.

答:
K. defaoite's comment is correct. In more detail, let $x=z^\beta/2$. Then
\begin{align}
z&=(2x)^{1/\beta}\\
dx&=\frac{\beta z^{\beta-1}}{2}dz\\
&=\frac{1}{\beta}2^{1/\beta}x^{1/\beta-1}dz\\
dz&=\frac{2}{\beta z^{\beta-1}}dx
\end{align}
Doing integration by substitution one has that
\begin{align}
\int_0^{\infty}\exp(-z^\beta/2)dz&=\int_0^\infty \exp(-x)\frac{1}{\beta}2^{1/\beta}x^{1/\beta-1}dx\\
&=\frac{1}{\beta}2^{1/\beta}\Gamma\left(\frac{1}{\beta}\right)
\end{align}
which is the right hand side of the inequality. It then follows immediately.

ContentLicense: CC BY-SA 4.0

Title: Gamma function integral inequality


答:
With $u=\frac {z^\beta}2$, we have $dz= \frac 1 \beta 2^{\frac 1 \beta}u^{\frac 1 \beta  -1}du$
$$\int_s^\infty \exp(-\frac{1}{2}z^\beta)dz=\frac{2^{\frac 1 \beta}}{\beta}\int_{2^{\frac 1 \beta}s^{\frac 1\beta}}^{+\infty}e^{-u}u^{\frac 1 \beta  -1}du\leq \frac{2^{\frac 1 \beta}}{\beta}\int_{0}^{+\infty}e^{-u}u^{\frac 1 \beta  -1}du=\frac{2^{\frac 1 \beta}}{\beta}\Gamma\left(\frac 1 \beta\right)
$$

Title: Gamma function integral inequality

Body:
In appendix 1 of the following paper, I found
\begin{align}
\int_s^\infty \exp(-\frac{1}{2}z^\beta)dz&<\frac{1}{\beta}2^{1/\beta}\Gamma\left(\frac{1}{\beta}\right)
\end{align}
however I can't figure out how they got that inequality. I tried writing out the gamma function as an integral and taking logs of both sides, but that got me nowhere.


问:
Set of prime numbers $q$ such that $\sum\limits_{p\leq\sqrt{q}}p=\pi(q)$
If we define the sum of consecutive prime numbers $p$ up to the square root of a given positive integer $n$ as $S(n)=\sum\limits_{p\leq\sqrt{n}}p$, and the prime counting function up to a given positive integer $n$ as $\pi(n)=\#\left\{p\leq n\right\}$, it can be showed that $S(n)\sim\pi(n)$ (for a proof, you can check http://vixra.org/abs/1911.0316). This led me to study the positive integers for which $S(n)=\pi(n)$.

If we define set $Q$ as the set of values of $n$ such that $\pi\left(n\right)=S\left(n\right)$, the first values are $Q=\{11,12,29,30,59,60,179,180,...\}$, for $\pi(n)=5,5,10,10,17,17,41,41,...$

It can be conjectured that the first value of $n$ with a concrete $\pi\left(n\right)=S\left(n\right)$ will be always a prime number. This conjecture is a corollary of the following

Conjecture 1.

It does not exist any squared prime number $p^{2}$ such that $\pi\left(p^{2}\right)=S\left(p^{2}\right)$ except of $p_{1}=2$. That is, $$\pi\left(p_{n}^{2}\right)\neq\sum_{k=1}^{n}p_{k}$$

If the Conjecture were false, we could have that, being $p_m$ and $p_{m+1}$ prime numbers such that $p_m<p_n^2<p_{m+1}$, then $S\left(p_{m}\right)=S\left(p_{m}+1\right)=S\left(p_{m}+2\right)=...=S\left(p_n^{2}-1\right)=S\left(p_n^{2}\right)-p_n$; and as  $\pi\left(p_n^{2}\right)=S\left(p_n^{2}\right)$, then $p_n^{2}\in Q$,and $p_n^{2}$ would be the first of a series of consecutive elements of $Q$ until $p_{m+1}$.

The conjecture has been tested and found to be true for the first thousands of primes.

If we define set $M$ as the set of values of $n$ such that $\pi\left(n\right)=S\left(n\right)$ and $n$ is some prime number, we find that $M=\{11,29,59,179,389,541,...\}$. You can see more terms in https://oeis.org/A329403. And if we define $p_{k}$ as the last prime number which is a term of $S\left(n\right)$, we can see that the first values of $k$ are $2,3,4,6,8,9,...$, which led me to formulate the following

Conjecture 2

Set $M$ has infinitely many elements

If the conjecture were true, as $M\subset Q$, the Conjecture implies that $\pi\left(n\right)$ intersects $S\left(n\right)$ infinitely many times, so $S\left(n\right)$ is not only asymptotically equivalent to $\pi\left(n\right)$: it is infinitely many times equal to $\pi\left(n\right)$. And it would show that the number of primes between $p_{n}^{2}$ and $p_{n+1}^{2}$ , on average, do not differ much from $p_{n+1}$.

Questions

(1) Any idea of how could be proved Conjecture 1, if it is already proved? Or maybe, with more computational power than mine, if anyone could verify if the conjecture supports itself also for big numbers, it would be great.

(2) With my computational power, I was able only to identify the first 19 prime numbers and $k$ values of the set (last one was $1756397$, for $k=216$). If anyone could calculate more of them to see if the conjecture supports itself also for big numbers, it would be great.

(3) Any idea of how to show that Conjecture 2 is true? I guess that it could be done with some theorem of the kind of Littlewood's oscillation theorem, but as an amateur mathematician obtaining such a result is beyond my actual knowledge.

Thanks in advance!

答:
My answer was wrong.

If $\zeta(s)$ has a zero of real part $> 3/4$ then $\sum_{p\le x}1-\sum_{p\le x^{1/2}} p$ changes of sign infinitely often. If $\zeta(s)$ has no zero of real part $> 3/4-\epsilon$ then $\sum_{p\le x}1-\sum_{p\le x^{1/2}} p$ still changes of sign infinitely often but due to more complicated reasons (similar to the prime number races theorem).

See https://mathoverflow.net/questions/383610/set-of-prime-numbers-q-such-that-sum-limits-p-leq-sqrtqp-piq-where?noredirect=1#383620


问:
Questions about locally Euclidean
Here is the definition of locally euclidean I was given:
A topological space $(X,τ)$ is said to be locally euclidean if there exists a positive integer n such that each point $x∈X$ has an open neighbourhood homeomorphic to an open ball about 0 in R^n with the euclidean metric. A Hausdorff locally euclidean space is said to be a topological manifold.

This is the first time I learned what locally euclidean means so I am wondering if I am on the right track with these questions below

(i)Prove that every non-trivial interval $(a,b)$, $a,b ∈ \Bbb{R}$, is locally euclidean.

Every open neighbourhood of elements of open intervals are just open intervals which are homeomorphic to any open interval in $\Bbb{R}$. So they are all homeomorphic to $(0-e,0+e)$.

(ii)  Let $S^1$ be the subspace of $\Bbb{R}^2$ consisting of all $x∈\Bbb{R}^2$ such that $d(x,0) = 1$, where $d$ is the Euclidean metric. Show that the space $S^1$ is locally Euclidean.

Every neighborhood of point of $S^1$ are just connected segment of the circle without the endpoints and we are given that these are homeomorphic to open intervals in $\Bbb{R}$ so they must be homeomorphic to $(0-e, 0+e)$.

(iii) Find an example of a locally Euclidean space which is not a topological manifold?

I am currently stuck on this one. This is the first time I learned what a topological manifold is and it seems like a lot of locally Euclidean space are Hausdorff.

答:
To help with the question you are stuck on consider:

The quotient space of the disjoint union of two copies of $R$, via the identification of $x$ in the first copy with $x$ in the second copy, for $x \neq 0$. This is known as the line with two origins.

Title: Questions about locally Euclidean


问:
Beginner Probability Question
This is my second day of stats & probability and I'm a little confused.

Here's the question:

The Abigail Construction Company is determining whether it should
  submit a bid for the construction of a new shopping mall. In the past,
  its main competitor, the Jared Construction Company, has submitted
  bids 60% of the time. When Jared does not submit a bid, the
  probability the Abigail will win the job is 70%. However, when Jared
  does submit a bid, the probability that the Abigail will win the job
  is only 40%. If Abigail wins a job, what is the probability that Jared
  submitted a bid?

What I've solved for thus far (it's not much):
J not - .2
  does - .7

I think the .6 in the beginning is irrelevant, but I could be wrong. How do I move forward?

P.S. If you know of a good place online to learn probability, please don't hesitate to post it!
Thanks!

答:
Let us consider how Abigail might get the job

Abigail can win if either of E1 or E2 occur

E1 = Jared submits and Abigail wins when Jared submits

E2 = Jared doesnot submit and Abigail wins when Jared Does not submit

P(JS) = Probability of Jared submitting = .6

P(~JS) = Probability of Jared not submitting = .4

P(A|JS) = Probability of Abigail winning when Jared submits = .4

P(A|~JS) = Probability of Abigail winning when Jared does not submit = .7

P(E1) = P(A|JS) * P(JS) = .6 * .4 = .24

P(E2) = P(A|~JS) * P(~JS) = .7 * .4 = .28

P(A) = P(E1) + P(E2)

P(A) = .52

P(JS|A) = P(E1) / P(A)

P(JS|A) = .24 / .52

P(JS|A) = .4615

ContentLicense: CC BY-SA 3.0

Title: Beginner Probability Question


问:
Ratio test for a series
Just trying to understand how the ratio test works. Suppose our infinite series ($n=1$) is

$\frac{7^n}{8^{n+1}}$ and the ratio test

$p = lim_{n->\infty} |\frac{a_{n+1}}{a_n}|$

Do we sub in any number for $n$ and calculate what they are in the series or do we just substitute $n+1$ into the infinite series for $a_{n+1}$ and calculate the limit?

答: $\displaystyle \frac {7^{n+1}}{8^{n+2}}\cdot\frac{8^{n+1}}{7^n}=\frac{7}{8}$ which is the limit as n approaches infinity. Since this is $<1$ the series $\displaystyle \sum_{n=1}^\infty \frac{7^n}{8^{n+1}}$ converges absolutely, in this case we know, to... $\displaystyle \frac 1 8\times \left(\frac{1}{1/8}-1\right)=\frac 7 8$

Title: Ratio test for a series


问:
Find $a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$
A sequence $\left\{a_n\right\}$ is defined as $a_n=a_{n-1}+2a_{n-2}-a_{n-3}$ and $a_1=a_2=\frac{a_3}{3}=1$

Find the value of $$a_1+\frac{a_2}{2}+\frac{a_3}{2^2}+\cdots\infty$$

I actually tried this using difference equation method.Let the solution be of the form $a_n=\lambda^n$
$$\lambda^n=\lambda^{n-1}+2\lambda^{n-2}-\lambda^{n-3}$$ which gives the cubic equation $\lambda^3-\lambda^2-2\lambda+1=0$. But i am not able to find the roots manually.

Body:
Let $$S=\sum_{n=1}^{\infty}\frac{a_n}{2^{n-1}}$$
We have,
$$S=\frac{1}{1}+\frac{1}{2}+\frac{3}{4}+P$$
Where $$P=\sum_{n=4}^{\infty}\frac{a_n}{2^{n-1}}$$
Using the given recurrence we have,
$$P=\sum_{n=4}^{\infty}\frac{a_{n-1}+2a_{n-2}-a_{n-3}}{2^{n-1}}$$
So we get,
$$P=\sum_{n=4}^{\infty}\frac{a_{n-1}}{2^{n-1}}+2\sum_{n=4}^{\infty}\frac{a_{n-2}}{2^{n-1}}-\sum_{n=4}^{\infty}\frac{a_{n-3}}{2^{n-1}}$$
By Change of variable for each summation we get
$$P=\sum_{k=3}^{\infty}\frac{a_k}{2^{k}}+2\sum_{k=2}^{\infty}\frac{a_k}{2^{k+1}}-\sum_{k=1}^{\infty}\frac{a_k}{2^{k+2}}$$
Which implies
$$P=\frac{1}{2}\sum_{k=3}^{\infty}\frac{a_k}{2^{k-1}}+\frac{2}{4}\sum_{k=2}^{\infty}\frac{a_k}{2^{k-1}}-\frac{1}{8}\sum_{k=1}^{\infty}\frac{a_k}{2^{k-1}}$$
$\implies$
$$P=\frac{1}{2}\left(\frac{a_3}{4}+P\right)+\frac{1}{2}\left(\frac{a_2}{2}+\frac{a_3}{4}+P\right)-\frac{S}{8}$$
Using $a_3=3,a_2=1$
$$P=\frac{3}{8}+\frac{P}{2}+\frac{1}{4}+\frac{3}{8}+\frac{P}{2}-\frac{S}{8}$$
Thus we get
$$\frac{S}{8}=\frac{3}{8}+\frac{1}{4}+\frac{3}{8}$$
$\implies$
$$S=8$$


问:
Prove that the tangent space has the same dimension as the manifold
I asked this question a couple of days ago. And I thought that I totally understood the question. However it turned out that I didn't, since the argument I constructed was proved to be wrong just now: Diffeomorphism preserves dimension

The original text from Guillemin and Pollack's book was:

"The dimension of the vector space $T_x(X)$ is, as you expect, the dimension $k$ of $X$. To prove this, we use the smoothness of he inverse $\phi^{-1}$. Choose an open set $W$ in $\mathbf{R}^N$ and a smooth map $\Phi': \mathbf{R}^N \rightarrow \mathbf{R}^k$ that extends $\phi^{-1}$. Then $\Phi'\circ\phi$ is the identity map of U, so the chain rule implies that the sequence of linear transformations

$\mathbf{R}^k\xrightarrow{d\phi_0}T_x(X)\xrightarrow{d\Phi_x'}\mathbf{R}^k$

is the identity map of $\mathbf{R}^k$. It follows that $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism, so $\dim T_x(X)=k$."

Just for everyone's information, in GP's book, the convention is that $\phi$ is a diffeomorphism from $\mathbf{R}^k$ to a k-dimensional manifold $X\subset \mathbf{R}^N$ ($\phi :\mathbf{R}^k\rightarrow X$).

My question is here:

How does $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism follow from previous argument? From the link given, it was shown that the existence of one-sided inverse does not guarantee that a linear map is an isomorphism. Thus I think there is a loophole in arguing "it follows that $d\phi_0 :\mathbf{R}^k \rightarrow T_x(X)$ is an isomorphism"...

Thanks a lot for your help!

答: That $d\phi_0$ has a left inverse implies that it is an injective linear map. The tangent space $T_x(X)$ is defined by Guillemin and Pollack to precisely be the image of $\mathbb{R}^k$ under the map $d\phi_0$, so it is also a surjection. Thus $d\phi_0$ is a linear isomorphism.</p>


答:
$$S_n=\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}=\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}-\sum_{i=0}^2\ (-1)^n \frac{x^{2n}}{2n!}$$
$$S_n=\frac{x^2-1}2+\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}$$ If you are aware of the complete and incomplete gamma function
$$\sum_{i=0}^n\ (-1)^n \frac{x^{2n}}{2n!}=\frac{e^{-x^2}\,\, \Gamma \left(n+1,-x^2\right)}{2 \Gamma (n+1)}$$

$$S_n=\frac{1}{2} \left(e^{-x^2}\frac{ \Gamma \left(n+1,-x^2\right)}{\Gamma
   (n+1)}+x^2-1\right)$$ which, for a given $n$ or a given $x$ can cancel.

As @bjorn93 did, let us use $x=10$ and $n=5$. $S_5$ will become negative as soon as $x>1.96259$. Checking for $x=2$, we have $-\frac{4}{15}$

Title: Prove that $0<\sum_{i=2}^n\ (-1)^n \frac{x^{2n}}{2n!}$ for $x>0$

Body:
So far, I've tried using induction:

Let P(n): $0<\sum_{i=2}^n\ (-1)^i \frac{x^{2i}}{2i!}$

Let $n=2$. Thus, P(2): $\frac{x^4}{4!}>0$ for $x>0$. Then, let P(k) be true (Induction Hypothesis). Now, prove the validity of P for n=k+1:

$P(k+1) : \sum_{i=2}^{k+1}\ (-1)^i\frac{x^{2i}}{2i!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{k+1}\frac{x^{2(k+1)}}{2(k+1)!} = \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}$

Let $k=2m-1$:

$\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} - (-1)^{k}\frac{x^{2(k+1)}}{2(k+1)!}=\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!}$ which is $>0$ since $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}>0$ and $\sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!} + (-1)^{2m}\frac{x^{4m}}{4m!} > \sum_{i=2}^k\ (-1)^i\frac{x^{2i}}{2i!}$ for $x>0$ according to the Induction Hypothesis.

I'm lost for the case where $k=2m$, I'd like some help there, please.


问:
Cardinality does not change when substracting a countable subset from an uncountable set
This comes from Exercise 11 of Section 2.1 of Kaplansky's "Set Theory and Metric Spaces".

Let $A$ be an uncountable set. Let $B \subseteq A$ be a countable set and let $B^c$ be its complement in $A$. Prove that there exists a bijection between $A$ and $B^c$.

I've found some related questions here: removing a countable subset from an infinite set doesn't change cardinality, and here: Altering an Infinite Set does not change cardinality, but none seems to be exactly the one I'm looking for.

Suposedly this problem can be solved without the axiom of choice, due to it being proposed before the axiom's introduction in the book.

It is quite easy to show that $B^c$ must be uncountable, but moreover I need to prove that it has the same cardinality as $A$. That doesn't look so simple. Here are some thoughts I've been considering:

Since $B^c$ can be split into some countable $C$ and uncountable $C^c$, and the same for $C^c$, this problem is equivalent to partitioning $A$ into a countable number of countable sets and its complement, which must be uncountable. And if we prove that it has indeed the same cardinality as $A$, we would've proved that any partition of $A$ must include a proper subset. It is a fact that proper subsets exist in $A$ due to it being infinite, but I don't think that's enough to say that it will be present on every partition. ¿Is there a way to prove this?

As a particular case, we obtain from this result that irrationals and trascendents have both the same cardinality as $\mathbb{R}$.

EDIT

So, following Berci's answer, I wrote this proof.

Since $B$ is countable and $A = B \cup B^c$ is uncountable it follows that $B^c$ is uncountable. Then it contains an uncountable infinite subset; call it $B_1$. By the same argument $B^c \backslash B_1$ must be uncountable. On the other hand, we have that $B \cup B_1$ is countable, so there exists a one to one correspondence between $B$, $B_1$ and $B \cup B_1$. And since $A \backslash B \cup B_1$ and $B \backslash B_1$ both have the same elements, it follows that there must exist a bijection between $A$ and $B^c$.

I think that's how it is suposed to be solved. I have some doubts about the conclusion though.

答:
It cannot be done in ZF. It has been shown to be consistent with ZF that there exists $x$ which is not Tarski-finite but is not Dedekind-infinite. That is, $x$ is not a bijective image of any bounded subset of $\Bbb N$ and $x$ has no countably-infinite subset.

If such an $x$ exists, let $A=(x\times \{0\})\cup (\Bbb N\times \{1\})$ and let $B=\Bbb N\times \{1\}$. Now if $f:A\to B^c=x\times \{0\}$ were a bijection then the image $f[B]$ would be a countably infinite subset of $x\times \{0\}$ which would imply that $\{y\in x: (y,0)\in f[B]\}$ is a countably infinite subset of $x$, a contradiction

.


问:
Show that the binomial formula $f\left(x+1\right)$ is correct (from Hoel)
This is from Hoel Probability book.
The question is: In calculating binomial probabilities, it is
convenient to calculate $f\left(x+1\right)$ from $f\left(x\right)$
by the formula $f\left(x+1\right)=k\left(x\right)f\left(x\right)$,
where $k\left(x\right)=\frac{\left[\frac{n-x}{x+1}\right]}{\frac{p}{q}}$.
Show that this formula is correct.

Accordingly, I should start from the binomial formula  $f\left(x\right)=\frac{n!}{\left(x+1\right)!\left(n-\left(x+1\right)\right)!}p^{x+1}q^{n-\left(x+1\right)}$ but I don't know how to proceed during the expassio of terms. Not sure if this is the right approach to answer the question.

Body: All you need is $(x+1)!=(x+1)x!, (n-x-1)!=\frac {(n-x)!} {(n-x)}$ and $p^{x+1}q^{n-(x+1)}=p^{x}q^{n-x} \frac p q$.


问:
For what integer values of $k$ does $a^{2}+b^{2}=kc^{2}$ have positive integer solutions?
The obvious case of $k=1$ clearly has solutions, for example, $(3,4,5)$ and $(5,12,13)$, among others. Trivially, it has infinitely many solutions by multiplying through by any number. It has infinitely many solutions aside from those that can be parametrized by $(2nm,n^{2}-m^{2},n^{2}+m^{2})$ for all $n,m\in\mathbb{Z}$. This is a well-known result.

The case of $k=2$ also has infinitely many solutions, among them $(1,7,5)$. As in the previous example I can simply multiply through by any number and get another valid triple. At some point just after Christmas I had worked out a parametrization (using the method of finding a rational point on the corresponding circle), but I cannot recall it off the top of my head. It was not much more complicated than for the case of $k=1$, however, and I sincerely doubt that I was the first one to have found it.

The case of $k=3$ is one for which I have been unable to find any solutions, but I cannot think of a proof that it has no solutions. I cannot find any rational points on the circle $x^{2}+y^{2}=3$, but that does not mean that those points do not exist.

The case of $k=4$ has solutions as well. Given $a^{2}+b^{2}=4c^{2}$, merely allow $d=2c$ and we return to the case of $k=1$ with the question of finding triples that satisfy $a^{2}+b^{2}=d^{2}$. This shows that the original question that I posed is equivalent to merely asking which squarefree integers $k$ have solutions.

I have similarly found solutions in the case of $k=5$, for example, $(11,2,5)$. Try as I might, I have found nothing for the cases of $k=6$ or $k=7$. The cases of $k=8$ and $k=9$ have solutions by a similar argument as the previous paragraph.

Initially, my guess was that there no solutions iff $k\cong3\mod4$ given my inability to find solutions in the $k=3$ and $k=7$ cases, until I decided to look at the $k=6$ case as well and found that I was similarly stumped. I feel like there's something else obvious that I'm overlooking, but whether it's solutions for the cases I can't find a solution for or some other connection between these numbers, I don't have a clue.

My apologies if this has been asked before, I couldn't find anything when I did a google search.

答:
The sum of two squares theorem tells you that for a given integer $k$, the equation
$$a^2+b^2=kc^2,$$
has a solution in positive integers if and only if $k$ does not have a prime factor $p\equiv3\pmod{4}$ with odd multiplicity. That is, if $k=\prod p^{v_p}$ is the prime factorization of $k$, then there exist solutions if an only if $v_p\equiv0\pmod{2}$ whenever $p\equiv3\pmod{4}$.


问:
Invariant subspaces of tensor product of SU(2)
Let $\varphi_n$ denote the standart irreducible representation of $SU(2)$ group with highest weight $n$.

I know that irreducible representations of $\varphi_2 \otimes \varphi_3 = \varphi_5 \oplus \varphi_3 \oplus \varphi_1$ (according to Clebsh-Gordan decomposition).

What will be the invariant subspace of $\varphi_2 \otimes \varphi_3$?

What will be the explicit form of the actions of irreducible reps on invariant subspaces?

答:
Working under the premise that you want to find, e.g. the submodule isomorphic to $\varphi_3$ inside the tensor product $\varphi_2\otimes\varphi_3$. Let the vectors
${\cal B}_2=\{x_0,x_1,x_2\}$ be a basis consisting of weight vectors of $\varphi_2$ (of respective weights 2,0,-2). Let similarly ${\cal B}_3=\{y_0,y_1,y_2,y_3\}$ be a basis of weight vectors of $\varphi_3$ (of respective weights 3,1,-1,-3). Then the space of vectors of weight three in the tensor product consists of the linear combinations
$$
a x_0\otimes y_1+bx_1\otimes y_0,
$$
where $a,b$ are arbitrary scalars. Calculate the effect of the raising operator (= a ladder operator that increases the weight) to such a linear combination. I dare not do that myself, because the details depend on how you normalized things and produced the bases ${\cal B}_2$ and ${\cal B}_3$ in the first place. You will find that - up to a scalar multiple - there is exactly one such linear combination that is annihilated by the ladder operator. That linear combination is then a high weight vector of weight three, and generates the copy $\varphi_3$.

You can similarly find (up to scalar multiple) a unique linear combination
$$
cx_0\otimes y_2+dx_1\otimes y_1+ex_2\otimes y_0
$$
of weight one vectors that is annihilated by the raising operator. Tha generates the remaining summand $\varphi_1$ inside $\varphi_2\otimes\varphi_2$.


问:
Nested equilateral triangles
Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.

Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.

Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$...

答:
Here I will show how you can draw the figure. The calculations that follow are somewhat similar to other answers.

We can build triangle $DEF$ by scaling down triangle $ABC$ and rotating it around its centroid $G$ by some angle $\alpha$.

Notice the hatched triangle $DHG$. This is a right triangle with sides:
$$GH = \frac13 AH =\frac{4}{12} AH$$
$$DG = \frac23 DY = \frac23 \times \frac58 AH = \frac{5}{12} AH $$
$$HD = \sqrt{DG^2 - GH^2} = \frac{3}{12} AH$$
It seems the designer of the problem intentionally used the ratio $\frac58$ to lead us to the $3-4-5$ right triangle!

Anyway, Now we are able to locate point $D$ on $BC$. It is at distance $\frac14 AH$ from point $H$ , the midpoint of BC. We can similarly locate points $E$ and $F$ , hence constructing triangle $DEF$.

Now, as for calculations, note that $\widehat{DEC}=120^o - \alpha$ and that $\widehat{DGH}=60^o - \alpha$. So $\widehat{DEC}=60^o+\widehat{DGH}$ . Therefore:
$$\cos \widehat{DEC} = \cos (60^o+\widehat{DGH}) = \cos60^o \cos \widehat{DGH} - \sin60^0 \sin \widehat{DGH}$$
$$= \frac12 \times \frac45 - \frac{\sqrt 3}{2} \times \frac35 = \frac{4-3\sqrt3}{10}$$
Note that when locating point $D$ we have two options, say, above and below $H$. The above calculations are for the option shown in the figure, where $D$ is selected below $H$. Alternatively, we could select $D$ to be above $H$, and respectively have $E$ below $K$. The interested reader can verify that in that case $\widehat{DEC}$ would be $\alpha$. And  $\; \cos\alpha \;$ can be calculated similar to above.
$$\cos \alpha = \cos (60^o-\widehat{DGH}) = \cos60^o \cos \widehat{DGH} + \sin60^0 \sin \widehat{DGH} = \frac{4+3\sqrt3}{10}$$

Title: Nested equilateral triangles


问:
Complex Gaussian Integral by change of variable
In this video (minute 5:20), they prove that
$$ I = \int_{-\infty}^{\infty}e^{i x^2}dx = e^{i\frac{\pi}{4}}\sqrt{\pi}$$
by deforming the integration contour in the complex plane from the real axis to a straight line crossing the origin at a $\frac{\pi}{4}$ angle. In other words, they perform the change of variable $ x = e^{i\frac{\pi}{4}}t$, and change the integration measure accordingly, i.e., $dx = e^{i \frac{\pi}{4}}dt$. Then, they claim that
$$ I = e^{i\frac{\pi}{4}}\int_{-\infty}^{\infty}e^{-x^2}dx,$$
where the resulting integral is a well known result.
I don't distrust the result (in fact, it's proven in this other question), but I'm concerned about the change of variable.
Why didn't they change the bounds of the integral to $\pm(1+i)\infty$? Is it always possible to proceed as they did?

P.S. The video is part of an online course on complex analysis by the MISiS

答:
In fact, video does not provide a proof thar this change is permissible. Because the function is even, $I = \int_{-\infty}^{\infty}e^{i x^2}dx =\frac{1}{2}\int_{0}^{\infty}e^{i x^2}dx$, it can be proved by means of consideration of integral along the following closed contour in the complex plane of the function $e^{i x^2}$:

$R$ is a big radius ($\to{\infty}$) and $r$ is a small radius ($\to0$). On the positive axis we get our initial integral; the integral along the line $y=ix$ is $\frac{i+1}{\sqrt2}\int_{R}^{r}e^{i\frac{1}{2} x^2(i+1)}dx=\exp(\frac{\pi{i}}{4})\int_{R}^{r}e^{- x^2}dx$.

The integral along a big circle can be evaluated by means of Jordan's lemma as $I_R=\int_{0}^{\frac{\pi}{4}}e^{iR^2e^{2i\phi}}iRe^{i\phi}d\phi<\int_{0}^{\frac{\pi}{4}}e^{-R^2{\sin(2\phi)}}Rd\phi<\frac{1}{2}\int_{0}^{\frac{\pi}{2}}e^{-R^2\frac{2\phi}{\pi}}Rd\phi=\frac{\pi}{4R}(1-e^{-R^2})\to0$ as $R\to\infty$. It is easy to evaluate the integral along a small circle $r$: $I_r<Ar $ ($A=const$) and $\to0$ as $r\to0$.

Given the fact that there is no singularities inside the chosen contour $\oint_C=0$,
and we get $\int_{0}^{\infty}e^{i x^2}dx+\exp(\frac{\pi{i}}{4})\int_{\infty}^{0}e^{- x^2}dx=0$, or $$\int_{0}^{\infty}e^{i x^2}dx=\frac{\sqrt{\pi}}{2}\exp(\frac{\pi{i}}{4})$$


问:
How many subset of $A=\{1,2,3...,35\} $ divisible by $5$.
How many subset of $A=\{1,2,3...,35\} $ has exeact 26 elements such that sum of all elements divisible by $5$.

I try set: $A(0),A(1),...,A(4)$ are set, which has 26 elements in $A$, satisfy sum of all elements divide $5$ remainder$0,1,2,3,4$.
We have $|A(0)|+|A(1)|+|A(2)|+|A(3)| +|A(4)|=\binom{35}{26}$

That's enought, if we can count each, but I can't to count continous.

答:
First we provide a brutal approach with generating functions. This is not meant to be insightful directly (though illustrates generating functions), but with numerics we can make a hint/observation at the very end. Denote
$$
A = \prod_{k=1}^{35} (1+x^kt)= (1+xt)(1+x^2t)(1+x^3t)\cdots(1+x^{35}t)=\sum c_{n,m}x^nt^m.
$$

Here $x^i$ keeps tracks of the use of the integer $i$ in the sum, and the variable $t$ keeps track of how many summands we use. Hence $c_{n,m}$ gives the number of ways to use $m$ many distinct numbers from $\{1,\ldots,35\}$ that sum to $n$. So for our problem seeking sums that are multiples of 5 with 26 summands, we look for the sum of coefficients
$$
\sum _k c_{5k,26}.
$$

Using SageMath(https://sagecell.sagemath.org/) does quick work for us:

```
var('x','t')
A = 1
for i in [1..35]:
    A = A* (1 + x^i*t)

m = A.expand().coefficient(t^26)
c = 0
for i in [0..150]:
    c = c + m.coefficient(x^(5*i))
c
---
output:
14121492
```

You can use this value to check your work.

Now to give you a hint/observation:
$$
\frac{35 \choose 26}{14121492} = 5.
$$

So perhaps you can make a guess on $|A(0)|,|A(1)|,|A(2)|,|A(3)|,|A(4)|$.

Hint 2.
We might have guess that $|A(0)| = |A(1)| =|A(2)| = |A(3)| = |A(4)|$, as $14121492$ goes in exactly 5 times into $35 \choose 26$. But how can we prove this? Well if we can establish bijections between these sets $A(i)$, then that shows their sizes are the same. Let us consider an element from $A(0)$, this is a 26-element set from $\{1,\ldots,35\}$ whose sum has remainder 0 modulo 5. Say they are
$$
a_1 , a_2 , \ldots, a_{26}.
$$
Now consider a new 26-element set where we add 1 to each of the elements above:
$$
a_1 +1 , a_2 +1 , \ldots, a_{26}+1
$$
What can we say about this new set? If we add these numbers up, we get a sum of $26 \equiv 1\mod 5$, which means this is an element from $A(1)$!. (Small detail, if $a_i+1$ exceeds 35, then we consider the number mod 35). This process is reversible by subtracting 1 from each element in an element of $A(1)$. Thus $|A(0)|=|A(1)|$. The rest can be shown similarly to show $|A(0)| = |A(i)|$ by adding $i$ to each entry like before.

Remark. This worked out nicely like this because we happen to have $gcd(26,5)=1$. If instead you are looking for sums that are divisible by 5 with using exactly 25 elements instead, then we won't have such nice simple bijections.

Body:
First we provide a brutal approach with generating functions. This is not meant to be insightful directly (though illustrates generating functions), but with numerics we can make a hint/observation at the very end. Denote
$$
A = \prod_{k=1}^{35} (1+x^kt)= (1+xt)(1+x^2t)(1+x^3t)\cdots(1+x^{35}t)=\sum c_{n,m}x^nt^m.
$$

Here $x^i$ keeps tracks of the use of the integer $i$ in the sum, and the variable $t$ keeps track of how many summands we use. Hence $c_{n,m}$ gives the number of ways to use $m$ many distinct numbers from $\{1,\ldots,35\}$ that sum to $n$. So for our problem seeking sums that are multiples of 5 with 26 summands, we look for the sum of coefficients
$$
\sum _k c_{5k,26}.
$$

Using SageMath(https://sagecell.sagemath.org/) does quick work for us:

var('x','t')
A = 1
for i in [1..35]:
    A = A* (1 + x^i*t)

m = A.expand().coefficient(t^26)
c = 0
for i in [0..150]:
    c = c + m.coefficient(x^(5*i))
c
---
output:
14121492

You can use this value to check your work.

Now to give you a hint/observation:
$$
\frac{35 \choose 26}{14121492} = 5.
$$

So perhaps you can make a guess on $|A(0)|,|A(1)|,|A(2)|,|A(3)|,|A(4)|$.

Hint 2.
We might have guess that $|A(0)| = |A(1)| =|A(2)| = |A(3)| = |A(4)|$, as $14121492$ goes in exactly 5 times into $35 \choose 26$. But how can we prove this? Well if we can establish bijections between these sets $A(i)$, then that shows their sizes are the same. Let us consider an element from $A(0)$, this is a 26-element set from $\{1,\ldots,35\}$ whose sum has remainder 0 modulo 5. Say they are
$$
a_1 , a_2 , \ldots, a_{26}.
$$
Now consider a new 26-element set where we add 1 to each of the elements above:
$$
a_1 +1 , a_2 +1 , \ldots, a_{26}+1
$$
What can we say about this new set? If we add these numbers up, we get a sum of $26 \equiv 1\mod 5$, which means this is an element from $A(1)$!. (Small detail, if $a_i+1$ exceeds 35, then we consider the number mod 35). This process is reversible by subtracting 1 from each element in an element of $A(1)$. Thus $|A(0)|=|A(1)|$. The rest can be shown similarly to show $|A(0)| = |A(i)|$ by adding $i$ to each entry like before.

Remark. This worked out nicely like this because we happen to have $gcd(26,5)=1$. If instead you are looking for sums that are divisible by 5 with using exactly 25 elements instead, then we won't have such nice simple bijections.


问:
Parametric equation gives dy/dt = sin(t) +3 and x(t)= 6t^2+ln(t) and asks for when the graph of the position will have a horizontal tangent
To my understanding if $\dfrac{dy}{dt}=0$ then the tangent is horizontal, but $\sin(t)+3$ will never equal $0$, therefore there is no horizontal tangent. Thanks!

答:
You cannot conclude so fast
$$x=6 t^2+\log (t)  \qquad \qquad \frac{dx}{dt}=12 t+\frac{1}{t}$$
$$\frac{dy}{dx}=\frac{\sin (t)+3}{12 t+\frac{1}{t}}$$ What happens when $t \to 0$ ?


问:
$Z(y^2-x^3-ax-b)$ is smooth if and only if $4a^3+27b^2\neq 0$
I want to show that $Z(y^2-x^3-ax-b)\subset \mathbb A^2$ is smooth if and only if $4a^3+27b^2\neq 0$. Let $f=y^2-x^3-ax-b$.

I know the following theorem:

If we show that $f$ is irreducible then $I(X)=(f)$.Then the Jacobi matrix is $(-3x^2-a,2y)$. If we know that $X$ is irreducible, then $X$ is smooth exactly when this matrix has rank $1$. The matrix has rank $1$ whenever both terms are not zero. I don't see how this is true only when $4a^3+27b^2= 0$.

Things I am confused about:

Why is $f$ irreducible?

Why is $X$ irreducible?

Why does the Jacobi have rank zero exactly when $4a^3+27b^2= 0$?

答:
Let $f = x^3 + ax + b$, let $F = y^2 - f(x) = y^2-x^3-ax-b$, and let $X = Z(F)$. (Sorry for changing your notation, but I'll just confuse myself if I don't have $X$ given by $y^2 = f(x)$.)

Irreducibility of $F$ basically follows from the fact $F$ is degree $2$ in $y$ and $x^3 + ax + b$ is not a square. If $F$ factors, then we can write it as a product of two monic, degree $1$ polynomials in $y$. Writing
$$
(y + g(x))(y + h(x)) = y^2 - x^3 - ax - b
$$
we find $h(x) = -g(x)$, so $y^2 - g(x)^2 = y^2 - x^3 - ax - b$ and hence $x^3 + ax + b = g(x)^2$. But $x^3 + ax + b$ is not a square in $k[x]$ (it has odd degree, for one thing), so this is impossible. (See here for more details.)

As mentioned in the comments, since $F$ is irreducible and $k[x,y]$ is a UFD, then the ideal $(F)$ is prime. Since prime ideals correspond to irreducible varieties (see here, for example), then $X = Z(F)$ is irreducible.

Note that $F_x = -f'(x)$ and $F_y = 2y$. Suppose that $(u,v)$ is a singular point of $X$, so $F(u,v) = F_x(u,v) = F_y(u,v) = 0$ by the Jacobian criterion. Then $0 = F_y(u,v) = 2v$, so $v = 0$. Thus $(u,v) = (u,0)$, and $u$ is a root of both $f'(x)$ and $F(x,0) = f(x)$. This implies that $u$ is a multiple root of $f$, hence the discriminant of $f$ is zero. The discriminant of the cubic $f(x) = x^3 + ax + b$ is $-(4a^3 + 27b^2)$, so the claim follows. Moreover, all the implications in this argument are reversible, so we get the iff statement.


问:
Proposition about Weierstrass's elliptic function, or $\wp$ function.
I cannot prove the proposition about Elliptic functions.

Proposition;

Every Elliptic function $f$ of order 2 whose pole set is contained in the lattice $\Lambda$ is written as $f=a+b \wp $, where $\wp$ is Weierstrass's elliptic function, i.e.

\begin{equation}
\wp (z)=\dfrac{1}{z^2} + \sum_{\omega \in \Lambda-\{0\}} \bigg( \dfrac{1}{(z-\omega)^2} - \dfrac{1}{\omega^2} \Bigg)
\end{equation}

In this page Direct construction of an arbitrary elliptic function of order $2$ with pole set contained in its lattice. , it is said that
for every $a \in \Lambda$, the Laurent series of $f$ about $a$ is written as
\begin{equation}
f(z)=\sum_{n=-2}^{\infty} c_n (z-a)^n \quad (c_{-2}\neq 0).
\end{equation}

And let $g:=f-c_{-2} \wp$.

The Laurent series of $g(z)$ about $a$ is written as
\begin{equation}
g(z)=\sum_{n=-1}^{\infty} c_{n} (z-a)^n.
\end{equation}

But I cannot understand why $c_{-2}$ disappears.

I have
\begin{equation}
g(z)=
\sum_{n=-2}^{\infty} c_n (z-a)^n -c_{-2}  \left( \dfrac{1}{z^2} + \sum_{\omega \in \Lambda-\{0\}} \bigg( \dfrac{1}{(z-\omega)^2} - \dfrac{1}{\omega^2} \Bigg) \right).
\end{equation}

I don't know what should I do in order to see the Laurent series of $g(z)$ about $a$. How should I deformate $g(z)$?

I would like you to give me some ideas.

答:
Suppose $f(z)=\sum\limits_{n=-2}^{\infty} c_n(z-a)^n$ is $f$'s Laurent series at $z=a$.

We know the first term in $\wp(z)$'s Laurent expansion around $z=a$ is $(z-a)^{-2}$, so therefore $f(z)$ and $c_{-2}\wp(z)$ share the same first term of their Laurent expansions, which means if we subtract one from the other that term cancels out, and you're only left with powers $(z-a)^n$ for $n\ge-1$.


问:
$\sum_i x_i^n = 0$ for all $n$ implies $x_i = 0$
Here is a statement that seems prima facie obvious, but when I try to prove it, I am lost.

Let $x_1 , x_2, \dots, x_k$ be complex numbers satisfying:

$$x_1 + x_2+ \dots + x_k = 0$$

$$x_1^2 + x_2^2+ \dots + x_k^2 = 0$$

$$x_1^3 + x_2^3+ \dots + x_k^3 = 0$$

$$\dots$$

Then $x_1 = x_2 = \dots = x_k = 0$.

The statement seems obvious because we have more than $k$ constraints (constraints that are in some sense, "independent") on $k$ variables, so they should determine the variables uniquely.  But my attempts so far of formalizing this intuition have failed.  So, how do you prove this statement?  Is there a generalization of my intuition?

答:
Rename the $x_i$ as $\alpha_i$ so that I can borrow the notation of this wikipedia article. Multiply the first row of the matrix in the article by $\alpha_1$, the second by $\alpha_2$, and so on. This gives a matrix with determinant

$$\left( \prod_i \alpha_i \right)\left(\prod_{i< j} (\alpha_i-\alpha_j) \right).$$

Since the homogeneous equation with these coefficients has a solution (the vector $(1,1,\cdots,1)$), the determinant must be zero. So either $\alpha_i=0$ for some $i$ or $\alpha_i=\alpha_j$ for some pair of distinct indices. In the first case, we reduce to the same problem with $k-1$ terms and can induct. In the second case, we can consider a slightly modified problem: identifying the $i$th and $j$th term, we have the same problem with $n-1$ variables, except the coefficient of $\alpha_j$ is $2$. But, since this still gives a nonzero solution to the corresponding homogeneous equation, we can again "induct" using the method described above. The number of variables must eventually decrease to one, from which we see $\alpha_1=0$, and hence $x_i=0$ for all $i$ in the original problem.


答:
Let $\Gamma$ be the set of all $26$-element subsets of $\mathbb{Z}_{35}$. Define the partition

$$ \Gamma=\Gamma_0\sqcup \Gamma_1\sqcup\Gamma_2\sqcup\Gamma_3\sqcup\Gamma_4 $$

where $\Gamma_k\subseteq\Gamma$ consists of those $A\in\Gamma$ with $\sum_{a\in A}a\equiv  k\bmod 5$.

There is a function $\tau$ on $\Gamma$ defined by $\tau(A)=\{a+1\mid a\in A\}$. You can show $\tau(\Gamma_k)=\Gamma_{k+1}$ (with the index interpreted $\bmod 5$). Therefore, each  $\Gamma_k$ has the same size, which must therefore be $1/5$th that of $\Gamma$.

In conclusion, $|\Gamma_0|=\frac{1}{5}\binom{35}{26}$.

Title: How many subset of $A=\{1,2,3...,35\} $ divisible by $5$.

Body:
How many subset of $A=\{1,2,3...,35\} $ has exeact 26 elements such that sum of all elements divisible by $5$.

I try set: $A(0),A(1),...,A(4)$ are set, which has 26 elements in $A$, satisfy sum of all elements divide $5$ remainder$0,1,2,3,4$.
We have $|A(0)|+|A(1)|+|A(2)|+|A(3)| +|A(4)|=\binom{35}{26}$

That's enought, if we can count each, but I can't to count continous.


问:
Integral on sphere and ellipsoid
Let $a,b,c \in \mathbb{R},$ $\mathbf{A}=\left[\begin{array}{*{20}{c}}
\mathbf{a}&{0}&{0}\\
{0}&\mathbf{b}&{0}\\
{0}&{0}&\mathbf{c}
\end{array}\right] , ~~\det A >1$

Let $~D = \{(x_1,x_2,x_3): x_1^2 + x_2^2 +x_3^2 \leq 1  \}~$ and

$~E = \left\{(x_1,x_2,x_3): \frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} + \frac{x_3^2}{c^2} \leq 1  \right\}~.$

Then for a compactly supported continuous function $f$ on $\mathbb{R}^3$, could anyone tell me which of the following are correct?

1.$\int_D f(Ax)dx = \int_E f(x)dx  $

2.$\int_D f(Ax)dx = \frac{1}{abc} \int_D f(x)dx $

3.$\int_D f(Ax)dx = \frac{1}{abc} \int_E f(x)dx $

4.$\int_{\mathbb{R}^3} f(Ax)dx = \frac{1}{abc} \int_{\mathbb{R}^3} f(x)dx $

答:
A function $~f~$ is said to be compactly supported if it is zero outside a compact set.

Let $~x=(x_1,x_2,x_3)\in\mathbb R^3~,$ be any arbitrary vector.
$$\therefore~~Ax=\left[\begin{array}{*{20}{c}}
\mathbf{a}&{0}&{0}\\
{0}&\mathbf{b}&{0}\\
{0}&{0}&\mathbf{c}
\end{array}\right]\left[\begin{array}{*{20}{c}}
{x_1}\\{x_2}\\{x_3}
\end{array}\right]=(ax_1,bx_2,cx_3)$$
Now $$\int_D f(Ax)dx =\iiint_{x_1^2+x_2^2+x_3^2\le1} f(ax_1,bx_2,cx_3)~dx_1~dx_2~dx_3\tag1$$
Putting $~~ax_1=y_1,~~bx_2=y_2,~~cx_3=y_3   \implies x_1=\dfrac{y_1}{a},~~x_2=\dfrac{y_2}{b},~~x_3=\dfrac{y_3}{c}$ $$\implies dx_1=\dfrac{dy_1}{a},~~dx_2=\dfrac{dy_2}{b},~~dx_3=\dfrac{dy_3}{c}$$
So $$x_1^2+x_2^2+x_3^2\le1\implies \dfrac{y_1^2}{a^2}+\dfrac{y_2^2}{b^2}+\dfrac{y_3^2}{c^2}\le1$$
So from $(1)$, $$\int_D f(Ax)dx =\iiint_{\frac{y_1^2}{a^2}+\frac{y_2^2}{b^2}+\frac{y_3^2}{c^2}\le1} f(y_1,y_2,y_3)~\dfrac{dy_1}{a}~\dfrac{dy_2}{b}~\dfrac{dy_3}{c}$$
$$=\dfrac{1}{abc}\int_E f(y) dy=\dfrac{1}{abc}\int_E f(x) dx$$ $($ as $\det A>1$, therefore $abc>1$ $)$

Thus option $(3)$ is correct but option $(1)$ and $(2)$ are not correct.

Similarly, we can show that $$\int_{\mathbb R^3} f(Ax)dx =\dfrac{1}{abc}\int_{\mathbb R^3} f(x) dx$$ Hence option $(3)$ and option $(4)$ are the only correct options.

Title: Integral on sphere and ellipsoid


问:
Show a graph with maximum degree 5 $(\Delta(G) =5 )$ is 2 colorable with components as paths or cycles for every color

Let $G$ be a graph with maximum degree 5 $(\Delta(G) =5 )$ Then there exists a coloring of $G$ in two colors(not necessarily proper) such that each connected component of the subgraphs induced by those colors is a path or a cycle

I already know $\chi(G)\leq \Delta(G)+1=6$, so I figured I'd take a 6 coloring of $G$ and merge color classes $1,2,3$ and $4,5,6$. That is a 2 coloring of $G$ in colors 1 and 2 respectively, but I don't know that each connected component of all the vertices of the same color is a path or a cycle, so that lead me to thinking this might not be true, However I can't think of a counter example. Any ideas?

答:
Pick the $2$-coloring which maximizes the number of edges with opposite-color endpoints.

Then each vertex $v$ has at most two neighbors of the same color. If $v$ had three neighbors of the same color, it would have at most two neighbors of the opposite color, and you could increase the number of edges with opposite-color endpoints by switching the color of $v$.

It follows that the subgraph induced by each color has maximum degree $2$, so each subgraph is the union of paths and cycles.


问:
If M is a finitely generated projective module then the Homomorphism ring is also projective
Ok so the problem is the following:
$\newcommand{\Hom}{\operatorname{Hom}}$

Given $M$ a finitely generated and projective left $R$-module, prove that $\Hom_{R}(M,R)$ seen as a right $R$-module is also projective.
Here $(fr)(m):=f(rm)$ for all $m\in M$, $f \in \Hom_{R}(M,R)$ and $r \in R$

My current attempt:

Since $M$ is finitely generated and proyective, define a surjective map
$\pi : R^n \to M$, where the map is defined by where it sends the basis of $M$.

Now as $M$ is proyective we have that $M \oplus \ker \pi \cong R^n$.
We also know that there exist a $Z(R)$-isomorphism $\varphi : \Hom_{R}(\bigoplus_{1}^n R,R) \rightarrow \bigoplus_1^n \Hom_R(R,R)$, where $Z(R)$ is the center of the ring.

On the other hand we also have a $Z(R)$-isomorphism
$$\phi: \Hom_{R}(M\oplus \ker\pi, R) \rightarrow \Hom_R(M,R) \oplus \Hom_R(\ker\pi,R).$$
We also have that
$$\Hom_{R}(M,R) \cong \Hom_{R}(R^n,R) \cong \Hom_R(\textstyle\bigoplus_1^nR,R)$$
and that
$$R^n \cong \Hom_R(R^n,R) \cong \Hom_R(M \oplus \ker\pi,R).$$

What I want to achive is to see that $\Hom_R(M,R)$ as a right $R$-module is a summand of a free group but for that I need to find the existence of a $R$-morphism not just a $Z(R)$ one, but I don't really know how to proceed, also I haven't used the fact that $M$ is left and $\Hom_R(M,R)$ is right.

Any help will be appreciated.

答: Since $M$ is finitely generated and projective, we have a split short exact sequence of the form $$0\rightarrow K\rightarrow F\rightarrow M\rightarrow 0$$  where $F$ is a free left $R$-module of finite rank. Since $\operatorname{Hom}_R(\cdot,R)$ is an additive functor, it preserves split short exact sequences. So $$0\rightarrow \operatorname{Hom}_R(M,R)\rightarrow \operatorname{Hom}_R(F,R)\rightarrow \operatorname{Hom}_R(K,R)\rightarrow 0$$ is also a split short exact sequence of right $R$-modules. But $\operatorname{Hom}_R(F,R)$ is a free right $R$-module. This shows $\operatorname{Hom}_R(M,R)$ is projective.


问:
Calculating the end point angle of an arc.
I am currently coding a program that draws a race track by combining multiple line segments that the user defines (which can be either a straight or a curve).

For my curve, I have the data for its radius, length and starting point. Using some maths explained here, I can also calculate the end point coordinates.

What would be a way to calculate the endpoint angle of the arc? It is important for me to know the end angle as the next line segment that connects with the arc will start with the same angle as the arc.

Example track Diagram

Edit

Problem Diagram

I am interested in finding the angle that a tangent intersecting the end point would make with the x axis.

答:
The angle an arc spans (in radians) is $$ \widehat{\rm angle} = \frac{\rm arclength}{\rm radius} $$

Then you simply add up all the angles of all the corners up to the one you are drawing to find where the orientation of the arc end.

When the final orientation is 360° you have completed one circuit of the track.

The more interesting problem is coming up with the arc endpoint coordinates from the standpoint coordinates $(x_1,\,y_1)$, the initial direction $\theta_1$, the radius $r$ and the arc length $s$.

A counter clockwise arc sweeps an angle $\varphi = s/r$ in radians, so the final direction of the track after the arc is $\theta_2 = \theta_1 + \varphi$.

The endpoint coordinates are found with some trigonometry

$$ \pmatrix{x_2 \\ y_2} = \pmatrix{x_1 -r \sin \theta_1 + r \sin \theta_2 \\ y_1 + r \cos \theta_1 - r \cos \theta_2 } $$

A clockwise arc sweeps an angle $\varphi = s/r$ in radians, so the final direction of the track after the arc is $\theta_2 = \theta_1 - \varphi$.

The endpoint coordinates are found with some trigonometry

$$ \pmatrix{x_2 \\ y_2} = \pmatrix{x_1 +r \sin \theta_1 - r \sin \theta_2 \\ y_1 - r \cos \theta_1 + r \cos \theta_2 } $$


问:
Symmetric, transitive and reflexive properties of a matrix
Say I had a relation
\begin{align}
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\end{align}
where $a,b,c,d \in \mathbb{R}$, where $X$ is related to $Y$ if and only if $\det(X) = \det(Y)$  (where $\det(A) = ad-bc$)

So I say it is reflexive since $xRx$ since $\det(X) = \det(X)$

However would it be Symmetric: $xRy$ implies $yRx$ (I would say yes, since to be related $\det(x)=\det(Y)$ so therefore $\det(y) = \det(x)$?)

Transitivity is a hard, $xRy yRb$ then $xRb$ (I would say yes since if $xRy$, $\det(x)=\det(y)$ and then $\det(y) = \det(b)$, so $xRb$ since $\det(b) = \det(x)$

And not anti symmetric since if $X=\{1, 2, 3, 4\}$ and $Y = \{4, 2, 3, 1\}$ then they would be $xRy$ but $X$ does not equal $Y$.

Am I on the right track with these?

答:
You're correct. Since the definition of the given relation uses the equality relation (which is itself reflexive, symmetric, and transitive), we get that the given relation is also reflexive, symmetric, and transitive pretty much for free.

To show that the given relation is not antisymmetric, your counterexample is correct. If we choose matrices $X,Y\in \left\{
\left[ \begin{array}{cc}
a & b \\
c & d
\end{array}\right] \text{ }
\middle| \text{ } a,b,c,d\in \mathbb{R}\right\}$
, where:
$$X=\left[ \begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}\right] \text{ and }
Y=\left[ \begin{array}{cc}
4 & 2 \\
3 & 1
\end{array}\right]$$

Then certainly $X$ is related to $Y$ since $\det(X)=1\cdot4-2\cdot3=-2=4\cdot1-2\cdot3=\det(Y)$. Likewise, since the relation was proven to be symmetric, we know that $Y$ is related to $X$. Yet $X\ne Y$.


问:
Application of Lehmann–Scheffé theorem in a sample with normal distribution

Let $Y_{1},Y_{2},\ldots,Y_{n}\sim N(0,\sigma^{2})$. Prove that $T(Y)=\frac{1}{n}\sum_{k=1}^{n}Y_{i}^{2}$ is the MVUE for $\sigma^{2}$.

My approach:  I need to prove that $T(Y)$ is the MVUE for $\sigma^{2}$, so I need to show that:

$T(Y)$ is a sufficient statistic for $\sigma^{2}$.

$T(Y)$ is a complete statistic.
Therefore, by the Lehmann–Scheffé theorem we can conclude that $T(Y)$ is the MVUE for $\sigma^{2}$.

Since that $$L(\sigma^{2})=\underbrace{\left( \frac{1}{\sqrt{\sigma^{2}} \sqrt{2\pi}}\right)^{n}}_{=h(\mathbf{y})}\underbrace{\exp \left\{-\frac{1}{2\sigma^{2}}\sum_{i=1}^{n}x_{i}^{2} \right\}\cdot \mathbf{1}_{\{x\in \mathbb{R}\}}(x_{i})}_{=g[T(\mathbf{y});\sigma^{2}]}$$
so by the Factorization's theorem we can conclude that $T(Y)$ is a sufficient statistic.

Now, for to prove that I need to show that: For all measurable function $g$ and for all $\sigma^{2}$ we have $$\mathbb{E}_{\sigma^{2}}g(T)=0 \implies \mathbb{P}_{\sigma^{2}}[g(T)=0]=1$$

But how can I prove that part? On the other hand, how would you use Rao-Blackwell's theorem to prove this result?

答:
First observe that Gaussian distribution belongs to the exponential family thus $T$ is not only Sufficient but Complete (and minimal) too.

To use Lehmann Scheffé lemma you have only to show that your estimator is unbiased for the variance:

$$\mathbb{E}[T]=\frac{1}{n} n\mathbb{E}[Y_1^2]=\sigma^2$$

that's all


问:
For which $q\in \mathbb{C}$ is the series $\displaystyle{\sum_n p(n)q^n}$ convergent?
Let $p:\mathbb{C}\rightarrow \mathbb{C}$, $z\mapsto a_kz^k+a_{k-1}z^{k-1}+\ldots +a_1z+a_0$ be a polynomial with coefficients $a_j\in \mathbb{C}$, $j=0, \ldots ,k$ and $a_k\neq 0$.

For which $q\in \mathbb{C}$ is the series $\displaystyle{\sum_n p(n)q^n}$ convergent?

We have that \begin{align*}\sum_n p(n)q^n&=\sum_n \left (a_kn^k+a_{k-1}n^{k-1}+\ldots +a_1n+a_0\right )q^n\\ & =a_k \sum_n n^k q^n +a_{k-1} \sum_n n^{k-1} q^n +\ldots +a_1 \sum_n n q^n + a_0 \sum_n q^n \end{align*}

So we have to check the convergence of a series of the form $\displaystyle{\sum_n n^jq^n}$.

Let $a_n=n^jq^n$.

Then from the ratio test we have that $$\left |\frac{a_{n+1}}{a_n}\right |=\left |\frac{(n+1)^jq^{n+1}}{n^jq^n}\right |=\left |\left (1+\frac{1}{n}\right )^jq\right |$$

To get convergence we want that this is smaller than $1$, right?

So we set $\left |\left (1+\frac{1}{n}\right )^jq\right |<1$ and solve for $q$ ?

答:
For $n\in\Bbb N$ we have $|p(n)|=|a_k|\cdot |n^k|\cdot |1+f(n)|$ where $\lim_{n\to\infty}f(n)=0.$ So $\lim_{n\to\infty}|p(n)|^{1/n}=1$ because $a_k\ne 0.$ So the  Cauchy-Hadamard Radius Formula tells us that the series converges if $|q|<1$ and diverges if $|q|>1.$

If $|q|=1$ and $k>0$ then  $|p(n)|\to\infty$ as $n\to\infty$ so the terms $|p(n)q^n|=|p(n)|$  do not $\to 0$ as $n\to \infty$ and the series diverges.

If  $|q|=1$ and $k=0$ then $p$ is the non-zero constant $a_0$, so $|p(n)q^n|=|a_0|$ does not $\to 0$ and the series diverges.

The proof of the  Cauchy-Hadamard Radius Formula is not hard nor long. It also shows that if a power series converges on a bounded open disc $D$ and $S=\overline S\subset D$ then the series converges uniformly on $S$. There is a nice presentation in Complex Calculus by Ahlfors.


问:
How do I go about graphing the region, $\Re(z^2)>1$ in the Complex Plane?
The questions asks to describe the region, $\Re(z^2)>1$, graphically in the complex plane. The math is fairly simple. If we let $z=x+iy$, then $z^2=(x+iy)(x+iy)\\z^2=x^2-y^2+i(2xy)$

Hence the $\Re(z^2)=x^2-y^2>1$

Graphically speaking, this function sketches a hyperbola in the real $xy$ plane. However, my confusion is that the question asks to represent $\Re(z^2)>1$ in the complex plane. How can I go about representing the real part of a function in the complex plane?

答: As a set, the complex plane is the same as $\mathbb{R}^2$. Your region is the "exterior" of the unit hyperbola" ( to the right of the right branch and to the left of the left one).


问:
norm of the resolvent inequality
Let $x$ be an element of a Banach Algebra. Let $\lambda \in \rho(x)$, where $\rho(x)$ is the resolvent of $x$.

Let ${d(\lambda, \sigma(x))}$ be the distance between $\lambda$ and $\sigma(x)$.

Show that $\| (\lambda-x)^{-1} \| \geq \frac{1}{d(\lambda, \sigma(x))}$.

Information I know:

$\rho(x)$ is an open set.

$\sigma(x)$ is compact.

I've seen many places using this inequality, but I cannot find a proof of it. And I'm not sure how to prove this by breaking down the definitions I know.

Thanks in advance!

答: Let $\mu \in \sigma (x)$. If $|\lambda -\mu| <\frac 1 {\|(\lambda -x)^{-1}\|}$ then $e+(\lambda -x)^{-1} (\mu-\lambda)$ is invertible and  the identity $(\mu -x) =(\lambda -x)[e+(\lambda -x)^{-1} (\mu-\lambda)]$ shows that $\mu-x$ is invertible. This contradicts the fact that $\mu \in \sigma (x)$. We have proved that $|\lambda -\mu| \geq \frac 1 {\|(\lambda -x)^{-1}\|}$ whenever $\mu \in \sigma (x)$. Taking infimum over all such $\mu$ we get the desired inequality.

Title: norm of the resolvent inequality


问:
Stochastic calculus of Gaussian white noise
I see a strange stochastic integral and don't know how to proceed?

Denote $\xi$ a Gaussian white noise,

Denote

$$I(n)=\int_t^{t+n\Delta t}ds\int_t^sd\xi$$
How to prove that $= \frac{1}{3}\Delta t^3$ and $= \frac{5}{6}\Delta t^3$?

Specifically, can anyone explain how to deal with these kind of integrals generally? Routine stochastic integral often only deals with Brownian motion. Any references and books are also greatly appreciated.

These equations are taken from a theoretical physics paper.

Many thanks!

答:
Welcome to MSE:
may you know that $(\Delta B(t))^2\sim \Delta t$ when $B(t)$ is brownian motion.I wish It helps you. so
$$\int_t^{t+n\Delta t}1ds=n\Delta t=o(\Delta t)$$and
$$\int_t^sd\xi\sim \sum \Delta B(t)=o(\sqrt{\Delta t})$$so It seems that $$I(n)=o(\sqrt{\Delta t}).o(\Delta t)=o(\Delta t^{\frac32})$$ can you proceed  now ?


问:
Solving the integral of $\frac {dx} {1+x^3}$

Compute integral $$\int_0^{\infty} \frac{dx}{1+x^3}$$.

I used the formula from complex variables by fisher (2.6 formula (9)) that $$\int_0^{\infty} \frac{dx}{1+x^a}=\frac {\pi} {a\sin(\pi/a)}$$ This means $$\int_0^{\infty} \frac{dx}{1+x^3}=\frac {\pi} {3\sin(\pi/3)}$$ Is that simply the answer or do I need to expand it to find all concrete values?

Additionally, from this answer here, the improper integral gives a different result:

$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$

is this result equal to the one I got above?

答:
Next, simplify
$$
F(x)=-\frac{1}{6}\ln|x^2-x+1|+\frac{1}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}+\frac{1}{3}\ln|x+1|
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|-\frac{1}{3}\ln\sqrt{|x^2-x+1|}
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln\left(\frac{|x+1|}{\sqrt{|x^2-x+1|}}\right).
$$
Then
$$\int_0^\infty \frac{dx}{1+x^3}=\lim_{X\rightarrow\infty}F(X)-F(0).$$
Compute the limit, and you are done.


问:
Cross product as factor in dot product
Given there are two vectors $w,v$ with $||w||=4$ , $||v||=1$ and $\phi=\frac{2\pi}{3}$

How do you transform the following expression into a form in which it can be computed with the given information?

$$
< w \times 5v,v-3w>
$$

The cross product as first factor is what puzzles me about this

答:
You should know that the dot product is distributive.
$\left<\mathbf{a}, \mathbf{b} + \mathbf{c} \right> = \left<\mathbf{a}, \mathbf{b} \right> + \left<\mathbf{a}, \mathbf{c} \right>$

So you should be able to obtain two dot products from this expression. Each of them would be what is called "scalar triple products."

How is the result of any cross product oriented with respect to the original vectors?

It might also help to be informed that scalars can be taken outside dot and cross products.

$\left<g\mathbf{a}, \mathbf{b} \right> = \left<\mathbf{a}, g\mathbf{b} \right> = g\left<\mathbf{a}, \mathbf{b} \right>  $

$(g\mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (g\mathbf{b}) = g(\mathbf{a} \times \mathbf{b}  )$

Title: Cross product as factor in dot product


问:
Polyhedra which can be perfectly split into self-similar pieces
A cube can be perfectly split into smaller equally sized cubes.  Similarly, a triangular prism can be perfectly split into smaller equally sized triangular prisms.

Is there a name for or list of the set of polyhedra which can be split into smaller self-similar pieces?

答:
You can find some non-trivial examples on my page
Polyhedron rep-tiles (rep-8)
Actually, there are thousands of such polyhedrons even for 8 equal pieces.</p>


问:
How do we prove the proposed expression is nonnegative?
So the expression which I am interested in is given by
\begin{align*}
f(b,\theta) = \frac{2b-1}{b\sqrt{(1-\theta)(2b-1)^{2} + \theta}} + \frac{\sqrt{(1-\theta)(2b-1)^{2} + \theta} - 1}{2b^{2}(\theta - 1)}
\end{align*}
where $b\in(0,1]$, $\theta\in\mathbb{R}_{>0}$ and $\theta\neq 1$.

It is part of my research project and I would like to know a good way to approach it.

Any hint is appreciated.

EDIT

According to WA, the solution is given by $0 < b \leq 1$ when $\theta > 0$.

Body:
Denote $A = (1-\theta)(2b-1)^2 + \theta$.

Since $(2b-1)^2 \le 1$, we have $A \ge (1-\theta)(2b-1)^2 + \theta (2b-1)^2 = (2b-1)^2$.

We have
\begin{align}
f(b, \theta) &= \frac{2b-1}{b\sqrt{A}} + \frac{\sqrt{A} - 1}{2b^2(\theta - 1)} \\
&= \frac{2b-1}{b\sqrt{A}} + \frac{A - 1}{2b^2(\theta - 1)(\sqrt{A} + 1)}\\
&= \frac{2b-1}{b\sqrt{A}} + \frac{2(1-b)}{b(\sqrt{A} + 1)}\\
&= \frac{2b - 1 + \sqrt{A}}{b\sqrt{A}(\sqrt{A} + 1)}\\
&\ge \frac{2b - 1 + \sqrt{(2b-1)^2}}{b\sqrt{A}(\sqrt{A} + 1)}\\
&\ge 0.
\end{align}
We are done.


问:
Find the expected value and variance of $f\left(x\right)=\left(\frac{1}{2}\right)^{x}x=1,2,3,\ldots$
Find the expected value and variance of $f\left(x\right)=\left(\frac{1}{2}\right)^{x}x=1,2,3,\ldots$ where $X$ is a discrete random variable. I try

$E\left[X\right]=\sum_{i=1}^{k}x_{i}f\left(x_{i}\right)=\sum_{i=1}^{\infty}x_{i}\left(\frac{1}{2}\right)^{x_{i}}=1\cdot\left(\frac{1}{2}\right)+2\cdot\left(\frac{1}{2}\right)^{2}+3\cdot\left(\frac{1}{2}\right)^{3}+\cdots$

$$E\left[X\right]=1\cdot\left(\frac{1}{2}\right)+2\cdot\left(\frac{1}{2^{2}}\right)+3\cdot\left(\frac{1}{2^{3}}\right)+\cdots$$
That sum converges to $2$, so $$E\left[X\right]=2$$
$Var\left(X\right)=E\left[\left(X-\mu\right)^{2}\right]=E\left[X^{2}\right]-E\left[X\right]^{2}$

$E\left[X^{2}\right]=\sum_{i=1}^{\infty}x_{i}^{2}\left(\frac{1}{2}\right)^{x_{i}^{2}}\approx0.7678$

$E\left[X\right]^{2}=4$

Not sure if those are the right steps in order to solve the problem.I know is wrong due to the negative resulting variance.

答: Should be $E(X^2)=\sum x_i^2 (\frac 1 2)^{x_i}$. Do not plug the function of x into the density.


问:
Ordered pairs $(m,n)$ such that ${\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m+c^m)}{m}\frac{(a^n+b^n+c^n)}{n}}$.
I am unable to proceed with the following problem, : If $a+b+c=0$ then find all the ordered pairs $(m,n)$ where $m,n\in\mathbb{N}$ such that
$$\boxed{\frac{a^{m+n}+b^{m+n}+c^{m+n}}{m+n}=\frac{(a^m+b^m+c^m)}{m}\frac{(a^n+b^n+c^n)}{n}}$$

I tried but I think I have no idea regarding this kind of problem although I tried a bit hard with all possible methods (which I can do/know) but that doesn't result any substantial thing which I can add and also it will not give a complete set of solutions, this trial approach provides a few solution(and quite laborious and too many limitations) for instance this well known stuff if $a+b+c=0$ then, $\frac{a^5+b^5+c^5}{5}=(\frac{a^3+b^3+c^3}{3})(\frac{a^2+b^2+c^2}{2})$ is one of the solution. Thanks for your attention.

答:
Let us first consider the case $m\le n$.

If $m=n$, then taking $(a,b,c)=(1,-1,0)$, we get $m=(1+(-1)^m)^2$ from which $m=4$ follows. However, when $m=n=4$, the equation does not hold for $(a,b,c)=(2,-1,-1)$.

If $m=1$, then taking $(a,b,c)=(2,-1,-1)$, we get $2^{n}=(-1)^{n}$ which is impossible.

So, in the following, $2\le m\lt n$.

If both $m$ and $n$ are odd, then taking $(a,b,c)=(2,-1,-1)$, we get
$$mn(2^{m+n-1}+1)=2(m+n)(2^{m-1}+(-1)^m)(2^{n-1}+(-1)^n)$$The LHS is odd while the RHS is even, which is impossible.

If both $m$ and $n$ are even, then taking $(a,b,c)=(1,-1,0)$, we get $(m-2)(n-2)=4$, but there are no solutions satisfying $2\le m\lt n$.

If exactly one of $m,n$ is odd, then taking $(a,b,c)=(2,-1,-1)$, we get
$$mn(2^{m+n}-2)=(m+n)(2^m+2(-1)^m)(2^n+2(-1)^n)\tag1$$Suppose here that $m$ is odd. Then, since we have $mn\gt m+n\ (\gt 0)$ (which is equivalent to $(m-1)(n-1)\gt 1$ which is true) and $$2^{m+n}-2\gt (2^m+2(-1)^m)(2^n+2(-1)^n)\ \ (\gt 0)$$ (which is equivalent to $2\gt 2^{m+1}-2^{n+1}$ which is true), the LHS of $(1)$ is larger than the RHS of $(1)$. So, $m$ has to be even to have
$$2^n\bigg(\underbrace{n(m2^{m}-2^{m}-2)-2^mm-2m}_{A}\bigg)+\underbrace{n(2^{m+1}-2m+4)+4m(2^{m-1}+1)}_{\gt 0}=0\tag2$$Now, suppose that $m\ge 4$. Then, we have
$$\begin{align}A=n\underbrace{(m2^{m}-2^{m}-2)}_{\gt 0}-2^mm-2m&\gt m(m2^{m}-2^{m}-2)-2^mm-2m
\\&=4m\bigg((m-2)2^{m-2}-1\bigg)
\\\\&\gt 0\end{align}$$implying that the LHS of $(2)$ is positive. So, we have to have $m=2$. Taking $(a,b,c)=(2,-1,-1)$, we have$$(n-6)2^{n-1}+2n+6=0\tag3$$Suppose here that $n\ge 7$. Then, the LHS of $(3)$ is positive. So, we get $n=3,5$ which are sufficient (see here, here, here).

In conclusion, considering the case $m\gt  n$, we see that the answer is
$$\color{red}{(m,n)=(2,3),(3,2),(2,5),(5,2)}$$


问:
Question Regarding the Axiom of Extensionality
Jech's text on Set Theory states the following:

If X and Y have the same elements, then X = Y :

∀u(u ∈ X ↔ u ∈ Y ) → X = Y.

The converse, namely, if X = Y then u ∈ X ↔ u ∈ Y, is an axiom of
  predicate calculus. Thus we have

X = Y if and only if ∀u(u ∈ X ↔ u ∈ Y).

The axiom expresses the basic idea of a set: A set is determined by
  its elements.

To check my understanding, are the following true?

(1) The axiom of extension states only that if two sets have the same members, then those two sets equal.

(2) It is, on the other hand, an axiom of the language of set theory (i.e., predicate calculus) that if two sets equal, then they have the same members.

A third question:

(3) It seems that the axioms of ZFC are very much independent of the language being worked with.  That is, the axioms of ZFC don't state we must be working in the language predicate calculus.  Then I take it there exist languages we could be working with that don't have the property that if two sets are equal, they have the same members (so that we could only conclude if two sets have the same members, they are the same and not the reverse).  Is this true?  Do mathematicians ever work in languages besides predicate calculus?

答:
Yes to (1) and (2).

As for (3): The language of set theory is $\{\in\}$. Therefore it seems to me that in any language in which you can express set membership you will necessarily be able to conclude "if two sets are equal, they have the same members". Of course, this corresponds to our intuition: two sets that are equal should contain the same elements.

Note that here the membership ranges over elements in the model. If you add elements to the sets that are not elements of the model then you can well have two sets that are equal in the model but not equal (do not contain the same elements) if seen from outside the model.

As for your last question: yes there are logical deduction systems other than first order logic (= predicate calculus). The language of sets $\{\in\}$ will still be $\{\in\}$ so that what I wrote above should still hold.


问:
Validity of Limits? Proof that $\lim_{x \to 1} \frac{x^{\alpha}-1}{x-1} = \alpha$ when $\alpha \in \mathbb{R}$
Synopsis

In Tao's Analysis 1, we are asked the following:

Let $\alpha$ be a real number, and let $f:(0,\infty) \to \mathbb{R}$
be the function $f(x) = x^{\alpha}$. Show that  $\lim_{x \to 1} \frac{f(x)-f(1)}{x-1} = \alpha$ using Exercise 10.4.2 and the comparison principle; you may need to consider right and left limits separately. The fact that the rationals are dense in the reals might also help.

Now, Exercise 10.4.2 proves the result for $f(x) = x^q$ where $q \in \mathbb{Q}$ and the comparison principle is given as follows:

Lemma 6.4.13 (Comparison principle) - Suppose that $(a_n)_{n=m}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ are two sequences of real numbers such that $a_n \leq b_n$ for all $n \geq m$. Then we have the inequalities $$\text{sup}(a_n)_{n=m}^{\infty} \leq \text{sup}(b_n)_{n=m}^{\infty}\\
\text{inf}(a_n)_{n=m}^{\infty} \leq \text{inf}(b_n)_{n=m}^{\infty} \\
\lim \text{sup}_{n \rightarrow \infty}a_n \leq \lim \text{sup}_{n \rightarrow \infty}b_n\\
\lim \text{inf}_{n \rightarrow \infty}a_n \leq \lim \text{inf}_{n \rightarrow \infty}b_n$$

It should be noted that Terence Tao builds the reals through Cauchy sequences of rational numbers, and defines real exponentiations in the following manner:

Definition 6.7.2 (Exponentiation to a real exponent) - Let $x> 0$ be real, and let $\alpha$ be a real number. We define the quantity $x^\alpha$ by the formula $x^{\alpha} = \lim_{n \to \infty} x^{q_n}$, where $(q_n)_{n=1}^{\infty}$ is any sequence fo rational numbers converging to $\alpha$.

Note that this exercise is given when we have not been taught L'Hopital's rule yet, and when we have yet even to prove that the derivative of $x^{\alpha} = \alpha x^{\alpha -1}$ for real $\alpha$. The previous exercise proved the statement for rationals and the exercise before that proved it for integers. This is not a question of computation but of generalization. Please take this into account.

My Attempt

I immediately thought of using double limits. Let $(q_k)_{k=1}^{\infty}$ be a sequence of rationals converging to $\alpha$. Then  $$ \lim_{x \to 1} \frac{x^{\alpha}-1}{x-1}  = \lim_{x \to 1}  \frac{\lim_{k \to \infty}(x^{q_k})-1}{x-1} = \lim_{k \to \infty} \lim_{x \to 1}  \frac{x^{q_k}-1}{x-1} = \lim_{k \to \infty} q_k = \alpha.$$ But this solution doesn't utilize the comparison principle or left and right limits or anything else Tao gives as hints except for the previous exercise. Furthermore, I'm not even sure if the limits are valid. Can any of you show me what's wrong with my solution (if it is wrong), or what Tao may have intended?

Update

Because of a helpful comment, I've realized that I can't abuse limits like I did above. But I'm out of ideas. I can't seem to think of anything for some reason. It's been bothering me for a hour. Any help would be appreciated!

答:
It sounds like they want you to prove it for the integers, then the rationals, then the real numbers.

If $\alpha$ is a positive integer $x^\alpha - 1 = (x-1)(x^{\alpha-1} + x^{\alpha-2} + \cdots + 1)$

$\lim_\limits{x\to 1} \frac {x^{\alpha} - 1}{x-1} = \alpha$

To cover the negative integers, lets assume $\alpha > 0$ and find $\lim_\limits{x\to 1} \frac {x^{-\alpha} - 1}{x-1}$

$x^{-\alpha} - 1 = -x^{\alpha}(x^{\alpha} - 1)$

$\lim_\limits{x\to 1} \frac {-x^{\alpha}(x^{\alpha} - 1)}{x-1} = \lim_\limits{x\to 1} -x^{\alpha}\lim_\limits{x\to 1}\frac {x^{\alpha} - 1}{x-1} = -\alpha$

Next the rationals... Suppose $\alpha = \frac {1}{q}$ with $q\in \mathbb Z^+$

$x^{\frac {1}{q}} - 1 = \frac {(x^{\frac {1}{q}} - 1)(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)} = \frac {x-1}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)}\\
\lim_\limits{x\to 1}\frac {x^{\frac {1}{q}} - 1}{x-1} = \frac {1}{q}$

Suppose $\alpha = \frac {p}{q}$

$x^{\frac {p}{q}} - 1 = \frac {x^p - 1}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1)}\\
\lim_\limits{x\to 1}\frac {x^{\frac {p}{q}} - 1}{x-1} = \lim_\limits{x\to 1}\frac {1}{(x^{q-1}{q} + x^{q-2}{q} + \cdots + 1))}\lim_\limits{x\to 1}\frac {x^p - 1}{x-1} = \frac {p}{q}$

We have proven our proposition is true for all $\mathbb Q \setminus \{0\}$

Now we get into the the hints.  Since the rationals are dense in the reals.  For all real $r$ there is a rational that is arbitrarily close to it.


问:
Stuck on this linear and homogenous PDE
$$\frac{1}{2}\frac{\partial^2w}{\partial y^2}+\sqrt{8x^2+25}\frac{\partial w}{\partial y}+4x^2w=0$$

"Find the general solution of the PDE in $w(x,y)$. State clearly any arbitrary functions that appear in general solutions."

Progression:

So far I tried using the method of separation, letting
$$w(x,y) = X(x)Y(y)$$
after subbing the partial derivatives into the original PDE, I divided the equation by $X(x)Y(y)$ and I got the following:

$$\frac{1}{2}\frac{Y''(y)}{Y(y)} + \sqrt{8x^2+25}\frac{Y'(y)}{Y(y)}+4x^2=0 $$

so usually I would let $X$ on one side and $Y$ on the other side of the equation that = a constant $\gamma$ and decompose into 2 separate ODEs but in this case, I can't find a way to separate $X$ and $Y$.

Sorry for the previous post I will delete it. This is not an assignment just a past year's exam question I was practicing. Thanks in advance for any help!

答:
Hint: Consider the simpler linear DE $\frac{\partial w}{\partial y}+x w=0$. This is equivalent to $$\frac{1}{w}\frac{\partial w}{\partial y}=\frac{\partial}{\partial y}\ln w=-x,$$

which we may immediately integrate w/r/t $y$ to obtain

$$\ln w(x,y)=-x y+C(x)\implies w(x,y)=A(x)e^{-x y}.$$

Note that this is basically equivalent to solving the linear ODE $\frac{dw}{dy}+xw=0$, treating $x$ as a constant, and then understanding that the coefficient of the general solution will depend on $x$ as well. Can you apply this insight to your DE?

Title: Stuck on this linear and homogenous PDE


答:
Let there be a function of a random variable $X$ denoted by $g(X)$. If you want to calculate the expected value you write:
$$E[g(X)] = \sum_{i=1}^{k} g(x_i)f(x_i)$$
In your case this $f(x)$ is the PDF. So if you want to find $E(X^2)$ you write:
$$E(X^2)=\sum_{i=1}^{k}x_i^2f(x_i) = \sum_{i=1}^{k}x_i^2\left(\frac{1}{2}\right) ^{x_i}$$
Hope this helps...

Body:
Find the expected value and variance of $f\left(x\right)=\left(\frac{1}{2}\right)^{x}x=1,2,3,\ldots$ where $X$ is a discrete random variable. I try

$E\left[X\right]=\sum_{i=1}^{k}x_{i}f\left(x_{i}\right)=\sum_{i=1}^{\infty}x_{i}\left(\frac{1}{2}\right)^{x_{i}}=1\cdot\left(\frac{1}{2}\right)+2\cdot\left(\frac{1}{2}\right)^{2}+3\cdot\left(\frac{1}{2}\right)^{3}+\cdots$

$$E\left[X\right]=1\cdot\left(\frac{1}{2}\right)+2\cdot\left(\frac{1}{2^{2}}\right)+3\cdot\left(\frac{1}{2^{3}}\right)+\cdots$$
That sum converges to $2$, so $$E\left[X\right]=2$$
$Var\left(X\right)=E\left[\left(X-\mu\right)^{2}\right]=E\left[X^{2}\right]-E\left[X\right]^{2}$

$E\left[X^{2}\right]=\sum_{i=1}^{\infty}x_{i}^{2}\left(\frac{1}{2}\right)^{x_{i}^{2}}\approx0.7678$

$E\left[X\right]^{2}=4$

Not sure if those are the right steps in order to solve the problem.I know is wrong due to the negative resulting variance.


问:
Show that any nonabelian group $G$ of order $12$ which contains a normal subgroup of order $4$ must be isomorphic to $A_4$.

Show that any nonabelian group $G$ of order $12$ which contains a normal subgroup of order $4$ must be isomorphic to $A_4$.

Hint. Show that $G$ is a split extension of a group of order $4$ by $\mathbb Z_3$, and, using the calculation of $Aut(\mathbb Z_2 \times \mathbb Z_2)$ given in Problem 14 of Exercises 4.6.13, show that here is only one such split extension which is nonabelian. (Exercise 4.6.13 tells us that $Aut(\mathbb Z_2 \times \mathbb Z_2) \cong S_3$.)

We know that $A_4 \cong (\mathbb Z_2 \times \mathbb Z_2) \rtimes_{\alpha} \mathbb Z_3$

Assume that $G$ is a group of order $12$ which has a normal subgroup $H$ of order $4$. Since $3\mid 12$ and is prime, we know that there exists a group of order $3$ in $G$. So we have an extension $f: \mathbb Z_3 \rightarrow G$ with kernel $H$. Since $H$ and $\mathbb Z_3$ are relatively prime, $f$ must be a split extension. Since split extensions are semidirect products, we just need to classify $H \rtimes_{\alpha} \mathbb Z_3$, right?

We know that $H \cong \mathbb Z_2 \times \mathbb Z_2$ or $\mathbb Z_4$.

Case 1:

$\mathbb Z_4 \rtimes_{\alpha} \mathbb Z_3$

We have $\alpha: \mathbb Z_3 \rightarrow Aut(\mathbb Z_4) \cong \mathbb Z_3$, and let $\mathbb Z_3 = \langle a \rangle$

$\bullet$ The homomorphism is trivial $\implies$ $\mathbb Z_4 \rtimes_{\alpha} \mathbb Z_3 \cong \mathbb Z_4 \times \mathbb Z_3$. Since both $\mathbb Z_4$ and $\mathbb Z_3$ are abelian, and the action is trivial; $\mathbb Z_3 \times \mathbb Z_3$ must also be abelian.

$\bullet$ $\alpha_1 : a \rightarrow a$

$\bullet$ $\alpha_2 : a \rightarrow a^2$

Let $\iota: H \rightarrow H \rtimes_{\alpha} K$, $\iota': H \rightarrow H \rtimes_{\alpha'} K$, $s: K \rightarrow H \rtimes_{\alpha} K$, and $s' : K \rightarrow H \rtimes_{\alpha'} K$

There is a proposition in our textbook that states:

Let $\alpha$ and $\alpha'$ be hoomorphisms of K into Aut(H). Then the following conditions are equivalent:

$\alpha$ is conjugate to $\alpha' \circ g$, with $g \in Aut(K)$

There is an isomorphism $\phi: H \rtimes_{\alpha} K \rightarrow H \rtimes_{\alpha'} K$ such that $im(\phi \circ \iota) = im \iota'$ and $im(\phi \circ s) = im s'$.

there is a commutative diagram (I cannot draw this here but this is on page 116 here).

I'm not sure if I'm using 1. in the proposition correctly, but this is what I said about $\alpha1$ and $\alpha2$:

We know that $e\alpha_2e^{-1} = \alpha_1(a)$, so they must be conjugate.

Case 2:

We need to classify $(\mathbb Z_2 \times \mathbb Z_2) \rtimes_{\alpha} \mathbb Z_3$

We have $\alpha: \mathbb Z_3 \rightarrow Aut(\mathbb Z_2 \times \mathbb Z_2) \cong S_3$

$\bullet$ If $\alpha$ is trivial, we have $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3 \cong \mathbb Z_2 \times \mathbb Z_6$. Since both $\mathbb Z_2$ and $\mathbb Z_6$ are abelian and the action is trivial, $\mathbb Z_2 \times \mathbb Z_6$ is also abelian.

We know that there are two elements in $S_3$ of order 3: (123) and (132).

So the other two homomorphisms must be:

$\bullet$ $\beta_1: a \rightarrow (123)$

$\bullet$ $\beta_2: a \rightarrow (132)$.

Now I want to show that $\beta_1$ and $\beta_2$ must be conjugate, but I'm stuck here.

My problem is that, in order to use the proposition, I need to have automorphisms, but since I'm using the fact that $S_3 \cong Aut(\mathbb Z_2 \times \mathbb Z_2)$, I'm not using automorphisms, I'm just using elements. So something like (for example) $\beta_1(13)$ makes no sense, right?

Basically, this was my plan:

I was going to show that $\beta_1$ and $\beta_2$ were conjugate. Since the problem is talking about nonabelian groups, the trivial homomorphisms of both cases don't really matter... since they both give abelian groups, right? So I would then try to prove that $(\mathbb Z_2 \times \mathbb Z_2) \rtimes_{\beta_1} Z_3 \cong \mathbb Z_4 \rtimes_{\alpha_1} \mathbb Z_3$, and then I would be done, right?

Thank you in advance

答:
You did a pretty nice work but I think you have a fatal "little" mistake there: it must be $\;\text{Aut}(\Bbb Z_4)\cong\Bbb Z_2\;$ and not isomorphic to $\,\Bbb Z_3\,$ ...!

With this you're done since the only possible homomorphism $\;\Bbb Z_3\to\Bbb Z_2\,$ is the trivial one, which yields the abelian group $\,\mathbb Z_4\times \Bbb Z_3\,$...

Thus, a non-abelian group of order $\;12\;$ with a normal subgroup of order $\,4\,$ must be an extension of $\,\Bbb Z_2\times \Bbb Z_2\,$ and thus it must be $\,A_4\,$ .


问:
Don't get how to apply the intermediate value theorem here
Note: I know how to prove this for $f(a) = f(a+100)$, but no clue how to do it for the value that's asked of me.

There's a continuous function $f(x): \mathbb R \to\ \mathbb R$

With these properties:

$f(x) ≤ 2 \ \text{for} \ x≥200$

$f(x) ≤ 1 \ \text{for} \ x≤100$

$f(x)=3 \ \text{for} \ x=150$

I'm supposed to show that there exists an $a$ for which the property $f(a)=(a+200)$ is true. I've tried applying the intermediate value theorem every which way, but at this point I don't know if that's the right way to go.

Body: Let $g(y)=f(y+200)-f(y)$. Then $g$ is continuous. We have $g(-50)=f(150)-f(-50)=3-f(-50)\geq 3-1=2>0$. On the other hand, $g(150)=f(350)-f(150)\leq 2-3=-1<0$. Hence, there is  a point $a$ between $-50$ and $150$ where $g(a)=0$.


问:
Will $y=x$ and $y=\sin^n \left(\frac{\pi x}{2} \right)$ meet in the interval $(0,1)$ for all $n \ge 2$?
I am currently learning the basics of real analysis, and still haven't learned the concepts and properties of pointwise convergence and uniform convergence of functions.

Still, I was thinking of a function $f_n : (0,1) \to \mathbb{R}$,
$$ f_n(x) = \sin^n \left(\frac{\pi x}{2} \right) $$
and if we took the $n$ to infinity, it would give
$$ \lim_{n \to \infty} f_n(x) = 0 $$

From that, I thought that there might exist some $f_n$ that $y=f_n(x)$ does not meet $y=x$.

I thought it this way since for every $x \in (0,1)$, $\lim_{n \to \infty}f_n (x) = 0$, I could use $\epsilon - N$. To be specific, for given $x \in (0,1)$, I have taken $\epsilon = x$, then exists a natural number $N$ such that the following holds:
$$ n \ge N \Rightarrow |f_n(x)| < x $$

I thought that this means that for that point $x$, $(x,f_n(x))$ is below $(x,x)$, and hence will not meet $y=x$.

Since I can take such $N$ for every point in $(0,1)$, I thought that $y=f_n(x)$ and $y=x$ will not meet for some large $N$.

But my instructor told me it will always meet since the function is pointwise convergent, but not uniformly convergent. I do not have enough knowledge of these concepts, so I am asking help for them.

My own guess to the failure of my conjecture(?) is that taking the large $N$ is impossible.
To take $N$, it has to be bigger or equal to all the other $N$ taken from each $x \in (0,1)$. In other words,
$$ N := \sup\{N_x : x \in (0,1)\}$$
where $N_x$ is the $N$ for each $x$.
I thought that the set is not upper-bounded hence unable to take $N$.

Thank you in advance for your considerate help.

答:
In this answer, I will provide a lower level (at least to a calculus level) proof that the answer to OP's post is yes. Further, I hope that the way I present the proof can serve as inspiration to anyone else that has a problem that requires chaining lots of inequalities together.

First, define the function

$$g_n(x)=\sin^n\left(\frac{\pi x}{2}\right)-x$$

Then note that for $n\geq 2$ (and $x=\frac{1}{3}$) we have that

$$\sin^n\left(\frac{\pi}{2}\cdot\frac{1}{3}\right)=\sin^n\left(\frac{\pi}{6}\right)=\frac{1}{2^n}<\frac{1}{3}\Rightarrow g_n\left(\frac{1}{3}\right)<0$$

Now, consider the sequence $x_n=1-\frac{1}{f(n)}$ where $f(n):\mathbb{N}\to\mathbb{R}$ is some increasing function such that $f(2)\geq 2$. We want to show that there exists such a function $f(n)$ such that $g_n(x_n)>0$. We choose not to fully spell out $f(n)$ at this point for two reasons: it is easier to write the following computations using simply $f(n)$ and using $f(n)$ provides a good case study on how one might go about proving similar problems. We have that

$$g_n\left(x_n\right)=\sin^n\left(\frac{\pi }{2}\cdot\left(1-\frac{1}{f(n)}\right)\right)-\left(1-\frac{1}{f(n)}\right)$$

But the trig term in this is just

$$\sin^n\left(\frac{\pi }{2}\cdot\left(1-\frac{1}{f(n)}\right)\right)=\sin^n\left(\frac{\pi}{2}-\frac{\pi}{2f(n)}\right)=\cos^n\left(\frac{\pi}{2f(n)}\right)$$

Using the Taylor Series approximation for $\cos(x)$, we have that

$$\cos^n\left(\frac{\pi}{2f(n)}\right)>\left(1-\frac{1}{2}\left(\frac{\pi}{2f(n)}\right)^2\right)^n=\left(1-\frac{\pi^2}{8f(n)^2}\right)^n$$

We may then use the binomial theorem to get that

$$\cos^n\left(\frac{\pi}{2f(n)}\right)>\sum_{k=0}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k=1+\sum_{k=1}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k$$

Plugging this all back into $g_n(x_n)$, we get that

$$g_n\left(x_n\right)=\sin^n\left(\frac{\pi }{2}\cdot\left(1-\frac{1}{f(n)}\right)\right)-\left(1-\frac{1}{f(n)}\right)$$

$$>1+\sum_{k=1}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k-\left(1-\frac{1}{f(n)}\right)$$

$$=\sum_{k=1}^n\binom{n}{k}\left(-\frac{\pi^2}{8f(n)^2}\right)^k+\frac{1}{f(n)}$$

Now, this is a complicated expression, but we can bound it by

$$>\frac{1}{f(n)}-\sum_{k=1}^n\binom{n}{k}\left(\frac{\pi^2}{8f(n)^2}\right)^k$$

Further, we have that

$$\sum_{k=0}^n\binom{n}{k}=(1+1)^n=2^n\Rightarrow \binom{n}{k}<2^n$$

Then the full expression can be bounded by

$$>\frac{1}{f(n)}-2^n \sum_{k=1}^n\left(\frac{\pi^2}{8f(n)^2}\right)^k>\frac{1}{(2n)!}-2^n \sum_{k=1}^\infty \left(\frac{\pi^2}{8f(n)^2}\right)^k$$

This is simply a convergent geometric series, which gives us

$$=\frac{1}{f(n)}-\frac{2^n\pi^2}{8f(n)^2-\pi^2}>\frac{1}{f(n)}-\frac{2^n 4^2}{8f(n)^2-4^2}$$

$$=\frac{1}{f(n)}-\frac{2^{n+1}}{f(n)^2-2}=\frac{f(n)^2-2-2^{n+1}f(n)}{f(n)(f(n)^2-2)}>\frac{f(n)^2-2^{2n+2}-2^{n+1}f(n)}{f(n)(f(n)^2-2)}$$

Now, recall that we are trying to show that $g_n(x_n)>0$. Thus, we simply need to pick $f(n)$ such that the numerator of the above expression is positive (since the denominator is always positive). This is easy (you may have noticed that we have simplified things in a particular way) as we may simply take $f(n)=2^{n+2}=2\cdot 2^{n+1}$. Then

$$f(n)^2-2^{2n+2}-2^{n+1}f(n)=4\cdot 2^{2n+2}-2^{2n+2}-2\cdot 2^{2n+2}=2^{2n+2}>0$$

We conclude that

$$g_n\left(1-\frac{1}{2^{n+2}}\right)>0$$

for all $n\geq 2$. Then by the mean value theorem, there exists $x_0\in \left(\frac{1}{3},1-\frac{1}{2^{n+2}}\right)$ such that $g_n(x_0)=0$ and we are done.


答:
In general, if $a$ is invertible and $y$ is not, then $$\|a-y\|\ge\frac{1}{\|a^{-1}\|}$$ This is because the set of invertible elements form an open set, and $a$ has a ball of radius at least $\|a^{-1}\|^{-1}$ containing invertible elements.

Applying this to $y=\mu-x$ and $a=\lambda-x$, where $\mu\in\sigma(x)$, $\lambda\in\rho(x)$, gives $$|\mu-\lambda|\ge\frac{1}{\|(\lambda-x)^{-1}\|}$$ The required formula is the limit as $\mu$ minimizes the distance $d(\lambda,\sigma(x))$

Title: norm of the resolvent inequality

Body:
Let $x$ be an element of a Banach Algebra. Let $\lambda \in \rho(x)$, where $\rho(x)$ is the resolvent of $x$.

Let ${d(\lambda, \sigma(x))}$ be the distance between $\lambda$ and $\sigma(x)$.

Show that $\| (\lambda-x)^{-1} \| \geq \frac{1}{d(\lambda, \sigma(x))}$.

Information I know:

$\rho(x)$ is an open set.

$\sigma(x)$ is compact.

I've seen many places using this inequality, but I cannot find a proof of it. And I'm not sure how to prove this by breaking down the definitions I know.

Thanks in advance!


问:
Does a product measure on a product space constructed from two sub-fields of the same space determine a measure on the underlying space?
Let $\mathcal{A}_1,\mathcal{A}_2$ be $\sigma$-algebras on $\Omega$. Let $P$ be a probability on $\mathcal{A}_1$ and let $Q$ be a Markov kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$. Set $K:=P\otimes Q$, so that $K$ is a probability measure on $\mathcal{A}_1\otimes\mathcal{A}_2$. Does there exist a probability measure $K'$ on $\sigma\left(\mathcal{A}_1,\mathcal{A}_2\right)$, that satisfies
$$
K'\left(D\cap E\right)=K\left(D\times E\right)
$$
for all $D\in\mathcal{A}_1$, $E\in\mathcal{A}_2$?</p>

答:
The answer is: no. $K'$ is not even necessarily consistent.

Set $\Omega:=\left\{0,1\right\}$, and $\mathcal{A}_1:=\mathcal{A}_2:=\mathcal{P}\Omega$.

Define

$$
P(0):=P(1):=\frac{1}{2}
$$

$$
Q\left(\omega,B\right):=P(B)
$$

Then $K=P^2$.

Set $D_1:=D_2:=E_1:=\left\{1\right\}$, $E_2:=\Omega$.

Then $D_1\cap E_1=D_2\cap E_2$, but

$$
K\left(D_1\times E_1\right)=\frac{1}{4}\neq\frac{1}{2}=K\left(D_2\times E_2\right)
$$


问:
Quadratic formula fails numerically at small "a" coefficients
I have a set of quadratic coefficients a, b, and c that need to be solved in order to determine the intersection between a parabola and a line on a plane. These coefficients have "arrived" from a different algorithm.

For context, the actual quadratic being solved is:
$$a = 4.3162933554 \times 10^{-11}$$
$$b = 1.0361118521 \times 10^{6}$$
$$c = -4.5813932360 \times 10^{5}$$
where $f(x)=ax^2+bx+c$.

My problem, however, is that when my parabola is so "stretched out" that it starts behaving like a line around the area of interest, the "a" coefficient becomes very small, and actually disregarding it completely and instead solving linearly for b and c works perfectly for the required intersection.

In this case what is actually happening with the formula is that the $-b+\sqrt(b^2-4ac)$ term is being calculated as 0.0 by the computer! (I've confirmed that this isn't specific to my implementation as my calculator does the same thing! Also, all of this is involving double floating point precision)

My question is, how can I approach finding this root in an optimal way? Also, are there any other cases where the formula may fail?

I've thought about disregarding the $a$ term and simply solving the equation linearly, but then at what point do you decide to do that? By extension, when should you solve a cubic equation quadratically?

答:
You can switch over to the modified solution formula, according to binomial formulas,
$$
x=\frac{-b+\sqrt{b^2-4ac}}{2a}=-\frac{2c}{b+\sqrt{b^2-4ac}}.
$$
The cancellation case in the first formula will give a numerically stable computation in the second formula, giving a value close to $-\frac{c}{b}$, the solution of the linear approximation.

In general, most of the cases with numerical issues when computing the roots, in the case of real roots, are avoided by first determining the non-cancelling variant of the numerator
$$
u=-b-{\rm sign}^+(b)\sqrt{b^2-4ac},~~~~{\rm sign}^+(0)=+1,
$$
and then computing the two roots as
$$
x_1=\frac{u}{2a},~~~ x_2=\frac{2c}{u}.
$$
After that, the stabilization discussion turns to the square root, to be able to keep computing it in the case of floating-point over- or underflow.


问:
Integral of a polynomial fraction: $\int \frac{4x^2-3x+2}{4x^2-4x+3}dx$
I have initially used long division to obtain: $$1+\frac{x-1}{4x^2-4x+3}$$
Now, I'm having difficulty solving $$\int \frac{x-1}{4x^2-4x+3}dx$$
Even if I do the $u$-substitution i.e. $u=4x^2-4x+3$ hence, $du=8x-4$ and $dx=\frac{1}{8x-4}du$, I can't seem to figure out how to deal with the $x-1$ in the numerator.
Any help regarding this will be appreciated. Thank you.

答: Hint: The substitution you can up with lets you evaluate $\int \frac {x-\frac1  2} {4x^{2}-4x+3}dx$. To evaluate $\int \frac 1 {4x^{2}-4x+3}dx$ write $4x^{2}-4x+3$ as $4(x-\frac1 2)^{2}+\frac 1 2$ and make the substiution $z=x-\frac  1 2$. Use the fact that $\int \frac 1 {z^{2}+c^{2}}dz=\frac 1  c \arctan \frac z c$ plus a constant.


答: Hint: Separate $x-1$ into $(x-1/2) -(1/2)$ and now perform these two integrals. The intuition behind this is that you can use your $u$-sub on the $x-1/2$ part as it is $(8x-4)/8$

Title: Integral of a polynomial fraction: $\int \frac{4x^2-3x+2}{4x^2-4x+3}dx$

Body:
I have initially used long division to obtain: $$1+\frac{x-1}{4x^2-4x+3}$$
Now, I'm having difficulty solving $$\int \frac{x-1}{4x^2-4x+3}dx$$
Even if I do the $u$-substitution i.e. $u=4x^2-4x+3$ hence, $du=8x-4$ and $dx=\frac{1}{8x-4}du$, I can't seem to figure out how to deal with the $x-1$ in the numerator.
Any help regarding this will be appreciated. Thank you.


问:
Identity in each group
Why must the identity be part of each group ? I was wondering about this in the context of "Algebra" by Michael Artin, where the author explains that there is essentially one group of order two.</p>

答:
The trivial group $\{e\}$ is the smallest possible group. It is also true that it is a subgroup of every possible group. Hence every group contains the identity.

Of course this a roundabout way of proving the statement, but you don't have to do it this way since identity being a part of every group is built into the definition of a group.

Edit: This, I realized, is not how you prove it, as it assumes that the identity is a part of any group. My answer therefore is wrong. You just have to work with the definition of a group and how that includes the existence of the identity.

Title: Identity in each group


问:
Inequality in Sobolev spaces that allows to prove weak differentiability
Let $f \in L^2([0,1])$.

Assume that for every $\varphi \in W^{1,2}([0,1])$ we know that
$$
\int_{0}^1 \varphi'(x) f(x) dx \leq \|\varphi \|_{W^{1,2}}.
$$
Does this imply that $f$ belongs to $W^{1,2}([0,1])$?

My guess is yes because we can define $f'$ as
$$
 \int_0^1 \varphi(x) f'(x) dx  = - \int_0^1 \varphi'(x) f(x) dx,
$$
but I am not able to formalize this idea.
Any hint is welcome, thank you

答:
In order to define a function $g \in L^2$ via the formula
$$\int_0^1 \phi \, g \, \mathrm d x = \int_0^1 \phi' \, f \, \mathrm{d}x$$
you need
$$\int_0^1 \phi' \, f \, \mathrm{d} x \le C \, \|\phi\|_{L^2}.$$
In this case, the right-hand side of the first formula is defined on $W^{1,2}$ and is continuous w.r.t. the $L^2$-norm. Since $W^{1,2}$ is dense in $L^2$, it extends uniquely to a continuous functional on $L^2$. By the Riesz representation theorem, this can be identified with an $L^2$-function $g$.


问:
Prove that $(K,\delta)$ is a compact metric space.
Let $(X, d)$ be a compact metric space. For $x ∈ X $ and $\epsilon > 0$, define

{$B_{\epsilon}(x) := {y ∈
X | d(x, y) < \epsilon}$}.

For $C ⊆ X$ and $\epsilon > 0$, define $B_{\epsilon}(C) := ∪_{x∈C}B_{\epsilon}(x)$.

Let K be the
set of non-empty compact subsets of X. For $C, C_0 ∈ K$, define $δ(C, C_0) = $inf{$\epsilon| C ⊆ B_{\epsilon}(C_0)$ and $C_0 ⊆ B_{\epsilon}(C)$}.

Assuming that it forms a metric space show that $(K, δ) $ is a compact metric space.

My attempt:

The metric space $(X,d)$ is compact so it is totally bounded.
Let $B=${$C_k$ } be a sequence of compact metric spaces in $K$. We need to show that the infinite set has a limit point.

$(X, d)$ is totally bounded so let $A_1$={$x_{1,1},x_{2, 1},....,x_{k_1,1}$} such that ${\cup {B_d(x_i,1) }} $ covers X for $i= (1,1),..,(k_1,1)$ and ${\cup {B_d(x_i,\frac{1}{2}) }} $ covers X for $i= (1,2),..,(k_2,2)$ ,so $A_2= x_{1,2},x_{2,2},...,x_{k_2,2}$

So we can generate a new sequence of compact metric space $A_1,A_2,...,A_n$ as mentioned above.Now,let {$\cup {(A_i)}$}=$A$ .We take the closure $cl(A)$ which is the closed subset of $(X,d)$.Hence it is compact so it is in $(K,\delta)$.

$(B_{\delta}(cl(A) \cap B)  /cl(A))\ne \phi$ for all $\epsilon>0$ is what we have to show to prove that $cl(A)$ is the limit point of the sequence {${C_k}$} to show that the sequence has bolzano weistrass property.

We pick an $\epsilon >0$ and proceed as ,

Now, $C_i \subset B_{\frac{1}{N_1}}(cl(A))$ where $\frac{1}{N_1} < \frac{\epsilon}{2}$ and $C_i \subset B_{\epsilon}(cl(A))$ where $C_i$ is any compact subset in the sequence {$C_k$} and $C_i$ can be chosen to be different from $cl(A)$.

Now, let $x \in Cl(A)$
then $x \in X$ and $x \in \cup B_d(x_i,\frac{1}{N_1})$ where $i=(1,N_1),(2,N_1),(3,N_1),...,(k_{N_1},N_1)$
then $d(x,x_i) < \frac{1}{N_1}$ and let $x_n \in C_i$ then $d(x_i,x_n) < \frac{1}{N_1}$ so $d(x,x_n) < \epsilon$ and the other case should also follow when $ x = x_i $ where $i=(1,N_1),(2,N_1),(3,N_1),...,(k_{N_1},N_1)$.
So $cl(A) \subset B_{\epsilon}(C_i)$.

Then $\delta(C_i,cl(A)) < \epsilon$ which allows me to conclude that $cl(A)$ is the limit point of the above mentioned seqeunce (as $\epsilon$ can be chosen arbitrarily).So by bolzano property I can claim the compactness.

Is my attempt ok? I have tried hard so that the notations are understandable. Thanks in advance.

答:
$X$ is compact, so it is totally bounded; for each $n\in\Bbb N$ let $D_n$ be a finite $2^{-n}$-net in $X$, i.e., a finite subset of $X$ such that $X=\bigcup_{x\in D_n}B_d(x,2^{-n})$, and let $\mathscr{D}_n$ be the family of non-empty subsets of $D_n$.

Let $\epsilon>0$; there is an $n\in\Bbb N$ such that $2^{-n}<\epsilon$. For $C\in K$ let

$$F=D_n\cap B_d(C,2^{-n})\in\mathscr{D}_n\,;$$

then $\delta(C,F)\le 2^{-n}<\epsilon$, so $\mathscr{D}_n$ is a finite $\epsilon$-net for $K$, and $K$ is totally bounded. To show that $K$ is compact, it only remains to show that $K$ is complete.

Let $\langle C_n:n\in\Bbb N\rangle$ be a Cauchy sequence in $K$. Let

$$H=\bigcap_{n\in\Bbb N}\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\in K\,,$$

and let $\epsilon>0$. If $\{n\in\Bbb N:C_n\nsubseteq B_d(H,\epsilon)\}$ is infinite, then $\bigcup_{k\ge n}C_k\nsubseteq B_d(H,\epsilon)$ for each $n\in\Bbb N$, so $\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\setminus B_d(H,\epsilon)$ is a non-empty compact set for each $n\in\Bbb N$, and

$$H\setminus B_d(H,\epsilon)=\bigcap_{n\in\Bbb N}\left(\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\setminus B_d(H,\epsilon)\right)\ne\varnothing\,,$$

which is absurd. Thus, for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $C_n\subseteq B_d(H,\epsilon)$ for all $n\ge n_\epsilon$.

Now suppose that $\{n\in\Bbb N:H\nsubseteq B_d(C_n,\epsilon)\}$ is infinite. There is an $m\in\Bbb N$ such that $\delta(C_k,C_\ell)<\frac{\epsilon}3$ whenever $k,\ell\ge m$. Fix $n\in\Bbb N$ such that $n\ge m$ and $H\nsubseteq B_d(C_n,\epsilon)$, and let $x\in H\setminus B_d(C_n,\epsilon)$. Let $k\ge n$. Then $\delta(C_k,C_n)<\frac{\epsilon}3$, so $C_k\subseteq B_d\left(C_n,\frac{\epsilon}3\right)$, and hence $B_d\left(C_k,\frac{\epsilon}3\right)\subseteq B_d\left(C_n,\frac{2\epsilon}3\right)$. But then

$$\begin{align*}
H&\subseteq\operatorname{cl}_X\left(\bigcup_{k\ge n}C_k\right)\subseteq\operatorname{cl}_X\left(\bigcup_{k\ge n}B_d\left(C_k,\frac{\epsilon}3\right)\right)\\
&\subseteq\operatorname{cl}_X\left(B_d\left(C_n,\frac{2\epsilon}3\right)\right)\subseteq B_d(C_n,\epsilon)\\
&\subseteq H\setminus\{x\}\,,
\end{align*}$$

which is absurd.

Thus, there is an $m_\epsilon$ such that $H\subseteq B_d(C_n,\epsilon)$ for all $n\ge m_\epsilon$.

Putting the pieces together, we see that $\delta(H,C_k)\le\epsilon$ for all $k\ge\max\{n_\epsilon,m_\epsilon\}$ and hence that $\langle C_n:n\in\Bbb N\rangle$ converges to $H$ in $\langle K,\delta\rangle$.


问:
The Exponential Group in a Banach Algebra: a proof of Rudin.
Define $G$  to be the group of invertible elements in a Banach algebra $A$ and $G_1$ the component of $G$ that contains $e\in G.$

Now, $\exp x=\sum_{n=0}^\infty \frac{x^n}{n!}$ and if $\Gamma$ is the subgroup generated by $\exp (A),$ a straightforward argument shows that $\Gamma$ is open: if we define $y=\sum_{n=0}^\infty \frac{-1}{n}(1-x)^n$ then the series converges absolutely whenever $\|1-x\|<1$ and a (at least for me) somewhat messy calculation shows that $\exp y=x$ so $\exp (A)$ contains a neighborhood of $e$, and since $\Gamma$ is a group, it follows that $\Gamma$ is open.

Unfortunately, Rudin does not do it this way. He cites the following theorem:

Theorem Suppose $A$ is a Banach algebra, $x\in A$, and the spectrum $\sigma(x)$ of $x$ does not separate $0$ from $\infty$. Then

(a) $x$ has roots of all orders in $A$,

(b) $x$ has a logarithm in $A$, and

(c) if $\varepsilon>0$, there is a polynomial $P$ such that $\|x^{-1}-P(x)\|<\varepsilon$.

to claim that $\exp(A)$ has nonempty interior in $G$ and uses this fact to prove that $\Gamma$ is open. I tried to do this as follows: take $y\in\exp(A)$ and note that as soon as $\|x-y\|<\|y^{-1}\|^{-1}$ then $\|e-y^{-1}x\|<1.$ It follows that $\sigma(y^{-1}x)\subseteq \{|\lambda-1|<1|\}$ and now the theorem applies to show that $y^{-1}x=\exp z$ for some $z\in A$ and so $x=y\exp z.$ Of course, unless $A$ is commutative, we can't conclude.

Since Rudin claims that $\exp(A)$ is open, as a passing remark, I guess it's easy, but I have not been able to prove it. I would like a hint to get me going in the right direction.

答:
$e=1$ has a neighborhood of invertible elements and, crucially, they are in $\exp(A)$, since $\log(1+x)$ is defined (as a series for example) for $x$ small enough.

(This is the extra bit required in your argument: $1-y^{-1}x\in\exp(A)$, not just $y^{-1}x$.)

Let $y\in\exp(A)$, then $y+x=y(1+y^{-1}x)\in\Gamma$ for $x$ small enough, by the above. Thus there is a neighborhood of $y$ inside $\exp(A)$, so $\exp(A)$ is an open subset.

(Note that $y^{-1}\in\exp(A)$.) You still have to figure out how small is small enough to make the argument rigorous.