问:
Shifting of axis of rotation of disc
Image for following question

Initially a disc is rotating about a point A on circumference with angular velocity $\omega$ . The point is then released and almost immediately point B is fixed on circumference. Show that the disc starts to rotate about B with angular velocity $\omega/3*(1+2*\cos \alpha)$ where $\alpha$ is angle subtended by AB at centre of disc.

This question is solved by conservation of angular momentum. $I\omega’ = I_0\omega + MV_{cm}*r*cos\alpha$

My queries are as follows -

When point A is released, the disc rotation would shift to its centre of mass, and because of this shift wouldn’t the $\omega$ change as Moment of Inertia has changed? (new MoI is 3x less.)

As this $\omega$ might have changed, so how can we take $I_0\omega$ in the calculation ?

Why this can’t be achieved by conserving energy ? (Is it because we are trying to minimise, rather ignore, effects of any reaction forces that might apply)

答:
The total angular momentum of the body about an axis through $B$ perpendicular to the disc comprises the angular momentum of the body about its centroid plus the moment of momentum, about that axis, of a particle of the same mass as the disc situated at the centroid.

Just before the point B is fixed, the disc is rotated with angular speed $\omega$ about the centroid, and the centroid has a velocity component perpendicular to the radius through B which is $r\omega\cos\alpha$.

Therefore, relative to axis $B$ the total angular momentum is $$I_o\omega+M(r\omega\cos\alpha)r$$

By conservation of angular momentum, this must equal $$I_B\omega'$$

Therefore we have$$\frac{3}{2}Mr^2\omega'=\frac{1}{2}Mr^2\omega+Mr^2\omega\cos\alpha$$
$$\implies\omega'=\frac{1}{3}\omega(1+2cos\alpha)$$

Note that conservation of energy does not apply because the act of fixing the axis at $B$ imparts an impulse to the disc.

Title: Shifting of axis of rotation of disc


问:
$10$ object in $2$ equal queues
I'm working on the following.

Ten objects shall be arranged in two queues of equal length. The order matters.

a) A and B must be in different queues.

b) A and B must be next to each other in the same queue.

Objects = $A,B,C,D,E,F,G,H,I,J$

Queues = $1,2$

a)
If we first place A, then there are 10 possible positions. If we then place B, then there are 5 possible positions. Then we will place the other eight objects in 8 possible positions. Then the same thing, but B is placed first, then A.

$$10\cdot 5 \cdot \binom{8}{8} \cdot 2=4032000$$

But if we place all then objects without restraints, we get a lower number. What am I doing wrong?

$$\binom{10}{10}=3628800$$

b) Calculate if A and B are next to each other in 1. Then the same thing for 2.

$$2\cdot \binom{5}{2}\cdot \binom{8}{8}=1612800$$

EDIT:

a) Removed the double counting. $$10\cdot 5 \cdot \binom{8}{8} =2016000$$

b) A and B can be placed next to eachother in each queue $2\cdot4$ possible ways (order matters). Then we've 2 queues, $2\cdot8$.

$$16\cdot \binom{8}{8} =645120$$

答:
For the part a) the answer is
$$2\binom84 (5!)^2=50\cdot 8!,
$$
where the first factor stays for the number of queues to place A, the second factor for the choice of the other objects joining A in the queue, and the third factor for the number of ways to permute both queues.

For the part b) the answer is
$$2\binom83 (4\cdot2!)3!5!=16\cdot8!,
$$
where the first factor stays for the number of queues to place A and B, the second factor for the choice of the other 3 objects joining A and B in the queue, the third factor for the number of ways to arrange A and B in the queue, the forth and fifth factors for the number of ways to permute the other objects in the two queues.

ContentLicense: CC BY-SA 4.0

Title: $10$ object in $2$ equal queues


问:
Find all $f(X)\in\mathbb{Z}[X]$ such that $\left(\frac{f(n)}{p}\right)=\left(\frac{n}{p}\right)$ [Legendre symbol and $p$ is a fixed prime]
Question: Let $f(X)\in\Bbb{Z}[X]$ be a non-constant polynomial such that $$\left(\frac{f(n)}{p}\right)=\left(\frac{n}{p}\right)$$ where $\left(\frac{\cdot}{p}\right)$ is the Legendre symbol and $p$ is a fixed odd prime. Find all such polynomials $f$.

Basic Observations: We know that $f(n+p)\equiv f(n)\pmod{p}$ for all $n\in\Bbb{Z}$. Then $$\left(\frac{f(n)}{p}\right)=\left(\frac{n}{p}\right)$$ if and only if $$\left(\frac{f(j)}{p}\right)=\left(\frac{j}{p}\right)\tag{1}$$ for all $j\in\{0,1,\ldots,p-1\}$. By Euler's criterion for quadratic residues, we get that $$\boxed{\left(\frac{f(j)}{p}\right)\equiv f(j)^{\frac{p-1}{2}}\pmod{p}}\qquad\boxed{\left(\frac{j}{p}\right)\equiv j^{\frac{p-1}{2}}\pmod{p}}$$ Therefore by condition $(1)$, we get that $$f(j)^{\frac{p-1}{2}}-j^{\frac{p-1}{2}}\equiv0\pmod{p}\;(\,\forall\,0\leq j\leq p-1)$$ Therefore we get that the polynomial $f(X)^{\frac{p-1}{2}}-X^{\frac{p-1}{2}}$ is in the ideal $(X^p-X)$ of $\Bbb{Z}/p\Bbb{Z}[X]$. In fact, this is an iff condition. Any $f$ satisfies the condition of the question if and only if $f(X)^{\frac{p-1}{2}}-X^{\frac{p-1}{2}}$ is in the ideal $(X^p-X)$ of $\Bbb{Z}/p\Bbb{Z}[X]$. I don't know how to proceed further to characterize all such polynomials. It is obvious that any polynomial of the form $aX^{2k+1}$ works, where $a$ is a quadratic residue modulo $p$. But what are all such polynomials? Any help would be highly appreciated. Thanks in advance!

Body:
$$f(x) = x\frac{p+1}{2}(x^{(p-1)/2}+1)g(x)^2+x\frac{p+1}{2}(1-x^{(p-1)/2}) h(x)^2+pu(x)+(x^p-x)v(x)$$
with $g,h$ having no zeros $\bmod p$.


问:
How to calculate conditional expectation with inequality condition
How to transition conditional expectation where the conditioned item is not an equation into a simpler form for calculation?

As a simple example, how to show that (from Introduction to Probability Models by S.M. Ross) $$E[R_1|R_1<R_2] = E[\min(R_1,R_2)]$$

where $R_1, R_2$ are continuous RVs following exponential distributions with rate $\lambda_1, \lambda_2$ respectively.

I did find:
How to calculate conditional expectation $E[X|X \geq 0]$?,
How to calculate conditional probability with inequality, and Conditional Expectation Multivariate Normal Distribution with inequality condition three questions, but it looks like they are either dealing with probabilities or not providing a concrete answer. I know there might not exist an explicit rule, but are there any general rules, or more examples or references?

Update: Please refer to the following image (Example 5.8 from the book). As is stated in @John Dawkins' answer, it holds when the two RVs follow a continuous distribution with finite mean.

Ross book Ex 5.8

答:
The assertion is true if $R_1$ and $R_2$ are i.i.d. with continuous distribution (and finite mean). In this case $P[R_1<R_2] = 1/2$, and
$$
\eqalign{
E[R_1\wedge R_2]
&= E[R_1; R_1<R_2]+E[R_2; R_2<R_1]\cr
&=2E[R_1; R_1<R_2] \qquad\hbox{(by symmetry)}\cr
&=E[R_1; R_1<R_2]/P[R_1<R_2]\cr
&=E[R_1\mid R_1<R_2]\cr
}
$$


问:
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1} + \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.

Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.

What I Tried: I thought of substituting $\sqrt{a - 1} = x$ , $\sqrt{a} = y$ .
This gives :- $$\rightarrow \sqrt[3]\frac{(x - y)^5}{(x + y)} + \sqrt[3]\frac{(x + y)^5}{(y - x)}$$
But I was not able to find any good factorisation for this. I even took some help from Wolfram Alpha and it gives me this :-

Another thing I thought of was to substitute only $\sqrt{a} = x$. This would give :-
$$\rightarrow \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} - x)^5}{(2x^2 - 1)} + \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} + x)^5}{1}$$

This looks more or less simpler to work with, but unfortunately I could not get any ideas.

Can anyone help me?

答: You can first get rid of the denominator in each fraction by multiplying above and below by the conjugate expression of the denominator. Then you will have 6th powers above, and you can get rid of the cubic roots.</p>


Body:
Simplify, $\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})}$.

What I Tried: I thought of substituting $\sqrt{a - 1} = x$ , $\sqrt{a} = y$ .
This gives :- $$\rightarrow \sqrt[3]\frac{(x - y)^5}{(x + y)} + \sqrt[3]\frac{(x + y)^5}{(y - x)}$$
But I was not able to find any good factorisation for this. I even took some help from Wolfram Alpha and it gives me this :-

Another thing I thought of was to substitute only $\sqrt{a} = x$. This would give :-
$$\rightarrow \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} - x)^5}{(2x^2 - 1)} + \sqrt[3]\frac{(\sqrt{(x + 1)(x - 1)} + x)^5}{1}$$

This looks more or less simpler to work with, but unfortunately I could not get any ideas.

Can anyone help me?

Body:
\begin{eqnarray*}
\sqrt[3]\frac{(\sqrt{a-1} - \sqrt{a})^5}{(\sqrt{a-1} + \sqrt{a})} + \sqrt[3]\frac{(\sqrt{a-1}
+ \sqrt{a})^5}{(\sqrt{a} - \sqrt{a-1})} = \frac{-(\sqrt{a-1} -\sqrt{a})^2 + (\sqrt{a-1} -\sqrt{a})^2}{\sqrt[3]{(\sqrt{a} - \sqrt{a-1})(\sqrt{a} + \sqrt{a-1})}} \\
= \frac{-(a-1+a-2 \sqrt{a(a-1)})+ a-1+a+2 \sqrt{a(a-1)}}{\sqrt[3]{a-(a-1)}}=4\sqrt{a(a-1)}
\end{eqnarray*}


问:
Does Lamé's proof of FLT hold for some n?
Despite the fact that Lamé's proof of FLT fails in general, is it correct that it holds for every n where the corresponding ring of integers of cyclotomic field is a UFD (which is all n with class numbers equal 1, like 3-22, for example)?

Thank you for your help.

答:
Your question is a bit imprecise, because it seems that Lamé has presented, at intervals of a few years, at least two "proofs" of special cases of FLT, see e.g. Catherine Goldstein historical note Gabriel Lamé et la théorie des nombres : « une passion malheureuse » ?  pp. 131 - 139 https://doi.org/10.4000/sabix.690.

His first communication, around 1840, was about the exponent 7, of which H. M. Edwards says in his book "A genetic approach to ANT" that [Lamé's arguments] are difficult, unmotivated and, worst of all, seem to be hopelessly tied to the case n =7. In his second, around 1847, after a reduction to the case of a prime exponent p, he made the famous mistake that $\mathbf Z[\zeta_p]$ is always a UFD. Later in his career Lamé came back to the particular cases $p=3,5$, with great care concerning the factoriality property (see Goldstein, op. cit.), but it does not seem that he went beyond these cases. Recall that $p$ is called a regular prime if $p$ does not divide the class number of $\mathbf Z[\zeta_p]$, and the smallest irregular prime is 37. So we can reasonably attribute to Kummer (and not Lamé) the theorem saying that FLT holds for regular primes (see e.g. Washington's "Introduction to cyclotomic fields", chap.1).


问:
Problem on $L^p$ space where $f(x)=x^{-1/n} \in L^p([1,\infty))$ is true or false

Prove or Disprove : $f(x)=x^{-1/n} \in L^p([1,\infty))$ and find $\|f\|_p $ if it exists .

I think if $\int_1^{\infty} \left(f(x)\right)^p\, dx <\infty $ then $f\in L^p([1,\infty]).$ Now $\int_1^{\infty} \left(x^{-1/n}\right)^p\, dx=\int_1^{\infty}x^{-p/n}\, dx $
Now what is $p/n > or <  1 $ here , where $1<p<\infty $

答: You think right. So if $p>n$ the integral converges, otherwise it diverges. In case it converges, it is easy to compute the integral. It equals $\frac{n}{p-n}$, so the norm is that number raised to the power $1/p$.


问:
Prove that a function $f$ is injective if and only if it respects intersection.
Let $f:X \to Y$ be a function. Prove that $f$ is injective if and only if $f(A\cap B)=f(A)\cap f(B),$ for any two sets $A, B \subseteq X.$

答:
We want to prove the following result.

Theorem. Let $f:X \to Y$ be a function. Hence, $f$ is injective if and only if $f(A\cap B)=f(A)\cap f(B),$ for any two sets $A, B \subseteq X.$

Sketch of the proof. Note that we want to prove an equivalence, i.e., a double implication. So, we must prove that
$$f \text{ is injective} \implies f(A \cap B) = f(A) \cap f(B)$$ and $$f(A \cap B) = f(A) \cap f(B) \implies f \text{ is injective}.$$

First, let’s recall a few definitions. One of the hypothesis is that $f$ is injective.

Definition. Let $A$ and $B$ be any sets and $f:A \to B$ a function. We say that $f$ is injective if
$$\forall x,y \in A, f(x) = f(y) \implies x = y.$$

The other hypothesis, states that the sets $f(A \cap B)$ and $f(A) \cap f(B)$ are equal.

Definition. Let $A$ and $B$ be any sets. We say that $A$ and $B$ are equal, and write $A = B$ if $A \subseteq B$ and $B \subseteq A.$

Proof. $\implies.$ Suppose that $f$ is injective.

Let $x \in f(A \cap B).$ Hence, it exists $y \in A \cap B$ such that $f(y) = x.$ Since, $y \in A \cap B$ we know that $y \in A$ and $y \in B.$ It comes that $f(y) \in f(A)$ and $f(y) \in f(B).$ Then, $x \in f(A)$ and $x \in f(B)$ which means that $x \in f(A) \cap f(B).$ Therefore, $f(A \cap B) \subseteq f(A) \cap f(B).$

Now, let $z \in f(A) \cap f(B).$ Then $z \in f(A)$ and $z \in f(B).$ From here, we get that there exists $a_1 \in A$ and $a_2 \in B$ such that $f(a_1) = z$ and $f(a_2) = z.$ Since $f$ is injective, then $a_1 = a_2$ (otherwise it would contradict our hypothesis). Then, we have that exists $a_1 \in A$ and $a_1 \in B$ such that $z = f(a_1).$ Hence, it exists $a_1 \in A \cap B$ such that $f(a_1) = z,$ and so $f(a_1) \in f(A \cap B).$ This means that $z \in f(A \cap B).$ Therefore, $f(A) \cap f(B) \subseteq f(A \cap B).$

Since, $f(A) \cap f(B) \subseteq f(A \cap B) \subseteq f(A) \cap f(B),$ we conclude that $f(A) \cap f(B) = f(A \cap B).$

$\Longleftarrow.$ Suppose that $f(A \cap B) = f(A) \cap f(B).$ Let $x, y \in A,$ such that $f(x) = f(y).$ Suppose that $x \neq y$ and let $X = \{x\}$ and $Y = \{y\}.$ Since $x \neq y,$ we know that $X \cap Y = \emptyset$ and therefore $f(X \cap Y) = \emptyset.$ Although, note that $f(X) = \{f(x)\} = f(Y)$ and therefore $f(X) \cap f(Y) = \{f(x)\} \neq \emptyset.$ But this is absurd, it contradicts our hypothesis. The absurd came from assuming that $x \neq y.$ Hence, $x = y.$ Therefore, by definition, $f$ is injective. $\square.$

Remark. Observe that in the $\implies$ part of the proof, we prove that $f(A \cap B) \subseteq f(A) \cap f(B)$ without using the hypothesis that $f$ is injective. That is because, for any two sets $A$ and $B$, $f(A \cap B) \subseteq f(A) \cap f(B).$


问:
What rule of inference has been used here?
Hi I'm struggling with this exercise

Exercise 3.4.10. Suppose that I and J are two sets, and for all α ∈ I ∪ J let Aα be a set.
If I and J are non-empty, show that (⋂α∈I Aα) ∩ (⋂α∈J Aα) = ⋂α∈I∪J Aα.

I know how to show that two sets are equal.

first let y be a arbitrary element.

if y∈ (⋂α∈I Aα) ∩ (⋂α∈J Aα)

then (y∈Aa for all a∈i) and (y∈Aa for all a∈j)

and then

I have to show that (y∈Aa for all a∈i U j). but this stage I can't deduce this statement from the above

statement using rule of inference that I know.

I can explain why this is valid inference in words,  but I can't explain by predicate calculus

please someone help me!

答:
We have, by definition of arbitrary intersection and union of two sets:$$\begin{align} y \in \bigcap_{\alpha \in I\cup J} A_\alpha & \iff \forall \alpha \in I\cup J (y \in A_\alpha)\\& \iff \forall \alpha (\alpha \in I\cup J \rightarrow y \in A_\alpha) \\ & \iff \forall \alpha(\alpha \in I \lor \alpha \in J \rightarrow y\in A_\alpha) \end{align}$$

You have shown $\forall \alpha(\alpha \in I \rightarrow y \in A_\alpha)$ and $\forall \alpha(\alpha \in U \rightarrow y \in A_\alpha)$.

We can prove this using the rule of inference called modus ponens, and by using proof by cases. We want to prove $\forall \alpha(\alpha \in I \lor \alpha \in J \rightarrow y\in A_\alpha)$, so let $\alpha$ be arbitrary. First suppose $\alpha \in I$. Apply modus ponens to this and the statement $\alpha \in I \rightarrow y \in A_\alpha$ (which was already proved) to conclude $y \in A_\alpha$ . Now suppose $\alpha \in J$ and apply modus ponens again, this time using $\alpha \in J \rightarrow y \in A_\alpha$. So $y \in A_\alpha$ holds for an exhaustive list of cases, and we have proven $\forall \alpha(\alpha \in I \lor \alpha \in J \rightarrow y\in A_\alpha)$.


问:
Reference for quickest detection (sequential hypothesis testing) problems
I am looking for a reference that covers the following special case of quickest detection (or sequential hypothesis testing) problem. I believe there should be a large literature on it either in mathematical finance or statistical decision theory.

Let  $\theta$ be unobserved. The decision maker (DM) would like to choose action $A_1$ if $\theta>0$ and $A_2$ otherwise. Before making decision, the DM can choose to wait to learn more about $\theta$, but this is costly. The DM is choosing when to stop learning and choose an action.

Thus, this is an optimal stopping time of the form
\begin{align}
\inf_{\tau} \mathbb{E}\left[g(\pi_{\tau}) - c\tau|\pi_0\right],
\end{align}
where $\pi_t$ is the DM's belief at time $t$ based on information DM received ($\pi_0$ is the prior belief) and $g(\pi)$ is the expected payoff of making a decision when the belief is $\pi$.

I am looking for books/notes (or accessible papers) that cover these types of problems. I am particularly, interested in the case where $\theta~\sim N(\mu,\sigma^2)$ and DM is observing process $dY_t=\theta dt + \nu dW_t,$ with $W_t$ a standard brownian motion.

答:
I think the most important reference on the Quickest Detection is the book is  "Quickest Detection" by H. Vincent Poor, Cambridge University Press.
Here you can find a list of the subject covered by the book
https://www.cambridge.org/core/books/quickest-detection/4A7686F64138F460CC10148FC58E52F8

Regards,

Andrea


问:
Proving Convexity of KL Divergence w.r.t. first argument?
I'd like to show that the KL divergence is convex w.r.t. its first argument, where the KL divergence is defined as

$KL(q|p) = \sum_x q(x) \log \frac{q(x)}{p(x)}$

This question suggests that I can show convexity using the log sum inequality, but thus far, I've only been able to show that $KL(\lambda q + (1-\lambda) r | p) \geq 0$. Can someone help me?

答:
Let me first rewrite $\mathrm{KL}$ slightly more conveniently: $$ \mathrm{KL}(q\|p) = \sum q(x) \log q(x) - \sum q(x) \log p(x).$$ The second of these terms is linear in $q$, so you only need to argue that $\varphi(q) := \sum q(x) \log q(x)$ is convex. This follows because the function $u \mapsto u \log u$ with $0 \log 0 := 0$ is convex on $\mathbb{R}_{\ge 0}$. One way to show this is the log-sum inequality: for any $u_1, u_2$ and $\lambda \in (0,1),$ take $a_1 = \lambda u_1, a_2 = (1-\lambda) u_2, b_1 = \lambda, b_2 = (1-\lambda)$, in which case the log-sum inequality tells us that $$ \lambda u_1\log u_1 + (1-\lambda) u_2 \log u_2 \ge (\lambda u_1 + (1-\lambda) u_2) \log \frac{(\lambda u_1 + (1-\lambda) u_2)}{(\lambda + 1 - \lambda)}. $$

Now we can use this inequality term by term in the sum in $\varphi$. Let $\lambda \in (0,1)$. Then $$ \varphi(\lambda q_1 + (1-\lambda)q_2) =  \sum (\lambda q_1(x) + (1-\lambda)q_2(x)) \log (\lambda q_1(x) + (1-\lambda)q_2(x)) \\ \le \sum \lambda q_1(x) \log q_1(x) +  (1-\lambda) q_2(x) \log q_2(x) \\ = \lambda \sum q_1(x) \log q_1(x) +  (1-\lambda) \sum q_2(x) \log q_2(x) \\ = \lambda \varphi(q_1) + (1-\lambda) \varphi(q_2). $$

(Formally, for each $x$, we're using the inequality above with $u_1 = q_1(x)$ and $u_2 = q_2(x)$.)

In fact pretty much the same argument applies to directly arguing the joint convexity of $\mathrm{KL}$ in $(p,q)$ - set $b_1 = \lambda v_1$ and $b_2 = (1-\lambda) v_2$ to show the joint convexity of $(u,v) \mapsto u \log \frac{u}{v},$ and then use the same term-by-term approach.


问:
Is there a cyclic code with the following parameters?
Is there exist a cyclic code over $F_{2^6}$ with parameters $[31,27,\geq5]$?
I tried to construct such code but I didn't succeeded?

答:
No. That's because there do not exist ANY cyclic codes over $K=\Bbb{F}_{64}$ with $(n,k)=(31,27)$.

Over the field $K$ the polynomial $p(x)=x^{31}-1$ factors as
$$
p(x)=(x-1)\prod_{j=1}^6 q_j(x),\qquad(*)
$$
where the factors $q_j(x), j=1,2,\ldots,6$, all have degree five. They are the six irreducible quintic factors of $x^{31}-1$ over the prime field $\Bbb{F}_2$. As $\gcd(5,6)=1$ an irreducible quintic remains irreducible over the degree six extension $K$.

A cyclic code of length $31$ has a generator polynomial $g(x)\in K[x]$ that is a factor of $p(x)$. If a rank $27$ cyclic code of length $31$ existed, we would need
$$\deg g(x)=31-17=4.$$

But the factorization $(*)$ implies that
$$
\deg g(x)\equiv \text{$0$ or $1$}\pmod 5,
$$
so this is impossible.

However, it is easy to construct a $(31,27,5)$-code with alphabet $K$. That will be an MDS-code, so we can simply shorten a Reed-Solomon code. If $\alpha$ is a generator of the multiplicative group of $K$, we can, for example, use $g(x)=(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4)\in K[x]$. It would generate a cyclic code of length $63=|K|-1$.
If we restrict the message polynomials $m(x)$ to degrees $\le 26$, we get the space
$$
\mathcal{C}=\{g(x)m(x)\mid m(x)\in K_{\le 26}[x]\},
$$
and that will work. Alternatively (and more or less equivalently), we can select some subset $S\subset K$ of $31$ elements. If $m(x)\in K[x]$, let's denote by $m(S)$
the vector $(m(s))_{s\in S}\in K^{31}$ gotten by evaluating the polynomial $m$ at all the points of $S$. Then the space
$$
\mathcal{C}=\{m(S)\mid m(x)\in K[x], \deg m(x)\le 26\}
$$
is a $27$-dimensional subspace of $K^{31}$. Because a degree $26$ polynomial can have at most $26$ zeros in $K$ (hence in $S$), the minimum Hamming weight of $\Bbb{C}$ is five.


问:
Show that a bounded integral operator on $L_2$ maps $L_2$ functions to continuous functions
Let $K(t,s)$ be a continuous function on $L_2([a,b])$, and let $A \in L(L_2[a,b])$ be the integral operator:
$$(Af)(t) = \int_a^b K(t,s)f(s)ds$$

I want to show that for any function $f \in L_2[a,b])$, the operator on the function $Af$ is always continuous.

I managed to show that $\{ Af \}_{f \in M}$ is pre-compact in $C[a,b]$ when $M$ is a set of continuous and bounded functions.

I also get that in this case $\overline {\{ Af \}_{f \in M}}$ is compact, and since $A$ is an integral operator then it's a compact operator.

I would have wanted to somehow represent $Af$ as a limit of continuous functions, but I am not sure exactly how to begin.

Help would be appreciated.

答:
Using Holder's inequality, you can directly verify that
\begin{align*}
|Af(t)-Af(s)|&=\left|\int\limits_a^b (K(t,z)-K(s,z))f(z)\, dz\right|\\
&\leq \int\limits_a^b |K(t,z)-K(s,z)||f(z)|\, dz\leq \|K(t,\cdot)-K(s,\cdot)\|_{L^2}\|f\|_{L^2}\\&\leq \max_{z\in [a,b]}|K(t,z)-K(s,z)|(b-a)^{1/2}\|f\|_{L^2}
\end{align*} Continuity now follows immediately from the continuity of $K$.


问:
existence of nonlinear transformation
do we have any transformation in form of  $ T : R^2 \longrightarrow R $ that holds
for each vector like $v$ in $ xy $ plane and any real constant like $ r$  this equality $$ T(rv) = rT(v) $$
but it isn't a linear transformation?

答: You can take, for instance,$$T(x,y)=\begin{cases}x&\text{ if }y=0\\0&\text{ otherwise.}\end{cases}$$

Title: existence of nonlinear transformation


问:
Equality for the Supremum of the mean square error
Given a statistical model $(X,\mathcal{F},(\mathbb{P}_\vartheta)_{\vartheta\in\Theta}$) and $\hat{\vartheta}$ be an estimator. Then (here the MSE of $\hat{\vartheta}$ in terms of $\vartheta$ is denoted as $R(\hat{\vartheta},\vartheta)$.

$$
\sup_{\vartheta\in \Theta}R(\hat{\vartheta},\vartheta)= \sup_\Pi \int_\Theta R(\hat{\vartheta},\vartheta)\Pi (d\vartheta)
$$
Where the supremum is taken over all a-priori distributions $\Pi$. Now what is clear to me ist that
$$
R(\hat{\vartheta},\vartheta)\leq \sup_\Pi \int_\Theta R(\hat{\vartheta},\vartheta)\Pi (d\vartheta)
$$
This can be shown using the Dirac-measure as one of the a-priori measures in $\Pi$. What is not clear to me is the other way, where the proof only says
$$
\text{It is clear that } \sup_{\vartheta\in \Theta}R(\hat{\vartheta},\vartheta)\geq \int_\Theta R(\hat{\vartheta},\vartheta)\Pi (d\vartheta)
$$
How can this be shown?

答:
I think applying Hölder's inequality proves the claim. ($p=\infty$ corresponds to supremum, and $q=1$ results in 1 integrating over the prior.) But I cannot see how "it is clear" either!

EDIT

On a second thought, it is clear. Suppose $f(x) < M$. Then $\int f(x)g(x)dx <  M \int g(x)dx $. Here, $M$ is the supremum of MSE, and $ \int \Pi(d\vartheta) = 1 $.


问:
A question about "nice" functions
Let $f:\mathbb R \rightarrow \mathbb R$ be a function such that $\lambda(I)=\lambda(f(I))$ for each interval $I \subseteq \mathbb R$. ($\lambda$ is Lebesgue measure here.) Let us call such functions ''nice'' functions.

Can we characterize nice functions?

For example;

$f(x)=x$ is a nice function.

$f(x)=2x$ is not a nice function. More generally if $m \in \mathbb R\setminus\{-1,1\}$ then $f(x)=mx$ is not a nice function.

Every translation is a nice function.

$$f(x)=
\begin{cases}
x+1; & x \in [-1,0) \\
x-1; & x \in [0,1) \\
x; & \text{elsewhere}
\end{cases}$$  is a nice function.

Further, let $\mathcal A$ be the family of nice functions. Is the following conclusion true?

$f \in \mathcal A$ iff there exists a partition $I_{\alpha}$ of $\mathbb R$ into pairwise disjoint intervals such that $f=ax+b_{\alpha}$ on $I_{\alpha}$ for $a=-1$ or $1$ and for suitable $b_{\alpha}$.

答:
What about this:

$$ f(x) = \left\{\begin{array}{ll}x+1 & \text{ if $x$ is rational}\\
x & \text{ otherwise}
\end{array}\right.$$

ContentLicense: CC BY-SA 3.0

Title: A question about "nice" functions


问:
Banach fixed point theorem with lip constant possibly attaining $1$
An important condition of Banach fixed point theorem is that the map is strict contraction with Lipschitz constant strictly less than 1. Then the convergence is exponentially fast. But what if we relax the condition to less and equal to 1. Are there still a convergence (albeit not at exponential speed)?</p>

答:
An alternative proof to Milo's theorem using variational ideas (original argument for Edelstein theorem):
Check that any two fixed points of $\mathrm{f}$ must be equal. Consider $Q(x):=d(x, f(x)) .$ This is continuous function on a compact space so it must admit a minimizer z. We can check that minimizer must be a fixed point (and thus unique by first sentence). For any sequence $\left\{f^{n}(x)\right\},$ there exists a subsequence $\left\{f^{n_{i}}(x)\right\}$ that converges to some limit $a$. By continuity of $f, f^{n_{i}+1}(x) \rightarrow f(a) .$ Meanwhile, $d\left(f^{n}(x), z\right)$ is a decreasing sequence lower bounded by 0 so it must converge to some limit $l$. This implies that $d\left(f^{n_{i}}(x), z\right) \rightarrow l, d\left(f^{n_{i}+1}, z\right) \rightarrow l$. By continuity of metric, $d(a, z)=l=$ $d(f(a), z)=d(f(a), f(z)) .$ This implies $a=z .$ So the subsequence converges to the unique fixed point. For any $n_{i}<n<n_{i+1}, d\left(f^{n}(x), z\right) \leq d\left(f^{n_{i}}, z\right) .$ This implies that the entire sequence converges to z.


问:
$(,) = x^2$ How to determine if injective or surjective?
I just learned to determine wether $f(x)=x^2$ is injective or surjective but how can I solve this when there are two variables? I've been struggling to get my head around this.
I know that for a function to be injective the following condition must be met:
$()≠()$ implies $≠$, or the contrapositive.
Surjective:
$∀ ∈ , ∃ ∈  \text{ such that } f(x) = y$.

How could I scale these definitions to handle two variables?
$(,) = x^2$
mapping from $\Bbb R \times \Bbb R$ to $\Bbb R$

Edit: I want to thank you all for your time and help. It's greatly appreciated! I can clearly see through these problems now.

答:
Let $f$ be the following function
\begin{align}
f \colon \mathbb{R} \times \mathbb{R} & \to \mathbb{R}\\
(x,y) & \mapsto x^2
\end{align}
for all $(x,y) \in \mathbb{R}^2.$

Problem. Is it $f$ injective? Is it $f$ surjective?

Sketch of the proof. Given a function $f$ we want to be able to show if $f$ is or not injective and if $f$ is or not surjective.

First, let’s recall a couple of definitions.

What does it mean to say that $f$ is injective?

Definition. Let $A$ and $B$ be any sets and $f:A \to B$ a function. We say that $f$ is injective if
$$\forall x, y \in A , f(x) = f(y) \implies x = y. \tag{1}$$

And what does it mean to say that $f$ is surjective?

Definition. Let $A$ and $B$ be any sets and $f:A \to B$ be a function. We say that $f$ is surjective if
$$\forall b \in B, \exists a \in A \colon f(a) = b. \tag{2}$$

Solving. Let’s start by verifying if $f$ is or not injective.

Let $0,1 \in \mathbb{R}.$ Then $(0,0),(0,1) \in \mathbb{R}^2.$ Since $0 \neq 1,$ then $(0,0) \neq (0,1).$ By definition of $f$ It follows that
$$f(0,0) = 0 ^2 = 0 = 0^2 = f(0,1).$$
Hence, $$\exists x,y,z,w \in \mathbb{R} \colon f(x,y) = f(z,w) \wedge (x,y) \neq (z,w).$$
But this is precisely the negation of $(1).$ So $f$ does not satisfies the condition $(1).$ Therefore, $f$ is not injective.

Now, let’s see if $f$ is or not surjective.

Since the domain of $f$ is $\mathbb{R}^2,$ note that for all $(x,y) \in \mathbb{R}^2$ we have that $x \in \mathbb{R}$ and, hence, $x^2 \geq 0.$ Then, $f(x,y) \geq 0$ for all $(x,y) \in \mathbb{R}^2.$ Which means that $\not\exists (x,y) \in \mathbb{R}^2 \colon f(x,y) < 0.$

This tells us that the condition $(2)$ is not satisfied by this function $f.$ Therefore, $f$ is not surjective.

(Also, note that $f$ is not injective.)


问:
Prove that $\sum_{i=0}^{\infty}(-1)^{3n} \dfrac{x^{4n}}{(5n)!}$ converges absolutely
I shall provide the whole problem. To prove is the absolute convergence of

$ \sum_{i=0}^{\infty}(-1)^{3n}  \dfrac{x^{4n}}{(5n)!}$

I decide to solve this problem using the Quotient Criteria and eventually end up with

$ \dfrac{\dfrac{(-1)^{n+1} x^{4(n+1)}}{(5(n+1))!}}{\dfrac{(-1)^{n}   x^{4n}}{(5n)!}}  = \dfrac{(-1)^{n+1} x^{4(n+1)} (5n)!}{(5(n+1))! (-1)^{n} x^{4n}} = - \dfrac{x^4 (5n)!}{(5(n+1))!}$

Now I just need to prove that $-\dfrac{x^4 (5n)!}{(5(n+1))!}$ converges, which is where I get stuck. It's obvious to me that it converges to $0$ since the bottom part of the equasion grows much faster than the top part.

How do I work with the $(5(n+1))?$

Did I make a mistake?

Body:
$-\dfrac{x^4 (5n)!}{(5(n+1))!}$ = $-\dfrac{x^4 (5n)!}{(5n+5)!}$  = $-\dfrac{x^4 (5n)!}{(5n+5)...(5n+1)(5n)!} $ = $-\dfrac{x^4}{(5n+5)...(5n+1)}$

$\lim_{n \to +\infty} \dfrac{x^4}{(5n+5)...(5n+1)} = 0$


问:
Inverse CDF for generating random points from U[0,1]
I read online that:

"The inverse CDF technique for generating a random sample uses the fact that a continuous CDF, F, is a one-to-one mapping of the domain of the CDF into the interval (0,1). Therefore, if U is a uniform random variable on (0,1), then X = F–1(U) has the distribution F."

From: Article

But I'm failig to undertand this, in the specific case that I have:

$\displaystyle F_{X}( x) =\frac{e^{x}}{1+e^{x}}$, and therefore $\displaystyle F^{-1}_{X}( x) =\log\frac{x}{1-x}$.

From the derativate I was able to get the PDF: $\displaystyle f_{X}( x) =\frac{e^{x}}{\left( 1+e^{x}\right)^{2}}$.

How are the two functions mapping one-to-one?

If I have a random point generator $\displaystyle U$ between $\displaystyle [ 0,1]$. How exactly should I be able to generate a random point in $\displaystyle f_{X}( x)$? This is what I'm not getting, like what should I do exactly, and why it works.

答:
WHY IT WORKS (more mind map than a real demo)

X is a random variable according to the CDF F(X) means that $P(X\leq x)=F_X(x)$. Let's introduce $u=F_X(x)$, and try to identify the corresponding distribution.

By assuming $F_X$ is strictly increasing (not always the case in case of "steps distribution")
$$X \leq x \Leftrightarrow F_X(X) \leq F_X(x) \Leftrightarrow U \leq u$$

Then,
$$P(U \leq u) = P(X \leq x)$$
By using $t=F_X(s)$ variable change, inducing $dt=f_X(s)ds$
$$P(U \leq u) = \int_{-\infty}^{x}f_X(s)ds=\int_{F_X(-\infty)=0}^{F_X(x)=u}1dt$$
Note: it's still working even in case of finite boundaries for X.

So, if X is randomly distributed according to the CDF $F_X$, then $U=F_X(X)$ is distributed according to a uniform distribution between $[0,1]$.

You can have exactly the symetric process from U a standard uniform distribution variable, and $X=F^{-1}_X(U)$:
$$U \leq u \Leftrightarrow F_X^{-1}(U) \leq F_X^{-1}(u) \Leftrightarrow X \leq x$$
With $s=F^{-1}_X(t)$, $ds= \frac{dt}{f_X(F_X^{-1}(t))}$:
$$P(X \leq x)=P(U \leq u)=\int_0^u1dt=\int_{F_X^{-1}(0)}^{F_X^{-1}(u)=x}f_X(F_X^{-1}(t))ds=\int_{F_X^{-1}(0)}^{F_X^{-1}(u)=x}f_X(s)ds$$

So you can see that if X is randomly distributed according to the strictly increasing CDF $F_X$, then $U=F_X(X)$ is uniformaly distributed in between $[0,1]$ and reciprocally.

So if you are able to identify the analytical $F_X^{-1}$ equation for any $F_X$ stricly increasing, you simply have to generate a random variable U according to a standard uniform distribution (e.g. between $[0,1]$), and then compute $X= F_X^{-1}(U)$. This new X variable will be distributed according to the CDF $F_X$.


问:
proof verification on separable metric space
let $(X,d)$ and $(Y,p)$ be metric spaces and $f:X\rightarrow Y$ continuous, if $A\subseteq X$  and $(A,d|_{A})$ is separable show that $$(f(A),p|_{f(A)})$$ is also separable.

Since $f$ is continuous and $A$ is countable then $f(A)$ is countable.
because $\overline{A}= X\Rightarrow B(x,\delta  )\cap X\neq \varnothing (x\in A)$

$\Rightarrow  f(B(x,\delta  ))\cap f(X)\neq \varnothing $ (can i do that?)

$f$ is continuous so $B(f(x),\varepsilon  )\cap Y\neq \varnothing $

答:
Hello I have been trying the same problem because i couldn't solve it in first and know I came up with a different solution which it maybe be correct could someone verify that? Thank you in advance.

Since $A$ is separable $\exists \, X\subset A : \overline{X} = A $ and $X$ is countable $\Rightarrow $ $f=cont$  $f(X)\subset f(A)$ $\Rightarrow $ $\overline{f(X)}\subseteq f(A)$ $ (1)$
it's enough to show $f(A)\subseteq \overline{f(X)}$

Since $\overline{X} = A $ $\Rightarrow  $ $\exists \ : x \in A$ and $x_n \in X : x_n\rightarrow x$ now since $f=cont$  we have that $f(x_n) \in f(X), f(x) \in f(A)$ and $f(x_n)\rightarrow f(x) $ with  $f(x_n) \in f(X)$  $\Rightarrow  f(x)\subseteq \overline{f(X)}  \Rightarrow  f(A)\subseteq \overline{f(X)} $ $(2)$ and we prove what we want.


问:
How to get to Catalan numbers recurrence relation from couting lattice paths
I'm currently taking a probability theory class and I've been stuck on a problem where I can find the solution using one method but I dont get how my TA got to the catalan numbers from counting, so heres how it goes (I didn't find any awnsers on HOW you get the recurrence relation i.e the reasonning that gets to it):

Suppose we place someone on a lattice $\mathbb{Z}^2_{\geq0}$ and we want to count how many ways he can get to$ (n,n)$,  such that $n\in \mathbb{N}_{\geq 1}$ and the the person cant pass trough the diagonal $y=x$ and he can only walk northward or westward on that lattice.

And here's my thoughts so far:

Attempt on a solution using recurrence relation

Let $C_k$ be the numbers of valid path from $(0,0)$ to $(k,k)| k\in \mathbb{N}_{\geq 1}$

Then $C_0 = 1$ (because there's only one way to stay where you are... you stay)

Moreover, if the person touch the diagonal at $(k,k)$, then he has $C_{n-k}$ ways to get to $(n,n)$ if he doesn't touch the diagonal again (because its as if (k,k) was the new origin... I'm not sure from here).

So by the product principle of counting, for every $C_k$ ways he can get to $(k,k)$ he has $C_{n-k}$ implies that he has $(C_{n-k})( C_k)$ ways to get to $(n,n)$.

However, my TA only wrote,

"let's consider the first time he touches the diagonal, then he has

$(C_0)(C_n) + (C_1)(C_{n-1}) + ... + (C_n)(C_0) = C_{n+1}$

So is my way of reasonning a good beginning and how do we get from his first sentence to that relation with a sum and $C_{n+1}$

Thank you so much

答:
Total Catalan paths from $(0,0)$  to  $(n+1,n+1)$ is equal to $C_{n+1}$

Let us split all Catalan paths in some mutually exclusive paths based on the first time, they meet diagonal $y=x$ after $(0,0).$

Let use take a Catalan path meeting $y=x$ after $(0,0)$ first time at $(r,r),\,r$ can be $1,2,,3\cdots,n+1.$ This path can be thought as starting from $(1,0)$ going to $(r,r-1)$ and never crossing $y=x-1$ then from $(r,r-1)$ to $(r,r)$ and finally from $(r,r)$ to $(n+1,n+1)$

Number of paths from $(1,0)$ to $(r,r-1)$ and not crossing $y=x-1$ is equal to $ C_{r-1}$

Number of paths from $(r,r-1)$ to $(r,r)$ is $1$

Number of paths from $(r,r)$ to $(n+1,n+1)$ is $C_{n+1-r}$

$\implies$ Number of paths meeting y=x after $(0,0)$ first time at $(r,r)$ is $C_{r-1}\times 1\times C_{n+1-r}$

$\implies C_{n+1}=\sum_{r=1}^{n+1} C_{r-1}C_{n+1-r}$


问:
How to show that any nontrivial valuation on the field of rational numbers is equivalent to some p-adic valuation?
I have started with a nontrivial valuation v on Q. Suppose V is the corresponding valuation ring. I want to show that V is equal to the valuation ring of some p-adic valuation for some prime p. Then by definition of equivalence this will hold. For this I use the fact that as v is nontrivial valuation so there must exist atleast one prime p such that 1/p is not in V. From this I can show that V⊆Vp where Vp is the valuation ring of the p-adic valuation on Q which is precisely the set of rationals with denominator  not divisible by p. But am stuck with the other way inclusion. Can anybody suggest to proceed furthur?</p>

答:
I don't know how you obtained $O_v\subset \Bbb{Z}_{(p)}$.

$O_v\subsetneq \Bbb{Q}$.

$v(1)=0$ gives that $v(n)\ge 0$ thus $\Bbb{Z}\subset O_v$.

There is some $p$ such that $v(p)>0$ because otherwise $O_v=\Bbb{Q}$.

$m_v\cap \Bbb{Z}$ is a prime ideal containing $p$ thus it is $(p)$ and hence $O_v$ contains $\Bbb{Z}_{(p)}$.

For any $a\in \Bbb{Q}-\Bbb{Z}_{(p)}$, $\Bbb{Z}_{(p)}[a]=\Bbb{Q}$, thus $O_v=\Bbb{Z}_{(p)}$.


问:
A simple but nonsense infinite sums rule
Assume $a>0$ and $0<r<1$ are given. The geometric series is known to be $\sum_0^{\infty} ar^n = \frac{a}{1-r}$.  Now differentiate w.r.t $r$ to obtain $\sum_0^{\infty}anr^n = \frac{ar}{(1-r)^2}$. So far we're cool.

Now take the two series $$S_1 = \sum_n \frac{1}{2^n}$$ and $$S_2 = \sum_n \frac{n}{2^n}.$$ One would trivially expect $S_2>S_1$, because obviously $S_2$ has an additional $n$ multiplied by each term. To my surprise, this is not the case! Invoking the result of the previous paragraph, you get
$$S_1 = \frac{1}{1-1/2} = 2$$
and
$$ S_2 = \frac{1/2}{1/4} = 2$$
so that $S_1 = S_2$.

Does this make any sense to you? Because it sure makes no sense to me. Or am I losing it?

答:
Yes, you are right $S_1 = S_2 = 2$ and it might be somewhat astonishing the first time. But first let's have a look at your derivation which needs to be somewhat revised. We can take $a=1$ and consider
\begin{align*}
S_1(r)=\sum_{n=0}^\infty r^n=\frac{1}{1-r}\tag{1}
\end{align*}
Derivation with respect to $r$ gives
\begin{align*}
\frac{d}{dr}S_1(r)&=\frac{d}{dr}\frac {1}{1-r}=\frac{1}{(1-r)^2}\tag{2}
\end{align*}
Note we have numerator $1$ and not $r$. On the other hand we also obtain
\begin{align*}
\frac{d}{dr}S_1(r)&=\frac{d}{dr}\sum_{n=0}^\infty r^n=\sum_{n=0}^\infty nr^{n-1}=\sum_{n=1}^\infty nr^{n-1}\\
&=\sum_{n=0}^\infty (n+1)r^n\\
&=\sum_{n=0}^\infty nr^n+\sum_{n=0}^\infty r^n\\
&=\sum_{n=0}^\infty nr^n+\frac{1}{1-r}\\
\sum_{n=0}^\infty nr^n& = \frac{d}{dr}S_1(r) - \frac{1}{1-r}=\frac{1}{(1-r)^2}-\frac{1}{1-r}\tag{3}
\end{align*}

We obtain from (1) - (3) by substituting $r=\frac{1}{2}$:
\begin{align*}
\color{blue}{S_1\left(\frac{1}{2}\right)}&=\frac{1}{1-\frac{1}{2}}\,\color{blue}{=2}\\
\color{blue}{S_2}&=\sum_{n=0}^\infty \frac{n}{2^n}
=\frac{1}{\left(1-\frac{1}{2}\right)^2}-\frac{1}{1-\frac{1}{2}}=4-2\,\,\color{blue}{=2}
\\
\end{align*}

We observe the term with $n=0$ in $S_2$ is zero, but the numerator $n$ with index $n>0$ is a compensation for it.


问:
Multplication operator self-adjoint
I have to show that the multiplication operator $M_{f}:D\to L^{2}(\mathbb{R})$ with $M_{f}g:=f\cdot g$ is self-adjoint if and only if $f:\mathbb{R}\to\mathbb{C}$ is real-valued, where $D$ denotes the natural domain (e.g. $g\in D$ if and only if $M_{f}g\in L^{2}(\mathbb{R})$)

In the proof of the direction $\Leftarrow$ i have the following problem:

So suppose $M_{f}$ is self-adjoint. Then we have for all $g,h\in D$
$$\langle M_{f}g,h\rangle = \langle g,M_{f}h\rangle.$$

which is equal to the statement:

$$\forall g,h\in D:\int_{\mathbb{R}}(f(x)-\overline{f}(x))\overline{g}(x)h(x)\,\mathrm{d}x=0$$

But then I can just say that $f(x)=\overline{f}(x)$ almost everywhere. How can I show that $f$ is real-valued on the whole domain?

答: Since you're working with $L^2$, all functions are identified up to sets of measure zero. So, you're good!

Title: Multplication operator self-adjoint


问:
Gaussian expectation of an exponential function
I am struggling to prove this,
$$
\int \mathcal{N}_\mathbf{x}(\mu,\Sigma)e^{a^T\mathbf{x}}d\mathbf{x} = e^{{a^T\mu}+\frac 12a^T\Sigma a}
$$</p>

答:
Complete the square in the exponent, that is, integrate the identity
$$
\mathcal{N}_\mathbf x(\mu,\Sigma)\cdot\mathrm e^{a^T\mathbf x}=\mathrm e^{a^T\mu+\frac 12a^T\Sigma a}\cdot\mathcal{N}_\mathbf x(\mu+\Sigma a,\Sigma).
$$


问:
Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1,\;a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$
Define  a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1$
$$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$
then what will be it's $\limsup (a_n)$ and $\liminf (a_n)$ and $\sup(a_n)$, $\inf(a_n)$.

I tried to determine the nature of $a_{2n+1}$ and $a_{2n}$. I got $a_{2n+1}$ is increasing and $a_{2n+1} \geq \text{something}$ where $a_{2n}$ is decreasing  and $a_{2n}\leq \text{something}$. So I could not draw any conclusion.

Can anyone please help me?

答:
I think the following will help the readers to understand the solution:

We know  $$\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt{|a_n| \times \frac {2}{|a_n|}}=\sqrt 2$$
$$\implies \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$
Given that $~a_1 = 1~$ and for $n\geq 1$
$$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$
When $~n~$ is even, $$a_{n+1} = \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$
Again when $~n~$ is odd, $$a_{n+1} = -\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\le-\sqrt 2$$
Which gives $~\lim\sup a_n=\sqrt 2~$ and $~\lim\inf a_n=-\sqrt 2~$.


问:
Find all integer solutions $(x,y)$ such that $2^x+2^y=x^2+y^2$
when i saw this question "if $2^x + 2^y = x^2 + y^2$, where $x$ and $y$ are nonnegative integers, what is the greatest possible value of $|x – y|$?" I wanted to figure out what are all the integer values that can satisfy that equation.

I thought on testing trivial values, starting with non-negative integers. I figured out that $x=y=2$ and $x=4,y=2$ would satisfy the equation, but testing integers one by one seems to be tricky.

答:
$$2^x+2^y=x^2+y^2$$
$$2^x-x^2=-(2^y-y^2)$$
For $x=0,1,2,3,4,5,\dots$, $2^x-x^2=1,1,0,-1,0,7,\dots$ and the difference only grows larger from there. We seek two values that are negatives of each other, as far apart as possible; those are $x=0$ and $x=3$, so the largest $|x-y|$ is $3$, attained by $x=0,y=3$:
$$2^0+2^3=9=0^2+3^2$$


问:
How can I prove that $\lim\limits_{m \to \infty}[(1 + \frac{1}{m})^m]^x = \exp(x)$ using polynomials?
I'm trying to prove that $\lim\limits_{m \to \infty}[(1 + \frac{1}{m})^m]^x = \exp(x)$ by expanding polynomials and comparing them, given that $\exp(x)$ is already defined as:

$$\exp(x) = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dotsc$$

I tried going about this by firstly expanding the said limit with binomial expansion as
$$\lim\limits_{m \to \infty}\left[\left(1 + \frac{1}{m}\right)^m\right]^x = \lim\limits_{m \to \infty} \left(\binom{m}{0} + \binom{m}{1} \frac{1}{m} + \binom{m}{2} \frac{1}{m^2} + \dotsb \right)^x $$

my main goal being to obtain polynomial $P(x)$ which I can compare with $\exp(x)$. But here I get stuck because, after expressing each binomial coefficient, I'm not sure what to do about the $m \to \infty$ or later on exponent $x$ for that matter.

I'd really appreciate your help.

答:
First note $$\left(1+\frac 1 m\right)^{mx}\\=1+mx\cdot \frac 1 m+\frac{mx(mx-1)}{2}\cdot \frac 1 {m^2}+\cdots+\frac{(mx)(mx-1)\cdots(mx-n+1)}{n!}\cdot\frac 1 {m^n}+\cdots\\=1+x+\frac{x(x-\frac 1 m)m^2}{2}\cdot \frac 1 {m^2}+\cdots+\frac{x(x-\frac 1 m)\cdots(x-\frac{(n-1)}m)m^n}{n!}\cdot\frac 1 {m^n}+\cdots\\=1+x+\frac{x(x-\frac 1 m)}{2}+\cdots+\frac{x(x-\frac 1 m)\cdots(x-\frac{(n-1)}m)}{n!}\cdot+\cdots$$
Now taking limits we get $$\lim_{m\to\infty}\left(1+\frac 1 m\right)^{mx}=1+x+\frac {x^2}2+\cdots+\frac{x^n}{n!}+\cdots=e^x$$

.${}{}{}$


问:
How to prove $f$ is alternating iff $f(v_1,\ldots ,v_k)=0$ whenever two of $v_1,\ldots,v_k$ are equal
My question comes from Loring Tu's An Introduction to Manifolds Second Edition.

The problem is

Let $f$ be a $k$-tensor on a vector space $V$. Prove that $f$ is alternating if and only if $f(v_1,\ldots,v_k)=0$ whenever two of the vectors $v_1,\ldots,v_k$ are equal.

The definition used for alternating functions is $\sigma f= (\text{sgn }\sigma)f$.

Proving the rightwards implication ("only if") is easy; I'm having trouble with the reverse direction.

I've tried expanding $f$ into its components using the basis $\{\alpha^{i_1}\otimes\cdots\otimes\alpha^{i_k}\}$ but wasn't able to take that anywhere. I was also considering somehow using the basis $\{\alpha^{i_1}\wedge\cdots\wedge\alpha^{i_k}\}$ for alternating functions, but couldn't think of a way to apply it.

I think my biggest hurdle on this problem is how to use the fact about the special case $v_a=v_b,a\ne b$ to prove anything about the cases where we don't have equivalent arguments.

If anyone has any tips or hints on how to progress, it would be greatly appreciated.

答:
Consider any $(v_i)_{i=1 \ldots k}$ and any $1 \leq j<l \leq k$ distinct integers.

Pose $w_j = w_l = v_j + v_l$ and $w_i = v_i$ for all the other values of $i$.

Now, $w_j = w_l$, so $f(w_1,\ldots,w_k)=0$. However, $$f(w_1,\ldots,w_k) = f(v_1,\ldots,v_j, \ldots, v_l, \ldots ,v_k) + f(v_1,\ldots,v_l, \ldots, v_l, \ldots ,v_k) + f(v_1,\ldots,v_j, \ldots, v_j, \ldots ,v_k) +f(v_1,\ldots,v_l ,\ldots, v_j ,\ldots, v_k) = f(v_1,\ldots,v_j ,\ldots, v_l, \ldots ,v_k) + f(v_1,\ldots,v_l, \ldots, v_j ,\ldots, v_k)$$

This shows it if $\sigma = (jl)$, can you finish from here?


问:
Using Cauchy-Euler method to solve $x^2 y''+2xy'-xy=0$
Given the following ODE
$$x^2 y''+2xy'-xy=0$$
I have to solve it using Cauchy-Euler's method. Writing
$$y=x^n$$
then
$$y'=n x^{n-1},\quad y''=n(n-1)x^{n-2}$$
and therefore
$$x^{2}\cdot n(n-1)x^{n-2}+2x nx^{n-1}-x\cdot x^n=0$$
$$\Rightarrow x^{n}(n^2+n-x)=0$$
which gives the equaiton
$$n^2+n-x=0$$
Solving the equation above, should the solution be
$$y(x)=x^{n}$$
where $n=\displaystyle\frac{-1\pm\sqrt{1+4x}}{2}$?

答:
Simplify by $x$ and rewrite the equation as$$x y''(x)+2 y'(x)=y(x)\tag{1}$$
set
$$y(x)=\sum_{n=0}^{\infty}a_nx^n$$
so we have
$$y'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}$$
and
$$y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$
Substitute in $(1)$
$$x\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+2\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=0}^{\infty}a_nx^n$$
and finally
$$\sum_{n=2}^{\infty}n(n-1)a_nx^{n-1}+\sum_{n=1}^{\infty}2na_nx^{n-1}=\sum_{n=0}^{\infty}a_nx^n$$
set equal the coefficients of the powers of $x$ in both sides
$$
\begin{cases}
2a_1=a_0,& a_1=\frac{1}{2}a_0\\
6a_2=a_1,& a_2=\frac{1}{12}a_0\\
12a_3=a_2,& a_3=\frac{1}{144}a_0\\
20a_4=a_3,& a_4=\frac{1}{2880}a_1\\
30a_5=a_4,& a_5=\frac{1}{86400}a_1\\
\ldots\\
a_n=\frac{1}{n!(n+1)!}
\end{cases}
$$
So the solution is, suppose $y(0)=y'(0)=1$,
$$y(x)=\sum _{n=0}^{\infty } \frac{x^n}{n! (n+1)!}$$
$$y(x)=\frac{I_1\left(2 \sqrt{x}\right)}{\sqrt{x}}$$
where $I_1(t)$ is the modified Bessel function of the first kind.


问:
Finding minimal distance between two curves using AM-GM inequality

Consider the following subsets of the plane: $C_{1}=\{(x,y)|x\gt 0,y=1/x\}$ and $C_{2}=\{(x,y)|x\lt 0,y=-1+1/x\}$ Given two points $P(x,y)$ and $Q(u,v)$ of the plane. Their distance $d(P,Q)$ is defined as:
$$d(P,Q)=\sqrt{(x-u)^2+(y-v)^2}$$
Show that there exists unique choice of $P_{0}\in C_{1}$ and $Q_{0}\in C_{2}$ such that $d(P_{0}, Q_{0})\le d(P,Q) \forall P \in C_{1}$ and $Q\in C_{2}$

Attempted:

Essentially all this problem asks for is to prove that there is a lower bound on the distance between the two subsets $C_{1}$ and $C_{2}$ of the plane and that, that minimal distance is achieved at unique $P_{0}$ and $Q_{0}$. To find the minimal distance, we can assert that, it has to be the length of the common normal. From the slopes of the two curves we can say, $P$ must be of the form $(x_{0},1/x_{0})$ and $Q$ of the form $(-x_{0},-1-1/x_{0})$. Now if we plug this into the expression for $d(P_{0},Q_{0})$, we get the following:

$$d(P_{0},Q_{0})=\sqrt{4x^2+1+\frac{4}{x}+\frac{4}{x^2}}$$

To minimise we can set the derivative of $d^2(P_{0},Q_{0})$ with respect to  $x$ so get that $2x^4-x-2=0$ which has only one positive solution in $(1,2)$ by the intermediate value theorem, which also proves the uniqueness of the choice of $P_{0}$ and consequently $Q_{0}$. However if I try to apply the AM-GM inequality, and rewrite the expression inside the square root sign as follows, I do not get the condition of equality in which all terms should be equal:

$$x^2+x^2+x^2+x^2+\frac{2}{x^2}+\frac{2}{x^2}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}\ge 10\cdot2^{1/5}$$

So how to go about it using the AM-GM inequality. Any hints are appreciated. Thanks.

答:
If the two coordinates are $(a,\frac1a)$ and $(-b,-1-\frac1b)$ for $a,b>0$, then the squared distance is $(a+b)^2 + (\frac1a + \frac1b + 1)^2$. We can make the substitution $a = e^u$ and $b = e^v$ to get the function
\begin{align}
f(u,v) &= (e^u + e^v)^2 + (e^{-u} + e^{-v} + 1)^2 \\
  &= 2 e^{-u-v}+2 e^{u+v}+e^{-2 u}+2 e^{-u}+e^{2 u}+e^{-2 v}+2 e^{-v}+e^{2 v}+1
\end{align}
which we optimize over all $u,v \in \mathbb R$.

This function is strictly convex, because each term is convex and some are strictly convex, which tells us that the minimizer (if it exists) is unique.

The minimum doesn't have to be achievable in general when optimizing a function over $\mathbb R^2$. However, we can argue that actually, $u,v \in [-2,2]$ (for instance), because $f(0,0)=13$, and one of the terms $e^{2u}, e^{2v}, e^{-2u}, e^{-2v}$ will exceed $13$ if $|u| > 2$ or $|v|>2$. So we can reduce the problem to minimizing $f$ over $[-2,2]^2$, and then a minimizer has to exist by the extreme value theorem.


答:
The slick $\log$ proof I alluded to:

If you know that $\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\dots$ for $|x|<1$, then
$$\left(1+\frac{1}{m}\right)^{mx}=\exp\left(mx\log\left(1+\frac{1}{m}\right)\right)=\exp(mx[1/m+o(1/m)])=\exp(x+o(1)),$$
which implies the claim.

Here's a more mechanical proof just in case.

Substitute $m=n/x$ to get the form $a_n=\left(1+\frac{x}{n}\right)^n$. Binomial expanding implies that $(a_n)$ is monotone increasing:
\begin{align*}
\left(1+\frac{x}{n}\right)^n &= \sum_{k\geq0}\frac{x^k}{k!}\cdot\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right) \\
&\leq\sum_{k\geq0}\frac{x^k}{k!}\cdot\prod_{i=0}^{k-1}\left(1-\frac{i+1}{n}\right)\\
&= \left(1+\frac{x}{n+1}\right)^{n+1}.
\end{align*}
Then as the factors $1-\frac{i}{n}<1$, it's clear that $\exp(x)$ is an upper bound for $(a_n)$. Hence $(a_n)$ converges to some limit that is at most $\exp(x)$. Fix some $m$, then for $n\geq m$, we have
$$a_n=\left(1+\frac{x}{n}\right)^n\geq\sum_{k=0}^m\frac{x^k}{k!}\cdot\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right).$$
You can see easily that for a fixed $k$, the product $\prod_{i=0}^{k-1}\left(1-\frac{i}{n}\right)$ converges to $1$ as $n\to\infty$. So taking the limit of the above inequality as $n\to\infty$ implies that
$$\lim_{n\to\infty}a_n\geq\sum_{k=0}^m\frac{t^k}{k!},$$
so now taking $m\to\infty$ implies the result.

Title: How can I prove that $\lim\limits_{m \to \infty}[(1 + \frac{1}{m})^m]^x = \exp(x)$ using polynomials?

Body:
I'm trying to prove that $\lim\limits_{m \to \infty}[(1 + \frac{1}{m})^m]^x = \exp(x)$ by expanding polynomials and comparing them, given that $\exp(x)$ is already defined as:

$$\exp(x) = \sum_{n=0}^{\infty} \dfrac{x^n}{n!} = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dotsc$$

I tried going about this by firstly expanding the said limit with binomial expansion as
$$\lim\limits_{m \to \infty}\left[\left(1 + \frac{1}{m}\right)^m\right]^x = \lim\limits_{m \to \infty} \left(\binom{m}{0} + \binom{m}{1} \frac{1}{m} + \binom{m}{2} \frac{1}{m^2} + \dotsb \right)^x $$

my main goal being to obtain polynomial $P(x)$ which I can compare with $\exp(x)$. But here I get stuck because, after expressing each binomial coefficient, I'm not sure what to do about the $m \to \infty$ or later on exponent $x$ for that matter.

I'd really appreciate your help.


问:
Fake proof that $\cot^4 x = -\tan^4 x$
What am I missing?

Consider:
$$ \int \frac{x^4}{\left(1-x^2\right)^{5/2}} dx$$

If we let $x = \sin \theta$ we get

$$ \int \frac{\sin^4 \theta}{\cos^5 \theta} \cos \theta d\theta$$

as $dx = \cos \theta d\theta$. We can simplify the above expression to

$$ \int \tan^4 \theta d \theta$$

However, if we let $x = \cos \theta$ we get

$$ -\int \frac{\cos^4 \theta}{\sin^5 \theta} \sin \theta$$

as $dx = - \sin \theta d\theta$.

We can simplify the above expression to

$$ -\int \cot^4 \theta d\theta$$

Now, the two simplified expressions are equal since they are equal to the original expression. So we have the following equality:

$$ \int \tan^4 \theta d\theta = -\int \cot^4 \theta d\theta$$
$$ \frac{d}{d\theta} \int \tan^4 \theta d\theta = \frac{d}{d\theta}\left( -\int \cot^4 \theta d\theta\right)$$

The derivative of the integral is the same as the function so

$$\tan^4 \theta = -\cot^4 \theta$$

答: The $\theta$s in the two susbtitutions are two variables, say $\theta$ and $\theta^\prime:=\pi/2-\theta$ so $\cot^4\theta^\prime=\tan^4\theta$. The extra $-$ sign follows from $d\theta^\prime=-d\theta$.

Title: Fake proof that $\cot^4 x = -\tan^4 x$


问:
A not-so-typical marble question
In a bag are 8 white marbles and 5 red. Two marbles are drawn from the bag.
What is the probability that both marbles are white if we know that one of them is white?
The answer is 0,412 but I don't know how to get there :(</p>

答:
Hint:

$$P(B \text{ happens given } A \text{ happens})=\frac{P(A\text{ and } B \text{ happen})}{P(A \text{ happens})}$$
In symbolic notation:
$$P(B|A)=\frac{P(A\cap B)}{P(A)}$$

Title: A not-so-typical marble question


问:
Let $p$ be a prime number and let $G$ be a $p$-group. Then $G$ contains a normal subgroup of order $p^k$ for every nonnegative $k\le r$
Let $p$ be a prime number, and let $G$ be a $p$-group: $|G|=p^r$ . Then $G$ contains a normal subgroup of order $p^k$ for every nonnegative $k\le r$

But are there any normal subgroup of order $p^n$ such that $n>r$ ?

How can we show that, I couldn't show.

答: No, since Lagrange's Theorem ensures that the order of any subgroup of a group divides the order of the group; $p^n\nmid p^r$ for $n>r$.


问:
Deriving the first moment of Collisionless Boltzmann Equation in Spherical Polar Coordinates
I am following these notes: Dynamics and Astrophysics of Galaxies.

After equation 6.37, we have:
\begin{equation*}
    p_r\,\frac{\partial f}{\partial r} + \frac{p_\theta}{r^2}\,\frac{\partial f}{\partial \theta} +
    \frac{p_\phi}{r^2\,\sin^2\theta}\,\frac{\partial f}{\partial \phi}
    -\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{p_\theta^2}{r^3}-\frac{p_\phi^2}{r^3\,\sin^2\theta}\right)\,\frac{\partial f}{\partial p_r}
    +\frac{p_\phi^2\,\cos\theta}{r^2\,\sin^3\theta}\,\frac{\partial f}{\partial p_\theta} = 0\,.\\
\end{equation*}

This is the Collisionless Boltzmann Equation in Spherical Polar Coordinates.

Then

We now multiply this by $p_r$ and integrate over all $(p_r,p_{\phi},p_{\theta})$ using
that $\mathrm{d}p_r\,\mathrm{d}p_\phi\,\mathrm{d}p_\theta = r^2\,\sin\theta\,\mathrm{d}v_r\, \mathrm{d}v_\phi\,\mathrm{d}v_\theta$ and using partial integration to deal
with the derivatives of f with respect to the momenta

\begin{align}\label{eq-spher-jeans-penult}
    \frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2_r})}{\partial r} + \frac{\partial (\sin\theta\,\nu\,\overline{v_r\,v_\theta})}{\partial \theta} & +
    \frac{\partial (\nu\,\overline{v_r\,v_\phi}/\sin\theta)}{\partial \phi}\\
    &
    +r^2\,\sin\theta\,\nu\,\left(\frac{\mathrm{d} \Phi}{\mathrm{d} r}-\frac{\overline{v_\theta^2}}{r}-\frac{\overline{v_\phi^2}}{r}\right) = 0\nonumber\,.
\end{align}

I would like to arrive to this result myself. Multiplying the SPC CBE by $p_r$ & going ahead with the suggested volume element, considering the first term in the above equation, not being worried about the integration limits gives us:

\begin{equation}
\int p_r \frac{\partial f}{\partial r} p_r r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
=\int v_r^2 \frac{\partial f}{\partial r} r^2 \sin \theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
=r^2 \sin \theta \int v_r^2 \frac{\partial f}{\partial r} \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}
\end{equation}

Using 6.32 from the mentioned notes,

\begin{equation}
\frac{\partial (r^2\,\sin\theta\,\nu\,\overline{v^2_r})}{\partial r} = \frac{\partial}{\partial r} \left( r^2 \sin\theta \int v^2_r f \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} \right)
\end{equation}

Which is not equal to what I have found just a line above.

What am I doing wrong?

Same question on Astronomy.SE and Physics.SE.

答:
After correspondence with the author of those notes, I arrived at the following conclusion.

When we do $\frac{\partial f}{\partial r}$, we require that all other (other than $r$) variables which f is dependent on, are held constant. That is: $\theta, \phi, p_r, p_{\theta}, p_{\phi}$ are constants.

Note that:

$(p_r, p_{\theta}, p_{\phi})=(\dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta})$

Keep in mind that:

$p_r$, ie $\dot{r}$ is held constant: it does not mean $r$ is held constant (in fact, we are differentiating with respect to $r$).

$p_{\theta}$, ie $r^2\sin^2\theta\dot{\phi}$ is held constant. This does not mean that either $r$, $\theta$ or $\dot{\phi}$ is held constant.

$p_{\phi}$, ie $r^2\dot{\theta}$ is held constant. We are not claiming this about $r$ or $\dot{\theta}$.

Writing the steps emphasizing this statement above:

$$\int d\vec{p} p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$

Writing out explicitly what we mean by $d\vec{p}$:

$$=\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r \frac{\partial f}{\partial r}  p_r \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$

We can move out the partial differentiation because all other term than $\frac{\partial f}{\partial r}$ is held constant: [THIS IS THE MAIN STEP]

$$=\frac{\partial}{\partial r}\int \mathrm{d}(p_r,p_{\theta},p_{\phi}) p_r^2 f \Bigr|_{\theta, \phi, p_r, p_{\theta}, p_{\phi}}$$

Changing from the $p$-nomenclature to the variables they represent:

$$\frac{\partial}{\partial r} \int \mathrm{d}(\dot{r}, r^2\dot{\theta}, r^2\sin^2\theta\dot{\phi})  v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$

ie

$$\frac{\partial}{\partial r} \int \mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) v_r^2 f \Bigr|_{\theta, \phi, \dot{r}, r^2 \sin^2 \theta \dot{\phi}, r^2\dot{\theta}}$$

This step I am unsure about (ie why I don't have a $dr$ and $d\theta$ term, for example), but we say:

$$\mathrm{d}(v_r, r \sin\theta v_{\phi},  r v_{\theta}) = r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi}$$

Rewrite integral

$$\frac{\partial}{\partial r} \int r^2 \sin\theta \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} r^2 \sin\theta \int \mathrm{d}v_r \mathrm{d}v_{\theta} \mathrm{d}v_{\phi} v_r^2 f$$
$$=\frac{\partial}{\partial r} \left( r^2 \sin\theta \nu \bar{v_r^2} \right)$$

As claimed by the notes.

Same answer on Astronomy.SE & Physics.SE.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答:
In subject $X$ the Fundamental Theorem of $X$ is always pretty important.

For example:

Fundamental Theorem of Finitely Generated Abelian Groups.

Fundamental Theorem of Calculus.

Fundamental Theorem of Arithmetic.

Fundamental Theorem of Algebra.

Fundamental Theorem of Galois Theory

Fundamental Theorem of.....

Title: Landmarks of subjects of mathematics


问:
Continuous Non negative martingale converging to 0
Is there any (non trivial) continuous non negative martingale which converges to 0?</p>

答: Sure, try $M_t=B_{t\wedge T}$ where $(B_t)$ is a Brownian motion starting from $B_0=1$ and $T$ is the first hitting time of $0$ by $(B_t)$.</p>


问:
Is the unit circle in $\mathcal{B}(\mathbb{R}^2)$
Is the unit circle in $\mathcal{B}(\mathbb{R}^2)$ ?

The definitions I am working with is that $\mathcal{B}(\mathbb{R}^2)$  is the $\sigma$-Algebra generated by open (equivalently closed, or clopen) rectangles. i.e sets of the form $(a_1,b_1) \times (a_2,b_2)$

Approach 1:

I think maybe I can get there in the limit of unions of lots and lots of rectangles. Here was my approach:

$S =(-1,1) \times (-1,1)$ and then we construct some kind of limit of squares to take away from the square, $S$, that leave us with just the circle?

Approach 2:

Consider $f:(\mathbb{R^2},\mathcal{B}(\mathbb{R}^2)) \to (\mathbb{R},\mathcal{B}(\mathbb{R}))$ by $f(x,y) = x^2 +y^2$

$f$ is clearly continuous and as such is Borel measurable. This means that the pre-image of any measurable set is measurable and hence $f^{-1}(0,1),$ the unit circle, is measurable?

Approach 2 seems to be correct but it feels too simple, are there any mistakes? Would approach 1 provide a fruitful answer?
Any help or even an alternative method of your own would be really helpful!

Happy holidays :)

答:
It is, I will try to make a visual proof, I will use the $\sigma-$alegbra generated by the closed rectangles, wich is the same as you said. Take the disk $B=B(0,1)$ now for $k\in \mathbb N$ imagine all the squares of the form $P_k=\{[n/2^k,(n+1)/2^k]\times[m/2^k,(m+1)/2^k]:m,n\in\mathbb Z\}$ (this is like a grid of squares of length $1/2^k$) and define $A_k$ as the union of all the squares in $P_k$ that are contained in the disk. Since any square in $P_{k}$ can be written as the union of four squares in $P_{k+1}$ you have that $A_k\subset A_{k+1}$, you have that $B=\bigcup A_k$.

To proove this take any point $p$ in $B$, since $B$ is open there is a ball $B_p$ of radius $\varepsilon$ centered in $p$ and contained in $B$, now take $k$ such that $\frac{1}{2^k}<\varepsilon$, then $p$ belongs to a square of $P_{k+1}$ contained in $B_p$:

Take a square $S$ of $P_{k+1}$ containing $p$ (since $P_{k+1}$ covers all $R^2$ the existance of such square is guaranteed) for any point $x$ in $S$ you have that $$d(x,p)\leq \frac{\sqrt{2}}{2^{k+1}}=\frac{1}{2^{k+1/2}}<\frac{1}{2^k}<\varepsilon$$
So $S\subset B_p\subset B$, hence for all $p\in B$ there exist $k$ such that $p\in A_k\subset B$.


问:
Writes in polar coordinates $x''-(1-x^2-(x')^2)x'+4x=0$.
I need to determine a periodic solution for :$x''-(1-x^2-(x')^2)x'+4x=0$.
We have the equivalent system:
$$\begin{cases}
       x'=y \\
       y'=(1-x^2-y^2)y-4x.
   \end{cases}
$$

We determined the stationary points for the equivalent system: $(0,0)$.

I saw that this solution is unstable.

But now I have to turn this system into polar coordinates. Unfortunately, I tried in all possibilities and I failed to bring the system to a beautiful shape depending on the polar coordinates.

I present what I tried:

Polar coordinates:
$\begin{cases}
       x(r,\theta)=r\cos(\theta) \\
       y(r,\theta)=r\sin(\theta)
   \end{cases}
$

We have that $x^2+y^2=r^2$ and $\tan(\theta)=\frac{y}{x}$. So $r'=\frac{xx'+yy'}{r}$ and $\theta'=\frac{xy'-x'y}{r^2}.$
So $$r'=\frac{xy+y[(1-x^2-y^2)y-4x]}{r}=\frac{xy+y^2-x^2y^2-y^4-4xy}{r}.$$
I tried to calculate, but I have no idea how to bring the system into polar coordinates to continue my work.
Thanks!

答: Note $$\theta'=\frac{(1-x^2-y^2)xy-4x^2-y^2}{r^2}=((1-r^2)\sin\theta-3\cos\theta)\cos\theta-1$$ and $$r'=\frac{xy+(1-x^2-y^2)y^2-4xy}r=-3r\sin\theta\cos\theta+(1-r^2)r\sin^2\theta$$ so $$(\log r)'=((1-r^2)\sin\theta-3\cos\theta)\sin\theta.$$ Thus $$\frac{(\log r)'}{\theta'+1}=\tan\theta.$$


问:
Equality in Triangle inequality in a normed vectorial space (not inner product space)
In a real inner product space $(E,<\cdot|\cdot>)$, if we note $||\cdot ||$ the associated norm, we have equality in the triangle equality, i.e. $||x+y||=||x||+||y||$, if and only if there exists $\lambda, \mu \in \mathbb{R}^+, \lambda x = \mu y$.

But is it still true in a normed vectorial space, where we can not associate a dot product ?

In fact in the proof for real inner product spaces, we use Cauchy-Schwarz, but since there is no dot product here, I can't see how I could prove it, if true.

Thanks in advance

答: Let $\|x\|_1 := |x_1| + |x_2|$ be the $\ell_1$ norm on $\mathbb{R}^2$. Then the triangle inequality attains equality for $x=(1,0)$ and $y=(0,1)$ despite $x$ and $y$ not being parallel.


问:
Evaluate $\lim_{x\to 0} \frac{\log(x^2)}{1/x}$
$\displaystyle \lim_{x\to 0} \frac{\log(x^2)}{1/x}=0$ (where the result is 0 because from hierarchy of infinities we know that $1/x$ goes to infinity faster than $\log(x^2)$)

Can I solve the limit in this way without using de l'hospital? To do: tell me if my idea is right.

答:
You didn't want to use L'Hopital's rule, but we can use it to check:

The top one's derivative is $\frac{2}{x}$ (assuming your $\log$ is the natural log, but it works either way because change of base adds a multiplicative constant)

The bottom one's derivative is $-\frac{1}{x^2}$.

$\frac{2}{x} * -x^2$ = $-2x$.

And we can plug in $x = 0$ to get $0$. So yes, it is correct!

Note that your "hierarchy of infinities" directly follows from the essence of L'Hopital's rule, so you're essentially reinventing the wheel here.


问:
Compute value of $\pi$ up to 8 digits
I am quite lost on how approximate the value of $\pi$ up to 8 digits with a confidence of 99% using Monte Carlo. I think this requires a large number of trials but how can I know how many trials?

I know that a 99% confidence interval is 3 standard deviations away from the mean in a normal distribution. From the central limit theorem the standard deviation of the sample mean (or standard error) is proportional to the standard deviation of the population $\sigma_{\bar X} = \frac{\sigma}{\sqrt{n}}$

So I have something that relates the size of the sample (i.e. number of trials) with the standard deviation, but then I don't know how to proceed from here. How does the "8 digit precision" comes into play?

UPDATE

Ok I think I am close to understand it. From CLT we have $\displaystyle \sigma_{M} = \frac{\sigma}{\sqrt{N}}$ so in this case $\sigma = \sqrt{p(1-p)}$ therfore $\displaystyle \sigma_{M} = \frac{\sqrt{p(1-p)}}{\sqrt{N}}$

Then from the Bernoulli distribution, $\displaystyle \mu = p = \frac{\pi}{4}$ therefore

$$\sigma_{M}=\frac{\sqrt{\pi(4-\pi)}}{\sqrt{N}}$$ but what would be the value of $\sigma_{M}$? and then I have $\pi$ in the formula but is the thing I am trying to approximate so how does this work? and still missing the role of the 8 digit precision.

答:
The usual trick is to use the approximation of 355/113, repeating multiples of 113, until the denominator is 355.  This version of $\pi=355/113$ is the usual implied value when eight digits is given.

Otherwise, the approach is roughly $\sqrt{n}$

Title: Compute value of $\pi$ up to 8 digits


问:
Tools to help make a stepping equation with a delay on the x axis
Context (Not important, feel free to skip ahead to the question):

I was interested (and thus started to investigate) in a vaccination proposition/question (https://youtu.be/M-jqf6i5OLc?t=990) regarding how long it would take to vaccinate people with the new AZ vaccine if we could only vaccinate 1 million per week and you had to wait 4 weeks until the second dose could be administered to the same patient. The following data denotes how long first doses take to be administered. Here's a fun/funny picture of my working out (nothing too complex, but cool to think about nonetheless)

Choice A: (give 1 million doses per week and forget about second doses)

Choice B: (give 1 million doses per week and then administer the second dose after 4 weeks)

Logistical problem: Only 1 million doses can be given per week

Question:

Here's the aforementioned data:

| X (weeks) | Y (doses in millions) |

|-|-|-|

|0|0|

|1|1|

|2|2|

|3|3|

|4|4|

|5|4|

|6|4|

|7|4|

|8|4|

|9|5|

|10|6|

|11|7|

|12|8|

|13|8|

|14|8|

|15|8|

|16|8|

It draws the following graph:

What kind of tricks are there to creating equations for steps of this kind. You can imagine that the pattern continues on as it does.

答:
The most common (and in many ways, the most preferable) way to do this is with a piecewise-defined function. We can express it as

$$y = \begin{cases}x - 4k, &\quad 8k \le x < 8k + 4, k \in \Bbb Z\\
4k+4, &\quad 8k+4 \le x < 8k+8, k \in \Bbb Z\end{cases}$$

By introducing the greatest integer function ($\lfloor z \rfloor$ is the greatest integer $\le z$), we can replace $k$ with $\left\lfloor \frac x8\right\rfloor$ to get
$$y = \begin{cases}x - 4\left\lfloor \dfrac x8\right\rfloor, &\quad x - 8\left\lfloor \dfrac x8\right\rfloor <4\\4\left\lfloor \dfrac x8\right\rfloor+4, &\quad x - 8\left\lfloor \dfrac x8\right\rfloor \ge4\end{cases}$$
This particular function has some symmetries that allow a simpler expression. If you move $\frac x2$ outside the bracket, you get
$$y = \dfrac x2 + \begin{cases}\dfrac x2 - 4\left\lfloor \dfrac x8\right\rfloor, &\quad x - 8\left\lfloor \dfrac x8\right\rfloor <4\\4\left\lfloor \dfrac x8\right\rfloor+4-\dfrac x2, &\quad x - 8\left\lfloor \dfrac x8\right\rfloor \ge4\end{cases}$$
which can be modified to
$$y = \dfrac x2 + \begin{cases}\dfrac x2 - 4\left\lfloor \dfrac {x+4}8\right\rfloor, &\quad x - 8\left\lfloor \dfrac {x+4}8\right\rfloor > 0\\4\left\lfloor \dfrac {x+4}8\right\rfloor-\dfrac x2, &\quad x - 8\left\lfloor \dfrac {x+4}8\right\rfloor \le 0\end{cases}$$
Or just $$y = \dfrac x2 + \left|\dfrac x2 - 4\left\lfloor \dfrac {x+4}8\right\rfloor\right|$$
While the last looks cleaner, it is only because the "piecewise" part was hidden in a pre-defined piecewise function, the absolute value. In general it is possible to use absolute value expressions to convert any piecewise defined expression to a single formula. However, this is usually not advisable. When given such an absolute value expression, it will usually be quite long. And even for this short equation, it is not clear from looking at it how this function will behave. The piecewise versions are all easier to interpret, particularly the first, which is clearly rising at slope $1$ for the first half of an interval of length $8$, and constant for the second half.

The second version is more straightforward to compute, since computer languages always come with a greatest integer (a.k.a "floor") function, and the piecewise part can be programmed with an if statement.


答: But why is the hierarchy ordered as stated? I think that's a proof in the wrong direction. With $y:=-\ln x$, note$$\lim_{x\to0^+}2x\ln x=2\lim_{y\to\infty}ye^{-y}=0$$because $\int_0^\infty ye^{-y}dy=1$ is a finite integral with continuous non-negative integrand. Then use that $x\ln(x^2)$ is odd.

Title: Evaluate $\lim_{x\to 0} \frac{\log(x^2)}{1/x}$

Body:
$\displaystyle \lim_{x\to 0} \frac{\log(x^2)}{1/x}=0$ (where the result is 0 because from hierarchy of infinities we know that $1/x$ goes to infinity faster than $\log(x^2)$)

Can I solve the limit in this way without using de l'hospital? To do: tell me if my idea is right.


问:
Approximation of a random variable by a sequence of simple random variables
It said in a probability book that any non-negative random variable $X$ can be approximated by a sequence of simple random variables (finite range) $X_1,X_2,\dots,X_n$ such that $lim_{n->\infty}X_n=X$.

Now the sequence of simple random variable is defined as :

$X_n(\omega)=k2^{-n} \text{where} k2^{-n} \leq X(\omega) \lt (k+1)2^{-n}$ \text{where} $k=0$ to $2^{2n}-1$

```<code>     = 0 otherwise
</code>```

I have some doubts regarding this construction :

i) why the first term in this sequence starts with a horizontal interval of length $2^{-1}$ with the number of intervals $2^{1}$. Is there any specific reason to it ? I understand that the range of $X_n$ spans over $2^n$ to make sure that $X_n$ converges to $X$ asap.

ii) Is this construction unique ?

答:
i) Why not? (But the conditions $k2^{-n}\leqslant X(\omega)\leqslant(k+1)2^{-n}$ in the definition of $X_n(\omega)$ should read $k2^{-n}\leqslant X(\omega)\lt(k+1)2^{-n}$.)

ii) No.


问:
Prove that $e^{\frac{x+y}{2}} \le \frac{e^x + e^y}{2}$
To prove:  $e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2}$

We observe:

$e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2} $

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le 1/2\sum_{n=0}^\infty \dfrac{x^n}{n!} + 1/2\sum_{n=0}^\infty \dfrac{y^n}{n!}$

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty 1/2\dfrac{x^n}{n!} + \sum_{n=0}^\infty 1/2 \dfrac{y^n}{n!} $

$\Leftrightarrow \sum_{n=0}^\infty  \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty \dfrac{x^n + y^n}{2\cdot n!}$

In essence, now we need to show that $ \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \dfrac{x^n + y^n}{2\cdot n!}$

I'm looking for an idea or a hint to help me solve this. I'm thinking about using something from the binomial theorem. Maybe I made a mistake as well.

答:
Another solution :

$e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2}$ comes directly from the convexity of $x\mapsto e^x$.

A convex function is a function such that for all $x,y$, and $\lambda \in [0,1]$, $f(\lambda x + (1-\lambda)y)\leqslant \lambda f(x) + (1-\lambda) f(y)$.

With $\lambda = 1/2$ you get your answer.

(and you can prove exp 's convexity, by deriving it two times)


答: Let $y=x+t$ then we have to prove $$e^{x+t/2}\le \frac{e^x+e^{x+t}}{2}$$ or $$\frac{e^t+1}{2}\ge e^{t/2}$$ $$\iff {(e^{t/2}-1)}^2\ge 0$$ which is true

Title: Prove that $e^{\frac{x+y}{2}} \le \frac{e^x + e^y}{2}$

Body:
To prove:  $e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2}$

We observe:

$e^{\frac{x+y}{2}} \le \dfrac{e^x + e^y}{2} $

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le 1/2\sum_{n=0}^\infty \dfrac{x^n}{n!} + 1/2\sum_{n=0}^\infty \dfrac{y^n}{n!}$

$\Leftrightarrow \sum_{n=0}^\infty \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty 1/2\dfrac{x^n}{n!} + \sum_{n=0}^\infty 1/2 \dfrac{y^n}{n!} $

$\Leftrightarrow \sum_{n=0}^\infty  \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \sum_{n=0}^\infty \dfrac{x^n + y^n}{2\cdot n!}$

In essence, now we need to show that $ \dfrac{(\dfrac{x+y}{2})^n}{n!} \le \dfrac{x^n + y^n}{2\cdot n!}$

I'm looking for an idea or a hint to help me solve this. I'm thinking about using something from the binomial theorem. Maybe I made a mistake as well.


Body:
$$\frac{e^x+e^y}{2}=e^{\frac{x+y}{2}} \cdot \frac{e^{\frac{x-y}{2}}+e^{\frac{y-x}{2}}}{2}$$
then,
$$\frac{e^x+e^y}{2}=e^{\frac{x+y}{2}} \cdot \cosh\left ( e^{\frac{x-y}{2}} \right )$$

as $\forall z, \cosh(z) \geq 1$, then

$$\frac{e^x+e^y}{2} \geq e^{\frac{x+y}{2}}$$


问:
$S$ and $G$ are roots of $x^2 −3x +1=0$. Find equation whose roots are $\frac{1}{S -2}$ and$\frac{1}{G-2}$
If S,G are roots of $x^2 −3x +1=0$, then find equation whose roots are $\frac{1}{S -2}$ , $\frac{1}{G-2}$.

So , the way my sir solved it is that

He took x = $\frac{1}{S -2}$ , then he got S = $\frac{2x+1}{x}$.

I am confused with is that we were told roots as $\frac{1}{S -2}$ and not only S for 2nd equation.

Why not just put $\frac{1}{S -2}$ in x and then solve ?

Body:
Note $SG=1,\>S+G=3$. Then

$$\frac1{S-2}\frac1{G-2}=\frac1{SG-2(S+G)+4}=-1
$$
$$\frac1{S-2}+\frac1{G-2}=\frac{S+G-4}{SG-2(S+G)+4}=1
$$
Thus, the equation with roots $\frac1{S-2}$ and $\frac1{G-2}$ is
$$x^2-x-1=0$$


问:
Negative value for area
I have a problem with the following example:Given the surfaces:$$x^2+y^2+z^2=5, x^2+z^2\ge (y-1)^2$$ find the area bounded with them. First problem appears when I try to sketch this in 3D. For the first one is sphere and that is OK, and the second one should be infinite cone right? (but not sure what part of it left or right, since it is on y axis).

Then for intersection I get $x^2+z^2=1$ and after cilindrical coordinates it is easy to see that $0\le r\le 1$ and $0\le \theta \le 2\pi$.
After that, I am calculating the area of cone and sphere seperatly. So for sphere i get $$\sqrt{5}\int\limits_{0}^{2\pi}d\theta \int\limits_{0}^{1}\frac{rdr}{\sqrt{5-r^2}}$$. But this integral has a negative value, so I am obviously mistaking somewhere but not sure where.

BTW i used that $y=\sqrt{5-x^2-z^2}$ from the equation of sphere and than plugged that into formula.

答:
It is easier to do it in spherical coordinates.

Your integral should be

$\displaystyle \int_0^{2\pi} \int_{cos^{-1}({\frac{2}{\sqrt5}})}^{\pi - cos^{-1}({\frac{1}{\sqrt5}})} 5 \sin \phi \, d\phi \, d\theta$

If you see from top ($z-$ axis) in $XY$ plane, this is how the projection looks. This sketch helps to understand the limits of integral.

EDIT: also given you were doing it in cylindrical coordinates, here is the integral in cylindrical coordinates with the correct limits. Please note $-1 \leq y \leq 2$ so you will have to split the integral into two parts, one for surface area above $XZ$ plane and one for below it. For $y = -1, r = 2 \,$ and for $\, y = 2, r = 1$

For above: $\displaystyle \sqrt{5} \int_{0}^{2\pi} \int_{1}^{\sqrt 5}\frac{r}{\sqrt{5-r^2}} \, dr \, d\theta$

For below: $\displaystyle \sqrt{5} \int_{0}^{2\pi} \int_{2}^{\sqrt 5}\frac{r}{\sqrt{5-r^2}} \, dr \, d\theta$

Title: Negative value for area


答: No, it's possible to have a flow of cost -n without having a Hamiltonian cycle. Consider for example a graph consisting of two disjoint cycles. To make the example less trivial, you could add some edges between them without introducing a Hamiltonian cycle. The two disjoint cycles still give a flow of cost -n.</p>

Title: Using Bellman-Ford to find a Hamiltonian cycle? (NP-complete)

Body:
Let $G(V,E)$ be a directed graph, where $V=\{a_1,\ldots,a_n\}$ is a set of vertices and $E$ is a set of ordered pairs of $V$, with $|V|=n$.

Now, let be $G(W,F)$ be a graph where $W$ is a set of vertices, such that $W=\{a_1,\ldots,a_{2n}\}$ with $|W|=2n$, and $F$ is a set of ordered pairs of $W$ defined as follows:
$$\forall i, j \in \{1, \ldots, n\} : (a_i,a_{j+n}) \in F \text{ if } (a_i,a_{j}) \in E \\
\forall i \in \{1, \ldots, n\} : (a_{i+n},a_i) \in F$$

Now we define the capacities:
$$c(a,b)=1 \text{ if } (a,b) \in F$$

And we define a cost:
$$\forall i \in \{1, \ldots, n\} : p(a_{i+n},a_i)=-1 \text{ if } (a_{i+n},a_i) \in F \\
p=0 \text{ else}$$

The source as $a_{n+1}$ and the sink as $a_{n+1}$.

Does the graph $G(W,F,c,p)$ have a minimum cost flow of $-n$ if and only if the graph $G(V,E)$ has a Hamiltonian cycle?

(Comment to an answer in the comments section)

Michael, for your matrix with capacities G(V,E) the matrix G(W,F) is:

```<code>cap = [[0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0],
       [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0]]

cost = [[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [-1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0]]
</code>```


问:
Calculate parameters of regression line
I'm learning statistics and trying to calculate the parameters of a regression line based on the data:

$(x,y)=(0,-1),(1,2),(2,9/2)$

Could you please tell me how would I do so? I'm not sure how to calculate the parameters based just on a table of values.

Thanks

答:
I'll meet you half way. Here is a complete run-through of doing linear regression by hand, with hints of how the method can be generalized. The compromise though, is I'll use different numbers than you provided. The machinery will work the same.

Starting with a simple example, consider the following (six) data points occurring as ordered pairs:
$$x_{i} : 0.6 , 1.8 , 2.8 , 3.6 , 4.2 , 5.6$$
$$y_{i} : 1.6 , 1.6 , 2.6 , 2.0 , 4.0 , 3.6$$

A scatter plot of $\left\{x_i,y_i\right\}$ would show a roughly linear arrangement of points, perhaps approximated the equation $$y = \frac{1}{2}x + 1 \:,$$ which could be discerned using a ruler. Of course, we want to do better than guessing, so begin with the most general form of a line in the $xy$-plane, namely $$f\left(x\right) = mx + b \:.$$ Note that some choice of $m$ and $b$ correspond to a  line that passes closer to the group of points $\left\{x_i,y_i\right\}$ than any other line.

The task now is to determine $m$ and $b$ based on the data given. To do this, suppose the line $f\left(x\right)=mx+b$ is sketched somewhere in the plane. Next, write the square of the vertical displacement from any given $y_i$ (up or down) to $f\left(x_i\right)$ for a single point: $$z_i^2 = \left(f\left(x_i\right) - y_i\right)^2 \:,$$ which can be done for all data points: $$F = \sum_{i=1}^N z_i^2 = \sum_{i=1}^N \left(mx_i + b - y_i\right)^2$$ In this form, we can guarantee a best-fitting line by finding the proper $m$ and $b$ that minimizes $F$. Taking the partial derivative with respect to $m$ and $b$ respectively, we have $$\frac{\partial F}{\partial m} = \sum_{i=1}^N x_i \cdot \left(m x_i + b - y_i\right) \hspace{2.54cm} \frac{\partial F}{\partial b} = \sum_{i=1}^N \left(mx_i + b - y_i\right) \:,$$ where setting $\partial F/\partial m = \partial F / \partial b = 0$ we get, after distributing the summation signs $$0 = m \sum_{i=1}^N x_i^2 + b \sum_{i=1}^N x_i - \sum_{i=1}^N x_i \: y_i \hspace{2.54cm} 0 = m \sum_{i=1}^N x_i + b \sum_{i=1}^N \left(1\right) - \sum_{i=1}^N y_i \:.$$ Letting $$X^2 = \sum_{i=1}^N x_i^2 \hspace{2.00cm} X = \sum_{i=1}^N x_i \hspace{2.00cm} Y = \sum_{i=1}^N y_i \hspace{2.00cm} XY = \sum_{i=1}^N x_i \: y_i \:,$$ the above is easily recognized as a system of two equations and two unknowns $m$ and $b$ (the rest are all numbers): $$0 = mX^2 + bX - XY \hspace{2.54cm} 0 = mX + bN - Y$$ Note that the variables $X^2$, $X$, $Y$, $XY$ don't follow ordinary algebra, that is $X \cdot X \neq X^2$, and so on. Solving for $m$ and $b$, we find $$m = \frac{N \cdot XY - X \cdot Y}{N \cdot X^2 - X \cdot X} \hspace{2.54cm} b = \frac{Y \cdot X^2 - X \cdot XY}{N \cdot X^2 - X \cdot X} \:.$$ Evidently, $m$ and $b$ are calculated in one iteration with many sub-steps.

For the example on hand, we have $N=6$, along with $$X^2 = 73.4 \hspace{2.00cm} X = 18.6 \hspace{2.00cm} Y = 15.4 \hspace{2.54cm} XY = 55.28 \:,$$ finally giving $$m = \frac{6 \cdot 55.28 - 18.6 \cdot 15.4}{6 \cdot 73.4 - 18.6^2} \approx 0.479 \hspace{2.54cm} b = \frac{15.4 \cdot 73.4 - 18.6 \cdot 55.28}{6 \cdot 73.4 - 18.6^2} \approx 1.082 \:,$$ corresponding to the best-fit line $$y = 0.479 \: x + 1.082 \:.$$

Title: Calculate parameters of regression line


问:
Show that the power series of complex function is defined and holomorphic

Let $(a_k)_{k\geq 0}$ be a sequence of complex numbers such that for
all $C>0$, $\lim_{k\to\infty}\mid a_k\mid C^k=0$. Show that
$$f(z)=\sum\limits^{\infty}_{k=0}a_k \exp(2i\pi kz)$$ is defined for
all $z$; that $f$ is entire, and that $f(z+1)=f(z)$ for all
$z\in\mathbb{C}$

I have proved the part $f(z+1)=f(z)$. To show that $f(z)$ converges and entire, I want to use the theorem

If $\{f_n\}^{\infty}_{n=1}$ is a sequence of holomorphic functions
that converges uniformly to a function $f$ in every compact subset of
$\Omega$, then $f$ is holomorphic in $\Omega$.

So I need to see what function it converges to, and $\sum\limits^{\infty}_{k=0}a_k e^{2i\pi kz}$ converges uniformly to that function in every compact subset of complex plane. But I am not sure how to proceed. Can you explain if this is the right idea, and if not, could you give me other suggestions? Thanks!

答:
It suffices to show that
$$ \tag{*}
 F(w) = \sum_{k=0}^\infty a_k w^k
$$
defines an entire function, because then $f(z) = F(e^{2\pi i z})$ is entire as the composition of entire functions.

Given $R > 0$ choose any $C > R$, then
$$
 |a_k w^k| \le |a_k| R^k = |a_k| C^k \left( \frac{R}{C} \right)^k  \le \left( \frac{R}{C} \right)^k
$$
for all $|w| \le R$ and all sufficiently large $k$, which (by the Weierstrass M-test) implies that the series $(*)$ converges uniformly on the disk $|w| \le R$.

In other words, the condition $\lim_{k\to \infty} |a_k | C^k = 0$ for all $C > 0$ implies (and actually is equivalent to) that the radius of convergence of the power series $(*)$ is infinity.

So your idea of proving uniform convergence on compact sets is correct, but (in the general case of arbitrary coefficients $a_k$) there is no closed form for the limit function. The Weierstrass M-test helps because it can be used to prove uniform convergence without knowing the limit function. This is closely related to the Cauchy criterion which can be used to prove convergence of a sequence without knowing the limit.


答:
I'll meet you the "other" half way:

Your data and fit should look like this:

Title: Calculate parameters of regression line

Body:
I'm learning statistics and trying to calculate the parameters of a regression line based on the data:

$(x,y)=(0,-1),(1,2),(2,9/2)$

Could you please tell me how would I do so? I'm not sure how to calculate the parameters based just on a table of values.

Thanks


问:
What are the polynomials $p \in \mathbb Z[x]$ such that $\vert p(z) \vert \le 1$ for all $z \in \mathbb U$?
The question is What are the polynomials $p \in \mathbb Z[x]$ such that $\vert p(z) \vert \le 1$ for all $z \in \mathbb U$? $\mathbb U$ stands for the complex numbers of modulus $1$.

Hint:

 Think to Fourier series

I wrote a solution below using the hint. However, any solution using a different approach is more than welcomed!

答:
Denote $\mathcal P = \{p \in \mathbb Z[x] \mid \vert p(x) \vert \le 1 \text{ for all } z \in \mathbb U\}$. As for
$$p(z) = \sum_{k=0}^n a_k z^k$$ we have
$$a_k = \frac{1}{2\pi} \int_0^{2 \pi} P(e^{it}) e^{-kt}\ dt$$ we can conclude that the coefficients of $P$ belong to $\{0, -1, 1\}$. Moreover for $p, q \in \mathcal P$, $pq \in \mathcal P$. Therefore if $p \in \mathcal P$, $p^2 \in \mathcal P$ from which we deduce that only one coefficient of $p$ can be non zero. Finally

$$\mathcal P = \{0\} \cup \{ \pm z^n \mid n \ge 0\}$$


问:
Clustering of vertices in an n dimensional cube
Consider the vertices of an n-dimensional cube. Distance between two vertices is measured as the minimum number of edges between the two vertices. Now consider a subset of these vertices.
If we call the total set of vertices as 'T' and the subset as 'S' then our objective is to partition 'S' into two sets 'A' and 'B' and for both of these sets find vertices $x_A$ and $x_B$ from T such that the sum total of the distance between $x_A$ and the vertices of A and $x_B$ and the vertices of B should be a minimum.

How to approach this question?

Given that we know the distance relation for every pair of vertex, is it possible to know the minimum distance through some simple calculation?

答:
To clarify, I am answering the following question:

For a subset $S$ of the vertex set of a graph $G$, let $G[S]$ denote the induced subraph on $S$.

Let $Q_n$ denote the $n$-th cube graph, and let $d(v,w)$ for $v,w \in V(Q_n)$ be the graph metric. Then your $T$ is just $V(Q_n)$, and you want to find a pair $(a,b)_{A\sqcup B} \in V(Q_n)^2$ so that:

"Fix a subset $S \subseteq V(Q_n)$, and let $A \sqcup B$ be a partition of $S$. Then:

$$\sum_{\alpha \in A} d(a,\alpha), \sum_{\beta \in B} d(b,\beta)$$

are both minimized, where the graph metric is taken over $Q_n[A]$ and $Q_n[B]$, respectively."

Notice that we only need be concerned with the distance in the induced subgraphs, because $a$ and $b$ must be in $A$ and $B$ respectively to achieve a minimum.

Then $a$ must be in the center of $Q_n[A]$, and the same for $b$.

To be specific, we define the eccentricity $\epsilon(v)$ of a vertex $v$ in $G$ to be:

$$\max_{w}\{d(v,w) : w \in G\}$$

The radius $r(G)$ of a graph is the minimum of all eccentricities, and the diameter $d(G)$ is the maximum.

The center of a graph Cent$(G) = \{v \in G : \epsilon(v) = r(G)\}$ are all the vertices of minimum eccentricities.

It follows that to minimize our formulae above, the vertices $a$ and $b$ must be in the respective centers of $Q_n[A]$ and $Q_n[B]$.

Provided $Q_n$ is not too large, this should be easy enough to compute with a program. I believe Mathematica has a built-in function to compute the center of the input graph.


答:
The substitution method your teacher is referring to goes as follows:

Let $y=\frac{1}{x-2}$ where $x=S, G$ are the roots of the given equation $x^2-3x+1=0$

Rearranging gives $x=\frac{1+2y}{y}$

Substituting this into the equation satisfied by $x$ gives $$(\frac{1+2y}{y})^2-3(\frac{1+2y}{y})+1=0$$

This simplifies to become $$y^2-y-1=0$$
This is the same as the answer obtained by Quanto using Vieta's formulas.

Title: $S$ and $G$ are roots of $x^2 −3x +1=0$. Find equation whose roots are $\frac{1}{S -2}$ and$\frac{1}{G-2}$

Body:
If S,G are roots of $x^2 −3x +1=0$, then find equation whose roots are $\frac{1}{S -2}$ , $\frac{1}{G-2}$.

So , the way my sir solved it is that

He took x = $\frac{1}{S -2}$ , then he got S = $\frac{2x+1}{x}$.

I am confused with is that we were told roots as $\frac{1}{S -2}$ and not only S for 2nd equation.

Why not just put $\frac{1}{S -2}$ in x and then solve ?


问:
Is $\mathfrak{so}(2n)$ contained in $\mathfrak{sl}(2n)$?
Let $\mathfrak{so}(2n) = D_n  =\left\{ \begin{pmatrix} A&B \\ C&-A^T \end{pmatrix}:  A,B,C\in M_n(K), B=-B^T, C=-C^T \right\}$ . Of course the matrices in $D_n$ have trace zero. This could lead me to conclude that $\mathfrak{so}(2n) \subseteq \mathfrak{sl}(2n)$, but this doesn’t seem right as for the inclusion of the respective Dynkin diagrams: in fact, it is not true that $D_n$ is contained in $A_n$. What am I missing here?

答: There is nothing wrong in what you did. The fact that a simple Lie algebra $\mathfrak g_1$ is a subalgebra of another simple Lie algebra $\mathfrak g_2$ doesn't imply the existence of a connection between their Dynkin diagrams (as your own example shows).


问:
Binomial sum with index as Arithmetic Progression

Consider a binomial sum of form,

$$ S = \sum_{k=1}^{\infty} \binom{n}{a_k}$$
Where,
$$ a_k  = a_o + (k-1) d$$ w/ $a_o \geq 0 $ $ d \geq 0$

An immediate simplification is that for terms with, $ a_k \geq n$ are zero. So, if there is a $j$ such that $n \in [a_j,a_{j+1} ]$ , then the sum's upper index turns to:

$$ S = \sum_{k=1}^{j} \binom{n}{a_k}$$

A brute force way to calculate would be to directly add them after that, but is there a simpler way to solve this?

The person who sent me this problem mentioned that I could use roots of unity to solve this.. but I'm not sure how that relates to here...

答:
The standard method to solve these involves the roots of unity filter (or finite Fourier analysis for the pretentious).

The idea is to look at the polynomial $f(x)=x^{d-a_0}(1+x)^n$. This has the property that the coefficient of $x^{dk}$ is $\binom{n}{a_0+dk}$. So if we could somehow extract only the coefficients of $x^{dk}$ and sum them, we would be done.

The trick is to consider $$\frac{f(1)+f(\omega)+\dots+f\left(\omega^{d-1}\right)}{d},$$ where $\omega=\exp(2i\pi/d)$ is the $d$-th root of unity. Can you see why this extracts only the coefficients of $x^{dk}$ in the sum?

To make this concrete, consider the following example:
$$\sum_{k\geq0}\binom{2021}{3k+1}.$$
Take the polynomial $f(x)=x^2(1+x)^{2021}$, so writing $\omega=\exp(2i\pi/3)$, the sum we want is
$$\frac{(1+1)^{2021}+\omega^2(1+\omega)^{2021}+\omega(1+\omega^2)^{2021}}{3}.$$
Now in general, at this point you'd just convert to polar form $1+\omega^m=re^{i\theta}$ and bash it out, but for $n=3$ you can take advantage of the neat trick that $1+\omega+\omega^2=0$ to make the calculation easier:
\begin{align*}
\frac{(1+1)^{2021}+\omega^2(1+\omega)^{2021}+\omega(1+\omega^2)^{2021}}{3} &= \frac{1}{3}\left[2^{2021}+\omega^2(-\omega^2)^{2021}+\omega(-\omega)^{2021}\right] \\
&= \frac{1}{3}\left[2^{2021}-\omega^{4044}-\omega^{2022}\right] \\
&=\frac{1}{3}\left[2^{2021}-2\right].
\end{align*}


问:
Solving $x^2 y''-2xy'=0$ using power series
Given the following ODE
$$x^2 y''-2xy'=0$$
I'm trying to solve it using power series. Since
$$y=\sum_{n=0}^{\infty} a_{n}x^{n}$$
then
$$x^{2}y''=\sum_{n=2}a_{n}n(n-1)x^{n}$$
and
$$2xy'=\sum_{n=1}^{\infty}2a_{n}nx^{n}=2a_{1}x+\sum_{n=2}^{\infty}2a_{n}nx^{n}$$
Therefore, in the equation, I get
$$-2a_{1}x+\sum_{n=2}^{\infty}(n^{2}-3n)a_{n}x^{n}=0$$
So
$$-2a_{1}=0\Rightarrow a_{1}=0$$
and
$$(n^{2}-3n)a_{n}=0,\quad\forall n\geq2$$
Since I'm looking for non-trivial solutions, $a_{n} \ne 0\Rightarrow n^{2}-3n=0$ which gives me $n=0$ or $n=3$.

I want to know if it's right and if so, what I'm supposed to do with these values of $n$.

答: This is an Euler equation, so there's an easier way to solve and one expects two solutions of the form $y=x^r$.  So what you've discovered is that $a_n$ is NOT zero only when $n=0$ or $n=3$.   So you're getting the right answer $a_0 x^0 + a_3 x^3.$


问:
Singular Values and Matrix Rank
I came across this problem and I'm not sure if the proof I have found is OK. Any suggestions or confirmation of correctness would be appreciated.

The Problem:

Show that the number of non-zero singular values of a matrix $A$ coincides with its rank

My Approach:

We have an arbitrary matrix $A_{m\times n}$. For every eigenvector $v_i$ of the modulus $|A|$ of $A$ (where $|A|^2=A^*A$, $|A|$ being $n\times n$), and its corresponding eigenvalue $\lambda_i$ ($\forall i=1,\,2,\,...,\,n$), we have $(|A|-\lambda_i \mathbb{I}_n)v_i=0$. For every singular value $\sigma_i$ of $A$ equal to zero, the corresponding eigenvalue $\lambda_i$ of $|A|$ is also zero. Therefore, from the previous expression, for every $\sigma_i=0$, we have:

$$|A|v_i=0$$

So the problem comes down to solving this homogeneous system. Because $|A|$ is self-adjoint, there always exists a basis of eigenvectors of $|A|$, which means that the geometric multiplicity of any eigenvalue of $|A|$ is always equal to its algebraic multiplicity. Therefore, $ \dim Ker|A| = n-r$, where $r$ is the number of non-zero eigenvalues of $|A|$, and therefore also the number of non-zero singular values of $A$.

Using the well known equality, $\dim Ker|A|=n-\text{rank}|A|$, and the previous result, we have:

$$n-r=n-\text{rank}|A|\rightarrow r=\text{rank}|A|$$

From the definition of $|A|$, for any $x \in \mathbb{C}^n$:

$$\mid \mid |A|x \mid \mid^2 =(|A|x,\,|A|x)=(|A|^*|A|x,\,x)=(|A|^2x,\,x)=(A^*Ax,\,x)=(Ax,\,Ax)=\mid \mid Ax \mid \mid^2 \rightarrow$$
$$\rightarrow\mid \mid |A|x \mid \mid=\mid \mid Ax \mid \mid$$

This means $|A|x=0\Leftrightarrow Ax =0$, $\forall x \in \mathbb{C}^n$, so $KerA=Ker|A|$. Then:

$$\text{rank}A=n-\dim KerA=n-\dim Ker|A|=n-(n-\text{rank}|A|)=\text{rank}|A|$$

Therefore, as $\text{rank}|A|=\text{rank}A$, we have:

$$r=\text{rank}|A|=\text{rank}A$$

Therefore, the number of non-zero singular values of an arbitrary matrix $A_{m\times n}$ is equal to the rank of such matrix.

答:
As mentioned by @user8675309, this is another possible approach:

The matrix $A$ can be decomposed with SVD in order to obtain $A=W \Sigma V^*$, with $W$ and $V^*$ being two square invertible matrices and $\Sigma$ being diagonal (with the singular values of $A$ in the diagonal), although not necessarily square. Then, $\text{rank}(A)=\text{rank}(W \Sigma V^*)=\text{rank}( \Sigma )$.

To prove $\text{rank}(W \Sigma V^*)=\text{rank}( \Sigma )$, we consider the case of two matrices $A$ and $B$, where $A$ is square invertible. Then, $\text{rank}(BA)\leq \text{rank}(B)$. Also, $\text{rank}(B)=\text{rank}(BAA^{-1})\leq \text{rank}(BA)\leq \text{rank}(B)$. Therefore, we have the expression:

$$\text{rank}(B)\leq \text{rank}(BA)\leq \text{rank}(B)$$

This means $\text{rank}(B)= \text{rank}(BA)$, so right multiplication by a square invertible matrix preserves rank. For left multiplication by square invertible matrices, we can take the transpose, because column rank is equal to row rank. We have:

$$\text{rank}(AB)=\text{rank}((AB)^T)=\text{rank}(B^TA^T)\leq \text{rank}(B^T)=\text{rank}(B)$$

Also, $$\text{rank}(B)=\text{rank}(B^T)=\text{rank}(B^TA^T(A^T)^{-1})\leq \text{rank}(B^TA^T)=\text{rank}((B^TA^T)^T)=\text{rank}(AB)\leq \text{rank}(B)$$

Therefore:

$$\text{rank}(B)\leq \text{rank}(AB)\leq \text{rank}(B)$$

This means $\text{rank}(B)= \text{rank}(AB)$, so left multiplication by a square invertible matrix also preserves rank.

The two previous results combined prove $\text{rank}(A)=\text{rank}(W \Sigma V^*)=\text{rank}( \Sigma )$.

Because the rank of $\Sigma$ is just the number $r$ of its nonzero rows (because it is diagonal), and $r$ coincides with the number of nonzero singular values, we can see that:

$$r=\text{rank}(A), \text{ where } r \text{ is the number of nonzero singular values of }A.$$

Title: Singular Values and Matrix Rank