问:
The point P lies onOV and is such that PNM is a right angle. Find the position vector of P
I dont understand how to find PN as I don't get where P lies I know that the vector $PNxMP=0 $ as it is perpendicular. The shade is a pyramid and And M and N are the midpoints of AB and BV respectively. vector $OA=6i $ $OC=3j$ and  $ OV=9k$

答:
Hint:

As Position vector of $A$ is $(6, 0, 0)$, of $V$ is $(0, 0, 9)$ and of $C$ is $(0, 3, 0)$, position vector of
$B$ $(6, 3, 0)$, $M$ $(6, \frac{3}{2}, 0), N(3, \frac{3}{2}, \frac{9}{2})$.

As point $P$ lies on $OV$, we assume its position vector as $P \ (0, 0, z)$. Using position vectors now find direction vectors $\vec {NM}$ and $\vec {NP}$. As they are perpendicular, equate their dot product to zero and find value of $z$.


问:
Expectation and Variance of a random variable as a function of sine and cosine
Let the random variables $X_1, X_2....X_n$ be independently Gaussian with mean, $\mathop{\mathbb{E}}[X_n]$ = $\mu sin(10n)$, and variance, $Var(X_n)$ = $\sigma^2(2+cos(10n))$. Does the sample mean $S_n=\frac{1}{n}\Sigma_{i=1}^nX_i$, converge to $\mu$ in probability. It's fact that, we use Chebyshev inequality to prove the convergence.

Chebyshev Inequality: $P(|S_n - \mathop{\mathbb{E}}[X_n]|\ge\epsilon)\le \frac{Var(S_n)}{\epsilon^2}$

As n->$\infty$, does RHS of inequality tends to 0  ?

答:
As n->∞, does RHS of inequality tends to 0 ?

obviously yes. $\mathbb{V}[S_n]=\frac{\mathbb{V}[X_n]}{n}$ thus in your RHS the numerator moves in $[\sigma; 3\sigma]$ while the denominator goes to $\infty$

ContentLicense: CC BY-SA 4.0


问:
Matrix inverses in function spaces
Consider the following two matrices, $D$ and $S$, defined in a function space:

$$D=\left(\begin{matrix}
0 & 1 & 0 & 0 & \cdots \\
0 & 0 & 2 & 0 & \cdots \\
0 & 0 & 0 & 3 & \cdots \\
0 & 0 & 0 & 0 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}\right)$$

$$S=\left(\begin{matrix}
0 & 0 & 0 & 0 & \cdots \\
1 & 0 & 0 & 0 & \cdots \\
0 & \frac12 & 0 & 0 & \cdots \\
0 & 0 & \frac13 & 0 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}\right)$$

I noticed that

$$DS=\left(\begin{matrix}
1 & 0 & 0 & 0 & \cdots \\
0 & 1 & 0 & 0 & \cdots \\
0 & 0 & 1 & 0 & \cdots \\
0 & 0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}\right)=I$$

However

$$SD=\left(\begin{matrix}
0 & 0 & 0 & 0 & \cdots \\
0 & 1 & 0 & 0 & \cdots \\
0 & 0 & 1 & 0 & \cdots \\
0 & 0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}\right)\neq I$$

Yet I was told that $AB=I\iff BA=I$. What is causing the discrepancy? Maybe I am wrong to say that $DS=I$, or maybe the proof that $AB=I\iff BA=I$ does not apply for infinitely sized matrices?

(Note: The proof I was told was that $AB=I\Rightarrow B=BI=B(AB)=(BA)B=IB\Rightarrow BA=I$.)

Context:

Some time ago, I learnt that the derivative and integral functions are both linear functions. More recently, I learnt that functions can be thought of as vectors and are in a function space. For the subset of continuous functions, we can use $x^0, x^1, x^2,\cdots$ as all continuous functions can be represented by an infinite series by Stone-Weierstrass Theorem.

I did some thinking with this definition of functions, and realized that the derivative was a linear transformation of functions, and could be represented by a matrix (represented by $D$ above). I also noticed that we the "pseudo-integral" (essentially the integral without the constant), represented by $S$ above, is also a linear transformation.

After playing around, I noticed that $DS=I$ (i.e. taking the pseudo-integral then the derivative of a function results in the function you started out with), while $SD\neq I$ (i.e. taking the derivative then the pseudo-integral of a function does not necessarily give the function you started out with). This seems to contradict the principle that $AB=I\iff BA=I$, so I wanted to inquire about this.

I haven't learnt functional analysis (or any sort of analysis beyond surface level) yet, so I am probably doing something very silly here and I won't understand much of the jargon.

答:
You found an example of two linear endomorphisms of a vector space such that one of them is a right inverse of the other one, but not a left inverse. It turns out that this is impossible for finite-dimensional vector spaces but, as you have noticed, not for infinite dimensional ones.

Another example is this: consider the space $\Bbb R[x]$ of all polynomials in one variable with real coefficients. If $p(x)\in\Bbb R[x]$, let $D(p(x))=p'(x)$ and let $P(p(x))$ be the primitive of $p(x)$ which maps $0$ into $0$. Then $D\circ P=\operatorname{Id}$, but $P\circ D\ne\operatorname{Id}$.

Title: Matrix inverses in function spaces


问:
Odds of Choosing Items out of a bag
Let's say there is a bag filled with a myriad of balls, labeled 1 through 7. There are an infinite number of each ball (i.e. odds don't diminish when one is removed). Edit A better phrasing than infinite-balls would be balls being replaced. /Edit

I get to draw 12 balls out of the bag.

What are the chances I'd end up with at least 1 of each number (and obviously some repeat pulls)? What are the chances I'd end up with all-but-1 of the 7 numbers (i.e. $\frac{6}{7}$)? Or $\frac{5}{7}$, or any lower combination?

(I'm not particularly gifted with Combinatorics. I was playing with smaller values, like 3-draws of 2-different numbers having $2^3$ possible options and $\frac{6}{2^3}$ of them fulfilling the condition. I tried 3-draws of 3-different numbers having $\frac{6}{3^3}$ fulfilling the condition. I tried 4 draws of 4-draws of 4-different numbers and got $\frac{24}{4^4}$ fulfilling the condition. Needless to say, I couldn't see a readily repeatable pattern to use on higher values :-( )

答:
The disadvantage with Stirling #s is that you need recourse to a table.

However, the question can easily be solved using inclusion-exclusion:

$Pr =\Large{{7^{12} - 7\cdot6^{12} + 21\cdot5^{12}-35\cdot4^{12} +35\cdot3^{12} - 21\cdot2^{12} + 7\cdot 1^{12}}\over 7^{12}}$

ContentLicense: CC BY-SA 3.0

Title: Odds of Choosing Items out of a bag


问:
Show that a strict positive homogeneous function of degree 0 has a minimum and maximum.
I'm stuck with the following question:

"Let $ f : \mathbb R^k \setminus {(0,0,...,0)} \rightarrow \mathbb R$
be a twice continuously differentiable and strictly positive homogeneous of degree 0. prove that it has a minimum and maximum at $ \mathbb R^k \ \setminus {(0,0,...,0)} $."

I've already shown that the Hessian of the function is neither positive definite nor negative definite.
I know that from Euler's identity, we have $ \langle\nabla f(a) , a \rangle = 0 $ for all $ a \in \mathbb R^k \setminus {(0,0,...,0)} $, however, I don't know how to show that it has a minimum or maximum.

I thought maybe I'll show a point that has a minimal/maximal value, however, I don't really have a point of thought about that.

答: First, homogeneous of degree 0 means $f(x)=f(x/\|x\|)$ with $x/\|x\|$ on the unit sphere $S^{k-1}=\{x\in\mathbb R^k|\|x\|=1\}$. Hence it suffices to restrict $f$ to this sphere without changing its image. But it's compact, so its image by any continuous function is also compact, thus $f$ attains maximum and minimum on this sphere (and thus all of $\mathbb R^k$).


问:
Is $x\log x$ Hölder continuous on $[0,1]$?
Consider the function $f:[0,1]\rightarrow \mathbb{R}$ given by

$$f(x)= x\log(x).$$

This function is not Lipschitz continuous at zero, but apparently Hölder continuous for any $\alpha<1$, i.e. there is a constant
$C_{\alpha}$ such that

$$\vert f(x)-f(y)\vert \le C_{\alpha} \vert x-y \vert^{\alpha},$$

but how can one show the Hölder continuity?

Please let me know if you have any questions or remarks.

答:
Same plan of attack as $x^\alpha$ as an example of an $\alpha$-Hölder continuous function, which should be reminiscent of proofs using e.g. mean value theorem and the sort.

Note $-x\log x\ge 0$ for $x\in[0,1]$ (with the obvious redefinition at zero).
Let $0\le x_0<x\le 1$ and consider $g:[x_0,1]\to\mathbb R$ defined by $$g(x) = -x\log x + x_0 \log x_0 + (x-x_0)\log(x-x_0).$$
Then
$$g'(x) = (-\log x - 1) - (-\log(x-x_0) - 1) = \log (x-x_0) - \log x\le 0,$$
which follows because of the monotonicity of $\log$ and $0<x-x_0\le x $. Thus $g$ is decreasing, with $g(x_0)=0$. This implies for any $0\le x<y\le e^{-1}$,  (note $-t\log t$ is increasing on $[0,e^{-1}]$)

$$ \frac{(-y\log y) - (-x\log x)}{-(y-x)\log(y-x)} = \frac{|y\log y - x\log x|}{|y-x|\lvert\log \lvert y-x \rvert \rvert} \le 1.$$
Therefore,
$$ \sup_{\substack{y\neq x\\x,y\in[0,e^{-1}]}} \frac{|y\log y - x\log x|}{|y-x|\lvert\log \lvert y-x \rvert \rvert} \le 1.$$
For points away from the origin, it is easy to check that the function is Lipschitz (in fact smooth). Therefore, it follows after some routine case checking (see end of answer) that for some $C>0$,
$$ \sup_{\substack{y\neq x\\x,y\in[0,1]}} \frac{|y\log y - x\log x|}{|y-x|\max(\lvert\log \lvert y-x \rvert \rvert,1)} \le C.$$
(The maximum is just to avoid 'irrelevant' considerations when $x$ and $y$ are far apart i.e. $|x-y|\to 1$. A function satisfying this type of inequality is said to be 'Log-Lipschitz'.)

This is in fact stronger than being $\alpha$-Hölder continuous: since
$ |r|\max(|\log |r||,1)\le C_\alpha |r|^\alpha$ for any $0<\alpha<1$, we immediately have
$$ \sup_{y\neq x} \frac{|y\log y - x\log x|}{|y-x|^\alpha} \le C_\alpha \sup_{y\neq x} \frac{|y\log y - x\log x|}{|y-x|\lvert\log \lvert y-x \rvert \rvert} \le CC_\alpha. $$
Indeed, setting $r=e^{-t}$ for $t> 0$, the above is the statement $\max(t,1)e^{-t} \le C_\alpha e^{-\alpha t} \iff \max(t,1) \le C_\alpha e^{(1-\alpha)t}$; so the statement holds with $C_\alpha = \frac1{1-\alpha}$.

(Routine case checks) for any $0<\epsilon<e^{-1}$ the function $x\log x$ is Lipschitz on $[\epsilon,1]$ with constant $\sup_{x\in [\epsilon,1]}| \log x +1|=\max(1,-\log\epsilon-1)=:C_1.$ (This is because the function has Lipschitz constant 1 on $[e^{-1},1]$.) Since we already have control for the region $0<x<y<e^{-1}$, this gives the required control for all $0\le x<y\le 1$ with $|x-y|<e^{-1}-\epsilon=:C_2.$ For the remaining regions of $(x,y)\in[0,1]^2$, it suffices to obtain control in the region $|x-y|>C_2$ where  we can crudely bound with $$\frac{|x\log x-y\log y|}{|x-y|\max(\lvert\log\lvert x-y\rvert\rvert,1)}\le \frac{\max_{x\in[0,1]}|x\log x|-\min_{x\in[0,1]}|x\log x|}{|x-y| }=\frac1{eC_2}.$$
Thus, for some fixed sufficiently small $\epsilon$, we can take  $$C=\max(1,C_1,\frac1{eC_2})=\max(1,-\log\epsilon-1,\frac1{1-e\epsilon}).$$

Setting $\epsilon=e^{-2}$ gives $C= 2/(1-e^{-1})\approx 1.58$. Numerically, the optimal choice of $\epsilon$ gives $C\approx 1.35$. but $C=1$ seems to be true even for the whole interval $[0,1]$.


问:
If $E=X-x$ find $\sqrt{X}-\sqrt{x}$
In the footnotes of my old (numerical methods and types of errors) lecture notes there is a question to find $\sqrt{X}-\sqrt{x}$ given that $E=X-x$. I am assuming that $E$ represents the error difference. The notes are old and not very clear but it mentions that
$\sqrt{X}-\sqrt{x}$ is proportional to $E/2\sqrt{X}$. Is there a way to show that?

答:
$$\sqrt X-\sqrt x=\frac{X-x}{\sqrt X+\sqrt x}=\frac E{2z}$$

where

$$z:=\frac{\sqrt X+\sqrt x}2.$$

When accuracy does not matter, you can use one of $\sqrt X$ and $\sqrt x$ for $z$.

Title: If $E=X-x$ find $\sqrt{X}-\sqrt{x}$


问:
Solution verification/clarification involving null space and row space for a set of planes
I was wondering if someone could verify my solution/work for this question alongside a few other clarifications I had.

a) Write the row space and the null space of the matrix $$\begin{bmatrix}1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9\\ \end{bmatrix}$$as a span of vectors.

After row reduction, I get:

$$\begin{bmatrix}1 & 2 & 3\\ 4 & 5 & 6 \\ 7 & 8 & 9\\ \end{bmatrix} \rightarrow \begin{bmatrix}1 & 0 & -1\\ 0 & 1 & 2 \\ 0 & 0 & 0\\ \end{bmatrix}$$

The basis of the row space just $\text{span}\{(1,0,-1),(0,1,2)\}$ and the null space is $\text{span}\{(1,-2,1) \}$.

b) Describe the space of solutions to the equations
\begin{cases}
      x_1+2x_2+3x_2=0 \\
      4x_1+5x_2+6x_3=0 \\
      7x_1+8x_2+9x_3=0
   \end{cases}

This is the same as the null space from part a, so it's just $\text{span}\{(1,-2,1) \}$.

c) Compare your answers from part (a) and (b). What do you observe and why?

The solution space for the given system is the same as the null space. This makes sense since the Null space is just the solution to the system $Ax=0$ and the coefficients of the given system is the same as the entries in the matrix.

d) Each equation in part (b) gives a plane. Describe the intersection of the 3 planes from part (b).

I'm not really sure how to describe this aside from the fact it's a scalar multiple of the null space of the the solution set.

e) Describe the span of the normal vectors of the 3 planes from the equations in part (b). Describe how it relates to part (a). What is the geometry of this span? How is it related to the intersection of the 3 planes from part (d)? Why?

I'm not 100% sure about this part. Would would I describe the span of the normal vectors? I know the normal vectors are $\{1,2,3\}$,$\{4,5,6\}$,$\{7,8,9\}$ but I am not sure what the span of the normal vectors are supposed to exactly describe.

The span would be something like $k_1\{1,2,3\}+k_2\{4,5,6\}+k_3\{7,8,9\}=(a,b,c)$ for some arbritary vector $\vec{v}=(a,b,c).$

If the mean the geometry of the null space, then that's a line but again, because I am not sure what the span of the normal vectors should describe, I can't think of what geometry I should get.

Similarly, I can't see how this is related to part d. Would it not just be the same set of vectors from part d?

答:
Your answers to a, b, c are fine.
For d, the three planes all intersect in a line, like pages of an open book.
For e, the span of the normal vectors is a plane--in particular, the plane perpendicular to the three planes represented by the equations.  The normal vectors span 2 dimensions, and they are orthogonal to the line of intersection, so their span is also orthogonal.</p>


问:
Prove that if $n^2$ is divided by 3, then also $n$ can also be divided by 3.

$n\in \Bbb N$

Prove that if $n^2$ is divided by 3, then also n can also be divided by 3.

I started solving this by induction, but I'm not sure that I'm going in the right direction, any suggestions?

答:
Proof by contraposition:

Suppose $ 3 \nmid n$, then $ n \equiv 1 (mod 3)$ or $ n \equiv 2 (mod 3)$

So $n^2 \equiv 1^2 \equiv 1 (mod 3) $ or $n^2 \equiv 2^2 \equiv 1 (mod 3) $

Therefore  $ 3 \nmid n^2$


问:
Given $e^{-5}\frac{5^n}{n!}=e^{-5}\frac{5^{n+1}}{(n+1)!},$ find the value of $n$
I do not know how to find the value of $n$ this is a past paper question I am trying to solve, any help is much appreciated. Thank you.

答:
Multiply both sides by $\frac{e^5(n+1)!}{5^n}$ simplify and get

$$5=n+1\rightarrow n=4$$

how do you know you must multiply both sides by...

I do not know. It is just a conclusion. To realize that you can rewrite you equation in the following way:

$$\frac{e^{-5}\cdot 5^n}{n!}=\frac{e^{-5}\cdot 5^n\cdot 5}{(n+1)\cdot n!}$$

As you can see, both sides contain

$$\frac{e^{-5}\cdot5^n}{n!}$$

thus this eliminates and it remains only

$$1=\frac{5}{n+1}$$

thus evidently, $n=4$


问:
Can a vector be sliced linearly?
There is a vector
A = [ 0, 1, 1, 2, 3, 5, 8, 13]
and another vector
B = [ 0, 1, 0, 0, 0, 0, 1, 1]
that signifies which indices/columns of A are to be extracted.

Is there a linear, vector operation that would yield
C = [1, 8, 13]?

答:
Let $$M=\begin{bmatrix}
 0&0&0 \\
 1&0&0 \\
 0&0&0 \\
 0&0&0 \\
 0&0&0 \\
 0&0&0 \\
 0&1&0 \\
 0&0&1
 \end{bmatrix}$$

where $M$ is obtained by decomposing $B$ into a sum of unit vectors and using the unit vectors to populate the columns of $M$.

Multiplying on the left by $A$ gives $C$.

Title: Can a vector be sliced linearly?


问:
Probability of first ball is red or second ball is red

A bag contains $2$ red and $6$ black balls. Two balls are drawn in succession without replacement from a bag. Then find the probability that the first  ball is red or the second ball is red.

What I have tried: Let $A: $ first ball is red and $B: $ second ball is red.

So $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Here $\displaystyle P(A)=\frac{2}{6}\times 1=\frac{1}{3}$

And $\displaystyle P(A\cap B)=\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}$

I did not understand. How do I solve $P(B)$? Help me please

答:
\begin{align}
P(B) &= P(A \cap B) + P(A^c \cap B)\\
&= P(A \cap B) + \frac{6}{8} \cdot\frac{2}{7}
\end{align}

Also, $P(A) = \frac{2}{2+6}$.

We do not have to compute $P(A \cap B)$ as it will cancel out.


答:
As long as you don't know what the first ball was, the second ball is just a uniformly random ball so the probability of $B$ is still $\frac13$.

If you don't find this convincing, you can check by considering the two possibilities: if $B$ occurs either both balls are red (probability $\frac1{15}$ as you already have) or the first ball is black and the second red (probability $\frac46\times \frac25=\frac{4}{15}$) so the total is $\frac5{15}$.

[edit: as Siong Thye Goh points out, the calculation in the question of $1/3$ is incorrect because there are $6$ black balls, not $6$ total balls.]

Title: Probability of first ball is red or second ball is red

Body:
A bag contains $2$ red and $6$ black balls. Two balls are drawn in succession without replacement from a bag. Then find the probability that the first  ball is red or the second ball is red.

What I have tried: Let $A: $ first ball is red and $B: $ second ball is red.

So $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Here $\displaystyle P(A)=\frac{2}{6}\times 1=\frac{1}{3}$

And $\displaystyle P(A\cap B)=\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}$

I did not understand. How do I solve $P(B)$? Help me please


答:
They are asking you to calculate the probability of the following cases:

$$\{RB,BR,RR\}$$

thus simply

$$1-\frac{6}{8}\times \frac{5}{7}+=\frac{13}{28}$$

that is the complement of $\mathbb{P}\{BB\}$


问:
Combination Probability: Chances of two particular people winning a raffle prize
So say there was an annual raffle. Each person receives one ticket only (i.e. you can't win more than one prize). If there are 510 ticketed people (including 1xFrank and 1xSally) and 32 prizes are drawn without replacement, what are the odds of Frank AND Sally BOTH winning a prize each?

Now, what are the chances of them both winning such an event two years in a row?

NOW, how would one suspect the raffle was rigged in some way? Is there a way to give quality assurance to these things?

I am completely out of my depth with this question but would appreciate a thoughtful answer. :)

答:
chances in the heading, odds in the question, I'll calculate the probability (Btw, chance is used loosely for probability, which is not the same as odds)

P(both win) = $\dfrac{32}{510}\dfrac{31}{509} = Z,\;say$

P(both win for two years in a row) = $Z^2$


问:
Mapping cylinder is a pushout

Mapping cylinder is a pushout where the definition of Mapping cylinder is the following: For given two topological spaces $X,Y$ and a continuous map $f:X\to Y$, a mapping cylinder $M_f$ is $X\times I\amalg Y/f(x)\sim (x,0)$ where $x\in X$.

i.e. I want to show that if $p:X\to X\times I$ by $x\mapsto (x,0)$ and $\varphi_1:Y\to M_f$ and $\varphi_2:X\times I\to M_f$ are induced maps and For given topological space $Z$ and two maps $\psi_1:Y\to Z$ and $\psi:X\times I\to Z$ such that the following diagram commute,

then there is a unique map $\varphi:M_f\to Z$ that makes the diagram commute. Could you help?

By the way, I'm not familiar with Categroy theory languages so the statement could be absurd

答: Consider the canonical inclusions $j_1 : Y \hookrightarrow Y \sqcup (X \times I)$ and $j_2 : X \times I \hookrightarrow Y \sqcup (X \times I)$. Now, from $\psi_1 : Y \to Z$ and $\psi_2 : X \times I \to Z$ one has a map $\psi : Y \sqcup (X \times I) \to Z$ so that $$\psi \circ j_1 = \psi_1 \quad \& \quad \psi \circ j_2 = \psi_2.$$ Additionally, if we assume that $\psi_1 \circ f = \psi_2 \circ p$, then $\psi$ identifies $j_1(f(x))$ with $j_2(x,0)$ for all $x \in X$, so $\psi$ factors through $M_f$: $$\varphi : M_f \to Z; \quad [z] \mapsto \psi(z)$$ (check that this is well-defined!).

Title: Mapping cylinder is a pushout


问:
When did we move from $\mathbb{Z}\left[\sqrt{d}\right]$ to the ring of integers $\mathcal{O}_{\mathbb{Q}\left[\sqrt{d}\right]}$ and why?
Gauss made great progress in number theory in $\mathbb{Z}$ by working in $\mathbb{Z}[i]$ (or equivalently $\mathbb{Z}\left[\sqrt{-1}\right]$), so much so that we call $\mathbb{Z}[i]$ the Gaussian integers now. And it was even known to the old mathematicians that solutions to Pell's equation $x^2 - dy^2 = 1$ could be better analysed by working in $\mathbb{Z}\left[\sqrt{d}\right]$.

But now in modern number theory we study much more the ring of integers $\mathcal{O}_{\mathbb{Q}\left[\sqrt{d}\right]}$. I find this confusing, as if we want to study Pell's equation with $d = 5$, we have that $\mathcal{O}_{\mathbb{Q}\left[\sqrt{5}\right]} = \mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ instead of $\mathbb{Z}\left[\sqrt{5}\right]$, which is not what we need. I was under the assumption that modern number theory usually tries to generalise its techniques but I don't see how this is a sensible generalisation and I don't see why the ring of integers is any more useful than just plain old $\mathbb{Z}\left[\sqrt{d}\right]$. So my question is:

Why is the ring of integers defined the way it is?

答: We want rings of integers of number fields to be Dedekind domains, which are roughly speaking rings where prime factorization holds at least on the level of ideals. Integral closure is necessary for that.</p>


问:
If the roots of the equation $x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$, prove that $(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$
My first post here

So I was doing some past year questions and this one popped out and I haven't been able to progress much upon it and I believe there's a trick which will get it done in no time.

If the roots of the equation
$x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$

we need to show that $$(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$$

Can I be helped with some hints to push me in the right direction to solve it?

Thanks in advance.

答:
Like Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity OR Problem based on sum of reciprocal of $n^{th}$ roots of unity

using Transformation of equations, let $1-x=y$

$$\implies(1-y)^n=1\iff  y^n-\binom n1y^{n-1}+\cdots+\binom n{n-1}(-1)^{n-1}y=0$$
whose roots are $y_k, 0\le k\le n-1$

Clearly, one of the them is $0,$ let $y_0=0$

So, the roots of $$y^{n-1}-\binom n1y^{n-2}+\cdots+\binom n{n-1}(-1)^{n-1}=0$$ will be $y_k, 1\le k\le n-1$

Finally use Vieta's formula


答:
You wrote "we have that $\mathcal{O}_{\mathbb{Q}\left[\sqrt{5}\right]} = \mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ instead of $\mathbb{Z}\left[\sqrt{5}\right]$, which is not what we need. "

This is not correct, we exactly need the ring of integers here, e.g., for most problems of algebraic number theory. The ring $\Bbb Z[\sqrt{5}]$ is not integrally closed, and hence not a Dedekind ring. So we cannot apply, among other things, the decomposition of ideals into prime ideals, which is crucial for many aspects.

Why do we need a decomposition for prime ideals here, and not for prime elements? Well, unfortunately the ring $\Bbb Z[\sqrt{5}]$  is not factorial, so we cannot proceed this way and we do need ideal decompositions.

Most rings appearing in number theory are no longer factorial, e.g. the rings $\Bbb Z[\zeta_n]$ for Fermat's equation $X^n+Y^n=Z^n$, but they are Dedekind rings, and this "recovers" some of the arguments we have for factorial rings - namely that $Z^n$ is an $n$-th power, so that writing
$$Z^n=X^n+Y^n=(X+Y)(X+\zeta_n Y)\cdots (X+\zeta_n^{n-1}Y)$$
as a product in the ring of integers of $\Bbb Z[\zeta_n]$ gives that the (coprime) factors are also $n$-th powers.
However, $\Bbb Z[\zeta_n]$ is only factorial for small $n$, so we can't argue this way.

Since $\Bbb Z[\zeta_n]$ is a Dedekind ring, we can write this as an ideal equation
$$
(Z)^n=(X^n+Y^n)=(X+Y)\cdot (X+\zeta_n=Y)\cdots (X+\zeta_n^{n-1}Y),
$$
and then at least make some progress.

Title: When did we move from $\mathbb{Z}\left[\sqrt{d}\right]$ to the ring of integers $\mathcal{O}_{\mathbb{Q}\left[\sqrt{d}\right]}$ and why?

Body:
Gauss made great progress in number theory in $\mathbb{Z}$ by working in $\mathbb{Z}[i]$ (or equivalently $\mathbb{Z}\left[\sqrt{-1}\right]$), so much so that we call $\mathbb{Z}[i]$ the Gaussian integers now. And it was even known to the old mathematicians that solutions to Pell's equation $x^2 - dy^2 = 1$ could be better analysed by working in $\mathbb{Z}\left[\sqrt{d}\right]$.

But now in modern number theory we study much more the ring of integers $\mathcal{O}_{\mathbb{Q}\left[\sqrt{d}\right]}$. I find this confusing, as if we want to study Pell's equation with $d = 5$, we have that $\mathcal{O}_{\mathbb{Q}\left[\sqrt{5}\right]} = \mathbb{Z}\left[\frac{1 + \sqrt{5}}{2}\right]$ instead of $\mathbb{Z}\left[\sqrt{5}\right]$, which is not what we need. I was under the assumption that modern number theory usually tries to generalise its techniques but I don't see how this is a sensible generalisation and I don't see why the ring of integers is any more useful than just plain old $\mathbb{Z}\left[\sqrt{d}\right]$. So my question is:

Why is the ring of integers defined the way it is?


问:
Minimize $(x+y)(x+z)$ with constraint without calculus

Let $x,y,z \in \mathbb R^+$ such that $xyz\cdot(x+y+z) = 1$

Find $\min\{(x+y)(x+z)\}.$

Using calculus, and Lagrange multipliers, I get:

$(x+y)(x+z) \ge2$ (with the equality occurring if and only if $y=z=1,\ x=\sqrt{2} - 1$).

But I want to solve it in an easy way, which doesn't need calculus.

How can I do it?
Thanks.

答: Hint: Recall the Heron's formula for triangle. Let $y+z=a, z+x=b, x+y=c$, then the problem becomes given the area to minimize $bc$.</p>


答:
Let A : First Ball is Red
B: Second Ball is Red.

You need to find:
$ P(A) P(B|A) + P(A)P(B^c | A) + P(A^c)P(B | A^c)$

These are respectively the probabilities for RR,RB,BR


问:
A coin is tossed 5 times and each flip is either heads or tails. How many possible outcomes contain more heads than tails?
The question is: A coin is tossed 5 times and each flip is either heads or tails. How many possible outcomes contain more heads than tails?

So my reasoning is that it can be:

5 heads = c(5,5) = 5!/5!(5-5)! = 5!/5! = 1

4 heads = c(5,4) = 5!/4!(5-4)! = 5!/4!1! = 5!/4! = 5

3 heads = c(5,3) = 5!/3!(5-3)! = 5!/3!2! = 5*4/2! = 10

so in total 16 possible outcomes. does this seem right?

答: There are $2^5=32$ equiprobable different outcomes. As $5$ is odd each outcome either has more heads or more tails. Due to symmetry both cases occur equally often. The answer therefore is $16$.


问:
Confirmation of correctness in proof regarding norm preserving operator
I just want to know if my solutions are correct for the following problem (euclidean norm assumed):

A linear transformation $T:\Bbb R^n \to \Bbb R^n$ is norm preserving if $|T(x)| = |x|$ for all $x \in \Bbb R^n$, and inner product preserving if $\left\langle T(x),T(y)\right\rangle = \left\langle x, y\right\rangle$.

$(a)$ Prove that $T$ is norm preserving if and only if $T$ is inner product preserving.

$(b)$ Prove that such a linear transformation is $1$-$1$ and $T^{-1}$ is of the same sort.

My solution for $(a)$ was: suppose $T$ is inner product preserving, then since $|x|=\sqrt{\left\langle x,x \right\rangle}$ we have that $|T(x)| = \sqrt{\left\langle T(x),T(x)\right\rangle}=\sqrt{\left\langle x, x\right\rangle} = |x|$ so that $T$ is norm preserving. Suppose now that $T$ is norm preserving, so we have the polarization identity:

$$\left\langle T(x), T(y)\right\rangle=\frac{|T(x)+T(y)|^2 - |T(x)-T(y)|^2}{4}$$

But by virtue of linearity $T(x)+T(y) = T(x+y)$ and $T(x)-T(y)=T(x-y)$ and so since $T$ preserves norms we have $\left\langle T(x),T(y)\right\rangle = \left\langle x,y\right\rangle$ so that $T$ is inner product preserving also.

My solution for $(b)$ was: Suppose that $T(x) = 0$, then we have $|T(x)| = 0$, but since $T$ is norm preserving $|x| = 0$ and this implies $x = 0$ so that $T$ is injective. Also by the rank-nullity theorem $T$ is surjective because $\dim \ker T = 0$ and so $\dim \operatorname{Im} T = \dim \Bbb R^n$. Now, because of that, given $y \in \Bbb R^n$ there's some $x \in \Bbb R^n$ such that $ y = T(x)$. In that case, consider the norm of $T^{-1}(y)$, because of the analysis that there's some $x$ such that $y= T(x)$ we will have:

$$T^{-1}(y) = x$$

Now recalling that $T$ preserves norms, we have that $|x| = |T(x)|$, but $T(x) = y$ so that $|x| = |y|$, so taking the norm on the above equation we finally have:

$$\left|T^{-1}(y)\right| = |x| = |T(x)| = |y|$$

Are my solutions correct? Thanks very much in advance!

答:
Just so that this question has an answer, I would make this comment into an answer.

You are absolutely correct in all your arguments. Where is your doubt?


问:
Boundary version of the fundamental lemma of calculus of variations
I was looking for a detailed version with proof of the following statement:
$$\int_{\partial D} (f\varphi  +X \cdot \operatorname{grad}\varphi ) \,\,\mathrm d a =0
~~~~\forall \varphi:D \to \mathbb R ~~~~~\Longrightarrow~~~~~ f=0,~X=0 ~~\text{on}~ \partial D .$$
Here $D$ is a three-dimensional domain.
It is some sort of boundary version of the fundamental lemma of calculus of variations with the derivatives of the test function.
A simple reference would be ok.
Thanks!

答:
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\n}{\mathbf{n}}$
$\newcommand{\pl}{\partial}$
$\DeclareMathOperator{\dv}{div}$
$\newcommand{\vp}{\varphi}$

For simplicity, let me assume that $D \subseteq \R^n$ is a bounded smooth domain, so that $\pl D$ is a closed submanifold. Let me also assume that $f \colon \pl D \to \R$ and $X \colon \pl D \to \R^n$ are smooth. Our assumption is that
$$
\int_{\pl D} f \vp + X \cdot \nabla \vp = 0
\quad \text{for all smooth } \vp,
$$
where $\vp$ is defined on all of $\overline{D}$ (or even $\R^n$) and $\nabla \vp$ is the full gradient.

Let us denote by $\n$ the unit outer normal to $\pl D$ and decompose $X$ and $\nabla \vp$ into their tangent and normal parts:
$$
X = X^\top + X^\perp \cdot \n,
\quad
\nabla \vp = \nabla_{\pl D} \vp + \pl_\n \vp \cdot \n,
$$
so that $X \cdot \nabla \vp$ becomes
$$
X \cdot \nabla \vp
= X^\top \cdot \nabla_{\pl D} \vp + X^\perp \cdot \pl_\n \vp.
$$
The first part can be integrated by parts over $\pl D$, yielding
$$
\int_{\pl D} X^\top \cdot \nabla_{\pl D} \vp
= - \int_{\pl D} \dv_{\pl D} X^\top \cdot \vp.
$$
Our main assumption then translates to the statement that for each smooth $\vp$ we have
$$
0 = \int_{\pl D} f \vp + X \cdot \nabla \vp
= \int_{\pl D} (f - \dv_{\pl D} X^\top) \cdot \vp + \int_{\pl D} X^\perp \cdot \pl_\n \vp.
$$
I claim that $\vp$ and $\pl_\n \vp$ can be chosen as arbitrary smooth functions on $\pl D$, so our starting assumption is equivalent to
$$
f = \dv_{\pl D} X^\top, \quad
X^\perp = 0
\quad \text{on } \pl D.
$$
To justify the claim, let us choose arbitrary smooth functions $a,b \colon \pl D \to \R$. The formula
$$
\vp(x+t\n) = a(x)+tb(x)
$$
gives us a well-defined function on a tubular neighborhood of $\pl D$, which can be extended elsewhere, if needed. Of course, $\vp = a$ and $\pl_\n \vp = b$ on $\pl D$.

What about curvature? The integration by parts formula $\int_{\pl D} X^\top \cdot \nabla_{\pl D} \vp = - \int_{\pl D} \dv_{\pl D} X^\top \cdot \vp$ is a consequence of  the formula $\int_{\pl D} \dv_{\pl D} V = 0$ valid for tangent fields $V$ on $\pl D$.
If in doubts, this can be further translated to an intrinsic formula (Gauss-Green, Stokes) for abstract manifolds, e.g. by translating fields and divergences into forms and differentials.

The mean curvature pops out when one applies $\dv_{\pl D} V$ (understood as $\operatorname{tr}(DV \circ (T_x \pl D)_\#)$) to a general (not tangent) vector field. The result is indeed
$$
\int_{\pl D} \dv_{\pl D} V = - \int_{\pl D} V \cdot \mathbf{h},
$$
where $\mathbf{h}$ is the mean curvature vector. See the definition of first variation here (sorry for unnecessary generality).


问:
A Euclidean algorithm problem
Suppose that, after running the Euclidean algorithm on two integers $a$ and $b$, we find that $r_n=3$, where $r_n$ is the last remainder in the Euclidean algorithm. Furthermore, we find that $r_{n-1}=6$ and $r_{n-2}=21$. What can we say about $a$ and $b$?

Obviously, by the definition of the Euclidean algorithm, $\gcd(a, b)=3$. However, that isn't sufficient (isn't strong enough for an iff condition). I tried for a long time to think of more conditionns, but only came up with a bunch of false conjectures. Can someone help me?

答:
As described in this answer, we can apply the Extended Euclidean Algorithm to $21$ and $6$ as follows:
$$
\begin{array}{r}
&&\overset{\substack{\hspace{-1in}\left\lfloor\frac{21}6\right\rfloor\hspace{-1in}\\\downarrow\\[3pt]\,}}{3}&\overset{\substack{\hspace{-1in}\left\lfloor\frac63\right\rfloor\hspace{-1in}\\\downarrow\\[3pt]\,}}{2}\\\hline
0&1&-3&7\\
1&0&1&-2\\
21&6&3&0\\
\end{array}
$$
Thus, the terminal three remainders, $(21,6,3)$, dictate and are dictated by the last two terms, $(3,2)$, in the continued fraction (along with the common factor of $3$). In fact, $(3;2)$ is the continued fraction for $\frac72=\frac{21}6$.

Title: A Euclidean algorithm problem


问:
Count the number of possible drawings
Take a square, ABCD. Add two points, E and F on AB such that AE=EF=FB. Now, add G, H on BC, I and J on CD, K and L on AD.

Now pick four pairs of the eight points, E,F,G,H,I,J,K,L. Draw the line segments between your picks, for example EG,FH, IL, JK. If we consider two such drawings equal if they can be rotated into each other, how many different drawings are there? What if we mandate that the drawing must have a line of symmetry?

答:
Start by arranging all the possible pairs in the following form:

\begin{pmatrix}
        \text{EF} & \text{EG} & \text{EH} & \text{EI} & \text{EJ} & \text{EK} & \text{EL} \\
         & \text{FG} & \text{FH}& \text{FI}& \text{FJ}& \text{FK}& \text{FL} \\
        && \text{GH}& \text{GI}& \text{GJ}& \text{GK}& \text{GL} \\
        &&& \text{HI}& \text{HJ}& \text{HK}& \text{HL}\\&&&& \text{IJ}& \text{IK}& \text{IL}\\&&&&& \text{JK}& \text{JL}\\&&&&&& \text{KL}\end{pmatrix}

Now we see that the number of pairing arrangements so that all $4$ pairs contain uncommon elements is $7\times 5\times 3\times 1=105$. Hence there must be $105$ such drawings. Notice that a drawing can be rotated to produce another pair drawing pattern from the other $104$ drawings if it is not symmetric about either the vertical (passing from the midpoints of $LK$ and $GH$) and horizontal (passing from the midpoints of $EF$ and $IJ$) axes, or the diagonal axes (lines $AC$ and $BD$). Also, a drawing can be rotated to produce 3 other drawing pattern from the other $104$ drawings if it is not symmetric about both the vertical (passing from the midpoints of $LK$ and $GH$) and horizontal (passing from the midpoints of $EF$ and $IJ$) axes, and the diagonal axes (lines $AC$ and $BD$).

Let us define diagonalizing as the process of rotating a drawing by $\pm45^\text{o}$ such that each node point is moved one unit forward (e.g. for $+45^\text{o}$ rotation, point $E$ becomes point $L$, $L$ becomes $K$, $K$ becomes $J$ and so on).

Notice that generally $\text{|EF|}\ne\text{|EL|}$, however diagonalizing allows a stretch or shrinkage for compensation.

I didn't know which of the cases you referred to so I listed all of them.

So if diagonalization and mirroring are not allowed then there are $35$ distinct drawings.

If diagonalization is allowed while mirroring is not, then there are $19$ distinct drawings.

If both diagonalization and mirroring are allowed, then there are $17$ distinct drawings.

Title: Count the number of possible drawings


问:
Is $x\log x$ Hölder continuous on $[0,1]$?
Consider the function $f:[0,1]\rightarrow \mathbb{R}$ given by

$$f(x)= x\log(x).$$

This function is not Lipschitz continuous at zero, but apparently Hölder continuous for any $\alpha<1$, i.e. there is a constant
$C_{\alpha}$ such that

$$\vert f(x)-f(y)\vert \le C_{\alpha} \vert x-y \vert^{\alpha},$$

but how can one show the Hölder continuity?

Please let me know if you have any questions or remarks.

答:
This can be seen by looking at the derivative:
$$
f'(x) = \log x + 1,
$$
which is controlled by the derivative of $f_\alpha(x) = x^\alpha$ (for any $0 < \alpha < 1$), which is $f_\alpha'(x) = \alpha x^{\alpha-1}$. Indeed, you should be able to check that $\left| \frac{f'(x)}{f_\alpha'(x)} \right|$ is bounded by some constant $C_\alpha > 0$ independent of $x \in (0,1)$ (but depending on $\alpha$).

Knowing that $f_\alpha$ is $\alpha$-Holder continuous, you can infer the same about $f$:
$$
|f(x)-f(y)| \le \int_x^y |f'(t)| dt \le C_\alpha \int_x^y |f_\alpha'(t)| dt = C_\alpha |f_\alpha(x)-f_\alpha(y)| \le C_\alpha C'_\alpha |x-y|^\alpha,
$$
of course also using monotonicity of $f_\alpha$.


问:
Invertible modules over a noetherian ring

Let $A$ be a commutative noetherian ring and let $M, N$ be two
finitely generated $A$-modules such that $M\otimes_A N\cong A$

I would like to know which are all such $A$-modules.

It seems to me that $M$ (and so $N$) has to be free and then since $M\otimes_A N\cong A$, it follows that they are both of rank $1$, i.e. isomorphic to $A$ itself.

My understanding of the problem

Using the noetherian hypothesis on $A$ we know that $M, N$ are also finitely presented and so we have two exact sequences \begin{equation*}A^{\oplus^{m_1}}\to A^{\oplus^{m_2}}\to M\to 0 \\ A^{\oplus^{n_1}}\to A^{\oplus^{n_2}}\to N\to 0 \end{equation*} We could try to tensor the first sequence by $N$ and the second one by $M$ to obtain \begin{equation*}N^{\oplus^{m_1}}\to N^{\oplus^{m_2}}\to A\to 0 \\ M^{\oplus^{n_1}}\to M^{\oplus^{n_2}}\to A\to 0 \end{equation*}but maybe this is a dead end. I don't have any further idea. Any hint/solution is appreciated.

答:
The modules $M$ satisfying this property are exactly the projective modules of (constant) rank $1$. In this case, you have automatically $N\simeq M^*$.

See for example this discussion on MO f or furhter information and references : https://mathoverflow.net/questions/104350/equivalent-definitions-of-invertible-modules


问:
Rayleigh Quotient with $2n \times 2n$ Matrices.
I need help understanding a problem.

For a bonus question in our problem set, we are told that $\bf{A}$ is a $2n \times 2n$ symmetric matrix of form $$\bf{A} = \begin{pmatrix} A_1 & A_3 \\ A_3 & A_2 \end{pmatrix}$$ where each entry is a $n \times n$ matrix, and that $A_1$ and $A_2$ are symmetric. We are then asked to prove that $$\lambda_\min (\bf{A}) \leq \min(\lambda_\min (A_1),\lambda_\min (A_2)).$$ We are asked to do this using the Rayleigh quotient $\frac{x^TAx}{x^Tx}$, which we've learned is bounded below and above by the smallest and largest eigenvalues of the matrix respectively, but is $x$ in this case supposed to be a vector of the form $\begin{pmatrix} X_1 \\ X_2 \end{pmatrix}$. If so, wouldn't this multiplication return a $n \times n$ matrix instead of a real value which can be bounded above and below by $\lambda$? It's all confusing me a bit, and I would appreciate it if someone could even just give me some definitions to work with, or some clarification on how the Rayleigh quotient could apply or work.

答:
Let $R(A,x)$ denotes the Rayleigh quotient. Then
$$
\lambda_{\min}(A)=\min_x R(A,x) \le \min_{x:x_{n+1}=\ldots=x_{2n}=0} R(A,x)=\lambda_{\min}(A_1).
$$
Similarly,
$$
\lambda_{\min}(A)=\min_x R(A,x) \le \min_{x:x_{1}=\ldots=x_{n}=0} R(A,x)=\lambda_{\min}(A_2).
$$

Here

$$
A=\left[
\begin{array}{ccc|ccc}
a_{1,1} & \cdots & a_{1,n} & a_{1,n+1} & \cdots & a_{1,2n}\\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{n,1} & \cdots & a_{n,n} & a_{n,n+1} & \cdots & a_{n,2n} \\ \hline
a_{n+1,1} & \cdots & a_{n+1,n} & a_{n+1,n+1} & \cdots & a_{n+1,2n}\\
\vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\
a_{2n,1} & \cdots & a_{2n,n} & a_{2n,n+1} & \cdots & a_{2n,2n} \\
\end{array}\right]
\quad\text{and}\quad x=\left[
\begin{array}{c}
x_{1} \\
\vdots  \\
x_{n}\\ \hline
x_{n+1} \\
\vdots  \\
x_{2n} \\
\end{array}\right].
$$


问:
Compute integral $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$
While solving a more general problem, I came across the following integral

$$I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$$

To start off, I defined $t:=x^2/2$ to get

\begin{align*}
I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2} \, dx &= 2\int_0^\infty x^{4n} e^{-x^2/2} \, dx \\
&= 2 \int_0^\infty (\sqrt{2t})^{4n} e^{-t} \frac{dt}{\sqrt{2t}} \\
&= 2 (\sqrt{2})^{4n - 1} \int_0^\infty t^{2n - 1/2} e^{-t} \, dt
\end{align*}

This seems to be a gamma function integral. I would start integrating by parts

\begin{align*}
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t} \, dt&=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt \\
    &=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt
\end{align*}

But how to proceed further?

I looked up in a table and the result should be

$$I=2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$$

PS: Alternative methods avoiding the gamma function are very welcomed! :)

EDIT: I think that what Yves suggested was the following

$$I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2}= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}\,dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2}\ \ 2\int_0^{+\infty}e^{-t} \, d(\sqrt{t})$$

Alrigtht! I see that $(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} = \Gamma\left(2n+ \frac 1 2 \right)$
NAIVE question though: how to see that $\int_0^{+\infty}e^{-t} \, d(\sqrt{t})=2 (\sqrt{2})^{4n - 1}$?

答:
You're really done after the first integration by parts: as $\displaystyle \Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\,dx,$

$\displaystyle 2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n -\frac{1}{2}} e^{-t} dt  =2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n + \frac{1}{2}-1} e^{-t} dt =2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right).$

In other words, your integral is the gamma function with $z = 2n+\frac{1}{2}$.

Regarding the other method, you have:

You've shown that if $\displaystyle I_n = \int_0^{\infty} t^{2n-\frac{1}{2}}\,{dt} $ then $I_n = \left(2n-\frac{1}{2}\right)I_{n-1}$ so that $$I_n =(2n-\frac{1}{2}) \cdot (2n-\frac{3}{2}) \cdot (2n-\frac{5}{2})\cdots\frac{3}{2} \cdot I_1 $$

Where $\displaystyle I_1 = \int_0^{+\infty}t^{\frac{1}{2}}e^{-t}\,dt$, then  $t = x^2$ shows that

$$\displaystyle I_1 = 2 \int_0^{\infty} x^2 e^{-x^2}\,{dx}. $$

Which evaluates to $\displaystyle \frac{1}{2}\sqrt{\pi}$ (see here). Therefore

$$\begin{aligned} I &= \sqrt{\pi} \cdot (2n-\frac{1}{2}) \cdot (2n-\frac{3}{2}) \cdot (2n-\frac{5}{2})\cdots\frac{3}{2} \cdot \frac{1}{2}\\& =\frac{\sqrt{\pi}}{2^{2n}}(4n-1) (4n-3) (4n-5)\cdots  3 \cdot 1 \\& =\frac{\sqrt{\pi}}{2^{2n}}\frac{(4n-1)(4n-2) (4n-3)(4n-4) (4n-5)\cdots 1}{(4n-2) (4n-4)  \cdots 2 }  \\& =\frac{\sqrt{\pi}}{2^{2n}}\frac{(4n-1)(4n-2) (2n-3)(2n-4) (2n-5)\cdots 3 \cdot \cdot 2 \cdot 1}{2^{2n} (2n-1) (2n-2) \cdots  1 }  \\& = \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}.\end{aligned}  $$

Therefore your original integral is $\displaystyle I =2 (\sqrt{2})^{4n - 1} \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}.$

This agrees with $\displaystyle I = 2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$ as $\displaystyle \Gamma\left(2n+ \frac 1 2 \right) = \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}$
(see here).


问:
Write down $n$ - tuple as a set
hello can you help me with that?

Let an $n$-tuple be defined as

$$ (a_1 , a_2 , \dots , a_n) = (( a_1 , a_2 , \dots , a_{n-1}),a_n) $$

write down $n$-tuple $ (a_1 , a_2 , \dots , a_n)$ as a set

I know that if we have $ (a_1,a_2,a_3) $ I can write it like this
$$  \{ \{a_1,a_2\}\{a_3\}\} = \{\{a_1,a_2\}\{ a_1,a_2,a_3 \}\} = \{ \{a_1\}\{a_1,a_2\}\{a_1,a_2,a_3\}\}    $$
and what i should do for $n$- tuple?

Title: Write down $n$ - tuple as a set

Body:
Hint:  Start with a pair $(a,b):=\{\{a\},\{a,b\}\}$, which is Kuratowski's now accepted definition.  Then generalize via the point of view that $(x_1,x_2,\dots,x_n)=(a,b)$, with $a=(x_1,\dots,x_{n-1}),b=x_n$, as you indicated.

So $(x_1,x_2,x_3)=((x_1,x_2),x_3)=\{\{(x_1,x_2)\},\{(x_1,x_2),x_3\}\}=\{\{\{x_1\},\{x_1,x_2\}\},\{\{\{x_1\},\{x_1,x_2\}\},x_3\}\}$.


答:
you can use the known formula (in any case very easy to prove) of the odd simple moments of the Standard Gaussian

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

Thus you integral results to me

$$\frac{\sqrt{2\pi}(4n)!}{4^n\cdot(2n)!}$$

To prove that this result match with the one you expected, just observe that

$$\frac{\sqrt{2\pi}(4n)!}{4^n\cdot(2n)!}=2^{2n+1/2}\cdot\Gamma\left(2n+\frac{1}{2}\right)$$

Your integral is the following

$$I=\sqrt{2\pi}\int_{-\infty}^{\infty} x^{2m}\phi(x)dx$$

where

$m=2n$

$\phi(x)$ is the density of a Standard Gaussian, that is a Gaussian distribution with mean zero and variance 1.

now you have only to apply the known formula for the simple odd moments I showed. The proof is available in many basic Statistics Textbook and it is very easy: it is enough to calculate the Taylor expansion of MGF and get the derivatives in $t=0$

Title: Compute integral $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$

Body:
While solving a more general problem, I came across the following integral

$$I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$$

To start off, I defined $t:=x^2/2$ to get

\begin{align*}
I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2} \, dx &= 2\int_0^\infty x^{4n} e^{-x^2/2} \, dx \\
&= 2 \int_0^\infty (\sqrt{2t})^{4n} e^{-t} \frac{dt}{\sqrt{2t}} \\
&= 2 (\sqrt{2})^{4n - 1} \int_0^\infty t^{2n - 1/2} e^{-t} \, dt
\end{align*}

This seems to be a gamma function integral. I would start integrating by parts

\begin{align*}
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t} \, dt&=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt \\
    &=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt
\end{align*}

But how to proceed further?

I looked up in a table and the result should be

$$I=2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$$

PS: Alternative methods avoiding the gamma function are very welcomed! :)

EDIT: I think that what Yves suggested was the following

$$I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2}= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}\,dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2}\ \ 2\int_0^{+\infty}e^{-t} \, d(\sqrt{t})$$

Alrigtht! I see that $(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} = \Gamma\left(2n+ \frac 1 2 \right)$
NAIVE question though: how to see that $\int_0^{+\infty}e^{-t} \, d(\sqrt{t})=2 (\sqrt{2})^{4n - 1}$?


问:
Given $e^{-5}\frac{5^n}{n!}=e^{-5}\frac{5^{n+1}}{(n+1)!},$ find the value of $n$
I do not know how to find the value of $n$ this is a past paper question I am trying to solve, any help is much appreciated. Thank you.

答:
A variant: simplify the common factors:
$$\mathrm e^{-5}\frac{5^n}{n!}=\mathrm e^{-5}\frac{5^n\cdot 5}{n!\,(n+1)}\iff 1=\frac 5{n+1}\iff n+1=5.$$


问:
$f_n\to f$ in $L^2$ implies $(f_n)^k\to f^k$ in the sense of distributions for all $k\in\mathbb{N}$?
I have a very naive question: Let $\{f_n\}_{n\in\mathbb{N}}\subset L^2(\mathbb{R})$ a sequence of functions satisfying $f_n\to f$ in $L^2$ for some $f\in L^2(\mathbb{R})$. Does this implies that, for all $k\in\mathbb{N}$, we have $$
(f_n)^k\to f^k \quad \hbox{in the sense of distributions }?
$$
Is this somehow direct? or maybe directly false? I have the feeling that it might be false since, for example, even on compact sets we could have functions belonging to $L^2$ such that they are not in $L^3$. However, with this naive intuition we could guess that the above convergence should hold for $k=1$, right?

答:
Hint

Let $\varphi $ is smooth. By Cauchy-Schwarz,
$$\left|\int_{\mathbb R}(f_n-f)\varphi ^{(k)}\right|\leq \|f_n-f\|_{L^2}\|\varphi ^{(k)}\|_{L^2}.$$


问:
Probability using a coin and a six-sided die
Assume a coin and a six-sided die. At the first move, you flip a coin. If you get a heads, you get to roll the die. If you get a tails, you have to flip the coin again. When you roll the die, if you get a $6$, the game is over. If you don't get a $6$, you have to flip the coin again. What is the probability of getting a $6$ on the die within $6$ moves? Each coin flip or die roll is a move.

I tried it with $4$ moves using brute force method with a tree diagram, and found an answer of $4$ in $60$, or $1$ in $15$. What I want to know is the general solution of this game within $n$ moves.

答:
We are going to compute the complimentary probability $p_n$ that there are no $6$ in $n$ moves. Note that there is a special case when we throw a head in the last move. If the move was not the last we were ought to throw a die but in this "defect" case it is not allowed.

Let us assume that $k$ heads was thrown in $n$ moves. In the normal case this means that there were $n-2k$ tails which amounts together to $n-k$ coins, which
we can permute in $\binom{n-k}k$ ways. In the defect case we will similarly obtain $n-2(k-1)-1=n-2k+1$ tails and $(n-2k+1)+(k-1)=n-k$ coins which we can arbitrarily permute (plus one additional coin which position is fixed). Plugging everything together we will obtain the probability:
$$
\binom{n-k}k \left(\frac56\right)^k\left(\frac12\right)^{n-k}+\binom{n-k}{k-1} \left(\frac56\right)^{k-1}\left(\frac12\right)^{n-k+1}.
$$

The total possibility can be computed by summing over all possible values of $k$:
$$
p_n=\frac1{2^n}\sum_{k=0}^{\left\lceil\frac n2\right\rceil}
\binom{n-k}k\left(\frac53\right)^k+\binom{n-k}{k-1}\left(\frac53\right)^{k-1},\tag1
$$
and $1-p_n$ will then give the probability to finish the game in $n$ moves.

Applying the formula $(1)$ to the case $n=4$ one obtains the value
$$
1-p_n=\frac{13}{72}
$$
which differs from your calculation but agrees with my check:
$$
\begin{array}{l|l}
XXXX& p\times\frac16\\
\hline
H6& \frac12\\
H\bar6H6&\frac12\frac56\frac12\\
TH6&\frac12\frac12\\
TTH6&\frac12\frac12\frac12\\
\hline
&\frac{13}{12}
\end{array}$$


问:
Integral involving integration by parts on second term
Can someone please step out the following integration by parts, I do not understand how to get to the solution.
$$\int\Big(\dfrac{e^x}{x}-\dfrac{e^x}{x^2}\Big)\space dx$$
How do I get to $\dfrac{e^x}{x}$ from this it says it is done by integration by parts on the second term but I am still confused, please help :)

答:
$$I=\int \left(\frac{e^x}{x}-\frac{e^{x}}{x^2}\right)dx$$
Do integration by parts on the first one and leave the second one. Take $e^x$ as second function.
$$I=\frac{1}{x} e^x+\int \frac{e^{x}}{x^2}dx-\int \frac{e^{x}}{x^2} dx=\frac{e^x}{x}+C$$

Now take second one first then, taking $e^x$ as first function, then
$$I=\int\frac{e^x}{x} dx+\frac{e^{x}}{x}-\int \frac{e^x}{x} dx= \frac{e^x}{x}+C$$


答:
No need for integration by parts.

Let $f(x)=\dfrac{1}{x}$.Then $f'(x)=-\dfrac{1}{x^2}$

Observe that now your integral is simply $\int e^x(f(x)+f'(x))dx$.

Also observe that $(e^xf(x))'=e^xf(x)+e^xf'(x)=e^x(f(x)+f'(x))$ which is your integrand.

Thus $\int e^x(f(x)+f'(x))dx=e^xf(x)+C=\dfrac{e^x}{x}+C$ which would be your solution.

Title: Integral involving integration by parts on second term

Body:
Can someone please step out the following integration by parts, I do not understand how to get to the solution.
$$\int\Big(\dfrac{e^x}{x}-\dfrac{e^x}{x^2}\Big)\space dx$$
How do I get to $\dfrac{e^x}{x}$ from this it says it is done by integration by parts on the second term but I am still confused, please help :)


问:
Why do we introduce a realisation of a generalized Cartan matrix?
When introducing the generalized Cartan matrix, one also introduces the the corresponding realisation. Why is this necessary?

I've read that if we didn't introduce the realisation, then we might not be sure to have linearly independence between the simpe roots in the finite dimensional case. However I'm not sure why? Here are my thoughts:

I'm thinking the trouble may occur when $\det(A)>0$ is removed. Then we may have $\det(A)=0$ so we have a non-zero vector $x\in \mathbb{C}^n$ st. $Ax=0$ or equivalently $\sum_{i=1}^nx_iA_{ij}=0$ for all $1\leq j\leq n$. This may somehow give a linear dependence between the roots. I know that $A_{ij}=2\frac{\langle \alpha_i,\alpha_j\rangle}{\langle\alpha_i,\alpha_i\rangle}$ for simple roots $\alpha_k$ ($1\leq k\leq n$) but I'm having a hard time convincing myself that  $\sum_{i=1}^nx_iA_{ij}=0$ gives $\sum_{i=1}^nx_i\alpha_i=0$?

答:
take for example $A = \begin{pmatrix} 2 & -2 \\ -2 & 2\end{pmatrix}$.
We cannot find any realization with $\{\alpha_1^\vee, \alpha_2^\vee\}\subset \mathfrak{h}$ and $\{\alpha_1, \alpha_2\}\subset \mathfrak{h}^*$  if $ \mathfrak{h}$ is a vector space of $\dim \mathfrak{h}=2$. In fact, you would get
$\langle\alpha_1^\vee,\alpha_1\rangle = -\langle\alpha_1^\vee,\alpha_2\rangle$ and $\langle\alpha_2^\vee,\alpha_1\rangle = -\langle\alpha_2^\vee,\alpha_2\rangle$. So, if $\alpha_1^\vee, \alpha_2^\vee$ are a basis, this forces $\alpha_2$ to be $-\alpha_1$. But we would like that both simple roots and simple coroots are linearly independent!

That's why you need to look in dimension $3$ instead.


问:
Discrepancy with absolute value in evaluating a triple integral
Evaluate

$$\iiint z^2 \,dx\,dy\,dz $$

over the region common to the sphere
$$x^2 + y^2 + z^2 = a^2$$
and the cylinder
$$x^2 + y^2 = ax.$$

Approach:

working in cylindrical co-ordinates, the integral should simply boil down to:-
$$\int_{-\pi / 2}^{\pi / 2} \int_0^{a\cos \phi} \int_{-\sqrt{a^2-s^2}}^{\sqrt{a^2-s^2}} z^2 \,dz\,d\phi$$

Which boils down to evaluating :

$$ \frac 2 3 \int_{-\pi / 2}^{\pi / 2}d\phi\int_0^{a\cos \phi} (a^2-s^2)^{3/2}s\,ds $$

Let $(a^2-s^2)=u^2$. Then, we get

$$ \frac 2 3 \int_{-\pi / 2}^{\pi / 2}\,d\phi \int_{a\left|\sin\phi\right|}^a u^4\,du \tag{E01} $$

Notice the $|\cdot|$ in the lower limit. (this was done as a result of $\sqrt{x^2}= |x|$ while changing the limits).  This will be the root cause of the trouble that follows:-

$$\frac{2a^5}{15} \int_{-\pi / 2}^{\pi / 2} (1-\left| \sin^5\phi \right|)\,d\phi = \frac{2a^5}{15} ( \pi - I)$$

Now, had the $| \cdot |$ not been present in $E01$, then $I$
would be $0$, since $\sin^5\phi$ is odd. So the answer would be $2\pi/15$.

But, the $|\cdot|$ means that $I$ would be something positive, which would mean the answer is something less than $2\pi/15$.

However, the given answer is: $2\pi/15$. Which made me question the existence of the $|\cdot|$.

Seems to be rather trivial that the $|\cdot|$ should be there.....The second scope for an error could be in determining the limits of the integral...But that seems fine too.

Where have I gone wrong then? Or do I just accept that the book "forgot" the $|\cdot|$ and I am indeed right?

答:
You are absolutely right, the absolute values should be there. Anyone who doubts is invited to write the integral as two times the integral over the upper half of the region (since the problem is symmetric with respect to $z$), so that you get $2 \int_0^{\pi/2}$ instead of $\int_{-\pi/2}^{\pi/2}$. Then there are no absolute values involved, and without question you do get a contribution from that $\sin^5 \phi$ term.

For what it's worth, Mathematica agrees: Integrate[z^2, {x, y, z} \[Element] ImplicitRegion[x^2 + y^2 + z^2 <= 1 && x^2 + y^2 <= x, {x, y, z}]]
returns
$$
\frac{2}{225}(-16+15\pi)
,
$$
or in other words
$$
\frac{2\pi}{15} - \frac{32}{225}
.
$$
(That's with $a=1$, obviously. To get the result for any $a>0$, just multiply by $a^5$, for dimensional reasons.)


问:
If systems of linear equations are equivalent, does the following computation hold?
Given the matrices $A, A', B$ and vectors $d, d', b$. Suppose the systems $Ax=d$ and $A'x=d'$ are equivalent and the matrices $A, A'$ have full rank. Furthermore assume that there exist solutions $y$ and $y'$ to $(A+B)y=d+b$ and $(A'+B)y'=d'+b$ and that $A+B$ and $A'+B$ have full rank.

Can we argue that because of the equivalency of the systems $Ax=d$ and $A'x=d'$ the equality $y=y'$ holds?

答:
No.

Take $A=2I_n$, $A'=I_n$, $d$ any nonzero vector, $d'=1/2. d$, $B=I_n$, $b=0$.

We end up with  the two systems $3y=d$ and $2y'=d'=1/2. d$.

They both have unique solutions $y=1/3.d$ and $y'=1/4.d$, which are distinct since $d\neq 0$.


问:
Probability of first ball is red or second ball is red

A bag contains $2$ red and $6$ black balls. Two balls are drawn in succession without replacement from a bag. Then find the probability that the first  ball is red or the second ball is red.

What I have tried: Let $A: $ first ball is red and $B: $ second ball is red.

So $P(A \cup B)=P(A)+P(B)-P(A \cap B)$

Here $\displaystyle P(A)=\frac{2}{6}\times 1=\frac{1}{3}$

And $\displaystyle P(A\cap B)=\frac{2}{6}\times \frac{1}{5}=\frac{1}{15}$

I did not understand. How do I solve $P(B)$? Help me please

答:
A bag contains 2 red and 6 black balls.

I did not understand. How do I solve P(B)?

Each individual ball has the same chance to be the first ball drawn.   Two among those eight are red.

Each individual ball has the same chance to be the second ball drawn.   Two among those eight are red.

The probability for obtaining two from two red balls when selecting two from eight balls is $\tbinom 22/\tbinom82$.

$\mathsf P(A)=1/4, \mathsf P(B)=1/4, \mathsf P(A\cap B)=1/28$ $$\mathsf P(A\cup B)=\dfrac{13}{28}$$

The probability for obtaining two from six black balls when selecting two from eight balls is $\tbinom 62/\tbinom 82$

$$\mathsf P((A\cup B)^\complement)=\dfrac{15}{28}$$


问:
Relation between L1 and L2 norms for the Fourier transform
If Fourier transform of $f(y)$ is $g(\beta)$, then $|f|_{L_y^2}=|g|_{L_\beta^2}$ by Plancherel's theorem. Is there a way to compare $|f|_{L_y^1}$ to $|g|_{L_\beta^2}$ somehow?

答:
No, unfortunately, if a function is in $L^1$, then the only thing one can say is that its Fourier transform is continuous and decaying at infinity, but not how fast it can decay.

Example: take a smooth compactly supported function $0≤\varphi ≤ 1$ and define
$$
f(x) = \frac{\varphi(x)}{|x|^{1-\varepsilon}}
$$
with $\varepsilon \in(0,1/2)$ so that $f∈ L^1$. Then
$$
\widehat{f}(x) = C_{\varepsilon}\,\widehat{\varphi} * \frac{1}{|x|^{\varepsilon}}
$$
is indeed continuous but only decays at infinity like $\frac{1}{|x|^{\varepsilon}}$ and so is not square integrable.

Similarly, if $f∈ L^2$, then $\widehat{f}$ is in $L^2$ but might not decay sufficiently at infinity to be in $L^1$ (take the same example but with $\varepsilon\in (1/2,1)$).


问:
Can sufficient statistics include indicators?
For example, Beta distribution is defined on $[0,1]$, and if we know that all the observations are in this range, then valid sufficient statistics would be $\prod x_i$ and $\prod (1-x_i)$. However, if we don't know if all the observations lie between $0$ and $1$, shouldn't we add the following indicators to the sufficient statistics:

$$I(x_{(1)} < 0)$$
$$I(x_{(n)} > 1)$$

or just $x_{(1)}$, $x_{(n)}$?

答:
In the Factorization theorem, the joint density is expressed in the form
$$f(\boldsymbol x) = h(\boldsymbol x) g(\boldsymbol T(\boldsymbol x) \mid \boldsymbol \theta),$$ where $h$ is a function of the sample $\boldsymbol x = (x_1, \ldots, x_n)$ but not of the parameters $\boldsymbol \theta = (\theta_1, \ldots, \theta_m)$, and $g$ is a function of both the sample and parameters, but only through some (possibly multivariate) function $\boldsymbol T$ of the sample; then this function $\boldsymbol T$ is our sufficient statistic.  The idea here is that we factor out the part of the density that does not depend on the parameters because that part is not informative of the value of the parameters, and the part that is informative is expressible in terms of some function of the sample that hopefully achieves some data reduction.

In light of this, you can immediately see that those indicators you mentioned do not depend on the parameters, so they are part of $h$ and not $\boldsymbol T$.  Including them as part of the sufficient statistic is unnecessary because there is no information about the parameters that is lost by omitting them.


问:
Finding a root of a fractional equation
I want to find a root of the following equation

\begin{equation}
\sum_{i=1}^m \frac{p_ia}{(1-xp_i)^2}=1
\end{equation}
where $x$ is variable with constraint $x>0$, $a$ and $p_i$ are some known positive number.

There might be many roots.
Is there any way to solve that equation?(or numerical method)

Any help would be appreciated!

答:
To simplify the problem, let us first change notations? Using $b_i=\frac a {p_i}$ and $c_i=\frac 1 {p_i}$, you want to find the zero's of function
$$f(x)=\sum_{i=1}^m \frac{b_i}{(x-c_i)^2}-1$$ which will show $2m$ roots among which some ( or none) would be real. Assuming, without any loss of generality, that the $c_i$'s are ordered $(c_1<c_2<\cdots<c_{m-1}<c_m)$ and different, there are potentially two roots between two asymptotes.

Suppose that your search for the roots between $c_k$ and $c_{k+1}$, amke the problem more continuous considering in this interval
$$g_k(x)=(x-c_k)^2(x-c_{k+1})^2 f(x)-(x-c_k)^2(x-c_{k+1})^2 $$ that is to say
$$g_k(x)=(x-c_k)^2(x-c_{k+1})^2\sum_{i=1}^{k-1} \frac{b_i}{(x-c_i)^2}+
b_k(x-c_{k+1})^2+$$
$$b_{k+1}(x-c_k)^2+(x-c_k)^2(x-c_{k+1})^2\sum_{i=k+2}^{m} \frac{b_i}{(x-c_i)^2}-(x-c_k)^2(x-c_{k+1})^2$$

At its boundaries
$$g_k(c_k)=b_k(c_{k+1}-c_{k})^2 \qquad \text{and} \qquad g_k(c_{k+1})=b_{k+1}(c_{k+1}-c_{k})^2$$ and you see that , in the range $c_k \leq x \leq c_{k+1}$, if  $b_{k}\,b_{k+1} <0 $,  function $g_k(x)$ looks like a straight line; otherwise  function $g_k(x)$ "looks more or less like a parabola".

If we are in the "nice" case, we can use as an estimate
$$x_0=\frac{b_{k+1}\,c_k- b_k\,c_{k+1}} {b_{k+1}-b_{k} }$$ and Newton method will converge like a charm.

To follow if you wish.

Title: Finding a root of a fractional equation


问:
How do I solve this normal line problem?
So I got this problem I've been struggling with.

''Determine the points of intersection between the parabola $x^2$ and its normal at the point $(a, a^2)$''.

And here is my try to solve it.enter image description here

As you can see I cant simplify the square root expression so I must have done something wrong.

答:
If $a=0$ the normal line is the $y$-axis, hence the unique intersection point is $(0,0)$.

For $a\ne 0$ the equation of the normal line at $(a,a^2)$ is $y=-\frac{1}{2a} x + (a^2+\frac 12)$,  hence the $x$-coordinate of the intersection points of the line with the parabola are the solutions of
$$x^2+\frac{1}{2a} x - (a^2+\frac 12)=0$$
One solution is obviously $x=a$ and using polynomial division we obtain:
$$x^2+\frac{1}{2a} x - (a^2+\frac 12)=(x-a)(x+a +\frac1{2a})$$

The intersection points are then $(a,a^2)$ and $(-(a +\frac1{2a}), (a +\frac1{2a})^2)$.


问:
FO-definability of the integers in (Q, +, <)
With $Q$ the set of rational numbers, I'm wondering:

Is the predicate "Int($x$) $\equiv$ $x$ is an integer" first-order definable in $(Q, +, <)$ where there is one additional constant symbol for each element of $Q$?

I know this is the case if multiplication is allowed.  I guess the fact that $<$ is dense would imply a negative answer to this question, via an EF game maybe; is there a similar structure with a dense order but with Int($x$) FO-definable?

答:
This is a very interesting question.

The answer is no.

First, I claim that the theory of the structure $\langle\mathbb{Q},+,\lt\rangle$ admits elimination of quantifiers. That is, every formula $\varphi(\vec x)$ in this language is equivalent over this theory to a quantifier-free formula. This can be proved by a brute-force hands-on induction over formulas. Allow me merely to sketch the argument. It is true already for the atomic formulas, and the property of being equivalent to a quantifier-free formulas is preserved by Boolean connectives. So consider a formula of the form $\exists x\varphi(x,\vec z)$, where $\varphi$ is quantifier-free. We may put $\varphi$ in disjunctive normal form and then distribute the quantifier over the disjunct, which reduces to the case where $\varphi$ is a conjunction of atomic and negated atomic formulas. We may assume that $x$ appears freely in each of these conjuncts (since otherwise we may remove it from the scope of the quantifier). If a formula of the form $x+y=z$ appears in $\varphi$, then we may replace all occurences of $x$ with $z-y$ and thereby eliminate the need to quantify over $x$ (one must also subsequently eliminate the minus sign after the replacing, but this is easy by elementary algebraic operations). We may do this even if $x$ appears multiply, as in $x+x+y=z$, for then we replace $x$ everywhere with $(z-y)/2$, but then clear both the $2$ and the minus sign by elementary algebraic manipulations. Thus, we may assume that equality atomic assertions appear only negatively in $\varphi$. All the other assertions merely concern the order. Note that a negated order relation $\neg(u\lt v)$ is equivalent to $v\lt u\vee v=u$, and we may distribute the quantifier again over this disjunct. So negated order relations do not appear in $\varphi$. The atomic order formulas have the form $x+y\lt u+v$ and so on. We may cancel similar variables on each side, and so $x$ appears on only one side. By allowing minus, we see that every conjunct formula in $\varphi$ says either that $a\cdot x\neq t$, or that $b\cdot x\lt s$ or that $u\lt c\cdot x$, for some terms $t,s,u$, in which $+$ and $-$ may both appear, and where $a$, $b$ and $c$ are fixed positive integer constants. By temporarily allowing rational constant coefficients, we may move these coefficients to the other side away from $x$. Thus, the assertion $\exists x\,\varphi(x,\vec t,\vec s,\vec u)$ is equivalent to the assertion that every $\frac{1}{c}u$ is less than every $\frac 1b s$. We may then clear the introduced rational constant multiples by multiplying through (which means adding that many times on the other side). Clearly, if such an $x$ exists, then this will be the case, and if this is the case, then there will be $x$'s in between, and so infinitely many, so at least one of them will be unequal to the $t$'s. This final assertion can be re-expressed without minus, and so the original assertion is equivalent to a quantifier-free assertion. So the theory admits elimination of quantifiers.

It now follows that the definable classes are all defined by quantifier-free formulas. By induction, it is easy to see that any such class will be a finite union of intervals, and so the class of integers is not definable.


问:
What is the disjunctive normal form of a formula
Given a formula A ∧ B, what is the disjunctive normal form of it?

Furthermore, given a formula A, what are the conjunctive and disjunctive normal forms of it?

答:
An expression in Disjunctive Normal Form is a disjunction of conjunctions of literals or their negations.

$A\land B$ is a disjunction of one conjunction of two literals.

Thus it is in Disjunctive Normal Form.


问:
Find rotation so that line doesn't intersect with circle
Hello math stackexchange community,

We have x as starting point, v as unit normalized direction, c as center of a circle and r as the radius of the circle.

To find if the line described as x + $\alpha$  v intersects the circle we can take the discriminant of the quadratic equation $||x + \alpha v - c||^2 < r^2$ and conclude if and how many collisions are there.

This is how it looks like

But how can we find the angle beta so that rotating v by beta would lead to line x + $\alpha$  rotate(v, beta) which doesn't intersect the circle?

Greetings, James!

答:
@Tanny Sieben Thank you! The idea to use the tangent is genius!

The angle between the tangent and the line pointing towards the center of the circle - $x + \alpha q$ - is $gamma = arcsin(r/d)$ where d is the distance from x to c and $q = \frac{c-x}{||c-x||^2}$ unit normalized direction vector.

The angle between $x + \alpha q$ and $x + \alpha v$ is $tau = arccos(q * v)$.

And $beta = gamma - tau$


问:
If a matrix has full rank, can it be inconsistent?
I read, that if a matrix is inconsistent you are able to bring it into a form where in one row there will only be zero entries, but the corresponding right hand side won't be zero, i.e. there exists a equation of the form $0x_1+0x_2+\cdots+0x_n=b_i\neq0$, which is an equality that can never be satisfied. But, if a matrix has full rank, there does not exist a row that does only contain zero elements, hence, if the matrix has full rank it cannot be inconsistent.

Is this correct?

Thank you very much in advance!

答: Matrices are not consistent or inconsistent, but rather systems of equations are.  If the  matrix $A$ corresponding to a system of equations $Ax=b$ is square, and has full rank, then the system will be consistent, since such a matrix corresponds to an isomorphism, and hence is surjective.


问:
Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series
Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...

答:
The functions $\sin$ and $\cos$, defined by the Taylor series have the following properties:

$$\begin{cases}
\sin'=\cos, & \sin 0=0\\
\cos'=-\sin, & \cos 0=1
\end{cases}$$

These properties are easy consequences of the definitions as power series, that can be differentiated term by term; moreover we know that the convergence radius (which is infinite, for $\sin$ and $\cos$) doesn't change when differentiating.

If we now consider the function
$$h(x)=(\sin x)^2 + (\cos x)^2$$
we have, by the chain rule,
$$
h'=2\sin\cdot\sin'+2\cos\cdot\cos'=2\sin\cdot\cos-2\cos\cdot\sin=0
$$
so $h$ is constant. Since $h(0)=0^2+1^2=1$, we have proved that, for all $x$,
$$
(\sin x)^2+(\cos x)^2=1
$$

How do you get $\cos(x+y)=\cos x\cos y-\sin x\sin y$? Just observe that when $f$ and $g$ are functions defined and differentiable over the whole real line, such that
$$f'=g,\qquad g'=-f,$$
these functions are uniquely determined as linear combinations of $\sin$ and $\cos$ by their values at $0$.

(Drawn from Lang's Introduction to mathematical analysis.)


问:
In classical computing, what is meant by a bit string function?
What is exactly meant by this function? I haven't found any clarification how exactly this function works:

$$f: \{0,1\}^n \rightarrow \{0,1\}^m$$

So it takes a n-bit string and turns it into an m-bit string but what is meant by a bit string and what are some examples of this function? Is a bit string 010101010101? This then gets put into the function and this turns into another bit string?

答:
Bit-strings just represent subsets of sets where a $1$ means the corresponding element of the set is included in the subset, and $0$ means it is excluded from your subset.

So if you have the following set: $\{2,3,5,8,13\}$ then the first bit in our bit-string corresponds to 2, the second to 3, the third to 5, the fourth to 8 and the fifth to $13$.

e.g. if you take a bit-string defined as $10110$, this represents a subset of our previous defined set as : $\{2,5,8\}$


问:
Alternative ways to show that a set of elements span a space?
Could we rigorously prove the following?

In a Hilbert Space $\mathcal{H}$, if $f, \phi_j \in \mathcal{H}$ for all $j \in J$, and $$\left<f, \phi_j\right> = 0 \text{  for all } j \in J \implies f = 0$$
then $\{\phi_j\}_{j \in J}$ spans $\mathcal{H}$.

This intuitively makes sense, but do we have formal proof?

P.S. This is from Ten Lectures on Wavelets Chapter 3, in the proof of proposition 3.2.1.

答:
Let $V := \displaystyle\bigoplus_{j \in J} \operatorname{span}\lbrace \phi_j \rbrace$. Your assumption implies $V^\perp = \lbrace 0 \rbrace$. Let $P: \mathcal{H} \rightarrow V$ be the linear projection onto $V$, i.e.
$$
(f - Pf, v) = 0 \quad \forall v \in V \quad \iff Pf = \operatorname*{argmin}_{v \in V} \lVert f - v \rVert.
$$
According to the projection theorem for Hilbert Spaces, this is well defined and the equivalence stated above holds. Note that $f$ in $\ker(P)$ iff $(f, v) = 0$ for all $v \in V$. So $\ker(P) = \lbrace 0 \rbrace$ according to our assumption.
Furtheremore, since $Pf \in V$, we have $PPf = Pf$ according to the $\operatorname{argmin}$ definition of the projection.
Note that for any $x \in \mathcal{H}$:
$$
x = \underbrace{(x - Px)}_{\in \ker(P) \implies 0} + \underbrace{Px}_{\in V}
$$
Therefore
$$
\mathcal{H} = \ker(P) \oplus V = \lbrace 0 \rbrace \oplus V = V
$$
and we are done.

$x - Px \in \ker(P)$ holds because of
$$
P(x-Px) = Px - PPx = Px - Px = 0.
$$


问:
Common tangent to two circles

Find the equations of the common tangents to the 2 circles:

$$(x - 2)^2 + y^2 = 9$$
  and

  $$(x - 5)^2 + (y - 4)^2 = 4.$$

I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.

答:
Like in my other answer, the equation of any tangent of $$(x-2)^2+y^2=3^2$$ will be $$x\cos t+y\sin t-(2\cos t+3)=0$$

Now if this has to be tangent of $(x-5)^2+(y-4)^2=4$ if the radius $=$ the distance from the center$(5,4)$

$$\dfrac{|5\cos t+4\sin t-(2\cos t+3)|}{\sqrt{\cos^2t+\sin^2t}}=2$$

$$\implies4\sin t+3\cos t-3=\pm2$$

Case $\#1:+\implies4\sin t+3\cos t=5$

Use Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ to solve for $\sin t,\cos t$

Case $\#2:-\implies4\sin t+3\cos t=1$

Use Weierstrass substitution to find $\sin t,\cos t$

Title: Common tangent to two circles


问:
Determine the Probability of no bin being empty when 10 balls are randomly tossed with equal probability into 6 bins
Suppose we have 10 items that we will randomly place into 6 bins, each with equal probability. I want you to determine the probability that we will do this in such a way that no bin is empty. For the analytical solution, you might find it easiest to think of the problem in terms of six events Ai, i = 1, . . . , 6 where Ai represents the event that bin i is empty, and calculate P(AC1 ∩···∩AC6)using DeMorgan’s law.

Analytical= 1- P(bin one empty) + P(bin 2 empty) + P(bin 3 empty) + P(bin 4 empty) + P(bin 5 empty) + P(bin 6 empty) – P(1 and 2 empty) – P(1 and 3 empty) – P(1 and 4 empty) – P(1 and 5 empty) – P(1 and 6 empty) – P(2 and 3 empty) – P(2 and 4 empty) – P(2 and 5 empty) – P(2 and 6 empty) – P(3 and 4 empty) – P(3 and 5 empty) – P(3 and 6 empty) – P(4 and 5 empty) – P(4 and 6 empty) – P(5 and 6 empty) + P( 1 2 3 empty) + P(1 2 4 empty) + P (1 2 5) empty + P (1 2 6 empty) + P( 1 3 4 empty) + P(1 3 5 empty) + P( 1 3 6 empty) + P(1 4 5 empty) + P(1 4 6 empty) + P(1 5 6 empty) – P(1 2 3 4 empty) – P(1235 empty) – P(1236 empty) – P(2345 empty) – P(2346 empty) – P(3456 empty) + P(12345 empty) + P(23456 empty) + P(34561 empty) + P(45612 empty) – P(123456 empty)

1-(6P(1 specific empty bin) – 15P(2 specific empty bins) + 20P(3 specific empty bins) – 15P(4 specific empty bins) +6*P(5 specific empty bins)

This is what I have so far but I don't know how to determine the individual probabilities above.

答:
Addendum added to examine the danger of overcounting from a different perspective.

This is a reaction to the answers of Math Lover and true blue anil.

I definitely regard Inclusion-Exclusion as the preferred approach.  Math Lover's answer completely covers that.

If you wish to consider the direct approach, then the Stirling Numbers of the Second Kind are dead on point.  This is exactly what true blue anil's answer details.

My answer is a manual exploration of what Stirling Numbers of the Second Kind represent,
how they are computed, and what their relevance is to this problem.

This Wikipedia Article
indicates that :

$\{ {n\atop k} \} = S(n,k) = $ the number of ways to partition a set of n objects into k non-empty subsets.

$\displaystyle S(n,k) = \frac{1}{k!}\sum_{i=0}^k (-1)^k \binom{k}{i} (k-i)^n.$

Suppose that you wanted to use the direct approach to the query's problem.  Further suppose that you wanted to manually derive each of the numbers, rather than relying on the given formulas.  This answer will illustrate the procedure, using the OP's constraints of $(10)$ balls going into $(6)$ rooms, and calculating the probability that each room gets at least one ball.

The probability will be expressed as

$$\frac{N\text{(umerator)}}{D\text{(enominator)}}
~~\text{where}~~ D = 6^{(10)} ~~\text{and}~~ N = N_1 + N_2 + \cdots + N_5.$$

$D = 6^{(10)}$, which was chosen for convenience, represents the total number of ways that each of the $10$ balls might be randomly distributed in the $6$ rooms.  Because of this choice, $N$ must be computed in a manner consistent with how $D$ is computed.  This means (for example) that [(ball-1 : room-1), (ball-2 : room-2)] must be considered distinct from
[(ball-1 : room-2), (ball-2 : room-1)].

There are 5 distinct ways that the 10 balls can be partitioned into 6 non-empty subsets of balls and then distributed into the various rooms.  These various ways are enumerated below.

$N_1 : ~~$ 5-1-1-1-1-1

This represents sending $5$ balls into one of the rooms, and $1$ ball into each of the remaining rooms.

There are $\binom{10}{5}$ ways of selecting $5$ balls into 1 unit.

Once this is done, you then have $6$ units that must be distributed into the 6 rooms.

There are $6!$ ways of doing this.

Therefore

$N_1 = \binom{10}{5} \times 6!.$

$N_2 : ~~$ 4-2-1-1-1-1

This represents sending $4$ balls into one of the rooms, $2$ balls into one of the other rooms, and $1$ ball into each of the remaining rooms.

There are $\binom{10}{4} \times \binom{6}{2}$ ways of

selecting $4$ balls into 1 unit and then selecting $2$ balls into another unit.

As before, there are $6!$ ways of distributing the units among the rooms.

Therefore

$N_2 = \binom{10}{4} \times \binom{6}{2} \times 6!.$

$N_3 : ~~$ 3-3-1-1-1-1

Warning - overcounting danger.

Superficially, one would suppose that the enumeration for $N_3$ would be:
$\binom{10}{3} \times \binom{7}{3} \times 6!.$

The best way to demonstrate why this is wrong, is by considering a specific example.

Consider sending the units $\{1,2,3\},\{4,5,6\}$ into Room-1 and Room-2, respectively.
In one instance, $\{1,2,3\}$ will be the first unit created, $\{4,5,6\}$ will be the second unit created, and the first unit will be sent to Room-1, while the second unit will be sent to Room-2.

In a different instance, $\{4,5,6\}$ will be the first unit created, $\{1,2,3\}$ will be the second unit created, and the first unit will be sent to Room-2, while the second unit will be sent to Room-1.

The overcounting danger is the price you pay for the convenience of using $(6!)$ to enumerate the number of ways of distributing the units into the rooms.  As long as you are aware of the danger, there is an easy fix.

By symmetry, because two of the larger units (more than 1 ball) have exactly the same size, you have to apply the scaling factor of $\frac{1}{2!}.$

Therefore

$N_3 = \binom{10}{3} \times \binom{7}{3} \times 6! \times \frac{1}{2!}.$

$N_4 : ~~$ 3-2-2-1-1-1

Again, the scaling factor of $\frac{1}{2!}$ will be needed,

because two of the larger units (more than 1 ball) have exactly the same size.

Therefore

$N_4 = \binom{10}{3} \times \binom{7}{2}  \times \binom{5}{2}
\times 6! \times \frac{1}{2!}.$

$N_5 : ~~$ 2-2-2-2-1-1

Here, the scaling factor of $\frac{1}{4!}$ will be needed,

because four of the larger units (more than 1 ball) have exactly the same size.

Therefore

$N_5 = \binom{10}{8} \times \binom{8}{2}  \times \binom{6}{2}  \times \binom{4}{2}
\times 6! \times \frac{1}{4!}.$

Putting this all together:

$$N ~=~ N_1 + N_2 + N_3 + N_4 + N_5$$

$$=~ 6! \times \left\{\left[\binom{10}{5}\right] ~+~
\left[\binom{10}{4} \binom{6}{2}\right] ~+~
\left[\frac{1}{2!}\binom{10}{3} \binom{7}{3}\right] ~+~
\left[\frac{1}{2!}\binom{10}{3} \binom{7}{2} \binom{5}{2}\right] ~+~
\left[\frac{1}{4!}\binom{10}{2} \binom{8}{2} \binom{6}{2} \binom{4}{2}\right]\right\}$$

$$=~ 6! \times \left[(252) + (3150) + (2100) + (12600) + (4725)\right]
~=~ 6! \times (22827) ~=~ (16435440).$$

Addendum

Examine the danger of overcounting from a different perspective.

Consider the subsets that are structured 4-2-1-1-1-1 :

against the subsets structured 3-3-1-1-1-1.

In order for the algorithm to work, each possible subset of the given structure must be counted exactly once.

With the 4-2-1-1-1-1, and the counting algorithm of
$\binom{10}{4} \binom{6}{2}$,

the specific collection of subsets

$\{1,2,3,4\}, \{5,6\}, \{7\}, \{8\}, \{9\}, \{10\}$,

will be counted exactly one time.

With the 3-3-1-1-1-1, and the counting algorithm of
$\binom{10}{3} \binom{7}{3}$,

the specific collection of subsets

$\{1,2,3\}, \{4,5,6\}, \{7\}, \{8\}, \{9\}, \{10\}$,

will be counted twice.

The first time that it will be counted is when $\{1,2,3\}$ is the first subset formed and $\{4,5,6\}$ is the second subset formed.

The second time that it will be counted is when $\{4,5,6\}$ is the first subset formed and $\{1,2,3\}$ is the second subset formed.

This explains the scaling factor of $\frac{1}{2!}.$

Examining the subsets structured as 2-2-2-2-1-1,

and the counting algorithm of
$\binom{10}{2} \binom{8}{2} \binom{6}{2} \binom{4}{2}$

the specific collection of subsets

$\{1,2\}, \{3,4,\}, \{5,6\}, \{7,8\}, \{9\}, \{10\}$,

will be counted (4!) times.

This is because with respect to the counting algorithm, any one of the four 2-element subsets might be created, first.  Then, any one of the three other 2-element subsets might be created second, and so forth.

This explains the scaling factor of $\frac{1}{4!}.$

One more example:

suppose that instead of forming 10 balls into 6 subsets,

you were forming 10 balls into only 4 subsets.

Consider the subsets structured as 3-3-2-2,

and the counting algorithm of
$\binom{10}{3} \binom{7}{3} \binom{4}{2}.$

How many different ways would the following collection of subsets be counted:

$\{1,2,3\}, \{4,5,6\}, \{7,8\}, \{9,10\}$?

The ordering of the creation of the two 3-element subsets could occur in $(2!)$ different ways.

Similarly, the ordering of the creation of the two 2-element subsets could also occur in $(2!)$ different ways.

This means that the correct enumeration for the structure of 3-3-2-2 would be

$\binom{10}{3} \binom{7}{3} \binom{4}{2} \times \frac{1}{2!} \times \frac{1}{2!}.$


问:
In a integral domain every prime element is irreducible
I'm trying to understand a proof of Hungerford's book which says that in a integral domain every prime element is irreducible:

I didn't understand why this implication $p=ab\implies p|a$ or $p|b$, is not the contrary $p=ab\implies a|p$ and $b|p$ ?

I'm a little confused

Thanks in advance

答: If $p = ab$, then in particular $p$ divides $ab$, because $ab = p \cdot 1$. Since $p$ is prime, it has to divide either $a$ or $b$.


问:
Proving that something is a topology on $X$
Let $(X,\mathcal{T})$ be a topological space. Let $A$ be a set such that $A \subseteq X.$ Show that the family $\mathcal{T}_{A} := \{U \cup (V \cap A) \mid U, V \in  \mathcal{T} \}$ is a topology on $X.$

The only problem I have is to show that $X \in \mathcal{T}_{A}$  and that $\emptyset \in \mathcal{T}_{A}.$

答:
$X = X \cup (\emptyset \cap A) \in \mathcal{T}_A$.

$\emptyset  =\emptyset \cup (\emptyset \cap A) \in \mathcal{T}_A$.

I don't see any issue there. We just use $X,\emptyset \in \mathcal{T}$ directly plus the definitions.


问:
Are $(1+p)^{p^{n-1}} - 1, (1+p)^{p^{n-2}} - 1$ divisible by $p^n$?
Is it true that $p^n$ divides $(1+p)^{p^{n-1}}-1$ if $p$ is an odd prime ? Furthermore, does $p^n$ divide $(1+p)^{p^{n-2}}-1$?

I tried to show the following, but made no progress:

$$ p^n \text{ divides } \binom{p^{n-1}}{k}p^k \text{ for each } k=1,\,2,\,\cdots,\,p^{n-1}$$

How to show this?

答:
LTE lemma is a good choice to prove it

If $p＝2$,then $p\mid 0$.
Let $p>2$ and  $p\mid (p+1)-1$, then we have
$$v_p((1+p)^{p^{n-1}}-1)＝v_p((1+p)-1)+v_p(p^{n-1})＝n$$


问:
Proof of acyclic models theorem in Vick's Introduction to Algebraic Topology
I am looking at the proof of the Acyclic Models theorem (5.3) in the book by James W. Vick: Homology Theory: an Introduction to Algebraic Topology.

My main issue is with the diagram on page 126:

He claims that the front face commutes due to the other faces commuting. First of all I am confused about what he means by front face. Intuitively, I would believe that the front face consists of the images of $M_\alpha$, but the context leads me to believe that it is actually the images of $X$ he is talking about when referring to the front face. Secondly, if the face with the $X$'s is indeed the one he means, I do not understand why commutativity of this face follows from commutativity of the others.

I will now give context about the diagram, but before reading this, I suggest having a look at the book yourself if you have the means to, since this will convey it more clearly.

$T$ and $T'$ are functors from a certain category $\mathcal{C}$ to the category of chain complexes. $T$ is free w.r.t. a set of models of $\mathcal{C}$.

$X$ and $M_\alpha$ are objects in $\mathcal{C}$ and more specifically, $M_\alpha$ comes from the set of models of $\mathcal{C}$.

$T_0$ and $T_0'$ are the corresponding functors to the category of abelian groups, defined as $(T(X))_0$ and $(T'(X))_0$ respectively.

$H_0$ is a functor which maps a chain complex to its zeroth homology group.

All vertical maps are different forms of $\pi$, a surjective map. More specifically, the quotient map from the zeroth cycle group of a chain complex into the zeroth homology group. Due to the nature of the functors $T$ and $T'$, these cycle groups are the same as the ambient group.

The diagonal maps are $T_0(f),T'_0(f), H_0(T(f)), H_0(T'(f))$, where $f$ is some map in $\operatorname{Hom}_\mathcal{C}(M_\alpha,X)$ whose existence is guarenteed if $T_0(X)\neq 0$. I do not think we can say anything on whether these maps are injective or surjective.

$\Phi$ is some natural transformation between the functors $H_0T$ and $H_0T'$.

Since $T$ is free w.r.t. the models, $T_0(X)$ will be free Abelian and generated by elements of the form $$ T_0(f)(e_\alpha)$$ with $e_\alpha$ in some $T_0(M_\alpha)$ for a model $M_\alpha$. $\phi$ acts on these generators by $$\phi(T(f)(e_\alpha))=T'(f)(\phi(e_\alpha))$$ and $\phi(e_\alpha)$ is a choice of element in $T'_0(M_\alpha)$ such that $\pi\circ \phi(e_\alpha) = \Phi\circ \pi(e_\alpha)$.

For completion's sake I add the definition of being free w.r.t. models:

Fix a category $\mathcal{C}$ and let $\mathcal{M}=\{M_\lambda\}_{\lambda\in\Lambda}$ be a collection of objects in $\mathcal{C}$. We call this collection the models of $\mathcal{C}$. A functor $T$ from $\mathcal{C}$ to the category of abelian groups is free with respect to the models $\mathcal{M}$ if there exists an element $e_\lambda\in T(M_\lambda)$ for each $\lambda$ such that for every object $X$ in $\mathcal{C}$
$$\{T(f)(e_\lambda)\mid \lambda\in\Lambda, f\in\operatorname{Hom}_\mathcal{C}(M_\lambda,X)\}$$ is a basis for $T(X)$ as a free abelian group.

So from this context I believe it is clear, he can only be talking about the face consisting of $X$'s as the front face, since if this is commutative, then the face consisting of $M_\alpha$'s will be commutative aswell, seeing as $M_\alpha$ is also an object of $X$.

I believe I have shown that all faces are commutative except for the face with the $X$'s. So my question is how does commutativity of the whole diagram follow from this.

答:
Consider the family of homomorphisms $\{T_0(f)\mid f \text{ occurs in some generator of }T_0(X)\}$. Then this set is jointly epimorphic. Indeed, suppose we have $g,h:T_0(X)\rightarrow Y$ homomorphisms for which $g\circ T_0(f) = h\circ T_0(f)$ for all $T_0(f)$ in the family. For a generator $T_0(p)(e_\alpha)$ of $T_0(X)$, we then have $$g(T_0(p)(e_\alpha)) = h(T_0(p)(e_\alpha))$$
since $T_0(p)$ lies in the family, now since $g$ and $h$ correspond on all generators, they must be equal.

Now this is precisely our case, for any $T_0(f)$ in the family, we have the above 3D diagram, where a priori all but the front face (with the $X$'s) commute. From this it must follow that $$\Phi_X\circ\pi\circ T_0(f) = \pi'\circ\phi^0_X\circ T_0(f)$$ by going through the five commutative diagrams. But now since the family is jointly epimorphic, we see that
$$\Phi_X\circ \pi= \pi'\circ \phi^0_X$$ i.e. that the front face commutes.


答:
In both cases, you'll want to use the Cauchy product, and the binomial theorem will be useful (for at least the first one), too. I leave the second one to you.

For the first one, $$\begin{align}\sin x\cdot\sin y &= \left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)\cdot\left(\sum_{n=0}^\infty\frac{(-1)^ny^{2n+1}}{(2n+1)!}\right)\\ &= \sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{(-1)^kx^{2k+1}}{(2k+1)!}\cdot\frac{(-1)^{n-k}y^{2(n-k)+1}}{(2(n-k)+1)!}\right)\\ &= \sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{(-1)^nx^{2k+1}y^{2(n+1)-(2k+1)}}{(2k+1)!(2(n+1)-(2k+1))!}\right)\\ &= \sum_{n=0}^\infty\frac{(-1)^n}{(2(n+1))!}\left(\sum_{k=0}^n\frac{(2(n+1))!x^{2k+1}y^{2(n+1)-(2k+1)}}{(2k+1)!(2(n+1)-(2k+1))!}\right)\\ &= \sum_{n=0}^\infty\frac{(-1)^n}{(2(n+1))!}\left(\sum_{k=0}^n\binom{2(n+1)}{2k+1}x^{2k+1}y^{2(n+1)-(2k+1)}\right)\\ &= -\sum_{n=0}^\infty\frac{(-1)^{n+1}}{(2(n+1))!}\left(\sum_{k=0}^n\binom{2(n+1)}{2k+1}x^{2k+1}y^{2(n+1)-(2k+1)}\right)\\ &= -\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right),\end{align}$$ whence $$-\sin x\cdot\sin y=\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right).$$

You can do some of the same sorts of manipulations to see that $$\cos x\cdot\cos y = 1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^n\binom{2n}{2k}x^{2k}y^{2n-2k}\right).$$ On the other hand, the binomial theorem shows us that for $n\ge 1$ we have $$(x+y)^{2n} = \sum_{j=0}^{2n}\binom{2n}jx^jy^{2n-j},$$ and splitting the right-hand side into two sums--one for even $j$ and one for odd $j$--gives us $$(x+y)^{2n} = \left(\sum_{k=0}^{n}\binom{2n}{2k}x^{2k}y^{2n-2k}\right)+\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right).$$ Can you put the pieces together and fill in the omitted steps/justifications?

Title: Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series

Body:
Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...


问:
Expected Value of Maximum Likelihood Estimator for $\operatorname{Beta}(\theta,1)$
I am having some difficulty finding the expected value of the MLE for the $\operatorname{Beta}(\theta,1)$ distribution.

Until now, I have found that the MLE for $\theta$ is: $$\hat{\theta} = -\frac{n}{\sum_{i=1}^n \ln X_i}$$

And I know that the expected value of $\hat{\theta}$ is: $$E[\hat{\theta}]=\frac{n}{n-1}\theta$$

But I really don't know how to calculate this expected value. I need to find the distribution of $\hat{\theta}$ to be able to calculate this expected value? Or exists another approach to this problem.

Furthermore, if I know that $T(X_1,...,X_n)=\sum_{i=1}^n\ln X_i$ is sufficient statistics for $\theta$, how can I find a non-biased estimator which is a function of the sufficient statistics for $\theta$? There is a method for that?

Thank you very much!

答:
But I really don't know how to calculate this expected value

Observe that

$$Y=-\log X \sim\exp(\theta)$$

Thus $\Sigma_y Y_i\sim \text{Gamma}(n,\theta)$

Is this enough for  you to derive your UMVU  Estimator?

Here is how to calculate your expectation using gamma distribution

$$\mathbb{E}[\hat{\theta}]=n\int_0^{\infty}\frac{1}{y}\frac{\theta^n}{\Gamma(n)}y^{n-1}e^{-n\theta}dy=n\theta\frac{\Gamma(n-1)}{\Gamma(n)}\underbrace{\int_0^{\infty}\frac{\theta^{n-1}}{\Gamma(n-1)}y^{(n-1)-1}e^{-n\theta}dy}_{=1}=$$

$$=\frac{n}{n-1}\theta$$


问:
Properties of closure on a topological space
Let $(X,t)$ be a topological space with $M,N \subseteq X$

I though that if $M\cap N\neq \varnothing  $ the it would be true that $CL(M)\cap CL(N)\subseteq CL(M\cap N)$ clearly this doesn't work. I figure it later.
But I did try to prove it, and now I want to understand where the mistake on my proof was. My proof went like this.

Let $M\cap N\neq \varnothing  $ and take $a \in CL(M)\cap CL(N) $
that means that for every open set $U$ such that $a \in U$ we have $U \cap M \neq \varnothing$
similarly, we have that for every open set $V$ that $a \in V$ we have $V \cap N \neq \varnothing$
I now just took the union of those 2 and said $M\cap U \cup V\cap N\neq \varnothing$ and consider $U \cup V=W$ and I said $(M\cap N)\cap W\neq \varnothing$ so $CL(M)\cap CL(N)\subseteq CL(M\cap N)$

My mistake was that I said that $(M\cap N)\cap W\neq \varnothing \equiv  CL(M\cap N)$ and that is wrong because I didn't show that for every open $W$ ?

答:
You'd have to start with an arbitrary neighbourhood of $a$ and show it intersects $M \cap N$. You didn't do that. You know the neighbourhood intersects $M$ and $N$ as well, but you're not guaranteed any point in $M \cap N$ in that neighbourhood.

E.g. If $M$ equals the set of rationals in $\Bbb R$ and $N$ the set of all irrationals plus $\{0\}$, then $M \cap N = \{0\}$ which equals its own closure while the closure of both $M$ and $N$ are $\Bbb R$ and we have $\Bbb R$ on the left hand side.


问:
Find limit of expressions where the numerator contains natural logarithm of 2 variables
find $\displaystyle \lim _{\left(x,y\right)\to (0,0)}\left(\frac{\ln(5xy+1)}{5x^8y^3+5xy} \right)$ where $x, y > 0$

I haven't touched math for quite a while. Normally I would try to factorize the numerator and the denominator so the denominator doesn't approach 0 but for this particular problem, I don't know how to start. Thank you for all the help you can provide.

答:
Hint:

$$\frac{\ln (5xy+1)}{5x^8y^3 + 5xy}= \frac{\ln(5xy+1)}{5xy} \times \frac{1}{x^7y^2 +1}  $$

Consider the first and second factors as separate limits. For the first one , let $5xy=t$. Then you get $$\lim_{t\to 0} \frac{\ln(1+t)}{t} $$ which is a well-known limit. The second one can be calculated just by substituting the values.


问:
Let $n \geq 4$. How many labeled trees are there with $n$ vertices where vertex with label 1 has degree 3?
So I'm a beginner to graph theory and really unsure how to approach this. I was thinking by counting Prufer codes, in which case you would want the second element in the Prufer code to be 3. So, it would be the number of Prufer codes where one of the labels are fixed at 3, so $n^{(n-1)-2}$ labeled trees?

答:
First of all, you should probably draw out the cases $n=4,5$ by hand, and you'd see that your guess is wrong.

I'll assume we are counting unrooted trees. In that case, think about what happens if we remove the node with label $1$. Then we have a forest of $3$ rooted trees (the root is the node connected to $1$). You can probably count these a number of ways, but I'll show how I'd do it with generating functions / the symbolic method.

The EGF for rooted labeled trees satisfies $T(z)=ze^{T(z)}$ (because $\mathcal T = \mathcal Z\star SET(\mathcal T)$). Let $\mathcal F^{(k)}$ be the class of rooted $k$-forests. Then
$$
\mathcal F^{(k)} = SET_k(\mathcal T) \implies F^{(k)}(z) = \frac{T(z)^k}{k!}
$$
Now we can use Lagrange inversion / the Lagrange–Bürmann formula to get $F_n^{(k)}$, the number of $k$-forests with $n$ nodes. $[z^n]$ denotes the $n$'th coefficient in a series expansion.
$$ \begin{split}
F_n^{(k)}
&= n![z^n] \frac{T(z)^k}{k!} \\
&= \frac{n!}{k!} [u^{n-1}]\frac 1n ku^{k-1} e^{nu} \\
&= \frac{(n-1)!}{(k-1)!} [u^{n-k}] e^{nu}\\
&= \binom{n-1}{k-1} n^{n-k}
\end{split} $$
Wonderful! So the number you seek is equal to $F_{n-1}^{(3)} = \binom{n-2}{2}(n-1)^{n-4}$.


问:
A question about Stolz-Cesàro theorem
Stolz-Cesàro theorem states that given some sequence $ \left(a_{n}\right)_{n\geq1} $ and a monotone strictly increasing sequence that diverge $ \left(b_{n}\right)_{n\geq1} $ , such that $$ \lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=L $$

Then the following holds:

$ \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=L $

Now, given a divergent sum $ \sum_{n=1}^{\infty}a_{n} $, we have that also  $ \frac{1}{n}\sum_{k=1}^{n}a_{k} $ diverge to $ \infty $ as we can see here

So consider $ a_{n}=\sum_{k=1}^{n}\frac{1}{k} $ and $ b_{n}=n $. Then

$$ \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\frac{\frac{1}{n+1}}{1}\underset{n\to\infty}{\longrightarrow}0 $$

But $$ \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k} $$

Does not converge to $0 $.

Hows is this possible consider that Stolz-Cesàro theorem is true? What am I missing?

Thanks in advance.

答:
Your mistake is here:

But $$ \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k} $$

Actually, $$ \lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac 1 k=0.$$

Title: A question about Stolz-Cesàro theorem


问:
Find the probability that the largest value $X_{\max}$ is greater than $0.92$
Let $X_1, X_2, \ldots , X_n \sim F$ with corresponding pdf $f(x) = 3x^2, \ 0 \leq x \leq 1$. If $ n = 10$ find the probability that the largest value $X_{\max}$ is greater than $0.92.$

To find this I have I was first asked to find the pdf for $X_{\max}$ which I found to
$$
f_{\max}(x) = nF^{n-1}(x)f(x) = n(x^3)^{n-1}3x^2 = 3nx^{3n-1}
$$
and then afterwards calculated
$$
P(X_{\max} \leq 0.92) = 1 - P(X_{\max} > 0.92) = 1 - \int_0^1 30x^{29} \, {\rm d}x = 1-1 = 0
$$
But I guess this is not right. Where did I do wrong?

TIA for any help.

答:
You got everything right except the very last part.  How did you conclude $$\Pr[X_{max} > 0.92] = \int_{\color{red}{x=0}}^{1} 30x^{29} \, dx?$$  It needs to be
$$\Pr[X_{max} > 0.92] = \int_{x=0.92}^1 30x^{29} \, dx.$$


Title: Find the probability that the largest value $X_{\max}$ is greater than $0.92$

Body:
Let $X_1, X_2, \ldots , X_n \sim F$ with corresponding pdf $f(x) = 3x^2, \ 0 \leq x \leq 1$. If $ n = 10$ find the probability that the largest value $X_{\max}$ is greater than $0.92.$

To find this I have I was first asked to find the pdf for $X_{\max}$ which I found to
$$
f_{\max}(x) = nF^{n-1}(x)f(x) = n(x^3)^{n-1}3x^2 = 3nx^{3n-1}
$$
and then afterwards calculated
$$
P(X_{\max} \leq 0.92) = 1 - P(X_{\max} > 0.92) = 1 - \int_0^1 30x^{29} \, {\rm d}x = 1-1 = 0
$$
But I guess this is not right. Where did I do wrong?

TIA for any help.

Body: $P(X_{max} >0.92)=\int_{0.92}^{1} 30x^{2}dx$. You took the integral from $0$ to $1$.