问:
How to prove this inequality? $4(a^2+b^2+c^2+d^2) \geq (a+b+c+d)^2$
How do you prove:
$$4(a^2+b^2+c^2+d^2)\geq (a+b+c+d)^2.$$ Sorry for the inconvenience because I'm new and the picture's size is to big to be uploaded.

答:
It's a highschool trick, let me show you below:

Suppose a b c d are all real number,

[result (1)] $a^2+b^2-2ab=（a-b)^2>=0,then a^2+b^2>=2ab$

expand the right-handside and apply  [result (1)]

$（a + b + c + d）^ 2=（a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2）+2ab+2ac+2ad+2bc+2bd+2cd$

$（a + b + c + d）^ 2\leq（a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2）+(a^2+b^2)+(a^2+c^2)+(a^2+d^2)+(c^2+b^2)+(d^2+b^2)+(c^2+d^2)=4（a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2）$


问:
Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$
I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots </p>

<p>$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$ </p>

答:
Given:

$(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$

Let a = $(x^2+5)$ which gives us a simple quadratic trinomial of the form $ax^2+bx+c=0.$

$a^2-15a+54=0$

Factoring this we get:

$(a-9)(a-6)=0$

$a-9=0$

$a-6=0$

$\implies a=9$ or $a=6$

$x^2+5=9$

$x^2+5=6$

$\implies x = \pm2$ or $\pm 1$

You can check that these solutions are correct by simply plugging values back in to see if you get $0$.

$a^2-15a+54=0 \implies$ $(9)^2 - 15(9) + 54 = 0 \implies 81-135+54=0$

OwnerDisplayName: Kalina Kama

ContentLicense: CC BY-SA 3.0


问:
Prove $ \forall x\in\mathbb{R} : |\sin{x}|<|x|$ geometrically
For $x\in(-\frac{\pi}{2}, -\frac{\pi}{2})\setminus \{0\}$ we can say that since the area of sector of circle is greater than the triangle (ABC) area I have $2|\sin{x}|<2|x|$. But what's about x in the 2nd and 3rd quadrant?
Look at the figure to easily understand what I am saying (Note: I have drawn a circle with a radius $>1$ to have a larger circle, but I am referring to a unit circle)

答: Hint: The inequality is trivial when $\lvert x\rvert>1$. (It is also not true when $x=0$.)

ContentLicense: CC BY-SA 4.0


问:
Find the derivative using the definition of derivative (limit).

Given $f(x)=\dfrac{5x+1}{2\sqrt{x}}$. Find $\dfrac{df(x)}{dx}=f'(x)$ using
the definition of derivative.

I have tried as below.

\begin{align*}
f'(x)&=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5(x+h)+1}{2\sqrt{x+h}}-\dfrac{5x+1}{2\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{\left(5(x+h)+1\right)\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x+h}-\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
\end{align*}

Now I can't find the limit. I confused how to simplify the limit. Anyone can give me  hint to solve it?

Note:

We were asked to find this derivative using
$$f'(x)=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$
instead of
$$f(x)=\dfrac{u(x)}{v(x)}\iff f'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}.$$

答:
You can expand $f(x)$ as $\frac{5x+1}{2\sqrt x}=\frac52\sqrt x+\frac1{2\sqrt x}$, then differentiate the two.

$$\begin{align}
f'(x)&=\left(\frac52\sqrt x\right)'+\left(\frac1{2\sqrt x}\right)'\\[1ex]
&=\lim_{h\to0}\frac{\frac52\sqrt{x+h}-\frac52\sqrt x}h+\lim_{h\to0}\frac{\frac1{2\sqrt{x+h}}-\frac1{2\sqrt x}}h
\end{align}$$

But if you don't already know or cannot directly use the fact that differentiation distributes over sums, you can instead expand the limand in your second line and regroup the terms in the numerator, then separate the limit into two. You would end up arriving at the same point either way:

$$\begin{align}
f'(x)&=\lim_{h\to0}\frac{\frac{5(x+h)+1}{2\sqrt{x+h}}-\frac{5x+1}{2\sqrt x}}h\\[1ex]
&=\lim_{h\to0}\frac{\frac52\sqrt{x+h}+\frac1{2\sqrt{x+h}}-\frac52\sqrt x+\frac1{2\sqrt x}}h\\[1ex]
&=\lim_{h\to0}\frac{\frac52\sqrt{x+h}-\frac52\sqrt x}h+\lim_{h\to0}\frac{\frac1{2\sqrt{x+h}}-\frac1{2\sqrt x}}h
\end{align}$$


问:
Prove that $c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right)$ holds on a smooth surface.

Problem: Given a infinitely small geodesic triangle $\triangle ABC$ on a smooth surface, denote the corresponding edges as $a,b,c.$ Prove: the area of $\triangle ABC$ (denote as $S$) and the Gauss curvature on $C$ ( denote as $K$) satisfies

$$c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right).$$

Since the triangle is infinitesimal, therefore it is naturally to use the first fundamental form and the Gauss-Bonnet theorem. However, I got stuck at how to handle the $\frac{KS}{3}$ in the equation. Since the coefficient is $\frac{1}{3}$, if using Gauss-Bonnet theorem, we would include new variables $\angle A,\angle B$. Could anyone help me out? Thanks a lot in advance!

答:
Comment @ robjohn : Can we check  for both signs of const $K?$

So for constant $K$ approximation we have $ \cos \tilde C$

$$= \cos (C+ \dfrac{KS}{3})= \cos (C\pm \dfrac{A+B+C-\pi}{3}) $$

K>0

For an equilateral spherical triangle $( a=b=c,A=B=C=\pi/2) $

$$\text{RHS}=\cos 2 \pi/3= -\dfrac12$$

$$ c^2= a^2+b^2+ab = 3 a^2 \to c= {\sqrt 3 a}, \text{ a contradiction }$$

K<0

For an equilateral  hyperbolic  spiky triangle $(A=B=C=\pi/6,a=b=c),$

$$ = \cos ( \dfrac{4C+A+B-\pi}{3}) = \cos (\dfrac{2C-A-B+\pi}{3}), $$

$$\text{RHS}=\cos \pi/3= \dfrac12$$

$$ c^2= a^2+b^2-ab = a^2 ,\; c=a.$$

Please comment what I am missing here in the K>0 approximation.


问:
Prime ideals of localization of absolutely flat modules
There is an exercise (chapter 3, Exercise 10, page number 44) in atiyah mc Donald.

$A$ is absolutely flat iff $A_m$ is a field for each maximal ideal $m$.

I actually proved this statement and it is not hard to do. But my question is where will maximal ideals map to?

We have the following, "There is an one-to-one correspondence between prime ideals of $A_m$ and prime ideals of $A$ not intersecting complement of $m$."

In the light of the above fact, in localization of absolutely flat ring at maximal ideal gives field, which has only two ideals. Now $0$ in $A$ corresponds to $0$ in $A_m$. So, To which ideal in $A_m$, $m$ corresponds to?

This might be some kinda trivial or silly question.  But I was clearly missing something and I couldn't figured it out. Please help.

Added:
Assume $A$ is an integral domain, if necessary.  So that $0$ will be prime.

答: If $A$ is an absolutely flat integral domain, then $A$ is a field, this is because absolutely flat rings are zero-dimensional, so every prime ideal is maximal. Thus there will be no $m$ to consider as $0$ is the only prime ideal in $A$.


问:
How to prove $\mathbb{Z}[i]$ is a ring but not a field?
Let i be the imaginary unit of $\Bbb{C}$ and $\mathbb{Z}[i]:= \{a + ib, ab \in\mathbb{Z} \} $

So the task is to prove that this a ring regarding multiplication and addition in $\mathbb{C}$ which is commutative, free of zero divisors and posseses a 1 element.
I already found out that it is a ring and that the 1 Element is (1,0) and 0 Element is (0,0), so in both cases $i= 0$. So I guess this is like proving a trivial Ring.
I am just confused because I am not sure about the $i$. How would I complete the proof? I am stuck because I don't know how to prove that it is free of zero divisors and not a field.
Thank you for the help.

答:
You can model $\mathbb Z[i]$ by using the matrix:
$$J=\begin{bmatrix}0&-1\\1&0\end{bmatrix} $$
Observe that $J^2=-I$, and observe that the map
$$a+bi \mapsto aI + bJ$$
is an embedding of $\mathbb Z[i]$ into $M_2(\mathbb Z)$.
Now, we can observe that, for example:
$$det (I+J) = det\begin{bmatrix}1&-1\\1&1\end{bmatrix}=2 $$
and thus $I+J$ is not an invertible matrix (in $M_2(\mathbb Z)$, since $2$ is not a unit in $\mathbb Z$), and therefore $\mathbb Z[i]$ is not a field.


问:
Clarification for two different methods on showing that a surface is smooth

Let $S = \{(x,y,z) \in \mathbf{R}^3 \mid 4-4x^2-2y^2, x^2+y^2 < 1 \}$. Show that $S$ is a smooth surface.

I'm going to propose two methods that I was taught on how to show that a surface is smooth.

$(1)$ A surface is smooth if all the partials are $C^1$ and the normal vector is non zero at each point i.e there exists a tangent plane at each point on the surface for which the normal is non zero.

Parameterizing $S$ as $\varphi(x,y) =(x,y,4-4x^2-2y^2)$ I have that $$D\varphi(x,y) = \begin{bmatrix} 1 && 0 \\ 0 && 1 \\ -8x && -4y \end{bmatrix}$$

from which it can be seen that the partials exist and are continuous.

Now I was thinking that it would satisfy to check whether the Jacobian determinant is non-singular to satisfy the condition for the normals, but the Jacobian determinant cannot be computed for non-square matrices so I guess I have to compute the cross product of the partials.

Computing $\frac{\partial \varphi}{\partial x} \times \frac{\partial \varphi}{\partial y} = (8x,4y,1) \ne0$. Which would imply that $S$ is smooth.

$(2)$ The surface is a smooth $2$-dimensional surface if rank$(D\varphi(x,y)) = 2.$

This can be clearly seen from $D\varphi(x,y)$ for example if I reduce the matrix to echelon form I have that $$D\varphi(x,y) = \begin{bmatrix} 1 && 0 \\ 0 && 1 \\ 0 && 0 \end{bmatrix}$$

so the rank is clearly $2$ and $S$ is a smooth $2$-dimensional surface.

What I would like to understand is how do these two statements coincide. So how are they equivalent? And also If someone could give me an idea of why the idea of looking at the Jacobian went wrong that would be appreciated.

答:
$\frac{\partial \varphi}{\partial x} \times \frac{\partial \varphi}{\partial y}$ is non zero means that the  basis vectors are linearly independent so u get a tangent palne which is  $\mathbf{R}^2$.

And when u compute the Jacobian and find it is having rank 2 (where u calculated) it means that the linear transformation  : U -> $\mathbf{R}^2$.

So both are same.
additional ref. Bar page 94


问:
Question about the Carleson's Theorm
Recently, I've been trying to understand the proof of Carleson's Theorem on Fourier Analysis given that $f\in L^2$

But I still find it quite hard to understand the relationship of the boundedness of Carleson operator and pointwise continuous of Fourier inversion series.

That is, how
$$\|\mathcal{C}f\|_2 \leq M\|f\|_2$$

where $\mathcal{C}$ is the Carleson Operator, would indicate

$$\lim_{N\rightarrow \infty}\int_{-N}^N \widehat{f}(\xi)e^{2\pi i x\xi} \,\mathrm{d}\xi=f(x)$$

Thank you in advance!

答:
The Carleson operator isn't bounded $L^2\rightarrow L^2$ or at least the proofs I've seen show the weaker bound
$$|\{x\mid Cf(x)>\lambda\}|\leq \lambda^{-2}||f||_{L^2}$$
which says that the Carleson operator is bounded $L^2\rightarrow L^{2,w}$, the latter pronounced weak $L^2$. Note that by Markov's inequality this would be implied by the $L^2$ bound.

Anyway, let $Lf(x)=\limsup_{N\rightarrow\infty}|f(x)-\int_{-N}^N e^{2\pi i \xi x}\hat{f}(\xi)\,d\xi|$ and fix $\epsilon>0$. We will show that $E_\epsilon=\{x\mid Lf(x)>\epsilon\}\rightarrow 0$ as $\epsilon\rightarrow 0$, which shows that $f(x)=\int_{-\infty}^\infty e^{2\pi i \xi x}\hat{f}(\xi)\,d\xi$ for almost every $x$. Let $g\in C^\infty_{c}(\mathbb{R})$ satisfy $||f-g||_{L^2}<\epsilon^2$ and notice that
$$E_\epsilon=\{x\mid \limsup_{N\rightarrow\infty}|f(x)-g(x)-\int_{-N}^Ne^{2\pi i \xi x}\hat{f}(\xi)-\hat{g}(\xi)\,d\xi|>\epsilon/2\}$$ where we have used that $g(x)=\int_{-\infty}^\infty e^{2\pi i \xi x}\hat{g}(\xi)\,d\xi$ for all $x$ because $g\in C^\infty_c(\mathbb{R})$ (this is just Fourier inversion). Then
$$|E_\epsilon|\leq |\{x\mid C(f-g)(x)>\epsilon/2\}|+ |\{x\mid |f-g|(x)>\epsilon/2\}|\leq 2(\epsilon/2)^{-2}\epsilon^4=2\epsilon^2$$
where we have used Chebyeshev's inequality to estimate the second term.

Title: Question about the Carleson's Theorm


问:
Show that $a^2b^2+b^2c^2+c^2a^2 \ge abc(a+b+c)$ For real $a,b,c$.
Show that $$a^2b^2+b^2c^2+c^2a^2 \ge abc(a+b+c)$$
For real $a,b,c$.

I tried to use:
$$\left(\sum_{i=1}^{3}a_{i}^{2}\right)\left(\sum_{i=1}^{3}b_{i}^{2}\right)\ge\left(\sum_{i=1}^{3}a_{i}b_{i}^{}\right)^{2}$$
$$(a^2+b^2+c^2)^2=(a^2+b^2+c^2)(a^2+b^2+c^2)\ge(ab+bc+ca)^2$$

But expanding and simplifying did not give the answer,I also used this approach:
$$\left(ab+bc+ca\right)^{2}=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc\left(a+b+c\right)$$

So we need to show that:
$$\left(ab+bc+ca\right)^{2}-2abc\left(a+b+c\right)\ge abc\left(a+b+c\right)$$

Or equivalently:
$$\left(ab+bc+ca\right)^{2}-3abc\left(a+b+c\right)\ge 0$$

But I don't know how to that.

Body: put $x=ab,y=bc,z=ac$ we have to prove $$x^2+y^2+z^2\ge xy+yz+zx$$ $$\iff \frac{1}{2}({(x-y)}^2+{(y-z)}^2+{(z-x)}^2)\ge 0$$


问:
Bound the minimum eigenvalue of a symmetric matrix via matrix norms
I'm reading a paper in which the authors prove an inequality of the following form:

$$\lVert H-H'\rVert_2 \leq \lVert H-H'\rVert_F \leq \epsilon \tag 1$$

Here $H$ and $H'$ are symmetric real matrices ($H'$ has all positive eigenvalues, if that matters), and the norms are the $L_2$ matrix norm and the Frobenius norm, respectively. With no justification the authors then claim:

$$\lambda_\text{min}(H) \geq \lambda_\text{min}(H') - \epsilon \tag 2$$

where $\lambda_\text{min}$ is the minimum eigenvalue of a matrix.

I can't see how to justify this, or even if (2) is even intended to be deduced from the (1). Here is the paper - the end of the proof of Lemma 3.2, page 6.

答:
This answer is based on this one. Below we will be working with some arbitrary inner product, and when we take the norm of a matrix, this means the operator norm associated with the vector norm we're using. We have:

Theorem. If $A$ and $B$ are real symmetric, then:

$$\lambda_\text{min} (A) \geq \lambda_\text{min} (B) - \lVert A-B\rVert$$
$$\lambda_\text{max} (A) \leq \lambda_\text{max} (B) + \lVert A-B\rVert$$

To prove this, the key is the expression $x^T Mx$, where $M$ is a symmetric matrix and $x$ has unit norm. We need two lemmas about this expression.

Lemma 1. For any matrix $M$ and any unit norm $x$:
$$-\lVert M\rVert \leq x^T Mx\leq \lVert M\rVert$$
Proof. Simple application of Cauchy-Schwartz and of the definition of an operator norm:
$$|x^TMx|\leq\lVert x\rVert \lVert Mx\rVert\leq \lVert x\rVert^2 \lVert M\rVert=\lVert M\rVert$$

Lemma 2. For any symmetric matrix $M$ and any unit norm $x$:
$$\lambda_\text{min}(M) \leq x^T M x \leq \lambda_\text{max}(M)$$
and the bounds are attained as $x$ varies over the unit sphere.

Proof. Let $M=P^TDP$ where $P$ is orthogonal and $D$ is diagonal. Then
$$x^TMx = (Px)^TD(Px)$$
As $x$ varies over the unit sphere, $Px$ varies also over the entire unit sphere, therefore the range of the latter expression above is simply the range of $y^TDy$ as $y$ ranges over the unit sphere. By the rearrangement inequality and some other simple arguments, the minimum is attained when $y$ is an eigenvector associated with $\lambda_\text{min}(M)$ and the maximum when $y$ is an eigenvector associated with  $\lambda_\text{max}(M)$.

Finally we can prove the theorem. For any unit norm $x$, we have

$$x^TAx = x^TBx + x^T(A-B)x$$

By applying Lemma 1 to the second term and Lemma 2 to the first term, the minimum of the left hand side is at least $\lambda_\text{min} (B)-\lVert A-B\rVert$. By Lemma 2, we know that the minimum of the left hand side is equal to $\lambda_\text{min} (A)$. A similar argument shows the other inequality in the theorem.


答:
\begin{eqnarray*}
a^2(b-c)^2+ b^2(c-a)^2+ c^2(a-b)^2 \geq 0.
\end{eqnarray*}
Divide by $2$ and rearrange.

Title: Show that $a^2b^2+b^2c^2+c^2a^2 \ge abc(a+b+c)$ For real $a,b,c$.

Body:
Show that $$a^2b^2+b^2c^2+c^2a^2 \ge abc(a+b+c)$$
For real $a,b,c$.

I tried to use:
$$\left(\sum_{i=1}^{3}a_{i}^{2}\right)\left(\sum_{i=1}^{3}b_{i}^{2}\right)\ge\left(\sum_{i=1}^{3}a_{i}b_{i}^{}\right)^{2}$$
$$(a^2+b^2+c^2)^2=(a^2+b^2+c^2)(a^2+b^2+c^2)\ge(ab+bc+ca)^2$$

But expanding and simplifying did not give the answer,I also used this approach:
$$\left(ab+bc+ca\right)^{2}=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc\left(a+b+c\right)$$

So we need to show that:
$$\left(ab+bc+ca\right)^{2}-2abc\left(a+b+c\right)\ge abc\left(a+b+c\right)$$

Or equivalently:
$$\left(ab+bc+ca\right)^{2}-3abc\left(a+b+c\right)\ge 0$$

But I don't know how to that.


问:
Polynomial equation root
I need to show whether:

$$\frac{1}{x}\times A - \left(\frac{1}{(x+b)(x-b)}-\frac{1}{(x+c)(c-x)}\right)$$

has a single root between $b$ and $c$, where $c>b>0$ and $A>0$. My numerical plots suggest so, but not 100% sure.

答:
From the discussion above and the limits, there exists at least one root where $P(x)$ crosses zero from below.

If the function $P(x)$ had multiple roots, then there exists an $x^\star\in\left[b,c\right]$ such that $P(x^\star)\leq 0$ and $P^\prime(x^\star)<0$ (i.e., when $P(x)$ crosses zero from above).

I show there is no such $x^\star$.

First off, we note that:
$$P^\prime\left(x\right)=\frac{2 x \left(-2 x^2 \left(b^2+c^2\right)+b^4+c^4+2 x^4\right)}{\left(b^2-x^2\right)^2 \left(x^2-c^2\right)^2}-\frac{A}{x^2}.$$

It follows that $P^\prime(x)<0$ if and only if

$$A>\overline{A}=\frac{2 x^3 \left(-2 b^2 x^2+b^4-2 c^2 x^2+c^4+2 x^4\right)}{\left(b^2-x^2\right)^2 \left(x^2-c^2\right)^2}.$$

Next, we see that
$$P\left(x,\overline{A}\right)=\frac{1}{2} \left(\frac{1}{(b+x)^2}+\frac{1}{(b-x)^2}+\frac{1}{(c-x)^2}+\frac{1}{(c+x)^2}\right)>0.$$

Since $P$ increases in $A$, it follows that $P\left(x,A\right)>0$,  $\forall A\geq \overline{A}$. Therefore,
$$ P^\prime\left(x\right)<0 \Longrightarrow P\left(x\right)>0 $$
and there is no root $x^\star$. In other words, every time the function decreases, it remains positive -- and it never crosses zero from above.

Correct?

Title: Polynomial equation root


问:
Bijection from ellipse to real line
I know that $f(x,y) = \cfrac{y}{x + 1}$ is a bijection from the unit circle with its south pole removed to the real line. I'm trying to find a similar bijection from the ellipse $S = \{(x,y): x^2 + 2y^2  =1 \}$  to the real line. The problem is that I only proved the first bijection mechanically (by showing that $f^{-1}(a) = \left(\cfrac{1-a^2}{1+a^2}, \cfrac{2a}{1 + a^2} \right)$ was its inverse) and didn't derive any of the two formulas myself - so I don't know what the bijection is actually "saying" nor consequently how to adapt it.

答: There is an obvious bijection between the circle $x^2+y^2=1$ and the ellipse $(z/a)^2+(w/b)^2=1$: $(z,w)=(ax,by)$. Composing that with the known bijection between the circle and line yields the desired result.

Title: Bijection from ellipse to real line


问:
Using Binomial Theorem for concert reservations
Q
A concert organiser observes that 3% of those making reservations for a concert do not show up for that concert. Based on this the organiser sells 100 seats for a venue with 97 seats. Use the binomial distribution to compute the probability that there will be a seat available for each person (with a reservation) who shows up for the concert.

Attempt

N = 100 seats
P = 0.97 (Probability of People Showing Up)
Q = 0.03
(Probability of People not Showing Up)

P(x < 97) = (97 + 0.1 = 100(0.97) / √100(0.97)(0.03)

I get syntax error so I'm guessing this is completely wrong?

答:
Set $X$ the rv "number of person not showing up at the concert", your probability is

$$\mathbb{P}[X\geq 3]=1-\mathbb{P}[X=0]-\mathbb{P}[X=1]-\mathbb{P}[X=2]$$

can you conclude by yourself?


问:
Linear Algebra eigenvector
Let $A$ be a $n\times n$ matrix. Let $v$ be an eigenvector of $A$ with corresponding eigenvalue $\lambda$, and let $w$ be and eigenvector of $A^T$ (the transposed of $A$) with corresponding eigenvalue $\alpha$. Show that if $\lambda\neq\alpha$, then $v$ and $w$ are orthogonal (Hint: compute $w^T Av$ in two different ways).

I am kind of lost and don't really know how to solve this.

答: Just to write them out, $Av=\lambda v$, $A^Tw=\alpha w.$ Note that this means that $(w^TA)^T=\alpha w^T.$ Now compute $w^TAv$ by $(w^TA)v$ first and then $w^T(Av)$ next and compare two expressions..

Title: Linear Algebra eigenvector


问:
The fourier transform preserves norms for a square-integrable function proof
Can you provide proof for "The Fourier transform preserves norms for a square-integrable function"?</p>

答:
The Schwartz functions are dense in $L^2$: because for $f\in L^2$ for $T$ large $\|f-f 1_{[T,-T]}\|_2$ is small and for $n$ large $\|f 1_{[T,-T]}-n e^{-\pi n^2 x^2}\ast f 1_{[T,-T]}\|_2$ is small ($\ast$ is the convolution).

The finite linear combination of Gaussian functions $e^{-a (x+b)^2+icx}$ are dense in the Schwartz functions thus in $L^2$ (do a convolution with $n e^{-\pi n^2 x^2}$ again and approximate the Riemann integral by a Riemann sum)

For such a finite linear combination of Gaussians functions let $(Wg)(y)=\int_{-\infty}^\infty e^{-2i\pi xy}g(x)dx,(W^* g)(y)=\int_{-\infty}^\infty e^{2i\pi xy}g(x)dx$, the known Fourier transform of the Gaussian gives that $W^* Wg =g$. Also it is easily seen that $\langle Wg,h\rangle=\langle g,W^*h\rangle$ thus $$\|Wg\|_2^2=\langle Wg,Wg\rangle=\langle g,W^* Wg\rangle=\langle g,g\rangle = \|g\|_2^2$$

Take a sequence $g_n$ of such  finite linear combination of Gaussians such that $\|f-g_n\|_2\to 0$ then its Fourier transform $Wf$ is defined as the limit in $L^2$ of the sequence $Wg_n$ (which is Cauchy by the previous result) so it is by definition that $\|f\|_2=\|Wf\|_2$.

It works the same way when replacing the Gaussians by any kind of functions for which you proved the Fourier inversion theorem (for example the functions such that $h,h'\in C^0\cap L^1$, from the Dirichlet kernel).


问:
Least squares intersection of three circles
Given is a triangle in the plane, with the coordinates of all three vertices known. I need to determine the location of a point $X$, for which the distances to all three triangle vertices are given (with some error). Therefore, $X$ is located at the intersection of the three circles around the triangle vertices, whose radii are the respective distances.

Since the distances are given with some error, the solution is not exact. How can the "best" intersection point be determined, possibly in a least-squares sense (e.g. minimizing the squared sum of deviations from the given distances)? Or is there a more reasonable way to define what is "best"?

答:
You want $x$ that satisfies the three equations
$$
\|v_i - x \|^2 = d_i^2,
$$
but such an $x$ doesn't exist in general. We can ignore that technicality, and obtain two linear equations from these three, by subtracting the second and the third from the first:
$$
(v_2 - v_1)^T x = \frac{1}{2}[(d_1^2 -d_2^2) + (v_2^2 - v_1^2)]
$$
$$
(v_3 - v_1)^T x = \frac{1}{2}[(d_1^2 -d_3^2) + (v_3^2 - v_1^2)],
$$
which can be written compactly in matrix form as
$$
V^Tx = w.
$$
Since the $v_i$ form a (presumably non-degenerate) triangle, $V$ is an invertible matrix, and we have a unique solution:
$$
x = V^{-T}w.
$$

I wasn't sure what this solution represented, but a friend supplied some insight. We can think of the triangle whose vertices are the $v_i$ as a triangle in a "weighted Delaunay triangulation". Unfortunately Wikipedia doesn't seem to have an entry on these yet. The dual diagram goes under various names, including "power diagram", "Laguerre diagram", "weighted Voronoi diagram", ...

Anyway, the point $x$ that we are finding is a vertex in the power diagram, where the weights on the $v_i$ are the $d_i$. It is the centre of the unique circle that intersects the circles $C(v_i,d_i)$ orthogonally. The caveat is that this 'circle' may have an imaginary radius.

In other words, if we take the three initial equations and add a third parameter, $h$, then the equations
$$
\|v_i - x \|^2 = d_i^2 + h,
$$
do admit a unique solution, where $h$ is the square of the radius of the orthogonal circle (some people call this the orthocircle, but its centre is not the usual orthocentre.)

I do not know whether this point is the same as you would find with Robert's optimization suggestion. Presumably this point can be characterized as an optimal solution of some functional.


问:
Prove $\sin(1/n)<1/n$ for all $n$
I need to prove $\sin(1/n)<1/n$ for all $n \in \Bbb N$ using mathematical induction.

Dont know how to start. Please help!

答:
Because of
\[ \sin(x)=x-\frac{x^3}{3!}+ \frac{x^5}{5!}-\frac{x^7}{7!} +\dots \]
As $x^n$ is monotone decreasing on $[0,1]$ in respect to $n$. Hence
\[ \sin(x)\leq x \] for $x$ in $[0,1]$.  As $\sin(x)\leq 1$ the inequality is even
true for all $x\in [0,\infty)$.

We can see that when $x-\sin(x)> 0$ holds for $x\in (0,1]$ this implies that $\sin\big(\frac{1}{n}\big) < \frac{1}{n}$ holds for all $n \in \mathbb{N}$.
At first we note that for $x=0$ $\sin(x)-x=0$. The derivative of
\[ x - \sin(x)\]
is $1-\cos(x)$. With the fundamental theoroem of calculus we know that
\[ x-\sin(x) = (1-\cos(\xi)) \cdot x \]
where $\xi \in (0,x)$. As $\cos (x) \leq 1$ and $x>0$ we have that $x-\sin(x)$ is positive and hence our inequality holds.

Title: Prove $\sin(1/n)<1/n$ for all $n$


答:
This is not going to be pretty, but it was not intended to so here goes:

To make it managable, let us define
$$
\begin{align}
a&:=5(x+h)+1\\
b&:=\sqrt{x+h}\\
c&:=5x+1\\
d&:=\sqrt x
\end{align}
$$
Then we have
$$
\begin{align}
f(x+h)-f(x)&=\frac a{2b}-\frac c{2d}\\
&=\frac{ad-cb}{2bd}\\
&=\frac{(ad-cb)(ad+cb)}{2bd(ad+cb)}\\
&=\frac{a^2d^2-c^2 b^2}{2bd(ad+cb)}
\end{align}
$$
where
$$
\begin{align}
a^2d^2-c^2 b^2&=(5h+(5x+1))^2x-(5x+1)^2(x+h)\\
&=5h(5h+2(5x+1))x-(5x+1)^2h
\end{align}
$$
or something like that (might need to double check). Hence you can divide the numerator by $h$ and both limits of numerator and denominator are non-zero and can be determined. Thus the limit of quotient will follow from that. Pretty tedious expressions!

Title: Find the derivative using the definition of derivative (limit).

Body:
Given $f(x)=\dfrac{5x+1}{2\sqrt{x}}$. Find $\dfrac{df(x)}{dx}=f'(x)$ using
the definition of derivative.

I have tried as below.

\begin{align*}
f'(x)&=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5(x+h)+1}{2\sqrt{x+h}}-\dfrac{5x+1}{2\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{\left(5(x+h)+1\right)\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x+h}-\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
\end{align*}

Now I can't find the limit. I confused how to simplify the limit. Anyone can give me  hint to solve it?

Note:

We were asked to find this derivative using
$$f'(x)=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$
instead of
$$f(x)=\dfrac{u(x)}{v(x)}\iff f'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}.$$


问:
(Multiplicative) inverse of $\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in\mathbb Q[\sqrt[3]{7}]$
Set $\mathbb Q[\sqrt[3]{7}] = \{F(\sqrt[3]{7}) \mid F ∈ Q[x]\}$ is a field (with the usual addition and the usual multiplication).

Calculate the (multiplicative) inverse of
$$\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in \mathbb Q[\sqrt[3]{7}]$$

Note: Application of the Euclidean algorithm to the polynomials $x^2 + 3x + 1$ and $x^3 - 7$ could help.

Attempt:

I know there is a multiplcative inverse $\beta$ with $\alpha\beta=1$, and that one should exist in $\mathbb Q[\sqrt[3]{7}]$, but do not know how I would go about expressing it in any form simpler than $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$.

How can I determine a simpler way to express the value of $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$?

答: Alternatively, $\alpha$ is a root of $x^3 - 3 x^2 - 60 x - 176$ and so its inverse is $\frac{1}{176}(\alpha^2-3\alpha-60)$.


问:
Conditions for a composite order to only have a single group
I just learned the theorem that for every prime number $p,$ there is just one group of order $p$ down to isomorphism (the cyclic group $\mathbb{Z}/p\mathbb{Z}$).

But interestingly, the converse of this statement is not true, as there is just a single group of order $15$.  This lead me to wonder: Is there a nice partial converse?

Consider a composite natural number $m \in \mathbb{N}$. I want to find the conditions for there to be multiple groups of order $m$. Since there is a cyclic group of every order, this is equivalent to finding a non-cyclic group of order $m.$  I have two conditions so far that imply multiple groups:

$m$ is even ($>2$)

Then $m=2k$ for some $k>1$. As the dihedral group $D_k$ has order $2k=m$ and is not cyclic, there are multiple groups of order m.

$p^2 \mid  m$ for some prime $p$

Then let $m = p^2k$ for some $k\in \mathbb{N}$. Then the group $(\mathbb{Z}/k\mathbb{Z}) \times (\mathbb{Z}/p^2\mathbb{Z})$ and the group $(\mathbb{Z}/k\mathbb{Z}) \times (\mathbb{Z}/p\mathbb{Z}) \times (\mathbb{Z}/p\mathbb{Z})$ both have order $p^2k = m$ and are not isomorphic.

But these two conditions definitely don't cover all cases. Is there a nice condition for a composite number m such that there are multiple groups of order $m$?

答: There is a characterization, but I wouldn't call it simple. If we define $f(n)$ to be the number of isomorphism classes of groups of order $n$, we see that $f(n) =1 $ if and only if $\gcd(n,\phi(n)) = 1$, where $\phi$ is the Euler totient function. More details about this can be found here (https://mathoverflow.net/questions/148731/for-which-n-is-there-only-one-group-of-order-n).


问:
permutation/combination problem
There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door and leave from another door?

I have done like this: They way of entering is 3 and exiting is also
  3, therefore the total way will be 3*3=9.

答: That’s the total number of ways in which he can enter and leave the room. The number of ways in which he can enter the room by one door and leave by another door, however, is only $3\cdot 2=6$: he has only $2$ choices of doors through which he may exit, since he must not leave by the door through which he entered.

Title: permutation/combination problem


问:
convergence of improper integral $\int_0 ^1 x(1-\log x)^{2a} dx$
I need to check whether the integral converges or not.

The integral is

$\int_0 ^1 x(1-\log x)^{2a} dx$

I see the integrand is continuous on the domain $(0,1)$, so I think if the limit at the endpoints exists then we can conclude that the integral converges, right??

The limit at the upper bound is zero, but the problem is to calculate the limit at the lower bound $x=0^+$.
I tried to calculate the limit using L'Hopital's rule, and get zero for any value of $a$, but not sure if this is true.

Any idea or suggestion!

答: Indeed, as $x \to 0$ you have that the function tends to zero. You can conclude that it is then bounded near $0$, so the integral $\int _0 ^\epsilon$ is well-defined and bounded by comparison for some $\epsilon > 0$. By your observations of continuity (well, bounded is enough) we clearly have that $\int_{\epsilon} ^1$ exists.


问:
Intersection of Sets and Probability
Is this equation True or False? $P(A ∪ B)$ = $P(A ∩ (B ∩ A^c$)).

I believe it to be False, but I am unsure on how to handle the two intersections. I think it is false because $P(A ∩ (B ∩ A^c$) is 0. I drew a Venn Diagram to see what $B∩A^c$ is. That intersected with $A$ is nothing. How does one tend to handle the two intersections with parentheses like that? If it was written as $A ∪ (B ∩ A^c)$ I know you can use Distribution Law. But how do we distribute intersection with another intersection?

答:
You've the right answer for the first prompt, but you've got the wrong reason.

I think it is false because $P(A \cap (B \cap A^\complement))$ is $0$. [You missed a closing parenthesis ')'.]

In fact, $P(A \cap (B \cap A^\complement))=0$ is a consequence of $P(\emptyset) = 0$, which is in the of the probability axioms, and $A \cap (B \cap A^\complement) = \emptyset$ is an observation from the Venn diagram to be proved.

As @angryavian points out, set intersections are commutative and associative, so you can easily prove that by some manipulation of the sets involved: you see $A$ and $A^\complement$ and you know that $A\cap A^\complement = \emptyset$, so when you rearrange the sets in "$A \cap (B \cap A^\complement)$", you try to rewrite it as "$A \cap (A^\complement \cap B)$" by applying the communtativity on $B \cap A^\complement$ first, and apply associativity to finish.

Title: Intersection of Sets and Probability


问:
How do you construct interval $(a, b)$ using just intervals of the form $(b, \infty)$?
My attempt is $(a, b) = (a, \infty) \setminus \cap_{n=1}^{\infty} (b - 1/n, \infty)$ with $n \in \mathbb{Z}^{+}$; $a,b \in \mathbb{R}$ and $a < b$. Is this correct?

答:
Yes, that works: the elements of $$\bigcap_{n\in\mathbb{N}} (b-{1\over n},\infty)$$ are exactly those real numbers which are $>b-{1\over n}$ for every $n\in\mathbb{N}$, which is to say $$\bigcap_{n\in\mathbb{N}} (b-{1\over n},\infty)=[b,\infty).$$ And $(a,\infty)\setminus [b,\infty)=(a,b)$ as desired.

The one subtlety here (which of course you avoided, but is worth mentioning for completeness) is that one needs to pay attention to the point $b$: e.g. $(a,\infty)\setminus (b,\infty)$ does not work since it gives $(a,b]$ instead.


问:
Relation between norm and variance of a random vector
I have been normalizing vectors for my work and there are generally two methods that I have been following. I assumed both the methods are equivalent until I found out they are not. The two methods are described below.

Take the square of the norm of the vector and divide this value by its length. To normalize, divide the vector by the square root of the above obtained value. This corresponds to the operation given below for any vector $\mathbf{x}$.
$$
E_{\mathbf{x}} = \frac{||\mathbf{x}||^2}{|x|}\hspace{2mm} \text{and}\hspace{2mm} \hat{\mathbf{x}} = \frac{\mathbf{x}}{\sqrt{E_{\mathbf{x}}}}
$$
$|\cdot|$ refers to dimension of the argument.

Use the variance function in the math library and divide the vector by the square root of its variance. Equivalent operations for vector $\textbf{x}$ are outlined below.
$$
\text{var}_{\textbf{x}} = \mathbb{E}(\textbf{x} - \mathbb{E}\textbf{x})^2 \hspace{2mm} \text{and}\hspace{2mm} \hat{\mathbf{x}} = \frac{\mathbf{x}}{\sqrt{\text{var}_{\mathbf{x}}}}
$$

The reason I believe both are equivalent is because in the norm case, we assume the origin to be at $\textbf{0}$ and we add the square of the distances from origin to each entry in the vector. In the variance case we move the origin to the mean of the random variable and then add the square of the distances taking the mean as origin.

Now I try to implement these two in python and following are the results.

```<code>import numpy as np
a = np.random.randn(1000)
np.linalg.norm(a) ** 2 / 1000
1.006560252222734
np.var(a)
1.003290114164144
</code>```

In these lines of code I generate 1000 length standard normal samples. Method 1 and method 2 give me equal values in this case. However when my samples have correlation, this is not the case.

```<code>import numpy as np
cov_matrix = 0.3*np.abs(1-np.eye(1000)) + np.eye(1000)
a = np.random.multivariate_normal(mean=np.zeros(1000), cov=cov_matrix)
np.linalg.norm(a) ** 2 / 1000
1.036685431728734
np.var(a)
0.6900743017090415
</code>```

I generate 1000 length multivariate normal random vector having covariance matrix with 1's along diagonals and 0.3 in all other off-diagonal entries. This is where I am confused. Method 1 and method 2 return different values.

Why is this the case? Why do both methods return same values in i.i.d. case and different values when the vector is correlated? Thanks.

答:
Your definitions are really vector-based operations implemented by matlab or Python (not the same as a probabilistic variance). So let me define them more clearly (I will use the matlab definitions, I assume Python definitions are the same).  You are dealing with $n$-dimensional random vectors $X=(X_1, ..., X_n)$.  Then:

Definition 1: $$E_X = \frac{\sum_{i=1}^n X_i^2}{n}$$

Definition 2: $$M_X = \frac{1}{n}\sum_{i=1}^n X_i$$

Definition 3: $$V_X = \frac{\sum_{i=1}^n (X_i-M_X)^2}{n-1}$$

Notice that $E_X, M_X, V_X$ are all random variables.

Observation 1:

If $\{X_i\}_{i=1}^{\infty}$ are i.i.d. with mean $E[X_1]$ and second moment $E[X_1^2]$ then by the law of large numbers
$$ \lim_{n\rightarrow\infty} E_X = E[X_1^2] \quad \mbox{(with prob 1)}$$
So in the special case when $\{X_i\}_{i=1}^{\infty}$ are i.i.d. Gaussian $N(0,1)$ then $E_X\rightarrow 1$ with prob 1.  Since $n=1000$ is "large" your numerical value $E_X=1.006560252222734$ makes sense. If you independently repeat the experiment you will get a new number for $E_X$ but, with high probability, it will still be very close to $1$. You would get similar results with $\{X_i\}_{i=1}^{\infty}$ any i.i.d. variables with $E[X_1^2]=1$, not necessarily having Gaussian distribution.

Observation 2:

In the special case when $\{X_i\}_{i=1}^{\infty}$ are i.i.d. Gaussian $N(0,1)$, then a surprising but classic result of statistics says that $(n-1)V_X$ has a  "chi-square" distribution with $n-1$ degrees of freedom.  In particular its mean is $n-1$ and its variance is $2(n-1)$.  So then $V_X = \frac{(n-1)V_X}{(n-1)}$ has mean $1$ and variance $\frac{2(n-1)}{(n-1)^2} = \frac{2}{(n-1)} \approx 0$ for large $n$.  So for $n=1000$ your numerical data $V_X = 1.003290114164144 \approx 1$ makes sense. If you independently repeat the experiment you will get a different numerical value for $V_X$ but, with high probability, it will still be very close to 1.

When you remove the i.i.d. assumption, so that you make the $\{X_i\}$ values correlated, then $V_X$ does not have the same distribution. That is why you get different numerical results in that case.


答: If you assume that a lecturer cannot exit from the door that he/she enters from then the correct answer is 3 * 2.</p>

Title: permutation/combination problem

Body:
There are 3 doors to a lecture room. In how many ways can a lecturer enter the room from one door and leave from another door?

I have done like this: They way of entering is 3 and exiting is also
  3, therefore the total way will be 3*3=9.


问:
For the following question show that locus is a circle.
Two circles touch one another internally at A, and a variable chord PQ of the outer circle touches the inner circle. Show that the locus of the incentres of the triangle APQ is another circle touching the given circles at A.

First I found the equation of the chord which is also the tangent to the smaller circle. Then I thought a lot on how to continue further;  but literally have no idea how to do it. Even if we use the formula of the incetre of a triangle then we have to find the length of the sides of the triangle which is damn tedious.
Please help me with shorter method.

答:
Let $H$, $K$ be the points where chords $AP$, $AQ$ intersect the inner circle (see figure). The homothety of centre $A$ carrying the outer circle to the inner circle also carries $P$ to $H$ and $Q$ to $K$, with
$$
{AP\over AH}={AQ\over AK}={R\over r},
$$
where $r$ and $R$ are the radii of inner and outer circle respectively.

Hence $HK$ is parallel to tangent $PQ$ and tangency point $T$ divides arc $HK$ into two equal parts.It follows that inscribed angles $\angle HAT$ and $\angle KAT$ are equal too and $AT$ is the bisector of $\angle PAQ$, so that incenter $I$ lies on $AT$.

Moreover, by the angle bisector theorem we have $IT:IA=PT:PA$, that is:
$$
{IT^2\over IA^2}={PT^2\over PA^2}={PH\cdot PA\over PA^2}=
{PH\over PA}={PA-HA\over PA}=1-{HA\over PA}=1-{r\over R}.
$$
Hence $AT/AI$ is a fixed ratio:
$$
{AT\over AI}={AI+IT\over AI}=1+{IT\over AI}=1+\sqrt{1-{r\over R}}
$$
and incenter $I$ is the image of $T$ under the homothety of centre $A$ and ratio given above. Hence its locus is a circle, the image of the inner circle under the same homothety.


问:
Basis for $\Lambda^k V^*$ the $k$-th exterior power of the dual
I have a quick algebra question :).

Say that $V$ is an $n-$dimensional vector space with basis $v_1,...,v_n$, and let $v_1^*, ..., v_n^*$ be the dual basis for $V^*$.

I am trying to prove that the $k-$th exterior power of $V^*$ : $\Lambda^k V^*$ is spanned by vectors of the form $v_{i_1}^*\wedge...\wedge v_{i_k}^*$ for $1 \leq i_1 < ... < i_k \leq n$. Where $\Lambda^k V^*$ is defined as the set of alternating $k-$forms $V^k \to \mathbb{R}$.

I looked up the proof on wikipedia and other sites, but they take it as a definition that $\Lambda^k V^*$ is the set of $\omega_1\wedge...\wedge\omega_n$ where the $\omega_i'$s are elements of $V^*$ (from which the claim follows easily).

However, I don't know how to get to that point. Is there a way to show that every $k$-form on $V$ can be expressed as the exterior product of $k$ $1-$forms? Or is there something easier that I am missing? I am a novice in the subject so any help would be appreciated.

Note: the definition of exterior product I use is that for $\alpha \in \Lambda^k V^*$, $\beta \in \Lambda^p V^*$, then:

$\alpha\wedge\beta(x_1,...,x_{k + p}) := \displaystyle\sum \text{sign}(\sigma) \alpha(x_{\sigma(1)},..., x_{\sigma(k)})\beta(x_{\sigma(k+1)}, ..., x_{\sigma(k+p)})$.

Thank you for your help :)

答:
The set of alternating $k$-forms is a vector space. Let's figure out its dimension.  If two alternating forms agree on all tuples of $(v_{i_1}, \ldots... v_{i_k})$ with $1 \leq i_1 < ... < i_k \leq n$, then, by the alternating condition, they agree on all tuples of distinct basis vectors (not necessarily ordered), and, since they both vanish on tuples with repeated vectors, they agree on all tuples of basis vectors, and hence, by multilinearity, on all tuples of vectors -- and so they are equal. This shows that the linear map sending the form to its values on all the tuples $(v_{i_1}, \ldots... v_{i_k})$ with $1 \leq i_1 < ... < i_k \leq n$  -- of which there are $\binom{n}{k}$ --  is  injective, so the dimension of the space of alternating forms is at most $\binom{n}{k}$.

On the other hand, the $\binom{n}{k}$  forms $v_{i_1}^*\wedge...\wedge v_{i_k}^*$ with $1 \leq i_1 < ... < i_k \leq n$ are alternating and linearly independent (each of them sends the ordered tuple $(v_{i_1},..., v_{i_k})$ with same indexes to $1$, all others to $0$, so a linear combination can only send everything to $0$ if all the coefficients are zero). This means that the dimension is exactly $\binom{n}{k}$ -- and the linearly independent set of the $v_{i_1}^*\wedge...\wedge v_{i_k}^*$s is actually a basis. QED.


问:
Isomorphicity of fields.
Given two irreducible polynomials $f$ and $g$.

What are some conditions under which we can conclude that the corresponding number fields are non-isomorphic?

For instance if $\deg(f) \neq \deg (g)$, then the corresponding number fields are non-isomorphic. For instance, if $\operatorname{disc}(f)$ and $\operatorname{disc}(g)$ are not in the same square class $\dfrac{\Bbb Q ^*}{\Bbb Q ^{*2}}$, then the corresponding number fields are non-isomorphic.

How can we realize that the number field corresponding to $x^5-pq=0$, is not isomorphic to the number field corresponding to $x^5-pq^2=0$?

答: There are many methods to distinguish two different fields using both more or less sophisticated methods by looking at invariants of the field  (degree, discriminant, local ramification, signature, etc.) For the particular case of $x^5 - pq$ and $x^5 - pq^2$, note that if the fields were the same there would be $\alpha, \beta \in K$ such that $\alpha^5 = pq$ and $\beta^5 = pq^2$, and so $\gamma = \beta/\alpha$ would satisfy $\gamma^5 = q$, and  $\delta = \alpha^2/\beta$ would satisfy $\delta^5 = p$. By degree considerations, this implies that the field $K$ is equal to $\mathbf{Q}(\gamma)$ and $\mathbf{Q}(\delta)$. The easiest way to distinguish these is by using local ramification at $p$ and $q$, but you can also use linear algebra...

Title: Isomorphicity of fields.


问:
How to compute signed angle between $n$-dimensional vectors?
I am looking for clarification on this question. The accepted answer says that measuring a signed distance is impossible because the plane spanned by two vectors $u$ and $v$ does not have a canonical orientation. But what does orientation mean here? Is it just picking a particular vector in the orthocomplement of $\text{span}\{u,v\}$ or must ond pick several such vectors?

I imagine that an orientation for the plane can be picked arbitrarily for any two vectors. Once it is defined, how would one go about computing the signed angle between them?

答:
Put aside $n$-dimensionality for a moment.  Let's look at a plane spanned by the vectors $a$ and $b$.  There is an unsigned angle between $a$ and $b$.  To compute the signed angle between $a$ and $b$, we must decide (for instance) which way is the direction of increasing angle and which is decreasing angle.

When presented with the usual coordinate plane, oriented in the usual way, the direction of increasing angles is anticlockwise and the direction of decreasing angles is clockwise.  Notice that to make sense of that claim, we have to know which side of the plane we look at it from.  If you are looking at the plane from the "wrong" side, we must swap "anticlockwise" and "clockwise" to get the correct directions.

So, to assign signed angles to vectors on a plane, we must know from which side to view the plane so that anticlockwise angles are positive.  So there are only two cases to resolve.

Let's temporarily take things down one dimension to see how much information we must specify to resolve these two cases.  We have a line in 3-space.  We want to know which direction on the line is positive and which negative.  At a distinguished point on the line (its origin), there are infinitely many (unit) vectors perpendicular to the line.  No one of these is sufficient to resolve the orientation of the line.  (To see this, treat the space as a metric space then rotate the entire space, with the line embedded in it, around the perpendicular vector by $\pi$ radians.  The orientation of the line has reversed, but nothing else has changed.)  Our experience with vectors in 3-space tells us that a linearly independent pair of vectors in the orthocomplement of the line is sufficient to pick an orientation by the "right-hand rule" once we assign an order to the list of two vectors.

The same thing happens for your plane in $n$-space.  You need to select $n-2$ linearly independent vectors, $V$, from the orthocomplement of your plane.  Then specify an order on the list of $n$ vectors $V \cup \{u,v\}$.  This ordered list will produce a positively or negatively oriented coordinate system and the plane will inherit this orientation in the ordered basis of $u$ and $v$ in whatever relative order they appear in the $n$-vector list.  Of course, if you swap $u$ and $v$ in that list the signs of both orientations (of the $n$-space and of the plane) simultaneously reverse, meaning positive angles remain positive and negative angles remain negative.  The freedom of choice is present in the $n-2$ vectors and their order in the list since exchanging any pair of these reverses the orientation on the space but does not change the order of $\{u,v\}$ in the plane, so reverses the signs of all signed angles in the plane.


问:
Let $a >0$. Find the area bounded by $x+y=a$ and $\sqrt{x}+\sqrt{y} = \sqrt{a}$.

Let $a >0$. Find the area bounded by $x+y=a$ and $\sqrt{x}+\sqrt{y} = \sqrt{a}$.

Graphing this it's a line $x+y=a$ on top of the curve $\sqrt{x}+\sqrt{y} = \sqrt{a}$. So solving for $y$ in both equations I have that $y= a-x$ and $y= (\sqrt{a}-\sqrt{x})^2$

So the integral becomes $$\int _0^a\int _{\left(\sqrt{a}+\sqrt{x}\right)^2}^{a-x}1 \ dy \ dx = \int _0^a\left(-2\sqrt{a}\sqrt{x}-2x\right) \ dx = -\frac{4}{3}\sqrt{a}a^{\frac{3}{2}}-a^2$$

but this seems to have a negative area, what might be happening here?

答:
Note that the area is
$$\int _0^a\int _{\left(\sqrt{a}-\sqrt{x}\right)^2}^{a-x}1 \ dy dx = 2\int _0^a\left(\sqrt{ax}-x\right)dx =\frac{a^2}{3}$$


问:
The diagonals of cyclic $ABCD$ meet at $P$, and $W$, $X$, $Y$, $Z$ are feet of perpendiculars from $P$ to the sides. Show $WX+YZ=XY+WZ$.

Let $ABCD$ be a cyclic quadrilateral whose diagonals meet at $P$. Let $W$, $X$, $Y$, $Z$ be the feet of $P$ onto $AB$, $BC$, $CD$, and $DA$ respectively. Show that $WX+YZ=XY+WZ$

I tried a lot but in vain. I literally have no idea how to begin and approach the problem. Probably, properties of circles and triangles will be used but I don't know from which to start with.
Please help.

答:
Diagram

As, $AZPW$, $PWBX$, $XPYC$, and $YPZD$ all are cyclic,
$\measuredangle PAZ = \measuredangle  PWZ = \measuredangle  PWX = \measuredangle PBX$.

Hence, $P$ lies on the angle bisector of the $\angle XWZ$.
Similarly, we can prove $P$ lies on the angle bisector of $\angle WXY, \angle ZYX, \angle YXW$.

Hence, $P$ is the incenter of $WXYZ$.

From Pitot's Theorem, we can conclude $WX + YZ = WZ + XY$. QED

This is my first answer in Math-StackExchange. Please forgive my technical mistakes :)


问:
Proper vector notation
I wrote a vector as $(a, b)$ in a maths exam but I was told it is not vector notation. However, I have seen this notation used all over this site and elsewhere. Is it correct or not?

答: I’ve also been told by my math teachers not to write vectors using the $(a, b)$ notation in order to avoid confusion, since that notation is often used to represent a single point on an $xy$-coordinate plane. Another answerer mentioned that it seems like something that varies by country/region, which makes sense—in the USA (the part that I’m from, at least) I was taught to write vectors as either $$ or $a\vec x + b\vec y$.

Title: Proper vector notation

Body: I’ve also been told by my math teachers not to write vectors using the $(a, b)$ notation in order to avoid confusion, since that notation is often used to represent a single point on an $xy$-coordinate plane. Another answerer mentioned that it seems like something that varies by country/region, which makes sense—in the USA (the part that I’m from, at least) I was taught to write vectors as either $<a, b>$ or $a\vec x + b\vec y$.


Title: Show that $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{2^{n}-1}>\frac{n}{2}$

Body:
Show that $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{n}-1}>\frac{n}{2}.$$

I used the fact that for all $k \ge 1:k <2^k$
$$\sum_{k=1}^{2^{n}-1}\frac{1}{k}>\sum_{k=1}^{2^{n}-1}\frac{1}{2^{k}}=\frac{1}{2}\left(\frac{1-\frac{1}{2^{2^{n}-1}}}{1-\frac{1}{2}}\right)=1-\frac{1}{2^{2^{n}-1}}.$$

But I don't know how to continue.

Body:
Hint

If $k\in\mathbb N$,

$$\frac{1}{2^k+1}+\frac{1}{2^k+1}+\frac{1}{2^k+2}+\cdots+ \frac{1}{2^{k+1}}>\frac{2^k}{2^{k+1}}=\frac{1}{2}.$$


问:
How to prove a differential entropy is not scale invariant?
For example, $S(X)=-E_X(\log(f_X))=-\int_{-\infty}^{+\infty}f_X(x)\log(f_X(x))dx$
A transformation of X changes the result:$S(aX)=S(X)+\log|a|$ and more in general $S(g(X))=S(X)+\int_{-\infty}^{+\infty}f_X(x)\log(|\frac{\partial g}{\partial x}|)dx$.

答: I believe the task is to show the general condition that differential entropy is not scale invariant. That the differential entropy of a random variable X is not the same as the differential entropy some function of the same random variable X. To prove the final line of the question.</p>


问:
Sum of square root of primes
I was playing around with prime numbers and a question came into my mind:

Let $S(n)$ denote the sum of square roots of primes from $2$ to the $n$th prime number.
Are there infinitely many numbers $n$ so that $\left\lfloor S(n) \right\rfloor$ is prime itself? (Where $\left\lfloor X \right\rfloor$ denotes the floor function.)
Please tell me if you had any ideas about it. I actually could do nothing. xD

答: If this problem is solvable, it would require advanced tools from analytic number theory. As for a heuristic: $S(n)$ is asymptotically $\sum_{k=1}^n \sqrt{k\log k} \sim \frac23n^{3/2}\sqrt{\log n}$; and I don't see any reason why $\lfloor S(n)\rfloor$ is more or less likely to be even, a multiple of $3$, or more generally divisible by any fixed prime. (This independence is borne out by numerical experiments.) Thus the prediction would be that $\lfloor S(n)\rfloor$ is just as likely to be prime as a random integer of the same size, which is about $1/\log S(n) \sim 2/(3\log n)$. In other words, I would predict that the number of $n\le x$ for which $\lfloor S(n)\rfloor$ is prime should be asymptotically $\frac23x/\log x$. (Numerical experiments also make this appear more likely than a constant times $x/\log^2x$.)</p>

Title: Sum of square root of primes


问:
$G$ is an abelian group and $x=a_1a_2...a_n$
Assume that $G$ is an abelian group of order $n$ and $a_1,a_2,...,a_n$ are its elements and we have $x=a_1a_2...a_n$.

If $G$ has more than one element such as $b\neq e$ and $b^2=e$, prove that $x=e$.

Answer:

Assume that $a\neq b$, $a^2=e$, $b^2=e$, $a\neq e$ and $b\neq e$.
$$(ab)^2=(ab)(ab)=a^2b^2=e$$
If $$ab=e\implies a=b^{-1}\implies a=b$$
But we assume that $a\neq b$. contradiction!

So, $ab \neq e$ and $(ab)^2=e$.

So, what should I do now?

STUCK!

答:
Note that the elements $x \in G$ such that $x^2 = e$ are a subgroup, call it $H$.

Every element in this group has order $2$. By the classification of finite abelian groups, we must have $H \cong (\mathbb{Z}/2 \mathbb{Z})^k$ for some $k$. The statement is clearly true for $H$ (the product of all elements is the identity).

Now consider the elements in $G \setminus H$. We can pair each element with their inverse to get product $e$, since no element is its own inverse.

Putting these two pieces together, you get that the product of all the elements is the identity.


问:
How many elements will be there in a group so that it has $4$ subgroups of order $3$?
I m thinking that a group of order $3$ will be of the form $\langle a \rangle=\{1,a,a^2=a^{-1} \}$, $ a^3=e$ therefore $a^2=a^{-1}$.
Therefore $\langle a \rangle=\{1,a,a^{-1} \}$.

Then there will be $3$ cyclic groups different then "$a$" ie "$b$", "$c$", "$d$" with no element same as any element of "$a$". So each group will have $2$ elements of order $3$. in total $4\cdot 2$ elements. Is this reasoning correct that all subgroups should be distinct?

答:
Generalizing this a bit: from the proof (1959) of James McKay of Cauchy's Theorem (the existence of an element of prime power order $p$, if $p$ divides the order of $G$) it follows that $$\#\{g \in G: o(g)=p\} \equiv -1 \text{ mod }p$$

If there are $l$ subgroups of order $p$, then $l=1$ or any pair of different subgroups of order $p$ have trivial intersection (apply Lagrange's Theorem on the intersection, being a subgroup of each of the subgroups of order $p$).
Hence, leaving out the identity element, there are $l(p-1)$ elements of order $p$. Combining this with the above formula yields $$l \equiv 1 \text{ mod } p$$

So in your case taking $p=3$, the minimal number of subgroups of order $3$ in a group is $4$, the next possibility would be $7$, $10$, ... etc.. As Hagen pointed out $G=C_3 \times C_3$ is the minimal example, and you can produce very large examples by looking at $G \times S$, where $S$ is a group with $3 \nmid |S|$.


问:
Find the derivative using the definition of derivative (limit).

Given $f(x)=\dfrac{5x+1}{2\sqrt{x}}$. Find $\dfrac{df(x)}{dx}=f'(x)$ using
the definition of derivative.

I have tried as below.

\begin{align*}
f'(x)&=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5(x+h)+1}{2\sqrt{x+h}}-\dfrac{5x+1}{2\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{\left(5(x+h)+1\right)\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
  &= \lim\limits_{h\to 0}
  \dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x+h}-\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
\end{align*}

Now I can't find the limit. I confused how to simplify the limit. Anyone can give me  hint to solve it?

Note:

We were asked to find this derivative using
$$f'(x)=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$
instead of
$$f(x)=\dfrac{u(x)}{v(x)}\iff f'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}.$$

答:
First, note that we don't need to worry about the derivative of $f$ when $x=0$, because your function isn't defined when $x=0$.

Using the Binomial expansion, as $h \to 0,$ which garuntees that $|\frac{h}{x}| < 1,$

$ \sqrt{x+h} = \sqrt{x} \sqrt{1 + \frac{h}{x}} = \sqrt{x} \left(1 + \frac{h}{x}\right)^\frac12 = \sqrt{x}\left(1 + \frac{\left(\frac12\right)}{1!\ } \left(\frac{h}{x}\right) + \frac{\left(\frac12\right)\left(\frac{-1}{2}\right) }{2!\ } \left(\frac{h}{x}\right)^2 + ...\right),$

i.e., $\ \sqrt{x+h} = \sqrt{x} + \frac{h}{2\ \sqrt{x} } + O(h^2).$

Substitute this into the last line of working in the question (but not the denominator), we get:

\begin{align*}
f'(x)&= \lim\limits_{h\to 0}
  \left(\dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\right)\\
&= \lim\limits_{h\to 0}
  \left(\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-(5x+1)\left(\sqrt{x}+ \frac{h}{2\ \sqrt{x} } + O(h^2)\right)}{2h\sqrt{x+h}\sqrt{x}}\right)\\
&= \lim\limits_{h\to 0}
  \left(\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x}-\frac52 h \sqrt{x} - \sqrt{x} - \frac{h}{2\ \sqrt{x} } }{2h\sqrt{x+h}\sqrt{x}} + O(\sqrt{h})\right).\\
\end{align*}

... and after some cancellation you will arrive at the answer.


答:
$\displaystyle\sum{k=1}^{2^n-1}\frac{1}{2^{k}}$ is an upper Riemann for the integral $\displaystyle\int_1^{2^n}\frac{\mathrm dx}x=\log(2^n)=n\log 2.$

Explain why $\:\log 2=\displaystyle\int_1^2\frac{\mathrm dx}x>\frac12$.


问:
A sequence of analytic functions with converging path integrals converges how?
Suppose you have a sequence $\{f_n\}_{n \in \mathbb Z^+}$ of analytic functions on a compact set $K$ such that for every smooth curve $\gamma$ in $K$ the sequence $\{\int_\gamma f_n\}_{n \in \mathbb Z^+}$ of path integrals over $\gamma$ converges. What can you say about the convergence of $f_n$?

Shouldn't the convergence should be uniform? Is the fact that $K$ is locally convex enough to allow you to define a primitive locally via a path integral and prove local uniform convergence?

答:
(First of all, analyticity is defined on open sets; I presume that functions are analytic in some domain $\Omega$ and we are to consider their behavior on a compact subset $K\subset\Omega$.)

The answer is negative in general. Take a sequence of holomorphic functions $F_n$ that converges pointwise to a non-holomorphic function (this can happen). Let $f_n=F_n'$. Then for any curve $\gamma$ we have $\int_\gamma f_n = F_n(b)-F_n(a)$ where $a,b$ are the endpoints of $\gamma$. Thus, $\int_\gamma f_n$ has a limit. However, we do not have locally uniform convergence of $f_n$.

Indeed, suppose that $f_n$ converge locally uniformly. Then so do $F_n$, because $F_n(z_0)$ converge at some fixed point $z_0$, and $F_n(z)-F_n(z_0)$ is found by integrating $f_n$. It follows that the limit $\lim F_n$ is holomorphic, contradicting our choice of $F_n$.


问:
Testing for similarity in two populations
The table shows the half-lives (in days) of methylmercury (CH203
3 ) the systems of six women and nine men.

Females: 56 , 45 , 76 , 97 , 43

Males: 85 , 56, 35 , 56 , 34 , 65 , 34 , 67 , 94

Test at the alpha = 0:05 if men and women metabolize CH203
3 at the same rate using two different suitable tests.

The two tests I have been thinking are testing the variance ratio using F distribution and two sample T test. For the two sample t test I am not sure if I should use pooled or non pooled data.

Also are there any other methods ?

答:
Here is a strip chart made in R, showing your data for women (bottom) and men.

```<code>f = c(56, 45, 76, 97, 43)
m = c(85, 56, 35, 56, 34, 65, 34, 67, 94)
stripchart(list(f,m), ylim=c(.5,2.5), meth="stack")
</code>```

The two samples do not seem to differ in location or spread.
It seems unlikely that any test is going to show a significant
difference for these data. Below we show results for
a Welch 2-sample t test and a nonparametric two-sample Wilcoxon test.

Welch t. A Welch two-sample t test shows a large P-value confirming
that there is no significant difference between sample means
63.4 for women and 58.4 for men.

```<code>t.test(f,m)

        Welch Two Sample t-test

data:  f and m
t = 0.39396, df = 8.0631, p-value = 0.7038
alternative hypothesis:
  true difference in means is not equal to 0
95 percent confidence interval:
 -24.01156  33.92267
sample estimates:
mean of x mean of y
 63.40000  58.44444
</code>```

More generally, if there is any doubt that the men and women
have the same population variances for such tests, you should
use the Welch two-sample test, which does not assume equal
variances.

It is not good practice to use an F-test to see whether variances differ significantly,
and then to use that result to 'branch' to either the Welch t test (if different) and the pooled t test (if not).

Wilcoxon rank sum test. From such small samples it is difficult to say whether populations are normal. The plots show no marked skewness or far outliers,
which might indicate non-normal populations. If prior experience
with such assays have shown non-normal data, you might want to
do a two-sample Wilcoxon (rank sum) test. This test does not
assume that data are normal but does assume that the two population
distributions are of the same shape (including roughly equal
variability) and are continuous (no ties).

For your data, a two-sample Wilcoxon test finds no significant
difference in location, but there is a warning message
that the P-value may not be exactly correct, on account of ties (obvious in the plots).

```<code>wilcox.test(f,m)

        Wilcoxon rank sum test with continuity correction

data:  f and m
W = 26, p-value = 0.6875
alternative hypothesis: true location shift is not equal to 0

Warning message:
In wilcox.test.default(f, m) :
  cannot compute exact p-value with ties
</code>```

Note: In order to make sure you get practice with various kinds of two-sample tests, it may be OK for
a textbook exercise to ask you to do two tests. However, in statistical practice, it is not OK to do multiple
tests and then to report the one that gives the smallest
P-value. Good practice is to choose the one test
that best fits the situation and to report its result.

Body:
Here is a strip chart made in R, showing your data for women (bottom) and men.

f = c(56, 45, 76, 97, 43)
m = c(85, 56, 35, 56, 34, 65, 34, 67, 94)
stripchart(list(f,m), ylim=c(.5,2.5), meth="stack")

The two samples do not seem to differ in location or spread.
It seems unlikely that any test is going to show a significant
difference for these data. Below we show results for
a Welch 2-sample t test and a nonparametric two-sample Wilcoxon test.

Welch t. A Welch two-sample t test shows a large P-value confirming
that there is no significant difference between sample means
63.4 for women and 58.4 for men.

t.test(f,m)

        Welch Two Sample t-test

data:  f and m
t = 0.39396, df = 8.0631, p-value = 0.7038
alternative hypothesis:
  true difference in means is not equal to 0
95 percent confidence interval:
 -24.01156  33.92267
sample estimates:
mean of x mean of y
 63.40000  58.44444

More generally, if there is any doubt that the men and women
have the same population variances for such tests, you should
use the Welch two-sample test, which does not assume equal
variances.

It is not good practice to use an F-test to see whether variances differ significantly,
and then to use that result to 'branch' to either the Welch t test (if different) and the pooled t test (if not).

Wilcoxon rank sum test. From such small samples it is difficult to say whether populations are normal. The plots show no marked skewness or far outliers,
which might indicate non-normal populations. If prior experience
with such assays have shown non-normal data, you might want to
do a two-sample Wilcoxon (rank sum) test. This test does not
assume that data are normal but does assume that the two population
distributions are of the same shape (including roughly equal
variability) and are continuous (no ties).

For your data, a two-sample Wilcoxon test finds no significant
difference in location, but there is a warning message
that the P-value may not be exactly correct, on account of ties (obvious in the plots).

wilcox.test(f,m)

        Wilcoxon rank sum test with continuity correction

data:  f and m
W = 26, p-value = 0.6875
alternative hypothesis: true location shift is not equal to 0

Warning message:
In wilcox.test.default(f, m) :
  cannot compute exact p-value with ties

Note: In order to make sure you get practice with various kinds of two-sample tests, it may be OK for
a textbook exercise to ask you to do two tests. However, in statistical practice, it is not OK to do multiple
tests and then to report the one that gives the smallest
P-value. Good practice is to choose the one test
that best fits the situation and to report its result.


问:
Prove that the triangle $ABC$ is a right triangle
Prove that the triangle with points $$A(a,b), B(c,d), C\left(h, d-\left(\frac{c-a}{d-b}\right)h+\left(\frac{c-a}{d-b}\right)c\right)$$

its sides comply with the Pythagorean theorem

I have not been able to show it,
some clue?

答:
We know that $A(a,b)$ and $B(c,d)$ do not form the hypotenuse as $C$ already has the x-coordinate $h$.

Assuming $AC$ to be the hypotenuse, products of slopes of $AB$ and $BC$ should be $-1$ for our assumption to be true.

Slope of $BC$ = $\frac{\frac{c-a}{d-b}\cdot{h-c}}{c-h}$

Slope of $AB$ = $\frac{d-b}{c-a}$

Slope of $AB$ $\cdot$ Slope of $BC$ = $-1$

Hence our assumption was true, and $ABC$ forms a right angled triangle with hypotenuse $AC$.


问:
Partition 100 people, 4 from each country into 4 groups with conditions
This is a problem from the $2005$ All-Russian Olympiad. Problem is as follows:

$100$ people from $25$ countries, four from each country, sit in a circle.
Prove that one may partition them onto $4$ groups in such way that no
two countrymen, nor two neighboring people in the circle, are in the
same group.

My approach was to try and find an algorithm that can monotonically reduce the number of bad states. However, I haven't been able to figure out a way to do this for every possibility. Does anyone have any hints on how to approach this problem?

I've tried to start with an arbitrary partition where each partition only has 1 person from each country, and preserve that with each move.

I also claim that there is always a country for which not all of its representatives' neighbors are in a single partition. Proof idea: Suppose not. WLOG, say that country $1$ has all of its representatives' neighbors in the first partition. So if $1_i$ is the representative from country $1$ in partition $i$, we have $1_1$ and its neighbor, WLOG from country $2$, is also in group $1$. So we have $1_1, 2_1$. Then $2_1$'s other neighbor must also be in partition $1$ or we are done. Continuing in this way, we find that all the representatives are in partition $1$, contradiction.

Using this, I try to decrease the number of "bad" pairs (i.e. neighbors or same country) in each partition monotonically. However, I haven't been able to figure out how to do so.

In particular I'm stuck on cases like the following: country $1$ has its representatives in partitions $1-4$ s/t the neighbors of $1_2,1_3,1_4$ are all in the first partition. $1_1$'s neighbors are in partition $1$ and another partition, say $2$. Trying to swap the placements of country $1$ representatives seems to only increase number of "bad" pairs, and trying to swap the neighbors may also increase the number of bad pairs.

Is there an observation that I'm missing here, or is this approach not the right way to go about it?

答:
Here's a two-step procedure partially borrowed from this question to find a partition. (Actually, maybe it is easier to read the other way around: my explanation here feels better than the one I posted there, so people confused about my answer to that question should read this answer instead.)

Number the people around the circle $1, 2, \dots, 100$.

Step 1: let's say that each country has two men and two women representing it. Obviously it is out of our hands where the men and women will sit.

Step 2: we give everyone a color that is either Red or Blue. We will make sure that from each country, the two men get different colors, and the two women get different colors. We will also make sure that any two people numbered $2k-1$ and $2k$ for some $k$ get different colors.

Graph-theoretically, this step just means two-coloring the union of even cycles, which can always be done. Each person has two constraints: they must be different from one of the people next to them, and they must be different from the person of their country of the same gender. So the graph of constraints is $2$-regular - a union of cycles. The cycles are all even because the constraints alternate between two different types, so each cycle has an even number of edges.

Step 3: we give everyone a shade that is either Light or Dark. We will make sure that from each country, the two Reds get different shades, and the two Blues get different shades. We will also make sure that any two people numbered $2k$ and $2k+1$ for some $k$ (or numbered $100$ and $1$) get different colors.

Graph-theoretically, this problem is identical to Step 2.

Now, the partition into four groups of $25$ is as follows: the Light Red, Dark Red, Light Blue, and Dark Blue groups. By the constraints we enforced in Step 2 and Step 3, each country has one of each color, and any two people sitting next to each other disagree either in color or in shade.


问:
For $n\geq 3$, does every horn $\Lambda^n_k\to N\mathcal C$ have a filler?
Let $N\mathcal C$ be the nerve of category. It's true that for $0<k<n$, every horn $\Lambda^n_k\to N\mathcal C$ can be extended to $\Delta^n$ in a unique way.

What about $k = 0$ or $n$? For $n \leq 2$, this certainly can't always be done (if $n=1$ this would imply $\mathcal C$ is the terminal category, and if $n=2$ that $\mathcal C$ is a groupoid).

However, for $n=3$, it seems that the horns can be filled even when $k = 0$ or $3$!
For instance, a horn $\Lambda^3_0\to N\mathcal C$ would correspond to three triangles $\sigma_1,\sigma_2,\sigma_3\in N\mathcal C_2$ fitting in the picture below:

The tetrahedron is already filled with $\sigma_0 := (d_0\sigma_3,d_0\sigma_1,d_0\sigma_2)$ as the $0$-th face.

Am I making any mistake? Does this pattern goes on for $n\geq 4$ i.e. in this case can every horn be filled?

答: You can fill in the three boundary edges of the triangle $\sigma_0$.  However, to fill in $\sigma_0$ itself (as well as the interior of the tetrahedron), you need to know that these three edges form a commutative diagram in your category.  This is not necessarily true.  For instance, if the $0$th object of the diagram is the initial object of your category, then you could put any three maps at all on the edges of $\sigma_0$, and the other three triangles will automatically commute, so you would be asserting that every triangular diagram in your category commutes!


问:
True/False: If $\sum_{n=1}^\infty a_{2n-1}+a_{2n}=0$ then $\sum_{n=1}^\infty a_n=0$
Prove or disprove: If $\sum_{n=1}^\infty a_{2n-1}+a_{2n}=0$ then $\sum_{n=1}^\infty a_n=0$

What I am seeing from this series is that it is the sequence of odd index elements added to the even index elements. This makes me think that it is true because the series are equal but I don't know how I would prove rigorously that this is true.

答: Try $a_n = \begin{cases}-1 \text{ if $n$ is odd} \\ 1 \text{ if $n$ is even}\end{cases}$.  Then $a_{2n-1} + a_{2n} = (-1) + 1 = 0$, so that $\sum\limits_{n = 1}^\infty (a_{2n-1} + a_{2n}) = \lim\limits_{N \to +\infty}\sum\limits_{n = 0}^N(a_{2n-1} + a_{2n}) = \lim\limits_{N \to +\infty} 0 = 0$, but the partial sum $\sum\limits_{n = 1}^N a_n = (-1) + 1 + (-1) + 1 + \dots = \begin{cases}-1 \text{ if $n$ is odd} \\ 0 \text{ if $n$ is even}\end{cases}$, so that $\sum_{n=1}^\infty a_n$ doesn't even exist.


问:
Prove that any linear operator on a finite dimensional inner product space , there exists a unique linear operator
Prove that for every linear operator $T$ on a finite dimensional inner product space $V$, there exists a unique linear operator $T^*$ on $V$ such that
$\langle T\alpha,\beta\rangle=\langle\alpha,T^*\beta\rangle$ for every $\alpha$ and $\beta$ in $V$.

Thanks for your time!O(∩_∩)O~

答:
This is how I've seen this proved before.
Let $y \in V$. Define a function $h_y: V \rightarrow F$ by $$h_y(x) = \langle T(x),y \rangle.$$ Then for $x_1, x_2 \in V$ and $a \in F$, we have
\begin{align}
h_y(ax_1 + x_2) &= \langle T(ax_1 + x_2), y \rangle \\
&= a \langle T(x_1), y \rangle + \langle T(x_2),y \rangle \\
&= a h_y(x_1) + h_y(x_2),
\end{align}
so our function $h_y$ is linear for all $y$. There's a theorem (6.8 from Friedberg, Insel and Spence Linear Algebra) that tells us there exists a unique vector $y' \in V$ such that
$$\langle T(x), y \rangle = \langle x, y' \rangle \; \forall x \in V.$$
Now define $T^*: V \rightarrow V$ by $T^*(y) = y'$, and we have
$$\langle T(x), y \rangle = \langle x, T^*(y) \rangle \; \forall x,y \in V.$$


问:
Solve $\int\frac{M\cos^2u-(M-1)\sec^2u}{\sqrt{1-M\sin^2u}}\,du$ without differentiating the solution
In this answer I said that "it is left as an exercise to show that"
$$\int\frac{M\cos^2u-(M-1)\sec^2u}{\sqrt{1-M\sin^2u}}\,du=(\tan u)\sqrt{1-M\sin^2u}+C\tag1$$
The original question asked for the derivation of a solution to an integral; the solution involves elliptic integrals. Because of the latter, I would ordinarily have the full power of elliptic integrals/functions at my disposal (i.e. Byrd and Friedman), but the OP said at the start

Yet again I might say I am very very rusty on my mathematics, but in this case I've never even heard of elliptic integrals up to the point I saw this paper.

And I did not want to scare the OP away by saying "oh, this is B&F XXX.XX, we have this and that". I wanted to be gentle.

However, I could not think of an easy way to justify the integral at the start of this question other than differentiating the solution I already knew. How could I solve $(1)$ from scratch in an elementary manner?

答: The integrand's numerator is$$\begin{align}M\cos^2u-(M-1)\sec^2u&=M-M\sin^2u+\sec^2u-M\sec^2u\\&=-M\sin^2u+\sec^2u-M\tan^2u\\&=-M\sin^2u+\sec^2u\cdot(1-M\sin^2u),\end{align}$$so the integrand is$$\frac{-M\sin u\cos u}{\sqrt{1-M\sin^2u}}\tan u+\sqrt{1-M\sin^2u}\sec^2u.$$Finish with the product rule.


问:
Standard free resolution of a Hopf algebra : ?$\exists$ an explicit chain homotopy?
Consider a Hopf algebra $H$ over $\mathbb C$ with $\varepsilon \colon H\to \mathbb C$ being its counit map. I am trying to figure out why the standard resolution of $\mathbb C$, seen as a right $H$-module via the counit map, is indeed a resolution. To my knowledge, this is well-known, from the book of Cartan and Eilenberg, where the standard resolution of an augmented algebra is considered. To be more precise, consider the chain complex

$$\cdots \longrightarrow H^{\otimes n+1} \longrightarrow H^{\otimes n}\longrightarrow\dots\longrightarrow H \otimes H\longrightarrow H \longrightarrow 0,$$ with differentials

\begin{align*}
 H^{\otimes n+1}  &\xrightarrow{d_{n+1}}  H^{\otimes n} \\
h_1\otimes \cdots \otimes h_{n+1} &\mapsto \varepsilon(h_1)h_2\otimes \cdots \otimes h_{n+1} + \sum_{i=1}^n(-1)^{i}h_1\otimes \cdots \otimes h_ih_{i+1} \otimes \cdots \otimes h_{n+1}.
\end{align*}

By $\otimes$, I mean $\otimes_{\mathbb C}$.
The augmented complex

$$\cdots \longrightarrow H^{\otimes n+1} \longrightarrow H^{\otimes n}\longrightarrow\dots\longrightarrow H \otimes H\longrightarrow H \longrightarrow \mathbb C \xrightarrow{\varepsilon} 0,$$
is a complex of right $H$-modules, where $H^{\otimes n}$ is a right $H$-module via the multiplication of an element of $H$ and the extreme right term in an elementary tensor of $H^{\otimes n}$.

My question is: how does one justifies that the augmented complex is exact? Is there an explicit chain homotopy to the identity chain map to zero? My guess (I am not an expert) was that the family $\{l_n: H^{\otimes n} \to H^{\otimes n+1}\}_{n\in \mathbb N}$  with  $l_0(1_{\mathbb C})=1_H$ and $l_n(h_1\otimes h_2\otimes \cdots \otimes h_n)=1_H\otimes h_1\otimes h_2\otimes \cdots \otimes h_n$ should work, i.e. that $d_{n+2}l_{n+1}+l_nd_{n+1}=\text{id}_{H^{\otimes n+1}}$ for all $n\in \mathbb N$, but what I obtain in some lower dimension computation is as follows

$(d_2l_1+l_0d_1 )(h_1)=\epsilon(h_1) \neq h_1$

$(d_3l_2+l_1d_2)(h_1\otimes h_2)= 1_H\otimes \varepsilon(h_1)h_2 \neq h_1 \otimes h_2$, etc.

答: You just need to put the 1 on the other side and to include a sign: the correct map is $l_n(h_1\otimes\cdots\otimes h_n) = (-)^n h_1\otimes\cdots\otimes h_n\otimes 1$.  For example, $l(d(h_1\otimes h_2)) = -\epsilon(h_1)h_2 \otimes 1 + h_1h_2 \otimes 1$ and $d(l(h_1\otimes h_2)) = \epsilon(h_1)h_2\otimes 1 - h_1h_2 \otimes 1 +h_1\otimes h_2$.


问:
How to solve this integral equation with Neumann series?
I am currently studying a course on mathematical methods for physicists. I am trying to solve the following integral equation:

$$y(x)=x+\int_0^1(1+xt)y(t)dt $$

with the Neumann series method. This exercise is taken from Arfken and Weber "Mathematical Methods for Physicists", seventh edition, chapter 21, exercise 21.3.2. I have already found the solution, which is $y(x) = -2$, with the separable-kernel tecnique. However, when trying to find a solution of the type:

$$ y(x) = \sum_{n=0}^{\infty}\lambda^ny_n(x)$$

I keep getting linear polynomials with "random" coefficients (in the sense that I don't find a pattern or recursive relation). Since the solution is a constant, I think this series should truncate in the first term. But $\lambda=1$, so all the orders are in fact the same. I don't really get what I should be doing. Any ideas? Thank you in advance.

答: Well, your linear polynomials are by no means "random". The integral operator takes a polynomial $a+b\,x$ to another polynomial $c+d\,x$ where $\begin{bmatrix}c \\d\end{bmatrix}=\begin{bmatrix}1 & \frac12\\\frac12 & \frac13\end{bmatrix}\begin{bmatrix}a \\b\end{bmatrix}$. Powers of that operator correspond to powers of the matrix. The eigenvalues of $\begin{bmatrix}1 & \frac12\\\frac12 & \frac13\end{bmatrix}$ are $\displaystyle\frac{4\pm\sqrt{13}}6$, and the powers of the matrix may grow like powers of the biggest eigenvalue, so the Neumann series can't converge for $\displaystyle\lambda\ge\frac6{4+\sqrt{13}}\approx0.78889744907202141376155746505900810750$.


问:
If $\mathfrak{h} \subset \mathfrak{gl}(V)$ is a Lie algebra and $V$ has a weight space decomposition, then $\mathfrak{h}$ is nilpotent.
Let $V$ be a finite dimensional vector space, over an algebraically closed field. Let $\mathfrak{h} \subset \mathfrak{gl}(V)$ be a Lie subalgebra, and assume that $V$ has a weigth space decomposition $$V= \bigoplus_{\lambda \ \in \mathfrak{h}^\ast }V^\lambda(\mathfrak{h}) \\ \text{where }  \ V^\lambda(\mathfrak{h})=\{v \in V: \forall \ H \in \mathfrak{h}, \ \exists \ n \in \mathbb{N} \text{ with } (H-\lambda(H)I)^nv=0  \}, $$ Such that each $V^\lambda(\mathfrak{h})$ is $\mathfrak{h}-$invariant. Then $\mathfrak{h}$ is nilpotent. Here's where I got so far:

By Engel's theorem, it suffices to prove that $\forall \ H \in \mathfrak{h}, \ H$ is nilpotent. Suppose this is not the case, then choose $H \in \mathfrak{h}$ which is not nilpotent, then it has a non-zero eigenvector $v \in V$, with eigenvalue $\mu$. Choose $\lambda \in \mathfrak{h}^\ast$ s.t. $v \in V^{\lambda}(\mathfrak{h}).$

Choose $k$ minimal with $(H-\lambda(H)I)^kv=0$, since $I$ and $H$ commute, we have:  $$0=(H-\lambda(H)I)^kv=\sum_{i=1}^k {n \choose i} \lambda(H)^{k-i}(-1)^iH^iv=\sum_{i=1}^k {n \choose i} \lambda(H)^{k-i}(-1)^i\mu^i v = (\mu-\lambda(H))^kv \\
\Rightarrow \lambda(H)=\mu \text{ and } k=1 \text{ by minimality of } k. $$ I think I should use the fact that the weight spaces are $\mathfrak{h}-$invariant to get a contradiction, but I'm stuck. Can someone help?

答: Hint: I don't think that you need any deep theorem in order to prove the claim. The assumption tells you that choosing an appropriate basis of $V$, the elements of $\mathfrak h$ are all represented by block diagonal matrices with each block given by some multiple of the identity plus a strictly upper triangular matrix. But it is easy to see directly that the direct sum of multiples of the identity and strictly upper triangular matrizes is a nilpotent Lie algebra, which implies the claim.


问:
Minimizing the maximum of $|x^2 - x - k|$
Let $k$ be a real number. Also, let $M(k)$ be the maximum value of $f(x) = |x^2 - x - k|$ in $0 \le x \le 1$. What are all real numbers $k$ which minimize $M(k)$?

答: Let $g(x)=x^2-x-k=(x-1/2)^2-k-1/4$. The graph of $g(x)$ is a parabola with the vertex at $(1/2, -k-1/4)$. Since $f(x)=|g(x)|$, the maximum of $f(x)$ at $[0,1]$ is attained when $x\in\{ 0,1/2,1\}$. We have $f(0)=f(1)=|k|$ and $f(1/2)=|k+1/4|$. So in order to minimize $M(k)$ we have to minimize $M’(k)=\max\{|k|, |k+1/4|\}$. It is easy to check that $M’(-1/8)=1/8$ and $M’(k)\ge 1/8$ for each $k$.


问:
Convergence of $\hat \beta_N - \hat \beta_{N-1}$ where $\hat \beta_N$ is the least squares solution of $Y_N = X_N\beta_N + \varepsilon_N$?
Suppose we are given a set of random observations $\{y_i,x_{i1},\dots,x_{ip}\}_{i=1}^N$. Based on these observations, we can form the multiple linear regression model in matrix form
$$
Y_N = X_N\beta + \varepsilon_N,
$$
where the error, conditioned on $X$, is normally distributed with mean zero and finite variance. I have used the subscript $N$ to explicitly indicate that we have used $N$ observations.

When we solve this model with least squares we find the estimated regression coefficient vector $\hat \beta_N$. We could also leave out the last observation and solve the model
$$
Y_{N-1} = X_{N-1}\beta + \varepsilon_{N-1},
$$
to find the estimated regression coefficients $\hat \beta_{N-1}$.

The explicit expression for the estimated regression coefficients is well-known. We have
$$
\begin{align}
\hat \beta_N &= \beta_N + (X_N^TX_N)^{-1}X^T \varepsilon_N, \\
\hat \beta_{N-1} &= \beta_{N-1} + (X_{N-1}^TX_{N-1})^{-1}X^T \varepsilon_{N-1}.
\end{align}
$$
As $N$ gets large intuitively it seems there will be negligible difference between the estmiated regression coefficients $\hat \beta_N$ ad $\hat \beta_{N-1}$. But can we quantify this precisely? Is it possible to show that $E(N) := \hat \beta_N - \hat \beta_{N-1}$ converges to zero in probability or almost surely as $N \to 0$? And in particular, how fast it converges to zero?

答:
(From the discussion on MO) I guess that the subscript $N$ in $\beta_N$ is a lapse.

It is well know that, conditionally on $X_N$, $\hat \beta_N$ is unbiased and $\mathrm{Cov}(\hat \beta_N) = \sigma^2 \big(X_N^T X_N\big)^{-1}$.

Now in order to have some convergence, one naturally needs additional (to the usual Gauss–Markov) assumptions. Say, if $x_i$ are iid and square integrable, then
$$
X_N^T X_N = \sum_{i=1}^N |x_i|^2 \sim n\,\mathrm{E}[|x_1|^2], n\to\infty,
$$
almost surely. As a result, $\hat \beta_n \overset{\mathrm P}\longrightarrow \beta$, $n\to\infty$, moreover, $\sqrt{n} |\hat\beta_n - \beta|$ is bounded in probability. (So the convergence rate is loosely $1/\sqrt{n}$.)


问:
Set of rationals isomorphic to $\omega \cdot \omega$
I have the following problem, but I don't know how to solve it:

Problem. Describe a set of rationals isomorphic to $\omega \cdot \omega$

答:
For $n\in \Bbb Z^+$ let $S_n\subset \Bbb Q\cap [n,n+1)$ be order-isomorphic to $\omega$ with $\min S_n=n$ and $\sup S_n=n+1.$ Now let $S=\cup_{n\in \Bbb Z^+}S_n.$

More generally, let $<'$ be a linear order on a non-empty countable set $B$. A jump in $B$ is an ordered pair $<x,y>$ of members of B with $x<'y,$ such that $no$ $z\in B$ satisfies $x<'z<'y.$ Augment $B$ to a larger ordered set $C$ by (i) inserting a copy of $\Bbb Q$ between $x$ and $y$ whenever $<x,y>$ is a jump in $B$, and (ii) adding a copy of $\Bbb Q$ below $\min_{<'}B$ if this min exists, and (iii) adding a copy of $\Bbb Q$ above $\max_{<'}B$ if this max exists. Then $C$ satisfies the premise of this (elementary) theorem of Georg Cantor:

A countably infinite linear order with no jumps, no min, and no max, is order-isomorphic to $\Bbb Q.$

So let $f:C\to \Bbb Q$ be an order-isomorphism. Then $f|_B$ is an order-isomorphism from $B$ to a subset of $\Bbb Q.$ Therefore any countable linear order (e.g. any countable ordinal) is order-isomorphic to a subset of $\Bbb Q.$

We can also use transfinite induction (without that Cantor theorem) to show that every countable ordinal embeds into $\Bbb Q.$


问:
Using the Rolle's method to prove function property
We had the following statement given to us, that, there are two functions $h$ and $j$, where $h,j : [a,b] \rightarrow \mathbb{R}$. And these functions are continuous and differentiable on this interval too. Now, what we have to prove is the following: $\frac{j(b)-j(a)}{h(b)-h(a)} = \frac{j^{'}(c)}{h'(c)}$. I am a little confused about how to do this question. I saw this post here:

Using Rolle's Theorem to Prove the Cauchy Mean Value Theorem.

But I do not know how to approach the question above.

答:
Firstly we assume that $h' \ne 0$ on $(a,b)$, $h,j$ are continuous on $[a,b]$ and $h,j$ have derivatives on $(a,b)$.

Now let's consider function $F(x)=h(x)-\lambda j(x)$, where $\lambda =\frac{j(b)-j(a)}{h(b)-h(a)}$. For $F$ holds Rolle's theorem, so $\exists c \in (a,b)$ such that $F'(c)=0$. It's easy to show, that $\lambda =\frac{j'(c)}{h'(c)}$.


问:
Integration of function using Hausdorff measure of Cantor Set
I am learning geometry on fractal shapes, along with how fractional calculus can relate to said geometry. At the moment I am trying to understand integration over a Hausdorff measure.

According to Falconer's The Geometry of Fractal Sets, the middle-third Cantor set has a $\log_3 2$-Hausdorff measure of $1$. I would like to describe this relation as $\int_{C} 1 \; \mathrm{d}H^{s}(x)=1$, where $C$ is the Cantor set, and $s = \log_3 2$.

My question is: How would the integral $$\int_{C} (e^x + x)\; \mathrm{d}H^{s}(x)$$ be evaluated?

I'm not entirely sure where to begin in evaluating this, since this is new to me. I would appreciate resources as well, but I think an answer to this problem will give me the tools I need to evaluate other functions on other fractal sets.

Thank you.

答:
I highly recommend the paper Evaluating Integrals Using Self-similarity by Robert Strichartz. Since Hausdorff measure on the Cantor set happens to be a self-similar measure, the ideas in that paper are applicable. In that paper, we learn  that the basic properties of the integral yield a self-similar identity for integration over the Cantor set $C$ with Hausdorff measure $\mu$:

$$\int_C f(x) \, d\mu =  \frac{1}{2}\int_C f\left(\frac{1}{3}x\right)d\mu + \frac{1}{2} \int_C f\left(\frac{1}{3}x+\frac{2}{3}\right)d\mu.$$

We can use this to evaluate the second term in your integral exactly:

$$
\begin{align}
  \int_C x \, d\mu &= \frac{1}{2}\int_C \left(\frac{1}{3}x\right)d\mu + \frac{1}{2} \int_C \left(\frac{1}{3}x+\frac{2}{3}\right)d\mu \\
 &= \frac{1}{6} \int_C x \, d\mu + \frac{1}{6} \int_C x \, d\mu + \frac{1}{3} \int_C 1 \, d\mu
\end{align}
$$

We can now solve for $\int_C x \, d\mu$ in terms of $\int_C 1 \, d\mu=1$ to get

$$\int_C x \, d\mu = \frac{1}{2}.$$

It's worth mentioning that this same technique can be applied to $\int_C x^n \, d\mu$ to derive a recursive technique to integrate $x^n$ for every integer $n$ and, therefore, any  polynomial.

Unfortunately,  I do not believe that the integrals of the transcental functions can be expressed  in closed form. They can be estimated numerically easily enough, though. For example, the picture below should convince you that

$$\frac{1}{4} \left(1+e^{2/9}+e^{2/3}+e^{8/9}\right) < \int_C e^x \, d\mu < \frac{1}{4} \left(\sqrt[9]{e}+\sqrt[3]{e}+e^{7/9}+e\right).$$

Taking that out to 20 iterations, I estimate that

$$1.75327920434 < \int_C e^x \, d\mu < 1.75327920485.$$


问:
How can I intuitively see independent events?
Here's a example with rolling a dice:

$A = \{1, 2\}$

$B = \{2, 3, 4\}$

$P(A \cap B) = P(\{2\})=1/6,$ and $P(A) = 1/3,$ and $P(B) = 1/2,$ so events $A$ and $B$ are independent events.

The definition of independence says, "if event $A$ doesn't influence event $B,$ then these events are independent".

If I add '$1$' to event $B$ and call it event $C,$ i.e., $C = \{1, 2, 3, 4\},$ then $A$ and $C $ are not independent events.

How can I understand this "event $A$ doesn't influence event $B$" more intuitively?

答:
The basic point is that independence of events $A$ and $B$ with positive probability implies $P(B)=P(B \mid A)$ and $P(A)=P(A \mid B)$.

In your example, 1s and 2s are a third of all and also a third of the rolls between 2 and 4 inclusive, so knowing that the roll is between 2 and 4 inclusive doesn't make it any more or less likely that the roll is a 1 or a 2. By contrast knowing that the roll is between 2 and 5 inclusive would make it less likely that it is a 1 or a 2, since 2s are only a quarter of the rolls between 2 and 5.


问:
Let $a >0$. Find the area bounded by $x+y=a$ and $\sqrt{x}+\sqrt{y} = \sqrt{a}$.

Let $a >0$. Find the area bounded by $x+y=a$ and $\sqrt{x}+\sqrt{y} = \sqrt{a}$.

Graphing this it's a line $x+y=a$ on top of the curve $\sqrt{x}+\sqrt{y} = \sqrt{a}$. So solving for $y$ in both equations I have that $y= a-x$ and $y= (\sqrt{a}-\sqrt{x})^2$

So the integral becomes $$\int _0^a\int _{\left(\sqrt{a}+\sqrt{x}\right)^2}^{a-x}1 \ dy \ dx = \int _0^a\left(-2\sqrt{a}\sqrt{x}-2x\right) \ dx = -\frac{4}{3}\sqrt{a}a^{\frac{3}{2}}-a^2$$

but this seems to have a negative area, what might be happening here?

答:
The method shown here is overkill, but I thought it would still be interesting to demonstrate. One could also consider changing coordinates from $(x,y)$ to $(u,v)$ with

$$\begin{cases}u=\sqrt x\\v=\sqrt y\end{cases}$$

Then $x+y=a$ transforms to the circle $u^2+y^2=a$ of radius $\sqrt a$ centered at the origin, while $\sqrt x+\sqrt y=a$ transforms to the line $u+v=a$ that passes through $(0,\sqrt a)$ and $(\sqrt a,0)$. The area element would be

$$\mathrm dx\,\mathrm dy=\left|\det\begin{bmatrix}u_x&u_y\\v_x&v_y\end{bmatrix}\right|\,\mathrm du\,\mathrm dv=4uv\,\mathrm du\,\mathrm dv$$

The bounded region in the $u$-$v$ plane is the segment of the circle in the first quadrant. To see this, pick any point in the original bounded region and see which point it maps to in the $u$-$v$ plane. (Plots shown below use $a=1$ and the point $(x,y)=\left(\frac13,\frac13\right)$, which maps to $(u,v)=\left(\frac1{\sqrt3},\frac1{\sqrt3}\right)$.)

Denote by $R$ the original bounded region, and by $S$ the transformed region in $(u,v)$ coordinates. Then the area of $R$ is

$$\iint_R\mathrm dx\,\mathrm dy=\iint_S4uv\,\mathrm du\,\mathrm dv=4\int_0^{\sqrt a}\int_{\sqrt a-u}^{\sqrt{a-u^2}}uv\,\mathrm dv\,\mathrm du=\frac{a^2}3$$


问:
Convergence of means and convergence in mean
Assume $(X_n)$ for $n \geq 0$ uniformly integrable and  $X_n \to X$ in probability. I want to show that $X_n \to X$ in $L^1$. I will probably have to achieve this with some bounding of $E[|X-X_n|]$. Convergence in probability implies convergence in distribution, and together with Skorokhod's theorem, this implies $E[X_n]\to E[X]$, but I don't think this can be used to bound $E[|X-X_n|]$. Any idea how to proceed?

答:
First, observe that $X$ is integrable, by applying Fatou's lemma to $\lvert X_{n_k}\rvert$, where $(X_{n_k})$ is such that $X_{n_k}\to X$ almost surely. Hence the sequence $(Y_n)_{n\geqslant 1}$ defined by $Y_n=\lvert X_n-X\rvert$ is uniformly integrable and converges in probability to $0$. In order to show the convergence in $\mathbb L^1$, recall that $\mathbb EY_n=\int_0^\infty\mathbb P(Y_n>t)dt$. Let $f_n(t):= \mathbb P(Y_n>t)$. Since $Y_n\to 0$ in probability, $f_n(t)\to 0$ for each positive $t$ which is a good start, but we cannot conclude yet (which is normal, as we did not use uniform integrability yet). Write for a fixed $R$
$$
\mathbb EY_n=\int_0^R\mathbb P(Y_n>t)dt+ \int_R^\infty\mathbb P(Y_n\mathbf{1}_{\{Y_n>R\}}>t)dt.
$$
The second term of the right hand side is smaller than $\mathbb E\left[Y_n\mathbf{1}_{\{Y_n>R\}}\right]$ hence
$$\tag{*}
\mathbb EY_n\leqslant \int_0^R\mathbb P(Y_n>t)dt+\sup_{k\geqslant 1}\mathbb E\left[Y_k\mathbf{1}_{\{Y_k>R\}}\right].
$$
As $n$ goes to infinity (for fixed $R$), $\int_0^R\mathbb P(Y_n>t)dt\to 0$ because $\int_0^R\mathbb P(Y_n>t)dt\leqslant R\mathbb P(Y_n>\varepsilon)+\varepsilon$
hence taking the $\limsup_{n\to\infty}$ in (*) and using uniform integrability finishes the proof.


问:
How to prove that a given matrix defines a Reed-Solomon code?
Assuming
\begin{equation}
\beta \space \epsilon \space \mathbb{F}_{q} \backslash \{0\}
\end{equation}
with
\begin{equation}
\beta^n = 1 \space and \space \beta^i \ne 1 \space \forall i \space \epsilon \space \{1,...,n-1\}\\
\beta^0 \ne \beta^1 \ne \beta^2 ... \ne \beta^{n-1}
\end{equation}

Given the parity check matrix of an [n,k] reed solomon code C

H:
\begin{pmatrix}
1 & 1 & 1 & ... & 1 & 1\\
1 & \beta & \beta^2 & ... & \beta^{n-2} & \beta^{n-1}\\
1 & \beta^2 & \beta^{4} & ... & \beta^{2n-4} & \beta^{2n-2}\\
\vdots & & & \ddots\\
1 & \beta^{n-k-1} & \beta^{2n-2k-2} & ... & \beta^{(n-k-1)\cdot(n-2)} & \beta^{(n-k-1)\cdot(n-1)}
\end{pmatrix}

How can I show, that

G:
\begin{pmatrix}
1 & \beta & \beta^2 & ... & \beta^{n-2} & \beta^{n-1}\\
1 & \beta^2 & \beta^{4} & ... & \beta^{2n-4} & \beta^{2n-2}\\
\vdots & & & \ddots\\
1 & \beta^{k-1} & \beta^{2k-2} & ... & \beta^{(k-1)\cdot(n-2)} & \beta^{(k-1)\cdot(n-1)}
\end{pmatrix}

is the generator matrix of C?

I know that
\begin{equation}
G \cdot H^\intercal = 0
\end{equation}
and that the product of (assuming i is a row of the matrix)
\begin{equation}
G_i \cdot H_{i+1} = 0
\end{equation}
as they have the same elements.

How can I show this for the remaining multiplications?
I assume I have to show that all multiplications end in n unique elements, but do not know how.

答:
The code that the OP is describing is a cyclic Reed-Solomon code whose generic codeword is the vector
$[C_0, C_1, \ldots, C_{n-1}]$ corresponding to the codeword polynomial $\displaystyle C(x) = \sum_{i=0}^{n-1}C_ix^i$ (often also simply called the codeword instead of codeword polynomial by those steeped in cyclic code theory), and the parity check matrix that the OP provides is equivalent to the statement that $$C(\beta^i)=0 \text{ for } i = 0, 1, 2, \ldots, n-k-1. \tag{1}$$
Thus, every codeword $C(x)$ has $\beta^i, i = 0, 1, 2, \ldots, n-k-1$ among its roots or zeroes, and so every codeword $C(x)$ is a multiple of the generator polynomial of the code given by
$$g(x) = \prod_{i=0}^{n-k-1}(x-\beta^i).$$

Now, with regard to the generator matrix of the cyclic Reed-Solomon code being described above, the form that the OP has written is a terrible one from the point of view of implementation as well as of understanding the theory, but no matter.  Let's ask whether each row of the hypothesized $G$ is a codeword. If $\displaystyle C(x) = \sum_{i=0}^{n-1}(\beta^m)^ix^i$ is the alleged codeword corresponding to the $m$-th row of the hypothesized $G$ (note that $1 \leq m \leq k$), does it satisfy $(1)$?  Well, for any $t, 0 \leq t \leq n-k-1$,
\begin{align}
C(\beta^t) &= \sum_{i=0}^{n-1}(\beta^m)^i(\beta^t)^i\\
&= \sum_{i=0}^{n-1}(\beta^{m+t})^i\\
&= \sum_{i=0}^{n-1}(\gamma)^i\\
&= \begin{cases} 0 & \text{if }~ t \neq n-m,\\
n & \text{if }~ t = n-m.\end{cases}
\end{align}
As $m$ ranges from $1$ to $k$, $n-m$ ranges from $n-1$ to $n-k$, and since $t$ is between $0$ and $n-k-1$, we see that $C(\beta^t) = 0$ for all $t, 0 \leq t \leq n-k-1$, and so $(1)$ holds: each row of the alleged $G$ is a codeword in the cyclic Reed-Solomon code. I will leave it to the OP to prove that the row space of $G$ has dimension $k$ so that $G$ is indeed a generator matrix of the $[n,k]$ cyclic Reed-Solomon code (i.e. that the rows of $G$ span the entire code, and not just a subcode.)


问:
Name of symbol of null set.
What is the name of the symbol we use for empty set?
Difference in notation of phi and void set? I found that symbols that we use for both are different but I know the name phi only. Not know the name of symbol of empty ser</p>

答: Although not mathematics, I thought I would add that in the APL programming language the empty vector is represented by the symbol ⍬, which is pronounced "zilde".

Title: Name of symbol of null set.


问:
if $\mathbf x$ is sampled randomly from a hypercube on $R^n$, what is the probability density for $|\mathbf x| = d$
if the vector $\mathbf x$ is sampled randomly from a uniform distribution on $[0, 1]^d$, what is the probability density function for $|\mathbf x|$? Is it easy to scale for $[0, n]^d$?</p>

答:
A partial answer.  If $X$ is uniformly distributed in $[0,1]$ then $X^2$ has pdf $$f(t) = \frac{1}{2 \sqrt{t}}$$
for $t \in [0,1]$.  Therefore the pdf of $| \mathbf{x}| = (\mathbf{x}_1^2 + \dotsc + \mathbf{x}_d^2)^{\frac{1}{2}}$ for $\mathbf{x} \in [0,1]^d$ is $$f_d(t) = 2t \, f^{\ast d}(t^2)$$ where $f^{\ast d}$ is the $d$-fold convolution of $f$.  It looks like this pdf gets complicated quickly.  For example already for $d=2$ using WA I get the following pdf:

$$
f_2(t) = \begin{cases}
\frac{\pi t}{2} & \textrm{if } t\in[0,1]\\[1ex]
t \left(\arcsin(t^{-1})-\arctan \sqrt{t^2-1}\right) & \textrm{if } t\in[1, \sqrt{2}]
\end{cases}
$$


问:
Symmetric Derivative is it right?
Let $ f $  be a function defined on some neighbourhood of a point $ x_0 $. If the following limit exists
$$ \lim_{h\to 0}{\frac{f\left(x_{0}+h\right)-f\left(x_{0}-h\right)}{2h}} $$
then $f$ is said to be symmetrically differentiable at $ x_{0} $  and the above limit is called the symmetric derivative of $ f$ at $ x_{0} $ and is denoted by $ f_{sym}'\left(x_{0}\right) $

Prove that if $ f $ has both right and left derivatives at $ x_0 $ then $ f$ is symmetrically differentiable at $ x_{0} $ (consequently, it would be proved that if $ f$ is differentiable at $ x_{0} $ then  $ f$ is symmetrically differentiable at $ x_{0} $)

So I wrote that
$$ f_{sym}'\left(0\right)=\lim_{h\to 0}{\frac{f\left(0+h\right)-f(0-h)}{2h}}=\lim_{h\to 0}{\frac{ 0}{2h}} = 0 $$
Is this proof right?

答:
$$ \lim_{h\to 0} \frac{f(x_0 + h) - f(x_0 - h)}{2h}=\lim_{h\to 0} \frac{f(x_0 + h)-f(x_0)}{2h}-\lim_{h\to 0} \frac{ f(x_0 - h)-f(x_0) }{2h}$$

Justifies the claim (after you fix it).

Title: Symmetric Derivative is it right?


问:
Conditions when $||x||_2 < ||y||_2 \rightarrow ||x||_1 < ||y||_1 $ holds?
Given that $x$ and $y$ are vectors in an n-dimensional vector space. Then when does the following identity hold ? i.e can we identify conditions in which the following identity holds ?
$$||x||_2 < ||y||_2 \rightarrow ||x||_1 < ||y||_1  $$

I am specially interested in the case when $x,y \in [0,1]^{n}$

Body: This is not true in general. Take $n=2$ and the points $ x=(1,0) $, $y=(3/5,3/5)$. We have $\|x\|_2 = 1$, $\|y\|_2 = \sqrt{18}/5 \approx 0.8485$ and $ \|x\|_1 = 1$, $\|y\|_1 = 6/5$.