答: The statement "If I have a sister, I have a sibling" is true. However. you are negating that statement which means that the statement "I have a sister and I do not have a sibling" should be false, which holds. The statement need not be true in order for it to be a valid negation.</p>

Title: Logic and implication negation

Body:
I am new to predicate logic and there is something I can't seem to understand.

Let's have an implication:
If I have a sister, I have a sibling.

Now, it's negation should be:
I have a sister and I do not have a sibling.

But I can't understand why the following is not a valid negation:
If I have a sister, I do not have a sibling.

It should be a negation because the first part of the implication happened and the second did not.


问:
Why are vector fields defined to be sections of the disjoint union of the tangent spaces? Isn't that overly complicated?
It is usual to define the tangent bundle $TM$ as the disjoint union of all tangent space and then to define vector fields as sections of
\begin{align}
TM=\bigsqcup_{p\in M}T_pM=\bigcup_{p\in M}T_pM\times\{p\}&\to M\\
(v,p)&\mapsto p
\end{align}
(I am reading John Lee's book, where tangent vectors are introduced as derivations, such that $T_pM\cap T_qM\neq\emptyset$.)

Why don't we simply define $TM=\bigcup_{p\in M}T_pM$ and say that a vector field is a map $\omega\colon M\to TM$ with $\omega_p\in T_pM$ for all $p\in M$.

Isn't this the essential property of a vector field: That is assigns to all $p\in M$ an element of $T_pM$?

An analogous question arises for Differential forms.

An attempt to clarify the question: I have to prepare a talk about the Maxwell equations in terms of differential forms and I have an audience who does neither know what a manifold is, nor what a tensor ist. I have 30 minutes for a crash course in this stuff. This means that I have to leave out as much information as possible, but I want everything I say to be correct. So one question I asked myself is: Do I miss out on something if I don't explain what disjoint unions and sections are and simply give them my definition above? Is there a good reason to define the tangent bundle as the disjoint union of tangent spaces instead of the ordinary union (besides wanting to introduce the notion of sections)? But this are questions I already asked myself before having to give this talk.

答:
This is just rewriting what has been written a few times, but when $M\subset \Bbb R^N$, then
$$TM = \{(x,v): x\in M, v\in T_xM\}\subset M\times\Bbb R^N.$$
There's your universe. For an abstract manifold, of course, this makes no sense, because there is no sensible thing to substitute for $\Bbb R^N$.

ContentLicense: CC BY-SA 4.0


问:
Any combination of nilpotent operators is still nilpotent?

If $N$ and $N'$ are nilpotent operator on a vector space and $N$ commutes with $N'$ then $N+N'$ is nilpotent.

$$(N+N')^r=\sum_{j=0}^r \binom{r}{j}N^j(N')^{r-j}=0$$ for some r large enough. but it seems if $N$ does not commute with $N'$, $N+N'$ is still nilpotent. So, Can I say that any combination of nilpotent operators is still nilpotent?

答:
For operators on a finite dimensional vector spaces you can actually characterize the sum of nilpotent operators (i.e. square matrices).

Lemma Let $A$ be an $n \times n$ matrix. Then $A$ is the sum of (at most three) nilpotent operators if and only if $\mbox{tr}(A)$.

Proof Since any nilpotent operator has trace $0$ the $\Rightarrow$ implication is obvious.

$\Leftarrow$.

Let $A=(a_{ij})$ be a matrix with $\mbox{tr}(A)=0$.

Then
$$A= B+C$$ where
$$B= \begin{bmatrix}
a_{11}& a_{11} & ...&a_{11}\\
a_{22}& a_{22} & ...&a_{22}\\
a_{33}& a_{33} & ...&a_{33}\\
...&...&...&...\\
a_{nn}& a_{nn} & ...&a_{nn}\\
\end{bmatrix}
$$
and $C_{ii}=0$ for all $1 \leq i \leq n$.

Next, $\mbox{tr}(A)=0$ implies that $B^2=0$.

Moreover, $C$ can be written in an obvious way as the sum of two nilpotent matrices:

$$C= \begin{bmatrix}
0& c_{12} & ...&c_{1n}\\
0& 0 & ...&c_{2n}\\
0& 0 & ...&c_{3n}\\
...&...&...&...\\
0&0 & ...&0\\
\end{bmatrix}+ \begin{bmatrix}
0& 0 & ...&0\\
c_{21}& 0 & ...&0\\
c_{31}&c_{32} & ...&0\\
...&...&...&...\\
c_{n1}& c_{n2} & ...&0\\
\end{bmatrix}
$$
i.e. the sum of a strictly upper triangular and a strictly lower triangular matrix.

\qed

Now your question becomes equivalent to : is any matrix with $\mbox{tr}(A)=0$ nilpotent? It is trivial to find invertible matrices with trace 0.


问:
Variance when changing from one random discrete variable to another
There 16 people divided to 8 pairs, each pair does the same task and has $p=0.8$ chance of completing it successfully, independent of the results of the other pairs

find Var(X), where X is the number of pairs which completed the task successfully

find Var(Y), where Y is the number of people which completed the task successfully

For (1) this is a classic binomial random variable so $$Var(X)= np(1-p) = 8 * (0.8)(1-0.8) = 0.128$$

As for (2), instinctively, I want to say that $Var(Y) = 2 * Var(X)$ because the $n=16$ instead of $8$ and the probability that a person completes the task remains $0.8$. But on the other hand  I thought that I can look at $Y$ as twice the number of pairs that completed the task so $$Var(Y)=Var(2\cdot X)=4\cdot Var(X) = 4 * 1.28 = 5.12$$ but I'm not sure which way is correct

答:
Part of your problem is that you have not precisely defined the following:

For a given pair of people, if their task is completed successfully, are both people considered to have completed the task?

Is it possible for exactly one person in a given pair to complete the task successfully?

To illustrate, I will describe a scenario consistent with my interpretation of your question.  Suppose the task is for each person to roll a fair ten-sided die numbered from $1$ to $10$ inclusive, and the task is successful if the sum of the numbers they roll is not divisible by $5$.  For example, if the numbers rolled are $4$ and $2$, then they succeeded, but if they roll $10$ and $5$ they failed.  You can check that the probability this occurs is exactly $p = 0.8$.  Then they can only both succeed or both fail.

In this case, the number of people succeeding can only be among the set $Y \in \{0, 2, 4, 6, \ldots, 16\}$, whereas the number of pairs succeeding can be among the set $X \in \{0, 1, 2, \ldots, 8\}$.  Moreover, $Y = 2X$ for all outcomes of $X$.

It is not true that $\operatorname{Var}[Y] = 2 \operatorname{Var}[X]$.  This is wrong.  For a scalar constant $c$, we have $$\operatorname{Var}[cX] = c^2 \operatorname{Var}[X],$$ and for $c = 2$, this means the variance of $Y$ is four times the variance of $X$.  The flaw in your first line of reasoning is that $Y$ is not itself binomial on $\{0, 1, 2, \ldots, 16\}$, because it is not possible to observe an odd number of people succeeding.  Therefore, $\operatorname{Var}[Y] \ne 16 p(1-p)$.


问:
What is the value of $\int_S \frac{\overline{f(z)}}{z-a}dz$ for an entire function $f(z)$, where $S$ is the unit circle.
Suppose that $f(z)$ is an entire function and $S$ be the unit circle, then prove that $$\int_S \frac{\overline{f(z)}}{z-a} dz=2\pi i \cdot\left\{\begin{aligned}\overline{f(0)}~\quad~|a|<1 \\
\overline{f(0)}-\overline{f(\bar{a}^{-1})} \quad |a|>1
        \end{aligned}\right.$$
How can I use the definition $\int_\gamma f(z)dz=\int_a^b f(\gamma(t))\gamma'(t)dt$ for a continuous function $f(z)$ on a piecewise smooth contour $\gamma:=\gamma(t),~a \leq \gamma \leq b$. Also how to use the analyticity of $f$ here?

答:
By noting that $\overline{z} = \frac{1}{z}$ for $z \in S$, we have

$$ I := \int_{S} \frac{\overline{f(z)}}{z-a} \, \mathrm{d}z
\stackrel{(z=w^{-1})}= \int_{-S} \frac{\overline{f(\overline{w})}}{\frac{1}{w}-a} \, \left(-\frac{\mathrm{d}w}{w^2} \right)
= \int_{S} \frac{\overline{f(\overline{w})}}{w(1-aw)} \, \mathrm{d}w. $$

Since $g(w) := \overline{f(\overline{w})}$ is an entire function, we may apply the residue theorem.

If $|a| < 1$, then the only pole inside of $S$ is $w=0$, and so,
$$ I = 2\pi i \, \underset{w=0}{\mathrm{Res}} \, \frac{g(w)}{w(1-aw)} = 2\pi i g(0). $$

If $|a| > 1$, then there are two poles, $w = 0$ and $w = \frac{1}{a}$, inside of $S$, and so,
$$ I = 2\pi i \biggl( \underset{w=0}{\mathrm{Res}} \, \frac{g(w)}{w(1-aw)} + \underset{w=a^{-1}}{\mathrm{Res}} \, \frac{g(w)}{w(1-aw)} \biggr) = 2\pi i \bigl( g(0) -  g(a^{-1}) \bigr). $$

Here, the following observation is useful:

If $P(z)$ and $Q(z)$ are analytic near $z = z_0$, $P(z_0) \neq 0$, and $Q(z)$ has a simple zero at $z = z_0$, then
$$ \underset{z=z_0}{\mathrm{Res}} \, \frac{P(z)}{Q(z)} = \frac{P(z_0)}{Q'(z_0)}. $$


问:
Reduction of Order - Confusion
Find the solution of the following initial value problem:

$y'' -y'\tan(t)+2y  =0$

$y(0) = 1$, $y'(0) = \sqrt[]{3}$

I've attempted this problem a couple of times and keep coming up with an incorrect answer. Anyone able help out?

ContentLicense: CC BY-SA 3.0

Title: Reduction of Order - Confusion

Body:
$y''-y'\tan(t)+2y=0$

$y''-\dfrac{\sin(t)}{\cos(t)}y'+2y=0$

$\cos(t)y''-\sin(t)y'+2\cos(t)y=0$

Note that it has trivial particular solution $y=\sin(t)$ .

Let $y=\sin(t)u$ ,

Then $y'=\sin(t)u'+\cos(t)u$

$y''=\sin(t)u''+\cos(t)u'+\cos(t)u'-\sin(t)u=\sin(t)u''+2\cos(t)u'-\sin(t)u$

$\therefore\cos(t)(\sin(t)u''+2\cos(t)u'-\sin(t)u)-\sin(t)(\sin(t)u'+\cos(t)u)+2\cos(t)\sin(t)u=0$

$\sin(t)\cos(t)u''+2\cos^2(t)u'-\sin(t)\cos(t)u-\sin^2(t)u'-\sin(t)\cos(t)u+2\sin(t)\cos(t)u=0$

$\sin(t)\cos(t)u''+(2\cos^2(t)-\sin^2(t))u'=0$

$\sin(t)\cos(t)u''=(\sin^2(t)-2\cos^2(t))u'$

$\dfrac{u''}{u'}=\tan(t)-2\cot(t)$

$\ln u'=-2\ln\sin(t)-\ln\cos(t)+c$

$u'=\dfrac{C_1}{\sin^2(t)\cos(t)}$

$u=C_1(\tanh^{-1}\sin(t)-\csc(t))+C_2$

$\therefore y=C_1(\sin(t)\tanh^{-1}\sin(t)-1)+C_2\sin(t)$

$y(0)=1$ :

$C_1=-1$

$\therefore y=1-\sin(t)\tanh^{-1}\sin(t)+C_2\sin(t)$

$y'=C_2\cos(t)-\tan(t)-\cos(t)\tanh^{-1}\sin(t)$

$y'(0)=\sqrt3$ :

$C_2=\sqrt3$

$\therefore y=1-\sin(t)\tanh^{-1}\sin(t)+\sqrt3\sin(t)$


问:
Definition of the field $F(\alpha)$
I have a little doubt. Let $F$ field and $\alpha\not\in F$. Then $F(\alpha):=\left\{a+b\alpha:a,b\in F\right\}$ by definition? or is it possible to prove it?

答:
You are exactly asking for what is a simple extension.

So better to refer to its definition, like in Wikipedia  simple extension.

Title: Definition of the field $F(\alpha)$


问:
Finding asymptotes to general functions
I stumbled upon this problem and didn't quite understand why the solution worked. I have two questions:

Why is it that the taylor series is enough to find the oblique asymptote? Specifically, why do the first two terms suffice? It seems like that may not work for all functions.

From pre-calculus I recall finding asymptotes of rational functions, but how do I find vertical, horizontal, and oblique asymptotes of a more general function (such as an exponential or square root, etc.)?

答: You got to be careful with those replacements in taylor series. You can only do that as long as your computations after the replacement stay within the interval of convergence. For the e-power, that's no big deal, since it converges kind of no matter what, but for arctanx for example, there are limitations. Nevertheless, using series is a clever way to find out about possible asymptotes</p>

Title: Finding asymptotes to general functions


问:
Is there an even faster way to find out when $y=\sin{5x} + \sin{x}$ cuts the x-axis for $0≤x≤\frac{π}{2}$?
The textbook method suggests using sum to product rules, but I know that someone on MSE will know a superior method! Please share?</p>

答: This is pure hindsight and I don't think it's "superior": when $x=\frac{\pi}{4}$ or $\frac{\pi}{3}, y=0$. Therefore $y=\sin 5x + \sin x$ cuts through the $x$-axis. But I really don't see the point of doing this.


问:
Explicit solution of a second order differential equation
Let $y:[0,1]\mapsto \mathbb{R}$. I'd like to know the existence of the explicit solution of the differential equation
$$
y^{\prime\prime}-xy^\prime-x^2y=3x-x^3
$$
with the boundary conditions $y(0)=y(1)=0$. I had tried to substitute $x=e^t$ but I could not change to the linear form. Could anyone give me a suggestion?</p>

Body:
Let $y=u+x$ ,

Then $y'=u'+1$

$y''=u''$

$\therefore u''-x(u'+1)-x^2(u+x)=3x-x^3$ with $u(0)=0$ , $u(1)=-1$

$u''-xu'-x-x^2u-x^3=3x-x^3$ with $u(0)=0$ , $u(1)=-1$

$u''-xu'-x^2u=4x$ with $u(0)=0$ , $u(1)=-1$


问:
automorphisms and field extension $E$ of $\mathbb{Q}$.
I want a hint. That is all I ask for. The question I am asked to prove is as follows:

Let $E$ be an extension field of $\mathbb{Q}$. Show that any automorphisn of $E$ acts as the identity on $\mathbb{Q}$.

Body: Hint: What is $\phi(0), \phi(1), \phi(n+1)),\phi(-n), \phi(n/m)$?


答:
We want to find $x$ in that range such that $\sin{5x}=-\sin{x}$ i.e. $x$ that satisfy either of the followings:

$$
\begin{align}
5x+x=k2\pi&\implies x=k\frac{\pi}{3}\\
\\
5x-x=(2k+1)\pi&\implies x=(2k+1)\frac{\pi}{4}
\end{align}
$$

Therefore the only solutions are $x=\frac{\pi}{3}$ and $x=\frac{\pi}{4}$

Title: Is there an even faster way to find out when $y=\sin{5x} + \sin{x}$ cuts the x-axis for $0≤x≤\frac{π}{2}$?

Body: The textbook method suggests using sum to product rules, but I know that someone on MSE will know a superior method! Please share?</p>


问:
Why does $\|(I+A^{-1}E)x\| = \|x-A^{-1}Ex\|$?
I was reading Lecture 15 of G.W. Stewart's Afternotes on Numerical Analysis, and he states the following theorem with proof:

"Let $A$ be nonsingular. If

$\|A^{-1}E\| < 1$, (15.3)

then $A+E$ is nonsingular.

To establish this result, we will show that under the condition (15.3) if $x \neq 0$ then $(A+E)x \neq 0$. Since $A$ is nonsingular $(A+E)x = A(I+A^{-1}E)x \neq 0$ if and only if $(I+A^{-1}E)x \neq 0$. But

$\|(I+A^{-1}E)x\| = \|x-A^{-1}Ex\| \geq \|x\| - \|A^{-1}E\|\|x\| = (1-\|A^{-1}E\|)\|x\| > 0$, which establishes the result."

What I don't understand about this proof is the step where he states that $\|(I+A^{-1}E)x\| = \|x-A^{-1}Ex\|$. This implies that $\|x+A^{-1}Ex\|=\|x-A^{-1}Ex\|$, and it's certainly not generally true for arbitrary vectors $a$ and $b$ that $\|a+b\| = \|a-b\|$.

In fact, I think that if we have that $\|a+b\| = \|a-b\|$, then this would imply that $a$ and $b$ are orthogonal, since then we would have $\|a+b\|^2 = \|a-b\|^2 \Rightarrow \|a\|^2+\|b\|^2 - 2\|a\|\|b\| \cos (\theta_1) = \|a\|^2+\|b\|^2 - 2\|a\|\|b\| \cos(\theta_2)$

by the Law of Cosines and because $\|b\|=\|-b\|$, where $\theta_1$ and $\theta_2$ must be supplementary angles since $\theta_1$ is the angle between $a$ and $b$ and $\theta_2$ is the angle between $a$ and $-b$. Solving this tells us that $\cos (\theta_1) = \cos (\theta_2)$, and, taking $0 \leq \theta_1, \theta_2 \leq \pi$, we have that $\theta_1 = \theta_2$. Combining this with the fact that $\theta_1 + \theta_2 = \pi$, since they are supplementary, we have that $\theta_1 = \theta_2 = \pi/2$, which implies that $a$ and $b$ are orthogonal.

All that said, I don't see why it should be the case that $x$ and $A^{-1}Ex$ are orthogonal. The only conditions that Stewart requires for this theorem are that $A$ be nonsingular and $\|A^{-1}E\| < 1$. I'm not sure if there's something else I'm missing or some other explanation for this. Any insight would be appreciated. Thank you!

答:
That is most likely a typo. It can be fixed easily-

$$\|(I+A^{-1}E)x\| = \|x\color{red}+A^{-1}Ex\| \geq \|x\| - \|A^{-1}E\|\|x\| = (1-\|A^{-1}E\|)||x|| > 0$$
and the inequality immediately following the modified step follows just from $\|My\| \le \|M\|\|y\|$ and
$ \|a+b\|\ge \|a\|-\|b\|$ which of course is just triangle inequality (in the form $\|a\| = \|(-b) + (a+b)\| \le \|{-b}\| + \|a+b\|$).


问:
How to find the maximum and minimum values of $\left|\sin 2x\right|+\left|\cos 2x\right|$
The expression is :$$\left|\sin2x\right|+\left|\cos2x\right|$$

For the maximum value I can ignore the absolute value and can get $$\sqrt2$$

But finding the minimum is what stumps me

I can understand it is $1$ by guessing but is there a rigorous approach ?

I didn't try differentiation because it can not be used on absolute value function .

Thank you in advance .

答:
Method 1 $2x=t$ we have $${(|\sin t|+|\cos t|)}^2=1+|\sin 2t|\ge 1$$

The conclusion is now obvious

Method 2 as $0\le |\sin t|,|\cos t|\le 1$ we must have $$|\sin t|+|\cos t|\ge |\sin^2 t|+|\cos^2 t|=\sin^2 t+\cos ^2t=1$$


答: If $||x+A^{-1}Ex||=||x-A^{-1}Ex||$ then we would have to have $$\big<x-A^{-1}Ex,x-A^{-1}Ex\big>=\big<x+A^{-1}Ex,x+A^{-1}Ex\big>$$ which means $$||x||^2-2\big<x,A^{-1}E\big>+||A^{-1}Ex||^2=||x||^2+2\big<x,A^{-1}E\big>+||A^{-1}Ex||^2$$ This is only true if $\big<x,A^{-1}E\big>=0$ i.e. $x$ and $A^{-1}Ex$ are orthogonal. However, since $$||x+y|| \geq \Big|||x||-||y||\Big|$$ is true for any $x,y\in \mathbb{R}^n$ I don't even see this this intermediate step is necessary.

Title: Why does $\|(I+A^{-1}E)x\| = \|x-A^{-1}Ex\|$?

Body:
I was reading Lecture 15 of G.W. Stewart's Afternotes on Numerical Analysis, and he states the following theorem with proof:

"Let $A$ be nonsingular. If

$\|A^{-1}E\| < 1$, (15.3)

then $A+E$ is nonsingular.

To establish this result, we will show that under the condition (15.3) if $x \neq 0$ then $(A+E)x \neq 0$. Since $A$ is nonsingular $(A+E)x = A(I+A^{-1}E)x \neq 0$ if and only if $(I+A^{-1}E)x \neq 0$. But

$\|(I+A^{-1}E)x\| = \|x-A^{-1}Ex\| \geq \|x\| - \|A^{-1}E\|\|x\| = (1-\|A^{-1}E\|)\|x\| > 0$, which establishes the result."

What I don't understand about this proof is the step where he states that $\|(I+A^{-1}E)x\| = \|x-A^{-1}Ex\|$. This implies that $\|x+A^{-1}Ex\|=\|x-A^{-1}Ex\|$, and it's certainly not generally true for arbitrary vectors $a$ and $b$ that $\|a+b\| = \|a-b\|$.

In fact, I think that if we have that $\|a+b\| = \|a-b\|$, then this would imply that $a$ and $b$ are orthogonal, since then we would have $\|a+b\|^2 = \|a-b\|^2 \Rightarrow \|a\|^2+\|b\|^2 - 2\|a\|\|b\| \cos (\theta_1) = \|a\|^2+\|b\|^2 - 2\|a\|\|b\| \cos(\theta_2)$

by the Law of Cosines and because $\|b\|=\|-b\|$, where $\theta_1$ and $\theta_2$ must be supplementary angles since $\theta_1$ is the angle between $a$ and $b$ and $\theta_2$ is the angle between $a$ and $-b$. Solving this tells us that $\cos (\theta_1) = \cos (\theta_2)$, and, taking $0 \leq \theta_1, \theta_2 \leq \pi$, we have that $\theta_1 = \theta_2$. Combining this with the fact that $\theta_1 + \theta_2 = \pi$, since they are supplementary, we have that $\theta_1 = \theta_2 = \pi/2$, which implies that $a$ and $b$ are orthogonal.

All that said, I don't see why it should be the case that $x$ and $A^{-1}Ex$ are orthogonal. The only conditions that Stewart requires for this theorem are that $A$ be nonsingular and $\|A^{-1}E\| < 1$. I'm not sure if there's something else I'm missing or some other explanation for this. Any insight would be appreciated. Thank you!


问:
Partial derivative confusion in lagrangian mechanics
An excerpt from an answer:

Consider the Lagrangian:

$L = (0.5)m(\dot{x}^2+\dot{y}^2 +\dot{z}^2)-U$.

The lagrange equation :

$\dfrac{\partial{L}}{\partial{x}}=\dfrac{d}{dt}\dfrac{\partial{L}}{\partial{\dot{x}}}$.

Evaluating this, we get:

$ -\frac{\partial{U}}{\partial{x}} = F_{x} $.

My issue with the evaluation:

The evaluation suggests $\frac{\partial{\dot x}}{\partial{x}}=0$ and $\frac{d}{dt}\left(\frac{\partial{U}}{\partial{\dot x}}\right)=0$, which unfortunately doesn't make sense to me.

How can we say in general that $\dot{x}$ is always not a function of $x$? We might have a situation where, say,$\dot{x} =ax+b$. How do we justify $\frac{\partial{\dot x}}{\partial{x}}=0$ for such a scenario?

If the answer is somehow "treat everything except $x$ as constant", then I will need some justification on what comes under "everything",especially $\dot{x}$, which might have a dependence on $x$ as I point out.

答:
It is a question of what you are given and what do you want to find in the problem. For example, you are given that the object is moving only due to a field. You need to find the position and velocity. Then the Lagrange equations will give you the relationship between position and velocity. There is no knowledge of the link before you solve the equation. So before you solve the equation, $x$ and $\dot x$ are independent variables.

What you are asking is a different question: what if there is an explicit relationship between position and velocity, such as the one you give in the example? Then there must be some unknown in the Lagrange equation. And that unknown is the force: what's the force that can move the object according to this constraint?

To show this by example, say a particle is moving under the action of the gravitational filed at the surface of the Earth. Then the force is $\vec F=-mg\hat y$, and you can write the potential as $U=mgy$. It's easy to calculate that $$v_y=\sqrt{v_y(0)^2-2gy}$$
Now what if $v_y=ay+b$? Then the answer is that you have the wrong potential. There is an extra force that will make the particle move that way. So using $U=mgy$ is incompatible with your problem.


答: Look at what this automorphism preserves in terms of the field structure on $E$. Can you "cover" $\mathbb{Q}$ by invariant elements?</p>

Title: automorphisms and field extension $E$ of $\mathbb{Q}$.

Body:
I want a hint. That is all I ask for. The question I am asked to prove is as follows:

Let $E$ be an extension field of $\mathbb{Q}$. Show that any automorphisn of $E$ acts as the identity on $\mathbb{Q}$.


问:
Expected projected length of radial vectors of n-sphere
Situation

In $n$-dimensional Euclidean space rests a unit $(n-1)$-dimensional sphere that is orthographically projected onto a $(n-1)$-dimensional plane. The topological definition of a sphere is used, i.e. only the surface points belong to a sphere.

Question

What is the expected projected length $\bar{x}_n$ of uniformly distributed radial unit vectors that start at the sphere center?

Solution strategy

One could calculate this by multiple integration using $n$-dimensional spherical coordinates but I hope there is a more elegant method. It seems there is a pattern that could be generalized to $n$ dimensions.

Solutions for $n=2$ and $n=3$

2-space (see drawing)

In 2-space (plane) a unit 1-sphere (circle) is projected onto a 1-plane (line).
The mean projected length from integration over a quadrant is $$\bar{x}_2=\frac{2}{\pi} =\frac{4}{2\pi}\int_{0}^{\frac{\pi}{2} } \sin(\alpha) \mathrm{d}\alpha$$

3-space

If we go 1 dimension higher, we arrive at the usual 2-sphere in 3-space. The radial vector is projected to a 2-plane. By integration over an octand we get for the mean projected length $$\bar{x}_3=\frac{\pi}{4} =\frac{8}{4\pi} \int_{0}^{\frac{\pi}{2} } \int_{0}^{\frac{\pi}{2} } \sin^2(\theta) \mathrm{d\theta} \mathrm{d}\phi$$

答:
Disclaimer: I used Mathematica for part of the problem.

As an expected value, this would be $$\mathbb{E}\left[ \sqrt{x_1^2+x_2^2 + ... + x_{n-1}^2} \right]$$

where $\left(x_1, x_2, ..., x_n\right)$ is a random point on the surface of the sphere. Since it is known that $x_1^2 + x_2^2 + ... + x_n^2 = 1$, this is equivalent to $$\mathbb{E}\left[ \sqrt{1 - x_n^2} \right] = \mathbb{E}\left[ \sqrt{1 - x_1^2} \right]$$

From here, $x_1$ can be chosen as $\frac{y_1}{\sqrt{\sum_{k=1}^n y_k^2}}$, where $y_k$ are chosen from $\mathcal{N}(0, 1)$. The expected value is then $$\mathbb{E}\left[ \sqrt{1 - \frac{y_1^2}{\sum_{k=1}^n y_k^2}} \right]$$

As an integral, this would be $$\int_0^{1} \left(1 - \mathbb{P}\left( \sqrt{1 - \frac{y_1^2}{\sum_{k=1}^n y_k^2}}<x \right)\right) dx$$

That inner probability is equal to $$\mathbb{P}\left( \frac{1-x^2}{x^2}\sum_{k=2}^n y_k^2 < y_1^2 \right)$$

Using the PDF of the chi-square distribution, this would be $$\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n-1}{2}\right)}\frac{1}{\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\infty}\int_{\frac{1-x^2}{x^2}z}^{\infty} z^{(n-1)/2-1}e^{-z/2}y^{1/2-1}e^{-y/2} dydz$$

Then the answer is (after some switching of bounds) $$1-\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\infty}\int_{0}^{\infty}\int_{\sqrt{\frac{z}{y+z}}}^{1}\left(z^{\frac{n-3}{2}}e^{-\frac{z}{2}}y^{-\frac{1}{2}}e^{-\frac{y}{2}}\right)dxdydz$$

This simplifies to $$\frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}\int_{0}^{\infty}\int_{0}^{\infty}\left(z^{\frac{n-2}{2}}e^{-\frac{y+z}{2}}y^{-\frac{1}{2}}\right)\sqrt{\frac{1}{y+z}}dydz$$

Mathematica then gives this as $$\frac{\pi^{\frac{3}{2}}}{\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{1}{2}\right)}\frac{\cos\left(\frac{n\pi}{2}\right)\Gamma\left(1-\frac{n+1}{2}\right)}{\left(\sin\left(\frac{n\pi}{2}\right)\Gamma\left(1-\frac{n}{2}\right)\right)^{2}}$$ for non-integer $n$. Taking the limit as $n$ approaches an integer and using Euler's reflection formula, this simplifies to $$\frac{\Gamma\left(\frac{n}{2}\right)^2}{\Gamma\left(\frac{n-1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)} \approx 1 - \frac{1}{2}n^{-1}-\frac{3}{8}n^{-2} + O(n^{-3})$$

Edit: Using that $\Gamma(k + \frac{1}{2}) = \frac{(2k-1)!!\sqrt{\pi}}{2^k} = \frac{(2k)!\sqrt{\pi}}{4^k}k!$ and $\Gamma(k) = (k-1)!$, splitting by odd and even cases, for odd $n$, it is $$\frac{\pi}{4^{n-1}}\cdot\frac{\left[\left(n-1\right)!\right]^{2}}{\left(\frac{n-3}{2}\right)!\left[\left(\frac{n-1}{2}\right)!\right]^{3}} = \frac{\pi(n-1)}{2^{2n-1}} \binom{n-1}{\frac{n-1}{2}}^2$$ while for even $n$, it is $$\frac{2^{2n-2}}{\pi}\frac{\left(\frac{n}{2}\right)!\left[\left(\frac{n}{2}-1\right)!\right]^{3}}{\left(n-2\right)!\left(n\right)!} = \frac{2^{2n+1}}{\pi}\frac{n-1}{n^{2}}\binom{n}{\frac{n}{2}}^{-2}$$


问:
Calculus inequality $\sqrt{x}\leqslant \frac14x+1$

Show that for all $x \geqslant 4$ the following inequality holds $$\sqrt{x}\leqslant \frac14x+1.$$
  Hint: $f(x) = \frac14x+1-\sqrt{x}$

So if we denote $f(x) = \frac14x+1-\sqrt{x}$.

We could just show that $f'(x) \geqslant0$ and that would satisfy the given inequality?

However $f'(x) = \frac14-\frac{1}{2\sqrt{x}}$ which doesn't hold when $x\geqslant 0$. What's the trick here?

答:
$$f(x) = \frac14x+1-\sqrt{x}$$

You get $f'(x) = \dfrac 14 - \dfrac{1}{2 \sqrt x}$ and $f''(x)=\dfrac{1}{4\sqrt{x^3}}$

So $f'(x)$

Is strictly increasing on the inverval $(0, \infty)$.

Has a vertical asymptote at $x=0$.

Passes through the point $(4,0)$.

So $f(x)$

Passes through the point $(0,1)$.

Is decreasing in the interval $0 \lt x \lt 4$.

Has a global minimum at $(4,0)$. (It follows that $f(x) \ge 0$.)

Is increasing on the interval $4 < x$.


问:
Terminology in graph theory [Directed graph]

Hi,

Have graph theorists agreed on the definitions of most of their basic objects? In particular, the definitions of "simple" and "elementary" path?

Sincerely,

P/s: I was kinda surprised learning that Germans have their own ways to define "path". And I also found that back in the 2010s, there were at least two schools of terminology for directed graphs.

(At least from what I read in [1])

[1]:

Note: There are two different definitions for "simple path". Here we follow the definition of Berge[1], Liu[2], Rosen[3] and others. A
"simple path" according to another group (Cormen et al[4], Stanat and
McAllister[5] and others) is a path in which no vertices appear more
than once.

答:
I think that there is more consistency these days than in the document you are citing, which is just over 20 years old.

A common set of definitions avoids "simple path" and "elementary path" entirely and uses the progression

walk (sequence of vertices and edges)

trail (no repeated edges)

path (no repeated vertices)

I would not be too surprised to encounter a paper which uses "path" to mean "trail" or "walk", but the above is what I would assume by default. Regarding the notions of "walk" or "trail", there is more confusion, because the middle ground where we allow repeated vertices but no repeated edges is very rarely necessary.

If you follow one of the standard textbooks by Bollobás, or Bondy and Murty, or Diestel, or West, you will have the right notion of "path". (Out of respect for all these authors I have listed their names in alphabetical order.)

It will probably take a long time before everyone agrees on this terminology, because graph theory is a disjointed field: in addition to graph theorists studying it, there are also network theorists with their own notation, and computer scientists with their own notation, and these groups talk to each other less often than they ought to.


问:
Is the scaling of the imaginary part a conformal map
Considering the following map from the complex plane z to Z:
$$  z = x+iy  \longrightarrow Z = x+iky $$

with $k$ a real value. Is this a conformal map? Such map would transform a circle into an ellipse for instance. Also, what would be the Jacobian of such transformation?
I guess it can be written $Z(z)= z + i (k-1) \sqrt{zz^*} \sin(\arg(z)))$ but I'm not sure how to derive this further.

答: By definition, a conformal map preserves angles locally.  What does your map do to the angle between the real axis and a diagonal line?</p>


问:
Show that the linear transformation of a measurable set is measurable
Let $E\subset\mathbb{R}^d$ be a Lebesgue measurable set and let $T\colon\mathbb{R}^d\to\mathbb{R}^d$ be a linear transformation.

Show that $T(E)$ is measurable and $m(T(E)) = \left|\det (T)\right|m(E)$.

I can't do it at all, I don't know how to start. Please help me!

答: One way to do this is to note that since $T$ is linear, if we define $\mu:=mT$ then $\mu $ is translation invariant so it is a positive multiple of the Lebesgue measure. That is, $mT=cm$. Now we can use $any$ measurable set $E$ to determine $c.$ Thus, if we take the standard cube $Q=[0,1]^n$ and apply $T$ we get a parallelopiped whose volume is |det T| and since $m(Q)=1$, it follows immediately that $c=|\det T|.$


答:
This is a fairly standard trick assuming that $\|\cdot \|$ is a submultiplicative norm.

Since $\|A^{-1}E\| < 1$, the matrix $D=\sum_{k=0}^\infty ((-A^{-1}E )^k $ is well defined
and since $D (I+A^{-1}E) = I$ we see that $D$ is invertible. Then we have
$DA^{-1} (A+E) = I$ and so $A+E$ is invertible with $(A+E)^{-1} = D A^{-1}$.


问:
How to integrate $\int \sqrt{x^2+a^2}dx$
$a$ is a parameter. I have no idea where to start</p>

Body:
Since $\sinh^2(x)+1
=\cosh^2(x)
$,
this suggests
letting $x = a \sinh(y)$.

$dx = a \cosh(y)$
and $x^2+a^2 = a^2(\sinh^2(y)+1)
= a^2 \cosh^2(y)$,
so $a \cosh(y) = \sqrt{x^2+a^2}$.

$\begin{align}
\int{\sqrt{x^2+a^2}}dx
&=\int a^2 \cosh^2(y) dy\\
&= a^2 \int (e^{2y}+2+e^{-2y})\,dy/4\\
&= (a^2/4) (e^{2y}/2 + 2y - e^{-2y}/2)\\
&=(a^2/4)(\sinh(2y)+2y)\\
&=(a^2/4)(2\sinh(y)\cosh(y)+2y)\\
&=(a^2/2)((x/a) (1/a)\sqrt{x^2+a^2}+\sinh^{-1}(x/a))\\
&=(x \sqrt{x^2+a^2})/2+(a^2/2)\sinh^{-1}(x/a)
\end{align}
$

I'll leave it at this - you can find $\sinh^{-1}$.


问:
Borel sigma algebras of subsets of $\mathbb{R}$
I'm trying to prove a result that seems to be used routinely, but the proof is skipped over.

Let $\mathbb{R}$ have the Euclidean metric, and this defines a topology. Let $A \subset \mathbb{R}$ be equipped with the subspace topology. Then,

$\mathcal{B}(A) = \{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$

If $A \in \mathcal{B}(\mathbb{R})$, then $\mathcal{B}(A) = \{ B \in \mathcal{B}(\mathbb{R}) : B \subset A\}$.

The proof for the $\subset$ direction of 1. is simple; note that $\{ B \cap A : B \in \mathcal{B}(\mathbb{R})\}$ is a sigma algebra, which contains all the open sets of $A$. There are also other questions on stackexchange which address this direction of the proof. However, I am stuck on the $\supset$ direction. Does anyone have a nice proof for this?

(2. is an immediate consequence of 1., as far as I can see)

答: The family $\mathcal S=\{X\subseteq\mathbb R:X\cap A\in\mathcal B(A)\}$ is a $\sigma$-algebra of subsets of $\mathbb R$, and it contains all open subsets of $\mathbb R$ (because their intersections with $A$ are open in the subspace topology of $A$). So $\mathcal S$ contains all of the Borel subsets of $\mathbb R$.


问:
Determinant of a 6x6 matrix
There's this part of my assignment which involves stochastic matrices and i've done most parts of it but there's one part which requires me to show that its eigenvalue is 1. The only way i can think of this is by calculating the det|A-$\lambda$I| and showing that equals to zero when $\lambda$=1. But really how do I calculate a determinant of a 6x6 matrices?
Given below is the stochastic matrice that i have found;

\begin{bmatrix}
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3} & 0 & 0 \\
    \frac{1}{2} & 0 & 0 & \frac{1}{3}& \frac{1}{2}& 0 \\
    0 & \frac{1}{2} & \frac{1}{3} & 0 & 0 & \frac{1}{2} \\
    0 & 0 & \frac{1}{3}& 0 & 0 & \frac{1}{2} \\
    0 & 0 & 0 & \frac{1}{3} & \frac{1}{2} & 0 \\
\end{bmatrix}

I don't think i'm suppose to compute this in a really long and complicated way but i also do have to find the eigenvector associated to eigenvalue 1. That's why I'm all confused about this. What would be the way to go about this?

答:
Subtract $I_6$ from the given matrix $M$, then find the reduced row-echelon form. We get
$$\left[\begin{matrix}1 & 0 & 0 & 0 & 0 & -1\\0 & 1 & 0 & 0 & 0 & -1\\0 & 0 & 1 & 0 & 0 & - \frac{3}{2}\\0 & 0 & 0 & 1 & 0 & - \frac{3}{2}\\0 & 0 & 0 & 0 & 1 & -1\\0 & 0 & 0 & 0 & 0 & 0\end{matrix}\right]$$
An eigenvector corresponding to $1$ is a vector in the nullspace of $M-I_6$. The above RREF shows that one such vector is $(1,1,3/2,3/2,1,1)^T$.

Title: Determinant of a 6x6 matrix


问:
Find the orthogonal vector of $f(x, y) = x^2y+y^3$ at a given point
Question: Which given vector is perpendicular to the following curve at the point (1, 2).$$f(x, y) = x^2y+y^3 = 10$$

My attempt: By taking the partial derivatives of this function, I was able to find that the gradient of the function is $(4, 13)$. Since the dot product of two vectors are zero if they are orthogonal with each other, I concluded the answer to be $<-13, 4>$. This turned out to be incorrect, however.

From my perspective, I imagine the gradient function to be similar to a slope function but in 3D and that it contains information regarding the function's rate of change at a given point.
But I've also seen pictures online that the gradient function is perpendicular to the function itself, which is quite confusing for me since the slope function is never perpendicular to the original function.

I would be much appreciated if someone could explain this to me. Thanks in advance!

答:
Think of the gradient $\nabla(f)$ as the ratio of change in $f$’s value to the change in position on $x-y$ plane. Now imagine what happen if you walk along the curve $f=10$ i.e. $\Delta f=0$

$$
\begin{align}
0&=\Delta f\\
\\
&=\lim_{\Delta x, \Delta y \to 0}{\nabla(f)}\cdot(\Delta x \hat{i} +\Delta y \hat{j})
\end{align}
$$

Therefore, $(-13,4)$ that you find is actually not the vector perpendicular to the curve but is actually a vector tangent to the curve at that point. The vector you are looking for is actually the one perpendicular to this one. Oh wait, is $\nabla(f)$ perpendicular to $(-13,4)$ ? :)


问:
Coin is tossed five times. Probability of getting head in first toss given that head was observed 3 times?
Let A: head observed on first toss
and B: 3 heads observed.

So far I have P(A)= 3/5 and P(B)=0.3125     [  5C3 x (0.5)^3 x (0.5)^2]

I tried to work out P(AnB) and I got 0.375   [ 4C2 x (0.5)^2 x (0.5)^2]

but when using the conditional probability formula I am not getting the right answer. Anyone know what I'm missing out or going wrong on?

答:
$P(A)$ is not $\frac35$, it's $\frac12$ since it concerns itself with a single toss. This is not needed for the conditional probability formula, though.

The real mistake is that $P(A\cap B)$ is a factor of $\frac12$ off; it should be $\frac6{32}$ since the first toss is fixed as a head.


问:
Compute iterated kernels for symmetric kernel $K(x,t)=\sum_{n=1}^\infty \frac{\sin(n\pi x) \sin(n \pi t)}{n}.$

Compute iterated kernls for symmetric kernel $$K(x,t)=\sum_{n=1}^\infty \frac{\sin(n\pi x) \sin(n \pi t)}{n}.$$

I am studying integral equations with symmetric kernels and there is a given problem. This problem is very unclear to me since there is no interval given, just the expression of the kernel.

My attempt so far: the expression looks very familiar to the form to the Mercer's theorem's expression of the kernel: $$K(x,t)=\sum_{n=1}^\infty \frac{\phi_n(x)\phi_n(t)}{\lambda_n}$$ assuming the values of the functions are real. It wouldn't be quite right to say that we have $\phi(x)=\sin(n\pi x)$ and $\lambda_n=n$ but in this case we could say that $$\phi_n(x)=\lambda_n\int_a^b K(x,t) \phi_n(t)dt \implies \sin(n \pi x)=n\int_0^1 K(x,t)\sin(n \pi t)dt.$$ assuming our interval is $[0,1].$ Is it possible to get $K(x,t)$ out of this equation?

What's the method to generate 'proper' kernel $K(x,t)$ out of the sum above? Any help would be highly appreciated.

答:
Your intuition is correct, kernel $K$ can be written as
$$K(x,t)=\sum_{n=1}^\infty \frac{\phi_n(x)\phi_n(t)}{\lambda_n}$$
with $\phi_n(x)=\sin(n\pi x)$ and $\lambda_n=n$.
This is no special theorem, we're just writing the definition of $K$.

So, assuming the integration interval is $[0, 1]$, and using the following notation for the inner product in $L^2([0,1])$:
$$\langle f, g\rangle = \int_0^1 f(x)g(x)dx$$
the square of $K$ is
$$K^2(x,t)=\sum_{n=1}^\infty \sum_{m=1}^\infty \frac{\phi_n(x)\phi_m(t)}{\lambda_n^2}\langle \phi_n, \phi_m\rangle$$
You can verify that $$\langle \phi_n, \phi_m\rangle=\int_0^1 \sin(n\pi u)\sin(m \pi u)du$$
is equal to $\frac 1 2$ if $n=m$, and is equal to $0$ otherwise.

Thus, $$K^2(x,t)=\frac 1 2 \sum_{n=0}^{+\infty}\frac{\phi_n(x)\phi_m(t)}{\lambda_n^2}$$
the $p$-th power of $K$ is given by
$$\boxed{K^p(x,t)=\frac 1 {2^{p-1}} \sum_{n=1}^\infty\frac{\phi_n(x)\phi_m(t)}{\lambda_n^p}=\frac 1 {2^{p-1}} \sum_{n=1}\frac{\sin(n\pi x) \sin(n \pi t)}{n^p}}$$

Edit: Following the comment from @Jean-Marie (merci!), an alternative form can be obtained by using the polylogarithm:
$$Li_p(z)=\sum_{n\geq 1}\frac {z^p}{n^p}$$
With this,
$$\begin{split}
K^p(x,t)&=\frac 1 {2^{p-1}} \sum_{n=1}\frac{\sin(n\pi x) \sin(n \pi t)}{n^p}\\
&= \frac 1 {2^p}\sum_{n=1}\frac{\cos(n\pi(x-t))-\cos(n\pi(x+t))}{n^p}\\
&= \frac 1 {2^p}\Re \left( \sum_{n=1}\frac{e^{in\pi(x-t)}-e^{in\pi(x+t)}}{n^p} \right)
\end{split}$$
$$\boxed{K^p(x,t) = \frac 1 {2^p} \Re \left( Li_p(e^{i\pi(x-t)}) - Li_p(e^{i\pi(x+t)}) \right)}$$
I'm not sure if that can be simplified further.

With $p=1$, $L_1(z) = -\ln(1-z)$, you get $$K(x,t)=\dfrac12\operatorname{ln}\left|\dfrac{\sin(\pi(x+t)/2)}{\sin(\pi(x-t)/2)}\right|$$


问:
eigenvalues and kernel
Say that A is a nxn matrice with eigenvalue $\lambda$. Let vector $\vec{v}$
∈ Ker((B − λIn)$^2$) but not an element of  Ker(B − λIn). (I have proved the contrary to be true though). How do i show that {$\vec{v}$ ,(B − λIn)$\vec{v}$} are linearly independent. I tried using

$c_1$$\vec{v}$ + $c_2$(B − λIn)$\vec{v}$ = 0 and expanding it in all sorts of ways but i can't show that $c_1$ and $c_2$ is 0. How do i show these two are linearly independent i must be missing something here.

答: If $c_1v+c_2(B-\lambda I)v=0$ then applying $B-\lambda I$ to both sides gives $$c_1(B-\lambda I)v+c_2(B-\lambda I)^2v=0$$ Since $v\in \text{ker}\Big[(B- \lambda I)^2\Big]$ we must have $c_1(B-\lambda I)v=0$. Now since $v\notin \text{ker}\Big[(B- \lambda I)\Big]$ we know $(B- \lambda I)v \neq 0$ telling us $c_1=0$ so original equation becomes $c_2(B-\lambda I)v=0$. Again, using the fact that $v\notin \text{ker}\Big[(B- \lambda I)\Big]$ we get $c_2=0$ as well.

Title: eigenvalues and kernel


问:
Logic and implication negation
I am new to predicate logic and there is something I can't seem to understand.

Let's have an implication:
If I have a sister, I have a sibling.

Now, it's negation should be:
I have a sister and I do not have a sibling.

But I can't understand why the following is not a valid negation:
If I have a sister, I do not have a sibling.

It should be a negation because the first part of the implication happened and the second did not.

答:
First note that
$$P\rightarrow Q\equiv\lnot P\lor Q$$
Here's how to properly negate the above statement
$$\lnot(P\rightarrow Q)\equiv\lnot(\lnot P\lor Q)\equiv P\land\lnot Q$$
Now let me show you why the following is not a negation of the original statement
$$P\rightarrow\lnot Q\equiv\lnot P\lor\lnot Q$$
Notice that this only negates $Q$, however we are seeking to negate the whole statement.
Also note that if $P\equiv\mbox{false}$, then we have
$$P\rightarrow Q\equiv P\rightarrow\lnot Q\equiv\mbox{true}$$
Which is clearly not a negation of the original statement. I hope this helps you understand.

Title: Logic and implication negation


答:
Since the other answer responded to the OP's question, I can point out the alternative approach.  Since (presumably) exactly 3 heads out of the 5 tosses were observed, the question is equivalent to the
following:

You have an urn with 5 balls, numbered 1 through 5.  You draw 3 of the 5 balls out, without replacement.  What is the probability that ball number 1 is among the 3 balls drawn.

Here, the numbers on each of the drawn balls can be construed to represent which of the 5 tosses came up heads.

Title: Coin is tossed five times. Probability of getting head in first toss given that head was observed 3 times?

Body:
Let A: head observed on first toss
and B: 3 heads observed.

So far I have P(A)= 3/5 and P(B)=0.3125     [  5C3 x (0.5)^3 x (0.5)^2]

I tried to work out P(AnB) and I got 0.375   [ 4C2 x (0.5)^2 x (0.5)^2]

but when using the conditional probability formula I am not getting the right answer. Anyone know what I'm missing out or going wrong on?


问:
What type of Hypergeometric series is this?
I am trying to find a closed form for the series

$$ \sum^\infty_{n=0} \frac{1}{n!} \frac{1}{n+1}(-z)^n {}_2F_2\left(m,n+1;\frac{1}{2},n+2; b z\right)$$

$m$ is a nonzero positive integer, and $b$, $z$ are positive real numbers. I to rewrite the sum as

$$ \sum^\infty_{n=0} \sum^\infty_{q=0}
 \frac{1}{n!} \frac{1}{q!}
 \frac{(m)_q}{(\frac{1}{2})_q}
\frac{(1)_{q+n}}{(2)_{q+n}}
(-z)^n (b z)^q$$

any idea what type of multi-variable hypergeometric function is the last equation?

答:
[Too long for a comment]. If you denote
$$ G(z)=\sum^\infty_{n=0} \frac{1}{n!} \frac{1}{n+1}(-z)^n {}_2F_2\left(m,n+1;\frac{1}{2},n+2; b z\right),$$
then certain combination of $G(z)$ and $G'(z)$ is a simpler single function:
$$\frac{d}{dz}\Bigl[zG(z)\Bigr]=e^{-z}{}_1F_1\left(m,\frac12;bz\right).$$
Therefore,
$$G(z)=\frac{1}{z}\int_0^ze^{-t}{}_1F_1\left(m,\frac12;bt\right)dt.$$
Also, for integer $m$ the $_1F_1$ function can be written in terms of error function.</p>


答:
$P[X_1 = 1 \text{ and } X_2+\cdots +X_5 =2] = {1 \over 2} \cdot \binom{4}{2} {1 \over 2^4}$.

$P[X_1+X_2+\cdots +X_5 = 3] = \binom{5}{3} {1 \over 2^5}$.

Hence the conditional probability is ${ \binom{4}{2} \over \binom{5}{3}} = {3 \over 5}$.


答: Direct answer:  Of the five tosses, three were heads and two were tails.  Since any of the five was equally likely to be the first, the probability that it was heads is 3/5.</p>


问:
What is the meaning of "quotient of polynomials in $\alpha,\beta,\gamma,\dots?$
I'm reading Artin's "Galois Theory", here:

A. Extension Fields.

If $E$ is a field and $F$ a subset of $E$ which, under the operations of addition and multiplication in $E$, itself forms a field, that is, if $F$ is a subfield of $E$, then we shall call $E$ an extension of $F$. The relation of being an extension of $F$ will be briefly designated by $F\subset E$. If $\alpha,\beta,\gamma,\dots$ are elements of $E$, then by $F(\alpha,\beta,\gamma,\dots)$ we shall mean the set of elements in $E$ which can be expressed as quotients of polynomials in $\alpha,\beta,\gamma,\dots$ with coefficients in $F$. It is clear that $F(\alpha,\beta,\gamma,\dots)$ is a field and is the smallest extension of $F$ which contains the elements $\alpha,\beta,\gamma,\dots$. We shall call $F(\alpha,\beta,\gamma,\dots)$ the field obtained after the adjunction of the elements $\alpha,\beta,\gamma,\dots$ to $F$, or the field generated out of $F$ by the elements $\alpha,\beta,\gamma,\dots$. In the sequel all fields will be assumed commutative.

What does he means by "quotient of polynomials in $\alpha,\beta,\gamma,\dots?$

答:
By "quotients of polynomials" Artin means rational functions of $\alpha,\beta,\gamma\dots$ like
$$\frac{c_1\alpha}{c_2\beta+c_3\gamma}$$
where the $c_i$ are elements of the smaller field $F$.


问:
Retract of $F_2$ onto $[F_2,F_2]$
I'm trying to show that the commutator subgroup  $[F_2,F_2]$ is not a retract of $F_2$. I was trying to do the proof by contradiction so I assume there is a retract $f: F_2 \to [F_2,F_2]$, then I know this is surjective and then I tried to apply abelianization to get the map $f^{ab}: \mathbb{Z}^2 \to {1}$ but this doesn't give me any contradictions since this map is surjective. Then I was thinking to apply the universal property of quotient groups somehow because I know $F_2/ [F_2,F_2] = \mathbb{Z}^2$ and that is a surjective map but I can't find anything that gives me a contradiction. I'm not sure where to go from here.

答: Abelianization does give you a contradiction; $[F_2, F_2]$ is a free group on countably many generators, so its abelianization is the free abelian group on countably many generators, and $f^{ab}$ can't be surjective.

Title: Retract of $F_2$ onto $[F_2,F_2]$


问:
$\mathbb{R}^{\mathbb{N}}$ Hilbert space with uncountable orthonormal basis?
Let's say we have a Hilbert space of sequences $u_n\in\mathbb{R}^{\mathbb{N}}$, equipped with some inner product $(\cdot,\cdot)$. Let's say there is a Hermitian operator $M$ acting on this space, with a continuous spectrum of eigenvalues $\lambda\in\mathbb{R}$. Because $M$ is Hermitian, its eigenvectors are automatically orthogonal to each other $(a^{\lambda},a^{\lambda'})=0$ for $\lambda\neq\lambda'$. The eigenvectors are assumed to be in the Hilbert space (i.e. have finite norm).

Owing to the existence of this Hermitian operator $M$, this Hilbert space has an uncountable orthonormal basis. I don't see any obstacles for such a scenario to occur, but it seems counter-intuitive for a sequence $u_n$ to have finite norm while also being orthogonal to an uncountable number of sequences.

Does this scenario make sense? Or is there something about $M$ and the Hilbert space which makes it impossible?

Edit: The reason I ask this question is that I seem to have come across a scenario in which this does in fact occur. The Hilbert space in question has the unusual inner product
$$(u,v)=\sum_{n=0}^{\infty}\frac{(2n+1)!!}{(2n)!!}u_nv_n.$$
And the Hermitian operator is
$$M_{nm}=(4n+3)\delta_{nm}+2n\delta_{n-1m}+(2n+3)\delta_{n+1m}.$$
You can check that it is Hermitian with respect to this inner product by noting that $\frac{(2n+1)!!}{(2n)!!}M_{nm}$ is symmetric. This operator can be written in the form $M=A+A^{\dagger}+3$, where
$$A_{nm}=2n\delta_{nm}+(2n+3)\delta_{n+1m}.$$
These operators obey the commutator
$$[A,A^{\dagger}]=2M.$$
This means in particular that
$$e^{cA}Me^{-cA}=e^{2c}M$$

Which implies that if $v$ is an eigenvector of $M$ with eigenvalue $\lambda$, then $e^{-cA}v$ is an eigenvector with eigenvalue $e^{2c}\lambda$, for any constant $c$. Thus providing a continuous spectrum of eigenvalues.

To find an eigenvector $v$, we may solve the recursive relation associated with the equation $Mv=\lambda v$, and find that there are normalizeable solutions for real $\lambda<0$, with asymptotic behavior
$$v_n\sim(-1)^n\frac{1}{\sqrt{n}}e^{-\sqrt{2n|\lambda|}},\quad n\to\infty$$

Hermiticity of $M$

Since this has been questioned a couple of times I will show that $M$ is Hermitian with respect to this inner product. As I mentioned it is equivalent to showing that $\frac{(2n+1)!!}{(2n)!!}M_{nm}$ is symmetric.

$$\frac{(2n+1)!!}{(2n)!!}M_{nm}=\frac{(2n+1)!!}{(2n)!!}(4n+3)\delta_{nm}+\frac{(2n+1)!!}{(2n-2)!!}\delta_{n-1m}+\frac{(2n+3)!!}{(2n)!!}\delta_{n+1m}$$

This is symmetric since

$$\frac{(2n+1)!!}{(2n-2)!!}\delta_{n-1m}=\frac{(2m+3)!!}{(2m)!!}\delta_{m+1n}.$$

Reviewing the definition of a Hermitian operator, it presumes it is a bounded operator, which we have found that this operator is not (assuming the rest of my calculations are correct). Perhaps the statement about orthogonality of eigenvectors is not true when the "Hermitian" operator is unbounded?

答:
Assuming you've defined your Hilbert space to be the space of sequences with finite norm with respect to your inner product, by rescaling appropriately, the Hilbert space of sequences you describe is isomorphic to $\ell^2(\mathbb{N})$, which is separable (admits a countable orthonormal basis), and hence which has the property that any set of orthonormal vectors in it must be at most countable. It follows that a bounded self-adjoint operator on $\ell^2(\mathbb{N})$ can have at most countably many distinct eigenvalues.

So your argument must break down somewhere, although I'm not sure where. Your calculations imply that the spectrum of $M$ is unbounded, so either those calculations are wrong or $M$ is unbounded. But I don't think even an unbounded self-adjoint operator on $\ell^2(\mathbb{N})$ can have uncountably many distinct eigenvalues. (I'm not even sure what an unbounded self-adjoint operator is, really.) So maybe $M$ is not self-adjoint, but I haven't checked this.


问:
Can we say anything about $\nu_p(x^n-y^n)$ if $p\nmid x-y$?
I know about the LTE lemma which says the following: If $p$ is a prime and $x,y,n$ are natural numbers with $p\nmid xy$ and $p\mid x-y$, then
$$ \nu_p(x^n-y^n)=\nu_p(x-y)+\nu_p(n),$$
where $\nu_p(n)$ is the multiplicity of $p$ in the prime factorisation of $n$.

I was wondering if we can make any statements about $\nu_p(x^n-y^n)$ if the conditions of the lemma do not hold; it is quite easy to reduce the case $p\mid xy$ to the case $p\nmid xy$, but I cannot immediately see how to make any statements about $\nu_p(x^n-y^n)$ in the case where $p\nmid xy$ and $p\nmid x-y$.

Can anyone offer any useful conclusions concerning $\nu_p(x^n-y^n)$ in this particular case or is it impossible to make any general statements here?

答:
(LTE in the form you stated it requires that $p$ be odd.)

As Greg Martin says in the comments, if $p \nmid xy$ then $p \mid x^n - y^n$ iff $\left( \frac{x}{y} \right)^n \equiv 1 \bmod p$. The set of $n$ for which this occurs consists of multiples of the order of $\frac{x}{y} \bmod p$, which can be any divisor of $p - 1$. If we write $d$ for the order, then information about $\nu_p(x^n - y^n)$ for $d \mid n$ can be obtained by applying LTE to $x' = x^d, y' = y^d$. The conclusion is that if $d \mid n$ (and $p$ is odd) then

$$\nu_p(x^n - y^n) = \nu_p(x^d - y^d) + \nu_p \left( \frac{n}{d} \right)$$

and if $d \nmid n$ then $\nu_p(x^n - y^n) = 0$.


问:
Finding the amount of prime numbers in $S_1, S_2,S_3,\ldots$
Let $\{a_n\}$ be a non-decreasing sequence of positive integers. For any positive integer $k,$ there are exactly $k$ terms in the sequence equal to $k.$ Let $S_n$ be the sum of first $n$ terms. How many prime numbers are there in the set $\{S_1,S_2,\ldots\}?$

My initial thought was to find a pattern of sorts, so I got
\begin{align*}
S_1 &= 1 \\
S_2 &= 3 \\
S_3 &= 5 \\
S_4 &= 8 \\
S_5 &= 11 \\
S_6 &= 14 \\
S_7 &= 18 \\
S_8 &= 22 \\
& \vdots \\
\end{align*}
I noticed that after a certain $S_n,$ all the numbers seemed to be composite no matter what. However, I wasn't quite sure how to prove this, so can somebody help me?

答:
Since the sequence was modified, the hint in the comment must be modified as well.

The $\dfrac {n(n+1)}2$-th term of the sequence is given by $\dfrac {n(n+1)(2n+1)}6$.

From this term to the $\dfrac {(n+1)(n+2)}2$-th term, the terms are given by $$\dfrac {n(n+1)(2n+1)}6 + k(n+1)=\frac {n+1}6(2n^2+n+6k)$$ for some integer $k$.

Since the above expression is always an integer, it cannot be prime for $n+1 >6$.


答:
Because the sequence is $(1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,…)$, the sum up to the end of a run of the same number is
$$S_{n(n+1)/2} = \frac{n(n+1)(2n+1)}{6}$$
which must be an integer. Therefore, the terms on the top of the fraction must contain a 2 and a 3 in their factorizations.

After $S_{n(n+1)/2}$, the following terms of $S$ are additions of $n+1$. If $n+1$ did not contribute in giving a 2 or a 3 to the fraction, then $n+1$ is a factor of each successive term, proving they are all composite. If it contributed a 2, then $(n+1)/2$ is a factor. A similar factor is found if it contributed a 3 or a 6. This proves that each term is composite whenever these factors are not equal to 1 or $n+1 \neq 1,2,3,6$. So one need only compute up to $S_{21}$.

Title: Finding the amount of prime numbers in $S_1, S_2,S_3,\ldots$

Body:
Let $\{a_n\}$ be a non-decreasing sequence of positive integers. For any positive integer $k,$ there are exactly $k$ terms in the sequence equal to $k.$ Let $S_n$ be the sum of first $n$ terms. How many prime numbers are there in the set $\{S_1,S_2,\ldots\}?$

My initial thought was to find a pattern of sorts, so I got
\begin{align*}
S_1 &= 1 \\
S_2 &= 3 \\
S_3 &= 5 \\
S_4 &= 8 \\
S_5 &= 11 \\
S_6 &= 14 \\
S_7 &= 18 \\
S_8 &= 22 \\
& \vdots \\
\end{align*}
I noticed that after a certain $S_n,$ all the numbers seemed to be composite no matter what. However, I wasn't quite sure how to prove this, so can somebody help me?


问:
automorphisms and field extension $E$ of $\mathbb{Q}$.
I want a hint. That is all I ask for. The question I am asked to prove is as follows:

Let $E$ be an extension field of $\mathbb{Q}$. Show that any automorphisn of $E$ acts as the identity on $\mathbb{Q}$.

答: Hint: $\phi(1) = 1 \implies \phi(n) =n$ for every $n \in \mathbb{Z}$.  Also, $\phi(n^{-1}) = \phi(n)^{-1}$</p>


问:
Rationalizing General Denominators
I've read on several answers how to rationalize a fraction of the form
$$
\frac{1}{\sqrt{a_{1}} + \dots +\sqrt{a_{n}}}
$$
by multiplying top and bottom by the numbers
$$\varepsilon_{1}\sqrt{a_{1}} + \dots +\varepsilon_{n}\sqrt{a_{n}},$$
where $\varepsilon_{i} \in \left\{1,-1\right\}$, avoiding the current denominator, that is, avoiding the number that corresponds to $\varepsilon_{i} = 1$, for all $i$. In fact, it is possible to do it restricting ourselves to the family
$$\sqrt{a_{1}} + \varepsilon_{2}\sqrt{a_{2}} + \dots +\varepsilon_{n}\sqrt{a_{n}},$$
that is, ignoring the terms where $\varepsilon_{1} = -1$. My questions are the following ones:

Is there a way of doing it involving less products (for n large)?

Can somebody give me a book or reference that deals with the more general case
$$
\frac{1}{\sqrt[m]{a_{1}} + \dots +\sqrt[m]{a_{n}}}
$$
or, more generally,
$$
\frac{1}{\sqrt[m_{1}]{a_{1}} + \dots +\sqrt[m_{n}]{a_{n}}},
$$
including an explicit algorithm and (if possible) a proof of its correctness?

Thank you in advance!

答:
Let $\alpha$ be the denominator. What you desire is the minimal polynomial of $\alpha$ in $\mathbb Q[x]$, which is the lowest degree polynomial $P$ with rational coefficients where

$$P(\alpha)=0$$

which means

$$\frac1\alpha=\frac{P(0)-P(\alpha)}{P(0)\alpha}=Q(\alpha)$$

is a polynomial in $\alpha$ and involves the minimum amount of products. There are many methods for finding such polynomials. If you wish you can even ask WolframAlpha.

The particular case where only square roots appear in the denominator can be seen here, where you can see several other methods, and generalizes to higher order radicals via using $\varepsilon_i$ being roots of unity. With some more theory, you will find that this factorization is minimal unless the products of powers of the terms are related e.g. $\sqrt[3]3+\sqrt[3]9$ has degree $3$ instead of degree $9$ (compare this to this).

As an example, the minimal polynomial of $\alpha=\sqrt2+\sqrt[3]3$ is

$$P(x)=x^6-6x^4-6x^3+12x^2-36x+1$$

which gives us the rationalization

$$\frac1{\sqrt2+\sqrt[3]3}=\frac1\alpha=36-12\alpha+6\alpha^2+6\alpha^3-\alpha^5$$

leaving expansion of the RHS (if it's what you want) to the reader.

Title: Rationalizing General Denominators


问:
A progression question with elements of geometry...
$ABCD$ is a square with area $1 cm^2$. Draw an isosceles right $\triangle CDE$, right angled at $E$. Draw another square $\square DEFG$, then draw an isosceles right $\triangle FGH$, right angled at $H$. This process is repeated infinitely so that no two figures overlap each other. Find the area of the entire figure.

Area of the first square is $1 cm^2$. Naturally, it's side is $1 cm$ and the side of the isosceles right $\triangle$ is equal to $\frac{1}{\sqrt2}$. And, it's area becomes:

$$\frac12 \times \left(\frac{1}{\sqrt2}\right)^2$$

And, the next square's area:
$$\left(\frac{1}{\sqrt2}\right)^2$$

I trust this question has to be answered using mathematical progression. But, I can't quiet keep my hands on simplifying the equation enough. Using the areas from above, I came up with:

$$1+\frac12\times\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2+\frac12\times\left(\frac{1}{\sqrt{2\sqrt2}}\right)^2+\left(\frac{1}{\sqrt{2\sqrt2}}\right)^2+\frac12\times\left(\frac{1}{\sqrt{2\sqrt{2\sqrt2}}}\right)^2...$$
$$\Rightarrow 1+\left(1+\frac12\right)\left(\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt{2\sqrt2}}\right)^2+\dotsb\right)$$
Now, squaring all, and taking out $\frac12$ out of the bracket from each,
$$1+\left(\frac32\right)\left[\frac12\left\{1+\left(\frac{1}{\sqrt2}\right)+\left(\frac{1}{\sqrt{2\sqrt2}}\right)+\dotsb\right\}\right]$$

答:
The area of the first square and triangle is $\frac54$ units and each succeeding such motif has linear dimensions $\frac1{\sqrt2}$ times that of the last, hence half the area. The area of the entire figure is thus
$$\frac54\left(1+\frac12+\frac14+\cdots\right)=\frac52$$


问:
Why every second even number have same number of even and odd Divisors.
Question: I was looking for natural numbers with same number of odd and even Divisors.Then I observed that, every second number in the list $2,4,6,8,10,12,14,...$ is such a number. Following list shows that, $2,6,10,14,...$ are even natural numbers that have same number of odd and even Divisors.

$2 \qquad 1 \qquad 1\quad\ast$

$4 \qquad 1 \qquad 2$

$6 \qquad 2 \qquad 2\quad\ast$

$8 \qquad 1 \qquad 3$

$10 \qquad 2 \qquad 2\quad\ast$

$12 \qquad 2 \qquad 4$

$14 \qquad 2 \qquad 2\quad\ast$

$16 \qquad 1 \qquad 4$

$18 \qquad 3 \qquad 3\quad\ast$

.........

How we can prove this in general? Is there is something special about this?

Note: I just copied the above list from  answer given here, https://math.stackexchange.com/a/48887/168676

答:
First of all, what property does every 2nd even number have? if we look at 2, 6, 10, 14, etc we see that they are all divisible by 2, but not by 4. The so the even numbers which dont have that property are all divisible by 4.

can we prove that all numbers with the property are going to be all numbers which are even and not divisible by 4? Yes, it isn't too difficult so I would recommend you give it a try, use the fundamental theorem of arithmetic. But if you dont want to try it, here's an outline of a proof:
Lets start by seeing why every number which is divisible by 2 but not by 4 will have this property.
Let $n\in\mathbb{N}$ such that $n$ is even but not divisible by 4. Using the fundamental theorem of arithmetic, we know that

$n=p_1^{a_1}\cdot p_2^{a_2}\cdot...\cdot p_k^{a_k}$

where each $p_i$ is a distinct prime number, and $a_i$ is some natural number. Since we know $n$ is even, we know one of those primes is 2, so WLOG let $p_1=2$ then we also know that $a_1=1$ as otherwise $n$ would be divisible by 4.

Now, how many numbers divide n? well if $m$ is a natural number that divides $n$, then $k=p_1^{b_1}\cdot p_2^{b_2}\cdot...\cdot p_k^{b_k}$

where $0\leq b_i\leq a_i$.
At this point it should be easy to see that the number of possible dividers will depend on the number of ways to assign all of the $b_i$s in particular, there are 2 options for $b_1$, and $a_i+1$ different options for each $b_i$. Meaning the total number of dividers will just be $(a_1+1)\cdot (a_2+1)\cdot...\cdot(a_k+1)$ However, for exactly half of these, $b_1=0$ which will be an odd divisor, and in the other half $b_1=1$ which will be an even divisor.

It should now be easy to see why odd numbers, and numbers divisible by 4 will not have the same number of even factors as odd factors.


问:
Proving integrability of the martingale $\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$
In trying to prove that $M_t:=\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$, I have hit a few roadblocks. I know how to show the martingale property as has been done many times in literature; but I haven't been able to properly justify that $\mathbb{E}|M_t|<+\infty$ so that I can finally conclude that it is a true martingale. Clearly $\textrm{e}^{-\frac12\lambda^2t}$ is bounded but the $\cosh$ isn't. How do I deal with the $\cosh$ and prove the integrability? Thanks!

答:
If $X \sim N(0,\sigma^{2})$ then $Ee^{cX} =e^{c^{2}\sigma^{2} /2} $ for any real number $c$. [In fact for any complex number $c$!]. Since $\cosh x=\frac {e^{x}+e^{-x}} 2$ we are done.  Ref:

https://en.wikipedia.org/wiki/Moment-generating_function

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjM8pqt_6btAhV1yDgGHZfpDi8QFjABegQIAhAC&url=https%3A%2F%2Fwww.le.ac.uk%2Fusers%2Fdsgp1%2FCOURSES%2FMATHSTAT%2F6normgf.pdf&usg=AOvVaw3QHSFjpCFrBgTFBxRwGAnK


问:
Can you prove that $f(x) = 0$ for every linear functional $f$ implies $x=0$ in a normed space without using Hahn-Banach?
In the context of another problem, I have to use the fact that if $f(x) = 0$ for all $f$, then $x=0$. I constructed my own proof of this using the Hahn-Banach theorem, which goes like this:

Proof: Assume that $x \neq 0$. Construct a functional $g$ in the following way: let $g(x) = 1$ and on $\text{span}(x)$ let $g$ be linear (this is well-defined on $\text{span}(x)$). By Hahn-Banach, $g$ can be continuously extended to the whole space, which contradicts the assumption that every functional is zero in $x$.

This is probably overkill for such a simple-looking statement, but I have no idea how to prove it in a more elementary way.

Edit:
I actually found a way to circumvent using Hahn-Banach (but not the axiom of choice). It goes like this:

Proof: Assume that $x \neq 0$. Let $(e_i)_{i \in I}$ be a basis of our vector space (where $I$ is some index set). Suppose that $x = \sum_{i \in I} x_i e_i$. Since $x \neq 0$, there exists an $x_i$ which is not zero, let's call its index $j$. Construct the following linear functional $g$: On $e_j$, let $g$ be 1, on every other basis vector, let $g$ be zero. In that case$$g(x) = g(\sum_{i \in I} x_i e_i) = \sum_{i \in I} x_i g(e_i) = x_j g(e_j) = x_j \neq 0 $$ which is a contradiction.

答:
As Ruy says in the comments, this is not a simple-looking statement at all. Without the Hahn-Banach theorem it's not clear that an arbitrary Banach space admits any nonzero continuous linear functionals, and in fact

The Hahn-Banach theorem is independent of ZF, and

Furthermore there are models of ZF containing Banach spaces with no nonzero continuous linear functionals (in other words, their continuous dual is trivial). See, for example, this MO answer, which describes a model of ZF in which $\ell^{\infty}/c_0$ (the Banach space of bounded sequences equipped with the sup norm, modulo the closed subspace of sequences converging to $0$) has trivial dual.


问:
Integral involving gaussian function
I would like to calculate the following integral:

$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\quad (x^2+y^2)^k\exp\left(-\dfrac{(x-x_0)^2+(y-y_0)^2}{a^2}\right)\,\mathrm dx\,\mathrm dy$$

Any clue on how to proceed?
Thanks

答:
You can use the fact that
$$(x^2+y^2)^k = \sum_{n=0}^k {k \choose n}x^{2n}y^{2(k-n)}$$
So
$$\mathrm{I}=\int_{\!-\infty}^{\!+\infty}\!\!\!\int_{\!-\infty}^{+\infty}dxdy(x^2\!\!+\!y^2)^k\! e^{-\frac{(x-x_0)^2+(y-y_0)^2}{a^2}}\!\!=\!\sum_{n=0}^k{k \choose n}\int_{-\infty}^{+\infty}\!\!\!x^{2n}e^{-\frac{(x-x_0)^2}{a^2}}dx  \!\!\int_{-\!\infty}^{\!+\infty}y^{2(k-n)}e^{-\frac{(y-y_0)^2}{a^2}}dy$$
One can work with it like with scaled plane moments of the normally distributed random variable. It is well known that $$\operatorname{E} \left[ X^p \right] =a^p \cdot (-i\sqrt{2}\mathrm{sgn}x_0)^p \; U\left( {-\frac{1}{2}p},\, \frac{1}{2},\, -\frac{1}{2}(x_0/a)^2 \right)$$ for a probability function $ \frac{1}{a\sqrt{2\pi}} e^{ -\frac{(x-x_0)^2}{2a^2} }$ but we have a scaled version (up to the multiplyer $\frac{1}{a\sqrt{\pi}}$).

So
$$\int_{-\infty}^{+\infty}\!\!\!x^{2n}e^{-\frac{(x-x_0)^2}{a^2}}dx=a\sqrt{\pi}\operatorname{E} \left[ X^{2n} \right]$$
and $$\int_{-\infty}^{+\infty}\!\!\!y^{2(k-n)}e^{-\frac{(y-y_0)^2}{a^2}}dx=a\sqrt{\pi}\operatorname{E} \left[ Y^{2(k-n)} \right]$$
So one can obtain:
$$\mathrm{I}=(-1)^k (a^2)^{k+1}\pi \sum_{n=0}^k{k \choose n}(\mathrm{sgn}x_0)^{2n}(\mathrm{sgn}y_0)^{2(k-n)}U\left(\!\!-\!n\!,\!\frac{1}{2}\!,\!-\frac{x_0^2}{a^2} \right)U\left(\!\!n\!-\!k\!,\!\frac{1}{2}\!,\!-\frac{y_0^2}{a^2} \right)$$
If $x_0=y_0=0$ then   $$\mathrm{E}\left[X^{2n}\right] =  \left(\frac{a}{\sqrt{2}}\right)^{2n}\,(2n-1)!! $$ and  $$\mathrm{E}\left[Y^{2(k-n)}\right] = \left(\frac{a}{\sqrt{2}}\right)^{2(k-n)}\,(2(k-n)-1)!! $$ So everything gets even simpler:
$$\mathrm{I}=\pi \left(\frac{a^2}{2}\right)^{k+1}\sum_{n=0}^k{k \choose n}(2n-1)!!(2k-2n-1)!!$$

Title: Integral involving gaussian function


问:
Sum of Indicators: $\sum_{n=1}^{\infty} \mathbb{1}\{X_1^2/\alpha^2 \geq n\} = \left(1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor\right)$
I was working on proving the following proposition:

Suppose $(X_n)$ is some iid sequence of random variables such that $\mathbb{E}(X_1)^2 < \infty$. Then for every $\alpha > 0$ we have

$$\mathbb{P}\left(\limsup_{n \to \infty}|X_n| \geq \alpha\sqrt{n}\right) = 0.$$

Proof. We wan't to invoke the Borel-Cantelli Lemma (B.C.), so since $X_n$ are iid we have

$$\sum_{n=1}^{\infty}\mathbb{P}(|X_n| \geq \alpha\sqrt{n}) = \sum_{n=1}^{\infty}\mathbb{P}(X_n^2 \geq \alpha^2 n) = \sum_{n=1}^{\infty}\mathbb{P}\left(\frac{X_1^2}{\alpha^2} \geq n\right).$$

Using the Monotone Convergence Theorem, we can represent the right-most term as

$$ \sum_{n=1}^{\infty}\mathbb{P}\left(\frac{X_1^2}{\alpha^2} \geq n\right) =  \sum_{n=1}^{\infty}\mathbb{E}\left[\mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\}\right] = \mathbb{E}\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\}.$$

Therefore,

$$\mathbb{E}\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = \mathbb{E}\left(1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor\right) \leq 1 + \frac{\mathbb{E}X_1^2}{\alpha^2} < \infty.  \tag{1}$$

So, we can invoke B.C. and are done. The bit about the indicator equality involving the floor function is given as a hint without going into detail.

Question.  How do we get the equality involving the floor in (1)? Specifically, I don't see how

$$\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = 1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor.$$

Could someone expand on this derivation please?

答:
I think there is a slight mistake. (I am interpreting $\left\lfloor x \right\rfloor$ as the greatest integer less than or equal to $x$).  If the sums are  from $0$ to $\infty$ then you get the value as stated, but otherwise you will get $\left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor$ instead of $\left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor +1$

Suppose $\left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor=k$. Then $k \leq \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor <k+1$. So  the only integers $n$ such that $\frac{X_1^2}{\alpha^2}\geq n$ are $n=1,2,...,k$. So when you add the indicators on LHS you see that $k$ terms are $1$ and the rest are zeros. Hence the sum is exactly $k$.


问:
What's the difference between Pick's formula and the Shoelace formula(Gausse's formula)?
So I was thinking of making a GUI in Python that calculates the area of a shape that the user draws. Therefore, I would need some sort of equation to find the area. I googled it and I found the Shoelace formula(made by Gauss himself); right here: https://en.m.wikipedia.org/wiki/Shoelace_formula

I read it and all of their examples are for polygons. But the user can draw circular shapes, so I did more research and found Pick's formula; right here: https://www.youtube.com/watch?v=vYmemgPa87I

And I think that Pick's formula is a lot easier to use. But what's the difference? Is there one better than the other or do they have their pros and cons? Thanks.

答:
But what's the difference?

Pick's theorem works only if the polygon vertices have integer coordinates.

Shoelace formula works with any real coordinates.

If you have a sequence of tuples corresponding to the consecutive points along the path the user drew, then the Shoelace formula is trivial to implement.

Let's say path is that sequence, with path[0] being the starting point, and path[len(path)-1] the final point.  Then, we can use the fact that in Python, path[len(path)-1] == path[-1]:

```<code>def pathArea(path):
    # Less than three points, and the area is zero.
    if len(path) < 3:
        return 0

    A = 0.0
    for i in range(0, len(path)):
        A += path[i-1][0] * path[i][1] - path[i][0] * path[i-1][1]

    return 0.5 * abs(A)
</code>```

Mathematically, the above uses
$$A = \frac{1}{2}\left\lvert \sum_{i=0}^{n-1} x_{i-1} y_{i} - x_{i} y_{i-1} \right\rvert$$
where $n$ is len(path), and $x_{-1} = x_{n-1}$ and $y_{-1} = y_{n-1}$ matching the final point in path.

When drawing, you can use line segments between the points, but note that you'll only want to add a new vertex if it differs from the previous one.  So, if self.currentPath is the path being drawn, and newX and newY are the current coordinates being drawn, use

```<code>    if (self.currentPath is None) or (len(self.currentPath) < 1):
        self.currentPath = [ (newX, newY) ]
    elif (self.currentPath[-1][0] != newX) or (self.currentPath[-1][1] != newY):
        self.currentPath.append( (newX, newY) ]
</code>```

so that the vertex list will not contain consecutive duplicate points.

Note that the above pathArea() function does not need the path to be closed, to start and end with the same point; there is an implicit line segment (polygon side) between the starting and the ending point.

If the user draws a closed figure with line segments poking out, drawing α instead of a nice circle, you will need to trim off the extra vertices.  Otherwise the actual area calculated is that of a fish shape, as an extra line segment is added between the starting and ending points.

To do that, you need to find vertex $S$ near the beginning of the path, and vertex $E$ near the end of the path, such that the line segment from the vertex $S$ to the next vertex (from $(x_S , y_S)$ to $(x_{S+1}, y_{S+1})$), intersects the line segment from the vertex preceding $E$ to vertex $E$ (from $(x_{E-1}, y_{E-1})$ to $(x_E , y_E)$); with $S < E$.  If you find the intersection, remove all initial vertices on the path up to $S$, and all vertices from $E$ to end, including both vertices $S$ and $E$, and prepend or append the intersection point to the path.

Body:
But what's the difference?

Pick's theorem works only if the polygon vertices have integer coordinates.

Shoelace formula works with any real coordinates.

If you have a sequence of tuples corresponding to the consecutive points along the path the user drew, then the Shoelace formula is trivial to implement.

Let's say path is that sequence, with path[0] being the starting point, and path[len(path)-1] the final point.  Then, we can use the fact that in Python, path[len(path)-1] == path[-1]:

def pathArea(path):
    # Less than three points, and the area is zero.
    if len(path) < 3:
        return 0

    A = 0.0
    for i in range(0, len(path)):
        A += path[i-1][0] * path[i][1] - path[i][0] * path[i-1][1]

    return 0.5 * abs(A)

Mathematically, the above uses
$$A = \frac{1}{2}\left\lvert \sum_{i=0}^{n-1} x_{i-1} y_{i} - x_{i} y_{i-1} \right\rvert$$
where $n$ is len(path), and $x_{-1} = x_{n-1}$ and $y_{-1} = y_{n-1}$ matching the final point in path.

When drawing, you can use line segments between the points, but note that you'll only want to add a new vertex if it differs from the previous one.  So, if self.currentPath is the path being drawn, and newX and newY are the current coordinates being drawn, use

    if (self.currentPath is None) or (len(self.currentPath) < 1):
        self.currentPath = [ (newX, newY) ]
    elif (self.currentPath[-1][0] != newX) or (self.currentPath[-1][1] != newY):
        self.currentPath.append( (newX, newY) ]

so that the vertex list will not contain consecutive duplicate points.

Note that the above pathArea() function does not need the path to be closed, to start and end with the same point; there is an implicit line segment (polygon side) between the starting and the ending point.

If the user draws a closed figure with line segments poking out, drawing α instead of a nice circle, you will need to trim off the extra vertices.  Otherwise the actual area calculated is that of a fish shape, as an extra line segment is added between the starting and ending points.

To do that, you need to find vertex $S$ near the beginning of the path, and vertex $E$ near the end of the path, such that the line segment from the vertex $S$ to the next vertex (from $(x_S , y_S)$ to $(x_{S+1}, y_{S+1})$), intersects the line segment from the vertex preceding $E$ to vertex $E$ (from $(x_{E-1}, y_{E-1})$ to $(x_E , y_E)$); with $S < E$.  If you find the intersection, remove all initial vertices on the path up to $S$, and all vertices from $E$ to end, including both vertices $S$ and $E$, and prepend or append the intersection point to the path.


问:
Find appropriate initial value of an exponential trend model
I was trying this exercise question but wasn't able to find a solution, so please help.

For the following trend model, $y={\mu}(x,{\beta})+{\epsilon = {\alpha-exp(-{\gamma}x)+{\epsilon}}}, {\alpha}>0, {\gamma}>0$,

where ${\beta}=({\alpha, {\gamma}})^T$. Then I was given a few given observations of x and y such as

(x1=1, y1=1.33;

x2=2, y2=1.55;

x3=4, y3=1.80;

x4=8, y4=1.96).

The question is to find appropriate initial values for $({\alpha0}, {\gamma}0)$.

I was trying to make a linear transformation for this model by using natural log but wasn't able to proceed as $log({\alpha}+exp(-{\gamma}x)+{\epsilon})$ does not give me too much insight...so please help, thanks in advance!

答:
I shall work with rational numbers to avoid losses of accuracy.

For the time being, just to simplify notations, replace $\gamma$ by $-\gamma$

Your equations are
$$\alpha -e^{-\gamma}=\frac {133}{100} \tag 1$$
$$\alpha -e^{-2\gamma}=\frac{31}{20} \tag 2$$
$$\alpha -e^{-4\gamma}=\frac{9}{5} \tag 3$$
$$\alpha -e^{-8\gamma}=\frac{49}{25} \tag 4$$

Now, subtracting
$$e^{-\gamma }-e^{-2 \gamma }=\frac{11}{50} \tag 5$$
$$e^{-2 \gamma }-e^{-4 \gamma }=\frac{1}{4} \tag 6$$
$$e^{-4 \gamma }-e^{-8 \gamma }=\frac{4}{25}\tag 7$$

Let $x=e^{-\gamma}$ and build the norm
$$\Phi=\left(x-x^2-\frac{11}{50}\right)^2+\left(x^2-x^4-\frac{1}{4}\right)^2+\left(x^4-x^8-\frac{4}{25}\right)^2$$ Expand
$$\Phi=x^{16}-2 x^{12}+\frac{58 x^8}{25}-2 x^6+\frac{109 x^4}{50}-2 x^3+\frac{47
   x^2}{50}-\frac{11 x}{25}+\frac{273}{2000}$$
$$\Phi'=16 x^{15}-24 x^{11}+\frac{464 x^7}{25}-12 x^5+\frac{218 x^3}{25}-6 x^2+\frac{47
   x}{25}-\frac{11}{25}$$ That is to say that we need to solution of
$$400 x^{15}-600 x^{11}+464 x^7-300 x^5+218 x^3-150 x^2+47 x-11=0$$ Plotting, we see that there is a single root close to $x=0.670$ that is to say $\gamma \sim-\log(0.670) \sim 0.400$.

Now, compute $(1)+(2)+(3)+(4)$ and solve for $\alpha$
$$\alpha=\frac{83}{50}+\frac{1}{4}\left(e^{-\gamma }+e^{-2\gamma }+e^{-4 \gamma }+e^{-8 \gamma }\right)$$ Replace $e^{-\gamma }$ by $0.670$ to get $\alpha\sim 2.00$.

Now, recompting the values, we find for the four rhs the values $$1.32968,1.55067,1.79810,1.95924$$

This almost looks like the solution.

Edit

If you want something much faster (probably in the mind of your teacher), you need $\alpha > 1.96$; so, say $\alpha_0=2$. Now, use $(5)$ which is a quadratic in $e^{-\gamma}$
$$x-x^2=\frac {11}{50} \implies x_\pm=\frac{5\pm\sqrt{3}}{10} $$
Now, look at $(6)$ for each of the values
$$x_+ \implies e^{-2 \gamma }-e^{-4 \gamma }=\frac{11 \left(39+10 \sqrt{3}\right)}{2500}\approx 0.24781$$ which is obviously fine.

So,
$$e^{-\gamma}=\frac{5\pm\sqrt{3}}{10}\implies \gamma=-\log \left(\frac{1}{10} \left(5+\sqrt{3}\right)\right)\sim 0.40$$


问:
Proof verification: $A, B$ are real symmetric commuting matrices then they share a common eigenvector.
I'm asked to prove if $A, B$ are real symmetric matrices that commute then they share a common eigenvector.

My approach: Since $A$ is real symmetric, it has an orthonormal eigenbasis. Now if all of these vectors are in the kernel of $B$ then $B$ has to be $0$ matrix. So WLOG lets take an eigenvalue $\lambda$ of $A$ and $W$ be the corresponding eigenspace for it, which is non trivial under $B$. As $B$ commutes with $A$, $A(Bv)= BA(v)=\lambda Bv$ amd hence $Bv \in W$ for any $v \in W$ so $B(W) \subset W$.

Now let us take the restriction of $B$ to $W$. We can now find a non zero eigenvector (perhaps with some other eigenvalue) $w'$ of $B$ in $W$ since $W$ does not lie in the kernel of $B$. Then this $w'$ is a common eigenvector for $A$ and $B$.

Is this proof correct? Did I use the symmetric information correctly? I don't think we needed both $A$ and $B$ to be symmetric here, since we are using that fact for only $A$.

答:
Your proof is incomplete as you didn't show why $B$ has an eigenvector in $W$. If you let $A$ be the identity matrix then you've shown that all real matrices have real eigenvectors. This is not true. For instance, the nontrivial rotation matrices have no real eigenvectors.

You need to use the fact that $B$ is symmetric to prove that $B|_W$ has an eigenvector. Indeed, you can just show that $B|_W$ is symmetric. Write $B$ in the orthonormal eigenbasis you specified at the start. Since it's orthonormal, this will still be symmetric. Without loss of generality put the eigenvectors in $W$ at the start of this basis. Then as $B[W] \subseteq W$, $B$ will contain a top left block corresponding to $B|_W$. Hence, $B|_W$ is symmetric so it has a real eigenvector.


问:
Confusion in the equivalence relation statement
In my classical analysis book, Chapter 1, it is written that:

Equivalence relation $(=)$ must be:

$E_1$ (reflexive) $a=a$

$E_2$ (symmetric) $a=b\ \Rightarrow b=a$

$E_3$ (transitive) $a=b, b=c \Rightarrow a=c$

QUESTION:

(i) It seems to me $E_1, E_2$ and $E_3$ are $(1)$ self evident axioms $(2)$ always true.

Are there any situations when any one or two out of $E_1,E_2,E_3$ being false?

(ii) If no, then should not the must be be replaced with are always?

答:
This is a common source of confusion for students. This textbook is not making statements about the usual equality operator (=) that you normally see in equations such as $1 + 1 = 2$. In fact, I think the authors made a poor choice by using the = symbol at all. They probably would have been better off using a tilde or another symbol which does not have a standard meaning.

Instead, this book is setting out requirements that any equivalence relation is required to satisfy. In other words, it's telling you what the phrase "equivalence relation" actually means.

To review definitions which you've probably seen earlier in the book:

The Cartesian product $X \times Y$ of two sets ($X$ and $Y$) is the set of all ordered pairs $(x, y)$, where $x$ ranges over all elements of $X$ and $y$ ranges over all elements of $Y$.

A binary relation (or sometimes just a "relation") is a subset of the Cartesian product. In other words, it's a set consisting of ordered pairs $(x, y)$ where $x \in X$ and $y \in Y$.

To make things easier to notate, relations are often given symbols. If you have some relation $R$, and its symbol is $\sim$, then you might write $x \sim y$ instead of writing $(x, y) \in R$ (but they both mean exactly the same thing). You might also refer to the relation as a whole as $\sim$ instead of $R$.

It may be the case that $X = Y$. If you want to have an equivalence relation, then this must be the case.

An equivalence relation is a particular kind of binary relation, which satisfies the requirements you have listed in your question. The actual equality relation obviously satisfies those requirements, but so does the relation "is parallel to" over the set of lines in the plane (when we take the convention that every line is parallel to itself). Therefore, the relation "is parallel to" is also an equivalence relation.


问:
Planar Geometry question
Suppose I have three points $p$, $f_1$, and $f_2$. I want to place a third point $f_3$ such that if you extend the line segments $pf_1$ and $pf_2$ into full lines, $f_3$ is going to be on the opposite side of $pf_1$ as $f_2$ is, and $f_3$ will also be on the opposite side of $pf_2$ from where $f_1$ is. What I want to show is that the point $p$ will be in the triangle formed by $f_1$, $f_2$, and $f_3$. I'm about 98% sure this is a correct statement, I'm just not sure how to go about proving its true beyond "drawing pretty pictures".

EDIT Clarification: If you take the line through $p$ and $f_1$, then you have partitioned the plane into two parts. The assumption is that $f_3$ must be in the partition that $f_2$ is not in.

答:
Here is an idea:

By rotating, translating, reflecting, and rescaling, we can assume that $p$ is the origin, $f_1=(1,0)$, and $f_2=(x_0,y_0)$ with $y_0>0$.

Case $1$: $x_0=0$.  Then $f_3=(x_1,y_1)$ is any point in the third quadrant.  The line from $f_3$ to $f_2$ has positive $y$-intercept, and the line from $f_3$ to $f_1$ has negative $y$-intercept, so that $p$ is inside the triangle formed by $f_1$, $f_2$, and $f_3$.

Case $2$: $x_0>0$.  Then $f_3=(x_1,y_1)$ is any point satisfying

$$\frac{y_0}{x_0}x_1<y_1<0$$

These are exactly the inequalities you need to show that the line from $f_3$ to $f_2$ has positive $y$-intercept, and the line from $f_3$ to $f_1$ has negative $y$-intercept, so that $p$ is inside the triangle formed by $f_1$, $f_2$, and $f_3$.

Case $3$: $x_0<0$.  Proceed as in Case $2$ using the inequalities:

$$y_1<0,\qquad y_1<\frac{y_0}{x_0}x_1$$

Title: Planar Geometry question


问:
How to get rank-in probabilities?
Suppose that we have n instances $X_1,\dots,X_n$ $(X_i \in \mathbb{R})$ uniformly sampled from the population without replacement.
We do not know the distribution of $X_i$ but know that the population size is N.

How to estimate the probabilities $Pr(pos(X_i) \geq T)$?

where $1\leq pos(X) \leq N$ indicates the position of $X_i$ in the ordered list of the entire instances in the population; the ordered list is obtained through sorting by the values of $X_i$ in descending order.
Here, $T$ $(1 \leq T \leq N)$ is a certain position in the ordered list.

答:
I'm going to rewrite the problem just a little bit, please let me know if I've changed the intent of your question.

We have a collection of $N$ random variables $X_1,\ldots,X_N$. We do not know the distribution of the $X_i$s but we do know the collection is independent and identically distributed, meaning all the $X_i$ follow the same distribution and the $X_i$ are mutually independent.

I'm going to make one assumption that I think you intended: the probability distribution of the $X_i$s is continuous which will imply $P(X_i=X_j)=0$ if $i\neq j$. This will give us that with probability $1$ all the $X_i$ will be different values. This helps avoid some extra complications with ties.

Here's the part I want you to reason out for yourself: for an arbitrary $i$ and $j$ such that $i\neq j$ what are $P(X_i< X_j)$ and $P(X_i > X_j)$? Take a pause here.

Once you accept that $P(X_i< X_j)=P(X_i > X_j)$, then you can see that they both must be $1/2$. Extending out this line of reasoning further, you'll see that all $N!$ orderings of the $N$ random variables are equally likely with probability $\frac{1}{N!}.$

With this insight, you can deduce that the probability that a specific random variable, say $X_i$, is ranked exactly $s$th is $\frac{(N-1)!}{N!} = \frac{(N-1)!}{N(N-1)!} = \frac{1}{N}$. The probability that $X_i$ is in the top $T$ can be found by summing these, so you get $\frac{1}{N}+\cdots + \frac{1}{N} = \frac{T}{N}$.

Title: How to get rank-in probabilities?


问:
Steps to calculate the tangential acceleration and normal acceleration of a vector
I am given $r(t) = \langle\cos(3) , 1 − \sin(3) , 3 \sin(3) + \cos(3)\rangle$ and $t= \pi /4$

I have calculated $r'(t)$ and  $r''(t)$.

I don't know what to do next to get either tangential or normal acceleration component. Any pointers?

答:
For any regular curve  $r(t)$, the velocity is $\dot r(t)$; this vector is clearly tangent to $r(t)$ for every $t$; the acceleration vector is evidently $\ddot r(t)$, but it is not in general tangent to $r(t)$.  We may find both the tangential and normal components of $\ddot r(t)$; first, we form the unit tangent vector field $T(t)$ to $r(t)$:

$T(t) = \dfrac{\dot r(t)}{\vert \dot r(t) \vert}, \tag 1$

where we have used the assumed regularity if $r(t)$ that is,

$\dot r(t) \ne 0, \tag 2$

in order to form $T(t)$, since (2) is equivalent to

$\vert \dot r(t) \vert \ne 0. \tag 3$

The tangential component of any vector field $v(t)$ along $r(t)$ is then

$v_T(t) = (v(t) \cdot T(t)) T(t); \tag 4$

in particular, taking

$v(t) = \ddot r(t), \tag 5$

we see that the tangential component of the acceleration is

$\ddot r_T(t) = (\ddot r(t) \cdot T(t)) T(t); \tag 6$

the component of acceleration normal to $r(t)$ is thus

$\ddot r_N(t) = \ddot r(t) - \ddot r_T(t) = \ddot r(t) - (\ddot r(t) \cdot T(t)) T(t); \tag 7$

indeed we see that

$\ddot r(t) = \ddot r_T(t) + \ddot r_N(t), \tag 8$

and

$\ddot r_N(t) \cdot T(t) = \ddot r(t) \cdot T(t) - (\ddot r(t) \cdot T(t)) T(t) \cdot T(t)$
$= \ddot r(t) \cdot T(t) - (\ddot r(t) \cdot T(t))  = 0.  \tag 9$

We note that $\ddot r_T(t)$ and $\ddot r_N(t)$ may both be expressed solely in terms of $r(t)$, that is, without explicit reference to $T(t)$.  We substitute (1) into (6):

$\ddot r_T(t) = \left (\ddot r(t) \cdot \dfrac{\dot r(t)}{\vert \dot r(t) \vert} \right ) \dfrac{\dot r(t)}{\vert \dot r(t) \vert} = \dfrac{\ddot r(t) \cdot \dot r(t)}{\vert \dot r(t) \vert^2}\dot r(t), \tag{10}$

and also into (7):

$\ddot r_N(t) = \ddot r(t) - \ddot r_T(t) = \ddot r(t) - \left (\ddot r(t) \cdot \dfrac{\dot r(t)}{\vert \dot r(t) \vert} \right ) \dfrac{\dot r(t)}{\vert \dot r(t) \vert}$
$= \ddot r(t) - \dfrac{\ddot r(t) \cdot \dot r(t)}{\vert \dot r(t) \vert^2}\dot r(t). \tag{11}$

I leave calculating $\ddot r_T(t)$ and $\ddot r_N(t)$ for the specific curve given in the text of the question itself to the sufficiently engaged amongst my readers; it is not difficult.


答:
I guess there's a small mistake in the end. Fix any $\omega \in \Omega$ . Then, $\frac{X_1(\omega)^2}{\alpha^2}$ is a finite quantity, so $\exists~ N_\omega \in \mathbb{N}$ such that $$\frac{X_1(\omega)^2}{\alpha^2} \geqslant  N_\omega\quad~\text{ and }~\quad \frac{X_1(\omega)^2}{\alpha^2} < N_\omega +1$$ and hence, $\left\lfloor\frac{X_1(\omega)^2}{\alpha^2}\right\rfloor = N_\omega$

Thus, for $n = 1 , 2 , \cdots , N_\omega$ , we have : $\mathbb{1}\left\{\frac{X_1(\omega)^2}{\alpha^2} \geq n\right\} = 1$ and otherwise, $0$ .

Therefore, $$\sum_{n=1}^{\infty}\mathbb{1}\left\{\frac{X_1(\omega)^2}{\alpha^2} \geq n\right\} = \sum_{n=1}^{N_\omega}\mathbb{1}\left\{\frac{X_1(\omega)^2}{\alpha^2} \geq n\right\} = N_\omega = \left\lfloor\frac{X_1(\omega)^2}{\alpha^2}\right\rfloor$$
Since this result is true for any $\omega \in\Omega$ , therefore : $$\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor$$

It is to be noted that even after this correction, Borel - Cantelli lemma works fine, so you'll obtain what you wanted. Hope it helps.

Title: Sum of Indicators: $\sum_{n=1}^{\infty} \mathbb{1}\{X_1^2/\alpha^2 \geq n\} = \left(1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor\right)$

Body:
I was working on proving the following proposition:

Suppose $(X_n)$ is some iid sequence of random variables such that $\mathbb{E}(X_1)^2 < \infty$. Then for every $\alpha > 0$ we have

$$\mathbb{P}\left(\limsup_{n \to \infty}|X_n| \geq \alpha\sqrt{n}\right) = 0.$$

Proof. We wan't to invoke the Borel-Cantelli Lemma (B.C.), so since $X_n$ are iid we have

$$\sum_{n=1}^{\infty}\mathbb{P}(|X_n| \geq \alpha\sqrt{n}) = \sum_{n=1}^{\infty}\mathbb{P}(X_n^2 \geq \alpha^2 n) = \sum_{n=1}^{\infty}\mathbb{P}\left(\frac{X_1^2}{\alpha^2} \geq n\right).$$

Using the Monotone Convergence Theorem, we can represent the right-most term as

$$ \sum_{n=1}^{\infty}\mathbb{P}\left(\frac{X_1^2}{\alpha^2} \geq n\right) =  \sum_{n=1}^{\infty}\mathbb{E}\left[\mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\}\right] = \mathbb{E}\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\}.$$

Therefore,

$$\mathbb{E}\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = \mathbb{E}\left(1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor\right) \leq 1 + \frac{\mathbb{E}X_1^2}{\alpha^2} < \infty.  \tag{1}$$

So, we can invoke B.C. and are done. The bit about the indicator equality involving the floor function is given as a hint without going into detail.

Question.  How do we get the equality involving the floor in (1)? Specifically, I don't see how

$$\sum_{n=1}^{\infty} \mathbb{1}\left\{\frac{X_1^2}{\alpha^2} \geq n\right\} = 1 + \left\lfloor\frac{X_1^2}{\alpha^2}\right\rfloor.$$

Could someone expand on this derivation please?


问:
How to prove $\lim\limits_{n\to\infty}\int^\pi_0 \pi f(nx)g(x)\ \mathrm {d}x = \int^\pi_0 f(x)\ \mathrm {d}x \int^\pi_0 g(x)\ \mathrm {d}x$
Question

Let $f$ be a continuous function on $\mathbb {R}$ such that $f(x + \pi) = f(x)$ for all $x \in \mathbb {R}$. Prove that, if $g$ is a continuous function on $[0, \pi]$, then $$\lim\limits_{n\to\infty}\int^\pi_0 \pi f(nx)g(x)\ \mathrm {d}x = \int^\pi_0 f(x)\ \mathrm {d}x \int^\pi_0 g(x)\ \mathrm {d}x.$$

This question just appeared in my final calculus examination this afternoon and I could not do it ):. I am just really bad at proofs. However, on reading the question, the condition of $f(x + \pi) = f(x)$ on $f$ tells me that $f$ must be some kind of periodic function and I see $\pi$ all around, so I am guessing we will have to make use of some trigonometry. Any intuitive explanations as to how the equality can be proven will be greatly appreciated :)

答:
Let $\epsilon>0$. Choose $N$ so that whenever $|x-y|<\pi/N$, we have $|f(nx)g(x)-f(nx)g(y)|<\epsilon/\pi$. This can be done since both $f$ and $g$ are uniformly continuous functions.

Now, for $n\ge N$

$$\eqalign{
\int_0^\pi \pi f(nx)g(x)\,dx&= \sum_{m=1}^n \int_{(m-1)\pi/n}^{m\pi/n}\pi f(nx)g(x)\,dx\cr
&\le\sum_{m=1}^n \Bigl[{\epsilon\over\pi}\cdot{\pi\over n}
+\int_{(m-1)\pi/n}^{m\pi/n} \pi f(nx) g(m\pi/n)\,dx\Bigr]\cr
&=\epsilon +\sum_{m=1}^n\Bigl[ g(m\pi/n) \int_{(m-1)\pi}^{m\pi} {\pi\over n}f(u)\,du\Bigr]\cr
&=\epsilon +\sum_{m=1}^n \Bigl[g(m\pi/n)\cdot{\pi\over n} \int_0^\pi f(x)\,dx\Bigr]\cr
&=\epsilon + \int_0^\pi f(x)\, dx \cdot\Bigl[\sum_{m=1}^n g(m\pi/n) \cdot{\pi\over n}\Bigr]
}$$

Since $\epsilon$ is arbitrary,
taking limits of both sides of the above gives
$$\limsup_{n\rightarrow\infty}\int_0^\pi \pi f(nx)g(x)\,dx \le \int_0^\pi f(x)\,dx \cdot\int_0^\pi g(x)\, dx.$$
In a similar manner, ones shows
$$
\liminf_{n\rightarrow\infty}\int_0^\pi \pi f(nx)g(x)\,dx \ge \int_0^\pi f(x)\,dx \cdot\int_0^\pi g(x)\, dx,
$$
whence the result follows.


问:
Continuous function that is only differentiable on irrationals
Can you help me finding a function $f : \mathbb{R} \rightarrow \mathbb{R}$ that is continuous in $\mathbb{R}$ and differentiable at $x$ iff $x \notin \mathbb{Q}$  ?

Thank you very much !

答:
Following David Mitra's comment :

Let's define the following Lebesgue-Stieltjes measure $m$ defined such that $m(E)=\sum_{q_{j}}\frac{1}{2^{j}}I_{q_{j}}(E)$, where $I_{q_{j}} = 1$ iff $q_{j} \in E$. Moreover, let's restrict $q_{j}$ to be the rationals that are in $E$ and are smaller than $\sup(E)$.  We can see that the $g(x) = \sum_{q_{j}}\frac{1}{2^{j}}H(x-q_{j})$, where $H$ is the Heaviside function, is such that $m(]a,b]) = g(b) - g(a)$.
Therefore, we know that $g$ is continuous at $x$ iff $m(\lbrace x \rbrace)= 0$. Thus, we can conclude that $g$ is continuous in $\mathbb{R}$. Finally, it's clear that $g$ is not differentiable on $\mathbb{Q}$, since the right and left limits of $g(x+h)-g(x)$ are not the same.

You think this is ok ?

BTW, thank you for all comments and for the beautiful answer regarding Weierstrass Function :)


问:
On expressing a subspace as intersection of kernels of linearly independent functionals
I tried to prove the following proposition

Let $X$ be a vector space over field $K$ of dimension $n<\infty$ and $X^*$ its algebraic dual. A subset $E\subseteq X$ is a subspace of dimension $p$ iff there exist linearly independent functionals $f_1,\ldots,f_{n-p}\in X^*$ such that $E=\bigcap_{i=1}^{n-p}f_i^{-1}(0)$.

I can prove $\Rightarrow$ by 1) taking complement dual basis corresponding to basis elements not in $E$, and 2) constructing an map $f$ from $X$ to $K^{n-p}$ as $f(x)=\bigl(f_1(x), \ldots\,f_{n-p}(x)\bigr)^T$ so $E$ is the kernel of $f$. But I was not able to proceed with $\Leftarrow$. For example, I tried to find $x_1,\ldots,x_{n-p}\in X$ such that they are independent and $f_j(x_i)=\delta_{ij}$ from independence of $f_1,\ldots,f_{n-p}$, but failed. So, can you please help me prove $\Leftarrow$ (you don't have to follow my unsuccessful attempt)? Thank you.

答:
First, find $x_1,\ldots,x_{n-p}\in X$ such that $f_j(x_i)=\delta_{ij}$. Applying the same construction of complementary basis$^*$ to independent $f_1,\ldots,f_{n-p}$ in dual space $X^*$, we got $g_1,\ldots,g_{n-p}$ which satisfies $g_j(f_i)=\delta_{ij}, i,j=1..n-p$. Taking $x_1,\ldots,x_{n-p}\in X$ to be the inverse images of the canonical mapping from $X$ to $X^{**}$, we have $f_j(x_i)=g_j(f_i)=\delta_{ij}$ from the definition of $g_j$'s.

Second, define $f$ from $X$ to $K^{n-p}$ as $f(x)=\bigl(f_1(x), \ldots\,f_{n-p}(x)\bigr)^T$, which is a linear transformation. We have, from the above construction, $$f(x_1)=(1,0,\ldots,0)^T\\
f(x_2)=(0,1,\ldots,0)^T\\
\dots\\
f(x_{n-p})=(0,0,\ldots,1)^T.$$
So, $f$ is surjective because $f\bigl(\sum\limits_{i=1}^{n-p}a_ix_i\bigr)=(a_1,a_2,\ldots,a_{n-p})$ for any $a_i\in K$. It can be seen that $E$ is the kernel of $f$, thus it is a subspace. From the rank-nullity theorem, ${\rm{rank\ of\ }}f+{\rm{nullity\ of\ }}f=n$ where ${\rm{rank\ of\ }}f=\dim K^{n-p}=n-p$, we get $\dim E={\rm{nullity\ of\ }}f=p$.

*: See, for example, Nathan Jacobson, "Lectures in Abstract Algebra, II. Linear Algebra", P53, or Erwin Kreyszig, "Introductory Functional Analysis with Applications", P114.


问:
$\{f_n\}$ be a sequence of continuous functions converging pointwise to $0$, show $\lim \int_0^1 f_n dx = 0$
This problem comes from Rudin's Real and Complex Analysis and is Exercise 2.10. Let $\{f_n\}$ be a sequence of continuous functions such that $0\leq f_n\leq 1$ and $f_n(x)\to0$ for all $x\in [0,1]$. We need to show that $$\lim_{n\to\infty}\int_0^1 f_n(x)dx = 0$$
In particular I am trying to show this without using Measure Theory or Lebesgue Integration.

What I have tried so far is to consider the sequence of functions $\{g_n\}$ defined by $$g_n(x)= \min\{f_1(x),\ldots f_n(x)\}$$
Then $\{g_n\}$ is a continuous, decreasing sequence of functions which converge pointwise to $0$. Since the sequence of continuous functions decreases to a continuous function, then we may apply Dini's Theorem to know that $g_n \to 0$ uniformly. Hence $$ \lim_{n\to 0}\int_0^1 g_n(x)dx = \int_0^1\lim_{n\to0}g_n(x)dx = 0$$
The only thing left to show is that $$ \lim_{n\to 0}\int_0^1 f_n(x)dx= \lim_{n\to 0}\int_0^1 g_n(x)dx $$
But this is where I am stuck. Again I am avoiding measure theory proofs.

答:
Isn't it a direct consequence of Lebesgue's Dominated convergence theorem ?

$(1)$ $f_n(x) \longrightarrow f(x) \equiv 0$ pointwise $\forall~x\in[0,1]$

$(2)$ For every $n\in\mathbb{N}$ , $|f_n(x)| < g(x) \equiv 1~~\forall~x \in[0,1]$

Therefore, by Lebesgue's DCT, $$\lim_{n\to\infty}\int_0^1 f_n(x)dx = \lim_{n\to\infty}\int_0^1 0~dx = 0$$