问:
Unilateral weighted shift operator and existence of unitary operator; spectrum of weighted shift

Let $(a_{n})$ be a bounded sequence of complex numbers and $\mathscr{H}$ be a Hilbert space with orthonormal basis $\{ e_{n} \}$.

Show that for every complex number $\lambda$ with $|\lambda| = 1$ there is a unitary (bounded) operator $U = U_{\lambda}$ such that $UAU^{-1} = \lambda A$, where $A$ is the unilateral weighted shift such that $A e_{n} = a_{n} e_{n + 1}$.

Also deduce that the spectrum of a weighted shift must be the union of concentric circles about $z = 0$.

I already showed that there is a unilateral weighted shift $A$ and that it is unique. Stuck on the unitary operator $U$.

答:
Let $Ue_n:=\lambda^ne_n$, where $\lambda=e^{i\theta}$, $\theta\in\mathbb{R}$.

Then its adjoint is $U^*e_n=\lambda^{-n}e_n$, since $\langle U^*e_n,e_m\rangle=\langle e_n,\lambda^m e_m\rangle=\bar{\lambda}^m\delta_{nm}$.

So $U^*Ue_n=\lambda^n \lambda^{-n}e_n=e_n=UU^*e_n$. Hence $U$ is unitary; moreover,

\begin{align}UAU^{-1}e_n&=UA(\bar{\lambda}^ne_n)\\
&= \bar{\lambda}^na_nUe_{n+1}\\
&=\bar{\lambda}^n\lambda^{n+1}a_ne_{n+1}\\
&=\lambda a_ne_{n+1}=\lambda Ae_n\end{align} Since they agree on all basis vectors, $UAU^{-1}=\lambda A$.


问:
$\sum_{k=1}^n [X_k-E(X_k)]$ converge in probability
$\{X_n\}$ are independent, $\sum_{k=1}^\infty Var(X_k)<\infty$,prove
$$\sum_{k=1}^n [X_k-E(X_k)]
$$converge in probability.

I have tried chebyshev inequality, but it's too loose.

答:
WLOG shift so that $\mathbb E[X_n]=0$ for all $n$. Let $S_n=\sum_{k=1}^nX_k$. Then we can just apply Chebyshev's inequality: for $1\leq m\leq n$, we have
\begin{align*}
\mathbb P(\lvert S_n-S_m\rvert>\epsilon)=\mathbb P\left(\left\lvert\sum_{k=m+1}^nX_k\right\rvert>\epsilon\right)&\leq\frac{1}{\epsilon^2}\operatorname{Var}\left(\sum_{k=m+1}^nX_k\right)\\
&\leq\frac{1}{\epsilon^2}\sum_{k>m}\operatorname{Var}(X_k).
\end{align*}
The RHS converges to $0$ as $m\to\infty$, since $\sum\operatorname{Var}(X_k)<\infty$, so we're done.

In fact, from one of many lemmas due to Levy, the independence of the $X_k$ implies that we in fact have almost sure convergence too.

ContentLicense: CC BY-SA 4.0


问:
Proving that surface is not elementary
I need help with proving or disproving that the following surface $x^2+y^2+z^2=a^2,a>0$ is elementary.
If I had another condition $z>0$ then the surface would be elementary because $rank g'=2$ where $g(x,y)=(x,y,\sqrt{a^2-x^2-y^2})$. But here I know that solution is that this set is not elementary, but not sure how to prove that using this definition with rank. Every help is appreciated.

答:
Here's the beginning of an approach:

Suppose we had such a homeomorphism $g : U \to \mathbb R^3$. Then, the component functions would satisfy
$$x(u, v)^2 + y(u, v)^2 + z(u, v)^2 = a^2,\text{ for all } u, v \in U.$$

The derivative $g'$ is of the form
$$g'(u, v) =  \begin{pmatrix}x_u(u, v) & x_v(u, v) \\ y_u(u, v) & y_v(u, v) \\ z_u(u, v) & z_v(u, v) \end{pmatrix}, $$
and we'd like to show that for all points $(u, v)$ this matrix cannot have rank 2. Note that differentiating the first equation with respect to $u$ and $v$ leads to two equations:
$$2 x x_u + 2y y_u + 2z z_u = 0,$$
$$2 x x_v + 2y y_v + 2z z_v = 0.$$

I claim that these two equations lead to the rank of $g'$ being smaller than 2. Perhaps you can show that there is at least one point $(u, v)$ where one of the 2 by 2 minors
$$x_u y_v - x_v y_u,\ x_u z_v - x_v z_u,\ y_u z_v - y_v z_u$$
vanishes.

Title: Proving that surface is not elementary


问:
Is there any analytic method for solving $\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$
Hi i am trying to solve the following two equations:
$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$
$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$
where $\dot{\theta}=\frac{d\theta}{dt}$ and $\dot{\phi}=\frac{d\phi}{dt}$ and $\ddot{\theta}=\frac{d^2\theta}{dt^2}$ and $\ddot{\phi}=\frac{d^2\phi}{dt^2}$. I am trying to solve this analytically and looked up the standard techniques but couldn't find how solve these. Basically i want $\phi$ in terms of $\theta$ that  is $\phi(\theta)$. Is there any way to do this?

答:
I think your system can be reduced to a Bernoulli Differential Equation: $y'(x)+p(x)\,y(x) = q(x) \, [y(x)]^n$.

$$$$

Rewrite $\dot{\phi}=\phi'(t)$ and $\ddot{\phi}=\phi''(t)$ by applying the chain rule for $\phi(\theta(t))$ and define $u(\theta)=\phi'(\theta)$ to get

$$u'(\theta) + 2 \cot(\theta) \, u(\theta) + \sin(\theta) \cos(\theta) \, [u(\theta)]^3 = 0 $$

$$$$

I will double-check this partial solution and write it down here step-by-step later.

$$$$

EDIT:

Considering the chain $\phi - \theta - t$, it's possible to write:

$$\frac{d\phi}{dt} = \frac{d\phi}{d\theta} \frac{d\theta}{dt}$$

$$\frac{d}{dt} \left[\frac{d\phi}{d\theta}\right] = \frac{d^2\phi}{dtd\theta}  = \frac{d^2\phi}{d\theta^2} \frac{d\theta}{dt}$$

$$\frac{d^2\phi}{dt^2} = \frac{d}{dt} \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] = \frac{d^2\phi}{dtd\theta} \, \frac{d\theta}{dt} + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} = \frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2}$$

Returning those results to your original system:

$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$

$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$

$$$$

$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} + 2\cot{\theta} \, \frac{d\theta}{dt} \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] =0$$

$$\frac{d^2\theta}{dt^2}- \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] ^2=0$$

$$$$

$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \frac{d^2\theta}{dt^2} + 2\cot{\theta} \, \frac{d\theta}{dt} \, \left[\frac{d\phi}{d\theta} \frac{d\theta}{dt}\right] =0$$

$$\frac{d^2\theta}{dt^2} = \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^2 \left[\frac{d\theta}{dt}\right]^2$$

$$$$

$$\frac{d^2\phi}{d\theta^2} \left[\frac{d\theta}{dt}\right]^2 + \frac{d\phi}{d\theta} \, \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^2 \left[\frac{d\theta}{dt}\right]^2 + 2\cot{\theta} \, \frac{d\phi}{d\theta} \, \left[\frac{d\theta}{dt}\right]^2 =0$$

$$$$

So you have a trivial equation, $[\theta'(t)]^2=0$, and

$$\frac{d^2\phi}{d\theta^2} + \sin\theta \cos\theta \, \left[\frac{d\phi}{d\theta}\right]^3 + 2\cot{\theta} \, \frac{d\phi}{d\theta} =0$$

$$$$

Define $u(\theta)=\phi'(\theta)$ to get a Bernoulli Differential Equation,

$$u'(\theta) + 2 \cot(\theta) \, u(\theta) + \sin(\theta) \cos(\theta) \, [u(\theta)]^3 = 0 \text{,}$$

which presumably has analytical solutions.


问:
Periodic point near Hyperbolic fixed point
This question is the last exercise of chapter 2 in Lan Wen`s Differential Dynamical system. (Exercise 2.12)

let $E$ a finite-dimensional normed vector space and $p \in E$ be a hyperbolic fixed point of $f$. Given any positive integer $m$, prove there is a neighborhood $V$ of $p$ such that any period point of $f$ in $V-{p}$ has a period greater than $m$.

here $U$ is open subset of $E$ and  $ f: U\longrightarrow E $ is $C^k$  and local Diffeomorphism.

答:
You can do it by contradiction.

Suppose this is not true. You will find a $M$ and a sequence of point $p_k \to p$ of periodic point converging to $p$ with period less than $M$.

Taking a sub-sequence, they can have the same period $m \leq M$ so they are fixed for $f^m$. Notice that $p$ is still a hyperbolic fixed point of $f^m$.

Now taking again a sub-sequence, they can approch $p$ in the same direction, that is $\frac{p-p_k}{\| p -p_k \|} \to v \in E$.

Computing the differential in the $v$ direction for $f^m$ with this sequence will give you a $0$ and this cannnot be since it is suppose hyperbolic.


问:
Continuous function from $(0,1) \to \mathbb{R}$ with minimun and maximum.

Construct a continuous function from $(0,1) \to \mathbb{R}$ with minimun and maximum.

I was convinced that $f(x) = \sin^2(\pi x)$ would have met these conditions, but it seems that this one doesnt have a minimum? Some cubic would also probably work here?

答: Hint: Try to construct a piecewise function by joining lines together at specific points. In this way, you can create a non-constant function with a maximum and a minimum.


问:
$Z(GL_n(\mathbb R)) = \{aI : a\neq0\} $
$Z(GL_n(\mathbb R)) = \{aI : a\neq0\} $

This article is the general case for $GL(n,k)$ where $k$ is a field. Could I prove it only with a basic linear algebra?

答:
Let $A \in Z(GL_n(\mathbb R))$.

Let $B$ be defined as $b_{ii}=i$ and $b_{ij}=0$ if $i\neq j$ (or any other diagonal matrix with pairwise distinct entries).

Then since $AB=BA$, for all $i \neq j$ we have

$$\sum_{k=1}^n a_{ik}b_{kj}=\sum_{k=1}^n b_{ik}a_{kj} \Rightarrow a_{ij}b_{jj}=b_{ii}a_{ij}\,.$$

Since $i \neq j$ we have $b_{ii} \neq b_{jj} \Rightarrow a_{ij}=0$.

This proves that $A$ is a diagonal matrix.

Now, use the fact that $A$ commutes with permutation matrices. Let $\sigma=(i,j)$ be any transposition and $P_\sigma$ the corresponding matrix.

Then $AP_\sigma=P_\sigma A$ implies $a_{ii}=a_{jj}$.

P.S. If you work over a finite field with less elements that the size of the matrix, instead of a matrix $B$ in the first part you need to consider a family $B_{i,j}$ of diagonal matrices so that $b_{ii} \neq b_{jj}$. AND, you need to study separately the case $k=F_2$.

ContentLicense: CC BY-SA 3.0

Title: $Z(GL_n(\mathbb R)) = \{aI : a\neq0\} $


答:
First, we must get rid of the middle man, $t$. In other words, we will make the following observations:

$$\dfrac{d\phi}{dt}=\dot\theta\dfrac{d\phi}{d\theta}$$ and
$$\dfrac{d^2\phi}{dt^2}=\dfrac{d}{dt}\dot\theta\dfrac{d\phi}{d\theta}=\ddot\theta\phi_{\theta}+\dot\theta^2\phi_{\theta\theta}.$$

Notice now how we have $\ddot\theta$ as $\dot\phi^2\sin\theta\cos\theta$. So the first equation becomes $$\dot\theta^2\sin\theta\cos\theta\cdot\phi^3_{\theta}+\dot\theta^2\phi_{\theta\theta}+2\cot\theta\dot\theta^2\phi_{\theta}=0$$ or $$\sin\theta\cos\theta\cdot\phi^3_{\theta}+\phi_{\theta\theta}+2\cot\theta\phi_{\theta}=0$$ This is the equation for $\phi(\theta)$. If you were not too picky about the parameter $t$, this could have been solved by [other answers on this post].
Now, notice that dividing through by $\phi_\theta$, we have
$$\sin\theta\cos\theta\cdot\phi^2_{\theta}+\dfrac{\phi_{\theta\theta}}{\phi_{\theta}}+2\cot\theta=0$$ Because we have a that second term as being $$\dfrac{d}{d\theta}\log(\phi_{\theta})$$ substitute $\phi=\int e^{\beta}\mathrm{d}\theta$. Thus,
$$\cos\theta\sin\theta e^{2\beta}+\beta_{\theta}+2\cot\theta=0$$ We're almost done. Divide through by $e^{2\beta}$, and subtract the beta-like terms to obtain
$$M+Le^{-2\beta}=\dfrac{d}{d\theta}e^{-2\beta}$$ where $M=2\sin\theta\cos\theta$ and $L=4\cot\theta$. This evidently takes the form
$$y'+ay=b$$ which has the general solution $$y=e^{-\int a}(k+\int b e^{\int a})$$ for some constant $k$, and so
$$e^{-2\beta}=e^{4\int \cot\theta\mathrm{d}\theta}\left[2\int \sin\theta\cos\theta \cdot e^{-4\int\cot\theta \mathrm{d}\theta}\mathrm{d\theta}+k\right]$$ which we know that
$$\phi =\int e^{\beta}\mathrm{d\theta}=\int \dfrac{e^{-2\int \cot\theta\mathrm{d}\theta}}{\sqrt{2\int \sin\theta\cos\theta \cdot e^{-4\int\cot\theta \mathrm{d}\theta}\mathrm{d\theta}+C_0}}\mathrm{d\theta}.$$

Edit A more complete solution would be where one notes that $e^{\int \cot \theta}=|\sin \theta|$ I'm ignoring constants because they're a pain, other than the one obtained by the linear solution. Then, $$\phi =\int \dfrac{1}{\sin^2\theta\sqrt{2\int \dfrac{\cos\theta}{\sin ^3\theta}\mathrm{d\theta}+C_0}}\mathrm{d\theta}$$ And since
$$\int \dfrac{1}{\sin^3\theta}d\sin \theta=-\dfrac{1}{2\sin^2\theta}$$ we have simply that
$$\phi = \int \dfrac{1}{\sqrt{C_0\sin^4\theta-\sin^2\theta}}d\theta=\arctan\left(\sqrt{C_0\sin^2\theta-1}\right)-\dfrac{\arctan\left(\dfrac{\sqrt{C_0\sin^2\theta-1}}{\sqrt{1-C_0}}\right)}{\sqrt{1-C_0}}$$

Please comment if I made any errors.

Title: Is there any analytic method for solving $\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$

Body:
Hi i am trying to solve the following two equations:
$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$
$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$
where $\dot{\theta}=\frac{d\theta}{dt}$ and $\dot{\phi}=\frac{d\phi}{dt}$ and $\ddot{\theta}=\frac{d^2\theta}{dt^2}$ and $\ddot{\phi}=\frac{d^2\phi}{dt^2}$. I am trying to solve this analytically and looked up the standard techniques but couldn't find how solve these. Basically i want $\phi$ in terms of $\theta$ that  is $\phi(\theta)$. Is there any way to do this?


问:
Help with $\lambda \in \Bbb C$ so that $f(z)= \lambda \cdot g(z)$ $\forall z \in \Bbb C$
Let $f$ and $g$ be two entire functions, s.t. $|f(z)|\leq |g(z)|$ $\forall z \in \Bbb C$. I wanna show that it exists $\lambda \in \Bbb C$, with $|\lambda |\leq 1$ so that $f(z)= \lambda \cdot g(z)$ $\forall z \in \Bbb C$.

Let's observe $h:= \frac{f}{g}$, with $g$ not equal to zero. By assumption $\vert h \vert \leq 1$ so all the singularities of h, which are isolated by the identity principle, are removable by the Riemann extension theorem.

Why can we conclude this?

答:
As pointed out in the comments, either $g$ vanishes identically (and in this case the conclusion is trivial, since both $g$ and $f$ are $0$) or it has isolated zeroes.

In the latter case, the quotient $h$ is in modulus $\le1$ and defines an holomorphic function on the whole plane minus a discrete set (the zeros of $g$). By Riemann thm, as pointed out, $h$ extends across these isolated singularities to an entire function which is then bounded.
Then by Liouville theorem $h$ is constant, that is, $\exists \lambda\in\Bbb C$ such that $0<|\lambda|\le1$ such that $h\equiv\lambda$.
Since $h=f/g$ you get the conclusion.


问:
Fourier Transform of Even/Odd Complex Functions
For any complex-valued $f \in L^1(\mathbb{R})$, let's define its Fourier transform $\hat{f}$ with the following convention
$$
\hat{f}(\omega) := \int_{\mathbb{R}} f(x) e^{-i \omega x} dx
$$

I would like a confirmation of the following:

$f$ even $\Rightarrow$ $\hat{f}$ even

$f$ odd  $\Rightarrow$ $\hat{f}$ odd

To prove for example the first statement, I would argue that
$$
\hat{f}(\omega)
= \int_{\mathbb{R}} f(x) e^{-i \omega x} dx
= \int_{\mathbb{R}} f(-x) e^{i \omega x} dx
= \int_{\mathbb{R}} f(x)  e^{i \omega x} dx
= \int_{\mathbb{R}} f(x)  e^{-i (-\omega) x} dx
= \hat{f}(-\omega)
$$
where the second equality is a consequence of the fact that the Lebesgue measure ($\lambda$) satisfies  $\lambda(A) = \lambda(-A)$, for every Borel set $A$ and the third is just using the hypothesis, i.e. that $f(x) = f(-x)$.

The second statement can be proved in the very same way.

Is the above correct?!?

I started having second thoguhts after seeing this heavily downvoted answer

Fourier transform of even/odd function
and especially its first comment.

答: Yes, you are correct. The corresponding parity of the Fourier transform follows from the parity of the function and the fact that the Lebesgue integral is invariant under reflection (which follows from the relative invariance of Lebesgue measure).</p>


问:
Finding $r$ given differential equations.
Suppose $y(x)=e^{rx}$. And we have following differential equations:

(1) $y'+y=e^x+e^{-x}$

(2) $y'+2y=e^{-x}$.

Then after substitution we get

(1) $re^{rx}+e^{rx}=e^x+e^{-x}$

(2) $re^{rx}+2e^{rx}=e^{-x}$

Is it safe to conclude that there is no such $r$ that satisfies (1) or (2)?

答:
These are linear ODE solve them by finding integrating function
The solution of $y'+Py_=Q$ is given by $$y=I(x)^{-1} \int I(x) Q(x) dx+ A I(x)^{-1}, I(x)=e^{\int p(x) dx}.$$
Here for
(1) $$y'+y=e^{x}+e^{-x}\implies y=e^{-x}\int e^{x}(e^{x}+e^{-x}) dx+C e^{-x} \implies y=e^{-x} (e^{2x}/2+1)+Ce^{-x} \implies y=e^{x}/2+Ae^{-x}$$


问:
Card Expected Value with Option to Skip Cards
In a normal deck of cards, you can either reveal the top card or guess whether that card is black. If you reveal the top card, you get to see what the card is and the game continues with one less card in the deck. If you were to make a guess, the game ends, and you get paid out $\$100$ if the card is black and $\$0$ if it's red.

What is the optimal strategy of this game and its expected value?

The lower bound has to be $0.5\times \$100 = \$50$ since you can just guess on the first card.

答:
There really is no optimal strategy: If you start with an equal number of black and red cards, you have a 50:50 chance of winning (hence an expected value of $50) no matter what you do (as long as you pick some card if the remaining deck gets down to all black). You can prove it by induction.

Given a deck consisting of $b$ black cards and $r$ red cards, the probability of winning with an optimal strategy, $P(b,r)$, satisfies the recursion

$$P(b,r)=\max\left\{{b\over b+r},{b\over b+r}P(b-1,r)+{r\over b+r}P(b,r-1) \right\}$$

That is, you can either guess the next card is black, in which case you win with probability $b/(b+r)$, or you can pass, in which case you'll be left with either one fewer black card hence probability of winning $P(b-1,r)$ or one few fewer red card hence probability of winning $P(b,r-1)$, with probabilities $b/(b+r)$ and $r/(b+r)$ respectively; optimality amounts to making the choice that gives the higher probability, i.e., taking the max. In particular, if the two choices are equal, then it doesn't matter which choice you make. And that's what we want to prove:

$${b\over b+r}P(b-1,r)+{r\over b+r}P(b,r-1)={b\over b+r}$$

Now it's clear that $P(1,0)=1$ and $P(0,1)=0$. This is the base case. So if we assume, inductively, that $P(b-1,r)=(b-1)/(b-1+r)$ and $P(b,r-1)=b/(b+r-1)$, then we have

$$\begin{align}
{b\over b+r}P(b-1,r)+{r\over b+r}P(b,r-1)
&={b(b-1)\over(b+r)(b-1+r)}+{rb\over(b+r)(b+r-1)}\\
&={b(b-1+r)\over(b+r)(b-1+r)}\\
&={b\over b+r}
\end{align}$$


问:
Optimal Strategy in Dice Game
In a recent interview I received the following question (an optimisation/strategy game)...which left me a bit stumped. The rules of play, you start with 0 points, then:

Roll three fair six-sided dice;

Now you have the option:

Hold: i.e. accept the values shown on your dice as the score for your turn. There is a caveat, if two or more dice show the same values, then all of them are flipped upside down - e.g. 1 becomes 6

OR

reroll the dice: You may choose to hold any combination of the dice on the current value shown (so you can choose to keep 1 dice the same and then reroll the other two). Rerolling costs you 1 point – so during the game and perhaps even at the end your score may be negative.
You can roll an infinite number of times...

My thoughts:

So clearly the best possible score is 18 and is achieved by rolling three 1s on the first roll
The reroll penalty prevents rolling forever to get 18.
If the value of the dice is greater than the expected value of rerolling them (accounting for the penalty), then you should stick...
I guess what I am asking is how do I work out the expected value of rerolling them (accounting for the penalty) and how does this fit into the optimal strategy...

答:
One important point is that, anytime in the game, your best action depends only on the dice on the table, and not on the number of penalties you've accumulated so far. Even if you have -20 in penalties, if your dice are bad, maybe you should reroll (and worry that the dice aren't fair after all).

Apart from that, I don't see much good properties on the problem, so to me it's a modeling problem and you have to compute it.

You start with an benefit function $f_0(\{x,y,z\})$ which is the value of your dice combination, and you compute successively the benefit if you also add the choice to reroll at most $i$ times : $f_{i+1}({x,y,z}) = \max f_i(\{x,y,z\}), (-1 + \mathbb E [f_i(\{X,y,z\})]), \dots$.

Since the $f_i$ are increasing and bounded, they converge (exponentially fast) to a $f_\infty$, and you can make your choice by comparing $f_\infty(\{x,y,z\})$ and $(-1 + \mathbb E [f_\infty(\{X,y,z\})])$, etc.

If you want the exact values of the expectation, you can see the transformation of $f_i$ to $f_{i+1}$ as a matrix multiplication (once you reached the point where the choice of the $\max$ will be stable), so you just have to find its stable point.

Title: Optimal Strategy in Dice Game


问:
Converting $r=(x^2+y^2+z^2)^{1/2}$ cartesian coordinates to spherical coordinates
Let $r=(x^2+y^2+z^2)^{1/2}$ be the radial distance from the origin expressed in cartesian coordinates.

I have been asked to express this in cylindrical coordinates.

Is $r=(x^2+y^2+z^2)^{1/2}$ the same as $r=x\hat{e}_x + y\hat{e}_y + z\hat{e}_z$

if yes then to convert to cartesian coordinate:

$R=\sqrt{x^2+y^2}=R$

$\theta=tan^{-1}(\frac{y}{x})=tan^{-1}===??$

$z=z$

i dont' understand what values we have put in?

答:
You have to use these $$\left\{\begin{aligned}x&=\rho \cos \varphi \\y&=\rho \sin \varphi \\z&=z\end{aligned}\right.$$
in order to convert from cartesian to cylindrical coordinates.

$r=(x^2+y^2+z^2)^{1/2}$ becomes $r=(\rho^2+z^2)^{1/2}$


问:
In how many ways can 4-digit numbers be formed using 2,2,8,8 if each digit is used once only?
In how many ways can 4-digit numbers be formed using 2,2,8,8 if each digit is used once only?

I'm confused as to how to solve this problem. If the question was "How many ways can 4-digit numbers be formed using 2 and 8 if each digit can be used any number of times?", the answer should be $2 \times 2 \times 2 \times 2 = 16$, since there a 2 possible digits (2 and 8) that can be used for each of the 4 positions.

However I'm not quite sure of how to solve the problem in the title.

答: If the number starts with a $2$, there are are three places we can put the other $2$. If the number starts with an $8$, there are three places we can put the other $8$. Since the number must start with a $2$ or an $8$, there are $\boxed{3+3=6}$ possibilities.


答:
The flaw in your reasoning is that your answer doesn't account for the exhaustion of digits, that is, if the first two places are filled with both the $2$'s, then there is only one way for the rest, which your solution counts as $2×2=4$.

You can rectify it as:

If the first two digits are filled with same digit, then total number of ways is $2×1×1×1=2$.

If the first two digits are distinct, then total number of ways is $2×2=4$.

Thus, total number of ways $=$ $\boxed{2+4=6}$.

ALTERNATIVE SOLUTION:

Think of it in pairs. The two $2$'s and $8$'s can form four pairs: $22$, $28$, $82$ and $88$. All the question remains is to find arrangement of two pairs, which can be done in ${}^4\text{P}_2=12$ ways. Since this is symmetrical for either $2$, the answer is $\boxed{6}$.

OR you can think of it, as fixing two $2$'s and filling the rest of places with $8$'s, giving the same result as above.

Hope this helps. Ask anything if not clear. Have a wonderful day ahead :)

Body:
In how many ways can 4-digit numbers be formed using 2,2,8,8 if each digit is used once only?

I'm confused as to how to solve this problem. If the question was "How many ways can 4-digit numbers be formed using 2 and 8 if each digit can be used any number of times?", the answer should be $2 \times 2 \times 2 \times 2 = 16$, since there a 2 possible digits (2 and 8) that can be used for each of the 4 positions.

However I'm not quite sure of how to solve the problem in the title.


问:
Weak convergence equivalent to weak convergence at $0$
Let $(X, \lVert \cdot \rVert)$ be a Banach space and $(T(t))_{t\geq0}$ a semigroup of linear operators , i.e for all $t, s \geq 0$ we have $T(t):X \to X$ is a linear operator, $T(t+s)=T(t)T(s)$ and $T(0)=$ Id.

For arbitrary but fixed $x\in X,~l \in X^{'}$ consider the mapping
$$
F_{l, x}: \mathbb{R}_+ \to \mathbb{R}, t \mapsto F_{l, x}(t):=l(T(t)x)
$$
Is continuity of $F_{l,x}$ at $t=0$, i.e.
$$
\underset{t\downarrow 0}{\lim} l(T(t)x) = l(x)
$$ enough for continuity everywhere?

[The motivation is 1. that weak continuity implies strong continuity, and 2. that there's the equivalent statement for strong continuity.]

答:
Yes because
$$
  l\big (T_{t+h}(x)\big ) - l\big (T_t(x)\big ) =   l\Big (T_{t}\big (T_h(x) - x\big )\Big ) =
  l'\big (T_h(x) - x\big ),
  $$
where
$
  l'=l\circ T_t.
  $


问:
Prove that every natural number is either even or odd using induction
It is a pretty basic question at first sight. It seems to be intuitively correct but I can't figure out a way to prove it. I think we have to use strong induction but even after sitting on this question for a long time, I still haven't got any leads.</p>

答:
The induction proof will be a lot easier to understand if the 'odd' and 'even' aspect is made algebraic from the outset.

Theorem

Every natural number is of the form $2k$ or $2k+1$ for some integer $k$.

Proof

The first natural number, $1$, is of the form $2\times 0+1$.

Suppose the theorem to be true for the natural number $n$. Then $n$ is either $2k$ or $2k+1$.

Then $n+1$ is either $2k+1$ or $2k+2=2(k+1)$. Hence the theorem is true for the natural number $n+1$.

By induction the theorem is true for all natural numbers.


问:
What is the Fourier series of $f(x) = \begin{cases} -2, & -4<x<-2 \\ x, & {-2<x<2} \\ 2, & {2<x<4} \end{cases}$
What is the Fourier series of $f(x) = \begin{cases}  -2, & -4<x<-2 \\  x, & {-2<x<2} \\  2, & {2<x<4}  \end{cases}$

I have already established that this function is an odd function and has a period of 8, however, I do not know how to move forward with this. Any explanation on what definition of $f(x)$ should I follow would be greatly appreciated.

答:
Hint.

If the function $f$ is odd on the symmetric interval $[-L,L]$, then you have the Fourier sine series

$$
f(x)\sim \sum_{n=1}^\infty a_n\sin (\frac{n\pi x}{L})
$$
where the coefficients are given by
$$
a_n=\frac{2}{L}\int_0^L f(x)\sin (\frac{n\pi x}{L})dx
$$

Notes.

In your example,

$$
\int_0^4 f(x)\sin (\frac{n\pi x}{4})dx
=\int_0^2x\sin (\frac{n\pi x}{4})dx
+\int_2^42\sin (\frac{n\pi x}{4})dx
$$


问:
Find the Taylor expansion of the function $f(z)=z/(z+1)^2$.
Find the Taylor expansion of the function  $f(z) = \frac{1}{(1+z)^2}$ about $z_0=i$

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

I know that
$\frac{1}{z+1}=\sum_{n=0}^{\infty}(-1)^n\frac{(z-i)^n}{(1+i)^{n+1}}$,

but I don't know how do $\frac{1}{(z+1)^2}$

答:
Hint

$$\frac{1}{(1+z)^2}=-\frac{\mathrm d }{\mathrm d z}\left(\frac{1}{1+z}\right).
$$


问:
Atiyah Ex5.29: Local ring of a valuation ring
Let $A$ be a valuation ring of a field $K$. Show that every subring of $K$ which contains $A$ is a local ring of $A$.

This problem is already asked and answered at mathoverflow. But I can't understand why $PA_P \subset M_B$ at step (b) of the answer.
Or there will be another way to show $B \subset A_P$. How can I prove it?

答: Recall that $P=A\cap M_B$. Since $M_B$ is an ideal of $B$, and $A_P\subset B$, every element of $PA_P$ is an element of $M_B$. </p>


问:
Chaos in finite field
Let's think about some finite field $\mathbb{F}$. Is it possible to construct a map

$x[n+1] = \mathcal{P}(x[n], x[n-1],...,x[n-k]), \ \ \  \forall x\in\mathbb{F} $

where $\mathcal{P}$ - polynomial with coeffs $\in\mathbb{F}$, so it's behavior  be non-periodic, but chaotic,

so x[n] be "jumping" randomly over $\mathbb{F}$?

(As $x[n+1] =1-2\cdot x[n]^2$ on [-1,1])

答: If $x[n+1]$ depends on $d$ of the previous values in the sequence (i.e. $x[n+1]=P(x[n],x[n-1],\ldots,x[n-d+1])$, the answer is negative. There are only $|\mathbb{F}|^d$ possible $d$-tuples of the values, so once any of them repeats (Dirichlet's principle), the subsequent values will start repeating too, making the function eventually-periodic.</p>

Title: Chaos in finite field


问:
Finding the length of the curve defined by $x(t)=t^2-2t$ and $y(t)=t^3-4t$ for $0\leq t\leq2$
So I was given the following prompt:

What is the length of the curve defined by the parametric equations $x(t)=t^2-2t$ and $y(t)=t^3-4t$ for $0\leq t\leq2$?

I understand the setup for the length formula, but I'm a bit confused about what the integral would look like for this equation.

Any help would be appreciated!

答:
$$L=\int_{0}^{2} \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} ~ dt$$
Here $dx/dt=2(1-t), dy/dt=3t^2-4$. We get
$$L=\int_{0}^{2} \sqrt{9t^4-20t^2-8t+20} ~dt, $$
which is not doable by hand, It can be done numerically and Mathematica gives $6.5105$.


问:
Length of the intersection between a sphere and a cylinder
Note: while writing this question I realized what I was missing so there is no question here, but I thought it is a nice exercise to share, and I'd like to see more solutions.

I am trying to calculate the length of the curve given by $$x^2+y^2+z^2=1, x^2+y^2=x$$

The curve is the intersection between a sphere and a cylinder, and we can  notice is is symmetric around the $xy$ and $xz$ planes, so I tried to calculate only the part where $y>0, z>0$ and multiply by 4.

For this part I used the parameterization: $\gamma(t)=(t, \sqrt{t-t^2}, \sqrt{1-t}), t \in [0, 1]$

And then the length should be $l=4\int_0^1 ||\gamma'(t)|| dt = 4\int_0^1 \sqrt{1+\frac{(1-2t)^2}{4(t-t^2)} +\frac{1}{4-4t}} dt = 4\int_0^1 \sqrt{\frac{t+1}{4t-4t^2}} dt$

And I'm not sure how to solve this integral but using an online calculator we can see the length is $\approx 7.64$

Different solutions, or suggestions for ways to solve this integral are welcome!

答:
This again requires elliptic integral, similar to your working.

In cylindrical coordinates, $x = r \cos t, y = r \sin t, z$.

Cylinder $x^2 + y^2 = x $ or $(x-\frac{1}{2})^2 + y^2 = \big(\frac{1}{2}\big)^2$ can be written in cylindrical coordinates as

$r = \cos t, -\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$.

At the intersection of sphere and cylinder,

$x = \cos^2 t, y = \cos t \sin t, z = \sqrt{1-r^2} = \sin t$

So $x'_t = -\sin2t, y'_t = \cos 2t, z'_t = \cos t$

Length of the curve = $ \displaystyle 2 \int_{-\pi/2}^{\pi/2} \sqrt{(x'_t)^2 + (y'_t)^2 + (z'_t)^2} \ dt = 4 \int_{0}^{\pi/2} \sqrt{1 + \cos^2t} \ dt \approx 7.64$

$\displaystyle \int_0^{\pi/2} \sqrt{1 + \cos^2t} \ dt $ can be written as,

$\displaystyle \sqrt2 \int_0^{\pi/2} \sqrt{1 - \frac{1}{2} \sin^2t} \ dt = \sqrt2 E(\frac{1}{\sqrt2}).$

$E(k) = \displaystyle \int_0^{\pi/2} \sqrt{1 - k^2 \sin^2t} \ dt$ is complete elliptic integral of the second kind as shared by Raffaele in the comment.


问:
Showing a multivariable function is differentiable at a point
So I've been asked a question:
"Show the function $f(x,y) = x^2 - xy + 3y^2$ is differentiable at the point $(1,1)$ using the definition of differentiability." The question also mentions that it wants us to indicate the linear and little-o terms, and prove the little-o terms are in fact little-o terms.

I've been given the answer already (see below) and understand most of what is going on but I'm not sure how we go from line 3 to line 4 (line 4 being the one that starts with $f(1,1)$).

I understand what happens after line 4, it's just the jump from line 3 to line 4 that I don't understand, where to these terms ($o(h)$, $A h$, and $f(1,1)$) come from?

Thanks in advance

Body:
From line 3 to line 4:

$f(1,1)=3$

$h=(h_1,h_2)$ and $Ah=h_1+5h_2$

$h_1^2-h_1h_2+h_2^2=o(h)$


问:
Normal between two lines
$$l_1:x=(-1,1,2)+t(1,-3,-2)$$
$$l_2:y=(1,1,1)+s(-2,6,-4)$$
Q: Find the equation of plane passing through $l_1$ and $l_2$.

My teacher wrote answer like this:

Since they're parallel, choose vector $u=(1,-3,-2)$ and choose normal vector $n=u$x$(x_0-y_0)=(-3,-5,6)$. Can someone please explain me how did he find $n=(-3,-5,6)$.

No matter how I choose vectors $x_0,y_0$ it gives $n=(-3,3,-6)$. Is there a way to obtain $n=(-3,-5,6)$?

Thanks!

答:
First of all, we know that the two lines l1,l2 are not parallels.

We know that because, by definition two lines are parallels if their direction vectors are parallels, that means that the two vectors must be linearly dependents (or the same).

In this case, the direction vector of l1 is v1 = (1, -3, -2) (the element of the equation that multiplies "t") and the direction vector of l2 is v2 = (-2, 6, -4). If we study these two vectors we see that they are not parallels because if we multiply v1 by -2 (to obtain the first component of v2, -2) we have as result 2*v1 = (-2, 6, 4) that is not v2, so we have proven that v1 and v2 are not linearly dependents so we have proven that l1 and l2 are not parallels.

To find the equation of the plane we need the normal vector of the plain.

By definition, the normal vector to a plane is a perpendicular vector to this plane.

If we choose l1 to find the normal vector, we need to find a perpendicular vector of

```<code>                               v1 = (1, -3, -2)
</code>```

Any perpendicular vector we choose will be fine. Your teacher has chosen n = (-3, -5, 6) but you can choose any perpendicular vector.

To understand how n is obtained, you can fix for example the first and second components of n = (-3, -5, x) and try to find the value of x knowing that it has to happend that

nv1' = 0 => (-3, -5, x) * (1, -3, -2)' = 0 => -3·1 + (-5)·(-3) + x·(-2) = 0 => -3+15-2x = 0

=> 12 - 2x = 0 => 12 = 2x => x = 6

And finally we have that n = (-3, -5, 6) and that's the way we obtain this vector

I hope it works to you, any doubt I will glad to answer if I know it ;)

Title: Normal between two lines

Body:
First of all, we know that the two lines l1,l2 are not parallels.

We know that because, by definition two lines are parallels if their direction vectors are parallels, that means that the two vectors must be linearly dependents (or the same).

In this case, the direction vector of l1 is v1 = (1, -3, -2) (the element of the equation that multiplies "t") and the direction vector of l2 is v2 = (-2, 6, -4). If we study these two vectors we see that they are not parallels because if we multiply v1 by -2 (to obtain the first component of v2, -2) we have as result 2*v1 = (-2, 6, 4) that is not v2, so we have proven that v1 and v2 are not linearly dependents so we have proven that l1 and l2 are not parallels.

To find the equation of the plane we need the normal vector of the plain.

By definition, the normal vector to a plane is a perpendicular vector to this plane.

If we choose l1 to find the normal vector, we need to find a perpendicular vector of

                               v1 = (1, -3, -2)

Any perpendicular vector we choose will be fine. Your teacher has chosen n = (-3, -5, 6) but you can choose any perpendicular vector.

To understand how n is obtained, you can fix for example the first and second components of n = (-3, -5, x) and try to find the value of x knowing that it has to happend that

nv1' = 0 => (-3, -5, x) * (1, -3, -2)' = 0 => -3·1 + (-5)·(-3) + x·(-2) = 0 => -3+15-2x = 0

=> 12 - 2x = 0 => 12 = 2x => x = 6

And finally we have that n = (-3, -5, 6) and that's the way we obtain this vector

I hope it works to you, any doubt I will glad to answer if I know it ;)


答: Pascal's triangle is there for a reason.  Number completely determined by which two of the 4 slots are assigned the digit "2" : $~\binom{4}{2}.$

Title: In how many ways can 4-digit numbers be formed using 2,2,8,8 if each digit is used once only?


问:
How many times can $C^{1}$ strictly convex functions intersect on a bounded interval?
The question  is related to question here but with a little of modification which makes the task a little difficult

Let $f:\mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$  be two strictly convex function of classe $C^{1}$ on an open interval $(a,b)$.

Assume $f$ and $g$ are not identical. such that the one-dimensional Hausdorff measure $\mathcal{H}^{1}(\{f(x)=g(x): x\in (a,b)\})=0$

Can we find an upper bound on how many times these two functions can intersect on $(a,b)$?

答:
No, there can be arbitrarily many intersections. In fact there can be infinitely many. Consider the function $h(x)=x^5\sin(\frac{1}{x})$ in $[-1,1]$ (in $0$ we define $h(0)=0$). That function can be checked to be $C^1$ and is $0$ in only countably many points.

Now consider the functions $f(x)=100x^2$ and $g(x)=100x^2+h(x)$ in $(-1,1)$. It is easy to see that their second derivatives are positive in the whole interval, and the two functions are equal in countably many points (so an infinite set but of measure $0$). So that´s a counterexample to the conjecture.


问:
Write $\int_1^e \frac{(\ln(\ln(x)))^3}{x}\,dx$ in terms of Gamma function.
Write $$\int_1^e \dfrac{(\ln(\ln(x)))^3}{x}\,dx$$ in terms of Gamma function.

I have tried $y=\ln(x)$ but not going to work.

I have tried to substitute $x=e^{e^y}$

but stuck, can you  give me good substitution? How to solve it?

答:
The substitution $y = \ln x$ works perfectly, no?
\begin{align*}
\int_1^e \frac{(\ln \ln x)^3}{x} dx = \int_0^1 (\ln y)^3 dy = \bigg[y(\ln^3(y) - 3\ln^2(y) + 6\ln(y) - 6)\bigg] \Bigg|_0^1 = -6
\end{align*}


问:
Proof of the existence of a holomorphic function from a simply connected domain to itselfs
Suppose that $U$ is simply connected domain in $\mathbb{C}$, and let $P$ be a point in $U$. Prove that for all $t \in \mathbb{R}$, there exists a  unique holomorphic function $f:U \to U$ such that $f(P)=P$ and $f'(P)=e^{it}$.

some of My idea:this is a problem  from 2011 UCLA PhD qualifications. Seeing the unique, I want to use the automorphism of the unit disk or Riemman mapping theorem, but I can not construct it. Thanks for your answer very much.

答:
By Riemman Mapping Theorm,there exists a unique conformal map $\phi$:$\mathbb{U}$ $\rightarrow$ $\mathbb{D}$

where $\phi$(P)=0,thus for all t$\in$ $\mathbb{R}$,denote h=$e^it$,consider the map:
{g=$\phi$$\circ$h$\circ$$\phi^-1$}\
then:g(P)=P and g'(P)=$\phi$'(0) *$e^it$ * 1/$\phi$'(0)=$e^it$ which satisfies.
Thanks for all of your hints!


答:
If you substitute $x=e^{e^y}$ your integral becomes

$$\int_{-\infty}^0 \dfrac{y^3}{e^{e^y}}\cdot e^{e^y}\cdot e^y\ dy = \int_{-\infty}^0 y^3 e^y dy$$

or with $z=-y$ and simplifying/switching limits

$$-\int_{0}^\infty (-z)^3 e^{-z} (-dz) = -\int_{0}^\infty z^3 e^{-z} dz = -\Gamma(4) $$

Title: Write $\int_1^e \frac{(\ln(\ln(x)))^3}{x}\,dx$ in terms of Gamma function.

Body:
Write $$\int_1^e \dfrac{(\ln(\ln(x)))^3}{x}\,dx$$ in terms of Gamma function.

I have tried $y=\ln(x)$ but not going to work.

I have tried to substitute $x=e^{e^y}$

but stuck, can you  give me good substitution? How to solve it?


问:
Strong inductive proof for recursive sequence given Explicit formula
I am stuck on this

Show that the recursively defined sequence $a_n=2·a_{n-1}+3·a_{n-2}, a_1=-1, a_2=13$ has the explicit formula $〖a_n=3^n+4(-1)^n〗$

I can prove the base cases no problem but I keep getting stuck on the transformation

so I have $a_{k+1}=2·a_{k} + 3·a_{k-1}$

=$2(3^k+4(-1)^k)+3( 3^{k+1})+4(-1)^{k-1})$

=$2*3^k+8*(-1)^k+3*3^{k-1}+12*(-1)^{k-1}$

=$$2*3^k+8*(-1)^k+3^k- 12(-1)^k$$
=$$3*3^k-4(-1)^k$$

How do I get that -4(-1)^k to turn into +$4(-1)^{k+1}$?
Did I goof the alegbra somewhere?

Body:
You want to show:

$$3^{n}+4(-1)^{n} = 2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}]$$

Simplifying from the left we have:

$$2[3^{n-1}+4(-1)^{n-1}] + 3[3^{n-2}+4(-1)^{n-2}] = 2[3^{n-1}] + 8[(-1)^{n-1}] + [3^{n-1}] + 12[(-1)^{n-2}] =$$

$$=3[3^{n-1}] - 8[(-1)^{n-2}] + 12[(-1)^{n-2}] = 3^{n} + 4(-1)^{n-2} = 3^{n} + 4(-1)^{n-2}$$


问:
What notation should I use for indexed family?
So I've just started learning a bit of topology and set theory and im getting confused on definitions of indexed family.

I've seen various websites and books describe an indexed family of sets like this

$\{A_{i}|i\in I \}$ is a function $S:I\longrightarrow A$ and we write $A_{i}$ instead of $S(i)$.

But on a post I saw on here I understood that an indexed family can contain repeated elements if this is so why do we use set builder notation to describe a family?
Shouldn't it be something more like $(U_i)_{i\in I}$ in this case the braces do not indicate a set.

Thanks in advance.

答: Unfortunately, both notations you describe are in use.  Some people use $\{A_{i}\mid i\in I \}$ (or interchangeably, $\{A_i\}_{i\in I}$) to denote an indexed family, and others use $(A_i\mid i\in I)$ or $(A_i)_{i\in I}$ (there are also other variations you will occasionally see, like $\langle A_i\rangle_{i\in I}$).  Personally, I strongly agree with you that the first notation doesn't make sense for an indexed family, since it should be describing a set, not a function.  So I would say that $(A_i)_{i\in I}$ denotes the indexed family itself whereas $\{A_i\}_{i\in I}$ denotes the set of terms of the indexed family, i.e. the image of the function $I\to A$.  However, this convention is not universal and some authors use $\{A_i\}_{i\in I}$ for the indexed family itself, or simply do not make a clear distinction between the indexed family and its image.


问:
Show that a vector that is orthogonal to every other vector is the zero vector
I have the following question, and I'd like to get some tips on how to write the proof. I know why it is, but I'm still not so great at writing it mathematically.

If $u$ is a vector in $\mathbb{R}^n$ that is orthogonal to every vector in $\mathbb{R}^n$, then $u$ must be the zero vector. Why?

I'm starting off like this, but I don't know if it's the right way to do it, or if it is and I just don't know how to continue.

\begin{align}
(\exists u\in\mathbb{R}^n)(\forall v\in\mathbb{R}^n)[\text{u is orthogonal to v}]&\iff u\cdot v=0\\
&\iff ?
\end{align}

From here, instinctively I want to divide both sides by $v$, but I don't know if there is such a thing as dividing a dot product.

答:
The dot product $\cdot$ is an unusual type of multiplication.  It takes two vectors in and produces a scalar out.  Imagine a mommy elephant and a daddy elephant giving birth to a giraffe.

There's no such thing as division in this context.  You need to do the proof by looking at components of the vectors.  $u=(u_1,u_2,\ldots, u_n), v=(v_1, v_2,\ldots, v_n)$, and $$u\cdot v=u_1v_1+u_2v_2+\cdots+u_nv_n$$

Then, following Jared's hint, you can try specific values for $v$ and learn things about the components of $u$.


问:
About a proof of fourier transformation of error function
In a proof of fourier transformation of error function, the formula
$$
\int_{-\infty}^{\infty} \frac{-i e^{-\pi^{2} t^{2}}}{\pi t} e^{i 2 \pi x t} d t=\frac{-i}{\pi} \int_{-\infty}^{\infty} i \pi \operatorname{sgn}(x-\tau) \frac{1}{\sqrt{\pi}} e^{-\tau^{2}} d \tau
$$
is used. How can I proof this equation? I cannot understand why $\operatorname{sgn}$ appears.

答:
Have a look at the Wikipedia page.  The argument is that
\begin{align}
\int_{-\infty}^{\infty} \frac{-i}{\pi} \frac{1}{t} e^{-\pi^2 t^2} e^{2\pi i xt} ~ dt = \frac{-i}{\pi}\mathcal F\Big( \frac{1}{t} \cdot e^{-\pi^2 t^2}\Big)(-x)
\end{align}
where $\mathcal F$ denotes the Fourier transform.  It then uses the convolution rule, $\mathcal F(f\cdot g) = \mathcal F(f) * \mathcal F (g)$,
\begin{align}
\int_{-\infty}^{\infty} \frac{-i}{\pi} \frac{1}{t} e^{-\pi^2 t^2} e^{2\pi i xt} ~ dt
&= \frac{-i}{\pi}\Bigg( \mathcal F\Big( \frac{1}{t}\Big) * \mathcal F \Big( e^{-\pi^2 t^2}\Big) \Bigg) (-x) \\
&= \frac{-i}{\pi}\int_{-\infty}^{\infty} \mathcal F\Big( \frac{1}{t} \Big)(-x-s) \mathcal F\Big( e^{-\pi^2 s^2} \Big)(s) ~ds
\end{align}
and then use the standard results, $\mathcal F(1/t) = -i\pi \operatorname{sgn}(x)$ and $\mathcal F( e^{-\pi^2 t^2}) = \frac{1}{\sqrt{\pi}} e^{-x^2}$.

Put these together and recognise $\operatorname{sgn(y)}=-\operatorname{sgn}(-y)$,
\begin{align}
\int_{-\infty}^{\infty} \frac{-i}{\pi} \frac{1}{t} e^{-\pi^2 t^2} e^{2\pi i xt} ~ dt
&= \frac{-i}{\pi}\int_{-\infty}^{\infty} -i\pi\operatorname{sgn}(-x-s) \cdot \frac{1}{\sqrt{\pi}}e^{-s^2} ~ds \\
&= \frac{-i}{\pi}\int_{-\infty}^{\infty} i\pi\operatorname{sgn}(x+s) \cdot \frac{1}{\sqrt{\pi}}e^{-s^2} ~ds \\
&= \frac{-i}{\pi}\int_{-\infty}^{\infty} i\pi\operatorname{sgn}(x-\tau) \cdot \frac{1}{\sqrt{\pi}}e^{-\tau^2} ~d\tau
\end{align}
where the last step resulted from a change of variable $\tau=-s$.


问:
proving that a function exits with derivative of $\cos(1/x)$
I have to prove that a function exists that in all real numbers is differentiable and its derivative is equal to:

$$
f(x)= \left\{
    \begin{array}{ll}
        \cos ( \frac{1}{x} ) & x \neq 0 \\
        0 & x=0
    \end{array}
\right.
$$

I've been trying to solve this for a while. Can someone give a hint to how I should approach this problem?

答: Hint: If $F(x)=-x^2\sin\left(\frac1x\right)$, then $F'(x)=\cos\left(\frac1x\right)-2x\sin\left(\frac1x\right)$.


问:
Can you make a sequence $(x_k)_{k \geq 1}$ of elements in the interior of $K$ that is dense in $K \subset \mathbb{R}^n$, which is convex and compact?
I'm reading a paper about approximators.
In the proof of Theorem 2 in the paper,
the authors mentioned that there is a sequence $(x_k)_{k \geq 1}$ of elements in the interior of $K$ that is dense in $K$ where $K \subset \mathbb{R}^n$ is convex and compact and has nonempty interior.

The example suggested in the paper is as follows:
"the set of vectors in the interior of $K$ that have rational coordinates, so by indexing its elements in an arbitrary way, we get a sequence that is dense in $K$."

Is it true?
It seems work intuitively, but I don't know how to show it in a rigorous way.

答:
If $K$ is convex and closed with non-empty interior, then $K$ is equal to the closure of its interior, $K= cl \ int \ K$.

Since $\mathbb Q^n$ is dense in $\mathbb R^n$ and $int \ K$ is open, the set $int \ K \cap \mathbb Q^n$ is dense in $int \ K$. Since the closure of $int\ K$ is $K$, these rational points are dense in $K$.


Body:
$$I=\int_{0}^{e} \frac{(\ln \ln x)^3}{x} dx$$ Let $\ln x=t$, then
$$I=\int_{0}^{1} \ln^3 t dt$$ Integrate it by parts taking $\ln^3 t$ as first function and 1 as second, sucessively, then
$$I=\ln^3t ~t|_{0}^{1}-\int 3\ln^2 t (1/t)t dt= -3 \int \ln^2 t dt$$ $$=-3[\ln^2 t|_{0}^{1} -\int_{0}^{1} 2 \ln t (1/t) t dt]=6\int_{0}^{1} \ln t dt=6(t\ln t-t)|_{0}^{1}=-6.$$


问:
$Z(GL_n(\mathbb R)) = \{aI : a\neq0\} $
$Z(GL_n(\mathbb R)) = \{aI : a\neq0\} $

This article is the general case for $GL(n,k)$ where $k$ is a field. Could I prove it only with a basic linear algebra?

答: We know that $G=GL_n(\mathbb{R})$ is generated by elementary matrices. Therefore $A$ is in $Z(G)$ if and only it commutes with all elementary matrices. By checking this condition, you will see that $A$ has to be diagonal.</p>

Title: $Z(GL_n(\mathbb R)) = \{aI : a\neq0\} $


问:
Number of points on elliptic curves over finite fields
I noticed that for elliptic curves of the form

$y^2≡x^3+a\pmod p$

sometimes the number of points is always $p+1$ for any choice of $a$. This seems to be the case for all $p ≡ 5 \pmod 6$.

Moreover, when this does not happen, i.e., for $p ≡ 1 \pmod 6$, it looks like there are exactly zero curves of such form where the number of points is $p+1$.

Can someone point me towards the right direction as to why this happens?

答:
I will explain the idea of why there are $ p+1$ points when $p\equiv 2 \pmod{3}$, $p$ prime odd.

First, observe that the map $x\mapsto x^3$  is a bijection on $k=GF(p)$, the field with $p$-elements, if $p\equiv 2 \pmod{3}$.

Hence the same is true for the map $f$ given by $x\mapsto x^3+1$ .

Now, there are $\frac{p-1}2$ numbers that are squares and not 0. This give $p-1$ points with $y$ coordinate non 0. There is also one with $y=0$, namely $(-1,0)$, and one point at infinity.


问:
Diagonalization of matrix and how to use it
I got this diagonalizable matrix $ H = Q\Lambda Q^T$ which is actually a Hessian of the function used in finding the optimal parameter $\theta$ ($\mu$ is regularization parameter and it is a scalar, if it matters, know that this equation came from first order optimality condition $H(\theta - \theta^*) + \mu\theta = 0$):
$$\theta = (H + \mu I)^{-1}H\theta^*$$
What I don't understand here is how they come from using property of the diagonalizable matrix H to the next result $$\theta = (H + \mu I)^{-1}H\theta^* = Q(\Lambda + \mu I)^{-1}\Lambda Q^T \theta^*$$
I understand they start like this :
$$\theta = (H + \mu I)^{-1}H\theta^* = (Q\Lambda Q^T + \mu I)^{-1}Q\Lambda Q^T \theta^*$$
But from here I don't understand how they reduce it to last term in the equation goal I showed one equation above.

答:
Since $H$ is a Hessian, it is symmetric, so that $Q$ is orthogonal and $QQ^T=I$.

Because they have the same inverse, these matrices are equal:

$$\left[Q(\Lambda + \mu I )Q^T\right]^{-1} = Q(\Lambda + \mu I )^{-1}Q^T $$

So, starting with: $$\theta = (H + \mu I)^{-1}H\theta^*$$
we have
$$\begin{aligned} \theta &= (Q\Lambda Q^T + \mu QQ^T)^{-1}H\theta^*\\
&=\left[Q(\Lambda + \mu I )Q^T\right]^{-1}H\theta^*\\
&= Q(\Lambda + \mu I)^{-1} Q^TH\theta^*\\
&= Q(\Lambda + \mu I)^{-1} Q^T Q\Lambda Q^T\theta^*\\
&= Q(\Lambda + \mu I)^{-1} \Lambda Q^T\theta^*.\\
\end{aligned}
$$


问:
$f(x_1,y_1)>f(x_2,y_2)$ and $g(x_1,y_1)>g(x_2,y_2)$ $\implies$ $x_1>x_2$ and $y_1>y_2$: what other conditions are needed?
in order to make the derivation in the title, what other conditions are needed, besides:
$\frac{\partial f}{\partial x} > 0$ and $\frac{\partial f}{\partial y} > 0$ and $\frac{\partial g}{\partial x} > 0$ and $\frac{\partial g}{\partial y} > 0$?

In single variable calculus, one only needs the monotonicity to make such derivations, but haven't seen anything analogous for multivariate functions.

thank you.

答:
Assume that your propierty holds for $f$. Then, the converse is also true: If $x_1\leq x_2$ or $y_1\leq y_2$ then $f(x_1,y_1)\leq f(x_2,y_2)$. So, for every couple $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2$ we have $f(x_1,y_1)\leq f(x_1,y_2)$ and $f(x_1,y_2)\leq f(x_2,y_2)$ (as $x_1\leq x_1$ and $y_2\leq y_2$).

We show that the inequalities above can't be strict: if  $f(x_1,y_1)<f(x_1,y_2)$ then $x_1<x_1$, which is a contradiction; if $f(x_1,y_2)<f(x_2,y_2)$ then $y_2<y_2$, which is also a contradiction. Then it must be $f(x_1,y_1)=f(x_1,y_2)$ and $f(x_1,y_2)=f(x_2,y_2)$, hence $f(x_1,y_1)=f(x_2,y_2)$. As this is true for every couple $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2$, $f$ is constant. Then the only functions verifying your property are, vacuously, the constant ones.


问:
Adjoint operator on a Hilbert space
I am working on this problem:

Consider the space $H=\{f:\mathbb{R}\to\mathbb{C} : \int_{-\infty}^{+\infty} e^{-|x|} |f(x)|^2 dx <\infty\}$, this is a Hilbert space with the inner product $<f,g >=\int_{-\infty}^{+\infty} e^{-|x|} f(x) \overline{g(x)} dx$.

Consider the operator $T:H\to H$ defined as $Tf(x)=f(x+1)$. Find the adjoint of T.

I have calculated $<Tf,g>$ by the substitution $x+1=y$. I split the integrals but I don't know how to reassemble them.
Could anyone help me? Thanks a lot.

答:
We start with
$$
\langle Tf,g\rangle = \int_{\mathbb R} e^{-|x|}f(x+1)g(x)dx.
$$
In order to get something that looks like $\langle f, h\rangle$,
we have to transform the integral to contain $e^{-|x|}f(x)$.
Now do a change of coordinates  $x\mapsto x-1$, then
$$\begin{split}
\langle Tf,g\rangle &= \int_{\mathbb R} e^{-|x-1|}f(x)g(x-1)dx\\
&=\int_{\mathbb R} e^{-|x|} e^{|x|-|x-1|}f(x)g(x-1)dx\\
&=\int_{\mathbb R} e^{-|x|} \cdot f(x) \cdot (e^{|x|-|x-1|} g(x-1))dx\\
&=\langle f, T^*g\rangle
\end{split}
$$
so
$$
(T^*g)(x)=e^{|x|-|x-1|} g(x-1)
$$

Title: Adjoint operator on a Hilbert space


问:
Is the conclusion wrong in this proof of Case 2 of FLT for n=3?
The classic proofs of some cases of Fermat´s last theorem are separated into two parts. These proofs use the reduced Fermat equation $z^n=x^n+y^n$.  The positive integers $x,y,z $ are coprime to each other. In Case$1$ they are coprime to the positive integer n, but in Case $2$, one of $x, y, z$ is divisible by $n$.

This attempt to prove Case $2$ for $n=3$ is inspired by the Theorems on FLT by the Norwegian mathematician Niels Henrik Abel. The proof is one of contradiction. The crucial point is the conclusion in the end of the proof.

Assume that (I) $z^3=x^3+y^3$ is true, when the positive integers $x,y,z $ are coprime to each other. Assume $z$ is the one which is divisible by $3$. Factorize (I) $z^3=x^3+y^3$ $\implies $ (II) $z^3=(x+y)((x+y)^2-3xy)$. The left side has $3$ factors $=3$. $ (x+y)^2-3xy$ has only $1$ factor $=3$ since $3xy$ has only $1$ factor $=3$  $\implies$ $x+y $  has $2$ factors $=3$.

Also, $x+y$ divides $z^3$  $\implies$ $x+y$  is coprime to $x$ and to $y$  $\implies$
If $x+y$  is not divisible by $3$, $x+y$  is coprime to $ (x+y)^2-3xy$.
Then $3$ is the only possible common factor between $x+y$ and $ (x+y)^2-3xy$.

$z^3=x^3+y^3$ $\iff $ $y^3=z^3-x^3$ $\iff $ $x^3=z^3-y^3$.
Factorizing of  $y^3=z^3-x^3$ and $x^3=z^3-y^3$  $\implies$

$y^3=(z-x)((z-x)^2+3xz)$  and  $x^3=(z-y)((z-y)^2+3yz)$.

Since $x$ and $y$ are not divisible by $3$ , the factors of these factorizations are coprimes and therefore cubes.
Let $z-x=b^3$ and $z-y=d^3$.  $b,d$ are positive coprime integers. Then

$ b^3+d^3=(z-x) + (z-y) =2z- (x+y)$  $\implies$ (III) $ b^3+d^3=2z- (x+y) $.
The right side has $1$ factor $=3$ $\implies$ the left side is divisible by $3$.

Factorizing of the left side $\implies$ $ b^3+d^3=(b+d)((b+d)^2-3bd)$. Then,
If one factor in the left side is divisible by $3$, the other factor is too $\implies$
The left side has $2$ factors $=3$ since $x+y$ has 2 factors $=3$ but $z$ only $1$ factor. That is a contradiction. $z^3\neq x^3+y^3$.

If $z$ has $2$ factors $=3$  $\implies$ $x+y $  in (II) $z^3=(x+y)((x+y)^2-3xy)$
has $5$ factors $=3$. That does not give a contradiction in (III) $ b^3+d^3=2z- (x+y)$.
But a contradiction is already reached for $z$ divisible by $1$ factor $=3$.

It is shown that $z^3\neq x^3+y^3$ $\iff $ $z^3\neq(x+y)((x+y)^2-3xy)$ when $z$ is divisible by $1$ factor $=3$. Then $z^3\neq(x+y)((x+y)^2-3xy)$ when $z$ is divisible by $2$ factors$=3$. If left side does not equal right side when both sides of the equation  is divided by $3$ factors $=3$, division of both sides by more factors $=3$ does not make them equal.

In the same way, inequality is shown when $x$ or $y$ divisible by $3$.

Case $2$ of FLT for n=3 is true.

答:
The proof is not correct. In line -10 you have used the factorisation of $b^3+d^3$ to prove that this expression is divisible by $9$. You then correctly deduce that $z$ is divisible by $9$.

You then claim that this is a contradiction but it isn't a contradiction, it just means that $9|z$ and $3^5|x+y$.

FLT for $n=3$ can be proved by elementary means - see for example Fermat's Affirmative Questions, The Mathematical Gazette, Vol 96, Number 535, March 2012.

The sort of equations you were using were used very successfully and with great ingenuity by Sophie Germain, for quite general values of $n$, but only for Case 1 of FLT.


问:
Integeration with one end infinite
Part1:For any natural number $n$ prove that $\int_0^1 \cos(nx)ln(x)dx$ exists and is finite.

Part2: Show that the above expression when n tends to inf is zero.

I have a proof for first part but I am not sure I just triangle inequality and the fact the the absolute value of cos is never more that 1 as follows:

$|\int_0^1 \cos(nx)ln(x)dx|\le\int_0^1 |\cos(nx)||ln(x)|dx \le \int_0^1 |ln(x)|$

But ln is always negative in that interval so we can easily show that the R.H.S is 1 so the left hand side shoud diverge.I don't know if this is completely ok.For the second part our master gave a hint that we should split the integeration interval into to intervals $[0,\frac{1}{n}]$ and $[\frac{1}{n},1]$ then the integral in the first part is easy to prove that it tends to zero when n tends to infinite but the second interval is hard to get conclusion any hints?

答:
This is just an attempt{and too long for a comment, hence posting it as an answer}. If the way I have done this is right, then we can see that the summation converges as k tends to $\infty$. However I haven't been able to prove the same when it's multiplied by $n^k$, but I hope this helps you and you figure it out from here.

If you do, kindly do share your further steps.

$$\int_0^1 \cos(nx)\ln(x)dx$$
Let us represent $\cos nx$ in terms of it's Taylor expansion.
Therefore,
$$\cos(nx) = \sum_{k=0}^{\infty} \dfrac{(-1)^k n^{2k} x^{2k}}{(2k)!}$$

Thus, $\int_0^1\cos (nx)\ln(x) $becomes
$$\int_0^1\ln(x)\sum_{k=0}^{\infty} \dfrac{(-1)^k n^{2k}x^{2k}}{(2k)!}$$
$$\sum_{k=0}^{\infty} \dfrac{(-1)^k n^k}{(2k)!} \int_0^1 x^{2k} \log(x) dx$$

Applying$\int UV=U\int V -\int U'\int V $
$$\sum_{k=0}^{\infty} \dfrac{(-1)^k n^k}{(2k)!} \Bigl(\frac{x^{2k+1}}{2k+1} \log(x)-\frac{x^{2k+1}}{(2k+1)^2}\Bigr)_0^1$$
$$=\sum_{k=0}^{\infty} \dfrac{(-1)^k n^k}{(2k)!} \Bigl(\frac{-1}{(2k+1)^2}\Bigr)$$

If we didn't have the $n^k$ term here, it's very obvious that this series converges. Since we do, I'm not sure of it's impact on the summation...

Title: Integeration with one end infinite


问:
Optimal Strategy in Dice Game
In a recent interview I received the following question (an optimisation/strategy game)...which left me a bit stumped. The rules of play, you start with 0 points, then:

Roll three fair six-sided dice;

Now you have the option:

Hold: i.e. accept the values shown on your dice as the score for your turn. There is a caveat, if two or more dice show the same values, then all of them are flipped upside down - e.g. 1 becomes 6

OR

reroll the dice: You may choose to hold any combination of the dice on the current value shown (so you can choose to keep 1 dice the same and then reroll the other two). Rerolling costs you 1 point – so during the game and perhaps even at the end your score may be negative.
You can roll an infinite number of times...

My thoughts:

So clearly the best possible score is 18 and is achieved by rolling three 1s on the first roll
The reroll penalty prevents rolling forever to get 18.
If the value of the dice is greater than the expected value of rerolling them (accounting for the penalty), then you should stick...
I guess what I am asking is how do I work out the expected value of rerolling them (accounting for the penalty) and how does this fit into the optimal strategy...

答:
The fact that you flip dice if they show the same value doesn't actually affect any expected values. You are just as likely to roll say $(1,1)$ (which gets flipped to $(6,6)$) as you are to roll $(6,6)$ in the game with no flipping. So I will essentially completely ignore this rule.

Now suppose our strategy is "re-roll a die if it comes up at most $x$, else hold the value". Then by the law of total expectation, your expected payoff $E$ for that particular die is
$$E=\frac{x}{6}(E-1)+\left(1-\frac{x}{6}\right)\frac{x+7}{2}\implies E=\frac{42-3x-x^2}{2(6-x)}.$$
Since $6$ is a small number, at this point you may as well just plug in the values of $x$, and you will find

$x$
$E(x)$

$0$
$3.5$

$1$
$3.8$

$2$
$4$

$3$
$4$

$4$
$3.5$

$5$
$1$

So our expected value is $12$ with the best strategy. We have two equally good strategies: either "hold values of three or more", or "hold values of four or more".

Title: Optimal Strategy in Dice Game


问:
What are the possibilities of a $4 \times 4 \times 4$ array that has two $0$s and two $1$s in each line?
So this is a $4$ by $4$ by $4$ cube consisting of $64$ ($1$ by $1$ by $1$) small cubes, which can have the value of $0$ or $1$.
What we call a line here is strictly straight, and not diagonal. So if we look at one side of the cube we have $4 \times 4$ perpendicular lines from each of the small cubes.

This way we can calculate, that the total lines of the cube are $4 \times 4 \times 3$ (4*4 from the 3 different dimensional sides). And out of the $64$ small cubes $32$ have the value $0$, and $32$ have $1$.
(Just to make sure everyone understand what i am talking about)

The main question is how many different possibilities are there, where you can't rotate one cube to the other. And how can you generate them mathematically?

Other question maybe:
How many are there where it is a different matrix in every aspect (you cant rotate the matrix, swap slabs).

答:
There are six columns.

Of the $6^3$ ways to fill the first three columns, calculate $2-sum$ to get the 4th column, and check whether it is all 0s and 1s,  and sums to 2.  You get 90 possible $4×4$ squares.

Of the $90^3$ ways to fill the first three layers, calculate the 4th layer and check whether it is valid.


问:
An affine $K$-algebra factored by a maximal ideal is contained in a finitely generated $K$-domain
I am trying to understand the proof of Proposition 1.2 from Gregor Kemper's "A Course in Commutative Algebra", which says that if $\varphi: A \rightarrow B$ is a homomorphism of the algebras $A,B$ over the field $K$ and if $B$ is an affine $K$-algebra and $\mathfrak{m}$ is a maximal ideal in $B$, then $\varphi^{-1}(\mathfrak{m})$ is a maximal ideal in $A$.

The step I'm having trouble with is when he uses Lemma 1.1,(b), which he proves previously and which states that an if an algebra $A$ over a field $K$ is itself a field and also $A$ is contained in an affine $K$-domain, then $A$ must be algebraic over $K$. This lemma is used for the $K$-algebra $B/\mathfrak{m}$ to conclude that $B/\mathfrak{m}$ is algebraic over $K$. I know that $B/\mathfrak{m}$ is a field because $\mathfrak{m}$ is maximal. But why is $B/\mathfrak{m}$ contained in an affine $K$-domain? Which is this finitely generated $K$-domain or how do we know $B$ is contained in one?

(A similar question was asked here for prime ideals instead of maximals, but got no answer.)

答: $B/\mathfrak{m}$ itself is an affine $K$-domain.  It is finitely generated as a $K$-algebra since $B$ is finitely generated as a $K$-algebra.  It is a domain since it is a field.


答:
Consider the function $F$ which is defined for $x\neq 0$ by
$$
F(x) =  - x^2 \sin \left( {\frac{1}{x}} \right) + 2\int_0^x {t\sin \left( {\frac{1}{t}} \right)dt}
$$
and by $F(0)=0$ for $x=0$. It is easy to check that $F'(x)$ exists and is $\cos(1/x)$ whenever $x\neq 0$. Now
$$
F'(0) = \mathop {\lim }\limits_{x \to 0} \frac{{F(x) - F(0)}}{{x - 0}} =  - \mathop {\lim }\limits_{x \to 0} x\sin \left( {\frac{1}{x}} \right) + \mathop {\lim }\limits_{x \to 0} \frac{2}{x}\int_0^x {t\sin \left( {\frac{1}{t}} \right)dt},
$$
and since
$$
\left| {x\sin \left( {\frac{1}{x}} \right)} \right| \le |x| \; \text{ and } \; \left| {\frac{2}{x}\int_0^x {t\sin \left( {\frac{1}{t}} \right)dt} } \right| \le \frac{2}{|x|}\int_0^{|x|} {tdt}  = |x|,
$$
we see that $F'(0) = 0$.

Title: proving that a function exits with derivative of $\cos(1/x)$

Body:
I have to prove that a function exists that in all real numbers is differentiable and its derivative is equal to:

$$
f(x)= \left\{
    \begin{array}{ll}
        \cos ( \frac{1}{x} ) & x \neq 0 \\
        0 & x=0
    \end{array}
\right.
$$

I've been trying to solve this for a while. Can someone give a hint to how I should approach this problem?


问:
Clarification about the measurability of a function
In a book I'm studying, it was stated that:

let $f:(\Omega,\mathcal A)\to (\Omega',\mathcal A')$ be measurable
where $\Omega'$ is a semi-metric space with semi-metric $d$, and that
$\mathcal A'$ is the sigma-algebra spanned by the open balls for this
semi-metric. If $x\in \Omega'$ is a fixed element, then $d(f,x)$ is measurable.

Question. Why is $d(f,x):(\Omega,\mathcal A)\to (\mathbb R,\mathcal B_{\mathbb R})$ measurable?

Should I explore the continuity of $d(f(t),x):\Omega'\to \mathbb R$ ? If it is indeed continous, then the composition $d(\cdot,x)\circ f$ is measurable.

*Def. $d$ is a semi-metric on some space $\Omega'$ as soon as:

$\forall x\in\Omega', d(x,x)=0$;

$\forall (x,y,z)\in \Omega'\times\Omega'\times \Omega', d(x,y)\leq d(x,z)+d(z,y)$.

答:
Well, $d(-,x)$ is continuous on $\Omega'$. Indeed, by the triangle inequality,

$$
d(y,x)-d(z,x)\leq d(y,z)
$$
and by symmetry, we get $|d(y,x)-d(z,x)|\leq d(y,z)$ so $d(-,x)$ is actually a contraction. Hence, $d(f,x)$ is a composition of a measurable map with a continuous function and hence, measurable.


问:
Is there even a point in defining the notion of a 'metric' (as opposed to a metric space), etc.?
When we define a new mathematical structure, we generally double up on definitions. We define structures (think: metric spaces, partially ordered sets, etc.) and also the ingredients that they're built up from (think: metrics, partial orders, etc.)

My question: are the ingredients even worth naming?

Consider that

(1) Structure-preserving maps go between structures, not ingredients, and

(2) Neither structures nor ingredients have the edge with respect to the ease with which new definitions can be created, e.g.

A partial ordering $\leq$ (carrier $X$) is linear iff either $x \leq y$ or $y \leq x$ for all $x,y \in X$.

A partially ordered set (carrier $X$, ordering $\leq)$ is linear iff either $x \leq y$ or $y \leq x$ for all $x,y \in X$.

(3) Structures are more general than ingredients because they're allowed to include multiple sorts. e.g. A vector space includes two distinct sorts, scalars and vectors.

So to repeat my question, are ingredients even worth naming? Why not just leave it at 'a metric space is an ordered pair $(X,d)$ such that...' without adding 'a metric is a function $d$ such that...'?

答:
The reason it's meaningful to define "metric" is that a space can have more than one metric, so we need a word to call those different functions.

The reason it's meaningful to define "metric space" is to distinguish those spaces that admit a metric from those that don't, without necessarily having to spell out what that metric is.


问:
Inverse Laplace of $\frac{K}{s(T s+1)}$
$$
y(t)=\mathcal{L}^{-1}\left[\frac{K}{s(T s+1)}\right]
$$

I know it should be:
$$
K\left(1-e^{-t / T}\right)
$$

But I get:

$$
y(t)=\mathcal{L}^{-1}\left[\frac{K}{s(T s+1)}\right] = [\text{partial fraction division}] = K \left( \mathcal{L}^{-1}\frac{1}{s} + \mathcal{L}^{-1}\frac{1-T}{sT+1} \right) = K\left( 1 + \frac{1}{T} \frac{1}{s + \frac{1}{T}} - \mathcal{L}^{-1}\frac{1}{T} T \frac{1}{s + \frac{1}{T}}  \right)= K \left( 1 + \frac{1}{T}e^{-1/T} - e^{-1/T} \right)= K \left( 1 + \frac{T-1}{T}e^{-1/T} \right)
$$

Where did it go wrong?

答: Your partial fraction decomposition is wrong. Note that $$\frac{K}{s(T s+1)} = \frac{K}{s} - \frac{KT}{sT+1} = \frac{K}{s} - \frac{K}{s+\frac{1}{T}} $$Now take the inverse transform which yields $$y(t) = Ku(t) - Ke^{-\frac{t}{T}}u(t) = Ku(t)(1-e^{-\frac{t}{T}})$$

Title: Inverse Laplace of $\frac{K}{s(T s+1)}$


问:
Is there any analytic method for solving $\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$
Hi i am trying to solve the following two equations:
$$\ddot{\phi} + 2\cot{\theta}{\dot\theta}{\dot\phi} =0$$
$$\ddot{\theta}-\sin\theta\cos\theta{\dot{\phi}}^2=0$$
where $\dot{\theta}=\frac{d\theta}{dt}$ and $\dot{\phi}=\frac{d\phi}{dt}$ and $\ddot{\theta}=\frac{d^2\theta}{dt^2}$ and $\ddot{\phi}=\frac{d^2\phi}{dt^2}$. I am trying to solve this analytically and looked up the standard techniques but couldn't find how solve these. Basically i want $\phi$ in terms of $\theta$ that  is $\phi(\theta)$. Is there any way to do this?

答:
Here is my attempt. Divide the first equation by $\dot\phi$, then we get
$$ \frac{d\ln\dot\phi}{dt}+2\frac{d\ln(\sin\theta)}{dt}=0.$$
We then obtain the integral of motion
$$\ln\dot\phi+2\ln\sin\theta=C.$$

Hence, we find $\dot\phi\sin^2\theta=e^C=B$.
Plugging this in the second equation, we find
$$ \ddot\theta-B^2\frac{\cos\theta}{\sin^3\theta}\dot\theta=0.$$
Multiplying this by $\dot\theta$, we find

$$\frac12\frac{d\dot\theta^2}{dt}+\frac b2\frac{d\sin^{-2}\theta}{dt}=0,$$
where $b=B^2$.
Then we get another integral of motion, a Hamiltonian $H$.
$$\frac{\dot\theta^2}2+\frac b{2\sin^2\theta}=H.$$
I believe this is right, but I don't have time to check it carefully now.
I hope this helps.


问:
$\epsilon - \delta$ definition of limit vs left and right hand definition of limit.
We initially learned in calculus that the limit of a function $f(x)$ at a point $x=c$ exists iff $$\lim\limits_{x \rightarrow c^+} f(x) = \lim\limits_{x \rightarrow c^-} f(x)$$
So what's not so rigorous about this definition that we are introduced to the new $\epsilon - \delta$ definition of limits in real analysis?

答:
If $f\colon A\subset\mathbb{R}\to \mathbb{R}$,  you have that $\lim_{x\to c}f(x)=l$ iff $\lim_{x\to c^{-}}f(x)=\lim_{x\to c^{+}}f(x)$.

Now, you defined $\lim_{x\to {c^{-}}}f(x)=l$ as:
$(\forall \varepsilon>0)( \exists \delta_{1}>0)(\forall x\in A)(-\delta_{1}<x-c<0\Rightarrow |f(x)-l|<\varepsilon)$

analogously you define the limit to the right, and with these two definitions you show that $\lim_{x\to c}f(x)=l$


问:
Minimum number of boxes we need to pack the items
We have $N$ items of the same size and we want to them in boxes. We have boxes of different sizes $\{M_1,\ldots , M_k\}$, how many we want from each size. Each box has be fully filled.

Design an algorithm that calculates the minimum number of boxes we need to pack the items.

We consider $M_1=1$ to have always fully boxes independent of $N$.

$$$$

It looks like Knapsack problem. So do we have to apply the algorithm of Knapsack problem for each box?

So we could sort the boxes in increasing order as for the size.

Then we starting packing items in that box, if it gets full we go to the next box.

We do that till no items are left.

Is the idea correct?

答:
We indeed can solve it as Knapsack problem - for each $n \leq N$ calculate $a_n$ - minimum number of boxes we need for $n$ items by recurrent
$$a_n = \begin{cases}
\infty, & n < 0 \\
0, & n = 0 \\
1 + \min_i a_{n - M_i}, & n > 0
\end{cases}$$

Greedy algorithm doesn't work, as usual: let $N = 10$, $M_2 = 6$, $M_3 = 5$. Then greedy algorithm uses one box of size $6$ and $4$ boxes of size $1$ for a total of $5$ boxes, while optimal solution is to use two boxes of size $5$.


问:
why is a regular open (closed) set called "regular"?
Certainly it doesn't have to have an obvious reason other than being sort of "simple" or "nice", but its resemblance to the "regular" in "regular semigroup" (https://en.wikipedia.org/wiki/Regular_semigroup) makes me wonder if they are indeed related.

I know that these two "regulars" are actually on different types of objects: the regular open set is certain type of set, while a regular element in a semigroup is an element $a$ having some algebraic property, i.e. $\exists x : axa = a$, but if we look at the closure/interior operation underneath the definition of a regular open/closed set (which are like algebraic operations as well), they are probably similar.

But how?

答:
General note

The adjective "regular," along with the adjective "normal," is probably one of the most overloaded adjectives in all of mathematics. Whenever mathematicians find a condition that makes things nice they are apt to try to find a good adjective to package it up, and early on especially, they didn't cast around very long: they had better things to do, like write groundbreaking math.

For those of us working in the second half of the 20th century and in the 21st century and beyond, we have to work a little harder while naming :)

Let me make a few general comments about the two notions that will set the stage for a comparison.  I will make the case that there is no close relationship between these two particular terms.

The set operations

but if we look at the closure/interior operation underneath the definition of a regular open/closed set (which are like algebraic operations as well), they are probably similar.

This here is related to the Kuratowski 14-set problem. If $k$ is the closure operator and $c$ is the complement operator on a topological space $X$, then the monoid generated by these two operations together has at most 14 distinct elements. For some spaces it will be smaller.

Now, if you alias the interior operator as $i=ckc$, you get a submonoid of that monoid with at most $7$ elements: $\{I, k, i, ki, ik, iki, kik\}$ where $I$ is the identity operator.  This is because the nontrivial kuratowski relation yields both $ikik=ik$ and $kiki=ki$ Again, for some spaces it will be smaller.

By definition, a subset $S$ is regular open if $S=ik(S)$, so it is a special class of subsets of $X$.

(von Neumann) regularity of semigroup elements

I say "von Neumann" regularity because even in ring theory the adjective "regular" is overloaded, and this is what we use to help distinguish this particular flavor.

I don't know if I can do justice to the notion of "regular element" of a semigroup in general. My main experience with them is through rings, but I suppose they are better known by semigroup theorists.

When for an $a$ there exists an $x$ such that $axa=a$, we can also say that "$a$ has an inner inverse" or sometimes "$x$ is a weak inverse of $a$."  Even though $ax$ and $xa$ aren't necessarily the identity, but they are at least idempotent, and you can "cancel" $x$ off of the left side of $a$ by using $a$ on the left, and so on.

Comparison

Now, you can already see that the semigroup notion is far departed from the set notion: the former discusses sets defined by a condition dictated by the topology, and the latter discusses the weak-invertibility of a binary operation. There is no  concept directly bridging the two.

What you may be picking up on is the relation of the latter to the former's operator monoid.  Indeed, in that monoid every element is idempotent and hence regular.


问:
Is exponential growth/decay always of higher order than logarithmic?
The book that I'm reading states that:

growth of $a^{n}$ is always higher/more than $(\log n)^b$. ($a,b$ constants; $a > 0$)

I'm a little confused as to what it means when $a<1$ (which makes it exponential decay I guess).

Does that mean the rate at which the exponential function is decreasing is faster than the rate which the logarithmic function is increasing at? (Imagine for example $b=5$ and $a=\frac{1}{1000}$)

答: It seems like the book might have had a typo because while it is true that an exponential always grows faster than any logarithm, that should only be true when $a > 1$


问:
Reduction of order
I have the equation $$x^2 y'' + (2x^2 - 3x) y' + 3y = 0$$ and the solution $$y_1 = x^3 e^{-2x}$$ Using reduction of order: $$y_2 = v y_1$$ I reduced the equation to $$v'' x^5 + v' x^4 = 0$$ so $v(x) = \ln x + C$. However, I thoroughly checked $$y_2 = x^3 e^{-2x} \ln x$$ and it does not work in the original equation. Any suggestions?

答:
If $y_(x)$ is a solution of $y''+P(x)y'+Q(x)y=0$ then it otther solution is given by
$$y_2(x)=y_1(x)\int \frac{e^{-\int P(x) dx}}{y_1^2(x)} dx$$
$$y_2(x)=y_1(x) \int \frac{e^{\int (-2+3/x) dx}}{x^6e^{-4x}}$$
$$y_2(x)=y_1(x) \int \frac{e^{-2x} x^3}{x^6e^{-4x}}dx$$
$$y_2(x)=y_1(x)\int \frac{e^{2x}}{x^3} dx=y_1(x) e^{2x}\left[\frac{-1}{2x^2}-\frac{1}{x}\right]+2y_1(x) \text{Ei}(2x).$$

EDIT: Yes instead of $\text{Ei}$ it should be leftas $\int \frac{e^{2x}}{x}dx$

Title: Reduction of order


问:
Fast paced but basic introduction to homological algebra
I'd like to learn some homological algebra. Being a physicist my abstract algebra background is not particularly strong so I find most of the usual books a bit forbidding. I recently tried Rotman's book "An introduction to homological algebra" and that's pitched at exactly the right level for me.

However Rotman's book walks a pretty long walk and getting to Tor or Ext takes forever. While that is surely interesting and conductive to a deeper understanding, for now I would prefer to have a much more superficial and instrumental understanding - being able to recognise/use homological algebra arguments where they emerge even if I don't grasp the subject at a deeper level or have not seen a proof of the big theorems.

To give an explicit example, I would like to be able to follow the argument in the proof of Lemma 6.22 of Kirillov "An introduction to Lie groups and Lie algebras", where Ext and some homological algebra are used to prove that $H^1(\mathfrak{g},V)=0$ for any representation $V$. Note that I am not especially interested in the result/proof itself, which surely can be reformulated to avoid the use of homological algebra, this is just an example of the kind of understanding of the subject I would like to get to.

Not being familiar with the subject I find it hard to devise by myself a shorter route through Rotman's book.

Can anyone recommend a homological algebra book or a set of notes which starts at about the same level as Rotman's but takes a quicker if less in depth route?

答:
Try Robert B. Ash's notes entitled Abstract Algebra: The Basic Graduate Year, especially Chapter 10 on homology and the accompanying Supplement. Also available in book form as Basic Abstract Algebra: For Graduate Students and Advanced Undergraduates. In the preface he writes:

I have attempted to communicate the intrinsic beauty of the subject.
Ideally, the reasoning underlying each step of a proof should be completely clear, but the
overall argument should be as brief as possible, allowing a sharp overview of the result.
These two requirements are in opposition, and it is my job as expositor to try to resolve
the conflict.

See also the end of the preface for his remarks on homological algebra:

In Chapter 10, we introduce some of the tools of homological algebra. Waiting until
the last chapter for this is a deliberate decision. Students need as much exposure as
possible to specific algebraic systems before they can appreciate the broad viewpoint of
category theory. Even experienced students may have difficulty absorbing the abstract
definitions of kernel, cokernel, product, coproduct, direct and inverse limit. To aid the
reader, functors are introduced via the familiar examples of hom and tensor. No attempt
is made to work with general abelian categories. Instead, we stay within the category of
modules and study projective, injective and flat modules.

In a supplement, we go much farther into homological algebra than is usual in the basic
algebra sequence. We do this to help students cope with the massive formal machinery
that makes it so difficult to gain a working knowledge of this area. We concentrate on
the results that are most useful in applications: the long exact homology sequence and
the properties of the derived functors Tor and Ext. There is a complete proof of the
snake lemma, a rarity in the literature. In this case, going through a long formal proof is
entirely appropriate, because doing so will help improve algebraic skills. The point is not
to avoid difficulties, but to make most efficient use of the finite amount of time available.


问:
Martingale proof
A sequence of rv $X_{n}$ i.i.d for $n \in \mathbb{N}$ with $S_{0}=0$ and $$S_{n} = \sum_{k=1}^{n}X_{k}.$$

with $k \leq n$ and $$\mathbb{E}[S_{k}\mid S_{n}] = \frac{k}{n}S_{n}$$

If $$M_{k} = \frac{S_{n-k}}{n-k},k=0,1,\dots,n-1$$ how can I show that $$\mathbb{E}[M_{k+1}\mid M_{0},\dots,M_{k}] =M_{k}$$

I am thinking that :

$$E[S_{n-k+1}\mid M_{0},\dots,M_{k}] = \mathbb{E}[X_{n-k+1}\mid M_{0},\dots,M_{k}]-S_{n-k}=a S_{n-k+1}-S_{n-k}$$

and find $a$. But is that right? And how can I find $a$?

答:
Preliminaries

$E[X_i|S_n]=a$ does not depend on $i$, if $i \le n$. Than since:

$na=\sum_{i\le n} E[X_i|S_n]=E[S_n|S_n]=S_n$

We have therefore that $E[X_i|S_n]=S_n/n$ and

$$E[S_k|S_n]=\frac{k}{n} S_n, k \le n  \ [1]$$

We also observe that:

$E[X_i|S_n,S_{n+1},..]=E[X_i|S_n,X_{n+1},...]=E[X_i|S_n] \ [2]$

because $X_{n+1}$ is independent of $X_i$. Analogously:

$$E[S_k|S_n,S_{n+1},..]=E[S_k|S_n], k\le n$$

The identity we want to show

Now we show:

$\mathbb{E}[M_{k+1}\mid M_{0},\dots,M_{k}] =M_{k}$

First observe:

$\mathbb{E}[M_{k+1}\mid M_{0},\dots,M_{k}]=1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n},\dots,S_{n-k}]$

And now apply first [2] and than [1]. All factors are going to simplify :).

In detail thanks to [2]:

$1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n},\dots,S_{n-k}]=1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n-k}]$

and thanks to [1]:

$1/(n-k-1)\mathbb{E}[S_{n-k-1}\mid S_{n-k}]=1/(n-k-1) \times (n-k-1)/(n-k) \times S_{n-k}=M_k$

Title: Martingale proof


答:
I don't know about you but I refer to the "metric" by name a lot even when it's the only one. It seems perverse to have something ubiquitous with no name. Imagine someone asks "What's $d$?" - what should you say? People would instantly adopt names for convenience anyway, why not standardize?

More mathematically perhaps, you often have e.g. two metrics on the same set. In your language, I'm only really supposed to talk about whole structures, but this situation clearly separates out the ingredients. Then one might reasonably talk about properties of these two metrics on this one set, not of these two unrelated structures.

Further, one might consider categorizations of possible metrics on a set. Here you are really intuitively dealing with a space of functions from the space to $\mathbb R$, not a space of structures. I would want to talk about "all possible complete metrics on $X$", not "all possible metric spaces of the form $(X,d)$". The former is more economical and efficient and in agreement with what you're doing, at least to my mind.

Title: Is there even a point in defining the notion of a 'metric' (as opposed to a metric space), etc.?

Body:
When we define a new mathematical structure, we generally double up on definitions. We define structures (think: metric spaces, partially ordered sets, etc.) and also the ingredients that they're built up from (think: metrics, partial orders, etc.)

My question: are the ingredients even worth naming?

Consider that

(1) Structure-preserving maps go between structures, not ingredients, and

(2) Neither structures nor ingredients have the edge with respect to the ease with which new definitions can be created, e.g.

A partial ordering $\leq$ (carrier $X$) is linear iff either $x \leq y$ or $y \leq x$ for all $x,y \in X$.

A partially ordered set (carrier $X$, ordering $\leq)$ is linear iff either $x \leq y$ or $y \leq x$ for all $x,y \in X$.

(3) Structures are more general than ingredients because they're allowed to include multiple sorts. e.g. A vector space includes two distinct sorts, scalars and vectors.

So to repeat my question, are ingredients even worth naming? Why not just leave it at 'a metric space is an ordered pair $(X,d)$ such that...' without adding 'a metric is a function $d$ such that...'?


问:
Is the argument below valid?
I have just started to learn Logic and want to understand is the following argument valid and how to formalize it in First-Order Logic:

```<code>Plato is Human
Plato has 5 fingers on his left hand
Plato has 5 fingers on his right hand

Socrates is Human
Socrates has 4 fingers on his left hand
Socrates has 4 fingers on his right hand
——
All humans have the same number of fingers on both hands
</code>```

And then:

```<code>All humans have the same number of fingers on both hands
Aristotle is Human
Aristotle has 3 fingers on his left hand
——
Aristotle has 3 fingers on his right hand
</code>```

Can I use Universal generalization to prove Argument 1?

答:
The first one is false because you can't prove the statement using given results. Just because it is true for those 2 people does not make it true for everyone.

The second one is true, because we are given the statement so you can get the results from it. In this case, because it is given that all humans have the same number of fingers on both sides, and Aristotle is human, he will have the same amount of fingers on both, which is 3.

Edit: I am assuming that the cases are separate. If they are not, then the second one would be false because the given statement was from the first one is false.

Title: Is the argument below valid?


答:
One could use epsilon-delta definition on the other two LHS, and RHS limits, and deduce that the original limit is less than epsilon.
(Pretty much what the guy above me said).</p>

Title: $\epsilon - \delta$ definition of limit vs left and right hand definition of limit.

Body:
We initially learned in calculus that the limit of a function $f(x)$ at a point $x=c$ exists iff $$\lim\limits_{x \rightarrow c^+} f(x) = \lim\limits_{x \rightarrow c^-} f(x)$$
So what's not so rigorous about this definition that we are introduced to the new $\epsilon - \delta$ definition of limits in real analysis?


问:
Grasping the practical usage of Central Limit Theorem
I've been reviewing the Central Limit Theorem in statistics and I'm having trouble intuitively grasping how it can be practically used. The theorem tells us, subject to constraints, that the distribution of the sample mean will be approximately normal, even if the population distribution is not. From what I understand, this remarkable fact now allows us to use inferential statistical methods that apply to the normal distribution to answer questions about the population distribution. But I struggle with intuitively grasping why this is possible? That is, if I want to answer some question about a random variable X of unknown distribution, how does working with the sample mean of X get me to the answer about X? Is there a step missing that allows me to infer properties of X from properties of the sample mean of X?

答: Many probability distributions can be entirely characterized by a few numbers. For example a gaussian distribution (i.e. bell curve) can be characterized by giving its mean and its variance. Most probability distributions have their mean as one of their parameters, and also when explaining your results to someone it's often effective to talk about what the expected value of something is going to be and with what confidence you can state that. So having the Central Limit Theorem show you how your sample mean relates to the true mean is useful in both characterizing the probability distribution and in explaining to people how confident you are in your result.</p>


问:
Continuity of norm of continuous random variable
Suppose I have a continuous random variable $X \in \mathbb{R}^n$. Let the norm function $f$ in $\mathbb{R}^{n}$ be the Euclidean norm, i.e. $f : \mathbb{R}^n \to [0,\infty)$ such that $f(x) = \sqrt{x^T x} \text{ } \forall x \in \mathbb{R}^{n}$.

Since $X$ is a continuous random variable, we know that $\mathrm{cdf}_{X}(\cdot)$ is a continuous function, which means that that $\mathrm{pdf}_{X}(\cdot)$ exists a.e. and is non-negative where defined.

Note: By $\mathrm{cdf}_{X}(x)$, I mean
$\mathrm{cdf}_{X}(x) = p\left(X_1\leq x_1, ..., X_n\leq x_n\right)$. Maybe in the real analysis community that's called a distribution function - I'm not sure.

My goal is to show that $Y \doteq f(X)$ is a continuous random variable, i.e. to show that $\mathrm{cdf}_{Y}(\cdot)$ is a continuous function.

Which would be a good way to show that? My intuition tells me that $\mathrm{pdf}_{Y}(\cdot)$ exists a.e. and is non-negative where defined, and that this would be sufficient to show that $\mathrm{cdf}_{Y}(\cdot)$ is continuous. However, I am having a hard time figuring out how to prove that.

Another way I thought of proving that would be perhaps showing how 1) the square of each component of $X$, $X_i^2$, is a continuous random variable, 2) showing how the sum $\sum_i^n X_i^2$ is a continuous random variable, and then 3) showing how $\sqrt{\sum_i^n X_i^2}$ is a continuous random variable. This seems like a much more roundabout, tedious way to go about it. I am also struggling a bit with showing this.

Any pointers or hints would be very much appreciated.

答:
As stated, you need to be careful with what you actually get by the a.e. existence of a derivative of your CDF. Consider $n>=2$ and $\tilde{X}$ a standard Gaussian on $\mathbb{R}^n$. Then $X=\frac{\tilde{X}}{\|X\|}$ defines a random variable which is uniformly distributed on the unit sphere. Hence, $\sqrt{X^T X}$ has a degenerate distribution.

However, it's quite obvious that the CDF of $X$ is continuous, since the area of $\mathbb{S}^n$ contained in a tube of width $\varepsilon$ tends to $0$ as $\varepsilon$ does so (that's simply continuity of measures).


问:
Prove that a sequence of functions converge uniformly
Assume $ f:[0,1]\to[0,1] $ is continuous, $ f\left(0\right)=0 $ and for any $ x \in (0,1] $ we have $ f(x)<x $.

Define $ f_1=f $ and for any $ n \in \mathbb{N} $ $ f_{n+1}=f\circ f_{n} $.

Prove that $ f $ comverge unifrmly in $[0,1]$

My intuition is that $ f_n $ converge uniformly to $ 0 $.
I tried to prove it using that for any $ x $, $ f_n(x) $ is decreasing and also $ f_n $ continuous for any $ n $. So by Dini's Theorem it sufficies to show that $ f_n $ converge pointwise to $ 0 $. But I couldnt see how to prove it.

Thanks in advance.

答:
For $x=0$ everything is clear I think. So you still have to show $\lim_{n\to\infty}f_n(x)=0$ pointwise for all $1\geq x>0$.

Let $1\geq x>0$ be fixded. As you already mentioned the sequence $f_n(x)$ is monotonically decreasing as
$$f_{n+1}(x)=f(f_n(x))\leq f_n(x).$$ Since $f$ is bounded from below by $0$ you have a bounded and monotonically decreasing sequence $f_n(x)$ and thus a convergent sequence. Now you still have to show that the limit is $0$:

Assume $\lim_{n\to\infty}f_n(x)=x_0\neq0$. It follows that $x_0>0$. Since $f$ is continuous it holds
$$f(x_0)=f(\lim_{n\to\infty}f_n(x))=\lim_{n\to\infty}f_{n+1}(x)=x_0.
$$ A contradiction to $f(y)<y$ for all $1\geq y>0$. Thus $\lim_{n\to\infty}f_n(x)=0$ pointwise for all $1\geq x>0$. Now the uniformly convergence follows from Dini's theorem, as you already correctly mentioned.


问:
Roots of $(az + b)^3 = c$, where $z \in \mathbb{C}$ and $a,b,c $ are real numbers greater than $0$.
Question: Find the roots of $(az + b)^3 = c$, where $z \in \mathbb{C}$ and $a,b,c $ are real numbers greater than $0$.

Attempt: I have solved for roots for questions of the type $z^n = a$, where $n \in \mathbb{N}$ and $a \in \mathbb{R}$, by considering the polar form $z_k = a^{1/n}e^{\pi/n(1 + 2k)i}$. However, I'm unsure how to approach this question. Any help?

答: Hint: first set $w = az + b$ and solve the system $w^3 = c$ for $w$ using the method you already know. Then can you find $z$?


答:
$G_Y(y)=P(Y\le y)=P(f(X)\le y)=P(X\le f^{-1} (y))=F_X(f^{-1}(y))=F_X(f^{-1} (y))$

This is continuous only when $f$ is a one-to-one function so it has an inverse. If f is decreasing then f inverse increases and the cdf makes sense. But if f is increasing then  f inverse decreases and the cdf of y makes no sense.

Title: Continuity of norm of continuous random variable

Body:
Suppose I have a continuous random variable $X \in \mathbb{R}^n$. Let the norm function $f$ in $\mathbb{R}^{n}$ be the Euclidean norm, i.e. $f : \mathbb{R}^n \to [0,\infty)$ such that $f(x) = \sqrt{x^T x} \text{ } \forall x \in \mathbb{R}^{n}$.

Since $X$ is a continuous random variable, we know that $\mathrm{cdf}_{X}(\cdot)$ is a continuous function, which means that that $\mathrm{pdf}_{X}(\cdot)$ exists a.e. and is non-negative where defined.

Note: By $\mathrm{cdf}_{X}(x)$, I mean
$\mathrm{cdf}_{X}(x) = p\left(X_1\leq x_1, ..., X_n\leq x_n\right)$. Maybe in the real analysis community that's called a distribution function - I'm not sure.

My goal is to show that $Y \doteq f(X)$ is a continuous random variable, i.e. to show that $\mathrm{cdf}_{Y}(\cdot)$ is a continuous function.

Which would be a good way to show that? My intuition tells me that $\mathrm{pdf}_{Y}(\cdot)$ exists a.e. and is non-negative where defined, and that this would be sufficient to show that $\mathrm{cdf}_{Y}(\cdot)$ is continuous. However, I am having a hard time figuring out how to prove that.

Another way I thought of proving that would be perhaps showing how 1) the square of each component of $X$, $X_i^2$, is a continuous random variable, 2) showing how the sum $\sum_i^n X_i^2$ is a continuous random variable, and then 3) showing how $\sqrt{\sum_i^n X_i^2}$ is a continuous random variable. This seems like a much more roundabout, tedious way to go about it. I am also struggling a bit with showing this.

Any pointers or hints would be very much appreciated.


问:
product of $L^2$ function and its $L^2$ derivative
Let $f$ be absolutely continuous on $[0,x]$ for all $x >0$. $f(0) = 0$ and $f,f' \in L^2([0,\infty))$. Prove that $\int\limits_0^x |ff'| dm\leq \frac{1}{2}(\int\limits_0^x |f'| dm)^2$. The problem is simple if we remove the absolute value, since $(f^2)' = 2ff'$, but because of the absolute value, I am not sure how to proceed.

答:
From absolute continuity, $f(t)=\int_0^tf'$, therefore $$\int_0^x|ff'|=\int_0^x\left|\int_0^tf'(s)f'(t)\,ds\right|\,dt\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt.$$ Changing the order of integration, we obtain that $$\int_0^x|ff'|\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt=\int_0^x\int_s^x|f'(s)f'(t)|\,dt\,ds,$$ and renaming the integration parameters in the last integral, we have $$\int_0^x|ff'|\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt=\int_0^x\int_t^x|f'(t)f'(s)|\,ds\,dt.$$ Hence, $$\begin{align*}2\int_0^x|ff'|&\leq\int_0^x\int_0^t|f'(s)f'(t)|\,ds\,dt+\int_0^x\int_t^x|f'(t)f'(s)|\,ds\,dt\\ &
=\int_0^x\int_0^x|f'(t)f'(s)|\,ds\,dt=\left(\int_0^x|f'|\right)^2.\end{align*}$$


问:
Extension of a semisimple Lie algebra with a semisimple Lie algebra is semisimple. Why?
I am looking for a proof, that the extension of a semisimple Lie algebra with a semisimple is again semisimple.

I know the Theorem of Weyl and the one of Levi, so I thought about maybe using them.

答:
In order for this question not to stay on the list of unanswered, here is the full answer:

Assume $L$ is a Lie algebra with a semisimple ideal $I$ such that $L/I$ is semisimple. Assume $S$ is a solvable ideal of $L$.

The ideal $(S + I)/I$ in $L/I$ is isomorphic to $S/(S\cap I)$ and thus solvable. But since $L/I$ is semisimple, this means that it must be $0$, which means that $S + I = I$ so $S\subseteq I$.

But now $S$ is a solvable ideal in the semisimple Lie algebra $I$, so we must have $S = 0$ which completes the proof.


问:
How many marbles must you take out in order to be certain of getting two of the same colour?
I came across this question as a part of a practice test I'm working on to prepare for a maths exam at school:

An urn contains 4 red, 6 blue, 3 green, 8 brown and 9 white marbles. The urn has a lid and you can't see into it. How many marbles must you take out in order to be certain of getting two of the same colour?

I think this problem would be solved by using probability theory but wasn't able to find a way to do it. My initial thinking was that since 3 is the smallest of the coloured marbles that if I picked 2 then it would give me at least two green marbles but then there's the issue of not getting 2 of green when picking.

答: This problem is pretty simple. Start out by taking one. Since you don't want them to be the same, next you take out a different color. Continue this until you get 5 marbles of all different colors. After this, no matter what you take, you will get a duplicate. Therefore, 5 is the most you can get with all different, so you would need 6. This is basically the Pigeonhole Principle.</p>


Title: Reduction of order

Body: I have the equation $$x^2 y'' + (2x^2 - 3x) y' + 3y = 0$$ and the solution $$y_1 = x^3 e^{-2x}$$ Using reduction of order: $$y_2 = v y_1$$ I reduced the equation to $$v'' x^5 + v' x^4 = 0$$ so $v(x) = \ln x + C$. However, I thoroughly checked $$y_2 = x^3 e^{-2x} \ln x$$ and it does not work in the original equation. Any suggestions?

Body:
$$x^2 y'' + (2x^2 - 3x) y' + 3y = 0$$
$$ y''+2y'-3 \dfrac {( x y' - y)}{x^2} = 0$$
$$y'+2y-3\dfrac y x=C_1$$
$$ {y'}-y(\dfrac 3 x-2)=C_1$$
$$\left( ye^{-\int(\dfrac 3 x-2)}\right)'=C_1e^{-\int(\dfrac 3 x-2)}$$
$$ \left( y\dfrac {e^{2x}}{x^3}\right)'=C_1\dfrac {e^{2x}}{x^3}$$
$$ y\dfrac {e^{2x}}{x^3}=C_1 \int \dfrac {e^{2x}}{x^3} dx+C_2$$
$$ y(x)= {e^{-2x}}{x^3} \left(C_1 \int \dfrac {e^{2x}}{x^3} dx+C_2 \right)$$
The intergal can't be calculated with elementary functions.


答:
You’re right, there seems to be a step missing. If $X_1,...,X_n$ are iid then the CLT says $X_1+...X_n\sim N(n\mu, n\sigma^2)$. It works because the unstated assumption is the sample is iid. You would only be able to tell the mean and variance, but not the type of distribution the sample came from. That’s about it to the “theorem”.

On the other hand, you’re saying that how can you tell anything about the individuals if you have $X_1+...X_n\sim (mean, variance)$, where $X_1$ is number of births and $X_n$ is cigarettes smoked in a day. You can’t.

Title: Grasping the practical usage of Central Limit Theorem

Body: I've been reviewing the Central Limit Theorem in statistics and I'm having trouble intuitively grasping how it can be practically used. The theorem tells us, subject to constraints, that the distribution of the sample mean will be approximately normal, even if the population distribution is not. From what I understand, this remarkable fact now allows us to use inferential statistical methods that apply to the normal distribution to answer questions about the population distribution. But I struggle with intuitively grasping why this is possible? That is, if I want to answer some question about a random variable X of unknown distribution, how does working with the sample mean of X get me to the answer about X? Is there a step missing that allows me to infer properties of X from properties of the sample mean of X?


问:
Green function of the Sturm Liouville problem
I am working on the following Sturm-Liouville problem

$Ly=-y^{''}$ with boundary conditions $y^{'}(0)=y(1)=0$

I have calculated the eigenvalues $(\frac{\pi}{2}+k\pi)^2$ but I don't know how to calculate the eigenfunctions and the green function.
Can someone help me? Thank you very much.

答:
Let $\varphi_{\lambda}(x)=\cos(\sqrt{\lambda}x)$ and $\psi_{\lambda}(x)=\sin(\sqrt{\lambda}(x-1))/\sqrt{\lambda}$. Then
$$
      \varphi_{\lambda}''=\lambda\varphi_{\lambda},\; \varphi_{\lambda}(0)=1,\; \varphi_{\lambda}'(0)=0 \\
      \psi_{\lambda}''=\lambda\psi_{\lambda},\;\psi_{\lambda}(1)=0,\;\psi_{\lambda}'(1)=1.
$$
The Green function solution of $(L-\lambda I)g=f$ subject to $g'(0)=0$, $g(1)=0$ is given by
$$
       g= \frac{\psi_{\lambda}(x)}{w(\lambda)}\int_0^x f(y)\varphi_{\lambda}(y)dy+\frac{\varphi_{\lambda}(x)}{w(\lambda)}\int_x^1f(y)\psi_{\lambda}(y)dy,
$$
where $\omega(\lambda)$ is the Wronskian of the solutions $\varphi_{\lambda}$ and $\psi_{\lambda}$ given by
$$
             \psi_{\lambda}'\varphi_{\lambda}-\varphi_{\lambda}'\varphi_{\lambda} \\=\cos(\sqrt{\lambda}(x-1))\cos(\sqrt{\lambda}x)+\sin(\sqrt{\lambda}x)\sin(\sqrt{\lambda}(x-1)) \\
 =\cos(\sqrt{\lambda}(x-1)-\sqrt{\lambda}x)=\cos(\sqrt{\lambda}).
$$
There are poles of order $1$ in the resolvent expression at $\sqrt{\lambda_n}=(n+1/2)\pi$ or $\lambda_n=(n+1/2)^2\pi^2$. These are the eigenvalues, and the eigenfunctions are $\psi_{\lambda_n}$.