问:
Finding the limit of a function defined by integral
I am trying (unsuccessfully) to find the following limit:

$$ \lim_{t \to \infty} \frac{1}{t^2}\int_0^t\ln(2e^x+x^2)dx $$

Some background - supposedly, this exercise is meant to be similar to what we've done in class, where we found the limit:
$$\lim \limits_{t \to 0} \frac{1}{t}\int _1^e \ln \left(1+\frac{t}{x}\right)dx$$

We started by simplifying the integral to:
$$ \int _1^e \ln \left( x+t \right) dx- \int _1^e \ln \left( x \right) dx$$

and using change of variables and the integral's additivity, arrived at the following:
$$ \int _e^{e+t} \ln(y)dy - \int _1^{1+t} \ln(x)dx $$

At this point we've used the mean value theorem, namely that there exist $c_t \in [e,e+t] $ and $d_t \in [1,1+t] $  such that:

$$ \int _e^{e+t} \ln(y)dy - \int _1^{1+t} \ln(x)dx = (e+t-e)\ln(c_t) - (1+t-1)\ln(d_t) = t\left(\ln(c_t) - \ln(d_t)\right) $$

and therefore our limit is:

$$\lim \limits_{t \to 0} \frac{1}{t} \int _1^e \ln \left(1+\frac{t}{x}\right)dx = \lim_{t \to 0} [\ln(c_t) - \ln(d_t)] = \ln(e) - \ln(1) = 1$$

Where we used the squeeze theorem and ln's continuity for the last steps.

I am unsure how to use the mean value theorem when $ t \to \infty $, I've tried using the change of variables $ y = \frac{1}{x} $ in order to turn the upper limit from $t$ to $\frac{1}{t}$ (which approaches zero similarly to what we've done in class), but this also means that the lower limit switches from $0$ to $\infty$ so I didn't see how this method would work.

I would appreciate any and all help/advice.

答:
This method does not involve mean value theorem:

$$ \lim_{t \to \infty} \frac{1}{t^2}\int_0^t\ln(2e^x+x^2)dx $$

Note that $2e^x+x^2$ is an increasing function. Therefore it follows that $\ln(2e^x+x^2)$ is an increasing function as well.

This would mean that if the upper limit tends to $\infty$, then the integral $\to \infty$ as well

Now note that the expression is of the form $\frac{\infty}{\infty}$

HINT 1:

Use L'Hopital's rule with Leibniz's rule

HINT 2:

 $$ \lim_{t \to \infty} \frac{1}{t^2}\int_0^t\ln(2e^x+x^2)dx = \lim_{t \to \infty} \frac{\ln(2e^t+t^2)}{2t} $$Apply L'Hopital's rule again

Regarding your last question, you could switch the limits using the fact that

$$\int_a^b f(x)dx=-\int_b^a f(x)dx$$


问:
Prove $<x, f'(x) \geq pf(x)$
Let f: A →ℝ, where A ⊆ ℝn is an open. We will say that f is homogeneous
of grade p over A if

f (tx) = t^(p).f(x)

for each t ∈ ℝ and x ∈ A for which    tx ∈ A. If such a function is differentiable
at x ∈ A, prove that

```<code><x, f'(x) > = pf(x)
</code>```

答:
$\frac{d}{dt} f(tx) = \frac{d}{dt} t^{p} f(x)$
Gives $pt^{p-1} f(x) = \langle x , \nabla f (tx) \rangle$

Setting t = 1 gives $pf(x) = \langle x, \nabla f(x) \rangle $

ContentLicense: CC BY-SA 4.0

Title: Prove $<x, f'(x) \geq pf(x)$


问:
Finding the basis and dimension of kerT and imT for a linear transformation
Let $F: \Bbb R[t]_2 \longrightarrow\Bbb R[t]_4$ given by $F\bigl(p(t)\bigr)= t^2p(t)$.

To find $\ker F$ I suppose I need to find the solutions for the following equation

$$F(a + bt+ct^2) = at^2 + bt^3 + ct^4 = 0.$$

The solution I find in the back of the book says $\ker F = \{0\}$. But how is that possible if the equation $at^2 + bt^3 + ct^4 = 0$ has other solutions apart from $0$?

What about $\operatorname{Im}F$? Do I need to find a generator set from $at^2 + bt^3 + ct^4$?

答:
The kernel is $\{0\}$ because $a_0+a_1t+a_2t^2\in\ker F\iff F(a_0+a_1t+a_2t^2)=0$. But$$F(a_0+a_1t+a_2t^2)=a_0t^2+a_1t^3+a_2t^4,$$which is the null polynomial if and only if $a_0=a_1=a_2=0$. The existence of roots of $a_0t^2+a_1t^3+a_2t^4$ doesn't matter here.

And $\operatorname{Im}F$ is the set of those elements of $\Bbb R[t]_4$ of the form $a_0t^2+a_1t^3+a_2t^4$. In other words,$$\operatorname{Im}F=\{P(t)\in\Bbb R[t]_4\mid P(0)=P'(0)=0\}.$$


问:
Rank of a (3x5) matrix
I got this matrix :

For me, since Column 1 = Column 3 = Column 4

It implies that the maximum rank of this matrix should be 2.
Since Column 2 is linearly independent of Column 5, the rank of this matrix is equal to 2.
This is coherent to the definition of the rank and to the fact that a (3x5) matrix has a rank inferior to 3.

My question is : Is my reasoning false ?

答:
Rank = 3

I removed the 3 linearly independent matrices where I should have removed only two.
It leads us to a 3x3 matrix with 3 linearly independant columns

Title: Rank of a (3x5) matrix


问:
How do you write ${r \over (r+1)!}$ as a difference?
We are asked to evaluate the following sum

$$\sum_{r=1}^\infty {r\over (r+1)!}$$

I thought it best to write the expression out as a difference and use the method of differences to find the sum.

However, how would one go about converting it into a difference? I lucked upon the solution, as it turns out:

$$
{r\over (r+1)!} \equiv {1 \over r!} - {1\over (r+1)!}
$$

But it was simply a result of me playing about and was not rigorous at all. What would've been a sure-fire method to reach this? More generally how should one approach expression like these when looking to convert them into differences, similar to how converting some expressions like ${1\over 4r^2-1 }$ into partial fractions can ease the solution?

答:
Well in case of factorials it is always a good idea to use the method of comparing coefficients.$$\frac{r}{(r+1)!}=\frac {A}{(r)!}-\frac{A}{(r+1))!}$$

The rest is just comparing coefficients.

The pointis to bring it of the form of $|f(r)-f(r+1)|$

In case of   $$\frac{1}{4r^2-1}=\frac{1}{4}\frac{1}{(r-1/2)(r+1/2)}=\frac{1}{4}(\frac{A}{r-1/2}-\frac{A}{r+1/2})$$

Sometimes it gets harder:say we have to find $$\sum r(r+1)(r+2)(r+3)$$

In this case we can  write $T_r=r(r+1)(r+2)(r+3)$$

$$=\frac{1}{5}r(r+1)(r+2)(r+3)(r+4-(r-1))$$

$$=\frac{1}{5}r(r+1)(r+2)(r+3)(r+4)-(r-1)(r)(r+1)(r+2)(r+3)=V(r-1)-V(r)$$

The summary is in case consecutive coefficients come trying to manipulate a constant term to the form $r+k-r$ sometimes works.

Exercise:Evaluate $$\sum_{r=1}^{n}\frac{1}{r(r+1)(r+2)}$$

Hint:

use partial fractions $$\frac{1}{r(r+1)(r+2)}=\frac{A}{r}-\frac{B}{r+1}+\frac{C}{r+2}$$

Although some times we have to think a bit differently as in this problem: sum of terms of series


问:
A suspicious way to conclude convergence
Let
$$S \triangleq \sum_{k=0}^\infty\,(-1)^k\quad.$$
On one hand, the sequence of partial sums alternates between $0$ and $1$, therefore, does not get arbitrarily close to any value, and $S$ can't converge. On the other hand, it's easy to see that
$$S = (-1)^0 + \sum_{k=1}^\infty\,(-1)^k = 1 + \sum_{k=0}^\infty\,-(-1)^k=1-S\quad\therefore\quad S = \frac12\quad,$$
so that $S$ is finite. One of these rationales has to be wrong, right?</p>

答: Your first argument is correct. The second contains a built-in assumption that $S$ exists, which is not the case.</p>

ContentLicense: CC BY-SA 3.0


问:
What is the $\lim _{n \to \infty} \sum_{k=0}^{n} \frac{\sqrt k }{n^{\frac32}}$?
What is $$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right)$$
Wolframalpha tells me its 2/3 but I don't understand why.

What I do know is that:
$$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right) < \lim _{n\to \infty }\left(\frac{n\sqrt{n}\:}{n^{\frac32}}\right) =1$$

Body:
By Stolz–Cesàro theorem we obtain

$$\frac{\sum _{k=1}^{n+1}\sqrt{k}-\sum _{k=1}^n\sqrt{k}}{\sqrt{(n+1)^3}-\sqrt{n^3}}=\frac{\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}=$$

$$=\frac{\sqrt{n+1}}{\sqrt{(n+1)^3}-\sqrt{n^3}}\cdot\frac{\sqrt{(n+1)^3}+\sqrt{n^3}}{\sqrt{(n+1)^3}+\sqrt{n^3}}=\frac{\sqrt{n+1}\left(\sqrt{(n+1)^3}+\sqrt{n^3}\right)}{3n^2+3n+1} \to \frac23$$


问:
Is this quadratic function monotonically increasing?

Given vector $x'$ and positive definite matrix $W$, let $$ f(x) := (x-x')^TW(x-x') $$ Show that $f$ is monotonically increasing, i.e. if $y-x' > x-x'$ component-wise, then $f(y) > f(x)$.

I do not know how to approach this. I do not know how to use the positive definiteness of $W$.

答:
As showed by Rammus, many more conditions are needed.

In fact, because f is differentiable, for it to be monotonically increasing, its derivative must also be nonnegative, that is:

$\nabla f(x) \ge 0$ for all vector $x$ in $range(f)$.

And this is also a sufficient condition.

So I guess by now you can find the needed conditions for your function $f$.


答:
No, the rank of the matrix in this case is 3. Firstly the matrix is a short-wide matrix $(m<n)$. So maximum rank is m at the most

The rank depends on the number of pivot elements the matrix. I would say that your statement "Column 1 = Column 3 = Column 4" is wrong. You can say that Columns 1, 2 & 3 are Linearly Dependent Vectors. If you reduce the matrix by performing row operations you will end to at this state

$$
\begin{bmatrix}
1 & 0 & 1 & 1 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1\\
\end{bmatrix}
$$

Now we can perform column operation and send either two from columns 1, 3 & 4 to $[0]$ but not all. Because one stays to be a Identity vector no matter what.

One more way to look at is Column Space of the matrix:

$$
C(Matrix) = c1 \begin{bmatrix} 5 \\5 \\4 \\ \end{bmatrix} + c2 \begin{bmatrix} 4 \\5 \\2 \end{bmatrix} + c3 \begin{bmatrix} 2 \\4 \\1 \\ \end{bmatrix}
$$

$$
rank(Matrix) = dim(C(Matrix))
$$

Therefore, rank is 3.

Title: Rank of a (3x5) matrix

Body:
I got this matrix :

For me, since Column 1 = Column 3 = Column 4

It implies that the maximum rank of this matrix should be 2.
Since Column 2 is linearly independent of Column 5, the rank of this matrix is equal to 2.
This is coherent to the definition of the rank and to the fact that a (3x5) matrix has a rank inferior to 3.

My question is : Is my reasoning false ?


问:
What happens to t-statistic in log transformation of linear regression?
Problem

Suppose that the logarithm of $Y$ is regressed on the logarithm of $X$, the fitted regression
being:

$$log \hat{Y} = \hat{\beta_1} + \hat{\beta_2} log X$$
Suppose $X^* = \mu X$, where $\mu$ is a constant, and suppose that $ log Y$ is regressed on log $X^*$
.
Determine how the regression coefficients are related to those of the original regression.
Determine also how the t statistic for $\hat{\beta_2}$ and $R^2$
for the equation are related to those in the
original regression.

My ideas:

We can substitute the $X^* = \mu X$ into the original regression and get that $log \hat{Y} = \hat{\beta_1} + \hat{\beta_2}log \mu + \hat{\beta_2}log X$. As $\mu$ is a constant the regression we stay the same, with exception of new intercept $\hat{\beta} := \hat{\beta_1} + \hat{\beta_2}log \mu $.

Also, as t-statistics is $t = \frac{\hat{\beta_2} - \beta_2}{s.e(\hat{\beta_2})}$, it seems that $t_{\text{stat}}$ will increase in $\mu$ times as $ s.e(\hat{\beta_2})$ decrease in $\mu$ times.

Can it somehow help me?

答:
You are almost right. The only thing, yo should've substituted $X = \frac{X^*}{\mu}$. As you correctly mentioned the slope coefficient stays the same. However, new intercept is $\hat{\beta_1} - \hat{\beta_2}log\mu$.

You may check that resiaduals for the original and new models are the same. Hence, RSS and $R^2$ stay unchanged.


问:
Find the points of discontinuities of the function $x \sin\left(\frac{\pi}{x^2+x}\right)$
Need to find the points of discontinuities of the function $f(x) = x \sin\left(\dfrac{\pi}{x^2+x}\right)$

Obviously at zero. I can’t take the one-sided limit ($\lim{x \to 0+}$ and $x \to 0-$) any way, sin is periodic, and at zero has a million oscillations.

In zero discontinuities, how to prove it. And how to get similar limits.

 sorry for lang. Frrrrom Rrrussia)

答:
Actually, you are able to take the limit at $0$. Note that $-1\leq\sin(y)\leq 1$ for any $y$, so we have:

$$-\vert x\vert \leq x\sin(\frac{\pi}{x^{2} + x})\leq \vert x \vert\text{ for }x\neq 0$$

We have the $x\neq 0$ condition because we wish to inspect the function's behavior at $x=0$. Then, we have:

$$\lim_{x\rightarrow 0}-\vert x\vert\leq\lim_{x\rightarrow 0}x\sin(\frac{\pi}{x^{2} + x})\leq \lim_{x\rightarrow 0}\vert x\vert$$

$$0\leq\lim_{x\rightarrow 0}x\sin(\frac{\pi}{x^{2} + x})\leq 0$$

$$\lim_{x\rightarrow 0}x\sin(\frac{\pi}{x^{2} + x}) = 0$$

To prove that the limit exists, you can do the same thing with left- and right-sided limits. You will get that they are both $0$. This method of evaluating limits is called the squeeze theorem and it is useful when you want to find the limit of a messy trig function.

However, this function is still not continuous at $x=0$ because it cannot be evaluated there.


答:
It's not true that you can't take the limit as $x \to 0$.

Actually, using the usual bound $$-1 \le \sin (\ \text{something} \ ) \le 1$$
you get
$$-|x| \le f(x) \le |x|$$
Hence by the squeeze theorem
$$\lim_{x \to 0} f(x)=0$$

Title: Find the points of discontinuities of the function $x \sin\left(\frac{\pi}{x^2+x}\right)$

Body:
Need to find the points of discontinuities of the function $f(x) = x \sin\left(\dfrac{\pi}{x^2+x}\right)$

Obviously at zero. I can’t take the one-sided limit ($\lim{x \to 0+}$ and $x \to 0-$) any way, sin is periodic, and at zero has a million oscillations.

In zero discontinuities, how to prove it. And how to get similar limits.

 sorry for lang. Frrrrom Rrrussia)


答:
Actually, $f$ is continuous at $0$, since, for each $f$ in the sdomain of $f$, $|f(x)|\leqslant|x|$, and therefore $\lim_{x\to0}f(x)=0=f(x)$. Being continuous has nothing to do with oscillationg.

And, in fact, your function is continuous at every point of its domain (assuming that the domain is $\Bbb R\setminus\{-1\}$ and that $f(0)=0$).


问:
Are hypergraphs more expressive than graphs?
I started studying hypergraphs theory some days ago.

I know that a hypergraph is a tuple $H = (X, E)$, in which $E \subseteq \mathcal{P}(X)$ and
is actually a generalisation of the notion of graph.

Though, I'm wondering why they're useful. I saw this example of this paper.
They explain how in the first sample I can't discern whether an author wrote more than one article, whereas in the second one (with the hypergraph representation) I can easily get this information.

But this is not true, right? I can always attach the information on the edges or nodes to compute that. In addition, from what I understood, I can always represent hyperedges $e \in E$ as cliques, right? Hence, I can always (?) transform an hypergraph to a graph. I must be wrong.

My questions are:

Is the notion of hypergraphs really necessary?

Do hypergraphs and graphs have the same expressivity?

Can I represent something with hypergraphs which I can not with graphs?

答:
If you try really hard, you can express anything with graphs, especially if you let yourself attach information to vertices or edges to help.

The way you're suggesting - replacing each hyperedge by a clique - is not the best. In this way, you can't distinguish the hyperedge $\{1,2,3,4,5\}$ from the three hyperedges $\{1,2,3\}$, $\{3,4,5\}$, and $\{1,2,4,5\}$. You have to have a graph with an edge-labeling saying which hyperedge each clique comes from, and that's awkward.

The standard way to represent hypergraphs as graphs is with the incidence graph. Given a hypergraph $(X,E)$, its incidence graph is the bipartite graph with vertices $X$ on one side, vertices $E$ on the other side, and an edge $xe$ if $x \in e$ in the hypergraph.

Via this representation, (multi-)hypergraphs are actually equivalent to bipartite graphs (with a designated "$X$" side and "$E$" side). Any multi-hypergraph gives a bipartite graph, and any bipartite graph gives a multi-hypergraph. Theorems about one can be turned into theorems about the other.

Sometimes we use hypergraphs anyway, because a concept is easier to express for the hypergraph than it is for the incidence graph. Many theorems about graphs have natural generalizations to hypergraphs, and representing them as incidence graphs is very unnatural.


答:
If you know about generalized harmonic numbers, we can do amazing things.
Let
$$S_n=\frac 1 {n^\frac 32} \sum_{k=0}^n \sqrt k=\frac {H_n^{\left(-\frac{1}{2}\right)}} {n^\frac 32}$$ The asymptotics being
$$H_n^{\left(-\frac{1}{2}\right)}=\frac{2 n^{3/2}}{3}+\frac{\sqrt{n}}{2}+\zeta
   \left(-\frac{1}{2}\right)+O\left(\frac{1}{n^{3/2}}\right)$$ then
$$S_n=\frac{2}{3}+\frac{1}{2 n}+\cdots$$

Title: What is the $\lim _{n \to \infty} \sum_{k=0}^{n} \frac{\sqrt k }{n^{\frac32}}$?

Body:
What is $$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right)$$
Wolframalpha tells me its 2/3 but I don't understand why.

What I do know is that:
$$\lim _{n\to \infty }\left(\frac{\sum _{k=1}^n\sqrt{k}\:}{n^{\frac32}}\right) < \lim _{n\to \infty }\left(\frac{n\sqrt{n}\:}{n^{\frac32}}\right) =1$$


问:
Relationship between transpose matrices and orthogonal complements
I have recently been introduced to the idea of orthogonal complements, and I am given a question to try to relate them. Suppose I have a subspace $V$ of $R^n$ and a square matrix A of order n, such that $Av$ $\in$ $V$ for all $v \in V$. How can I make use of this to show that for an orthogonal complement of $V^⊥$, $A^Tw \in V^⊥$ for all $w \in V^⊥$? I am at a complete loss... Does anyone have any advice or tips?

In addition, suppose that $V = span\{u_1,u_2,...,u_k\}$. We have another matrix $B$ such that $u_1,u_2,...,u_k$ form the rows of $B$. How is it that the nullspace of $B$ is equal to $V^⊥$?

Thank you!

答:
If I understand you correctly, you are asked you show the following:

Let $V$ be a subspace of $\mathbb{R}^n$ and $A$ be an $n \times n$ matrix such that for all $v \in V$ we have $Av \in V$. Then for all $w \in V^{\perp}$, we have $A^{T} w \in V^{\perp}$.

PROOF: Let $v \in V$ $w \in V^{\perp}$. We need to show that $(A^T w) \cdot v = 0$

$$(A^T w) \cdot v=(A^T w)^T v=w^T A v=w \cdot (Av) = 0,$$

where we used that $w$ is orthogonal to $Av$ since $A \in V$ when $v \in V$.


答:
The function is defined when $x^2+x\ne0$, so the domain is $\Bbb R-\{0,-1\}$. It is continuous over its domain. It would be wrong to say that the function is "discontinuous" at $0$ since $0$ does not belong to the domain of the function, so $f(0)$ is not defined.

We can however talk about the limit of the function at $0$.$$L=\lim_{h\to0}h\sin\left(\frac\pi{h^2+h}\right)$$ Note that $\sin$ is bounded within $[-1,1]$ irrespective of what is inside $\sin$. So as $h\to0$, the function approaches $0$, giving $L=0$. Alternatively note that $|f(x)|\le|x|$ and use the squeeze theorem to get $0\le\lim |f(x)|\le\lim|x|=0$.


问:
A question based on metric space $d(x_i , y_i) $= $\sum_{i\geq 1 }|x_i -y_i| 2^{-i}$
This question was asked in my analysis quiz and I was unable to solve some options although I tried at home also.

Question: Let X={$(x_i)_{i\geq 1}: x_i \in$ {0,1} for all$ i\geq 1 $} with metric $d(x_i , y_i)$ = $\sum_{i\geq 1 }|x_i -y_i| 2^{-i}$. Let $f: X \to [0,1]$ be the function defined by $f(x_i)_{i\geq 1} = \sum_{i\geq 1}x_i 2^{-i}$. Which of the following options are correct:

1 f is continuous.

2 f is onto

3 f is 1-1

4 f is open

I have proved 1. But I don't know how to approach 2, 4 . I proved 3 right . BUt answer doesn't have it as correct.

I hav egiven a lot of time on it. I request you to tell me 2,3,4.

Waiting for your help!!

答:
Well, firstly, you need correct your $f$, it has to be $f(x) = \sum_{i} x_i 2^{-i}$ , i guess.

I think I encountered this one when learning contruction of $\mathbb{R}$.

2) Surjectivity is easy. In fact, if you look closely, this is in someway the formal the question of writing a real number in binary system.

So, loosely, we have for $1 > x \ge 0$:

$ x= \sum_{n \ge 1} \frac{ \left[ 2^nx \right]- 2\left[ 2^{n-1}x \right]}{2^n}$

3)This is wrong. Remind yourself that we have:
$\frac{1}{2} = \sum_{n \ge 1} \frac{1}{2^{n+1}}$

4) Open? I have no idea what this means. Sorry.

Update for part 3
Consider the two following sequence:
$$ x= (1,0,0,0,0 ,....)$$
and
$$ y= (0,1,1,1,.....)$$

You can see that $f(x) = \frac{1}{2}$ and $f(y) = \sum_{n \ge 1 } \frac{1}{2^{n+1}}= \frac{1}{2}$

Thus $f(x)=f(y)$ , while $x \ne y$


答:
Given $f(0)=0$, $f(x)$ will be continuous as the left and right limit abiut $x=0$ exist by sandwich theorem:$$ -x\le x\sin \frac{\pi}{x^2+x} \le x ~~~~(1)$$.

But by sandwich theorem (1) left lim is 1 and right limit is -1 about $x=-1$
the limit does not exist.So $f(x)$ is essentially discontinuous at $x=-1$.


问:
If $f (x) <g (x) $ for every $x \in [a, b]$, then there exists $δ> 0$ such that $f + δ ≤ g$.
Question:
Prove that if $f, g: [a, b] → \mathbb{R}$ are continuous and $f (x) <g (x) $ for  every $x \in [a, b]$, then there exists $δ> 0$ such that $f + δ ≤ g$.

I have tried to take the $f-g$ function and apply something like intermediate value theorem, but couldn't conclude anything.  I've seen a solution that uses covers, but we haven't seen that yet and would like to know if there is an easier way to fix it.

答:
let $h=g-f>0$ on $[a,b]$. Since $h$ is continuous on a closed interval, it has a global minimum (extreme value theorem).

then let $\delta=\min_{x\in[a,b]}{h(x)}$

Then, by definition $\delta>0$ and $\forall x\in[a,b], f(x)+\delta\leq g(x) $


答:
Suppose not. Then for every $\delta>0$ there is a point $x\in[a,b]$ where $g(x)-f(x)<\delta$. In particular, there is a sequence $x_n\in[a,b]$ with $g(x_n)-f(x_n)<1/n$.

The sequence $x_n$ doesn't necessarily have a limit, but by Bolzano-Weierstrass there is a convergent subsequence $x_{n_i}\to x\in[a,b]$. Now $g(x_{n_i})-f(x_{n_i})\to 0$, so $g(x)=f(x)$.

Title: If $f (x) <g (x) $ for every $x \in [a, b]$, then there exists $δ> 0$ such that $f + δ ≤ g$.

Body:
Question:
Prove that if $f, g: [a, b] → \mathbb{R}$ are continuous and $f (x) <g (x) $ for  every $x \in [a, b]$, then there exists $δ> 0$ such that $f + δ ≤ g$.

I have tried to take the $f-g$ function and apply something like intermediate value theorem, but couldn't conclude anything.  I've seen a solution that uses covers, but we haven't seen that yet and would like to know if there is an easier way to fix it.


答:
The idea is that you want to prove that there exists some constant $\delta$ such that $g(x)-f(x) \geqslant \delta >0$ for all $x \in [a,b]$. Your only hypothesis are the continuity of $f$ and $g$, with $g >f$, over the set $[a,b]$.

But there is a well-known property of continuous functions over closed bounded interval : if $h : [a,b] \to \mathbb{R}$ is continuous, then $h$ is bounded and its infimum and supremum are maximum and minimum.

Thus, if $h = g-f$, then $h$ is continuous over $[a,b]$ and has a minimum :
\begin{align}
\exists c \in [a,b],~ \forall x \in [a,b],~ h(x) \geqslant h(c)
\end{align}
Consequently,
\begin{align}
\exists c \in [a,b],~ \forall x \in [a,b],~ g(x) \geqslant h(c) + f(x)
\end{align}
By assumption, $h(c) = g(c)-f(c) >0$, and so the result follows by defining $\delta = h(c)$.


答:
The Mean Value Theorem says
$$
\tfrac32k^{1/2}\le\overbrace{(k+1)^{3/2}-k^{3/2}}^{\frac32\kappa^{1/2}\text{ for }k\lt\kappa\lt k+1}\le\tfrac32(k+1)^{1/2}
$$
Summing, we get
$$
\frac32\sum_{k=0}^{n-1}k^{1/2}\le n^{3/2}\le\frac32\sum_{k=1}^nk^{1/2}
$$
or equivalently
$$
\frac23n^{3/2}\le\sum_{k=1}^nk^{1/2}\le\frac23n^{3/2}+n^{1/2}
$$
Divide by $n^{3/2}$ and take the limit.


问:
Derivative Calculation $\Delta(x\cdot\nabla u) = \color{red}{2}\Delta u + x \cdot (\nabla(\Delta u))$
$\Delta(x\cdot\nabla u) = \Delta u + x \cdot (\nabla(\Delta u))$

Does this hold for $x\in\Bbb R^n$ and $u : \Bbb R^n \to \Bbb R$ ? If it does, why does it?

I calculated the above formula by writing out all the variables from $x_1$ to $x_n$ and guessing that it must be correct. But I am not familiar with 'direct' calculations using $\nabla$ and $\Delta$, so I need some help. Thank you!

答:
Since there is no standard writing of the derivative of a vector field I don't know of a way to do this in a coordinate free manner. I'm coming to the following result which is different from your formula.

The $j$ in the calculation is fixed, and $\delta_{ij}$ is the Kronecker delta. I write $x_i$ for the $i^{\rm th}$ component of $x$ and $u_{.i}$ for the partial derivative of the scalar function $u$ with respect to $x_i$. Then
$$\eqalign{
{\partial\over\partial x_j}\left(\sum_i x_i u_{.i}\right)&=
\sum_i\bigl(\delta_{ij} u_{.i}+x_i u_{.ij}\bigr)\cr
 &=u_{.j}+\sum_ix_i u_{.ij}\ .\cr}$$
Doing this a second time we obtain
$$\eqalign{
{\partial^2\over\partial x_j^2}\left(\sum_i x_i u_{.i}\right)&=
u_{.jj}+{\partial\over\partial x_j}\sum_ix_i u_{.ij}\cr
&=u_{.jj}+\sum_i\bigl(\delta_{ij}u_{.ij}+ x_i u_{.ijj}\bigr)\cr
&=2u_{.jj}+\sum_i x_i u_{.ijj}\ .\cr}$$
Summing over $j$ we find that in fact
$$\Delta(x\cdot \nabla u)=2\Delta u+x\cdot\nabla(\Delta u)\ .\tag{1}$$
Note that in the one-dimensional case we have
$$\bigl(x u'(x)\bigr)''=\bigl(u'(x)+x u''(x)\bigr)'=2u''(x)+x\bigl(u'(x)\bigr)''\ ,$$
as obtained in $(1)$.


问:
Show that $\|x\|_{1}=\sum_{i=1}^{n} x_{i} \text { and }\|x\|_{\infty}=\max _{i=1, \ldots, n} x_{i}$ is not defined not strictly convex space.
Show that the vector space $\mathbb{C^n}$ with the either of the norms $$
\|x\|_{1}=\sum_{i=1}^{n} |x_{i}| \text { and }\|x\|_{\infty}=\max _{i\in\{1, \ldots, n\}} |x_{i}|
$$ is not strictly convex.

I am trying to think about a counter example and thus show that $$||U+V|| \neq ||U||+||V||$$
for $v=tU$ or $u=tV$ for $t\in \mathbb{R}$ but I couldnt find any counter example.

Body: $x_1=(1,0)$ , $x_2=(1,1)$ for both norm, I guess ?


答:
A norm is strictly convex if and only if the unit ball does not contain intervals.
Take $x_1=(1,0,\dots,0)$ and $x_2=(0,1,\dots,0)$. Then for every $0\leq t\leq 1$
the vector $v_t=tx_1+(1-t)x_2=(t,1-t,\dots,0)$ satisfies $\|v_t\|_1=1$, so that norm is not strictly convex. Now take $y_1=(1,1,\dots,0), y_2=(1,0,\dots,0)$. Then for every $0\leq t\leq 1$ the vector $u_t=ty_1+(1-t)y_2=(1,t,\dots,0)$ satisfies $\|u_t\|_{\infty}=1$, so both unit balls contains intervals in their boundaries.

Title: Show that $\|x\|_{1}=\sum_{i=1}^{n} x_{i} \text { and }\|x\|_{\infty}=\max _{i=1, \ldots, n} x_{i}$ is not defined not strictly convex space.

Body:
Show that the vector space $\mathbb{C^n}$ with the either of the norms $$
\|x\|_{1}=\sum_{i=1}^{n} |x_{i}| \text { and }\|x\|_{\infty}=\max _{i\in\{1, \ldots, n\}} |x_{i}|
$$ is not strictly convex.

I am trying to think about a counter example and thus show that $$||U+V|| \neq ||U||+||V||$$
for $v=tU$ or $u=tV$ for $t\in \mathbb{R}$ but I couldnt find any counter example.


问:
What is the set containing only integers congruent to 89 modulo 17?
What is the set containing only integers congruent to 89 modulo 17</p>

答:
$$89\equiv 4 \pmod {17}$$ So we want the set of all integers $x$ such that $\;x\equiv 89 \pmod{17}\iff x \equiv 4\pmod{17}$. That can be written in a number of ways, e.g.:

$$\{x\mid x\equiv 4 \pmod{17}\} \quad \text{or}\quad \{x \mid x = 17k + 4;\;\; k\in \mathbb Z\}$$


答:
I would not say "writing out the expression as a difference", but more "turning it into more familiar terms". There is no general method for this.

In $\dfrac r{(r+1)!}$ we don't see a simplification, because we need $r+1$ at the numerator. So we can use an artifice and write

$$\dfrac r{(r+1)!}=\dfrac{r+1-1}{(r+1)!}=\frac1{r!}-\frac1{(r+1)!}$$ and voilà, by chance. We now can work out the partial $e$ series.

Title: How do you write ${r \over (r+1)!}$ as a difference?

Body:
We are asked to evaluate the following sum

$$\sum_{r=1}^\infty {r\over (r+1)!}$$

I thought it best to write the expression out as a difference and use the method of differences to find the sum.

However, how would one go about converting it into a difference? I lucked upon the solution, as it turns out:

$$
{r\over (r+1)!} \equiv {1 \over r!} - {1\over (r+1)!}
$$

But it was simply a result of me playing about and was not rigorous at all. What would've been a sure-fire method to reach this? More generally how should one approach expression like these when looking to convert them into differences, similar to how converting some expressions like ${1\over 4r^2-1 }$ into partial fractions can ease the solution?


问:
A realization of a group defined by semidirect product
Consider the group $G=C_3\rtimes C_9$, where $\rtimes$ denotes the semidirect product and $C_n$ denotes the cyclic group with $n$ elements.

Can this group elements be describe using matrices with respect to regular matrix multiplication? In not, is there an easy way to understand its elements? Thanks.

答:
There is only isomorphism class of nontrivial semidirect products $C_9 \rtimes C_3$ (where nontrivial means not isomorphic to the direct product), and such a group is defined by the presentation $\langle x,y | x^9=y^3=1, y^{-1}xy=x^4 \rangle$.

This group can be represented by $3 \times 3$ complex matrices, with
$$x \to \left(\begin{array}{ccc}0&0&1\\w&0&0\\0&1&0\end{array}\right),
\ \ \ \ y \to \left(\begin{array}{ccc}w&0&0\\0&w^2&0\\0&0&1\end{array}\right),$$
where $w$ is a primitive cube root of $1$.


答:
You get the same result by applying the formula for a geometric series:

\begin{array}{ccc}
S &=& a + ar + ar^2 + \dots \\
rS &=& ar + ar^2 + ar^3 + \cdots
\end{array}

Hence $S - rS = a$ and so $S = \frac{a}{1-r}$. If you put $a=1$ and $r=-1$ then you get $S=\frac{1}{2}$. The problem is that $S$ is only well-defined if $|r|<1$. If $|r| \ge 1$ then $S$ is not well-defined. You can't do arithmetic with something that is not well-defined. Your final answer does not make sense because you have been manipulating objects that do not make sense.

You can do all sorts of crazy things when working with infinity. For example: $\infty + \infty = \infty$, so dividing through by $\infty$ gives $1+1 =1$.

Title: A suspicious way to conclude convergence

Body:
Let
$$S \triangleq \sum_{k=0}^\infty\,(-1)^k\quad.$$
On one hand, the sequence of partial sums alternates between $0$ and $1$, therefore, does not get arbitrarily close to any value, and $S$ can't converge. On the other hand, it's easy to see that
$$S = (-1)^0 + \sum_{k=1}^\infty\,(-1)^k = 1 + \sum_{k=0}^\infty\,-(-1)^k=1-S\quad\therefore\quad S = \frac12\quad,$$
so that $S$ is finite. One of these rationales has to be wrong, right?</p>


问:
Confusion about Change of Variables in Integration
Suppose I want to calculate

$$\int_{0}^{2\pi} f(x) \, dx$$

for some continuous $f: \mathbb{R} \to \mathbb{R}$. Then if I make the change of variable $u = \sin x$, I get:

$$\int_{u(0)}^{u(2\pi)} f(u(x)) \cdot \, u'(x) \, dx = \int_{\sin 0}^{\sin 2\pi} f(\sin x) \, \cos x \, dx = \int_{0}^{0} f(\sin x) \, \cos x \, dx = 0.$$

Obviously, something's going wrong here -- it's not true that $\int_{0}^{2\pi} f(x) \, dx = 0$ for all $f$. But I can't seem to figure out where the flaw is! As far as I'm aware, $u = \sin x$ is a perfectly legal change of variable. Can anyone shed some light on what's going on? Any suggestions would be greatly appreciated.

答: In order to make a change of variable, you need your function diffeomorphism, so it needs to be a one to one function on $[0,2\pi] $ and the $\sin$ function is not, and the change of variable you made was $u=\sin{x} $, which means that $du=\cos  x dx$ so $\int \frac{f(\arcsin(u))}{\cos(\arcsin(u))}du=\int  \frac{f(\arcsin(u))}{\sqrt{1-u^2}}du$


问:
Restriction of a faithful representation of a von Neumann algebra
Suppose that $M$ is a von Neumann algebra represented on a Hilbert space $H$. Suppose $H_0$ is a subspace of $H$ invariant under $M$ so that we get a representation $\pi: M \rightarrow B(H_0) $. Will $\pi$ be faithful? I do not much about von Neumann algebras. Is representation of $M$ on $B(H)$ is faithful?

答:
No. Take $H=M=\mathbb C\oplus\mathbb C$, $H_0=\mathbb C\oplus0$.

Not sure what you mean by the last question.


问:
Simple model-theoretic arguments in set theory
Consider the following simple proof that PA has models with nonstandard numbers.

The type $\{ x > n \mid n \in \mathbb{N} \}$ is finitely realizable, so there is an $M \models \mathrm{PA}$ which fully realizes it. Any realization of this type is larger than all the standard natural numbers. QED

Arguments of this kind, and analogous ones via omitting types instead of realizing them, are the bread and butter of model theory. My question is this. Are there any obvious reasons why simple arguments of this kind can't prove the basic results of undecidability?

Naively, "PA doesn't prove its own consistency" is equivalent to "there is $M \models \mathrm{PA}$ which contains a proof of 0=1". (Of course, such a proof would be nonstandard.) I tried for a bit to prove this by a type-realization argument like the one above, but couldn't. Are there known obstacles to such a strategy?

Relatedly, could there be a simple type realizing/omitting argument to construct an $M \models \mathrm{PA}$ which contains some sentence $\varphi$ such that, in M, there are no proofs of $\varphi$ nor $\lnot\varphi$? This wouldn't quite show incompleteness of PA (as $\varphi$ could be nonstandard)  but it would be interesting to me.

In general, I guess I'm looking for a bit more of a connection between basic computability and basic model theory -- even if the connection is, "here are good reasons why simple methods from the latter don't work in the former".

答:
The point is that any technique for showing that (say) $\mathsf{PA}$ is incomplete must use something rather special about $\mathsf{PA}$, such as its computable axiomatizability. This is because there are in fact complete consistent extensions of $\mathsf{PA}$; the most obvious example is true arithmetic $\mathsf{TA}=Th(\mathbb{N};+,\times)$, but there are more exotic examples as well.

Basic model theory does provide us with quite flexible tools for building models with various sructural properties (compactness, Lowenheim-Skolem, omitting types). However, these tools are too flexible for establishing independence results: since they apply to all theories without finite models, including complete ones, they can't possibly be used to establish incompleteness.


问:
How to do a double integral for a parallelogram?

So I found that the Jacobian is equal to $1$. Then I found the bounds and then I found the integral by plugging in $u$ and $v$ for $x$ and $y$ and I integrated this:

$$\int _6^{10}\int _1^3(u+3v)dudv=208.$$

But $208$ is not the correct answer. What part of my setup is wrong? Thanks.

答: You exchanged the order of integration. $y=v$ ranges from $1\to3$ and $x+2y=u$ ranges from $6\to10$.


问:
If $\frac{dy}{dx}= 1-x-xy$, find the solution
This is a really short question, so I can’t write much here. Basically, I am not able to separate the variables, and I don’t know how to reduce it to that form either. What should I substitute in order to solve it?</p>

答:
We need to solve $$y'=1-x-xy$$
where $y'=\frac{dy}{dx}$ and $y=y(x)$.

Now, we  can see that $$y'+xy=-x+1$$
Let, $\mu(x)=\exp\left(x \right)=e^\frac{x^{2}}{2}$
Multiply both sides by $\mu(x)$ and then substitute $e^{x^{2}/2}x=\frac{d}{dx}\left(e^{x^{2}/2}\right)$  So, we have $$ \frac{d}{dx}\left(e^{x^{2}/2}y\right)=-e^{x^{2}/2}(x-1) \implies \int \frac{d}{dx}\left(e^{x^{2}/2}y\right) dx =\int -e^{x^{2}/2}(x-1)dx$$ Solving we get, $$e^{x^{2}/2}y=-e^{x^{2}/2}+\sqrt{\pi/2} \text{erfi} \left(\frac{x}{\sqrt{2}} \right)+c$$
Therefore, $$\boxed{ y(x)=\frac{1}{e^{x^{2}/2}}\left( -e^{x^{2}/2}+\sqrt{\pi/2} \text{erfi} \left(\frac{x}{\sqrt{2}} \right)+c\right)}$$


问:
Why is this set of polynomials not a vector space?
Why is the set of polynomials $M=\{ f(x)=2\lambda+7\lambda x +4\lambda x^2 + x^3 | \lambda \in $R$\}$ not a vector space over $R$?

答: Hint: Does the zero polynomial belong to $M$?


答: It is not a vector space under the usual addition of polynomials as addition and the usual multiplication with real numbers as scalar mutiplication, in other words, it is not a subspace of the usual vector space of real polynomials. Indeed, the zero polynomial is not in $M$, the sum of any two elements of $M$ is not in $M$, and for any $c\ne 1$, the $c$-fold multiple of any element of $M$ is not in $M$.

Title: Why is this set of polynomials not a vector space?

Body: Why is the set of polynomials $M=\{ f(x)=2\lambda+7\lambda x +4\lambda x^2 + x^3 | \lambda \in $R$\}$ not a vector space over $R$?


答: You can easily check that if $\lambda_1,\lambda_2\in M$, then $\lambda_1+\lambda_2\notin M $


问:
I did all explanation. Can you just teach me how to calculate this interior product?
Orient the unit sphere $S^n$ in $\Bbb R^{n+1}$ as the boundary of the closed unit ball.

Show that an orientation form on $S^n$ is

$w=\sum _{i=1}^{n+1}(-1)^{i-1}x^i dx^1∧...∧dx^i∧...∧dx^{n+1}$

I think that

I need to calculate i$_{X}(dx^1 ∧···∧dx^{n+1})$.

Let's write $X=\sum x^i\frac{\partial}{\partial x^i}$

Then I obtain that
$i_{\sum x^i\frac{\partial}{\partial x^i}}(dx^1 ∧···∧dx^{n+1})$

After here, How to calculate this? I have my final exam tommorow. So I need to learn this calculation. Please show me. Thank you so much:)

答:
Using the results that for 1 form $ \omega \in \Omega^1(M) $ you have $i_X(\omega) = \omega(X)$ and in general for any two forms $\omega,\eta $ you have $ i_X(\omega\wedge \eta) = (i_X\omega)\wedge \eta + (-1)^{\deg(\omega)}\omega\wedge (i_X\eta) $. Using these two it is easy to prove for 1 forms $ \omega_1,\omega_2,...,\omega_k \in \Omega^1(M) $ you have (the hat implies that is omitted)
$$ i_X(\omega_1\wedge\omega_2\wedge...\wedge \omega_k) = \sum_{i=1}^k (-1)^{i-1} \omega_i(X)(\omega_1\wedge\omega_2\wedge...\hat{\omega_i}...\wedge \omega_k) $$
So you have your answer.</p>


答:
$$\frac{dy}{dx}= 1-x-xy$$
$$\frac{dy}{dx}+xy= 1-x$$
$$\frac{dy}{dx^2}\dfrac {dx^2}{dx}+xy= 1-x$$
$$2\frac{dy}{dx^2}+y= \dfrac 1x-1$$
$$2\frac{dy}{dv}+y= \dfrac 1 {\sqrt v}-1$$
$$2y'+y= \dfrac 1 {\sqrt v}-1$$
Where $v= x^2$.

Can you solve this one ?

Title: If $\frac{dy}{dx}= 1-x-xy$, find the solution

Body: This is a really short question, so I can’t write much here. Basically, I am not able to separate the variables, and I don’t know how to reduce it to that form either. What should I substitute in order to solve it?</p>


问:
Proving a determinant equal a 0
How can I prove that this type of determinant is equal to zero. I couldn't figure out how to start this test or any counter-test. If p is one, the determinant is null. How to begin?

$
\begin{equation}
A = \begin{bmatrix}
(1 - p)a_1 & a_2 & \dots  & a_n \\
a_1 & (1 - p)a_2 & \dots & a_n  \\
\vdots  & \vdots & \ddots &  \vdots \\
a_1 & a_2 & \dots & (1 - p)a_n \\
\end{bmatrix}
\end{equation}
$

答:
The determinant is $(-1)^n a_1 \cdots a_n p^{n-1}(p-n)$ and so is zero iff $a_k=0$ for some $k$ or $p=0$ or $p=n$.

This follows from the matrix determinant lemma, since $A=D+B$ where $D$ is a diagonal matrix and $B$ has rank $1$.

Title: Proving a determinant equal a 0


问:
Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$
Here is my attempt:

Let $(x, y)$ be any such point.

Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.

Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$

Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.

Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$

Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?

答: If you had actually checked each candidate point for equality of distance from the coordinate axes and $(4,2)$, you would have seen that only $(2,2)$ and $(10,10)$ satisfy the conditions of the original question.


问:
Definiton of inner product on $L^2$ with positive definite symmetric matrix
Consider a measure space $\left(\Omega,\mathcal A,\mu\right)$. Now one can define an inner product $$\langle f,g\rangle_Q:=\int_\Omega\int_\Omega f(x)g(y)q(x,y)\mu\left(dx\right)\mu\left(dy\right),$$ where $q$ is symmetric and positve definite in the sene that $\forall A_1,\ldots A_n\in\mathcal A$ the matrix $Q$ with elements $q_{i,j}=\int_{A_i}\int_{A_j}q(x,y)\mu(dx)\mu(dy)\ge0$ is symmetric and positve definite.

Does this definition have to be understood in the following way. We take $q_{ij}$ as given and now construct $q(x,y)$ such that above equation for every measurable set holds true? I am not sure if $q$ is now well defined (i.e unique).

I really appreciate some enlightening on this defintion.

答: The $q$ is given. The matrix with entries $q_{ij}$ is used to establish a property that $q$ satisfies.


答:
All solutions must lie in the same quadrant as $(4,2)$. If any solution is in a different quadrant, it's distance from both axes is less than distance from $(4,2)$. Can you see why?

 The vector joining any point in a different quadrant cuts the axes before reaching the point $(4,2)$. The length of perpendicular dropped on the closest axis from the solution point is less than the length of this vector (along hypotenuse).

Title: Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$

Body:
Here is my attempt:

Let $(x, y)$ be any such point.

Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.

Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$

Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.

Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$

Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?


问:
prove that $(1 + x)^\frac{1}{b}$ is a formal power series
How to prove that $(1 + x)^\frac{1}{b}$ (where $b$ is an integer) can be expressed as a formal power series without using Binomial theorem?</p>

答:
The coefficient recurrence arises from the obvious first-order differential equation, namely

$$\rm\ \frac{y'}y\ =\ (log\ y)'\:=\ \bigg(\frac{log(1+x)}b\bigg)'\:=\ \frac{1}{b\ (1+x)}\ \ \ \Rightarrow\ \ \ y\: =\ b\ (1+x)\ y'$$

Therefore $\displaystyle\rm\ \ y\ =\ \sum_{k\ge 0}\ a_k x^k\ =\ b\ (1+x)\ \sum_{k\ge 0}\ (k+1)\ a_{k+1}\ x^k\:,\ \: $ which, after algebra, yields

$$\rm a_{k+1}\ =\ \frac{1-b\:k}{b\:(k+1)}\ a_k,\ \ \ a_0 = 1$$

As a check, note that it yields the binomial formula for $\rm\ b = 1/n,\ y = (1+x)^n\ $ since then

$$\rm \frac{a_{k+1}}{a_k}\ =\ \frac{n-k}{k+1}\ =\ \frac{n\choose k+1\:}{n\choose k\:}$$

and $\rm\ a_k = 0\ $ for $\rm\ k > n\ $ since the above implies $\rm\ a_{n+1} = 0\ $ hence $\rm\ a_{n+i} = 0\ $ for all $\rm\:i>0\:.$


问:
Multivariate Berry-Esseen/ Please help!
I've got a problem with understanding Berry-Esseen inequality for random vectors. You see, I keep coming across various forms of this theorem, all assuming a unit covariance matrix $I$, though it's seldom the case. I thought that maybe we can multiply the initial centered vector $X$ (or sum of vectors, if you please), for which we have the following:
$$
\sqrt{n} \left( X - \mu \right) \to N(0, \Sigma),
$$
– we can multiply our $X$ by some matrix $L : L \Sigma L^{\top} = I$, and then the new vector $LX$ will converge to $N(0, I)$ with the very same speed (and by speed I mean the same upper bound in Berry-Esseen).

Is it true? If not, why? And could you cast some light on what happens when the initial distribution is discrete - because in continuos case I can at the very least suggest that the abovementioned multiplication can be interpreted as a linear transformation, and therefore there will be change of density transformation with the determinant of Jacobian, and so on.

Please, help, this question drives me bonkers.

答:
You can multiply you random vector with $\Sigma^{-1/2}$, where $\Sigma=E[XX^T]$ is the covariance matrix.

Let $\tilde X = \Sigma^{-1/2} X$, Then $$E[\tilde X\tilde X^T] = \Sigma^{-1/2} E[XX^T]\Sigma^{-1/2} = \Sigma^{-1/2}\Sigma\Sigma^{-1/2} = I.$$

Title: Multivariate Berry-Esseen/ Please help!


问:
Probability questions : roll a fair die and compose a box with x colored balls
We roll a fair die and compose a box with 11 colored balls. Of these, the number of green balls is equal to the number obtained by rolling the dice. We then proceed by extracting a ball from the box. To calculate

the probability of getting a green ball out of the box Answer

the probability that there are 4 green balls in the box, if the ball
we have extracted is green Answer

We now proceed by extracting 2 balls from the box with putting them inside again. To calculate

the probability of pulling at least one green ball out of the box.
Answer

ps. i actually don't know where to start.

答:
Firstly, since we are rolling a fair die, we are likely to get an integer from 1 to 6 with equal probability (1/6 for each). Hence the probability of having 1 green ball in the box is 1/6, same as the probability of having 2, 3, 4 , 5 or 6 green balls in the box.

Hence,
P(Getting a green ball out of the box)
= (1/6)(1/11) + (1/6)(2/11) + (1/6)(3/11) + (1/6)(4/11) + (1/6)(5/11) + (1/6)(6/11)
= (1/6)*(21/11)
= 7/22

For the second part, we are conditioning P(4 green balls in the box) on P(Getting a green ball out of the box).
P(4 green balls in the box)= P(rolling a 4 on dice) = 1/6
P(P(4 green balls in the box ∩ getting a green ball out of the box) = (1/6)*(4/11) = 2/33

P(4 green balls in the box | Getting a green ball out of the box) = (2/33)/(7/22) = 4/21

For the third part,
P(Getting at least 1 green ball) = 1 - P(Getting no green balls)

P(Getting no green balls)
= (1/6)(10/11) + (1/6)(9/11) + (1/6)(8/11) + (1/6)(7/11) + (1/6)(6/11) + (1/6)(5/11)
= (1/6)*(45/11)
= 15/22

P(Getting at least 1 green ball) = 1 - 15/22 = 7/22.


问:
Solving the functional equation $f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy)$
Problem: find all functions $f:\mathbb{R}\to\mathbb{R}$ such that
$$f\left(x^2+f(y)\right)=f(x)^2+y^4+2f(xy),\ \ \ \forall x,y\in\mathbb{R}\text.$$

答:
It can be shown that the only function $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \left( x ^ 2 + f ( y ) \right) = f ( x ) ^ 2 + y ^ 4 + 2 f ( x y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is $ f ( x ) = x ^ 2 $. It's straightforward to check that this is indeed a solution. We prove that it's the only one.

General Properties of $ f $:

Let $ a = f ( 0 ) $ and $ b = f ( 1 ) $. By letting $ y = 0 $ and $ y = 1 $ in \eqref{0}, we respectively get the following:
$$ f \left( x ^ 2 + a \right) = f ( x ) ^ 2 + 2 a \text ; \tag 1 \label 1 $$
$$ f \left( x ^ 2 + b \right) = f ( x ) ^ 2 + 2 f ( x ) + 1 \text . \tag 2 \label 2 $$
Substituting $ - x $ for $ x $ in \eqref{1} and comparing with \eqref{1} itself, we have $ f ( - x ) ^ 2 = f ( x ) ^ 2 $. Then, substituting $ - x $ for $ x $ in \eqref{2} and comparing with \eqref{2} itself we get $ f ( - x ) = f ( x ) $; i.e. $ f $ is an even function. Letting $ x = 0 $ in \eqref{0} we have
$$ f \big( f ( y ) \big) = y ^ 4 + a ^ 2 + 2 a \text . \tag 3 \label 3 $$

Showing $ f ( 0 ) = 0 $ and $ f ( 1 ) = 1 $:

We can use \eqref{1} and \eqref{3} together to get
$$ f \left( x ^ 4 + 2 a x ^ 2 + a ^ 2 + a \right) = f \left( x ^ 2 + a \right) ^ 2 + 2 a \\
= f ( x ) ^ 4 + 4 a f ( x ) ^ 2 + 4 a ^ 2 + 2 a = f \Big( f \big( f ( x ) \big) \Big) + 4 a f ( x ) ^ 2 + 3 a ^ 2 \\
= f \left( x ^ 4 + a ^ 2 + 2 a \right) + 4 a f ( x ) ^ 2 + 3 a ^ 2 \text , $$
which by letting $ x = \sqrt { \frac 1 2 } $ yields $ 4 a f \left( \sqrt { \frac 1 2 } \right) ^ 2 + 3 a ^ 2 = 0 $, and that implies $ a \le 0 $. We can then use \eqref{1} and the fact that $ f $ is even to get
$$ f \left( \sqrt { - \frac a 2 } \right) ^ 2 + 2 a = f \left( \frac a 2 \right) = f \left( - \frac a 2 \right) = f \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 2 + 2 a \text . $$
On the other hand, substituting $ f ( x ) $ for $ x $ in \eqref{2} we can use \eqref{3} and see that
$$ f \left( f ( x ) ^ 2 + b \right) = \Big( f \big( f ( x ) \big) + 1 \Big) ^ 2 = x ^ 8 + 2 ( a + 1 ) ^ 2 x ^ 4 + ( a + 1 ) ^ 4 \text . $$
As we know that $ f \left( \sqrt { - \frac a 2 } \right) ^ 2 = f \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 2 $, we can conclude that
$$ \left( \sqrt { - \frac a 2 } \right) ^ 8 + 2 ( a + 1 ) ^ 2 \left( \sqrt { - \frac a 2 } \right) ^ 4 = \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 8 + 2 ( a + 1 ) ^ 2 \left( \sqrt { - \frac { 3 a } 2 } \right) ^ 4 \text , $$
which can be simplified to $ a ^ 2 \left( 5 a ^ 2 + 4 ( a + 1 ) ^ 2 \right) = 0 $, and thus $ a = 0 $. Now, putting $ x = 1 $ in \eqref{1} we have $ b ^ 2 = b $. But if $ b = 0 $, putting $ x = 0 $ in \eqref{2} leads to a contradiction. Hence $ b $ can't be equal to $ 0 $, and we must have $ b = 1 $.

Proving $ f ( x ) = x ^ 2 $:

Using \eqref{0}, \eqref{1}, \eqref{3} and $ a = 0 $, we have
$$ f \left( x ^ 4 + y ^ 4 \right) = f \Big( x ^ 4 + f \big( f ( y ) \big) \Big) = f \left( x ^ 2 \right) ^ 2 + f ( y ) ^ 4 + 2 f \big( x ^ 2 f ( y ) \big) \\
= f ( x ) ^ 4 + f ( y ) ^ 4 + 2 f \big( x ^ 2 f ( y ) \big) \text , $$
which by symmetry yields
$$ f \big( x ^ 2 f ( y ) \big) = f \big( y ^ 2 f ( x ) \big) \text . \tag 4 \label 4 $$
Now, putting $ y = 1 $ in \eqref{4} and using \eqref{3}, $ a = 0 $ and $ b = 1 $ we get $ f \left( x ^ 2 \right) = f \big( f ( x ) \big) = x ^ 4 $, and thus $ f ( x ) = x ^ 2 $ for all $ x \ge 0 $. As $ f $ is even, we conclude that $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $.


问:
construct an orthogonal family of support projections by Zorn's lemma
Let $M$ be a von Neumann algebra and $NS(M)$ be its normal state space.

Set $P_0(M):=\{s_{\phi}: \phi \in NS(M)\}$. There is a conclusion: For any projection $p\in M$, it follows from Zorn's lemma that there is an orthogonal family $(P_i)_{i\in I}$ in $P_0(M)$ satisfying $p=\sum_{i\in I} P_i$. Where $s_{\phi}$ is the support projection of the normal state $\phi$.

I tried to construct a partial orederd set as following :

Set $S_p=\{J：p=\sum_{j\in J}P_j, P_j\in P_0(M), P_iP_j=0 \quad \text{for any }i\neq j\}$, $S_p$ is ordered by inclusion.  But I cannot check every chain in $S_p$ has a upper bound in $S_p$.

答:
Consider the following set:
$$S_p:=\left\{ J \subseteq P_0(M)\ \middle|\ \sum_{P\in J}P \le p,\quad P\cdot Q = 0 \text{ for all $P,Q$ in $J$} \right\}$$
ie $S_p$ consists of all sets of mutually orthogonal support projections that sum up to something less than $p$ (this sum is to be understood to converge in the ultra-weak topology or in SOT or in whatever von Neumann topology you like to look at). Give this an order relation
$$J\leq I :\iff \sum_{P\in J} P \leq \sum_{P\in I} P,$$
ie $J$ is less than $I$ if the sum over $I$ is closer to the upper bound $p$.

Now for any chain $J_i$ in $S_p$ define $J:=\bigcup_i J_i$.All projections in $J$ will remain mutually orthogonal and their sum will remain less than the upper bound $p$. Hence $J\in S_p$ and obviously $J\ge J_i$ for all $i$. By Zorn's Lemma there are maximal elements of $S_p$.

Whats left is to check that any maximal element of $S_p$ must necessarily sum up to $p$. Suppose that $J$ is a maximal element and denote with $P_J=\sum_{P\in J}P$. Then $p-P_J$ is an orthogonal projection, if it isn't $=0$ then there must be some support projection $P'\leq p-P_J$, you can then check that $J\cup\{P'\}\geq J$, contradicting maximality.


问:
How to integrate $f(x)$?
I've been asked to integrate $\int{f(x)}^2 dx$ between the ranges of $L$ and $-L$.

I'm stuck! I understand how to integrate a constant or a function as in $x^2$ or something, but the $f(x)$ format is not something I'm familiar with. Please could someone explain this to me and perhaps guide me?

This is not the actual question I'm trying to solve. I think it might help if I put the above in context, so here's the question I'm trying to solve (I'm not asking anyone to do my homework, this isn't homework...):

Show that for a periodic function, defined over the integral $x=-L$ to $x=L$:

$$\frac1{L} \int_{-L}^L{f(x)}^2 dx = \frac{a_0^2}{2} + \sum_{n=1}^{\infty}(a_n^2 + b_n^2)$$

where $a_0$, $a_n$ and $b_n$ are Fourier coefficients of $f(x)$.

答:
First, you don't know anything about $f$ other than that it's periodic and has Fourier coefficients $a_n,b_n$. This means the exercise is not about integrating a particular function, like $x^2$ or $\cos(x)$, it's about integrating a general Fourier series.

So, second, what can we do? All we know is that $f(x)$ is periodic of period $2L$ and has Fourier coefficients. So really the only thing you can do is write down a Fourier expansion of $f$ and integrate its square:
$$f(x) = \sum_{n=0}^\infty a_n\cos(2\pi nx/L) + b_n\sin(2\pi nx/L),$$
so
\begin{align*}f(x)^2 &= \sum_{n=0}^\infty\sum_{m=0}^\infty\bigg(a_n\cos(2\pi nx/L) + b_n\sin(2\pi nx/L)\bigg)\bigg(a_m\cos(2\pi nx/L) + b_m\sin(2\pi nx/L)\bigg)\\
&=\sum_{n,m}\bigg(\mbox{product of the two factors}\bigg)\\
&= \mbox{etc ...}
\end{align*}
I'll leave the details for you to work out. Don't forget the most important theorem of Calculus II: $$\int_{-1}^1 \cos(2n\pi x)\sin(2m\pi x)\ dx = \delta_{nm} = \begin{cases}0, & n\neq m\\ 1, & n=m.\end{cases}$$

Third, for a general perspective, note that we can regard $f$ as an element of $L^2[-L,L]$, which has an orthonormal basis of integral-frequency cosines and sines. The Fourier coefficients of $f$ are the coordinates of $f$ with respect to this basis. Since the inner product is integration, your integral is just the square of the norm of $f$:
$$\int_{-L}^L f^2\ dx = \|f\|^2_{L^2}.$$
This suggests that, indeed, your integral will evaluate to the sum of the squares of the Fourier coefficients of $f$.

Title: How to integrate $f(x)$?


问:
Prove that distance between the two sets is $0$
I need to show that distance between the sets $y =0$ and $y = e^{-x}$ is $0$ under $d_2$ metric

Now let $A = \{(x,0)|x \in \mathbb{R}\}$

$B = \{(x,e^{-x})|x \in \mathbb{R}\}$

$d(A,B) = \text{inf} \{d(x,y)|x \in A, y \in B\}$

Now, suppose I pick the sequence $x_n =  (n,0) \in A$ and $y_n = (n,e^{-n}) \in B$, then

$d(x,y) = \sqrt{0 + e^{-2n}}  \to 0$ , as $n \to \infty$

Hence, there exist a sequence $(x_n,y_n)$ such that $d(x_n,y_n)\to  0$, therefore $d(A,B) = 0$

Is this proof correct ?

Please tell me if there are any errors.

Thank you.

答:
Indeed, you have :

$$\forall n\in \mathbb{N}, 0\leq \inf\{d(x,y)|x\in A, y\in B \}\leq d(x_n,y_n) $$

And by passing to the limit in the inequality, you get :
$$0\leq \inf\{d(x,y)|x\in A, y\in B \}\leq0 $$

So $d(A,B)=0$


问:
Solve a coupled second-order ODE
In a mechanics problem, I am trying to solve for the motion of two particles with position vectors $\boldsymbol{r_1}$ and $\boldsymbol{r_2}$.

I have written the equation's of motion in the form
$$\left( \begin{array}{c} \ddot{\boldsymbol{r_1}} \\ \ddot{\boldsymbol{r_2}} \end{array} \right) = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{array}{c} \boldsymbol{r_1} \\ \boldsymbol{r_2} \end{array} \right)$$
where $a, b, c, d$ are constants.

As the problem suggests, I have found the two real eigenvalues, $\lambda_1, \lambda_2$, and eigenvectors, $\boldsymbol{u_1}, \boldsymbol{u_2}$, corresponding to $\lambda_1$ and $\lambda_2$ respectively. The problem then says to use these to solve for $\boldsymbol{r_1}$ and $\boldsymbol{r_2}$, but I don't know where to go from here. How do I solve a system of this form?

答:
Let $r(t)=e^{\lambda t}\mathbf{u}$. Then one has
$$ \lambda^2e^{\lambda t}\mathbf{u}=A e^{\lambda t}\mathbf{u}\text{ or }A\mathbf{u}=\lambda^2\mathbf{u} $$
where
$$ A=\left( \begin{matrix} a & b \\ c & d \end{matrix} \right). $$
Thus $\lambda^2$ is an eigenvalue of $A$. Let $\mu_1,\mu_2$ be two eigenvalues of $A$ with corresponding eigenvectors $\mathbf{u_1,u_2}$. Then
$$ \lambda_1^2=\mu_1,\lambda_2=\mu_2^2, A\mathbf{u}_1=\lambda_1^2\mathbf{u}_1, A\mathbf{u}_2=\lambda_2^2\mathbf{u}_2$$ Then the general solution $\mathbf{r}(t)$ can be expressed as
$$ \mathbf{r}(t)=(c_1e^{\sqrt{\lambda_1}t}+c_2e^{-\sqrt{\lambda_1}t})\mathbf{u}_1+(c_3e^{\sqrt{\lambda_2}t}+c_4e^{-\sqrt{\lambda_2}t})\mathbf{u}_2.$$
Here $\mu_1\neq\mu_2$ is assumed. If $\mu_1=\mu_2$, you can treat similarly.

Title: Solve a coupled second-order ODE


问:
A problem on number theory involving concepts of primes, divisibility and factorials
Prove that for any positive integers $m$ and $n$, there exists set on $n$ consecutive positive integers each of which is divisible by a number of the form $a^m$ where $a$ is any integer.

To be honest, I'm not even sure what the question is asking. I mean should I write $a=1$ and move on, or is there an actual non trivial solution.

Any hint/ help is appreciated!

答:
It's a Chinese Remainder Theorem problem.  Take the system

$$x\equiv 0 \pmod{2^m}$$

$$x+1 \equiv 0\pmod{3^m}$$

etc.

$$x+n-1 \equiv 0 \pmod{p_k^m}$$

where $p_k$ is the $k$th prime.


问:
decide for what $n \in \mathbb{N}: 3^n \equiv n \mod 13$ is valid.
It is not enough to try every option mod 13
because

3^1≡1 (mod 13) not true

3^2≡2 (mod 13) not true

3^3≡3 (mod 13) not true

3^4≡4 (mod 13) not true

3^5≡5 (mod 13) not true

3^6≡6 (mod 13) not true

3^7≡7 (mod 13) not true

3^8≡8 (mod 13) not true

3^9≡9 (mod 13) not true

3^10≡10 (mod 13) not true

3^11≡11 (mod 13) not true

3^12≡12 (mod 13) not true

3^13≡0 (mod 13) not true (a^(prime number)≡a(mod (prime number)

I dont know how to find those numbers, there is no repetition...

答:
I try to show some results numerically:

$3^3\equiv 1\ mod (13)$

⇒$3^4\equiv 3 \ mod (13) $

⇒$3^{16}\equiv 3^4 \ mod (13)\equiv 3 \ mod (13)\equiv (13+3=16) \ mod (13) $

In this case $n=16$

$3^{29}=3^{16}\times 3^{13}\equiv (16\times 3=48) \ mod (13)\equiv 9 \ mod (13) $

⇒ $\frac {3^{29}}{3^2}=3^{27} \equiv 1  \ mod (13)\equiv (2\times 13 +1=27) \ mod (13)$

Or you may simply argue:

$(3^3)^3=3^{27}\equiv 1 \ mod (13)\equiv 27 \ mod (13)$

in this case $n=27$

I found by python $n= 16, 27, 35, 55, 66 , \cdot\cdot\cdot$

You can justify all these by similar argument.


问:
What are the prerequisites to Jech's Set theory text?
I'm looking for a book to self-study axiomatic set theory, and heard this was a classic.  What are the main prerequisites for this text? My knowledge of set theory isn't too great. Probably the only time I came across nontrivial set theory was when I read the proof that every nonzero ring has a maximal ideal (Zorn's Lemma).</p>

答: Hrbacek & Jech, Introduction to Set Theory, is a substantially gentler introduction than Jech, Kunen, or Halbeisen, but it still gets as far as discussing $\diamondsuit$ and Martin’s Axiom. (I’m not familiar with the others that have been mentioned so far.)


问:
Deriving Marginal Distribution from Two Independent Beta Distributions
I have two independent random variables:

$X$ is Beta($a, b$), $x \in [0, 1]$

$Y$ is Beta($a+b, \gamma$), $y \in [0, 1]$

$$f(x,y) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1}(1-x)^{b-1}\frac{\Gamma(a+b+\gamma)}{\Gamma(a+b)\Gamma(\gamma)}y^{a+b-1} (1-y)^{\gamma-1}$$

I want to perform the transformations for $U = XY$, $V=Y$.
$$U=XY \rightarrow U=XY \rightarrow X=U/V, \qquad u \in [0,1]$$
$$V=Y \rightarrow Y=V, \qquad v\in[0,1]$$

$$\det{J}= \frac{1}{v}*1-0=\frac{1}{v}$$
So I have,

$$f(u, v) = \frac{\Gamma(a+b+\gamma)}{\Gamma(a+b)\Gamma(\gamma)} (u/v)^{a-1}(1-(u/v))^{b-1}v^{a+b-1} (1-v)^{\gamma-1} * \frac{1}{v}, \qquad 0 \le u \le v \le 1$$

I want to find the marginal distribution of $u$.

$$f_{U}(u) = \frac{\Gamma(a+b+\gamma)}{\Gamma(a)\Gamma(b)\Gamma(\gamma)} u^{a-1} \int^1_{u} (1-(u/v))^{b-1}v^{a+b-1} (1-v)^{\gamma-1} \frac{1}{v} dv$$

QUESTION: However, I am unsure how to go about this integral. How can I proceed with this integration find the marginal distribution for $U$?

答:
The joint density results to me

$$f(u,v-u,1-v)=\frac{\Gamma(a+b+\gamma)}{\Gamma(a)\Gamma(b)\Gamma(\gamma)}u^{a-1}(v-u)^{b-1}(1-v)^{\gamma-1}$$

This is a Dirichlet distribution with parameters $(a,b,\gamma)$ thus it is immediate to derive the 3 marginal laws:

$$ \bbox[5px,border:2px solid black]
{
U\sim Beta(a; b  +\gamma)
\qquad (1)
}
$$

$$ \bbox[5px,border:2px solid red]
{
(V-U)\sim Beta(b; a  +\gamma)
\qquad (2)
}
$$

$$ \bbox[5px,border:2px solid blue]
{
(1-V)\sim Beta(\gamma; a+b)
\qquad (3)
}
$$

(3) is an obvious result, knowing that $V\sim Beta(a+b;\gamma)$


问:
Derive the formula for the Arc Length in polar coordinates
I want to prove the formula:

$\frac{ds}{d\theta } =\sqrt{\rho^{2} +\left( \acute{\rho } \right)^{2}  } $

One solution is to start with rectangular coordinates

$x=\rho .\cos \left( \theta \right)  \  \  \  and\  \  \  \  y=\rho .\sin \left( \theta \right)  \  \  \  ,\  \rho =f\left( \theta \right)$

$\frac{dx}{d\theta } =\acute{\rho } .\cos \left( \theta \right)  -\rho .\  \sin \left( \theta \right)  $

$\frac{dy}{d\theta } =\acute{\rho } .\sin \left( \theta \right)  +\rho .\  \cos \left( \theta \right) $

$\frac{ds}{d\theta } =\sqrt{\left( \frac{dx}{d\theta } \right)^{2}  +\left( \frac{dy}{d\theta } \right)^{2}  } =\sqrt{\rho^{2} +\left( \acute{\rho } \right)^{2}  } \  $

Now I want to reach the same result using only polar variables.
With reference to attached figure, the arc length could be approximated by a sum of very small line segments

$\left( \Delta s\right)^{2}  =\rho^{2} +\left( \rho +\Delta \rho \right)^{2}  -2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)$

$\left( \frac{\Delta s}{\Delta \theta } \right)^{2}  =\left( \frac{\rho +\Delta \rho }{\Delta \theta } \right)^{2}  +\left( \frac{\rho }{\Delta \theta } \right)^{2}  -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)  }{\left( \Delta \theta \right)^{2}  } $

$\left( \frac{\Delta s}{\Delta \theta } \right)^{2}  =\left[ \frac{\rho +\Delta \rho }{\Delta \theta } -\frac{\rho }{\Delta \theta } \right]^{2}  +\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  }{\left( \Delta \theta \right)^{2}  } -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)  }{\left( \Delta \theta \right)^{2}  }$

$\left( \frac{\Delta s}{\Delta \theta } \right)^{2}  =\left( \frac{\Delta \rho }{\Delta \theta } \right)^{2}  +\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  }{\left( \Delta \theta \right)^{2}  } -\frac{2\cdot \rho \cdot \left( \rho +\Delta \rho \right)  \cdot \cos \left( \Delta \theta \right)  }{\left( \Delta \theta \right)^{2}  }$

Now  $\  \  \  \  \  \left( \lim_{\Delta \theta \rightarrow 0} \  \frac{\Delta \rho }{\Delta \theta } \right)^{2}  =\left( \frac{d\rho }{d\theta } \right)^{2}  =\left( \acute{\rho } \right)^{2} $

I couldn’t figure out how the remaining expression could be equal to $\rho^{2} $

Could this or a different approach lead to the same formula?

答:
Keep the $\Delta$ terms up to second order only,

\begin{align}
  \cos (\Delta \theta) & \approx 1-\frac{(\Delta \theta)^2}{2}   \\
  (\Delta s)^2
  & \approx
  \rho^{2}+(\rho +\Delta \rho)^{2}-2\rho (\rho +\Delta \rho) \cos ( \Delta \theta ) \\
  & \approx
  [\rho^{2}-2\rho (\rho +\Delta \rho)+(\rho +\Delta \rho)^{2}]+
  \rho (\rho +\Delta \rho) (\Delta \theta)^2 \\
  & \approx
  [\rho-(\rho+\Delta \rho)]^2+\rho^2 (\Delta \theta)^2+
  \require{cancel} \cancel{\rho^2 (\Delta \rho)(\Delta \theta)^2} \\
  & \approx (\Delta \rho)^2+\rho^2 (\Delta \theta)^2 \\
\end{align}

Take $\Delta \theta \to 0$, the metric becomes

$$ds^2=d\rho^2+\rho^2 d\theta^2$$

that is

$$
\left(
  \frac{ds}{d\theta}
\right)^2=\rho^2+
\left(
  \frac{d\rho}{d\theta}
\right)^2$$


问:
Core of a set and its gauge
Let $A$ be a convex absorbing set in a linear space $X$ with gauge $p$. I want to show that

$\{x\in X: p(x)<1\}=$core of $A$.

We have $p(x)=\inf\{\lambda>0: x \in \lambda A\}$ and core of $A$ is the set of all $a$ in $A$ such that for all $x\in X\setminus \{a\}$ there exists  $0<r<1$ with the line segment
$[a, (1-r)a+rx)$ contained in $A$.

Any hint is appreciated.

答:
If $a\in core(A)$ then $p(a)<1$.

Maybe part of the difficulty is that you are using an equivalent but more inconvenient definition of $core(A)$. Let me use the following definition: $a\in core(A)$ if and only if for all $d\in X$ there is $r>0$ such that $a+sd\in A$ for all $s\in [0,r]$.

Since $A$ is absorbing $0\in core(A)$.
Now let $p(a)<1$. Then $\frac1\lambda a \in A$ for some $\lambda\in (0,1)$. Hence we can write
$$
a = \lambda \cdot  \frac1\lambda a + (1-\lambda)0.
$$
That is, $a$ lies on the line between a point in $core(A)$ and a point in $A$. Then it follows easily that $a\in core(A)$: Let $d\in X$. Then there is $r>0$ such that $0+sd\in A$ for all $s\in [0,r]$. Taking the convex combination as above it follows that
$$
a+(1-\lambda)sd = \lambda \cdot  \frac1\lambda a + (1-\lambda)( 0+sd) \in A.
$$
Hence $a\in core(A)$.

The latter argumentation can be used to prove that the convex combination of a point in the closure of $A$ and a point in the algebraic (topological, relative) interior of $A$ is again such an interior point.

Title: Core of a set and its gauge


答:
Point lies on the bisector $(t,t)$
$$(t-4)^2+(t-2)^2=t^2$$

$$t^2-12 t+20=0\to t=2;\;t=10$$

Points are $(2,2);\;(10,10)$

The other possibility, $A$ on the other bisector $(t,-t)$, leads to the equation:
$$(-t-2)^2+(t-4)^2 =t^2$$
which has no real solutions.
$$$$
$$
$$


问:
Does A exists such that for all $f$, $||f||_1 \le A \cdot ||f||_2$?
Let $E = C^0([0, 1], \mathbb{R})$, is the statement $\exists A \in \mathbb{R}^*, \forall f \in E, ||f||_1 \le A \cdot ||f||_2$ true ? With $||f||_1 = \int_0^1 |f(x)|dx$ and $||f||_2 = (\int_0^1 (f(x))^2 dx)^{1/2}$

答:
Let $(E,\langle\cdot,\cdot\rangle)$ be a real vector space with a scalar product $\langle \cdot,\cdot\rangle$. For $x$ and $y$ vectors in $E$, the Cauchy-Schwarz inequality  states that
\begin{align}
|\langle x,y\rangle | \leqslant \|x\|\|y\|
\end{align}
In $E=\mathcal{C}^0([0,1])$, the scalar product is defined by
\begin{align}
\int_0^1 f(t)g(t)\mathrm{d}t
\end{align}
The $\|\cdot\|_2$ and $\|\cdot\|_1$ norms are defined by
\begin{align}
{\|f\|_2} &= \sqrt{\int_0^1 f(t)^2 \mathrm{d}t} & \|f\|_1 = \int_0^1 |f(t)|\mathrm{d}t
\end{align}
Thus, if $f \in E$, you can apply the CS inequality to $x= |f|$ and $y = 1$, and
\begin{align}
\|f\|_1 = \int_0^1|f| = \int_0^1 |f|\times 1 \underset{CS}{\leqslant} \left\||f|\right\|_2 \times \|1\|_2
\end{align}
You have to check that $|f|$ and $f$ have the same $\|\cdot\|_2$ norm to conclude.


问:
A Bernoulli trials problem
Two parents decide to have children until they have 3 children of the same gender one after another (3 in a row). If p(boy)=p(girl)=1/2, how many children are they expected to have?

I have tried to draw a tree diagram to analyse the problem, but i can't generate a useful expression for the probability that a 'success' will happen on the nth trial.

So far P(3)=1/4, P(4)=1/8, P(5)=1/8, P(6)=3/32, P(7)=3/64, P(8)=5/128 and P(9)=1/32.

I have read a similar article on this forum about finding E(x) and Var(x) for 2 consecutive successes, but generating an expression here is far more difficult. In the case of first 2 consecutive successes in k trials, every success must be followed by a failure until the (k-1)th trial. Here not every success needs to be followed by a failure, though 2 consecutive successes must be followed by a failure until the (k-2)th trial.

many thanks,
Yun Fei

答:
They have a kid, cost $1$. We will add this at the end.

After the first kid, the parents are in one of two states: State A if the two preceding children are of the same sex (so they are almost there); State B is there is a previous child, but they are not in State A.

Let $a$ be the expected number of additional children if they are in State A, and $b$ the expected number of additional children if they are in State B.

If they are in State A, then with probability $\frac{1}{2}$, they get their three in a row, and breeding is over.  And with probability $\frac{1}{2}$ they go into State B, and their expectation becomes $b$. The cost in either case is $1$ more child. Thus
$$a=\frac{1}{2}(1)+\frac{1}{2}(1+b).$$

If they are in State B, then with probability $\frac{1}{2}$ they get a child of the same sex as the previous one, and enter State A. With probability $\frac{1}{2}$ they get a child of the opposite sex, and stay in State B. Thus
$$b=\frac{1}{2} (1+a)+\frac{1}{2}(1+b).$$

Solve. We get $b=6$. Add the initial $1$ mentioned in the first paragraph. The mean number of children is $7$.

Remarks: $1.$ We can also express the expectation as a series, by calculating the probability that breeding ends after $3$ trials, $4$, and so on. The seres can be summed explicitly. But the conditional expectation approach we used above is much smoother, and is of wide applicability.

$2.$ I would prefer a phrasing in terms of coin tossing, since in fact boy births and girl births are not equally likely. And the tacit independence assumption that we made is also not correct.

Title: A Bernoulli trials problem


问:
A function on space of $n\times n$ real matrices in euclidean space $\Bbb{R}^{n\times n}$
This question was asked in my analysis assignment and my answers are wrong. So, I am asking it here.

Let $M_n(\mathbb{R})$ denote the space of all $n\times n$ real matrices identified with the Euclidean space $\mathbb R^{n^2}$. Fix a column vector $x\neq 0$ . Define $f:M_n(\mathbb{R}) \to \mathbb{R}$ by  $f(A)= (A^2 x ,x)$ . THen

A) $f$ is linear

B) $f$ is differentiable

C) $f$ is continuous but not differentiable

D) $f$ is unbounded

I marked A , C but answers are B,D so Can you please tell  how to prove B, D and disprove A,C.

I shall be really thankful for your help.

Waiting for your help!!

答:
I am going to assume $(\cdot,\cdot)$ is the usual inner product.

A. If $f$ was linear, then as stated in the comments you should get $f(2A)=2f(A)$, but since theres a square on the 2 there you will get a coefficient mismatch.

B. You can open the inner product explicitly and you will see that you get a polynomial in the coefficients of $A$, of 2nd degree. Alternatively, you can claim both $A \mapsto A^2$ is differentiable and you compose it with a multiplication by vectors on the left and the right to get $x^\top A^2 x$ which are constants, so it's still differentiable

D. You can look at the sequence of matrices $(A_k = k I_n)_{k \in \mathbb N}$. You get that
$$f(A_k) = k^2 (I_nx,x)=k^2\| x\|^2$$
which tends to infinity as $k\rightarrow \infty$, so $f$ can't be bounded.


问:
How to calculate simple stochastic integral?
Can you suggest how can I calculate $\int_0^t e^{cs}dW_s$ where $W_s $ is Brownian motion and $c$ is constant?

答: Write $X_t=\int_0^te^{cs}\,dW_s$, deduce that $dX_t=e^{ct}dW_t$ and use Ito's Lemma.


问:
Prove that $n^{\log_2(3)}\geq n\log_2(n)$
How may I prove the following claim:

$$n^{\log_2(3)}\geq n\log_2(n)$$

Any hint please?

答:
HINT

We have that

$$n^{\log_2 3}=2^{\log_2 n\cdot \log_2 3} =3^{\log_2 n}$$

$$n^{\log_2 3}\ge n \log_2(n) \iff 3^{\log_2 n}\ge n \log_2(n) \iff \frac{3^{\log_2 n}}{\log_2(n)} \ge n$$

then let $n=2^x$ to obtain $f(x)=\left(\frac32\right)^x-x\ge 0$.


答:
If you write $a_{ij}$ for the elements of the matrix $A$ (its "coordinates" in $\mathbb R^{n^2}$), and similarly $x_i$ for the coordinates of the vector $x\in \mathbb R^n$, then you can write the explicit form of $f:\mathbb R^{n^2}\to \mathbb R$ as
$$f(a_{ij}) = \left\langle A^2x,x\right\rangle = \sum_{i,j,k}x_i a_{ij} a_{jk} x_k$$
which is a polynomial in the coordinates $a_{ij}$ with coefficients depending on the $x_i$. This is differentiable, which proves B and disproves C.

edit for comments:

In components, the inner product is
$$\langle v,w\rangle = \sum_k v_k w_k,$$
and matrix multiplication is described by
$$(AB)_{ij} = \sum_k a_{ik}b_{kj},$$
that is, the $(i,j)$th element of the matrix $AB$ is given by the above number. Similarly, matrices multiply with vectors to give a vector, whose $i$th component is
$$(Av)_i = \sum_k a_{ik}v_k.$$
Now put all this together!

Title: A function on space of $n\times n$ real matrices in euclidean space $\Bbb{R}^{n\times n}$

Body:
This question was asked in my analysis assignment and my answers are wrong. So, I am asking it here.

Let $M_n(\mathbb{R})$ denote the space of all $n\times n$ real matrices identified with the Euclidean space $\mathbb R^{n^2}$. Fix a column vector $x\neq 0$ . Define $f:M_n(\mathbb{R}) \to \mathbb{R}$ by  $f(A)= (A^2 x ,x)$ . THen

A) $f$ is linear

B) $f$ is differentiable

C) $f$ is continuous but not differentiable

D) $f$ is unbounded

I marked A , C but answers are B,D so Can you please tell  how to prove B, D and disprove A,C.

I shall be really thankful for your help.

Waiting for your help!!


问:
Finding a concrete Doob decomposition

Let $X_1, X_2, ...$ be i.i.d. random variables such that $\mathbb{E}[X_n^2] < \infty$ and $\mathbb{E}[X_n] = 0$ for all $n$. Let $M = \sum_{k = 1}^n X_k$. We consider everything relative to the filtration $\mathcal{F}$ generated by the $X_i$ and want to calculate predictable process $\langle M \rangle$ such that $M^2 - \langle M \rangle$ is Martingale.

So the requirement $\mathbb{E}[X_n^2] < \infty$ allows us to use Doob's decomposition theorem, which means that $\Delta \langle M \rangle_n = \mathbb{E}[\Delta(M_n^2) | \mathcal{F}_{n - 1}] = \mathbb{E}[X_n^2 | \mathcal{F}_{n - 1}] - 2M_{n - 1} \mathbb{E}[X_n | \mathcal{F}_{n - 1}]$. If we can evaluate this and determine $\langle M \rangle_1$ we know $\langle M \rangle_n$ for all $n$, but I struggle with both of these. We have not used the requirement $\mathbb{E}[X_n] = 0$ yet. The only way I can think of using this, is that this implies that $M$ is martingale. I don't know why that is relevant though. Can we even determine $\langle M \rangle_1$ instead of leaving that arbitrary? The definition of martingales does not require anything of the first element of the stochastic process.

答:
Suppose $Y=(Y_n)$ is an adapted sequence of integrable random variables, and you seek the Doob decomposition $Y_n = K_n+V_n$ in which $K$ is a martingale and $V$ is predictable. Observe that you would then have $\Bbb E[Y_n\mid\mathcal F_{n-1}] = K_{n-1}+V_n$ for $n= 1,2,\ldots$. Taking differences: $\Bbb E[Y_n\mid\mathcal F_{n-1}]-Y_{n-1} = V_n-V_{n-1}$. Therefore (assuming $V_0=0$)
$$
V_n=\sum_{k=1}^n \left(\Bbb E[Y_k\mid\mathcal F_{k-1}]-Y_{k-1}\right)
$$
for $n=1,2,\ldots$. In the same way,
$$
K_n=Y_0+\sum_{k=1}^n \left(Y_k-\Bbb E[Y_k\mid\mathcal F_{k-1}]\right).
$$

Now you can apply the discussion above to $Y_n=M_n^2$, and your $\langle M\rangle$ would be my $V$.

Title: Finding a concrete Doob decomposition


问:
Meaning of equation $dx=\sum_{A}\omega_Ae_A$.
I am reading some notes about Riemannian Structures. In definition of moving frame I see blow text and can't understand what $dx$ is:

By a moving frame in $U\subseteq \mathbb{R}^N$ we mean a choice of smoothly varying orthonormal bases $\{e_1(x),\ldots, e_N(x)\}$ for all $\mathcal{T}_xU, x\in U$. Taking exterior derivatives we obtain $$dx = \sum_A \omega_A e_A , de_A=\sum_B \omega_{BA}e_B$$
  where $\omega_A$'s and $\omega_{BA}$'s are 1-forms.

答:
Think of $x\colon U\hookrightarrow\mathbb R^n$. Then you can think of $dx$ both as an $\mathbb R^n$-valued $1$-form on $U$ and as a fancy way of thinking of the identity map on $T_xU\cong\mathbb R^n$ (as a tensor field of type $(1,1)$ on $U$).

Note that $\{e_A\}$ and $\{\omega_A\}$ are dual bases. We are writing, more pedantically, the identity map as
$$dx=\sum_A \omega_A\otimes e_A, \quad\text{i.e., for any }v\in T_xU,\quad
v=dx(v)=\sum_A \omega_A(v)e_A\,.$$


答:
Note that $\ln_2(3)=1.5849...>\frac{3}{2}$ and that $\ln(2)=0.693>\frac{1}{2}$. We also have that

$$\sqrt{100}-2\ln(100)=10-4\ln(10)=10-4(2.302...)>10-4\cdot 2.5=0$$

Also for $n\geq 100$ we have

$$\frac{d}{dn}\bigg(\sqrt{n}-2\ln(n)\bigg)=\frac{\sqrt{n}-4}{2n}\geq \frac{10-4}{2n}>0$$

Since

$$\sqrt{100}-2\ln(100)>0\text{ and }\frac{d}{dn}\bigg(\sqrt{n}-2\ln(n)\bigg)>0$$

we may conclude that for $n\geq 100$

$$\sqrt{n}-2\ln(n)>0$$

Manipulating this we get

$$\sqrt{n}>2\ln(n)=\frac{\ln(n)}{\frac{1}{2}}>\frac{\ln(n)}{\ln(2)}=\ln_2(n)$$

$$n^{\ln_2(3)}>n^{\frac{3}{2}}>n\ln_2(n)$$

To finish the proof, we simply check that the inequality is true for all $1\leq n\leq 99$ (which it is). We conclude

$$n\ln_2(n)\leq n^{\ln_2(3)}$$

for all $n\in\mathbb{N}$.

Title: Prove that $n^{\log_2(3)}\geq n\log_2(n)$

Body:
How may I prove the following claim:

$$n^{\log_2(3)}\geq n\log_2(n)$$

Any hint please?


问:
Is the Oval ( based on Ptolemy inequality) known?
It has a property of enclosing quadrilaterals so the ratio of their diagonals product  and sum of their opposite sides pair  products is constant $(e<1)$. The curve is from a family defined by the Ptolemy Inequality

In order to rope in the Ptolemy Inequality Oval took three points on a unit generating circle radius $ a=1 $ and the fourth one outside the circle

$$(-1,0),(0,-1),(1,0),(x,y)$$

as particular vertices of a non-cyclic  quadrilateral. The ratio $e$ defines its equation.

$$ \dfrac{\sqrt 2 \sqrt{x^2+(1+y)^2}}{\sqrt{y^2+(x+1)^2} + \sqrt{y^2+(x-1)^2}} =e<1 \tag 1 $$

Special case  $e=1$ is the circle enclosing cyclic quadrilaterals that have the property given by Ptolemy theorem. A set of non-cyclic quadrilaterals can be inscribed in this oval shape. In this drawing $ e=0.95; $

Some shapes for Other $e$ values

Further simplification yields a fourth degree algebraic curve:

$$\left(-a^4-2 a^3 y+a^2 \left(2 \left(e^2-1\right) x^2-2 y^2\right)-2 a y \left(x^2+y^2\right)-\left(x^2+y^2\right)^2\right)+\frac{\left((a+y)^2+x^2\right)^2}{2 e^2}=0$$

答:
Just a comment because I want to insert a picture

You can find it here https://mathcurve.com/courbes3d/crepe/crepe.shtml . Sorry if the link is in french .

Hope it helps you !


问:
Show that $\mathbb{M}$ is a probability measure and that under that measure the random variable $X$ has normal distribution $N(a,1)$
We define a measure $\mathbb{M}$ which is absolutely continuous w.r.t $\mathbb{P}$, and has the Radon–Nikodym derivative $\mathbb{M}/\mathbb{P}$ = $^{−^{2}/2}$. I need to show that $\mathbb{M}$ is a probability measure and that under the measure $\mathbb{M}$ the random variable  has normal distribution
(,1). I tried to show that $\mathbb{M}(\Omega)=1$, with $\Omega$ the unvivers using the definition from the course :$\mathbb{M}(\Omega)=\int_{\Omega} d\mathbb{M}/d\mathbb{P}~d\mathbb{P}=\int_{\Omega} e^{aX-a^{2}/2} d\mathbb{P}$. But then it seems like I can work with the expectation (I don't know why) and make appear a $1/\sqrt{2\pi}$. I don't know how to pass from $d\mathbb{P}$ to $dx$. Can I have some help please.

答:
For $\ X\ $ to be $\ \mathcal{N}(a,1)$-distributed with respect to $\ \mathbb{M}\ $, you must have
\begin{align}
\mathbb{M}\big(X\le x\big)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-\frac{x^2}{2}+ax-\frac{a^2}{2}}dx\\
&=\int_{X\le x}d\mathbb{M}\\
&=\int_{X\le x}e^{aX-\frac{a^2}{2}}d\mathbb{P}\\
&=\int_{-\infty}^xe^{ax-\frac{a^2}{2}}dF_\mathbb{P}(x)\ ,
\end{align}
where $\ F_\mathbb{P}\ $ is the distribution of $\ X\ $ with respect to $\ \mathbb{P}\ $.  For this to be true for all $\ x\ $, $\ F_\mathbb{P}\ $ must be the standard normal distribution.  If you make that assumption you should then be able to prove the required results (and it will be impossible to prove that $\ X\ $ is $\ \mathcal{N}(a,1)$-distributed with respect to $\ \mathbb{M}\ $ without that assumption).


问:
No proper dense open subsets of $\mathbb{R}^n$?
Let $U$ be a dense open subset of $\mathbb{R}^n$, a positive integer, and suppose that $X:=\mathbb{R}^n-U$ is $0$-dimensional; in the sense of the Lebesgue covering dimension.  Then is $U$ homeomorphic to $\mathbb{R}^n$ if and only if $U=\mathbb{R}^n$?

答: What about $$\mathbb{R}^2 \setminus \lbrace (x,0), x \geq 0 \rbrace \quad \quad ?$$


问:
If $\int {yy''dx} = 3xy$ is it possible to find y'?
So one of the people I know on discord gave me this problem to solve for fun.
I've tried using integration by parts on the left side and it didn't do anything.
The integral transforms into the ODE $yy''- (3xy)'=0$ but it's non-linear and I have no idea how to solve that. Would a series solution work?
Notice that I'm looking for y' not y so maybe there's some trick there?
Any help is appreciated. Thank you for reading.
Here, $y'$ and $y''$ denote the first and second order derivatives of $y$ w.r.t. $x$.

答:
There is impossible to directly integration of both sides , you have to face directly to solve $yy''-(3xy)'=0$ .

$yy''=3xy'+3y$

$y''=\dfrac{3xy'}{y}+3$

you can consider as the ODE of the type http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=416


问:
Solutions to the functional equation $f(f(x))=f(x+1)$
I am trying to find all solutions to $f(f(x))=f(x+1)$ with $f:\mathbb{R}\rightarrow \mathbb{R}$.

I have found that $f(x)=x+1$ is a solution and $f(x)=c$ for any $c \in \mathbb{R}$ and
I think that there are no other rational (polynomial) solutions.

Are there any other (perhaps non-continuous) solutions?

答:
There are no other polynomial solutions - if $f$ has degree $n$, then $f(f(x))$ has degree $n^2$, which is greater than $n$ if $n > 1$. And if $f$ has degree of at most $1$, then $f(x) = ax + b$, and we can find that solutions are only $a = 0$ and $a = 1, b = 1$ - which corresponds to $f(x) = c$ and $f(x) = x + 1$.

There are many discontinuous solutions even without AC. For example, $$f(x) = \begin{cases}0,\ x \in \mathbb{Z}\\ \frac{1}{2}, x \notin \mathbb{Z} \end{cases}$$.

There are also some other continuous solution. For example, $f(x + n) = |x|$ where $n \in \mathbb{Z}$ and $-\frac{1}{2} \leq x < \frac{1}{2}$ (any number has unique such representation): $f(f(x + n)) = f(x) = x$ and $f(x + n + 1) = x$. Alternative form of the same function: $f(x) = \rho(x, \mathbb Z) = \min\limits_{n \in \mathbb Z} |x - n|$.


答:
There are in fact uncountably many non-continuous solutions:

You can take $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(x)=0$, if $x \in \mathbb{Q}$, and $f(x)=y$, if $x \notin \mathbb{Q}$, for every irrational number $y \in \mathbb{R}$.

This works since there are uncountably many irrational numbers and the sum of an irrational number with a rational number is irrational.

Title: Solutions to the functional equation $f(f(x))=f(x+1)$

Body:
I am trying to find all solutions to $f(f(x))=f(x+1)$ with $f:\mathbb{R}\rightarrow \mathbb{R}$.

I have found that $f(x)=x+1$ is a solution and $f(x)=c$ for any $c \in \mathbb{R}$ and
I think that there are no other rational (polynomial) solutions.

Are there any other (perhaps non-continuous) solutions?


问:
Difference in values of centroid for a Kite with different approaches
Suppose a Kite with coordinates as (0.5,0.5), (1/3,1/3), (0.5, 0) and (1,0)

Centroid abscissa=1/4*(0.5+1/3+0.5+1)=7/12

Consider 2 triangles with coordinates (0.5,0.5), (1/3,1/3), (1,0) and (1/3,1/3), (0.5, 0) and (1,0)

Now, they both have equal areas and both their Centroid Abscissa are 11/18 each.

Thus now Abscissa of COM of the body would be 11/18

Why is it different in different approaches? What point I might be missing?

答:
You are computing the centroid of a triangle or quadrilateral as the arithmetic mean position of their vertices. But only in the case of a triangle (or for some special quadrilaterals) this is the same as the centroid of the area of the figure.

Your reasoning, then, gives you the centroid of the area of the kite, which is in fact the midpoint of the centroids of those two triangles composing it. But this is not the centroid of the kite vertices.

To obtain the centroid of the kite vertices from the triangles, you should consider your triangles but with the vertices they have in common (let's name them $A$ and $B$) having half the mass of the other vertex. In that case the centers of mass of both triangles vertices have abscissa $7/12$.


问:
How to prove using Ferrer's diagrams?
Using Ferrer's diagrams, show that the number of partitions of $n$ into parts of size 1 or 2 is equal to the number of partitions of $n$ into two parts.</p>

答:
It isn’t: it’s equal to the number of partitions of $n$ into at most two parts. The concept that you want is the conjugate of a partition.

Added: Here’s the Ferrers diagram of a partition whose parts all have sizes $1$ or $2$:

$$\begin{align*}
&\bullet\bullet\\
&\bullet\bullet\\
&\bullet\\
&\bullet\\
&\bullet
\end{align*}$$

And here is the Ferrers diagram of its conjugate partition:

$$\begin{align*}
&\bullet\bullet\bullet\bullet\bullet\\
&\bullet\bullet
\end{align*}$$

Notice that it has two parts. Here’s another example:

$$\begin{align*}
&\bullet\bullet\\
&\bullet\bullet\\
&\bullet\bullet\\
&\bullet\bullet\\
&\bullet\bullet
\end{align*}$$

And its conjugate:

$$\begin{align*}
&\bullet\bullet\bullet\bullet\bullet\\
&\bullet\bullet\bullet\bullet\bullet\end{align*}$$

And one more example:

$$\begin{align*}
&\bullet\\
&\bullet\\
&\bullet\\
&\bullet\\
&\bullet
\end{align*}$$

And its conjugate

$$\bullet\bullet\bullet\bullet\bullet$$

Title: How to prove using Ferrer's diagrams?


答:
There are an infinite number of non-continuous solutions which can be created by combining the two solutions you gave. For example, define $A_k$ as the set of algebraic numbers of degree $k$ or less. That is, $A_k$ is the set of algebraic numbers whose minimal polynomial has degree $k$ or less. Now, if $\alpha\in A_k$ then $\alpha+1\in A_k$ and if $\alpha\not\in A_k$ then $\alpha+1\not\in A_k$. Now, define

$$f(x)=\begin{cases}
      x+1 & x\in A_k \\
      c & x \in \mathbb{R}/A_k
   \end{cases}$$

(for any constant $c$). For any particular $x$ we have

$$f(x)\in A_k \Leftrightarrow x\in A_k$$

Thus, $f(x)$ satisfies the functional equation.


问:
Berry-Esseen vs concentration inequalities
I am trying to understand a bit more how concentration inequalities work but I am having some trouble. Indeed there is this Berry–Esseen theorem that implies that if $Z_n = (X_1+\ldots +X_n)\sqrt{n}/\sigma$ then then $Z_n\to g\sim \mathcal{N}(0,1)$ and quantitatively we have something like $\lvert P\{Z_N≥t\}−P\{g≥t\}\rvert≤\frac{C}{\sqrt{N}}$. Also it seems that this bound is tight. But the Hoeffding bound can bound the tail by something that looks like $e^{-t^2}$ like in the exponential. Intuitively these this seem a bit counter-intuitive and I don't get it.

答:
Well. I'm not sure what you asked.

Berry-Essen and Hoeffding deal with different objects. $Z_n$, which you provided, describe how the mean of $(X_1,X_2,...,X_n)$ distributes around $\mathbb{E}(X)$. And Berry-Essen tells you the speed at which these distributions converge to a normal law
.
While Hoeffding provides an estimation of how quick the mean $\frac{X_1+...+X_n}{n}$ converges to $\mathbb{E}(X)$.

So, one (Hoeffding) gives you a local view on the concentration while other (Berry-Essen) gives you a global estimation. And intuitively, the bound for a global estimation is less tight.

P/s: Of course, we can tout Hoeffding somehow to get some estimation on the convergence in law, but I think that is not the point. Hope this help.


问:
Are weak homotopy equivalences always homotopic?
I was wondering whether two weak equivalences between spaces $X$ and $Y$, are homotopic, or not. If yes, a proof / reference would be very welcome. If not, a counterexample would be interesting!
More generally, I am interested in the case of any category with a notion of weak equivalences / homotopies between weak equivalences, but I think the archetypal case of topological spaces is a good way to start.

Have tried a few things so far, but nothing worth mentioning yet.

答: A counterexample is given by considering the antipodal map and the identity map $S^1\to S^1$. Both are weak equivalences, but certainly not homotopic.


答: No, the identity map $S^2 \to S^2$ and the antipodal map are self-homotopy equivalences of $S^2$, yet the two maps are not homotopy equivalent.

Title: Are weak homotopy equivalences always homotopic?

Body:
I was wondering whether two weak equivalences between spaces $X$ and $Y$, are homotopic, or not. If yes, a proof / reference would be very welcome. If not, a counterexample would be interesting!
More generally, I am interested in the case of any category with a notion of weak equivalences / homotopies between weak equivalences, but I think the archetypal case of topological spaces is a good way to start.

Have tried a few things so far, but nothing worth mentioning yet.


问:
Showing a function of Bernoulli random variable is a martingale
Let $X_{0},X_{1},\cdots$ be i.i.d Bernoulli random variables with parameter $p$ (i.e., $\mathbb {P}(X_{i}=1)=p$, $\mathbb {P}(X_{i}=0)=1-p$). Define $S_{n}=\sum_{i=1}^{n} X_{i}$ where $S_{0}=0$. Define
\begin{align*}
Z_{n}=\left(\frac {1-p}{p}\right)^{2S_{n}-n},\quad n=0,1,2,\cdots
\end{align*}
Let $\mathcal {F}_{n}=\sigma(X_{0},X_{1},\cdots,X_{n})$. I want to show that $Z_{n}$ is a martingale with respect to this filtration. When I check if $\mathbb {E}(Z_{n+1}\vert \mathcal {F}_{n})=Z_{n}$, I have the following:
\begin{align*}
\mathbb {E}(Z_{n+1}\vert \mathcal {F}_{n})=\mathbb {E}\left( \left(\frac {1-p}{p}\right)^{2S_{n+1}-n-1}\vert\mathcal {F}_{n}\right)=\mathbb {E}\left( \left(\frac {1-p}{p}\right)^{2S_{n}-n}\cdot \left(\frac {1-p}{p}\right)^{2X_{n+1}-1}\vert\mathcal {F}_{n}\right)=\left(\frac {1-p}{p}\right)^{2S_{n}-n}\mathbb {E}\left(\left(\frac {1-p}{p}\right)^{2X_{n+1}-1}\vert\mathcal {F}_{n}\right)=\left(\frac {1-p}{p}\right)^{2S_{n}-n}\mathbb {E}\left(\frac {1-p}{p}\right)^{2X_{n+1}-1}=\left(\frac {1-p}{p}\right)^{2S_{n}-n}\cdot\left(\frac {1-p}{p}\right)^{2p-1}
\end{align*}
I don't know how to show that $\left(\frac {1-p}{p}\right)^{2p-1}=1$ or is there anything wrong with my argument?

答:
Your computation of $\Bbb{E} \left(\frac{1-p}{p} \right)^{2X_{n+1}-1}$ is wrong. Note that it is not equal to $\left(\frac{1-p}{p}\right)^{2\Bbb{E}[X_{n+1}]-1}$, which is what you have written. It is in fact equal to (using the formula $E[f(X)] = \sum_{k} P(X=k)f(k)$):
$$
\Bbb{E} \left(\frac{1-p}{p} \right)^{2X_{n+1}-1} = \left(\frac{1-p}{p}\right)^{2(0)-1} P(X_{n+1} = 0) + \left(\frac{1-p}{p}\right)^{2(1)-1} P(X_{n+1} = 1) \\
= \frac{p}{1-p}(1-p) + \frac{1-p}{p}(p) = p+1-p = 1
$$

and therefore the result of your computation is just $\left(\frac{1-p}{p}\right)^{2S_n - n}$, which shows one part of the martingale conditions. You still need to show that the random variable $Z_i$ has finite expectation, and is adapted to the required filtration. Both of these are easy exercises.

EDIT : To see a use of this martingale, let us imagine we are playing a simple game , where you are tossing coins that turn heads with probability $p$, and you are counting the number of successes, which is $S_n$. The above martingale provides great control on $S_n$. Furthermore, if we were , say losing /gaining a rupee for each time we won/lost (refer to Gambler's ruin) then the above martingale(or a variant) , along with stopping time results , will give us probabilties for various functions of the game (the most common being : if we start with $n$ dollars and wish to get to $N>n$ dollars before we go bankrupt while playing this game, what is the probability of that happening?)


问:
Reference request: theorem concerning random addition mod a positive integer
I am wondering if anyone can provide reference to a work providing roughly the following statement.

Let $N>2$ be a positive even integer, and $x_0$ be a nonnegative integer less than $N$.  Further, consider a partition of the $N$ elements less than $N$ into two equally sized "classes", $A$ and $B$, so that for every $y \in \mathbb{N}$, we have $y \mod N$ falls into either the $A$ or $B$ class (i.e. is an element of the set $A$ or $B$).

I am looking for a theorem that says, roughly, for uniformly randomly chosen $y \in \mathbb{N}$, the probability that $x_0 + y \mod N$ is in either class $A$ or $B$ is equal.  This is a somewhat trivial statement, but I would like a reference nonetheless.

答:
Consider a charge on $(\mathbb{N},2^{\mathbb{N}})$ of the form $\mathsf{P}(C):=T(\delta_C)$, where $T$ is a Banach limit on $\ell^{\infty}$ and $(\delta_C)_n=\mathbf{1}_C(n)$ (see this question).

Let $H_C:=\{y\in \mathbb{N}:(x_0+y \mod N)\in C\}$. Then
$$
\mathsf{P}(H_A)=\frac{1}{2} \quad\text{and}\quad \mathsf{P}(H_B)=1-\mathsf{P}(H_A)=\frac{1}{2}
$$
because the Césaro limit of the sequence $\delta_{H_A}$ is $1/2$.


问:
Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$
I was looking back in my junk, then I found this:

$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$

What is the value of $xyz$?

A) $5$

B) $4$

C) $1$

D) $-4$

E) $-5$

It's pretty easy, any chances of solving this question? I already have the
answer for this, but I didn't fully understand.

Thanks for the attention.

答:
\begin{align} x + z + y &= u=5 \tag{1}\label{1} ,\\ x^2 + z^2 + y^2 &= v=21 \tag{2}\label{2},\\ x^3 +z^3 + y^3 &= w=80 \tag{3}\label{3}. \end{align}

What is the value of $xyz$?

Surprisingly, the Ravi substitution
works in this case, despite that
not all the numbers $x,y,z$ are positive,
and hence, the corresponding triangle is "unreal".

So, let
\begin{align}
a &= y + z
,\quad
b = z + x
,\quad
c = x + y
\tag{4}\label{4}
,\\
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
\tag{5}\label{5}
,
\end{align}

where the triplet $a, b, c$ represents the sides of a triangle
with semiperimeter $\rho$, inradius $r$ and circumradius $R$.

Then

\begin{align}
x + z + y &= \rho
\tag{6}\label{6}
,\\
x^2 + z^2 + y^2 &= \rho^2-2(r^2+4rR)
\tag{7}\label{7}
,\\
x^3 + z^3 + y^3 &= \rho(\rho^2-12rR)
\tag{8}\label{8}
,\\
xyz&=\rho\,r^2
\tag{9}\label{9}
.
\end{align}

Excluding $rR$ from \eqref{7}-\eqref{8},
we get

\begin{align}
r^2&=
\tfrac16\,\rho^2-\tfrac12\,v+\tfrac13\,\frac w{\rho}
\tag{10}\label{10}
,
\end{align}

and from \eqref{9} we have the answer

\begin{align}
xyz&=
\tfrac16\,\rho^3-\tfrac12\,v\rho+\tfrac13\,w
=
\tfrac16\,5^3-\tfrac12\,21\cdot5+\tfrac13\,80
=-5
\tag{11}\label{11}
.
\end{align}

As a bonus, we can find that

\begin{align}
rR &= \tfrac1{12}\,\frac{\rho^3-w}{\rho}
\tag{12}\label{12}
\end{align}

and $x=\rho-a,\ y=\rho-b,\ z=\rho-c$ are the roots of cubic equation

\begin{align}
x^3-\rho\,x^2+(r^2+4rR)\,x-\rho r^2&=0
\tag{13}\label{13}
,\\
\text{or }\quad
x^3-\rho\,x^2+\tfrac12\,(\rho^2-v)\,x-\tfrac13\,w-\tfrac16\,\rho\,(\rho^2-3v)&=0
\tag{14}\label{14}
,\\
x^3-5\,x^2+2\,x+5&=0
\tag{15}\label{15}
.
\end{align}

One of the roots of \eqref{15} is

\begin{align}
x &= \tfrac53+
\tfrac23\,\sqrt{19}\,\cos\Big(\tfrac13\,\arctan(\tfrac9{25}\,\sqrt{331})\Big)
\approx  4.253418
\tag{16}\label{16}
,\\
\text{the other two are }\quad
y&\approx -0.773387
,\quad
z\approx  1.519969
\tag{17}\label{17}
.
\end{align}


答:
Taking $\log_{n}$ of both sides, we wish to prove:

$$\log_{2}(3) \geq \log_{n}(\log_{2}(n)) = \frac{\ln(\ln(n)) - \ln(2)}{\ln(n)}$$

Note that this restricts the domain to $n\in(1, \infty)$. We will prove the case where $n\in(0,1]$ later.

Thus, we want to find the global maximum of $f(n) = \frac{\ln(\ln(n)) - \ln(2)}{\ln(n)}$. This will use calculus. Taking the derivative, we obtain:

$$f'(n) = \frac{\ln(n)\cdot \frac{d}{dn}(\ln(\ln(n)) - \ln(2))- (\ln(\ln(n)) - \ln(2))\cdot \frac{d}{dn}(\ln(n))}{\ln(n)^{2}}$$

$$f'(n) = \frac{\frac{1}{n} - \frac{\ln(\ln(n)) - \ln(2)}{n}}{\ln(n)^{2}} = \frac{1 + \ln(2)-\ln(\ln(n))}{n\ln(n)^{2}}$$

Now we wish to find the critical points of $f(x)$. Since both the function and its derivative are continuous on their domain ($n\in(1,\infty)$, the same domain as $\ln(\ln(n))$), we only need to find the points where $f'(x) = 0$. We have:

$$\frac{1 + \ln(2)-\ln(\ln(n))}{n\ln(n)^{2}} = 0$$

$$1 + \ln(2)-\ln(\ln(n)) = 0\ (n\neq 0, 1)$$

$$\ln(\ln(n)) = 1 +\ln(2)$$

$$\ln(n) = 2e$$

$$n = e^{2e}$$

Testing $f'(x)$ on both sides of $n=e^{2e}$, we find that $f'(n) < 0$ for $n > e^{2e}$ and $f'(n) > 0$ for $1 < n < e^{2e}$. Note that this covers the whole domain of $f(x)$. Thus, $n = e^{2e}$ is where the global maximum is.

Then, $f(e^{2e}) = \frac{1}{2e}\approx 0.1839 < \log_{2}(3) \approx 1.5849$. Thus, the inequality is true for all $n\in (1,\infty)$. To prove that it is true in $n\in(0, 1]$, just note that the RHS of the original inequality is $\leq 0$ and the LHS is $\geq 0$ for $n$ in that interval.

Thus, the inequality is true for all positive $n$.


问:
Extending nullhomotopic function to boundary of sphere question
Suppose $f:\mathbb{S}^n\rightarrow Y$ is a continuous map null homotopic to a constant map $c$. In other words: $F: f\simeq c$ , where $c(x)=y$

Now, we may extend $f$ to a continuous map $g: D^{n+1}\rightarrow Y$ by defining

$g(x)= y$ if $0\leq ||x||\leq \frac{1}{2}$ and $F(\frac{x}{||x||},2-2||x||)$ if $\frac{1}{2}\leq ||x||\leq  \frac{1}{2}$

Now, on $||x||=\frac{1}{2}$, $F(\frac{x}{||x||},2-2||x||)=F(\frac{x}{||x||},1)=c=y$.

Hence $F$ is continuous by the gluing lemma.

I was wondering as to what the intuition is for constructing such a function, what geometrical clues allow one to define such a function $F$

答:
Here is the general case:

Let $f\colon (X,x_0) \to (Y,y_0)$ and $g\colon (S^n,s) \to (X,x_0)$ be basepoint preserving maps. Let $(Z,x_0)$ be the space that arises from $X$ by attaching a $D^{n+1}$ disk via $g$. Then the following are equivalent:

$f_*[g] = 0$ in $\pi_n(Y,y_0)$

$f$ extends to a map $Z\to Y$.

Since $f_*[g] = [f\circ g]$ is nullhomotopic in $\pi_n(Y,y_0)$, there is a homotopy $$H\colon S^n\times[0,1]\to Y$$ between $f\circ g$ and $\operatorname{const} = \{*\}$, i.e. $$H(\cdot, 0) = f\circ g,\ H(\cdot,1) = \{*\}$$

As Tyrone mentioned, since $H$ is constant on $S^n\times\{1\}$, we factor through a map $\widetilde{H}\colon S^n/S^n\times\{1\} \to Y$ via

$$S^n\times[0,1] \xrightarrow{q} S^n/S^n\times\{1\} \xrightarrow{\widetilde{H}} Y$$

Observe that $ S^n/S^n\times\{1\} \simeq D^{n+1}$, therefore you have

$$\widetilde{H}\colon D^{n+1} \to Y$$

since $$\widetilde{H}\vert_{S^n} \equiv f$$ the map $\widetilde{H}$ extends $f$ to a map $Z\to Y$.