问:
Roulette strategy
I came up with a strategy, which imitates the martingale (doubling) strategy , but it seems that there is a good probability to gain profit, which I want to compute.

So this is how we start:

We do not bet on black/red, but we chose columns, or dozens. Let's say we start with 100 chips.

Bet $1$ chip in e.g. a column. The probability to win is $12/37$. If you win, start over again, while if you lose:

Bet again $1$ chip to a column. Why again $1$? This is because we will still have profit if we win ( I will have won 3 chips, and lost 2 ). Again, if win go to step $1$, if lose:

Bet $2$ chips in a column. Each time we bet the lowest amount of funds, such that we have profit in case we win. In this case, if we win we will have played $4 $ chips , and won $6$, which gives the minimum  profit($2 $ chips).

Continuing with this strategy, one has to bet in columns/dozens, the following sequence of chips (each time he loses), in order to have the minimum profit :
$$1,1,2,3,4,6,9,14,21,31,47\dots$$
So, here is my question:

Using this strategy, and starting with $100$ chips, what is the probability of gaining $+50\%$, and what is the probability of doubling?
You continue until you have doubled, and stop if you have less chips than they are required to put the next bet.

*We have to consider that when we gain some profit, e.g. we have reached $140$ chips, we are able to bet more chips of the above sequence. With $100$ chips we can lose up to 9 times, and then bet $31$ chips. But with $140$ chips we can lose one more time , and then bet $46$.

Due to the complexity of the problem, I would also appreciate a numerical solution, but I do not have the knowledge to run a simulation.

Anyone can figure out a solution using probability theory?

答:
We either end as winners with 150 or 151 chips or as losers with between 0 and 50 chips (if we lose early, with between 0 and 33 chips;  in general the upper limit is $\frac13$ of the last peak). In a fair game this would mean a winning probability between $\frac12$ and $\frac35$ (just by the relation of up and down movements $+50:-50$ to $+50:-67$). Even this rough estimate (that does not take into account the bank advantage) matches well with Sharkoe's simulation.

A more precise calculation for $p_{k,d}$, the probability to reach $150$ when starting with $k$ and trying to win at least $d$ in the first sequence

$p_{k,d}=0$ if $2k<d$

$p_{k,1}=1$ if $k\ge 150$

$p_{k,d}=\frac{12}{37}p_{k+2b,1}+\frac{25}{37}p_{k-b,d+b}$ with $b=\lceil \frac d2\rceil$

This allows us to compute the probybility exactly as a fraction. However, numerator and denominator have about $785$ digits, so the numerical value from rounding that fraction
$$ p_{100,1}\approx0.56097114279613511032732301110367086534$$
should be good enough for us.

Code in PARI/GP (with memoization for values $p_{k,1}$, $100\le k\le 150$), edited to use less memory):

```<code> Start=100;Target=150
 A=vector(Target-Start+1,n,-1);
 getP(k,d)={local(pkd);
    if(2*k<d, 0, if(k>=Target, 1,
       pkd=if(d==1&&A[k-Start+1]>=0,
             A[k-Start+1],
             12/37*getP(k+2*ceil(d/2),1)+25/37*getP(k-ceil(d/2),d+ceil(d/2))
          );
       if(d==1,A[k-Start+1]=pkd);
       pkd))
    }
 getP(Start,1) +.0
</code>```

The new code after editing does not reflect it, but the "worst" case among the cases occuring is $p_{39,62}\approx0.1842$.

Title: Roulette strategy

Body:
We either end as winners with 150 or 151 chips or as losers with between 0 and 50 chips (if we lose early, with between 0 and 33 chips;  in general the upper limit is $\frac13$ of the last peak). In a fair game this would mean a winning probability between $\frac12$ and $\frac35$ (just by the relation of up and down movements $+50:-50$ to $+50:-67$). Even this rough estimate (that does not take into account the bank advantage) matches well with Sharkoe's simulation.

A more precise calculation for $p_{k,d}$, the probability to reach $150$ when starting with $k$ and trying to win at least $d$ in the first sequence

$p_{k,d}=0$ if $2k<d$

$p_{k,1}=1$ if $k\ge 150$

$p_{k,d}=\frac{12}{37}p_{k+2b,1}+\frac{25}{37}p_{k-b,d+b}$ with $b=\lceil \frac d2\rceil$

This allows us to compute the probybility exactly as a fraction. However, numerator and denominator have about $785$ digits, so the numerical value from rounding that fraction
$$ p_{100,1}\approx0.56097114279613511032732301110367086534$$
should be good enough for us.

Code in PARI/GP (with memoization for values $p_{k,1}$, $100\le k\le 150$), edited to use less memory):

 Start=100;Target=150
 A=vector(Target-Start+1,n,-1);
 getP(k,d)={local(pkd);
    if(2*k<d, 0, if(k>=Target, 1,
       pkd=if(d==1&&A[k-Start+1]>=0,
             A[k-Start+1],
             12/37*getP(k+2*ceil(d/2),1)+25/37*getP(k-ceil(d/2),d+ceil(d/2))
          );
       if(d==1,A[k-Start+1]=pkd);
       pkd))
    }
 getP(Start,1) +.0

The new code after editing does not reflect it, but the "worst" case among the cases occuring is $p_{39,62}\approx0.1842$.


问:
Weak convergence to strong convergence in a Hilbert space (Fredholm Alternative)
This question is related to the post here. It is regarding the closed range argument as in part (ii) of the theorem (Fredholm Alternative) below.

I have a question regarding, how the author arrives at "then (5) implies $u_{k_j}\to u$" in the proof, which is bolded below. It is not clear to me, how one can go from weak convergence to strong convergence like this.

I hope someone could clarify the argument here for me. Thank you in advance.

(PDE Evans, Appendix D, Theorem 5)

THEOREM 5 (Fredholm Alternative). Let $K : H \to H$ be a compact linear operator. Then

(i) $N(I-K)$ is finite dimensional,

(ii) $R(I-K)$ is closed,

(iii) $R(I-K)=N(I-K^*)^\perp$,

(iv) $N(I-K)=\{0\}$ if and only if $R(I-k)=H$,

and

(v) $\dim N(I-K)=\dim N(I-K^*)$.

proof of (ii):  We next claim there exists a constant $\gamma > 0$ such that $$\|u-Ku\|\ge \gamma \|u\| \quad \text{for all }u\in N(I-K)^\perp. \tag{4}$$ Indeed, if not, there would exist for $k=1,\ldots$ elements $u_k \in N(I-K)^\perp$ with $\|u_k\|=1$ and $\|u_k-Ku_k\|<\frac 1k$. Consequently, $$u_k-Ku_k \to 0. \tag{5}$$ But since $\{u_k\}_{k=1}^\infty$ is bounded, there exists a weakly convergent subsequence $u_{k_j} \rightharpoonup u$. By compactness $Ku_{k_j} \to Ku$, and then (5) implies $u_{k_j} \to u$. We therefore have $u \in N(I-K)$ and so $$(u_{k_j},u)=0 \quad (j=1,\ldots).$$ Let $k_j \to \infty$ to derive a contradiction to (4).

答:
Observe that
$$\|u_{k_j}-u_{k_\ell}\|\leq \|u_{k_j}-Ku_{k_j}\|+\|Ku_{k_j}-Ku\|+\|Ku-Ku_{k_\ell}\|+\|Ku_{k_\ell}-u_{k_\ell}\|\rightarrow 0,$$ which shows that $(u_{k_j})$ is Cauchy. Since $H$ is complete, this must converge. Additionally, the strong limit must match the (unique) weak limit, which is $u$.

EDIT: Alternatively, as pointed out by @DavidMitra in the comments, you can avoid showing that it's Cauchy by explicitly showing strong convergence to $Ku$, via the same sort of argument as before.

Calculate
$$\|u_{k_j}-Ku\|\leq \|u_{k_j}-K u_{k_j}\|+\|Ku_{k_j}-Ku\|\rightarrow 0,$$ then argue as above.

This is easier, but my mind went to Cauchy first for some reason.

ContentLicense: CC BY-SA 4.0


问:
Show that $O_{3}(\mathbb{R})$ spans $\mathcal{M}_3(\mathbb{R})$.
It was given as an assignment problem.

Given a real vector space $Y,$ let $T: \mathcal{M}_3(\mathbb{R})\rightarrow Y$ be a linear operator such that $T$ restricted to $O_{3}(\mathbb{R}),$ the set of orthogonal matrices, is a constant mapping. Show that $T \equiv 0 .$ Can the result be extended to dimension $n \neq 3 ?$

My attempt:

Suppose the constant is c then, T(I)=c and T(-I) is c. $\implies$ T(0) = c $\implies$ c is 0.

So T is identically zero on $O_{3}(\mathbb{R})$.

If I show $O_{3}(\mathbb{R})$ spans $\mathcal{M}_3(\mathbb{R})$ then we are done.
I don't know where to start, I was trying to use SVD but couldn't succed.

答:
Another proof: we note that the function $\langle  A,B \rangle = \operatorname{tr}(AB^T)$ defines an inner-product over $M_n(\Bbb R)$. It suffices to show that relative to this inner product, $O_n(\Bbb R)^\perp = \{0\}$.

Indeed, suppose that $A \in O_n(\Bbb R)^\perp$. The existence of the real polar decomposition ensures that there exists a $U \in O_n(\Bbb R)$ for which $AU^T$ is positive semidefinite. However, since $A \in O_n(\Bbb R)^\perp$, we must have
$$
\operatorname{tr}(AU^T) = \langle A,U \rangle = 0.
$$
On the other hand, $AU^T$ is positive semidefinite, which means that $\operatorname{tr}(AU^T) \implies AU^T = 0$. Finally, $U$ is invertible, so $AU^T = 0 \implies A = 0$.


问:
Use Cauchy's Residue Theorem to solve for $\displaystyle{\frac{1}{2\pi i} \oint_{|z|=1} \frac{\cos \left(e^{-z}\right)}{z^2} dz}$
I know how to solve for the basic about this.  But in this problem

$\displaystyle{\frac{1}{2\pi i} \oint_{|z|=1} \frac{\cos \left(e^{-z}\right)}{z^2} dz}$,

I don't know how to start.  Can somebody help me or guide me about this? Or just give me a hint.  Thanks.  I would appreciate the help.  Thanks.

答:
First, recall the following form of Cauchy's integral formula. Let $f$ be a function holomorphic on some neighborhood of unit disk, then
$$f^{(n)}(w)=\frac{n!}{2i\pi}\oint_{|z|=1}\frac{f(\zeta)}{(\zeta-w)^{n+1}}d\zeta.$$

The above theorem is sometimes only mentioned for $n=0,$ but it is true more generally. Once you have the above formula, you observe that $f(z)=\cos(e^{-z})$ is holomorphic in the disk, and taking $w=0,$ you obtain
$$f'(0)=\frac{1}{2i\pi}\oint_{|z|=1}\frac{f(z)}{z^2}dz.$$

This is what was mentioned in the comment above. Therefore, you job reduces to finding the derivative $\frac{d}{dz}\vert_{z=0}\cos(e^{-z}),$ which is a straightforward computation using chain-rule.


问:
Standard Brownian Motion process
I would appreciate the help in the following

$\{x(t): 0\leq t \leq 1\}$ Standard Brownian motion.

$ t^*=\inf\{t\in[0,1]:x(t)=\underset{0\leq s \leq 1}{\sup} x(s) \}$
random variable , $\{y(t): 0\leq t \leq 1\}$ Standard Brownian motion are independent,
$ \mathbb{E}(t^*)<\infty $ then

$\mathbb{E}\left[(y(t^*))^2\right] =\mathbb{E}(t^*)$

I would like to know if this result is true and if it is true, how can I prove it?

答:
This question is not really about Brownian motion.

You see, $t^*$ and $y$ are independent, which makes:
$$\mathbb{E}( y(t^*)^2 |t^*=t )= \mathbb{E}( y(t)^2 |t^*=t)= t$$
Hence forth what follows.

P/s: $t^*$ and $y$ are independent

Title: Standard Brownian Motion process


问:
Prove $\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$
$a,b,c >0$ and $a+b+c=3$, prove
$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3$$

I try to apply AM-GM

$$\left(\frac{a+1}{a+b} \right)^{\frac25}+\left(\frac{b+1}{b+c} \right)^{\frac25}+\left(\frac{c+1}{c+a} \right)^{\frac25} \geqslant 3\cdot \sqrt[3]{\left(\frac{a+1}{a+b} \right)^{\frac25}\left(\frac{b+1}{b+c} \right)^{\frac25}\left(\frac{c+1}{c+a} \right)^{\frac25}}$$
Thus it remains to prove
$$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1   $$ with the condition $a+b+c=3.$

 But I found the counter example for $$\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \geqslant 1   $$ :(

答:
Only a partial answer.

Assuming $a\le b\le c$, then we have $0<a\le1$ and $1\le c<3$. Now we have two cases, $b\le1$ and $b>1$. The case $b\le1$ is easy to deal with.

Assuming $0<a\le b\le1\le c<3$, we have
\begin{align}
\left(\frac{a+1}{a+b} \right)\left(\frac{b+1}{b+c}\right)\left(\frac{c+1}{c+a} \right) \ge 1
&\iff (a+1)(b+1)(4-a-b)\ge(a+b)(3-a)(3-b)\\
&\iff {\left(2-a-b\right)} {\left(1-a\right)} {\left(1-b\right)}\ge0.
\end{align}

Now assuming $0<a\le1\le b \le c<3$. Below is not an answer but only an analysis.

We want to know why this case is difficult, and why $\frac25$ is important.

Using the series expansion of $\exp$, and let $A = \operatorname{diag}\left(\ln\left(\frac{a+1}{a+b} \right),\ln\left(\frac{b+1}{b+c}\right),\ln\left(\frac{c+1}{c+a} \right)\right)$, then we have
\begin{align}
LHS
&= \sum_{n=0}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!}\\
&= 3 + \frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3 +R_3,
\end{align}
where
$R_3 = \sum_{n=4}^\infty\left(\frac25\right)^n\frac{\operatorname{tr}A^n}{n!} \ge 0$, since $e^x-(1+x+x^2/2+x^3/6)$ is positive for any $x\in\mathbb R$.

Then we can prove the inequality if we have
$$\frac25\operatorname{tr}A + \frac12\cdot \left(\frac25\right)^2\operatorname{tr}A^2+ \frac16\cdot \left(\frac25\right)^3\operatorname{tr}A^3\ge\!\!\!?\;0,$$
which can be simplified to
$$75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge\!\!\!?\;0.\tag{1}$$

Numerical results suggest that using the 3 first terms is enough to prove the inequality. Note that in the first case where $b\le1$, using the first term $\operatorname{tr}A$ is enough (and what we did in the first part is in fact proving $\operatorname{tr}A\ge0$), that's why that case is easy.

So,

Why the case $b\ge1$ is difficult?

Because we have 2 more terms, $\operatorname{tr}A^2$ and $\operatorname{tr}A^3$, to deal with.

Why $\frac25$ is important?

Because $\frac25$ gives the coefficients 75, 15, and 2, which makes $75\operatorname{tr}A + 15\operatorname{tr}A^2+ 2\operatorname{tr}A^3\ge0$.

ContentLicense: CC BY-SA 3.0


问:
What is a Fourier Symbol
I was wondering what exactly the term Fourier symbol, or sometimes said the symbol of the Fourier transform means. Also, are any books that cover this topic in detail?

Edit:
After the comments, I have come to realise I mean to ask either: what is the symbol of the Fourier transform operator, or also known as Fourier multiplier.

Are there are good recommended texts that cover the topic of symbols of operators, and the Fourier transform, well?

答:
Probably you are interested in the symbol $P$ of of a pseudo-differential operator $T$ on $\mathbb R^n$ (or other...), the latter described by (up to choices)
$$
Tf(y) \;=\; \int_{\mathbb R^n} e^{i\langle y,\xi\rangle}\; P(\xi,y) \Big(\int_{\mathbb R^n} e^{-i\langle \xi,x\rangle} \; f(x)\;dx\Big)\;d\xi
$$
This generalizes (in one way) the idea that Fourier transform converts constant-coefficient differential operators to multiplications by polynomials $P(\xi)$.

G. Folland's books (both Tata intro to PDEs and fuller textbook on PDEs) give good settings for and intros to simple cases of pseudo-differential operators. Many other sources (such as L. Hormander's books on PDE) treat general cases, which may obscure points of interest for a person new to the ideas.

Title: What is a Fourier Symbol


问:
Taylor expanding a vector function of many variables
How does one Taylor expand a vector function of many variables? The question arises in the context of deriving the geodesic deviation in Newtonian gravity, where we need to subtract as follows, and the answer given was: $$\overline{g}(\overline{x}+\overline{N}, t) - \overline{g}(\overline{x}, t) = (\overline{N} \cdot \overline{\nabla})\overline{g} + \omicron(N^{2}).$$

I wanted to know how this obtained.

答:
Analogous to the single-variable Taylor expansion
$$
f(x) = f(a) + f'(a)(x-a)+O(x^2)
$$
we have the multivariate expansion
$$f(\mathbf{x})=f(\mathbf{a})+\nabla f(\mathbf{a})(\mathbf{x-a})+\ldots$$
Simply substitute in $\mathbf{x\rightarrow x+N}$ and $\mathbf{a \rightarrow x}$ into the second equation to obtain
$$
f(\mathbf{x+N})=f(\mathbf{x})+(\nabla f(\mathbf{x}))\cdot \mathbf{N}+\ldots
$$
and the result follows.

OwnerDisplayName: Nihar Karve


答:
How does one Taylor expand a vector function of many variables?

$$\vec{f}=\vec{f}(\vec x)$$

taylor :

$$\vec{f}(\vec x+d\vec x)=\vec{f}(\vec x)+\frac {\partial \vec f}{\partial \vec x}\,d\vec{x}$$

where :

$$\frac {\partial \vec f}{\partial \vec x}\,d\vec{x}=(d\vec{x}\cdot \vec \nabla)\,\vec f(\vec x)$$

edit:
for
$$\vec f=\left[ \begin {array}{c} f_{{1}} \left( x_{{1}},x_{{2}} \right)
\\ f_{{2}} \left( x_{{1}},x_{{2}} \right)
\end {array} \right]
$$
is:
$$\frac {\partial \vec f}{\partial \vec x}= \left[ \begin {array}{cc} {\frac {\partial }{\partial x_{{1}}}}f_{{1}
} \left( x_{{1}},x_{{2}} \right) &{\frac {\partial }{\partial x_{{2}}}
}f_{{1}} \left( x_{{1}},x_{{2}} \right) \\ {\frac {
\partial }{\partial x_{{1}}}}f_{{2}} \left( x_{{1}},x_{{2}} \right) &{
\frac {\partial }{\partial x_{{2}}}}f_{{2}} \left( x_{{1}},x_{{2}}
 \right) \end {array} \right]
$$

the Jacobi matrix.

Title: Taylor expanding a vector function of many variables

Body:
How does one Taylor expand a vector function of many variables? The question arises in the context of deriving the geodesic deviation in Newtonian gravity, where we need to subtract as follows, and the answer given was: $$\overline{g}(\overline{x}+\overline{N}, t) - \overline{g}(\overline{x}, t) = (\overline{N} \cdot \overline{\nabla})\overline{g} + \omicron(N^{2}).$$

I wanted to know how this obtained.

OwnerDisplayName: Eli


答:
The fact that $\overline{g}$ is a vector function is inconsequential here. Consider a component-wise expansion of $\overline{g}(\overline{x}+\overline{N}, t)$ as: $$g_{i}(\overline{x}+\overline{N}, t) = g_{i}(\overline{x}, t) + \sum_{j=1}^{n}(x_{j}+N_{j} - x_{j})\frac{\partial}{\partial N_{j}}g_{i}(\overline{x}+\overline{N}, t) + \omicron(N^{2}).$$

So, $$\overline{g}(\overline{x}+\overline{N}, t) = \overline{g}(\overline{x}, t) + \sum_{j=1}^{n}N_{j}\frac{\partial}{\partial N_{j}}\overline{g}(\overline{x}+\overline{N}, t) + \omicron(N^{2}).$$

The second term in the expansion can be interpreted as a dot product. It has a summation over $j$, which can be simplified by recognising that the summation over the partial derivatives is the gradient operator, while that over $N_{j}$ is the vector $\overline{N}$. So, the second term becomes: $\overline{N} \cdot \nabla$, and hence, the subtraction in your question yields the above dot product.

Hope this answers your question.

OwnerDisplayName: PhutureFysicist


问:
Integrating $1/\theta$ over a range
My textbook is telling me:

$$\int_{x}^{1} \frac{1}{\theta} d\theta = {|\log x|}$$

While I arrive at:

$$\int_{x}^{1} \frac{1}{\theta} d\theta = |\log(1)| - {|\log(x)|} = - {|\log(x)|}$$

Why am I wrong?

$0 < x < 1$

Edit: Here is a pic from the textbook

答:
The answer in the book is not wrong but is over-complicated.

The integral of $1/\theta$ in the positive real line is $\ln(\theta)$. Therefore

$$\int_x^1 1/\theta d\theta = \ln(1)-\ln(x)= -\ln(x)$$

However since $0<x<1$ (I suppose) we have that $\ln(x)$ is negative and so $-\ln(x) = |\ln(x)|$

Title: Integrating $1/\theta$ over a range


问:
Hensel's lemma exercise
I read this thread and the author starts by stating a problem:

Suppose $p>2$ is a prime. Show that $x^p+y^p=z^p$ has a solution in $\mathbb{Z}_p^{\times}$ if and only if there exists an integer $a$ such that $p\not\mid a(a+1)$ and $$(a+1)^p=a^p+1\pmod{p^2}.$$

They claim that this is a good exercise in Hensel's Lemma and they will not prove it there. Coincidentally, I first came in touch with Hensel's lemma today, and do not know how to prove this result. Can someone help me please?

Thanks in advance!

答:
If there is a solution in $\Bbb{Z}_p^\times$ then there is a solution in $\Bbb{Z}/(p^2)^\times$ which is $z^p=x^p +y^p \bmod p^2$ which gives a solution $Z^p=X^p+1 \bmod p^2$.

Since $$(Z+p^n c)^p= Z^p \bmod p^{n+1}$$
then $Z\bmod p$ determinates $Z^p \bmod p^2$, thus $Z^p = X^p+1\bmod p^2$ implies $(X+1)^p = X^p+1\bmod p^2$.

Conversely given $b=X^p+1$ and $n\ge 1$, if there is $Z_n$ such that $Z_n^p=b\bmod p^{n+1}$ then there is $Z_{n+1}\equiv Z_n \bmod p^n$ such that $Z_{n+1}^p = b\bmod p^{n+2}$, thus if $Z_1$ exists then $Z=\lim_{n\to \infty} Z_n$ is a solution to $Z^p=b$.

Title: Hensel's lemma exercise


问:
Partial Differentiation using chain rule
I am trying to express $\frac{\partial^2u}{\partial y^2}$ and $\frac{\partial^2u}{\partial x\partial y}$ of the function $u=f(x,y,g(x,y))$ with partial derivative notation of $f$ and $g$.

However I am having a hard time doing so because I am a bit confused with the notation.

I know that $\frac{\partial u}{\partial y}=\frac{\partial f}{\partial y}+ \frac{\partial f}{\partial g}\frac{\partial g}{\partial y}$. But the problem is that I am not sure how to express $\frac{\partial^2u}{\partial y^2}$. If I differentiate the $\frac{\partial u}{\partial y}$ by $y$, I am thinking I will get $\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 f}{\partial y^2}+ ...$ , but I am not quite sure how the second term of the right hand side ($\frac{\partial f}{\partial g}\frac{\partial g}{\partial y}$) will look like. Similarly, I have trouble in expressing $\frac{\partial^2u}{\partial x\partial y}$ using partial differentiation notation of $f$ and $g$.

How should it be done?

Thanks.

答:
First lets start by $\frac{\partial u}{\partial y}$.

You already evaluated this one right:

\begin{equation}
\left( \frac{\partial u}{\partial y} \right)_x = \left( \frac{\partial f(x,y,g)}{\partial y} \right)_x = \left( \frac{\partial f}{\partial y} \right)_{g,x} + \left( \frac{\partial f}{\partial g} \right)_{x,y} \left(\frac{\partial g}{\partial y}\right)_x
\end{equation}

For the next one, we want:

\begin{equation}
\left( \frac{\partial }{\partial y} \left( \frac{\partial u}{\partial y} \right)_x \right)_{x}
\end{equation}

We can evaluate this using the chain rule:

\begin{equation}
\left( \frac{\partial }{\partial y} \left( \frac{\partial u}{\partial y} \right) \right)_{x} =  \left(\frac{\partial}{\partial y} \left( \frac{\partial f}{\partial y} \right)_{g,x} \right)_x + \left( \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial g} \right)_{x,y}  \right)_x  \left(\frac{\partial g}{\partial y}\right)_x + \left( \frac{\partial}{\partial y} \left(\frac{\partial g(x,y)}{\partial y}\right)_x \right)_x \left( \frac{\partial f}{\partial g} \right)_{x,y}
\end{equation}

Lets evaluate term by term:

Note that the variables we are keeping constant are also an indicative that the final derivative will depend on them, so for the terms that contains $f$, the first and the second, you can replace the partial derivative with this operator:
\begin{equation}
\left(\frac{\partial}{\partial y} \right)_x = \left( \frac{\partial }{\partial y} \right)_{x,g} + \left( \frac{\partial }{\partial g} \right)_{x,y} \left( \frac{\partial g}{\partial y} \right)_{x}
\end{equation}
And for the last derivative you simply get:

\begin{equation}
\left( \frac{\partial}{\partial y} \left(\frac{\partial g(x,y)}{\partial y}\right)_x \right)_x \left( \frac{\partial f}{\partial g} \right)_{x,y} = \frac{\partial^2 g}{\partial y ^2} \left( \frac{\partial f}{\partial g} \right)_{x,y}
\end{equation}

If do this calculation you will get:

\begin{equation}
2 \frac{\partial }{\partial g} \left( \frac{\partial f}{\partial y} \right) \frac{\partial g}{\partial y}+ \frac{\partial^2 f}{\partial y^2} + \left(\frac{\partial g}{\partial y}\right)^2 \frac{\partial^2 f}{\partial g^2} +\frac{\partial f}{\partial g} \frac{\partial^2 g}{\partial y^2}
\end{equation}

Where in this last equation I have not stated the variables kept constant, but you may identify them easily.
With this headstart, I think you can easily do the other derivative, which I won't put here cause it is too much to type in latex.
Any question just ask!


问:
Infimum in a Sequence
Let $a_n$ be a sequence such that
$$ \inf{a_n} = 2 $$
and
$$ \lim{a_n} = 5 $$

Does exist $n_0$ such that $a_{n_0} = 2$?

I'm not sure how to approach this question, I believe that $2$ is in $a_n$, since there only exists one partial limit, and so if $2$ wasn't in $a_n$, then $\inf{a_n} > 2$ (since $a_n$ must be increasing?).

I'm not sure how I could justify myself, or if my argument is correct.

答: Hint: for large enough $n$ we have $a_n>3$. Consider the finitely many elements of the sequence before that point.

Title: Infimum in a Sequence


问:
What is the general equation of a cubic polynomial?
I had this question:

"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$"
and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.

I took the general equation of a cubic equation, which was: $x^3 +bx^2/a + cx/a + d/a$.

Through simultaneous equations, I found what $b/a, c/a, d/a$ should equate to for my unknown cubic polynomial. Am I supposed to make $b/a, c/a, d/a$ all positive, then substitute it into the general formula?

Any help would be greatly appreciated, thanks.

答:
Let $a$ be a root of $x^3+2x+1=0$ and $b=a^2$ be  a root of the required equation

So, $a^3+2a+1=0\implies a\cdot b+2a+1=0\implies a=-\frac1{b+2}$

As $a$ be a root of $x^3+2x+1=0$, put this value of $a$ in $x^3+2x+1=0$

On simplification, I get $(b+2)^3-2(b+2)^2-1=0\iff b^3+4b^2+4b-1=0$

So, the required equation will be $y^3+4y^2+4y-1=0$


问:
Spectrum of an unbounded weighted shift operator on a Hilbert space
Consider the Hilbert space $H=\ell^2(\mathbb{Z}\times\mathbb{Z})$ with an orthonormal basis $\{e_{i,j}\}_{i,j\in\mathbb{Z}}$ and fix a complex number $\mu$ such that $|\mu|\in(0,1).$ Define the operator $T$ on $H$ by
$$T(e_{i,j})=\mu^i e_{i,j+1}$$
Then $T$ is a densely defined unbounded normal operator on $H$ with spectrum $\mathbb{C}^{\mu}\colon =\{\lambda\in\mathbb{C}\mid |\lambda|\in\mu^{\mathbb{Z}}\}$ where $\mu^{\mathbb{Z}}\colon=\{\mu^i\mid i\in\mathbb{Z}\}$. It is clear that $T$ cannot be bounded as the indices $i,j$ runs over the set of all integers. My questions are the following:

$1$. I suspect the spectrum of $T$ is $\textit{contained}$ in $\mathbb{C}^{\mu}$ and not equal to $\mathbb{C}^{\mu}$. Clearly, $\mu^i$ is in the spectrum for all $i\in\mathbb{Z}$ and real $\mu\in(0,1)$. I can't see why $\lambda$ is in the spectrum with $|\lambda|=|\mu|^i$ when $\mu$ is complex number such that $0<|\mu|<1.$

$2$.If $U$ is a unitary operator on $H$ which commutes with $T$, then how are the spectra of $T$ and $UT$ related? are they equal?

Any suggestion will be highly appreciated.

答:
Let
$$
H_k=\overline{\operatorname{span}}\{e_{k,j}:\ j\in\mathbb Z\}.
$$
Then $H=\bigoplus_kH_k$. More importantly, each $H_k$ is invariant for $T$, so $T=\bigoplus_k T_k$, where $T_k=T|_{H_k}$. Then $\sigma(T)=\overline{\bigcup_k\sigma(T_k)}$.

On $H_k$, we have
$$
T\sum_j x_je_{k,j}=\sum_j \mu^kx_je_{k,j+1}.
$$
That is, $T_k=\mu^k\,B$, where $B$ is the bilateral shift. Thus
$$\sigma(T_k)=\mu^k\,\sigma(B)=\mu^k\,\mathbb T=|\mu|^k\,\mathbb T.
$$
Thus
$$
\sigma(T)=\overline{\{\lambda\in\mathbb C:\ \exists k\in\mathbb Z, |\lambda|=|\mu|^k\}}.
$$
The closure is definitely needed, as $0\in\sigma(T)$.

As for your second question, the answer is "not related". For instance  if $T=I$, you have $\sigma(T)=\{1\}$ and you are free to choose $\sigma(UT)=\sigma(U)$.


问:
Set theory and bijective functions
I was presented with this statement in my discrete mathematics course. So far i can't figure out if it's true or false.

Let there be a set $U$. Also, let there be a set $A$ such that $A⊆U$.

For any set $B$, function f is defined as follows:

$f: \mathcal{P}(U) → \mathcal{P}(U)$

$f(B)= A ∩ B$

Prove: $f$ is a bijective function if and only if $A=U$.

答: lulu's comment gives you everything you need, but if you like I can furnish an answer based on said comment.  If $f$ is bijective then it must be injective.  However note that by construction $f(A^c) = \emptyset$; since $f(\emptyset) = \emptyset$ then we must have $A^c = \emptyset \Rightarrow A = U$.  On the other hand if $A = U$ then clearly $f$ is bijective.

Title: Set theory and bijective functions


答:
Your mistake is that the antiderivative of $\dfrac 1x$ is not $|\log x|$ but $\log x$. Then

$$\int_x^1\frac{dx}x=\log 1-\log x=-\log x$$ and in the particular case that $\log x <0$, this is also $|\log x|$.

The traditional solution is

$$\int_a^b\frac{dx}x=\log|b|-\log|a|.$$

I claim that it is technically more correct to write

$$\int_a^b\frac{dx}x=\log\frac ba.$$

Title: Integrating $1/\theta$ over a range

Body:
My textbook is telling me:

$$\int_{x}^{1} \frac{1}{\theta} d\theta = {|\log x|}$$

While I arrive at:

$$\int_{x}^{1} \frac{1}{\theta} d\theta = |\log(1)| - {|\log(x)|} = - {|\log(x)|}$$

Why am I wrong?

$0 < x < 1$

Edit: Here is a pic from the textbook


问:
Limit of a function vs sequence as $\lim_{n\rightarrow \infty}n\cos (\frac{\pi }{2}+\pi n)$
I know that $\lim_{n\rightarrow \infty}n\cos (\frac{\pi }{2}+\pi n)$ does not exist. However, I am not sure about the sequence
$(n\cos (\frac{\pi }{2}+\pi n))_{n=1}^\infty$. By evaluating this sequence, I get indeterminate form $\infty\cdot0$. Intuitevly, I believe that the answer is $0$, because $0$ can be treated as a solid (?). For instance, $\lim_{n\rightarrow \infty}n\cdot0=0$. So, how can I interpret these results?

答:
We have
$$
n \cos(n \pi + \frac{\pi}{2})=n(\cos(n\pi)\cos(\frac \pi 2) -\sin(n\pi)\sin(\frac{\pi}{2}))=0\to0.
$$


问:
How to prove that the points $E, F, G$ are collinear

Given $\triangle ABC$ and $\odot(ABC) $, $AG$ and $AD$ are the angle bisectors of $\angle A$ .  $CD$ and $AB$ intersects at $E$. $DB$ and $AC$ intersects at $F$. Prove that $E, F, G$ are collinear.

I found that $DC=DB$. By Menelaus theorem I need to prove $$\frac{BG} {CG} \cdot \frac{CF} {AF} \cdot \frac{EA} {EB} =1$$ and do I need to use the angle bisector theorem also?

答:
This is a Projective solution. Refer this if you are new to projective. Also the point $H$ in your Diagram, wasn't necessary, so I removed it and introduce a new point called $H$.

Solution: Define $H = DA\cap BC$.

By $lemma$ $2$ from the handout I linked, we get that it is enough to show that $(H,G;C,B)=-1$.

By $lemma$ $5$ from the handout I linked, we get that if we show that $\angle AGH=90$ and $AG$ bisects $\angle BAC$, then we are done. But as $AD, AG$ are angle bisectors we have $\angle HAG =90$ and by question it's given that $AG$ bisects $\angle BAC$. And we are done.


问:
Continuity of a function of product spaces
Let $f: X \times Y \rightarrow Z $ such that.

$ \forall x_0 \in X , f_{x_0} : Y \rightarrow Z, y\mapsto f(x_0,y)$ is continuous.

$ \forall y_0 \in Y , f_{y_0} : X \rightarrow Z, x \mapsto f(x,y_0)$ is continuous.

How i can prove that $f$ is continuous?

Thanks in advance.

答:
It isn't. Let $f:\Bbb R\times \Bbb R\to \Bbb R$ be the function

$(x,y)\mapsto\begin{cases}
0 &\text{ , if }x=y=0\\
\frac{xy}{x^2+y^2} &\text{ , else }
\end{cases}$

For any fixed $x_0$ the function $f_{x_0}$ is continuous. Since $f$ is symmetric in $x$ and $y$, $f_{y_0}$ is continuous for any $y_0$. But $f$ is not continuous in $x=y=0$. Consider the sequence $(x_n,y_n)=(1/n,1/n)$. It converges to $(0,0)$ along the diagonal and $f(x_n,y_n)=1/2$. But $f(\lim_{n\to\infty}(x_n,y_n))=f(0,0)=0$.


问:
Kolmogorov: question on definition of Independence from his book
I'm reading Kolmogorov Foundations of the Theory of Probability, Page 9, which explains Independence and have questions.  Kolmogorov provides the following pictured Definition I for independence.  The definition is similar to what more modern books provide with exception to the inclusion of the experiment number.

The definition is confusing because an experiment $U^{(n)}$ is a decomposition of events.  A decomposition is a set of disjoint sets such that when union is taken, the result is $E$.  Kolmogorov typically uses $q_1, q_2, ..., q_n$ typically to denote elementary events (of the sample space).  From what I read, the definition implies that the intersection of an element from each experiment's decomposition.

My questions, is above explanation correct?  If so, Definition I is $= 0$ because an intersection of elementary events $A^{(1)}_{q_1} \cap A^{(1)}_{q_2} \cap ....\cap A^{(n)}_{q_n} = P(0) = 0$.  I think I have read something wrong, because there should be the possibility that these events intersect

Appreciate any guidance on understanding of Kolmogorov's grammatical and symbolic explanation of independence mentioned here.

答:
$A_{q_i}^{(i)}$ isn't an elementary event - it's some set of events.

Each experiment $\mathfrak{A}^{(i)}$ is a decomposition of $E$ into sets $A_j^{(i)}$. Sets say $A_{q_1}^{(1)}$ and $A_{q_2}^{(2)}$ are from different decompositions, so they can intersect.


答:
By definition
$$\forall \epsilon>0 \mbox{, } \exists N \mbox{ s.t. } \forall n\ge N \mbox{, }|a_n-5|<\epsilon$$
so given $\epsilon=1$,
$$\exists N \mbox{ s.t. } \forall n\ge N \mbox{, }|a_n-5|<1$$
so for all of these $n$, $4<a_{n}$ so there exist only other $N-1$ possible terms that are under $4$.

This means that $\inf a_n = \inf_{n<N}a_n=\min_{n<n_0}a_n=2 $ becouse of the finiteness of the set on which we are taking the $\inf$, so there exists by definition an element $a_{n_0}$ with $n_0<N$ such that $a_{n_0}=2$.

Title: Infimum in a Sequence

Body:
Let $a_n$ be a sequence such that
$$ \inf{a_n} = 2 $$
and
$$ \lim{a_n} = 5 $$

Does exist $n_0$ such that $a_{n_0} = 2$?

I'm not sure how to approach this question, I believe that $2$ is in $a_n$, since there only exists one partial limit, and so if $2$ wasn't in $a_n$, then $\inf{a_n} > 2$ (since $a_n$ must be increasing?).

I'm not sure how I could justify myself, or if my argument is correct.


问:
"L'Hôpital's rule" vs. "L'Hospital's rule"?
I know this is not strictly a mathematical question, and I considered putting it on Linguistics SE, but I decided that seeing as this is most probably a mathematical history question, it would be better placed here on math SE.

My question is:

Why is "L'Hôpital's rule" often referred to as "L'Hospital's Rule" in english mathematical literature?

I am aware that the translation from French to English of "L'Hôpital" is "The Hospital", but I haven't seen any cases of other french names which correspond to proper nouns being translated into english, so why the special case here?

Again, sorry if this is completely the wrong place to put this question, and moderators feel free to migrate this question to a more appropriate SE board if one exists, but as I said, I believe this to be the most appropriate board.

答:
The thermometer scale, which is called now the Delisle scale, is from a french name "de l'Isle" (von Isulen).  It's just that english mangles the assorted addons.

Even in street names, the road in brisbane, called Swann road, was formerly swann's road.


问:
Is the Gauss norm for $p$-adic power series a supremum of values taken over the unit disk?
Let $A$ be the ring of convergent power series over $\mathbf{Q}_p$ in one variable. That is, elements of $A$ are formal series $$f(t) = \sum_{m=0}^\infty a_m t^m$$ such that the coefficients $a_m$ tend to zero in $\mathbf{Q}_p$ as $m \rightarrow \infty$.

We define the Gauss norm on $A$ by setting
$$\Vert f(t) \Vert = \sup_m |a_m|$$
It is obvious that for any point $z \in \mathbf{Z}_p$, we have $|f(z)| \leq \Vert f(t) \Vert$.

Failed attempt (see edit below): In fact, this upper bound should be attained i.e. there should be some $z \in \mathbf{Z}_p$ with $|f(z)| = \Vert f(t) \Vert$.

How do we show the existence of such a $z$?

Edit: So it turns out the above is not true. The supremum is not always necessarily attained. Here is a revised question that is hopefully more true.

Let's replace $\mathbf{Q}_p$ with its completed algebraic closure $\mathbf{C}_p$. The definition of convergent power series and Gauss norms for $\mathbf{C}_p$ are unchanged. Now we have

$$\sup_{z \in \mathcal{O}_{\mathbf{C}_p}} |f(z)| \leq \Vert f(t) \Vert$$

Can we upgrade this inequality to an equality?

I have dropped the requirement that the upper bound on the right side be attained by some $z \in \mathcal{O}_{\mathbf{C}_p}$.

答:
$$\|f\| = \sup_n |a_n|_p = \sup_{b\in O_K} |f(b)|_p, \qquad O_K=\bigcup_{n\ge 1}\Bbb{Z}_p[\zeta_{p^n-1}]$$ Proof: $\sup_{b\in O_K} |f(b)|_p\le \|f\|$, and $\|f\| = |p^m|_p$ means that $f=p^m g+p^{m+1}h$ where $g\in \Bbb{Z}_p[x],h\in A\cap \Bbb{Z}_p[[x]]$ and $g\bmod p $ is a non-zero, this polynomial has finitely many roots in the field $O_K/(p)$ thus for some $b\in O_K$, $g(b) \in O_K^\times$ and $|f(b)|_p = |p^m g(b)|_p = \|f\|$.

We can replace $K$ by any extension of $\Bbb{Q}_p$ with an infinite residue field.


问:
Series expansion of $x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$ at $x=1$
In various texts it is stated that a good approximation of,

$x\sqrt{x^2-1} - \ln(x + \sqrt{x^2-1} )$,

about $x=1$ is given by,

$\frac{4\sqrt{2}}{3} (x-1)^{3/2}$.

Plotting the graphs this is indeed the case, however, I am not sure how this approximation is derived. Entering the original equation in Wolfram Alpha it does list the above approximation as the first term in the Puiseux series expansion of the original function. I am not familiar with these and looking online I can't seem to find a way into calculating a Puiseux series for the relevant expression.

Any suggestions or pointers for how to go about this would be greatly appreciated :)

答:
The Approximation in the Question

I recognized this as the result of an integral that uses the trigonometric substitution $x=\sec(\theta)$, so I took the derivative of ${\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)}$ and got $2{\textstyle\sqrt{t^2-1}}$. Thus,
$$
{\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)}
=2\int_1^x{\textstyle\sqrt{t^2-1}}\,\mathrm{d}t\tag1
$$
For $t\approx1$, $t^2=(1+(t-1))^2\approx1+2(t-1)=2t-1$; therefore, $\sqrt{t^2-1}\approx\sqrt{2t-2}$. Thus,
$$
\begin{align}
\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)
&\approx2\int_1^x\sqrt{2t-2}\,\mathrm{d}t\tag2\\[6pt]
&=\textstyle\frac{4\sqrt2}3(x-1)^{3/2}\tag3
\end{align}
$$

A Better Approximation

We can get a better approximation if we note that
$$
\begin{align}
\textstyle2\sqrt{t^2-1}
&=\textstyle2\sqrt{2(t-1)+(t-1)^2}\tag{4a}\\
&=\textstyle2\sqrt2\sqrt{t-1}\sqrt{1+\frac{t-1}2}\tag{4b}\\
&\approx\textstyle2\sqrt2\sqrt{t-1}\left(1+\frac{t-1}4\right)\tag{4c}\\
&=\textstyle2\sqrt2\sqrt{t-1}+\frac{\sqrt2}2\sqrt{t-1}^3\tag{4d}
\end{align}
$$
Therefore,
$$
\begin{align}
\textstyle x\sqrt{x^2-1}-\log\left(x+\sqrt{x^2-1}\right)
&\approx2\sqrt2\int_1^x\sqrt{t-1}\,\mathrm{d}t+\frac{\sqrt2}2\int_1^x\sqrt{t-1}^3\,\mathrm{d}t\tag5\\[6pt]
&=\textstyle\frac{4\sqrt2}3(x-1)^{3/2}+\frac{\sqrt2}5(x-1)^{5/2}\tag6
\end{align}
$$

Graphs


问:
An inequality in probability theory
I am trying to understand a paper but am stuck at some point. However I could understand it if I proved the following result, but I am not sure whether it acutally is true.

$\mathbf{Conjecture:}$ let $Z_1.\, \dots, Z_n $ be iid $\sim Z \in L^2$, with $\mathbb{E}Z=0$. Then
$$
\mathbb{E}[\sqrt{\sum_{i=1}^n Z_i^2}] \leq \sup_{\lambda\in \mathbb{R}^n : ||\lambda||_2 = 1} \mathbb{E}[|\sum_{i=1}^n \lambda_i Z_i|]
$$

I think something like this is being us in the paper, but I cannot see whether it is true. Any help?

答:
This is false.  Take $n=2$ and take $Z_1, Z_2$ i.i.d. Gaussian with mean 0 and variance 1. By symmetry, the following expression is the same for all $(\lambda_1, \lambda_2) \in \mathbb{R}^2$ that satisfy $\lambda_1^2 + \lambda_2^2=1$:
$$ E[|\lambda_1 Z_1 + \lambda_2Z_2|] $$
Thus this is the same for $(\lambda_1, \lambda_2) = (1,0)$ and
$$ E[|\lambda_1 Z_1 + \lambda_2Z_2|] = E[|Z_1|] = \sqrt{\frac{2}{\pi}}\approx 0.79788456  \quad \mbox{(whenever $\lambda_1^2+\lambda_2^2=1$)}$$
On the other hand
$$ E\left[\sqrt{Z_1^2 + Z_2^2} \: \right] = \sqrt{\frac{\pi}{2}} \approx 1.2533141$$

Edit:
To give one proof of the symmetry observation: Consider the rotation matrix
$$ A = \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$
Define
$$ \begin{bmatrix}X_1\\ X_2\end{bmatrix} = A\begin{bmatrix}Z_1\\Z_2\end{bmatrix}$$
Then $X_1, X_2$ are again i.i.d. Guassian mean 0, variance 1.  So for any measurable function $f(x)$ (including $f(x)=|x|$) and any constant $\theta \in \mathbb{R}$:
\begin{align}
E\left[f\left([\cos(\theta) \: \sin(\theta)] \cdot \begin{bmatrix}Z_1\\Z_2\end{bmatrix}\right)\right] &= E\left[f\left([\cos(\theta) \: \sin(\theta)] \cdot A\begin{bmatrix}Z_1\\Z_2\end{bmatrix}\right)\right]\\
&=E\left[f\left([1 \: \: 0] \begin{bmatrix}Z_1\\Z_2\end{bmatrix}\right)\right]\\
&=E[f(Z_1)]
\end{align}
where the first equality uses the fact that random vector $[Z_1;Z_2]$ has the same distribution as random vector $A[Z_1;Z_2]$; the second equality uses the fact that
$$  [\cos(\theta) \: \sin(\theta)] \cdot A = [1 \: \: 0]$$

Title: An inequality in probability theory


问:
$P(|X|\ge\lambda a)\ge (1-\lambda)^2a^2$ for $0\le \lambda \le 1$
If $E(X^2)=1$ and $E(|X|)\ge a >0$, then $P(|X|\ge\lambda a)\ge (1-\lambda)^2a^2$ for $0\le \lambda \le 1$.

I can see from the well known inequality $E(|X|) \le E(|X|^2)^{1/2}$ that it must be the case that $a\le 1$. But what to do next I'm not sure.

答:
Note that
$$a\le E(|X|)=E(|X|\cdot 1_{|X|< \lambda a})+E(|X|\cdot 1_{|X|\ge \lambda a})\le \lambda a +E(|X|\cdot 1_{|X|\ge \lambda a}).\tag{1}$$

By Cauchy-Schwarz inequality and noting that $E(X^2)=1$, we have

$$E(|X|\cdot 1_{|X|\ge \lambda a})\le \big(E(X^2)\big)^{\frac{1}{2}}\cdot \big(E(1_{|X|\ge \lambda a})\big)^{\frac{1}{2}}=\big(P(|X|\ge \lambda a)\big)^{\frac{1}{2}}.\tag{2}$$
Combining $(1)$ and $(2)$, the conclusion follows.


问:
Proving a polynomial splits fully in a certain field
I have a polynomial $f(x)=x^6+3$, and define $a$ a root of $f(x)$. Then construct the field $\mathbb{Q}(a)$. I need to prove that $g(x)=x^3+3$ splits completely over $F(a)$. It hints at the fact that $a^2$ is a root of $g(x)$ and $a^3$ is a root of $x^2+3$. However I'm at a complete loss of how to use this information. I can gather that $(x-a^2)(x^2+a^2x+a^4)$ is a factorization of $g(x)$, but I don't think this is the right approach.

Any hints would be appreciated, thanks in advance.

答: Let $h(x)=x^2+a^2x+a^4$. The discriminant of $h$ is $D=a^4-4a^4=-3a^4$. We have $(a^5)^2=(a^2a^3)^2=a^4(a^3)^2=-3a^4$. Implies that $D$ has a root in $\mathbb{Q}(a)$. This implies that the roots of $h(x)$ are in $\mathbb{Q}(a)$.


问:
In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

In a rectangle $ABCD$ where $AB = 6$ , $BC = 3$ , point $P$ is chosen on $AB $ such that $\angle APD = 2 \angle CPB$. Find $AP$.

What I Tried: Here is a picture :-

You can see I tried angle-chasing, but that really did not lead me to any useful info. Also I tried Pythagoras theorem, taking $AP = k$ and $PB = (6 - k)$ gives :-
$$k^2 + 9 = AP^2$$
$$(6 - k)^2 + 9 = PC^2$$
And that is $3$ variables in $2$ equations, so I need to find another value before trying this.

I think the fact that $AD = \frac{AB}{2}$ should be used somehow, maybe by some construction or something. Also in Geogebra I saw that $\angle APD = 65.7^\circ$ , $\angle CPB = 32.85^\circ$ , which are $2$ odd-looking angles, and $AP$ came to be $1.35$ .

Can anyone help me how is this the case? Thank You.

答:
Let $a=AP$ be the unknown to be found, we are trying to write an equation that $a$ should satisfy.

Let $Q$ be the point of intersection of the angle bisector of $\hat P$ in $\Delta APD$ with the opposite side $AD$ in this triangle. So $PQ$ divides this angle $\hat P$ in two angles of measure $x$, notation as in the OP. Then we have successively:

$PD$ is $\sqrt{3^2+a^2}$, Pythagoras.

$\Delta APQ\sim\Delta BPC$, so we obtain a formula for $AQ$ in terms of $a$, since $AQ:a=BC:(6-a)$. Explicitly, $AQ=3a/(6-a)$.

This gives also $QD=AD-AQ=3-AQ$ in terms of $a$. Explicitly, $QD=(18-6a)/(6-a)$.

We write now the angle bisector theorem in $\Delta APD$:
$$
\frac{QD}{QA}=\frac {PD}{PA}\ .
$$
this is the needed equation. Making things explicit, the resulted equation is:
$$
\frac{18-6a}{3a}=\frac{\sqrt{9+a^2}}{a}\ .
$$
We simplify with $a>0$, get $6-2a=\sqrt{9+a^2}$. So $6-2a>0$ (i.e. $a<3$), and
$$
(6-2a)^2 = 9+a^2\ .
$$
This is $a^2-8a+9=0$, the two roots are $4\pm 7$, and we pick the one $<3$, which is $4-\sqrt 7$.


问:
Evaluate the infinite sum of $\sum_{n=0}^{\infty}\frac{1}{3^n(n+1)}$
Would someone be able to show which well-known power series works with this? And help evaluate it? The sum is  $\sum_{n=0}^{\infty}\frac{1}{3^n(n+1)}$.

答:
Use the geometric series:

$$ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$

Now, to get $ \frac{1}{i+1}$ in denominator sum, we can get that factor by integrating both side w.r.t.x

$$ -\ln(1-x)= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1 } +C$$

Now, we need put $x=\text{something}$ to make this sum match with the one in the question, I'll let you finish the problem.

Note: I've used the property that you can interchange integrals and summations in the domain of summation in the integration step.


问:
Find, in radians the general solution of cos 3x = sin 5x
I am studying maths as a hobby. I have come across this problem:

Find a general solution for the equation cos 3x = sin 5x

I have said, $\sin 5x = \cos(\frac{\pi}{2} - 5x)$

so

$\cos 3x = \sin 5x \implies 3x = 2n\pi\pm(\frac{\pi}{2} - 5x)$

When I add $(\frac{\pi}{2} - 5x)$ to $2n\pi$ I get the answer $x = \frac{\pi}{16}(4n +1)$, which the book says is correct.

But when I subtract I get a different answer to the book. My working is as follows:

$3x = 2n\pi - \frac{\pi}{2} + 5x$

$2x = \frac{\pi}{2} - 2n\pi$

$x = \frac{\pi}{4} - n\pi = \frac{\pi}{4}(1 - 4n)$

but my text book says the answer is $\frac{\pi}{4}(4n + 1)$

Is the book wrong?

答:
If you write $m$ in place of $n,$ you reached at $\dfrac{\pi(1-4m)}4$

We $$\dfrac{\pi(1-4m)}4=\dfrac{\pi(1+4n)}4\iff m=-n$$

In our case $m$ is any integer  $\iff n=-m$ also belong to the same infinite set of integers

In their case $n$ is so.


答:
No, the two are equivalent. In particular, if $m$ = $-n$, then $$\dfrac{\pi}{2}(1 - 4m) = \dfrac{\pi}{2}(4n + 1),$$
so all that's really happened is tha tyou've listed the solutions in a different order.

Title: Find, in radians the general solution of cos 3x = sin 5x

Body:
I am studying maths as a hobby. I have come across this problem:

Find a general solution for the equation cos 3x = sin 5x

I have said, $\sin 5x = \cos(\frac{\pi}{2} - 5x)$

so

$\cos 3x = \sin 5x \implies 3x = 2n\pi\pm(\frac{\pi}{2} - 5x)$

When I add $(\frac{\pi}{2} - 5x)$ to $2n\pi$ I get the answer $x = \frac{\pi}{16}(4n +1)$, which the book says is correct.

But when I subtract I get a different answer to the book. My working is as follows:

$3x = 2n\pi - \frac{\pi}{2} + 5x$

$2x = \frac{\pi}{2} - 2n\pi$

$x = \frac{\pi}{4} - n\pi = \frac{\pi}{4}(1 - 4n)$

but my text book says the answer is $\frac{\pi}{4}(4n + 1)$

Is the book wrong?


答:
Consider the classic series
$$ \sum_{n=0}^{\infty} t^{n} = \frac{1}{1-t} $$
and integrate both sides. This leads to
\begin{align}
\int_{0}^{t} \sum_{n=0}^{\infty} u^n \, du &= \int_{0}^{t} \frac{du}{1-u} \\
\sum_{n=0}^{\infty} \left[ \frac{u^{n+1}}{n+1} \right]_{0}^{t} &= \left[ - \ln(1-u) \right]_{0}^{t} \\
\sum_{n=0}^{\infty} \frac{t^{n+1}}{n+1} &= - \ln(1-t)
\end{align}
and finally
$$ \sum_{n=0}^{\infty} \frac{t^n}{n+1} = - \frac{\ln(1-t)}{t}. $$

Letting $t$ be a value within the range of convergence gives the desired series.

Title: Evaluate the infinite sum of $\sum_{n=0}^{\infty}\frac{1}{3^n(n+1)}$

Body: Would someone be able to show which well-known power series works with this? And help evaluate it? The sum is  $\sum_{n=0}^{\infty}\frac{1}{3^n(n+1)}$.


问:
What is $E[(\hat \beta_1 - \beta_1)X_1]$ where $\beta_1$ is a linear regression coefficient and $\hat \beta_1$ is its least squares estimate?
Define the linear regression model:
$$
Y_i = \beta_0 + \beta_1 X_i + \varepsilon_i, \quad \quad i = 1,\dots,n.
$$
Let $\hat \beta_0$ and $\hat \beta_1$ be the estimates of $\beta_0$ and $\beta_1$ when we solve the regression model with least squares. $\hat \beta_0$ and $\hat \beta_1$ are dependent on the $X_i$'s and $Y_i$'s since they were computed using them.

I am wondering what is the expected value of $\hat \beta_1 - \beta_1$ multipled by one of the $X_i$'s, say $X_1$ for example. That is, what is $E[(\hat \beta_1 - \beta_1)X_1]$?

答:
$$
\mathsf{E}\!\left[(\hat{\beta}_1-\beta_1)X_j\right]=\mathsf{E}\!\left[\frac{\sum_{i=1}^n(X_i-\bar{X})\varepsilon_iX_j}{\sum_{i=1}^n (X_i-\bar{X})^2}\right].
$$
If $\mathsf{E}[\varepsilon_i\mid X_1,\ldots,X_n]=0$,
$$
\mathsf{E}\!\left[(\hat{\beta}_1-\beta_1)X_j\right]=0.
$$


问:
I have a problem in proof of $\lim\limits_{x\to \infty} \frac{1}{x}=0$
Can we say that $x>0$ when $\lim\limits_{x\to \infty} \frac{1}{x}=0$, because x approaches to $\infty$?

Is the following proof correct?

step1:
$$|\frac{1}{x}-0|<\epsilon,M<x,0<x$$
step2:
$$\frac{1}{x}<\epsilon,M<x$$
step3:
$$\frac{1}{\epsilon}<x,M<x $$
step4:
$$\frac{1}{\epsilon}<M$$

答:
By the definition of a limit, we must have

$$\forall\epsilon>0:\exists M:\forall x> M:\left|\frac1x-0\right|<\epsilon.$$

This is true, because with $M=\dfrac1\epsilon$, we do have

$$\forall x>M:\left|\frac1x-0\right|<\epsilon.$$

So for all $\epsilon$ we can find an $M$ that fits.


问:
Proving a series converges IFF its sequence converges
My question is this:

Let {$x_k$} be a sequence of real numbers. Let $a_k = x_k - x_{k+1}$ for all natural numbers $k$. I want to prove that $\sum_{k=1}^{\infty}{a_k}$ converges IFF $x_k$ converges.

I believe that I have the reverse direction (if $x_k$ converges then $\sum_{k=1}^{\infty}{a_k}$ converges). For this, I (roughly) said that since $x_{k+1}$ is a subsequence of $x_k$ and $x_k$ converges, then $x_{k+1}$ converges as well. I said that the difference of two converging sequences will also converge, therefore $a_k$ converges. Then if $a_k$ converges, then $\sum_{k=1}^{\infty}{a_k}$ converges as well. My first question is if this sounds correct.

Meanwhile, I don't understand how to prove the forward direction. I tried looking at the sequence of partial sums of the series, but I found myself making improper assumptions. So I tried to find a counterexample instead, but I couldn't find one of those either. Any hints on this part? Thank you!

答:
We know that $\frac{1}{n}\to0$ but $\sum_{n \ge 1} \frac{1}{n}$ isn't convergent. So $a_n$ is convergent doesn't imply that $\sum a_n$ is convergent. Here, it suffices to notice that
$$
\forall N\ge0, \sum_{k=0}^N x_{k}-x_{k+1}=x_0-x_{N+1}.
$$


问:
How to integrate $ \int \frac{x}{\sqrt{x^4+10x^2-96x-71}}dx$?
I read about $ \int \dfrac{x}{\sqrt{x^4+10x^2-96x-71}}dx$ on the Wikipedia Risch algorithm page. They gave an answer but I don't understand how they got it.

答:
The problem is $\int\frac{x\,\mathrm{d}x}{y}$ where $y^2=x^4+10x^2-96x-71$ is a polynomial with integer coefficients and no double roots. There is a rarely-known “Chebyshev substitution” which decides if integrals like this one are elementary functions.

His criterion is the following: the $\mathbb{Z}[x]$-continued fraction expansion of $y$ contains a quadratic coefficient after the constant (e.g. $y=[c_0;c_1,\ldots,c_n,c_{n+1},\ldots]$ where $c_0,c_{n+1}$ are quadratic and $c_1,\ldots,c_n$ are linear) iff the integral is an elementary function. In this case “$w=$ the numerator of $[c_0;c_1,\ldots,c_n]$” is a good substitution.

So we first work out the continued fraction expansion of $y$: $$y=x^2+5-\frac{48}{x-1}\mathbin{\raise{-11mu}{+}}\frac{6}{x+5}\mathbin{\raise{-11mu}{+}}\frac{24}{x-4}\mathbin{\raise{-11mu}{-}}\frac{3}{x-4}\mathbin{\raise{-11mu}{+}}\frac{24}{x+5}\mathbin{\raise{-11mu}{+}}\frac{12}{x-1}\mathbin{\raise{-11mu}{-}}\frac{48}{x^2+5+y}.$$ Now $c_1,\ldots,c_6$ are linear, but $c_7$ is quadratic, so the integral is an elementary function (or, a pseudoelliptic integral). We have
$$ [c_0;c_1,\ldots,c_6]=\frac{x^8+20x^6-128x^5+54x^4-1408x^3+3124x^2+10001}{x^6+15x^4-80x^3+27x^2-528x+781}=\frac{w}{z}.$$
For these polynomials we observe that $w^2-y^2z^2=2^{16}\cdot3^7$ and $w'=8xz$, so
$$\int\frac{x\,\mathrm{d}x}{y}=\int\frac{xz\,\mathrm{d}x}{yz}=\frac18\int\frac{\mathrm{d}w}{\sqrt{w^2-2^{16}\cdot3^7}}=\frac18\log\left(w+\sqrt{w^2-2^{16}\cdot3^7}\right)+C.$$

This method is found in Chebyshev's paper “On the integration of differentials that contain a square root of a third or fourth degree polynomial”, available in French here.


问:
What can I say about a map multiplication ?(2)
I have asked this question here What can I say about a map multiplication?, here is the question and its answer:

The question:

If I know the following:

If I have the following homomorphism $f: \mathbb Z_{p^a} \rightarrow \Bbb Z_{p^b}$ defined by multiplication by $n/d$ where $n = p^b$ and $d = \gcd(p^a, p^b).$

And I know the following:

if $a\geq b$ then $f$ is onto.

if $a \leq b$ then $f$ is 1-1.

How can I use this piece of information to conclude something about this function: multiplication by $rn/d$ for $r= p^s x$ and $0 \leq r < d$? when is this map 1-1 and when it is onto?

Could anyone help me in that please?

Its Answer:

Let us handle the natural generalisation of the problem you present. Consider $m, n \in \mathbb{N}^{\times}$ together with the notation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for any $r \in \mathbb{Z}$. Let $d\colon=(m; n)$ denote the greatest common divisor of the two numbers and also consider $m'\colon=\frac{m}{d}, n'=\frac{n}{d}$ (the fractions exist since $d \neq 0$).

Let us fix $k \in \mathbb{N}^{\times} \cap n'\mathbb{Z}$ and let $f \in \mathrm{End}_{\mathbf{Gr}}(\mathbb{Z})$ be the additive group endomorphism given by $f(r)=kr$ for arbitrary $r \in \mathbb{Z}$. It is obvious that $f[\mathbb{Z}]=f\left[\langle m \rangle\right]=\left\langle f(m)\right\rangle \leqslant n\mathbb{Z}$, since $f(m)=mk \in mn'\mathbb{Z}=m'dn'\mathbb{Z}=m'n\mathbb{Z} \leqslant n\mathbb{Z}$, which means that $f$ induces a quotient morphism $g \in \mathrm{Hom}_{\mathbf{Gr}}\left(\mathbb{Z}_m, \mathbb{Z}_n\right)$, described by $g\left(\overline{r}\right)=\widehat{f(r)}=\widehat{kr}$, where $\overline{\bullet}$ denotes classes modulo $m$ respectively $\widehat{\bullet}$ refers to classes modulo $n$.

By virtue of the general properties of quotient morphisms, we have $\mathrm{Ker}g=f^{-1}\left[n\mathbb{Z}\right]/m\mathbb{Z}$. It is easy to see that $f^{-1}\left[n\mathbb{Z}\right]=\frac{n}{(k; n)}\mathbb{Z}$, whence $\mathrm{Ker}g=\frac{n}{(k; n)}\mathbb{Z}/m\mathbb{Z} \approx \mathbb{Z}_{\frac{m(k;n)}{n}} \ (\mathbf{Gr})$. In particular, $g$ is injective if and only if $\frac{m(k; n)}{n}=1$, relation equivalently expressed as $m(k; n)=n \Leftrightarrow m \mid n \wedge (k; n)=\frac{n}{m}$.

As to the image, we have that $\mathrm{Im}g=\left(\mathrm{Im}f+n\mathbb{Z}\right)/n\mathbb{Z}=\left(k\mathbb{Z}+n\mathbb{Z}\right)/n\mathbb{Z}=(k; n)\mathbb{Z}/n\mathbb{Z}=\frac{n}{(k; n)}\mathbb{Z} \ (\mathbf{Gr})$. Therefore, $g$ is surjective if and only if $(k; n)=1$, which entails $n'=1$ and thus $n \mid m$. The surjectivity of $g$ is thus attained if and only if $n \mid m$ and $(k; n)=1$.

Here is my question now:

I have a case that contradicts the required conditions for $g$ to be being surjective
If we took $m=2^4, n = 2^3 $ and $r = 2$ that case will not be surjective. Am I correct in that example or no? could anyone help me in that?

答: From what I can tell, the assertion is that the group homomorphism $f : \mathbb Z \to \mathbb Z$ defined by $f(r) = kr$ induces a surjective group homomorphism $g(r + m \mathbb Z) = kr + n \mathbb Z$ if and only if $n \,|\, m$ and $\gcd(k, n) = 1.$ For the case that $m = 2^4,$ $n = 2^3,$ and $r = 2,$ we have that $$k = \frac{rn}{\gcd(m, n)} = \frac{2^4}{2^3} = 2$$ so that $\gcd(k, n) = 2.$ Ultimately, there is no contradiction because $\gcd(k, n) \neq 1.$


问:
A definite multiple integral

$$\int_0^1\int_\sqrt[3]{x}^1 4\cos(y^4)\,\mathrm dy\,\mathrm dx$$

What I got was

$\sin(1)x+\cos(x^2) dx$ and now I am stuck.

I suddenly froze. Could someone help me? Haven't done calculus for a long time.

答:
Note that this is a definite integral, and hence the result should be a number and not a function.

First, note that the inner integral is quite tough to compute. It's not obvious what an anti-derivative of $\cos(y^4)$ should be. Therefore, it is sometimes convenient to change the order of integration, i.e. in this exercise integrate with respect to $x$ first and then with respect to $y$:

$$
\int_0^1\int_\sqrt[3]{x}^1 4\cos(y^4)\,\mathrm dy\,\mathrm dx=\int_0^1\int_0^{y^3}4\cos(y^4)\,\mathrm dx\,\mathrm dy.
$$
All you have to consider here is why the upper and lower bounds of the integrals are as they are. To this end, observe that in the first integral we have $0\leq x\leq 1$ and $\sqrt[3]{x}\leq y\leq 1$ which is equivalent to $0\leq y\leq 1$ and $0\leq x\leq y^3$. Now, this should be an easy task to integrate.

Title: A definite multiple integral


问:
Is $f(n)=\sqrt{n}$ the only function from $\Bbb{N}_0$ to $[0,\infty)$, with $f(100)=10$ and $\sum_{k=0}^n\frac{1}{f(k)+f(k+1)}=f(n+1)$?

Let $f:\mathbb{N_0} \rightarrow \left[0,\infty\right)$ be a function such that

a) $f(100)=10$

b)$\dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}=f(n+1)$ for all $n\geq 0$.

Evaluate $f(n)$.

Clearly, $f(n)=\sqrt{n}$, satisfies the conditions, but my question is, is it the only function possible?

I don't see how using the value at $n=100$ would prove the function to be $\sqrt{n}$, however if instead of $100$, $f(1)=1$ was given, then we can prove by induction that $\sqrt{n}$ would be the only function by starting from the first term and finding the next term.

答:
Using (b) with $n$ and then (b) with $n-1$ and subtracting, we get
\begin{align*}
f(n+1) - f(n) &= \bigg( \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)} \bigg) \\
&\qquad{}- \bigg( \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n-1)+f(n)} \bigg) \\
&= \frac1{f(n)+f(n+1)}.
\end{align*}
Therefore $f(n+1)^2 - f(n)^2 = 1$ for all $n\ge0$, which immediately fixes all the values given any single value.


答:
$$\sin 5x = \cos (\frac{\pi}{2}-5x)= \cos 3x $$

$$3x=\frac{\pi}{2}-5x+2k\pi$$

$$x=\frac{\pi}{16}+\frac{k\pi}{4}=\frac{\pi}{16}(1+4k)$$

or

$$3x=-(\frac{\pi}{2}-5x)+2k\pi$$

$$x=\frac{\pi}{4}-k\pi$$

$$x=\frac{\pi}{4}+k\pi =\frac{\pi}{4}(1+4k)$$

where $k\in Z$

writing $-k\pi$ or $k\pi$ does not change the solution set. Because $-k$ is the opposite of $k$ in integers.


答:
(Fill in the gaps as needed. If you're stuck, show your work and explain why you're stuck.)

Ignore condition (a) for now (This just gives us a specific solution).

Claim: Prove that condition $b$ is equivalent to

$ \frac{1}{ f(0) + f(1) } = f(1)$, and

$ f(n)^2 - f(n-1)^2 = 1$ for $n\geq 2$

Corollary:  The family of solutions that satisfies $(b)$ is given by

$$ \begin{align} f(0) & = a, \\
f(1) & = \frac{1}{2} ( \sqrt{ a^2 + 4} - a ), \\
f(n) & = \sqrt{ f(1)^2 + n-1 } \end{align}. $$

Corollary: Hence, using condition $a$, since $f(100) = 10$, so $f(1) = 1$ and $ f(0) = a = 0$. Thus, the solution is uniquely $ \sqrt{n}$.

Notes

If we just care about condition $b$, then for $ n \geq 1$,  $\sqrt{n} \geq f(n) > \sqrt{n-1}$. You can show that by showing that the range of $ \frac{1}{2} \sqrt{a^2 + 4} - a ) $ on $[0, \infty)$ is $ (0, 1]$.

Title: Is $f(n)=\sqrt{n}$ the only function from $\Bbb{N}_0$ to $[0,\infty)$, with $f(100)=10$ and $\sum_{k=0}^n\frac{1}{f(k)+f(k+1)}=f(n+1)$?

Body:
Let $f:\mathbb{N_0} \rightarrow \left[0,\infty\right)$ be a function such that

a) $f(100)=10$

b)$\dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}=f(n+1)$ for all $n\geq 0$.

Evaluate $f(n)$.

Clearly, $f(n)=\sqrt{n}$, satisfies the conditions, but my question is, is it the only function possible?

I don't see how using the value at $n=100$ would prove the function to be $\sqrt{n}$, however if instead of $100$, $f(1)=1$ was given, then we can prove by induction that $\sqrt{n}$ would be the only function by starting from the first term and finding the next term.


问:
Condition of quotient space to satisfy T1 axiom
Here are the definitions that were given in Munkres's general topology:

Definition 1: A topological space satisifes the T1 axiom if every finite point set is closed.

Definition 2: Let $X$ be a topological space, and let $X^*$ be a partition of $X$ into disjoint subsets whose union is $X$. Let $p:X \to X^*$ be the surjective map that carries each point of $X$ to the element of $X^*$ containing it. In the quotient topology induced by $p$, the space $X^*$ is called a quotient space of $X$.

Question: I'm currently stuck on one statement that says in order for $X^*$ to satisfy the T1 axiom, one simply requires that each element of the partition $X^*$ be a closed subset of $X$. I'm not quite sure how is this true. Can anyone explain this to me? Thank you!

答:
By definition the quotient topology consists of all $U\subset X^*$ such that $p(U)^{-1}$ is open in $X$, so also for every closed $C\subset X^*$, we have that its preimmage is closed.

Knowing this, if we get an element $x^*\in X^*$ we have that it is a class of elements of $X$, in other terms it means $x^*=A\subset X$, so if we show that $A$ is closed, then by definition $\{ x^* \}$ is closed in $X^*$.

This is what we wanted, if fact a finite union of closed subset is closed.


问:
$\oint_{\gamma}(2z-3\bar z+1)\,dz$ where $\gamma$ is the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
$\oint_{\gamma}(2z-3\bar z+1)\,dz$ where the contour $\gamma$ is the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ traversed clockwise.

My difficulty is how to translate the ellipse formula to analytical path .
for circle it's easy ($e^{zi}$), but I don't know how to do so for ellipse.

答:
$$ z(t) = 2 \cos t + i3 \sin t$$

Explanation:

$$ z(t) = x(t) + i y(t)$$

Now, for the ellipse for a trignomeetric parameterization: $x=2 \cos t$ $y=3 \sin t$, plug that same thing into the complex function in $t$


问:
Derivative of Vectorization of matrix products w.r.t. a matrix
Suppose $\lambda \in l\times1 $， $ y \in l\times 1$，$A \in l\times mn $， $L \in m\times r $，$R \in n\times r $

$f=1/2 \parallel L\parallel_{F}^{2} +  \lambda^{T} (y-A \text{vec}(LR^{T}) )$

I want to calculate the minimum of $f$.

So, how can I calculate

$\frac{\partial f}{\partial L} $ and $\frac{\partial f}{\partial R} $

I am confused with the derivative.
Can someone tell me ?

答:
$\def\p#1#2{\frac{\partial #1}{\partial #2}}$
Use a colon to denote the trace/Frobenius product
$$A:B \;=\; \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} \;=\; {\rm Tr}(A^TB)$$
This product can also be applied to vectors by treating them as rectangular matrices (set $\,n={\tt1}$) in which case it corresponds to the usual dot product.  The cyclic property of the trace allows the terms in such products to be rearranged in many equivalent ways, e.g.
$$\eqalign{
A:BC &= B^TA:C = AC^T:B \\
B:C &= B^T:C^T = C:B \\
B:C &= {\rm vec}(B):{\rm vec}(C) \\
}$$
Note that terms on the RHS and LHS of the colon must have identical dimensions.

Use ${\rm devec}$ to denote the reverse of the vec operator
$$A={\rm devec}(a) \quad\iff\quad a={\rm vec}(A)$$
For ease of typing, define the matrix
$$M = {\rm devec}(A^T\lambda)$$
Write the objective function using the above definitions,
then calculate its differential.
$$\eqalign{
 f &= \tfrac 12L:L - \lambda:\Big(A\,{\rm vec}(LR^T)-y\Big) \\
   &= \tfrac 12L:L - A^T\lambda:{\rm vec}(LR^T) - \lambda:y \\
   &= \tfrac 12L:L - M:LR^T - \lambda:y \\
df &= L:dL - M:(dL\,R^T+L\,dR^T) \\
   &= (L-MR):dL - M^TL:dR \\
}$$
Holding $R$ constant (i.e. setting $dR=0$) yields the gradient with respect to $L$, while holding $L$ constant yields the gradient wrt $R$.
$$\eqalign{
\p{f}{L} = L-MR, \qquad \p{f}{R} = -M^TL \\
}$$


问:
Square LP. Consider the LP with $A$ square and nonsingular. Minimize $c^Tx$.

Square LP. Consider the LP
$$\text{minimize }c^Tx\\
\text{subject to } Ax \le b$$
with $A$ square and nonsingular. Show that the optimal value is given by
$p^* = c^TA^{-1}b$ if $A^{-1}c \le 0$, and $p^{*} = -\infty$ otherwise.

In the solution the problem is transformed to $$\text{minimize }c^TA^{-1}y\\
\text{subject to } y \le b$$

by the change of variables $y=Ax$. Then, it says if $A^{-T}c \le 0$  the optimal solution is $y = b$, with $p^* = c^TA^{-1}b$. Otherwise, the LP is unbounded below.

The question is how the condition $A^{-1}c \le 0$ was derived (I don't understand the underlying logic)? Even if that condition is satisfied why does it imply that the optimal solution is $y = b$, with $p^* = c^TA^{-1}b$?

答:
Let $d=A^{-T}c$, then your objective function is now

$$\min d^Ty=\sum_{i=1}^ n d_i y_i$$

subject to $$y \le b$$

If one of the term $d_i > 0$, we can set $y_i \to -\infty$ while fixing the other $y_j$ and the overall objective function will go to $-\infty$.

If each of the term $d_i \le 0$, then we would want $y_i$ to be as large as possible to reduce the quantity.


问:
Likelihood of sum of dice roll is exactly 1 million
A board game is set up such that there is a number line with squares numbered 0..1 million. You roll a standard 6 sided die and move forward the number of spaces that you roll. Eventually you will either land on, or pass the 1 millionth spot. What is the probability you land exactly on the 1 millionth spot?

In other words, what is the probability that after a number of rolls the sum is exactly 1 million

答:
You can also analyze this problem via generating functions.

Let $P_{n,k}$ be the probability that you land on square $n$ after exactly $k$ turns, and let $P_n$ be the probability that you ever land on square $n$. So we're trying to find $P_{10^6}$. Also, note that $P_n=\sum_{k=0}^\infty P_{n,k}$: every number on the die is positive, so we can only land on a given square at most once.

Now, let $p(z)=\frac{1}{6}z+\frac{1}{6}z^2+\frac{1}{6}z^3+\frac{1}{6}z^4+\frac{1}{6}z^5+\frac{1}{6}z^6$ be the probability-generating function of the distribution of an individual die roll. Then the probability-generating function for the distribution of your position after $k$ turns is given by $\left[p(z)\right]^k$. That is, we have

$$
\left[p(z)\right]^k=\sum_{n=0}^\infty P_{n,k} z^n \, .
$$

Combining this formula with the above infinite sum, it follows that the generating function for the sequence $P_n$ is:

$$
f(z)=\sum_{k=0}^\infty \left[p(z)\right]^k=\frac{1}{1-p(z)}
$$

by summing the geometric series.

We can study the asymptotic behavior of this sequence by examining the poles of $f(z)$: that is, the complex roots of the polynomial $1-p(z)$. As we can see:

The strictly smallest-magnitude pole is $z=1$.

The next-smallest poles are roughly at $0.551685 \pm 1.253349i$, and so they have magnitude $\rho \approx 1.36939$.

Since the smallest-magnitude pole is at $z=1$, the sequence $\{P_n\}$ has a nonzero limit $A$, where the principal part of $f(z)$ at $z=1$ is $\frac{A}{1-z}$. Applying the residue formula gives

\begin{align*}
A&=\lim_{z \to 1} (1-z)f(z)\\
&=\lim_{z \to 1}\frac{1-z}{1-p(z)}\\
&=\frac{-1}{-p'(1)}&&\text{(by L'Hopital's rule)}\\
&=\frac{1}{p'(1)} \, .
\end{align*}

Since $p'(1)=\frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}+\frac{6}{6}=\frac{7}{2}$, we recover the limit $\lim_{n \to \infty} P_n=\frac{2}{7}$ as in the other answers.

Since the next-smallest poles have magnitude $\rho$, we have $P_n=\frac{2}{7}+O(\rho^{-n})$. For $n=10^6$, $\rho^{-10^6} \approx 10^{-136527}$, so the error in the approximation $P_{10^6} \approx \frac{2}{7}$ is too small to care about.


Body:
For $n\geq 1$,
\begin{align}
f(n+1) &= \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}\\
& = f(n)+\dfrac{1}{f(n)+f(n+1)}\\
\implies f(n+1)^2&=1+f(n)^2 =n+f(1)^2\\
\implies f(n) &= \sqrt{n-1+f(1)^2}.
\end{align}
Further, we have $$10 = f(100)=\sqrt{99+f(1)^2}\implies f(1)=1.$$

Finally, we also derive $$f(1)=\dfrac{1}{f(0)+f(1)}\implies f(0)=0.$$

Hence, we conclude that $f(n)=\sqrt{n}.$


问:
Curvature of a curve whose unit tangent creates a constant angle with z-axis
I have this question:
Consider a curve $\gamma$(s) and its projection to the plane $\beta$(s), i.e
$$\gamma(t) = \begin{bmatrix} f(s)\\ g(s) \\ h(s) \end{bmatrix},  \beta(t) = \begin{bmatrix} f(s) \\ g(s) \\ 0 \end{bmatrix} $$
Where s is the arc length parameter for $\gamma$.

Is s also the arc length parameter for $\beta$?

Let $T$ be the unit tangent vector of $\gamma$. Prove that the angle between $T(s)$ and the z-axis is at least $\frac{\pi}{4}$ iff $|h'(s)|\leq \frac{1}{\sqrt 2}$.

Assume $|h'(s)|\leq \frac{1}{\sqrt 2}$ holds for all $s$. Assume there is a constant C so that $$\kappa_{\gamma}\leq C$$ is satisfied everywhere. Prove that the curvature $$\kappa_{\beta}\leq 100C.$$

It would be great if someone could help me with this problem because I have been working on it for quite some time and have no idea how to arrive at the conclusion for part 3. Thank you in advance!

答:
Here is the answer to your third question:

As $s$ is arc length along $\gamma$ we have
$$\kappa_\gamma(s)=|\dot\gamma\times\ddot\gamma|=\sqrt{(\dot g\ddot h-\ddot g\dot h)^2+(\dot h\ddot f-\ddot h\dot f)^2+(\dot f\ddot g-\ddot f\dot g)^2}\ .$$
Along $\beta$ the variable $s$ is no longer arc length; therefore we have to use the formula
$$|\kappa_\beta(s)|={|\dot f\ddot g-\ddot f\dot g|\over(\dot f^2+\dot g^2)^{3/2}}\ .$$

By assumption $\dot h^2(s)\leq{1\over2}$. As $\dot f^2(s)+\dot g^2(s)+\dot h^2(s)\equiv1$ it follows that $\dot f^2(s)+\dot g^2(s)\geq{1\over2}$, so that we can immediately deduce
$$|\kappa_\beta(s)|\leq\sqrt{8}\>\kappa_\gamma(s)$$
for all $s$.


问:
Prove the inequality: $(y+z)*x^4+(x+z)*y^4+(x+y)*z^4\le((x+y+z)^5)/12$ for positive reals.
The inequality is homogeneous, but I am not sure if that helps.
Also, I was thinking about considering a concave function and apply Jensen or Karamata on LHS.</p>

答: Hint. Since the inequality is homogeneous, you may assume wlog that $x+y+z=1$. Thus, we want to prove that for positive real numbers $x,y,z$ that satisfy $x+y+z=1$, the following inequality holds $$\frac1{12}\geqslant \sum x^4\cdot (1-x)$$ Define now $f(x)=x^4\cdot (1-x)$, for $x\in[0;1]$ and notice that $f''(x)=4\cdot(3 - 5 x)\cdot x^2$. Hence, $f$ is convex on $\left[0; \frac35\right]$ and concave on $\left[\frac35; 1\right]$. You are left with some case work: you might assume wlog $x\leqslant y\leqslant z$ and have to see what happens if, (case $1$) $x\leqslant y\leqslant z\leqslant \frac35$ or (case $2$) $x\leqslant y\leqslant \frac35\leqslant z$... But this is (more or less) smooth :)


问:
Sum of $\frac{\Lambda(n)}{\log(n)}$ over divisors
I am wondering if $\sum_{d\mid n}\frac{\Lambda(d)}{\log(d)}$ (which evaluates to $1/k$ for $d=p^k$ and $0$ otherwise) has any interesting significances or bounds.

For $n = p_1^{k_1}\dots p_s^{k_s}$, we would have

$$\sum_{d\mid n}\frac{\Lambda(d)}{\log(d)} = \sum_i H_{k_i}$$
where $H_k$ is the $k$-th harmonic number, but is there a good way to approximate this?

Here is the plot for $(n, \sum_{d\mid n}\frac{\Lambda(d)}{\log(d)})$ for $1\le n \le 50000$:

答:
We can begin by computing

\begin{align*}
\sum_{n<N}\sum_{d|n}\frac{\Lambda(d)}{\log(d)}&=\sum_{d<N}\sum_{n<N/d}\frac{\Lambda(d)}{\log(d)}\\
&=\sum_{d<N}\left[\frac{N}{d}\right]\frac{\Lambda(d)}{\log(d)}\\
&=\sum_{p<N}\left[\frac{N}{p}\right]+O\left(\sqrt{N}\right)\\
&=\sum_{p<N}\frac{N}{p}+O(\pi(N))\\
&=\sum_{p<N}\frac{N}{p}+O\left(\frac{N}{\log(N)}\right)\\
&=N\left(\log(\log(N))+M+O\left(\frac{1}{\log(N)}\right)\right)+O\left(\frac{N}{\log(N)}\right)
\end{align*}

Here, the last estimate is from Merten's third theorem with "simple" error bounds (i.e not dependent on PNT). Thus,

$$\frac{1}{N}\sum_{n<N}\sum_{d|n}\frac{\Lambda(d)}{\log(d)}=\log(\log(N))+M+O\left(\frac{1}{\log(N)}\right)$$

without PNT


问:
$\sigma$-field generated by Borel sets
I what sense in the snippet is the $\sigma$-field $\cal M$ on $M$ generated by all the functions $$m\to m(C)$$
for $C$ Borel;how does $\cal M$ look like ?

答:
$M$ is the set of all counting measures on $(\mathbb{R},\mathcal{B}) $, where $\mathcal{B}$ is the Borel $\sigma$-field.

For each $C \in \mathcal{B}$, define a function $f_C : M \rightarrow \mathbb{R}$ such that $f_C(m)=m(C)$.

Let $\mathcal{M}$ be the $\sigma$-field generate by the functions in $\{ f_C : C \in \mathcal{B}\}$.

It means that  $\mathcal{M}$ be the $\sigma$-field generate by the sets in $\{ f_C^{-1}(D) : C \in \mathcal{B} , D\in \mathcal{B} \}$, in other words, $\mathcal{M}$ be the $\sigma$-field generate by the sets in $\{ \{m \in M : m(C) \in D \} : C, D \in \mathcal{B} \}$.

Using the properties of $\sigma$-field and Borel $\sigma$-field, we can simplify to:  $\mathcal{M}$ be the $\sigma$-field generate by the sets in $\{ \{m \in M : m(C) > a \} : C \in \mathcal{B} \textrm{ and } a \in \mathbb{R}\}$.

Remark 1: Let us prove, step by step, that
$$\{ f_C^{-1}(D) : C, D \in \mathcal{B} \}= \{ \{ m \in M : m(C) \in D \} : C, D \in \mathcal{B} \}$$

Note that, given any $ C, D \in \mathcal{B}$,

$$m \in  f_C^{-1}(D)  \Leftrightarrow (m \in M  \textrm{ and }  f_C(m) \in D)
\Leftrightarrow (m \in M  \textrm{ and }  m(C) \in D)$$

So, for all $ C, D \in \mathcal{B}$, $$f_C^{-1}(D)  = \{ m \in M : m(C) \in D \}  \tag{1}$$  It follows that
$$\{ f_C^{-1}(D) : C, D\in \mathcal{B}  \}= \{ \{ m \in M : m(C) \in D \} : C, D \in \mathcal{B} \}$$

Remark 2: Let us prove, step by step, that
\begin{align*}
\sigma(\{ \{ m \in M : m(C) \in D \} : & \, C, D \in \mathcal{B} \})= \\
&=\sigma(\{ \{m \in M : m(C) > a \} : C \in \mathcal{B} \textrm{ and } a \in \mathbb{R}\})
\end{align*}

First, let $\Gamma= \{(a,+\infty) \subseteq \mathbb{R} : a \in  \mathbb{R}\}$.
Note that $\mathcal{B}$ is generated by $\Gamma$, that is  $\mathcal{B}=\sigma(\Gamma)$.

So, we must prove:
\begin{align*}
\sigma(\{ \{ m \in M : m(C) \in D \} : & \,C\in \mathcal{B} \textrm{ and } D\in \sigma(\Gamma) \})= \\ &=\sigma(\{ \{m \in M : m(C) \in D \} : C \in \mathcal{B} \textrm{ and } D \in \Gamma \})
\end{align*}

It means, using $(1)$:
$$ \sigma(\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \})= \sigma(\{ f_C^{-1}(D) : C \in \mathcal{B}, D \in \Gamma \}) $$

Clearly,
$$\{ f_C^{-1}(D) : C \in \mathcal{B}, D \in \Gamma \} \subseteq \{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \}$$

So, we have,

$$\sigma(\{ f_C^{-1}(D) : C \in \mathcal{B}, D \in \Gamma \}) \subseteq \sigma (\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \})$$

Now, let us write $\Psi =\sigma (\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \Gamma \})$.

For each  $C\in \mathcal{B}$, consider  $\Sigma_C = \{E \subseteq \mathbb{R}: f_C^{-1}(E) \in \Psi\}$. Using the properties on the inverse image and the fact that $\Psi$ is a $\sigma$-algebra, it is easy to prove that $\Sigma_C$ is $\sigma$-algebra. It is also clear that $\Gamma \subset \Sigma_C$. So $\sigma(\Gamma) \subseteq \Sigma_C$.

It means that, for each $D\in \sigma(\Gamma)$ , $f_C^{-1}(D) \in \Psi$.
It means, for all  $C\in \mathcal{B}$, $$\{f_C^{-1}(D) : D\in \sigma(\Gamma)\} \subseteq \Psi$$
So
$$\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \} = \bigcup_{C\in \mathcal{B}}\{f_C^{-1}(D) : D\in \sigma(\Gamma)\} \subseteq \Psi$$

Since $\Psi$ is a $\sigma$-algebra, we have

$$\sigma (\{ f_C^{-1}(D) : C\in \mathcal{B}, D\in \sigma(\Gamma) \}) \subseteq \Psi$$

Title: $\sigma$-field generated by Borel sets


问:
How do I prove $\mu(\{x\in X: f(x)>c\} \cap A) \leq \frac{1}{c}\int_A f(x) d\mu(x)$.?
Let $f$ be a measurable function and $A$ a measurable set in the measure space $(X,M, \mu)$. Prove that for every $c>0$ the following inequality holds

$\mu(\{x\in X: f(x)>c\} \cap A) \leq \frac{1}{c}\int_A f(x) d\mu(x)$.

I know

$\mu(\{x\in X: f(x)>c\} \cap A)= \int_{X} \mathbb{X}_{\{x\in X: f(x)>c\} \cap A\}}(y) d \mu(y) $

Can anyone suggest how do I solve the given result?

答:
Hint: Notice that $\chi _{\{x:f(x)>c\}}=\chi _{\{x:\frac{f(x)}{c}>1\}}\leq 1 <\frac{f(x)}{c}$ if $x$ is in $\{x\in X: f(x)>c\}.$
Also $\chi _{A\cap B}=\chi _A\cdot \chi _B.$


问:
Finding the limit when denominator = 0
I'm having trouble solving this limit:

$$\lim_{x \to -2^-} \frac{1}{(x + 2)^2}$$

I can't find a way to rationalize the denominator. Also, is there a way to do it without plugging in -2.001 and stuff or graphing it?

EDIT:

I realized after asking this question that it doesn't matter if you take the above limit from the right, left, or both. It's always $+\infty$. Here's an equation that gives $+\infty$ from the right and $-\infty$ from the left: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$

How do I (algebraically) determine if it is positive or negative?

答:
You asked for an algebraic solution and that's well-covered in other answers, but I'd like to offer a conceptual way of reasoning through it.  Let's start with your revised example:
$$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$
As $x$ gets close to $3$, the numerator is getting close to $-1$ and the denominator is getting close to $0$.  Something nonzero (rather, not tiny-approaching-zero) divided by something tiny-approaching-zero is going to be big (far from zero), suggesting a limit of $\infty$ or $-\infty$ (that is, suggesting that the limit does not exist, but may be of the specific $\infty$ or $-\infty$ kind of does-not-exist).

Now, how do we tell whether it's $\infty$, $-\infty$, or neither?  Well, the numerator of the fraction is getting close to $-1$, so it's negative.  The denominator of the fraction is getting close to zero, but specifically as $x\to 3^+$, $x>3$, so $x-3>0$ and the denominator is positive.  The fraction is the quotient of a negative number and a positive number, so it's negative and
$$\lim_{x \to 3^+} \frac{x - 4}{x - 3}\to-\infty.$$

Title: Finding the limit when denominator = 0


问:
$i^{th}$ root(s) of unity
If we define $S:=\{z\in\mathbb{C}:z^n=1\}$ (i.e. the $n^{th}$ roots of unity), then $|S|=n$ (i.e. we have $n$ of them). We can even go as far as to say: $$S=\{z_k:k\in\mathbb{N}\cap[1,n]\}=\{e^{2\pi i/k}:k\in\mathbb{N}\cap[1,n]\}$$
Plotting these points on an Argand diagram shows the above point, namely that the plot looks symmetric across the real axis, and has rotational symmetry order $n$. It follows that, if we let $n\to\infty$, the plot of points tends towards being a circle centred on the origin, radius $1$.

But $z^i$, and even $z^{1/i}$, are both (possibly) well-defined ideas, so can we talk about the $i^{th}$ roots of unity? How many of them are there? What geometric interpretation is there when we talk about them, and can we analogously look at the $2i^{th}$ roots, etc.?

I know that $z^{1/i}:=e^{(1/i)\log{z}}$, so $z^{1/i}=1$ means that we search for solutions to $(1/i)\log{z}=0 \Rightarrow \log{z}=0$.

I guess a similar question is: what geometric/insightful interpretation is there to the fact that $i^i=e^{(-\pi+4k)/2}$ (for $k\in\mathbb{Z})$?

Edit: turns out that I can't multiply things by $0$ correctly... mistake fixed, thanks.

答:
One considers not so much the cyclotomic numbers $cis(2\pi m/n)$, but rather the table of chords $C(m/n) = 1-cis(2\pi m/n)$.  $cis() = cos() + i sin()$.

It's a pretty straight-forward proof that the intersection between the set of spans of the cyclotomics, and the set of fractional numbers (Z/Z), is the set of reals.

Using the further item that the product of numbers $C(m/n)$ for 0

So if you determine that the chord of an angle is 4-sqrt(13), then it must be irrational.

Title: $i^{th}$ root(s) of unity


问:
Finding remainder using Fermat's Little Theorm

Use Fermat's Little Theorem to find the remainder of $5^{15}$ divided by $1337$.

I know Fermat's Little Theorem, but failed to understand how to use it in this.

output.944

Body:
$$5^3\equiv-66\pmod{191}$$
$$5^4\equiv-330\equiv52,5^5\equiv5\cdot52\equiv69,5^6\equiv5\cdot69\equiv-37,5^7\equiv5\cdot(-37)\equiv6$$

$$5^{15}\equiv5(5^7)^2\equiv5\cdot6^2\equiv-11\pmod{191}\ \ \ \ (1)$$

$$5^3\equiv-1\pmod7\implies5^{15}=(5^3)^5\equiv(-1)^5\pmod7\ \ \ \ (2)$$

Use Chinese Remainder Theorem


答:
As stated, the result is wrong. Take for example $f$ defined on $A=[0,1]$ constant and equal to $-1$, $c=1/2$ and $\mu$ the Lebesgue measure.

The result is correct if you suppose $f$ non negative as you then have:

$$\begin{aligned}
\int_A f d\mu &= \int_{A \cap \{f(x) \gt c\}} f d\mu + \int_{A \cap \{f(x) \le c\}} f d\mu\\
&\ge \int_{A \cap \{f(x) \gt c\}} f d\mu\\
&\ge c \mu(\{x \in X \mid f(x)\gt c\} \cap A)
\end{aligned}$$

and therefore the desired result.

Title: How do I prove $\mu(\{x\in X: f(x)>c\} \cap A) \leq \frac{1}{c}\int_A f(x) d\mu(x)$.?

Body:
Let $f$ be a measurable function and $A$ a measurable set in the measure space $(X,M, \mu)$. Prove that for every $c>0$ the following inequality holds

$\mu(\{x\in X: f(x)>c\} \cap A) \leq \frac{1}{c}\int_A f(x) d\mu(x)$.

I know

$\mu(\{x\in X: f(x)>c\} \cap A)= \int_{X} \mathbb{X}_{\{x\in X: f(x)>c\} \cap A\}}(y) d \mu(y) $

Can anyone suggest how do I solve the given result?


答:
Assuming $c>0$:
$$c\cdot \mu(\{x\in X: f(x)>c\} \cap A)=c\cdot  \int_{X} \mathbb{X}_{\{x\in X: f(x)>c\} \cap A\}}(y) d \mu(y)=$$
$$=c\cdot \int_{X} \mathbb{X}_{\{x\in X: f(x)>c\}}(y)\mathbb{X}_A(y) d \mu(y) =\int_{A} c\cdot\mathbb{X}_{\{x\in X: f(x)>c\}}(y) d \mu(y)\le$$
$$\le\int_{A\cap \{x\in X: f(x)>c\} } f(y) d \mu(y)$$
The result can be extended to what you wrote if $f\ge 0$ on $A$.


问:
restricted of the entropy function is concave and upper semi continuous
Let $T:X \to X$ be a topologically mixing subshifts of finite type and $f:X\to \mathbb{R}$ be a continuous function. It is well known that the entropy function $\mu \mapsto h_{\mu}(T)$ is concave. I want to show that the following function is concave
$$ H(t)=\sup\{h_{\mu}(T), \mu \in \mathcal{M}(X, T), \int f d\mu=t\},$$
where $\mathcal{M}(X, T)$ is the space of invariant measure.

My attempt: for every $\mu_{1}, \mu_{2} \in \mathcal{M}(X, T)$, let $\mu=\lambda \mu_{1}-(1-\lambda)\mu_{2}$ for some $\lambda \in (0,1)$.

Since the entropy is concave, $h_{\mu}(T)\geq \lambda h_{\mu_{1}}(T)+(1-
\lambda)h_{
\mu_{2}}(T).$

Let $a=
\int f d\mu_{1}$ and $\int f d\mu_{2}=b.$
$$\sup \{h_{\mu}(T): \mu \in \mathcal{M}(X,T), \int f d\mu=\lambda a+(1-\lambda)b\}\geq \sup \{\lambda h_{\mu_{1}}(T)+(1-
\lambda)h_{
\mu_{2}}(T): \mu \in \mathcal{M}(X,T), \int f d\mu=\lambda a+(1-\lambda)b\}$$

I don't know how to get the result.

Edit: Is $H(t)$ upper semi continuous?

答:
You wrote all the right ingredients in your post.

You want to prove that $H$ is concave in $t$, so you want to prove that for any $s\in [0,1]$ and $a,b\in \mathbb{R}$ you have $H((1-s)a + sb) \geq (1-s) H(a) + s H(b)$.

For a real number $l\in \mathbb{R}$ let $\mathcal{M}_l$ be the set of invariant probability measures $\mu$ such that $\int fd\mu = l$, this set could be empty.

Fix $a,b \in \mathbb{R}$ and $s\in [0,1]$. Since $H(t)$ is defined by a sup, you can take two sequences of probability measures,  $(\mu^a_n)_{n\in \mathbb{N}}$ and $(\mu^b_n)_{n\in \mathbb{N}}$ such that for each $n\in \mathbb{N}$, $\mu^l_n \in \mathcal{M}_l$ and  $H(l) = \lim_{n\to +\infty} h_{\mu^l_n}(T)$, for $l=a,b$.

Define $\nu^s_n = (1-s) \mu^a_n + s \mu_n^b$. Since $\int f d\nu^s_n = (1-s) a + sb$, we have that $\nu^s_n \in \mathcal{M}_{(1-s) a + s b}$ for every $n\in \mathbb{N}$.  Also $h_{\nu^s_n}(T) = (1 -s) h_{\mu^a_n}(T) + s h_{\mu^b_n}(T)$. In particular, since $H(t)$ is defined by the supremum,
$$
H((1-s)a + sb) \geq \lim_{n\to +\infty} h_{\nu^s_n}(T) = \lim_{n\to + \infty} (1 - s ) h_{\mu^a_n}(T) + s h_{\mu^b_n}(T) = (1-s)H(a) + sH(b).
$$


问:
Determining the correct upper bound for an integral in polar coordinates

This seems super easy. But i am just a little bit stuck here. Haven't done much calculus recently. Can someone help me out real quick?

Thank you in advance!

答:
Hint:

$$z=\sqrt{x^2+y^2}\implies 18=x^2+y^2+z^2=2(x^2+y^2)\implies x^2+y^2=9$$

Well now: do you remember what $\,r\,$ is in polar coordinates...?


问:
Elliptic Curve has Canonical Bundle $K_E = \mathcal{O}_E$
Let $E$ be an elliptic curve over a field $k$. I'm looking for a proof that for the canonic divisor $K_E$ we have $K_E = \mathcal{O}_E$? RR says $h^0(K_E) = h^0(\mathcal{O}_E)+deg(K_E) +g-1= h^0(\mathcal{O}_E)=1$ since $E$ elliptic. I don't see wy this already imply $K_E = \mathcal{O}_E$.

Body:
It might be helpful to simply $\bf{copy\text{-}paste}$ materials from Silverman's book.
Fix a field $K.$
Restrict ourselves to char $K\neq 2$ case with $e_{1}\neq e_{2}\neq e_{3}:$
$E:y^{2}=(x-e_{1})(x-e_{2})(x-e_{3})$ in $\Bbb{P}^{2}_{K}.$
To show $$\Omega^{1}_{E}\cong \mathcal{O}_{E},$$
we need to show canonical divisor class is same as the divisor class of $0.$
As canonical divisor class is given by $\text{div}(\omega)$ with any $0\neq\omega\in\Omega^{1}_{E}.$
Consider $dx/y.$ In fact $\text{div}(dx/y)=0,$ where
$$\text{div}(dx)\overset{(1)}{=}(e_{1},0)+(e_{2},0)+(e_{3},0)-3(\infty)\overset{(2)}{=}\text{div}(y).$$
To show (1), consider $dx=d(x-e_{1})$ firstly and we have
$$\text{ord}_{(e_{1},0)}(d(x-e_{1}))=\text{ord}_{(e_{1},0)}(x-e_{1})-1.$$
Admit $\text{ord}_{(e_{1},0)}(x-e_{1})=2.$ Then $\text{ord}_{(\infty)}(x-e_{1})=-2$ and $\text{div}(x-e_{i})=2(e_{i},0)-2(\infty).$ Moreover, we have $\text{ord}_{(\infty)}(d(x-e_{1}))=-3$ and (2).

To show $\text{ord}_{(e_{1},0)}(x-e_{1})=2,$
we show $x-e_{1}\in \mathfrak{m}_{(e_{1},0)}^{2}.$
Here $$\mathfrak{m}_{(e_{1},0)}=(x-e_{1},y),\,\,(x-e_{1},y)^{2}=((x-e_{1})^{2},(x-e_{1})y,y^{2}).$$
Consider $$-(x-e_{1})(e_{3}-e_{1})(e_{2}-e_{1})=y^{2}-((x-e_{1})^{2}-(e_{2}+e_{3}-2e_{1})(x-e_{1}))(x-e_{1})\equiv 0\text{ mod }(x-e_{1},y)^{2}$$ and thus $(x-e_{1})\in (x-e_{1},y)^{2}.$


问:
How many 5-digit palindromes are divisible by 9?
What I already have,

Palindrome in form XYZYX, where X can’t be 0.

Divisibility rule of 9: sum of digits is divisible by 9. So, we have 2(X+Y)+Z = 9M.

The first part is divisible by 9 if and only if X+Y is divisible by 9. So, we have 10 pairs out of 90. And each such pair the total sum is divisible by 9 when Z is also divisible by 9. There are 2 such Zs: 0, 9. So, there are 20 divisible palindromes.

If (X+Y) mod 9 = 1, then 2(X+Y) mod 9 = 2; and in order for the total sum to be divisible by 8, Z must have the remainder of 1 when divided by 9. There is 1 such Z: 1. And again, we have 10 xy pairs with the given remainder. So, this case yields 10*1 = 30 more palindromes.

Same logic as on previous step applies to the case when 2(X+Y) mod 9 = 2.

So, total number of divisible palindromes = 80?

When using this method, I only get 80 numbers of 5-digit palindromes that are divisible by 9(?) i dont think im doing this method correctly, can someone show me whats going on here

答:
As you have rightly mentioned, we need to figure out all tuples of $(X,Y,Z)$ such that $$2(X+Y)+Z\bmod 9 = 0 \implies 2(X+Y)\bmod 9 = (9-Z)\bmod 9.\tag{1}$$
Note that for any value of $2(X+Y)\bmod 9=0,1,\ldots,8$, the corresponding value of $(X+Y)\bmod 9$ is unique. Also, for any given $k=0,1,\ldots,8$, the tuples that satisfy $(X+Y)\bmod 9=k$ are given by
$$(k, 9) \text{ and } (X,k-X\bmod 9) \text{ for } X=1,2,\ldots,9.$$ So, there are $10$ tuples $(X,Y)$ corresponding to any given value of $2(X+Y)\bmod 9$.

Further, for any given value of $2(X+Y)\bmod 9$ in $(1)$, the corresponding value of $Z$ is also unique except when $2(X+Y)\bmod 9 = 0$. When $2(X+Y)\bmod 9 = 0$, $Z$ can either be $0$ or $9$. Consequently, with $2(X+Y)\bmod 9=1,2,\ldots,8$, there are $80$ palindromes divisible by $9$, and with $2(X+Y)\bmod 9=0$, there are $20$ palindromes divisible by $9$.

Hence, the total number of $5-$digit palindromes divisible by $9$ is $100$.


问:
Why does rotation by a quaternion require multiplying two times?
Given a vector $p$, to rotate it by a quaternion $q$, we use the formula:

$$p' = q p \hat{q}$$

where $\hat{q}$ is the conjugate of $q$. But if we use rotational matrices, then it's just

$$p' = Rp$$

While it's clear why the matrices work this way, I just cannot develop any intuition on quaternion rotation formula. I realise the formula does work, but still my guts feel as if it was “rotate $p$ by $q$ and then compose it with the inverse rotation of $q$”. Maybe there is some simple explanation?

答:
I will try to provide a simple intuition. Suppose q = $\hat i$,$q^{-1}=-\hat i$  and p = $\hat j$.

Now, $q*p = \hat k$ will give you a 90-degree rotation anticlockwise.

Similarly, $q^{-1}*p$ will give you a 90-degree rotation clockwise.

So we start thinking if these operations are done in succession wouldn't these operations cancel out each other?

The answer to this is as follows -

$q*p$ is not the same as $p*q$. So when you express your operations as $q*(p*q^{-1})$ you are actually performing a 180-degree rotation. To see this let us compute the result
$\hat j* - \hat i = -\hat k, -\hat k*\hat i =-j$

Now let us perform $q^{-1}*(q*p) = -\hat i * \hat k = \hat j$ (cancel out operation)

The double cover property is important because, in the case of 2d, the addition of a single imaginary axis did the job. But in the 4d case, we have 3 extra imaginary axes that have to express 6 degrees of freedom. Hence in order to have closure on the 6 DoF we need to have the double cover.


问:
Prove that $X$ and $Y$ follow a normal distribution
Let two random variables $X$ e $Y$ with the following joint distribution function:
$$
F(x,y)=\Phi(x)\Phi(y)[1+\alpha(1-\Phi(x))(1-\Phi(y))]
$$
with $|\alpha|\leq 1$ and $\Phi(\cdot)$ the distribution function of a standard normal distribution. Prove that the marginal distributions corresponding to $X$ and $Y$ are standard normal.

I'm having some problems proving this statement. I only have to prove it for $X$ because for $Y$, the procedure is the same.

To prove for $X$, I'm using the fact that $f_{X}(x)=\displaystyle\int_{-\infty}^{+\infty}f(x,y)dy$ and that $f_{X,Y}(x,y)=\frac{\partial^2F}{\partial x \partial y}(x,y)$.

The problem with this approach is that I obtain a very long expression and when I try to integrate, I obtain terms like $\displaystyle\int\Phi(y)f(y)dy$ that i do not know how to resolve. I'm not sure if I have some errors calculating $f_{X,Y}$ or if this is the incorrect approach to prove this.

Any suggestions?

答: Since $$F(x,y):=P(X\le x, Y\le y),\tag{*}$$ a quick way to obtain the marginal distribution for $X$ is to let $y\to\infty$ in (*); the RHS will then converge to $P(X\le x)$.


问:
show that there exists a set with $x_{1},.....,x_{n}$ as its elements
Let $x_{1},....,x_{n}$ be sets for $n\in\mathbb{N}$ and $n\ge 1$.

a) show that there exist a set with $x_{1},....,x_{n}$ as its elements

b)show that $x_{1}\cup x_{2}\cup...\cup x_{n}$ is a set.

My first thought was to do it like that

we can form a set $\{x_{1},x_{2}\}$ which is a set by axiom of pairs and a set $\{x_{n}\}$

then
$\{x_{1},x_{2}\}\cup \{x_{n}\}$ is also a set by the union axiom but $\{x_{1},x_{2}\}\cup \{x_{n}\}=\{x_{1},x_{2},...x_{n}\}$

is this okay as the answer to a)?

what should be done in b)?

I know that for example $x_{1}\cup x_{2}:=\bigcup\{x_{1},x_{2}\}$

does it mean that

$x_{1}\cup....\cup x_{n}=\bigcup\{x_{1},....,x_{n}\}$

but how to show that this is a set?

答: For part b) all you need to do is use the set you obtain from part a) and apply the union axiom to it to prove $\bigcup\{x_1,\cdots,x_n\}$ is a set.</p>


问:
$\oint_{\gamma}(2z-3\bar z+1)\,dz$ where $\gamma$ is the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
$\oint_{\gamma}(2z-3\bar z+1)\,dz$ where the contour $\gamma$ is the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ traversed clockwise.

My difficulty is how to translate the ellipse formula to analytical path .
for circle it's easy ($e^{zi}$), but I don't know how to do so for ellipse.

答:
Forget explicit parameterization of $\gamma$, just use Stoke's theorem. In particular, use the version stated in complex coordinates.

Let $E$ be the ellipse bounded by $\gamma$. Since $\gamma$ walks around $E$ in clockwise direction, it is "negative" to the orientation of $\partial E$, the boundary of ellipse. Apply Stoke's theorem in complex coordinates, we have

$$\int_\gamma (2z - 3\bar{z} +1 ) dz = \int_{-\partial E}(2z - 3\bar{z} + 1) dz
= -\int_E d(2z - 3\bar{z} + 1) \wedge dz\\
= 3\int_E d\bar{z} \wedge dz
= 6i \int_E \frac{d\bar{z}\wedge dz}{2i}$$
In terms of Cartesian coordinates,

$$\frac{d\bar{z}\wedge dz}{2i}
= \frac{d(x-iy) \wedge d(x+iy)}{2i} = dx \wedge dy$$
is simply the area element. Since ellipse $E$ has semi-major/minor axes $3$ and $2$, we have:

$$\int_\gamma (2z - 3\bar{z} +1 ) dz = 6i\verb/Area/(E) = 6i(6\pi) = 36\pi i$$

For comparison, let us redo the computation in Cartesian coordiantes.

We can parametrize $E$ as

$$[0,2\pi] \ni \theta\quad\mapsto\quad (x,y) = (2\cos\theta,\color{red}{-}3\sin\theta) \in \mathbb{R}^2 \sim \mathbb{C}$$

Since $\gamma$ walks around $E$ in clockwise direction, the sign in front of $\sin\theta$ is negative instead of positive. Plug these into original integral, it becomes

$$\begin{align} &\int_0^{2\pi} (2(2\cos\theta - 3\sin\theta i) - 3(2\cos\theta + 3\sin\theta i) + 1)(-2\sin\theta - 3\cos\theta i) d\theta\\
= &\int_0^{2\pi} -(2 + 41\cos\theta)\sin\theta + (30\sin^2\theta + 6\cos^2\theta - 3\cos\theta)i d\theta\end{align}$$
Throwing away terms which clearly don't contribute, we get

$$\begin{align}\int_\gamma(2z - 3\bar{z} +1 )dz
&= i\int_0^{2\pi}(30\sin^2\theta + 6\cos^2\theta)d\theta\\
&= i(30\pi + 6\pi) = 36\pi i\end{align}
$$
Same number $36\pi i$ we obtained before.


问:
Fourier transform and energy of a convolution with a delta comb
Hi guys i have to find the fourier transform of the convolution:
$$ sinc(t/2T)*\sum\limits_{n-\infty}^{+\infty} (-1)^{n}\delta(t - nT) $$

i was thinking of express the summatory as : $$\sum\limits_{n-\infty}^{+\infty} (-1)^{n}\delta(t - nT) = \mathrm{III}_{2T}(t) -\mathrm{III}_{T}(t-1)$$
so we have:

$$ sinc(t/2T)*[\mathrm{III}_{2T}(t) -\mathrm{III}_{T}(t-1)]$$

So the convolution in the fourier domain became a product:
$$ rect(f2T)\mathrm{III}_{1/2T}(t)-2rect(f2T)\mathrm{III}_{1/T}(t)e^{ipif} $$

Well is this a good result and how i can find the energy from this formula  ?

答:
$rect(f2T)\mathrm{III}_{1/2T}(t)$ is a single Dirac delta so you are almost done. I think you meant  $2\mathrm{III}_{2T}(t) - \mathrm{III}_{T}(t)$ where $$\mathrm{III}_T(t)=\sum_n \delta(t-nT)=|T|^{-1} \mathrm{III}_1( t/|T|)$$ whose $e^{-2i\pi f t}$ Fourier transform is $$\mathrm{III}_1( |T| f)=|T|^{-1}\mathrm{III}_{1/|T|}( f)$$

Alternatively you could replace $\sum\limits_{n-\infty}^{+\infty} (-1)^{n}\delta(t - nT)$ by its Fourier series (which converges only in the sense of distributions) and change the order of summation obtaining a series of $sinc \ast e^{iat} = \widehat{sinc}(a) e^{iat}$.

Your function is $2T$ periodic, so with the energy you meant $\int_0^{2T} |g(t)|^2dt$ ?


问:
Prove by induction that for any integer $n \geq 7, 3^n \leq n!$, where $n! = n(n-1)(n-2)...$
This is a practice problem, and but I am having trouble what I should do. I am new to proof writing, so I am a little bit uncomfortable with induction.

Question: Prove by induction that for any integer $n \geq 7, 3^n \leq n!$, where $n! = n(n-1)(n-2)...$.

My Attempt:

Suppose $P(n)$ is the assertion: $n \geq 7, 3^n \leq n!$. First we need to prove that the assertion for the base case is true. That means showing that when $n = 7, 3^n \leq 7!$, so

\begin{align*}
3^7 &\leq 7! \\
2187 &\leq (7)(6)(5)(4)(3)(2)(1) \\
2187 &\leq 5040
\end{align*}

This confirms that the inequality holds for $n = 7$

Assume that the assertion $P(n)$ is true for $n = k$, such that $P(k)$ is true. That means to assume that $3^k \leq k!$ (Inductive Hypothesis) is true

Then by making the inductive step, by using the inductive hypothesis, it follows that $P(k + 1)$ is true. $$3^{k+1} \leq (k + 1)!$$

We want to show that if $P(k + 1)$ follows $P(k)$, and the conclusion is true along with the base case, then this will prove that $P(n)$ is true

\begin{align*}
3^{k+1} &\leq (k + 1)! \\
(3)^k (3)^1 &\leq (k + 1)k! \\
\end{align*}

This is where I am lost, and do not know how to proceed.

Any help is appreciated.

答:
Consider the following for any $k\geq 7$:

$$3^{k+1} = 3\times 3^k \leq 3 (k!)\leq k(k!) = (k+1)!.$$
Here, the first inequality holds from $P(k)$ and the second inequality is because $k\geq7>3$.

Hence, we showed that when $P(k)$ holds, $P(k+1)$ also holds.