答:
If you simplify the original expression using the definitions
$$\cosh x =\frac{e^x+e^{-x}}{2} \quad \sinh x =\frac{e^x-e^{-x}}{2}$$

you'll see that the second factor is simply $e^x$. Then
$$y(x)=e^{2x}\cdot e^x=e^{3x}$$
from which it follows
$$y''(x)=9e^{3x}=9y(x)$$


问:
Why does $3+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n),\quad n\ge1223$?
Let $P$ denote $\text{primes}$, and $\pi(x)$ denote $|P| \le x$.

Here's my first question: Why does

$$3+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n),\quad n\ge1223$$

And similarly:

$$2+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n),\quad 11\le n\lt1223$$

I know that it is a fairly weak statement as of yet, but I can't find this (or a statment similar enough) anywhere, which seems weird, as it seems to be a interesting way of defining $\pi(x)$.

If anyone has any explanation on why this is the case, please share.

Thanks in advance!

Edit:

If the statment is redefined as $j+ (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right) = \pi(n)$, is there a way to determine $j$?

Edit 2:

As the accepted answer indicates:

$$j + (-1)\left(\left\lfloor\sum_{k=1}^{|\{x\in P,\;x\le n\}|} \frac{P_k}{1-P_k}\right\rfloor\right),\quad j>0,\; e^{e^{j-1}} < n < e^{e^{j}} = \pi(n)$$

答:
Note that $$-\frac {P_k}{1-P_k}=1+\frac1{P_k-1}>1+\frac1{P_k}$$
and the sum of prime reciprocals diverges.
Therefore, for $n$ big enough, your $\sum$ will be bigger than $\pi(n)+r$ for any $r$, thus disproving your conjecture. (The divergence is slow, however, for the $r$th steps you have to go up to $n$ in the order of $e^{e^{r}}$).</p>

ContentLicense: CC BY-SA 3.0


问:
If $a+b+2 \sqrt{ab}= 2(b-a)$, Prove that $\frac{b}{a}=9$
$a$ and $b$ are both Real numbers (different than $0$)

I saw this problem in the math Olympiads of my home country

What I’ve tried so far:

Factoring $a+b+2 \sqrt{ab}$ to $(\sqrt{a} +\sqrt{b})^{2}$

Expanding the whole equation

And thank you for your help

答:
Divide the equation by $a$, let $r=b/a$. This gives
$$1+r+2\sqrt r = 2(r-1)$$
set $x=\sqrt r>0$  and we see a quadratic $$1+x^2 + 2x = 2(x^2-1) \iff 0 = x^2 - 2x -3 = (x-3)(x+1)$$
the positive solution is $x=3$, so $\frac ba = r = x^2 = 9$.

ContentLicense: CC BY-SA 4.0


答:
Dividing $a+b+2\sqrt{ab}=2(b-a)$ by $a$ gives
$$1+b/a+2\sqrt{b/a}=2b/a-2.$$
Hence, $3-b/a+2\sqrt{b/a}=0$. This is a quadratic equation in $\sqrt{b/a}$.

Title: If $a+b+2 \sqrt{ab}= 2(b-a)$, Prove that $\frac{b}{a}=9$

Body:
$a$ and $b$ are both Real numbers (different than $0$)

I saw this problem in the math Olympiads of my home country

What I’ve tried so far:

Factoring $a+b+2 \sqrt{ab}$ to $(\sqrt{a} +\sqrt{b})^{2}$

Expanding the whole equation

And thank you for your help


答:
Divide throughout by $a$.

So,

$\begin{align}1+ \frac{b}{a} + 2\sqrt{\frac ba} = 2 \frac ba -2 \Rightarrow \frac ba - 2\sqrt{\frac ba} = 3 \end{align}$

$\Rightarrow\sqrt{\frac ba} = 3  $ or $-1$.

As $-1$ is not possible, $\sqrt{\frac ba} = 3$

or $\frac ba = 9$


问:
Density of $\{ u \in \mathcal{C}_{0}^{\infty}(\mathbb{R}^n) \: : \: | \nabla u | \le 1 \}$ in $\operatorname{Lip}_{1}(\mathbb{R}^n)$
Let $\mathcal{C}_{0}^{\infty}$ denote the space of $\mathcal{C}^{\infty}$ functions with compact support and Let $\operatorname{Lip}_{1}(\mathbb{R}^n)$ be the space of Lipschitz continuous functions having Lipschitz constant equal to 1.

Is it true that the space $\{ u \in \mathcal{C}_{0}^{\infty}(\mathbb{R}^n) \: : \: | \nabla u | \le 1 \}$ is dense in $\operatorname{Lip}_{1}(\mathbb{R}^n)$? And if the answer is yes, how can I prove it?

Any help is appreciated.

答:
The Lipschitz functions need to have compact support. Otherwise, the function
$f(x):=\sin x$ is a counterexample as it cannot be approximated uniformly with functions with compact support.

If $f$ is a Lipschitz function with compact support, i.e., $f\in C^{0,1}_c(\mathbb R^n)$, then the convolution with the standard mollification kernel $\rho_\epsilon * f$ is a smooth function with compact support. Since $D (\rho_\epsilon * f) = \rho_\epsilon * (Df)$, the resulting function is 1-Lipschitz as well. In addition,
$$\begin{split}
|f(x) - \int_{\mathbb R^n} \rho_\epsilon(x-y)f(y)dy|
&= |\int_{\mathbb R^n} \rho_\epsilon(x-y)(f(x)-f(y))dy|\\
&\le 1\cdot \int_{\mathbb R^n}\rho_\epsilon(x-y)|x-y|dy \\
&\le \epsilon \int_{\mathbb R^n}\rho_\epsilon(x-y)dy=\epsilon,
\end{split}$$
since the support of $\rho_\epsilon$ is contained in a ball of radius $\epsilon$ and $\int_{\mathbb R^n}\rho_\epsilon(y)dy=1$.
Hence $\rho_\epsilon * f \to f$ uniformly.


问:
Vector Related problem
If $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ are unit vector such that ${\left| {\overrightarrow a  - \overrightarrow b } \right|^2} + {\left| {\overrightarrow b  - \overrightarrow c } \right|^2} + {\left| {\overrightarrow c  - \overrightarrow a } \right|^2} = 9$.

Then the value $\left| {2\overrightarrow a  + 5\overrightarrow b  + 5\overrightarrow c } \right| = \_\_\_\_\_\_$

My approach is as follow

${\left| {\overrightarrow a  - \overrightarrow b } \right|^2} + {\left| {\overrightarrow b  - \overrightarrow c } \right|^2} + {\left| {\overrightarrow c  - \overrightarrow a } \right|^2} = {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} - 2\overrightarrow a .\overrightarrow b  + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} - 2\overrightarrow b .\overrightarrow c  + {\left| {\overrightarrow c } \right|^2} + {\left| {\overrightarrow a } \right|^2} - 2\overrightarrow c .\overrightarrow a  = 9$

$\left( {\overrightarrow a .\overrightarrow b  + \overrightarrow b .\overrightarrow c  + \overrightarrow c .\overrightarrow a } \right) =  - \frac{3}{2}$

How do I proceed from here

答:
For convenience we shall take $a=\vec{a},b=\vec{b},c=\vec{c}$

by your work$$a\cdot b+b\cdot c+c\cdot a+\frac{3}{2}=0$$ $$\implies|a+b+c|^2=0$$ $$|a+b+c|=0$$ $$\implies a+b+c=0$$ $$|2a+5b+5c|=|2a-5a|=3$$

Title: Vector Related problem


答:
We have $(b-a) = (\sqrt{b} - \sqrt{a})(\sqrt{b} + \sqrt{a})$

So with your step we get:

\begin{align*}&(\sqrt{b} + \sqrt{a})^2 = 2(\sqrt{b} - \sqrt{a})(\sqrt{b} + \sqrt{a}) \\ \iff &(\sqrt{b} + \sqrt{a}) = 2(\sqrt{b} - \sqrt{a}) \\ \iff & 9 = \frac{b}{a} \end{align*}


答:
Notice that the left side of the equation is actually equal to a square
$$(\sqrt{a} + \sqrt{b})^2 = 2(b-a)$$

Now apply the usual $c^2 - d^2 = (c+d)(c-d)$

$$(\sqrt{a} + \sqrt{b})^2 = 2(\sqrt{a} + \sqrt{b})(-\sqrt{a} + \sqrt{b})$$

and finally

$$\sqrt{a} + \sqrt{b}= 2\sqrt{b} - 2\sqrt{a}$$

from which

$$\frac{\sqrt{b}}{\sqrt{a}} = 3$$

And hence your claim that $\frac{b}{a} = 9$.


问:
Why is the subspace topology same as the Order topology?
Hi i am reading Topology by Munkres and there in the proof of theorem 27.1 it is written that

Given $a<b$, let $\mathcal A$ be a covering of $[a,b]$ by sets open in $[a,b]$ in the subspace topology(which is the same as the order topology).

I can't understand why is the subspace topology same as the order topology in this case? Is this always true? Btw i know how subspace topology and order topologies are defined. If $(X,\tau)$ is a topological space and $A$ be any subset of X then we define $\tau_A$ to be the set with subsets as : $\tau_A=\{U\cap A | U\in \tau\}$. And order topology is defined in a different way.So how can they be same here? For reference i am attaching the screenshot where i have highlighted the part where this is mentioned.

答:
I haven't done topology in a while, but verifying this seems straight-forward by comparing basis elements.

Let $O$ be a basis element of the order topology. Then,

if $O = [a,x), O = (-\infty, x) \cap [a,b]$

if $O = (x,y), O = (x,y) \cap [a,b]$

if $O = (x,b], O = (x,\infty) \cap [a,b]$

(Here $a < x, y < b$.) So $O$ is a basis element in the subspace topology in every case. Conversely, let $U$ be a basis element of the subspace topology. Then $U = R \cap [a,b]$, where $R = (x,y)$ is a basis element of the topology of $X$. (Here $x$ may be $-\infty$ and $y$ may be $\infty$.)

if $x < a$ and $y \notin [a,b]$, then $U = [a,b]$ or $U = \emptyset$

if $x < a$ and $y \in [a,b]$, then $U = [a,y)$

if $x \in [a,b]$ and $y > b$, then $U = (x,b]$

if $x \in [a,b]$ and $y \le b$, then $U = (x,y)$

again, $U$ is a basis element of the order topology in all cases.


答:
The conditions are that either $(a<0\land b<0)$ or $(a>0\land b>0)$. If the former, then $$\begin{cases}a<0\\ b<0\\ a+b+2\sqrt{ab}=2(b-a)\end{cases}\Leftrightarrow\begin{cases}a<0\\ b<0\\-(\sqrt{-a}-\sqrt{-b})^2=2(\sqrt{-a}-\sqrt{-b})(\sqrt{-a}+\sqrt{-b})\end{cases}\Leftrightarrow\\\Leftrightarrow\begin{cases}a<0\\ b<0\\(\sqrt{-a}-\sqrt{-b})(3\sqrt{-a}+\sqrt{-b})=0\end{cases}\Leftrightarrow \begin{cases}a<0\\ b<0\\ a=b\end{cases}\lor\begin{cases}a<0\\ b<0\\ \sqrt{-b}=-3\sqrt{-a}\end{cases}\Leftrightarrow\begin{cases}a<0\\ a=b\end{cases}$$

If they are both positive, $$\begin{cases}a>0\\ b>0\\ a+b+2\sqrt{ab}=2(b-a)\end{cases}\Leftrightarrow\begin{cases}a>0\\ b>0\\ (\sqrt a+\sqrt b)^2-2(\sqrt b-\sqrt a)(\sqrt a+\sqrt b)=0\end{cases}\Leftrightarrow\\\Leftrightarrow\begin{cases}a>0\\ b>0\\ (\sqrt a+\sqrt b)(3\sqrt a-\sqrt b)=0\end{cases}\Leftrightarrow\begin{cases}a>0\\ b>0\\ \sqrt a=-\sqrt b\lor 3\sqrt a=\sqrt b\end{cases}\Leftrightarrow\\\Leftrightarrow \begin{cases}a>0\\ b>0\\ b=9a\end{cases}$$

So, for non-zero real numbers the identity holds if and only if either $b=a<0$ or $b=9a>0$.


问:
How can I justify the substitution of a complex variable in the following integral?
$ \displaystyle I(k) = \int_{0}^{\infty} e^{(i-1)x}x^{k} \mathrm{d}x \tag*{} $

I want to know how can we justify the substitution $ (1-i)x=t $ while the upper and lower limits of the integral remains the same.

答:
$$\int_0^\infty e^{(i-1)x}x^k\,dx=\int_0^\infty (\cos x+i\sin x)e^{-x}x^k\,dx.$$

The antiderivative is of the form $(\cos(x)\,P(x)+\sin(x)\,Q(x))\,e^{-x}$ where $P,Q$ are polynomials. Hence the limit $x\to\infty$ is zero, and only the bound $x=0$ brings a contribution.


问:
Show that if for every sequence $(x_{n})$ converging to zero, $(f(x_n))$ is limited, then $f$ is a limited functional
Let $\mathbb{X}$ a normed vector space. If $f\colon \mathbb{X}\to\mathbb{K}$ is a linear functional such that for all sequence $(x_{n})$ that converges to zero we have that $(f(x_{n}))$ is bounded. Show that $f$ is a bounded functional. Could someone help me, please? I was unable to come up with ideas for this exercise.

答: Let assume by contradiction that a functional like that can be unbounded, then $\exists(x_n)_{n\in\mathbb{N}}$ a bounded sequence of elements of $\mathbb{X}$ such that $(f(x_n))_{n\in\mathbb{N}}$ is unbounded (we can assume $\lim_{n\to+\infty}|f(x_n)|=+\infty$, if it is not strictly true just consider a subsequence of $(x_n)_{n\in\mathbb{N}}$ with this proprierty and rename that $(x_n)_{n\in\mathbb{N}}$), let now consider $(y_n)_{n\in\mathbb{N}}$ such that $y_n=\frac{x_n}{\sqrt{1+|f(x_n)|}}\forall n\in\mathbb{N}$. Now, $(x_n)_{n\in\mathbb{N}}$ is buonded and $(f(x_n))_{n\in\mathbb{N}}$ goes to $+\infty$, so $(y_n)_{n\in\mathbb{N}}$ is infinitesimal and $f(y_n)=\frac{f(x_n)}{\sqrt{1+|f(x_n)|}}$, so $\lim_{n\to+\infty}|f(y_n)|=+\infty$, contradiction.


问:
You're given $51$ natural numbers between $[1,100]$. Show that there are two of these numbers whose sum is $101$.

You're given $51$ natural numbers between $[1,100]$. Show that there are two of these numbers whose sum is $101$.

This seems to be an application of the pigeonhole principle? I started considering different sets as the boxes, but this didn't quite lead anywhere. How should I approach this kind of problem?

Body: Hint: Consider pairs $(1,100),(2,99),\cdots ,(50,51).$


问:
Uniform convergence of the series $\sum_{n\ge 1}\frac{ne^{-nx}}{1+n^2}$
I'm triying to prove the uniform convergence of the series $$\sum_{n\geq 1} \frac{ne^{-nx}}{1+n^2}$$ for $x\in (0,\infty)$, I tried to use the M-test many times, but only can acomplish this for $x\in[a,\infty)$, $a>0$. Can be an error in the exercise?

答:
If the series was uniformly convergent on $(0,+\infty)$ then for each $\varepsilon>0$ there would be an $N$ such that
$$
 \sum\limits_{n = N}^\infty  \frac{ne^{ - nx} }{1 + n^2 }  < \varepsilon
$$
for all $x>0$. Note however, that for $N\geq 1$
with $x=1/N$,
$$
\sum\limits_{n = N}^\infty  {\frac{{ne^{ - n\frac{1}{N}} }}{{1 + n^2 }}}  > \sum\limits_{n = N}^{2N} {\frac{{ne^{ - n\frac{1}{N}} }}{{1 + n^2 }}}  > e^{ - 2} \sum\limits_{n = N}^{2N} {\frac{n}{{1 + n^2 }}}  > e^{ - 2} \frac{{2N^2 }}{{1 + (2N)^2 }} > 0.05.
$$
Thus, if $\varepsilon<0.05$, there cannot be such a positive $N$.


问:
Infinite Limit Involving a Factorial
$$\lim_{x\to\infty} \left(\frac{x+1}{(x!)^{\frac{1}{x-1}}}\right) $$

I'm pretty sure that this approaches infinity, yet I'm having trouble proving it. Hope someone can help, thanks.

答:
As by Stirling's Approximation for  $\lim_{n\to\infty},$

$n!\approx\sqrt{2\pi}\left(\dfrac{n^{n+\dfrac12}}{e^n}\right)$

$$\lim_{x\to\infty}\dfrac{x+1}{(x!)^{\dfrac1{x-1}}}$$

$$=\dfrac{\lim_{x\to\infty}(x+1)\lim_{x\to\infty}e^{\dfrac1{1-\dfrac1x}}}{\lim_{x\to\infty}(\sqrt{2\pi})^{\dfrac1{x-1}} \lim_{x\to\infty}x^{1+\frac1{2x}}}$$

$$=\dfrac{e\cdot\lim_{x\to\infty}\left(1+\dfrac1x\right)}{\lim_{x\to\infty}x^{1/2x}}$$

Use this  $\lim_{x\to\infty}x^{\dfrac1x}=1$

Title: Infinite Limit Involving a Factorial


问:
Is there an integral domain containing an element having no factorization into irreducible elements?
I know that the ring of algebraic integers can be an answer because this ring has no irreducible elements.

So, is there an integral domain with nonempty set of irreducible elements and an element having no factorization into irreducibles?

答: Consider $R=\{P \in \mathbb{Q}[X],\, P(0) \in \mathbb{Z}\}$. $R$ has irreducible elements, such as $X^2+1$. However, $X$ has no factorisation into a product of irreducible elements. Indeed, any factorisation of $X$ is of the form $n \cdot \frac{X}{n}$, and $\frac{X}{n}$ is never irreducible (it can be written as $\frac{X}{(n+1)!} \frac{(n+1)!}{n}$ and none of the factors are units).


问:
Show that $P\left(\bigcap_{i=1}^n E_i^c\right) \le e^{-\sum\limits_{i=1}^n P(E_i)}$ given $(E_i)$ are independent

$E_1,\ldots,E_n$ are independent. We want to show that $$P\left(\bigcap_{i=1}^n E_i^c\right) \le e^{-\sum\limits_{i=1}^n P(E_i)}$$
More generally, for a countable sequence $(E_i)_{1}^\infty$ of independent events, show that
$$P\left(\bigcap_{i=1}^\infty E_i^c\right) \le e^{-\sum\limits_{i=1}^\infty P(E_i)}$$

I'm trying to use Boole's inequality and the fact that $e^{-x} \ge 1-x$ for all $x\in\mathbb R$.

To use Boole's inequality, I first invoked De-Morgan's Law to do the following:
$$P\left(\bigcap_{i=1}^n E_i^c\right) = P\left(\left(\bigcup_{i=1}^n E_i\right)^c\right) = 1 - P\left(\bigcup_{i=1}^n E_i\right)$$

Then I tried to put $x = P\left(\bigcup_{i=1}^n E_i\right)$  and use $e^{-x} \ge 1-x$, but sadly I didn't get to the desired inequality. What am I missing? I did note that $$P\left(\bigcup_{i=1}^n E_i\right) \le \sum_{i=1}^nP\left(E_i\right)$$ but it didn't produce the right inequality.

Once I'm able to prove it for $n$, putting $n\to\infty$ should do the rest of the job. Since probability measures are continuous, we know we can switch $P$ with $\lim_{n\to\infty}$.

Thanks for the help!

答: Independence tells us that $$P\left(\bigcap_{i=1}^n E_i^c\right) = \prod_{i=1}^n P(E_i^c) = \prod_{i=1}^n (1 - P(E_i)),$$ which is the expression to which you should start applying all your inequalities.


问:
Let $(A,B) \in S_n(\mathbb R)^2$ $\forall X \ne 0, (AX|X) \ge 0 $ Show that $ sp(AB) \subset \mathbb R $
Let $(A,B) \in S_n(\mathbb R)^2$
$\forall X \ne 0, (AX|X) \ge 0 $
Show that $ sp(AB) \subset \mathbb R $

My work

I took an eigen value of $AB$ to show that it was equal to its conjugate

To do that I considered the product $ \bar{X}^T ABX$ with X the vector associated to the value $\lambda $

It is on the one hand equal to $\lambda \bar{X}^T X$ and on the other hand to $(BA\bar{X})^T X$. However I am blocked here. I don’t know where to use the inner product from here.

答:
$A$ is positive semidefinite so it has a positive semidefinite square root $A^{1/2}$. We get

$$\sigma(AB) =\sigma(A^{1/2}A^{1/2}B) = \sigma(A^{1/2}BA^{1/2}) \subseteq \Bbb{R}$$

since $A^{1/2}BA^{1/2}$ is symmetric.


问:
Calculating $E(1+M_t)^2\int_0^t M_s^2 ds$ and $E(1+M_t)^2\int_0^t M_s^2 dM_s$
Let $ M= (M_t)_{t\geq 0}$ be a Standard Brownian Motion. I would like to calculate

$$E\left[(1+M_t)^2\int_0^t M_s^2 ds\right]$$ and $$E\left[(1+M_t)^2\int_0^t M_s^2 dM_s\right].$$

For the first one, I tried to use Fubini's theorem, but I get a strange result. I think I'm calculating wrongly
$$
E\int_0^t M_t^2M_s^2 d s
$$

For the first one, I use Itô's formula:
$$\int_0^t M_s^2 dM_s = \frac{1}{3}M_t^3 -\int_0^t M_s ds,$$
ant then again Fubini...

Am I doing it right? Is there other way? Thanks

答:
You seem to be doing it correctly. In the first case, you can use the Tonelli theorem rather than the Fubini theorem, because the integrand of the double-integral $(1+M_t)^2M_s^2$ is non-negative a.s., therefore we may interchange integrals, and the first one equals :
$$
E\left[(1+M_t)^2\int_0^t M_s^2 ds\right] = \int_0^t E\left[(1+M_t^2)M_s^2\right]ds
$$

To evaluate the inner integral is a standard exercise in independence and stationarity of increments, as mentioned in a comment : indeed, \begin{align}
(1+M_t^2)M_s^2 & = M_s^2 + ((M_t-M_s)+M_s)^2M_s^2 \\& = M_s^2 + (M_t-M_s)^2M_s^2 + M_s^4 + 2(M_t-M_s)M_s^3
\end{align}

and therefore by linearity of expectations :
\begin{align}
E[(1+M_t^2)M_s^2] & = \color{blue}{E[M_s^2]} + \color{green}{E[(M_t-M_s)^2M_s^2]} + \color{red}{E[M_s^4]} + \color{fuchsia}{E[2(M_t-M_s)M_s^3]} \\
\\ & = \color{blue}{s} + \color{green}{E[(M_t-M_s)^2]E[M_s^2]} + \color{red}{E[N(0,s)^4]} +  \color{fuchsia}{E[2(M_t-M_s)]E[M_s^3]} \\
&= \color{blue}{s} + \color{green}{(t-s)s} + \color{red}{3s^2}+\color{fuchsia}{0}
\\ &= \underline{\mathbf{s + ts + 2s^2}}
\end{align}

and now $\int_0^t (s+ts+2s^2)ds$ is easy to calculate.

Key takeaway : Apply Fubini while looking out for Tonelli (which makes things so much easier!), and use independence of increments for taking expectations of polynomials of Brownian motion evaluated at various points.

For the second one, you have to apply the Ito formula as you have done, in order to simplify $\int_0^t M_s^2 dM_s$, with $f(x) = \frac{x^3}{3}$. (The use of the formula is justified since the initial process is Brownian motion, obviously an Ito process, and $f$ is smooth)
\begin{align}
\frac{M_s^3}3 = \int_0^t M_s^2dM_s + \int_0^t M_sds
\end{align}

and therefore $\int_0^t M_s^2 dM_s = \frac{M_s^3}{3} - \int_0^t M_sds$. Whereupon the second question simplifies :
$$
E\left[(1+M_t^2)\int_0^t M_s^2dM_s\right] = E\left[(1+M_t^2)\left(\frac{M_s^3}{3} - \int_0^t M_sds\right)\right]\\
= E\left[(1+M_t^2)\frac{M_s^3}{3}\right] - E\left[(1+M_t^2)\int_0^t M_sds\right]$$

Now, for the first one, write $M_t = (M_t-M_s)+M_s$, expand out and use independence and stationarity of increments.

For the second one, we need to use Fubini, because a non-negative integrand is not present. However, it is not difficult to show that $$\int_0^t E\left[|(1+M_t^2)||M_s|\right]ds < \infty$$

using independence of increments : $$(1+M_t^2)M_s = M_s + M_s(M_t-M_s)^2 + (M_s)^3 + 2(M_s)^2(M_t-M_s)$$ and therefore by the triangle inequality and independence $$E\left[|(1+M_t^2)||M_s|\right] \leq E[|M_s|] + E[|M_s|]E[|M_t-M_s|^2]+E[|M_s|^3] + 2E[|M_s^2|]E[|M_t-M_s|] $$

Now, one can use scaling, reduce to the standard normal and show that the RHS is a polynomial in $s$ and $t$, whose integral from $0$ to $t$ is then obviously finite.

Once we use Fubini, then again independence of increments, as shown in part $1$, completes the answer.

On other ways to evaluate these : I am not so sure. The stage is still a very basic one, in the sense that we aren't using anything beyond the definition of the Brownian motion, and the Ito formula. I thought some kind of Stochastic Fubini might work for the second part (it is a theorem that allows an exchange of an Ito and a Lebesgue integral) but I think it overcomplicates the situation, so doesn't really count.


问:
Order of the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$
Back to easier problems for a bit...

I have been told that it is possible to find the order of the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ without considering automorphisms of $\mathbb{Z}_{720}$.  Here's what I have:

We have that $\mathrm{Aut}(\mathbb{Z}_{720}) \cong U(720)$.  Then since $720=2^4 \cdot 3^2 \cdot 5$ we have $U(720)=U(16) \oplus U(9) \oplus U(5)$.  Some simple computations show that $U(16)$ is not cyclic, while $U(9)$ and $U(5)$ are, so $U(720) \cong U(16) \oplus \mathbb{Z}_6 \oplus \mathbb{Z}_4$.  Since $\gcd(6,4) \ne 1$, $\mathbb{Z}_6 \oplus \mathbb{Z}_4$ is not cyclic, so this, coupled with the fact that the noncyclic $U(16)$ of order 8 would have a cyclic subgroup of order 4 at the most, gives us that the largest cyclic subgroup of $\mathrm{Aut}(\mathbb{Z}_{720})$ is of order $6$.

Is this correct?

I didn't look at automorphisms of $\mathrm{Aut}(\mathbb{Z}_{720})$ per se, but I did consider elements of $\mathrm{Aut}(\mathbb{Z}_{720})$ up to isomorphism...  Is there a way to derive this result just from looking at the structure of $\mathbb{Z}_{720}$?

Thanks.

答: Once you have found the structure $\def\Z{\Bbb Z}\Z_4\oplus\Z_2$, you can apply the structure theorem for fintely generated Abelian groups (or more elementarily a few applications ot the Chinese remainder theorem) to conclude that $(720)\cong\Z_2\oplus\Z_2\oplus\Z_4\oplus\Z_{12}$. But if you only need to find the order of the largest cyclic subgroup, it is easier to use that the order of an element $g$ in a direct product of groups $G_i$ is the least common multiple of the orders of its components $g_i\in G_i$. In particular if the groups $G_i$ are Abelian (so that the set of occurring orders is closed under least common multiples) the maximal possible order is the least common multiple of the maximal orders in the groups $G_i$. For $U(720)$ the latter number is $\operatorname{lcm}(4,6,4)=12$. </p>


问:
Cauchy's Functional Equation $ f ( x + y ) = f ( x ) + f ( y ) $ with the Additional Assumption $ f \left( x ^ { n + 1 } \right) = x ^ n f ( x ) $

Assume that $ n $ is a given positive integer. Determine all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f ( x + y ) = f ( x ) + f ( y ) \tag 0 \label 0 $$
forall $ x , y \in \mathbb R $, and
$$ f \left( x ^ { n + 1 } \right) = x ^ n f ( x ) \tag 1 \label 1 $$
for all $ x \in \mathbb R $.

I've recently encountered two posts (here and here) where the problem was somehow similar to the above one; one was in fact equivalent to the above question for $ n = 3 $, and the other one was similar to the case $ n = 1 $. While those posts asked for solutions with additional regularity assumptions like continuity and differentiability, I found out that \eqref{0} and \eqref{1} themselves were enough for showing that all solutions are of the form $ f ( x ) = a x $ for some constant $ a \in \mathbb R $, and posted answers to them not using the additional regularity conditions.

Let's consider the case $ n = 2 $. Letting $ a = f ( 1 ) $ and using \eqref{1} together with \eqref{0} and its well-known consequences, we have
$$ \left( x ^ 2 + 2 x + 1 \right) \big( f ( x ) + a \big) = ( x + 1 ) ^ 2 f ( x + 1 ) = f \left( ( x + 1 ) ^ 3 \right) \\
= f \left( x ^ 3 \right) + f \left( 3 x ^ 2 \right) + f ( 3 x ) + f ( 1 ) = 3 f \left( x ^ 2 \right) + \left( x ^ 2 + 3 \right) f ( x ) + a \text . \tag 2 \label 2 $$
Similarly, by calculating $ f \left( ( x - 1 ) ^ 3 \right) $ in two different ways, we get
$$ \left( x ^ 2 - 2 x + 1 \right) \big( f ( x ) - a \big) = - 3 f \left( x ^ 2 \right) + \left( x ^ 2 + 3 \right) f ( x ) - a \text . \tag 3 \label 3 $$
Adding \eqref{2} and \eqref{3} (and dividing by $ 2 $) we get
$$ \left( x ^ 2 + 1 \right) f ( x ) + 2 a x = \left( x ^ 2 + 3 \right) f ( x ) \text , $$
which shows that $ f ( x ) = a x $.

For $ n = 1 $, the problem can be solved more easily, as we only need to consider $ f \left( ( x + 1 ) ^ 2 \right) $ (see my answer to the related post). For $ n = 3 $, the "$ f \left( ( x \pm 1 ) ^ 2 \right) $ technique" above gives us an equation containing $ f \left( x ^ 2 \right) $ and $ f ( x ) $, and then we can use this equation and continue similar to the case of $ n = 1 $ (see my answer to the related post).

It seems that as \eqref{1} takes the multiplicative structure of $ \mathbb R $ into account in a very specific way, it could be true than for any $ n $ the solutions are linear. But the problem with my above approach for $ n \ge 4 $ is that using the expressions for $ f \left( ( x \pm 1 ) ^ { n + 1 } \right) $, there will be more than one term of the form $ f ( x ^ m ) $ ($ m > 1 $) in the equation, and I don't know how to handle that. Is there any other way to get around this problem? Or is there in fact a nonlinear solution for larger $ n $?

答:
The general thing of what you're asking is the following:

If $f$ is a Cauchy function and $f(x^n)=x^kf(x^{n-k})$ for positive integers $n,k$, then $f$ is linear.

The proof is as follows:

Put $x\rightarrow x+q$ where $q$ is an arbitrary rational number. Then let's expand $f((x+q)^n)$:

First of all (although it is clear) note that:

For a Cauchy function $f$, $f(qx)=qf(x)$ and $f(q)=qf(1)$ hold for all rational numbers $q$.

$$f((x+q)^n)=(x+q)^kf((x+q)^{n-k})$$

$$f((x+q)^n)=f\bigg(\displaystyle\sum_{t=0}^{n}{\binom{n}{t}\cdot x^tq^{n-t}}\bigg)=\displaystyle\sum_{t=0}^{n}{\binom{n}{t}\cdot q^{n-t}f(x^t)}$$

On the other hand:

$$(x+q)^kf((x+q)^{n-k})=\bigg(\displaystyle\sum_{i=0}^{k}{\binom{k}{i}\cdot x^iq^{k-i}}\bigg)\bigg(\displaystyle\sum_{j=0}^{n-k}{\binom{n-k}{j}\cdot x^jq^{n-k-j}}\bigg)$$

So finally:

$$\displaystyle\sum_{t=0}^{n}{\binom{n}{t}\cdot q^{n-t}f(x^t)}=\bigg(\displaystyle\sum_{i=0}^{k}{\binom{k}{i}\cdot x^iq^{k-i}}\bigg)\bigg(\displaystyle\sum_{j=0}^{n-k}{\binom{n-k}{j}\cdot q^{n-k-j}f(x^j)}\bigg)$$

Now assume these two as polynomials on $q$. Because these two polynomials are equal to each other for each rational number $q$, they are also equal for all real values $q$. Hence, the coefficients are equal in both sides.

By checking the coefficients of $q^{n-1}$ we'll have:

$$nq^{n-1}f(x)=q^k(n-k)q^{n-k-1}f(x)+kxq^{k-1}\cdot q^{n-k}f(1)$$

By simplifying this:

$$nf(x)=(n-k)f(x)+kxf(1)$$

And finally $f(x)=xf(1)$ for all $x\in\mathbb{R}$.

P.S. The following is also true (the proof is very much like this so please do it yourself:

If $f$ is a Cauchy function and $f(x^n)=x^kf(x)^{n-k}$ for positive integers $k,n$, then $f$ is linear.


问:
Does the following lower bound improve on $I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)}$, where $q^k n^2$ is an odd perfect number?
Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Define the abundancy index
$$I(x)=\frac{\sigma(x)}{x}$$
where $\sigma(x)$ is the classical sum of divisors of $x$.

Since $q$ is prime, we have the bounds
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1},$$
which implies, since $N$ is perfect, that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$

By considering the negative product
$$\bigg(I(q^k) - \frac{2(q-1)}{q}\bigg)\bigg(I(n^2) - \frac{2(q-1)}{q}\bigg) < 0,$$
since we obviously have
$$\frac{q}{q-1} < \frac{2(q-1)}{q},$$
then after some routine algebraic manipulations, we arrive at the lower bound
$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q-1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$

Now, a recent MO post improves on the lower bound for $I(n^2)$, as follows:
$$I(n^2) > \bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg)$$

Added February 3, 2021 - 8:10 PM (Manila time) Here is a quick way to show the improved lower bound for $I(n^2)$.  We have
$$I(n^2)=\frac{2}{I(q^k)}=\frac{2q^k (q - 1)}{q^{k+1} - 1}=\frac{2q^{k+1} (q - 1)}{q(q^{k+1} - 1)}=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1}}{q^{k+1} - 1}\bigg)$$
$$=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{(q^{k+1} - 1) + 1}{q^{k+1} - 1}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1} - 1}\bigg)$$
$$>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)$$
since $q^{k+1} > q^{k+1} - 1$.  But, of course, we obtain
$$I(n^2)>\bigg(\frac{2(q-1)}{q}\bigg)\bigg(1 + \frac{1}{q^{k+1}}\bigg)=\bigg(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\bigg).$$
QED.

Repeating the same procedure as above, we have the negative product
$$\Bigg(I(q^k) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg)\Bigg(I(n^2) - \left(\frac{2(q-1)}{q}\bigg)\bigg(\frac{q^{k+1} + 1}{q^{k+1}}\right)\Bigg) < 0.$$

This implies, after some algebraic manipulations, that
$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}.$$

But WolframAlpha says that the partial fraction decomposition of the new lower bound is given by
$$\frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}} = \frac{3q^2 - 4q + 2}{q(q - 1)} + \frac{2(q - 1)}{q^{k+2}} - \frac{q}{(q - 1)(q^{k+1} + 1)}.$$

So essentially, my question boils down to:

QUESTION: Is it possible to produce an unconditional proof (that is, for all $k \geq 1$ and for all special primes $q \geq 5$) for the following inequality?
$$\frac{2(q - 1)}{q^{k+2}} > \frac{q}{(q - 1)(q^{k+1} + 1)}$$

MY ATTEMPT

I tried to ask WolframAlpha for a plot of the above inequality, it gave me the following GIF image:

So it does appear that the inequality is unconditionally true, which would mean that the new lower bound for $I(q^k) + I(n^2)$ improves on the old.  Is it possible to prove this analytically?

And lastly:  Based on this answer to a closely related question, since we appear to have obtained an improved lower bound for $I(q^k) + I(n^2)$, can we then say that there is indeed an integer $a$ such that $k \leq a$?

答:
Suppose to the contrary that there exists an integer $k \geq 1$ and a (special) prime $q \geq 5$ such that
$$\frac{q}{(q-1)(q^{k+1}+1)} \geq \frac{2(q-1)}{q^{k+2}}.$$
This inequality is equivalent to
$$q^{k+3} \geq 2(q-1)^2 (q^{k+1}+1) = 2q^{k+1} - 4q^{k+2} + 2q^{k+3} + 2q^2 - 4q + 2,$$
which in turn is equivalent to
$$0 \geq q^{k+3} - 4q^{k+2} + 2q^{k+1} + 2q^2 - 4q + 2 = q^{k+2} (q - 4) + 2q^{k+1} + 2q(q - 2) + 2 > 0,$$
a contradiction.

We therefore conclude that
$$\frac{2(q-1)}{q^{k+2}} > \frac{q}{(q-1)(q^{k+1}+1)}$$
for all integers $k \geq 1$ and all (special) primes $q \geq 5$.

Hence, the new lower bound
$$I(q^k) + I(n^2) > \frac{q^{k+2}}{(q - 1)(q^{k+1} + 1)} + \frac{2(q-1)(q^{k+1} + 1)}{q^{k+2}}$$
does indeed improve on the old lower bound
$$I(q^k) + I(n^2) > 3 - \frac{q-2}{q(q - 1)} = \frac{3q^2 - 4q + 2}{q(q - 1)}.$$

It remains to be seen whether this implies that there does exist an integer $a$ such that $k \leq a$, if $q^k n^2$ is an odd perfect number with special prime $q$.


问:
What construction gives the syntax of an algebraic structure?
Let's say I have some algebraic structure $M$ (let's use a monoid for simplicity). I have a specific subset $S \subset M$. The free monoid on $S$ encodes what elements of $M$ I can get by combining elements of $S$.

I'm interested not only on what elements of $M$ I can get, but also on how I can get them. In this case the set of binary trees with elements of $S$ at their leaves.

So, If the Free construction gives me the semantics, what construction gives me the syntax?

答: In this particular case you could just look at the free magma (non-associative object with binary operation) on $S$. In general contained in any algebraic theory $\mathcal T$ is a subtheory $\mathcal T’$ with the same operations but no axioms, and you could look at the free $\mathcal T’$-object on a subobject.


问:
Evaluate $\int \arcsin \left(\sqrt{x}\right)dx$.

Evaluate:
$$\int \arcsin \left(\sqrt{x}\right)dx$$

I am really stuck here. I know that I have to use $u-$substitution and set $u = \sqrt{x}$, to get $\left(2\int \arcsin \left(u\right)\cdot \:udu \right)$ but how can I continue from here? Could I use integration by parts and use $u = \arcsin(u), v'=u$?

答: Yes, you can use integration by parts:\begin{align}\int\arcsin(u)u\,\mathrm du&=\frac12\arcsin(u)u^2-\frac12\int\frac{u^2}{\sqrt{1-u^2}}\,\mathrm du\\&=\frac12\arcsin(u)u^2-\frac12\left(\int\frac{u^2-1}{\sqrt{1-u^2}}\,\mathrm du+\int\frac1{\sqrt{1-u^2}}\,\mathrm du\right)\\&=\frac12\arcsin(u)u^2+\frac12\int\sqrt{1-u^2}\,\mathrm du-\frac12\arcsin(u).\end{align}And now you can compute $\int\sqrt{1-u^2}\,\mathrm du$ doing $u=\sin(\theta)$ and $\mathrm du=\cos(\theta)\,\mathrm d\theta$.


问:
Number of different ways you can place positive integers into an array of size N such that the product of the integers is K
To find the number of ways to place positive integers into an array of size $N$ such that the product of the numbers is $K$. So the problem boils down to finding the numbers of ways to place the prime factors of $k$ into $N$ slots. For Example, if $N = 2$ and $K = 24$

Prime Factors of $24$:  $2^3 \times 3^1$

So we have to find the number of ways of placing three 2's and one 3 in 2 slots. I saw the problem in the discussion forums and came to know that this can be done with the stars and bars concept. It is stated that the number of ways of placing three 2's and one 3 in 2 slots is equal to placing three 2's in 2 slots multiplied by placing one 3 in 2 slots.

placing three 2's in 2 slots (stars and bars) = $\binom{3+2-1}{3}$ = $\binom{4}{3}$

placing one 3 in 2 slots (stars and bars) = $\binom{1+2-1}{1}$ = $\binom{2}{1}$

So the total number of ways = $ \binom{4}{3} $ * $\binom{2}{1}$ = 4 * 2 = 8

The parts which I didn't understand is

how the number of ways of placing three 2's and one 3 in 2 slots is equal to placing three 2's in 2 slots multiplied by placing one 3 in 2 slots

why can't we apply the stars and bars concept to all of the prime factors like placing 4 (three 2's and one 3) into 2 slots.

答:
Because these choices are independent. You can divide the three twos as either $(2\times 2\times 2,-)$, $(2\times 2, 2)$, $(2, 2\times 2)$ or $(-, 2\times2\times 2)$. You can put the three either as $(3,-)$ or $(-,3)$. Now you can take any of the options for twos and combine it with any of the options for three, to get $4\times 2$ possibilities. E.g., taking the first option for each gives $(2\times2\times 2\times 3,1)$ (since there are no factors of $2$ or $3$ in the second slot), whereas taking the third and second gives $(2,2\times2\times3)$.

If you place four things in two slots, there are $5$ ways to do this, but this doesn't distinguish which things go where, just how many. So splitting as three things in the first slot is just counted as one way by this method, but actually this could correspond to $(2\times2\times2, 3)$ or $(2\times 2\times3,2)$, which are different.


问:
Manipulating the triangle inequality to compare two sides
With $\mathbf x$ and $\mathbf y$ two vectors, the triangle inequality can be written as:

$$
|\mathbf x|-|\mathbf y| \le |\mathbf x + \mathbf y|\le |\mathbf x|+|\mathbf y| \tag{A}
$$

which can be broken down into a system of inequalities (labelled as (1) and (2)):

\begin{equation}
\begin{cases}
|\mathbf x|-|\mathbf y| \le |\mathbf x + \mathbf y| \\
|\mathbf x + \mathbf y|\le |\mathbf x|+|\mathbf y|
\end{cases}
\end{equation}

then, with the assumption of $|\mathbf y|>0$, (1)-(2) yields:

$$
|\mathbf x| \le |\mathbf x + \mathbf y| \tag{B}
$$

But is this in fact true? Is (B) a direct consequence of (A) when $|\mathbf y|>0$? or is (B) is invalid because (1)-(2) is not allowed as it is not obvious if the subtraction preserves the inequality?

This question arose during my attempt to this problem, where I build triangles from 3 points on a line in order to try and show which point is the last one (to no avail so far).

答: (B) does not follow from (1)-(2). For a very simple counter-example, work in the 1-dimensional space of real numbers and take x=3, y=-2.


答:
Substitute $y=\arcsin\sqrt x$. Then, $dx= \sin 2y \>dy$ and
$$\int \arcsin \sqrt{x}\>dx
=\int y \sin 2y \>dy \overset {IBP}= -\frac12 y \cos 2y +\frac14\sin2y+C
$$

Title: Evaluate $\int \arcsin \left(\sqrt{x}\right)dx$.

Body:
Evaluate:
$$\int \arcsin \left(\sqrt{x}\right)dx$$

I am really stuck here. I know that I have to use $u-$substitution and set $u = \sqrt{x}$, to get $\left(2\int \arcsin \left(u\right)\cdot \:udu \right)$ but how can I continue from here? Could I use integration by parts and use $u = \arcsin(u), v'=u$?


问:
How to calculate $\int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta$
How to calculate:

$$ \int_0^{2\pi} \sqrt{1 - \sin^2 \theta}\;\mathrm d\theta $$

答:
\begin{align}
\int_0^{2\pi} \sqrt{1 - \sin^2 \theta} d\theta &= \int_0^{2\pi} \sqrt{\cos^2 \theta} d\theta \\ &= \int_0^{2\pi} | \cos \theta | d\theta
\\ &= 4 \int_0^{\frac{pi}{4}} \cos \theta d\theta
\\ &= 4
\end{align}</p>


问:
Geometrical proof for property of chord of contact with a circle
Let their be a circle with center O, and radius r. Let their be a point P outside the circle at a distance d from O. Let a variable line passing through P cut the circle in 2 points, A and B. We define a point C between A and B such that PC is harmonic mean of PA and PB. Prove that the locus of C is chord of contact of P with respect to the circle.

I can do it through coordinate geometry, I just wanted to find a geometrical method. I tried looking at its inversion, and trying to find relations but I couldn't.
Any help is appreciated. Thanks.

答:
Another approach. For simplicity let p has distance 2R from the center of circle with center at O and radius R. If PC is tangent at C We have A, B and C coincident and we may write:

$PC=PA=PB$

For any position of A and B we have:

$PC^2=PA\times PB$

$PC=\frac {PA\times PB}{PC}=\frac {2\times PB\times PA }{2 PC=PA+PB}$

Now we evaluate the position of point H which is the foot of altitude of right angled triangle PCO.

$CP^2=OP^2-CO^2=(2R)^2-R^2=3R^2$

We have area of this triangle as:

$S=\frac{PC \times CO}2=\frac{R^2\sqrt 3}2$

$CH=\frac{R^2\sqrt 3}{2\times 2R}=\frac{R\sqrt 3}4$

In triangle PCO we can write:

$CH^2=PH\times OH$

$PH(OP-PH)=\frac{3R^2}4$

OR:

$2R\times PH-PH^2=\frac{3R^2}4$

$4PH^2+3R^2-8PH\times R=0$

Solving this equation for PH we get:

$PH=\frac {3R}2=\frac{6R^2}{4R}=\frac{2(R)\cdot (3R)}{R+3R}$

Or:

$PH=\frac{2 PA\times PB}{PA+PB}$

Which means H has similar particularity as C and CH is part of chord CC'.


问:
Is this how we write radian
When $\theta$ = Arc Length / radius of circle.

Then it is theta = metre/metre .Then , we say theta has the unit radians.

So why is m/m radian?

答:
SI system has seven base units plus two supplementary units: radian and steradian. The latter two don't add to the dimensions, and are dimensionless.

Okay, angle is dimensionless quantity, so why should we assign an dimensionless unit to it? Simply because, if we do not, then we won't be able to distinguish between simple numbers and angles.

Angles are different from normal numbers, and hence need a distinction. This can be done by some symbol or so. This symbol is itself designated as an unit.

Hope this helps. Ask anything if not clear :)

Title: Is this how we write radian


问:
Is my claim true about the maximum and minimum of the given function?
I have this three-variable function
$$f(x,y,z)=x^9 \cos (2 x y+x z)+\left(5 x^4+1\right) \cos (x y+x z)+\left(5 x^4+3\right)\\ \qquad+\left(4 x^3+1\right) \cos (x y+2 x z)+x^3 \cos (x z)+\left(2 x^2+7\right) \cos (2 x y+2 x z)\\ \qquad+\left(x^2+5\right) \cos (x y)+\left(2 x^2+7\right) \cos (2 x z)\\+\left(3 x^7+5 x^2+1\right) \cos (x y-x z)+(x+1) \cos (2 x y)$$
for $x,y,z>0$.

Now, since all the coefficients of the functions $\cos$ are positive, can I claim that the inequality $f_{-1}\leq f(x,y,z)\leq f_{+1}$ holds, where $f_{\pm1}$ are the function $f(x,y,z)$ when all the $\cos$ are $+1$ and $-1$?

答: Yes, considering that the variables only take positive values, setting all the co-sines to 1 will provide an upper bound and setting all the co-sines equal to -1 will provide a lower bound.</p>


问:
Guides/tutorials to learn abstract algebra?
I recently read up a bit on symmetry groups and was interested by how they apply to even the Rubik's cube. I'm also intrigued by how group theory helps prove that "polynomials of degree $\gt4$ are not generally solvable".
I love set theory and stuff, but I'd like to learn something else of a similar type. Learning about groups, rings, fields and what-have-you seems like an obvious choice.

Could anyone recommend any informal guides to abstract algebra that are written in (at least moderately) comprehensible language? (PDFs etc. would also be nice)

答:
You can also follow Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996.I am sure this will be of great help to start with.

Also follow some useful recommendation here.


问:
Help with a revolution solid
Suppose I have an area in the cartesian system formed by the $y$ axis and a given function $y=f(x)$. How do I evaluate the volume of the solid formed by completely revolving this area around the $y$ axis (given a known interval $x\in[a,b]$)? Assuming that you can't invert the function to a form like $x=f(y)$. Thank you in advance.

答:
As described in the Wikipedia Solid of revolution article, the volume of the solid formed by rotating the area between the $y$ axis and $y = f(x)$ and lines $x = a$ and $x = b$ is
$$V = 2 \pi \int_a^b x \lvert f(x) \rvert dx$$

For $f(x) \ge 0$, this is called shell integration,
$$V = 2 \pi \int_a^b x \; f(x) \; dx$$

Title: Help with a revolution solid


问:
Inequality $\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} \leq \sqrt{3}$.
Recently I answered this Question which required a proof for the upper and lower bound of a given function. While seemingly a calculus question, I realised that it can be transformed into an inequality problem, and my solution is detailed there. However, it seems that we can improve upon the upper bound substantially, from $2$ to $\sqrt{3}$. This leads us to the following claim:

Let $x,y,z \in \mathbb{R^+},xyz=1$. Prove that $\dfrac{1}{\sqrt{1+2x}} + \dfrac{1}{\sqrt{1+2y}} + \dfrac{1}{\sqrt{1+2z}} \leq \sqrt{3}$.

Equivalently, by the substitution $x=\dfrac{b}{a}, y=\dfrac{c}{b}, z=\dfrac{a}{c}$, it suffices for us to prove that for all $a,b,c \in \mathbb{R^+}$, $\sqrt{\dfrac{a}{a+2b}} + \sqrt{\dfrac{b}{b+2c}} + \sqrt{\dfrac{c}{c+2a}} \leq \sqrt{3}. $ But after thinking for a while, I still have no clue on how to proceed.

Any help will be much appreciated.

答:
Your inequality doesn't true because the maximal value of k for which the inequality
$$\dfrac{1}{\sqrt{1+kx}} + \dfrac{1}{\sqrt{1+ky}} + \dfrac{1}{\sqrt{1+kz}} \leqslant \frac{3}{\sqrt{k+1}},$$
is true for all positive real numbers $x,\,y,\,z$ satisfy $xyz=1$ is $k = \frac54.$


问:
worked conservative question. Can someone double check?
Can you please check this question for me because my answer is different to my friends and they say it's wrong...

Given the equation: $G = (x^3 -3xy^2)\vec{i} +(y^3-3x^2y)\vec{j} +2\vec{k}$ is conservative, find the potential function for the fields that are conservative.

My answer
$f: \nabla f=F \\
fx=x^3-3xy^2 \\
fy=y^3-3x^2y \\
fz=z$

$\int(x^3-3xy^2)dx
=x^4/4-3x^2y^2/2 +h(y, z) \\
fy \text{ or } f'y= -3x^2 y +hy(y, z) \\
fy=y^3-3x^2y= -3x^2y+hy(y, z) \\
hy(y,z)=y^3$

$\int y^3 dy = y^4/4 +gz$
$fz=0$ because no $z$ term ^
$\int 0dz =c$
$F= x^4/4-3x^2y^2/2 +y^4/4+c$ <-potential function
my other friend solution was $F=x^4/4 -3x^2y^2/2 +z^2/2+y^4/4$.

答: If we accept that the vector field $F = (x^3 -3xy^2)\vec{i} +(y^3-3x^2y)\vec{j} +2\vec{k}$ is conservative, so there is a function, say $f$ such that, $\nabla f=F$. So  $$f_x=x^3-3xy^2,~~f_y=y^3-3x^2y,~~f_z=z$$ Assume that $f_x=x^3-3xy^2$, so $$f(x,y,z)=\frac{x^4}4-\frac{3}2x^2y^2+h(y,z)$$ Then $f_y=-3x^2y+h_y(y,z)$ but $f_y=y^3-3x^2y$, so we get $$-3x^2y+h_y(y,z)=y^3-3x^2y\Longrightarrow h_y(y,z)=y^3\to h(y,z)=\frac{y^4}4+g(z)$$ Therefore, we have up to now: $$f(x,y,z)=\frac{x^4}4-\frac{3}2x^2y^2+\frac{y^4}4+g(z)$$ Now take a differentiation with respect to $z$, we have: $$f_z=g'(z)$$ but $f_z=z$, so $g'(z)=z$ and then $g(z)=\frac{z^2}2$. Hence $$f(x,y,z)=\frac{x^4}4-\frac{3}2x^2y^2+\frac{y^4}4+\frac{z^2}2$$</p>


问:
Is $\lnot\lnot\forall x(A(x)\vee\neg A(x))$ a theorem of intuitionistic logic?
I have a short question. I know that $\neg\neg(A\vee\neg A)$ is a theorem of intuitionistic logic. What about $\neg\neg\forall x(A(x)\vee\neg A(x))$?

答:
No, it is not. To prove this, we can give a countermodel (I'll use Kripke semantics).

For a world $v$ in a Kripke model $M$, the meaning of $M,v\Vdash\lnot\lnot\forall x(A(x)\lor \lnot A(x))$ is: for all worlds $v'\geq  v$, there exists a world $v''\geq v'$ such that for all worlds $w\geq v''$ and all $x\in M_{w}$, $A(x)$ holds at $w$ or for all worlds $w'\geq w$, $A(x)$ does not hold  at $w'$.

In other words, there is a dense set above $v$ of worlds $v''$ with the property that the truth value of $A(x)$ does not ever "flip" for any element of any world above $v''$.

For our countermodel, take the frame to be $\mathbb{N}$ with its usual order, $M_n = \{0,\dots , n\}$, and interpret $A$ in $M_n$ as $M_n\setminus\{n\}$. Then $M,0\not\Vdash\lnot\lnot\forall x(A(x)\lor \lnot A(x))$, since for any world $n$, the truth value of $A(n)$ "flips" when moving from $n$ to $n+1$.


问:
Is this operator a finite rank operator on $L^2$?
Is the operator $T: L^2([0,1]) \rightarrow L^2([0,1])$ defined as:
$$Tu(t) = t u(t)$$
a finite rank operator?

More in general: how can I understand if an operator $T$ on $L^2$ is a finite rank operator? Thanks!

答:
Let $u_n(t):=t^{n-1}$ for $ n \in \mathbb N.$ Then $u_n \in  L^2([0,1])$ for all $n$ and

$$Tu_n=u_{n+1}.$$ This shows that the functions $t,t^2,t^3,...$ are contained in $im(T).$

But $t,t^2,t^3,...$ are linearly independent in $L^2([0,1])$ , thus $T$ is not a finite rank operator.


问:
What rules were applied to get this solution when differentiating
I've come across this derivation that looks fairly simple yet I cannot get it right.We are given solutions but I dont understand it.The function looks like this

$$f(x) = \cos y \space \ln x + y\sqrt{x^2 +3} $$

Now since $\cos y$ and $\ln x$ are multiplied and $y$ and the square term as well than we use the multiplication rule. The given solution looks like this

$$ -\sin y*y' \ln x + \cos y \frac 1x + y'\sqrt{x^2 +3} + y\space  \frac 1 2 \space \frac {1}{x^2 +3}\space 2x$$

The part with the first multiplication makes sence, we derivate sin and ln x, we leave y as it is since x is the differation variable but how does one get to  $$ \frac 1 2 \space \frac {1} {x^2 +3}\space 2x  $$ from the root?  If we would to get ride of the root by making it $$(x^2 + 3)^{1/2} $$ than we would have to use the power rule, which is bring 1/2 infront than we have to substract 1 from 1/2 so that is -1/2. But how does the 2x appear? Im aware that the $x^2 + 3$ under the root were also derivated but why?I thought the chain rule only applied to multiplication.A litle insight would be great,thanks!

答:
There's a typo in the book.

$$\frac{d}{dx}(y\sqrt{x^2+3}) = y'\sqrt{x^2+3} + \frac{y}{2} \frac{1}{\color{blue}{\sqrt{x^2+3}}}2x$$

Now for the derivative of $\sqrt{x^2+3}$

$$\frac{d}{dx}\sqrt{x^2+3} = \left[\frac{d}{d(\color{red}{x^2})} \sqrt{\color{red}{x^2}+3}\right]\times\left[\frac{d(\color{green}{x^2})}{d\color{green}x}\right] = \frac{1}{2\sqrt{x^2+3}}\times (2x)$$


问:
Are the following function 1st derivatives continuous at (0,0)?
Apologies: I've had to rephrase this question quite a few times

Suppose the following function:
$$f(x,y)=\frac{2x^3y^2}{x^4+y^2}$$
$$f(0,0)=0$$
Show it is differentiable/not differentiable at $(0,0)$

I attempted this question first by applying the definition of differentiability which yielded the following expression:
$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to(0,0)}\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}$$

If the above equals 0, then the function is infact differentiable at $(0,0)$. I can't seem to find a counterexample which proves that it is discontinuous so I attempted to prove its continuity by applying the $\epsilon-\delta$ definition of limits. Unfortunately, no matter whether I express the function in terms of polarcoordinates or not, I cannot bound the following
$$|\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}| < \epsilon$$
for all $\epsilon$.

I'd be greatly appreciative for a solution to this kind of problem!

答:
Credits to Kavi Rama Murthy for pointing out the $|2x^2y|<x^4+y^2$ inequality for me.

Note that:

$(x^2-y)^2 \geq 0 \implies x^4 - 2x^2y + y^2 \geq 0 \implies x^4 + y^2 \geq 2x^2y$

$(x-y)^2 \geq 0 \implies x^2 - 2xy + y^2 \geq 0 \implies x^2 + y^2 \geq 2xy \implies x^2+y^2 \geq xy \implies \sqrt{x^2+y^2} \geq \sqrt{xy}$

So it turns out the function is differentiable at (0,0). If $0 < \sqrt{x^2+y^2} < \delta$ then:

$$|\frac{2x^3y^2}{(x^4+y^2)\sqrt{x^2+y^2}}| = |\frac{(2x^2y)(\sqrt{xy})(\sqrt{xy})}{(x^4+y^2)\sqrt{x^2+y^2}}| = |\frac{2x^2y}{x^4+y^2}||\frac{\sqrt{xy}}{\sqrt{x^2+y^2}}||\sqrt{xy}| \leq (1)(1)\sqrt{xy} < \delta = \epsilon
$$

Thus the definition of limits is satisfied therfore $f(x,y)$ is differentiable at $(0,0)$


问:
Topological understanding of real Analysis
I am almost done with my introductory class in Real Analysis for my undergrad and I am having a bit of difficulty grasping the idea of Basic Topology in R and applying it in Real Analysis. I studied alongside my course using 'Understanding Analysis 2ed' by Abott, but I am still lost... For instance I get what Open, Closed and Compact sets are, but when it comes to using it for proofs and in relation to continuity, I have no idea how to use the concepts at all. How did you guys approach it? Are there maybe online lectures that are good?

Thanks!

答:
The way I see it you have two options :

Forget about the word topology for the time being. Understand open and closed sets in $\mathbb{R}$ (or $\mathbb{R}^d$) via the definition of open balls, that is a set $A\subset\mathbb{R}^d$ is open if for all $x\in \mathbb{R}^d$ there exists $r>0$ such that $B_r(x) \subset A$, where $B_r(x)$ is the open ball of radius $r$ around $x$

$$
B_r(x):=\{y\in\mathbb{R}^d~:~|x-y|<r\}.
$$

Take the definition of a continuous function as the "$\epsilon,\delta$" definition which you will of learnt in your Analysis course. Continue along using these definitions / concepts, until you realise you need topology, then take a course in it.

If you have some spare time learn topology now. If you want to just get a quick idea of the point of it I recommend this fantastic $<1.5$ hour lecture : https://www.youtube.com/watch?v=7G4SqIboeig&list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_&index=1&ab_channel=TheWE-HeraeusInternationalWinterSchoolonGravityandLight . In short : given a set $M$ we can define a topology $\mathcal{O}$ on $M$, which must satisfy three conditions (look them up). Once this CHOICE of $\mathcal{O}$ is made we call $(M,\mathcal{O})$ a topological space. The elements of $\mathcal{O}$ are subsets of $M$ and are called the open sets of $M$.

So the notion of open sets on $\mathbb{R}$ that you first learn in Analysis is ONE way of viewing open sets, this coincides with equipping $\mathbb{R}$ with whats called the Standard Topology, sometimes written $\mathcal{O}_{standerd}$.

Note people frequently work with the standard topology and hence don't explicitly say that they are working with it.

The whole point of topology is to add additional structure to your space which allows you to talk about continuity. Since the continuity of a function is directly related to the notions of open and closed sets we need to state which topologies we are working with. A function may be continuous with respect to one topology but not continuous with respect to another topology! Look up the topological definition of continuity.


问:
Sum of edge weights in cycles of a directed graph
I am studying for an exam in graph theory and algorithms, and am stuck on the following problem:

Let $G=(V,E)$ be a directed graph with edge weights $c:E(G) \rightarrow \mathbb{R}$.

Show that the following two statements are equivalent:

(1) $c$ is conservative, that is, there is no cycle with negative weight in $(G,c)$.

(2) There is a function $\pi:V(G) \rightarrow \mathbb{R}$ such that the relation $c(e) + \pi(v) - \pi(w) \geq 0 $ holds for all edges $ e = (v,w) \in E(G)$.

My attempted solution so far: Assuming (2) is true, look at any cycle in $G$, let the cycle have $k$ edges and $k$ unique vertices. We label the edges $1,...,k$ and the vertices $1,...,k$, such that $e_i = (v_i,v_{i+1})$ for $1 \leq i \leq k-1$ and $e_k = (v_k,v_1)$. (2) then yields

$$\pi_1 + c_1 \geq \pi_2$$
$$\pi_2 + c_2 \geq \pi_3$$
$$...$$
$$\pi_{k} + c_k \geq \pi_1.$$

Combining all these, we get $\pi_1 + c_1 + c_2 +...+c_k \geq \pi_1$, so the sum of weights in any cycle must be greater than or equal to zero, implying (1). So if I did this right, (2) $\Rightarrow$ (1) should be ok.
This is where I am stuck: how do I prove (1) $\Rightarrow$ (2), or $\neg$(2) $\Rightarrow$ $\neg$(1)? (2) has to hold for all edges, so I cannot just look at cycles. I am unsure how to approach this. A second part of the problem concerns proving that a suitable $\pi$ can be computed in $O(nm)$ time complexity, with $n$ the number of vertices in $G$ and $m$ the number of edges. I sense that if I can grasp the first part, that second part would also be within grasp.

答:
Your solution to the first part is correct. As for the second part, let's add a new vertex $s$ to $G$, along with a directed edge $sv$ with zero cost for all $v \in V$. (*) For each vertex $v$, let $\pi(v)$ be the length of a shortest $s-v$ path in $G$. You should try showing that this $\pi$ has the desired property; more precisely, if $vw$ was an edge with $c(vw) + \pi(v) < \pi(w)$, then a walk consisting of a shortest $s-v$ path and then the edge $vw$ would contain a negative cycle.

Such a function $\pi$ is sometimes called a feasible potential (with respect to $c$). We can define another cost function $c'$ on $G$ by letting $c'(vw) = c(vw) + \pi(v) - \pi(w)$. Then the condition on $\pi$ means exactly that $c'$ is non-negative. The interesting observation is that for any walk in $G$ starting with some vertex $u$ and ending with $v$, the cost of the walk according to $c'$ is the cost according to $c$ plus $\pi(v) - \pi(u)$. In other words, the cost of every $u-v$ walk changes by the same amount. This means that shortest paths according to $c$ are the same as the shortest paths according to $c'$!

Using this observation, we can find the shortest path between all pairs of vertices in $G$ by first computing a feasible potential and then using Dijkstra's algorithm with the cost function $c'$ for each source vertex $v$ to find shortest paths starting from $v$. This is called Johnson's algorithm. Note that this is faster than running the Bellman-Ford algorithm for each source vertex (and if the graph is sparse, then it is even faster than the Floyd-Warshall algorithm).

(*) Instead of this, we could just take a source vertex from each connected component (in the undirected sense) of $G$ and do the same thing component-wise, but I think this is slightly cleaner.


问:
Combining two pdf
I would appreciate some insights on this question I've had this past days.

I'll try to phrase the problem as best as I can. I'll try to make some sense of it by considering a sales context.

Consider some sales time horizon $t\in \left [ 0,\infty  \right ]$. The sales over time $X$ for some product given some condition $y$ ( day of the week for example ) follows a clearly defined pdf.
Let's imagine that the sales over time $X$ for the same product given a condition $z$ ( weather ) also follows a clearly defined pdf.

If we have $p_{X}(t|y)$ and $p_{X}(t|z)$ can we build a third pdf i.e. $p_{X}(t|y,z)$ using this two pdf. This combined pdf would give us the probability of purchase for some day of the week and some specific weather condition ?

For simplicity we can assume that $p_{X}(t|y)$ and $p_{X}(t|z)$ belong to the same family i.e. they are both normal or binomial distributions or whatever they may be but they have an analytical form.

I've tried to explore a solution with the Bayes Theorem but I don't seem to get anywhere.

答:
There are different ways to construct the third pdf; it may not be unique.

Just need to make sure $\int{f_{X|Y,Z}(t|y,z)f_Z(z)dz}=f_{X|Y}(t|y)$ and $\int{f_{X|Y,Z}(t|y,z)f_Y(y)dy}=f_{X|Z}(t|z)$.

Title: Combining two pdf


问:
A paradox? Or a wrong definition?
Let $A$ be a commutative ring with $1 \neq 0$. Then writing $V(1) = V((1))$, we have $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \sqrt{(1)} = (1)$.

But then $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \{x : x \in \mathfrak{p} \text{ for all prime ideals } \mathfrak{p} \supseteq (1)\} \supseteq \text{Set}$, where the last containment is justified since there is no prime ideal that contains $(1) = A$, so $x$ can be anything by vacuous truth.

I had a similar confusion like this in my topology course, but the answer that I heard from my teacher was that "of course, $A$ is the universe that we consider," so $\bigcap_{\mathfrak{p} \in V(1)}\mathfrak{p} = \{x \in A : x \in \mathfrak{p} \text{ for all prime ideals } \mathfrak{p} \supseteq (1)\} = (1)$.

But then why "pure" definition from (naive) set theory fails and forces us to consider $A$ as the ambient space?

答:
Consider the defining formula for the class $\bigcap\cal S$, where $\cal S$ is a collection of sets:

$$\bigcap\mathcal S=\{x\mid\forall S\in\mathcal S, x\in S\}$$

When $\cal S=\varnothing$, the assumption is satisfied by every object in the universe. And as we know, the collection of all objects in the universe is not a set. Therefore $\bigcap\varnothing$ is not a set, it doesn't "exist" in the sense that we want it to exist.

But if we agree that the ambient space is $A$ itself, our ring, then this is just all the objects which are in $A$, that is to say $A$ itself. Since $A$ exists, it is a set, it's fine.

When we agree on an ambient space, we actually say that whatever formula is used to pick whatever elements, they will always be members of that ambient space. So if we agree that $A$ is our ambient space, and $\cal S$ is a collection sets, then we actually write:

$$\bigcap\mathcal S=\{x\in A\mid\forall S\in\mathcal S, x\in S\}$$

Now the empty intersection would be $A$ itself, which we assume is a set.

Also see:

intersection of the empty set and vacuous truth

Unary intersection of the empty set

Title: A paradox? Or a wrong definition?


问:
Proving each subsequent term equal to one another in an equation
Q. The maximum value of $\cos x_1\cdot \cos x_2 \cdot \cos x_3...\cos x_n$ under restrictions $0\le x_1,x_2,x_3,...,x_n \le \frac{\pi}{2}$ and $\cot x_1 \cdot \cot x_2 \cdot \cot x_3...\cot x_n=1$.

MY ATTEMPT:

From question, we get $$\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n=\sin x_1 \cdot \sin x_2 \cdot \sin x_3...\sin x_n ...(1)$$
Using AM,GM inequality, we get:
$$\frac{\cos x_1+\cos x_2+\cos x_3+...+\cos x_n}{n}=\sqrt[n]{\cos x_1\cdot \cos x_2 \cdot \cos x_3...\cos x_n}\,\,\,...(2)$$
$$\frac{\sin x_1 +\sin x_2+\sin x_3+...+\sin x_n}{n}=\sqrt[n]{\sin x_1 \cdot \sin x_2 \cdot \sin x_3...\sin x_n}\,\,\,...(3)$$

The RHS term in equations $(2)\,\,,(3)$ are equal by using equation $(1)$.
So we get:
$$\cos x_1+\cos x_2+\cos x_3+...+\cos x_n=\sin x_1 +\sin x_2+\sin x_3+...+\sin x_n$$
Now i notice that if i somehow prove that $\sin x_1=\cos x_1,\,\,\sin x_2=\cos x_2,...,\,\,\sin x_n=\cos x_n$, so i get every angle as '$\frac\pi 4$'
$$\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n=\biggl(\frac{1}{\sqrt 2}\biggr)^n=\biggl(\frac{1}{ 2^\frac{n}{2}}\biggr)$$;
Which is the answer. But i am unsure how to prove such thing or is it even possible with how much i have attempted. Also if possible, give any other solution to the question which is shorter/interesting.

答:
I have another solution to the question, if anyone is interested:

We know from question:
$$\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n=\sin x_1 \cdot \sin x_2 \cdot \sin x_3...\sin x_n ...(1)$$
Let $$x=\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n$$
$$x^2=(\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n)^2$$
$$x^2=(\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n)(\cos x_1 \cdot \cos x_2 \cdot \cos x_3...\cos x_n)$$
$$x^2=(\cos x_1 \cdot \sin x_1 \cdot \cos x_2 \cdot\sin x_2\cdot \cos x_3 \cdot \sin x_3 ...\cos x_n \cdot \sin x_n)$$
$$x^2=\frac{1}{2^n}(\sin 2x_1 \cdot \sin 2x_2 ...\sin 2x_n)$$
Maximum value of $(\sin 2x_1 \cdot \sin 2x_2 ...\sin 2x_n)=1$
$$x=\biggl(\frac{1}{ 2^\frac{n}{2}}\biggr)$$


问:
Questions on doubling map D(x)
For the doubling map D on [0, 1):
$$D(x)=\begin{cases}
      2x & 0\leq x< \frac{1}{2} \\
      2x-1 & \frac{1}{2}\leq x <1
\end{cases}$$

List all points whose orbits end up landing on 0 and are thereby
eventually fixed.

Let x ∈ [0, 1) and suppose that x is given in binary form as $a_0, a_1, a_2 ...$ where each $a_j$ is either 0 or 1. First give a formula for the binary representation of D(x). Then explain why this causes orbits
of D generated by a computer to end up eventually fixed at 0.

I have no idea how to solve this can anyone provide any hints?

答:
Can you guess how the binary representation behaves under the doubling map?
Recall that $x \in [0,1)$ has binary representation $a_1,a_2,a_3,\dots$ (I drop the $a_0$ term for a more convenient notation) if and only if
$$ x = \sum_{k \geq 1} a_k 2^{-k}.$$
Note that $x < 1/2$ if and only if$^*$ $a_1 = 0$ (and, hence, $x \geq 1/2$ if and only if $a_1 = 1$).
Thus, if $x = (a_1,a_2,\dots)$, then what is $D(x)$ in binary?
This can then be used to tackle both questions.
Try it yourself first and only check the spoiler below if you get stuck.

$^*$Here, I implicitly assume that we take the finite representation if both a finite and an infinite one exist (as is the case for $x=1/2$).

 Solution: Plugging the binary representation into the formula for the doubling map shows that $D(a_1,a_2,\dots) = (a_2,a_3,\dots)$.
 So, the doubling map shifts the sequence to the left, always dropping the first entry.
 In particular, if a point $x$ admits a finite binary representation, i.e. where $a_k = 0$ for all $k \geq n$, then $D^n(x) = 0$.
 Conversely, if $x$ does not admit a finite binary representation, then it admits an infinite binary representation in which there is a non-zero entry $a_{n_k}$ with $n_k \geq k$ for all $k \in \mathbb{N}$.
 In this case, $D^{n_k}(x) = (a_{n_k},a_{n_k+1},\dots) \ne 0$ for all $k \in \mathbb{N}$, so $x$ does not eventually get mapped to 0.
 Lastly, of course, a computer always works with finite representations.

Title: Questions on doubling map D(x)


问:
newton's method on $\frac{1}{x}$
I was trying to use newton's method to approximate $\frac{1}{x}$, by finding the solution to $\frac{1}{x}=0$,I got $$x_{k+1}=x_k-\frac{\frac{1}{x_k}-n}
{-\frac{1}{x_k^2}} = 2x_k-nx_k^2$$
I was trying to find where it is convergent to the root, $0$ and $1$ both don't work, and I did not manage to find any other candidate.

can any of you help me find where it converges?

I'm pretty sure this isn't a very efficient way to approximate it, are there any other better ways?

答:
You have
$$x_{k+1}-\frac{1}{n} = -n(x_k^2-2\frac{1}{n}x_k + \frac{1}{n^2}) =-n(x_k-\frac{1}{n})^2$$
So, for all $k\in \mathbb{N}$
$$x_k-\frac{1}{n} = (-1)^k n^k (x_0-\frac{1}{n})^{2k}$$

The sequence converges if and only if the initial value of the sequence $x_0$ satisfies the following condition:
$$\sqrt{n}|x_0-\frac{1}{n}|<1$$
or
$$\frac{1}{n} - \frac{1}{\sqrt{n}}< x_0 < \frac{1}{n} + \frac{1}{\sqrt{n}}$$

Title: newton's method on $\frac{1}{x}$


问:
Is this how we write radian
When $\theta$ = Arc Length / radius of circle.

Then it is theta = metre/metre .Then , we say theta has the unit radians.

So why is m/m radian?

答:
"Only angles have a symbol. Not like refractive index which have no unit".

I see.

In optics, the refractive index (also known as refraction index or
index of refraction) of a material is a dimensionless number that
describes how fast light travels through the material. It is defined
as

$ n=\frac {c}{v},\ $ where $c$ is the
speed of light in vacuum and $v$ is the phase velocity of light in the
medium.

We could create a unit for refractive index, like $refrix$. If so, we would say, "the refractive index of light in Perspex is $1.52\ refrix$."

But we don't need to. We can just say, "the refractive index of light in Perspex is $1.52$". And it's unambiguous what we mean here, because there's no other commonly-utilised English term for "ratio of the speed of light in a vacuum and the phase velocity of light through the medium".

However, unfortunately we have two different commonly-utilised ratios that describe an angle between two straight lines that meet at a point (First, extend the lines so that they have the same length. Then draw the circle). One is:

$$\frac{\text{arc length}}{\text{radius}}.$$

The other is $$\frac{\text{arc length}}{\text{radius}} \times \frac{180}{\pi}.$$

We must distinguish between these two ratios because they are different and they are both commonly-used, therefore we give them different names. If we didn't give different names, then it wouldn't be clear which one we are referring to/talking about.

Title: Is this how we write radian


问:
Inconsistent notation fro Del-Operator
It has recently come to my attention that the operations carried out with a Del operator do not match their notation.

That is, the divergence or curl are written as

$\nabla \cdot \vec{v}$ and $\nabla \times \vec{v}$.

The Del operator is a vector operator, and has been expressed as above given that it outputs divergence and curl in Cartesian coordinates; however, this notation is also used in curvilinear coordinates. This troubles me because that is not really the case for curvilinear coordinates (if i carry out the dot product or the cross product literally I will get the wrong answer), yet we still use vector products notation and we are not carrying out those operations at all but rather carrying out integrals.

Can I carry out a dot-product in cylindrical coordinates and arrive to the correct form of the divergence? Or the vector notation does not represent the actual divergence and curl, and it is just customary.

Thank you for your time and answers.

答:
No, it is still the dot product, even in other coordinate systems. The inner product is actually a $(0,2)$ tensor, and thus is coordinate invariant. Just because $(a^1,a^2)\boldsymbol{\cdot}(b^1,b^2)=a^1b^1+a^2b^2$ in the standard basis does not mean that this is how we compute the dot product in other bases. E.g, for two vectors with components $(u^1,u^2)~;~(v^1,v^2)$ with respect to the polar coordinate system, the dot product is
$$(u^1,u^2)\boldsymbol{\cdot}(v^1,v^2)=u^1 v^1 \cos(u^2-v^2)$$
There are similar constructions for the cross product.

Unfortunately, there is no easy formula to take the dot product between vectors expressed via some arbitrary coordinate system without first having to convert back to the standard basis.

Title: Inconsistent notation fro Del-Operator


问:
For what complex $z$ does $f(z) = \frac12((-1+i\sqrt{3})z+(-2\sqrt{3}-18i))$ represent a rotation around $c$?

The function
$$f(z) = \frac{(-1 + i \sqrt{3}) z + (-2 \sqrt{3} - 18i)}{2}$$
represents a rotation around some complex number $c$. Find $c$.

I don't know how to proceed, help would be appreciated.

Thank you :)

答: Hint: A rotation around $c$ is given by $(z-c)w+c$. Compare this is $f(z)=az+b$ to find $w$ and $c$.


问:
Show that $\bigcup_{n=1}^{\infty} (0, 2+\frac{1}{n}] = (0, 3]$
Show that: $$ \bigcup_{n=1}^{\infty}  (0, 2+ \frac{1}{n}] = (0, 3]$$

The above statement seems quite obvious but i'm unsure how one would prove it.

So far i've done this:

$A_n := (0, 2+ \frac{1}{n}]$  for $n \in N$

$A_{n+1} \subset A_n $    $ ,\forall$ $ n \in N$

So $\bigcup_{n=1}^{\infty} A_n = A_1 = (0, 3] $

I'm not sure if this is the way i'm meant to answer the question.

答: Yes your proof is perfect. Considering that each $A_{n+1} \subset A_{n}$ for all natural numbers, the largest 'parent' set that we can observe from this union is when we can choose the smallest possible natural number: $n=1$. Thus $A_{1}=\bigcup_{n=1}^{\infty} A_n = (0,3]$.


问:
Show that equation has no roots $ \in \mathbb{Z} $
Given $ P(x) := 7x^5-23x^2 +2x -8,$ show that $ P(x) $ has no roots in $ \mathbb{Z}. $

My (primitive solution);

Assume that it exists atleast one root in $ \mathbb{Z}. $ We can then write $ 7x^5 -23x^2+2x = x(7x^4-23x+2)=8.$

$ 8 $ is then a product of 2 factors, and both must divide $8.$ We have the set of $ D_8 = \{\pm 1,\pm2,\pm4,\pm8 \}$. So we can test for every $ x \in D_8 $ and see that none of them yields 8.

Questions;

Is this a solid proof and are there any other way to show it? And does "Integral root" mean a root in $ \mathbb{Z} $?

答:
Yes, integral root means an integer root.

The question has many answers already at this site, so let me give you just a comment, how to solve such questions in general: there is also the so-called Rational Root Theorem, which makes it easy to verify that $p(x)$ has no integral and no rational root. By the way, $P(x)$ is even irreducible over $\Bbb Q$.


答:
You are quite right to assume that Arzela Ascoli would help you here. The statement goes as follows:

$L_{A,B}$ is compact, if it is closed and satisfies the two conditions of equicontinuity and pointwise boundedness.

Inside the definition of $L_{A,B}$ you are given two conditions - one similar to continuity, one similar to boundedness. You might want to try using the first to show equicontinuity and the second to show pointwise boundedness.

It remains to show that $L$ is closed. What would happen is you have a sequence of functions $(f_n)_{n\in\mathbb{N}}\subseteq L_{A,B}$ that converges? Can you show the fact that the limit function $f$ is also inside $L_{A,B}$? Can you show that $f$ satisfies both conditions?

Title: Is the set $L_{A,B}=\{f\in C([0,1]): |f(x)-f(y)|\leq A(x)|x-y|,\ |f(x)|\leq B(x),\ x,y\in [0,1]\}$ compact?

Body:
Givene $A,B\colon [0,1]\to [0,\infty)$ , is the set $L_{A,B}=\{f\in C([0,1]): |f(x)-f(y)|\leq A(x)|x-y|,\ |f(x)|\leq B(x),\ x,y\in [0,1]\}$ compact??
Note $(C([0,1]),d_\infty)$ when $d_\infty(f,g)=\sup \{|f(x)-g(x)|: x\in [0,1]\}$ Arzela Ascoli I guess?


问:
Guides/tutorials to learn abstract algebra?
I recently read up a bit on symmetry groups and was interested by how they apply to even the Rubik's cube. I'm also intrigued by how group theory helps prove that "polynomials of degree $\gt4$ are not generally solvable".
I love set theory and stuff, but I'd like to learn something else of a similar type. Learning about groups, rings, fields and what-have-you seems like an obvious choice.

Could anyone recommend any informal guides to abstract algebra that are written in (at least moderately) comprehensible language? (PDFs etc. would also be nice)

答:
I really think that Isaacs book Algebra: A Graduate Course introduces the group theory in detail without omitting any proof. It may sound difficult because of the adjective "Graduate" but I do not think that the explanations are that difficult to follow for undergraduates as long as they know how to write proofs.

The best freebies for algebra in my opinion is Milne's website (http://www.jmilne.org/math/). Not every note is complete, but his excellent notes tell you which books to buy to corroborate them.


问:
Hartshorne Example 5.2.2
Let $X = \operatorname{Spec} A$, $Y \subseteq X$ closed subscheme defined by ideal $I \unlhd A$, and let $i : Y \rightarrow X$ denote the inclusion morphism.

Then, Hartshorne claims that $i_{*}\mathcal{O}_{Y} \cong (A/I)^{\sim}$, the sheaf associated to the  $A$-module $A/I$.

I am having trouble proving this rigorously.

Firstly, as $Y$ is a closed subscheme, it follows $i^{\sharp} : \mathcal{O}_{X} \rightarrow i_{*}\mathcal{O}_{Y}$ is surjective. Therefore, we have

$i_{*}\mathcal{O}_{Y} \cong \mathcal{O}_{X} /\mathcal{I} $

where $\mathcal{I} :=$ ker $i^{\sharp}$.

Hence, I think a good idea is to prove that
$\mathcal{O}_{X} /\mathcal{I} \cong (A/I)^{\sim}$.

But I have not been able to come up with any argument that I find convincing. For example, I looked at the surjective map

$\mathcal{O}_{X} \rightarrow (A/I)^{\sim} $

and tried to prove its kernel is $\mathcal{I}$, but I couldn't come up with a rigorous answer.

Maybe my problem is that I don't really understand what the map $(i,i^{\sharp})$ is?

Any help would be much appreciated!

答:
Following the discussion in the comment section, I will try to answer my own question. My exposition may not be fully rigorous, but I think it contains the right ideas following hints given in the comments.

We have

$(Y,\mathcal{O}_{Y}) =$ $(Spec A/I, \mathcal{O}_{Spec A/I }$).

Recall,

$\mathcal{O}_{Y}(U) = \{s : U \rightarrow \coprod_{p \in U} (A/I)_{p} : \cdots  \}$ (such that "locally looks like fractions").

Also, note that

$i_{*}\mathcal{O}_{Y}(U) = \mathcal{O}_{Y}(U \cap Y) = \{s : U' \rightarrow \coprod_{p \in U'} (A/I)_{p} : \cdots  \}$

where $U' := U \cap Y \subseteq Y$ is open in $Y$.

Then, note that for principle open $D(f) \subseteq X$,

$i_{*}\mathcal{O}_{Y}(D(f)) = \mathcal{O}_{Y}(D(f)\cap Y) = \mathcal{O}_{Spec A/I}(D([f]))$.

But,

$\mathcal{O}_{Spec A/I}(D([f])) \cong (A/I)^{\sim}(D(f))$

via

$(A/I)^{\sim}(D(f)) \rightarrow \mathcal{O}_{Spec A/I}(D([f])) $

$s \mapsto s'$

where if $\pi : A \rightarrow A/I$ denotes the projection map, then

$s'(\pi(p)) := s(p)$.

Thus, this isomorphism on the principle open sets yields an isomorphism of the sheaves.

Title: Hartshorne Example 5.2.2


问:
$L^2$ product of three functions two of which are orthogonal
Let $f,g,h \in L^2(0,1) \cap L^\infty(0,1)$. Suppose that $f,g$ are orthogonal in $L^2$.
Is it true that $$\int_{0}^1 fgh \ dx = 0 \ ?$$ I would say yes because
$$|\int_{0}^1 fgh \ dx| \le \Vert h \Vert_\infty| \int_0^1 fg \ dx| = 0, $$
but I feel that I'm missing something

答: I think it is false just choose $f(x)=1$ , $g(x)=1$ when $x\leq 1/2$ and $g(x)=-1$ when $x>1/2$ , so clearly they are orthogonal in $L^2$ and choosing $h=g$ or some other function would do the job


答:
The absolute values in your inequality are misplaced (outside the integral rather than inside). The version with absolute values outside the integral can be shown to be false with the functions
$$\begin{align}
f(x)&=1\\
g(x)&=\sin(2\pi x)\\
h(x)&=\sin(2\pi x)
\end{align}$$
as then $f$ and $g$ are orthogonal. Thus the RHS of the inequality is $0$, but the LHS is positive as it is an integral of a positive function over a set of positive measure.

This example is then a counterexample to your claim.

Title: $L^2$ product of three functions two of which are orthogonal

Body:
Let $f,g,h \in L^2(0,1) \cap L^\infty(0,1)$. Suppose that $f,g$ are orthogonal in $L^2$.
Is it true that $$\int_{0}^1 fgh \ dx = 0 \ ?$$ I would say yes because
$$|\int_{0}^1 fgh \ dx| \le \Vert h \Vert_\infty| \int_0^1 fg \ dx| = 0, $$
but I feel that I'm missing something


问:
Finding an example of a bounded sequence in a complete metric space such that the sequence has no partial limit
I'm working through an analysis text and I've come across this exercise:

Give an example of a complete metric space $X$ and a bounded sequence $\left(x_{n}\right)$ in $X$ such that the sequence $\left(x_{n}\right)$ has no partial limit.

The reason that I'm stuck with this one is because I'm shaky about what it means for a sequence to be bounded. Does this mean that for any positive integers $n$ and $m$, the set of all possible $|x_{n}-x_{m}|$ is a bounded set of real numbers?

If so, I'm not sure how to relate the boundedness of a sequence to the completeness of a metric space.

I know that a metric space $X$ is said to be complete if every Cauchy sequence in $X$ converges to a point in $X$... what does this have to do with bounded sequences?!? Thanks in advance for any help.

EDIT: I'm going to add a couple of definitions.

A sequence is said to be bounded if there exists a constant $C\gt 0$ such that for any positive integers $n$ and $m$ the inequality $|x_{n}-x_{m}|\lt C$ holds.

A point $x$ in a metric space $X$ is said to be a partial limit of a sequence $\left(x_{n}\right)$ if for every $\epsilon\gt 0$ there are infinitely many values of $n$ for which $x_{n}\in B\left(x,\epsilon\right)$.

答:
Fix any point $y_0\in X$, and define a sequence $(x_n)$ to be bounded if there exists a constant $C>0$ such that $d(x_n,y_0)\leq C$ for all $n$. This definition does not depend on $y_0$, since if you choose any other $y_1\in X$, then by the triangle inequality,
$$|d(x_n,y_0)-d(x_n,y_1)|\leq d(y_1,y_0),$$
where the right-hand-side is a constant, so the sequence $(x_n)$ is bounded "with respect to $y_0$" if and only if it is bounded "with respect to $y_1$".

Hint for the exercise: Consider a space with no limit points.


问:
Is this operator on $L^2$ compact?

Let $T: L^2([0,1]) \rightarrow L^2([0,1])$ be a linear and bounded
operator defined as: $$Tu(t) = \log(1+t)u(t)
 \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space
 t\in[0,1]$$

Is $T$ compact? Justify the answer.

Hi, I'm having a lot of issues trying to understand if an operator in $L^2$ is compact or not. May u help me by telling me what is the optimal procedure in order to understand if a given operator in $L^2$ is compact? Many thanks!

答:
There is no "optimal procedure", as there are lots of ways to express an operator.

One common way to tell if an operator is compact, is to show that it is a norm limit of finite-rank operators.

On the opposite direction, the spectrum of a compact operator is always countable.

For a multiplication operator like yours, it is an easy-to-establish fact that the spectrum of an operator of the form $u\longmapsto f\,u$ is the closure of the image of $f$. That means that, over $[0,1]$, whenever your operator is multiplication by a continuous function its spectrum will be uncountable.

Title: Is this operator on $L^2$ compact?


问:
Let $M$ be a maximal ideal of $R$, then $M+(x)$ is a maximal ideal of $R[x]$.
Let $M$ be a maximal ideal of $R$, then $M+(x)$ is a maximal ideal of $R[x]$.

(I forget where the exercise come from, so I'm not sure whether $R$ has an identity.)

My attempt is as follows:

Let $N$ be an ideal in $R[x]$ such that $M+(x) \subseteq N \subseteq R[x]$. Let $N' = \{a \in R : a = f(0) \mbox{ for some } f \in N\}$, then $N'$ is an ideal in $R$ such that $M \subseteq N' \subseteq R$.

Let $a, b \in N'$, then $a = f(0)$ and $b = g(0)$ for some $f, g \in N$. Let $r \in R (\subseteq R[x])$. Since $N$ is an ideal in $R[x]$, we have $f-g, rf$, and $fr$ in N. It follows from
$$ a-b = (f-g)(0), ra = (rf)(0), \mbox{ and } ar = (fr)(0) $$
that $a-b, ra$, and $ar$ in $N'$.
Let $a \in M$, then $a+x \in M+(x) \subseteq N$. So we obtain that $a \in N'$.

Since $M$ is a maximal ideal in $R$, we have $N' = M$ or $N' = R$.

Note that $N \subseteq N'+(x)$. If $N' = M$, then $N = M+(x)$.

But I cannot prove the another case. Can somebody help me?

Thank you.

答:
I forget where the exercise come from, so I'm not sure whether  has an identity.

Almost certainly identity is assumed, because if not it is not even immediately clear what $R[x]$ means. (If it means $\{\sum r_i x^i\mid r_i\in R \text{ (finite support)}\}$, then how can one even be sure if $x\in R[x]$?)

Let $N' = \{a \in R : a = f(0) \mbox{ for some } f \in N\}$, then $N'$ is an ideal in $R$ such that $M \subseteq N' \subseteq R$.

Uhh... aren't you already almost done right after this step?  By maximality of $M$, either $N'=R$, whence $N'+(x)=R+(x)=R[x]$ or else $N'=M$ whence $N'+(x)=M+(x)$.

What you did is equivalent to what most people would do: compute $R[x]/(M+(x))$.

By the second isomorphism theorem, $R[x]/(M+(x))\cong \frac{R[x]/(x)}{(M+(x))/(x)}\cong\frac{R}{M}$, so that $M$ is maximal in $R$ iff $M+(x)$ is maximal in $R[x]$.


问:
$\ln(a+b)\leq\ln(a)+\ln(b)$
Suppose $a,b>0$ are integers. Do we then have that
$$
\ln(a+b)\leq \ln(a)+\ln(b)?
$$

I think, since this is equivalent to $a+b\leq ab\Leftrightarrow 1\leq b-\frac{b}{a}$

it holds for $1<b<a$ only and, moroever, it is a strict inequality for these values.

答:
You are correct. This indeed fails if $a=1$ and $b\in\mathbb{N}$, as $\log(1)=0$, so the inequality becomes
$$\log(1+b)\leq\log(b)$$
which is clearly false. However, if we strengthen the premise to $a,b>1$, then this inequality holds.

Your strengthening to $1<b<a$ omits the equality case $a=b=2$, as
$$\log(2+2) = \log(2\cdot 2) = \log(2)+\log(2)$$

Title: $\ln(a+b)\leq\ln(a)+\ln(b)$


问:
Understanding this Poisson Queueing Process
Consider $N$ busses, each which breaks down independently at a rate of $\mu$.  Once a bus breaks down it is sent to a repair shop which can only repair one bus at a time and takes an exponentially distributed amount of time with parameter $\lambda$ to repair it.

I'm trying to find the equilibrium distribution of the number of busses in service.

Could some one explain to me how to set this up?  Am I supposed to do a Q matrix?  Is $\mu$ intended to be determinate, so that the first time around all buses break down at the same time, and then it takes an exponentially distributed amount of time with parameter $\lambda$ for the first bus to exit the repair shop?  Then that would mean it would take twice the random variable for the next one to exit, and so on, but then as soon as one exists it comes back in $\mu$ time.  Like I said I'm just lost as to how to set this up.

Thanks

答:
The first paragraph describes some transition rates on the space $\{0,1,\ldots,N\}$ for the number of busses in service, which are as follows:

For every $0\leqslant n\leqslant N$, the rate $n\to n-1$ is $\mu n$.

For every $0\leqslant n\leqslant N-1$, and the rate $n\to n+1$ is $\lambda$.

The rate $N\to N+1$ is $0$.

From here, one can determine the stationary distribution by the usual procedure.


问:
Let $\Gamma _g(f)=\int_{\mathbb R}fg$. Prove that $\|\Gamma _g\|=\|g\|_{L^{p'}}$.
Let $p>1$ and $p'$ s.t. $\frac{1}{p}+\frac{1}{p'}=1$. Let $g\in L^{p'}$ and $\Gamma _g\in(L^p)'$ defined by $$\Gamma _g(f)=\int_{\mathbb R}fg.$$
Define $$\|\Gamma _g\|:=\sup_{\|f\|_{L^p}=1}|\Gamma _g(f)|.$$
How can I prove that $\|\Gamma _g\|=\|g\|_{L^{p'}}$ ? If $p=2$, then I have that  that $|\Gamma _g(g/\|g\|_{L^{p'}})|=\|g\|_{L^{p'}}$ and thus $$\|\Gamma _g\|=\|g\|_{L^{p'}}.$$

But if $p>1$ and $p\neq 2$, I don't see how I can do because $g\notin L^p$ a priori. Any idea ?

答:
Hint

If $f=g^{\frac{p'}{p}}$, then $f\in L^p$ and $fg=g^{p'}$.


问:
Convergence of stochastic process
I'm stucked with this exercise. Hope someone can help me

$X_t = 3+a_t$ with $a_t \sim (0,\sigma^2)$. $a_t$ is iid.

$Y_n=\frac{1}{\sqrt{n+2}} \sum_{t=1}^{n} X_t $

Does $Y_n$ converge in distribution? What distribution?

答:
You have

\begin{align}
Y_n &= \frac{1}{\sqrt{n+2}}\sum_{i=1}^nX_i \\
    &= \frac{1}{\sqrt{n+2}} \sum_{i=1}^n(3 + a_i) \\
&= \frac{3n}{\sqrt{n+2}} + \frac{1}{\sqrt{n+2}}\sum_{i=1}^n a_i \\
\end{align}

Because $a_i$ are iid and follows a normal distribution $N(0,\sigma^2)$, then $\sum_{i=1}^n a_i$ follow the normal distribution $N(0,n\sigma^2)$. Hence, $Y_n$ follows the normal distribution
$$Y_n = N(\frac{3n}{\sqrt{n+2}} ,\frac{n}{n+2}\sigma^2 )$$
with mean of $\frac{3n}{\sqrt{n+2}}$ and variance $\frac{n}{n+2}\sigma^2$.

When $n \rightarrow +\infty$, $Y_n$ still has a normal distribution of variance $\sigma^2$ but its mean tends to $+\infty$.

Title: Convergence of stochastic process


Title: solve $\log(-0.7x-9)=1+5\log(5)$. I get $370$ whereas textbook solution says $-44655.7143$

Body:
I'm to solve $\log(-0.7x-9)=1+5\log(5)$ using a calculator. I arrived at $370$ when I'm told the solution is $-44655.7143$.

My working:

$$\log(-0.7x-9)=1+5\log(5)$$
$$\log(-0.7x-9)=1+\log(25)$$
$$\log(-0.7x-9)=\log(25)+\log(10)$$
$$\log(-0.7x-9)=\log(250)$$
$$-0.7x-9=250$$
$$-0.7x=259$$
$$x=\frac{259}{0.7}$$
$$x=370$$

Where did I go wrong, and how can I arrive at $-44655.7143$?

Body:
There are two errors in your calculations. First, $5^5 \ne 25$...And in the third line before the end you are missing a sign. It should be, using your steps,
$$\log(-0.7x-9)=1+5\log(5)$$
$$\log(-0.7x-9)=1+\log(3125)$$
$$\log(-0.7x-9)=\log(3125)+\log(10)$$
$$\log(-0.7x-9)=\log(31250)$$
$$-0.7x-9=31250$$
$$-0.7x=31259$$
$$x=-\frac{31259}{0.7}$$
$$x\approx - 44655.7 $$


问:
Manipulating the triangle inequality to compare two sides
With $\mathbf x$ and $\mathbf y$ two vectors, the triangle inequality can be written as:

$$
|\mathbf x|-|\mathbf y| \le |\mathbf x + \mathbf y|\le |\mathbf x|+|\mathbf y| \tag{A}
$$

which can be broken down into a system of inequalities (labelled as (1) and (2)):

\begin{equation}
\begin{cases}
|\mathbf x|-|\mathbf y| \le |\mathbf x + \mathbf y| \\
|\mathbf x + \mathbf y|\le |\mathbf x|+|\mathbf y|
\end{cases}
\end{equation}

then, with the assumption of $|\mathbf y|>0$, (1)-(2) yields:

$$
|\mathbf x| \le |\mathbf x + \mathbf y| \tag{B}
$$

But is this in fact true? Is (B) a direct consequence of (A) when $|\mathbf y|>0$? or is (B) is invalid because (1)-(2) is not allowed as it is not obvious if the subtraction preserves the inequality?

This question arose during my attempt to this problem, where I build triangles from 3 points on a line in order to try and show which point is the last one (to no avail so far).

答:
No, we cannot subtract inequalities in the following way:
$$
a \leq b\\
c\leq d\\
\implies a-c\leq b-d $$
For example, take $a=3,b=4,c=1,d=3$ to see that it's wrong. I think this is what you did to obtain what you have. So
$$|\mathbf x|\leq |\mathbf x+\mathbf y| $$
is wrong.
A counterexample in $\mathbb{R}^2$: $\mathbf x=(1,0),\mathbf y=(-1,0)$.


问:
How to find the kernel of $\pi$
Define a map $\pi : \mathbb{R}^2 \to \mathbb{R} $  by $\pi((x,y))=x$ .Prove that $\pi$ is a  homomorphism and  find the kernel  of $\pi$

My attempt :  To show that $\pi$ is homomorphism

Proof :  Take  $(a, b) , (c,d) \in \mathbb{R}^2$ ,  we have  $\pi \left((a,b) + (c,d)\right) = \pi(a+c,b+d) = a+c= \pi(a,b) + \pi(c,d) $ $\implies$ $\pi$ is homomorphism

Im  confused about the kernel i,e how to find the kernel  of $\pi?$

答:
Hint.

You need to find all $(x,y)\in\mathbb{R}^2$ such that $\pi((x,y))=0$.

Now it is time to review what a "kernel" means.

Notes.

Given a (group) homomorphism $f:G\to H$, the kernel of $f$ is all the elements $g$ in $G$ such that $f(g)=e_H$ where $e_H$ is the identity element of $H$.

In your problem:

$f=\pi$

$G=\mathbb{R}^2$

$H=\mathbb{R}$

$e_H=0$

Title: How to find the kernel of $\pi$


问:
How to prove $G(X)=Y$ has a solution $X$ for any $Y$ $\iff$ $A$ and $B$ have no common eigenvalues.

Let $A,B\in \Bbb C^{n\times n}$ and $G:\Bbb C^{n\times n} \to \Bbb C^{n\times n}$ defined by $$G(X)=XA-BX$$
Please show that $G(X)=Y$ has a solution $X$ for any $Y$ $\iff$ $A$ and $B$ have no common eigenvalues.

By Show that $B$ must equal $0$ if $A$ and $C$ have no common Eigenvalues, I know how to prove $\Leftarrow$ part. For $\Rightarrow$, since $G$ is linear and by the assumption, $G$ is surjective, $G$ is invertible. Hence, $A$ and $B$ are not similar and they have different characteristic polynomials. But this does imply they have no common eigenvalues. Appreciate any suggestion.

答:
Here is a proof. As you note, $G$ is surjective linear map over $\Bbb C^{n \times n}$. Because $\Bbb C^{n \times n}$ is finite-dimensional, $G$ is also injective, which is to say that $G(X) = 0 \implies X = 0$.

On the other hand, suppose for the purpose of contradiction that $A,B$ have a common eigenvalue $\lambda$. Let $x \in \Bbb C^n \setminus \{0\}$ be such that $A^Tx = \lambda x$, and let $y \in \Bbb C^n \setminus \{0\}$ be such that $By = \lambda y$. Let $X = yx^T$. Verify that $X \neq 0$, but $G(X) = 0$, contradicting the injectivity of $G$.


答:
The projection mapping $\pi(x,y)=x$ has kernel $\{0\}\times\Bbb R$, since for all $x,y$:

$x=\pi(x,y) = 0$.

Title: How to find the kernel of $\pi$

Body:
Define a map $\pi : \mathbb{R}^2 \to \mathbb{R} $  by $\pi((x,y))=x$ .Prove that $\pi$ is a  homomorphism and  find the kernel  of $\pi$

My attempt :  To show that $\pi$ is homomorphism

Proof :  Take  $(a, b) , (c,d) \in \mathbb{R}^2$ ,  we have  $\pi \left((a,b) + (c,d)\right) = \pi(a+c,b+d) = a+c= \pi(a,b) + \pi(c,d) $ $\implies$ $\pi$ is homomorphism

Im  confused about the kernel i,e how to find the kernel  of $\pi?$


问:
Question 6 of the 2nd round of the 15th Iran mathematical olympiad.
I came across this question the other day on an old Iran MO. The question goes as follows,

Let $P$ be the set of all points in $R^n$ with rational coordinates. For points $A,B ∈ P$, one can move from $A$ to $B$ if the distance $|AB|$ is $1$. Prove that every point in $P$ can be reached from any other point in $P$ by a finite sequence of moves if and only if $n ≥ 5$.

I have tried a few things such as polar coordinates and vectors but didn't make a huge amount of progress. Any advice or a solution would be greatly appreciated.

答:
We first show that not every point can be reached from $(0,0)$ for $n \le 4$. That it can be done for $n \ge 5$ will be shown afterwards. That this solves the general problem follows from the symmetry of the "move from A to B" definition.

Part 1:  $n \le 4$

Let the rational units be defined as

$$RU_n=\{r=(r_1,\ldots,r_n) \in P| \text{ satisfying } |r|^2=r_1^2+\ldots+r_n^2=1\}.$$

A move can be made from $A\in P$ to $B \in P$ iff $B-A \in RU_n$, where the difference is made component-wise. That means the points in $P$ that can be reached from $(0,0)$ in a finite series of moves are exactly those that can be expressed as a finite sum

$$x=\sum_{i=1}^k x_i, \text{ where } \forall i \in \{1,\ldots,k\}: x_i \in RU_n. \tag{*}\label{eq0}$$

We'll now show no component of a rational unit can have a denominator divisible by $4$ (in reduced form) if $n \le 4$.

Assume otherwise, then there exists a $r \in RU_n$ and writing $\forall i=1,\ldots,n: r_i=\frac{a_i}{b_i}$ in reduced form, we assume w.l.o.g that $b_1$ contains (one of) the highest powers of $2$ among the $b_i$, so $b_1=2^kb'_1, 2\nmid a_1, 2 \nmid b'_1, k \ge2 $. Note that $d=lcm (b_1,\ldots,b_n)$ is then of the form $d=2^ko$ with odd $o$. Using the definition of rational units we get

$$\sum_{i=1}^n \frac{a_i^2}{b_i^2}=1$$

and if we multiply this with $d^2$ we get

$$ \sum_{i=1}^n c_i^2=d^2, \text { where } c_i=a_i\frac{d}{b_i}\in \mathbb Z. \tag{1} \label{eq1}$$

As we noted above $2\nmid a_1$ and since $\frac{d}{b_1}=\frac{o}{b'_1}$ is also odd, so is $c_1$! Since $n \le 4$, we can add $0^2$ terms to the left side of \eqref{eq1} to fill up to 4 summands if necessary, sowie finally get

$$\sum_{i=1}^4 c_i^2 = d^2=16d'^2 \tag{2} \label{eq2}$$

Considering \eqref{eq2} mod 4, and remembering that $0$ and $1$ are the only possible quadratic residues mod 4, the only solutions are $c_1^2\equiv c_2^2 \equiv c_3^2 \equiv c_4^2 \equiv 0 \pmod 4$ and $c_1^2\equiv c_2^2 \equiv c_3^2 \equiv c_4^2 \equiv 1 \pmod 4$. Since we know that $c_1$ is odd, only the second option is possible, so all the $c_i$ are odd.

Considering \eqref{eq2} now mod 16, and remembering that the only odd quadratic residues mod 16 are $1$ and $9$, we see that \eqref{eq2} is impossible to fullfil with all odd $c_i$ mod 16, because the only options are $4\times 1, 3\times 1 + 1\times 9, 2\times 1 + 2\times 9,1\times 1 + 3\times 9$ and $4\times 9$, which are $\equiv 4, 12, 4, 12, 4 \pmod {16}$ in that order, so never $0$.

Ok, so we proved that no component of a rational unit has a denominator that is a multiple of $4$. But \eqref{eq0} shows that each component of a point that is reachable from $(0,0)$ is the sum of components of rational units. But such a sum can never be $\frac14$. Because if it were, then

$$\frac14=\sum_{i=1}^k \frac{a_i}{b_i},$$

where $\frac{a_i}{b_i}$ is the reduced form of a component of a rational unit, so $4\nmid b_i$ and hence $4\nmid lcm(b_1,\ldots,b_k)$. Multiplying the above equation with $lcm(b_1,\ldots,b_k)$ gives an integer result on the right hand side, but not an integer on the left hand side, as $lcm(b_1,\ldots,b_k)$ cannot "cancel" the $4$ in the denominator.

So any point in $P$ where one cordinate is $\frac14$ cannot be reached from $(0,0)$ for $n \le 4$.

This concludes the proof for part 1.

Part 2: $n \ge 5$

We use an old theorem from number theory, Lagrange's four-square theorem, that every natural number can be represented as the sum of four integer squares.

That means for every integer $k \ge 1$ we have for integers $a_k, b_k, c_k, d_k$ such that

$$a_k^2 + b_k^2+ c_k^2 + d_k^2 = 4k^2 -1,$$

and thus

$$\left(\frac{1}{2k}\right)^2 + \left(\frac{a_k}{2k}\right)^2 + \left(\frac{b_k}{2k}\right)^2 + \left(\frac{c_k}{2k}\right)^2 + \left(\frac{d_k}{2k}\right)^2  = 1.$$

That means for $n\ge 5$ there exists a rational unit $r_1$ with $\frac1{2k}$ as first component (if $n > 5$, just fill up with zeros). Since the definition of the rational unit only uses the square of each component, changing any number of signs in a rational unit makes the resulting number also a rational unit. Let $r_2$ be the rational unit obtained from $r_1$ by flipping all the signs, expcept on the first component. That means

$$r_1+r_2=(\frac1k,0,0,\ldots).$$

Obviously $((-r_1) + (-r_2)=(-\frac1k,0,0,\ldots)$, and similar by permutating the components rational units can be found such that the sum of 2 of them produces the zero vector except for a given component, where the component is $\frac1k$ or $-\frac1k$.

But each vector from $p\in P$ can be obtained as a finite sum of vector with only one component.

$$p=(p_1,\ldots,p_n) = (p_1,0,\ldots,0) + (0,p_2,0,\ldots) + \ldots + (0,\ldots,0,p_n), \tag{3}\label{eq3}$$

and if $p_1=\frac{a}{b}, b > 0$, then

$$(p_1,0,\ldots,0)=\sum_{i=1}^{|a|}(\pm\frac1b,0,\ldots,0),$$

with the signs before the fraction chosen the same as $a$. The same can of course be done for the other summands in \eqref{eq3}. But the summands thus calculated are exactly of the type that we proved are the sum of 2 rational units, so each $p=(p_1,\ldots,p_n) \in P$ is the sum of a finite number of rational units, so can be reached from $(0,0)$.

This concludes part 2 of the proof.


问:
Evaluate the orthogonal complement of a rank-1 updated orthogonal matrix.
Assume $V \in \mathbb{R}^{r\times r}$ and $\hat{V} = [V ,a]$. $V$ is an orthogonal matrix. $a \in \mathbb{R}^{r}$. I want to evaluate $\hat{V}^\perp$. Here is my attempt:

$\hat{V}^\perp = Null(\hat{V}^T) $, $Null$ is the nullspace.

$Null(\hat{V}^T) = \{ \nu: [V ,a]^T\nu = 0 \} = \{\nu: V_1^T\nu =0,\dots,V_r^T\nu=0,a^T\nu = 0 \}$, where $V_i$ is the ith column of $V$.

$V_i^T\nu = 0$ if $ \nu = 0 \text{ or } \nu = \sum_{1, j \ne i}^{r-1} \alpha_j V_j$ since $V$ is an orthogonal matrix. $a^T\nu = 0$ if $\nu = 0$ or $ \nu = a^{'}$. where $a'$ is an orthogonal vector to $a$

Now, I am not sure how to write the basis of $Null(\hat{V}^T)$ in a rigorous way?

答:
Since $V$ is orthogonal, $V^T$ is orthogonal as well, in particular of full rank and its null space is trivial, that is $V^\perp = \{0\}$. Augmenting $V$ by an additional vector can only make $V^\perp$ smaller, since in addition to being orthogal to a basis (which is only possible for $\nu=0$) you are adding an additional orthogonality constraint.

In conlusion, $$\hat V^\perp = \underbrace{V^\perp}_{=\{0\}} \cap a^\perp = \{0\}.$$


答:
Let $X=L^{\infty}(\mathbb{R})$, let $f_n \in X$ and define

$$
f_n(x) = \begin{cases}
1 & x \in [n, n+1] \\
0 & \text{else}
\end{cases}
$$
The sequence $(f_n)_{n=0}^{\infty}$ is bounded by 1, but has no convergent subsequence.

Title: Finding an example of a bounded sequence in a complete metric space such that the sequence has no partial limit

Body:
I'm working through an analysis text and I've come across this exercise:

Give an example of a complete metric space $X$ and a bounded sequence $\left(x_{n}\right)$ in $X$ such that the sequence $\left(x_{n}\right)$ has no partial limit.

The reason that I'm stuck with this one is because I'm shaky about what it means for a sequence to be bounded. Does this mean that for any positive integers $n$ and $m$, the set of all possible $|x_{n}-x_{m}|$ is a bounded set of real numbers?

If so, I'm not sure how to relate the boundedness of a sequence to the completeness of a metric space.

I know that a metric space $X$ is said to be complete if every Cauchy sequence in $X$ converges to a point in $X$... what does this have to do with bounded sequences?!? Thanks in advance for any help.

EDIT: I'm going to add a couple of definitions.

A sequence is said to be bounded if there exists a constant $C\gt 0$ such that for any positive integers $n$ and $m$ the inequality $|x_{n}-x_{m}|\lt C$ holds.

A point $x$ in a metric space $X$ is said to be a partial limit of a sequence $\left(x_{n}\right)$ if for every $\epsilon\gt 0$ there are infinitely many values of $n$ for which $x_{n}\in B\left(x,\epsilon\right)$.


问:
Prove or disprove that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is a principal ideal in $\mathbb{Z}[X]$

Prove or disprove that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is a principal ideal in $\mathbb{Z}[X]$

Assume that the ideal $\Big((X+1)^2,(2X+1)(X^2-2)\Big)$ is principal. Then exist generator $f(X)\in \mathbb Z[X]$ such that: $$\Big((X+1)^2,(2X+1)(X^2-2)\Big)=f(X)\mathbb Z[X]$$
However I don't have idea how to (dis)prove that $f$ exist.

答:
Division with remainder gives

$2x^3+x^2-4x-2 = (2x-3)(x^2+2x+1) + 1$.

Thus $1$ lies in the ideal and so the ideal equals $\Bbb Z[x]$.


问:
Why does $\frac{dy}{dx} \propto y$ imply $x_1 - x_2$ is constant for $\frac{y_2}{y_1}$ constant where $(x_1, y_1)$ and $(x_2, y_2)$ can vary?
The time taken for the activity of a radioactive sample to halve is called the half-life. This is a constant.

More generally, given the curve $y=Ae^{-bx}$, given $y_1$ and $y_2 = ky_1$, then $\frac{y_2}{y_1}$ is always the same number for all choices of $y_1$.

I can justify this by writing $y_1 = Ae^{-bx_1}$ and $y_2 = Ae^{-bx_2}$ and then using $y_2 = ky_1$ to get $x_1 - x_2 = \frac{1}{b} \ln k$, which is a constant.

But is there anyway to justify this starting from the fact that we have a curve that satisfies $\frac{dy}{dx} \propto y$?

答:
$\frac{dy}{dx} \propto y \Rightarrow \frac{dy}{dx} = ky \ $   ($k$ is a constant)

Now you're provided with 2 arbitrary points $(x_1,y_1)$ and $(x_2,y_2)$.

So,

$\begin{align}&\int_{y_1}^{y_2}\frac{dy}{y} = k\int_{x_1}^{x_2} dx \\\Rightarrow \ & \ln\frac{y_2}{y_1} = k(x_2- x_1) \text{ or }  \boxed{x_1 - x_2 = \frac1k\ln\frac{y_1}{y_2}}\end{align}$

Now, $k$ and  $\frac{y_1}{y_2}$ are constats. So, $\boxed{x_1- x_2}$ is also constant