问:
How do we get the formula of the sequence?
A woman is walking home with distance $d$ and speed $v$.

The dog is happy and runs at speed $\frac{3v }{ 2}$ always between woman and House back and forth.

(i) At what distances $(d_n)_{n\geq 1}$ do women and dogs meet?

(ii) Determine the total path length of the dog with the help of $(d_n)_{n\geq 1}$.

$$$$

EDIT:

Let $x$ be the distance between the woman and the house and let $s$ be the distance between the dog and the house.

At the beginning we have $x_0=d$ and $s_0=0$.

It holds that $x=d-vt$ and $s=\frac 32v t$.

The woman and the dog meet at $x=s=d_1$ for $t=t_1$.

From $x=s$ we get $d-vt=\frac{3}{2}vt \Rightarrow \frac{5}{2}vt=d \Rightarrow t=\frac{2d}{5v}$, so $t_1=\frac{2d}{5v}$.

So we have that $d_1=x(t_1)=d-v\cdot \frac{2d}{5v}=d- \frac{2d}{5}=\frac{3d}{5}$.

After the meeting of the woman and the dog, the dog goes back to the house and then again to the woman.

At time $t_1$ the dog is at $s_1=\frac{3d}{5}$.

The dog goes in direction of the house, so $s=\frac{3d}{5}-\frac{3v}{2}t$.

When $s=0$ the dog arrived at the house, so $t=\frac{2d}{5v}$.

Then the dog goes back in direction of the woman, so $s= \frac{3}{2}vt$ with $t>\frac{2d}{5v}$.

At time $t_1$ the woman is also at $x_1=\frac{3d}{5}$.

The woman continues in diection to the house, so $x=\frac{3d}{5}-vt$.

For the second meeting, it holds $x=s=d_2$.

We get \begin{equation*}\frac{3d}{5}-vt=\frac{3}{2}vt \Rightarrow \frac{3d}{5}=\frac{5}{2}vt \Rightarrow t=\frac{6d}{25v}\end{equation*} So  they meet at \begin{equation*}d_2=x(t_2)=\frac{3d}{5}-v\cdot \frac{6d}{25v}=\frac{9d}{25}\end{equation*}.

So the $n$-th meeting is at $d_n =\left (\frac{3}{5}\right )^nd$.

Is that correct so far?

For the second question do we take the sum of all the terms $d_n$ ? Or should we take twice each term $d_n$ since the dog goes back and forth?

答:
The distance of the woman from home is given by $d-vt$, and that of the dog by $d-\frac32vt$, until it reaches it, at time $t_1=\frac23\frac dv$. It comes back to its mistress with equation

$$\frac32v(t-t_1)=\frac32vt-d.$$

The moment they meet is obtained by solving

$$\frac32vt_2-d=d-vt_2$$ or $$t_2=\frac25\frac dv,$$ at a distance $\frac35d$ of home.

Now this repeats proportionally, substituting $\frac35d$ for $d$, and we have a geometric progression of common ratio $\frac35$.

The distances of the meetings are given by the partial sums of the sequence

$$\frac25d,\frac25\frac35d,\frac25\frac9{25}d,\cdots\frac25\frac{3^n}{5^n}d,\cdots$$

which indeed converges to $d$.

The dog travels $\frac32$ times more than the woman, hence $\frac32d$.


问:
Cubes inscribed into a sphere
Suppose n ≥ 3 unit cubes are inscribed into a sphere, such that every
three of them have a common vertex. Prove that all n cubes have a common vertex.

Could anyone please help me the approach?

Thanks in advance.

P.S. I edited out what I have tried since it was pointed out my approach was wrong.

答:
Suppose four unit cubes A, B, C and D are inscribed in a common sphere. This means that not only are they the same size, they also have the same centre.

Suppose:

B, C, D share a common vertex P.

A, C, D share a common vertex Q.

If P and Q coincide, then obviously all four cubes share a common vertex. The same is true if P and Q are diametrically opposite each other because every vertex is diametrically opposite another vertex of the same cube. Therefore we can assume that P and Q are distinct non-opposite vertices.

Then the two cubes C and D share two distinct non-opposite vertices (that is to say an edge or a face diagonal), as well as their centre. The only way this is possible is for them to fully coincide. Since the triplet A, B, and C share a common vertex, this vertex must also be shared by D.

For larger n you can use induction using essentially the same proof. The base case, n=4, is above. For induction, suppose it is true for n-1. Given cubes $C_1,...C_n$ satisfying the requirements. First exclude $C_1$ to get a vertex P that is shared by the other cubes. Then exclude $C_2$ to get a vertex Q that is shared by the other cubes. If P and Q coincide or are opposite, then we are done. If not, then $C_3$ and $C_4$ share both vertices P and Q, so must coincide. Then exclude $C_4$ to find a vertex that must be shared by all the cubes.

ContentLicense: CC BY-SA 4.0

Title: Cubes inscribed into a sphere


问:
About Uniform Convergence of $\sum_{n=1}^\infty\frac{\sin nx}{n}$ on $[0,2\pi]$

Is $\sum_{n=1}^\infty\dfrac{\sin nx}{n}$ uniform convergent on $[0,2\pi]$?

I think it is not. However, I could not prove it by Cauchy's criterion.

答:
A related problem.

Hint: Notice that the series is the Fourier series of the function

$$ -\frac{\pi+x}{2}. $$

Now, use the following result:

Theorem: The Fourier series of a $2\pi$-periodic continuous and piecewise smooth function converges uniformly.

ContentLicense: CC BY-SA 3.0


问:
What is the definition of the abelian group denoted by $\mathbb{Q}^*$ with $\times$?
My text states that $\mathbb{Q}^*$ with $\times$ is an abelian group.

I am unable to locate the definition of the set $\mathbb{Q}^*$. That is, I cannot find the definition of Q star, maybe because it is difficult to search for this notation.

This occurs in item (iv) of the first example at https://dec41.user.srcf.net/h/IA_M/groups/1_1.

It occured to be this may relate to the the non-zero rationals or some subset of the quaternions, but I cannot find a definition of this notation.

答:
For any ring $R$ the notation $R^{\times}$ or $R^{\ast}$ usually denotes the group of units i.e. all those $x \in R$ s.t. $\exists y \in R: xy = 1$.

In the case of $\mathbb{Q}$ this is just $\mathbb{Q} \setminus \{0\}$


答:
The answer is that $\mathbb{Q}^\star$ is the set of all nonzero rational numbers.

$\mathbb{Q}^* = \mathbb{Q} \setminus \{0\}$

https://proofwiki.org/wiki/Symbols:Q/Non-Zero_Rational_Numbers

Title: What is the definition of the abelian group denoted by $\mathbb{Q}^*$ with $\times$?

Body:
My text states that $\mathbb{Q}^*$ with $\times$ is an abelian group.

I am unable to locate the definition of the set $\mathbb{Q}^*$. That is, I cannot find the definition of Q star, maybe because it is difficult to search for this notation.

This occurs in item (iv) of the first example at https://dec41.user.srcf.net/h/IA_M/groups/1_1.

It occured to be this may relate to the the non-zero rationals or some subset of the quaternions, but I cannot find a definition of this notation.


问:
What does George E. Martin mean by "The belief that geometries can be classified by their symmetry groups is no longer tenable"?
In the preface to George E. Martin's Transformation Geometry: An Introduction to Symmetry, he writes (emphasis mine)

Transformation geometry is a relatively recent expression of the successful venture of bringing together geometry and algebra. The name describes an approach as much as the content. Our subject is Euclidean geometry. Essential to the study of the plane or any mathematical system is an understanding of the transformations on that system that preserve designated features of the system.

...

The belief that geometries can be classified by their symmetry groups is no longer tenable. However, the correspondence for the classical geometries and their groups remains valid. Undergraduates should not be expected to grasp the idea of Klein's Erlanger program before encountering at least the projective and hyperbolic geometries. Therefore, although the basic spirit of the text is to begin to carry out Klein's program, little mention of the program is made within the text.

What does he mean in the bolded statement? From the Erlangen program Wikipedia page, what I understand is that there are geometric objects that have "the same symmetries" (I don't know how else to put it), but are nevertheless distinct.

Comments would be appreciated!

答:
I suppose that he means that we now know that the symmetry group is not sufficient to capture the space. The idea of the Erlanger Program is that the group of symmetries determines what are the geometric figures and geometric properties within that space. If you have two subsets of the space that can be transformed into each other with a symmetry, then they are instances of the same type of geometric figure, and if a property is invariant under symmetry transformations, then that is a meaningful property for that given geometry.

The Lens example given by Jason De Vito in the comments illustrates how the symmetry group doesn't characterize the space


问:
use Schwarz inequality to get $\int_x^{+\infty }f(t) dt \leq \frac{1}{1+x^2}$
$f(x)\ge 0,\forall x\in \Bbb R$. If

$$\int_{-\infty}^{+\infty}f(x) dx =1, \int_{-\infty}^{+\infty}xf(x) dx =0,\int_{-\infty}^{+\infty}x^2f(x) dx =1. $$

Show that , for every $x\gt0$,
$$\int_x^{+\infty }f(t) dt \leq \frac{1}{1+x^2}$$

I have a method: proof by contradiction

If $a\gt0$ such that $\int_a^{+\infty }f(t) dt \gt \frac{1}{1+a^2}$, then

$$ \int_{a}^{+\infty}xf(x) dx \gt \dfrac{a}{1+a^2},\int_{a}^{+\infty}x^2f(x) dx \gt\frac{a^2}{1+a^2}. $$

so

$$\int_{-\infty}^a f(x) dx \lt -\dfrac{a^2}{1+a^2}, \int_{-\infty}^a xf(x) dx \lt -\dfrac{a}{1+a^2},\int_{-\infty}^a x^2f(x) dx \lt\frac{1}{1+a^2}. $$

$$ \int_{-\infty}^a (ax+1)^2f(x) dx \lt a^2 \frac{1}{1+a^2}-2a\frac{a}{1+a^2}+\frac{a^2}{1+a^2}=0. $$

I heard that the inequality is related to probability theory，and it can't be improved, how to cite a function to get this?  Are there any other methods, use Schwarz inequality?

答:
What does this have to do with probability?

It is true that this falls under the realm of probability theory, because if $X$ is a real valued random variable with density $f(x)$ i.e. $P(X \in (a,b)) = \int_a^b f(x)dx$ for all $a<b$, then :

$\int_{-\infty}^{\infty} f(x)dx = 1$ and $f(x) \geq 0$ assert that $f$ is a well-defined density and $X$ a well defined random variable.

$\int_{-\infty}^\infty xf(x)dx = 0$ is equivalent to $E[X] = 0$.

$\int_{-\infty}^{\infty} x^2f(x)dx = 1$ is equivalent to $E[X^2] = 1$,which when combined with the previous statement , gives $Var(X) = E[X^2] - E[X]^2 = 1-0=1$.

$\int_{x}^{\infty} f(t)dt = P(X \geq x)$ for $x>0$, which we write as $P(X - E[X] \geq x)$ for reasons becoming clear later.

$\frac{1}{1+x^2} = \frac{Var(X)}{Var(X)+x^2}$.

Thus, the inequality in probabilistic terms, is $P(X - E[X] \geq x) \leq \frac{Var(X)}{Var(X) + x^2}$ for every $x >0$.  This is called the Cantelli inequality. It is tight, by using the random variable $X$ given by a Bernoulli random variable with any parameter $p$, and taking its density.

A proof without using the Cauchy-Schwarz inequality

I will translate the probabilistic proof into a proof without such  terminology. Fix $f$ and $x >0$. We have , for any $u \geq 0$ :
$$
\int_{x}^\infty f(t)dt \leq \int_x^\infty f(t)dt + \int_{-\infty}^{-x} f(t)dt \\= \int_{x+u}^\infty f(t-u)dt + \int_{-\infty}^{-x-u} f(t-u)dt
$$

and now we introduce $x+u$ into both integrals. By domination , since $\frac{t^2}{(x+u)^2} \geq 1$ for every $|t| \geq x+u$:
$$
\int_{x+u}^\infty f(t-u)dt \leq \frac{1}{(x+u)^2} \int_{x+u}^{\infty} t^2f(t-u)dt \\
\int_{\infty}^{-x-u} f(t-u)dt \leq \frac{1}{(x+u)^2} \int_{-\infty}^{-x-u} t^2f(t-u)dt
$$

and adding them gives:
$$
\int_{x+u}^\infty f(t-u)dt + \int_{-\infty}^{-x-u} f(t-u)dt\\ \leq \frac{1}{(x+u)^2} \int_{x+u}^{\infty} t^2f(t-u)dt + \frac{1}{(x+u)^2} \int_{-\infty}^{-x-u} t^2f(t-u)dt \\
\leq \frac 1{(x+u)^2} \int_{-\infty}^{\infty} t^2f(t-u)dt =\int_{-\infty}^{\infty} (t+u)^2f(t)dt
 $$

which upon expansion and simplification becomes $$
\int_x^\infty f(t)dt \leq \frac{1 + u^2}{(x+u)^2}
$$

for every $u \geq 0$. So the best bound is when the right hand side is minimized i.e. the $u \geq 0$ for which $\frac{1+u^2}{(x+u)^2}$ is minimal. This can be done e.g. using differentiation : differentiating w.r.t $u$ gives $\frac{2(xu-1)}{(x+u)^3}$, which is zero when $u=\frac 1x$. One can check the second derivative is positive at this point. At $u = \frac 1x$ we have $\frac{1 + u^2}{(x+u)^2} = \frac{1}{1+x^2}$ , as desired.

A proof using the Cauchy Schwarz inequality

The Cauchy-Schwarz (-Буняковский) inequality, in the context of integrals, states that :

If $f,g$ are square-integrable functions on $(a,b)$ for any $a<b$ possibly infinite, then $fg$ is integrable on $(a,b)$ and $$
\left|\int_{a}^{b} f(x)g(x)dx \right| \leq \sqrt{\int_{a}^b f(x)^2dx} \sqrt{\int_{a}^{b} g(x)^2dx}
$$

To use this correctly, we must play around a little with the start, use Cauchy-Schwarz, and then correct the changes.

So let's start with $f$ as in the question and $x>0$. Our intuition for the change of variable in the next step is derived from the previous proof : recall the minimizer $\frac 1x$.

We write :
$$
\int_{x}^\infty f(t)dt = \int_{x+\frac 1x} f\left(t - \frac 1x\right)dt \\
\leq \int_{x+\frac 1x}^\infty f\left(t-\frac 1x\right)\frac{t}{x + \frac 1x} dt
$$

where the inequality follows because $\frac{t}{x+\frac 1x} \geq 1$ whenever $t \geq x + \frac 1x$, so a domination of integrand occurs.

Finally, for the Cauchy Schwarz split :
$$
\int_{x+\frac 1x}^\infty f\left(t-\frac 1x\right)\frac{t}{x + \frac 1x} dt  = \int_{x+\frac 1x}^{\infty} \left[t\sqrt{f\left(t-\frac 1x\right)}\right]\left[\frac{\sqrt{f(t-\frac 1x)}}{(x+\frac 1x)}\right] dt
$$

applying it (on the second line : the first line is just what we know from before) :
$$
\int_{x}^\infty f(t)dt \leq \int_{x+\frac 1x}^\infty f\left(t-\frac 1x\right)\frac{t}{x + \frac 1x} dt \\ \leq \sqrt{\int_{x+\frac 1x}^{\infty} t^2f\left(t-\frac 1x\right)dt}\sqrt{ \int_{x+\frac 1x}^{\infty} \frac{f\left(t - \frac 1x\right)}{(x+\frac 1x)^2}dt}\\ = \frac 1{x+\frac 1x} \sqrt{\int_{x}^\infty (t +1/x)^2 f(t)dt} \sqrt{\int_x^\infty f(t)dt}
$$

take the $\sqrt{\int_x^\infty f(t)dt}$ to the other side and square both sides to get :
$$
\int_x^\infty f(t)dt \leq \frac{\int_{x}^{\infty} (t+\frac 1x)^2f(t) dt}{(x+\frac 1x)^2} \leq \frac{\int_{-\infty}^{\infty} (t+\frac 1x)^2 f(t)dt}{(x+\frac 1x)^2}
 \\
= \frac 1{(x+\frac 1x)^2} \left(1 + \frac 1{x^2}\right) = \frac 1{1+x^2}$$

after simplification, as desired.


问:
Equality of Floors of some Partial Sums
Let $S_n=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\cdots+\frac{1}{n!}$ denote the $(n+1)^{st}$ partial sum in the series expansion for $e=\sum_{k\ge 0}\frac{1}{k!}$. I want to prove that $\lfloor n\cdot(S_n+1/n!)\rfloor=\lfloor n\cdot S_n\rfloor$ for $n\ge 3$. That is, $$\left\lfloor\frac{n}{0!}+\frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{n!}\right\rfloor=\left\lfloor\frac{n}{0!}+\frac{n}{1!}+\frac{n}{2!}+\cdots+\frac{n}{n!}+\frac{n}{n!}\right\rfloor.$$ I've tried to write out some kind of induction argument and use the fact that $\lfloor x+y\rfloor\ge\lfloor x\rfloor +\lfloor y\rfloor$, but I just get really messy expressions that don't seem to lead anywhere. I'm not even sure if induction is the right approach here. Any help is greatly appreciated, thanks in advance!

答:
Assume $n \geq 3$.

$S_n$ is a rational number and in lowest terms its denominator is $n!$. Therefore $nS_n$ has lowest terms denominator $(n-1)!$, so $\lceil nS_n \rceil - n S_n \geq \frac{1}{(n-1)!}$. If equality held in this expression then we would have $nS_n + \frac{1}{(n-1)!} = m$ for integer $m$, so $n! S_n + 1 = (n-1)!m$. But $n! S_n$ and $(n-1)!$ are both even numbers, a contradiction. Therefore $\lceil nS_n \rceil - n S_n >\frac{1}{(n-1)!}$.

Since $nS_n$ and $n(S_n + \frac{1}{n!})$ differ by $\frac{1}{(n-1)!}$, this implies that $\lfloor n S_n \rfloor = \lfloor n(S_n + \frac{1}{n!}) \rfloor$. QED

Side note: Using the remainder of the Taylor series for $e^x$ at $x=1$, $0 < e - S_n \leq \frac{e}{(n+1)!}$ (Lagrange remainder). This easily implies that $\lfloor n S_n \rfloor = \lfloor ne \rfloor$.

Title: Equality of Floors of some Partial Sums


问:
Why does this trigonometric identity problem work?

Why do I multiply by 1 + sin (theta)? If I multiply on one side, shouldn't I multiply on the other side too? Sorry for the weird formatting. I'm still learning how to use MathJax.

答:
You are multiplying by 1 since
$$\frac{1+\sin(\theta)}{1+\sin(\theta)}=1$$
And
$$x\cdot1=x\cdot\frac{y}{y}=x$$
And from there on you are just rewriting your fraction.


问:
Ordinary differential equations with double resonance
I want to know what is the definition of "resonance, double resonance" in

ordinary differential equations with double resonance

for exemple this : what it means the probleme is resonant in infity ?

Please,

Thank you.

答:
Apologies but I don't have time to explain in more detail. The following link has a gentle explanation: https://www.google.co.uk/url?sa=t&source=web&cd=4&ved=0CEQQFjAD&url=http%3A%2F%2Fwww.math.pku.edu.cn%3A8000%2Fvar%2Fpreprint%2F562.pdf&ei=nR6rUfr2FMeiO8eNgJgD&usg=AFQjCNF8AqsnYZp0r3yj98-2Q2CKpuJnww&sig2=BalUlpIsDbxM4BUJn6gNZQ

The following is my original answer.

Write $L$ for a linear differential operator. For example, the equation $\ddot{y}+\omega^2y = f(t)$ can be written as
$$Ly = f(t) \qquad \text{where}\qquad L\equiv \frac{\mathrm d%2}{\mathrm d t^2} + \omega^2$$

Eigenfunctions $y_n$ such that $Ly_n=\lambda_n y_n$ are interesting because if someone gives you the equation $Ly = \sum_n c_n y_n(t)$ you can solve it by
$$y(t) = \sum_n \frac{c_n}{\lambda_n}y_n(t)\quad\text{because}\quad L\sum_n \frac{c_n}{\lambda_n}y_n = \sum_n \frac{c_n}{\lambda_n} L y_n = \sum_n \frac{c_n}{\lambda_n} \lambda_n y_n = \sum_n c_n y_n$$

But this goes wrong when $\lambda_n=0$ for one of the forcing eigenfunctions, since you can't divide by zero. Therefore, forcing with zero-eigenvalue solutions is special, and called resonance. A particular example is a forced equation $Ly=f$ with $Lf=0$. For example, a $\sin (\omega t)$ forcing for the above example of $L$ is forced.

You can often fix this by multiplying by the independent variable $t$. This is because if $Lf=0$ then $$L(tf) = t L(f) + \text{something with }f,f',\ldots = \text{something}$$ in general. (To see this, just expand $L$ as a sum of derivative operators.) However, if the "something" is zero, because $tf$ is also a solution to $L(tf)=0$, then you can try $L(t^2f)$ and so on. It might be fun for you to figure out how this all works.

This is called 'resonance' because $tf$ grows faster than $f$; for example, $\sin(\omega t)$ forcing is a small, constant forcing but it leads to the max. of $|y|$ growing like $t$ over time.

An example is given by $Ly = y''$, which satisfies $L(1)=0$ and $L(t)=0$ but then $L(t^2)=2\propto 1$. That is, $L$ strips off the two $t$s we had to add to $1$. Then we can solve $Ly=1$ by $y=\frac{1}{2}t^2$.

Anyway, I don't know what 'double resonance' means; it might be this, or not!


答:
$$\frac ab=\frac cd\iff ad=bc.$$

This directly leads to

$$1-\sin^2\theta=\cos^2\theta.$$

Title: Why does this trigonometric identity problem work?

Body: Why do I multiply by 1 + sin (theta)? If I multiply on one side, shouldn't I multiply on the other side too? Sorry for the weird formatting. I'm still learning how to use MathJax.


问:
Limit of ratio goes to zero
I  am trying to show, if a sequence $\{a_n\}$ satisfies
$$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}=0$$
then
$$\lim_{n\to \infty}a_{n}=0.$$

I figured there exists an $\epsilon_N$ such that the following holds

$$|a_{N+1}|< \epsilon_N |a_N|$$

and $\lim_{n\to \infty}\epsilon_{N}=0.$

Since $\lim_{n\to \infty}a_{n}\ne  \infty$, then $\lim_{N\to \infty}\epsilon_{N}*|a_{N}|=0.$

One can conclude $\lim_{N\to \infty}|a_{N+1}|=0$. Is this correct? Is there another way to show this?

答:
Take $N\in\Bbb N$ such that $n\geqslant N\implies\left|\frac{a_{n+1}}{a_n}\right|<\frac12$. Then

$|a_{N+1}|\leqslant\frac12|a_N|$;

$|a_{N+2}|\leqslant\frac1{2^2}|a_N|$;

$\vdots$

$|a_{N+k}|\leqslant\frac1{2^k}|a_N|$

and therefore $\lim_{k\to\infty} a_{N+k}=0$, from which it follows that $\lim_{n\to\infty}a_n=0$.

Title: Limit of ratio goes to zero


问:
$\max\{n_1,n_2\} < \max\{m_1,m_2\}$ implies $n_1n_2 > m_1m_2?$
Suppose we have two sets of natural number $n_1,n_2$ and $m_1,m_2$ such that $$m_1+m_2 = n_1 + n_2$$ and$$\max\{n_1,n_2\} < \max\{m_1,m_2\}.$$ From here can we claim that $$n_1n_2 > m_1m_2?$$

Body:
WLOG assume $n_1 \le n_2, m_1 \le m_2$, and $m_2=n_2 +d, d>0$, then $m_1=n_1+n_2-(n_2+d)=n_1-d$

$$n_1 n_2 - m_1 m_2 = n_1 n_2 - (n_1-d)(n_2+d)\\
= n_1 n_2 -(n_1n_2+n_1d-n_2d-d^2)\\
= d(n_2-n_1)+d^2 >0$$


问:
If $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x)\neq x$ for all $x$, must it be true that $f(f(x))\neq x$ for all $x$?
Let $f: \Bbb R → \Bbb R$ be a continuous function such that $f(x)=x$ has no real solution  . Then is it true that $f(f(x))=x$ also has no real solution ? </p>

答:
If $f(f(c))=c$ and $f(c)=r$ then $f(r)=c$.

Since $c\neq r$ it follows that there is an $s$ between $r$ and $c$ such that $f(s)=s$ (in the case $c<r\Rightarrow f(r)-r=c-r>0$ and $f(c)-c=r-c<0$...) $\Rightarrow\Leftarrow$.


问:
find the coordinates of the vectors $(u,v)$ in the ordered basis $B=\{a,b\}$?
let $a= (x_1,x_2)$  and $b=(y_1,y_2)$ be  vectors in $\mathbb{R}^2$ such that $$x_1y_1 +x_2y_2=0$$ , $$x_1^2+ x_2^2 =y_1^2 +y_2^2 =1$$

It is given that $B=\{a, b\}$ is  basis for $\mathbb{R}^2  $

find the coordinates  of the vectors  $(u,v)$ in the ordered basis  $B=\{a,b\}$

My attempt : I take $(u,v)=   c_1 a + c_2b= (c_1x_1 +c_2y_1 , c_1x_2 +c_2y_2)$

so the coordinates are  $c_1x_1 +c_2y_1=u$ and  $v= c_1x_2 +c_2y_2$

答: That basis is an orthnormal basis, and therefore the coordinates of $(u,v)$ with respect to that basis are $\bigl\langle(u,v),a\bigr\rangle(=ux_1+vx_2)$ and $\bigl\langle(u,v),b\bigr\rangle(=uy_1+vy_2)$. That is$$(u,v)=(ux_1+vx_2)a+(uy_1+vy_2)b.$$


问:
Binary matrix, Permutation and Rank

Let $B$ be a square 0-1 matrix and there exists no permutation matrices $P_1$ and $P_2$ such that $\mathrm{tr}(P_2^TBP_1)=n$. Then show that $B$ does not have full rank.

Somehow I think the proof is related to being able to write matrix as a linear combination of permutation matrices.

答:
This isn't true. Consider e.g. $n=2$ and $B=\pmatrix{0&1\\ 1&0}$. Using the tracial property, we have $\operatorname{tr}(P^TBP)=\operatorname{tr}(PP^TB)=\operatorname{tr}(B)=0\ne n$, but $B$ has full rank.

The statement can be corrected by requiring that $\operatorname{tr}(BP)\ne n$ for all permutation matrices $P$ (or equivalently, $\operatorname{tr}(P_2^TBP_1)\ne n$ for all permutation matrices $P_1$ and $P_2$, because $\operatorname{tr}(P_2^TBP_1)=\operatorname{tr}(BP_1P_2^T)$ and $P_1P_2^T$ is a permutation matrix). This means $\sum_ib_{i\sigma(i)}\ne n$ for all permutations $\sigma$. Since $B$ is a $0-1$ matrix, $\sum_ib_{i\sigma(i)}\ne n$ only if $b_{i\sigma(i)}=0$ for some $i$. Therefore $\det(B)=\sum_\sigma\operatorname{sgn}(\sigma)\prod_ib_{i\sigma(i)}=0$, i.e. $B$ is singular.

Title: Binary matrix, Permutation and Rank


问:
Gamma Integral with Complex Factor on Exponential
I've set out to work out the following gamma-type integral where the exponential function has a complex factor multiplying the argument:
$$
\int _{0}^{\infty} x^{z-1}e^{-\alpha x} \, dx, \quad \alpha \in \mathbb{C}.
$$
This particular integral has its value reported in many places to be
$$
\frac{\Gamma(z)}{\alpha ^{z}}
$$
and what I've been attempting to do is simply reproduce this result. However, I'm running into a bit of a problem trying to work out the path integral once the substitution $u = \alpha x$ is made, since
$$
\int _{0}^{\infty}x^{z-1}e^{-\alpha x} \, dx = \frac{1}{\alpha ^z} \lim _{t \rightarrow \infty} \int _{0}^{\alpha t} u ^{z-1}e^{-u} \, du.
$$
This last integral is, as far as I understand, a path integral along a line that crosses the origin in the complex plane, and it shouldn't be simply $\Gamma (z)$, right? I mean, I may be getting this wrong, but the definition of $\Gamma (z)$ involves an integration over the real line, and not some slanted line in the complex plane.

I attempted to separate the integral in two by taking first the integration along the real line and then another one along the imaginary line, but it didn't get me anywhere. Am I missing something very obvious here? I initially thought this would be a rather straightforward result. Any and all help is appreciated! If you see any mistakes in my calculations please feel free to correct them!

答:
By the Cauchy integral theorem, an integral of $f(u) = u^{z-1} e^{-u}$ on a path from $0$, on to $\alpha t$, on an arc to $|\alpha|t$ on the real line, and finally back to $0$, is zero, because the integrand is holomorphic over the whole region. Therefore
$$
\int_{0}^{\alpha t} f(u) \, du = - \int_{\textrm{arc}} f(u) \, du - \int_{|\alpha| t}^0 f(u) \, du
$$
The next step is to argue using the estimation lemma that in the limit $t \to \infty$, the integral along the arc vanishes -- however, this may not be true for all values of $\alpha$. After that, the result follows immediately.

To apply the estimation lemma, note that for any point $u = R e^{i \theta}$, where $R = |\alpha| t$,
$$
\left| u^{z-1} e^{-u} \right| = R^{z-1} e^{-R\cos(\theta)},
$$
and then show that
$$
\lim_{R \to \infty} \max_{u \textrm{ on arc}} \left| f(u) \right| = 0.
$$


问:
Does this notation describe an interval or a set?
The MathWorld page on Stieltjes integrals here contains the following:

"...consider the Riemann sum $$\sum _{i=0}^{n-1} f\left(\xi _i\right)
   \left(g\left(x_{i+1}\right)-g\left(x_i\right)\right)$$ with $\xi _i \in[x_i,x_{i+1}]$."

I am not 100% sure of the meaning of $\xi _i \in[x_i,x_{i+1}]$ in this context. Does it mean that $\xi _i$ is contained within the closed interval $[x_i,x_{i+1}]$, or does it mean that $\xi _i$ is equal to either $x_i$ or $x_{i+1}$?

In this context, the square brackets would ordinarily denote an interval. But then one would write $\xi _i = [x_i,x_{i+1}]$, no? And, surely, if it means either/or one would use curly brackets for the list $x_i,x_{i+1}$? So, the expression as written on MathWorld is half one thing, half another...

Given that I'm trying to learn about Stieltjes integrals from first principles, I'd like to be sure.

答: This notation $\xi_i \in [x_i,x_{i+1}]$ means it belongs to that closed interval, i.e. $\xi_i$ is in that interval. So the notation you are reading is fine, your understanding of it seems to be off. One would not write $\xi_i = [x_i,x_{i+1}]$ since this does not make sense, a number cannot be equal to an interval, it usually belongs to an interval which is what they wrote.


问:
Posterior for Gamma prior and Gamma likelihood with known shape
Following "Introduction to Mathematical Statistics" of Hogg et al. (Exercise 11.2.2) I am trying to calculate the posterior distribution of $\theta$ with the following information given:

Let $X_1,X_2,...,X_{10}$ be a random sample of size $n=10$ from a gamma distribution with $\alpha = 3$ and $\beta=\frac{1}{\theta}$. Suppose we believe that $\theta$ has a gamma distribution with $a=10$ and $b=2$.

I tried to calculate the posterior in general form first and then substitute the variables with the concrete values above. For the Prior I have:

$$ f_{\Theta}(\theta)=\frac{b^a}{\Gamma(a)} \cdot \theta^{a-1} \cdot e^{-b\theta} $$
And for the likelihood:

$$ f_{X\mid \Theta}(x\mid\theta)=\frac{\beta^\alpha}{\Gamma(\alpha)} \cdot x^{\alpha-1} \cdot e^{-\beta x}$$
Therefore for the posterior the following should be true:

$$ f_{\Theta \mid  X_1,X_2,...,X_n}(\theta\mid x_1,x_2,...,x_n)\propto \prod_{I=1}^{n}\frac{\beta^\alpha}{\Gamma(\alpha)} \cdot x_i^{\alpha-1} \cdot e^{-\beta x_i} \cdot \frac{b^a}{\Gamma(a)} \cdot \theta^{a-1} \cdot e^{-b\theta}$$
Now removing all terms independent of $\theta$ my result is the following:
$$f_{\Theta \mid  X_1,X_2,...,X_n}(\theta\mid x_1,x_2,...,x_n)\propto \beta^{n\alpha} \cdot e^{-\beta\sum_{I=1}^{n}x_i}\cdot\theta^{a-1}\cdot e^{-b\theta}$$

Inserting $\beta=\frac{1}{\theta}$ and aggregating:
$$f_{\Theta \mid  X_1,X_2,...,X_n}(\theta\mid x_1,x_2,...,x_n)\propto \theta^{a-n\alpha-1} \cdot e^{-\frac{\sum_{I=1}^{n}x_i}{\theta}-b\theta}$$

Finally I am struggling to relate this to any common distribution. Also taking the following hint from the exercise into account: "Can the posterior distribution be related to a chi-square distribution?"
Can anybody help?

答:
I think it is only a matter of parametrizing the gamma density. If you consider $\beta,b$ as the scale parameter, Gamma density is  the following

$$f_X(x)=\frac{1}{\Gamma(\alpha)\beta^\alpha}x^{\alpha-1}e^{-x/\beta}$$

Thus the likelihood is

$$p(\mathbf{x}|\beta)\propto\theta^{30}\cdot e^{-\theta \Sigma_i X_i}$$

The prior is

$$\pi(\theta)\propto \theta^9\cdot e^{-\theta/2}$$

Thus the posterior is

$$\pi(\theta|\mathbf{x})\propto\theta^{39}\cdot e^{-\theta(1/2+\Sigma_i X_i)}$$

Now we immediately recognize that the posterior is a $Gamma\Big[40;\frac{2}{2 \Sigma_i X_i+1}\Big]$

Obviously it is related with a $\chi_{(80)}^2$ which can be obtained by a simple linear transformation of the Posterior


答: Lets $z^T$ be the transpose of the column vector $z$ and set $(u,v)=z$. Since the basis is orthonormal the projection matrix onto each vector is $aa^T$ and $bb^T$ and so we have $z = (aa^T)z + (bb^T)z = a(a^Tz) + b(b^Tz)$ and since $a^Tz$ is the dot product $a \cdot z$, it is a scalar and commutes with $a$ and $b$ so we can write this as $(a \cdot z)a + (a \cdot z)b = (ux_1 + vx_2)a + (uy_1 + vy_2)b$ and we are done.

Title: find the coordinates of the vectors $(u,v)$ in the ordered basis $B=\{a,b\}$?

Body:
let $a= (x_1,x_2)$  and $b=(y_1,y_2)$ be  vectors in $\mathbb{R}^2$ such that $$x_1y_1 +x_2y_2=0$$ , $$x_1^2+ x_2^2 =y_1^2 +y_2^2 =1$$

It is given that $B=\{a, b\}$ is  basis for $\mathbb{R}^2  $

find the coordinates  of the vectors  $(u,v)$ in the ordered basis  $B=\{a,b\}$

My attempt : I take $(u,v)=   c_1 a + c_2b= (c_1x_1 +c_2y_1 , c_1x_2 +c_2y_2)$

so the coordinates are  $c_1x_1 +c_2y_1=u$ and  $v= c_1x_2 +c_2y_2$


问:
moments of a Geometric brownian motion using Ito's lemma
Can someone explain me what is wrong in my derivation of the formula for the moments of a GBM using Ito's lemma (I am not interested in other methods) ?
\begin{equation}
dX=\mu Xdt+\sigma XdW_t
\end{equation}
setting $Y=X^n$ we have
\begin{equation}
\frac{dY}{Y}=(n\mu+\frac{1}{2}n(n-1)\sigma^2t)dt+ n \sigma dW_t
\end{equation}
at this point my book says "the expectation of $Y=X^n$ follows directly:
\begin{equation}
E[X(t)]^n=X(0) \exp{(n\mu+\frac{1}{2}n(n-1)\sigma^2)}t
\end{equation}
however, when I try to do this step something goes wrong. I start solving for  $Y_t$ in the usual way (taking the logarithm and then computing the differential) and so I obtain
\begin{equation}
\begin{split}
d\log Y_t& =\frac{1}{Y}dY-\frac{1}{2Y^2}(dY)^2 \\
& = \frac{1}{Y} \biggl( (n\mu+\frac{1}{2}n(n-1)\sigma^2t)Ydt+ n \sigma Y dW_t \biggr)-\frac{1}{2Y^2}n^2 Y^2 \sigma^2 dt \\
& = [ (n\mu+\frac{1}{2}n(n-1)\sigma^2t)-\frac{1}{2}n^2 \sigma^2 ]dt + n \sigma dW_t
\end{split}
\end{equation}
As you can see I obtain a $-\frac{1}{2}n^2 \sigma^2$ in the drift that I should not obtain

Thank you for your help

答:
You have
$$
\Bbb E(e^{σ·W_t}) = e^{\frac12σ^2·t}
$$
so that the extra term in the solution formula for $Y_t$ cancels again when taking the expectation.


问:
Absolute error from root in false position method
Suppose I know that $f(x_1)$ and $f(x_2)$ have opposite signs, so $f(x)=0$ has a root $x\in[x_1,x_2]$. Then using the false position method, I have a guess for the root
$$x_3=\frac{f(x_2)x_1-f(x_1)x_2}{f(x_2)-f(x_1)},$$
and I can iterate on either $[x_1,x_3]$ or $[x_3,x_2]$ depending on the sign of $f(x_3)$.

Usually we terminate the process when $|f(x_n)|<\epsilon$ for some specified $\epsilon$.

How can I pick $\epsilon$ so that I am certain that my guess for the root $x_n$ is within $\delta$ of the true value of the root, i.e. $|x_n-x|<\delta$?

I am not sure how to pick such an $\epsilon$ when we don't even know the true value $x$ of the root. When $\delta$ is sufficiently small, something like $\epsilon=\delta f'(x)$ could work, but obviously this requires that you (a) know the true value of the root and (b) know the derivative of the function, two assumptions that are definitely not true in general.

答:
A more robust criteria for termination which does not have the issues you point out would be to use an estimate of the derivative, since we expect to have

$$f(x_n)\approx f'(x)(x_n-x),\quad|x_n-x|\approx\left|\frac{f(x_n)}{f'(x)}\right|,\quad f'(x)\approx\frac{f(a)-f(b)}{a-b}\tag{1, 2, 3}$$

where $a<x<b$. As you may notice, this simply ends up becoming the estimate

$$|x_n-x|\approx|x_n-x_{n+1}|\tag4$$

as Claude Leibovici mentions.

Another strategy would be to instead use a better estimate of the slope. Instead of using the endpoints of your interval, of which one side is very inaccurate, you could instead use the last two computed points, replacing $f'(x)$ with

$$f'(x)\approx\frac{f(x_{n+1})-f(x_n)}{x_{n+1}-x_n}\tag5$$

which, in the case of twice differentiable functions with non-vanishing second derivative at the root, can be shown to lead to an overestimate of the absolute error (which is desirable).

Below a graphical demonstration of this is shown. As can be seen, every iteration of false position gives a point on the right of the root. Likewise, if you estimate the slope using the last two computed points, you get an estimate of the root on the left side. The difference between the last computed point and this one is an upper bound on the absolute error.

Hence one can conclude that in most instances one should eventually have

$$|x_{n+1}-x|\stackrel<\simeq\left|\frac{f(x_{n+1})}{f(x_{n+1})-f(x_n)}(x_{n+1}-x_n)\right|\tag6$$

This approach is not flawless however, as it can easily lead to premature termination. Early on one may have the last two computed points be nearly vertical, or even pointing in the wrong direction. A much safer strategy would then be to use an anti-stalling method, such as the Illinois method, or along the lines of what was presented so far in this answer:

Try the usual false position twice.

Try using $(5)$ to compute the next estimate of the root instead of the usual false position.

Use bisection if the previous step gives an estimate outside of your current bounds or if the length of the bracketing fails to halve.

Repeat steps 1, 2, and 3 until your bracketing interval is sufficiently small.

In this way you can be certain that your bracketing interval shrinks and that the estimated absolute error is always an over-estimate of the real absolute error.


答:
Your first equality $x_1y_1 + x_2y_2$ is exactly the definition of $(x_1, x_2)$ and $(y_1, y_2)$ are orthogonal.

Your second equality tells you that both vectors have a norm equal to 1.

Hence, they form an orthonormal basis, and the formula giving the coordinates of any vector in this basis is very simple: you simply have to compute the scalar product of your vector and of the vectors of the basis. No "correction" is needed since your basis is orthogonal, and no "renormalization" is needed since your basis vectors have already normalized.

Finally;

$(u, v) = \left<(u, v), a\right> a + \left<(u, v), b\right> b = (ux_1 + vx_2)a + (uy_1 + vy_2)b$.

As a result, the coordinates of $(u, v)$ in the new basis  are $(ux_1 + vx_2, uy_1 + vy_2)$


问:
Proving continuity of $f(x)=\sqrt{1−x^3}$ on $[0,1]$
I have to prove the continuity of the function with $\epsilon, \delta$. For this I have to prove that the function is continuous at every $x \in [0, 1]$ by proving:

$$\forall 0 < \epsilon, \exists 0 < \delta, \forall a \in [0, 1]: |x - a| < \delta \implies \left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right| < \epsilon.$$

This is where I am right now:

\begin{align}
\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|&= \frac{\left|\sqrt{1 - x^3} - \sqrt{1 - a^3}\right|\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{x^3 - a^3}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&= \frac{(x - a)(x^2 + ax + a^2)}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3} + \sqrt{1 - a^3}\right|}\\
&\leq \frac{3\delta}{\left|\sqrt{1 - x^3 + 1 - a^3}\right|}
\end{align}

Any hint is welcome.

答:
If $a\not=1$, then you can continue with

$${3\delta\over\sqrt{1-x^3}+\sqrt{1-a^3}}\lt{3\delta\over\sqrt{1-a^3}}$$

so it suffices to let $\delta=\epsilon\sqrt{1-a^3}/3$ to make the epsilon-delta proof work for $a\in[0,1)$. But for $a=1$ you need a separate argument to show that there is a $\delta$ for which $|x-1|\lt\delta\implies|\sqrt{1-x^3}|\lt\epsilon$.  I'll let you think this over first; please leave a comment if you'd like more help.


问:
Souslin operation $\mathcal{\Gamma}$ contains countable union and intersection
This is exercise 25.5 II) in Kechris Descriptive Set Theory:

Denoting by $\Gamma_\sigma$, $\Gamma_\delta$ the class of sets that are respectively countable unions or countable intersections of sets in $\Gamma$, show that if $X\in\Gamma$, then $\Gamma_\sigma\cup\Gamma_\delta\subseteq\mathcal{A}\Gamma$.

First off all I have a misunderstanding with the exercise, which might be some irrelevant concern on my end, so you make nothing wrong if you skip this part, if you are only interested in the actual question.

In which sense is $\Gamma$ a Souslin scheme/collection of subsets of a set $X$?
The given definitions are as follows:

Let $(P_s)_{s\in\mathbb{N}^{<\mathbb{N}}}$ be a Souslin scheme on a set $X$, i.e., a familiy of subsets of $X$ indexed by $\mathbb{N}^{<\mathbb{N}}$. The Souslin operation $\mathcal{A}$ applied to such a scheme produces the set

$\mathcal{A}_sP_s=\bigcup_{x\in\mathcal{N}}\bigcap_{n} P_{x|n}$.

Given any collection $\Gamma$ of subsets of a set $X$ we denote by $\mathcal{A}\Gamma$ the class of sets $\mathcal{A}_sP_s$, where $P_s\subseteq X$ are in $\Gamma$.

So we just assume that $\Gamma$ is such a collection of subsets of a set $X$ (with the additional assumption that $X\in\Gamma$)? With other words $\Gamma\subseteq\mathrm{Pow}(X)$ (power set of $X$).
Because the task does not specify this further. In particular which set it is a collection of subsets from. Anyways...

As far as the actual exercise goes:

I want to show $\Gamma_\sigma\cup\Gamma_\delta\subseteq\mathcal{A}\Gamma$.

To show $\Gamma_\delta\subseteq\mathcal{A}\Gamma$, I have to find for $G\in\Gamma_\delta$ a Souslin scheme $(P_s: s\in\mathbb{N}^{<\mathbb{N}})$, such that $\mathcal{A}P_s=G$.

$G$ is a countable intersection of elements in $\Gamma$, so it is most natural to take a Souslin scheme which contains the (countable many) sets which get intersected.
I struggle alot to grasp what is happening here.

I need to associate an countable subset of $\Gamma$ (which consists of the sets which get intersected to get the set $G$) with an element $x$ of $\mathcal{N}=\mathbb{N}^{\mathbb{N}}$, this is my first issue.
It seems I have to convert this into a Souslin scheme such that $\bigcap_n P_{x|n}=G$. This is the second issue.

Can you help me out?
Thanks in advance.

答:
If $B_0, B_1, B_2, \ldots \in \Gamma$ then $\bigcup_n B_n \in \mathcal{A}\Gamma$, and also $\bigcap_n B_n \in \mathcal{A}\Gamma$.

The $\Gamma$ is understood to be some collection of subsets of some $X$, so a subset of $\mathscr{P}(X)$, really. He does asume $X \in \Gamma$.

This can be seen by constructing "degenerate cases" of Suslin schemes:

For the intersection, set $A_s = B_{|s|}$, for all $s \in \Bbb N^{<\Bbb N}$, where $|s|$ denotes the length, as usual. For every $x \in \mathcal{N}$ the sequence $A_{s|n}$ just becomes the sequence $B_n$ and the whole union reduces to a single term $\bigcap_n B_n$.

For the union, we let the constant sequences (say with value $n$) in $\mathcal{N}$ give rise to an intersection equal to $B_n$ and all others only to at most subsets of that. So the union over $\mathcal{N}$ will be precisely $\bigcup_n B_n$.

E.g. set $A_s$ to be $B_{l(s)}$ for $s \in \Bbb N^{<\Bbb N}$, where $l(s)$ is the last term of $s$ (first term would also work BTW) and we also set $A_\emptyset = X$ (as a "neutral" first term in the intersection). As promised the constant $x \in \mathcal N$ just give rise to the set $B_n$, for that constant. So $\bigcup_n B_n \subseteq \mathcal{A}_s(A_s)$ and the reverse inclusion is immediate, so $\bigcup_n A_n \in \mathcal{A}\Gamma$.

So simple Souslin schemes give unions and intersections. In this case we don't "abuse" the option of taking actual uncountable unions.


答:
First of all, $\Gamma$ is not a Suslin scheme. Your confusion arises from the use of the symbol $\mathcal{A}$ as a class operator: $\mathcal{A}\Gamma$ is the class of all sets obtained by applying the Suslin operation to a scheme $\{P_s\}$ with all $P_s$ in $\Gamma$.

Now to the question proper, intersections are immediate: Assume $Q_n$ for $n\in\mathbb{N_0}$ are sets in $\Gamma$. If you define that $P_s := Q_{|s|}$ (where $|s|$ is the length of $s$), then $\mathcal{A}_sP_s$ will equal $\bigcap_n Q_n$ (note that all the terms in the big union appearing in the $\mathcal{A}$ operation are the same).

By a similar reasoning, we can obtain the union of the $Q_n$, this time repeating a set in the intersections. Using the assumption that $X\in\Gamma$ allows for a slightly simpler solution. Define
$$\begin{align*}
P_{()} &:= X\\
P_s &:= Q_{s(0)},
\end{align*}
$$
where $()$ denotes the empty sequence. This choice ensures that that all terms in the intersections inside the Suslin operation are equal and the final result is the countable union.

Title: Souslin operation $\mathcal{\Gamma}$ contains countable union and intersection

Body:
This is exercise 25.5 II) in Kechris Descriptive Set Theory:

Denoting by $\Gamma_\sigma$, $\Gamma_\delta$ the class of sets that are respectively countable unions or countable intersections of sets in $\Gamma$, show that if $X\in\Gamma$, then $\Gamma_\sigma\cup\Gamma_\delta\subseteq\mathcal{A}\Gamma$.

First off all I have a misunderstanding with the exercise, which might be some irrelevant concern on my end, so you make nothing wrong if you skip this part, if you are only interested in the actual question.

In which sense is $\Gamma$ a Souslin scheme/collection of subsets of a set $X$?
The given definitions are as follows:

Let $(P_s)_{s\in\mathbb{N}^{<\mathbb{N}}}$ be a Souslin scheme on a set $X$, i.e., a familiy of subsets of $X$ indexed by $\mathbb{N}^{<\mathbb{N}}$. The Souslin operation $\mathcal{A}$ applied to such a scheme produces the set

$\mathcal{A}_sP_s=\bigcup_{x\in\mathcal{N}}\bigcap_{n} P_{x|n}$.

Given any collection $\Gamma$ of subsets of a set $X$ we denote by $\mathcal{A}\Gamma$ the class of sets $\mathcal{A}_sP_s$, where $P_s\subseteq X$ are in $\Gamma$.

So we just assume that $\Gamma$ is such a collection of subsets of a set $X$ (with the additional assumption that $X\in\Gamma$)? With other words $\Gamma\subseteq\mathrm{Pow}(X)$ (power set of $X$).
Because the task does not specify this further. In particular which set it is a collection of subsets from. Anyways...

As far as the actual exercise goes:

I want to show $\Gamma_\sigma\cup\Gamma_\delta\subseteq\mathcal{A}\Gamma$.

To show $\Gamma_\delta\subseteq\mathcal{A}\Gamma$, I have to find for $G\in\Gamma_\delta$ a Souslin scheme $(P_s: s\in\mathbb{N}^{<\mathbb{N}})$, such that $\mathcal{A}P_s=G$.

$G$ is a countable intersection of elements in $\Gamma$, so it is most natural to take a Souslin scheme which contains the (countable many) sets which get intersected.
I struggle alot to grasp what is happening here.

I need to associate an countable subset of $\Gamma$ (which consists of the sets which get intersected to get the set $G$) with an element $x$ of $\mathcal{N}=\mathbb{N}^{\mathbb{N}}$, this is my first issue.
It seems I have to convert this into a Souslin scheme such that $\bigcap_n P_{x|n}=G$. This is the second issue.

Can you help me out?
Thanks in advance.


问:
Continuous, differentiable, concave functions and the Euclidean norm of the gradient
Let $f \colon \Omega \subseteq \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$ be a continuous differentiable function over $\Omega$. Suppose that the function $f$ is concave, and fix two points $ \mathbf{x} = (x_1, \dots, x_n), \mathbf{y} = (y_1, \dots, y_n) \in \Omega $.

$(1) $ If $x_i \leq y_i$ for all $i = 1, \dots, n$ and $\Omega = \mathbb{R}^n $, does it hold $ \parallel \nabla_{\mathbf{x}} f \parallel \geq \parallel \nabla_{\mathbf{y}} f \parallel $?

$(2) $ If $f(\mathbf{x}) \leq f(\mathbf{y})$ and $\Omega = \mathbb{R}^n $, does it hold $ \parallel \nabla_{\mathbf{x}} f \parallel \geq \parallel \nabla_{\mathbf{y}} f \parallel $?

$(3) $ Does points $(1)$-$(2)$ hold, if we consider proper subsets $\Omega $ of $\mathbb{R}^n $?

答:
partial answer:
(1) is false. Take a concave function which is symmetric about the origin (e.g. $f=-\|x\|^2$). if $0\leq x_i$ and $\|\nabla f(\mathbf{x})\|>\|\nabla f(0)\|$, then due to symmetry we'd get $-x_i\leq 0$ but $\|\nabla f(-\mathbf{x})\|<\|\nabla f(0)\|$.


问:
Integrating $\int_0^\infty \frac{\sin (ax)}{x^3}dx$
I am trying to integrate this integral,

$$I_3 = \frac{ -4 }{ \pi} \int_0 ^{\infty} \frac{d\lambda}{\lambda^4} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda) \left(-p_3 \lambda\cos\left(p_3 \lambda\right)\right)\cdot\sin\left(p_4 \lambda\right) \\
    = \frac{4p_3 }{ \pi} \int_0 ^{\infty} \frac{d\lambda}{\lambda^3} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda) \cos\left(p_3 \lambda\right)\sin\left(p_4 \lambda\right)$$
I changed this trigonometric product to a sum of  $\sin\left(\left(dp_4+cp_3+bp_2+ap_1\right)\lambda\right)$ where $a,b,c,d= \pm1$ But then integral takes the form $\int_0 ^{\infty} \frac{d\lambda}{\lambda^4} \sin\left(k\lambda\right)$ Which seems divergent to me. Is there any other way to integrate this?

Background: I am trying to calculate $$ D_2(p_3,p_4) =\frac{-4}{\pi} \int \frac{d\lambda}{\lambda^4} \cdot \sin(p_1 \lambda)\cdot \sin(p_2 \lambda)
 \left(\sin\left(p_3 \lambda\right)-p_3 \lambda\cos\left((p_3 \lambda\right)\right)\cdot\left(\sin\left(p_4 \lambda\right)-p_4 \lambda\cos\left((p_4 \lambda\right)\right)$$ which has a finite value according to Appendix C of this paper and $I_3$ is one of the four terms after expanding it. If you can suggest any other simpler way to calculate $D_2$, that would also be great.

Thanks in advance :)

答:
$$I(a)=\int_0^\infty\frac{\sin(ax)}{x^4}dx$$
we can analyse its converge by the following: $u=ax\Rightarrow x=u/a,\,dx=du/a$
$$I(a)=\int_0^\infty\frac{\sin(u)}{(u/a)^4}\frac{du}a=a^3\int_0^\infty\frac{\sin(u)}{u^4}du$$
now we need to look at what happens as the integrand tends to zero, to do this I will split it into two parts:
$$J=\int_0^\epsilon\frac{\sin(u)}{u^4}du+\int_\epsilon^\infty\frac{\sin(u)}{u^4}du$$
the problem is this first integral, for a small enough $\epsilon$ it is fair for us to say $\sin u\approx u$ so:
$$\int_0^\epsilon\frac{\sin u}{u^4}du=\int_0^\epsilon\frac1{u^3}du\to\infty$$


问:
How can I construct a variation problem whose solution is $f^{(3)}=0$?
For example, for the variation problem
$$
\min_{f\in H([a,b])} \int_a^b f''^2(x) \mathrm{d} x
$$

Based on Euler-Lagrange, we can obtain
$$
f^{(4)}=0
$$

Now I need to find a variation problem whose solution is
$$
f^{(3)}=0
$$

However, it seems really hard to construct such a variation problem. Something like $\min \int f' f'' \mathrm{d} x$ and $\min \int f f^{(3)} \mathrm{d} x$ cannot work.

Could anyone please help me with that?

答:
The Euler-Lagrange equation corresponding to the stationary values of the functional
$$
I[f] = \int_a^b \mathcal{L}(x, f, f', \dots, f^{(k)}) \, \text d x
$$
reads
$$
\frac{\partial \mathcal L}{\partial f} - \frac{\text d}{\text d x}\frac{\partial \mathcal L}{\partial f'} + \dots + (-1)^k \frac{\text d^k}{\text d x^k}\frac{\partial \mathcal L}{\partial f^{(k)}} = 0\, .
$$
One notes that the differential equation $f^{(2p)} = 0$ with even-order derivative is deduced from all the Lagrangian functions of the form $$
\mathcal{L} = f^{(k)}f^{(2p-k)}, \qquad k=0, \dots, 2p.
$$
However, it is difficult to perform the same operation for odd-order derivatives, say differential equations of the form $f^{(2p+1)} = 0$. A famous example is the case of the damped harmonic oscillator equation
$$
f'' + 2\zeta\omega f' + \omega^2 f = 0 \, ,
$$
for which one may introduce an $x$-dependent Lagrangian to account for the damping term $2\zeta\omega f'$. Alternatively, one may consider a quasi-variational framework and introduce dissipation functions, or make use of further tricks (see other answers).


问:
calculus and exponential functions
$0\leq t\leq3$, $t$ (measured in hours)

$$G_1(t)=4e^{t/2}$$ Diabetic
$$G_2(t)=8-4e^{-t/2}$$ Non diabetic-

Relating to the question but the whole question isn't important, there are six parts and I cant seem to get two of them right.

d) When does the diabetic patient have a glucose level of $10$? (G corresponds to glucose levels)
My working out:

$$G_1(t)=4e^{t/2}$$
Let $G_1=10$
$$10=4e^{t/2}$$
$$10/4=e^(t/2)$$
$$ln(10/4)=t/2$$
$$t= 2ln(10/4)$$
$$t \approx 1.8$$ (1 hour and 80 min so becomes two hours etc)
however this isn't the answer :( it is actually 1 hour and 50 minutes!

f) At which point in the experiment has the diabetic patient twice the glucose levels of the non diabetic patient?

I know $$2G_1=G_2$$
$$2(4e^{t/2})=8-4e^{-t/2}$$

BUT i CANT SEEM TO FIND $t$

help please, thankyou in advance!

答:
Part $1$ is good. See the comment of Alex Becker. $1.8$ hours is one hour and $0.8$ of another hour which consists of $60$ minutes. $0.8$ of $60$ is $0.8 \cdot 60 = 48$. Therefore $1.8$ hours is $1$ hour and $48$ minutes.

Part $2$

EDIT:
You must set (!)
$$
G_1 = 2G_2
$$
From
$$
4e^{\frac{t}2} = 2(8 - 4e^{\frac{-t}2})
$$
you get, by multiplying the equation by $e^{\frac{t}2}$,
$$
4e^t = 16 e^{\frac{t}2} - 8
$$
or
$$
e^t - 4e^{\frac{t}2} + 2 = 0
$$
Now substitue $e^{\frac{t}2} = x$ and use $ (e^{\frac{t}2})^2 =e^{\frac{t}2} \cdot e^{\frac{t}2} = e^t$ to get
$$
x^2 -4x + 2 =0
$$
Tell me if you can proceed from here. It looks like no real solution is there.

If we analyse a bit the function for diabetic patient we can see that it is exactly $4$ at $t=0$ and monotonically increases to $4e^{\frac32}=17.93$ at $t=3$.

On the other hand, the function for non-diabetic patient is exactly $4$ at $t=0$ and monotonically increases to $8-4e^{-\frac32}=7.11$ at $t=3$.

Title: calculus and exponential functions


问:
The limit and asymptotic analysis of $a_n^2 - n$ from $a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}$
I came up with the following question which is the follow up of
How to prove that for $a_{n+1}=\frac{a_n}{n} + \frac{n}{a_n}$ , we have $\lfloor a_n^2 \rfloor = n$?

Problem: Let $a_1 = 1,\quad a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n},\quad n\ge 1$.

Prove that $\lim_{n\to \infty} (a_n^2 - n) = \frac{1}{2}$;

Give the asymptotic analysis of $a_n^2 - n - \frac{1}{2}$.

Edit (2021/02/16) I also posted in https://mathoverflow.net/questions/384047/asymptotic-analysis-of-x-n1-fracx-nn2-fracn2x-n-2

For 1), I use the mathematical induction to prove the claim
$$n + \frac{1}{2} - \frac{2}{n} < a_n^2 < n + \frac{1}{2} + \frac{13}{4n} + \frac{13}{8n^2} + \frac{157}{16n^3}, \quad n \ge 22. \tag{1}$$
However, we need to verify it for $n = 22$ (a computer is required).
Are there simpler solutions?

$\color{blue}{\textbf{Edit}}$ 2021/02/15:

For 1), there is a solution in [1] (I know it from @haidangel's post The variation of a Ukrainian Olympiad problem: 10982). The authors proved that
$\frac{n^2}{n-1/2} \le a_n^2 \le \frac{(n-1/2)^2}{n-3/2}$ for all $n\ge 3$.

[1] Yuming Chen, Olaf Krafft and Martin Schaefer, “Variation of a Ukrainian Olympiad Problem: 10982”,
The American Mathematical Monthly, Vol. 111, No. 7 (Aug. - Sep., 2004), pp. 631-632

For 2), I have no idea currently. I want to find something like: for example, for the recurrence relation
$b_0 = 1, b_{n+1} = b_n + \frac{1}{b_n}, n\ge 0$,
we have $b_n \sim \sqrt{2n} + \frac{\sqrt{2}}{8\sqrt{n}}\ln n + o(\frac{\ln n}{\sqrt{n}})$. (Thank @Diger for pointing out the mistake. See the comment.)

About how to construct the claim (1): I want to find $d_n, c_n$ such that, for sufficiently large $n$,
$$n + \frac{1}{2} - d_n < a_n^2 < n + \frac{1}{2} + c_n.$$
To use the the mathematical induction, we need
$$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2
< \frac{n + \frac{1}{2} + c_n}{n^2} + \frac{n^2}{n + \frac{1}{2} - d_n} + 2
< n + 1 + \frac{1}{2} + c_{n+1},$$
$$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2
> \frac{n + \frac{1}{2} - d_n}{n^2} + \frac{n^2}{n + \frac{1}{2} + c_n} + 2
> n + 1 + \frac{1}{2} - d_{n+1}$$
which results in
$$c_{n+1} - \frac{c_n}{n^2} > \frac{n + \frac{1}{2}}{n^2} + \frac{n^2}{n + \frac{1}{2} - d_n} + \frac{1}{2} - n,$$
$$c_n < \frac{n^2}{n - \frac{1}{2} - d_{n+1} - \frac{n + \frac{1}{2} - d_n}{n^2}} - n - \frac{1}{2}.$$
We first choose $d_n$, then determine $c_n$.
For example, $d_n = \frac{2}{n}$ and $c_n = \frac{13}{4n} + \frac{13}{8n^2} + \frac{157}{16n^3}$.

答:
Since $a_1=1>0$ it is clear that $a_n>0$ for all $n$. Squaring gives $$a_{n+1}^2 = \frac{a_n^2}{n^2} + \frac{n^2}{a_n^2} + 2$$ and defining $a_n^2=nb_n$, this recurrence becomes $$b_{n+1}=\frac{b_n}{n(n+1)} + \frac{n}{(n+1)b_n} + \frac{2}{n+1}$$
with $b_1=1$. Now suppose $$1\leq b_n \leq 1+\frac{1}{n}+ \frac{2}{n^2}$$
which is true for $b_2=2$, $b_3=4/3$ and $b_4=(13/12)^2$ and continue inductively, i.e.
$$b_{n+1}\geq \frac{1}{n(n+1)} + \frac{n}{(n+1)(1+1/n+2/n^2)} + \frac{2}{n+1} = 1 + \frac{3n+2}{n(n+1)(n^2+n+2)}\geq 1$$
and also $$b_{n+1}\leq \frac{1+1/n+2/n^2}{n(n+1)} + \frac{n}{n+1} + \frac{2}{n+1} \\ = 1 + \frac{1}{n+1} + \frac{2}{(n+1)^2} - \frac{1-2/n-3/n^2-2/n^3}{(n+1)^2} \leq 1 + \frac{1}{n+1} + \frac{2}{(n+1)^2}$$
whenever $n\geq 4$. Taking the limit on both sides it follows $$\lim_{n\rightarrow \infty} b_n = 1 \, .$$ Next we formally write $b_n$ as an asymptotic expansion $$b_n = 1 + \sum_{k=1}^\infty \frac{c_k}{n^k}$$ and insert it into $$(n+1)b_{n+1}b_n - \frac{b_n^2}{n} - 2b_n - n = 0$$ which gives after some extensive algebra
$$0 = 2c_1 - 1 \\
+ \sum_{m=1}^\infty \frac{1}{n^m} \left\{ \sum_{k=0}^m (c_{m-k} + c_{m+1-k}) \sum_{l=0}^k \binom{-l}{k-l} c_l + \sum_{l=0}^{m+1} \binom{-l}{m+1-l} c_l - \sum_{k=0}^{m-1} c_k c_{m-1-k} - 2c_m \right\} \, .$$
Setting the coefficient of each power $n^{-m}$ to zero, iteratively gives a linear equation for $c_{m+1}$ ($m=1,2,3,...$) in terms of $c_0=1$ and $c_k$ ($k=1,2,...,m$). The coefficient of $n^0$ already gives $c_1=1/2$.

The first higher coefficients read: $c_2=5/8, c_3=13/16, c_4=155/128, c_5=505/256$. The denominator seems to follow a power of $2$ pattern.


问:
Proof check $f(x)=\frac{2x}{x-4}$ diverges as $x\to4$ using $\epsilon-\delta$
My nephew asked me to consider  $\lim_\limits{x\to 4 }\left(\frac{2x}{x-4}\right)$ for a problem set he was doing.

I explained that it diverged by substituting $x=4+\delta$ and letting $\delta\to 0$.

I thought I'd try to do it formally using $\epsilon-\delta$ definition of divergence for practice as I haven't done one in years and I find them difficult.

Could someone please check the following as there's likely an error?

I used this definition:
$\forall M>0, \exists\delta >0:|x-a|<\delta\implies f(x)>M$

For the problem at hand:

$\lim_\limits{x\to 4 }\left(\frac{2x}{x-4}\right)$

We require:

$\forall M>0, \exists\delta >0:|x-4|<\delta\implies \frac{2x}{x-4}>M$

Some scratchwork is necessary

1st manipulate $\frac{2x}{x-4}>M$ as follows

$\left|\frac{2x}{x-4}\right|>|M|\implies \frac{|2x|}{|x-4|}>M$

Note that we may assume that $\delta\leq3\implies 1\leq x \leq7$

So $|2x|\geq2$

$\frac{|2x|}{|x-4|}\geq\frac{2}{|x-4|}>M$

So we may choose $\delta=\frac{2}{M}$

Proof:
$\forall M>0 \exists \delta=\frac{2}{M}:|x-4|<\delta=\frac{2}{M}\implies \frac{2x}{x-4}>M$

Thank you

答:
It is nearly fine. The only problems that I see are:

The final value for $\delta$ should be $\min\left\{\frac2M,3\right\}$, not just $\frac2M$.

What that actually proves is that $\lim_{x\to4^+}\frac{2x}{x-4}=\infty$. That is, what you actually proved is that, for any $M>0$, if you take the $\delta$ from above, then$$4<x<4+\delta\implies\frac{2x}{x-4}>M$$and that is enough to prove that that limit doesn't exist (in $\Bbb R$). But $\lim_{x\to4^-}\frac{2x}{x-4}=-\infty$.


问:
Draw the set $M_{4}$={(x, y) ; |xy|< 1/4}
I want to draw the set $M_{4}$={(x, y) ; |xy|< 1/4}

My attempt is that I evaluate the expression of |xy|<1/4 depending on the values of x an y, i.e depending on the quadrant.

For the second quadrant I get the following:

x < 0 $\implies$ |x| = -x

y $\ge$0  $\implies$ |y| = y

$\implies$ |xy| =-xy < $\frac{1}{4}$ $\implies$ y > - $\frac{1}{4x}$

Which does not make any sense as the set is enclosed by the lines y=-$\frac{1}{4}$ y = $\frac{1}{4x}$, and thus for the second quadrant I should get y < - $\frac{1}{4x}$. I don't understand where I go wrong, any help would be much appreciated!

答: An idea: draw first the two hyperbolas $\;y=\pm\cfrac1{4x}\;$ , and then "bound" the points with both coordinates different from zero belonging to $\;M_4\;$ by them. Of course, the above involves assuming $\;x\neq0\;$, but together with $\;y=0\;$ (both axis are contained in $\;M_4\;$) you shall get what you need.


问:
How to find the primitive function of $\frac{1}{5+2\sin x-\cos x}$
I am trying to find the primitive function of $\displaystyle\int_{}^{}\frac{dx}{5+2\sin x-\cos x}$. I've got $$\int_{}^{}\frac{dx}{5+2\sin x-\cos x}=\frac{\textrm{arctan}\left(\frac{3\tan\frac{x}{2}+1}{\sqrt{5}}\right)}{\sqrt{5}}$$ However the plot of this function isn't continuous - exactly these points $\pi+2k\pi$ make the problem. It means that this primitive function doesn't exist on whole $\mathbb{R}$, but just on some small interval. So what I have to do in this case? And how the whole primitive function looks like?

Thank you

答:
Remember that on each interval of the type $(\pi+2k\pi, x+(2k+2)\pi)$ your antiderivative is actually.

$$\int_{}^{}\frac{dx}{5+2\sin x-\cos x}=\frac{\textrm{arctan}\left(\frac{3\tan\frac{x}{2}+1}{\sqrt{5}}\right)}{\sqrt{5}} +C$$

Now, at $x=\pi+2k\pi$ your function has a jump of $\frac{\pi}{\sqrt{5}}$, so if you chose your constant as

$$\int_{}^{}\frac{dx}{5+2\sin x-\cos x}=\frac{\textrm{arctan}\left(\frac{3\tan\frac{x}{2}+1}{\sqrt{5}}\right)}{\sqrt{5}} + k \frac{\pi}{\sqrt{5}} \,;\, x \in (\pi+2k\pi, x+(2k+2)\pi)$$

your antiderivative becomes continuous on $\mathbb R$... To make it more accurate, let

$$ g: \mathbb R \backslash \{ \pi+ 2k \pi \} \,;\, g(x)= k \frac{\pi}{\sqrt{5}} \,;\, x \in (\pi+2k\pi, x+(2k+2)\pi) \,,$$

Then $g$ is locally constant, and

$$\frac{\textrm{arctan}\left(\frac{3\tan\frac{x}{2}+1}{\sqrt{5}}\right)}{\sqrt{5}} +g(x) +C \,,$$

is an antiderivative of your function on  $\mathbb R \backslash \{ \pi+ 2k \pi \}$ which has removable discontinuities at $ \{ \pi+ 2k \pi \}$, thus it can be extended to a continuous function on $\mathbb R$.


问:
Finding $\mathbb{E}(s^T)$ for simple symmetric random walk on $\mathbb{Z}$

Let $X_n$ be a simple, symmetric random walk on $\mathbb{Z}$ with
  $X_0=0$. Let $$ T=\inf\{n\ge 1 : X_n=0\} $$ Compute $\mathbb{E}(s^T)$
  for fixed $s\in(0,1)$.

Apologies if this has been asked before, it seems like the sort of question that may have but not that I've found.

We have already shown that the walk is null recurrent.

I thought to use the identity
$$
P_{i,j}(s)=\delta_{ij}+F_{i,j}(s)P_{j,j}(s)
$$
setting $i=j=0$ and rearranging
$$
\mathbb{E}(s^T)=F_{0,0}(s)=\frac{P_{0,0}(s)-1}{P_{0,0}(s)}
$$
where
$$
P_{0,0}(s)=\sum^{\infty}_{n=0}s^n\mathbb{P}(X_n=0)=\sum^{\infty}_{k=0}\Big(\frac{s}2\Big)^{2k}{2k\choose k}
$$
and I don't see how this is useful or that it even converges.

Preferably I would like an answer that doesn't use Catalan numbers (although that may be useful for answer checking)

Thank you

Body:
First, note that
$$
\mathsf{P}(T=2n)=\frac{2^{-2n}}{2n-1}\binom{2n}{n}.
$$
Thus, for $s\in(0,1)$,
$$
\mathsf{E}s^T=\sum_{k\ge 1}\frac{(s/2)^{2n}}{2n-1}\binom{2n}{n}=1-\sqrt{1-s^2}.
$$

To see the last equality, notice that
$$
A\equiv\sum_{n\ge 1}\frac{(s/2)^{2n}}{2n-1}\binom{2n}{n}=\left(\frac{s}{2}\right)^2\sum_{n\ge 1}\frac{(s/2)^{2n-2}}{2n-1}\binom{2n}{n}
$$
and
$$
\int_0^s \left(\frac{x}{2}\right)^{2n-1}\, dx=\frac{s(s/2)^{2n-2}}{2n-1}.
$$
Consequently, since
$$
\sum_{n\ge 1}\binom{2n}{n}x^{2n-2}=\frac{4}{\sqrt{1-x^2}\left(1+\sqrt{1-x^2}\right)},
$$
we get
$$
A=\frac{s}{4}\int_0^s\frac{4}{\sqrt{1-x^2}\left(1+\sqrt{1-x^2}\right)}\,dx=1-\sqrt{1-s^2}.
$$


问:
$C^*$-algebraic tensor product preserves strict inclusions?
Let $I$ be a proper closed ideal of a $C^*$-algebra $A$ and let $B$ be a unital (for simplicity) $C^*$-algebra. We have a natural inclusion
$$I \otimes B \subseteq A \otimes B$$
Here, the tensor products are the spatial ones (minimal tensor product).

Can it ever happen that $I \otimes B = A \otimes B?$

I tried the following:

Let $a \in A \setminus I$. We want to show that $a \otimes 1_B \notin I \otimes B$. It is easy to see that $a \otimes 1_B \notin I \odot B$ (algebraic tensor product), but how can we ensure $a \otimes 1_B$ is not a limit of elements in $I \odot B$?

答:
Considering the quotient map $q: A\to A/I$, one has that
$$
  q\otimes \text{id} : A\otimes B \to (A/I)\otimes B
  $$
is a well defined, nonzero map, vanishing on $I\otimes B$.  Therefore $I\otimes B\subsetneq A\otimes B.$


答:
You are right to consider the set in four different quadrants.

For $xy<0$, you have $-xy<\frac14$.

For $x>0$ and $y<0$,
$$
y>\frac{-1}{4x}
$$

For $x<0$ and $y>0$,
$$
y<\frac{-1}{4x}
$$

Notes.

The mistake you make is that for $x<0$ and $y>0$, you should have
$$
-xy<\frac14\Rightarrow y\color{red}{<}\frac{-1}{4x}
$$ because you are dividing the quantity $-x$ on both sides and it is a positive quantity. So the direction of the inequality does not change.

Title: Draw the set $M_{4}$={(x, y) ; |xy|< 1/4}

Body:
I want to draw the set $M_{4}$={(x, y) ; |xy|< 1/4}

My attempt is that I evaluate the expression of |xy|<1/4 depending on the values of x an y, i.e depending on the quadrant.

For the second quadrant I get the following:

x < 0 $\implies$ |x| = -x

y $\ge$0  $\implies$ |y| = y

$\implies$ |xy| =-xy < $\frac{1}{4}$ $\implies$ y > - $\frac{1}{4x}$

Which does not make any sense as the set is enclosed by the lines y=-$\frac{1}{4}$ y = $\frac{1}{4x}$, and thus for the second quadrant I should get y < - $\frac{1}{4x}$. I don't understand where I go wrong, any help would be much appreciated!


问:
Evaluating the double integral $\int_0^\infty d a \int_0^\infty d b\ \frac{ \sin(x a) \sin( y b ) }{a+b}$.
Consider the double-integral
$$
F(x,y) := \int_0^\infty d a \int_0^\infty d b\ \frac{ \sin(x a) \sin( y b ) }{a+b} \ .
$$
By playing around in Mathematica I have come to believe that the above evaluates to
$$
F(x,y) \stackrel{?}{=} \frac{\pi}{2(x+y)} \ .
$$
How would one prove this result?

EDIT: I've been able to evaluate the $a$-integral, showing that the above is equal to
$$
F(x,y) = \int_0^\infty db\ \sin(by) \bigg[ \text{Ci}(bx) \sin(bx) + \frac{1}{2} \cos(bx) \big( \pi - \text{Si}(bx) \big) \bigg] \ ,
$$
where $\text{Ci}$ and $\text{Si}$ are the cosine integral and sine integral functions, respectively. From here I cannot make progress.

There is a symmetry in the order of integration (ie. the above would look the same if I did the $b$-integral first). It is curious that the function seems to depend on $x+y$: I don't see how this dependence falls out of the definition of $F$.

答:
$$F(x,y)=\int_0^\infty\int_0^\infty\frac{\sin(xa)\sin(yb)}{a+b}da\,db$$
$u=xa,da=du/x$ and $v=yb,db=dv/y$ so:
$$F(x,y)=\int_0^\infty\int_0^\infty\frac{\sin u\sin v}{u/x+v/y}\frac{du}{x}\frac{dv}{y}=\int_0^\infty\int_0^\infty\frac{\sin u\sin v}{yu+xv}du\,dv$$
makes the sines simpler but we are probably going to have to use another system I think

$$F(x,y)=\int_0^\infty\int_0^\infty\frac{\sin(xa)\sin(yb)}{a+b}da\,db$$
we could try $u=a+b, v=a-b$ we just have to make sure the jabobian is non-zero:
$$da\,db=\left|\frac{\partial(a,b)}{\partial(u,v)}\right|du\,dv$$
we can work out that:
$$a=\frac{u+v}{2}$$
$$b=\frac{u-v}{2}$$
and so:
$$\left|\frac{\partial(a,b)}{\partial(u,v)}\right|=\left|\begin{matrix}\frac12 & \frac12 \\ \frac12 &-\frac12\end{matrix}\right|=-\frac12$$
$$da\,db=-\frac12du\,dv$$
so our integral will become:
$$F(x,y)=-\frac12\iint\limits_S\frac{\sin\left(\frac x2(u+v)\right)\sin\left(\frac y2(u-v)\right)}{u}dudv$$
Now you need to separate $u,v$ using compound angle rule and work out the domain $S$


Title: Evaluating the double integral $\int_0^\infty d a \int_0^\infty d b\ \frac{ \sin(x a) \sin( y b ) }{a+b}$.

Body:
Consider the double-integral
$$
F(x,y) := \int_0^\infty d a \int_0^\infty d b\ \frac{ \sin(x a) \sin( y b ) }{a+b} \ .
$$
By playing around in Mathematica I have come to believe that the above evaluates to
$$
F(x,y) \stackrel{?}{=} \frac{\pi}{2(x+y)} \ .
$$
How would one prove this result?

EDIT: I've been able to evaluate the $a$-integral, showing that the above is equal to
$$
F(x,y) = \int_0^\infty db\ \sin(by) \bigg[ \text{Ci}(bx) \sin(bx) + \frac{1}{2} \cos(bx) \big( \pi - \text{Si}(bx) \big) \bigg] \ ,
$$
where $\text{Ci}$ and $\text{Si}$ are the cosine integral and sine integral functions, respectively. From here I cannot make progress.

There is a symmetry in the order of integration (ie. the above would look the same if I did the $b$-integral first). It is curious that the function seems to depend on $x+y$: I don't see how this dependence falls out of the definition of $F$.

Body:
$$I=\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin ax \sin bx}{a+b} da db =
\int_{0}^{\infty} \int_{0}^{\infty} \sin ax \sin by ~ da ~ db ~ e^{-(a+b)t}~dt$$
$$I=\int_{0}^{\infty} dt\left(\int_{0}^{\infty} \sin ax ~e^{-ta} da \int_{0}^{\infty}\sin by ~  db ~ e^{-tb}\right)$$
$$I=\int_{0}^{\infty} \frac{xy}{(t^2+x^2)(t^2+y^2)}dt= \frac{xy}{y^2-x^2} \int_{0}^{\infty}\left(\frac{1}{t^2+x^2}-\frac{1}{t^2+y^2}\right)=\frac{\pi}{2(x+y)}.$$
Note that:
$$J=\int \sin ax ~e^{-ta} da=\Im \int_{0}^{\infty} e^{-(t-ix)}=\frac{x}{t^2+x^2}$$


问:
Uniform distribution solution verification

An insurer offers a travelers insurance policy. Losses under the policy are uniformly distributed on the interval $[0, 5]$.

The insurer reimburses a policyholder for a loss up to a maximum of $4$.

Determine the cumulative distribution function, $F$, of the benefit that the insurer pays a policyholder who experiences exactly one loss under the policy.

My attempt:

Let $Y$ represent the payment RV. Then $Y = x$ if $0\le x<4$ and $Y = 4$ if $4\le x<5$. So the cumulative distribution function is
$$F_Y(x) = \begin{cases}
\displaystyle\int_0^x t\; dt = \frac{x^2}{2} \;\;\text{if $0\le x<4$}\\
\displaystyle\int_4^x 4\; dt = 4(x-4) \;\;\text{if $4\le x<5$}
\end{cases}$$
This is not the correct answer. Can someone please suggest where I have gone wrong?

答:
$$F_B(b) =
\begin{cases}
0,  & \text{if $b<0$ } \\
\frac{b}{5},  & \text{if $0\leq b<4$ } \\
1, & \text{if $b\geq 4$ }
\end{cases}$$

Observe that your CDF has "a jump" in $b=4$ thus the random varaible $B$ is not absolutely continuous and for this reason it does not admit a density

Uniform rv in $[0;5]$ has the following density $f_X(x)=\frac{1}{5}\cdot\mathbb{1}_{[0;5]}(x)$ thus the CDF can be calculated with integrals in the following way

$$F_B(b)=\frac{1}{5}\int_0^b dt=\frac{b}{5}$$

The second part of the CDF is

$$\frac{4}{5}+\frac{1}{5}\int_4^5 dt=1$$


问:
Cosine rule to find a side of a triangle
I would like to ask you for help for the following problem:

The sides of a $\triangle ABC$ are $BC=2,AB=5,AC=\sqrt{22}$ and point $D$ lies on $AC$, such that $BD=3$. Find $CD$.

 $CD=\dfrac{\sqrt{22}}{2}$

I have just studied the Cosine rule and I think I am supposed to use it. I don't see how, though. To find $CD$ we can use the Cosine rule in $\triangle BCD:CD^2=BC^2+BD^2-2.BC.BD.\cos\measuredangle CBD$ but I don't see how to find $\cos\measuredangle CBD$. I also found $\cos\gamma=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{1}{4\sqrt{22}}=\dfrac{\sqrt{22}}{88}.$

答:
First, use the cosine rule on triangle ACB to find $\measuredangle ACB$.

Then use the cosine rule on triangle BCD to find the length of $CD$.


问:
Why does every $ m \times n$ matrix of rank $r$ reduce to $(m \times r)$ times $(r \times n)$?
How can I prove the following statement?

Every $m \times n$ matrix of rank $r$ reduces to $(m \times r)$ times $(r \times n)$:

$A = $ (pivot columns of $A$) (first $r$ rows of $R$) = (COL)(ROW).

[ Source: Gilbert Strang, Introduction to Linear Algebra, question $56$, section $3.2$. ]

I think that the $R$ is reduced row echelon form. I believe there is a brief elegant proof.

答:
Let $m\times n$ matrix $A=\left[a_1, \ldots, a_r, a_{r+1},\ldots,a_n\right]$, without loss of generality (I guess you can switch the columns), let first $r$ columns are its pivot columns, denote as $P$, and the left $n-r$ columns are free columns, denote as $F$.

Its reduced echelon matrix $R=\left[\begin{array}{c c}
I_{r\times r}&F^0_{r\times(n-r)} \\
0_{(m-r)\times r}& 0_{(m-r)\times (n-r)}
\end{array}\right] $.

Since $R$ is derived from row eliminations, it’s equivalent to left multiply a series of elementary matrices, which are $m\times m$ and invertible.

So we have

$A=ER$, $E$ is an invertible $m\times m$ matrix.

Rewrite the equation above as
$[P,F] = E\left[\begin{array}{c c}
I & F^0\\
0 & 0
\end{array}\right]
$,

so, $P=E\left[\begin{array}{c}
I\\
0\end{array}\right],F=E\left[\begin{array}{c}
F^0\\
0\end{array}\right]$.

We want to prove $A=P[I,F^0]$, that is, $A$ equals the pivot columns of $A$ times of first $r$ rows of $R$.

Since $P[I,F^0]=[PI,PF^0]=[P, PF^0]$, the first $r$ columns are pivot columns themselves. left we want to show $PF^0=F$, the free columns.

Since $P=E\left[\begin{array}{c}
I\\
0\end{array}\right]$, then

$PF^0 = E\left[\begin{array}{c}
I\\
0\end{array}\right]F^0=E\left[\begin{array}{c}
F^0\\
0\end{array}\right]=F$. It's indeed the free columns. Proved.


答:
Your definition of divergence is wrong, so I didn't read any further.

To prove: $\lim_{x \to 4} \frac{2x}{x-4}$ does not exist.

In order to get the correct epsilon-delta definition of divergence, you must specify the correct epsilon-delta definition of convergence, and then negate the definition.

The epsilon-delta definition of convergence is:

There exists a limit $L$ such that:

$\forall \epsilon > 0 ~~\exists ~\delta > 0$ such that

when $0 < |x - 4| < \delta$, then $|f(x) - L| < \epsilon,$

where $f(x) = \frac{2x}{x-4}$.

In order to prove divergence, you must therefore prove that:

For any proposed limit $L$,

a specific $\epsilon > 0$ can be found such that

regardless of how small you make $\delta$, where $0 < \delta$,

you will be able to find at least one $x$ such that both
$0 < |x - 4| < \delta~$ and $~|f(x) - L| \geq \epsilon.$

It is sufficient to demonstrate that regardless of the proposed value of $L$, setting $\epsilon = 1$ will work.

With $L$ some fixed value, and $\epsilon = 1$, regardless of the  positive value of $\delta$ chosen, you can arbitrarily decide that you are going to look for a (positive) $\delta_1 < \delta$

and then choose $x = 4 + \delta_1.$

Then, you will have that the numerator of $f(x)$ will be

greater than $8$, and

the denominator of $f(x)$ will be $\delta_1$.

Therefore, $f(x)$ will be greater than $\frac{8}{\delta_1}.$

Since you can choose $\delta_1$ as small as you want, you can choose $\delta_1$ small enough

so that $[f(x) - L] > 1 = \epsilon.$

This means that regardless of your proposed value of $L$, and regardless of your choice of (positive) $\delta$, you will be able to find at least one $x$ such that $0 < |x - 4| < \delta$ and

$|f(x) - L| > 1.$

Title: Proof check $f(x)=\frac{2x}{x-4}$ diverges as $x\to4$ using $\epsilon-\delta$

Body:
My nephew asked me to consider  $\lim_\limits{x\to 4 }\left(\frac{2x}{x-4}\right)$ for a problem set he was doing.

I explained that it diverged by substituting $x=4+\delta$ and letting $\delta\to 0$.

I thought I'd try to do it formally using $\epsilon-\delta$ definition of divergence for practice as I haven't done one in years and I find them difficult.

Could someone please check the following as there's likely an error?

I used this definition:
$\forall M>0, \exists\delta >0:|x-a|<\delta\implies f(x)>M$

For the problem at hand:

$\lim_\limits{x\to 4 }\left(\frac{2x}{x-4}\right)$

We require:

$\forall M>0, \exists\delta >0:|x-4|<\delta\implies \frac{2x}{x-4}>M$

Some scratchwork is necessary

1st manipulate $\frac{2x}{x-4}>M$ as follows

$\left|\frac{2x}{x-4}\right|>|M|\implies \frac{|2x|}{|x-4|}>M$

Note that we may assume that $\delta\leq3\implies 1\leq x \leq7$

So $|2x|\geq2$

$\frac{|2x|}{|x-4|}\geq\frac{2}{|x-4|}>M$

So we may choose $\delta=\frac{2}{M}$

Proof:
$\forall M>0 \exists \delta=\frac{2}{M}:|x-4|<\delta=\frac{2}{M}\implies \frac{2x}{x-4}>M$

Thank you


问:
Proof of closed form Hofstadter G-Sequence
I'm working through a discrete maths text book and was stumped as to how to prove the closed form solution of the Hofstadter G-Sequence

$a(0) = 0$ and $a(n) = n - a(a(n-1)), n \geq 1$

The closed form is $a(n) = \lfloor(n+1)\mu\rfloor$  where $\mu$ is the inverse of the golden ratio.

The text book says to prove the following two statements on the way to proving this.

Show that $n\mu -\lfloor n\mu \rfloor + n\mu^2 - \lfloor n\mu^2 \rfloor = 1$ for all n > 0

Show that $\lfloor(1+\mu)(1-\alpha)\rfloor + \lfloor\alpha+\mu\rfloor = 1$ for all $\alpha \in R, 0 \leq \alpha < 1\space and\space \alpha \ne 1-\mu$

I'm relatively certain I was able to confirm those those two statements but I'm not sure where to apply them when I start on the main proof. I think it's just my weakness with algebra, particularly how to deal with the floor that's holding me up.

答:
Our goal is to prove $$\lfloor \mu  (n+1)\rfloor  = n-\lfloor \mu  \lfloor \mu  n+1\rfloor \rfloor. $$

If you can prove  $\lfloor \mu  \lfloor \mu  n+1\rfloor \rfloor  = \lfloor (1-\mu ) (n+1)\rfloor $ then you should be able to prove the above without too much effort.

ContentLicense: CC BY-SA 2.5


问:
Central elements in the Dihedral Group of order $ 2n.$
Continuing my independent journey through "Abstract Algebra" (see this previous question for context and notation), I am working on:

If $n = 2k$ is even and $n \geq 4$, show that (a) $z = r^k$ is an element of order 2 which (b) commutes with all elements of $D_{2n}$ (Dihedral group of a regular n-gon). Show also that (c) $z$ is the only non-identity element of $D_{2n}$ which commutes with all elements of $D_{2n}$.

(a)  $z^2 = (r^k)^2 = r^{2k} = r^n = 1$, so the order of $z$ divides 2.

$\quad\quad$  Does the hypothesis that $n \geq 4$ come into play here?

(b)  Elements in $D_{2n}$ are of the form $r^i$ or $sr^i$, $0 \leq i \leq n - 1 $

$\quad\quad$  For elements $r^i$, we have: $zr^i = r^kr^i = r^{k + i} = r^{i + k} = r^ir^k = r^iz$,

$\quad\quad$  so $z$ commutes with these elements.

$\quad\quad$ For elements $sr^i$, I am a little confused (yet still trying to remember what needs to be shown!):

$\quad\quad$  $zsr^i = r^ksr^i = (r^ks)r^i = (sr^{-k})r^i = sr^{-k + i}$

$\quad\quad$ Yet $sr^iz = sr^ir^k = sr^{k + i}$

$\quad\quad$ The two don't seem equal to me, so I must be missing something?

(c) The only thing that comes to mind is the argument that the identity in a group is unique,

$\quad\quad$  but I'm having a hard time generalizing/extending that concept to this group.

Thanks in advance for any hints!

答:
(a) The fact that the order is exactly $2$ (and not merely divides $2$) now follows simply because the order of $r$ is exactly $n$, and $0\lt k\lt n$, so $z=r^k\neq 1$.

No, the hypothesis that $n\geq $ does not affect the fact that $z$ is of order $2$. This holds in all dihedral groups. That hypothesis has to do with (c) (for example, if you take $D_4$, then every element commutes with every other element, as you should have little trouble verifying, so that the assertion in (c) would be false then).

(b) Remember part (a)! You proved that $z^2 = e$, so $z=z^{-1}$. That means that $r^k = (r^k)^{-1} = r^{-k}$. You have $z(sr^i) = sr^{-k+i} = sr^{-k}r^i$; but since $r^{-k}=r^k$, then...

(c) You have already shown that every element commutes with $r^k$. Now you need to show that if $s^jr^i$, $0\leq j\leq 1$, $0\leq i\leq n-1$ commutes with everything, then either $s^jr^i = r^k$, or $s^jr^i = 1$.

Show that if $0\lt i\lt n$, and $r^i$ commutes with $s$, then $i=k$. That will show that the only power of $r$ that commutes with everything is $r^k=z$ (this is where you will need that $n\geq 4$).

Now think about elements of the form $sr^i$. If they commute with everything, then they must commute with both $s$ and with $r$. So, for example, you must have that $s(sr^i) = s^2r^i = r^i$ is equal to $(sr^i)s = s^2r^{-i} = r^{-i}$; when do you have $r^i = r^{-i}$? So that whittles down the possible elements that can commute with everything. And now, what happens when you compute $r(sr^i)$ and $(sr^i)r$? When, if ever, do you get equality?


问:
Which of the below would be less likely to achive
I would like to ask if I were to "design" a game where the less likely of the recognized dice combinations is the better, which would be the "best" of all?

The theme is a railyard. Building a ready-to-depart train is the goal. Six standard dice are used. Wagons are the numbers 1, 2, 3, 4, 5 and number 6 is the engine. Without an engine, only a loose consist can be scored. Regardless of probability, a loose consist will be inferior to any train.

Loose consist: 123, 234, 345, 1234, 2345, 12345.

Single-headed trains: 3456, 23456, 123456.

Double-headed trains: 34566, 234566.

Triple-headed-trains do not count as there should be more wagons than engines in a train.

An intermodal is(?) the (planned) highest scoring combination, which can be single or double-headed, with four (and no less than four) wagons. In ascending order, these are 665555, 655555, 664444, 644444, 663333, 633333, 662222, 622222, 661111, 611111.

My question is, whether an intermodal is more or less likely than a single or double-headed train? Which should be the "ultimate train" in the game? An intermodal (engine(s) + identical wagons) or a train (engine(s) + sequence)?

Kindly help me out.

答:
Hint:

The probability of a certain combination of a given length $n$ is proportional to number of ways to permute it:

$$
\frac{n!}{n_1!n_2!\cdots n_6!},
$$
where $n_i$ is the number of dice showing $i$ in the sequence.

Therefore the probability of a doubly headed intermodal is proportional to:
$$
5\frac{6!}{2!4!}=75,
$$
where the factor $5$ counts the ways to choose the wagon number, whereas the probability of the double headed train is proportional to
$$
\frac{6!}{2!}=360.
$$

One observes that the probability of an intermodal is 4.8 times less than that of the double-headed train.


问:
Confusion with Notation - What does $t = s0$ or $t=s1$ mean?
Reference: Blackburn, Modal Logic, Section 1.3, Ex 1.3.2:

$B$ is a set of strings of $0$s and $1$s, and $R_1,R_2$ are two relations defined on $B$ by (i) $sR_1t$ iff $t=s0$ or $t=s1$ (ii) $sR_2t$ iff $t$ is a proper initial segment of $s$.

What does the notation $s0$ and $s1$ mean? I couldn't find this in the book.

答: $t$ and $s$ are bit strings.  $s0$ is the bit string $s$ with an extra 0 bit appended.  $s1$ is the bit string $s$ with an extra 1 bit appended.


问:
Number of possible passwords with same characters
We have a password with length of 6 characters, out of a possible 27 letters, 10 digits and 8 special characters.

My questions are:

A: How many possible combinations there are?

B: How many combinations have at least two same characters?

C: How many combinations have at least one special character?

D: How many combinations have exactly two same characters (and others are different characters)?

The answer to A is $(27+10+8)^6$ but B, C and D are a mystery to me. Could I have some advice on the problem?

答:
B:  Consider the opposite event, that no characters are repeated.  Choose the first character.  Once having chosen the first character, choose the second character noting that it can not be the same as the first.  Then choose the third character noting it cannot be either of the first two characters, etc...  Apply rule of product and conclude.

 $45\times 44\times 43\times 42\times 41\times 40$, the falling factorial written commonly as $~_{45}P_6$, as $P(45,6)$, or my preference being $45\frac{6}{~}$

Using this count, subtracting it away from the result from A gives the count we are interested in

 $45^6-45\frac{6}{~}$

C: Again, consider the opposite event, the number of sequences with no special characters.  The same logic used in A can be used here to count the opposite event.  The same logic used in B taking the opposite event away from the overall total will work here as well.

 $45^6 - 37^6$

D: Pick the character which is to be repeated.  Pick the locations of that character.  Then, from left to right, pick what characters occupy the remaining positions noting that we may not repeat for these.

 $\binom{6}{2}45\frac{5}{~}$


答:
You're right about $A$: the total number is $45^6$. For $B$ you could count the ones that have all characters different, so $45 \times 44 \ldots \times 40$ and subtract that from the total. For C you can count those without a special character and subtract that amount from the total (so $45^6- 37^6$).

$D$ is the trickiest one: there you can first choose the duplicated character in $45$ ways (all are possible), and then the two positions among the $6$ where they will occur (in $\binom{6}{2} = 15$ ways) and then the $4$ remaining positions go like the different characters under B, with starting with $44$ options, so we get $45 \times 15 \times (44 \times 43 \times 42 \times 41)$ ways.

Title: Number of possible passwords with same characters

Body:
We have a password with length of 6 characters, out of a possible 27 letters, 10 digits and 8 special characters.

My questions are:

A: How many possible combinations there are?

B: How many combinations have at least two same characters?

C: How many combinations have at least one special character?

D: How many combinations have exactly two same characters (and others are different characters)?

The answer to A is $(27+10+8)^6$ but B, C and D are a mystery to me. Could I have some advice on the problem?


问:
initial value problem: y'' + 4y = f(t) , y(0)= y'(0)=0. f(t) = { 0 if t <3; t if t >3}
Solve the initial value problem:

$$y'' + 4y = f(t) , y(0)= y'(0)=0. $$
where
$$ f(t) = \begin{cases} 0 &t < 3 \\ t & t > 3\end{cases} $$

I've solved for the homogeneous equation, $y'' + 4y = 0,$ to get the general solution,
$y = A\cos(2t) + B \sin(2t)$ for some arbitrary constants $A$ and $B$.

But, now where do I go next?

Thanks!

Body:
I would use a Laplace transform:

$$\scr{L}(y''+4y)=\scr{L}f(t)$$

$$s^2Y(s)+4Y(s)=F(s)$$

$$Y(s)=\frac{F(s)}{s^2+4}$$

$$y(t)=\scr{L}^{-1} \frac{f(s)}{s^2+4}$$

The Laplace transform of $f(t)$ is $\frac{1}{s^2}e^{-3s}$.

$$y(t)=\scr{L}^{-1}\frac{1}{s^2(s^2+4)}e^{-3s}$$

Using partial fraction decomposition:

$$\frac{1}{s^2(s^2+4)}=\frac{1}{4}(\frac{1}{s^2}-\frac{1}{s^2+4})$$

$$\frac{1}{4}\scr{L}^{-1}\{\frac{1}{s^2}-\frac{1}{s^2+4}\}=\frac{1}{4}t-\frac{1}{8}\sin{2t}\space u(t)$$

$$\scr{L}^{-1}e^{-3s}=\delta{(t-3)}$$

The property of convolution states that:

$$f(t) * g(t)=\scr{L}\{f(s)g(s)\}$$

$$y(t)=\int_{-\infty}^\infty\delta{(\tau-3)}(\frac{1}{4}t-\frac{1}{4}\tau-\sin{(2(t-\tau))})u(t-\tau)d\tau$$
$$y(t)=(\frac{1}{4}t-\frac{3}{4}-\frac{1}{8}\sin(2t-6))u(t-3)$$


问:
Why does the image of an interval being unbounded imply there exists a sequence with the image of the sequence greater than n?
I am trying to understand a specific step in the proof of the Extreme Value Theorem.

The step is showing that the image of a continuous function over a compact interval, $f:[a,b] \rightarrow \mathbb{R}$, is bounded.

The proof assumes the image, $f([a,b])$ is unbounded to get a contradiction.

Why is the following true?

$$f([a,b]) \text{ unbounded } \implies \exists \ (x_n)_{n \in \mathbb{N}} \in [a,b] \text{ such that } f(x_n) \geq n$$

答: That is the definition of unbounded.  Intuitively, it says how ever far away you look, there is a point farther.  If there were not such a point, the image would be bounded above by the first $n$ where it fails.  You must be considering functions with positive range or you should have absolute value signs around $f(x_n)$ to consider the case where $f$ is unbounded negative.


问:
find angle x in isosceles triangle
Considering the attached image, is that possible to compute for $\angle x$ in the isosceles triangle below, with no further known values?

$\\$I have found $\angle B$ and $\angle C$ are $50^o$. So $\angle B$ is divided by $\overline{BP}$ to $10^o$ and $40^o$. But I can't find how the $\angle C$ is divided by $\overline{CP}$. Also its clear that $\overline{BP}$ is perpendicular to leg $\overline{AC}$.

答:
Construct equilateral triangle $ABX$ as in the picture.

Note that the concave angle $BXA$ equals $300^\circ = 2\cdot 150^\circ = 2\angle BPA$, hence $P$ lies on the circle with center $X$ and radius $XA=XB$. It follows that $\angle PXB = 2\angle PAB = 40^\circ$.

On the other hand, $AB=AX=AC$, so $A$ is the circumcenter of triangle $BXC$, hence $\angle CXB = \frac 12 \cdot \angle CAB = \frac 12 \cdot 80^\circ = 40^\circ$.

Since $\angle PXB = 40^\circ = \angle CXB$, points $X, P, C$ are collinear.

Since $AC=AX$, we have $\angle PCA = \angle AXP = 2\cdot \angle ABP = 20^\circ$, hence $$\angle APC = 180^\circ - \angle CAP - \angle PCA = 180^\circ - 60^\circ - 20^\circ = 100^\circ.$$

Title: find angle x in isosceles triangle


问:
Computing coefficients in the linear basis functions for 2D FEM
I am currently looking to write a GFEM code but to do this I need to truly understand the construction of a standard FEM code in 2D.

For the example we will be taking triangluar elements (K) with coordinates $(x_1,y_1),\ (x_2,y_2),\ (x_3,y_3)$. Now we know that the linear hat function can be written in the following way:

\begin{equation} \phi_i = a_i + b_ix + c_iy \end{equation}

where for every node, N_i we will have
$$ \phi_i(N_j) = \begin{cases} 1, \ i=j \\ 0,\ i\neq j \end{cases}$$.

Now this makes sense and I have been using it in my research until now. The issues being is how do we compute the next part? (which I have always taken for granted)

We can then find that the coefficients are:
$$ a_i = \frac{x_jy_k - x_ky_j}{2|\mathbf{K}|},\quad b_i = \frac{y_j-y_k}{2|\mathbf{K}|},\quad c_i = \frac{x_k-x_j}{2|\mathbf{K}|}  $$,

where $|\mathbf{K}|$ is the area of the triangle,

$$ det(K) = \begin{vmatrix} x_1 & x_2 & x_3\\ y_1 & y_2 & y_3 \end{vmatrix} $$.

Applying the condition above gives:
$$ 1 = a _1 + b_1x_1 + c_1y_1 \\  1 = a _2 + b_2x_2 + c_2y_2 \\ 1 = a _3 + b_3x_3 + c_3y_3,$$
but I am unsure how to proceed from here as these equations are independent of one another.
If someone could help me get derive the formulas for $a_i$ etc that would be great!

Thanks in advance.

答:
The condition is: $$\phi_i(N_j) = \delta_i^j$$ where $\phi_i$ is the hat function on the triangle defined by the nodes $(x_i,y_i)$, with $i=1,2,3$.

Look WLOG at the $\phi_1$: you have $\phi_1(x_1)=1, \phi_1(x_2)= \phi_1(x_3)=0$, i.e. the following system needs to be solved for $(a_1,b_1,c_1)$.

$\begin{bmatrix}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{bmatrix} \begin{bmatrix} a_1 \\ b_1 \\ c_1 \end{bmatrix}= \begin{bmatrix} 1\\0\\0 \end{bmatrix}$

The determinant of the matrix is equal to $2|K|$, by elementary geometry. Then, the solution can be obtained pretty easily by Cramer's rule: $$a_1 = \frac{x_2 y_3 - x_3 y_2}{2|K|}$$

$$b_1=\frac{y_2-y_3}{2|K|}$$

$$c_1=\frac{x_3-x_2}{2|K|}$$

The argument above can be easily extended in order to obtain the general formula for $a_i,b_i,c_i$.


问:
Space-time white noise grows at infinity?
I am reading some introduction texts on SPDE's and I often find the phrase "noise grows at infinity" (for instance in Gubinelli & Hofmanova's paper on $\Phi^4$ https://arxiv.org/abs/1804.11253, page 4) what does this mean?

Referring to noise as space-time white noise which can formally be understood as a centred Gaussian process $\xi$ with covariance function $\mathbb{E}[\xi(t,x)\xi(s,y)]=\delta(t-s)\delta^d(x-y)$ how am I supposed to make sense of this blow-up?

Rigorously defined $\xi$ is not a (classical) function, but only a distribution so point-wise evaluation doesn't even make sense. However, say I were to interpret is as a function or I would perhaps look at its mollification how could I see that it grows at infinity? And does this mean spacial infinity or as time goes to infinity or both?

I would be very happy with some heuristics or a even a formal argument.

答: One needs a suitable way to quantify the growth of a temperate Schwartz distribution $T$. We know the definition of being in $\mathscr{S}'$ somehow (in a rather mysterious way) says $T$ grows at most like a polynomial at infinity, i.e., like $|x|^{\alpha}$. The question is how to make this quantitative and in particular find the $\alpha$, or rather the infimum of $\alpha\in\mathbb{R}$ such $T$ "grows at most as $|x|^{\alpha}$". One way to find a good definition of this growth exponent is to do an inversion $x\mapsto x/|x|^2$ so this becomes an exponent for a singularity at a point (the origin) and then use for instance the Steinman scaling degree. See the article "On-shell extension of distributions" for a definition and use for extension of distributions. Finally, once you have the definition, aim for a Theorem of the form 1) if $\alpha>\alpha_0$, then with probability one, the random distribution $\xi$ grows at most like $|x|^{\alpha}$, and 2) if $\alpha<\alpha_0$ then  the event $\xi$ grows at most like $|x|^{\alpha}$ has probability zero. This is a bit similar to well known results about Brownian motion being Holder $(\frac{1}{2})^{-}$.


问:
A group with six elements which are given partially by relations.
In a textbook I saw the following Group
$$
  G = \{ 1, x, x^2, y, xy, x^2y \}
$$
and it was said that it is the $S_3$, surely the $S_3$ is a model of this group, but when I set $x^3 = y$ (in $S_3$ the relation would be $x^3 = 1$) I get that
$$
  G = \{ 1, x, x,^2, x^3, x^4, x^5 \}
$$
and this group is cyclic, hence commutative, so could not be the $S_3$. I cannot see were this would be a contradiction to $ G = \{ 1, x, x^2, y, xy, x^2y \}$ so I think this group has at least two models, is this true? So why did my textbook said the group $S_3$ is $G = \{ 1, x, x^2, y, xy, x^2y \}$?

答:
Your list of expressions for group elements has the nice property that if the expression $u \cdot v$ is in the list, then so are $u$ and $v$. Such a list turns the Cayley graph of the group into a directed acyclic graph with a single source (the identity), or perhaps better a spanning tree with direction. The multiplication in the group is given by the (labeled) edges omitted from the spanning tree.

A wonderful way to create such a directed graph is to decide on a “reduction ordering” on expressions. Many expressions will yield the same group element, so we desire a way to choose the “simplest” expression. If $u\cdot v$ is simplest, then both $u$ and $v$ should be simplest (otherwise why not simplify them).

Such a reduction ordering determines a special set of relations that suffice to actually simplify expressions. These relations are called a “rewriting system” and they are the natural thing: if $u$ is not a simplest form (it equals $v$ where $v$ is simplest), but every proper sub-expression of $u$ is a simplest, then we need the rule $u \mapsto v$.

Doing this for your list of expressions finds the rules $x^3 \mapsto ?$, $y^2 \mapsto ?$ and $yx \mapsto ?$. All other missing edges of the Cayley graph can be found from these by a sort of “translation” or “substitution”.

I believe these are the only possibilities:

$\begin{array}{ccc|c}
x^3 & y^2 & yx & G \\ \hline
1 & 1 & xy & C_6 \\
1 & 1 & x^2y & S_3 \\
y & 1 & xy & C_6 \\
1 & x & xy & C_6 \\
1 & x^2 & xy & C_6
\end{array}$

As you can see the choices are limited, and not independent. Additionally $C_6$ appears in several disguises.


问:
Exponential Stability in Lyapunov Razumikhin Theorem
Is there an exponential stability version of the Lyapunov-Razumikhin Theorem for retarded functional differential equations? i.e. analogous to ODEs, given a FDE $\dot{x}(t)=f(x_t)$ does the following imply global exponential stability?

For a positive definite $V:\mathbb{R}^n\to\mathbb{R}^+$ with $c_1||x||^p\leq V(x)\leq c_2||x||^p$ it holds,
$$\dot{V}(x_t) \leq -\alpha V(x(t)), \quad\text{if } V(x(t+\theta))\leq \beta V(x(t)) \quad\forall \theta \in [-\tau,0], \quad \text{where }\alpha>0, \beta>1.$$
Where $x(t)\in\mathbb{R}^n$ and $x_t(\cdot)= x(t+\cdot)\in \mathcal{C}\left([-\tau,0]\right)$.

答:
In case someone is interested in it: The answer to my above question is yes and a proof can be found here. Furthermore, the rate of decay is at least $$\gamma = \min\left(\alpha, \log( \beta)/\tau\right)/p.$$

However, the proof is quite a lot more complicated as for ODEs or Krasovskii functionals.


问:
English sentences to Aristotle (Categorical) logic
I am studying Aristotle's logic (Categorical logic AEIO). There are rules on how you can form arguments: Premise 1, Premise 2, and Conclusion: All A is B; All B is C; then "All A is C" is valid. Everything is clear if we have simple Arguments.

How to translate the next sentence in Categorical Logic: " The number of negative claims in the premises must be the same as the number of negative claims in the conclusion. If the argument follows this rule then it is valid otherwise it is not valid"?

Something like:

```<code>(C is Negative) and (P1 is negative) and not (P2 is negative)
(C is Negative) and not (P1 is negative) and (P2 is negative)
(C is Positive) and not (P1 is negative) and not (P2 is negative)

All provided argument is Follow rules
All Follow rules is Valid argument
The provided argument is Valid argument
</code>```

It is not clear how to formulate such rules in Categorical logic.

答:
The first examples is not very clear.

The second one looks like:

Every argument that Follow the rules is a Valid argument (this is an A proposition: $\text {Afv}$).

The provided argument Follows the rules.

Therefore: The provided argument is a Valid argument.

The argument is valid but it is not, strictly speaking, a syllogism, because the second premise is not a Categorical proposition. "The provided example" refers to an individual and not to a general term.

Its logical form is:

$\forall x (Fx \to Vx), Fp \vDash Vp.$

The rule needed are: Universal instantiation and Modus Ponens.

Conclusion: not every valid argument can be formulated as a Syllogism.

Syllogism is a proper part of predicate logic: the part called Monadic predicate logic.


问:
Connecting the two definitions of $e$
There are two different ways I've been taught to understand the meaning of $e$.

$e$ can be defined as the total growth from continuously compounding interest in a single period. To make the growth rate the same as the dollar amount, consider I start with $\$1$. Then, $a_n$ represents the amount I have after compounding $n$ times.
\begin{align}
a_n&=1\cdot\left(1+\frac{1}{n}\right)^n\\[5pt]
e=a_{\max}&=\lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n
\end{align}
As we increase the number of times we compound the interest, the final amount (in the case that we start with $\$1$), approaches $e$.

All exponential functions increase at a rate proportional to their current value. For example, consider the derivative of $2^x$:
\begin{align}
\frac{d\left(2^x\right)}{dx}&=\lim_{h \to 0}\frac{2^{x+h}-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x2^h-2^x}{h}\\[5pt]
&=\lim_{h \to 0}\frac{2^x\left(2^h-1\right)}{h}\\[5pt]
&=2^x\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}
\end{align}
From this we can see that the rate at which $2^x$ changes is proportional to its current value, with a proportionality constant of $$\lim_{h \to 0}\frac{\left(2^h-1\right)}{h}.$$ For $2^x$, this constant happens to be $\ln(2)$ (why is this the case?). $e$ can be defined as the base where this proportionality constant is $1$; therefore, the rate $e^x$ grows at is exactly its current value.

My question is this: how can the same constant have two (apparently) separate definitions? Is there a way of looking at $e$ that will combine these two definitions, and make it obvious as to why they are true? Also, why is the proportionality constant of any exponential functions growth rate the natural log of its base? Is there a way to understand why this is the case intuitively?

答:
Let's start with your second definition.  We want to find the derivative of $f(x) = a^x$, for some $a>0.$  Using the definition of derivative, we get

\begin{align}
f'(x) &= \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}\\
&= \lim_{h\to 0}\frac{a^{x+h} - a^{x}}{h}\\
&=a^{x}\lim_{h\to 0}\frac{a^{h} - 1}{h}.
\end{align}

Now, as you said, one way to define $e$ is to define it such that $$\lim_{h\to 0}\frac{e^{h} - 1}{h} = 1.\tag{$*$}$$  Now, I'm going to be somewhat informal here.  Suppose we have $$\frac{e^{h} - 1}{h} = 1.$$  Then, doing some manipulation: \begin{align}
\frac{e^{h} - 1}{h} &= 1\\
e^{h} - 1 &= h\\
e^{h} &= h + 1\\
e &= (h+1)^{1/h}
\end{align}

Now, by taking limits we have that
\begin{equation*}
\lim_{h\to 0}\frac{e^{h} - 1}{h} = 1 \iff e = \lim_{h\to 0}(h+1)^{1/h}.
\end{equation*}

Taking the new limit, $e = \lim_{h\to 0}(h+1)^{1/h}$, and setting $n = 1/h$, we note that $h\to 0 \implies n\to \infty$, and we can change variables to get $$e = \lim_{n\to\infty}\left(1 + \frac{1}{n}\right)^{n},$$
which matches your first definition.

As to understanding where the natural log comes from, i.e. why we have $$\lim_{h\to 0}\frac{a^{h}-1}{h} = \ln(a),$$ this is a straightforward consequence of $(*)$:
\begin{align}
\lim_{h\to 0}\frac{a^{h} - 1}{h} &= \lim_{h\to 0}\frac{e^{\ln(a^{h})}-1}{h}\\
&=\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h}\\
&=\frac{\ln(a)}{\ln(a)}\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h}\\
&=\ln(a)\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h\ln(a)}.\\
\end{align}
Now, set $k = h\ln(a)$, and note that because $a > 0$, we have that $h\to 0\implies k \to 0$ as well, and so we have by $(*)$ that:
\begin{align}
\lim_{h\to 0}\frac{a^{h} - 1}{h} &=\ln(a)\lim_{h\to 0}\frac{e^{h\ln(a)}-1}{h\ln(a)}\\
&=\ln(a)\lim_{k\to 0}\frac{e^{k} - 1}{k}\\
&=\ln(a)\cdot 1\\
&=\ln(a).
\end{align}

Title: Connecting the two definitions of $e$


问:
Deducing finiteness of the integrand from finiteness of integral
Suppose we know that:
$$
\int_{[-\pi, \pi]^d} \, g(k) \,  h^2(k) \, dk  \leq\int_{ [-\pi, \pi]^d} \, h(k) \, dk < \infty.
$$
where $h(k) = \sin(\frac{k_1}{2})^2 +  \sum_{i=2}^{d} \sin(k_i)^2.$
Assuming $d > 2$, is it possible from this to deduce that,
$$
\int_{[-\pi, \pi]^d} \, g(k)  < \infty?
$$

答: Not in general. Take the case $d=1$ and $g(k) = 1/h(k) =1/\sin^2(k)$.