问:
Show that $u$ is increasing if $u''=a(t)u+b(t)u'$
All functions are defined in an open interval $I$, and $a,b$ are continuous. The problem asked me to show that $u$ is monotonically increasing in $\{t\geq t_0:t\in I\}$, provided that $u''=a(t)u+b(t)u'$, and $u(t_0)=0$, $u'(t_0)=1$, $a(t)>0$ for all $t$.

I have so far found that:
$$u''(t_0)= a(t_0)u(t_0)+b(t_0)u'(t_0)=b(t_0) $$ Also I wrote the problem as $$u'(t)=1+\int_{t_0}^t a(s)u(s)+b(s)u'(s)ds $$ I tried exploring the methods $u(t)>u(s)$ if $t>s$, and even $u'(t)>0$ for $t\geq t_0$, but goes nowhere. I'm don't see more directions to take not sure how to go from here, any idea is appreciated
I have added the original problem came from some chapters I found out somewhere, but without author, problem and chapter are added.

答:
This is a visual argument, that I think leads to a visual intepretation of the argument already given by Hagen Von Eitzen .

If we convert the differential equation to a system of first order equations we can visualize the associated vector field.

The system is not autonomous and therefore the vector field depends on time but:

when $v=0$, it equals to $a(t)u \hat{v}$ and it points in the first quadrant, because $a>0$;

when $u=0$, the $u$ component is time independent and positive (equal to zero only in the origin). The $v$ component changes with time ;

In the figure I plotted a visualization of the vector field at an arbitrary time t.

It visually follows (and I think the argument can be made rigorous following arguments similar to the ones given by Hagen von Eitzen) that if we start in the first quadrant, we cannot escape it. Therefore $v=u'$ will remain always positive, i.e. the function is monotonous.


问:
Integrating a 2-form
Integrate the 2-form $\omega=\frac1x dy\wedge dz-\frac1y dx\wedge dz$ over the following surface:

The top half of the unit sphere using the following parametrization.

$(r,\theta)\to (r\cos\theta, r\sin\theta,\sqrt{1-r^2})$, where $0\leq \theta\leq 2\pi$ and $0\leq r\leq 1$.

I am currently working through "A Geometric Approach to Differential Forms" by David Bachman, which gives a rather informal introduction to the topic, and this is a task, which I am unable to solve.

The reason I struggle here is that I have the following formula:

$\int_M \omega =\int_R \omega_{\phi(x,y)}\left(\frac{\partial\phi}{\partial x}(x,y),\frac{\partial\phi}{\partial y}(x,y)\right) dx\wedge dy$.

And I now want to integrate the 2-form above, which is part $dy\wedge dz$ and $dx\wedge dz$.

So I am confused on how to apply this formula in this case, where we have "$dx\wedge dz-dy\wedge dz$" and not just "$dx\wedge dy$". Can you help me out?

Thanks in advance.

答:
I now solved this as follows.

I calculate the integrals seperatly. So first I calculate the integral for $\frac{1}{x} dy\wedge dz$

Then
$\int_M \frac{1}{x}dy\wedge dz=\int_R \frac{1}{r\cos(\theta)} \left(\left\langle \cos(\theta),\sin(\theta),\frac{-r}{\sqrt{1-r^2}}\right\rangle,\left\langle-r\sin(\theta),r\cos(\theta),0\right\rangle\right)dr\wedge d\theta$

Where $\langle\cdot\rangle$ just notes a vector.

Now to respect $dy\wedge dz$ it seems I have to calculate the determinant given by the y- and z-coordinates. Hence

$\int_R \frac{1}{r\cos(\theta)}\begin{vmatrix}\sin(\theta)& r\cos(\theta)\\ \frac{-r}{\sqrt{1-r^2}}&0\end{vmatrix}dr\wedge d\theta=\int_R \frac{r}{\sqrt{1-r^2}}dr\wedge d\theta=\int_0^{2\pi}\int_0^1 \frac{r}{\sqrt{1-r^2}}dr\,d\theta=2\pi$

And with a similar calculation we achiev

$\int_M -\frac{1}{y}dx\wedge dz=2\pi$.

So the result is $2\pi+2\pi=4\pi$, which is correct according to the solutions in the named book.

One question remains: How does the integral "remember" the coordinates we started with after employing the formula for the parametrization.
In the first integral I respected y- and z-coordinate.

In the second integral I take the determinant with regards to the x- and z-coordinate, which leads to

$\begin{vmatrix}\cos(\theta)&-r\sin(\theta)\\ \frac{-r}{\sqrt{1-r^2}}&0\end{vmatrix}$.

Since the solution is correct, I assume that this is the right way to do it, but I do not know why we have to take these specific determinants, as the formula is the same in both cases. So how is $dx\wedge dz$ respectively $dy\wedge dz$ respected in the formula?

Thanks in advance.

ContentLicense: CC BY-SA 4.0

Title: Integrating a 2-form


问:
Convergence of $\int_{-\infty}^\infty \frac{1}{1+x^6}dx$
Okay, so I am asked to verify the convergence or divergence of the following improper integrals:

$$\int_{-\infty}^\infty \frac{1}{1+x^6}dx$$

and

$$\int_1^\infty \frac{x}{1-e^x}dx$$

Now, my first attempt was to use comparison criterion with $$\int \frac{1}{x^2}$$

and conclude that both of the improper integrals converge given that they are smaller than the general term $\frac{1}{x^2}$.

Is it the right path? Also, are the antiderivatives of the improper integrals given easy to find?

Thanks.

答:
For the second, note that:

$\int_1^\infty \frac{x}{1-e^x}dx \leq \sum_2^{\infty} \frac{n}{1 - e^{n}}$,

so we can apply the ratio test:

$\lim_{n \to \infty} \frac{(1 -e^n)(n+1)}{(1-e^n)n}$

using L'Hopital once we get:

$\lim_{n \to \infty} \frac{1 - e^n(n+2)}{1 - e^{n+1}(n+1)}$ =

$\lim_{n \to \infty}\frac{1}{e^{n+1}(n+1)-1} + \lim_{n \to \infty}\frac{e^n(n+2)}{1 -e^{n+1}(n+1)}$ =

$\lim_{n \to \infty}\frac{e^n(n+2)}{1 -e^{n+1}(n+1)} = $

(using L'Hopital and simplifying)

$\lim_{n \to \infty} \frac{n+3}{en + 2e} = $

(using L'Hopital again)

$\frac{1}{e} < 1$

so the series, and therefore the integral converges.

ContentLicense: CC BY-SA 3.0


问:
Expanding Dawson's integral in a series of Hermite functions
We want to show that
$$F(x)=\sqrt{\pi } \sum
   _{n=0}^{\infty }
   \frac{(-1)^n H_{2
   n+1}(x)}{2^{3 n+3} \Gamma
   \left(n+\frac{3}{2}\right)}$$
where $F(x)$ is the Dawson Integral ($F(x)=\exp \left(-x^2\right)
   \int_0^x \exp
   \left(u^2\right) \, du$) and $H_n(x)$ is the $n$  th Hermite polynomial.

Since $F(x)$ is odd, we can write $F(x)=\sum _{n=0}^{\infty }
   c_{2 n+1} H_{2 n+1}(x).$ where
$$c_{2 n+1}=\frac{\int_{-\infty
   }^{\infty } \exp
   \left(-x^2\right) F(x) H_{2
   n+1}(x) \, dx}{2^{2 n+1} (2
   n+1)! \sqrt{\pi }}$$
Employing the generating function for the Hermite polynomials we can write
$$\int_{-\infty }^{\infty } \exp
   \left(-(x-t)^2\right) F(x)
   \, dx=\sum _{n=0}^{\infty }
   \frac{t^n \int_{-\infty
   }^{\infty } F(x) \exp
   \left(-x^2\right) H_n(x) \,
   dx}{n!}$$
How do we proceed to get the coefficient of $t^n$ ($n$ odd) on the left hand side ?

答:
We will proceed to evaluate the integral appearing above in $c_{2 n+1}$.
The Dawson Integral can be expressed in the form
$$F(x)=\int_0^{\infty } \exp
   \left(-y^2\right) \sin (2 x
   y) \, dy$$
Hence,
$$\int_{-\infty }^{\infty } \exp
   \left(-x^2\right) F(x) H_{2
   n+1}(x) \,
   dx=\int_0^{\infty } \exp
   \left(-y^2\right)
   \int_{-\infty }^{\infty }
   \exp \left(-x^2\right) H_{2
   n+1}(x) \sin (2 x y) \, dx
   \, dy$$
To evaluate the $x$-integral on the right hand side we invoke the generating function for the Hermite polynomials.
$$\exp \left(-x^2\right) \sin (2
   y x) \exp \left(2 x
   t-t^2\right)=\sum
   _{n=0}^{\infty }
   \frac{H_n(x) t^n \exp
   \left(-x^2\right) \sin (2 y
   x)}{n!}$$
Integrating both sides with respect to $x$ from $-\infty$ to $\infty$ we obtain
$$\int_{-\infty }^{\infty } \exp
   \left(-(x-t)^2\right) \sin
   (2 y x) \, dx=\sum
   _{n=0}^{\infty } \frac{t^n
   \int_{-\infty }^{\infty }
   H_n(x) \exp
   \left(-x^2\right) \sin (2 y
   x) \, dx}{n!}$$
The integral on the left is readily evaluated as $\sqrt{\pi } \exp
   \left(-y^2\right) \sin (2 t
   x y)$. Expanding this expression in a power series in $t$ we obtain the realtion
$$\sqrt{\pi } \exp
   \left(-y^2\right) \sum
   _{n=0}^{\infty }
   \frac{(-1)^n (2 y t)^{2
   n+1}}{(2 n+1)!}=\sum
   _{n=0}^{\infty } \frac{t^{2
   n+1} \int_{-\infty
   }^{\infty } H_{2 n+1}(x)
   \exp \left(-x^2\right) \sin
   (2 y x) \, dx}{(2 n+1)!}$$
where we have recognized that the integral vanishes for even values of $n$.
Equating coefficient of like powers of $t$ we have
$$\int_{-\infty }^{\infty } H_{2
   n+1}(x) \exp
   \left(-x^2\right) \sin (2 y
   x) \, dx=(-1)^n \sqrt{\pi }
   \exp \left(-y^2\right) (2
   y)^{2 n+1}$$
We then obtain
$$\int_{-\infty }^{\infty } \exp
   \left(-x^2\right) F(x) H_{2
   n+1}(x) \, dx=(-1)^n
   \sqrt{\pi } 2^{2 n+1}
   \int_0^{\infty } \exp
   \left(-2 y^2\right) y^{2
   n+1} \, dy$$
The integral on the right hand side evaluates to $(-1)^n \sqrt{\pi } 2^{n-1} n!$
Finally, we find the desired coefficient
$$c_{2 n+1}=\frac{(-1)^n
   n!}{2^{n+2} (2 n+1)!}$$
which can be shown to be equivalent to that given in the original problem statement


问:
Is it possible to Calculate This integral using fourier transform?
I am trying to calculate $$\displaystyle\int\limits^{\cssId{upper-bound-mathjax}{\infty}}_{\cssId{lower-bound-mathjax}{-\infty}} \dfrac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}\,\cssId{int-var-mathjax}{\mathrm{d}x}$$
for $a,b >0$

I recognize that I can split it to multiply of the derivative of $\frac{ln(x^2+a^2)}{2}$ and $\frac{ln(x^2+b^2)}{2}$ but I am stuck on calculate transform Fourier of the $ln$ function

Body:
Why not to use
$$\dfrac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}=\frac1 {a^2-b^2}\Bigg[\frac {a^2}{x^2+a^2}-\frac {b^2}{x^2+b^2} \Bigg]$$


问:
Differential equation of all second-order central curves
I am trying to solve the task: Write the differential equation of all second-order central curves whose principal axes coincide with the x and y axes. The answer is $x(y''y+y'^2)-y'y=0$.
I tried to differentiate hyperbola and ellipse formulae $$\frac{x^2}{a^2}\pm\frac{y^2}{b^2}=1$$ $$Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$$ but it doesn't work.
I tried to integrate the answer $$\frac{y''}{y'}+\frac{y'}{y}=\frac{1}{x}$$ It's a inhomogeneous equation. Integrate homogeneous part first $$\frac{y''}{y'}=-\frac{y'}{y}$$ $$ln(y')=-ln(y)+ln(C)$$ $$y'=\frac{C}{x}$$ $$\frac{y^2}{2}=C_1x+C_2$$ $$C_1'x+C_2'=0$$ $$C_1'=\frac{1}{x}$$ $$C_2'=-1$$ $$C_1=ln(x)+ln(\bar{C_1})$$ $$C_2=-x+\bar{C_2}$$ $$\frac{y^2}{2}=x(ln(x)+ln(C_1))-x+C_2$$ but I can't differentiate this into the answer either. Any help will be appreciated.

Body:
$$ax^2+by^2=1$$
$$ax\;dx+by\;dy=0\quad\implies\quad y'=\frac{dy}{dx}=-\frac{ax}{by}\tag 1$$
$$d(y')=-\frac{a}{by}dx+\frac{ax}{by^2}dy $$
$$y''=\frac{d(y')}{dx}=-\frac{a}{by}+\frac{ax}{by^2}y'\tag 2$$
From $(1)$ and $(2)$ :
$$\frac{a}{b}=-\frac{yy'}{x}=\frac{y''}{-\frac{1}{y}+\frac{xy'}{y^2}}$$
$$-yy'(-\frac{1}{y}+\frac{xy'}{y^2})=xy''$$
$$yy'-xy'^2=xyy''$$
$$\boxed{x(yy''+y'^2)=yy'}$$


问:
If two topologies have the same continuous maps into and out of them, must they be the same?
Let $X$ be a set and let $T_1,T_2$ be topologies on $X$. Assume that for any topological space $Y$, the following conditions hold:

(1) $f:Y\rightarrow X$ is continuous wrt $T_1$ iff $f:Y\rightarrow X$ is continuous wrt $T_2$

(2) $f:X\rightarrow Y$ is continuous wrt $T_2$ iff $f:X\rightarrow Y$ is continuous wrt $T_1$

Must $T_1,T_2$ be the same topologies?

答:
We only need (2) to see this: suppose $O \in \mathcal{T}_1$. Let $Y=\{0,1\}$ with topology $\{\emptyset, \{0\}, Y\}$, so called Sierpiński space. Define $f_O: X \to Y$ by

$$f_O(x) = \begin{cases}
       0 & x \in O\\
       1 & x \notin O\\
\end{cases}$$

and note that $f_O:(X,\mathcal{T}_1) \to Y$ is continuous as $f_O^{-1}[\{0\}]=O$ is open and $\{0\}$ is the only non-trivial open subset of $Y$.

So by assumption (2) applied to $Y$ we get that $f_O$ is also continuous from $(X,\mathcal{T}_2)$ and so $O=f_O^{-1}[\{0\}] \in \mathcal{T}_2$, so that $\mathcal{T}_1 \subseteq \mathcal{T}_2$. The same idea proves the reverse inclusion from $(2)$ as well.

Bonus exercise: can you also show it only from assumption (1) for some fixed $Y$ as well?


问:
Why can't we construct a Vitali Set?
We have an existence proof, but why is it hard to construct a concrete example of one?</p>

答:
If by "concrete" you mean not invoking the Axiom of Choice, then as pointed out in the comment, this is impossible.

This suggests that the whole thing boils down to defining "concrete". I'd guess you want it to at least be fixed: the Axiom of Choice provides an arbitrary set satisfying the requirements, and thus the proof of the existence of Vitali sets does not pin down a unique set. However, if a unique Vitali set could be constructed, such a constructive process would not refer to the Axiom of Choice, since it is against the very nature of constructiveness(!). This is impossible.

There might be some other definitions of "concreteness", which may or may not lead to a "concrete Vitali set".

Title: Why can't we construct a Vitali Set?


问:
Prolongation of the solution

Given
$$\begin{cases}y'=2(3x^2-1)\sqrt{y} \\y(1)=9 \end{cases}$$

the exercises asks me to determine if the solution can be extended to all $\mathbb R$.

The function $f(x,y):=2(3x^2-1)\sqrt{y}$ is $\mathcal C^{\infty}(\mathbb R\times [0,+\infty))$ and the solution of the Cauchy's problem is, assuming $y\ne 0$
$$y(x)=\pm (x^3-x+3)^2 (\text{only the positive solution is acceptable}).$$
The solution is defined in the most wide interval in which the function is not zero, so for a negative $x_{\alpha}$, the solution is defined in $(x_{\alpha},+\infty)$.

The derivative is also zero in $x_{\alpha}$ zero, since $y'(x_{\alpha})=2(3x_{\alpha}^2-1)\sqrt{y(x_{\alpha})}=0$, hence we can extend the solution with the constant function $y\equiv 0$.

Is the prolongation unique? I think that there are infinite possible prolongations of the solution to all $\mathbb R$, but I'd like to formalize this.

答:
It is a little less intuitive to think in negative time direction than in positive, as one has to also mentally reverse the vector fields.

We know $x_α=-\frac2{\sqrt3}\cosh(u_α)<-1<-\frac1{\sqrt3}$, where $\cosh(3u_α)=\frac{27\sqrt3}{2}$. Thus the factor $3x^2-1$ and with it the vector field is positive for $x<x_α$, $y>0$. Solutions in that region grow away from the zero axis, can not return there. Or seen in the other direction, the solution set contracts towards the zero axis when looking into the past, so once zero is reached, there is no getting away.

On the other hand, the solution segments with positive values are unique, so there is no other path to reach the initial point than the one you computed.

Title: Prolongation of the solution


问:
Did I solve the limit $\lim_{x \to 1} (\ln(x) \cdot \ln(1-x))$ correctly?
$\lim_{x \to 1} (\ln(x) \cdot \ln(1-x))$
I tried variable changing to change the limit to $\infty$ and then using the common limit $$\lim_{x \to \infty} \frac{\ln(x)}{x}=0$$ as below:

$x=X + 1 \implies \lim_{X \to 0} (\ln(X+1)\cdot\ln(-X))$

Then

$X = 1/y \implies \lim_{y \to \infty} (\ln((1/y)+1)\cdot\ln(-1/y))$

Then I multiplied and divided by $(\frac{1}{y}+1)$ and $(\frac{-1}{y})$

And then I had this

$\lim_{y \to \infty} ((1/y)+1)(-1/y)\cdot\frac{\ln(\frac{1}{y}+1)}{\frac{1}{y}+1}\cdot\frac{\ln(\frac{-1}{y})}{\frac{-1}{y}}=0$

Did I do it right?
Note: I knew the limit was 0 before I tried to solve it so it explains why I did these precise steps.

答:
Setting $t=1-x$, we have
$$
\lim_{x\to 1^-} \log(x)\log(1-x) =
\lim_{t\to 0^+} \log(1-t)\log(t)
= \lim_{t\to 0^+} \frac{\log(1-t)}{t}\cdot t\log(t)
$$
If you know that the following two (standard) limits exist, and what they are:
$$
\lim_{t\to 0^+}\frac{\log(1-t)}{t} \quad \text{ and } \quad \lim_{t\to 0^+} t\log t,
$$
then you can conclude.


问:
Direct Proof: $A\subseteq\bigcap_{n\lt\omega}f^n[A]\cup\bigcup_{n\lt\omega}f^n[A]\setminus f^n[B]\cup \bigcup_{n\lt\omega} f^n[B]\setminus f^{n+1}[A]$
In the course of trying to prove a larger claim, I want to demonstrate directly that $A = P \cup H_0 \cup H_1 \cup H_2 ... \cup \  K_0 \cup K_1 \cup K_2 ...$

where $P = \bigcap_{n \lt \omega}f^n[A] = \bigcap_{n \lt \omega}f^n[B]$,

$H_n = f^n[A] \setminus f^n[B]$, and $K_n = f^n[B] \setminus f^{n+1}[A]$.

Here is the extra information that is relevant:

$f:  A \xrightarrow{1-1} B$, and the notation $f^n$ simply refers to $n$ iterations of the function $f$ (with $f^0$ representing $I_A$)

$A \supseteq B \supseteq f[A] \supseteq f[B] \supseteq f^2[A] \supseteq f^2[B] \supseteq...$

Previously, I have also proven that $P$, $H_m$ and $K_n$ are all pairwise disjoint.

Seeing as the one direction is fairly trivial, I have condensed the desired proof into the following form:

$A \subseteq \bigcap_{n \lt \omega}f^n[A] \ \cup \ \bigcup_{n\lt \omega}f^n[A] \setminus f^n[B] \  \cup \ \bigcup_{n\lt \omega} f^n[B] \setminus f^{n+1}[A] $

I have successfully demonstrated  $A \subseteq \bigcap_{n \lt \omega}f^n[A] \ \cup \ \bigcup_{n\lt \omega}f^n[A] \setminus f^n[B] \  \cup \ \bigcup_{n\lt \omega} f^n[B] \setminus f^{n+1}[A] $. However, I could only do so using a proof by contradiction.

Specifically, I assumed $A \nsubseteq \bigcap_{n \lt \omega}f^n[A] \ \cup \ \bigcup_{n\lt \omega}f^n[A] \setminus f^n[B] \  \cup \ \bigcup_{n\lt \omega} f^n[B] \setminus f^{n+1}[A] $ and arrived at a contradiction.

My set up for this was $\exists x \in A \Big(x \notin P \land x \notin \bigcup_{n\lt \omega} H_n \land x \notin \bigcup_{n\lt \omega} K_n \Big )$

$x \notin P \leftrightarrow \exists x \in \omega \Big (x \notin f^n[A] \Big) \land \exists n \in \omega \Big (x \notin f^n[B]\Big)$

$x \notin \bigcup_{n\lt \omega} H_n \leftrightarrow \forall n \in \omega \Big ( x \notin f^n[A] \Big ) \lor \forall n \in \omega \Big (x \in f^n[B] \Big) $

$x \notin \bigcup_{n\lt \omega} K_n \leftrightarrow \forall n \in \omega \Big ( x \notin f^n[B] \Big ) \lor \forall n \in \omega \Big (x \in f^{n+1}[A] \Big) $

The contradictions that arise from these three simultaneous statements are straightforward.

I am really at loss on how to prove this claim directly.

I've played around with it quite a bit using a working example of $A=\mathbb N$, $B=\mathbb N \setminus \{2,3\}$ and $f$ being the natural number square function (where $n \mapsto n^2$)...where, for example, $P$ equals $\{0,1\}$

I have some visual intuition about how these sets partition $A$:

If anyone could offer some input on how to formulate a direct proof, I would greatly appreciate it. Cheers~

答:
Let

$$C_n=\begin{cases}
f^{n/2}[A],&\text{if }n\text{ is even}\\
f^{(n-1)/2}[B],&\text{if }n\text{ is odd}\,,
\end{cases}$$

so that $C_0=A$, $C_1=B$, and $C_n\supseteq C_{n+1}$ for all $n\in\omega$. For each $x\in A\setminus P$ let $n(x)=\min\{k\in\omega:x\notin C_k\}$; clearly $n(x)>0$, and

$$\begin{align*}
x\in C_{n(x)-1}\setminus C_{n(x)}&=\begin{cases}
f^{k-1}[B]\setminus f^k[A],&\text{if }n(x)=2k\\
f^k[A]\setminus f^k[B],&\text{if }n(x)=2k+1
\end{cases}\\
&=\begin{cases}
K_{k-1},&\text{if }n(x)=2k\\
H_k,&\text{if }n(x)=2k+1\,.
\end{cases}
\end{align*}$$

Thus,

$$A\setminus P\subseteq\bigcup_{n\in\omega}H_n\cup\bigcup_{n\in\omega}K_n\,,$$

and the result follows.


答: Since 10 ≡ -1 (mod 11), you can just raise both sides to the power of $n$.</p>

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
Finding the Riemann $\zeta$ function by adelic integration
I am referring to Tao's blog post about Tate's thesis. Introduce the adeles $\mathbb A$ of $\mathbb Q$ and the adelic Mellin transform
$$Z(s) = \int_{\mathbb A^\times} = g(x) |x|^s d^\times x.$$

Here, $g = \prod_v g_v$ is a product over places $v$ of $\mathbb Q$ and $g_v$ is a self-dual function on $\mathbb Q_v$, more precisely the Gaussian $g(x) = \exp(-\pi x^2)$ for the real place and the characteristic function of integers $\mathbb 1_{\mathcal O_p}$ at finite places $p$. I would like to justify that
$$Z(x) = \pi^{-s/2} \Gamma(s/2) \zeta(s), $$
as in the blog's equation (24), directly from properties of the adeles (and not the Euler product of $\zeta$). In particular, Tao emphasizes that we can use an explicit fundamental domain for $\mathbb A^\times / \mathbb Q^\times$, viz.
$$\mathbb A^\times = \bigsqcup_{k \in \mathbb Q^\times} k \left( J := \mathbb R_+ \prod_p \mathbb Z_p^\times \right).$$

Cutting by classes modulo $\mathbb Q^\times$, we get
$$Z(s) = \int_{\mathbb A^\times / \mathbb Q^\times} \left( \sum_{k \in \mathbb Q^\times} g(kx) \right) |x|^s d^\times x$$
where we recognize the adelic analogue of the theta function used in the derivation of the classical functional equation.

However, I do not understand how using the decomposition above we would find another way to recover $Z(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s)$.

答:
$$Z(s)= \sum_{k\in \Bbb{Q}_{>0}}\int_{(k\Bbb{\hat{Z}^\times)\times R^*}} g(x)|x|^s d^\times x =
  \sum_{k\ge 1} \int_{(k\Bbb{\hat{Z}^\times)\times R^*}}g(x) |x|^s d^\times x $$ $$=\sum_{k\ge 1} |k|_{fin}^s \int_{\Bbb{\hat{Z}^\times\times R^*}}g(x)|x|^s d^\times x  = \sum_{k\ge 1} k^{-s} \int_{-\infty}^\infty e^{-\pi t^2} |t|^s \frac{dt}{t}= \zeta(s)\pi^{-s/2}\Gamma(s/2)$$
Where $\int_{\Bbb{Z}^\times} d^\times x$ is normalized to $1$


问:
Rigorous proof that $E(X|X+Z) = E(Y|Y+Z)$ when $X, Y, Z$ are independent and $X\overset{d}{=} Y$
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, and let $X, Y, Z:\Omega\rightarrow[-\infty,\infty]$ be independent random variables, such that $X, Y$ are $\mathbb{P}$-integrable and identically distributed. By symmetry considerations, intuitively $E(X|X+Z) = E(Y|Y+Z)$. Can this intuition be given a valid formal statement? For instance, is it true that $E(X|X+Z) \overset{d}{=} E(Y|Y+Z)$? If so, how can this be proved rigorously?

答:
I will show that
$$
\mathbb{E}(X|X+Z) \overset{d}{=} \mathbb{E}(Y|Y+Z)\tag{*}
$$
by proving the more general claim:

Claim

Let $(\Omega_i, \mathcal{F}_i, \mathbb{P}_i)$ be a probability space for $i \in \{1,2\}$, and let $U^i_1, U^i_2:\Omega_i\rightarrow[-\infty,\infty]$ be random variables. Write $U^i = (U^i_1, U^i_2)$. Assume that

$U^i_1$ is $\mathbb{P}_i$-integrable.

$U^1 \overset{d}{=} U^2$.

$U^1_2 \overset{d}{=} U^2_2$.

Then $\mathbb{E}_{\mathbb{P}_1}(U^1_1|U^1_2) \overset{d}{=} \mathbb{E}_{\mathbb{P}_2}(U^2_1|U^2_2)$.

We can obtain $(*)$ from the claim by setting
$$
\begin{align*}
\Omega_1 &= \Omega_2 = \Omega,\\
\mathcal{F}_1 &= \mathcal{F}_2 = \mathcal{F},\\
\mathbb{P}_1 &= \mathbb{P}_2 = \mathbb{P},\\
U^1_1 &= X,\\
U^1_2 &= X + Z,\\
U^2_1 &= Y,\\
U^2_2 &= Y + Z.
\end{align*}
$$

Note that since (a) $X, Z$ are independent, (b) $Y, Z$ are independet, (c) $X \overset{d}{=} Y$, we have $(X,Z) \overset{d}{=}(Y,Z)$, and therefore $X+Z \overset{d}{=} Y+Z$ and $(X, X+Z) \overset{d}{=} (Y, Y+Z)$.

Proof of the claim

By definition of conditional expectation, there are Borel functions $\alpha_1, \alpha_2:[-\infty,\infty]\rightarrow[-\infty,\infty]$ such that $\mathbb{E}_{\mathbb{P}_i}(U^i_1|U^i_2) = \alpha_i\circ U^i_2$. Hence, $\alpha_i\circ U^i_2$ is $\mathbb{P}_i$-integrable. Hence, $\alpha_i$ is $\mathbb{P}_{U^i_2}$-integrable.

For every Borel set $A \subseteq \mathbb{R}$, define $g_A:\mathbb{R}^2\rightarrow\mathbb{R}$ by $g_A(x,y) = x\mathbb{1}_A(y)$. Then $U^i_1\mathbb{1}_{\{U^i_2\in A\}} = U^i_1\big(\mathbb{1}_A\circ U^i_2\big) = g_A\circ U^i$.

Then, for every Borel set $A \subseteq \mathbb{R}$ we have
$$
\begin{align*}
\int_A \alpha_i\ d\mathbb{P}_{U^i_2} &= \int_{\{U^i_2 \in A\}}\alpha_i\circ U^i_2\ d\mathbb{P}_i\\
&= \int_{\{U^i_2 \in A\}}\mathbb{E}_{\mathbb{P}_i}(U^i_1|U^i_2)\ d\mathbb{P}_i\\
&= \int_{\{U^i_2 \in A\}}U^i_1\ d\mathbb{P}_i\\
&= \int U^i_1\mathbb{1}_{\{U^i_2 \in A\}}\ d\mathbb{P}_i\\
&= \int g_A\circ U^i\ d\mathbb{P}_i\\
&= \int g_A\ d\mathbb{P}_{U^i}.
\end{align*}
$$

Since $U^1 \overset{d}{=} U^2$, we have $\mathbb{P}_{U^1} = \mathbb{P}_{U^2}$, and therefore $\int_A \alpha_1\ d\mathbb{P}_{U^1_2} = \int_A \alpha_2\ d\mathbb{P}_{U^2_2}$ for every Borel set $A \subseteq \mathbb{R}$. Since $U^1_2 \overset{d}{=} U^2_2$, we have $\mathbb{P}_{U^1_2} = \mathbb{P}_{U^2_2}$, and therefore for every Borel set $A \subseteq \mathbb{R}$ we have $\int_A \alpha_1 - \alpha_2\ d\mathbb{P}_{U^i_2} = 0$, which implies that $\alpha_1 - \alpha_2 = 0$ $\mathbb{P}_{U^i_2}$-a.s., i.e. that $\alpha_1 = \alpha_2$ $\mathbb{P}_{U^i_2}$-a.s.

Now, let $A \subseteq \mathbb{R}$ be a Borel set. Then
$$
\begin{align*}
\mathbb{P}_i\big(\mathbb{E}_{\mathbb{P}_i}(U^i_1|U^i_2) \in A\big) &= \mathbb{P}_i\big(\alpha_i\circ U^i_2 \in A\big)\\
&= \mathbb{P}_{U^i_2}(\alpha_i \in A)\\
&= \mathbb{P}_{U^i_2}\big(\{\alpha_i \in A\}\cap\{\alpha_1 = \alpha_2\}\big).
\end{align*}
$$

It follows that $\mathbb{P}_1\big(\mathbb{E}_{\mathbb{P}_1}(U^1_1|U^1_2) \in A\big) = \mathbb{P}_2\big(\mathbb{E}_{\mathbb{P}_2}(U^2_1|U^2_2) \in A\big)$ for every Borel set $A \subseteq \mathbb{R}$. In other words, $\mathbb{E}_{\mathbb{P}_1}(U^1_1|U^1_2) \overset{d}{=} \mathbb{E}_{\mathbb{P}_2}(U^2_1|U^2_2)$, Q.E.D.

Remark

The above proof shows that if $(\Omega_1, \mathcal{F}_1, \mathbb{P}_1) = (\Omega_2, \mathcal{F}_2, \mathbb{P}_2)$ and if $U^1_2 = U^2_2$ $\mathbb{P}$-a.s., then $\mathbb{E}(U^1_1|U^1_2) = \mathbb{E}(U^2_1|U^2_2)$ $\mathbb{P}$-a.s. ($\mathbb{P} = \mathbb{P}_1 = \mathbb{P}_2$.)

Indeed, under these conditions define
$$
\begin{align*}
N_0 &= \{U^1_2 \neq U^2_2\},\\
N_i &= \big\{U^i_2 \in \{\alpha_1 \neq \alpha_2)\}\big\},\quad i \in \{1,2\}\\
N &= N_0 \cup N_1 \cup N_2.
\end{align*}
$$
Then
$$
\begin{align*}
\mathbb{P}(N_0) &= 0,\\
\mathbb{P}(N_i) &= \mathbb{P}_{U^i_2}(\alpha_1 \neq \alpha_2) = 0,\quad i \in \{1,2\}\\
\mathbb{P}(N) &= 0,
\end{align*}
$$
and for every $\omega \in \Omega\setminus N$ we have
$$
\big(\mathbb{E}(U^1_1|U^1_2)\big)(\omega) = (\alpha_1\circ U^1_2)(\omega) = (\alpha_2\circ U^1_2)(\omega) = \big(\mathbb{E}(U^2_1|U^2_2)\big)(\omega).
$$

In particular, $\mathbb{E}(X|X+Y) = \mathbb{E}(Y|X+Y)$ $\mathbb{P}$-a.s.


问:
Orders of rotations in higher dimension
Do we have a general method to figure out the order of the group of rotations in a higher dimension?

My first attack is on 3-d. We can figure out the order of the group of rotations of a regular solid(like tetrahedron or dodecahedron) by considering the axis of rotation.

But for higher dimensions, if we use the same method for dimension 4, we need to consider the rotations around the axis and faces. I guess we can shorten this process by taking a subgroup.

Is there a more efficient way to think about this? Thanks in advance!

答:
I assume you mean the rotational symmetry group of a polytope.

The best technique might be the orbit-stabilizer theorem. This says that given any choice of feature that the group acts on, the size of the group is the size of the stabilizer times the size of the orbit. For illustration, let's consider a cube and a triangular prism in three dimensions, with various features.

CUBE, $\,24=3\times8=2\times12=4\times6$ symmetries using any of the following "features":

vertices: $8$ vertices (the orbit), and $3$ rotations fixing a given vertex (the stabilizer)

edges:  $12$ edges (the orbit) and $2$ rotations fixing a given edge (the stabilizer)

faces: $6$ faces (the orbit) and $4$ rotations fixing a given face (the stabilizer)

axes: there are $3$ axes thru faces, $2\times4$ rotations fixing a given axis (the stabilizer)

inscribed tetrahedra: there are $2$ inscribed tetrahedra, each fixed by $12$ rotations

You can decide for yourself which of the above is most practical.

TRIANGULAR PRISM, $6$ symmetries, assuming triangular faces and long rectangles.

vertices: $6$ vertices (orbit) but only the identity rotation fixing it (stabilizer)

long edges: $3$ long edges (orbit) and $2$ rotations fixing one of them (stabilizer)

short edges: $6$ triangle edges (orbit), only identity fixing it (stabilizer)

Similar reasoning for faces.

Exercise. Show the octahedron also has $24$ rotational symmetries.

Now let's try something in four dimensions: the $24$-cell. First take the integer lattice $\mathbb{Z}^4\subseteq\mathbb{R}^4$, and notice opposite points are exactly $2$ units apart by the Pythagorean theorem, so we can take the permutations of the points $(\pm1,0,0,0)$ and the combinations of $(\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2},\pm\frac{1}{2})$ (the midpoints of line segments from the origin the the hypercube antipodes away from it) to get vertices of the $24$-cell. The key is that any rotation which fixes a vector restricts to a rotation of its orthogonal complement, which has one lower dimension; this allows essentially an inductive argument.

There are $24$ vertices. The vector $(1,0,0,0)$'s orthogonal complement contains $6$ points of the $24$-cell, the octahedron with coordinate axis vertices. Therefore the $24$-cell has a total of $24^2=576$ symmetries.

Title: Orders of rotations in higher dimension


问:
Cardinality of Cartesian Product of finite sets.
If $a = \{1,2,3\}$ and $b = \{a,b,c\},\;$ FIND $\;n(a\times b)$

Or is it impossible to multiply these sets?

What will be the answer?

答:
Let's use capital letters for sets: so let $$A = \{1, 2, 3\},\;\;\text{ and} \;\; B = \{a, b, c\}$$ and $n(A) = |A| = 3,\;n(B) = |B| = 3$.

Then the Cartesian-product $\,A\times B\;$  is a set of all ordered pairs $$A \times B= \{(a_i, b_j)\mid a_i \in A, b_j \in B\}.$$

In this case, $$A \times B = \{(1, a), (1, b), (1, c), (2, a),(2, b), (2, c), (3, a), (3, b), (3, c)\}$$

In general, for two sets $P, Q$, $$\;|P\times Q| = |P| \times |Q|$$

So, if $n(A \times B) = |A\times B|,\;$ then $n(A \times B) = n(A) \times n(B) = 3 \times 3 = 9$


问:
Why does Cantor's diagonal argument not work for rational numbers?
If we map every integer to a string that represents a rational number, and produce a number different from all the ones listed, we are essentially following Cantor's algorithm. But why does it not apply? Is it because we can't be certain that the number produced is a rational number?</p>

答:
Despite replying to an old post, I would like to supplement things that that I missed out after reading all replies.

Fact:

$\pi$ is irrational

Cantor's method has 1 property and I will respect it

The generated new number must be different

Cantor's contradiction makes sense because the new number is

Real number

Resulted from the method, which means its different from all enumerated real numbers

The contradiction will makes sense to $\mathbb{Q}$ if the new number is

Rational number

Resulted from the method, which means its different from all enumerated rationals

Here's the key. The very method can give this number--$\pi$,

Rational number (No, it isn't)

Resulted from the method (Yes, it is different from all since its irrational)

This is how you conclude why is the generated number not necessarily rational, because the method gives a lot of freedom in picking the number


问:
What is the necessary condition on $f$ such that the DE $dy/dx=f(y)$ has a solution?
In my book, it is written that we shall assume throughout the discussion that follows that $f$ and its derivative $f'$ are continuous functions of $y$ on some interval $I$.

I am guessing that this assumption is made in order that the autonomous DE $dy/dx=f(y)$ has a solution. However, I feel like the continuity of $f$ is enough to guarantee that the DE has a solution since this would imply that $dy/dx$ is continuous on some $I$. What am I missing?

答:
Continuity of $f'$ guarantees uniqueness of solution $\begin{cases} \frac{dy}{dx} = f(y) \\ y(x_0) = y_0\end{cases}$, take a look at Picard–Lindelöf theorem. However, only continuity of $f$ guarantees only existence of solution, not uniqueness; it is Peano theorem.

For example, consider problem $\begin{cases} \frac{dy}{dx} = 2\sqrt{y} \\ y(0) = 0\end{cases}$. It has two solutions $y = 0$ and $y = x^2$, because $f(y) = 2\sqrt{y}$ is continuous near zero and $f'(y)$ isn't. However, $\begin{cases} \frac{dy}{dx} = 2\sqrt{y} \\ y(1) = 1\end{cases}$ has unique solution $y = x^2$, at least in some neighbourhood of $(x = 1, y=1)$.


问:
A p-adic Fourier transform
Consider the field of $p$-adic numbers $\mathbb Q_p$. Define the character $\chi(u p^n) = e(p^n)$ for all $n \in \mathbb Z$ and all unit $u$. In particular it is trivial on integers. This allows to define a Fourier transform and I am interested in computing the one of $\mathbb{1}_{\mathcal O_p}$ (characteristic function of the $p$-adic integers $\mathcal O_p$.

I tried to cut by valuations, without success (the measure is normalized so that $\mathcal O_p$ gets measure one). Let $y = p^{-n} v$ for $n \in \mathbb Z$ and $v$ a unit:
\begin{align}
\int_{\mathbb Q_p} e_p(-x y) \mathbb{1}_{\mathcal O_p}(x) dx & = \int_{\mathcal O_p} e_p(-x y) dx \\
& = \sum_{k \geq n} \int_{\mathcal p^k O_p^\times} e_p(-x y) dx + \sum_{0\leq k < n} \int_{\mathcal p^k O_p^\times} e_p(-x y) dx
\end{align}

Now I can write $x = p^k u \in p^k \mathcal{O}_p^\times$, and get
\begin{align}
& \sum_{k \geq n} \int_{\mathcal O_p^\times} e_p(-p^{k-n} uv) p^{-k} du + \sum_{0 \leq k < n} \int_{\mathcal p^k O_p^\times} e_p(-p^{k-n} uv) p^{-k} du \\
& = \sum_{k \geq n}  p^{-k} \mathrm{vol}(\mathcal O_p^\times) + \sum_{0 \leq k < n} e(p^{k-n}) p^{-k} \mathrm{vol}(\mathcal O_p^\times)
\end{align}

When $n \leq 0$, i.e. $y$ is integer, I indeed get 1 as expected. However when it is not the case I don't know how to compute this sum (and, expectedly, get 0).

答:
For $a,b\in \Bbb{Z}_p$ and $k\ge 0$ an integer $$\exp(-2i\pi ab/p^k)=\exp(-2i\pi (ab\bmod p^k)/p^k)$$ is well-defined. Then

$$\int_{\mathbb Z_p} \exp(-2i\pi x y)dx=\lim_{n\to \infty}\frac1{p^n} \sum_{a=0}^{p^n -1} \exp(-2i\pi a y)= 1_{y\in \Bbb{Z}_p}$$

Title: A p-adic Fourier transform


问:
Probability distribution vs. probability mass function (PMF): what is the difference between the terms?
Consider a discrete case. PMF is the probability each value of random variable gets. So, for example, X ~ Poisson(2). I plot these probabilities (below), so I can say that I show the PMF of X. But on the other hand I show the distribution of X. For example, I can say whether the distribution I have is symmetrical or not. So, what is the difference between probability distribution and PMF terms (in discrete case)? Below I also bring the definitions from Wikipedia, but it is not helpful either.

Many thanks!

A probability mass function (pmf) is a function that gives the probability that a discrete random variable is exactly equal to some value.

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment.

答:
Probability Mass Function.

I would say pmf of a discrete random variable is a graph or a table or a formulae that specifies the proportion or probabilities associated with each possible value the random variable can take.

It is a function that gives the probability that a discrete random variable is exactly equal to some value.


问:
$f$ differentiable. Show there's $c\in(0,\infty)$ so that $f^{\prime}(c)=0$
let $f : \mathbb{R} \longrightarrow \mathbb{R}$ be differentiable for all $x \in \mathbb{R}$, and let $L\in\mathbb{R}$ so that $f(0)=L$ and $\underset{x\rightarrow\infty}{\lim}f\left(x\right)=L$.

prove there's $c\in(0,\infty)$ so that $f^{\prime}(c)=0$.
I get that I'm supposed to use Rolle's theorem, but I can't find 2 x values so $f(x_1)=f(x_2)$.

答:
Take any $x_0>0$. If there is another $x_1\ne x_0$ with $f(x_0)=f(x_1)$ we are done by Rolle's theorem. If there is no such $x_1$ argue using the hypothesis $f(x)\overset{x\to\infty}{\rightarrow} L$ why $x_0$ has to be an extremum of $f$, hence $f'(x_0)=0$.

Edit: Wlog assume $f(x_0)>L$. If there is some $0<x<x_0$ with $f(x)>f(x_0)$ then by the intermediate value theorem there is an $x_1\in(0,x)$ with $f(x_1)=f(x_0)$ and we are done. Assume there is some $x>x_0$ with $f(x)>f(x_0)$. Since $f(x)\to L$ there has to be a $y>x$ with $f(y)<f(x_0)$. So again we get an $x_1\in(x,y)$ with $f(x_1)=f(x_0)$.


问:
Problem with Poisson integral
I have heat equation
$$
u_t=u_xx \\
u(x,0)=xe^{\frac{-x^2}{3}}
$$
I solve by the Poisson integral
$$
u(x,t)=\frac{1}{2\sqrt{\pi t}} \int xe^{\frac{-x^2}{3}}e^{\frac{-(y-x)^2}{4t}} dx
$$

I worked with him (see image), but I have no idea how to finish it. Can u help me with it? Thanks
enter image description here

答:
$$A=\frac{(y-x)^2}{4 t}+\frac{x^2}{3}=\left(\frac{1}{4 t}+\frac{1}{3}\right) x^2-\frac{ y}{2 t}x+\frac{y^2}{4 t}$$ Complete the square to get
$$\left(\frac{1}{4 t}+\frac{1}{3}\right)\Bigg[\left(x-\frac{3 y}{4 t+3}\right)^2 +\frac{12 t y^2}{(4 t+3)^2}\Bigg]$$ Lat
$$a=\left(\frac{1}{4 t}+\frac{1}{3}\right)\qquad b=\frac{3 y}{4 t+3}\qquad c=\left(\frac{1}{4 t}+\frac{1}{3}\right)\frac{12 t y^2}{(4 t+3)^2}$$  there are simplifications to be done and you face
$$\int x e^{-a(x-b)^2-c}\,dx=e^{-c}\int x e^{-a(x-b)^2}\,dx$$ Let $x=b+y$ to face two usual inetgrals.

Title: Problem with Poisson integral


问:
Is $A$ a perfect square?
Consider
$$A=a^2+2ab^2+b^4-4bc-4b^3,$$
where $a,b,c\in\mathbb{Z}$ and $b\neq0$ such that $b|a$ and $b|c$, so $b|A$.

Now I want to know that Is $A$ a perfect square?

答:
No, $A$ is not a perfect square for such $a,b,c$ values. As a counterexample, take $a=b=c=1$, for which $A=-4$.

EDIT: The answer below corresponds to the original question "Can $A$ be a perfect square?".

Yes, it can be 0. Let $a=bn$ and $c=bm$. Then:
$$\begin{align}
A&=a^2+2ab^2+b^4-4bc-4b^3\\
&=b^2n^2+2b^3n+b^4-4b^2m-4b^3\\
&=b^2 \big((n+b)^2-4(m+b)\big)
\end{align}$$
So any values that make $(n+b)^2-4(m+b)$ a square are valid ones. So, for instance, any $n=-b$ and $m=-b$ are enough, but you can find many others.

Title: Is $A$ a perfect square?


问:
Why dynamical systems must be affine with respect to noise in order to apply kalman filtering
I know that to apply Kalman Filtering for state estimation purposes one first assumption is that we can model the system as:
$$x(k+1) = F(k)x(k) + G(k)u(k) + v(k) $$
$$y(k) = H(k)x(k) + w(k) $$

I am trying to find a nice/simple explanation of why it is important that the $v(k)$ and $w(k)$ appear in the previous equations in an affine form (i.e., in a linear way). This is an assumption that all the books of state estimation do when first explain the basics of Kalman Filtering. However, none of the books explain why this is so important. What would it happen if the noise would appear in a non-affine way. E.g.:
$$x(k+1) = F(k)x(k) + G(k)u(k)v(k) $$
$$y(k) = H(k)x(k)w(k) $$

I know that it is still possible to use kalman filters but the equation become more involved since the relationship between the noise and the state/input of the system is nonlinear. However, where/how can I explain in simple terms why this (often hidden) assumption is so important?

The equations above are applied to a discrete linear system but the same goes for nonlinear systems and when therefore a (more complex) Extended Kalman Filter (EKF) must be applied. Indeed, the same affinity assumption is made:

$$x(k+1) = f(x(k),u(k),k) + v(k) $$
$$y(k) = h(x(k),k) + w(k) $$

Thanks in advance

答:
I don't have a background in stochastic processes, so I am not completely sure if time varying noise covariances could maybe pose some nuances in the optimal estimation result. However, in Kumar, P. R., and Pravin Varaiya. "Stochastic systems: estimation, identification and adaptive control." (1986), page 93 defined the general state space model considered by a Kalman filter as

\begin{align}
x_{k+1} &= A_k x_k + B_k u_k + G_k w_k, \\
y_k &= C_k x_k + H_k v_k,
\end{align}

with $A_k$, $B_k$, $G_k$, $C_k$ and $H_k$ possibly time-varying, known matrices of appropriate dimensions. The variables $w_k$ and $v_k$ are zero mean possibly multivariate normally distributed with covariances $Q$ and $R$ respectively. These covariances are not time-varying. This is equivalent to using $w_k'$ and $v_k'$ instead of $G_k w_k$ and $H_k v_k$ respectively, where $w_k'$ and $v_k'$ have time-varying covariances $G_k Q\,G_k^\top$ and $H_k R\,H_k^\top$ respectively.

For the normal Kalman filter result it is required that the model is linear. So terms containing $x_k\,w_k$ would not be allowed. However, $G_k u_k v_k$ could also be interpreted as using $G_k'=G_k u_k$ as the time-varying matrix multiplied by $v_k$, since $u_k$ is known.

For the extended Kalman filter the matrices $A_k$, $B_k$, $G_k$, $C_k$ and $H_k$ can be estimated from the dynamics

\begin{align}
x_{k+1} &= f(k,x_k,u_k,w_k), \\
y_k &= h(k,x_k,v_k),
\end{align}

by taking the appropriate partial derivatives and substituting in the current (estimated) values (for the stochastic variables $w_k$ and $v_k$ I think one should use their expected value, so zero, when evaluating these matrices).


问:
A winning wager that loses over time
This problem was posted in Scientific American (vol. 321.5, Nov 2019, p. 73), and it was troubling.

The game:

We flip a fair coin.
If we flip heads we gain 20% of our bet
If we flip tails we lose 17% of our bet.
Starting bankroll: $100

Stipulations: We must bet all the chips we have, cannot reload, and we must play at least 10 flips.

Note- There's no minimum bet, in the sense that we can keep playing this game with an infinitesimally small chip stack. Since we're always betting a fraction of our bankroll, there's no risk of ruin.

The expected value is net positive. Using the first flip as an example:
EV = (0.5 * $20) - (0.5 * $17) = + $1.50

For 10 flips considering all outcomes:
EV = +16.05%
https://docs.google.com/spreadsheets/d/15nXStFnsEHFU938erWaCKDVMmFdaOVeHNJ-taQZLcJs/edit?usp=drivesdk

However, there's another way to look at this. When we win our bet is multiplied by 1.2. When we lose our bet is multiplied by 0.83.

Let's say we win one, lose one: 1.2*0.83 = 0.996

If we win 5 flips and lose 5 flips in any order:
1.2 x 1.2 x 1.2 x 1.2 x 1.2 x 0.83 x 0.83 x 0.83 x 0.83 x 0.83 = 0.9802, for a net loss of -$1.98.

Losses "hit harder" than wins. From this perspective it looks like a disadvantageous game.

Put technically, the geometric mean is less than 1, which means we expect our bankroll to shrink on average.

In this game the Kelly Criterion says we should risk no more than 7.5%, however we're forced to risk more than double that, 17%. It's a well known principle that betting more then twice Kelly will result in a shrinking bankroll. This is just another way to say the Geometric mean of growth is less than 1.

Scientific American claims the casino will make money over time. They claim that this holds true even if you flip 100 or 10,000 times or more.

I can't wrap my head around the fact that a +EV bet could lose over time in a game with no risk of ruin. I suspect this has to do with "volatility drag". While I can understand the math, I can't intuitively wrap my head around it.

So my question is, should you play this game?

答:
S Spring has it right.  I've also detected a pattern for figuring out how much one must win back after such a percentage loss, but expressed as a fraction. In this case, one needs a win of seventeen 83rds. :-)

The "trick" is that after the initial bet, the amount that you are taking a percentage of has changed. Have added a pic of a table that shows the percentages of gains required for "getting even" after a particular percentage loss.

My fractional trick is merely a pattern that I noticed. That 17/83 of mine is S Spring's .2048.

I'm sure I'm not the first to have noticed this "fractional trick", but haven't been able to Google up anything related to it. I'm certainly no mathematician and am embarrassed to be here in the presence of those who are. Though, would love to be pointed to a URL that might say something more about this fractional trickage.

Title: A winning wager that loses over time


问:
Independent Spaces
What does it mean for spaces to be independent?

答:
That a vector space $V$ and $W$ are (linearly) independent means you can not write any element of $V$ in terms of a linear combination of elements of $W$ and also consider the same condition replacing $V$ and $W$.

As an example, see the impossibility of writing a non-zero vector on the $x$-axis in terms of some elements on the $y$-axis.

Title: Independent Spaces


问:
Tough Number Theory mod problem
If the product of the factors of $30^{12}$ that are congruent to 1 mod 7 can be expressed as $2^{a} \cdot 3^{b} \cdot 5^{c},$ find $a+b+c$.

I tried using mod rules to simplify but I got a very strange answer which was divisible by 7 and did not fit the answer criteria. I noted that 15 is 1 mod 7 and even 8 is 1 mod 7, but I do not know how to go ahead from there.

EDIT: I also noted that 6 is -1 mod 7 and 30^12 is 1 mod 7. But that didn't help either.

答:
With Toby Mak's great insight I give a  short answer with the least amount of calculation below.

Claim: $a+b+c = (2\cdot 13^2 + 2\cdot 13+5)\cdot 18 = 6642$.

Proof: Suppose the number of $3^{12}$'s divisors congruent to $1 \pmod 7$ is $k$. First we note $2^6 \cdot 3^6 \cdot 5^6$ is one of them, with the sum of exponents equal to $18$. For any other such divisor of the form $2^x 3^y 5^z$, we have $2^{12-x} 3^{12-y} 5^{12-z} \equiv 1 \pmod 7$, therefore we have $(k-1)/2$ pairs of divisors such that the exponents sum to $36$,  plus a lonesome divisor $2^6 \cdot 3^6 \cdot 5^6$, henceforth $a+b+c=36(k-1)/2 + 18 = 18k$. (We also know $\color{blue}{a=b=c=6k}$.)

Next, for each of the $13\cdot 13$ pairs of $x,y, 0\le x,y \le 12 $, there are $2$ solutions of $z, 0\le z \le 12$ such that $5^z \equiv 2^{-x} 3^{-y} \pmod 7$, except when $2^x 3^y \equiv 1 \pmod 7$ there are $3$ solutions.

Finally, for each of the $13x$'s$, 0\le x \le 12$ there are $2$ solutions of $y$ such that $3^y \equiv 2^{-x} \pmod 7$, except when $x=0,3,6,12,15$ there are $3$ solutions. Therefore, $k=2\cdot 13^2 + 2 \cdot 13 + 5$ and we are done.

Title: Tough Number Theory mod problem


问:
$X$ taken from $N(0,\theta)$ .Doubt related to MLE and Sufficient statistic
Let $X$ be a single observation from $N(0,\theta).\ \ (\theta=\sigma^2)$

Writing joint density.
$L(x;\theta)=\dfrac{e^{-\frac{x^2}{2\theta}}}{\sqrt{2\pi\theta}}=\dfrac{1}{\sqrt{2\pi\theta}}\cdot e^{-\frac{x^2}{2\theta}}$

We see that $T(X)=X^2$ sufficient for $\theta$

$(a)$ Is $X$ is sufficient statistic?

Since the original sample is always sufficient we conclude that $X$ is sufficient.

$(b)$ Is $|X|$ is sufficient statistic?

$|X|=\sqrt{X^2}$

I am stuck in this one. So please guide me here.

$(c)$ Is $X^2$ is sufficient statistic?

Yes, that's what we get from Neymann Factorization theorem.

$(d)$ What is the maximum likelihood estimator of $\sqrt{\theta}$

When $X \sim N(\mu,\sigma^2)$ MLE of $\sigma^2$is $\dfrac{\sum_{i}(X_i-\bar{X})^2}{n}$

in our case, we have $\dfrac{X^2-0}{1}$

using invariance property MLE of $\sqrt{\theta}$ is $\sqrt{X^2}=X$

B part is where I am stuck and I want to know if I calculated everything right? Where I am going wrong? Any tips to keep in mind and please correct me.

答:
(a) is wrong. You found with factorization theorem that $T=X^2$ is sufficient for $\sigma^2$. $X$ is not sufficient because it is NOT a monotonic function of $T$

(b)

$$T_1=|X|=\sqrt{X^2}$$

In this case, $|X|$ is a monotonic function of $T$ thus $T_1$ is sufficient too for $\sigma^2$

(c) this is wrong too...

Easy find that $\hat{\theta}=X^2$ (observe that MLE must be function of the sufficient estimator, if it exists)

Now using invariance property

$$\hat{\sigma}_{ML}=\sqrt{\hat{\theta}}=|X|$$


问:
What subjects should I study to learn about eigenfunctions? What good textbooks would you recommend for learning the subject?
I googled eigenfunction and look it up in wikipedia, but still I do not know where I should start to learn the subject. I have two questions, and allow me to repeat the title of this question.

What subjects should I study to learn about eigenfunctions?

What good textbooks would you recommend for learning the subject?

Thank you for any answers.

答:
This is not a very good answer to your question, but based on the article it looks like you might actually want to study partial differential equations. Eigenfunction expansion is a nice technique to help understand these equations, in exactly the same way eigenvectors let us understand linear transformations (some PDEs actually are just linear transformation, in a sense).

Since you want to understand PDE involved in population dynamics, I would suggest you look at a more applied book on PDEs. I unfortunately don't know any. A quick search for "Applied PDE Book" gave this as the number one on Amazon: http://www.amazon.com/Applied-Partial-Differential-Equations-Edition/dp/0130652431

This book appears to cover the topics you need to understand the paper, based on a very casual glance of both the paper and the contents of the book. I don't know the level of difficulty of the text, so be prepared to review or learn new things in linear algebra or analysis.

Many of the comments suggest learning functional analysis. I initially wanted to recommend this, but I'm afraid that if you go out and pick up a typical book on functional analysis, it will be too abstract and difficult to transfer to your paper. A sufficiently advanced PDE book should include the techniques from functional analysis you need.


问:
Banach contraction theorem with composition

Problem 5. Let $(\mathcal{X},\mathrm{d}_\mathcal{X})$ be a non-empty complete metric space. Suppose that $f,g:\mathcal{X}\longrightarrow\mathcal{X}$ are two Banach's contractions of $\mathcal{X}$. Prove that there always exists a unique point $x_0\in\mathcal{X}$ such that $f\big(g(x)\big)=x_0$.

Dear all,
I am revising for my exam in two weeks by doing old exams. This was a problem in an old exam, but I don't know how to prove this. I guess that f and g need to have the same fixed point in order for this to be true, so I tried to prove this, but I didn't come far. Could anyone please help me with solving this problem?
Thanks in advance!

答:
They do not need to have the same fixed point—the composition is also a contraction!

I.e., if there exists $λ , μ \in [0,1)$ such that, for all $x,y \in \mathcal X$,

$$d\big(f(x),f(y)\big) \leq λ d\big(f(x),f(y)\big)$$

and

$$d\big(g(x),g(y)\big) \leq μ d\big(g(x),g(y)\big),$$

then, for all $x,y \in \mathcal X$,

$$d\big(fg(x),fg(y)\big) \leq λ \,d\big(g(x),g(y)\big) \leq λ μ \,d\big(x,y\big) $$

This should suffice! I.e., apply the CMT after composing, not before.

Picture:


问:
Relative homology of disconnection (if $X=A\cup B$, then $H_n(X,A)\cong H_n(B)$ for all $n$)
From Rotman's book on algebraic topology (Exercise 6.2):

If $X=A\cup B$ is a disconnection (i.e., $A$ and $B$ are nonempty open sets with empty intersection), then $H_n(X,A)\cong H_n(B)$ for all $n\ge0$.

I think this should probably use the previous exercise, which used Mayer-Vietoris to show that $H_n(X)=H_n(A)\oplus H_n(B)$, but since I can't actually "divide" both sides by $H_n(A)$, I'm not sure how to continue.

I also thought about using the long exact sequence for relative homology. However, I was unable to bring $B$ into this sequence.

Finally, I thought about going back to the definitions of $H_n$ as $\ker\partial_n/\text{im}~\partial_{n+1}$. But this seemed a bit messy so, if I'm being honest, I didn't really pursue this track any further.

Does anyone have any suggestions for me? (Given that this exercise appears in Chapter 6, I feel like it should use Mayer-Vietoris or excision, and would like to find a solution using one of those techniques, if possible; but of course any hints are appreciated.)

答:
You can use the excision axiom: Excise $A$ (which is possible because $\bar A = A \subset A = \text{int}(A)$ to get
$$H_n(X,A) \approx H_n(X \setminus A, A \setminus A) = H_n(B,\emptyset) = H_n(B).$$


问:
Crank-Nicolson method for solving parabolic problem in 1D
Condiser the following initial-boundary value problem for $u = u(x,t),$
$$u_t - au_{xx} = f(x,t) \text { for } 0 < x< L \text { and } 0 < t < T,$$

along with boundary and initial conditions. This is a model of heat conduction in $1$D.

We know that the implicit Euler method is first-order accurate in time because the first-order backward difference approximates $u_t(x_p, t_n)$ only to first order in $\Delta t$, that is,

$$\dfrac {u(x_p,t_n) - u(x_p,t_{n-1})}{\Delta t} = u_t(x_p,t_n) + O(\Delta t).$$

However, viewed as an approximation to $u_t(x_p,t_{n-1/2})$, the backward difference becomes a central difference (with step-size $\frac{1}{2}\Delta t$), and is therefore second-order accurate by:
$$\dfrac {u(x_p,t_n) - u(x_p,t_{n-1})}{\Delta t} = u_t(x_p,t_{n-1/2}) + O(\Delta t^2).$$

Suppose $U_p^{n} \approx u(x_p,t_n).$ Then $u_t(x_p,t_n) \approx \dfrac{U_p^{n} - U_p^{n-1}}{\Delta t}.$

Furthermore, $u_{xx}(x_p,t_n) \approx  \dfrac{U_{p+1}^{n-1/2} - 2U_p^{n-1/2} + U_{p-1}^{n-1/2}}{\Delta x ^2} $ where $U_p^{n-1/2} = \dfrac{U_p^n + U_p^{n-1}}{2} \approx u(x_p, t_{n-1/2}) $

When viewed as an approximation to $u_t(x_p,t_{n+1/2})$, the forward difference (as employed in the explicit Euler) becomes a central difference (with step-size $\frac{1}{2}\Delta t$), and is therefore second-order accurate by:
$$\dfrac {u(x_p,t_{n+1}) - u(x_p,t_n)}{\Delta t} = u_t(x_p,t_{n+1/2}) + O(\Delta t^2).$$

My question is the following:

Can we, in the Crank-Nicolson method, approximate $u_t(x_p,t_n)$ by $\dfrac{U_p^{n+1} - U_p^{n}}{\Delta t}$ and $u_{xx}(x_p,t_n) \text { by }$ $$\dfrac{U_{p+1}^{n + 1/2} - 2U_p^{n + 1/2 } + U_{p-1}^{n + 1/2}}{\Delta x^2}  \text { where } U_p^{n + 1/2} = \frac{1}{2}(U_p^n + U_p^{n+1}).$$

Will this make any difference except that the index $n$ will be from $1 \text { to } N$ in the first case and $0 \text { to } N - 1$ in the latter ?

答:
You seem to be discussing the implicit midpoint method,
$$
U^{j+1}=U^j+ΔtF\left(\tfrac12(U^{j+1}+U^j)\right).
$$
Which could also be implemented as a backwards Euler step of half the step size
$$
U^*=U^j+\frac{Δt}2F(U^*)
$$
followed by a forward Euler step
$$
U^{j+1}=U^*+\frac{Δt}2F(U^*).
$$
Yes, that is a second order method (in time), the same as the implicit trapezoidal method
$$
U^{j+1}=U^j+\frac{Δt}2(F(U^{j+1})+F(U^j))
$$
that is usually associated as the time evolution in C-N.


问:
Compact & Hausdorff but not Orientable
Imagine $\Bbb{R}^3$ with this equivalence relation $(x,y,z) \sim (x',y',z') \iff  \exists t \in \Bbb{R}^3 - \{0\}:  x'=tx, y'=ty, z'=tz$.

In this case how can I show that $\Bbb{R}^3/\sim$ is compact and hausdorff but it isn't orientable?

Thank you in advance!

答:
This space is the same as $\mathbb R \mathbb P^2 \cup \{ 0 \}$.

But let us try to see it more directly. Let us call your space $X$ (and let us also forget the origin, since it just makes the computations messier). So $X= \mathbb R^3 - \{ 0\} / \sim $.

First, I claim that there is a map $\pi: \mathbb R^3 -\{0\} \to S^2$ and a map $\Pi:S^2 \to X$ such that the following diagram commutes:
$$
\begin{array}{cccc}
\mathbb R^3 - \{ 0\} & \xrightarrow{\pi} & S^2 \\
&\searrow & \downarrow \\
&& X
\end{array}
$$

Indeed, $\pi$ is defined by $x \mapsto x/|x|$, and $\Pi$ is defined by sending $x$ to its equivalence class. Note that it is a $2:1$ map.

This proves that $X$ is compact, since it is the image of a compact space. It also proves Hausdorffness, because $\Pi$ is locally a diffeomorphism (explain why and why this is enough).

So $X$ can be identified with $S^2/\mathbb Z_2$, where $\mathbb Z/2$ acts by $x \mapsto -x$.

Here's a sketch of the non-orientability: the group action of $\mathbb Z / 2$ is not orientation-preserving on $S^2$, then use this old math.stachexchange answer.

Title: Compact & Hausdorff but not Orientable


答: Subspaces $W_1,...,W_k$ of $V$ are said to be independent if the only combination $$w_1+\cdots+w_k=0$$ with each $w_j\in W_j$ is $w_1=\cdots=w_k=0$. (This definition should be a bit earlier on the page.)

Title: Independent Spaces

Body: What does it mean for spaces to be independent?


问:
Prove that $( X,τ)$ is homeomorphic to a subspace of $\Bbb{R}$ if and only if $X$ is countable
"Let $(X,τ)$ be a discrete topological space. Prove that $( X,τ)$ is homeomorphic to a subspace of $\Bbb R$ if and only if $X$ is countable."

How do I get started on this one? I know that R has both countable and uncountable subspace, and you can only have a homeomorphism if the cardinality is the same. Now sure where to go from there.

答:
If $X$ is countable, you can simply enumerate the elements of $X$ (say, $x_i$) and map $x_i$ to $i \in \mathbb{N}$.  This is (almost tautologically) a bijection, and it's a homeomorphism since $\mathbb{N}$ is discrete.

For the other direction, you need to prove that if $Y \subset \mathbb{R}$ is an uncountable set, then it is not discrete. If $Y$ were uncountable and discrete, then for each $y \in Y$ there exists an open set $U_y$ such that $U_y \cap U_{y'} = \emptyset$ for $y \neq y'$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, each $U_y$ contains an element of $\mathbb{Q}$, say $q_y$, and all these $q_y$ are distinct. But then $\mathbb{Q}$ is uncountable, contradiction.


问:
Differentiability in $(0,0)$
Let $f(x,y)=e^{\sqrt{x^2+y^2}}$. I would like to define the points for which $f$ is non-differentiable. Is it sufficient argument that $f$ is non-differentiable in $(0,0)$ because $\sqrt x$ is non-differentiable in $(0,0)$, or should I prove it by definition? Or maybe there is a quick way to prove it without using a definition?

答:
Your function is differentiable at $0$ because $$f(x,y)-f(0,0)-\partial _xf(0,0)x-\partial _yf(0,0)y=e^{\sqrt{x^2+y^2}}-1\to0$$
if $(x,y)\to (0,0)$.

Title: Differentiability in $(0,0)$


问:
generator for a vector space
I stumbled upon an exercise where I had to determine a value k inside S :(A1,A2,A3), with A1,A2 and A3 all being 2x2 matrices, so that S generates a vector space. (k being atleast once in every matrix)

I normally solve to find the basis by checking linear independence and getting the value for k that way. But here since there are only 3 matrices inside S that are 2x2 I get a 4x3 matrix for which I can't check de determinant so I'm kinda clueless at this point how to solve for k.

sorry if my words don't seem logic, I'm not familiar with every English term for these kind of exercises.

答:
Let $M$ be your $4 \times 3$ matrix. You want to know whether $M$ has full rank $3$. If you want to use determinants, you can look at $\det(M^TM)$. Indeed, $M^TM$ will be a $3 \times 3$ matrix which will be invertible if and only if $M$ has full rank $3$.

This follows from $\ker M = \ker (M^TM)$ so one matrix is injective if and only if the other one is.

Title: generator for a vector space


问:
How to solve the Cauchy problem?
I have a task that looks like
$$ \frac{\partial u}{\partial t} = 2\frac{\partial^2 u}{\partial x^2} + \frac{\partial u}{\partial x} + u + 3e^{t},\, u\left(x, 0\right)=\sin{x}$$

Please help, thanks :)

Title: How to solve the Cauchy problem?

Body:
Some details:
$$\mathcal{L}^{-1}\left(\ \frac{3}{(s-1)^2}\right)=3te^t$$
$$\mathcal{L}^{-1}\left(\ \frac{1}{s^2+2s+2}\right)=e^{-t}\sin t$$
$$\mathcal{L}^{-1}\left(\ \frac{s}{s^2+2s+2}\right)=e^{-t}(\cos t-\sin t)$$


问:
Roots of polynomials with a specific coefficient
Please prove in the following polynomial
$$z^m-az^{m-1}-b=0 \hspace{1cm} z \in \mathbb{C} \hspace{1cm} m \in \mathbb{N}$$
if $$|a|>2$$ then holds at least one of the roots in the$$|z|>1$$
I tried to use the following relation but I did not succeed.
$$\sum_{ i=1}^m {z_i} = -\frac{-a}{1}=a$$
If you have an idea to prove it, please say. thank you

答:
Let $m > 1$ be an integer, let $a,b\in\mathbb{C}$ be such that $|a| > 2$, and let $z_1,...,z_m\in \mathbb{C}$ be the roots (repetitions allowed) of the equation
$$z^m-az^{m-1}-b=0$$
We want to show $|z| > 1$ for some $z\in\{z_1,...,z_m\}$.

If $b=0$, then $z=a$ is a root, and we're done.

So assume $b\ne 0$.

Then $z_1,...,z_m$ are all nonzero.

Suppose $|z| \le 1$ for all $z\in\{z_1,...,z_m\}$.
\begin{align*}
\text{Then}\;\;&
z\in\{z_1,...,z_m\}
\\[4pt]
\implies\;&
z^m-az^{m-1}-b=0
\\[4pt]
\implies\;&
z^{m-1}(z-a)=b
\\[4pt]
\implies\;&
|z|^{m-1}|a-z|=|b|
\\[4pt]
\implies\;&
|z|^{m-1}(|a|-|z|)\le|b|
\\[4pt]
\implies\;&
|z|^{m-1}(2-1) < |b|
\\[4pt]
\implies\;&
|z|^{m-1}|z| < |b|
\\[4pt]
\implies\;&
|z|^m < |b|
\\[4pt]
\implies\;&
|z| < |b|^{\large{{\frac{1}{m}}}}
\\[4pt]
\implies\;&
\prod_{i=1}^m |z_i| < |b|
\\[4pt]
\implies\;&
\left|\prod_{i=1}^m z_i\right| < |b|
\end{align*}
contradiction.


问:
Generalisation of an expression.
I have been solving matrices questions lately and this pattern keeps showing up:

Given matrix $ P = \begin{bmatrix}
1 & 0 & 0\\
4 & 1 & 0\\
16 & 4 & 1\\
\end{bmatrix}$ and $I$ be identity matrix of order 3. If $[Q] = [q_{ij}]$ is a matrix such that $P^{50}-Q= I$ then $\frac{q_{31}+q_{32}}{q_{21}} = ?$

While solving I found this pattern for a general matrix of the type:
$$L = \begin{bmatrix}
1 & 0 & 0\\
a & 1 & 0\\
b & a & 1\\
\end{bmatrix}$$
As

$$L^n = \begin{bmatrix}
1 & 0 & 0\\
a×n & 1 & 0\\
K & a×n & 1\\
\end{bmatrix}$$

Where $K = (L_{31})_{{n-1}}+a^2×(n-1) + b$

If I have to figure out $L_{50}$ using this algorithm, I need to find $L_{49}, L_{48}$ and so on which basically defeats the purpose. How do you reiterate the value of $L_{n-1}$ so that I can use directly the value of $b$ to calculate $L_{31}$ of any power of matrix?

答:
By recursion you actually show that
$$P^n=\begin{bmatrix} 1 & 0 & 0 \\ 4n &
1 & 0 \\ 16n + \frac{(n-1)n}{2}4^2 & 4n &
1\end{bmatrix} ,$$
so you necessarily have
$$Q=\begin{bmatrix} 0 & 0 & 0 \\ 4n &
0 & 0 \\ 16n + \frac{(n-1)n}{2}4^2 & 4n &
0\end{bmatrix} $$
where $n=50$ (but we will substitute later), so we obtain
$$\frac{q_{31}+q_{32}}{q_{21}}= \frac{16n + \frac{(n-1)n}{2}4^2 + 4n}{4n} = 4 + 2(n-1) + 1 =103.$$

Title: Generalisation of an expression.


答:
This is an application of the triangle inequality. You are working in $\mathbb{C}$, so we have
$$z^m-az^{m-a}-b = (z-\alpha_1) \cdots (z-\alpha_m) = 0$$
And $\alpha_1, \cdots, \alpha_m$ are the roots. Now suppose $|\alpha_i| < 1$ for all $i$. Note that
$$b = \alpha_1 \cdots \alpha_m.$$
But we have
$$z^m-az^{m-1} = b$$
which implies
$$|a||z^{m-1}| - |z^m| \leq |z^m-az^{m-1}| = |b|$$
i.e.
$$(2-|z|)|z^{m-1}| \leq |b|$$
Now take $z = \alpha_0$ to be the root with the largest modulus. Then
$$(2-|\alpha_0|)|\alpha_0|^{m-1} \leq |\alpha_1| \cdots |\alpha_m|$$
Now every term on the left-hand side is greater than or equal to every term on the right, which $(2-|\alpha_0|) > 1$ strictly greater than any term on the right. This is a contradiction.

Title: Roots of polynomials with a specific coefficient

Body:
Please prove in the following polynomial
$$z^m-az^{m-1}-b=0 \hspace{1cm} z \in \mathbb{C} \hspace{1cm} m \in \mathbb{N}$$
if $$|a|>2$$ then holds at least one of the roots in the$$|z|>1$$
I tried to use the following relation but I did not succeed.
$$\sum_{ i=1}^m {z_i} = -\frac{-a}{1}=a$$
If you have an idea to prove it, please say. thank you


问:
Why bending some rules is meaningful, while bending others is fruitless?
This might be a purely philosophical question, but still...

Up to some points in geometry, Euclid's axioms were accepted, even though the 5th one was causing headache.

Then someone comes up and say "hey, what if we consider it's wrong ?", and bam!, here come spherical and hyperbolical geometries. Such theories ave a meaning and can even be used in physics to describe our universe.

In another fiels, mathematicians have known for quite some time that we can't make the square root of a negative number.

Then someone comes up and say "hey, what if we consider $\sqrt{-1}$ has a solution ?" and bam!, here come complex numbers, very useful in describing a lot of things, in particular waves in physics.

What is so special about these "rules" (5th axiom of Euclid and "no squared real number squared is negative") that we can suddenly consider them as false and gets something useful out of it ?

It seems to me, if I say "hey, let's consider $\pi=4$" or "hey, let's consider $x=x+1$ has a solution in some new set of numbers", it might bring some new theories, but they will be a complete waste of time.

Is there a reason why it is so ?

答:
An extension makes sense (and can in the best case turn out to be very useful), if it respects the structures we currently have and creates no contradictions. Consistency is necessary to create anything meaningful.

The complex numbers are , for example , useful because they form an algebraically closed field.

Another very important example is the system ZFC , which is a much stronger theory than the PA that are contained in ZFC.

Because you mentioned $\pi=4$ , actually in the USA, this was defined by law (!) but I do not exactly remember where and when. But such ideas , as you correctly pointed out, have no merit whatsoever.


问:
Is $A$ a perfect square?
Consider
$$A=a^2+2ab^2+b^4-4bc-4b^3,$$
where $a,b,c\in\mathbb{Z}$ and $b\neq0$ such that $b|a$ and $b|c$, so $b|A$.

Now I want to know that Is $A$ a perfect square?

答:
In general, A is obviously not a perfect square. A counterexample is when a = 3, b=1 and c=1 in which case A = 8. However, there are many values of a, b, and c where A is a perfect square. One special case is when b = 1. If b = 1, and you let a + 1 be the hypotenuse of a right triangle whose sides are natural numbers and let $c = \frac{E^{2}}{4}-1$, where E is an even leg of the right triangle, then you can find infinitely many solutions to meet the conditions of your equation. For example, take the Pythagorean triple 5, 4, 3.

a + 1 = 5 which means a = 4

b = 1

$c = \frac{4^{2}}{4}-1=3$

You can view a list of Pythagorean triples here.

Title: Is $A$ a perfect square?


问:
Showing the sum of n independent Bernoulli trials follows Binomial distribution

Suppose $U_1,...,U_n$ are independent Bernoulli random variables taking values $0$ or $1$ with $Pr(U_i=1) = \theta$ and $Pr(U_i=0)=1-\theta$

Let $S = U_1+...+U_n$ be the number of $U_i$ equal to one.

Show $S \sim Binomial(n, \theta)$

I'm unclear as to what I need to show in order to achieve this. Would it be enough to show:

$(1)$ That the sample of space of $S$ is the same as a Binomial variable with such parameters.

$(2)$ The probability of each element of the sample space and hence the probability distribution of $S$ is also the same as a Binomial variable with such parameters.

Or is there something specific way to show that $S$ is a Binomial random variable?

答:
The easiest proof is to use Moment Generating Function and its properties

$$ \bbox[5px,border:2px solid black]
{
MGF_{U_i}=(1-\theta)+\theta e^t
\qquad (1)
}
$$

$$ \bbox[5px,border:2px solid black]
{
MGF_S=\Pi_i MGF_{U_i}=[(1-\theta)+\theta e^t]^n
\qquad (2)
}
$$

We immediately recognize in (2) the MGF of a binomial $B(n;\theta)$

To realize that, take 2 iid Bernulli rv's.

Observe that $S=U_1+U_2$ can take values in $\{0;1;2\}$ with probablities

$P(S=0)=P(U_1=0;U_2=0)=(1-\theta)^2$

$P(S=1)=P(U_1=0;U_2=1)+P(U_1=1;U_2=0)=2\theta(1-\theta)$

$P(S=2)=P(U_1=1;U_2=1)=\theta^2$

Using a compact notation you get

$$P(S=s)=\binom{2}{s}\theta^s(1-\theta)^{2-s}$$

$s=0;1;2$

that is a binomial $B(2;\theta)$


问:
Find the modular residue of this product..
Please help me solve this and please tell me how to do it..

$12345234 \times 23123345 \pmod {31} = $?

edit: please show me how to do it on a calculator not a computer thanks:)

Body:
As $10^1\equiv 10\pmod{31},$

$10^2=100\equiv7,$

$10^3\equiv10\cdot7\equiv8,$

$10^4\equiv49\equiv18,$

$10^5=10^2\cdot10^3\equiv 7\cdot8\equiv25,$

$10^6=(10^3)^2\equiv8^2\equiv2,$

$10^7=10^4\cdot10^3\equiv18\cdot8\equiv20,$

$$12345234=4+3\cdot10+2\cdot10^2+5\cdot10^3+4\cdot10^4+3\cdot10^5+2\cdot10^6+1\cdot10^7$$
$$\equiv4+3\cdot10+2\cdot7+5\cdot8+4\cdot18+3\cdot25+2\cdot2+1\cdot20\pmod{31}$$


答:
The reason these rules were bend was because the original set of rules did not suffice for every purpose. Euclid's postulates were designed and deducted specifically for plane geometry. It is then reasonable that if one wants to do geometry on a curved space, the set of rules have to be adapted to suit this new situation.

As Peter mentions, it is mostly useful to extend an existing structure in such a way that there are no contradictions in the extended structure. However, one can also modify the existing structure in order to obtain a new structure which has maybe a bit other implications and philosophical interpretations.

It is hard to know beforehand bending which rules gives you meaningful insights, and which not, but experimenting and trying new things without knowing the outcome has been characteristically of science since day one.

Title: Why bending some rules is meaningful, while bending others is fruitless?

Body:
This might be a purely philosophical question, but still...

Up to some points in geometry, Euclid's axioms were accepted, even though the 5th one was causing headache.

Then someone comes up and say "hey, what if we consider it's wrong ?", and bam!, here come spherical and hyperbolical geometries. Such theories ave a meaning and can even be used in physics to describe our universe.

In another fiels, mathematicians have known for quite some time that we can't make the square root of a negative number.

Then someone comes up and say "hey, what if we consider $\sqrt{-1}$ has a solution ?" and bam!, here come complex numbers, very useful in describing a lot of things, in particular waves in physics.

What is so special about these "rules" (5th axiom of Euclid and "no squared real number squared is negative") that we can suddenly consider them as false and gets something useful out of it ?

It seems to me, if I say "hey, let's consider $\pi=4$" or "hey, let's consider $x=x+1$ has a solution in some new set of numbers", it might bring some new theories, but they will be a complete waste of time.

Is there a reason why it is so ?


问:
If $(X,d)$ is a compact metric space then is $A$ a closed subset of $K$?
Since $(X,d)$ is compact it is totally bounded so we will have a finite set $A_n$ such that $\displaystyle\bigcup_{x \in A_n} B_d\left(x,\frac{1}{n}\right)$ covers $(X,d)$. Now we take $A$ as the union of these$A_n$. Is $A$ closed?

答:
Think about an interval $[a,b]\subset \mathbb{R}$ with the usual metric. You can choose each $A_n$ as a subset of the rational numbers, so $A$ also will be a subset of $\mathbb{Q}\cap [a,b]$.

The interior of $A$ is empty and the closure of $A$ is $[a,b]$. Hence $A$ is neither open or closed.


问:
Line segments intersecting at the nine-point center
Can you provide a proof for the following claim:

Claim.Given any $\triangle ABC$. The tangent lines to the circumscribed circle of $\triangle ABC$ are constructed at vertices $A$,$B$,$C$ . Let point $D$ be the intersection point of tangent lines that passes through vertices $A$ and $C$ , point $E$ the intersection point of tangent lines that passes through vertices $A$ and $B$ , point $F$ the intersection point of tangent lines that passes through vertices $B$ and $C$ , and let $H_1$,$H_2$,$H_3$ be the orthocenters of  $\triangle ACD$, $\triangle AEB$ and $\triangle BFC$ respectively. Then line segments $AH_3$,$BH_1$ and $CH_2$ concur at the nine-point center of $\triangle ABC$.

GeoGebra applet that demonstrates this claim can be found here.

I don't know how to start the proof. All I know is that nine-point center of the triangle lies in the middle of the line segment whose endpoints are orthocenter and circumcenter , but I don't know how to use that fact.  Any hints are welcomed.

答:
In fact we can say more : $N_9$ is the midpoint of segments $AH_i$.

First we show that $BOCH_2$ is a rhombus. $\triangle BEC$ is isosceles so $H_2$ lies on its symmetry axis. So $BH_2C$ is isosceles as is $BOC$. Hence $OH_2$ is perpendicular to $BC$ at its midpoint $M$. Also $\angle OBC = 90 -A = 90 - \angle EBC = \angle H_2CB$.

Next we show $AHH_2O$ is a parallelogram. $AH \perp BC \Rightarrow AH || OH_2$. It is known that $2OM = AH$. So $AH = OH_2$.

We conclude $AH_2$ is bisected at intersection of diagonals of parallelogram $AHH_2O$. But midpoint of $OH$ is $N_9$. We're done.


问:
Homeomorphic image of $D^m$ is closed in $\mathbb{R}^n$
Suppose $A \subset \mathbb{R}^n$. Define
$$D^m=\{x \in \mathbb{R}^{m+1} : ||x|| \leq 1 \}$$

Suppose $A$ and $D^m$ are homeomorphic , then is it necessary that $A$ is closed subset of $\mathbb{R}^n$

I thought that homeomorphism sends open sets to open sets, but it is not the case. Hence I cannot use that. I wonder if this statement is true. Can I get some hint?

答:
Yes, since $D^m$ is compact $A$ is also compact. But $\Bbb R^n$ is Hausdorff, so compact subsets are closed.

The compactness argument is really necessary here, take for example $[0,+\infty)$ and $[0,1)$. These subsets (with the subspace topology) are homeomorphic but the first is closed in $\Bbb R$ while the second is not.


答:
Why don't you try this: $10 \equiv -1 \ (\text{mod} \ 11) \Longrightarrow (10)^{n} \equiv (-1)^{n} \ (\text{mod} \ 11)$

And for using induction assume it's clear that for $n=1$, the result holds. Now assume that is true for $n=k$. That is $(10)^{k} \equiv (-1)^{k} \ (\text{mod} \ 11)$, and prove it for $n=k+1$.


问:
Are all integer fractions rational?
Any repeating decimal can be written as a fraction $\frac{a}{b}$ where $a$ and $b$ are integers. But is the reverse true. Will any fraction $\frac{a}{b}$ where $a$ and $b$ are integers produce a quotient with a repeating decimal?</p>

答:
The answer is yes.  Every rational number has a repeating decimal expansion.

It's an easy exercise to prove this (hint: consider trailing 9s as above).

Title: Are all integer fractions rational?


答: $D^m$ is compact, and so is every set homeomorphic to $D^m$. In $\mathbb R^n$ every compact is closed.

Title: Homeomorphic image of $D^m$ is closed in $\mathbb{R}^n$

Body:
Suppose $A \subset \mathbb{R}^n$. Define
$$D^m=\{x \in \mathbb{R}^{m+1} : ||x|| \leq 1 \}$$

Suppose $A$ and $D^m$ are homeomorphic , then is it necessary that $A$ is closed subset of $\mathbb{R}^n$

I thought that homeomorphism sends open sets to open sets, but it is not the case. Hence I cannot use that. I wonder if this statement is true. Can I get some hint?


问:
Prove or disprove that the matrix is invertible
Let $P$ be a $n \times n $ matrix with integer entries. Let $q$ be a non integer and $Q = P + qI$ where $I$ is identity matrix.

Prove or disprove that $Q$ is invertible.

This is easy if $P$ is a $2 \times 2$ matrix. In this case it is easy to see that determinant of $Q$ is non zero. Let $P$ be a matrix

\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix}

then we see that $\det Q = (a + q)(d + q) - bc$. If $q$ is an irrational then the determinant is obviously non zero. If $q$ is a rational then let $q = \frac mn$ where gcd of $m$ and $n$ is $1$. Now it is easy to see that if
$$
(an+m)(dn+m) = n^2 bc
$$
then any prime which divides $n$ doesn't divide $m$ which implies that it doesn't divide LHS as well. So determinant of $Q$ is non zero so we conclude that $Q$ is invertible.

Now how to prove for higher dimension matrices?

答:
You basically want to know whether $-q$ is an eigenvalue of $P$. This is true if and only if $k_P(-q) = 0$, where $k_P(x) = \det(x I-P)$ is the characteristic polynomial of $P$. Since $P$ has integer coefficients, we have $k_P \in \Bbb{Z}[x]$ and it is a monic polynomial with $\deg k_P = n$ so clearly if $q$ not an algebraic integer or is an algebraic integer of degree $> n$ then $-q$ cannot be a zero of $k_P$ so $P+qI$ is invertible.

On the other hand, if $q$ is algebraic of degree $\le n$, then let $p(t) = c_0 + c_1 t + \cdots + c_{k-1}t_{k-1} + t^k \in \Bbb{Z}[x]$ be the minimal polynomial of $-q$ with $k \le n$. You can check that the $n \times n$ matrix
$$P = \begin{bmatrix}0 & 0 & \cdots & 0 & -c_0 & 0 & 0 & \cdots & 0\\ 1 & 0 & \cdots & 0 & -c_1 & 0 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 & -c_2 & 0 & 0 & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\  0 & 0 & \cdots & 1 & -c_{k-1}  & 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 & 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & \cdots & 0 & 0 & 0 & 1 & \cdots & 0\\  \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 0 & 0 & 0 & 0 & \cdots & 1 \end{bmatrix}$$
has $k_P(x) = p(x)(x-1)^{n-k}$ so $k_P(-q)=0$ and therefore $P+qI$ is not invertible.

For your $2 \times 2$ example, your analysis is wrong, for example $\sqrt{2}$ is an algebraic integer with minimal polynomial $x^2-2$ so for $$P = \begin{bmatrix} 0 & 2 \\ 1 & 0 \end{bmatrix}$$
we have that $P+\sqrt{2} I$ is not invertible. Also for any $P$ and $q = \frac12$ (which is not an algebraic integer), we have
$$\det \left(P+\frac12 I\right) = \left(a+\frac12\right)\left(d+\frac12\right) - bc \in \Bbb{Z}+\frac14$$
so it is nonzero and hence $P+\frac12 I$ is invertible. Similarly, for any $P$ and $q = \sqrt[3]{2}$ which is an algebraic integer of degree $3$, you can check that $P+\sqrt[3]{2} I$ is invertible.


问:
Prove that $f(x)=e^{\frac{1}{x}}$ is uniformly continuous.
Problem: Prove that $f(x)=e^{\frac{1}{x}}$ is uniformly continuous on $(a,1)$ for $a \in (0,1)$.

My solution: By using the Lagrange theorem for $x,y \in (a,1)$ we have
$$\frac{f(x)-f(y)}{x-y}=f'(c)=e^{-\frac{1}{c^2}},$$
where $c \in (x,y)$. This yields

$$\vert e^{\frac{1}{x}}-e^{\frac{1}{y}} \vert < e^{-1} \vert x-y \vert.$$

So, $\forall\varepsilon > 0$, $\exists \delta =\varepsilon e >0$ that $\forall x,y \in (a,1)$, if
$$\vert x-y \vert < \delta$$
then
$$\vert f(x) -f(y) \vert = \vert e^{\frac{1}{x}}-e^{\frac{1}{y}} \vert < e^{-1} \vert x-y \vert <e^{-1}\delta=\varepsilon.$$
Therefore $f$ is uniformly continous on (a,1).

Question: In my solution, I have not used "$x,y>a$". That means, $f$ is uniformly continous on $(0,1)$ but we can easily prove that $f$ is not uniformly continous on $(0,1)$. Could you show me why my solution is wrong?

Body:
$$
(e^{1/x})'=-\frac{1}{x^2}e^{1/x},
$$
and hence there exists a $c\in(x,y)$, such that
$$
e^{1/x}-e^{1/y}=(x-y)\cdot \frac{1}{c^2}e^{1/c}.
$$
Thus
$$
|e^{1/x}-e^{1/y}|=|x-y| \frac{e^{1/c}}{c^2}\le |x-y| \frac{e^{1/a}}{a^2}.
$$


问:
Convergence of $\int_0^n f(x)\, dx - [\frac{1}{2} f(0) + f(1) +\dots + f(n-1) + \frac{1}{2} f(n)]$
If $f(x)$ is increasing, strictly concave, and twice continuously differentiable for $x > 0$, does the difference
$$
d_n = \int_0^n f(x)\, dx - [\tfrac{1}{2} f(0) + f(1) + f(2) +\dots + f(n-1) + \tfrac{1}{2} f(n)]
$$
converge to a finite positive limit when $n\to +\infty$?

The error form provided by the trapezoidal rule is not helpful since it just asserts that $d_n = -n f''(\xi_n)/12\geqslant 0$, for some $0 < \xi_n < n$, or that
$$
d_n\leqslant -\tfrac{1}{12}[f''(\xi_1) + f''(\xi_2) +\dots + f''(\xi_n)]
$$
with  $k-1 < \xi_k < k$. Since $f(x)$ is increasing and concave, then $f''(x)$ is negative, bounded, and increasing for $x > 0$, but then again, how to assure the converge of $\sum_{k=1}^n f''(\xi_n)$ when $n\to +\infty$?

Body:
Consider
$$
f(x)=-\frac{1}{(x+1)(x+2)}.
$$
Clearly $f$ is increasing and concave in $[0,\infty)$.

Then
$$
\frac{1}{2}f(0)+f(1)+\cdots+f(n-1)+\frac{1}{2}f(n)=\cdots=-\frac{3}{4}+\frac{1}{n+1}-\frac{1}{2(n+1)(n+2)}\\=-\frac{3}{4}+\frac{1}{n}+{\mathcal O}(n^{-2})
$$
while
$$
\int_0^n f(x)\,dx=\int_0^n \frac{dx}{x+2}-\int_0^n\frac{dx}{x+1}=\log(n+2)-\log 2-\log(n+1)\\=-\log 2-\log\Big(1+\frac{1}{n+1}\Big)\\=-\log 2-\frac{1}{n}+{\mathcal O}(n^{-2})
$$
Thus
$$
d_n\to -\frac{1}{2}+\log 2>0.
$$


问:
How to find the orthogonal projection of a matrix onto a subspace?
$
\newcommand{\<}{\left \langle}
\newcommand{\>}{\right \rangle}
$

I have fought 2 vectors(?!) in basis which are

$M_1 = \begin{bmatrix}
    1 & 0\\
    0 & -1\\
    \end{bmatrix}$ and
$M_2 = \begin{bmatrix}
    0 & 1\\
    1 & 0\\
    \end{bmatrix}$

However, I do not know how to do the orthogonal projection of a matrix onto the subspace. Can anyone please help me to solve this problem number 2?

Consider the inner product space $\mathit M_{22}$ with the standard inner product
$\ = tr(AB^T)$

(1) Find an orthogonal basis for the subspace $\mathit W = \{A \in M_{22}|A^T = A$ and $\mathit trA=0\}$
and explain the reason.

(2) For $\mathit A = \begin{bmatrix}
    0 & 1\\
    1 & 1\\
    \end{bmatrix} in \space M_{22}$, find $\mathit B \in W$ such that $\|\mathit A - B\|$ ≤ $\|\mathit A - C\|$ for any $\mathit C \in W$.

答:
Since you have an orthogonal basis $\{M_1, M_2\}$ for $W$, the orthogonal projection of $A$ onto the subspace $W$ is simply
$$B= \left\langle A,\frac{M_1}{\|M_1\|}\right\rangle \frac{M_1}{\|M_1\|} + \left\langle A,\frac{M_2}{\|M_2\|}\right\rangle \frac{M_2}{\|M_2\|}.$$

Do you know how to prove that this orthogonal projection indeed minimizes the distance from $A$ to $W$?


问:
Is Multiplication A System?
My textbook says:

Clock arithmetic and modular systems were built upon ordinary numbers and involved familiar operations such as addition, subtraction, multiplication, and division.

So multiplication is a part of a system, but does that mean it is a system? It then states that an abstract system likely has properties such as: closure, commutative, associative, identity, inversive, and distributive. It does not define what these properties are. It also does not state what the requirements are to be classified as a system. There is no direct definition of a "system"

Then there is a question: "Is multiplication a system? Explain." What information do I need to provide to sufficiently answer this question?

I don't understand how to identify the properties of a system. What possible properties could a system have? Are a certain number of properties required in order to be classified as a system?

答:
Multiplication is both a system and an operation.
Multiplication is used as an operation within the properties which define systems. As http://en.wikipedia.org/wiki/Multiplication states, Multiplication has properties which include: a commutative property, distributive property, identity element, zero element, negation and an inverse element. Thus, multiplication is also a system. Multiplication is often used within other systems such as the modular arithmetic system, clock arithmetic, etc.</p>

Title: Is Multiplication A System?


问:
Prove by induction: for every integer $n > 0$, $a_n = 2 \cdot 3^n + n^2-6$
I am working on what should be a simple induction proof exercise. However, I'm not sure if I am failing to simplify an expression after substitution correctly, or whether a prior step is incorrect.

The exercise:

Let $a_1 = 1$ and for every integer $n > 0$ let $$a_{n+1} = 3a_n - 2n^2 + 2n + 13$$

Use induction to prove that for every integer $n > 0$ $$a_n = 2 \cdot 3^n + n^2 - 6$$

My steps:

Let $P(n)$ be the statement that for every integer $n > 0$, $a_n = 2 \cdot 3^n + n^2 - 6$

For $n = 1$, $a_1 = 2 \cdot 3^1 + 1^2 - 6 = 1$. Thus, $P(1)$ is true.

Suppose that $P(k)$ is true. That is, $a_k = 2 \cdot 3^k + k^2 - 6$.

Then (using substitution to replace $a_n$ in the given definition for $a_{n+1}$), $$a_{k+1} = 3(2 \cdot 3^k + k^2 - 6) - 2k^2 + 2k + 13 = (2 \cdot 3^{k+1}) + k^2 + 2k -5$$

This is not the needed result.

Is my error in the simplification of the expression in step 4? Or is a previous step (also) incorrect?

Body: It is the needed result if you notice $$(2\cdot 3^{k+1}) + k^2 + 2k - 5 = (2\cdot 3^{k+1}) + k^2 + 2k +1 - 6 = (2\cdot 3^{k+1}) + (k+1)^2 - 6$$


问:
for what value of $r$ does the sequence and series converge
Suppose $a_n = \dfrac{4^n}{n^3} + (r^2)^n $ and $$S=\sum_{n=0}^\infty\frac{4^n}{n^3+(r^2)^n}$$
I think that the denominator $> 0$ as $n\to\infty$, but that doesn't seem right. please help me with this problem: when is this sequence and series converge?
i divided the numerator and denminator by $4^n$ but i'm unable to proceed forward to find a general vale of r for which the sequence and series converge

答:
You can write $$a_n=\frac{1}{\frac{n^3}{4^n}+\left(\frac r2\right)^{2n}},$$ so that as $n\to\infty$ we find that $\frac{n^3}{4^n}\to 0.$ On the other hand, if $|r|<2,$ then $$\left(\frac r2\right)^{2n}\to 0$$ as well, so that then $a_n\to\infty.$ However, for $|r|\ge 2,$ we find that $$\left(\frac r2\right)^{2n}=1$$ or $$\left(\frac r2\right)^{2n}\to \infty,$$ and convergence of $a_n$ follows in either of the last cases.

If follows that for the series $S$ to exist, you must have $|r|>2.$ However, that may be insufficient for convergence of the series. For more delicate analysis, note that $$a_n<\left(\frac{2}{r}\right)^{2n}$$ for large $n,$ so that upon applying the root test we see that $S$ exists exactly whenever $|r|>2.$


问:
Differentiability in $(0,0)$
Let $f(x,y)=e^{\sqrt{x^2+y^2}}$. I would like to define the points for which $f$ is non-differentiable. Is it sufficient argument that $f$ is non-differentiable in $(0,0)$ because $\sqrt x$ is non-differentiable in $(0,0)$, or should I prove it by definition? Or maybe there is a quick way to prove it without using a definition?

答:
It is sufficient to prove that one of the partial derivatives, with respect to either $x$ or $y$, does not exist when the other variable is zero.  Set $y=0$.  Then using the Taylor's expansion for the exponential function,
\begin{align}
\left. \frac{\partial f}{\partial x}\right \rvert_{x,y=0}&=\lim_{x\to0}\frac{e^{\sqrt{x^2}}-1}{x} \\
&=\lim_{x\to0} \Big( \frac{|x|}{x}+\frac{1}{2!}x + o({x^2} ) \Big)
\end{align}
which (accepting the limit of the sum is the sum of the limits in this case) has limit $\pm1$ depending on the sign of $x$, so $f$ is not differentiable at $(0,0)$.

Title: Differentiability in $(0,0)$


问:
If $S$ is a simple subgroup of a finite group $G$ such that $NS=SN$ where $N$ is any subnormal subgroup of $G$, then $S\subseteq N_G(N)$
I’m looking for a proof of the following statement, or any hint/help on how to attempt one:

If $S$ is a simple subgroup of a finite group $G$ such that $NS=SN$ where $N$ is any subnormal subgroup of $G$, then $S\subseteq N_G(N)$ for every subnormal subgroup $N$ of $G$.

答:
Let $N \lhd \lhd G$. We prove by induction on the subnormal depth of $N$ in $G$ that $S \le N_G(N)$. This is clear for subnormal depth $1$, i.e. when $N \lhd G$.

Otherwise, there exists $K \lhd G$ with $N \le K$ such that the subnormal depth of $N$ in $K$ is less than in $G$. If $S \le K$ then the result follows by induction, and otherwise, since $S$ is simple, we have $S \cap K = 1$.

Now $SN=NS$ implies that $NS$ is a subgroup of $G$ and so $N^S \le NS \cap K = N(S \cap K) = N$, and hence $S \le N_G(N)$.


问:
Number of solutions to $x^k \equiv h \pmod {q^n}$
Could someone please give me a hint/solution to
the question, say $q$ is a prime and $(q,h)=1$, then
$$
x^k \equiv h \pmod {q^n}
$$
has at most $k$ solutions $1 \leq x < q^n$?

Thanks!

答:
Let $p$ be an odd prime. Then any positive power $q$ of $p$ has a primitive root. That is, there is a $g$ such that as $i$ travels from $1$ to $\varphi(q)$  the powers $g^i$ travel through the $\varphi(q)$ reduced residue classes modulo $q$. In group-theoretic terms, $g$ generates the group of multiplicative units modulo $q$. Here as usual $\varphi$ is the Euler $\varphi$-function.

Suppose that $h$ and $q$ are relatively prime, where $q$ is a power of an odd prime.  We will show that the congruence $x^k \equiv h\pmod{q}$, if it has a solution, has precisely $\gcd(k,\varphi(q))$ solutions modulo $q$. As a consequence, the congruence has at most $k$ solutions modulo $q$.

Suppose that $x^k\equiv h\pmod{q}$. Any such $x$ is congruent to $g^i$ for a suitable $i$, where $1\le i\le \varphi(q)$. Let $h\equiv g^j\pmod{q}$. Then $x^k\equiv h\pmod{q}$ if and only if
$$g^{ik}\equiv g^j\pmod{q}.$$
The above congruence holds if and only if
$$ik\equiv j\pmod{\varphi(q)}.$$
But the linear congruence $yk\equiv j\pmod{\varphi(q)}$ has a solution if and only if $\gcd(k,\varphi(q))$ divides $j$. and if it has a solution, it has precisely $\gcd(k,\varphi(q))$ solutions.

Remark: As pointed out by Zev Chonoles, the result does not hold for powers of $2$ greater than or equal to $2^3$. These numbers do not have a primitive root. The result does hold for $q$ of the form $2p^n$, for $p$ and odd prime, since these $q$ do have a primitive root.


问:
Suppose $\gcd(a,y)=s$ and $\gcd(b,y)=t$. Prove that $\gcd(\gcd(a,b),y)=\gcd(s,t)$.
All I have so far is that $$s|a, s|y, t|b, \text{ and } t|y.$$ I also know

$$\gcd(\gcd(a,b),y)=\gcd(a,b,y)=\gcd(a,gcd(b,y))$$

by the associative property of gcd. It would suffice to show $$\gcd(a,b,y)=\gcd(gcd(a,y),\gcd(b,y)).$$
I'm just not sure how to prove it. Thanks for your help.

答:
We have :
$\gcd(a,y) = s$. And $\gcd(b,y) = t. $
so :
$$\gcd(\gcd(a,b),y)= \gcd(a,b,y,y)$$
$$ =\gcd(\gcd(a,y),\gcd(b,y))$$
Which is the same as :
$$=\gcd(s,t)$$


问:
how did they arrive to the rule of addition of vectors?
I'm an A2 igsce math student and I'm taking mechanics for the first time in math this session.

I wanted to ask about how they arrived to the rule of addition of vectors. How did they know that if we add the X's and Y's of two vectors they would get a third vector which has exactly the same direction and magnitude of the force that could replace these two vectors or forces.

I'm convinced that it's correct and can feel or see the direction a point will accelerate in if two certain forces are applied to it. And can feel how the resultant force tends to get closer to the bigger force and how if two equal forces are applied with the same angle the resultant force is going to be exactly between them.

So I'm sure who invented the vectors had the same feelings and visions too but how did he arrive at this simple method to get such fascinating and exact results, not only did he manage to get the direction but the magnitude of the resultant too!

答:
Vectors don't exist in reality, they are just a mathematical model that works to describe some aspects of reality. Physics is not mathematics and not all propositions can be derived simply by logic(even if this fact seems to be true also in maths in a certain sense). So I think that many experiments have been made to grasp the nature of forces and how they "sum" to each other.
You seem to be attached to an extremely linear model of history of mathematics and physics:there isn't "who invented vectors". Human mind is not a formal system and the thought that the mathematical and physical history follow the linear path of a modern book is pure illusion.</p>


问:
proving limits exist
Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.

I basically need to determine how large $n$ has to be to imply

$$\frac{3n+24}{n^3-6}<\epsilon$$

the idea is to upper bound the numerator and lower bound the denominator.
For example, since $3n + 24 ≤ 27n$, it suffices for us to get $\frac{27n}{n3-6} < ε$.

it is inferred, thereafter, that all we need is $\frac{n^3}2 ≥6$ or $n^3 ≥12$ or $n>2$.
this is where I have a problem. I don't understand how we got to the point where all we need is  $\frac{n^3}2≥6$.

答: The most dominant term in numerator is $4n^3$ and the most dominant term in denominator is $n^3$ so the limit is $4$.

Title: proving limits exist


问:
Is $\|M^s\|\leq \|M\|^s$ for every $s\in \mathbb{R}$?
Let $M$ be a self-adjoint operator on a complex Hilbert space $H$.

Let $s\in \mathbb{R}$. Is
$$\|M^s\|\leq \|M\|^s\;?$$

Since $M$ is selfadjoint, then by the Spectral Theorem we have
$$
M^s=\int_{\sigma(M)}\lambda^s\,dE(\lambda).
$$
Then
$$
\|M^s\|\leq\int_{\sigma(M)}|\lambda|^s\,\|dE(\lambda)\|.
$$

答:
Consider $$M = \pmatrix{2 & 0 \\ 0 & 3}, \quad M^{-1} = \pmatrix{\frac12 & 0 \\ 0 & \frac13}$$
so $\|M\| = 3$ and $\|M^{-1}\| = \frac12$ so
$$\|M^{-1}\| \le \|M\|^{-1}$$
yields $\frac12 \le \frac13$ which is not true.

For positive $s > 0$ the claim is true since the function $x \mapsto x^s$ is continuous and increasing so
$$\|M^s\| = \max_{\lambda \in \sigma(M^s)}|\lambda| = \max_{\lambda\in\sigma(M)}|\lambda^s| = \max_{\lambda\in\sigma(M)}|\lambda|^s = \left(\max_{\lambda\in\sigma(M)}|\lambda|\right)^s = \|M\|^s$$
so even equality holds.


问:
Classifying 'Normally Perfect' Fields
A field is said to be perfect if it has no non-separable algebraic extensions. Aside from being very useful as ground fields, they have a nice classification: they are exactly the fields of characteristic $0$ and the fields of characteristic $p>0$ that are closed under $p^{th}$ roots.

Along these lines I am wondering if there is a nice characterization of fields that are 'normally perfect' - fields with no non-normal algebraic extensions.

Just as a field being perfect is equivalent to it having no non-separable finite extensions, a field being 'normally perfect' is equivalent to it having no non-normal finite extensions, so it is easy to see that all finite fields are 'normally perfect'. It is also a trivial fact that all algebraically closed fields and all real closed fields are 'normally perfect'. But beyond this I'm having a difficult time seeing what fields could be 'normally perfect' or finding nice necessary conditions for these fields.

答:
Not a complete answer, but it might interest you (and anyway, it's too long for a commment)

If $F$ is normally perfect, Let $G$ be its absolute Galois group, and let $H$ be a closed subgroup of $G$. Then $F_{sep}^H/F$ is spearable, and normal by assumption. In other words, this extension is Galois, hence $H$ is normal.

Now  let us assume that the following is true:

Claim (?) If $H$ is a non normal subgroup of $G$, then its closure is non norml.

In this case, every subgroup of $G$ is normal. But groups whose all subgroups are normal are classified (https://en.wikipedia.org/wiki/Dedekind_group); they are either abelian, or of the form $E_2\times A\times Q_8$, where $A$ is a an abelian torsion group where are elements have odd order, and $E_2$ is a $2$-elementary abelian group.

An consequence of Artin Schrier theorem ( https://kconrad.math.uconn.edu/blurbs/galoistheory/artinschreier.pdf) implies that the torsion part of $G$ is $2$-elementary abelian, so $Q_8$ does not appear and $A$ does not appear either.

Hence, in any case $G$ is abelian. Conversely, if $G$ is abelian, any separable extension of $F$ is Galois.

Hence, if the claim above is true, a field $F$ is perfect and normally perfect if and only if its absolute Galois group is abelian.

This would solve your question in zero characteristic.

Anyway, indepedently of the veracity of the claim, fields $F$ of characteristic zero with abelian absolute Galois group give you another family of examples. You can take for example $F=F_0((t_1))((t_2))\cdots((t_n)),$$ where $F_0$ has charactertstic zero.

Title: Classifying 'Normally Perfect' Fields


问:
A Question about Doctoral Theses in Mathematics
This is most definitely a soft question, which I'm sure may get some negative attention, and perhaps even be voted closed. However, I genuinely would like to generate answers on this matter as it concerns anyone that decides to get a doctorate in mathematics. (Since I believe this is the best forum to do so, here goes...)

Question: Does a doctoral thesis in mathematics have to contain more than an abstract, a proposition and proof of a new and exciting result?

I strongly believe that brevity is beauty to mathematicians and see no problem with a thesis that contains a lengthy abstract (~350 words) to generate enthusiasm and then jumps right into the heart of a novel and far-reaching result. The proof provided is extremely condensed in the spirit of Zagier, leaving almost all details that can be found elsewhere to be found elsewhere, in the works cited. If the result has merit, I see no reason why it can't be submitted or published in such a manner. To support my stance, I offer the following condensed theses:

David Rector, "An Unstable Adams Spectral Sequence", MIT (1966), 9 pgs.

Burt Totaro, "Milnor K-Theory is the simplest part of algebraic K-theory", Berkeley (1989), 12 pgs.

Herman Buvik, A New Proof of Torelli's Theorem, NYU (1962), 12 pgs.

Eva Kallin "A non-local function algebra" Berkeley (1963), 13 pgs.

Edmund Landau "Neuer Beweis der Gleichung $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} = 0$", Berlin University (1899), 14 pgs.

Barry Mazur, "On Embeddings of Spheres", Princeton (1959), 26 pgs.

John F. Nash "Non-Cooperative Games", Princeton (1950), 27 pgs.

Kevin Walker, "An Extension of Casson's Invariant to Rational Homology Spheres", Berkeley (1989). 29 pgs.

Although I'd appreciate to hear from anyone worth giving his two cents, I'm specifically eager to hear from those senior members of the community.

Thanks!

答:
[Disclaimer: I’m a fairly junior member of the community — in my 3rd year of postdoc-ing.]

Yes, there are certainly plenty of excellent short theses in maths.  There’s nothing necessarily wrong with a short thesis; it doesn’t have to contain any more than what you say.

However, I’d suggest being quite careful about the issue.  In a small thesis based tersely around a key result, that result has to bear a lot more weight than if it were part of a larger thesis along with a bit more general exploration of the area.  If I go to a restaurant and the portions are tiny, then that can be fine, but the food had better be really good.  And this isn’t just silly prejudice: a longer thesis really can exhibit aspects of a candidate’s knowledge and capabilities that a single tersely-proved result, even an excellent one, doesn’t attest to.

Finally, the audience of your thesis is your PhD committee.  If the certain faculty member is on your committee, or close to people who are, then it’s not good to antagonise them.  What will be read by a wider audience is determined by what’s on your webpage, on the arXiv, or in journals.  If you make sure to make your perfect, terse exposition the more readily accessible one in these ways, then perhaps it may be easier to bite your tongue and write the thesis that will be more acceptable to your thesis committee.


答:
If $\frac{n^3}2\ge 6$ then
$$n^3-6\ge n^3-\frac{n^3}2=\frac{n^3}2 $$
is your desired monomial lower bound for the denominator.

In fact, we are allowed to be wasteful and may begin right away with, say, assuming $n>1000$.
That makes $3n+24<3.024 n$ and $n^3-6>0.999999994n^3$ and so the error term certainly $<\frac{3.025}{n^2}$. But that doesn't matter for the sole purpose of proving the limit where any other bound of the form $<\frac C{n^2}$ (or even weaker) is good enough.

Title: proving limits exist

Body:
Prove $\lim\frac{ 4n^3+3n}{n^3-6}= 4$.

I basically need to determine how large $n$ has to be to imply

$$\frac{3n+24}{n^3-6}<\epsilon$$

the idea is to upper bound the numerator and lower bound the denominator.
For example, since $3n + 24 ≤ 27n$, it suffices for us to get $\frac{27n}{n3-6} < ε$.

it is inferred, thereafter, that all we need is $\frac{n^3}2 ≥6$ or $n^3 ≥12$ or $n>2$.
this is where I have a problem. I don't understand how we got to the point where all we need is  $\frac{n^3}2≥6$.


答:
Just look at the highest power of $n$ upside and downside (and divide everything by that power of $n$):

$$\lim_{n\to\infty}\frac{4n^3+3n}{n^3-6}= \lim_{n\to\infty}\frac{4+\frac{3}{n^2}}{1-\frac{6}{n^3}}=\frac{4+\frac{3}{\infty}}{1-\frac{6}{\infty}}=\frac{4+0}{1-0}=\boxed{4}$$

Title: proving limits exist