问:
Inequality with Gamma.
Consider the function $$f(x)=e^{\frac{x^2}{2}}\int^\infty_0e^{-t^2}t^{2n}\cos(2xt)dt,\quad x\in\Bbb R$$

I find a lot difficultes to prove $$\forall (x,n)\in\Bbb R^+\times\Bbb N:|f(x)|\leq C\Gamma(n+\frac{1}{2})$$
for some const $C$.
Thanks for any help.

My attempt:

Put $$f(x)=\frac{1}{\Gamma(n+\frac{1}{2})}e^{\frac{x}{2}}\int^\infty_0 e^{-t^2}t^{2n}\cos(2\sqrt{x}t)dt.$$

Let $x>0$, by the change of variable $t=\sqrt{n}s$ we get $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_0 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds. $$
Now we apply Laplace's method wich says:\
\textbf{Theorem} Consider the integral  $\int^b_a f(t)e^{-n g(t)}dt$ where $g$ is of class $\mathcal{C}^2$ on $[a,b[$ (here $a<b\leq+\infty)$ and $f$ is a continuous function on  $[a,b[$ such that

$f(a)\not=0$

$\forall t\in[a,b[:g'(t)>0$

$g'(a)=0$

$\forall t\in[a,b[:g''(t)>0$

Then $$\int^b_a f(t)e^{-n g(t)}dt\sim\sqrt{\frac{\pi}{2g''(a)}}\frac{e^{-ng(a)}f(a)}{\sqrt{n}},\quad n\to+\infty.$$

So we can write $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}\Big(n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds+n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds\Big).$$
Put $$I_1=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds$$
with $g(s)=s^2-2\ln(s), \quad s\geq 1$ and $f(s)=\cos(2\sqrt{x}s)$, by the previous theorem we have
$$I_1\sim \sqrt{\frac{\pi}{8}}n^{n}\sqrt{n}\frac{e^{-n}}{\sqrt{n}}\cos(2\sqrt{x})$$
(we have $g'(s)=2(s-\frac{1}{s})>0$ and $g''(s)=2(1+\frac{1}{s^2})>0$)
Since $n!\sim(\frac{n}{e})^{n}\sqrt{2\pi n}$ we obtain $$I_1\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$.

Using the same method we obtain $$I_2=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$

Now by using the well-known result  $\frac{n!}{\sqrt{n}}\sim \Gamma(n+\frac{1}{2})$

we get $$\forall x>0,\qquad f(\frac{x}{n})\sim \cos(2\sqrt{x})$$

So, $\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow  \Big|f(\frac{x}{n})\Big|\leq C$ for some constant $C>0$.

Therefore if we replace $x$ by $nx$ we obtain:\

$\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow  \Big|f(x)\Big|\leq C$

And hence $$\sup_{(n,x)\in \{n\in\Bbb N; n\geq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$

It remains the case $(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+$ and it is easy because if $n\leq n_0$ we have

$$f(x):=\Big(\frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})}\Big)e^{\frac{-x}{2}}L^{\frac{-1}{2}}_{n}(x)\leq C' e^{\frac{-x}{2}}P(x) $$
for some constant $C'>0$ independent of $n$ and $P$ is polynomial with $\text{deg}(P)=n\leq n_0$.

Since $\lim_{x\to+\infty}e^{\frac{-x}{2}}P(x)=0$ we obtain $$\sup_{(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$

Finally: $$\sup_{(n,x)\in\Bbb N\times\Bbb R_+}\Big|f(x)\Big|<+\infty. $$

The case $x=0$ is easy because $\Psi(0)=c$.

答:
According to 22.10.15 in [1], we have
$$H_{2n}(x) = (-1)^n\mathrm{e}^{x^2}\frac{2^{2n+1}}{\sqrt{\pi}}\int_0^\infty
\mathrm{e}^{-t^2}t^{2n} \cos (2xt) \mathrm{d} t$$
where $H_{n}(x)$ is the Hermite polynomial satisfying
$$\int_{-\infty}^\infty \mathrm{e}^{-x^2}H_n(x) H_m(x) \mathrm{d} x = \sqrt{\pi} 2^n n!\delta_{nm}, \
n, m = 0, 1, 2, \cdots$$

Thus, we have
$$\mathrm{e}^{x^2/2}\int_0^\infty \mathrm{e}^{-t^2}t^{2n} \cos (2xt) \mathrm{d} t
= \frac{1}{2} \mathrm{e}^{-x^2/2}\Gamma(n + 1/2) (-1)^n \frac{n!}{(2n)!} H_{2n}(x)$$
where we have made use of $\Gamma(n + \tfrac{1}{2}) = \frac{(2n)!}{n!} \frac{\sqrt{\pi}}{2^{2n}}$.

The question is: Is there absolute constant $C$ such that
$$\mathrm{e}^{-x^2/2} \frac{n!}{(2n)!} |H_{2n}(x)| \le C ?$$
Theorem 1 in [2] says the answer is NO: It holds that
$$\max_{x> 0} \mathrm{e}^{-x^2/2} \frac{n!}{(2n)!} |H_{2n}(x)|
> C_1 n^{1/6}$$
for some absolute constant $C_1$.
The same estimate is given in Theorem 1 in [3] where
$p_n(x) = (\sqrt{\pi}\, 2^n n!)^{-1/2}H_n(x)$.

Reference

[1] M. Abramowitz and I. A. Stegun (Eds.), “Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables”, 1964.

[2] Ilia Krasikov, “New Bounds On The Hermite Polynomials”, https://arxiv.org/pdf/math/0401310.pdf

[3] "Estimates of the Hermite and the Freud polynomials", Journal of Approximation Theory,
Volume 63, Issue 2, November 1990, Pages 210-224.

Title: Inequality with Gamma.


问:
Finding all radial solutions for biharmonic equation $\Delta^2u=0$ in $n$ dimensions
I need help finishing the proof:

Let $u(x) = v(|x|)$ be a solution of the problem $\Delta^2u = 0$. Where $\Delta^2u = \Delta(\Delta(u))$ is the Bilaplacian operator.

We have by standart calculus that
$$\frac{\partial}{\partial x_i}v(|x|) = v'(|x|)\frac{x_i}{|x|} \implies \nabla v(|x|) = v'(|x|)\frac{x}{|x|}.$$
$$\frac{\partial^2}{\partial x_i^2}v(|x|) = v''(|x|)\frac{x_i^2}{|x|^2}+v'(|x|)\left[\frac{1}{|x|}+\frac{x_i^2}{|x|^3}\right] \implies \Delta v(|x|) = v''(|x|)+v'(|x|)\frac{(n-1)}{|x|}.$$

and repeating the same calculations with $\Delta v$, using $r$ in place of $|x|$ (to shorten the notation) will result in:

$$\Delta^2 v(r) = v^{(4)}(r)+2(n-1)\frac{v'''(r)}{r}+(n-3)(n-1)\frac{v''(r)}{r^2}-(n-3)(n-1)\frac{v'(r)}{r^3}$$

As $v(r)$ is solution of the biharmonic equation, multiplying the relation above by $r^{n-1}$ and rearranging the therms will give

$$0 = r^{n-1}\Delta^2 v(r) = \left(v^{(4)}(r)r^{n-1}+(n-1)v'''(r)r^{n-2}\right)+(n-1)\left(v'''(r)r^{n-2}+(n-2)v''(r)r^{n-3}\right)-(n-1)\left(v''(r)r^{n-3}+(n-3)v'(r)r^{n-4}\right)$$
$$ = \frac{d}{dr}\left(v'''(r)r^{n-1}\right)+(n-1)\frac{d}{dr}\left(v''(r)r^{n-2}\right)-(n-1)\frac{d}{dr}\left(v'(r)r^{n-3}\right)$$
$$=\frac{d}{dr}\left(v'''(r)r^{n-1}+(n-1)v''(r)r^{n-2}-(n-1)v'(r)r^{n-3}\right) = 0$$

Therefore there's a constant $c_1$ such that

$$v'''(r)r^{n-1}+(n-1)v''(r)r^{n-2}-(n-1)v'(r)r^{n-3} = c_1$$
$$\frac{d}{dr}\left(v''(r)r^{n-1}\right)-(n-1)v'(r)r^{n-3} = c_1$$

How do I continue from here? The calculations I'm doing are the same used to find the radial solution of the laplace equation. On this point with just the laplacian we would be able to solve the resulting ODE easily. But in this case it seems more complicated. Is there a product I'm missing that could help solving the remaining equation?

Thanks in advance.

答:
We can define $\Delta$ in a bit different way and it will help you to find a general solution. If the system is spherically symmetric we have to deal only with one variable and $f(\vec{r})=f(r)$.

$\Delta{f(r)}=div(\vec\nabla{f(r)})$; gradient has only one component: $\vec\nabla{f(r)}=\vec{e_r}\frac{\partial}{\partial{r}}f(r)$

We can get the expression for $\Delta{f(r)}$ in $n$ dimensions in different ways; for instance, from the Stokes theorem for an arbitrary radial vector $\vec{A(r)}$

$div({A(r)}\vec{e_r})\sim\lim_{\delta{r}\to0}\frac{1}{r^{n-1}\delta{r}}\left((r+\delta{r})^{n-1}A(r+\delta{r})-r^{n-1}A(r)\right)=\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}A(r)\right)$

If we take $\vec{A(r)}=\vec\nabla{f(r)}$ we get (accurate to constant factors)
$$\Delta{f(r)}\sim\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)$$

(you can check this expression, for instance, for $n=3$)

And
$$\Delta\Delta{f(r)}\sim\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}\left(\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)\right)\right)=0$$

Integrating step by step we get

$$r^{n-1}\frac{\partial}{\partial{r}}\left(\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)\right)=C_1$$

$$\frac{1}{r^{n-1}}\frac{\partial}{\partial{r}}\left(r^{n-1}\frac{\partial}{\partial{r}}f(r)\right)=\frac{C_1}{2-n}r^{2-n}+C_2$$  and so on (you can easily do it yourself).

Finally, I got (please, check it!)
$$f(r)=\frac{C_1}{2(2-n)(4-n)}r^{4-n}+\frac{C_2}{3n}r^{2}+\frac{C_3}{2-n}r^{2-n}+C_4$$

The uncertainties at points $n=2$ and $n=4$ could be disclosed and give  additional terms $\sim\log(r)$

ContentLicense: CC BY-SA 4.0


问:
Prove that the degrees lie in a range
Let $G=(V,E)$ be a graph with $|V|=n$ and $|E|=m$ prove that
$$
\min_{u\in V} \{d(u)\}\leq 2\frac{m}{n}\leq \max_{v\in V} \{ d(v)\}
$$
now my first intuition is to assume that $\min\limits_{u\in V} \{d(u)\} =0$ holds because it is not a connected graph. And my second assumption is to use $\max\limits_{v\in V} \{ d(v)\} =n -1$ (connected to everything, except it self) , and from another exercise we got

$$
d(v)\geq \frac{n-1}{2}
$$

which could be interpreted as
$$
n-1\leq 2d(v)
$$
which would help in getting the right side, the only thing baffling me at the moment is the $\frac{m}{n}$ part, any intuition on this?

答:
How about we just summarize all the degrees? We know:

$$\min_{v \in V} d(v) \le d(u) \le \max_{v \in V} d(v).$$

Applying this to all elements in the sum $2m = \sum_{u \in V} d(u)$, we get that

$$n \min_{v \in V} d(v) \le \sum_{u \in V} d(u) \le n\max_{v \in V} d(v),$$

so

$$\min_{v \in V} d(v) \le \frac{2m}{n} \le \max_{v \in V} d(v).$$

ContentLicense: CC BY-SA 3.0

Title: Prove that the degrees lie in a range


问:
Equation with two variables?
im studying for an exam and i came up across this problem

find all integers $m,n$ that satisfy the equation

$\ (m+2n+1)(m-n+2)=5 $

i tried to open up the parenthesis and then finding out one of the variables and then substituing it into the first equation, would that be correct?
thank you all!

ps. sorry if it's an easy question!
pps the answers would be  $m=2 ,n= -1$ or $m= -4 , n = -1 $

Title: Equation with two variables?

Body:
$$(m + 2 n + 1) (m - n + 2)=5$$
As $m,n$ are integers we can have
$$
\begin{cases}
m + 2 n + 1=1\\
 m - n + 2=5\\
\end{cases}
$$
Or
$$
\begin{cases}
m + 2 n + 1=5\\
 m - n + 2=1\\
\end{cases}
$$
or
$$
\begin{cases}
m + 2 n + 1=-1\\
 m - n + 2=-5\\
\end{cases}
$$
Or
$$
\begin{cases}
m + 2 n + 1=-5\\
 m - n + 2=-1\\
\end{cases}
$$


问:
Why the set of all $n{×}n$ PSD matrices is not a subspace of all $n{×}n$ symmetric matrices?
I know this question is kind of basic, but I am pretty confused about that.

PSD is also symmetric, how can it not belongs to the symmetric's subspace? I have no idea to prove it.

答:
Because the set of positive semidefinite matrices is not closed under multiplication by scalars. Suppose $A$ is a positive definite matrix, and hence positive semidefinite, and $\lambda > 0$ is an eigenvalue. If we multiply $A$ by $s < 0$, we get a negative definite matrix since:

$$Ax = \lambda x \implies (sA)x = (s \lambda) x.$$

So the problem is not that the set of positive definite matrices is not contained in the symmetric subspace, but that it is not itself a vector space.


问:
What does it mean that a category satisfies right Ore condition?
What is the right Ore condition and what does it mean that a category satisfies right Ore condition?
Also can you please give an example of a category which satisfies this condition?</p>

答:
It would be great if you could supply us with more context here. Are you interested in a category $\mathcal{C}$ satisfying the right Ore condition or rather a collection of morphisms $S\subset \text{Mor}(\mathcal{C})$ to satisfy the right Ore condition? These are somewhat different things.

Following nlab, a category $\mathcal{C}$ satisfies the right Ore condtion if for every two morphisms $A\to B$ and $C\to B$, there exists an object $D$ and morphisms $D\to A$ and $D\to C$ yielding a commutative square. Any abelian category satisfies this condition and $D$ is simply the pullback of $A\to B\leftarrow C$. More generally, any additive category having all kernels also has all pullbacks and thus satisfies the right Ore condition.

A slightly different animal is the following: Let $\mathcal{C}$ be a category and let $S\subseteq \text{Mor}(\mathcal{C})$ be a collection of morphisms. Then $S$ is said to satisfy the right Ore condition if for any morphism $A\to B$ and any morphism $s\colon C\to B$ with $s\in S$, there exists an object $D$ together with morphisms $D\to C$ and $t\colon D\to A$ such that $t\in S$ and $ABCD$ is a commutative square.

In fact, $S$ is called a right multiplicative system if it is closed under compositions, satisfies the right Ore condition as above and one more axiom (see the definition of a right multiplicative system here).

The point of a right multiplicative system $S$ in a category $\mathcal{C}$ is that you can build a new category $\mathcal{C}[S^{-1}]$ obtained by declaring all morphisms in $S$ to become isomorphisms. Thanks to the structure on $S$, the category $\mathcal{C}[S^{-1}]$ is relatively easy to understand.

A decent example of the latter is the following: Let $\mathcal{A}$ be an abelian subcategory and let $\mathcal{S}$ be a Serre subcategory, i.e. given a short exact sequence $$A\to B\to C,$$ $B\in \mathcal{S}$ if and only if $A,C\in \mathcal{S}$. Define a system $S\subseteq \text{Mor}(\mathcal{A})$ by those morphisms $s$ such that $\ker(s),\text{coker}(s)\in \mathcal{S}$. Then $S$ is both a left and right multiplicative system. In fact, the localized category $\mathcal{A}[S^{-1}]$ is again abelian.


问:
Is there a dot product with which the following linear operator becomes Hermitian
Given the linear operator

$A \in L(M_2(\mathbb{C}))$

$A \begin{bmatrix}a & b \\ c & d \end{bmatrix}=\begin{bmatrix}a-b & -a+b \\ d & -c \end{bmatrix}$

Is there a dot product where the operator becomes Hermitian? ($A^*=A$)

I don't know how to prove that there exists one, but I don't know how to construct a dot product either. Any help is appreciated.

答:
An hermitian operator is diagonalizable in an orthogonal basis.

If your operator is not diagonalizable, then it is not hermitian with respect to any inner product.

If it is diagonalizable, pick any basis of eigenvalues and construct an inner product which makes it an orthogonal basis.

Rejoyce..


问:
Sums of Binomials
I want to understand how to sum binomials and I can not figure out how this comes together:

$$\sum _{l=0}^k \binom{x}{l} \binom{y}{k-l}=\binom{x+y}{k}$$

答:
This will be a duplicate, but the combinatorial argument is:

Choose a team of $k$ from $x$ girls and $y$ boys by having $l$ girls and $k-l$ boys and summing over possible $l$

Choose a team of $k$ from from $x$ girls and $y$ boys by treating them as $x+y$ children

Title: Sums of Binomials


问:
Enlarging a probability space
The book I'm reading states:
Let $\mathbb{F}$ be some filtration on $(\Omega, \mathcal{G}, \mathbb{Q})$ and $\eta$ a random variable such that

it is independent of $\mathbb{F}$ and

it has unit exponential law under $\mathbb{Q}$ (i.e. the pdf $f_\eta(x)=e^{-x}$)

The statement I'm confused about is as follows
"Of course $\eta$ [...] [is] given on a common probability space $(\Omega, \mathcal{G}, \mathbb{Q})$ and thus a suitable enlargement of the underlying probability space may be required to produce $\eta$."

My question is:
Why does $(\Omega, \mathcal{G}, \mathbb{Q})$ not suffice and how would such an enlargement look like?

答:
If $\Omega$ is a finite set and $\mathbb Q$ is the uniform measure on it it is obvious that you cannot construct any continuous random variable on this space.
Consider $(\Omega, \mathcal G, \mathbb  Q) \times (\Omega', \mathcal G', \mathbb  Q')$ where $(\Omega', \mathcal G', \mathbb  Q')$  is any probability space on which there is an exponetnial (1) random variable $\xi$. Define $\eta (\omega, \omega')=\xi(\omega')$. Then $\eta$ has the required properties.

Title: Enlarging a probability space


问:
Orthogonality of discrete sine functions
can somebody provide a proof of the attached result ? (excerpt from a thesis document). It seemed obvious but I got lost in the calculations. Many thanks
Gerald

But discrete $\sin$ functions that we consider are orthogonal when $x_j\ne y_j$,
$$\sum_{k=1}^{2R-1}\sin\left(\pi k\frac{x_j+R}{2R}\right)\sin\left(\pi k\frac{y_j+R}{2R}\right)=R\delta(x_j,y_j),$$
where $\delta(x,y)=0$ when $x\ne y$ and $\delta(x,y)=1$ when $x=y$.

答:
There is also a direct proof using the geometric partial sum

$$ \sum_{j=0}^N q^j = \frac{q^{N+1}-1}{q-1}. $$

I will prove the statement of the following form

$$ \sum_{j=1}^N \sin(\frac{\pi n j}{N+1})\sin(\frac{\pi n^\prime j}{N+1}) = \frac{N+1}{2}\delta_{nn\prime}$$

for $n, n^\prime \in \{1, \dots, N\}$. First, we note that $\sin(0)=0$, therefore we can include $j=0$ in the sum. Next we relate the $\sin$ to the exponential function $\sin(x) = (e^{ix} - e^{-ix})/2i$. Thus, we write

$$
\sum_{j=1}^N \sin(\frac{\pi n j}{N+1})\sin(\frac{\pi n^\prime j}{N+1})
 \\= \frac{-1}{4}\sum_{j=0}^{N}\left[
   \exp(i\pi \frac{n+n^\prime}{N+1}j)
 - \exp(i\pi \frac{n-n^\prime}{N+1}j)
 - \exp(-i\pi \frac{n-n^\prime}{N+1}j)
 + \exp(-i\pi \frac{n+n^\prime}{N+1}j)
\right].
$$

Each of the addends can be evaluated using the geometric partial sum:
$$\sum_{j=0}^{N}\exp(\pm i\pi \frac{n+sn^\prime}{N+1}j) = \frac{\exp(\pm i\pi(n+sn^\prime)) - 1}{\exp(\pm i\pi \frac{n+sn^\prime}{N+1}) -1}$$
where $s\in\{-1, +1\}$. Strictly speaking, we have to treat the case $n+sn^\prime=0$  separately, as the denominator becomes $0$. In this case the sum is trivially $N+1$ as we add only ones, agreeing with the limit of $n+sn^\prime\rightarrow 0$ using l'Hopital.

We distinguish two cases:

The case if both $n$ and $n^\prime$ are even/odd is simple: $n+sn^\prime$ is even and therefore $\exp(\pm i\pi(n+sn^\prime))=1$; the terms with $s=+1$ vanish, the terms with $s=-1$ both give $\delta_{nn^\prime}(N+1)$.

If one of $n$ and $n^\prime$ is even and the other is odd, $n+sn^\prime$ is odd and therefore $\exp(\pm i\pi(n+sn^\prime))=-1$. The terms compensate, as
$$\frac{1}{\exp(+ix)-1} + \frac{1}{\exp(-ix)-1} = -1$$

Q.E.D


问:
Continuous injection of $[a,b]$ is strictly monotonic from Zorich's book
I am reading Zorich's book "Analysis I" and ran into the following statement:

Let $f:[a,b]\to \mathbb{R}$ and $f$ is continuous on $[a,b]$. Then $f$ is injective if and only if $f$ is strictly monotonic.

Proof: $\Leftarrow$ is quite obvious.

The most interesting part is $\Rightarrow$. Suppose that $f$ is not strictly monotonic. Then we can find three points $x_1<x_2<x_3$ from $[a,b]$ such that $f(x_2)$ does not lie in between $f(x_1)$ and $f(x_3)$. In this case $f(x_3)$ lies in between $f(x_1)$ and $f(x_2)$ or $f(x_1)$ lies in between $f(x_2)$ and $f(x_3)$.

Question: Suppose $f$ is not strictly monotonic, i.e. $f$ is not strictly increasing and is not strictly decreasing. Then we can find points $x_1,x_2,x_3,x_4\in [a,b]$ such that $x_1<x_2$ with $f(x_1)\geq f(x_2)$ and $x_3<x_4$ with $f(x_3)\leq f(x_4)$.

I have spent smth like couple of hours trying to derive that exists three points with the above properties as in Zorich's book but I failed.

I'd be thankful if you can help me, please!

答: Proceed by contradiction and assume $f:[a,b]\to\Bbb R$ is a continuous injection but $f$ is not strictly monotonic. Choose $x_1,x_2,x_3\in [a,b]$ such that $x_1<x_2<x_3$ but $f(x_1)< f(x_2)>f(x_3)$. Now choose any point $$c\in (f(x_1),f(x_2))\cap (f(x_2),f(x_3))$$ so that the Intermediate Value Theorem gaurantees there exists $x_4\in(x_1,x_2)$ and $x_5\in(x_2,x_3)$, necessarily distinct because $(x_1,x_2)\cap(x_2,x_3)=\emptyset$, and such that $f(x_4)=f(x_5)=c\;$ which contradicts the fact that $f$ is an injection. $\it{reductio\;ad\;absurdum}$


问:
if a function $f$ is a Regular-monomorphism for two functions $g,h :Y\to Z$ and its an epimorphism, then it's an isomorphism.
My take: let $f:X\to Y$ be a regular monomorphism for $g,h: Y\to Z$:
by assumption, $g\circ f=h\circ f$, and for every other $x,y:A \to Y$ such that $g(x)=h(y)$, there exists a unique function from $A\to X$ call it $u_x$ and $u_y$ for each $x,y$. (We know they are one and the same but that's not needed at the moment. I don't think)
Hence, $g(f(u_x))=h(f(u_y))$ we know that $f(u_x)=f(u_y)=x=y$; therefore, $(gf)u_x=(hf)u_x$.
Therefore, $gf=hf=I_y$. (i think this is wrong. $gf$ cant be a function from $Y \to Y$)
Seems like everything i have so far is wrong. (This is wrong)

Any Ideas on how to prove it?

I haven't proven surjectivity, or injectivity. I am trying to build The Elementary Theory of the Category of Sets from scratch.

Before you ask me any further questions like: why "sets" why "functions." It's in the book. I didn't come up with the question or the wording. I have 4 axioms and 10 definitions (so far) and this turns out to be an exercise following the fact that every equalizer is monic. That's pretty much all I have. I have injectivity, but no surjectivity (it is "yet to be proven," as the author says).

I am working, as it seems by the author's absolutely repitative insistence, in the category of sets; however, not everything set-theoretic is workable so far (like 1-1 and onto $\iff$ surjective).

答:
Note that since $f$ is an epimorphism $g=h$. Thus, we see that $1_Y$ fits the equalizer diagram, so that there exists $k : Y \rightarrow X$ such that $f \circ k = 1_Y$. This implies that
$f \circ k \circ f  = f = f \circ 1_X$, and since $f$ is a monomorphism, we get that $k\circ f= 1_X$. Thus we showed that $k$ is an inverse of $f$.


问:
Finding the limit for a rational expression involving factorials
The task is to find
$$L=\lim_{n\rightarrow \infty} \frac{1}{2n}\sqrt[n]{\frac{(2n+1)!}{n!}}.$$

I say we resort to Sirling's approximation $k!\approx \sqrt{2\pi k} (\frac{k}{e})^k$:

$$ L \approx \lim_n \frac{1}{2n}\frac{\bigg(\sqrt{2\pi (2n+1)}(\frac{2n+1}{e})^{2n+1}\bigg)^{1/n}}{\bigg(\sqrt{2\pi n}(\frac{n}{e})^{n}\bigg)^{1/n}}\approx \frac {4n^2/e^2}{2n^2/e} \approx \frac {2}{e}.$$

however, as usual there is a solution manual that disagrees and says $L=2$. I'd appreciate it if you could help me figure out what's wrong.

答:
You can take logs to get the expression $$\frac{1}{n}\sum_{i=0}^{n}\log(n+1+i)-\log(2n)$$ whose limit is same as that of $$\frac {1}{n}\sum_{i=1}^{n}\log(n+i)-\log (2n)=\frac{1}{n}\sum_{i=1}^{n}\log(1+i/n)-\log 2$$ and you can see the limit as $$\int_{0}^{1}\log(1+x)\,dx-\log 2=\log 2 - 1$$ and thus $L=2/e$.

The approach avoids the Stirling approximation and one needs a little care when using Stirling.

You may also try the standard result that if $a_n>0$ and $a_{n+1}/a_n\to L$ then $a_n^{1/n}\to L$. Here use $a_n=b_n^n$ where $b_n$ is the expression under limit in question.


问:
Expectation of positive supermartingale.
Suppose that $\{(X_n, B_n), n \ge 0\}$ is a positive supermartingale and v is a stopping time. Show

$E(X_0) \ge E(X_v1[v<\infty])$. This is exercise 10.51 in textbook a probability path.

答:
Let $\tau$ be a stopping time. By the optional sampling theorem, $(X_{n \wedge \tau})_{n \in \mathbb{N}}$ is a supermartingale and so

$$ \mathbb{E}(X_{n \wedge \tau}) \leq \mathbb{E}(X_0). \tag{1}$$

Since

$$\lim_{n \to \infty} X_{n \wedge \tau} = X_{\tau} \quad \text{on $\{\tau<\infty\}$},$$

and $X_{n \wedge \tau} \geq 0$, it follows from Fatou's lemma that

$$\mathbb{E}(X_{\tau} 1_{\{\tau<\infty\}}) \leq \liminf_{n \to \infty} \mathbb{E}(X_{n \wedge \tau}1_{\{\tau<\infty\}}) \leq \liminf_{n \to \infty} \mathbb{E}(X_{n \wedge \tau}) \stackrel{(1)}{\leq} \mathbb{E}(X_0).$$


问:
Can 2 matrices performing the same linear transformation be unequal?
If $\Bbb R^m,\Bbb R^n,\Bbb R^o,\Bbb R^p$..., are all different vector spaces, and $A,B,C,D,E$ are transformation matrices.

And there are 2 linear transformations performing the same mapping
$$\phi:\Bbb R^n\to \Bbb R^m,\\\gamma:\Bbb R^n\to \Bbb R^m$$
$$\phi(x)=Ax,\\\gamma(x)=Bx$$

Now let there be 3 other transformations:
$$\phi_1:\Bbb R^n \to \Bbb R^o,\\\phi_2:\Bbb R^o \to \Bbb R^p,\\\phi_3:\Bbb R^p \to \Bbb R^m$$ And
$$\phi_1(x)=Cx,\\\phi_2(x)=Dx,\\\phi_3(x)=Ex$$

Let $$\gamma(x)=\phi_3(\phi_2(\phi_1(x)))\\ \Longrightarrow B=EDC$$

Now if both $\phi$ and $\gamma$ perform the same linear transformation, wrt. the same bases. will $A$ and $B$ be equivalent? Is there a simple way to prove this?

Another way to probably frame this question: 'Is a matrix performing a particular linear transformation always unique?'

Based on the answers I've gotten, if I can equate $Ax=Bx$, can I say $A$ and $B$ are equivalent matrices? Regardless of what they are composite of?

答:
Suppose that $A$ and $B$ are different. Then $B=A+M$.

Since $Ax=Bx$ we have that $Ax=(A+M)x$

Then $Ax=Ax+Mx$

$0=Mx$ for all $x$

$M$ is a zero matrix. So $B=A$


答:
The answer is yes: The matrix representation of a linear transformation with respect to particular bases is unique. To see this, let $\mathcal{B}$ and $\mathcal{C}$ be bases for $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively, and $T: \mathbb{R}^n \rightarrow \mathbb{R}^m$ a linear transformation. I assume you know that there exists a matrix representation of $T$ with respect to $\mathcal{B}$ and $\mathcal{C}$, so we only show uniqueness. Suppose $M$ and $M'$ are matrices such that for all $x \in \mathbb{R}^n$

$$[T(x)]_{\mathcal{C}} = M[x]_{\mathcal{B}} = M'[x]_{\mathcal{B}}.$$

We get $(M-M')[x]_{\mathcal{B}} = 0$ for all $x$, which means that $M-M' = \mathbf{0} \implies M = M'$.

EDIT based on addendum to the question: Notice that it is important that the relation holds for all $x$ in order to conclude that $M-M' = \mathbf{0}$. If you have $Ax = Bx$ for all $x$, we have $A = B$. You should make sure you understand why that is.

Title: Can 2 matrices performing the same linear transformation be unequal?

Body:
If $\Bbb R^m,\Bbb R^n,\Bbb R^o,\Bbb R^p$..., are all different vector spaces, and $A,B,C,D,E$ are transformation matrices.

And there are 2 linear transformations performing the same mapping
$$\phi:\Bbb R^n\to \Bbb R^m,\\\gamma:\Bbb R^n\to \Bbb R^m$$
$$\phi(x)=Ax,\\\gamma(x)=Bx$$

Now let there be 3 other transformations:
$$\phi_1:\Bbb R^n \to \Bbb R^o,\\\phi_2:\Bbb R^o \to \Bbb R^p,\\\phi_3:\Bbb R^p \to \Bbb R^m$$ And
$$\phi_1(x)=Cx,\\\phi_2(x)=Dx,\\\phi_3(x)=Ex$$

Let $$\gamma(x)=\phi_3(\phi_2(\phi_1(x)))\\ \Longrightarrow B=EDC$$

Now if both $\phi$ and $\gamma$ perform the same linear transformation, wrt. the same bases. will $A$ and $B$ be equivalent? Is there a simple way to prove this?

Another way to probably frame this question: 'Is a matrix performing a particular linear transformation always unique?'

Based on the answers I've gotten, if I can equate $Ax=Bx$, can I say $A$ and $B$ are equivalent matrices? Regardless of what they are composite of?


问:
Why is $\langle \operatorname{grad} f, X\rangle_g$ independent of the metric on a Riemannian manifold?
Let $(M,g)$ be a Riemannian manifold and let $f \in C^{\infty}(M)$. Let $X$ be a smooth vector field on $M$. In smooth local coordinates $(x^i)$ on $M$, we can write $g = g_{ij} dx^i \otimes dx^j$ as well as $X = X^i \partial_{x^i}$. Now I have computed that $\langle\operatorname{grad} f, X \rangle_g =  X^i\frac{\partial f}{\partial x^i}$ as follows. If we let $g^{ij}$ be the inverse matrix of $g_{ij}$ We have
$$\begin{eqnarray*} \langle\operatorname{grad} f, X \rangle_g &=&  \left\langle g^{ij} \frac{\partial f}{\partial x^i}\partial_{x^j} , X^k\partial_{x^k} \right\rangle_g\\
&=& g^{ij} \frac{\partial f}{\partial x^i}X^k\left\langle \partial_{x^j},\partial_{x^k} \right\rangle_g\\
&=& g_{jk}g^{ij}  \frac{\partial f}{\partial x^i}X^k \\
&=& g_{kj}g^{ji}\frac{\partial f}{\partial x^i}X^k \hspace{1cm} (\text{by using that $g_{ij}$ is a symmetric matrix})\\
&=& X^i\frac{\partial f}{\partial x^i}. \end{eqnarray*}$$

My question is: Is there any geometric reason as to why $\langle \operatorname{grad} f, X \rangle_g$ should be independent of the metric $g$?

答:
Any Riemannian manifold can be alternatively described by gauge fields on a flat Euclidean manifold (it may take several gauge fields on overlapping subsets to completely describe the manifold, but it can be done).

Let the basis vectors of the flat Euclidean space be $e_1, e_2, \ldots$ with corresponding basis covectors $e^1, e^2, \ldots$.  There is the usual partial derivative operator $\partial = e^1 \partial_1 + e^2 \partial_2 + \ldots$.

There is a gauge field $\underline h$, which is a linear map.  $\underline h^{-1}$ maps a vector $X$ to its gauge invariant counterpart $\underline h^{-1}(X)$.  The adjoint map $\overline h$ maps a derivative $\partial f$ to its gauge invariant counterpart $\overline h(\partial f)$.

Only operations on gauge-invariant quantities are, in general, geometrically meaningful.  So only $\overline h(\partial f) \cdot \underline h^{-1}(X)$ is a meaningful quantity.  But the definition of the adjoint means that $\overline h(A) \cdot \underline h^{-1}(B) = \underline h \underline h^{-1}(A) \cdot B = A \cdot B$ for any $A$ and $B$.

Thus, the dependence of the dot product on the gauge field cancels out, but only when one quantity depends on $\underline h^{-1}$ (that is, it's a vector) and the other depends on $\overline h$ (that is, it's a covector).  This is a time-honored shortcut to getting geometrically meaningful results without having to pay attention to the gauge.  It says that, for instance, the dot product of a basis vector and its corresponding basis covector (say $e_i \cdot e^i$, no summation implied) should always be $1$, regardless of gauge.

In other words, if you started with everything being vectors and then defined covectors, you would pick up a factor of the metric $\underline g = \overline h^{-1} \underline h^{-1}$ to do so, but bringing that covector back to being a vector just picks up $\overline g^{-1}$, so the metric cancels itself out.


问:
Calculating power series from function
I am trying to determine the power series of $f(x)=\frac{e^{x}-1}{x}$ (suppose $f(0)=0$). I know that $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$ and $\frac{1}{x}=\sum_{k=0}^\infty(-1)^k(x-1)^k$ for $(1-x)<1$, but I can't just combine them can I? I tried combining them and then checked the definition of the n-th derivative at $x=0$ of a power series to check if it matched the expected derivative, but it didn't.

Title: Calculating power series from function

Body:
Recall that the Malaurin series for $e^x$ is
$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$
Hence,
$$\begin{align}
&e^x-1=\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\\
\implies &\frac{e^x-1}{x}=\frac{1}{1!}+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\cdots\\
\end{align}$$
So
$$\frac{e^x-1}{x}=\sum_{r=0}^\infty\frac{x^r}{(r+1)!}$$


问:
Why in an inconsistent axiom system every statement is true? (For Dummies)
I would like to know if someone can explain in a somehow down to earth (almost logic free) way why is it true that in an axiom system where there is some statement $P$ such that $P$ and its negation $\lnot P$ are true, then every statement in the system is true?

I'm not sure if this can be done, but basically since I don't know any formal logic at all, I'm interested in seeing if at least the argument can be conveyed in an intuitive way, or if the idea can be explained without talking about first or second order logic and using symbols like $\top$, $\bot$, and $\vdash$.

This previous question is like the formal version (which I don't understand) so maybe my question can be thought of as a version for dummies of that question.

Thanks a lot in advance.

答: "inconsistent" means that if you ask the same question twice, you get a different answer. To say that you get the answer "true" every time is "consistent". The definition of "negative proposition" (the opposite of a proposition) is meant for the purpose of getting a negative result in the event the proposition gets a positive. So if we want to make a system which is "consistent" and gives the same result for a proposition and its negation, we cannot do it within the accepted meaning of the definitions "consistent" and "negation"</p>

ContentLicense: CC BY-SA 2.5


答:
Though direct construction from the development of the exponential is the easy way, we can go for the explicit evaluation of the derivatives. To make the computation manageable, we start from

$$xy=e^x-1,$$

then

$$y+xy'=e^x,\\2y'+xy''=e^x,\\3y''+xy'''=e^x,\\\cdots$$ and setting $x=0$,

$$y_0=1,\\2y'_0=1,\\3y''_0=1,\\\cdots$$

Title: Calculating power series from function

Body: I am trying to determine the power series of $f(x)=\frac{e^{x}-1}{x}$ (suppose $f(0)=0$). I know that $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$ and $\frac{1}{x}=\sum_{k=0}^\infty(-1)^k(x-1)^k$ for $(1-x)<1$, but I can't just combine them can I? I tried combining them and then checked the definition of the n-th derivative at $x=0$ of a power series to check if it matched the expected derivative, but it didn't.


问:
Lines in a rectangle with a specific point $P$

I came across this interesting problem that might teach me something new.

On the above diagram, $ABCD$ is a rectangle and point $P$ such that $PB = 4\sqrt{2}$, $PC = 4$ and $PD= 3$.

Our goal is to find $PA$. If this was a problem that has good description I would just search up it, however, this is not easy to describe to a search engine, so here's my actual question: How do you find a length of a given line that is in a rectangle connected to some point inside the rectangle and knowing how long each of the other lines connecting the rectangle's corner with the specific point? (Hopefully you didn't have a headache reading this)

I suspect there is some sort of formula to figure this out, so any help would be greatly appreciated!

答:
Hint

Draw parallel lines from point $P$ to the all four sides of the rectangle. Apply the Pythagorean Theorem four times and solve the system of equations.

 You should obtain: $PB^2+PD^2=PA^2+PC^2$


问:
Find $\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz$
Find $\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz$ where $\gamma$ is a linear pieces parameterization of closed polygonal chain $[w_0,w_1,w_2,w_3,w_0]$ with vertices $w_0=1+i, w_1=1-i, w_2=-1+i,w_3=2$. Consider the cases  $z_0=\frac 12, z_0=-1$.

I know that such task I should do by using the Cauchy's integral theorem. However, on the basis of the theorem itself, I have no intuition yet how to go about such tasks. Anyone would like to guide me how to think about a task so that I can do the next task one myself?

答:
The case $z_0=-1$. There is an open and connected set $D$ with the following properties:

the convex hull of $[w_0,w_1,w_2,w_3,w_0]$ is contained in $D$ and $z_0 \notin D.$ Hence the function $\frac{(\sin z)^2}{(z-z_0)^2}$ is holomorphic on $D$. Cauchy's theorem gives then:

$$\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz=0.$$

The case $z_0=1/2.$ Let $f(z)= \sin^2 z$. Since $z_0$ is an element of the interior of the convex hull of $[w_0,w_1,w_2,w_3,w_0]$, we get by Cauchy's integral formula for the first derivative, that

$$\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz= 2 \pi i f'(1/2).$$


答:
Let's draw 4 perpendecular lines on all four edges from point P. The lines meet the edges at points L, R, U, D. I will call the heights: $h_{L}$,  $h_{R}$, $h_{U}$, $h_{D}$

$$   PB^2 - PL^2 = LB^2 = CR^2 = CP^2 - PR^2    $$
Thus:
$$ 32 - 16 = PL^2 - PR^2   $$

Similarly:

$$ PA^2 - PD^2 = ...= PL^2 - PR^2 $$

Thus, according to the two previous equations:
$$ PA^2 - 9 =  16 $$
$$PA = 5$$

Title: Lines in a rectangle with a specific point $P$

Body:
I came across this interesting problem that might teach me something new.

On the above diagram, $ABCD$ is a rectangle and point $P$ such that $PB = 4\sqrt{2}$, $PC = 4$ and $PD= 3$.

Our goal is to find $PA$. If this was a problem that has good description I would just search up it, however, this is not easy to describe to a search engine, so here's my actual question: How do you find a length of a given line that is in a rectangle connected to some point inside the rectangle and knowing how long each of the other lines connecting the rectangle's corner with the specific point? (Hopefully you didn't have a headache reading this)

I suspect there is some sort of formula to figure this out, so any help would be greatly appreciated!


问:
How to prove the equality $\int^{na}_{ma}\frac{\ln(x-a)}{x^2+a^2}\,dx = \int^{\frac{a}{m}}_{\frac{a}{n}}\frac{\ln(x+a)}{x^2+a^2}\,dx$?
How to prove the equality
$$\int^{na}_{ma}\frac{\ln(x-a)}{x^2+a^2} \, dx= \int^{\frac{a}{m}}_{\frac{a}{n}}\frac{\ln(x+a)}{x^2+a^2}\, dx,$$

where $a, m, n$ are strictly positive numbers such that $mn=m+n+1$?

答:
$x=a^2/u$ should help you. For example, $dx/(x^2+a^2)=-du/(a^2+u^2)$. I believe the constraint comes into play when consisting the extra terms in the logarithm.

However, I'm not quite sure about the $\pm$ signs in the logarithm; are these correct?

This substitution is motivated by the observation that you go from $am\to a/m$ via $x\to a^2/x$.


问:
Determinant and matrices operations
My teacher told that in a determinant if we do some operations on a row or column then the coefficient of that should be 1 otherwise the whole determinant should be multiplied with that quantity.
I am confused over his sentence... If we take for instance if C1 goes to xC1 - C2 then column 1 should only be multiplied with "x" Right?... How come the whole determinant is getting multiplied with "x"???</p>

答:
You may know that $\det(XY) = \det(X)\det(Y)$. You can represent the row and column operations by multiplying the matrix by another matrix from left (for row operations) or from right (for column operations). If you multiply your matrix by another matrix with determinant equal to one, the product has the same determinant as the original matrix.

The operation which you ask about can be represented by multiplying a matrix from right by this matrix:$$
Y=\begin{bmatrix}
x&0&0&0\\
-1&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}
$$
And you see that $\det(Y)=x$, not 1. Therefore it scales the determinant by factor $x$.

If your operation was C1 going to C1-xC2, the matrix would be:$$
Y=\begin{bmatrix}
1&0&0&0\\
-x&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}
$$
In this case, $\det(Y)=1$, so it doesn't change the determinant.

Title: Determinant and matrices operations


问:
Relation between a set $A$ and the interior of its closure
Let $A$ be a subset of a metric space $(X , d)$.

I know that if $A$ is open then it is contained in the interior of its closure. But if $A$ is not open then is it true that A is not contained in the interior of its closure. There are examples of non open sets which are not contained in the interior of their closure. For example any non empty finite subset of $\mathbb{R}$ (with
usual metric) satisfies this statement as the interior their closure is empty. However, is this true for any non open subset of a general metric space.

Any help please. Thanks in advance.

答: If $D\subseteq X$ is dense, then $\overline{D}=X$ (by definition) and so $int(\overline{D})=X$. Therefore $D\subseteq int(\overline{D})$. So all you need is to take some non-open dense subset, e.g. $\mathbb{Q}\subseteq\mathbb{R}$.


问:
How does $(0,y) = (y,0)\cdot(0,1)$?
I was reading a script and cannot understand how does $(0,y) = (y,0)\cdot(0,1)$ in the second line. How can we say this? The section is  "rebuilding the complex numbers from ground up" and this is a multiplication of 2 tuples.

答:
In terms of the representation $(a,b) = a+bi$, we obtain

$(y,0)\cdot (0,1) = y\cdot i= yi =(0,y)$.

Title: How does $(0,y) = (y,0)\cdot(0,1)$?


答:
If we write complex numbers as an ordered pair of real numbers $(a,b)$ then the rules of arithmetic are

$(a,b)+(c,d)=(a+c,b+d)$

$(a,b)\cdot(c,d)=(ab-bd,ad+bc)$

This example justifies the second, and presumably there were earlier sections saying we want something like

A distributive property

Real scaling such as $(a,0)\cdot (c,d)=(ac,ad)$

A square root of $-1$ so $(0,1)\cdot(0,1)=(1,0)$

and the second gives $(y,0)\cdot(0,1)=(0,y)$

but we would need to see the context to be sure

Title: How does $(0,y) = (y,0)\cdot(0,1)$?

Body: I was reading a script and cannot understand how does $(0,y) = (y,0)\cdot(0,1)$ in the second line. How can we say this? The section is  "rebuilding the complex numbers from ground up" and this is a multiplication of 2 tuples.


问:
How to perform a push forward change of measure.
Im a bit confused with the push forward formula https://en.wikipedia.org/wiki/Pushforward_measure

Let $\rho$ be a probability density associated to a probability measure $\mu$ on $\mathbb{R}^d$, i.e "$\rho(x)dx=d\mu(x)$". Let $f:\mathbb{R}^d\to\mathbb{R}^d$ and denote $f_{\#}\mu$ be the push forward of $\mu$ by $f$, and $f_{\#}\rho$ the associated density pushed forward.

Then $\textbf{is it true that}$ for any integral

$$
\int_{\mathbb{R}^d} \big( g \circ f_{\#}\rho \big) (x) df_{\#}\mu(x)  =\int_{\mathbb{R}^d} \big( g\circ \rho \big) \rho(x)dx
$$

答: What does $f_\sharp\rho$ mean? Is it the density of $f_\sharp\mu$ with respect to the Lebesgue measure? In the general case there does not necessarily exist such a density, e.g. $f:x\mapsto0$. Even if $f_\sharp\rho$ exists, your equation has no reason to hold. Take for instance $d=1$, $\mu$ equal to the Lebesgue measure on $(0,1)$ and $f:x\mapsto2x$, so that $\rho(x)=1_{(0,1)}(x)$, $f_\sharp\mu$ is half of the Lebesgue measure on $(0,2)$ and $f_\sharp\rho(x)=\frac12\times1_{(0,2)}(x)$. Let $g:\mathbb R\to\mathbb R$ be any nonnegative measurable map such that $g(x)=0$ on $[1,2]$ and $g(x)>0$ on $(0,1)$. Then in your equation, the left-hand side is positive whereas the right-hand side is zero.


问:
Does the family of compact subsets of a Polish space belong to the Effros Borel structure?
Since the answer to my previous question (see Effros Borel structure
) turned out to be negative, I don't know how to solve the following problem:

Let $X$ be a Polish space and let $\mathcal{F}(X)$ be the space of all closed subsets of $X$ equipped with the Effros Borel structure. Is it necessary true that $\mathcal{K}(X)$, the set of all compact subsets of $X$, is Borel as a subset of $\mathcal{F}(X)$?

Trivially, the answer is positive if $X$ is compact. However, the case I am interested in is $X = \mathbb{U}$, where $\mathbb{U}$ stands for the universal Urysohn space (which is not compact).

答:
Yes, $\mathcal K(X)$ is indeed Borel in $\mathcal F(X)$.

In fact when $X$ is locally compact and $\mathcal F(X)$ has the Fell topology, $\mathcal K(X)$ is $\mathbf{\Pi}^0_3$ at worst. To see that recall that a subspace of a metrizable space is compact iff it is complete and totally bounded. Since we are only dealing with closed sets we need to worry only about being totally bounded. Let $D\subseteq X$ be a dense countable set, then $F\in\mathcal F(X)$ is totally bounded iff for all $n\in\Bbb N$ there is a finite $A\subseteq D$ such that $F\subseteq\overline{\bigcup_{a\in A}B_{1/n}(a)}$, which means that we can write:

$$\mathcal K(X)=\bigcap_{n\in\Bbb N}\bigcup_{A\in[D]^{<\omega}}\left\{F\in\mathcal F(X)\mid F\subseteq \overline{\bigcup_{a\in A}B_{1/n}(a)}\right\}.$$

(I'm taking closure of the union of balls even though it wouldn't be needed for total boundedness to make sure that the set is in the Borel $\sigma$-algebra, being the complement of a generating set)


问:
Question about Rank-Nullity Theorem Proof
I understand the proof except for the very last part that says that the argument also shows these vectors are distinct.

How exactly does the argument show this. I need this to be super detailed otherwise I will go crazy if I don't fgure it out. I'm ci fused about the very last two sentences.

答:
I guess you are confused as to how they show the vectors in $S=\{T(v_{k+1}), \dots,T(v_{n})\}$ to be linearly independent.

To show that $S$ is linearly independent, assume that $\sum_{j=k+1}^nb_jT(v_j)=0$. We need to show that each $b_j$ is zero. By linearity of $T$, we find that $$T(\sum_{j=k+1}^nb_jv_j)=\sum_{j=k+1}^nb_jT(v_j)=0.$$
Hence $\sum_{j=k+1}^nb_jv_j\in \ker(T)=N(T)$. By construction, we have a basis $\{v_1, \dots ,v_n\}$ of $V$ such that $\{v_1, \dots , v_k\}$ is a basis of $\ker(T)=N(T)$. Hence $\sum_{j=k+1}^nb_jv_j\in \ker(T)=N(T)$ implies that $\sum_{j=k+1}^nb_jv_j$ can be written as $$\sum_{j=k+1}^nb_jv_j=\sum_{i=1}^kc_iv_i.$$
Rewriting we get that $$\sum_{j=k+1}^nb_jv_j-\sum_{i=1}^kc_iv_i=0.$$ Looking at the last expression, this precisely is a linear combination of the vectors $\{v_1, \dots ,v_n\}$ which is zero. Since $\{v_1, \dots ,v_n\}$ is a basis, the only such combination is the zero combination. It follows that all coefficients (thus all $b_j$ and all $c_i$) are zero. In particular, the $b_j$'s are zero as required.


问:
How to build the Mathematics for outage probability.
I am reading a paper that calculates outage probability. I had reached upto following:
Step 1: The outage probability is defined as $P_o = Pr\left(\gamma \geq \gamma_{th}\right)$-----(1), where $\gamma$ is SNR and $\gamma_{th}$ is SNR threshold. Then i have the expression of SNR that i had substituted into definition of SNR which is given in (1). Thus the step 2 is obtained as follows.

Step 2: $P_o = Pr\left(\frac{P_t|h|^2}{\alpha|f|^{2}|g|^{2}P_{t}+\sigma^2}\geq\gamma_{th}\right)$-----(2), where $h,f,g $ are rayleigh random variable, $\sigma^2$ is variance of additive white gaussian noise (AWGN) and $P_t$ is transmit power. After some simple manupulation, i reached upto the step 3.

Step3: $P_o = Pr\left(|h|^{2}\geq\alpha|f|^{2}|g|^{2}\gamma_{th}+\frac{\sigma^2\gamma_{th}}{P_t}\right)$------(3). After step 3, the step 4 is given as

Step4: $P_o = \int_0^{\infty}\int_{\frac{\gamma_{th}(\alpha P_tx+\sigma^2)}{P_t}}^{\infty}f_{|h|^{2}}(y)f_{|f|^2|g|^2}(x)dydx$------(4)
I am not getting how Eq. (4) is obtained. Any help please.
Note: I apologize for the very long question that I had asked. However I had written so that every one understood my query. And I just want how Eq. (4) is obtained from Eq. (3)

答:
I think that what’s going on is as follows:

First (and check the paper for notation): I think $f_{|h|^2}(y)$ designates the probability density function for a Rayleigh distribution of scale parameter $|h|$.

So the inner integral with respect to $dx$ is accumulating the portion of $P_0$ for a given value of $h$. The lower bound of this integral is the smallest valid value of $h$.

And then the outer integral with respect to $dy$ is accumulating over all values of $h$.

In more detail:
$$\begin{align*}
  P_0 &= Pr\!\left(
    |h|^2 \geq \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t}
  \right)
\\&=
  \lim_{\delta y \to 0}
  \sum_{k=0}^{\infty}
    Pr\!\left(
      |h|^2 \geq
      \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t}
      \,\Bigg{|}\,
      h = y_k
    \right)
    Pr\Bigl(h \in [y_k,y_{k+1})\Bigr)
    &\text{where $y_k = k\delta y$}
\end{align*}$$
by the law of total probability. Intuitively, we tile the positive number line into intervals of width $\delta y$.  Those are the possible values for $h$.  Now $Pr\Bigl(h \in [y_k,y_{k+1})\Bigr) \approx f_{|h|^2}(y) \delta y$ and the approximation becomes perfect as $\delta y \to 0$ so the sum becomes an integral
$$
P_0 = \int_{0}^{\infty}
      Pr\!\left(
      |h|^2 \geq
      \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t}
      \,\Bigg{|}\,
      h = y
    \right)
   f_{|h|^2}(y) \, dy
$$

I think that something similar is going on with the inner integral with respect to $x$, so that
$$\begin{align*}
  Pr\!\left(
      |h|^2 \geq
      \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t}
      \,\Bigg{|}\,
      h = y
    \right)
  &=
  \lim_{\delta x \to 0}
  \sum_{\ell=0}^{\infty}
    Pr\!\left(
      |h|^2 \geq
      \alpha|f|^2|g|^2\gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t}
      \,\Bigg{|}\,
      h = y, |f||g| = x_{\ell}
    \right)
    Pr\Bigl( |f||g| \in [x_{\ell}, x_{\ell+1}) \Bigr)
    & \text{where $x_{\ell} = \ell \delta x$}
\\&=
  \int_{\alpha x \gamma_{th} + \frac{\sigma^2\gamma_{th}}{P_t}}^{\infty}
     f_{|f|^2|g|^2}(x) \, dx
\end{align*}$$

Hence
$$\begin{align*}
  P_0 &= \int_{0}^{\infty}
      \left(
        \int_{\frac{\gamma_{th}(\alpha P_t x + \sigma^2)}{P_t}}^{\infty}
          f_{|f|^2|g|^2}(x) \, dx
      \right)
   f_{|h|^2}(y) \, dy
\\&=
  \int_{0}^{\infty}
  \int_{\frac{\gamma_{th}(\alpha P_t x + \sigma^2)}{P_t}}^{\infty}
    f_{|h|^2}(y)
    f_{|f|^2|g|^2}(x)
    \, dy dx
\end{align*}$$
where the last step is from rearranging the order of the integration and bringing the factor $f_{|h|^2}(y)$ inside the inner integral.


问:
Differentiate a vector-valued function means to differentiate each of the components separately
If we have $f: \mathbb{R^n} \to \mathbb{R^m} $.

We can write $f$ in terms of its components $f = (f_1, f_2, ..., f_m)$

We want to calculate the $partial \, derivative$ with respect to the $j^{th}$ component of the variable $x$ at the point $x_0$ ($i.e.$ calculate $\frac {\partial f} {\partial {x_j}} (x_0)$)

It is said that (by Terence Tao in his book Analysis 2, chapter 6.3), to differentiate a vector-valued function one just has to differentiate
each of the components separately.

$i.e. \qquad $ $\frac {\partial f} {\partial {x_j}} (x_0)$ = ($\frac {\partial {f_1}} {\partial {x_j}} (x_0)$, $\frac {\partial {f_2}} {\partial {x_j}} (x_0)$, ..., $\frac {\partial {f_m}} {\partial {x_j}} (x_0)$)

I have tried to prove this statement but failed to have any progress. Could you please give me a suggestion for a rigorous proof of this statement ?

答:
Note that:\begin{align*}
\frac{\partial f}{\partial x_j}(x_0)&=\lim_{t\to 0}\frac{f(x_0+te_j)-f(x_0)}{t}\\&=\left(\lim_{t\to 0}\frac{f_1(x_0+te_j)-f_1(x_0)}{t},\dots,\lim_{t\to 0}\frac{f_m(x_0+te_j)-f_m(x_0)}{t}\right)\\
&=\left(\frac {\partial {f_1}} {\partial {x_j}} (x_0), \frac {\partial {f_2}} {\partial {x_j}} (x_0), ..., \frac {\partial {f_m}} {\partial {x_j}} (x_0)\right)
\end{align*}
In the second equality I used the fact that a limit in $\Bbb R^m$ may be computed componentwise.


问:
What does the discriminant of an algebraic number field mean intuitively?
If $E/F$ is a finite extension of fields and $\alpha_1,\ldots, \alpha_n$ is a basis
of $E/F$, the discriminant of $\{\alpha_1,\ldots, \alpha_n\}$ is $$\det(\operatorname{Tr}_{E/F}(\alpha_i\alpha_j))$$

I want to know if there is any other interpetration of this quantity? What information is actually conveyed by this quantity?

答:
The discriminant has a geometric interpretation as squared volume. The analogy to have in mind is that the pairing $(x,y) \mapsto {\rm Tr}_{E/F}(xy)$ in a finite extension of fields $E/F$ is analogous to the pairing $(v,w) \mapsto v \cdot w$ for vectors $v$ and $w$ in ${\mathbf R}^n$.  The trace pairing is an $F$-bilinear mapping $E \times E \rightarrow F$ while the dot product is an ${\mathbf R}$-bilinear mapping ${\mathbf R}^n \times {\mathbf R}^n \rightarrow {\mathbf R}$.

If an $n$-dimensional box in ${\mathbf R}^n$ has a vertex at the origin and its edges coming out of the origin are $v_1, \dots, v_n$, then the traditional way of describing the volume uses the matrix with the $v$'s as the columns: the volume is $|\det A|$ where $A = [v_1 \cdots v_n]$ is the $n \times n$ matrix whose $j$th column is $v_j$. The product $A^\top{A}$ has $(i,j)$ entry $v_i \cdot v_j$, and its determinant is $\det(A^\top{A}) = \det(A^\top)\det(A)$, which is $\det(A)^2$. Therefore the volume of the box is $\sqrt{\det(A^\top{A})} = \sqrt{\det(v_i \cdot v_j)}$. So we have two formulas for the volume of that box:
$$
{\rm volume} \ = \det[v_1 \cdots v_n] = \sqrt{\det(v_i \cdot v_j)}.
$$
Therefore
$$
{\rm volume}^2 \ = \det(v_i \cdot v_j).
$$

Replacing $n$-dimensional Euclidean space with a finite extension of fields $E/F$ of degree $n$ and a basis $v_1,\dots,v_n$ of ${\mathbf R}^n$ with a basis $e_1,\dots,e_n$ of $E/F$, then the analogue of $\det(v_i \cdot v_j)$ is $\det({\rm Tr}_{E/F}(e_ie_j))$, which is the discriminant of the basis.

As one example of this analogy at work, any two bases of $E/F$ have their discriminants related by a nonzero square factor (namely the square of the determinant of the change-of-basis matrix), and this is analogous to the squared volume of two boxes in ${\mathbf R}^n$ being related to each other by the square of the determinant of the change-of-basis matrix. The algebraic calculations underlying both results are very similar (some may even say it's the same calculation).

As another example of the analogy at work, $\det(v_i \cdot v_j)$ and $\det({\rm Tr}_{E/F}(e_ie_j))$ both make sense even if the $v_i$'s and $e_i$'s are not bases, and in that case the numbers are $0$. If the $v_i$'s are linearly dependent then the "box" they span is lower than $n$-dimensional, so it makes geometric sense that $\det(v_i \cdot v_j)$ is $0$ because the $n$-dimensional volume of the box is $0$. If $v_1,\dots, v_n$ is a basis then $\det(v_i \cdot v_j)$ is not $0$ since the the box spanned by the $v_i$'s is $n$-dimensional, so the box has nonzero volume. Is $\det({\rm Tr}_{E/F}(e_ie_j))$ nonzero if $e_1,\dots,e_n$ is a basis of $E/F$? This algebraic question is more subtle than the situation with vectors in ${\mathbf R}^n$. If the trace function ${\rm Tr}_{E/F}$ is not identically zero on $E$ then $\det({\rm Tr}_{E/F}(e_ie_j)) \not= 0$ when the $e_i$'s are a basis, but if the trace is identically $0$ then of course $\det({\rm Tr}_{E/F}(e_ie_j)) = 0$ for any $e_i$'s at all. For number fields, or more generally any fields of characteristic $0$, the trace is not identically $0$ since ${\rm Tr}_{E/F}(1) = [E:F] \not= 0$ in $F$. More generally, the trace is not identically $0$ if and only if $E/F$ is a separable field extension.


问:
Understanding descent on elliptic curves
I'm trying to understand the proof for finding the rank of an elliptic curve in the from $Y^2 = F(X)$ in the case where $F(X)$ has 3 rational roots.
I believe the book Lectures on Elliptic Curves by Cassels covers this in Chapter 15 entitled The weak finite basis theorem.

The proof treats the cases where F(X) has 3 rational roots, one rational root and  no rational roots together, however I'm only interested in the first case.

My main question is: in this specific case what is $\Theta$?

My second question is how do I interpret (iii) in this case in the homomorphism definition? What do points of the form $(a,0)$ map to?

Many thanks for your help.

答:
When $F(x)=(x-r_1)(x-r_2)(x-r_3)$, $\mathbb{Q}[\Theta]$ is $\mathbb{Q}^3$ and $\Theta$ is $(r_1,r_2,r_3)$.

For your second question, take for instance $a=r_1$ (in general, $a$ is some $r_i$), then the image of $(a,0)$ is the class in $\mathcal{M}$ of any $x=(\lambda,r_2-r_1,r_3-r_1)$ such that its norm (aka the product of these three elements) is the square of a nonzero rational.


问:
Reordering rows in an array for minimum total euclidean distance
I am looking for an algorithm that can find the shortest path between a set of vertices in 5 dimensions given a single source.

For example if I have a 10 x 5 array, where each row corresponds to a point in 5 dimensional space. How do I reorder the rows such that travelling from top to bottom of the array results in the minimum total Euclidean distance travelled.

My end goal is to perform this operation in python.

I feel this must be a well defined problem but my research has not led me anywhere conclusive. It seems to me that a solution would incorporate graph theory - a minimum spanning tree. I'd be very grateful for any suggestions.

答:
we can use the concept of  hierarchical clustering using single linkage. At each step, you connect two group of nodes that have minimum distance which is defined as the minimum distance that exist between a point from one group to the point from another group. Just google it and you can see its concept is what you are looking for. Python Sklearn package provides this algorithm. It creates clusters in form of chains. If you continue to completely merge all points into one cluster, you will get what you are looking for.

A very simple example:
Ler's say your points have $1$ dimensuon. Data is $1, 2, 7,9, 20$

It first groups $1$ and $2$. Then, it will group $7$ and $9$. Then, it will group $1,2,7,9$ together since $2$ and $7$ has smallest single-linkage between-cluster distance. And, finally it adds 20 to the group.

At each stage of merging, just find the pair that makes the chain- i.e. leads to merging two clusters together. Doing so upto the end and you will get what you want. (You can either write the algorithem yourself which is easy, or modify sklearn hierarichal clustering function to get what you want)

EDIT
So, after thinking more, I realized this algorithm doesn't work since merging two close groups doesn't necessarily occur at the end of one chain and the beginning of one chain. In other words, if $A \to B \to C$ is a preferable traversing path in one group and $D \to E \to F$ is another, merging these two chains might happen at $B \to E$. So, even rearranging the points doesn't help as it violates the idea of having the smallest distance from one point to the next.

Could anyone confirm if this is a NP-hard problem if n gets really large?


问:
Set theory task
Let F be a set of sets.
We say a set is n-pretty  (it's writen in my text book, I don't know the exact name)(n is fixed nat. number) if
$$\forall X (X \in F \iff \forall Y (Y \subseteq X \land |Y| <= n \implies Y \in F)$$.
And we say that a set is finite-pretty if:
$$\forall X (X \in F \iff \forall Y (Fin(Y) \land Y \subseteq X \implies Y \in F)$$

So the task is to find if every n-pretty set is finite-pretty

and if every finite-pretty set is n-pretty.

I think that every finite-pretty set is n-pretty. Since it's true for every finite set, it will be true for every set with cardinality n. But I can't seem to  prove it in terms of set theory. And what about the other direction ? Any tips?

答:
It's indeed true that all finite-pretty $F$ is $n$-pretty for any $n \in \Bbb N$ because (trivial set theory) $|Y| \le n \implies Y \in \textrm{Fin}(Y)$. This is almost true by definition: a set is finite iff there is some $n \in \Bbb N$ such that its cardinality is $=n$ or $\le n$, definitions vary by text...

The reverse fails in general: the powerset of $\Bbb N$ without the singletons is a boring example of a $2$- and $3$-pretty set that is not finite pretty.

Title: Set theory task


问:
How is it possible that we can solve some equations numerically, but can't solve them analytically?
For example, there is the Abel–Ruffini theorem that states there is no analytical solution to polynomial equations of degree five or higher, mentioning $x^5-x-1=0$ as the simplest example. Now if we take $$x^5-x-a=0$$ we can calculate $x$ numerically for every value of $a$ without any trouble. So clearly all the information necessary is contained in the equation. But despite this, we are not able to provide a formula like $x(a)=\dots$.

How is it possible that we have all the information necessary, but we still can not shape it into an explicit formula? What property is it that makes some equations solvable analytically, and others not?

Body:
You cannot obtain the analytical solution of the equation but you can obtain explicit asymptotic expressions of $x$.

To simplify the notations, let $a=b^5$ and we shall try to solve
$$x^5-x-b^5=0$$ assuming $b>0$. Using high-order methods (order $n$) we shall have
$$x_{0}=b+\frac{b}{5 b^4-1}$$
$$x_{1}=b+\frac{5 b^5-b}{25 b^8+1}$$
$$x_{2}=b+\frac{25 b^9+b}{125 b^{12}+25 b^8+5 b^4-1}$$
$$x_{3}=b+\frac{125 b^{13}+25 b^9+5 b^5-b}{625 b^{16}+250 b^{12}+50 b^8-5 b^4+1}$$
$$x_{4}=b+\frac{625 b^{17}+250 b^{13}+50 b^9-5 b^5+b}{3125 b^{20}+1875 b^{16}+500 b^{12}+6   b^4-1}$$
$$x_{5}=b+\frac{3125 b^{21}+1875 b^{17}+500 b^{13}+6 b^5-b}{15625 b^{24}+12500 b^{20}+4375 b^{16}+375 b^{12}+35 b^8-7 b^4+1}$$ Suppose that we do this forever.

Using $b=3$, some results

$$\left(
\begin{array}{ccc}
 n & x_{n} & \text{decimal representation} \\
 0 & \frac{1215}{404} & 3.007425742574257425742574 \\
 1 & \frac{246645}{82013} & 3.007389072464121541706803 \\
 2 & \frac{100137870}{33297277} & 3.007389162783491274676905 \\
 3 & \frac{6254765325}{2079799117} & 3.007389162671713933610637 \\
 4 & \frac{33012650893518}{10977179575985} & 3.007389162671662103584418 \\
 5 & \frac{13403136063083787}{4456734841451944} & 3.007389162671662635777060 \\
6 & \frac{680209145067512871}{226179289833989464} &3.007389162671662634787269 \\
7 & \frac{2209319270264084987187}{734630322436023080017} & 3.007389162671662634787290 \\
8 & \frac{896983610363648795857113}{298259906465446908872354} & 3.007389162671662634787294
\end{array}
\right)$$


问:
difference between Sobolev space and bounded variation space
I learned Sobolev spaces & bounded variation spaces. I read this sentence:

Sobolev spaces include functions such that $\int |f^{(1)}(x)|^2 dx < \infty$, but do not include functions that are piecewise smooth. The space of bounded variation $\{f: \int |f^{(1)}(x)|dx < \infty\}$ include such functions.

So the only difference in two equations is square, I am not sure how this makes a big difference. Can anyone elaborate on this difference with example of step function $f(x)=0$ if $x<0$, $f(x)=1$ if $x\ge0$ for $x \in [-1,1]$?

答:
An illuminating way to think about Sobolev spaces is that functions from such spaces are integrable, don't have jumps, are differentiable almost everywhere and their derivative is integrable. This is given a precise meaning using the weak derivative:
$$
  f = Dg
  \qquad \stackrel{def}{\Longleftrightarrow} \qquad
  \int_\Omega f \varphi = -\int_\Omega g \varphi'
  \quad \forall \varphi \in C^\infty(\Omega)
  \quad \forall \Omega \text{ compact}
$$
If $f, g \in C^1(\mathbb{R})$, this is just integrating by parts, but you can also introduce removable signularities and the weak derivative will still work. Notice however, that if $g$ had a jump, such a function $f$ wouldn't exist. Then the Sobolev spaces are defined as follows:
$$
  W^{k,p}(\Omega) = \Big\{
    f \in L^p(\Omega)
    \; \Big| \;
    D^n f \in L^p(\Omega) \;\; \forall n \in \{ 1, ... k \}
  \Big\}
$$
The space you describe in your question by “$\int |f^{(1)}(x)|^2 \mathrm{d}x < \infty$” is the Sobolev space $W^{1,2}$. However, what you're missing is that this is the weak derivative – therefore it “automatically forbids” any jumps in $f$. The square in the integral isn't important at all for your question, the space $W^{1,1}$ is characterized by $\int |D^1 f(x)| \, \mathrm{d}x < \infty$, without the square. But it's still the weak derivative.

On the other hand, the space $BV$ of (not necessarily continuous) bounded variation functions is characterized by the fact that they allow for removable discontinuities and jumps [src], and by the fact that they have a derivative almost everywhere. The difference here is that this derivative is the classical derivative of a function. If $f \in BV$ has a jump in a point, it just doesn't have a derivative there. In other words, the integral $g(x) = \int_0^x f^{(1)}(x') \mathrm{d}x'$ would be equal to the function $f$ piecewise shifted by a constant, so that it doesn't have jumps anymore.

This also explains, why one would want to use the weak derivative instead of the almost-everywhere derivative: it doesn't lose track of the jumps, which can be quite important eg. in physical models.

On a final note, this wouldn't happen if we were talking about continuous functions with bounded variation. The Sobolev space $W^{1,p}$ could be alternatively described as (a.e.) absolutely continuous functions with suitable growth conditions. Since the space of continuous bounded-variation functions is a proper subset of absolutely continuous functions.


问:
Let $E = C[0, 1]$ and $A = \{f \in E \mid f(x) > 0 \text{ for all } 0 \leqslant x \leqslant 1 \}$. Is $A$ open when the norm of $E$ is the L1-norm?

Let $E = C[0, 1]$ and $A = \{f \in E \mid f(x) > 0 \text{ for all } 0 \leqslant x \leqslant 1 \}$. Is $A$ open when the norm of $E$ is the L1-norm?

Let $f \in A$. Now I have $B(f,r)= \{g \in A \mid d(f,g) < r \}.$ Let $g \in B(f,r)$ now I think it would satisfy to show that $g$ outputs only positive values in order to show that it belongs to $A$ which would imply that $B(f,r) \subset A$? Using the  $L1$ norm I have $$d(f,g) = |f(x) -g(x)|$$ but I cannot seem to show from here that $g$ would be positive?

答:
Let $\epsilon>0$. If $f \in A$ is a continuous positive function, we know that there exists $M$ such that $f(x)\leq M$.  Thus, we know that there exists $\delta$ such that $$\int_{1-\delta} ^1 f \leq \epsilon/2M$$. Define the function $g$ as
$g(x)= f(x)$ if $x \leq 1-\delta$ and connect $f(1-\delta)$ to zero via a line with domain $[1-\delta,1]$, this function is continuous. Now, note that $d(f,g) < \epsilon$ and that $g(x)=0$. Thus, any open ball containing$f$ is not contained in $A$.


问:
Differential equations (SO ODE)
My problem is based on the next differential equation:

$\frac{1}{2}\sigma^2 P^2 V'' +rPV' - rV + GP-C = 0.$

Have someone idea how to solve it.
Thanks

答:
I guess you want to solve for $V(t)$, so that all coefficients are constants since $V$ depends on $t$.

Put $V(t)=e^{\lambda t}$ in $\frac{1}{2}\sigma^2 P^2 V'' +rPV'-rV=0$ and solve for $\lambda$. You will probably find two solutions. Then you need to find any solution to the whole ODE. You can find a constant solution if you use $V=D=const$ and plug it into the ODE and solve for D. The general solution, will be $V(t)=c_1 e^{\lambda_1 t}+c_2 e^{\lambda_2 t}+D$, where $c_1, c_2 \in \mathbb{R}$.

Title: Differential equations (SO ODE)


问:
On 'proving' that $\langle x,y_1,\ldots,y_n | [x, y_1 \cdots y_n] = 1 \rangle \simeq \mathbb Z^2$.
As the title says, I want to prove that $G:= \langle x,y_1,\ldots,y_n \mid [x, y_1 \cdots y_n] = 1 \rangle \simeq \mathbb Z^2$ algebraically.

Edit: well, this is embarrassing. That is not what I want, because it is not true (and that is why I asked this in the first place, it seemed quite suspicious). So:

Where have I messed up in my proof?

Edit': I have found the error in what follows, it is now added as an answer.

Consider a regular $2(n+1)$-gon $P$ whose edges are labeled  $x,y_1,\ldots,y_n,x^{-1},y_n^{-1},\ldots,y_1^{-1}$, say, clockwise. The quotient $X$ of $P$ that identifies the edges of $\partial P$ according to this labeling gives a torus: we can "fold" $P$ so that each edge $y_i$ matches the edge labeled with its inverse. This produces a tube whose boundary corresponds to the edges labeled $x$ and $x^{-1}$ respectively, and the orientation is coherent with the torus (i.e. we do not get a Klein bottle). Hence $\pi_1(X) \simeq \mathbb Z^2$. On the other hand, a straightforward analysis via van Kampen with $U = X \setminus \{[p]\}, p \not \in \partial P$ and and $V$ (the projection of) a disk around $p$ yields $ \pi_1(X) \simeq G$.

答:
For completion's sake: the space $X$ is indeed a torus, the mistake was thinking that puncturing $X$ would retract to a wedge of $n+1$ copies of $S^1$ hence yielding the claimed presentation.

That is not the case; we have $U \simeq  S^1 \vee S^1$ which agrees with the fact that $\pi_1(X) \simeq \mathbb Z^2$.


问:
Integrability of an "almost" CR-structure defined thanks to an almost contact metric structure.
Let $(M,\eta,\xi,\varphi)$ be an almost contact metric manifold, that is:

$M$ is a smooth manifold

$\eta$ is a contact 1-form on $M$, i.e $\eta$ is a 1-form and $\mathrm{d}\eta|_{\ker \eta}$ is non-degenerate

$\xi$ is the Reeb vector field of $\eta$, that is $\xi \in \ker \mathrm{d}\eta$ and $\eta(\xi) \equiv 1$

$\varphi$ is a section of $End(TM)$ with $\varphi^2 = -\mathrm{id} + \eta\otimes\xi$ and $\varphi (\xi) = 0$

One can think of this manifold as an "almost" (I don't know if this is the right terminology) CR manifold: let $H = \ker \eta$ be the contact distribution and $J = \varphi|_{H}$, which is a section of $End(H)$. Then $J$ is a complex structure on the contact distribution $H$. For $(M,H,J)$ to be a CR manifold, $J$ needs to be integrable on $H$, that is, in the complexified bundle $(H^{\mathbb{C}},J)$ with the usual holomorphic/anti-holomorphic decomposition $H^{\mathbb{C}} = H^{1,0} \oplus H^{0,1}$, one needs to have $[H^{1,0},H^{1,0}] \subset H^{1,0}$.

I remember having read a paper where it was stated that such an "almost" CR structure is integrable and leads to a CR manifold if and only if the Nijenhuis tensor of $\varphi$, defined by
$$
N(X,Y) = -[\varphi X,\varphi Y]+ \varphi[\varphi X,Y] + \varphi[X,\varphi Y] -  \varphi^2[X,Y]
$$
satisfies a certain condition. I think this condition was something like $N$ taking values proportional to $\xi$.

The problem is I cannot find this paper anymore, and some long research on the Internet and on arXiv had been unproductive. Moreover, I remember that, in this paper, this was only stated and wasn't proved.

My question is the following: does anybody have a reference about this fact?

Edit

Here are some more lines to be more precise about my question.
Regarding only $J$, it is clear that $(M,H,J)$ is integrable if and only if the Nijenhuis tensor of $J$ indentically vanishes. My question is more about how to translate this condition to a condition on the full tensor $\varphi$ and not just its restriction?

A - possibly abusive - parallel can be done with almost Kaehler geometry: given an almost Kaehler manifold $(M,g,J)$, there are three equivalent conditions that assure the integrability of the structure, that is $(M,g,J)$ is a Kaehler manifold. It may happen that some condition is more relevant in some context than the others: in a Riemannian setting, it is easier to show some parallel condition than some symplectic ones, etc. Here, I try to find a condition on $\varphi$ without applying any restriction on it; moreover, I am almost sure I have already read such a condition!

答:
I finally found the conditions for $\varphi|_H$ to be integrable: first, one needs to have a compatibility condition relying $\eta$ and $\varphi$:
\begin{align}
\forall X,Y \in \Gamma(H),~ \mathrm{d}\eta(\varphi X, Y) &= - \mathrm{d}\eta(X,\varphi Y), & \mathrm{d}\eta(\varphi X,\varphi Y) = \mathrm{d}\eta(X,Y),
\end{align}
and secondly, the Nijenhuis tensor $N_{\varphi}$, restribted to $H$, has to take values in $\mathbb{R}\xi$. Here is a proof:

Notice that $\varphi\left(\varphi^2+1\right) = 0$, and hence (the $\mathbb{C}$-linear extension of) $\varphi$ is diagonalizable in $TM\otimes \mathbb{C}$, with
$$
TM\otimes \mathbb{C} = \mathbb{C}\xi \oplus H^{1,0}\otimes H^{0,1},
$$
with $H^{1,0} = \left\{X - i\varphi X \mid X \in H \right\}$ and $H^{0,1} = \overline{H^{1,0}} = \left\{X+i\varphi X \mid X \in H \right\}$.

For $V$ a section of $TM\otimes \mathbb{C}$, we have the decomposition
$$
V = \eta(V)\xi + V^{1,0} + V^{0,1}.
$$
A direct computation shows that $V+ i \varphi V = \eta(V)\xi + 2 V^{0,1}$. It follows that $$V \in H^{1,0} \iff \eta(V) = 0 \text{ and } V^{0,1} = 0 \iff V + i \varphi V = 0.$$
Now, if $X$ and $Y$ are in $H$, a direct computations shows that, writing $V = [X-i\varphi X,Y-i\varphi Y]$,
$$
V = [X,Y] - [\varphi X,\varphi Y] -i [\varphi X,Y] -i [X,\varphi Y]
$$
Hence, using the fact that $\eta\circ \xi=0,~\varphi \xi = 0,~ \varphi ^2 = -\mathrm{id} + \eta\otimes \xi$, and that $\eta([A,B]) = -\mathrm{d}\eta(A,B)$ for $A,B \in H$:

\begin{align}
V + i\varphi V &= \left(-\varphi^2[X,Y] + \varphi[\varphi X,Y] + \varphi[X,\varphi Y] - [\varphi X,\varphi Y] - \mathrm{d}\eta(\varphi X,\varphi Y)\otimes \xi\right) \\
&+ i\varphi \left(-\varphi^2[X,Y]+ \varphi[\varphi X,Y] + \varphi[X,\varphi Y] - [\varphi X,\varphi Y] \right) \\
&+ \left[\left(\mathrm{d}\eta(\varphi X,\varphi Y) - \mathrm{d}\eta(X,Y) \right) - i\left(\mathrm{d}\eta(\varphi X,Y) + \mathrm{d}\eta(X,\varphi Y) \right)\right]\otimes \xi \\
&= \left(N_{\varphi}(X,Y) - \mathrm{d}\eta(\varphi X,\varphi Y)\otimes \xi \right) + i\varphi N_{\varphi}(X,Y) \\
&+ \left[\left(\mathrm{d}\eta(\varphi X,\varphi Y) - \mathrm{d}\eta(X,Y) \right) - i\left(\mathrm{d}\eta(\varphi X,Y) + \mathrm{d}\eta(X,\varphi Y) \right)\right]\otimes \xi.
\end{align}
This equation yields the following:
\begin{align}
V \in H^{1,0} & \iff \left\{\begin{array}{rcl}N_{\varphi}(X,Y) - \mathrm{d}\eta(\varphi X,\varphi Y)\otimes \xi &=& 0 \\
\varphi N_{\varphi}(X,Y) &=& 0 \\
\mathrm{d}\eta(X,Y) - \mathrm{d}\eta(\varphi X , \varphi Y) &=& 0 \\
\mathrm{d}\eta(\varphi X,  Y) + \mathrm{d}\eta(X,\varphi Y) &=& 0 \end{array}\right. \\
&\iff \left\{\begin{array}{rcl} N_{\varphi}(X,Y) &=& \mathrm{d}\eta(\varphi X,\varphi Y) \otimes \xi \\
N_{\varphi}(X,Y) & \in & \mathbb{R}\xi  \\
\mathrm{d}\eta(\varphi X, \varphi Y) &=& \mathrm{d}\eta(X,Y) \\
\mathrm{d}\eta(\varphi X, Y) &=& -\mathrm{d}\eta(X,\varphi Y)
\end{array} \right. .
\end{align}
One can easily show, thanks to the definition of $N_{\varphi}$ and the fact that $\eta\circ \varphi = 0$, that under the third condition ($\mathrm{d}\eta(\varphi X,\varphi Y) = \mathrm{d}\eta(X,Y)$), the two first conditions are equivalent. Finally:
$$
[H^{1,0},H^{1,0}] \subset H^{1,0} \iff\forall X,Y \in \Gamma(H),~\left\{\begin{array}{rcl} \mathrm{d}\eta(X,Y) &=& \mathrm{d}\eta(\varphi X,\varphi Y) \\
\mathrm{d}\eta(\varphi X,Y) &=&-\mathrm{d}\eta(X,\varphi Y) \\
N_{\varphi}(X,Y) &\in& \mathbb{R}\xi\end{array} \right. .
$$
One can also remark that the two first conditions are equivalent for vector fields tangent to $H$. This yields this final equivalence:
$$
\varphi|_H \text{ is integrable } \iff \left\{\begin{array}{l} \mathrm{d}\eta \text{ is of type } (1,1) \text{ with respect to } \varphi \\  N_{\varphi}  \text{ restricted to } \Gamma(H) \text{ takes values in }  \mathbb{R}\xi \end{array}\right.
$$


问:
Find $\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz$
Find $\int_{\gamma}\frac{(\sin z)^2}{(z-z_0)^2}dz$ where $\gamma$ is a linear pieces parameterization of closed polygonal chain $[w_0,w_1,w_2,w_3,w_0]$ with vertices $w_0=1+i, w_1=1-i, w_2=-1+i,w_3=2$. Consider the cases  $z_0=\frac 12, z_0=-1$.

I know that such task I should do by using the Cauchy's integral theorem. However, on the basis of the theorem itself, I have no intuition yet how to go about such tasks. Anyone would like to guide me how to think about a task so that I can do the next task one myself?

答:
Since $-1$ is outside the region of $\Bbb C$ bounded by the range of the loop (see the picture below),$$\int_\gamma\frac{\sin^2z}{z+1}\,\mathrm dz=0.$$

Concerning $\frac12$, we have\begin{align}\int_\gamma\frac{\sin^2z}{z-1/2}\,\mathrm dz&=2\pi i\operatorname{Ind}_\gamma\left(\frac12\right)\frac{\sin'(1/2)}{1!}\\&=-2\pi i\cos\left(\frac12\right).\end{align}


答: Let $f_n(x)=nx$ if $0 \leq x \leq \frac  1n$ and $f_n(x)=1$ if $\frac 1 n < x \leq 1$. Let $f(x)=1$ for all $x$. Then it is easy to check that $(f_n)$ is  sequence in $A^{c}$ conveging in the $L_1$ norm to $f$ but $f$ belongs to $A$. It follows that $A^{c}$ is not closed, so $A$ is not open.

Body:
Let $E = C[0, 1]$ and $A = \{f \in E \mid f(x) > 0 \text{ for all } 0 \leqslant x \leqslant 1 \}$. Is $A$ open when the norm of $E$ is the L1-norm?

Let $f \in A$. Now I have $B(f,r)= \{g \in A \mid d(f,g) < r \}.$ Let $g \in B(f,r)$ now I think it would satisfy to show that $g$ outputs only positive values in order to show that it belongs to $A$ which would imply that $B(f,r) \subset A$? Using the  $L1$ norm I have $$d(f,g) = |f(x) -g(x)|$$ but I cannot seem to show from here that $g$ would be positive?


问:
Solving general non-homogenous wave equation with homogenous boundary conditions
I have been given the following PDE to solve:
$$
u_{tt}-u_{xx}=g(t)\sin x \;\;\; (t,x)\in (0,\infty)\times(0,\pi)\\u(0,x)=u_t(0,x) \;\;\;x\in(0,\pi)\\
u(t,0) = u(t,\pi) =0 \;\;\; t>0
$$

This shouldn't be too hard. I have to solve it by separating variables so I write $u(t,x)=v(t)w(x)$ and substituting this into the pde it's clear that $w(x)=\sin x$ and then I have the following ODE for $v$:
$$
v'' +v = g(t) \;\;\;\;\;: \;\;\;v(0)=v'(0)=0
$$

But I don't know how to solve this for a general $g(t)$. The solution they have given us says that:
$$
v(t) = \int_{0}^{t}g(t-\tau)\sin \tau \;\;d\tau
$$
I know this is something to do with Fourier series but I'm not sure why this applies in this case?

Any help welcome :)

答:
Okay I think I have solved it but its not simple:

By looking at the ode we guess a solution of the form:

$$
v(t) = A(t)\cos(t)+B(t)\sin(t)$$
and then we differentiate twice. However, because we picked such a general form, we have a degree of freedom still to play with so we set:
$$
A'\cos t +B'\sin t = 0
$$
With this equation and with the ODE we derive the following system of simultaneous equations:
$$
B' \cos t - A' \sin t =g\\
A' \cos t + B\ \sin t =0
$$

We rearrange and solve for $A', B'$ and then integrate to get:

$$
A(t) = -\int_{0}^{t} g(s)\sin(s) \;ds\\
B(t) = \int_{0}^{t}g(s)\cos(s) \;\;ds
$$

Now we substitute this into our formula for $v$ and apply the compound angle formula for $\sin$ and then change variables $\tau = t-s$ and this gives us the solution:

$$
v(t) = \int_{0}^{t}g(t-\tau)\sin(\tau) \;\;d\tau
$$


问:
Spivak, Ch 2 Prologue, Problem 12d: is there a systematic way to find two irrational numbers whose sum and multiplication result in rational number?
The question in Spivak is: are there two irrational numbers whose sum and product are both rational? Below is my solution, and my question is if there is a more systematic way of finding the two irrational numbers.

Assume:
$$a,b \in Q'\tag{1}$$
$$a+b =r_1\tag{2}$$
$$ab = r_2\tag{3}$$
$$r1, r2 \in Q$$
Then,
$$a=r_1-b \rightarrow (r_1-b)b = r_2 \rightarrow b^2-r_1b+r_2=0$$
$$\triangle = r_1^2-4r_2 \geq0 \rightarrow r_1^2 \geq 4r_2$$
So, at this point we know the relationship between $r_1$ and $r_2$ necessary to satisfy (2) and (3), but this includes rational and irrational options for $a$ and $b$ and we only want irrational ones.

Is the only way to proceed here to make guesses?

For example, $r_1=0 \rightarrow r_2 \leq 0$ and we have $a+b=0 \rightarrow a=-b\ and\ ab=r_2$. We can now guess an $r_2$ and hope we find irrational values for $a$ and $b$. $$r_2 = -3 \rightarrow -b^2=-3 \rightarrow b^2=3 \rightarrow b=\pm \sqrt{3}$$.So, we have two solutions for $a,b$: $a=-\sqrt{3}, b=\sqrt{3}$ and $a=\sqrt{3}, b=-\sqrt{3}$.

We could likewise guess different values for $r_1$ and $r_2$ and obtain other solutions.

Is this the only way to obtain these solutions or is there a more systematic way?

答:
How about
$$a+\sqrt b$$
and $$a-\sqrt b$$
where $a,b\in\mathbb{Q},\sqrt b\notin \mathbb{Q} $ ? That seems the simplest answer to me.


问:
how to determine a transformation matrix for transforming x and y with specific equation?
I need to determine a matrix that can be used to:

a)  transform $x$ and $y$ using equations

$$
\left\{
\begin{array}{c}
x'=3x + 4y \\
y'=-x + 2y \\
\end{array}
\right.
$$

Then I need to:

b) Transform a triangle $(0,0),(1,0),(1,1)$ using this transformation

c) What are the coordinates of point $(10,10)$ in the basis of transformation?

Could I have some advise? Thanks in advance.

答:
$\left\{
\begin{array}{c}
x'=3x + 4y \\
y'=-x + 2y \\
\end{array}
\right. \Rightarrow \left\{
\begin{array}{c}
x'=(3,4)(x,y)\\
y'=(-1,2)(x,y) \\
\end{array}
\right.$

$\left\{
\begin{array}{c}
x'=(3,4)(x,y)\\
y'=(-1,2)(x,y) \\
\end{array}
\right. \Rightarrow \begin{pmatrix}  3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} x` \\ y` \end{pmatrix}$

$\begin{pmatrix}  3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 0 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

$\begin{pmatrix}  3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 3 \\ -1 \end{pmatrix}$

$\begin{pmatrix}  3 & 4 \\ -1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix}=\begin{pmatrix} 7 \\ 1 \end{pmatrix}$

as itcan be seen, image of triangle is also a triangle (because kernel of matrix is $\{0\}$)


问:
Question on next prime function
The function $P(n)$ gives the smallest prime larger or equal to n. Example: $P(3)=3,P(4)=5.$

Show that: equation $P(x)^2-P(x^2)\equiv 4\pmod6$ have only one solution as $x=3$?

Source code Pari GP

```<code>for(x=1,100000,if((nextprime(x)^2-nextprime(x^2))%6==4,print([x])))
[3]
</code>```

答:
If $x>3$, $P(x)$ can only have the form $6k+1$,$6h+5$ or
$$P(x) \equiv \pm 1 \pmod 6\iff (P(x))^2 \equiv 1 \pmod 6$$
From the equation, we can deduce  that
$$1-P(x^2)\equiv 4 \pmod 6 \iff P(x^2)\equiv 3 \pmod 6 $$
Hence, $P(x^2)$ is divisible by 3 $\implies$ contradiction.

So, $x \le 3$ and the sole solution is $x=3$.

Title: Question on next prime function


问:
I don't understand what the correct form of this inequality is
I have the following inequality:

$$
(C+Bk)(1-m)+B(m+f)(1-a)>0
$$

I follow the following steps:

$$
(C+Bk)(1-m)>-B(m+f)(1-a)
$$

$$
C(1-m)>-B(m+f)(1-a)-Bk(1-mm)
$$

$$
\frac{C}{B}(1-m)>-(m+f)(1-a)-k(1-mm)
$$

$$
\frac{C}{B}>\frac{-(m+f)(1-a)-k(1-m)}{1-m}
$$

$$
\frac{-C}{B}<\frac{(m+f)(1-a)+k(1-m)}{(1-m)}
$$

Is this correct? But what if I do the opposite, that is:

$$
B(m+f)(1-a)>-(C+Bk)(1-m)
$$

$$
B(m+f)(1-a)>-C(1-m)-Bk(1-m)
$$

$$
B((m+f)(1-a)+k(1-m))>-C(1-m)
$$

$$
B((m+f)(1-a)+k(1-m))>-C(1-m)
$$

$$
B>-C\frac{(1-m)}{((m+f)(1-a)+k(1-m))}
$$

$$
\frac{B}{-C}<\frac{(1-m)}{((m+f)(1-a)+k(1-m))}
$$

They are clearly describing different things so I feel like I'm doing something wrong, but I cannot figure out what and which result is the correct one. Sorry, not a mathematician, I hope this isn't a very stupid question. C and B are positive(>0), while the other variables are between 0 and 1.

答: All your inequalities are correct  if you assume that  $(m+f)(1-a)+k(1-m)$ is also positive but there is no contradiction. Under your assumptions both the inequalities say that  a negative number is less than a positive number (so they say nothing!)


答:
Both inequalities say that a negative number is smaller than a positive number which is true. This follows from your original equation which also has a positive number on the left and zero on the right. The original inequality doesn't give any information it just says a positive number is greater than zero (except $m=0$ and $a=0$). So anything you follow from that has to be true but you can't learn anything from it. No variable is restricted in any way; it doesn't matter what values you assign to the variables the inequality is true.

You can see that they say the same thing by applying $1/x$ to both sides. Normally this would result in the sign switching, but because one number is negative and the other is positive 1/negative is still negative and 1/positive is still positive, so the sign doesn't switch and you get the same equation for both.

Title: I don't understand what the correct form of this inequality is

Body:
I have the following inequality:

$$
(C+Bk)(1-m)+B(m+f)(1-a)>0
$$

I follow the following steps:

$$
(C+Bk)(1-m)>-B(m+f)(1-a)
$$

$$
C(1-m)>-B(m+f)(1-a)-Bk(1-mm)
$$

$$
\frac{C}{B}(1-m)>-(m+f)(1-a)-k(1-mm)
$$

$$
\frac{C}{B}>\frac{-(m+f)(1-a)-k(1-m)}{1-m}
$$

$$
\frac{-C}{B}<\frac{(m+f)(1-a)+k(1-m)}{(1-m)}
$$

Is this correct? But what if I do the opposite, that is:

$$
B(m+f)(1-a)>-(C+Bk)(1-m)
$$

$$
B(m+f)(1-a)>-C(1-m)-Bk(1-m)
$$

$$
B((m+f)(1-a)+k(1-m))>-C(1-m)
$$

$$
B((m+f)(1-a)+k(1-m))>-C(1-m)
$$

$$
B>-C\frac{(1-m)}{((m+f)(1-a)+k(1-m))}
$$

$$
\frac{B}{-C}<\frac{(1-m)}{((m+f)(1-a)+k(1-m))}
$$

They are clearly describing different things so I feel like I'm doing something wrong, but I cannot figure out what and which result is the correct one. Sorry, not a mathematician, I hope this isn't a very stupid question. C and B are positive(>0), while the other variables are between 0 and 1.


问:
Don't linear combinations of partial derivatives entirely determine Jacobian?
I'm learning multivariable calculus for the first time, having just taken a course on theoretical linear algebra and group theory, and the instructor motivated the Jacobian definition of a function derivative by giving an example of how, if we defined a multivariable derivative purely in terms of the existence of partial derivatives, we could get functions that are not continuous at a point, but for which the derivative still exists, which violates what we want from a definition of a derivative (as a stronger condition than continuity), and so we instead take the "existence of a linear map" definition instead.

But doesn't the existence of enough partial derivatives completely determine the Jacobian? For example, if you had $\frac{\partial f}{\partial x_i}$ for all components $0 \le i \le n$ for $f(x): \mathbb{R}^n \to \mathbb{R}$, don't you automatically guarantee the existence of the Jacobian
$$\begin{bmatrix} \frac{\partial f}{\partial x_1} \frac{\partial f}{\partial x_2} \cdots \frac{\partial f}{\partial x_n} \end{bmatrix}$$ by just using the partial derivatives you've shown exist. And does the existence of such a Jacobian satisfy the "existence of a linear map" condition to show that $f$ is differentiable, as a multivariable function?

My intuition is breaking down because I intuitively imagine a function as differentiable if the limits from all possible paths exist, and are the same. Aren't said limits entirely specified by these set of partial derivatives, from which you can get any direction derivative by a linear combination? And so in some sense, you've specified the datum for the "total derivative" (ie. $Df(x)$) when specifying all the partial derivatives?

答:
Existence of the Jacobian matrix at $c$ only needs, by definition, the partial derivatives at $c$. It is also true that a function is differentiable at $c$ if and only if the Jacobian exists at $c$ and $\lim_{\lvert h\rvert\to0}\frac{\left\lvert f(c+h)-f(c)-\nabla_c f\cdot h\right\rvert}{\lvert h\rvert}=0$.

However, it might still not be the case that, while the Jacobian at $c$ exists, $$\lim_{\lvert h\rvert\to0}\frac{\left\lvert f(c+h)-f(c)-\nabla_c f\cdot h\right\rvert}{\lvert h\rvert}=0$$

For instance, consider $$f(x,y)=\begin{cases}0&\text{if }x\le 0\lor y\le x^2\lor y\ge 4x^2\\ (x^2+y^2)^{1/4}\exp\left(\frac{1}{(y-x^2)(y-4x^2)}+\frac49x^{-4}\right)&\text{if }x>0\land x^2<y<4x^2\end{cases}$$

Then $f$ is $C^\infty$ on $\Bbb R^2\setminus\{0\}$, $\lim_{t\to 0} \frac{f(tv)-f(0)}{t}=0$ for all $v\in\Bbb R^2$, $f$ is continuous, but $f$ is not differentiable at $0$ because $$\limsup_{h\to 0}\frac{\lvert f(h)-f(0)-\nabla_0f\cdot h\rvert}{\lvert h\rvert}\ge\limsup_{t\to 0^+}\frac{\lvert f(t,5t^2/2)\rvert}{\lvert (t,5t^2/2)\rvert}=\limsup_{t\to 0^+}\left(t^2+\frac{25}{4}t^4\right)^{-1/4}=\infty$$

Also, $(x^2+y^2)^{-1/2}f(x,y)$ is again $C^\infty(\Bbb R^2\setminus\{0\})$, has all the directional derivatives equal to $0$ at $0$ and it isn't continuous at $0$.

Of course, by the total differential theorem, existence of the partial derivatives in a neighbourhood of $c$ and continuity of the partial derivatives at $c$ is sufficient to guarantee differentiability at $c$.


答:
After dealing with the small cases we can assume that $x \geq 3$.
Observe that the equation reduces to, $$P(x)^2-P(x^2) \equiv 0 \pmod 2$$ and $$P(x)^2-P(x^2) \equiv 1 \pmod 3$$
now the former one is obvious as we are assuming $x \geq 3$. (just note that prime number larger than $2$ is always odd).

Now we solve the equation,
$$P(x)^2-P(x^2) \equiv 1 \pmod 3$$

Recall that a square number is always congruent to either $1$ or $2$ modulo $3$.
\

$\textbf{Case 1:}$ $P(x)^2 \equiv 0 \pmod 3$

In this case, as $P(x)$ is a prime number divisibe by $3$. We observe that $P(x)= 3 \implies x = 2 \text{ or } 3 $.

$\textbf{Case 1:}$ $P(x)^2 \equiv 1 \pmod 3$

In this case $P(x)^2 \equiv 1 \pmod 3 \implies P(x^2) \equiv 0 \pmod 3 $
Now as in last case, we observe that $P(x^2) = 3$. But this is impossible.

Thus we conclude that the given equation has only two solutions. Namely, $x=2 \text{ and } x=3$

Title: Question on next prime function

Body:
The function $P(n)$ gives the smallest prime larger or equal to n. Example: $P(3)=3,P(4)=5.$

Show that: equation $P(x)^2-P(x^2)\equiv 4\pmod6$ have only one solution as $x=3$?

Source code Pari GP

```<code>for(x=1,100000,if((nextprime(x)^2-nextprime(x^2))%6==4,print([x])))
[3]
</code>```


问:
Ordered triples that sastify the system.
Find all the real system solutions:

$$2x+x^2y=y$$
$$2y+y^2z=z$$
$$2z+z^2x=x$$

My attempt

Trivial solution : (0,0,0)

We have:

$$\frac{1}{y}=\frac{1-x^2}{2x}$$

$$\frac{1}{z}=\frac{1-y^2}{2y}$$

$$\frac{1}{x}=\frac{1-z^2}{2z}$$

Then:

$$x+\frac{1}{x} + y+\frac{1}{y} + z+\frac{1}{z}=0$$

How to procede?

答:
$$y=\dfrac{2x}{1-x^2}$$  immediately commands me to use $x=\tan A$

to find $$y=\tan2A,z=\tan(2\cdot2A),x=\tan(2\cdot4A)$$

$$\implies\tan8A=\tan A$$

$\implies8A=n\pi+A$ where $n$ is any integer


问:
Solving algebraic equation (compound interest)
Task: Say you have an initial capital $c_0$ on your bank account and after each year the bank gives you $p \%$ of your hole capital $c$ what you directly put to your bank account. Additionally, you deposit another $t\$$ on your bank account at the end of each year after the bank paid you the interest. After how many years will your capital exceed $b$, i.e. when does $c > b$ hold?

My recurrence formula approach:
$$
f: \mathbb{R} \to \mathbb{R};\space\space f(x) = \begin{cases}(1+\dfrac{p}{100})\cdot c_0 + t, \space \text{if }\space x = 1\\[10pt](1+\dfrac{p}{100})\cdot f(x-1) + t, \space\text{else }\\\phantom{x}\end{cases}
$$

Where $f(x)$ describes $c$ after $x$ years.

Solution: I solved it using telescoping, since it is a rather simple recurrence relation:
$$
f(n) = p\cdot f(n-1) + t = p\cdot(p\cdot f(n-2) + t) + t = \ldots = p^n \cdot c_0 + \sum\limits_{k =0}^{n-1}p^k \cdot t \\= p^n \cdot c_0 + \dfrac{1-p^{n}}{1-p}\cdot t.
$$
Notice that for simplicity we define $p := \dfrac{\text{percentage}}{100} + 1$, in contrast to the $p$ we used above, which was the "raw" percentage.

Now I'm trying to solve for $n$, which doesn't seem that easy to me. Moreover, is there a function that better represents my problem? I just started diving into analysis a few days ago so my knowledge is very little. I'm thankful for any helpful comment!

答:
In your recurrence relation, your $p$ should rather be $1 + \frac{p}{100}$ by your notation (since a factor larger than one means that the capital accumulates interest). Let's write $i = \frac{p}{100}$. If you put these into $f(n)$, the accumulated value at time $n$, we get
$$
f(n) = (1+i)^n c_0 + \frac{(1+i)^n - 1}{i} t = (1+i)^n \left[ c_0 + \frac{1}{i} \right] - \frac{t}{i} .
$$
Solving $f(n) = b$, we can make $n$ the subject and thus solve for $n$:
$$
(1+i)^n \left[ c_0 + \frac{1}{i} \right] - \frac{t}{i} = b .
$$
Rounding $n$ up to the nearest integer then gives you the first time period where the accumulated capital exceeds $b$.

EDIT: To solve for $n$, first make $n$ the subject by moving all the terms without $n$ to one side wher possible:
$$
(1+i)^n = \frac{b + \frac{t}{i}}{ c_0 + \frac{1}{i}} .
$$
Take logarithms (any base will do) to get
$$
n \log(1+i) = \log \left(  \frac{b + \frac{t}{i}}{ c_0 + \frac{1}{i}} \right)
$$
Thus, we have isolated $n$ and we can solve
$$
n = \log \left(  \frac{b + \frac{t}{i}}{ c_0 + \frac{1}{i}} \right) / \log(1+i) .
$$


问:
Question about smooth functions and their signs with given initial conditions

Problem Statement

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a smooth function (infinitely differentiable) where $f(x)\geq 0$ for all $x\in\mathbb{R}$, $f(0)=0$, and $f(1)=1$. Show that there is some $n\in\mathbb{N}$ and an $x_0\in\mathbb{R}$ such that $f^{(n)}(x_0)<0$.

Proof Attempt

We can consider two cases:

Case (1): There exists a $z<0$ such that $f(z)>0$. By the MVT, there exists a point $z_0\in(z,0)$ such that $$f'(z_0)=\frac{f(z)-f(0)}{z-0}=\frac{f(z)}{z}<0 $$ so we are done.

Case (2): For all $x<0$, $f(x)=0$. In this case, $f'(x)=0$ for all $x<0$, which implies that, by left sided derivatives and the fact that $f'$ is smooth, $f'(0)=0$. It then follows that,  since $f'(x)=0$ on $(-\infty,0]$, we have $f^{(n)}(x)=0$ for all $x\in(-\infty,0]$.

For $n=1$, define $z_1$ as the real number in $(0,1)$ such that $$f'(z_1)=\frac{f(1)-f(0)}{1-0}=1$$For $n\in\mathbb{N}$, define $z_{n+1}$ by the real number in $(0,z_n)$ such that $$f^{(n+1)}(z_{n+1})=\frac{f^{(n)}(z_n)-f^{(n)}(0)}{z_n-0}=\frac{f^{(n)}(z_{n})}{z_n}>0$$ whose existence is given by the mean value theorem. This defines the sequence $(z_n)$, which converges to $0$ and is strictly decreasing.

Edit 2: Continuing, we see that by definition, we have $$f^{(n+1)}(z_{n+1})=\frac{f^{(n)}(z_n)}{z_n}>f^{(n)}(z_n)$$ for any $n\in\mathbb{N}$ with $n\geq 2$, since $0<z_n<1$, thus the sequence $f^{(n)}(z_n)$ is strictly increasing and always greater than or equal to $1$. Now, $f^{(n)}(0)=0$, but since $(z_n)$ can get arbitrarily close to $0$ and $f^{(n)}(z_n)\geq 1$ for all $n\in\mathbb{N}$, we have $f^{(n)}(0)=0<1\leq f^{(n)}(z_n)$ for large enough $n$.

This is has almost shown that $f$ is not differentiable for some value of $n$, but I am not quite sure that it has exactly shown that. If it has, then we are done, as Case (2) cannot happen. I would like help either formalizing this or fixing this if it is not true.

This is where my argument, to me, seems to break down. Intuitively, the slope of the secant lines between $f^{(n)}(z_n)$ and $f^{(n)}(0)$ is increasing without bound as $n$ increases. My idea was to create a sequence of points which converged to $0$, but broke the differentiability at $0$. I would greatly appreciate any tips in proceeding, critiques of my proof outline, or alternative proof methods. Thank you in advance.

Edit 1: In Case (2), I am not explicitly making any assumptions that would lead to a contradiction, but why is this proof going in that direction? Either Case (2) cannot happen, which I find intuitively true, but my intuition often fails me, or one of my assumptions in the beginning of Case (2) is false.

Edit: Just to close this off, one of my faulty assumptions was that I assumed the sequence I constructed converged to $0$, which it doesn't have to.

答:
This is actually a very hard question. So hard, that it appeared in the 2018 Putnam Exam.
There are lots of solutions to this and I do not know which one suits your knowledge best. So just choose.</p>


问:
Transitive groups
Someone told me the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle is $A_6$ itself.

(1) What does it mean to be a "transitive subgroup?"  I know that a transitive group action is one where if you have a group $G$ and a set $X$, you can get from any element in $X$ to any other element in $X$ by multiplying it by an element in $G$.  Is a transitive subgroup just any group that acts transitively on a set?  And if so, does its transitiveness depend on the set it's acting on?

(2) Why is $A_6$ the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle?

Thank you for your help :)

答:
Here is another way to look at it:

Let $H$ be a transitive subgroup of $A_6$ containing a 5-cycle and a 3-cycle.  Then $5$ divides $|H|$, and also $6$ divides $|H|$ since $H$ acts transitively.  Thus $30$ divides $|H|$, and if $H$ is not all of $A_6$, by simplicity $H$ must have order $30$ or $60$.  $30$ is impossible, as such a group has a (normal) subgroup of order $15$, and every group of order $15$ is cyclic.  Yet there is no element of order $15$ in $A_6$.  $60$ is also impossible, as every such subgroup of $A_6$ is a copy of $A_5$, which is the point stabilizer of one of $\{1,2,3,4,5,6\}$, and thus not transitive.  Thus $H$ is $A_6$.

EDIT - Here is an argument that if $H$ has order $60$ and contains a 3-cycle, then it is a point stabilizer in $A_6$.  First, every subgroup of $A_6$ of order $60$ is isomorphic to $A_5$, and thus $H$ is generated by its order 3 elements. We can assume up to conjugation that the 5-cycle in $H$ is $(12345)$.  Since all order 3 elements in $A_5$ are conjugate, all order 3 elements in $H$ are 3-cycles. If no 3-cycle in $H$ contains a $6$, then $H$ cannot be transitive, since it is generated by these 3-cycles.  Thus, assuming $H$ is transitive, and by considering appropriate powers of $(12345)$, we can also assume $H$ contains the 3-cycle $(126)$.  But then $(12345)(126)=(16)(2345)$ has order $4$, and $A_5$ has no elements of order $4$.

Title: Transitive groups


问:
How can you prove that a conditional bivariate Gaussian is a univariate Gaussian?
Is there a way to prove that a bivariate Gaussian becomes a univariate Gaussian when conditioned on one of the two variables?</p>

答:
Consider the density of a jointly Gaussian vector $(X,Y)$

$$f_{XY}(x,y)=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2)}}\exp\left\{ -\frac{1}{2(1-\rho^2)}\left[ \frac{(x-\mu_X)^2}{\sigma_X^2}-2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}   +\frac{(y-\mu_Y)^2}{\sigma_Y^2} \right]  \right\}=$$

$$=\frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2)}}\exp\left\{- \frac{1}{2(1-\rho^2)} \frac{(x-\mu_X)^2}{\sigma_X^2}  \right\}\cdot\exp\left\{ -\frac{1}{2(1-\rho^2)}\left[ \frac{(y-\mu_Y)^2}{\sigma_Y^2}-2\rho\frac{(x-\mu_X)(y-\mu_Y)}{\sigma_X\sigma_Y}    \right]  \right\}=$$

Now sum and subtract the following quantity in the argument of the second exp

$$\frac{\rho^2(x-\mu_X)^2}{\sigma_X^2}$$

...and after some (easy) algebraic manipulations you get

$$f_{XY}(x,y)=f_X(x)\cdot f_{Y|X}(y|x)$$

Where

$$X\sim N(\mu_X;\sigma_X^2)$$

$$(Y|X=x)\sim N\left[\mu_Y+\rho\frac{\sigma_y}{\sigma_X}(x-\mu_X);\sigma_Y^2(1-\rho^2)   \right]$$

Observe that if you set $\rho=0$ (uncorrelation) the joint density becomes

$$f_{XY}(x,y)=f_X(x)\cdot f_Y(y)$$

meaning that, in this model, uncorrelation and independence are equivalent.


答:
Using resultants we can write down all solutions over $\Bbb C$, and then look for the real solutions. They are
$$
(x,y,z)=(0,0,0),(i,i,i),(-i,-i,-i)
$$
and
$$
y=\frac{x(x^4 - 20x^2 + 15)}{4},\quad
z=\frac{x( - 3x^4 + 62x^2 - 87)}{8},
$$
where $x$ is a root of the polynomial
$$
x^6 - 21x^4 + 35x^2 - 7=0.
$$
This polynomial has six real roots! It is cubic in $y=x^2$.

Title: Ordered triples that sastify the system.

Body:
Find all the real system solutions:

$$2x+x^2y=y$$
$$2y+y^2z=z$$
$$2z+z^2x=x$$

My attempt

Trivial solution : (0,0,0)

We have:

$$\frac{1}{y}=\frac{1-x^2}{2x}$$

$$\frac{1}{z}=\frac{1-y^2}{2y}$$

$$\frac{1}{x}=\frac{1-z^2}{2z}$$

Then:

$$x+\frac{1}{x} + y+\frac{1}{y} + z+\frac{1}{z}=0$$

How to procede?


问:
Example of dual of vector in $\mathbb{R}^3$
Assuming the basis of $\mathbb{R}^3$ is $e_1, e_2, e_3$ and the corresponding dual basis vectors are $\theta^1, \theta^2, \theta^3$ (by the standard identification $\theta^i(e_j)=\delta^i_j$).

What is the dual of $2e_1$? My guess would be $2\theta^1$, however, I'm unsure of whether the coefficient $2$ would become $\frac{1}{2}$ under the isomorphism from $\mathbb{R}^3$ to its dual space.

Any clarification would be appreciated, thanks!

Edit: I've realised I am referring to the outcome of applying a musical isomorphism (given all the necessary structure of a pseudo-Riemannian manifold (i.e. metric) required). In this case, the isomorphism is conventionally known as its flat, $$\flat:TM\to T^*M$$
where $M=\mathbb{R}$.

答:
There is not a dual linear form of a vector.

You can complement $2 e_1$ into a basis $\{2 e_1, f_2, f_3\}$. Providing that $f_2,f_3$ belong to the subspace spanned by $\{e_2,e_3\}$, the dual basis will then be $\{1/2 \theta_1, f_2^*, f_3^*\}$.

However again, there is not a notion of the dual of a single vector.