问:
Is (Inverse) Fourier Transform still a "sum" of sinusoids?
I'm trying to wrap my mind around the FT and specifically the IFT:

$$x(t) = \frac{1}{2T}\int_{-\infty}^{\infty}x(\omega)e^{i\omega t}d\omega
$$

I understand that they are generalization of the Fourier Series to non-periodic functions, where you take the period to be infinite, and the interval to be 0.

My question is - does it still hold that what you get is, sort-of, a "sum" of sinusoids? The $x(\omega)$ representing the frequency-domain, are the magnitudes of the different frequencies, but how do you get rid of the imaginary part $i$ when expanding $e^{i\omega t}$ using Euler's formula?

Specifically, is there a way to (numerically approximate) the rect function (time domain) by "sampling" of frequencies from the sinc function (frequency domain) and appropriate sinusoids? (To be honest, the sinc function itself is already somewhat similar to the rect, so I think some approximation is possible...)

答:
I believe the rectangle function $\Pi(x)$ can be expressed as a nested series of sinusoids as illustrated in formula (1) below where the parameter $f$ is the evaluation frequency and assumed to be a positive integer. When evaluating formula (1) below (and all formulas derived from it) the evaluation limit $N$ must be selected such that $M(N)=0$ where $M(x)$ is the Mertens function. Formula (2) below contains a closed form representation of the two nested sums over $k$ in formula (1) below. Formulas (1) and (2) below are based on this answer I posted to my own question "Does this formula correspond to a series representation of the Dirac delta function $\delta(x)$?" on MathOverflow.

(1) $\quad\Pi(x)=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\frac{1}{\pi}\sum\limits_{n=1}^N\mu(n)\left(\sum\limits_{k=1}^{f\ n}\frac{\cos\left(\frac{2 \pi k}{n}\right) \left(\sin\left(\frac{2 \pi k \left(x+\frac{1}{2}\right)}{n}\right)-\sin\left(\frac{2 \pi k \left(x-\frac{1}{2}\right)}{n}\right)\right)}{k}-\frac{1}{2} \sum\limits_{k=1}^{2\ f\ n} \frac{\sin\left(\frac{\pi k \left(x+\frac{1}{2}\right)}{n}\right)-\sin\left(\frac{\pi k \left(x-\frac{1}{2}\right)}{n}\right)}{k}\right)$

(2) $\quad\Pi(x)=\underset{\underset{M(N)=0}{N\to\infty}}{\text{lim}}\frac{1}{4 \pi i}\sum\limits_{n=1}^N\mu(n)\left(\log\left(1-e^{\frac{i \pi (2 x-3)}{n}}\right)-\log\left(1-e^{\frac{i \pi (2 x-1)}{n}}\right)-\log\left(1-e^{\frac{i \pi (2 x+3)}{n}}\right)+\log\left(1-e^{\frac{i (2 \pi x+\pi )}{n}}\right)+\log\left(1-e^{\frac{i \pi (1-2 x)}{n}}\right)-\log\left(1-e^{\frac{i \pi (3-2 x)}{n}}\right)+\log\left(1-e^{-\frac{i \pi (2 x+3)}{n}}\right)-\log\left(1-e^{-\frac{i (2 \pi x+\pi)}{n}}\right)-\log\left(1-e^{\frac{i \pi (2 x-1)}{2 n}}\right)+\log\left(1-e^{\frac{i \pi (2 x+1)}{2 n}}\right)+\log\left(1-e^{\frac{i \pi (1-2 x)}{2 n}}\right)-\log\left(1-e^{-\frac{i \pi (2 x+1)}{2 n}}\right)\right)$

Figure (1) below illustrates the reference function $\Pi(x)$ in blue and formulas (1) and (2) for $\Pi(x)$ in orange and green respectively where formula (1) is evaluated at $f=8$ and formulas (1) and (2) are both evaluated at $N=39$.

Figure (1): Illustration of formulas (1) and (2) for $\Pi(x)$ (orange and green)

Figure (2) below illustrates the reference function $\Pi(x)$ in blue and formula (2) for $\Pi(x)$ evaluated at $N=39$ and $N=101$ in orange and green respectively.

Figure (2): Illustration of formula (2) for $\Pi(x)$ evaluated at $N=39$ and $N=101$ (orange and green)

Figures (1) and (2) above illustrate formulas (1) and (2) above evaluate at an offset from the reference function $\Pi(x)$, and Figure (2) above illustrates the magnitude of this offset decreases as the magnitude of the evaluation limit $N$ increases. This offset is given by $-\frac{3}{4}\sum\limits_{n=1}^N\frac{\mu(n)}{n}$ which corresponds to $-0.0378622$ at $N=39$ and $-0.0159229$ at $N=101$. Since $-\frac{3}{4}\sum\limits_{n=1}^\infty\frac{\mu(n)}{n}=0$, formulas (1) and (2) above converge to the reference function $\Pi(x)$ as $N\to\infty$ (and as $f\to\infty$ for formula (1)).

Formula (3) below approximates the Fourier transform $\int_{-\infty}^\infty\Pi(x)\ e^{2 i \pi x y}\,dx=\text{sinc}(\pi y)$ with the finite integral $\int_{-X}^X\Pi(x)\ e^{2 i \pi x y}\,dx$ where the finite integral is evaluated using a modified version of formula (1) above which compensates for the offset of formula (1) from the reference function $\Pi(x)$ which was explained in the preceding paragraph. The Fourier transform of formula (1) needs to be evaluated at finite limits because of the idealization of the Fourier transform of the $\sin$ function as a pair of Dirac delta ($\delta$) functions. The offset of formula (1) from the reference function $\Pi(x)$ explained in the preceding paragraph needs to be compensated for in the evaluation to eliminate an oscillation in the result which is otherwise very noticeable at modest values of the evaluation limit $N$.

(3) $\quad\text{sinc}(\pi y)=\underset{\underset{M(N)=0}{X,N,f\to\infty}}{\text{lim}}\sum\limits_{n=1}^N\mu(n)\left(\frac{3 \sin(2 \pi X y)}{4 \pi n y}+\frac{n}{\pi} \left(\sum\limits_{k=1}^{f\ n}\frac{e^{-2 i \pi X y} \sin\left(\frac{\pi k}{n}\right) \cos\left(\frac{2 \pi k}{n}\right) \left(k \left(1+e^{4 i \pi X y}\right) \sin\left(\frac{2 \pi k X}{n}\right)+i n y \left(-1+e^{4 i \pi X y}\right) \cos\left(\frac{2 \pi k X}{n}\right)\right)}{\pi k (k-n y) (k+n y)}-\frac{1}{2}\sum\limits_{k=1}^{2\ f\ n}\frac{4 \sin\left(\frac{\pi k}{2 n}\right) \left(k \cos(2 \pi X y) \sin\left(\frac{\pi k X}{n}\right)-2 n y \sin(2 \pi X y) \cos\left(\frac{\pi k X}{n}\right)\right)}{\pi k \left(k^2-4 n^2 y^2\right)}\right)\right)$

Figure (3) below illustrates formula (3) above evaluated at $X=35$, $N=39$, and $f=8$ in orange overlaid on the reference function $\text{sinc}(\pi y)$ in blue. Note the evaluation of formula (3) in orange is so accurate it pretty much hides the underlying evaluation of the reference function $\text{sinc}(\pi y)$ in blue.

Figure (3): Illustration of formula (3) for $\text{sinc}(\pi y)$ orange


问:
Hi I need help with this exercise to my mechanic physics classes
A) A traffic light with a mass of 140 kg is hung up as shown in the figure above. Calculate the forces in
cables AB and AC

I got answer 1.33 kN, which is the correct solution. I used trigonometry and and decomposed the forces from the ropes.
Excercise a

B) A parade will pass the traffic light, and the local municipal engineer proposes to install one
extra cable DE with a length of 20 m to have a greater free height below point A. The cable section BD is
now 10.05 m long. Is this possible when the maximum tension the cables can withstand is 6 kN?

I have done the a part, but I am going to include it here. The answer of a is 1.33 kN
We have given that the weight of the traffic light is 140 kg. I don't understand how am I supposed to calculate the forces in b). I have gotten two new robes so there needs to be more forces in this system. Also the new robe is making an angle. The answer is supposed to be above 6 kN, but I am not getting it. Could someone give a tip please?

Exercise B

答:
Comment

A sketch is necessary to be drawn. The following schematic sketch is given to help get started in posing the question as I understood it. It is not fully correct. It may be later deleted.

To push up the traffic light the engineer suspends two smaller lights as shown at 20 m distance apart without extra cable length but with same total weight. (It is 16.012 m below top line of posts if calculated). The cable is position is elevated to black from red at start.

When a 20 meters rope is added and points B,C are brought nearer to the posts, lights are pushed further up as in green position.

The calculation of forces is done by drawing triangle of static forces.

ContentLicense: CC BY-SA 4.0


问:
Which of the sets are equal?
Which of these sets are equal?

$A= \{ z \in \mathbb{C}; \frac{1}{z}+\frac{1}{\overline{z}}\leq2\}$ ---------> $x^2-x+y^2 \geq 0 $

$B= \{ z \in \mathbb{C}; \frac{1}{z}+\frac{1}{\overline{z}}< 2\}$ ---------> $x^2-x+y^2 > 0 $

$C= \{ z \in \mathbb{C}; |z-\frac{1}{2}|\geq\frac{1}{2}\}$ ---------> $x^2-x+y^2 \geq 0 $

$D= \{ z \in \mathbb{C}; |z-\frac{1}{2}|>\frac{1}{2}\}$ ---------> $x^2-x+y^2 > 0 $

Well, I tried to solve this problem by writing in Cartesian coordinates (look above). The the solution manual only states "B = D. No other pair is equal."

Why are A and C not equal?

答: We have $0\in C$ but $0\notin A$.

Title: Which of the sets are equal?

Body: We have $0\in C$ but $0\notin A$.


问:
Pointwise and uniform convergence of $f_n(x)=xe^{-nx}$
Let $f_n:[0, +\infty)\rightarrow \mathbb{R}$, $f_n(x)=xe^{-nx}$.

Show that $(f_n)_n$ converges pointwise and calculate $$f(x):=\lim_{n\rightarrow +\infty}f_n(x)\ \ \  \forall x\in [0,+\infty)$$
Does $f_n$ converge also uniformly to $f$ on $[0, +\infty)$ ?

$$$$

I have done the following:

We have that \begin{equation*}f(x)=\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}xe^{-nx}=0 \text{ since } x\geq 0\end{equation*}
So the sequence of functions converges pointwise on $[0, +\infty)$ to $0$.

Is that the way to show that $(f_n)_n$ converges pointwise? Or do we have to prove that in an otherway because after that question it is asked to calculate the limit?

As for the uniform convergence:

Let $\epsilon> 0$ arbitrary.

It holds that $\displaystyle{\lim_{y\rightarrow +\infty}ye^{-y}=0}$.

So there is a $M>0$ such that $0<ye^{-y}<\epsilon$ for all $y>M$.

For each $n\in \mathbb{N}$ with $n>M$ we get $\frac{1}{n}<\frac{1}{M}$.

We have that \begin{equation*}|f_n(x)|=\left |xe^{-nx}\right |=\left |\frac{1}{n}\cdot nxe^{-nx}\right |=\left |\frac{1}{n}\right |\cdot \left |nxe^{-nx}\right |\ \overset{y:=nx}{=}\ \left |\frac{1}{n}\right |\cdot \left |ye^{-y}\right |<\frac{1}{M}\cdot \epsilon\end{equation*}
So we get also uniform convergence on $[0, +\infty)$ to $0$.

Is everythig correct?

Or do we not need to use $\displaystyle{\lim_{y\rightarrow +\infty}ye^{-y}=0}$ ?

答:
Your proof of pointwise convergence is fine.

Concerning uniform convergence, note that $f_n'(x)=e^{-nx}(1-nx)$ and that it follows from this that $f_n$ is increasing on $\left[0,\frac1n\right]$ and decreasing on $\left[\frac1n,\infty\right)$. Therefore, $\max f_n=f_n\left(\frac1n\right)=\frac1{ne}$. Since $\lim_{n\to\infty}\max f_n=0$, the convergence is uniform.


问:
Average rank of a sequence of iid N(0,1) random variables.
Suppose we have $X_1, ..., X_n$ are iid N(0,1), and the rank is defined as below
$$
r(X_i) = \sum_{i \neq j} I(X_j \geq X_i)
$$
What is the expected rank?

I get that it is $(n-1)/2$ but Mood in "On the Asymptotic Efficiency of Certain Nonparametric Two-Sample Tests" states it to be (n+1)/2.

Am I missing something here?

答:
There is a very strong symmetry along the variables that we can exploit; because all variables are iid the probability that the rank is uniformy distributed among the possibilities.

There are $n$ possible ranks, between $0$ and $n-1$. Hence the average rank is $E[r(X_i)] = \sum_{i=0}^{n-1} \frac{i}{n} =  \frac{n(n-1)}{2n} = \frac{n-1}{2}$


问:
$\lim_{x \to 0} (e^x - 1)/x^8$ .
I want to know how to solve this limit problem with Maclaurin's series expansion of only $e^x$ in the neighborhood of $x=0$.

Title: $\lim_{x \to 0} (e^x - 1)/x^8$ .

Body:
$$\frac{e^x-1}{x^8}=\frac{1+x+o(x^2)-1}{x^8}.$$

You can conclude.


问:
Complex Analysis:Show that $a \in \mathbb{Z}$ and that there is $\theta \in \mathbb{R}$ so that $f(z)=e^{i\theta}z^{a}$.
Let $f$ be analytical on ann $A(0;r,R)$ and suppose there is $a \in \mathbb{R}$ such that $|f(z)|=|z|^{a}$, $\forall z \in A$. Show that $a \in \mathbb{Z}$ and that there is $\theta \in \mathbb{R}$ so that $f(z)=e^{i\theta}z^{a}$.

I have the impression that using Schwarz's lemma forms $ f(z)=e^{i\theta}z^{a} $, but I cannot develop it. I can't think of results that prove  $  a \in \mathbb{Z} $. Could you help me with ideas?

答:
(I use $r$ for $\vert z\vert$, so I will change the domain to $A(0;R_1,R_2)$)

Write $f(re^{i\phi})=R(r,\phi)e^{i\psi(r,\phi)}$. By assumption $R(r,\phi)=r^a$. Thus the Cauchy-Riemann equations give us \begin{align*}
\frac{\partial\psi}{\partial\phi}(r,\phi)&=a\\
\frac{\partial\psi}{\partial r}(r,\phi)&=0\\
\end{align*}
Hence $\psi(r,\phi)=a\phi+\theta$ for some constant $\theta\in\Bbb R$. We then get for $z=re^{i\phi}$: $$f(z)=f(re^{i\phi})=r^ae^{i(a\phi+\theta)}=e^{i\theta}z^a$$ in $A(0,R_1,R_2)\setminus\{x\in \Bbb R\mid R_1<x<R_2\}$. We know that $f$ is actually defined on the whole annulus $A(0;R_1,R_2)$ which is not possible if $a\notin\Bbb Z$.

Edit: The Cauchy-Riemann equations for the function in polar coordinates as above are:\begin{align*}
r\frac{\partial R}{\partial r}&=R\frac{\partial\psi}{\partial\phi}\\
\frac{\partial R}{\partial \phi}&=-rR\frac{\partial\psi}{\partial\phi}
\end{align*}
Now plug in $R(r,\phi)=r^a$.


问:
Showing that if $N \le G$ is finite minimal normal and every simple homomorphic image is abelian, then N is elementary abelian
This is part 2 of this question.  Unsurprisingly, I'm having some difficulty with it.  Hints are much appreciated.  Here's the setup again:

"Let $N$ be a finite minimal normal subgroup of a group $G$, and suppose that $N$ has the property that every simple homomorphic image is abelian.  Show that all nonidentity elements of $N$ have equal prime order."

As a hint I'm given "show that for some prime $p$, the subset $S=\{x \in N \,|\, x^p=1\}$ contains a nonidentity element."

Basically as far as I've gotten is showing that $N$ has prime power order:

Let $\phi: G \to H$ be a nontrivial homomorphism.  Since $\phi(N)$ is simple and abelian, it is a cyclic group of prime order, say $p$.  If $x \in N$, then $|\phi(x)|$ divides $|x|$, so $|x|$ is a multiple of $p$.  Furthermore, it is a power of $p$ since if it had some other prime factor, say $q$, then $q$ divides $|N|$ and so by Cauchy's Theorem we have an element of order $q$, which $p$ does not divide.  Therefore $|N|$ is a power of $p$.

How we prove that it cannot have elements of prime power order greater than $p$ is what is eluding me.  I will continue thinking and check back tomorrow to see where I have gone astray...

Note: Before even posting I now realize that's not even a valid argument: elements of order $q$ could exist, but they must lie in $\ker(\phi)$...  Should I think further about this?

答:
The trick is to first show that $N$ is abelian and after that to show that all non-identity elements have order $p$ for some fixed prime $p$.

Note that any non-trivial group will have a simple quotient, so the condition that any simple quotient is abelian means that $N'$ is a proper subgroup of $N$. But $N'$ is a characteristic subgroup of $N$, so it is normal in $G$, and by assumption of minimality of $N$, this means that $N' = \{1\}$ so $N$ is abelian.

Now, since $N$ is abelian, any Sylow subgroup is characteristic, so by minimality, we conclude that $N$ must be a $p$-group for some prime $p$.

But in any abelian $p$-group the set of elements of order dividing $p$ is a non-trivial characteristic subgroup, so by minimality, this subset is $N$ itself, which precisely means that $N$ is elementary abelian.

ContentLicense: CC BY-SA 3.0


问:
Computing the integral $ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}$
Compute the following integral
$$ \int_{1}^{\infty} \frac{x \,\mathrm{d} x}{\sqrt{2x^6+2x^3+1}}.$$
This integral obviously converges. However, is it computable?

答:
A bit more Mathematica legerdemain: If we substitute $v=2x^3+1$, we obtain the form

$$I=\frac{1}{3 \sqrt[6]{2}}\int_3^\infty \frac{1}{\sqrt[3]{v-1}}\frac{dv}{\sqrt{1+v^2}}.$$
Writing the integration range as $\int_3^\infty = \int_1^\infty -\int_1^3$, Mathematica yields
$$I= \frac{2\pi}{3\sqrt{3}}{_2}F_1\left(\frac13,\frac56,1,-1\right)-  \frac12 F_1\left(\frac23;\frac12,\frac12;\frac53;-1+i,-1-i\right)$$
where ${_2}F_1(a,b;c;x)$ is the Gauss hypergeometric function and $F_1(a;b,c;d;x,y)$ is the Appell hyper-geometric function referenced in J.G.'s answer. Note that this means

\begin{align}
\int_0^\infty \frac{x\,dx}{\sqrt{2x^6+2x^3+1}} &= \frac{2\pi}{3\sqrt{3}}{_2}F_1\left(\frac13,\frac56,1,-1\right),\\
\int_0^1 \frac{x\,dx}{\sqrt{2x^6+2x^3+1}} &= \frac12 F_1\left(\frac23;\frac12,\frac12;\frac53;-1+i,-1-i\right)
\end{align}

Mathematica can't simplify the first term further, but there's so many identities for hypergeometric functions that further simplification seems plausible. The Appell F1 term, however, seems rather impenetrable.


问:
Question about Ito integral
I was wondering if Ito integral:

$\int_0^T B(t)dB(t) $

is Gaussian (in which B(t) is Brownian Motion)??

Thank you so much, I appreciate any help ^^

Title: Question about Ito integral

Body:
As a simple and brief description, $\int_0^T B(t)dB(t)=\displaystyle\sum\limits_{j=0}^{n-1}B(t_j)\Delta B(t_j)$ you can expand sigma as below
$$\sum\limits_{j=0}^{n-1}B(t_j)\Delta B(t_j)=\\
\sum\limits_{j=0}^{n-1}B(t_j)(B(t_{j+1})- B(t_j))=\\
\frac12\sum\limits_{j=0}^{n-1}2B(t_j)B(t_{j+1})- 2B^2(t_j)=\\
\frac12\sum\limits_{j=0}^{n-1}B^2(t_{j+1})-B^2(t_j)-B^2(t_{j+1})+2B(t_j)B(t_{j+1})- B^2(t_j)=\\
\frac12\sum\limits_{j=0}^{n-1}B^2(t_{j+1})-B^2(t_j)-(B^2(t_{j+1})-2B(t_j)B(t_{j+1})+B^2(t_j))=\\
\frac12\sum\limits_{j=0}^{n-1}B^2(t_{j+1})-B^2(t_j)-\frac12\sum\limits_{j=0}^{n-1}(B^2(t_{j+1})-2B(t_j)B(t_{j+1})+B^2(t_j))=\\
\frac12\sum\limits_{j=0}^{n-1}(B^2(t_{j+1})-B^2(t_j))-\frac12\sum\limits_{j=0}^{n-1}(B(t_{j+1})-B(t_j))^2=\\
\underbrace{\frac12\sum\limits_{j=0}^{n-1}(B^2(t_{j+1})-B^2(t_j))}_{telescopic}-\frac12\sum\limits_{j=0}^{n-1}(\underbrace{B(t_{j+1})-B(t_j)}_{\Delta B(t_j)})^2=\\
\frac12(B^2(T)-B^2(0))-\frac12\underbrace{(\Delta B(t_0)^2+\Delta B(t_1)^2+\cdots+\Delta B(t_{n-1})^2)}_{(\Delta t_0+\Delta t_1+...+\Delta t_{n-1})=T-0}=\\
\frac12(B^2(T)-B^2(0))-\frac12 T$$ if $B(0)=0$ then $$
\int_0^T B(t)dB(t)=\displaystyle\sum\limits_{j=0}^{n-1}B(t_j)\Delta B(t_j)=\frac12 B^2(T)-\frac12 T$$


问:
Topology counter example
-We known that every subspace of T2 space is T2space..
but If we reverse it
-if A is subspace of X and A was a T2 space need to be space X T2space also.?
If the statement is wrong plz given example.</p>

答:
No, a single point $\{x\}$ in any space $X$ is Hausdorff, regardless of what separation axioms $X$ satisfies.

For a more sophisticated example you can take any Hausdorff space $A$ and add a point to it $X:=A\sqcup\{*\}$ with the following topology: $U\subseteq X$ is open if and only if $U=X$ or $U\subseteq A$ is open. Forget about Hausdorfness, the special $\{*\}$ is not even closed in $X$. But $A\subseteq X$ is Hausdorff.

The statement isn't even true if we assume that $X$ has all proper subsets Hausdorff, as the trivial topology on $\{1,2\}$ shows.

Title: Topology counter example


问:
Are these graph Hamilton or not?
I have read it a lot about that but I am still confusing how to find out if it is Hamilton graph or not. I have check here some similar examples but I was not able to  understand it.

I know that a graph to be Hamilton it needs to do from one "starting" point we put  to do all the graph until the end which is our "starting" point.

I have created those on a paper please help understand if those are Hamilton or not.

For these Graphs

Here I have created 1 and 2 as you can see graphs. As I have seen, if a graph has inside another circuit, then it is not a Hamilton. Is this right? As a result if there are similar like those with in, inside a "circle lets call it" then it is not a Hamilton graph?

And, for these Graphs

With 3 and 4, I am thinking to do a start point in an edge and to do all the graph until I end again on the "start point" but I see it that it is not happening. So my question is, are those Hamilton or not? My thought is that those 3 and 4 graphs are not Hamilton. If these are not Hamilton, how to understand it? (is this the way?). In the other hand if we want to make those Hamilton (if those are not) what should exactly need to change?

Is these Graph Hamilton? I guess it is.

答:
The OP wants a Hamiltonian cycle, that is easy, for instance computing (using Mathematica):

```<code>g = Graph[{1 \[UndirectedEdge] 2, 1 \[UndirectedEdge] 3,
   1 \[UndirectedEdge] 4, 2 \[UndirectedEdge] 3,
   3 \[UndirectedEdge] 4, 3 \[UndirectedEdge] 5,
   2 \[UndirectedEdge] 5, 4 \[UndirectedEdge] 6,
   5 \[UndirectedEdge] 6, 4 \[UndirectedEdge] 7,
   6 \[UndirectedEdge] 7, 6 \[UndirectedEdge] 8,
   7 \[UndirectedEdge] 8, 5 \[UndirectedEdge] 8 },
  VertexCoordinates -> {{0, 2}, {1, 2}, {1/2, 3/2}, {0, 1}, {1,
     1}, {1/2, 1/2}, {0, 0}, {1, 0}}]
HighlightGraph[g,FindHamiltonianCycle[g][[1]]]
</code>```

For an arbitrary graph h, simply compute:

```<code>FindHamiltonianCycle[h]
</code>```

If such a cycle exists, it will be returned.  If not, the set of edges will be empty.

Here's a more complicated case:

Title: Are these graph Hamilton or not?

Body:
The OP wants a Hamiltonian cycle, that is easy, for instance computing (using Mathematica):

g = Graph[{1 \[UndirectedEdge] 2, 1 \[UndirectedEdge] 3,
   1 \[UndirectedEdge] 4, 2 \[UndirectedEdge] 3,
   3 \[UndirectedEdge] 4, 3 \[UndirectedEdge] 5,
   2 \[UndirectedEdge] 5, 4 \[UndirectedEdge] 6,
   5 \[UndirectedEdge] 6, 4 \[UndirectedEdge] 7,
   6 \[UndirectedEdge] 7, 6 \[UndirectedEdge] 8,
   7 \[UndirectedEdge] 8, 5 \[UndirectedEdge] 8 },
  VertexCoordinates -> {{0, 2}, {1, 2}, {1/2, 3/2}, {0, 1}, {1,
     1}, {1/2, 1/2}, {0, 0}, {1, 0}}]
HighlightGraph[g,FindHamiltonianCycle[g][[1]]]

For an arbitrary graph h, simply compute:

FindHamiltonianCycle[h]

If such a cycle exists, it will be returned.  If not, the set of edges will be empty.

Here's a more complicated case:


问:
If $N \lhd G$ and $A \lhd B \le G$ then $AN \lhd BN$?
The content of the question is in the title. I'm asking this in relation to a proof that (assuming the above conditions): $$BN/AN \cong B/A(B \cap N)$$
Which I believe follows directly from an application of the second isomorphism theorem for groups.

Body:
$AN \subseteq BN$ is clear since $A \subseteq B$.

Then we must show that $xANx^{-1} \subseteq AN$ for $x \in BN$.

$x=bn$ so show that $bn(an_{1})n^{-1}b^{-1} \in AN.$

Using the normality of $N$ in $G$ and of $A$ in $B$ as well the extensive use of the generalized associative law for groups,

$bn(an_{1})n^{-1}b^{-1} = b(na)n_{1}n^{-1}b^{-1}=b(an_{2})(n_{3}b^{-1})= (ba)n_{2}(n_{3}b^{-1})=(a_{2}b)n_{2}(b^{-1}n_{4})=a_{2}(bn_{2})(b^{-1}n_{4})=a_{2}(n_{5}b)(b^{-1}n_{4})=(a_{2}n_{5})(b^{-1}b)n_{4}=(a_{2}n_{5})n_{4}=a_{2}(n_{5}n_{4})=a_{2}n_{6} \in AN$.


问:
Prove the existence of a principal submatrix of order $r$ in $M\in\Bbb F^{n\times n}, M=-M^T,\ \operatorname{rank}(M)=r$
Let $M$ be a skew-symmetric matrix of $\operatorname{rank}(M)=r$, prove that there exists a principal submatrix of order $r$.

I have a solution for the version which doesn't require the submatrix to be principal (this is not my solution)
take away all but $r$ linearly independent columns of $M$, call the matrix $P$
since $\operatorname{rank}M=r$ this is possible
then if $P$ is $r \times r$, done, invertible

if $P$ is $n \times r$ where $n < r$ then $\operatorname{rank}(M)\leqslant n < r$ so this case can't happen

so assume $P$ is $n \times r$ where $n > r$,
now we look at $P^T = -P$,
$\operatorname{rank}(P) = \operatorname{rank}(-P) = r$

then you can take away all but $r$ columns of $P^T$ obtaining $Q^T=$ taking all but $r$ rows of $P$ obtaining $Q$
then $Q$ is $r \times r$ and all columns are linearly independent

So I am looking for a solution which proves the existence of a principal matrix.

答:
Here's a proof that holds over any field $\mathbb F$ for a skew symmetric (or indeed a regular symmetric) matrix.

$M\mathbf x = \mathbf 0 =-M^T\mathbf x =M^T\mathbf x\implies \mathbf x^T M= \mathbf 0^T$

having rank $r$, rank-nullity tells us $\dim\ker\big(M\big) = n-r$.   Build a basis for the nullspace.

$\big\{\mathbf x_1, ..., \mathbf x_{n-r}\big\}$.  Now apply the basis extension algorithm to the ordered set of standard basis vectors.  i.e. append $\mathbf e_1$ to the prior set if it is linearly independent, and discard it otherwise.  Then consider $\mathbf e_2$ and so on.

The result is we have a basis for $\mathbb F^n$ given by

$\big\{\mathbf e_{\sigma_{(1)}}, ..., \mathbf e_{\sigma_{(r)}},\mathbf x_1, ..., \mathbf x_{n-r}\big\}$

collect these in a matrix

$B:= \bigg[\begin{array}{c|c|c|c|c|c|c}
\mathbf e_{\sigma_{(1)}} & \cdots &  \mathbf e_{\sigma_{(r)}}& \mathbf x_1 &\cdots & \mathbf x_{n-r}
\end{array}\bigg]$

if it's easier, you can write this as

$B:= P \bigg[\begin{array}{c|c|c|c|c|c|c}
\mathbf e_{1} & \cdots &  \mathbf e_{r}& P^T\mathbf x_1 &\cdots & P^T\mathbf x_{n-r}
\end{array}\bigg]$

for some permutation matrix $P$

Finally, effect a congruence transform, i.e.

$B^T M B = \begin{bmatrix} C_{r\times r} &\mathbf {0}\\ \mathbf {0}& \mathbf {0}_{n-r \times n-r} \end{bmatrix}$

since $B$ is invertible, we have

$r=\text{rank}\big(M\big)=\text{rank}\big(B^T M B\big)= \text{rank}\big(C_{r\times r}\big)$

and $C_{r\times r}$ is a principal sumbatrix of $M$ as desired.


问:
What does this mean and how to solve it?
The equation described on the chord $3x + y + 5 = 0$ of the circle $x^2 + y^2 = 16$ as a diameter, is:

(A) $x^2 + y^2 + 3x + y - 2 = 0$

(B) $x^2 + y^2 + 3x + y - 22 = 0$

(C) $x^2 + y^2 + 3x + y + 1 = 0$

(D) $x^2 +y ^2 + 3x +y - 11 = 0$

答:
They want the equation of the red circle in the picture below.

I agree that is badly written.

I had said:

Given line $3x+y+5=0$ and circle $x^2+y^2=16$ call $A$ and $B$ their intersection points. Find the equation of the circle having $AB$ as diameter.


问:
Generating functions combinatorical problem
In how many ways can you choose $10$ balls, of a pile of balls containing $10$ identical blue balls, $5$ identical green balls and $5$ identical red balls?

My solution (not sure if correct, would like to have input):

Define generated function:

$$\begin{align}
A(x) & =(x^0+x^1+x^2+...+x^{10})(x^0+x^1+x^2+...+x^5)(x^0+x^1+x^2+...+x^5) \\
& ={1-x^{11} \over 1-x}\cdot \left({1-x^6 \over 1-x}\right)^2 \\
& =(1-x^{11})(1-2x^6+x^{12}) \cdot {1 \over (1-x)^3} \\
& =(1-2x^6-x^{11}+x^{12}+2x^{17}-x^{23})\cdot \sum_{n=0}^∞ {n+2 \choose 2} \cdot x^n.
\end{align}$$

We look for the coefficient of $x^{10}$, so we get:

$$a_{10}=1 \cdot {10+2 \choose 2}-2\cdot {4+2 \choose 2}=36.$$

This seems incorrect (sadly I'm terrible in 'ordinary combinatorics' so I'm not sure how to calculate this 'regularly'). I would love to get input and hints.

Thanks in advance.

答:
Here’s one elementary approach. As you see, it confirms your result.

Let $b,g$, and $r$ be the numbers of blue, green, and red balls chosen to make up a set of $10$ balls. You’re looking for the number of solutions in non-negative integers to the equation $$b+g+r=10\;,\tag{1}$$ subject to the condition that $g\le 5$ and $r\le 5$. (You also have to have $b\le 10$, but that imposes no additional constraint when the sum is to be $10$.) Without the upper bounds this is a standard stars-and-bars problem whose solution is

$$\binom{10+3-1}{3-1}=\binom{12}2\;.$$

However, this count includes solutions with too many green or red balls. Let $g'=g-6$; then there is a bijection between solutions to $$b+g'+r=4\tag{2}$$ in non-negative integers and solutions to $(1)$ for which $g>5$. Thus, we need only count solutions to $(2)$ to get the number of solutions to $(1)$ with $g>5$. This is another stars-and-bars problem, and the answer is

$$\binom{4+3-1}{3-1}=\binom62\;.$$

Similarly, there are $\dbinom62$ solutions to $(1)$ that have $r>5$. There are no solutions that exceed the upper limits on both $g$ and $r$, so the number of solutions to $(1)$ that satisfy all of the conditions is

$$\binom{12}2-2\binom62=36\;.$$


问:
Notation of Holder space used in stochastic analysis
Usually the Holder space is denoted by the sdandard notation $C^{k,\alpha}$ or $C^{k+\alpha}$ where $\alpha \in (0,1]$ and $k$ is an integer, but when I read some materials on SPDE, they use the notation $C^{\gamma}$ where $\gamma > 0$. What is the precise meaning of this notation? I am very confused when $\gamma$ takes integer value.

答:
For the purpose of this question having an answer, I recreate my comments here. In the SPDE literature (especially in singular SPDE) it is usual to define $C^\gamma$ to be the space of functions that are $\lceil \gamma \rceil - 1$ times differentiable whose $(\lceil \gamma \rceil - 1)$-th derivative is $(\gamma - \lceil \gamma \rceil + 1)$-Holder continuous.

In particular, if $\gamma$ is an integer then this space is larger than the space of $\gamma$ times continuously differentiable functions.


问:
$f \in L^p \iff \sum_{k\in\mathbb{Z}}2^{kp}\lambda_f(2^k)<\infty$
The problem is,

$$f \in L^p \iff \sum_{k\in\mathbb{Z}}2^{kp}\lambda_f(2^k)<\infty$$

I have the solution for $\Rightarrow$, but I don't understand a determined step.

Let $f\in L^p$,

\begin{align*}
\sum_{\mathbb{Z}} \lambda_f(2^k)2^{kp}&=\sum_{\mathbb{Z}}\lambda_f(2^k)\frac{p}{1-2^{-p}}\int_{2^{k-1}}^{2^k} \alpha^{p-1}\,d\alpha\\
&\le \sum_{\mathbb{Z}}\frac{p}{1-2^{-p}}\int_{2^{k-1}}^{2^k}\lambda_f(\alpha) \alpha^{p-1}\,d\alpha \quad(*)\\
&=\frac{p}{1-2^{-p}}\int_{0}^{\infty}\lambda_f(\alpha) \alpha^{p-1}\,d\alpha\\
&=\frac{p}{1-2^{-p}}\int|f|^p < \infty
\end{align*}

I couldn't understand the step $(*)$. I probably can't understand it because I'm having a little difficulty understanding exactly what it would be

$$\int\lambda_f(\alpha)\,d\alpha$$

Recalling that $\lambda_f(\alpha)$ is the distribution function, given by $\lambda_f(\alpha)=\mu(\{x:|f(x)|>\alpha\})$.

答:
Hint

If $\alpha \in [2^{k-1},2^k]$, then

$$|f(x)|>2^k\implies |f(x)|>\alpha ,$$
and thus
$$ \lambda _{f}(2^k)\leq \lambda _f(\alpha ).$$

Edit

Therefore $$\alpha ^{p-1}\lambda_f(2^k)\leq \alpha ^{p-1}\lambda _f(\alpha ),$$
for all $\alpha \in [2^{k-1},2^k]$, and thus $$\lambda _f(2^k)\int_{2^{k-1}}^{2^k}\alpha ^{p-1}\,\mathrm d \alpha \leq \int_{2^{k-1}}^{2^k}\lambda _f(\alpha )\alpha ^{p-1}\,\mathrm d \alpha .$$


问:
Zeta in summations
Does anyone know if there is any reference that discuss the following kind of summations?

$$S = \displaystyle\sum_{n=0}^\infty a_n \zeta(n+2)$$

I have read Srivastava's article https://www.sciencedirect.com/science/article/pii/0022247X88900133. But none deals with such series. Especially when:

$a_n = \frac{2}{(n+2)(n+3)}$. Which leads to

$S = \frac{\zeta(2)}{3} + \frac{\zeta(3)}{6} + \frac{\zeta(4)}{10} + \cdots$

Have you seen this one before?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
S & \equiv \bbox[5px,#ffd]{2\sum_{n = 0}^{\infty}
{\zeta\pars{n + 2} \over \pars{n + 2}\pars{n + 3}}} =
2\sum_{n = 2}^{\infty}{\zeta\pars{n} \over n\pars{n + 1}}
\\[5mm] & =
2\sum_{n = 2}^{\infty}\zeta\pars{n}
\pars{\int_{0}^{1}x^{n - 1}\,\,\dd x}
\pars{\int_{0}^{1}y^{n}\,\dd y}
\\[5mm] & =
2\int_{0}^{1}\int_{0}^{1}
\bracks{\sum_{n = 2}^{\infty}\zeta\pars{n}
\pars{xy}^{n - 1}}y\,\dd x\,\dd y
\\[5mm] & =
2\int_{0}^{1}\int_{0}^{1}
\bracks{-\gamma - \Psi\pars{1 - xy}}y\,\dd x\,\dd y
\end{align}
$\ds{\gamma}$ is the Euler-Mascheroni Constant and $\ds{\Psi}$ is the Digamma Function.

The relation between $\ds{\zeta\ \mbox{and}\ \Psi}$ can be seen in
$\ds{\color{black}{\bf 6.3.14}}$ of
$\mbox{A & S Table}$.

Then,
\begin{align}
S & \equiv \bbox[5px,#ffd]{2\sum_{n = 0}^{\infty}
{\zeta\pars{n + 2} \over \pars{n + 2}\pars{n + 3}}}
\\[5mm] & =
2\int_{0}^{1}
\bracks{-\gamma -
{\ln\pars{\Gamma\pars{1 - y}} \over - y}}y\,\dd y
\\[5mm] & =
2\bracks{-\,{1 \over 2}\,\gamma +
\int_{0}^{1}\ln\pars{\Gamma\pars{1 - y}}\dd y}
\\[5mm] & =
\bbx{\ln\pars{2\pi} - \gamma} \approx 1.2607 \\ &
\end{align}

I used the well known result
$\ds{\int_{0}^{1}\ln\pars{\Gamma\pars{1 - y}}\,\dd y
= {\ln\pars{2\pi} \over 2}}$

Title: Zeta in summations


问:
How to interpret a two-sided graph?

What does the "right" Y-axis represent?

Taken from here: https://istheservicedown.co.uk/status/virgin-media/2654675-bristol-bristol-england-united-kingdom

答:
Looking around the provided site it looks like the blue curve represents the number of reports for Virgin Media outages in the UK, whereas the green curve is the number of reports specific to Bristol.  The units on the left are for the blue (country-wide) curve, and the units on the right are for the green (local) curve.  For example, consider the excerpted images below.  Note that in each case the blue curve matches the graph of total Virgin Media outages, whereas the green curve (and corresponding scale on the right axis) change depending on locality.

Virgin Media outages in the UK:

Virgin Media outages in the Bristol:

Virgin Media outages in the London:

Title: How to interpret a two-sided graph?


问:
Find the supremum and infimum of a set X
Let the set $X=\{\frac{s}{s^2+2}, s\in\mathbb{Z}\}$ I have to find the supremum and the infimum. I have done in this way:

I can observe that $X=\{\frac{s}{s^2+2}, s\in\mathbb{N}\}\cup \{\frac{-t}{t^2+2}, t\in\mathbb{N}\}$ and now I study separetely the infimum and supremum of the two sets that give me in the union the set X.

1)$\{x_s=\frac{s}{s^2+2}, s\in\mathbb{N}\}$:
since $\frac{s}{s^2+2}>\frac{s+1}{(s+1)^2+2}$ $\forall s\geq 1$ then my sequence is eventually decreasing and so $\exists \lim_{s\to\infty} x_s=0=inf$. The sup is instead given by $\frac{s}{s^2+2}$ evaluated at s=1, so $sup=\frac{1}{3}$.

2) $ \{x_t=\frac{-t}{t^2+2}, s\in\mathbb{N}\}$:

since $\frac{-t}{t^2+2}>-\frac{t+1}{(t+1)^2+2}$ $\forall t\geq 1$ then my sequence is eventually increasing and so $\exists \lim_{t\to\infty} x_t=0=sup$. The inf is instead given by $\frac{-t}{t^2+2}$ evaluated at t=1, so $inf=\frac{-1}{3}$.

So finally $supA=max\{\frac{1}{3},0\}=\frac{1}{3}$ and then $infA=max\{\frac{-1}{3},0\}=\frac{-1}{3}$.

TO DO: check my solving.

答:
hint

Let $$f(x)=\frac{x}{x^2+2}$$

$ f $ is an odd function.

$$(\forall x\ge 0)\; \;\; f'(x)(x^2+2)^2=2-x^2$$

From here the maximum of $ X $ will be $ \max X =f(1) = f(2)  =\frac 13$ .

By symmetry, $\min X =-\max X $.


问:
Find $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for $f(x) = a e^x + b \cos(\pi x /2)$?
I am reading through some exercises of the chapter I am reading about numerical integration. I stumbled upon the following question:

Find the formula $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for functions of the form $f(x) = a e^x + b \cos(\pi x/2)$

I know that for any polynomial of degree $\leq n$ its integral can be exactly estimated (that is, is equal to) by the formula
$$
\int_a^b f(x) \,dx = \sum_{i=0}^{n} A_i f(x_i)
$$
where $A_i = \int_a^b \ell (x)\, dx$ (Lagrange fundamental polynomial) for $n+1$ distinct nodes $x_i \quad 0 \leq i \leq n$.

However, the $f$ in the exercise is clearly not a polynomial. So no idea how to find such a formula that is exact as with numerical integration for polynomials. Can you help?

答:
Integrating explicitly and equating to 0 the coefficients multiplying $a$ and $b$ individually one gets the linear system:

$e-1=A_0e^{x_0}+A_1e^{x_1}$ [1]

$\frac{2}{\pi}=A_0cos(\pi x_0/2)+A_1cos(\pi x_1/2)$ [2]

If we have $x_0$, $x_1$ and $A_0,A_1$ that solve [1] and [2] we are done.

Let's take $x_0$ and $x_1$ arbitrary. Since this system is linear in $A_0,A_1$ one can simply invert the matrix of coefficients and solve for these:

$\begin{pmatrix}
  A_0 \\
  A_1
\end{pmatrix}$  =$\begin{pmatrix}
  e^{x_0} & e^{x_1}\\
  cos(\pi x_0/2) & cos(\pi x_1/2)
\end{pmatrix}^{-1}$ $\begin{pmatrix}
  e-1 \\
  2/\pi
\end{pmatrix}$

It is simple to find $x_0$ and $x_1$ such that the system is invertible checking that the determinant is different from zero.


问:
Topology induced by Basis B
I have recently started studying topology on my own. I have a confusion regarding defination of topology of set X induced by basis $\mathfrak{B}$. ${\tau} = \{ U: x\in U \implies \exists B :x \in B \in \mathfrak B, B \subset U   \}$

Here defination of topology $\tau$ means i need to include all sets such that x belongs to some B in basis. It means i need to include B, B+atleast 1 element in X .... X

Does it mean finally $\tau$ becomes powerset of X? please correct if wrong

答:
It means that the topology is defined by all sets $U$ that satisfy the property that each of its elements has at least a basis element $B$ that pass through that point and is entirely included in that set $U$.

Note that there is a typo in what you write. The first inclusion symbol must be a "belongs to".

Title: Topology induced by Basis B


答:
Let $f(s) = s/(s^2 + 2)$.  Then clearly for all $s \in \mathbb Z$, $$f(-s) = -s/((-s)^2 + 2) = -s/(s^2 + 2) = -f(s).$$  So if $f(n) = \sup X$, we must have $f(-n) = \inf X$.

Next, observe that $f(s) > 0$ if $s > 0$, so $f(s) < 0$ if $s < 0$, and $f(s) = 0$ if $s = 0$.  So it suffices to consider $s \in \mathbb Z^+$ and $\sup X$ only.

Finally, consider the quotient $$\frac{f(s+1)}{f(s)} = \frac{(s+1)(s^2 + 2)}{s(s^2 + 2s + 3)} = \frac{s^3 + s^2 + 2s + 2}{s^3 + 2s^2 + 3s} = 1 - \frac{(s-1)(s+2)}{s(s^2 + 2s + 3)}.$$  When $s > 0$, the denominator of the second term is always positive.  Since the numerator of the second term is zero if $s = 1$ (we ignore the negative root since $s > 0$), and is positive for $s > 1$, it follows that this ratio is equal to $1$ only when $s = 1$, and is strictly less than $1$ if $s > 1$.  Therefore, for $s \in \mathbb Z^+$, we see $f(s)$ is greatest when $s = 1$, and $\sup X = f(1) = \frac{1}{3}$, from which it follows that $\inf X = -\frac{1}{3}$.

Title: Find the supremum and infimum of a set X

Body:
Let the set $X=\{\frac{s}{s^2+2}, s\in\mathbb{Z}\}$ I have to find the supremum and the infimum. I have done in this way:

I can observe that $X=\{\frac{s}{s^2+2}, s\in\mathbb{N}\}\cup \{\frac{-t}{t^2+2}, t\in\mathbb{N}\}$ and now I study separetely the infimum and supremum of the two sets that give me in the union the set X.

1)$\{x_s=\frac{s}{s^2+2}, s\in\mathbb{N}\}$:
since $\frac{s}{s^2+2}>\frac{s+1}{(s+1)^2+2}$ $\forall s\geq 1$ then my sequence is eventually decreasing and so $\exists \lim_{s\to\infty} x_s=0=inf$. The sup is instead given by $\frac{s}{s^2+2}$ evaluated at s=1, so $sup=\frac{1}{3}$.

2) $ \{x_t=\frac{-t}{t^2+2}, s\in\mathbb{N}\}$:

since $\frac{-t}{t^2+2}>-\frac{t+1}{(t+1)^2+2}$ $\forall t\geq 1$ then my sequence is eventually increasing and so $\exists \lim_{t\to\infty} x_t=0=sup$. The inf is instead given by $\frac{-t}{t^2+2}$ evaluated at t=1, so $inf=\frac{-1}{3}$.

So finally $supA=max\{\frac{1}{3},0\}=\frac{1}{3}$ and then $infA=max\{\frac{-1}{3},0\}=\frac{-1}{3}$.

TO DO: check my solving.


问:
To construct jump discontinuous function
$$f(x)=\begin{cases}
\frac{\tan kx}{x}, & x<0\\
3x + 2k^2, &x\ge 0
\end{cases}
$$

Hi! I'm trying to construct a function  from this problem with jump discontinuity but from my knowledge the variable $x$ in the denominator with limit approaching $0$ for the left side of it would would result $x=0$ making it discontinuous with a vertical asymptote hence infinite discontinuity?
How can I compute a value for $k$ to keep it as discontinuous with a jump?

答:
$$ \lim_{x \rightarrow 0^-} \frac{ \tan(kx)}{x} = \lim_{x \rightarrow 0^-} \frac{ \frac{k\sin(kx)}{\cos(kx)}}{kx} = \lim_{x \rightarrow 0^-} \frac{ \sin(kx)}{kx} \cdot \lim_{x \rightarrow 0^-}\frac{k}{\cos(kx)} = k$$

Using the known $\frac{\sin(kx)}{kx}$ limit

Thus we need to check for what values of $k$ we have a continuity, or in other words the two functions have the same output at $x=0$:

$$ k = \lim_{x \rightarrow 0^+} 3x + 2k^2 = 2k^2 \\ k = 2k^2 \\ k = 0, \frac{1}{2}$$

Thus, if you want a function with a jump continuity you need to pick some values that are not $0.5, 0$


问:
Solving $2^x=4x$

$2^x=4x$

I cant seem to solve this equation. The furthest I have been able to come is
$x-\log_2(x)=2$, but I can't figure how to solve. When I graph $2^x$ and $4x$ they intersect at $x=4$ and $x=0.31$, so I know it is possible to solve.

答:
Classic problem:

$$ x = -\frac{W\left(-\frac{\log (2)}{4}\right)}{\log (2)}, {\rm or}\ -\frac{W_{-1}\left(-\frac{\log
   (2)}{4}\right)}{\log (2)}$$

where $W$ is the ProductLog function.

Numerical evaluation of these analytic solutions:  $.309907$ and $4$.

Title: Solving $2^x=4x$


问:
How to prove the lower semi-continuity of this functionnal on $H^1(\Omega)$
I am studying the following application which goes from $H^1(\Omega)$ to $\mathbb{R}$ with $\Omega$ a bounded regular subset of $\mathbb{R}^3$ :

$$H : u \mapsto \int_{\Omega} \left(|\nabla u|^2 - a \cdot  \nabla u \right) \ \mathrm{d} x$$

where $a$ is a fixed vector in $\mathbb{R}^3$. I would like to prove that $H$ verifies the lower semi-continuity definition.

I know 3 equivalent ways of defining lower semi-continuity, denoting $X=H^1(\Omega)$ :

for all $u_0 \in X$, for all $t<H(u_0)$, there exists a neighbor of $u_0$ such that :

$$\forall u \in X, H(u) \geq t$$

for all real $\alpha$, the set $\{x \in X \ | \ H(u) \leq \alpha \}$ is closed in X.

the epigraph $\{(u,\alpha) \in X \times \mathbb{R}, \ | \ H(u)\leq \alpha  \}$ is closed in X.

As I am very unfamiliar with this notion, I'm not sure which definition should I use to show the lower semi-continuity of $H$.

Any advices are welcomed ! Maybe we'll have to change $X$ to $H^1_0(\Omega)$ to use Poincaré inequality and change the norm.

答:
Your functional is actually well-defined on $\mathrm{H}^1(\Omega)$. It is also strongly lower-semicontinuous with respect to the topology of this space. Indeed,
define $f(z):= \vert z \vert^2 -a \cdot z$ for $z \in \mathbb{R}^{3}$. You then have $H(u)=\int_{\Omega}f(\nabla u)$.

In order to show that $H$ is strongly l.s.c on $\mathrm{H}^1(\Omega)$, I use the following characterization (valid on metric spaces) : $H$ is l.s.c if and only if for all $u \in \mathrm{H}^1(\Omega)$ and for all sequence $(u_n) \in \mathrm{H}^1(\Omega)^{\mathbb{N}}$ such that $u_n \rightarrow u$ in $\mathrm{H}^1(\Omega)$, we have $H(u) \leq \liminf H(u_n)$.

Thus, take $u \in \mathrm{H}^1(\Omega)$ and $(u_n) \in \mathrm{H}^1(\Omega)^{\mathbb{N}}$ such that $u_n \rightarrow u$ in $\mathrm{H}^1(\Omega)$. There is nothing to prove if $\liminf H(u_n)=+\infty$. If $\liminf H(u_n)<+\infty$, we can pick a subsequence $(u_{n_k})$ of $(u_n)$ such that $$\nabla u_{n_k} \rightarrow \nabla u \ \ a.e \ \ \text{and} \ \ H(u_{n_k}) \rightarrow \liminf H(u_n), \ \ \text{when} \ \ k \rightarrow + \infty.$$

Since for almost any $z$, we have $f(z) \geq -a \cdot z$, we can use the Fatou lemma to write
$$ \liminf \int_{\Omega} (f(\nabla u_{n_k})+a \cdot \nabla u_{n_k} ) \geq \int \liminf (f(\nabla u_{n_k})+a \cdot \nabla u_{n_k})=\int_{\Omega} (f(\nabla u)+ a \cdot \nabla u),$$
because $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ is continuous. Since $\nabla u_{n_k} \rightarrow \nabla u$, we also have
$$\liminf \int_{\Omega} (f(\nabla u_{n_k})+a \cdot \nabla u_{n_k} ) \leq \lim \int_{\Omega} f(\nabla u_{n_k}) + \int_{\Omega} a \cdot \nabla u.$$
We end up with
$$ H(u)=\int_{\Omega} f(\nabla u) \leq \lim \int_{\Omega} f(\nabla u_{n_k})=\lim H(u_{n_k})=\liminf H(u_n),$$
and this completes the proof.

Note that since $f$ is convex, the functional $H$ is convex. This implies that $H$ is also weakly lower semicontinuous on $\mathrm{H}^1(\Omega)$.


问:
Determinant of associated real matrix from a complex one
Given $X\in \operatorname{GL}(n, \mathbb{C}) $, let $X_r\in \operatorname{GL}(2n, \mathbb{R}) $ be the real matrix obtained by substituting to each complex entry $a+ib$ the matrix $\begin{pmatrix} a & & - b \\ b & & a \end{pmatrix}$.
Does someone can give me the details of the proof that $\det(X_r) =|\det(X)|^2$?
The only thing I have tried is to use the fact that the determinant is a wedge product but it seems not work.

答: Since $|a+ib|^2=\det\begin{pmatrix} a & & - b \\ b & & a \end{pmatrix}$,$$\begin{align}\det X\cdot\overline{\det X}&=\sum_{\sigma,\,\sigma^\prime\in S_n}\varepsilon_\sigma\varepsilon_{\sigma^\prime}\prod_{j=1}^n\prod_{k=1}^nX_{j\sigma(j)}\overline{X_{k\sigma^\prime(k)}}\\&=\sum_{\sigma,\,\sigma^\prime}\varepsilon_{\sigma\circ\sigma^\prime}\prod_j(XX^\dagger)_{j(\sigma\circ\sigma^\prime)}\\&=\sum_{\sigma\in S_{2n}}\varepsilon_\sigma\prod_{j=1}^{2n}(X_r)_{j\sigma(j)}\\&=\det X_r.\end{align}$$


问:
If the group if abelian, must all automorphisms be outer automorphisms?
I am having some thought about this question.

Clearly, if the group $G$ is abelian, then $\mathrm{Inn}(G) = {e}$.  But what about $\mathrm{Aut}(G)$ and $\mathrm{Out}(G) = \mathrm{Aut}(G)/\mathrm{Inn}(G)$?  Must $\mathrm{Aut}(G) = \mathrm{Out}(G)$?  Take $G = \mathbb{Z}_2 \times \mathbb{Z}_2$ as the example.  We know that $\mathrm{Aut}(G)$ consists of 6 bijective functions.  Must $\mathrm{Out}(G)$ also consist of 6 bijective functions?

Different example:

$G = D_4$

$G$ is not abelian and noncyclic.  It consists of $4$ elements in $\mathrm{Inn}(G)$ and $8$ elements in $\mathrm{Aut}(G)$.  Must $\mathrm{Out}(G)$ consist of 2 elements?  I found that there are 4 functions instead of 2.

答:
Can you see that you have correctly answered your question?

For abelian group $G$, we know $\,\operatorname{Inn}(G) \cong e.\;$ So, we have that $$\operatorname{Out}(G)\cong \operatorname{Aut}(G)/\operatorname{Inn}(G)\cong \operatorname{Aut}(G)$$


答:
The equation has indeed the easy integer solution $x=4$ which can be found by inspection. Graphing reveals a second solution (and it is possible to formally prove that there are no other real roots).

If you heard of the Taylor development, as the second root is not that large,
you can develop $e^{x\ln2}$ to the second order and solve

$$1+x\ln2 +\frac{(x\ln2)^2}2=4x.$$

The smallest solution of this quadratic is $x=0.30935\cdots$, which is close the searched answer.

Even better, you can use the known approximation $0.31$ and write

$$e^{(0.31+y)\ln 2}=e^{0.31\ln 2}e^{y\ln 2}=4(0.31+y).$$

Solving as above,

$$y=-9.3067447651\cdots\,\cdot10^{-5}$$ and $$x=0.3099069323807\cdots,$$ where all decimals are exact !

Title: Solving $2^x=4x$

Body:
$2^x=4x$

I cant seem to solve this equation. The furthest I have been able to come is
$x-\log_2(x)=2$, but I can't figure how to solve. When I graph $2^x$ and $4x$ they intersect at $x=4$ and $x=0.31$, so I know it is possible to solve.


问:
Twice differentiable but not infinitely differentiable
I want an example of a function satisfying:

(1) $f:\mathbb R\to\mathbb R$ twice continuously differentiable

(2) $f''(x)$ $\ge 0$ for every x$\in \mathbb R$

(3)$\int\limits_0^x{f(t)dt}$ is NOT infinitely differentiable with respect to x.

I find it very difficult to get a function satisfying conditions 2 and 3. If I try $e^t$ it violates 3. If I try a function involving $t^5$sin(1/t), it violates condition 2. Please help me with a correct example. Thank you.

答:
How can $$G(t)=\int_0^t[1-s^2] ds = t(1-\frac13t^2)$$ be differentiable at both $1$ and $-1$ and yield $g(s)$? As a hobbyist, I have to work it out explicitly.
Since $G(-1)=-\frac23$ and $G(1)=\frac23$, we define
$$G(t)=\begin{cases}
t(1-\frac13t^2), &|t|\le 1,\\
-\frac23, &t<1,\\
\frac23, &t>1
\end{cases}$$
Piecewise differentiation yields $g(t)=g(s)$.
Then differentiating $$f(x)=\begin{cases}
\frac{x^2}{12}(6-{x^2}), &|x|\le 1,\\
-\frac23x, &x<1,\\
\frac23x, &x>1
\end{cases}$$
yields $G(x)=G(t)$. Thanks!


问:
prove that $3x - x^3 < 2 \cdot \sin(\pi/2 \cdot x)$ $\forall x \in (0, 1)$
I faced problems of such kind before and usually I would just chech that on the ends of segment values of compared functions are equal and then find their derivatives and make sure that derivative of one function is always greater on some interval or segment. However, in this case it just didn't work. I found derivative but on some interval $A \subset (0, 1)$ one is greater than the other and on another interval $B \subset (0, 1)$ everything is vice versa.

Do you have any other meaningful approaches to this problem?

答:
Let's prove another inequality, first: $\sin$ is a concave function in $[0,\pi]$, so we have (by Jensen's inequality) $$(1-x)\,\sin0+x\,\sin\frac\pi6=\frac{x}2\le\sin\left((1-x)\cdot0+x\cdot\frac\pi6\right)=\sin\frac\pi6x$$ for $x\in[0,1]$, with equality only in the endpoints of the interval. Moreover, both sides of the inequality take values in $[0,1/2]$. In that interval, the function $g(y)=3y-4y^3$ is monotone increasing (derivative $3(1-4y^2)\ge0$), and the well-known triplication formula says $$g(\sin\phi)=\sin3\phi,$$ so we have
$$g(x/2)\le\sin\frac\pi2x.$$ Multiplying both sides by $2$, we arrive at $$3x-x^3\le2\sin\frac\pi2x$$ on $[0,1]$, and as it's clear from the derivation, there's equality only at the endpoints of the interval.


问:
Argand Diagram - Quadrants help
I forgot the rules of adding angles when it comes to argand diagrams.

In the first quadrant, you add 90 degrees to whatever angle you get, what about Q2 Q3 Q4 ?

This picture will explain what i mean :

notice how 18.43 was added with 180 .. can someone list the rules for the other quadrants as well ?

what if i have -3i -3j for example ?

Thanks

答: The decision to add 180 degrees to the inverse tangent is based on the sign of the denominator "inside" the inverse tangent.  Add 180 degrees only if denominator < 0.  Also, for this idea to work, use angles (in degrees) in (-180, 180] instead of the more typical [0, 360).   </p>

Title: Argand Diagram - Quadrants help


问:
Showing the orthogonal subspace to the Range of A is the null space of the complex conjugate of A
I have matrices $A \in \mathbb{C}^{m \times n}$

My goal is to show that $R^{\perp}(A) = N(A')$.

Conceptually I get that the orthogonal space to the range of the matrix is going to be the null space because otherwise it'd be in the row space but I'm having a hard time expressing it in math.

I haven't done too much complex matrix work so I'm not sure what pitfalls I need to be wary of

答: The column space (range) of $A$ is the row space of $A^T$. Thus $v\in R^\perp(A)\iff\forall i,v^*r_i=[0]\iff v^T\bar{r_i}=0$ where $r_i$ is a row vector of $A^T$. This gives $\bar{A^T}v=0\iff v\in\ker(\bar{A^T})$.


问:
Relations between Bochner spaces $L^p(0,T;L^q)$
Let $(X,\|\cdot\|_X)$ be a Banach space. The function $u=u(t,x)$ belongs to Bochner space $L^p(0,T;X)$ if the norm
$$ \|u\|_{L^p(0,T;X)} = \left(\int_0^T \|u(t,\cdot)\|_X^p \mathrm{d}t\right)^{1/p} \text{ for } p<\infty, $$
$$ \|u\|_{L^\infty(0,T;X)} = \mathrm{ess}\sup_{t\in[0,T]} \|u(t,\cdot)\|_X $$
is finite.

Now let $X=L^q(\Omega)$ for bounded $\Omega\subset\mathbb{R}^n$. I was wondering, if there are some embeddings between spaces $L^{p_1}(0,T;L^{q_1}(\Omega))$ and $L^{p_2}(0,T;L^{q_2}(\Omega))$, depending on $p_i$ and $q_i$.

From the interpolation in Lebesgue spaces we have this result, moreover from the boundedness of $[0,T]$ and $\Omega$ we obviously have
$$L^p(0,T;L^{q_1}(\Omega)) \subset L^p(0,T;L^{q_2}(\Omega)) \text{ for } q_1>q_2$$
and
$$ L^{p_1}(0,T;L^q(\Omega)) \subset L^{p_2}(0,T;L^q(\Omega)) \textrm{ for } p_1>p_2. $$

However, is it possible that
$$ L^{p_1}(0,T;L^{q_1}(\Omega))\subset L^{p_2}(0,T;L^{q_2}(\Omega)) $$
for some $p_1>p_2$ and $q_1<q_2$?

In other words, can we compensate the integrability over one variable by the higher integrability of the other? In particular I am interested in the inclusion of $L^2(0,T;L^2(\Omega))$ and $L^1(0,T;L^\infty(\Omega))$.

EDIT: it seems that there's no relation whatsoever. In particular if $\Omega=[0,T]=[0,1]$ there are simple examples for no such relation between $L^2(0,T;L^2(\Omega))$ and $L^1(0,T;L^\infty(\Omega))$: take $f(t,x)=\frac{x}{\sqrt{t}}$ and $g(t,x)=\frac{t}{x^{1/4}}$ - one of them is in $L^1(0,T;L^\infty)$ but not in $L^2$ and the second one vice versa. I believe these examples can be easily modified to other cases.

I'm not deleting this post in case somebody would have a similar problem in the future.

答:
No, this is not true, as long as $L^{q_1}(\Omega)\setminus L^{q_2}(\Omega)\neq\emptyset$.

You can see it considering a function $f\in L^{q_1}(\Omega)\setminus L^{q_2}(\Omega)$ and setting $u(t,x)=f(x)$.


答:
You can just set for instante $x_0=0$ and $x_1=1$ and compute $A_0,A_1$ such that $$
\int_0^1 e^x dx = A_0 e^0 + A_1 e^1, \quad \int_0^1 \cos \frac{\pi x}{2}\, dx = A_0 \cos \frac{\pi \cdot 0}{2} + A_1 \cos \frac{\pi \cdot 1}{2}.
$$

This way you obtain the quadrature
$$
Q(f)=\frac{2}{\pi} f(0) + \frac{e \pi -2-\pi }{e \pi } f(1).
$$

If you require the formula to be exact to an extended set of functions, you'll be able to remove the initial arbitrarity in the choice of $x_0,x_1$. For instance, if you take $x_0 \approx 0.221611$, $x_1 \approx 0.79518$, $A_0 \approx 0.513634$, $A_1 \approx 0.486366$, the formula will be exact for functions of the form
$$
f(x)= a e^x + b \cos \frac{\pi x}{2} + c \sin \frac{\pi x}{2} + d.
$$

Title: Find $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for $f(x) = a e^x + b \cos(\pi x /2)$?

Body:
I am reading through some exercises of the chapter I am reading about numerical integration. I stumbled upon the following question:

Find the formula $\int_0^1 f(x) \,dx \simeq A_0 f(x_0) + A_1 f(x_1)$ that is exact for functions of the form $f(x) = a e^x + b \cos(\pi x/2)$

I know that for any polynomial of degree $\leq n$ its integral can be exactly estimated (that is, is equal to) by the formula
$$
\int_a^b f(x) \,dx = \sum_{i=0}^{n} A_i f(x_i)
$$
where $A_i = \int_a^b \ell (x)\, dx$ (Lagrange fundamental polynomial) for $n+1$ distinct nodes $x_i \quad 0 \leq i \leq n$.

However, the $f$ in the exercise is clearly not a polynomial. So no idea how to find such a formula that is exact as with numerical integration for polynomials. Can you help?


问:
Complex Analysis:Show that $a \in \mathbb{Z}$ and that there is $\theta \in \mathbb{R}$ so that $f(z)=e^{i\theta}z^{a}$.
Let $f$ be analytical on ann $A(0;r,R)$ and suppose there is $a \in \mathbb{R}$ such that $|f(z)|=|z|^{a}$, $\forall z \in A$. Show that $a \in \mathbb{Z}$ and that there is $\theta \in \mathbb{R}$ so that $f(z)=e^{i\theta}z^{a}$.

I have the impression that using Schwarz's lemma forms $ f(z)=e^{i\theta}z^{a} $, but I cannot develop it. I can't think of results that prove  $  a \in \mathbb{Z} $. Could you help me with ideas?

答: The principal branch of $g(z)=z^a$ is holomorphic in $\mathbb C\setminus (-\infty,0].$ We then have both $f,g$ holomorphic in $U=A(r,R)\setminus (-\infty,0],$ an open connected set. Note both $f,g$ are nonzero in $U.$ Thus in $U$ we have $f/g$ holomorphic, with $|f/g|=1.$ It follows that $(f/g)(U)$ is not open. By the open mapping theorem, $f/g = C$ in $U,$ where $C$ is a constant of modulus $1.$ Therefore $f(z)/C = z^a$ in $U.$ It follows that $z^a$ extends to be holomorphic in $A(r,R).$ But we know for that to happen $a$ must be an integer.


问:
Albers projection is area-preserving
I read that the Albers projection (https://en.wikipedia.org/wiki/Albers_projection) is area-preserving. The projection is:

$$x=\frac{\sqrt{C-2n\sin \phi}\sin n \lambda}{n}$$
$$y=\rho_0 - \frac{\sqrt{C-2n\sin \phi}\cos n \lambda}{n}$$

So the Jacobean matrix is:

$$\dfrac{d(x, y)}{d(\phi, \lambda)} = \begin{bmatrix}
        \frac{-\sin n \lambda \cos \phi}{\sqrt{C-2n\sin \phi}} & \frac{\cos n \lambda \cos \phi}{\sqrt{C-2n\sin \phi}} \\
        \sqrt{C-2n\sin \phi}\cos n \lambda & \sqrt{C-2n\sin \phi}\sin n \lambda \\
        \end{bmatrix}$$

And the Jacobean has determinant

$$-\sin^2 n \lambda \cos \phi - \cos^2 n \lambda \cos \phi = -\cos \phi$$

However, for the projection to be area-preserving, it must have determinant $\pm1$. So where have I gone wrong with my calculations?

答:
For the projection to be area-preserving it does not have to have determinant ±1.

See this response to a similar question is a very good explanation https://math.stackexchange.com/q/33184.

Basically,you have to consider the "area element" of the coordinate system.
If you compare the jacobean you have calculated to the surface area element in spherical coordinates - you will find they are the same as long as you align your axis.
( Surface Element in Spherical Coordinates contains proofs that the area element is $^2 sin()$ and in our case r=1)

I hope this helps

Title: Albers projection is area-preserving


问:
Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$
For $n$ normal number, the book solved it like this:

If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.

If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .

But I didn't solve it like this.

My logic is that since 3 is a prime number then $(2n+1)2^{4n+5} = 3 \pmod{7}$ means either $2^{4n+5} = 3\pmod{7}$ and $ 2n + 1 = 1\pmod{7} $ or the other way around.

But since there is no $n$ value that can make $2^{4n+5} = 3\pmod{7}$ then it means:

$ 2n + 1 = 3 \pmod{7}$ and by calculating we find in the end $n = 7k' + 1$

And $ 2^{4n+5} = 1\pmod{7}$ and by calculating we find that $n = 3k''$

So it means that $n = 7k' + 1$ AND $n = 3k''$

So, I know that my solution is faulty but can anyone explains to me the right solution or point out where I went wrong?

答:
Problem: Find $n$ where $(2n+1)2^{4n+5} \equiv 3 \bmod{7}$

My approach:

Starting at $n=0,1,2...$, all $\bmod 7$, knowing that $2^3\equiv 1$, we can see that  $2^{4n+5}\equiv 2^{n+2}$ cycles through $\{4, 1, 2\}$. Since $4^{-1}\equiv 2$, the implied $3$-cycle of $required$ values for $(2n+1)$ to satisfy is thus $\{2\cdot 3, 1\cdot 3, 4\cdot 3\} = \{\color{red}6,\color{green}3,\color{blue}5\}$, each of which can be found at one point in the $7$-cycle of its values $\{1, \color{green}3, \color{blue}5, 0, 2, 4, \color{red}6\}$.

So we need one of $n \underset{(3,7)}{\equiv} (0,6),(1,1),(2,2)$ which can be worked through the Chinese Remainder Theorem to give $n \equiv \{6,1,2\} \bmod 21$.

Your approach fails because of your assumption that only $1\times 3 = 3\times1 \equiv 3 \bmod 7$, whereas also $2\times 5 \equiv 4\times 6 \equiv 3 \bmod 7$; a prime modulus implies all coprime numbers have a multiplicative inverse.

The book's approach is incomprehensible, and $n \equiv \{\color{red}9,1,2\} \bmod 21$ is wrong.


问:
System of congruences with $x$ and $y$
Solve the system of congruence relations:

$$5x+17y\equiv 1 \pmod{24}$$

$$6x+10y\equiv 2\pmod{24}$$

Can someone help me? I usually use Chinese theorem, but here that is not possible.

答:
Subtract both equations to make $x$ appear.

$\begin{cases}5x+17y\equiv 1\pmod{24}\\6x+10y\equiv 2\pmod{24}\end{cases}\iff
\begin{cases}5x+17y\equiv 1\pmod{24}\\x\equiv 7y+1\pmod{24}\end{cases}$

We can now substitute in the first equation to solve for $y$.

$5x+17y\equiv 35y+5+17y\equiv 4y+5\equiv 1\pmod{24}\iff 4y\equiv 20\pmod{24}$

Now all coefficients are divisible by $4$ including the modulo, therefore we can simplify it to $$y\equiv 5\pmod 6$$

Note: do not make the mistake to go $\, y\equiv 5\pmod{24}\, $, this is not true, you cannot divide by $4$ because $\gcd(24,4)\neq 1$, but you can reduce the equation to a lower modulo when this one is also divisible.

We continue by setting $\, y=6n+5\, $ for some integer $n$.

$\begin{cases}y\equiv 6n+5\pmod{24}\\x\equiv 7y+1\pmod{24}\end{cases}\iff
\begin{cases}y\equiv 6n+5\pmod{24}\\x\equiv 42n+36\pmod{24}\end{cases}\iff
\begin{cases}y\equiv 6n+5\pmod{24}\\x\equiv 18n+12\pmod{24}\end{cases}$

You can verify the the solutions are cycling for $n\ge4$, you get for $n=0,1,2,3$

$$\{(12,5),(6,11),(0,17),(18,23)\}$$

Title: System of congruences with $x$ and $y$


问:
Flipping coins- percentages of heads vs tails
If I flip a coin multiple times and count the number of time it fell on heads and the number of times it fell on tails and keep a track of them. In how many flips on average will the delta between percentage of heads and percentage of tails will be less than 0.1%?</p>

答:
If I flip a coin "$2$" times... On average the delta between percentage of heads and percentage of tails will be less than $0.1 \%$. Thank you.

As It can be $\{HH \}$ or $ \{TT \}$ or $\{HT\}$ or $\{TH \}$

I went through a lot of sleeplessness because of this question randomly popping up somewhere I don't even remember now.


问:
Operator norm and eigenvalues
I am trying to understand if bounding the operator norm (or L2 induced norm) of the difference of two matrices can be used to bound the eigenvalues (specifically, the min eigenvalue) of matrix products.

Formally, given full-rank matrices $Q \in \mathbb{R}^{n \times d}$ and $\hat{Q} \in \mathbb{R}^{n \times d}$. If eigenvalues of $Q^TQ$ are known (say $\lambda_1, \cdots, \lambda_d$) and that $||Q - \hat{Q}||_{2\rightarrow 2} = \epsilon$ (L2 induced norm or operator norm.) Can we say anything about the eigenvalues of the matrix $\hat{Q}^TQ$ in relation to $\lambda_1, \cdots, \lambda_n$?

My use case is that I know $\min_{1\leq i\leq d} \lambda_i > 0$ since $Q^TQ$ is PD. I am wondering if I can say anything about whether the minimum eigenvalue of $\hat{Q}^TQ$ will be negative. I understand that $\hat{Q}^TQ$ need not be symmetric and hence, the eigenvalues could be complex. From numerical experiments (using a symmetric $Q$ and $\hat{Q} = Q + \epsilon I$ with varying $\epsilon$ ensuring eigenvalues are always real) , I see that for sufficiently large $\epsilon$, the minimum eigenvalue becomes negative but I am curious if we can establish a relationship between $\epsilon$ and the minimum eigenvalue of $\hat{Q}^TQ$.

答: One way would be to define the perturbation $\epsilon V = (\hat Q - Q)^T Q$, so that you are interested in finding eigenvalues of $Q^T Q + \epsilon V$. For small $\epsilon$ you can approximate eigenvalues and eigenvectors using perturbation theory, it is a standard tool in quantum mechanics (https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics), https://en.wikipedia.org/wiki/Eigenvalue_perturbation#:~:text=In%20mathematics%2C%20an%20eigenvalue%20perturbation,to%20changes%20in%20the%20system. )

Title: Operator norm and eigenvalues


问:
Show this set can't be precompact via counterexample
Let $I=[0,1]$ and $B=\{u\in\mathcal{C}^{1}(I,\mathbb{R})\mid \|u'\|_{L^{2}(I)}\leq1\}$, then it can be easily shown $B$ is an equicontinous family of functions. Now, how  do we find a counterexample, i.e. a sequence $u_{n}$ which doesn't have converging subsequences, thus deducing it is not relatively compact?

答: I'm not sure about this, but can't you just take the sequence of constant functions $u_n\equiv n $?


问:
Differentiation - Limits Equal (Possibly MVT, Rolle's or L'Hopital)
Got a quick question from a past exam paper.

If $f:R \rightarrow R$ is differentiable, and $f$ is such that $\lim_{ x\rightarrow \infty } f(x)=\lim_{ x \rightarrow -\infty} f(x)=0$ and there is a point $c$ such that $f(c)>0$, then prove that $f$ has is bounded above, and attains its maximum at a point $x_0$ where $f'(x_0)=0$.

P.S - Sorry about the formatting, I'm a bit new to this :|

答:
The idea is that as long as you consider a sufficiently large compact set you can apply Weierstrass Theorem.

the fact that the function goes to $0$ with $|x| \to \infty$ implies that for each $\varepsilon > 0$ there exists $M > 0$ such that $|f(x)| < \varepsilon$ for all $x$ such that $|x| > M$.

If you now take, for example, $\varepsilon = f(c)$ you are allowed to look for the maximum in $[-M,M]$, where Weierstrass Theorem holds, proving boundedness and the existence of a maximum.

The fact that $f'(x_0) = 0$ is a classical result, I guess you do not have doubts about this.


问:
Probability of winning specified amount of money
We play a game in which we toss a coin - if we have head we get 1\$ otherwise, we lose 1\$. Let's say we do it 50000 times. What's the probability of losing at least 10\$?
I started by calculating the excepted value of one time:
$$
x_{1} = 1
\\ x_{2} = -1
\\EX= \frac{1}{2}\cdot1 - \frac{1}{2}\cdot 1 = 0
$$
So the excepted value in the whole game is $0$.
Now I don't really know what to do. Do I need to use Chebyshev's inequality? Is it OK if $EX=0$?

答:
The key word here is the binomial distribution. The standard presentation is that if you flip a $p$-coin $N$ times, and count the number $X$ of heads, then the distribution is
$$\mathbb P(X=k)=\binom{N}{k}p^k(1-p)^{N-k}.$$
We write $X\sim\operatorname{Bin}(N,p)$ to indicatte that $X$ has this distribution.

Your profit $X$ after $50000$ tosses isn't quite $\operatorname{Bin}(50000,1/2)$ though, since we subtract when the coin comes up tails. If $Y\sim\operatorname{Bin}(50000,1/2)$, then it should be clear that $X=2Y-50000$.

So then $X\leq-10\iff Y\leq24995$. Then
$$\mathbb P(Y\leq24995)=\sum_{k=0}^{24995}\mathbb P(Y=k)=\sum_{k=0}^{24995}\binom{50000}{k}\cdot\left(\frac{1}{2}\right)^{50000}.$$
This sum is obviously not tractable by hand. Most scientific calculators would have some kind of function that computes the CDF for a binomial random variable. You should get $0.484$ as the probability, which seems reasonable -- just a smidgen under half.


问:
Proving that there exists a solution to $f\bigl(x+\frac1n\bigr)-f(x)=0$

Given $f:[0,1]\to\mathbb R$ which is continuous, $f(0)=f(1)$, and $n\in\mathbb N^*$ (doesn't include $0$), show that $f\bigl(x+\frac1n\bigr)-f(x)=0$ has a solution.

This was shown to me by an acquaintance who told me that if we consider $h(x)=f\bigl(x+\frac1n\bigr)-f(x)$, then the fact that $\sum_{k=0}^{n-1}f\bigl(\frac kn+\frac1n\bigr)-f(\frac kn)=0$ somehow tells us that there are two opposite terms in that sum, and, as such, we can apply the IVT to $h(x)$.

Clearly the sum is null because it is in the form $f(1) + (f(a)-f(a)\text{ same argument a})+(f(b)-f(b)\text{ same argument b})+...-f(0)$ and $f(1)=f(0)$, not because there somehow exists two different in argument terms that cancel $f(a)+f(b)=0$.

In any case, we require a $h(a)$ and a $h(b)$ such that $h(a)h(b)<0$, not in terms of $f$.

But, from what I have understood, this is the correction given by their professor, so maybe I am missing something?

EDIT: Ignore the following nonsense...

I thought of going about it by contradiction, but the use of a limit leaves me undecided as to whether this counts as a solution.

The negative of the proposition is $\exists n\in\mathbb N^*\,|\,\forall x\in\mathbb R,\,f\bigl(x+\frac1n\bigr)-f(x)\neq0$, i.e $f\bigl(x+\frac1n\bigr)<f(x)$ or $>$. In both cases, if we take $n\to\infty$, we get $f(x)<f(x)$, which is impossible.

Any comments? Thank you!

答:
$$\sum_{k=0}^{n-1} \left(f\left(\frac{k+1}{n}\right) - f\left(\frac{k}{n}\right)\right) = f(1) - f(0) = 0$$ (it's a telescoping sum).  If any $f\left(\frac{k+1}{n}\right) - f\left(\frac{k}{n}\right) = 0$ you're done.  They can't all be positive or the sum would be positive, and similarly they can't all be negative.  So if none are $0$, there are
two consecutive ones that have opposite sign, say $f((k+1)/n) - f(k/n) < 0$ and $f((k+2)/n) - f((k+1)/n) > 0$.  Now apply the Intermediate Value Theorem to $h(x) = f(x+1/n) - f(x)$ on $[k/n, (k+1)/n]$.


问:
Trigonometric Identities, are they injective or well-defined?
Proving trignometric indentities is quite simple, when you know certain properties of $e^{ix}$. Therefore identites such as $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$ and $\sin(x+y)=...$ by simply using Eulers Identity and comparing Real Parts to each other and Imaginary Part to each other!

My question: What guarantees me to say that if the Imaginary/Real Part of $z$ is equal to Imaginary/Real Part of $w$ then they are actually eqaul. To be more precise, is it defined as a function ? Then this function has to be injective. A quick example plotting the function $x^2$ on $2$and$-2$ will give me the number 4 but this doesnt tell me that $-2=2$ since x^2 is not injective. Thanks

答:
You ask

What guarantees me to say that if the Imaginary/Real Part of $z$ is
equal to Imaginary/Real Part of $w$ then they are actually equal?

That is essentially the definition of complex numbers as pairs of real numbers. It has nothing to do with trigonometric identities, or the injectivity of any functions of real or complex variables.


问:
understanding the conditional expectation
Given a sigma field $\mathfrak{g}$ and a random variable $X$. A conditional expectation of $X$ is defined as $E(X|\mathfrak{g})$ and this conditional expectation is $\mathfrak{g}$ measurable.Now define an indicator function $\mathfrak{1}_G$ where $G\in\mathfrak{g}$. Consider the expression $E(X|\mathfrak{g})\cdot1_G$. If I have an element $l\in\mathfrak{g}$ and $l\not\subseteq G$, then the  expression $(E(X|\mathfrak{g})\cdot1_G)(l)$ means  $E(X|l)\cdot0$. Am I correct on this?

edit:
changed $E(X|\mathfrak{g})\cdot1_G(l)$ to $(E(X|\mathfrak{g})\cdot1_G)(l)$

clarifying:

$(E(X|\mathfrak{g})\cdot1_G)(l)$

the $\cdot$ notation here means multiplication

答:
Since $E(X \mid \mathfrak{g})$ is a $\mathfrak{g}$-random variable, I think it would be better to write
$$(E(X \mid \mathfrak{g}) \cdot 1_G)(l) = E(X \mid \mathfrak{g})(l) \cdot 0.$$
As far as I know, something like "$E(X \mid l)$" is not defined.


问:
Is it surjective because of $K \subset H$?
Let $G$ be a group and $H,K$ normal subgroups of $G$ s.t. $K \subset H$. If I define the following map: $\phi:G /K \to G/H, gK \mapsto gH$ then the claim is that it is surjective. Is it because we have that $K \subset H$ and hence $G /H \subset G/K$?

答:
More precisely, this is because of the $3^{ rd }$ isomorphism theorem: $H/K$ is a normal subgroup of $G/K$, and
$$(G/K)\big/(H/K)\simeq G/H.$$

What can be said of $G/H$ as compared to $G/K$ is that every $H$-coset is a union of $K$-cosets.


问:
About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime
The number $30$ has a curious property:

All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers!

I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not find any.

Are there any such numbers $\gt 30$?

答:
Let $n>30$ be such a number. Observe that $p^2<n$ implies $p|n$.
Thus $2|n$, $3|n$, $5|n$ follow directly from $n>25>9>4$.
Thus $n$ is a multiple of $30$ and $n>30$, which implies $n\ge 60>49$ and hence $7|n$.
Up to now, $n$ is divisible by the four smallest primes.

Lemma. The product of four consecutive primes is greater than the square of the next prime:
$$\tag1p_kp_{k+1}p_{k+2}p_{k+3}> p_{p+4}^2.$$

Proof:
Using Bertrand's postulate, we have $p_{k+2}>\frac{p_{k+3}}2$, and $p_{k+4}\le 2p_{k+2}$, hence $(1)$ follows from
$$p_kp_{k+1}p_{k+2}p_{k+3}>3\cdot 5\cdot \frac{p_{k+3}}{2}p_{k+3}> 4p_{k+3}^2>p_{k+4}^2$$
at least if $p_k\ge 3$. The case $p_k=2$ is verified directly: $2\cdot3\cdot 5\cdot 7>11^2$. $_\square$

Now if $n$ is divisible by  four consecutive primes $p_k,\ldots,p_{k+3}$, we conclude from $(1)$ that $n$ is also divisible by the next prime, because otherwise $p_{k+4}^2<n$ is relatively prime to $n$. Hence $n$ is divisible by the four consecutive primes $p_{k+1},\ldots,p_{k+4}$.
By induction, we conclude that $n$ is divisible by all primes, but that is of course absurd.
Hence there is no such number beyond $30$.


答:
The fact that the application is surjective is trivial, since taken $gH\in G/H$, then  $\phi(gK)=gH$. The tricky part is proving that this is actually a well posed application. And this can be done as follows:
$$aK=bK\Rightarrow ab^{-1}\in K\subset H \Rightarrow   ab^{-1}\in H \Rightarrow aH=bH \Rightarrow \phi(aK)=\phi(bK)$$
Now you have just to prove that is an homomorphism (trivial), find its kernel and then apply first isomorphism theorem.

Title: Is it surjective because of $K \subset H$?

Body: Let $G$ be a group and $H,K$ normal subgroups of $G$ s.t. $K \subset H$. If I define the following map: $\phi:G /K \to G/H, gK \mapsto gH$ then the claim is that it is surjective. Is it because we have that $K \subset H$ and hence $G /H \subset G/K$?


答:
We model the coin tosses as a binomial distribution.

Let X be the number of heads, then
$$\text{X} \sim Bn( 50000, p=0.5)$$
Then we know the net value of the whole game is
$$\text{net value} = 1(\text{X}) - 1(50000-\text{X})=2\text{X} - 50000$$
To know the probability of loosing at least \$10:
$$\mathbf{P}(\text{net value} \leq -10) = \mathbf{P}(2\text{X} - 50000 \leq -10)=
\mathbf{P}(\text{X} \leq 24995)  ≈ 0.483947$$
Various methods can be used to calculate this final value. I chose to use wolfram alpha as a calculator but for a more laborious method  you could approximate to the normal distribution and then use a table.

Title: Probability of winning specified amount of money

Body:
We play a game in which we toss a coin - if we have head we get 1\$ otherwise, we lose 1\$. Let's say we do it 50000 times. What's the probability of losing at least 10\$?
I started by calculating the excepted value of one time:
$$
x_{1} = 1
\\ x_{2} = -1
\\EX= \frac{1}{2}\cdot1 - \frac{1}{2}\cdot 1 = 0
$$
So the excepted value in the whole game is $0$.
Now I don't really know what to do. Do I need to use Chebyshev's inequality? Is it OK if $EX=0$?


问:
Manifolds with boundary
Let $M$ be the left half space $\{(x^1,\ldots, x^n) \in \mathbb R^n\mid x^1 \le 0\},$ with orientation form  $dx^1 \wedge \cdots \wedge dx^n.$ Show that an orientation form for the boundary orientation on
$\partial M=\{(0,x^2, \ldots , x^n) \in\mathbb R^n\}$ is $dx^2\wedge \cdots \wedge  dx^n.$

答:
It is enough to show that $\omega=dx^2\wedge..\wedge dx^n$ in not degenerated on the boundary.
Let $x=(0,x^2,..,x^n)$ an element of the boundary. $e_i, i\geq 1$ the vector whose only non zero coordinate is its $i$ coordinate which is one is tangent to the boundary at $x$, $\omega_x(e_2,..,e_n)=1$.

Title: Manifolds with boundary


问:
$f$ is Lipschitz if and only if there exists $L\geq0$ such that $|f'(x)|\leq L$
Let $f:[a,b]\to\mathbb{R}$ be absolutely continuous.

Prove that $f$ is Lipschitz if and only if there exists $L\geq0$ and a set $E\subset[a,b]$ such that, $m(E)=0$ and $f$ is differentiable at each $x\in[a,b]\setminus E\quad$  with $|f'(x)|\leq L$.

My attempt:

For $\Rightarrow$ direction:

Since $f$ is absolutely continuous on $[a,b]$, it is differentiable a.e on $[a,b]$

And notice:

$\begin{align}
|f'(x)|&=\left|\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\\
&\leq\lim\limits_{h\to0}\left|\frac{f(x+h)-f(x)}{h}\right|\\
&\leq \lim\limits_{h\to0}\left| \frac{L|(x+h)-x|}{h}\right|\\
&=L \quad\text{ Where, $L$ is the Lipschitz constant}
\end{align}$

Is that correct?

And for the other side of the implication I would really appreciate your help

答:
Important: this proof only works when $f$ is differentiable in $\mathbb{R}$.

The reverse implication can easily be seen with the Mean value theorem. Assuming $|f'(x)|\leq L$ (being $L\geq 0$) for any $x\in\mathbb{R}$, using the mean value theorem we know that, given $x,y\in\mathbb{R}$ (we will assume $x<y$), there exists $c\in(x,y)$ such that
$$\frac{f(x)-f(y)}{x-y}=f'(c).$$
Using absolute values, and using that $|f'|\leq L$ we get that
$$\frac{|f(x)-f(y)|}{|x-y|}\leq |f'(c)|\leq L,$$
so we end up with
$$\boxed{|f(x)-f(y)|\leq L\cdot|x-y|}$$
and we conclude $f$ is Lipschitz


问:
Litterature on Dynamic graph theory
I was wondering if anyone knows any good articles or papers or books on graph theory that deals with changing graphs and not just static ones.

So far I've only found qualitative descriptions of developing a dynamic graph theory and no actual derivations.

Particularly I'm wondering about dynamical systems represented as graphs.

Any help would be appreciated and thanks in advanced.

答:
Since this question has been posted, there have been many works in this area. Many of them are quite empirical, but let me mention the following more mathematical works:

a nice algorithmic-oriented introduction to the field, that surveys various approaches and sees temporal graphs mostly as sequences of graphs over time.

some works on computational complexity of time-varying graphs, showing that some dynamics families (similar to graph families) may be defined that induce different computational complexity classes. They study many important algorithmic problem, in particular (but not only) reachability and path-related ones.

some algebraic approaches defining semi-rings on temporal networks, which make it easy to compute their key (temporal and structural) properties, without resorting to the usual time-slicing approach.

our work on Stream Graphs and Link Streams focusing on fine-grained dynamics of links and nodes. Its key feature is to see graphs as special cases of stream graphs, when there is no dynamics. It extends the basics of graph theory (like density, paths, or cliques, e.g.) in a way consistent with graph concepts.

There are plenty of other references that would be worth mentioning, of course. But I think these four already provide a representative panel of modern approaches to dynamic graphs, with mathematical grounds.

Title: Litterature on Dynamic graph theory


问:
Proof that f = {(x,n) e R x Z | n =< x < n + 1} is a function.
I have to prove that f = {(x,n) e R x Z | n =< x < n + 1} a function is.

I started with saying (x,n1) and (x,n2) are both elements of the function. This means n1 =< x < n1 + 1 (1) and n2 =< x < n2 + 1 (2). I substracted n1 from (1) and n2 from (2) this gives 0 =< x - n1 < 1 (3) and 0 =< x - n2 < 1 (4) when I substract (3) from (4), this gives 0 =< n1 - n2 < 0 (5). This is impossible since a number can't be both greater than or equal to 0 and smaller than 0. I wanted to work towards n1 - n2 = 0. Can someone help me find the mistake I made?

Thanks!

答:
We cannot subtract inequalities!

Consider for example $1<2,\ 0<2$, if you subtract them, you would receive $1<0$ which is false.

It's not working because addition of inequalities does work, and subtraction is $a-b=a+(-b)$, while multiplication by $-1$ reverses the inequality to be subtracted.

If we carefully add appropriate forms of the inequalities instead:
$$0\le (x-n_1)<1\\ -1<(n_2-x)\le 0$$
then we do get $-1<(n_2-n_1)<1$, implying $n_2-n_1=0$ as these are integers.


问:
Division and linearity condition

A rational number a/b with (a, b) = 1 is called a reduced fraction. If
the sum of two reduced fractions is an integer, say (a/b) + (c/d) = n,
prove that |b| = |d|.

I am having trouble to understand the solution i found of this exercise:

Note that (a/b) + (c/d) = (ad + bc)/(bd). This is an integer implies
bd|(ad + bc), which by linearity implies d|bc and b|ad. However, since
(a, b) = (c, d) = 1, this implies that d|b and b|d. Hence |b| = |d|

I don't understand how does the linearity apply here, i mean, it does not make sense to me to use it here. The only things i could conclude is that:

$ ad = r$ mod $(bd)$

$bc = bd - r$ mod $(bd)$

Let me know if i am wrong, but linearity is not this?: if $a|m, a|n => a|(k_{1}n + k_{1}m)$

I mean, the solution used in the contrary way, the right side implying the left side.

But, even if what the autor of the solutions did is right, why does " bd|(ad + bc), which by linearity implies d|bc and b|ad"?

答:
$bd|ad+bc\implies d|\color{blue}{ad+bc}.$

Since $d|\color{green}{ad}$, it follows that the linear combination $(\color{blue}{ad+bc})-\color{green}{ad}=bc$ is also divisible by $d$.

A similar argument shows that $b|ad$.

Title: Division and linearity condition


问:
partial derivative of cost function using chain rule
I need to compute the partial derivative of the cost function of a neural network, with respect to $z_1$ and $z_2$. The cost function is defined below:

In my attempt to do this:

I calculated both partial derivatives to be essentially the same, I'm not sure this is correct? if its not, could someone point me to where ive gone wrong. thanks - i apolgise for using pictures!

答:
$$c=\ln(e^{z_1}+e^{z_2})-z_y$$
$$\frac{\partial c}{\partial z_1}=\frac{e^{z_1}}{e^{z_1}+e^{z_2}}$$
$$\frac{\partial c}{\partial z_2}=\frac{e^{z_2}}{e^{z_1}+e^{z_2}}$$
remember with partial derivatives: it is the same as taking a derviative but you treat all other varables as constants so for each one we can say:
$$c=\ln(e^{z_1}+a_1)-a_2$$
$$\frac{dc}{dz_1}=\frac{e^{z_1}}{e^{z_1}+a_1}$$
now substitute in your other variables as the "constants"

The mistake you made in each step was the following:
$$\frac{\partial u_2}{\partial z_1}=\frac{\partial u_2}{\partial z_2}=0$$
Other than this your answer is correct :)


答:
If $f$ is constant, then we are done.  So assume $f$ is nonconstant and, without loss of generality, assume
$$
f(0) = f(1) = 0.
$$
This, together with the nonconstancy of $f$, implies that $f$ either attains a strictly positive maximum at a point interior to $[0, 1]$ or attains a strictly negative minimum at a point interior to $[0, 1]$.  For definiteness, assume $f$ attains a strictly positive maximum at a point $\overline{x}$ interior to $[0, 1]$.

Introduce the function
$$
g(c) = (\max f) - c, \quad c \geq 0.
$$
Then
$$
g(\overline{x}) = 0.
$$
For $c$ close enough to $\max f$, but strictly smaller, the equation
$$
g(c) = 0
$$
had at least two solutions (we prove this by applying the Intermediate Value Theorem twice, once on $[0, \overline{x}]$ and once on $[\overline{x}, 1]$).  Among these solutions, let $x_{c,-} < \overline{x} < x_{c,+}$ be the closest two to $\overline{x}$.  Let $h(c)$ be the distance between these two.  It has a removable discontinuity at $c = \overline{x}$, which we fill by putting
$$
h(\overline{x}) = 0,
$$
extending $h$ to being a continuous function for $c \in [0, \mbox{some positive }\epsilon]$.

Picking a positive integer $n$ large enough to have
$$
{1 \over n} < \epsilon,
$$
we see that there exists a $c$ such that
$$
h(c) =  {1 \over n }.
$$

P.S. Realizing I may have misread the quantifier for $n$, hence the Community wiki.

Title: Proving that there exists a solution to $f\bigl(x+\frac1n\bigr)-f(x)=0$

Body:
Given $f:[0,1]\to\mathbb R$ which is continuous, $f(0)=f(1)$, and $n\in\mathbb N^*$ (doesn't include $0$), show that $f\bigl(x+\frac1n\bigr)-f(x)=0$ has a solution.

This was shown to me by an acquaintance who told me that if we consider $h(x)=f\bigl(x+\frac1n\bigr)-f(x)$, then the fact that $\sum_{k=0}^{n-1}f\bigl(\frac kn+\frac1n\bigr)-f(\frac kn)=0$ somehow tells us that there are two opposite terms in that sum, and, as such, we can apply the IVT to $h(x)$.

Clearly the sum is null because it is in the form $f(1) + (f(a)-f(a)\text{ same argument a})+(f(b)-f(b)\text{ same argument b})+...-f(0)$ and $f(1)=f(0)$, not because there somehow exists two different in argument terms that cancel $f(a)+f(b)=0$.

In any case, we require a $h(a)$ and a $h(b)$ such that $h(a)h(b)<0$, not in terms of $f$.

But, from what I have understood, this is the correction given by their professor, so maybe I am missing something?

EDIT: Ignore the following nonsense...

I thought of going about it by contradiction, but the use of a limit leaves me undecided as to whether this counts as a solution.

The negative of the proposition is $\exists n\in\mathbb N^*\,|\,\forall x\in\mathbb R,\,f\bigl(x+\frac1n\bigr)-f(x)\neq0$, i.e $f\bigl(x+\frac1n\bigr)<f(x)$ or $>$. In both cases, if we take $n\to\infty$, we get $f(x)<f(x)$, which is impossible.

Any comments? Thank you!


问:
Global conformally flat coordinates in 2d spacetimes
Let $(M,g)$ be a 2 dimensional pseudo-Riemannian manifold that is topologically a disc. Is it possible to construct a global coordinate system in which the metric is conformally flat? I.e. coordinates $(t,x)$ which cover the whole manifold such that the line element takes the form

$ds^2=\Omega^2(t,x)(-dt^2 + dx^2)$

for some conformal factor $\Omega$.

答:
Yes, every disk (topologically) is conformally flat. String theorists would say that the disk has "no moduli".

If it were not possible to flatten a disk to the standard round form by Weyl transformations, there would have to exist a family of inequivalent solutions. The family would be labeled by some parameters, "moduli". It's pretty clear that they would have to be continuous. So there would have to exist new "disks" that are conformally inequivalent to the standard disk but that are infinitesimally close to it.

It's impossible because if one defines a slightly perturbed disk by describing its boundary via polar coordinates
$$ r(\phi) = r_0 + \epsilon(\phi), $$
where we also use $z=r \exp(i\phi)$, it is always possible to find a holomorphic function of $z$ such that the curve given by the equation above is mapped to the standard $r=r_0$ round form. Just Fourier-expand $\epsilon(\phi)$ and redefine $z$ by a Laurent series whose coefficients are the Fourier coefficients I just mentioned. One can give details. But in the formula
$$z'=z+\sum_{n=0}^\infty \delta_n z^n$$
one has enough coefficients $\delta_n$ to adjust all the Fourier modes of $\epsilon(\phi)$. They're essentially the same things. Because every two (topologically) disks may be connected by continuous transformations (not necessarily preserving anything beyond topology), it's possible to divide the proof of their conformal equivalence to the pairs of nearby manifolds.

Moduli and conformal isometries of Riemann surfaces

Much more generally, if there is a Riemann surface with $g$ handles and $b$ boundaries and $c$ crosscaps (boundaries whose opposite points are identified, and therefore help to create an unoriented but possibly closed manifold), the number of real moduli $\mu$ can be calculated from the equation
$$\frac{1}{3} ( s - \mu ) = \chi = 2 - 2g - b - c $$
where $\chi$ is the Euler character, calculable by the explicit formula above, and $s$ is the real dimension of the conformal isometry group of the manifold; usually $s$ vanishes for a sufficiently high value of $g$ etc. It's the Riemann-Roch index theorem which may be extended to the Atiyah-Singer theorem etc.

So for example, the disk has an $SU(1,1)$ isometry group - isomorphic to the $SL(2,R)$ conformal isometry of its conformally equivalent surface, the half-plane - which has $s=3$ and the formula above becomes
$$ \frac{1}{3} ( 3 - 0  ) = 2-1 = 1 $$
For a sphere, $s=6$ and $\chi=2$ since $g=b=c=0$. For a torus, $s=2$ from the $U(1)^2$ group and $\mu=2$ as well because $\chi=2-2=0$. Similarly for a Möbius strip, $b=c=1$ and $\chi=0$, matching $s=\mu=1$. For a Klein bottle, $b=0$, $c=2$, and $s=\mu=1$ as well. All other topologies of Riemann surfaces have $s=0$ (no isometries) so the formula above simplifies to  $\mu = 6g+b+c - 6$ for the number of real moduli.

Indefinite signature

If your word "pseudo-Riemannian" really meant that you have an indefinite signature, like in special relativity, then it is not correct to say that such manifolds ever  have the "topology of a disk". In particular, seemingly compact sets of points cut into the Minkowski space are not open sets - a pretty basic problem. It's because open sets require that together with every point, you must include a whole neighborhood of all points whose distance is smaller than $\epsilon$.

That's not possible in a Minkowski-like space because all lightlike-separated points - that can be arbitrarily far in the coordinate space - have proper distance from the original point equal to zero, so it is smaller than $\epsilon$. Topologies work differently in the Lorentzian-signature manifolds and they're really very messy.

In string theory - following the procedures in physics in general - the relevant calculations are always done with the Euclidean, positive-definite-signature manifolds, and only the final results are extrapolated back to the Minkowski-signature spacetime. That's perfectly OK and one may show that the resulting theory satisfies all required consistency conditions. This "messy Lorentzian topologies" case is a textbook example of the fact that the calculations in the Wick-rotated framework are more well-behaved than the calculations that someone would naively try to do directly in the original, indefinite-signature spacetime.


问:
Words counting problem
What is the number of words, which are made by shifting all lower case letters in the english alphabet and none of them contains any of the four subwords (null, one, two, three)?</p>

答:
Let us focus on a simpler example. Suppose, that we are not allowed only one subword, say, 'null'. Without that restriction, we can form $26^{26}$ words (assuming a word length of $26$ and that we are allowed repetitions).

However, we want to exclude the word 'null' from our list. Now, the word 'null' can appear in any one of the following positions at the start of the word, in the middle or towards the end. There are $23$ such possible places where the word 'null' can appear. For each one of the $23$ places, we can fill the remaining places in $26^{22}$ ways.

Thus, the number of words where the word 'null' appears is $23 . 26^{22}$. Therefore, the number of words where 'null' does not appear is: $26^{26} - 23 . 26^{22}$

Title: Words counting problem


答:
Since $f$ is absolutely continuous, we have that
$$ f(x) - f(y) = \int_x^y f'(t) \, dt .$$
Then
\begin{align}
\text{$f$ is Lipschitz with constant $L$} &\Leftrightarrow
-L (y-x) \le \int_x^y f'(t) \, dt \le L (y-x) \text{ for all $x < y$} \\
&\Leftrightarrow  \int_x^y (L - f'(t)) \, dt \ge 0 \text{ and } \int_x^y (L + f'(t)) \, dt \ge 0\text{ for all $x < y$}
\end{align}
Also
$$ \int_x^y g(t) \, dt \ge 0 \text{ for all $x < y$} \Leftrightarrow g \ge 0 \text{ a.e.} $$
Hence $f$ is Lipschitz with constant $L$ if and only if $L - f'(t)$ and $L + f'(t)$ are greater than or equal to zero for almost every $t$.

Title: $f$ is Lipschitz if and only if there exists $L\geq0$ such that $|f'(x)|\leq L$

Body:
Let $f:[a,b]\to\mathbb{R}$ be absolutely continuous.

Prove that $f$ is Lipschitz if and only if there exists $L\geq0$ and a set $E\subset[a,b]$ such that, $m(E)=0$ and $f$ is differentiable at each $x\in[a,b]\setminus E\quad$  with $|f'(x)|\leq L$.

My attempt:

For $\Rightarrow$ direction:

Since $f$ is absolutely continuous on $[a,b]$, it is differentiable a.e on $[a,b]$

And notice:

$\begin{align}
|f'(x)|&=\left|\lim\limits_{h\to0}\frac{f(x+h)-f(x)}{h}\right|\\
&\leq\lim\limits_{h\to0}\left|\frac{f(x+h)-f(x)}{h}\right|\\
&\leq \lim\limits_{h\to0}\left| \frac{L|(x+h)-x|}{h}\right|\\
&=L \quad\text{ Where, $L$ is the Lipschitz constant}
\end{align}$

Is that correct?

And for the other side of the implication I would really appreciate your help


答: Count all, subtract those that contain the forbidden words. To count, e.g. the permutations that don't contain one, consider that as a "superletter" and count the possible words containing it. Need count those containing twone, and so on. Use the idea of inclusion-exclusion to exclude the forbidden ones.

Title: Words counting problem

Body: What is the number of words, which are made by shifting all lower case letters in the english alphabet and none of them contains any of the four subwords (null, one, two, three)?</p>


问:
What is the probability that it is raining?
A bus comes around your house every day.
When it rains, there is a 20% probability that it is going to be late.
When it doesn't rain, there is a 10% probability that it's going to be late.
The probability that it will rain is 40%.
Today, the bus was late. What is the probability that it was raining?
(By the way, the correct answer is 0.571, but I couldn't figure out how to get there.)</p>

答:
According to Bayes' Rule,

$$ P[A|B] = \frac{P[B|A]\cdot P[A]}{P[B]} $$

This gives us

$$ P[\text{Rain}|\text{Late}] = \frac{P[\text{Late}|\text{Rain}]\cdot P[\text{Rain}]}{P[\text{Late}]} $$

Of these, we know

$$ P[\text{Late}|\text{Rain}] = 0.2 , $$
$$ P[\text{Rain}] = 0.4 . $$

So, we still need $P[\text{Late}]$.

For this, we can use the fact that

$$ P[A] = P[A \cap B]+P[A \cap B^C] . $$

This gives us

$$ P[\text{Late}]=P[\text{Late} \cap \text{Rain}]+P[\text{Late} \cap \text{No Rain}] $$

$$ =P[\text{Late} | \text{Rain}]\cdot P[\text{Rain}]+P[\text{Late} | \text{No Rain}]\cdot P[\text{No Rain}] $$

$$ = 0.2 \cdot 0.4 + 0.1 \cdot (1-0.4) $$
$$ = 0.14 $$

Substituting these values all into the expression we found above, we have:

$$ P[\text{Rain}|\text{Late}] = \frac{P[\text{Late}|\text{Rain}]\cdot P[\text{Rain}]}{P[\text{Late}]} $$

$$ = \frac{0.2 \cdot 0.4}{0.14} $$
$$ \approx 0.571 $$


问:
Bold 1 with subscripts - How should this be read?
I am struggling to understand the notation i.e. what does the bold 1 refer to?

$\mathbf{1}_{A\cap B}(x)=\mathbf{1}_{A}(x)\times\mathbf{1}_{B}(x)$

答:
It's presumably the indicator function
\begin{align*}
{\bf 1}_S(x) &= \begin{cases}
1 & \text{if $x\in S$}; \\
0 & \text{if $x\not\in S$}.
\end{cases}
\end{align*}
It's also (and more commonly, at least in my experience) denoted by $\chi_S$.


答:
It's the indicator function.
$$
\textbf{1}_A(x)=\left\{\begin{array}{ll}1&\text{ if }x\in A\\0&\text{ if }x\notin A\end{array}\right.
$$

Title: Bold 1 with subscripts - How should this be read?

Body:
I am struggling to understand the notation i.e. what does the bold 1 refer to?

$\mathbf{1}_{A\cap B}(x)=\mathbf{1}_{A}(x)\times\mathbf{1}_{B}(x)$


答:
We use the rules of conditional probability.
Generally speaking:
$$\mathbb{P}(\text{A}|\text{B}) = \frac{\mathbb{P}(\text{A} \text{ and } \text{B})}{\mathbb{P}(\text{B})}$$
Where $\mathbb{P}(\text{A}|\text{B})$ means "probability of A given B".

So here we have the probability that it is raining is
$$\frac{\text{probability that it is raining and the bus was late}}{\text{probability that the bus was late}}$$
The probability that the bus was late is $0.4 \times 0.2 + 0.6 \times 0.1$. (The sum of the case the bus is late and it rains and the bus is late and it does not rain.

The probability that the bus was late and it is raining is $0.4 \times 0.2$

So the answer is  $0.4\times0.2/(0.4\times0.2+0.6\times0.1) = 0.571$

Title: What is the probability that it is raining?

Body:
A bus comes around your house every day.
When it rains, there is a 20% probability that it is going to be late.
When it doesn't rain, there is a 10% probability that it's going to be late.
The probability that it will rain is 40%.
Today, the bus was late. What is the probability that it was raining?
(By the way, the correct answer is 0.571, but I couldn't figure out how to get there.)</p>


问:
Can I always write group elements $L,R$ as $VA$ and $VA^{-1}$ respectively, for some $V,A$?
My first question is, given two different elements $L,R$ in the Lie group $SU(N)$ can one always make write
$$L=V A, \quad \quad R=VA^\dagger,$$
for some $V, A \in SU(N)$ ?

Secondly, how does this generalise to general Lie (or even non-Lie) groups. Is it always possible to decompose in the above fashion (with $A^\dagger$ generalised to $A^{-1}$)?

答:
I first claim that $A$ and $V$ exist iff $R^{-1}L$ is the square of some element in the group.

First, if $A$ and $V$ exist, then $R^{-1}L = AV^{-1} VA = A^2$ is a square.  Conversely, if $R^{-1}L$ is a square, then there is a solution to $A$ to the equation $A^2 = R^{-1} L$.  Setting $V = LA^{-1}$ works.  That is, $VA = LAA^{-1} = L$ while $VA^{-1} = L(A^2)^{-1} = L(R^{-1}L)^{-1} = R$.

Having established the claim, note that in any connected compact Lie group $G$, the exponential map is surjective.  Given $g\in G$, we may write $g = \exp(X)$ for some $X$ in the Lie algebra of $G$.  Then $\exp(1/2  X)^2 = \exp(X) = g$, so $g$ is a square.  Hence, in a connected compact Lie group, you can always find $V$ and $A$.

On the other hand, in some groups, not every element is a square.  The smallest is $\mathbb{Z}/2\mathbb{Z}$, where the non-trivial element is not a square.


问:
Prove $\sin \alpha+\sin \beta+\sin \gamma \geq\sin 2\alpha+\sin 2\beta+\sin 2\gamma $
Prove that $\sin \alpha+\sin \beta+\sin \gamma \geq\sin 2\alpha+\sin 2\beta+\sin 2\gamma $ where $\alpha$ $,\beta$ $,\gamma$ are the angles of a triangle</p>

答:
$\sin A+\sin B+\sin C=\dfrac{a+b+c}{2R}=\dfrac{S}{rR}$

$\sin 2A+\sin 2B+\sin 2C=2(\sin A\cos A+\sin B\cos B+\sin C\cos C)$

$=\dfrac{a\cos A+b\cos B+c\cos C}{R}=\dfrac{2S}{R^{2}}$

$\sin A+\sin B+\sin C\geq \sin 2A+\sin 2B+\sin 2C$

$\Longleftrightarrow \dfrac{S}{rR}\geq \dfrac{2S}{R^{2}}$ $\Longleftrightarrow$ $R\geq 2r$ : Euler formula