问:
A function F from X to Y is continuous if and only if F is continuous to F(X)
Today I read in my topology book that a function F from X to Y is continuous if and only if F is continuous to F(X). My question is this: Let K be a closed subset of X and Suppose there is a continuous function F from K to (-1,1). So (-1,1) is F(K). Let Y be [-1,1]. Because that function F from X to Y is continuous if and only if F is continuous to F(X), The origin of [-1,0] should be a closed subset of K. But the origin of [-1,0] is exactly the same as the origin of (-1,0] which isn't necessarily closed. I obviously missing here an important detail.(The reason I ask this question is because of the Tietze extension theorem.)

By the way, how can I write with mathematic notations? It would really help me to write more formally and clear.

答:
It's useful to think of the "nice pullbacks for open sets" definition of continuity: $f:X \to Y $ is continuous if for every open set $U \subset Y$, $f^{-1}(U)$ is an open set in the $X$ (where here $f^{-1}$ denotes the set inverse, i.e.,
$$
f^{-1}(U) = \{x \in X \mid f(x) \in Y \}.
$$

Now suppose that $F : X \to F(X) \subset Y$ is continuous. Then for any open set $U$ in $F(X)$, we have $F^{-1}(U)$ is open.

Now suppose that $U'$ is an open set in $Y$. I'd like to show that $F^{-1}(U')$ is open in $X$.

Well, $K = U' \cap F(X)$ is open in $F(X)$ by the definition of the subset topology. So $F^{-1}(K)$ is open, by continuity from $X \to F(X)$.

And $F^{-1}(K) = F^{-1}(U')$ [Why? I leave you to work that out...]. So $F^{-1}(U')$ is open in $X$ for an arbitrary open set $U' \subset Y$. Hence $F: X \to Y$ is also continuous.

All this can be converted into the closed-sets definition for continuity, and you can use to see where your mistake is (if you so choose). The key error is, I suspect, what Brian Scott pointed out, namely, that you need to better understand how the subspace topology is defined.


问:
Matlab algorithm takes as input an integer $n$, and generates a random $n$ by $n$ symmetric positive definite and strictly diagonally dominant matrix.
I could write an algorithm to generate a random symmetric positive definite matrix, but how to edit it to also have this matrix to be strictly diagonally dominant?

Here is my code:

function $A$ = generateSPDmatrix$(n)$
% Generate a dense n x n symmetric, positive definite matrix

$A = rand(n,n)$; % generate a random n x n matrix

% construct a symmetric matrix

$A = 0.5*(A+A')$;

% since $A(i,j) < 1$ by construction and a symmetric diagonally dominant matrix is symmetric positive definite, which can be ensured by adding nI

$A = A + n*eye(n)$;

end

答:
As you have recalled it, a real symmetric (more genrally Hermitian) diagonally dominant matrix ${\displaystyle A}$ with real non-negative diagonal entries is positive semidefinite (see (https://en.wikipedia.org/wiki/Diagonally_dominant_matrix))

Here is a Matlab program giving a random matrix as you desire it, in the same spirit as you.

(I think the comments I have given are enough for following the instructions; remember that operator "sum" on a matrix gives the row vector of the sums column by column):

```<code>n=5;
rng('shuffle');A=rand(n);
B=A+A';% or A*A'
C=diag(diag(B)); % extraction of the diagonal part of B
D=B-C;
m=max(sum(D)); % m is the biggest sum in columns (or in rows) of D
E=D+m*eye(n),% the zero diagonal of D is filled with values m
eig(E), % all eigenvalues are positive
</code>```

ContentLicense: CC BY-SA 4.0

Body:
As you have recalled it, a real symmetric (more genrally Hermitian) diagonally dominant matrix ${\displaystyle A}$ with real non-negative diagonal entries is positive semidefinite (see (https://en.wikipedia.org/wiki/Diagonally_dominant_matrix))

Here is a Matlab program giving a random matrix as you desire it, in the same spirit as you.

(I think the comments I have given are enough for following the instructions; remember that operator "sum" on a matrix gives the row vector of the sums column by column):

n=5;
rng('shuffle');A=rand(n);
B=A+A';% or A*A'
C=diag(diag(B)); % extraction of the diagonal part of B
D=B-C;
m=max(sum(D)); % m is the biggest sum in columns (or in rows) of D
E=D+m*eye(n),% the zero diagonal of D is filled with values m
eig(E), % all eigenvalues are positive


问:
Automorphism group of a Cayley graph
Let $G$ be a group. Let $\Gamma = \Gamma(G,X)$ be the Cayley graph of $G$ defined with respect to a generating set $X$. I want to show that $G\cong \text{Aut}(\Gamma)$. Note that by $\text{Aut}(\Gamma)$ I am not referring to the automorphism group of the underlying undirected graph, but rather the detailed graph where each edge is directed and labelled with the appropriate generator.

For example, in the following directed and labelled graph, there is only one non-trivial automorphism: the one where I send $1$ to $4$. Indeed, the rest of the automorphism is uniquely determined by describing the image of a single vertex under the automorphism.

I tried to follow this post, but I was a little confused. My questions are the following:

How are the elements of $\text{Aut}(\Gamma)$ defined? Since it is different from the usual definition of a graph isomorphism, I wasn't sure how to go about making this definition.

Why is it easy to see that $T_h\in\text{Aut}(\Gamma)$? (I suppose the answer to this question depends on how $\text{Aut}(\Gamma)$ is defined.)

答:
The vertices of the Cayley graph $\Gamma$ are the elements of $G$, there is an edge $(g,gs)$ for every $s\in X$ (where $X$ is the generating set) and $g\in G$. The edge $(g,gs)$ is labelled by the generator $s$. An automorphism of $\Gamma$ is a permutation of the vertex set which  induces a permutation of the edge set that also preserves the labels of the edges. (It sends any edge to an edge with the same label.)

Call such a function $\phi:G\to G$. The fact it preserves edges means $(\phi(g),\phi(gs))$ must be an edge for all $g\in G,s\in S$, and it must have the same label $s$, which means the second coordinate $\phi(gs)$ must be the first one, $\phi(g)$, times $s$. That is, $\phi(gs)=\phi(g)s$ for all elements $g\in G,s\in S$. You can do the same idea for the edges $(gs^{-1},g)$ labelled by $s$ in order to show $\phi(gs^{-1})=\phi(g)s^{-1}$ too.

Since all $g\in G$ are products of elements of $S$ and their inverses, by induction $\phi(g)=\phi(e)g$. That is, any labelled automorphism $\phi$ of $\Gamma$ is simply left multiplication $T_h(g):=hg$ by a group element $h=\phi(e)$, and conversely (which follows simply from the associative property).

Title: Automorphism group of a Cayley graph


问:
Integral question help me?
I am in the middle of solving a diff equation and I have  to solve $\int\dfrac{1}{\cos x}e^{\tan x}\,dx$? I am thinking about integration by parts but it is a very long way.is there another short way? Thank you :)

**Edit: The diff equation which is linear of first order that  I have to solve is (ysinx-1)dx +cosxdy=0

Edit again,I first find y=c*e^[-int(tgx)] then I find y' and then I replace both of them in y' +y*tgx=1/cosx..And here I have a c'=____ and then I find the integral of c' which is c and finally replace it in y=c*e^[-int(tgx)].The problem is that here I have this difficult integration..

答:
Your setup is incorrect, so this integral never comes up.  Your ODE is,
$$
y' + y\tan x = \sec x
$$
which can be solved using an integrating factor to turn the LHS into a total derivative. This turns the ODE into,
$$
\cos x (y \sec x)' = \sec x
$$
The integration is thus $\int dx \sec^2 x$ which is just $\tan x$.  This gives $y(x)=\sin x$ which can be easily seen to solve the ODE.</p>

ContentLicense: CC BY-SA 3.0

Title: Integral question help me?


问:
Find the tangent to the graph of the function which is perpendicular to the line
Find the tangent to the graph of the function $f (x) = \ln (x-1) ^ 2$ which is perpendicular to the line $p$: $y = \frac{2}{5}x +2$

I decided $kp = 2/5$ and planted in $kn = -1/kp = -5/2 \to kt = -5/2$. Then I derived the function $f'(x) = 2/(x-1)$.

Substituted into the equation
$kt = -5/2 = 2/(x-1) \to x=1/5$.
And put in the original function. And determined $y=\ln(16/25)$

Is this procedure correct and how do I express the tangent equation?

答:
So we want to find a tangent to $f(x) = \ln ((x-1)^2)$ that has the slope of $\frac{-1}{2/5} = -\frac{5}{2}$. Indeed the derivative you got was correct, so we just need to find the point $(x, f(x))$ for which we have a tangent with the slope $-\frac{5}{2}$. So

$$ f'(x) = -\frac{5}{2} = \dfrac{2}{x-1} \implies x=\frac{1}{5} $$

So the point $\left(\frac{1}{5}, \ln \left(\frac{16}{25}\right)\right)$ has a tangent to $f(x)$ with the slope $-\frac{5}{2}$, So

$$ y -  \ln \left(\frac{16}{25}\right) = -\frac{5}{2}\left(x-\frac{1}{5}\right) $$


问:
Contour integration around pole and essential singularity
I've got the following integral:

$$
\frac{1}{2\pi i}\oint_\limits{|z|\,=\,R} \frac{z^5}{z-1} e^\frac{2}{z} dz,
$$

where $R$ - sufficiently large number.

I've tried to evaluate it with residues. The integrand has a simple pole at $z = 1$ and an essential singularity at $z=0$.

Evaluation of the residue about the pole is trivial: it equals to $e^2$. I tried to deal with the singularity:

$$
\frac{z^5}{z-1} e^\frac{2}{z} = z^4 \frac{1}{1-\frac{1}{z}} e^\frac{2}{z} = \left[z^4 + z^3 + z^2 + z + 1 + \frac{1}{z} + ... \right]\left[1 + \frac{2}{z} + \frac{2}{z^2} + \frac{4}{3z^3} + \frac{2}{3z^4} + \frac{4}{15z^5} + ... \right].
$$

After term-by-term multiplication and collecting terms with $\frac{1}{z}$ I got the residue $\frac{109}{15}$. Therefore, the integral equals to $e^2 + \frac{109}{15}$.

But I am told it's wrong, and I can't find a problem for a long time. Can someone tell, is that solution wrong? Very grateful in advance for your help and time. Thanks!

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{R\ \mbox{is a}\ \underline{\it sufficiently\ large\ number}}$:
\begin{align}
&\bbox[5px,#ffd]{{1 \over 2\pi\ic}
\oint_{\verts{z}\ =\ R}\,\,\,
{z^{5} \over z - 1}\expo{2/z}\,\dd z}
\\[5mm] \stackrel{z\ \mapsto\ 1/z}{=}\,\,\,&
{1 \over 2\pi\ic}\bracks{%
-\oint_{\verts{z}\ =\ 1/R}\,\,\,
{1/z^{5} \over 1/z - 1}\expo{2z}
\,\pars{-\,{\dd z \over z^{2}}}}
\\[5mm] = &
{1 \over 2\pi\ic}
\oint_{\verts{z}\ =\ 1/R}\,\,\,
{\expo{2z} \over z^{6}\pars{1 - z}}\,\dd z
\\[5mm] = &
{1 \over 2\pi\ic}\bracks{2\pi\ic\,{1 \over \pars{6 - 1}!}
\,\lim_{z \to 0}\,\,\totald[5]{}{z}
\bracks{z^{6}\,{\expo{2z} \over z^{6}\pars{1 - z}}}}
\\[5mm] = &
{1 \over 120}
\,\ \underbrace{\lim_{z \to 0}\,\,\totald[5]{}{z}
\pars{{\expo{2z} \over 1 - z}}}_{\ds{872}}\ =\
\bbx{109 \over 15} \approx 7.2667
\end{align}


问:
How can you find the limit of this: $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$?
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$

I would start with expanding it with $* \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}} \;$ but I don't know how to progress from there.

I can't use L'Hospital's rule. I also have 3 other exercises that are like this but if I can see one solved, I think I will be able to do the other ones as well.

Thanks

答: As $x\to0$, famously $\frac{\sin x}{x}\to1$ and $\frac{1-\cos x}{x^2}=\frac12$, so $\frac{x\sin x+1-\cos x}{x^2}=\frac32$ and$$\frac{x^2}{\sqrt{1+x\sin x}-\sqrt{\cos x}}=\frac{x^2}{1+x\sin x-\cos x}(\sqrt{1+x\sin x}+\sqrt{\cos x})=\frac23\cdot2=\frac43.$$


答:
hint

After multiplying by the conjugate as you done, the denominator becomes
$$1+x\sin(x)-\cos(x)=$$

$$2\sin(\frac x2)\Bigl(x\cos(\frac x2)+\sin(\frac x2)\Bigr)$$

The function can written as

$$\frac{x}{2\sin(\frac x2)}\frac{1}{\cos(\frac x2)+\frac 1x\sin(\frac x2)}$$
$$×(\sqrt{1+x\sin(x)}+\sqrt{\cos(x)})$$

the limit is then
$$1×\frac{1}{1+\frac 12}×2=\frac 43$$

Title: How can you find the limit of this: $\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$?

Body:
$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$

I would start with expanding it with $* \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}} \;$ but I don't know how to progress from there.

I can't use L'Hospital's rule. I also have 3 other exercises that are like this but if I can see one solved, I think I will be able to do the other ones as well.

Thanks


答:
Hint:

Multiplying numerator and denominator with $\sqrt{1+x\sin x}+\sqrt{\cos x}$, you obtain
$$\frac{x^2}{1+x\sin x-\cos x}\cdot\bigl(\sqrt{1+x\sin x}+\sqrt{\cos x}\mkern1.5mu\bigr). $$
You can rewrite the fraction as
$$\frac 1{\cfrac{1-\cos x}{x^2}+\cfrac{\sin x}x},$$
in which each term in the denominator corresponds to a standard limit.


问:
Question about zero sets and measure theory
If $Z\subset\mathbb{R}$ is a zero set we know that for any $\epsilon>0$ there is a countable covering of $Z$ by open intervals $(a_i,b_i)$ such that:
$$\sum_i^\infty b_i-a_i<\epsilon$$
This definition was pretty much ripped straight from Pugh. Now in an excerise in Pugh I'm supposed to show that the statement:

$Z$ is the zero zero set

Is equivalent to:

For each $\epsilon>0$ there is a countable covering of $Z$ by closed intervals $[a_i, b_i]$ with total length $\sum b_i - a_i < \epsilon$.

Now whats throwing me through a loop here is the use of closed intervals. I know that $Z$ can be covered by open intervals, but does that somehow allow me to also cover it with closed intervals? Am I missing something completely obvious?

答:
Suppose you have an open cover of $Z$ by intervals $(a_i,b_i)$. Then the closed intervals $[a_i,b_i]$ also cover $Z$ since $(a_i,b_i)\subseteq[a_i,b_i]$ for each $i\in\mathbb{N}$, and the length of each closed interval is the same as the length of the corresponding open interval, so:
$$\sum_{i=1}^\infty\ell\big([a_i,b_i]\big)=\sum_{i=1}^\infty\ell\big((a_i,b_i)\big)<\varepsilon.$$
Conversely, suppose you have a cover of $Z$ by closed intervals $[a_i,b_i]$ such that the sum of the lengths $\sum_{i=1}^\infty\ell\big([a_i,b_i]\big)<\varepsilon/2$. For each $i\in\mathbb{N}$, consider the open interval $(a_i',b_i')\equiv(a_i-\varepsilon/2^i,b_i+\varepsilon/2^i)$. Then $[a_i,b_i]\subseteq(a_i',b_i')$ for each $i$, and so the intervals $(a_i',b_i')$ cover $Z$, and the sum of the lengths of these intervals is less than $\varepsilon$. Hence, covering by closed sets is equivalent to covering by open sets.


问:
What is the exact value of
What is $ \sum_{n=1}^{\infty} \frac{n}{2^{n}}$ equal to?

I know by the ratio test this converges absolutely, thats not what I'm looking for in this case, I want the actual value, is there a way aside from numerical approximation to get this value?

The value obtained from the ratio test is $\frac{1}{2}$ now I just need to know what this approximates to, Wolfram says 2 but how would one prove this?

Thanks

答:
We can use a trick on the geometric series.
$$\sum^\infty_{n=0}x^n=\frac{1}{1-x}$$
We can take the derivative with respect to $x$ of both sides
$$\sum^\infty_{n=0}nx^{n-1}=\frac{1}{(1-x)^2}$$
Let $x=\frac{1}{2}$
$$\sum^\infty_{n=0}\frac{n}{2^{n-1}}=\frac{1}{(1-\frac{1}{2})^2}$$
$$\sum^\infty_{n=0}\frac{n}{2^n}=2$$

Title: What is the exact value of


问:
Chromatic 8 graphs
I am looking for chromatic number 8 graphs that use the least number of edges preferably. Also my graph shouldn't be k8.

I don't want the Mycielski graph.

suggest any online database or resources if they exist.

答:
You can check out the Database of Interesting Graphs on The House of Graphs.

Here you can search, for example, for a graph with chromatic number equal to $8$, but clique number less than $8$ (so that the graph does not contain a $K_8$). This gives three results [1] [2] [3], none of which come from the Mycielski construction.

Title: Chromatic 8 graphs


答:
Let's say we have
$$
f(r) = \sum_{n=0}^{\infty}r^{n}
$$
Of course, in order for this to make sense we require ${|r|<1}$. So our function is only defined on ${(-1,1)}$. Now, let's take the derivative:
$$
\frac{df}{dr}=\frac{d}{dr}\sum_{n=0}^{\infty}r^n
$$
Now here's the question: can you interchange the sum and derivative operator here? In other words, is it true that
$$
\frac{d}{dr}\sum_{n=0}^{\infty}r^n =^{?} \sum_{n=0}^{\infty}\frac{d}{dr}r^n
$$
in this case - the answer is yes. But this is not always true for a general series involving functions. I'd suggest researching the conditions necessary for doing such interchange for future reference, but for now you can take my word in this case it's okay to do so.

So we get
$$
\frac{df}{dr}=\sum_{n=0}^{\infty}\frac{d}{dr}r^n = \sum_{n=0}^{\infty}nr^{n-1}
$$
The cool thing is, we know an alternative representation for ${f(r)}$. That is
$$
f(r) = \sum_{n=0}^{\infty}r^n = \frac{1}{1-r}
$$
This tells us
$$
\frac{df}{dr}=\sum_{n=0}^{\infty}nr^{n-1} = \frac{d}{dr}\left(\frac{1}{1-r}\right)
$$
Using this knowledge, are you now able to show that
$$
\sum_{n=0}^{\infty}n\frac{1}{2^n} = 2
$$
?

Title: What is the exact value of

Body:
What is $ \sum_{n=1}^{\infty} \frac{n}{2^{n}}$ equal to?

I know by the ratio test this converges absolutely, thats not what I'm looking for in this case, I want the actual value, is there a way aside from numerical approximation to get this value?

The value obtained from the ratio test is $\frac{1}{2}$ now I just need to know what this approximates to, Wolfram says 2 but how would one prove this?

Thanks


问:
Find MLE given a mixed density
Given i.i.d. data $x_1,...,x_n$ from a r.v. $X$ with density

$$f(x) = \begin{cases} \frac{1}{4} \theta_1 &\text{if } x=1  \\
\frac{1}{4}(1-\theta_1) &\text{if } x=2 \\
\frac{3}{4}\frac{1}{\theta_2}e^{-(x-3)/ \theta_2} &\text{if } x \geq 3 \ , \end{cases} $$

find the MLE of $\theta =(\theta_1, \theta_2)$.

$f(x)$ needs to be normalized, and this results in $1=\int_3^{\infty}\frac{1}{\theta_2}e^{-(x-3)/ \theta_2}dx$, which is only satisfied for $\theta_2=1$, Is this then also the MLE for $\theta_2$?

Edit: The density is defined with respect to the measure $\mu=\mu_0 1\{x < 3\}+\mu_1 1\{x \geq 3\}$ where $\mu_0$ is counting measure and $\mu_1$ is length measure.

答:
Say you have $k_1$ observations where $x_i=1$ and $k_2$ observations with $x_i=2$ and the rest are $x_i\geq 3$. Then the likelihood will be $$L(\theta_1,\theta_2)=\left(\frac{\theta_1}{4}\right)^{k_1}+\left(1-\theta_1\right)^{k_2}\frac14^{k_2}+\prod_{I}\frac{3}{4}\frac{1}{\theta_2}e^{-(x_i-3)/ \theta_2}$$ where in the last sum we have $I$ including all the indices for which $x_i\geq 3$.

The maximization is simple since the discrete and the continuous parts "do not have anything in common".

Title: Find MLE given a mixed density


Body:
I'm going to write it out.

$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}}$

$\lim\limits_{x \to \; 0} \frac{x^2}{\sqrt{1+x\sin(x)} \; - \; \sqrt{\cos(x)}} * \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}$

$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{1+x\sin(x) \; - \; \cos(x)}$

$\lim\limits_{x \to \; 0} \frac{x^2*\sqrt{1+x\sin(x)} \; + \; x^2*\sqrt{\cos(x)}}{x^2*\frac{\sin(x)}{x} \; x^2*\frac{1- \cos(x)}{x^2}}$

$\lim\limits_{x \to \; 0} \frac{\sqrt{1+x\sin(x)} \; + \; \sqrt{\cos(x)}}{\frac{\sin(x)}{x} \; +\frac{1- \cos(x)}{x^2}}$

$\frac{\sqrt{1+0}+1}{1+\frac{1}{2}} = \frac{2}{\frac{3}{2}} = \frac{4}{3}$


问:
Closed form of $I=\int_{0}^{c}{t^\kappa\ e^{-t}\cos{\left(t\right)} dt}$
Let $c,\kappa\in\mathbb{R}_+^\ast$. I'm trying to simplify the following integral in order to compute it fast (so Monte-Carlo is out of the question) : $$I=\int_{0}^{c}{t^\kappa\ e^{-t}\cos{\left(t\right)} dt}$$

If I try using complex numbers, I end up with the following :

$$I=\frac{1}{2}\int_{0}^{c}{t^\kappa e^{-t}\left(e^{it}+e^{-it}\right)dt}=\frac{1}{2}\left(\int_{0}^{c}{t^\kappa e^{-t\left(1+i\right)}dt}+\int_{0}^{c}{t^\kappa e^{-t\left(1-i\right)}dt}\right)$$

My major mostly focuses on probability & stats so I'm not sure and/or confortable with what to do next.
A u-substitution would replace the upper bound $c$ with a complex number (that's not allowed, right ?).

If the gamma cumulative distribution function were expanded to complexe parameters, I could have computed it this way on R (with help from the pgamma function) :

$$I=\frac{\Gamma\left(\kappa+1\right)}{2}\left(\left(\frac{1-i}{2}\right)^{\kappa+1}F_{\Gamma\left(\kappa+1,\frac{1-i}{2}\ \right)}\left(c\right)+\left(\frac{1+i}{2}\right)^{\kappa+1}F_{\Gamma\left(\kappa+1,\frac{1+i}{2}\ \right)}\left(c\right)\right)$$

But unfortunately that's not the case. Any insight or suggestions on what to do next would be very much appreciated...

答: To see how the expression given by @DavidG.Stork is derived, we first write the exponential as a series to get \begin{align}\int t^\kappa e^{-t(1\pm i)}\,dt&=\int t^\kappa\sum_{n=0}^\infty\frac{(-t(1\pm i))^n}{n!}\,dt\\&=C+\sum_{n=0}^\infty\frac{(-1\mp i)^nt^{n+\kappa+1}}{n!(n+\kappa+1)}\\&=C+t^{\kappa+1}\sum_{n=0}^\infty\frac{((-1\mp i)t)^n}{n!(n+\kappa+1)}\\&=C-\frac{\Gamma(\kappa+1,(1\pm i)t)}{(1\pm i)^{\kappa+1}}\end{align} where $$\Gamma(a,z)=-\sum\limits_{n=0}^\infty\frac{(-1)^nz^{n+a}}{n!(n+a)}=z^aE_{1-a}(z)$$ is the incomplete gamma function. Thus\begin{align}2I&=\int_0^ct^\kappa e^{-t(1+i)}\,dt+\int_0^ct^\kappa e^{-t(1-i)}\,dt\\&=\left(-\frac{\Gamma(\kappa+1,(1+i)c)-\Gamma(\kappa+1)}{(1+i)^{\kappa+1}}\right)+\left(-\frac{\Gamma(\kappa+1,(1-i)c)-\Gamma(\kappa+1)}{(1-i)^{\kappa+1}}\right)\\&=\Gamma(\kappa+1)\left((1+i)^{-(\kappa+1)}+(1-i)^{-(\kappa+1)}\right)-c^{\kappa+1}(E_{-k}((1+i)c)+E_{-k}((1-i)c))\end{align} and since $(1+i)^{-(\kappa+1)}+(1-i)^{-(\kappa+1)}=2^{-(\kappa+1)/2}\cos(-(\kappa+1)\pi/4)$ the result follows.


问:
Intersection of subspace which projector do not commute
I am a little bit confused by the following.

Consider two subspace $U$ and $V$ of a given vector space. I associate to them $\Pi_U$ and $\Pi_V$ their projectors.

In the case $[\Pi_U,\Pi_V]=0$; the projector on $U \cap V$ is simply $\Pi_{U \cap V} = \Pi_U \circ \Pi_V = \Pi_V \circ \Pi_U$

But when the two projector do not commute, $\Pi_U \circ \Pi_V$ might not even be a projector.

My question is:

Is there a physical meaning for intersection of subspaces which projector do not commute ? Does that mean that the intersection is necesseraly $\{0\}$ ? If so I couldn't manage to show it.

答:
The projector onto $U$ is not unique!  It depends on the complement of $U$ chosen to be the kernel of $\Pi_U$.

However, given any $U$ and $V$, there are always choices of $\Pi_U$ and $\Pi_V$ which commute!

EDIT.  Here is how to get commuting $\Pi_U$ and $\Pi_V$.

Find a subspace $U_1⊆U$ such that $U=(U∩V) ⊕ U_1$,

Find a subspace $V_1⊆V$ such that $V=(U∩V) ⊕ V_1$,

Prove that $U+V = U_1 ⊕ (U∩V) ⊕ V_1$,

Denoting by $E$ the ambient space, find a subspace $W⊆E$ such that $E=(U+V) ⊕ W$,

Prove that $E = U_1 ⊕ (U∩V) ⊕ V_1 ⊕ W = U ⊕ V_1 ⊕ W = U_1 ⊕ V ⊕ W$,

Let  $\Pi_U$ be the projection onto  $U$  along (i.e. with kernel) $V_1 ⊕ W$,

Let  $\Pi_V$ be the projection onto  $V$  along  $U_1 ⊕ W$.

Check that  $\Pi_U$ and $\Pi_V$ commute!


问:
How to derive the recursive equation for number of binary trees with 'n' nodes?
Now I know that the number of binary trees with n nodes is given by (2n!/(n+1)! n!)

How can I derive this using the basic definition of a binary tree, which says that a binary tree is either an empty tree, or a tree with two children which are also trees?

I thought I could use induction, but I couldn't go beyond the base case:

Base case: f(0) = 1    (There is 1 binary tree with 0 nodes).

Also, I am not supposed to know the term (2n!/(n+1)!), I am supposed to derive it. I don't think induction helps for this. From what I know, induction is used to prove something you know is a fact, not derive that fact.

答:
If $f(n)$ is the number of unlabelled binary trees on $n$ nodes, then $f(0)=1=f(1)$. The definition then yields a recurrence: to get a binary tree with $n+1$ nodes, start with a root node and hang from it a left subtree with $\ell$ nodes and a right subtree with $r$ nodes, where $0\le\ell,r\le n$ and $\ell+r=n$. For a given $\ell$ there are $f(\ell)$ possible left subtrees and $f(n-\ell)$ possible right subtrees, and every binary tree with $n+1$ nodes is obtained uniquely in this way, so

$$f(n+1)=\sum_{\ell=0}^nf(\ell)f(n-\ell)\,.$$

This answer shows (with somewhat different notation) how to use generating functions to derive a closed form for $f(n)$.


问:
How to find number of arrangements of 18 exams with a specific condition?
Student should take $18$ exams. Three of those are Biology, Chemistry and Physics, no two of which can be taken consecutively. What's the number of arrangements in which student can take these exams.

I figured that the way to solve this is to take the number of arrangements that don't fulfill this condition from the total number of $18!$.

Here I get stuck because we can consider arrangements in which we treat all three as a whole and find that there will be $15! \times 6$, accounting for permutation within their group. Or you can also choose groups of $2$, but my problem is that by doing this some of the arrangements accounted for in the previous part will also be considered.

What's the best way to tackle this problem?

答:
There are different methods we could employ here.

Method 1:  We use the Inclusion-Exclusion Principle.

If there were no restrictions, there would be $18!$ possible arrangements of the exams.  From these, we must exclude those arrangements in which at least two of the Biology, Chemistry, and Physics exams are consecutive.

Two of the three exams are consecutive:  There are $\binom{3}{2}$ ways to choose two of the three subjects Biology, Chemistry, and Physics to be examined consecutively.  Place those two exams in a block.  Then we have $17$ objects to arrange, the block of two consecutive exams and the other $16$ exams. The objects can be arranged in $17!$ ways.  The exams within the block can be arranged in $2!$ ways.  Hence, there are
$$\binom{3}{2}17!2!$$
such arrangements.

However, if we subtract these $17!2!$ cases from the total, we will have subtracted too much since we will have subtracted each case in which all three exams are consecutive twice, once when we designated the first two of those three consecutive exams as the consecutive exams and once when we designated the last two of those three consecutive exams as the consecutive exams.  We only want to subtract such cases once, so we must add them back.

All three of the exams are consecutive:  Place the Biology, Chemistry, and Physics exams in a block.  We now have $16$ objects to arrange, the block of three exams and the other $15$ exams.  The $16$ objects can be arranged in $16!$ ways.  The three exams can be arranged within the block in $3!$ ways.  Hence, there are
$$16!3!$$
such arrangements.

By the Inclusion-Exclusion Principle, the number of admissible arrangements of the exams is
$$18! - \binom{3}{2}17!2! + \binom{3}{3}16!3!$$

Method 2:  We arrange the Biology, Chemistry, and Physics exams among the other $15$ exams so that no two of the Biology, Chemistry, and Physics exams are consecutive.  We can arrange the other $15$ exams in $15!$ ways.  This creates $16$ spaces in which we can place the Biology, Chemistry, and Physics exams, $14$ spaces between successive exams, one at the beginning, and one at the end.  To ensure that no two of the Biology, Chemistry, and Physics exams are consecutive, we must choose three of these $16$ spaces in which to place the Biology, Chemistry, and Physics exams, then arrange these three exams in the selected spaces.  This can be done in $\binom{16}{3}3!$ ways.  Hence, there are
$$15!\binom{16}{3}3!$$
admissible arrangements of the exams.


答:
The intuition of subtracting from the total number is correct.
Now we need to compute how many arrangements there are that don't fullfill the condition.

So, let's count how many arrangements where Chemistry is direcly after Biology.
That is
$$16!\times 17$$
where the 17 is given by the position the group "BC" has in the arrangements.

We can do the same for any ordered pair, which there are $6$.
So we have a total of
$$16!\times 17\times 6$$

But we counted some arrangements too many times.
For example, if an arrangement contains the sequence $BCT$ it gets counted because it contains $BC$ and it gets counted again as it contains $TC$, so we counted it $2$ times.
It follows that we have to add again every arrangement where the three subjects are one after another.
In total, they are
$$15!\times 6\times 16$$

So the result should be
$$18!-16!\times 17\times 6 +15!\times 6\times 16=16!\times (18\times 17-17\times 6+6)=16!\times 35\times 6$$

Title: How to find number of arrangements of 18 exams with a specific condition?

Body:
Student should take $18$ exams. Three of those are Biology, Chemistry and Physics, no two of which can be taken consecutively. What's the number of arrangements in which student can take these exams.

I figured that the way to solve this is to take the number of arrangements that don't fulfill this condition from the total number of $18!$.

Here I get stuck because we can consider arrangements in which we treat all three as a whole and find that there will be $15! \times 6$, accounting for permutation within their group. Or you can also choose groups of $2$, but my problem is that by doing this some of the arrangements accounted for in the previous part will also be considered.

What's the best way to tackle this problem?


问:
how to show that curve lies on a sphere
Let $\gamma$ be a parameterization by arc length of a curve $C$. Prove that,

if there are: $X_0 \in \mathbb{R}^3$, and $\alpha,\beta \in [a,b] \to
\mathbb{R}$,  differentiable on $[a, b]$, such that $\alpha(s)N(s) +
\beta(s)B(s) +\gamma(s) = X_0$  for all $s \in [a,b]$, then $C$ lies
on a sphere.

答:
Given that

$X_0 = \gamma(s) + \alpha(s)N(s) + \beta(s) B(s), \tag 1$

we have

$X_0 - \gamma(s) = \alpha(s)N(s) + \beta(s) B(s), \tag 2$

whence

$\Vert X_0 - \gamma(s) \Vert ^2 = (X_0 - \gamma(s)) \cdot X_0 - \gamma(s))$
$= (\alpha(s)N(s) + \beta(s) B(s)) \cdot (\alpha(s)N(s) + \beta(s) B(s)) = \alpha^2(s) + \beta^2(s), \tag 3$

since

$N(s) \cdot N(s) = 1 = B(s) \cdot B(s) \tag 4$

and

$N(s) \cdot B(s) = 0. \tag 5$

We next differentiate (1) with respect to $s$, observing that $X_0 \in \Bbb R^3$ is fixed:

$0 = \dfrac{d}{ds}(X_0)$
$= \dot{\gamma}(s) + \dot{\alpha}(s)N(s) + \alpha(s) \dot N(s) + \dot{\beta}(s) B(s) + \beta(s) \dot B(s), \tag 6$

and recalling that

$\dot \gamma(s) = T(s), \tag 7$

and the Frenet-Serret equations

$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s), \tag 8$

$\dot B(s) = -\tau(s) N(s), \tag 9$

we transform (6) into

$0 = \dfrac{d}{ds}(X_0)$
$= T(s) + \dot{\alpha}(s)N(s) + \alpha(s) (-\kappa(s) T(s) + \tau(s) B(s)) + \dot{\beta}(s) B(s) - \beta(s) \tau(s) N(s)$
$ = T(s) + \dot{\alpha}(s)N(s) - \alpha(s) \kappa(s) T(s) + \alpha(s) \tau(s) B(s) + \dot{\beta}(s) B(s) - \beta(s) \tau(s) N(s)$
$= (1 - \alpha(s) \kappa(s)) T(s) + (\dot \alpha(s) - \beta(s) \tau(s))N(s) + (\dot \beta(s) + \alpha(s) \tau(s))B(s); \tag {10}$

by virtue of the orthogonality of $T(s)$, $N(s)$ and $B(s)$ we may infer that

$\alpha(s) \kappa(s) = 1, \tag{11}$

$\dot \alpha(s) = \beta(s) \tau(s), \tag{12}$

$\dot \beta(s) = -\alpha(s) \tau(s); \tag{13}$

thus,

$\alpha(s) \dot \alpha(a) + \beta(s) \dot \beta(s) = \alpha(s) \beta(s) \tau(s) - \beta(s) \alpha(s) \tau(s) = 0, \tag{14}$

which is easily seen to imply that

$\dfrac{d}{ds}(\alpha^2(s) + \beta^2(s)) = 0, \tag{15}$

and we conclude that $\alpha^2(s) + \beta^2(s)$ is constant; returning to (3), we see that this implies that $\gamma(s)$ lies in the sphere of radius $\sqrt{\alpha^2(s) + \beta^2(s)}$ centered at $X_0$.

A Few Further Remarks:  First, we observe that (11) may be written

$\alpha(s) = \dfrac{1}{\kappa(s)}, \tag{16}$

which shows not only that

$\alpha(s) > 0, \tag{17}$

but also that $\alpha(s)$ is the radius curvature of $\gamma(s)$ at $s$.  Second, (16) yields

$\dot \alpha(s) = -\dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{18}$

and combining this with (12) we find

$\beta(s) \tau(s) = -\dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{19}$

or, in the event $\tau(s) \ne 0$,

$\beta(s) =  -\dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)} = \dfrac{d}{ds} \left ( \dfrac{1}{\kappa(s)} \right )\left ( \dfrac{1}{\tau(s)} \right ). \tag{20}$

In light of (16) and (20) we see that (2) may be expressed in the form

$X_0 - \gamma(s) = \dfrac{1}{\kappa(s)} N(s) - \dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)} B(s)$
$= \dfrac{1}{\kappa(s)} N(s) + \dfrac{d}{ds} \left ( \dfrac{1}{\kappa(s)} \right )\left ( \dfrac{1}{\tau(s)} \right )B(s), \tag{21}$

which in accord with (3) shows that the radius of the sphere, $\Vert \gamma(s) - X_0 \Vert$, satisfies

$\Vert \gamma(s) - X_0 \Vert^2 = \dfrac{1}{\kappa^2(s)} + \dfrac{\dot \kappa^2(s)}{\kappa^4(s) \tau^2(s)}. \tag{22}$

This formula has appeared elsewhere on this site, see for example this question.

From (20),

$\dot \beta(s) = -\dfrac{\ddot \kappa(s)(\kappa^2(s) \tau(s)) - \dot \kappa(s) (\kappa^2(s) \tau(s))'}{\kappa^4(s) \tau^2(s)}$
$= -\dfrac{\ddot \kappa(s)(\kappa^2(s) \tau(s)) - \dot \kappa(s) (2\kappa(s) \dot \kappa(s) \tau(s) + \kappa^2(s) \dot \tau(s))}{\kappa^4(s) \tau^2(s)}$
$= -\dfrac{\ddot \kappa(s)(\kappa^2(s) \tau(s)) - 2\kappa(s) \dot \kappa^2(s) \tau(s) - \kappa^2(s) \dot \kappa(s) \dot \tau(s)}{\kappa^4(s) \tau^2(s)}.  \tag{23}$

We note that (11) implies that

$\alpha(s) > 0; \tag{24}$

if

$\beta(s) = 0, \tag{25}$

then (12) yields

$\dot \alpha(s) = 0, \tag{26}$

that is, $\alpha(s)$ is a positive real constant; in light of (25) we also have

$\dot \beta(s) = 0, \tag{27}$

and then via (13) we see that

$\tau(s) = 0, \tag{28}$

which is sufficient that $\gamma(s)$ lie in a plane (see my answer to this question, click here); now (2) becomes

$X_0 - \gamma(s) = \alpha(s)N(s), \tag {29}$

from which, in accord with (3),

$\Vert X_0 - \gamma(s) \Vert^2 = \alpha^2(s), \tag{30}$

or

$\Vert X_0 - \gamma(s) \Vert = \alpha(s) = \dfrac{1}{\kappa(s)}, \tag{31}$

whence since $\alpha(s)$ and $\kappa(s)$ are constant, indicates that $\gamma(s)$ lies in the sphere or radius $\alpha(s)$ centered at $X_0$; when combined with the fact that $\gamma(s)$ lies in a plane, allows the conclusion that $\gamma(s)$ in fact lies in a circle of radius $\alpha(s)$ centered at $X_0$.

We have seen above (ca. (28)) that $\gamma(s)$ lies in a plane;

Need to re-save in .txt file!

End of Remarks.

Title: how to show that curve lies on a sphere


问:
What is the rank of the set of functions from $\omega$ to $\omega$?
My stab was to say each member is a set of pairwise distinct ordered pairs of $\omega$ into $\omega$, so each function has rank $\omega + 1$ , therefore the entire set of functions has rank $\omega+ 2$

Does this work?

答: This is the right idea but slightly off.  Each ordered pair of elements of $\omega$ has finite rank, so each function has rank $\omega$, not $\omega+1$ (it has no elements of rank $\omega$).  So, the set of all such functions has rank $\omega+1$.


问:
10 ball as put in 7 boxes randomly. What is the probability that at least one box is empty?
My solution:
First, put one ball at each box and then distribute the remaining 3 balls in 7 boxes.

$p=1-\frac{\binom{9} {3} }{\binom{16}{10}}$

Is this correct?

Or?

$p=1-\frac{\binom{10} {7} 7!7^3}{7^{10} }$

答:
Let's have a look at your two attempts and then discuss a third option.

In the first one you use the formula for combinations with repetition. Let $x_i\in \mathbb{Z}_{\geq 0}$ be the number of balls that end up in the $i$-th box. Indeed there are $\displaystyle\left(\!\!\binom{7}{10}\!\!\right) = \binom{16}{10}$ possible assignments of values to $x_1,x_2,\ldots,x_7$ such that $x_1+x_2+\ldots+x_7 = 10$. But these assignments are not equally probable. For example, there is only one way to get the assignment $x_1=10, x_2=\ldots=x_7=0$. Namely, every ball has to go into the first box. That happens with probability $7^{-10}$. On the other hand, there are many ways to get the assignment $x_1=4, x_2=\ldots=x_7=1$, $\binom{10}{4} \cdot 6!$ to be exact. Each of these ways is equally probable; it happens with probability $7^{-10}$. With this in mind, you could classify the assignments that result in no box being empty. Such an assignment...

could have one 4 and six 1s. There $7 \cdot \binom{10}{4}\cdot 6!$ ways to get such an assignment.

could have one 3, one 2 and five 1s. $\#\text{ways} = 7\cdot \binom{10}{3} \cdot 6 \cdot \binom{10-3}{2} \cdot 5!$

or it could have three 2s and four 1s. $\#\text{ways} = \binom{7}{3}\cdot \binom{10}{2} \cdot \binom{10-2}{2} \cdot \binom{10-2-2}{2} \cdot 4!$.

Dividing the sum of the above by $7^{10}$ gives you the probability that no box ends up empty.

$ $

Let's now have a look at your second attempt. You calculated the number of ways that no box ends up empty to be $\binom{10}{7}\cdot 7! \cdot 7^3$. So you choose seven of your balls, you put them in a different box each, and then you are left with three balls which you can put anywhere. This is on the right track but has one problem. Consider the scenario where four balls end up in one box. Which of the four was included in your initial choice of seven balls? It could be any one of them! So you have counted this scenario $4$ times - one for each of those four balls. Similarly, if you end up with a box with three balls and another with two, you have counted that scenario $3\cdot 2 = 6$ times. And if you end up with three boxes with two balls each, you have counted that scenario $2^3=8$ times. So, once again, you start by choosing seven balls, putting each into a different box, and you are left with three balls. You have to consider what scenario will result in each way you put those three into boxes, and divide by the number found above. This way you'll avoid double-counting.
$$\displaystyle\binom{10}{7}\cdot 7! \cdot \left(\frac 74 + \frac{7 \cdot \binom{3}{2} \cdot 6}{6} + \frac{\binom{7}{3}\cdot 3!}{8}\right)$$
Dividing this by $7^{10}$ you will get the same result as before.

$ $

Another way to attack the problem is to use the inclusion-exclusion principle.
Let $p_i$ be the probability that $i$ given boxes end up empty.
Then $p_i = \dfrac{(7-i)^{10}}{7^{10}}$. The probability that at least one of the boxes does end up empty is
$$\sum_{i=1}^7(-1)^{i+1}\binom{7}{i}p_i.$$


问:
Proving the limit exists when point tends to the cluster point of a set
Let c be a cluster point of A $\subset \mathbb{R}$ and $f: A \rightarrow \mathbb{R}$ be a function. Suppose that for every sequence in {$x_{n}$} in A, such that $\lim x_{n} = c$, the sequence $\{f(x_{n})\}_{0}^{\infty}$ is Cauchy. Prove that $\lim _{x\rightarrow c}f(x)$ exists.

I did not know how to approach this question, but I found a skeleton answer here:

Why do I need to show uniqueness?

I still don't understand how we show that the limit exists though.

答:
$\Bbb R$ is complete, so for each sequence $\sigma=\langle x_n:n\in\Bbb N\rangle$ in $A$ converging to $c$, the Cauchy sequence $\langle f(x_n):n\in\Bbb N\rangle$ converges to some $\ell_\sigma\in\Bbb R$. Your first step should be to show that if $\sigma$ and $\tau$ are two sequences in $A$ converging to $c$, then $\ell_\sigma=\ell_\tau$.

HINT: If $\sigma=\langle x_n:n\in\Bbb N\rangle$ and $\tau=\langle y_n:n\in\Bbb N\rangle$, consider the interlaced sequence $\langle x_0,y_0,x_1,y_1,x_2,y_2,\ldots\rangle$.

Once you’ve done that, it’s straightforward to verify that $\lim\limits_{x\to c}f(x)=\ell$, where $\ell$ is the common value of the limits $\ell_\sigma$.


问:
Calculating probability that a certain amount of time goes by without the event occurring
Can anybody help me with this question that I stumbled upon in my statistics book today?

Mia breaks glasses at the rate of four per week.Let T be the time, in weeks, between successive breakages of glasses. Then, $f(t)=4e^{-4t}$ for $t\geq0$. What is the probability that a week goes by without Mia breaking any glasses?

I know I'm supposed to use integration to find the probability, I just dont know how to go about this question.
Any help with this question would be much appreciated!

答:
Follow @BrianTung's Comment to obtain the CDF of $T,$
which is $F_T(t) = 1 - e^{-4t},$ for $t > 0.$
Then the probability that a given week goes by
without Mia breaking a glass is $P(T > 1) = 1 - P(T\le 1).$

In R, an exponential density function is dexp and an
exponential CDF is pexp, where the parameter for your problem is rate $\lambda = 4.$

```<code>1 - pexp(1, 4)
[1] 0.01831564
</code>```

In the left panel of the figure below, the desired area is to the right of $1$ under the
density curve. The CDF is shown in the right panel.

R code for figure:

```<code>par(mfrow=c(1,2))
 curve(dexp(x, 4), 0, 2, lwd=2, ylab="PDF",
       main="Density of EXP(rate = 4)")
  abline(h=0, col="green2");  abline(v=0, col="green2")
  abline(v = 1, lwd=2, lty="dotted")
 curve(pexp(x,4), 0, 2, lwd=2, ylab="CDF",
       main="CDF of Exp(rate-4)")
  abline(v=1, col="green2"); abline(h=0:1, col="green2")
  abline(h = 1, lwd=1, lty="dotted")
par(mfrow=c(1,1))
</code>```

Body:
Follow @BrianTung's Comment to obtain the CDF of $T,$
which is $F_T(t) = 1 - e^{-4t},$ for $t > 0.$
Then the probability that a given week goes by
without Mia breaking a glass is $P(T > 1) = 1 - P(T\le 1).$

In R, an exponential density function is dexp and an
exponential CDF is pexp, where the parameter for your problem is rate $\lambda = 4.$

1 - pexp(1, 4)
[1] 0.01831564

In the left panel of the figure below, the desired area is to the right of $1$ under the
density curve. The CDF is shown in the right panel.

R code for figure:

par(mfrow=c(1,2))
 curve(dexp(x, 4), 0, 2, lwd=2, ylab="PDF",
       main="Density of EXP(rate = 4)")
  abline(h=0, col="green2");  abline(v=0, col="green2")
  abline(v = 1, lwd=2, lty="dotted")
 curve(pexp(x,4), 0, 2, lwd=2, ylab="CDF",
       main="CDF of Exp(rate-4)")
  abline(v=1, col="green2"); abline(h=0:1, col="green2")
  abline(h = 1, lwd=1, lty="dotted")
par(mfrow=c(1,1))


问:
Star Trek Discovery and the origin of "The Burn"
This maths puzzle was inspired by watching the latest season of Star Trek Discovery.

Warning: minor spoilers for season 3 of Star Trek Discovery ahead.

Many years ago, something caused hundreds to starships across the galaxy to explode. For the longest time it was thought this had occurred at exactly the same time throughout the galaxy, despite how odd this may seem. But one day Michael Burnham, chief science officer of USS Discovery, recovered the black boxes of two of the destroyed ships and noticed that they hadn't quite been destroyed at the same time.
"Hang on" -- she thought -- "if I can get my hands on a third black box, I might be able to pinpoint the location of the source of the signal that caused the ships to explode."

How can she locate the origin of the signal? What is the minimum number of black boxes required?

Some clarifications:

We assume that some signal caused the spaceships to instantly detonate when it reached them.

We assume it was sent from somewhere in the physical realm and that it spread through space at a constant velocity uniformly in every direction.

One might be tempted to assume this velocity is the speed of light; however if that were the case it would have been noticed long before. Therefore the speed of propagation is taken as an unknown.

Each black box contains the spatial coordinates of the ship and the time at which it received the signal. Note however that we don't know at what time the signal was originated.

Don't think for a second I'm taking relativity into account :)

答:
Attempt at a solution

After thinking about this problem for some time I believe I have come up with a (partial) solution.

We can think of each black box as representing a point in a four-dimensional space, with time being the fourth coordinate. Then the signal can be thought of as a 4D cone, with the vertex at the source of the signal, and the surface representing the points in spacetime at which other ships receive the signal. We know that the cone has a circular spherical base (we assume the velocity of propagation is uniform in every direction) and that it is aligned with the time axis, so it is a so-called right circular cone. It helps to visualize the problem as having two spatial dimensions plus one in time (see image).

After some digging I found the following equation form of a circular cone with its apex at the origin, aligned with the vector $\vec d$ and aperture $2\theta$:
$$
F(\vec u) = (\vec u \cdot \vec d)^2 - (\vec d \cdot \vec d)(\vec u \cdot \vec u)(\cos\theta)^2 = 0
$$

We know that our cone is aligned with the time axis, so $\vec d = [0\, 0\, 0\, 1]^T$, however we don't know the location of the apex, so we will replace $\vec u$ with $\vec u - \vec x$, where $\vec x$ are the spacetime coordinates of the source, and $\vec u$ of a black box:
\begin{align}
F(\vec u - \vec x) &= ((\vec u - \vec x) \cdot \vec d)^2 - (\vec d \cdot \vec d)((\vec u - \vec x) \cdot (\vec u - \vec x))(\cos\theta)^2\\
&= (u_t - x_t)^2 - ||\vec u - \vec x||^2(\cos\theta)^2 = 0
\end{align}

Now we have one equation and five unknowns: three spatial coordinates, one temporal coordinate, and the cone's aperture $\theta$, which is effectively the velocity at which the signal propagates. This suggests we need at least five black boxes to uniquely solve the problem.

Caveats:

I am casually extending the above formula to four dimensions. Not sure if that works.

Not just any 5 points will do: if, for instance, these 5 points lie on a line in 4D space, the system of equations will be underdefined.

I don't know whether this could be solved with fewer than five points.


问:
Can one prove Pappus hexagon theorem with cross ratios alone
Can one prove Pappus theorem knowing that projections don't change the cross ratio and that $(A,B;C,D) = (A,B;C,E) \iff D=E$?

I was reading this in exercise 6 they say we can prove Pappus theorem with projections and cross ratio, but they don't quite do it.

Pappus theorem is this: Given $A,B,C$ colinear in $r$ and $D,E,F$ colinear in $s$, then $X = AE \cap BD, Y =BF \cap CE, Z= CD \cap AF$ then $X,Y,Z$ are colinear. They hint use to define $P = BF \cap DC$ and $Z' = XY \cap CD$.

答:
This is a proof using cross ratios from S. Dobos Cross ratio in use, The Mathematical Gazette,
Vol. 95, No. 534 (November 2011), pp. 444-453:


问:
Does Norm imply Inner Product? If not, what's a counter-example?
I'm currently working my way through the book "Linear Algebra Done Right" by Sheldon Axler.

In the book, he defines a norm the following way.

Let $V$ be an inner product space and let $\langle u, v\rangle$ denote the inner product of $u$ and $v$. Then, the Norm of any vector $v \in V$ is defined as $||v|| = \sqrt{\langle v, v\rangle}$.

However, I was digging through Wikipedia and I discovered the concept of a "Normed Vector Space", i.e. a vector space on which a norm is defined. Inner product spaces seemingly constitute a subset of normed vector spaces.

My question is this - how is it possible to have a normed vector space that is not an inner product space if it seems like the definition of a norm depends on the presence of an inner product?

答: Not quite. If we have an inner product, ee can use it to define a norm. But there are norms that are induced by no inner product. For instance,$$\|(x,y)\|=|x|+|y|.$$


问:
Does this linear combination of vector norms describe a convex set?
Consider the surface identified by the following implicit equation
\begin{equation}
|\vec{x}|+|\vec{y}|-|\vec{z}|-|\vec{x}+\vec{y}-\vec{z}|=0,
\end{equation}
with $(\vec{x},\vec{y},\vec{z})\in\mathbb{R}^{3N}$ and $|\cdot |$ being the Euclidean norm in $\mathbb{R}^N$.
Is such a surface convex in $\mathbb{R}^{3N}$? Are there some regions of $\mathbb{R}^{3N}$ in which the restriction of the surface to the region is convex?. Observe that this surface can be obtained by intersecting the following cones:
\begin{align}
t_x&=|\vec{x}|, \quad t_y=|\vec{y}|, \\
t_z&=|\vec{z}|, \quad t_x+t_y-t_z=|\vec{x}+\vec{y}-\vec{z}|,
\end{align}
where now $(t_x,\vec{x},t_y,\vec{y}, t_z, \vec{z})\in \mathbb{R}^{3(N+1)}$.

答:
You say that the case $N=1$ is "trivially convex".

I'm sorry but I don't agree at all.

I have been computing by hand at first, working at the elimination of the different absolute values, octant by octant, obtaining, as you say, one linear relationships, but not relationships with inequalities. Therefore the 3D set is a surface (S), not a volume, made of 8 planar sectors (6 with a 90° opening, 2 with a 60° opening) assembled as a double umbrella (symmetrical with respect to the origin).

Therefore (S) is not eligible to be a convex set: the midpoint of 2 points of (S), is - most often - not in (S). If you restrict (S) to one of these 8  planar sectors, you have a convex set...

Then, I have written a Matlab program (see below) that has given this, confirming my computations:

```<code>clear all;close all;hold on;axis equal
f= @(x,y,z)(abs(x)+abs(y)-abs(z)-abs(x+y-z));
a=3; N = 100; W=4;L='linewidth';
plot3([-a,a],[0,0],[0,0],L,W); % x axis
plot3([0,0],[-a,a],[0,0],L,W);
plot3([0,0],[0,0],[-a,a],L,W);
plot3([a,-a],[0,0],[a,-a],L,W);
plot3([0,0],[a,-a],[a,-a],L,W);
u = linspace(-a,a, N);
[X, Y, Z] = meshgrid(u, u, u);
D = f(X, Y, Z);
isosurface(X,Y,Z,D,-0.01);% levelset of f for value v=-0.01
view([-126,12])
xlabel('x');
ylabel('y');
zlabel('z');
</code>```

I f we consider now the region defined by $|x|+|y|-|z|-|x+y-z|<0$ we will get the interior of the umbrellas, with "nice level sets". See for example below the level set

$$|x|+|y|-|z|-|x+y-z|=-0.5$$

Body:
You say that the case $N=1$ is "trivially convex".

I'm sorry but I don't agree at all.

I have been computing by hand at first, working at the elimination of the different absolute values, octant by octant, obtaining, as you say, one linear relationships, but not relationships with inequalities. Therefore the 3D set is a surface (S), not a volume, made of 8 planar sectors (6 with a 90° opening, 2 with a 60° opening) assembled as a double umbrella (symmetrical with respect to the origin).

Therefore (S) is not eligible to be a convex set: the midpoint of 2 points of (S), is - most often - not in (S). If you restrict (S) to one of these 8  planar sectors, you have a convex set...

Then, I have written a Matlab program (see below) that has given this, confirming my computations:

clear all;close all;hold on;axis equal
f= @(x,y,z)(abs(x)+abs(y)-abs(z)-abs(x+y-z));
a=3; N = 100; W=4;L='linewidth';
plot3([-a,a],[0,0],[0,0],L,W); % x axis
plot3([0,0],[-a,a],[0,0],L,W);
plot3([0,0],[0,0],[-a,a],L,W);
plot3([a,-a],[0,0],[a,-a],L,W);
plot3([0,0],[a,-a],[a,-a],L,W);
u = linspace(-a,a, N);
[X, Y, Z] = meshgrid(u, u, u);
D = f(X, Y, Z);
isosurface(X,Y,Z,D,-0.01);% levelset of f for value v=-0.01
view([-126,12])
xlabel('x');
ylabel('y');
zlabel('z');

I f we consider now the region defined by $|x|+|y|-|z|-|x+y-z|<0$ we will get the interior of the umbrellas, with "nice level sets". See for example below the level set

$$|x|+|y|-|z|-|x+y-z|=-0.5$$


问:
How to find the limit of this function
We have the function $$\dfrac{\sqrt{n^4 + 100}}{4n}$$

I think the best method is by dividing by $n$, but I have no idea what that yields, mainly because of the square root.

答:
Perhaps another method I overlooked, stupidly enough:

$$ \dfrac{\sqrt{n^4 + 100}}{4n} = \dfrac{n^4 + 100}{4n^5 + 400n} = \dfrac{ n^4 + 100}{4n \cdot n^4 + 400n} = \dfrac{101}{404n} = \dfrac{1}{4n}$$

I'm not sure it is correct, I played around a bit with the algebra.

Title: How to find the limit of this function


问:
Categories and quivers / directed graphs associated
I am currently working on associating categories and quivers.

a) Let $C$ be an arbitraty category. I need to show that one can associate with $C$ a quiver $Q_C$ by sending an object $X$ in $C$ to a vertex and a morphism $f:X\longrightarrow Y$ to a directed arrow between vertices $X$ and $Y$. This seems so obvious to me that I don’t really know how to prove it explicitly. Can somebody help me here?

b) I was wondering whether the converse is also true, whether it is possible to associate a category with any quiver. I know that a quiver is a directed graph, ie. a collection of vertices joined by directed arrows. I think the answer is No, because if we take a quiver where two vertices $X$ and $Y$ are not joined by an arrow, this would mean that we don’t have a morphism between those objects. Or because if we take a quiver where there is no arrow from a vertex $X$ to itself, this would mean we don’t have an identity morphism for this object.  Is that the right argumentation or am I completely wrong?

答:
We can define a functor from the category of small categories to the category of quivers, which takes a category (viewed as a quiver with two operations – composition and identity – subject to equations expressing associativity and unit laws) to its underlying quiver, which is the operation you describe. The definition is obvious, because we're just forgetting some extra structure defined on top of the quiver (this sort of functor is called forgetful).

There is also a functor in the opposite direction, which takes a quiver, and freely adds composites for any pair of edges sharing a source and target, and adds an identity edge for each vertex. So, while a quiver can not necessarily be equipped with the structure of a category, we can always "complete" it to a category. This functor is related to the one we described above: it is left adjoint to the forgetful functor from quivers to small categories. This demonstrates that the construction of a category from a quiver is a free construction. (Intuitively, this says that the two operations are inverse in an appropriate categorical sense.)


问:
If $(B_t)$ is a Brownian motion, then $P(\sup_{t\in [0,\infty )}B_t=0)=P(\sup_{t\in [0,\infty )}B_t<\infty )$.
If $(B_t)$ is a Brownian motion, how can I prove that $$P(\sup_{t\in [0,\infty )}B_t=0)=P(\sup_{t\in [0,\infty )}B_t<\infty )\ \ ?\tag{1}$$

I can't use the fact that $P(\sup_{t\in [0,\infty )}B_t=\infty)=1 $ (because we use $(1)$ to prove this). The fact that  $$P(\sup_{t\in [0,\infty )}B_t=0)\leq P(\sup_{t\in [0,\infty )}B_t<\infty ) $$
is fine. For the converse inequality, I tried to use that $(tW_{1/t})$ is a Brownian motion, but I can't conclude. any idea ? :-)

答:
Let $X:=\sup B_t$, for all $0<a<b$, by scaling property, we imply that:
$$P(X<a)=P(X<b)$$
Hence, conclusion


问:
What is the solution of $x^3+x=1$?
According to Wolfram|Alpha, the solution of $x^3+x=1$ is approximate $0.68233$ or exactly this monstrosity:

$x_0=\frac{\sqrt[3]{\frac{1}{2}(9+\sqrt{93})}}{3^{\frac{2}{3}}}-\sqrt[3]{\frac{2}{3(9+\sqrt{93})}}$

$x^3+x=1$ is so simple, that I refuse to believe that this ugly construct is the simplest way.  Am I right?

Title: What is the solution of $x^3+x=1$?

Body: $$  \left( \frac{1}{2} \left( 1 + \sqrt{\frac{31}{27}} \right) \right)^{1/3} + \left( \frac{1}{2} \left( 1 - \sqrt{\frac{31}{27}} \right) \right)^{1/3}  $$


问:
What does 'Unique Common Irreducible Component' mean?
Im working through some stuff on representations of $S_n$. I came across a certain result and I am not sure exactly what it means.

We have some representations $V, W$ of $S_n$ (they are defined precisely but I dont think that is needed) and we are able to prove that
$$\dim \text{Hom}_{S_n}(V, W) = 1 $$
then they say "Let $Z$ be the unique common irreducible component of $V$ and $W$". I am not sure exactly what this means. Are we able to extract some dimension 1 representation from $V$ and $W$ since their hom space has dimension 1?

答:
Let $V_{\lambda}$ be the irreducible representations of $S_n$ (over $\mathbb{C}$ a priori, but in fact they're all defined over $\mathbb{Q}$) and write $V, W$ as a direct sum $V \cong \bigoplus_{\lambda} n_{\lambda} V_{\lambda}, W \cong \bigoplus_{\lambda} m_{\lambda} V_{\lambda}$ where the $n, m$ denote multiplicities. Schur's lemma gives

$$\dim \text{Hom}_{S_n}(V, W) = \sum_{\lambda} n_{\lambda} m_{\lambda}$$

(exercise), so this dimension is equal to $1$ iff $n_{\lambda} = m_{\lambda} = 1$ for a unique irreducible representation $V_{\lambda}$ (meaning $V$ and $W$ both have a component isomorphic to $V_{\lambda}$ with multiplicity $1$, which is therefore unique) and $n_{\lambda'} m_{\lambda'} = 0$ otherwise (meaning $V$ and $W$ do not have a component isomorphic to $V_{\lambda'}$ in common).

$Z$ refers to $V_{\lambda}$. In terms of the unique-up-to-scale nonzero intertwining operator $f : V \to W$, $f$ factors canonically as a composite $V \to Z \to W$, where $Z \cong \text{im}(f)$.


问:
Is 'x=+1 or x=-1' $\iff$ '(x+1)(x-1)=0' correct?
I am quite certain that 'x=+1 or x=-1' $\impliedby$ '(x+1)(x-1)=0' is true (you use this principle when solving quadratic equations).

However, I am less secure in affirming that  'x=+1 or x=-1'$\implies$ '(x+1)(x-1)=0' is also true (if it is true, then you can say 'x=+1 or x=-1' $\iff$ '(x+1)(x-1)=0' overall as suspected in my main question).

I am not sure if this latter statement is correct because when you substitute x=+1 in the first bracket (x+1)and x=-1 in the second bracket (x-1) of the expression (x+1)(x-1) the result becomes (+1+1)(-1-1) = -4 which clearly is not 0. This is different to when you substitute x=-1 in the first bracket and x=+1 in the second bracket where the result becomes (-1+1)(1-1) = 0.

Hence, it appears that actually 'x=+1 or x=-1' does not imply '(x+1)(x-1)=0' (because there are cases where it equals 4 instead). Therefore,  only 'x=+1 or x=-1' $\impliedby$ '(x+1)(x-1)=0' is true and this is our end statement.

Am I right in believing this?

答:
When you write $ P \; or Q \; $, this means that we have may be $ P $, may be $ Q$ and may be both.

for example
$$x>0 \; or\; x <2$$

But in your case, you cannot have $ P: x=1$ and $ Q:x=-1$.

You should write $$x=1 \;xor\;x=-1\iff (x-1)(x+1)=0$$
"xor" is the exclusive disjunction.


问:
Trying to get the idea of what the second fundamental form of a hypersurface is
Let $M \subset \mathbb{R}^n$ be a submanifold of codimension $1$. Suppose it is given by $M = f^{-1}(\{ 0\})$ for some $f: \mathbb{R}^n \to \mathbb{R}$. I was wondering if someone could give me an idea of what exactly the second fundamental form of $M$ measures, in particular what does it mean if the second fundamental form vanishes identically on $T_p M$ for some $p \in M$?

I vaguely understand that it measures curvature and I had the following idea which is possibly incorrect.

Suppose $\gamma: [-1, 1] \to M$ be a smooth function such that $\gamma(0) = p$.  Is it then true that:
the second fundamental form vanishes identically on $T_p M$ if and only if given any such $\gamma$ the second derivative of $f \circ \gamma(t)$ vanishes at $t =0$?

I would appreciate if someone could explain me the intuition behind the second fundamental form, and also if I my idea above is incorrect or not.. Any comments appreciated. Thank you!

答:
The second fundamental form of a hypersurface refers to different -equivalent- notions: sometimes it refers to a scalar bilinear form, sometimes, to an endomorphism, and sometimes, to a vector valued functions on the tangent space. These notions are equivalent but some authors chose to talk only about one or another. I will here present the idea behind the endomorphism notion.

Suppose $M \subset \mathbb{R}^n$ is an orientable hypersurface, and let $\nu$ be a normal vector field, that is a -smooth- function $\nu : p \in M \mapsto \nu(p) \in \mathbb{R}^n$ such that $\|\nu(p)\| = 1$ and $\nu(p) \perp T_pM$.

Suppose $M$ is just a linear -or affine- hyperplane. Then the function $\nu$ is constant, equal to the normal of the hyperplane. Thus, there is no local change in $\nu(p)$.

Suppose $M$ is the unit sphere. Then $\nu(p)=p$ because the normal of a point in the unit sphere is directed by this point. Locally, $p \mapsto \nu(p)$ has the same variation than $p$.

These two examples shows that the infinitesimal change in $\nu(p)$ gives some information on the local geometry of $M$: if $M$ is flat around $p$ (like an hyperplane), $\nu$ does not change around $p$, and if $M$ is convex (like a sphere) around $p$, $\nu$ changes.

The second fundamental form, defined in terms of endomorphisms and usually denoted by $S$, and also known as the shape operator or weingarten operator, is the derivative of the function $p\mapsto \nu(p)$. More precisely, if $\nabla$ denotes the covariant derivative -or Levi-Civita connexion- of $M$, and if $p\in M$, then $S(p)$ is defined to be the linear transform $v \in T_pM \mapsto \nabla_{v}\nu(p) \in T_pM$. In terms of derivation along a curve, if $\gamma : I \to M$ is a curve such that $\gamma(0) = p$ and $\gamma'(0) = v$, then
$$ S(p)v = {\nu(\gamma)'(0)}^{\perp}$$
where $\perp$ refers to the orthogonal projection onto $T_pM$ in $\mathbb{R}^n$.

Geometrically, $S(p)v$ measures how much the hypersurface $M$ is curved at $p$ in the direction of $v$. If it is vanishing, then $M$ looks flat at $p$ in the direction of $v$.

A property of $S$ is that it is a symmetric endomorphism of $T_pM$, that is $\langle S(p)v,w\rangle = \langle v,S(p)w\rangle$. Hence, $S(p)$ is a diagonalisable endomorphism with othonogonal eigenvectors: this says that there exist orthogonal eigen-directions in $T_pM$ for $S(p)$. They are the principal directions in which $M$ is curved at $p$. Moreover, if $S(p)$ is symmmetric -definite- positive, then $M$ is -strictly- convex at $p$.

The bilinear definition of the second fundamental form is the bilinear form associated to the symmetric endomorphism $S$, that is $II_p(u,v) = \langle S(p)u,v\rangle$. They have the exact same properties.

Note that $S$ is an extrinsic property of $M$: as an abstract manifold in itslef, $M$ does not have any notion of shape operator at all. It depends on the geometry of the ambiant space (here $\mathbb{R}^n$). One can remember this by remembering that $S$ is defined by a normal vector field on $M$ and then, refers to an ambiant space in which $M$ is embedded.

Edit There was an edit in the question stating that $M$ is defined to be the zero level-set of a smooth function $f : \mathbb{R}^n \to \mathbb{R}$, and asking if the second fundamental form $S$ is zero at $p\in M$ if and only if for every smooth path $\gamma : I \to M$ with $\gamma(0)=p$ one has $\left(f\circ \gamma\right)''(0)=0$. The answer is no for the simple reason that if $\gamma : I \to M$ is a smooth function, then $f\circ \gamma : I \to \mathbb{R}$ is always the null function (because $M = f^{-1}(\{0\})$!). Consequently this condition is always true.

For example, $\mathbb{S}^n \subset \mathbb{R}^{n+1}$ is defined as $f^{-1}(\{0\})$ with $f(x) = \|x\|^2 -1$. Any path $\gamma : I \to \mathbb{S}^n$ has $\|\gamma\|^2 = 1$ by definition, thus $f\circ\gamma$ is the constant null function.


问:
Can't argue with success? Looking for "bad math" that "gets away with it"
I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").

One example would be "cancelling" the 6's in

$$\frac{64}{16}.$$

Another one would be something like

$$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;.$$

Yet another one would be

$$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;.$$

Note that I am specifically not interested in mathematical fallacies (aka spurious proofs).  Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for all cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.

答:
I'm not sure where it originates from, but I first came across this wonderful example in the "Why I should not teach first-year analysis" section at the end of Samir Siksek's Introduction to Abstract Algebra lecture notes (having ignored the warning about irrepairable damage to my soul):

Let $\int = \int_0^x$. We wish to solve the integral equation $f + \int f = 1$. Obviously, we begin by factoring out the $f$, to obtain $f(1 + \int) = 1$.
Dividing both sides by $1 + \int$ then gives $f = (1 + \int)^{-1}$, and the binomial theorem allows us to rewrite that as \begin{align*}f(x) &=1 - \int 1 + \iint 1 - \iiint 1 + \cdots\\&= 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots \\ &= \sum\limits_{n=0}^\infty \dfrac{(-x)^n}{n!}\\&= e^{-x}.\end{align*}
Substituting that back into the original equation we note that $$e^x + \int_0^xe^{-y}dy = e^x - e^x + 1 = 1,$$ as expected.


问:
Laurent series for $\frac{1}{(z-1)(z+1)}$ centered at $z = 1$
practice problem got us stuck -- please help us!!! Thank you..<3

Find a Laurent series for $\frac{1}{(z-1)(z+1)}$ centered at $z=1$ and specify the region in which it converges.

What we did was looking at the region $|z-1|>2$, where the function is holomorphic. For this region we can use the formulae for Laurent series and $c_k=\frac{1}{2\pi i}\int_\gamma\frac{(w-1)^{-k-2}}{w+1}dw$. We are stuck at for which values of $k$ the integral goes to 0 and for other values of $k$ how to evaluate the integral...

Thanks!

答: Hint: partial fractions and geometric series.</p>


问:
How to create a 4-sided die and an 8-sided die such that the sum follows the same distribution as the sum of 2 standard 6-sided dice?
We know that the sum of two standard 6-sided dice follows:
$$P(X_1+X_2=2)=\frac{1}{36},\dots, P(X_1+X_2=6)=\frac{5}{36}$$ $$P(X_1+X_2=7)=\frac{6}{36}, P(X_1+X_2=8)=\frac{5}{36},\dots, P(X_1+X_2=12)=\frac{1}{36}$$
How to design a 4-sided die and an 8-sided die and the numbers on the dice and the probability of getting each face is chosen by the designer such that the sum of the dice has the same distribution as the sum of the above 2 standard dice?

答:
Mimic the generating function approach used for Sicherman dice.  You want a product $$p(x)q(x)=(x+x^2+x^3+x^4+x^5+x^6)^2=[x(x+1)(x^2+x+1)(x^2-x+1)]^2$$
with $p(0)=q(0)=0$, $p(1)=4$, and $q(1)=8$.  But $p(1)q(1)=36$, so $p(1)=4$ implies $q(1)=9 \not= 8$.  You can do it with a 4-sided die and a 9-sided die:
\begin{align}
p(x) &= x(x+1)^2 = x+2x^2+x^3 \\
q(x) &= x(x^2+x+1)^2(x^2-x+1)^2 = x+2x^3+3x^5+2x^7+x^9
\end{align}
That is, $(1,2,2,3)$ and $(1,3,3,5,5,5,7,7,9)$.

There are exactly two other possibilities, based on where you put the two copies of factor $x^2-x+1$.

I had missed the part about also deciding the probability of each face.  Here is such a solution for a 4-sided die and an 8-sided die:
\begin{matrix}
\text{face} &1 &2 &3 &4 &5 &6 &7 &8 \\
\hline
\text{die 1} &1/6 &1/3 &1/3 &1/6 \\
\text{die 2} &1/6 &0 &1/6 &1/6 &1/6 &1/6 &0 &1/6 \\
\end{matrix}


问:
How to prove that if a number is divisible by two other numbers, then it is divisible by their product
I would like to prove if $a \mid n$ and $b \mid n$ then $a \cdot b \mid n$ for $\forall n \ge a \cdot b$ where $a, b, n \in \mathbb{Z}$

I'm stuck.

$n = a \cdot k_1$

$n = b \cdot k_2$

$\therefore a \cdot k_1 = b \cdot k_2$

EDIT: so for fizzbuzz it wouldn't make sense to check to see if a number is divisible by 15 to see if it's divisible by both 3 and 5?

答:
If $a$ and $b$ are relatively prime, then $x a + y b = 1$ for some integers $x, y$ (Bezout's identity).
Then
$$
n = k_1 a \overset{\times yb}{\Leftrightarrow} (yb)n = (yk_1) ab\\
n = k_2 b \overset{\times xa}{\Leftrightarrow} (xa)n = (xk_2) ab
$$
Adding the two equations together we get $(xa + yb)n = (yk_1 + x k_2)ab \Leftrightarrow n = k (ab)$, where $k = yk_1 + xk_2$ is an integer. This proves that $ab \mid n$.


问:
What is the rank of the set of functions from $\omega$ to $\omega$?
My stab was to say each member is a set of pairwise distinct ordered pairs of $\omega$ into $\omega$, so each function has rank $\omega + 1$ , therefore the entire set of functions has rank $\omega+ 2$

Does this work?

答: HINT. $rank(S)=\sup \{rank(T)+1:T\in S\}.$


问:
Find the center and the radius of convergence of this complex series.
$$\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!} (z - \frac{1}{2}\pi)^{2n}$$
$$\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(2(n+1))!} (z - \frac{1}{2}\pi)^{2n}$$

So I know a power series is $\sum_{n=0}^{\infty} a_n(z-z_0)^n = a_0 + a_1(z-z_0) + a_2(z-z_0)^n...$ and that the radius is going to equal $\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n}\right |$ and if that value converges to 0, the radius is \infty.

So $\lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n}\right | = \left | \frac{(-1)^n}{(2n)!)}\right | \cdot \left | \frac{(2n+2)!}{(-1)^{n+1}}\right |$

So since the absolute value of (-1) is 1, we can say that portion is just equal to $1^n$ and they cancel.

Can someone help me finish the rest? IS the numerator just left with $(2n+1)(2n+2)$ so does this limit approach infinity?

What about this next problem .How do I get this into the correct form?

$$\sum_{n=0}^{\infty} (\frac{(2-i)}{(1+5i)}) z^n$$

答:
You offer two power series, but then write as if there were only one. It is rather confusing what you mean by including the second series. But I'll join you in ignoring its existence in favor of the first. There are a couple issues your use of the Ratio test here.

The first is that in the Ratio test $a_n$ refers to the coefficient of the $(z - c)^n$ term, not the term of index $n$ in however the series is expressed, as you have interpreted it here. Therefore
$$a_n = \begin{cases}\dfrac{(-1)^{n/2}}{n!}, &n \text{ even}\\0,&n\text{ odd}\end{cases}$$
And $\left|\frac{a_{n+1}}{a_n}\right|$ thus flips back and forth between being $0$ and being undefined, so the Ratio test fails.

However this is actually a simple problem to overcome in this case. Let $w = \left(z -\frac \pi 2\right)^2$. The series becomes $$\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}w^n$$ which is a power series to which the Ratio test can successfully be applied. But you have to be careful. When you find the radius of convergence $R$, that is the radius of the power series in $w$, not the power series in $z -\frac \pi 2$. The radius of convergence for the $z$ power series will be $\sqrt R$. (Do you see why?)

The second issue is simpler: you made a mistake in applying the formula

$$\left|\frac{a_{n+1}}{a_n}\right|\ne \left | \frac{(-1)^n}{(2n)!)}\right | \cdot \left | \frac{(2n+2)!}{(-1)^{n+1}}\right |$$
Examine it carefully to spot the problem.

Your second question is also unclear. What more "correct form" are you looking for? It is in the traditional form for a power series already.


问:
How would you approach finding the limit of the convergent sum 1/sqrt(n^2+k), k = 0 to k = n
I solved this question by using a wonky method and would like to hear different approaches to the solving the problem.

The task was to find out if the $S_n=\sum\limits_{k=1}^n \frac{1}{\sqrt{n^2+k}}$ converges and if so, what the sum converges to.

My wonky approach:
Found out that it convergences by creating a bigger sum that's easier to analyze. If the bigger sum converges => the smaller one does as well.

Bigger sum (n:terms): $n*\frac{1}{\sqrt{n^2+1}} = \frac{n}{n}*\frac{1}{\sqrt{1+\frac{1}{n^2}}} = \frac{1}{\sqrt{1+\frac{1}{n^2}}} $ which becomes 1 when n --> inf

Then I noticed that if you created a new sum with that smallest number in summation, it's limits is also 1

Then I saw that the last term converges aswell to 1

And therefor $ 1 \leq \lim Sn \leq 1$ <=> $\lim Sn = 1$

答:
Essentially you are correct, though you want to tidy a few things, such as the title which starts at $k=0$ while the question starts at $k=1$.

You have essentially said (using the easier $n^2+k > n^2$) $$\frac{1}{\sqrt{1+\frac1n}}=\sum\limits_{k=1}^n \frac{1}{\sqrt{n^2+n}} \le \sum\limits_{k=1}^n \frac{1}{\sqrt{n^2+k}} \le \sum\limits_{k=1}^n \frac{1}{\sqrt{n^2}}= 1$$

and the limits of the left-hand and right-hand expressions are both $1$ as $n\to \infty$, so the limit of the middle expression $S_n$ sandwiched between them is also $1$.


问:
Complete randomness/disorder and determinism
Wikipedia page about randomness says that "complete disorder" and "true randomness" are impossible according to Ramsey Theory and Cristian S. Calude. I don't understand it.

https://en.wikipedia.org/wiki/Randomness

So if I randomly select numbers from an infinite set of numbers e.g. from a set of complex numbers, will there always be some formula or pattern or a way to predict successive numbers from the previous ones?

There are no nondeterministic theories, patterns, methods or elements in mathematics?
What with quantum mechanics in Copenhagen way of understanding it?
What with probability theory and stochastic process?

Thanks!

答:
You're essentially right. Ramsey theory says that, in any sufficiently big system, there will be some small (often very small compared to the whole system) parts that look orderly.  That doesn't contradict randomness or disorder in the system as a whole.

For example, if I flip a coin 100 times to produce a sequence of "heads" and "tails", you can pick out 50 of those terms that are all heads or all tails. Those 50 terms look very orderly, but that doesn't contradict the randomness of the whole sequence of 100 terms. It doesn't mean that the coin was biased or that I cheated.

In nontrivial theorems of Ramsey theory (as opposed to the pigeonhole principle used in the preceding paragraph), the orderly part of the system is generally much smaller, not half of the whole system (50 out of 100 coin flips) but maybe the logarithm (or log log, or even smaller) of the system size.


问:
how many empty sets are there?
Would I be correct in saying that in the category of sets, the "class of sets that are isomorphic to the empty set is a proper class"?

In other words, there are LOTS of initial objects in the category of sets, but they're all related to each other via unique isomorphisms?

Similarly, for any category $C$, if $X$ is an object of $C$, then the full subcategory of objects isomorphic to $X$ is a proper class (ie, there are LOTS of them)?

EDIT: Let me elaborate a bit. I'm not entirely sure if I should be considering ZFC or ETCS, etc. I'm asking this question from the point of view of algebraic geometry and its use of category theory. I mean, I know people say that the class of say, all elliptic curves over $\mathbb{C}$ is a proper class, but the class of $\mathbb{C}$-isomorphism classes of elliptic curves is a set. That would seem to imply that every isomorphism class is a proper class, and hence this would seem to say that the isomorphism class (ie, bijection class) of the empty set in the category of sets should also be a proper class, right?

Lana's answer with the extensionality axiom seems to make sense, in that there is only ONE empty set, so that the isomorphism class of the empty set is a singleton set. On the other hand, the isomorphism class of a set with 1 element should be a proper class right?

Generalizing a bit, would it be true that the isomorphism class of any initial (or terminal) object of a category be a singleton set? It seems to me that the isomorphism class of, say $\text{Spec }\mathbb{Z}$ in the category of schemes is a proper class, simply because there are many different presentations of rings isomorphic to $\mathbb{Z}$?

will

答: By extensionality axiom you get that there exists exactly one empty set, and that's all folks!</p>

Title: how many empty sets are there?


问:
Show that $f_ng_n \to fg$ in $B(X)$.

If $f_n \to f$ and $g_n \to g$ in $B(X)$, show that $f_ng_n \to fg$ in $B(X)$.

Note that given a set $X$, $B(X)$ is the vector space of all bounded, real-valued functions $f: X \to \mathbb{R}$ and we supply $B(X)$ with the sup-norm $||f||_{\infty} = \sup\limits_{x \in X}|f(x)|$.

Given that  $f_n \to f$ and $g_n \to g$ in $B(X)$; which is precisely $||f_n-f||_{\infty} \to 0$ and $||g_n-g||_{\infty} \to 0$. Thus we can find some $N \in \mathbb{N}$ such that for every $\epsilon > 0$, $$||f_n-f||_{\infty} < \frac{\epsilon}{2||g_n||_{\infty}} \ \ \ and \ \ \ ||g_n-g||_{\infty} < \frac{\epsilon}{2||f||_{\infty}}, \ \ \ \forall n\geq N. $$ Then we may compute the following: $$||f_{n}g_{n}-fg||_{\infty} \leq ||f_{n}g_{n}-fg_{n}||_{\infty} + ||fg_{n}-fg||_{\infty} \leq ||f_{n}-f||_{\infty}||g_n||_{\infty}+||g_{n}-g||_{\infty}||f||_{\infty}.$$ Therefore, $||f_{n}g_{n}-fg||_{\infty} < \epsilon$ so  $||f_{n}g_{n}-fg||_{\infty} \to 0$. It follows that $f_ng_n \to fg$ in $B(X)$.

Is this a correct argument for this problem? There was a lot of attention to detail when setting up and performing the triangle inequality, so I just want to be sure.

答:
First, since $f_n\to f$, in the $\infty-$ norm, then for $\varepsilon=1$, there exists an $N$, such that for $n>N$,
$$
\|f_n-f\|<1\quad\Longrightarrow\quad \|f_n\|< \|f\|+1.
$$
Hence
$$
\|f_n\| \le M=\max\{\|f\|+1, \|f_1\|,\ldots,\|f_N\|\}.
$$

Therefore
$$
\|f_ng_n-fg\|\le \|f_ng_n-f_ng\|+\|f_ng-fg\|\le \|g\|\|f_n-f\|+\|f_n\|\|g_n-g\|\\ \le
\|g\|\|f_n-f\|+M\|g_n-g\|\to 0.
$$

Title: Show that $f_ng_n \to fg$ in $B(X)$.


问:
How to determine the rotation angles of a telescope
I have two control buttons on my telescope, one to increase/decrease the right ascension (rotation around z-axis) and one to increase/decrease the declination (rotation around y-axis). My first target has a right ascension of 10 deg and declination of 20 deg. First I command a 20-deg rotation around the y-axis and then a second rotation of 10 deg about the z-axis. Now my telescope (original pointing direction to (1, 0, 0)) points to
$$ (0.925416 \hspace{3mm} 0.163176 \hspace{3mm} 0.342020)$$
The second target has a right ascension of 20 deg and declination of 40 deg. If I repeat the same matrix multiplications as before (20-deg rotation around the y-axis followed by a rotation of 10 deg about the z-axis) I am not ending up exactly at right ascension of 20 deg and declination of 40 deg. This is the result:
$$ (0.712860 \hspace{3mm} 0.291390 \hspace{3mm} 0.637905)$$
But I should point to:
$$ (0.719846 \hspace{3mm} 0.262003 \hspace{3mm} 0.642788)$$

My question is, how do I calculate the correct rotation angles for the third and the fourth rotation? I could numerically solve for the unknown rotation angles, but I wonder if there exists an analytical solution where I do not need a root finder. I am grateful for any help.

答:
I assume that when your telescope is pointing to $(1,0,0)$, it is pointing at the equator on the local meridian. If so, the declination axis is aimed at the point due east: $(0,1,0)$ and the polar axis is pointing at the north celestial pole: $(0,0,1)$.

The matrix for rotating $\alpha$ in right ascension is
$$
R_\text{A}(\alpha)=
\begin{bmatrix}
\cos(\alpha)&\sin(\alpha)&0\\
-\sin(\alpha)&\cos(\alpha)&0\\
0&0&1
\end{bmatrix}
$$
The matrix for rotating $\delta$ in declination is
$$
R_\text{D}(\delta)=
\begin{bmatrix}
\cos(\delta)&0&\sin(\delta)\\
0&1&0\\
-\sin(\delta)&0&\cos(\delta)
\end{bmatrix}
$$
However, since matrix multiplication is not commutative, $R_\text{D}$ is only valid for rotating when the scope is aimed on the local meridian. As the right ascension changes, the declination axis is rotated. So to see the effect of rotating $\delta$ on the declination axis after your telescope has rotated a total of $\alpha$ in right ascension, you would use the matrix
$$
R_\text{D}(\delta\vert\alpha)=R_\text{A}(-\alpha)\,R_\text{D}(\delta)\,R_\text{A}(\alpha)
$$
$R_\text{D}(\delta\vert\alpha)$ undoes the action of rotating $\alpha$ in right ascension, performs the rotation of $\delta$ in declination, then restores the action of rotating $\alpha$ in right ascension.

So, as you computed:
$$
(1,0,0)\,R_\text{D}\!\left(20^{\large\circ}\right)R_\text{A}\!\left(10^{\large\circ}\right)=(0.925417, 0.163176, 0.342020)
$$
But then, using $R_\text{D}\!\left(20^{\large\circ}\vert10^{\large\circ}\right)$
$$
(1,0,0)\,R_\text{D}\!\left(20^{\large\circ}\right)R_\text{A}\!\left(10^{\large\circ}\right)R_\text{D}\!\left(20^{\large\circ}\vert10^{\large\circ}\right)R_\text{A}\!\left(10^{\large\circ}\right)=(0.719846, 0.262003, 0.642788)
$$
To repeat the process, you would use $R_\text{D}\!\left(20^{\large\circ}\vert20^{\large\circ}\right)$, since you have rotated a total of $20^{\large\circ}$ in right ascension.


问:
Solution to complex differential equation $f'(z)=\frac{h(z)}{z^2+1}$
I'm trying to solve a problem in complex analysis: Q 4E of the 2012 IB mathematical Tripos paper 4 section I. This is the link to the question:
https://www.maths.cam.ac.uk/undergrad/pastpapers/files/2012/PaperIB_4.pdf

$Let \; h:\mathbb{C}\to \mathbb{C}$ be holomorphic with $h(i) \neq h(-i)$.

Does there exist a holomorphic function $f$ defined in $|z|<1$ for which $f'(z)=\frac{h(z)}{z^2+1}$ ?

Does there exist a holomorphic function $f$ defined in $|z|>1$ for which $f'(z)=\frac{h(z)}{z^2+1}$ ?

I'm not sure how to even begin solving this problem. Any help would be greatly appreciated.

答:
Let $\gamma$ be a curve contained in $|z|<1$. It is easy to see that $\frac{h(z)}{z^2+1}$ is holomorphic in $\text{int}(\gamma)$, and thus by Cauchy's theorem
$$\oint_\gamma \frac{h(z)}{z^2+1}dz=0$$
Since this is true for every closed curve, $\frac{h(z)}{z^2+1}$ has a primivite $f$ inside the open unit disc (by the fundamental theorem of line integrals), and $f$ is holomorphic

Now, let $\gamma$ be a curve contained in $|z|>1$ encircling the origin. By the residue theorem,
$$\oint_\gamma\frac{h(z)}{z^2+1}dz=2\pi i\left(\text{Res}_{z=i}\frac{h(z)}{z^2+1}+\text{Res}_{z=-i}\frac{h(z)}{z^2+1}\right)$$

Since $\frac{h(z)}{z^2+1}=h(z)\frac{1}{(z+i)(z-i)}$, the two residues are $\frac{h(i)}{2i}, \frac{h(-i)}{-2i}$. Thus:
$$\oint_\gamma\frac{h(z)}{z^2+1}dz=\pi (h(i)-h(-i))\neq 0$$
Thus, by the fundamental theorem for line integrals, $\frac{h(z)}{z^2+1}$ does not have a primitive there.


问:
What is the radius of convergence of a power series in two variables?
What is the radius of convergence of a power series in two real variables?

If I were to fix one of the variables (i.e. make it a real constant), then would the radius of convergence simply be related to the nearest singularity?

答:
A power series in several complex variables doesn't have a radius of convergence. In general, the domain of convergence is a logaritmically convex Reinhardt domain, but not necessarily a ball. Consider for example the series
$$\sum_{k=0}^\infty z^k w^k \qquad\text{and}\qquad \sum_{j=0}^\infty \sum_{k=0}^\infty z^j w^k$$
respectively.</p>


问:
Isometry group G is a compact?
Hello everyone I found the following statement in my lecture but I'm not sure why it's true.

Let $(X,d) $ a metric space (compact). Hence we can say that the isometry group G is a compact in $(C(X,X),d_\infty)$ i.e $G$ is a family of closed and equicontinuous applications.

I thought that using Arzelà-Ascoli theorem would help but it leads nowhere. Thanks in advance for your help.

答:
Hint: Let $f_n$ be a sequence of isometries. Consider $h_n(x)=d(x,f_n(x))$
show that the family $h_n$ is absolutely bounded and equicontinuous.

Ascoli implies the existence of a subsequence $h_p$ which converges towards $h$; $h_p(x)=d(x,f_p(x))$ is a Cauchy sequence implies that $(f_p(x))$ is Cauchy and $f_p(x)$ converges towards $f(x)$. Show that $f$ is an isometry.

Title: Isometry group G is a compact?


问:
What does it mean for the partial derivatives of $f(x, y$) to be $0$ in the outline of the method of exact equations?
In the outline of the method of exact equations given in the textbook below the authors write  $\partial f/\partial y \equiv 0$ and $\partial f/\partial x \equiv 0$ and denotes these as $M(x, y)$ and $N(x, y)$ and I'm very much confused by this. Does this mean $M(x, y)$ and $N(x, y)$ have to equate to $0$? Doesn't make sense with the examples given later in the text. I'm guessing the answer is a basic fact that I'm overlooking.

Outline of the method (Differential Equations by Simmons and Krantz)

答:
Note that when you're solving an exact differential equation, you're looking for a solution in the form $f(x,y)=c$. Since this is the solution of the differential equation, it's derivative with respect to $x$ will be our original equation. Hence, when you implicitly take the derivative of $f(x,y)=c$ you get the total derivative
$$
\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=0.
$$
For exact differential equations, we represent the two partial derivatives as $M$ and $N$ respectively. The reason your textbook says that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are identically zero is because if we were to implicitly take the partial derivative of $f(x,y)=c$ with respect to $x,$ we can easily see the result would be $\frac{\partial f}{\partial x}=0.$ However, the partial derivative here (given by $M(x,y)$ in the differential equation) is still a function, for which all legitimate values of $x$ and $y$, the result is zero.


问:
Find a expression using pgf $G_X$ for the probability that $X$ is divisible by 4.
A random variable $X$ has probability generating function $G_X$.

(a) Find a simple expression using $G_X$ for the probability that $X$ is even. Hint: consider the value of $G_X(-1)$.

(b) Suggest a similar expression for the probability that $X$ is divisible by 4 -- be creative about what values of the generating function you might evaluate!

I had no problem with solving a simpler question in part (a), but am really struggling with part (b).

答:
In part (a) you used two numbers such that both are square roots of 1. Can you extend this to something that involves 4? Hint: you need 4 numbers.

 Find a, b, c, d, such that $$aG(i)+bG(-1)+cG(-i)+dG(1) = \sum_{4|n} p(n). $$


问:
Proving a Sobolev norm inequality
I need to show that $\|u\|_{H^1}^2\leq \|u\|_{L_2}\|u\|_{H^2},\forall u\in H^2\cap H^1_0$ using the Helmholtz equation: \begin{align*}-\Delta u+u&=f,\text{ in }\Omega\\u&=0,\text{ on }\partial\Omega\end{align*}

My attempt:

From the Helmholtz equation I got the bilinear form $a(u,v)=(\nabla u,\nabla v)+(u,v)$ where $(\cdot,\cdot)$ means $L_2$ inner product. From another proof, I know that $(u,v)\leq \|u\|_{H^{-1}}\|v\|_{H^1},\forall u\in H^{-1}, v\in H^1$ and so consequntly I said that $(u,u)\leq \|u\|_{L_{2}}\|u\|_{H^2}$ since both the $L_2$ and $H^2$ norms are greater than or equal to $H^{-1}$ and $H^1$ norms respectively.

Then I thought of getting a similar bound for the first term $(\nabla u,\nabla v)$ in the bilinear form and using the fact that this bilinear form is coercive to finally get to the desired inequality. But I cannot figure out how to get that upper bound.

Is this the correct way to prove the result? If so, how to get the upper bound for that first term?

答:
If $u,v\in H^2(\Omega)\cap H_0^1(\Omega)$, then
$$
\int_{\Omega}u\Delta v\,dx=-\int_{\Omega}\nabla u\cdot\nabla v\,dx.
$$
Hence
$$
\int_{\Omega}u\Delta u\,dx=-\int_{\Omega}|\nabla u|^2\,dx
$$
and thus
$$
\|u\|_{L^2}\|u\|_{H^2} \ge \int_{\Omega}u(u-\Delta u)\,dx=\int_{\Omega}(u^2+|\nabla u|^2)\,dx=\|u\|_{H^1}^2
$$

which implies tha

Title: Proving a Sobolev norm inequality


问:
If $L/K$ is a field extension then $\operatorname{char}(L) = \operatorname{char}(K)$
I need to show that If $L/K$ is a field extension then $\operatorname{char}(L) = \operatorname{char}(K)$. This is what I've tried:

Well, if $\operatorname{cha}(K) = p$, then $1_{K} + 1_{K} + \dots 1_{K} = 0$ (sum of $p$'s $1_{K}$). But $1_{K} = 1_{L}$ since $L/K$ is a  field extension. So $\operatorname{char}(L)=p$.

Is this right? Thank you!

答:
For every ring with identity $R$, there is a unique ring homomorphism $\chi_R\colon\mathbb{Z}\to R$, which sends $n$ to $n\mathbf{1}$ (the integral multiple of the identity; I'll use $\mathbf{1}$ to denote the identity in $R$, for better clarity).

There are two cases: either $\ker\chi_R=\{0\}$ or $\ker\chi_R=n\mathbb{Z}$, for some (unique) $n>0$. In the former case the characteristic of $R$ is $0$; in the latter case the characteristic is $n$, because $n\mathbf{1}=0$, but $m\mathbf{1}\ne0$ for every $m$ with $0<m<n$.

Thus if $R$ is a subring of $S$, the characteristic of $R$ and $S$ must be the same, because by uniqueness of $\chi_R$ we have that $\chi_R$ is the restriction to $R$ of $\chi_S$.


问:
Counterexample to domain additivity of uniformly continuous function
What I'm looking for is a counter-example to the following question:

Let $A,B\subseteq\mathbb{R}$ be closed sets and $f$ a real-valued function of a real variable such that its restrictions to $A$ and $B$ respectively are uniformly continuous. Is $f$ uniformly continuous on $A\cup B$?

Those are my considerations so far:

$A$ and $B$ must be unbounded, otherwise, they would be compact and so would be their union. Thus, since $f$ is surely continuous on $A\cup B$, $f$ would be uniformly continuous as well.

$A$ and $B$ cannot be unbounded intervals, otherwise, we can prove again (by an $\epsilon-\delta$ argument) that $f$ is uniformly continuous on $A\cup B$.

$A$ and $B$ cannot have the discrete topology (like $\mathbb{N}\subseteq\mathbb{R}$), otherwise uniform continuity is immediate.

This is easily achieved by considering infinite unions instead of finite ones, so I'm not interested in that either.

Could somebody provide any hint on what I could try? Also, the counter-example can be as complex as needed (supposing I didn't miss something very trivial) so feel free to do your "worst".

答: Let $A=\mathbb N$, let $B=\{n+2^{-n-1}:n\in\mathbb N\}$, and let $f$ be identically $0$ on $A$ and identically $1$ on $B$.


问:
Prove that $\lim_{}{x_{n}}=L\iff L=x_{1}+\sum_{k=1}^{\infty }{(x_{k+1}-x_{k})}$
Let $(x_{n})$ be a sequence of real numbers. Show that $\lim_{}{x_{n}}=L\iff L=x_{1}+\sum_{k=1}^{\infty }{(x_{k+1}-x_{k})}$.

I've tried using the fact that the sequence $(x_{n})$ can be expressed as a series of partial sums, taking for example:

\begin{equation}
a_{1}=x_{1}\\
a_{n+1}=x_{n+1}-x_{n}\\
\Rightarrow a_{1}+\cdots+a_{n}=x_{1}+(x_{2}-x_{1})+\cdots+(x_{n+1}-x_{n})=x_{n}
\end{equation}

But I dont' know where to go next to prove the $\Rightarrow $ and $\Leftarrow $ statements. I'm thankful for any kind of help.

答:
You are almost there.

As you almost wrote,
$s_n
=x_{1}+\sum_{k=1}^{n-1}{(x_{k+1}-x_{k})}
=x_{n}
$
and since
$\lim_{n \to \infty} s_n
=\lim_{n \to \infty} x_n
=L
$.

Since
$\lim_{n \to \infty} s_n
$
is the definition of
$x_1+\sum_{k=1}^{\infty}(x_{k+1}-x_k)
$,
you are done.


问:
$\lVert A^{-1} \rVert \ge \frac{1}{\lVert A-B \rVert}$ for norm $\lVert•\rVert$ and nonsingular (matrix) $A$ and singular $B$?
I have a question. For any norm $\lVert•\rVert$ and any nonsingular matrix $A$ and singular matrix $B$,
$$\lVert A^{-1} \rVert \ge \frac{1}{\lVert A-B \rVert}$$

How can we show that?

答:
It is not true for any square matrix norm. For example, if $\|\cdot\|_1$ is a norm, so is $\|\cdot\|_a=a\|\cdot\|_1$, for every $a>0$, and observe that for sufficiently small $a>0$,
we have $\|A^{-1}\|_a\|A-B\|_a=a^2\|A^{-1}\|_1\|A-B\|_1<1$.

However, it is true for every induced norm, i.e., norm of the form
$$
\|E\|=\sup_{\|x\|=1}\|Ex\|,
$$
where $\|\cdot\|$ is an arbitrary norm in $\mathbb R^n$.

Proof. Let $\|\cdot\|$ be an arbitrary norm in $\mathbb R^n$. Since $B$ is singular, there exists an $x_0\ne 0$, such that $Bx_0=0$. In fact, we may choose $x_0$, so that $\|x_0\|=1$.
So
$$
\|I-A^{-1}B\|=\sup_{\|x\|=1}\|(I-A^{-1}B)x\| \ge \|(I-A^{-1}B)x_0\|=\|x_0\|=1.
$$
But a property of induced norms is $\|EF\|\le \|E\|\|F\|$. Hence
$$
\|A^{-1}\|\|A-B\|\ge\|A^{-1}(A-B)\|=\|I-A^{-1}B\|\ge 1
$$
and thus
$$
\|A^{-1}\|\ge \frac{1}{\|A-B\|}.
$$


答:
Assume $\|A^{-1}\| < \frac1{\|A-B\|}$. Then using the submultiplicativity of the norm we have

$$\|I-A^{-1}B\| = \|A^{-1}(A-B)\| \le \|A^{-1}\|\|A-B\| < 1$$

so $A^{-1}B$ is invertible with inverse $\sum_{n=0}^\infty (I-A^{-1}B)^n$ which converges since it converges absolutely. Hence $B$ is invertible as well, which is a contradiction.

Title: $\lVert A^{-1} \rVert \ge \frac{1}{\lVert A-B \rVert}$ for norm $\lVert•\rVert$ and nonsingular (matrix) $A$ and singular $B$?

Body:
I have a question. For any norm $\lVert•\rVert$ and any nonsingular matrix $A$ and singular matrix $B$,
$$\lVert A^{-1} \rVert \ge \frac{1}{\lVert A-B \rVert}$$

How can we show that?


问:
The number of operations appear in the multiplication of two upper triangular matrices
Imagine $A$ and $B$ are both $n\times n$ upper triangular matrices. How many operations exist in the $AB$?

I know that multiplication of two upper triangular matrices is upper triangular and operations that appear in finding the main diagonal is $n$, but I can't find how many operations appear in the $j>i$ matrix elements, can someone help me?

答:
The $\ i^\text{th}\ $ row or column of an $\ n\times n\ $ upper triangular matrix contains $\ n-i+1\ $ entries that could be non-zero. If all those entries in both factors are in fact non-zero, and you compute the $\ ij\ $ entry of the product (for $\ i\le j\ $) by taking the dot product of the $\ i^\text{th}\ $ row of the first factor with the $\ j^\text{th}\ $ column of the second, then that operation will require $\ n-j+1\ $ multiplications and $\ n-j\ $ additions.  The total numbers, $\ M\ $, of multiplications, and, $\ A\ $, of additions, are therefore given by
\begin{align}
M&=\sum_{i=1}^n\sum_{j=i}^n(n-j+1)\\
&=\frac{n(n+1)(n+2)}{6}\\
A&=\sum_{i=1}^n\sum_{j=i}^n(n-j)\\
&=\frac{(n-1)n(n+1)}{6}\ .
\end{align}
Since multiplications are significantly more computationally expensive than additions, it doesn't make much sense to add these together to get the total number of "operations".

You should also be made aware that there are matrix multiplication algorithms that require asymptotically fewer than the $\ O(n^3)\ $ operations needed for the straightforward method of multiplication treated above, so the above results are only valid for that specific method.


问:
$f\in C^2[-1,1]\text{ then }\exists \xi\in(-1,1)\text { such that }\displaystyle\int_{-1}^1 x^2f(x)\mathrm dx=2f(0)+\frac15f''(\xi)$
I am trying to solve the following problem :

$f\in C^2[-1,1]\text{ then }\exists \xi\in(-1,1)\text { such that }\displaystyle\int_{-1}^1 x^2f(x)\mathrm dx=2f(0)+\frac15f''(\xi)\tag*{*}$

If we write $$f(x)=f(0)+xf'(0)+\frac{x^2}2f''(\zeta)\text{ for some }\zeta\in(-1,1)$$

then $\displaystyle\int_{-1}^1 x^2f(x)\mathrm dx=\frac23f(0)+\frac15f''(\zeta)$

But I can't able to find the form of $(*)$

I've also tried expanding $x^2f(x)$ around $0$:

$$x^2f(x)=\left.x^2f(x)\right|_{x=0}+x\cdot\left.(x^2f(x))'\right|_{x=0}+\frac{x^2}2\cdot\left.(x^2f(x))''\right|_{x=\zeta},\text{ for some }\zeta\in(-1,1)$$

but this doesn't lead to $(*)$

So my question is how to proceed with this problem.

答:
Your conclusion is not always true. Consider the function $f(x)=1, \forall x\in [-1,1]$.

We have $\displaystyle \int_{-1}^1x^2f(x)dx=\int_{-1}^1x^2dx=\frac{2}{3}$ and $\displaystyle 2f(0)+\frac{1}{5}f''(\xi)=2$ for any $\xi\in (-1,1)$


问:
Can we prove a right continuous real-function has left limits everywhere?
càdlàg function is basically a function defined on the real numbers (or a subset) that is everywhere right-continuous and has left limits everywhere.
Can right-continuity be the sufficient condition for a function to be a càdlàg function? In other words can we derive left limit on domain from right continuity?</p>

答:
NO.

Consider the function
$$
f(x)=\left\{
\begin{array}{ccc}
\sin(1/x) & \text{if} & x<0, \\
0 & \text{if} & x\ge 0.
\end{array}
\right.
$$
Then $f$ is right continuous everywhere, but has no left limit at $x=0$.


问:
Placing $18$ numbered teams into $3$ groups.

Problem

There are $18$ football teams. Each of them randomly gets placed into $3$ possible groups
such that each group contains exactly $6$ teams. The teams are numbered from $i = 1, \ldots,18$.

Without loss of generality consider group $1$.
Let $X_{\min}$ be a random variable which maps group $1$ to the lowest team number contained in it.

Find the distribution of $X_{\min}$.

Approach

Let's look at an example: Denote group $1$ as $G_1$ and say $G_1 = \{8,4,3,9,5,12\}$. The smallest number in this set is $3$. We can easily observe that $X_\min = 3$ if and only if all teams with numbers $k < 3$ were assigned to either group $2$ or group $3$.

Let $A_i$ be the event that team $i$ gets assigned to group $1$. Obviously $P(A_i) = \frac{1}{3}$.
Then the probability that some team $i$ does not get assigned to group $1$ is $P(\overline{A_i}) = 1 - P(A_i) = \frac{2}{3}$.

So in general we get
$$
    P(X_\min = k) = P(A_k \cap \overline{A_{k-1}} \cap \cdots )
$$

My question: can we write $P(X_\min = k) = P(A_k) \cdot P(\overline{A_{k-1}}) \cdot \ldots$

If not (which I think), how can I calculate this probability?

Maybe looking at arbitrary sets $\{k, x_2, x_3, x_4, x_5, x_6\}$, then choosing $5$ numbers from the remaining upper $18-k$ numbers to get all such $\binom{18 - k}{5}$ sets, so we are left with the binomial distribution:
$$
    P(X_\min = k) = \binom{18 - k}{5} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^5
$$

I feel like I'm missing something.

答:
Isn't it the same as: you pick 6 different numbers from $1, 2, \ldots, 18$ and find the distribution of the smallest?

So $$\Pr(X_\min = k) = \frac{{18-k \choose 5}}{{18 \choose 6}}, \forall 1\leqslant k \leqslant 13.$$