问:
Bell Numbers and Equating Coefficients
I need help solving the following exercise:

Let B(n) be the Bell - Numbers of $[n]$ and $$\exp(x) := \sum_{n\geq 0} \frac{x^n}{n!}$$ the exponential function.

Prove by equating the coefficients and using
$$ B(n) = \sum\limits_{a_1+\cdots+a_k=n, a_i \geq 1} \frac{1}{k!} \binom{n}{a_1 \cdots a_k}$$ that for $n \geq 0$
$$\sum\limits_{n \geq 0} B(n)/n! \cdot x^n = \exp(\exp(x) - 1)$$

I tried to play around with the equation but haven't gotten anywhere yet.

For the left side I am not sure how to handle the range of the 2nd sum:

$$\sum\limits_{n \geq 0} B(n)/n!\cdot x^n = \sum\limits_{n \geq 0} \frac{\sum\limits_{a_1+\cdots+a_k=n, a_i \geq 1}\frac{1}{k!}\binom n {a_1 \cdots a_k}}{n!}\cdot x^n$$

How do I get this simplified and converted to a polynomial form?

I know that
$$\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}
= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
But how do I proceed for $\exp(\exp(x)-1) = e^{e^x-1}$?

Thank you in advance!

答:
We are interested in expansion of the term $\exp(\exp(x)-1)$, now $$\exp(x) - 1 = \sum_{n=1}^\infty \frac{x^n}{n!}$$ and since $\displaystyle \exp(z) = \sum_{m=0}^\infty \frac{z^m}{m!}$ we should start by considering $\displaystyle \left(\sum_{n=1}^\infty \frac{x^n}{n!}\right)^m$ for $m=2,3,\ldots$, here is a table just for $m=2$, extrapolation will give all other $m$:

$$\begin{array}{c|l}
\sum_{n=1}^\infty \frac{x^n}{n!} & \frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\ldots \\
\times \frac{x^1}{1!} & \frac{x^2}{1!1!}+\frac{x^3}{1!2!}+\frac{x^4}{1!3!}+\frac{x^5}{1!4!}+\frac{x^6}{1!5!}+\ldots \\
\times \frac{x^2}{2!} & \frac{x^3}{1!2!}+\frac{x^4}{2!2!}+\frac{x^5}{2!3!}+\frac{x^6}{2!4!}+\frac{x^7}{2!5!}+\ldots \\
\times \frac{x^3}{3!} & \frac{x^4}{1!3!}+\frac{x^5}{2!3!}+\frac{x^6}{3!3!}+\frac{x^7}{3!4!}+\frac{x^8}{3!5!}+\ldots \\
\vdots & \cdots
\end{array}$$

and so we find $$\exp\left(\sum_{n=1}^\infty \frac{x^n}{n!}\right) =  \frac{1}{0!} + \sum_{n=1}^\infty \left(\frac{1}{1!} \sum_{a=n}\frac{x^{a}}{a!} + \frac{1}{2!} \sum_{a+b=n}\frac{x^{a+b}}{a!b!} + \frac{1}{3!} \sum_{a+b+c=n}\frac{x^{a+b+c}}{a!b!c!} + \ldots \right)$$

So you can try to see whether this is equal to the strange multinomial coefficient you have written.

Now I will show how to use the above formula to compute Bell numbers,

First the 0th Bell number (coefficient of $x^0$) is just 1/0! = 1. The first Bell number is the coefficient of $x^1$, and the only term which contributes this is 1/1! x^1/1! so B_1 = 1 as well. A more interesting one is B_3:

All contributions to $x^3$ terms come from $\sum_{n=1}^\infty \frac{1}{1!} \sum_{a=n}\frac{x^{a}}{a!} + \frac{1}{2!} \sum_{a+b=n}\frac{x^{a+b}}{a!b!} + \frac{1}{3!} \sum_{a+b+c=n}\frac{x^{a+b+c}}{a!b!c!}$ we could remove the "..." part since a+b+c+d+.. can never be 3 (since all the numbers must be $\ge 1$).

Now consider each sum one at a time,

{a = 3}

{a = 2,b = 1}, {a = 1, b = 2}

{a = 1,b = 1,c = 1}

so we have the $x^3$ coefficient being: $1/1! x^3/3! + 1/2! (x^3/1!2! + x^3/2!1!) + 1/3! x^3/1!1!1! = 5/6 x^3$ so the third bell number is $5$.

Title: Bell Numbers and Equating Coefficients


问:
Approximation of alternating series $\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$
$\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$

I am asked to find the no. of terms needed to approximate the partial sum to be within  0.0000001 from the convergent value of that series. Hence, I tried find the remainder, $R_n = S-S_n$ by the improper integral

$\int_{n}^{\infty} \frac{ (-1)^n(0.55)^{2n-1}} {(2n-1)!} dn $

However, I don't know how to integrate this improper integral so is there other method to solve this problem?

答:
Hint:

$\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+ \dots$

Your expression is simply $\sin (0.55)$.

ContentLicense: CC BY-SA 3.0


问:
Nice ODE that has wildly behaving solutions
Can we find an ODE
$$ f^{(n)}(x) + a_{n-1}(x)f^{(n-1)}(x)+  \ldots + a_0(x) f(x) = b(x),$$

with coefficients $a_i(x), b(x)$ in $C^\infty(\mathbb{R})$, bounded in codomain, whose solutions $f(x) \in Map(\mathbb{R},\mathbb{R})$ behave wildly?

I'm looking for some minimal examples. So hopefully $n$ can be as low as possible. It'd be better if $b \equiv 0$, and all the $a_i$'s are simple enough as polynomials. The more chaotic $f$ is the better.

EDIT: A comment below suggests there's none by Picard's theorem.

If there's none, would you please refer to a theorem that guarantees nicety? What are some minimal setups for bad solutions to exist for good DEs?

答:
The answer is no. Your equation is linear. The solutions to $n$-th order linear equations (with $b = 0$) are an $n$-dimensional vector space. See any standard differential equations book for this. You haven't specified how "bad" you would like solutions to be, but a typical minimal requirement, but not sufficient, for chaos is 3rd order nonlinear.

Edit: one of the common examples comes from the Rossler system, see Wikipedia, which is a system of 3 first order equations. From them you can derive a really horrible looking 3rd order equation, but this is not usually done because the system is more agreeable for study. The solution trajectories are chaotic and elegant.

ContentLicense: CC BY-SA 4.0


问:
Sangaku: to prove one of the intangents is parallel to $BC$
Given an acute triangle $\triangle ABC$ whose incircle is $I(r)$. Let $O(R)$ be the circle through $B$ and $C$ and which touches $I(r)$ interiorly. Show that the circle $P(p)$ which is tangent to $AB$, $AC$ and $O(R)$ (externaly) is such that one intagent line from $P(p)$ and $I(r)$ is parallel to side $BC$.

I have seen a couple solutions for this online but I never really understood them and they all seem wrong to me. For example, this is the problem 2 in here. The solution they give doesn't ring a bell. When they say that $\angle ACB - \angle ADE = \angle AED - \angle ABC$ (which is correct) and then they claim that  $\angle ADE < \angle ACB$ and $\angle AED > \angle ABC$ imply that $\angle ADE=\angle ABC$ it just sounds wrong. They could just use the congruence of $ADE$ and $AFG$ but instead they use this confuse argument.

And things get worse after the co-axial system. They give a huge jump and conclude that $BCD'E'$ is cyclic. This is best solution I have found for this problem but I just can't agree with it.

答:
The proof of the theorem is simply wrong.  It incorrectly asserts that the points of tangency of $\odot O$ with $\odot I$ and $\odot P$ are located at $D'$ and $E'$, respectively.  This is impossible, because $DE$ is tangent to $\odot I$ and $\odot P$ at $D'$ and $E'$, respectively, thus if $\odot O$ is tangent to $\odot I$ and $\odot P$ at these same points, $\odot O$ is tangent to line $DE$ at two distinct points--impossible, unless $D' = E'$ which implies $\triangle ABC$ is isosceles.  Refer to the figure:

To be precise, the claim presented in the theorem is true; however, the proof is incorrect.  If one is not convinced, we can draw the diagram on the Cartesian plane and explicitly calculate the relevant coordinates.


问:
Find Taylor series for these related series.
So I want to find the Taylor series of:

$$\frac{1}{(3-z)^2}$$

so it sort of looks like this:

$$\frac{1}{(3-z)} = \frac{1}{3} \cdot \frac{1}{(1 - \frac{z}{3})} = \frac{1}{3} (1 + \frac{z}{3} + \frac{z^2}{9} + ...)  = \sum_{n=0}^{\infty} \frac{z^n}{3^{n+1}}$$

Can I take the derivative? But what can I take the derivative of?

I am told the answer is:

$$\sum_{n=0}^{\infty} \frac{nz^{n-1}}{3^{n+1}}$$ but I'm not sure how to get there?

答:
The main idea behind why differentiation term by term is justified:

The series $f(z)=\frac 13\sum_{n=0}^\infty \left(\frac{z}{3} \right)^n$ has radius of convergence $R=3$.  This means the series converges absolutely when $|z|<3$. Take any $z$ with $|z|<3$. Then there exists $r>0$ such that $|z|\leq r <3$. Then by the Weierstrass M-test, we have
$$
\left|\frac{z}{3} \right|^n\leq \left(\frac{r}{3} \right)^n
$$
so $f_N(z)=\frac 13\sum_{n=0}^N \left(\frac{z}{3} \right)^n$ converges to $f(z)$ uniformly and absolutely. Similar methods show that $f_N'$ converges uniformly.

Finally, the uniform convergence of $f_N \to f$ and $f_N'$ to some function for $|z|\leq r$ guarantees that
$$
f'(z)=\lim_{N\to \infty} f_N'(z).
$$
In other words, we can bring the derivative operator inside.


问:
Average number of head streaks in $N$ coin tosses
You throw $N$ times a fair coin. What is the expected number of subsequences of $k$ consecutive heads?

For instance, with $N=6$ and $k=2$, for the outcome HTHHHH, we would get $3$ such subsequences.

I know the solutions to two related problems, but did not find this one being asked:

what is the expected number $N$ needed to get $k$ consecutive heads when throwing coins (see here)

what is the probability of having at least $k$ consecutive heads for $N$ throws (see here)

答: The number of possible sub-sequences is  (n-k+1). Multiply these by the probability for each, p = (.5)^k. So, for n=6, k=2, E [k consecutive heads] = (6-2+1)(.25) = 1.25.</p>


问:
Can a given symmetric matrix be written as a linear combination of the identity and a rank-$1$ matrix?
Given an invertible symmetric matrix $M$, is there a way to determine whether it can be written as $$M = a v v^T + b \mathrm{I}$$ for some scalars $a,b \neq 0$ and a vector $v$? I'm given matrix $M$ and want to find $a, b$ and $v$.

答:
If $M$ is an $n\times n$ matrix of the form $M=avv^T+bI$, then $M$ is clearly symmetric and we may always suppose that
$v^Tv=1$, by replacing $v$ with $v'=(v^Tv)^{-1/2}v$, and $a$ with $a'= (v^Tv)a$.

Assuming henceforth that $v^Tv=1$, observe that
$$
  Mv=avv^Tv+bv = (a+b)v,
  $$
and that, for every vector $u$ perpendicular to $v$, we have
$$
  Mu=avv^Tu+bu = bu.
  $$
Therefore  the spectrum of $M$ coincides with  $\{a+b,b\}$, where the multiplicity of $a+b$ is $1$ (provided $a\neq 0$), and the multiplicity of $b$
is $n-1$.

Conversely, every symmetric matrix whose spectrum consists of exactly two points, say $\lambda _1$ and $\lambda _2$, with multiplicities
1 and $n-1$, respectively, may be written as
$$
  M=avv^T+bI,
  $$
where $b$ is the spectral value with multiplicity $n-1$, namely $b=\lambda _2$,  and  $a=\lambda _1-\lambda _2$.  Finally, $v$ is any unit
eigenvector associated to $\lambda _1$.


问:
A special trick needs a special name
What do we call the process $(-1)^n$ that we use to alternate $+$ and $-$? Do Mathematicians call it something like "the alternator"?

答:
I'm pretty sure that we refer to anything containing $(-1)^n$ as alternatingly signed; this passes also on objects characterized by the presence of $(-1)^n$ (see, for instance, alternating series), or by its elements being limited to $\{-1,0,1\}$ (e.g. alternating sign matrices).

Also, when multiplying by $(-1)^n$, we very often refer to it by expressions like "the alternating factor".

Unless you're looking for some other, more specific use, I would say that "alternating (something)" is a pretty standard expression.

Title: A special trick needs a special name


问:
Calculate two vectors given their norms and angle

For two vectors $\mathbf{u,v}$ in $\mathbb{R}^n$ euclidean space, given:

$\|\mathbf{u}\| = 3$

$\|\mathbf{v}\| = 5$

$\angle (\mathbf{u,v})=\frac{2\pi}{3}$

Calculate the length of the vectors

$4\mathbf{u}-\mathbf{v}$

$2\mathbf{u}-7\mathbf{v}$

I'm not sure how to approach this with the given information

With the formula for the angle between the two vectors being

$$\cos \theta=\frac{\mathbf{u\cdot v}}{\|\mathbf{u}\|\cdot\|\mathbf{v}\|}$$

I already have the denominator, but how do I get the point product of u and v in this case?

A point to start would be most appreciated


问:
Prove that $\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+2}\right)\cdot\ldots\cdot\left(1-\frac{1}{2n}\right)=\frac{1}{2}$
Prove the following equality:$$\left(1-\frac{1}{n+1}\right)\left(1-\frac{1}{n+2}\right)\cdot\ldots\cdot\left(1-\frac{1}{2n}\right)=\frac{1}{2}$$
$$\text{for all }\;n\in\mathbb{N}\;.$$

Could you tell me if my proof is correct?

Is there another way to prove it?

Does it exist a simpler proof?

We are going to prove by induction the following equality:

$\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n}\right)=\dfrac12\;.\quad\color{blue}{(*)}$

For $\;n=1\;$ we get that

$\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n}\right)=$

$=\left(1-\dfrac12\right)=\dfrac12$.

Now we suppose that $(*)$ is true and prove that

$\left(1-\dfrac{1}{n+2}\right)\left(1-\dfrac{1}{n+3} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)=\dfrac12\;.$

It results that

$\left(1-\dfrac{1}{n+2}\right)\left(1-\dfrac{1}{n+3} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)=$

$=\dfrac{\left(1-\dfrac{1}{n+1}\right)\left(1-\dfrac{1}{n+2} \right)\cdot\ldots\cdot\left(1-\dfrac{1}{2n+2}\right)}{\left(1-\dfrac{1}{n+1}\right)}=$

$=\dfrac{\dfrac12\left(1-\dfrac{1}{2n+1}\right)\left(1-\dfrac{1}{2n+2}\right)}{1-\dfrac{1}{n+1}}=$

$=\dfrac{\dfrac12\left(\dfrac{2n}{2n+1}\right)\left(\dfrac{2n+1}{2n+2}\right)}{\dfrac{n}{n+1}}=$

$=\dfrac12\cdot\dfrac{\color{red}{2}\color{blue}{n}}{\color{green}{2n+1}}\cdot\dfrac{\color{green}{2n+1}}{\color{red}{2}(\color{brown}{n+1})}\cdot\dfrac{\color{brown}{n+1}}{\color{blue}{n}}=$

$=\dfrac12\;.$

Hence, by induction, the equality $(*)$ is true for all $n\in\mathbb{N}\;.$

答:
HINT

Notice that each denominator cancels the next numerator, whence we get

$\begin{align}
&\left(1 - \frac{1}{n+1}\right)\left(1 - \frac{1}{n+2}\right)\cdot\ldots\cdot\left(1 - \frac{1}{2n}\right) =\\
&=\left(\frac{n}{n+1}\right)\left(\frac{n+1}{n+2}\right)\cdot\ldots\cdot\left(\frac{2n-1}{2n}\right) = \frac{n}{2n}=\frac{1}{2}\;.
\end{align}$


问:
On the proof that $x^2+y^2-z^2=1$ is a topological manifold
Let $S=\{(x,y,z)\in\mathbb{R}^3\ |\ x^2+y^2-z^2=1\}$ and, for every $\theta \in \mathbb{R}$: $$L_\theta = \{(x,y,z)\in \mathbb{R}^3\ |\ (x-z)\cos\theta=(1-y)\sin\theta,\ (x+z)\sin\theta=(1+y)\cos\theta\}$$
One can easily prove that $L_\theta=\{(\sin2\theta,-\cos2\theta,0)+t(\cos2\theta,\sin2\theta,1)\ |\ t,\theta\in \mathbb{R}\}$ for all $\theta \in \mathbb{R}$ and that $\bigcup_{\theta \in \mathbb{R}} L_\theta=S.$

This proves that the map $\sigma:\mathbb{R}^2\rightarrow S, \ (\theta,t)\mapsto (\sin2\theta+t\cos 2\theta, -\cos 2\theta+t\sin2\theta, t)$ is well defined. I'm convinced of the fact that one can find a family $\{U_i\}_{i \in I}$ of open sets of $\mathbb{R}^2$ such that the restriction of $\sigma$ to $U_i$ (which I'll indicate with $\sigma_i:U_i\rightarrow \sigma(U_i)$) is a homeomorphism for all $i \in I$ and such that $\bigcup_{i\in I}\sigma(U_i)$, but I'm having trouble coming up with such an atlas. Does this lead to some viable solution? How would one define the charts?

答:
$x^2+y^2=1+z^2\subset \Bbb{R}^3$ is isomorphic to the cylinder $X^2+Y^2=1\subset \Bbb{R}^3$ through

$$(x,y,z) \to (\frac{x}{\sqrt{1+z^2}},\frac{y}{\sqrt{1+z^2}},z),\qquad (X,Y,Z)\to  (X\sqrt{1+Z^2},Y\sqrt{1+Z^2},Z)$$

You need two charts to cover the cylinder $$\phi_1(u,v) = (\cos u,\sin u,v),\qquad \phi_2(u,v) = (-\cos u,-\sin u,v),\qquad (u,v)\in (-\pi,\pi)\times \Bbb{R}$$
$\phi_i^{-1}\phi_j$ is clearly smooth because $\phi_1(u,v)=\phi_2(u\pm\pi,v)$


问:
Intuition for multivariable functions
Normal equations are easy enough to perceive as geometrical shapes or volumes when defined explicitly in the form $y=f(x)$.

But how do I understand equations like $g(x+y)=e^yg(x)+e^xg(y)$? I can make some sense of it algebraically, with it being a definition of a function and $x$ and $y$ being two variables. But how is it intuitively represented?

答:
$\def\e{\varepsilon}
\def\To{\rightarrow}$Note that $g$ in
\begin{align*}
g(x+y) &= e^y g(x)+e^x g(y) \tag{1}
\end{align*}
is not a multivariable function.
It is a function of one variable.
This is an equation (called a functional equation) for which the unknown is a function, namely the function $g$.
That is, to solve this equation involves not finding an $x$- or $y$-value for which the equation is satisfied but to find a function for which the equation holds for all $x$, $y$.

The geometric meaning of (1) is clear:
the value of $g$ at $x+y$ is related to its value at the points $x$ and $y$ as well as on $x$ and $y$ directly through (1).
Let us examine a simpler relationship implied by (1),
\begin{align*}
g(2x) = 2e^x g(x).\tag{2}
\end{align*}
This states that the height of the function at $2x$ is $2e^x$ times the height of the function at $x$.
(Let $x=1$ so
$g(2) = 2e g(1)$.
This states that the height of the function $g$ at 2 is $2e$ times the height of the function at 1.)
A relationship such as (1) is extremely restrictive and almost completely specifies the function $g$.

A good intuition comes from recurrence relations.
Suppose that
\begin{align*}
f(0) &= a \\
f(n+1) &= r f(n)
\end{align*}
where $n$ is a nonnegative integer.
This is the recurrence relation for the geometric sequence.
Note the similarity with (1):
the value of $f$ at $n+1$ is related to its value at $n$.
The geometric sequence could instead be defined by
\begin{align*}
f(0) &= a \\
f(m+n) &= r^m f(n),
\end{align*}
where $m,n$ are nonnegative integers, in which case the similarity to (1) is even more transparent.

Addendum

Let us agree that we wish to solve (1) on the reals.
Let $x=y=0$.
Then
$g(0) = 2g(0)$.
Thus, $g(0) = 0$.
Consider

\begin{align*}
g(x+\e) - g(x)
    &= (e^\e-1)g(x) + e^x g(\e). \tag{3}
\end{align*}
This implies that if $\lim_{\e\To0}g(\e)=0$, that is, if $g$ is continuous at $x=0$, that $g$ is continuous for all reals.
In fact, if we assume that $g$ is differentiable at $x=0$ this implies that $g$ is differentiable for all reals since
\begin{align*}
g'(x) &= \lim_{\e\To0} \frac{g(x+\e)-g(x)}{\e}
    = g(x) + e^x g'(0). \tag{4}
\end{align*}

We can begin graphing $g$ by point plotting using (2).
Suppose, without loss of generality, that $g(1) = e$.
Since $g(2x) = 2e^x g(x)$,

\begin{align*}
g(2) &= 2e^1g(1)
    = 2e^2 \\
g(4) &= 4e^4 \\
\vdots
\end{align*}
Similarly, since $g(x/2) = e^{-x/2}g(x)/2$,
\begin{align*}
g(1/2) &= \frac{e^{-1/2}}{2}g(1)
    = \frac{e^{1/2}}{2} \\
g(1/4) &= \frac{e^{1/4}}{4} \\
\vdots
\end{align*}
See the figures below.
(Values of $g$ for $x<0$ may be found by setting $y=-x$ in (1), with the result
$g(-x) = -e^{-2x}g(x)$.)

How to find the form of $g$ for a generic $x$-value?
There are many ways to approach this.
Here we solve the differential equation (4).
Applying the condition that $g(0) = 0$ we find the solution
$g(x) = x e^x g'(0).$
If $g(1) = e$ we find
$$g(x) = x e^x.$$
(Note that this is equivalent to the condition that $g'(0) = 1$.)
The graph of this function is given in the figures below as a solid line.
(One can also transform this problem as indicated by @Christian Blatter. This will result in a well-known functional equation.
It should be noted that the conditions that $g$ be continuous or differentiable at $0$ are not necessary and that generally the solution may be quite pathological.)

Figure 1. Plot of $g(2^k)$ for $k=0,1,\ldots$.

Figure 2. Plot of $g(2^{-k})$ for $k=0,1,\ldots$.

Title: Intuition for multivariable functions


问:
Find a limit $\lim_{x\rightarrow0}\frac{\int_{\cos(\sqrt{x})}^{1}\frac{\cos(t)}{1-t}dt}{\sqrt{x}}$
Find
$$\lim_{x\rightarrow0}\frac{\int_{\cos(\sqrt{x})}^{1}\frac{\cos(t)}{1-t}dt}{\sqrt{x}}$$
Substitute $y = \sqrt{x}$, by Mean Value Theorem for integrals, set $L = \lim_{y\rightarrow0}\frac{\frac{\cos(c)}{1-c}(1-\cos(y))}{y} = [\cos(y) \leq c \leq 1 \rightarrow 1- c \leq1-\cos(y)] = \lim_{y\rightarrow0}\frac{(\frac{\cos(c)\dot(1-c)}{1-c})}{y} \leq L$. We know that $0 < \cos(c) \leq \cos(1)$, thus by bounds $L \geq \lim_{y\rightarrow0}\frac{\cos(c)}{y} = +\infty$.

To me, the interval of integration is shrinking yet the denominator inside tends to approach $0$. How to solve it properly?

答:
You have to take limit as $x \to 0$ through positive values since $\sqrt x$ is involved. In the following I take limits as $y \to 0$ through positive values.

$\lim_{ y\to 0} \frac {\int_{\cos y}^{1} \frac {\cos t} {1-t} dt}y =\lim_{ y\to 0}\frac {\sin y \cos (cos y)} {1-\cos y}$ by L'Hopital's Rule. Noting that $\cos (\cos y) \to \cos 1$ as $ y \to 0$ and $\frac {\sin y } {1-\cos y}=\frac {2 \sin (y/2) \cos (y/2)} {2\sin^{2}(y/2)}=\cot (y/2)\to \infty$ we see that the givem limit is $\infty$.


答:
All head streaks either end $\ldots HT$ or are at the far end ending $\ldots H$ and all such cases are head streaks

The expected number of times $TH$ appears is $\frac{n-1}{4}$ and the probability the final throw is an $H$ is $\frac12$

So add these together to get $\dfrac{n+1}{4}$ expected head streaks

Title: Average number of head streaks in $N$ coin tosses

Body:
You throw $N$ times a fair coin. What is the expected number of subsequences of $k$ consecutive heads?

For instance, with $N=6$ and $k=2$, for the outcome HTHHHH, we would get $3$ such subsequences.

I know the solutions to two related problems, but did not find this one being asked:

what is the expected number $N$ needed to get $k$ consecutive heads when throwing coins (see here)

what is the probability of having at least $k$ consecutive heads for $N$ throws (see here)


问:
Probability that a bird is shot down out of n birds
Assume we have $n$ birds and $10$ hunters. A hunter chooses a bird randomly and independently shoots it down with a probability of $0.6$

(Note that hunters can choose and shoot at the same bird).

I want to find the probability that the $i$-th  bird is shot down.

$P(i$-th bird is shot down by some  hunter$) = 1 - P(i-$th bird is not shot down by all hunters$) = 1 - (1 - P($all hunters shoot down bird $i))$

This gives us $P(i$-th bird is shot down$) = 1 - 1 - (\frac{1}{n} \cdot 0.6)^{10}$

However, I don't think this is correct because logically it should be $P(i$-th bird is shot down$) = 1 - (1 - \frac{1}{n} \cdot 0.6)^{10}$

Where did I go wrong in breaking down the probabilities?

答: P(bird is not shot down by all hunters) does not equal 1-P(all hunters shoot down bird). It equals P(all hunters fail to hit bird) which leads to your correct finall formula.</p>


问:
Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$
$a_1 = 1 $ ,  $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$

It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.

The problem is to actually find the limit (it is $0$) because if I take arithmetic of limits:

$ \lim_{x \to +\infty} a_{n+1} =  \frac{\sqrt{1+ \lim_{n \to +\infty} a_n^2+1}}{\lim_{n \to +\infty} a_n}$

$g = \frac{\sqrt{1+g^2+1}}{g}  \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$

答:
If you have already proved that the limit exists and is greater than or equal to $0$, then assume by way of contradiction that

$$\lim_{n\to\infty}a_n=L>0$$

Then following your work (which is all valid since $L>0$) we get

$$L^2+1=\sqrt{L^2+1}$$

$$L^4+2L^2+1=L^2+1$$

$$L^4+L^2=0$$

$$L^2(L^2+1)=0$$

This has solutions $L=0$, $L=i$, and $L=-i$. Since $L$ is a positive real number, we have reached a contradiction. Thus, $L$ is a non-negative non-positive real number. Of course, the only such number is $L=0$. We conclude

$$\lim_{n\to\infty}a_n=0$$


答:
The expression in the right hand side inspires me to use $$a_m=\tan(b_m),-\dfrac\pi2\le b_m<\dfrac\pi2$$

$m=1\implies b_1=?$

$$a_{n+1}=\tan(b_{n+1})=\cdots=\tan\dfrac{b_n}2=\cdots=\tan\dfrac{b_{n+1-r}}{2^r},0\le r\le n$$

Set $r=n$ and then $n\to\infty$

Title: Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$

Body:
$a_1 = 1 $ ,  $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$

It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.

The problem is to actually find the limit (it is $0$) because if I take arithmetic of limits:

$ \lim_{x \to +\infty} a_{n+1} =  \frac{\sqrt{1+ \lim_{n \to +\infty} a_n^2+1}}{\lim_{n \to +\infty} a_n}$

$g = \frac{\sqrt{1+g^2+1}}{g}  \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$


问:
Find the limit of trigonometric function
$$
\lim_{x\to+\infty}(\frac{2}{\pi}\text{arctan}\;x)^{x}
$$

Even though I've expanded via Taylor series, I can't figure out

答:
This is of the form $\left[1^\infty\right]$, which one may address by taking logarithms to obtain the more familiar $[\infty \cdot 0]$ form.

Let $L = \lim_{x \rightarrow \infty} \left( \frac{2}{\pi} \arctan x \right)^x $.  By the continuity of the logarithm, we may exchange the logarithm and the limit, then apply a familiar property of logarithms.
\begin{align*}
\ln L &= \ln \lim_{x \rightarrow \infty} \left( \frac{2}{\pi} \arctan x \right)^x  \\
    &= \lim_{x \rightarrow \infty} \ln \left( \left( \frac{2}{\pi} \arctan x \right)^x \right)  \\
    &= \lim_{x \rightarrow \infty} x \ln \left( \frac{2}{\pi} \arctan x \right)  \\
\end{align*}
Then, as is typical for an application of l'Hopital's rule, we continue.
\begin{align*}
\ln L &= \lim_{x \rightarrow \infty} \frac{\ln \left( \frac{2}{\pi} \arctan x \right)}{1/x}  \\
    &\overset{l'H}{=} \lim_{x \rightarrow \infty} \frac{\frac{2}{\pi (x^2 + 1)}}{-1/x^2}  \\
    &= \lim_{x \rightarrow \infty} \frac{-2x^2}{\pi (x^2 + 1)}
\end{align*}
We perhaps already recognize the value of the limit.  Perhaps we continue using l'Hopital's rule again.  Alternatively, this is a rational function so we can proceed by factoring out "the big part".
\begin{align*}
\ln L &= \lim_{x \rightarrow \infty} \left( \frac{x^2}{x^2} \cdot \frac{-2}{\pi(1 + x^{-2})}   \right)  \\
    &= \frac{-2}{\pi(1+0)}  \\
    &= -2/\pi  \text{.}
\end{align*}
So $L = \mathrm{e}^{-2/\pi}$.


答:
You can prove it directly like this
$$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n} < \frac{\sqrt{1+a_n^2+\frac{1}{4}a_n^4}-1}{a_n} = \frac{1+\frac{1}{2}a_n^2-1}{a_n} = \frac 12 a_n$$
Therefore $a_n \to 0$, $n\to \infty$.


问:
Estimating PDF of ratio of random variables
I have a step in my project in which I have two random variables $U$ and $V$ with their distributions, and I need to calculate the quantity $\frac{U}{U+V}$. Unfortunately, this ratio does not have an analitically tractable distribution.

My question is: is it possible to compute the PDF of the ratio by taking samples from the numerator, taking samples from the denominator, computing the ratio and then using the histogram of the ratios as the estimate for the PDF?

Thank you very much!

答:
Graphical comment: Using R.

```<code>u = runif(10^6); v = runif(10^6)
x = u/(u+v)
hist(x, prob=T, br=50)
</code>```

You can derive the PDF of $X = \frac{u}{U+V}$ by using
a bivariate transformation.

Body:
Graphical comment: Using R.

u = runif(10^6); v = runif(10^6)
x = u/(u+v)
hist(x, prob=T, br=50)

You can derive the PDF of $X = \frac{u}{U+V}$ by using
a bivariate transformation.


答:
Let $\dfrac{2\arctan x}\pi=1+y$

$x=\tan\left(\dfrac\pi2+\dfrac{y\pi}2\right)=-\cot?$

We have $$\left(\lim_{y\to0^+}(1+y)^{1/y}\right)^{-\lim_{y\to0}\dfrac y{\cot\dfrac{y\pi}2}}$$

The inner limit converges to $e,$ right?

What about the limit in the exponent?

Title: Find the limit of trigonometric function

Body:
$$
\lim_{x\to+\infty}(\frac{2}{\pi}\text{arctan}\;x)^{x}
$$

Even though I've expanded via Taylor series, I can't figure out


问:
$\|f*g\|_q\leq \|g\|_q \|f\|_1$ and $\|f*g\|_\infty\leq \|g\|_q \|f\|_{q^{'}}$, $(1/q+1/q^{'}=1)$?
Now I'm reading the Young inequality. It says that if $f \in L^p(R)$, $g \in L^q(R)$, $1\leq p,q\leq \infty$, $1/p+1/q\geq 1$. Then how could we have the following inequalities:

$$\|f*g\|_q\leq \|g\|_q \|f\|_1$$

$$\|f*g\|_\infty\leq \|g\|_q \|f\|_{q^{'}},\quad (1/q+1/q^{'}=1)$$

$f*g$ means the convolution

The two inequalities seems not obvious, the second one is similar with the holder inequality, but actually not.

Could someone give me some hints or helpful links?
Thanks a lot!

答:
The second inequality is easy. The first one is tricky, and was already proved by robjohn here as an answer to a different question. I just answer this for the sake of completeness because I couldn't find a proof of the second inequality here.

2) For the second inequality, I assume $1<q<\infty$ as the cases $q=1$ and $q=\infty$ are trivial. First note that
$$
|(f\ast g)(x)|=\Big|\int f(x-y)g(y)dy\Big|\leq \int |f(x-y)||g(y)|dy
$$
then  by Hölder and variable change
$$
\leq \left(\int|f(x-y)|^{q'} \right)^\frac{1}{q'} \left(\int|g(y)|^{q} \right)^\frac{1}{q}=\left(\int|f(y)|^{q'} \right)^\frac{1}{q'} \left(\int|g(y)|^{q} \right)^\frac{1}{q}
$$
whence
$$
\|f\ast g\|_\infty \leq \|f\|_{q'}\|g\|_q.
$$

1) The first inequality is a particular case of Young's inequality
$$
\|f\ast g\|_s\leq \|f\|_p\|g\|_q\qquad\mbox{when}\quad \frac{1}{p}+\frac{1}{q}=1+\frac{1}{s}, \quad 1\leq p,q,s\leq \infty
$$
which is proved by robjohn in this thread. This does not cover the cases when $p,q,$ or $s$ are infinite. But $s=\infty$ is the case above, and your inequality has $p=1$, which yields $s=q$.


问:
Let $G,K$ be groups with $H\le G$. It is not always possible to extend a homomorphism $\sigma:H\to K$ to $G$
This is Exercise 4.2 of Roman's, "Fundamentals of Group Theory: An Advanced Approach". According to this MSE search and this Approach0 search, it is new to MSE.

The Question:

Let $G$ and $K$ be groups with $H\le G$. Show that it is not always possible to extend a homomorphism $\sigma:H\to K$ to $G$.

Thoughts:

This looks, prima facie, like a job for the extreme cases, like $K$ (and hence $H$) being trivial, although I haven't found anything; the solution seems more elusive than that.

My second idea came in the form of supposing $G$ and $K$ are finite, then employing some trick - I don't know what yet - with the orders of the groups involved, like Lagrange's Theorem for the subgroup $H$ in $K$, aiming to get $\lvert \sigma'(G)\rvert\nmid \lvert K\rvert$ somehow, for some potential extension $\sigma':G\to K$. This seems feasible at first but, after a while, I have doubts.

Besides, I'm not entirely sure whether the extension is from $H$ to $G$ or from $K$ to $G$ (if you see what I mean). The word "extension" is defined only once prior to the exercise ibid. but it is in an entirely different context. For completeness, here is that definition:

Let $G$ be a group. We refer to subgroups $H$ and $K$ of $G$ for which $H\le K$ as an extension.

Please help :)

答:
The division-by-$2$ map $2\mathbb{Z}\rightarrow\mathbb{Z},\,k\mapsto k/2$ is a bona fide group homomorphism, but it cannot be extended to a group homomorphism $\mathbb{Z}\rightarrow\mathbb{Z}$, for such an extension $\tilde{f}$ would have to satisfy
$$1=f(2)=\tilde{f}(2)=2\tilde{f}(1),$$
yet no such integer $\tilde{f}(1)$ exists.


问:
Segment above another segment
I received a task with multiple exercises and this is the only one I don't know how to solve.
So, we represent a segment in the Cartesian space by an instance of the

class Segment {
public Point3D A,B; // the coordinates of the endpoints of the segment
}

We say that a segment s1 is above another segment s2 if there is a vertical line 'l' that intersects both segments, such that the l ∩ s1 is above l ∩ s2. The following picture illustrates the situation when segment s1 is above segment s2:
Click

So I have to create a method that returns

0 if s1 and s1 intersect,

1 if s1 is above s2,

2 if s2 is above s1, and

3 otherwise.

答: Project the segments onto the $xz$ plane. If these projected segments intersect, then either one of the segments is above the other or they intersect.

Title: Segment above another segment


问:
Do third derivatives commute?
How does one prove that $$\partial_{i}\partial_{j}\partial_{k} = -\partial_{j}\partial_{i}\partial_{k}?$$

The fact that they permute cyclically follows directly from the symmetry of second derivatives. How and why does this hold? Why aren't third derivatives commutative?

答:
They are. When your functions are "nice" enough (in this case, for example, $C^3$ or more), third derivatives commute (with no change of any sign). And this easily follows from second derivatives' commutativity (Schwarz's theorem). For example:

$$ \partial_j\partial_k f=\partial_k\partial_j f \;\Rightarrow\; \partial_i(\partial_j\partial_k f)=\partial_i(\partial_k\partial_j f) $$

(Note that there's no multiplication, only composition of differential operators.)

Title: Do third derivatives commute?


问:
A particle travels at velocity $|t - 1|$ at time $t$.
Question: A particle travels at velocity $|t - 1|$ at time $t$. Divide the interval $[0, 2]$ into $2k$ intervals of equal length where $k\in\mathbb{N}$. Find the minimum velocity of the particle on each of these time intervals.

My approach to this question is to experiment with the numbers. For example, if $k = 1$, if $k = 2$, if $k = 3$, etc... And I think that you would have to find the derivative of the velocity of the particle to find the minimum velocity.

But I am not really sure how to approach from here.

Any help will be appreciated!

答:
Hint

For $t < 1,$ the particle's speed is decreasing, as it approaches $t = 1$, which implies that
you want which end of the interval for minimum velocity?

For $t > 1,$ everything is symmetrically the opposite.

No need to take derivatives.


问:
Fabius function and equivalent
The Fabius function $F$ can be defined on $[0,1]$ by

$F(0)=0$

$F(1)=1$

on $[0,\frac{1}{2}]$ $F'(x)=2F(2x)$

on $[\frac{1}{2},1]$ $F'(x)=2F(2(1-x))$

It's a known example of a not analytic $C^\infty$ function.

The Fabius function can also be defined as the CDF of the random variable $X$ such that
$$X=\sum_{i=1}^\infty 2^{-i}U_i$$
where $U_i$ are independent random variables uniform on $[0,1]$.

Is it possible to find a usual function $f$ such that
$$\lim_{x\rightarrow 0}\frac{F(x)}{f(x)}=1$$

Some functions like $f(x)=e^{-\frac{1}{x^2}}$ could be good candidates, but I don't really know how to find such a function, or given a function $f$, how to compute the limit.

答:
Based oh this answer, the asymptotic of the Fabius function for $x\to0^+$:
$$\small F(x)\sim \frac{1}{2^{7/12} \sqrt{\pi \, x}}\, \exp \left(\frac{1}{\log 2} \left(\gamma _1+\frac{\gamma_0^2}{2}-\frac{\pi ^2}{12}-q-\frac{q^2}{2}\right)\right),\quad q=W_{-1}(-x \log2),$$
where $W_{-1}(x)$ is the non-principal real-valued branch of the Lambert W-function, and $\gamma_n$ are the Stieltjes constants (in particular, $\gamma_0$ is the Euler–Mascheroni constant $\gamma$).

Title: Fabius function and equivalent


问:
Two questions about $a^{2^{n-1}} \mod {2^{n}}$ and Fermat's Little Theorem
This is a question that comes in two parts.

First, prove that if $a^{2^{n-1}}\not\equiv1 \pmod {2^{n}}$, then $a$ is even.

Second, prove that for any odd prime $p$, if $a$ is odd and $a^{2^{n-1}(p-1)}\not\equiv1 \pmod 2^{n}p$, then $p|a$.

For the first one, I realized I can try proving the contrapositive, and then use induction. Then what should I do next? Fermat's Little Theorem?

For the second one, can you split the modulus into $2^{n}$ and $p$? I think this might be a good method. Answers and ideas are welcome!

答:
For the first question, use https://mathworld.wolfram.com/EulersTotientTheorem.html or https://mathworld.wolfram.com/CarmichaelFunction.html

and https://en.m.wikipedia.org/wiki/Proof_by_contradiction

For the second, if $p\nmid a\iff(a,p)=1$

$$a^{p-1}\equiv 1\pmod p$$

If $(a,2)=1,$ using Carmichael or Totient function,

$$a^{2^{n-1}}\equiv1\pmod{2^n}$$

$$\implies a^{[p-1,2^{n-1}]}\equiv1\pmod{[p,2^{n-1}]}$$


问:
Cardinality of the set of functions that maps from $\mathbb{N}$ to $\{1,2,3\}$.
I know that from the other direction, the cardinality of the set of function that maps from  $\{1,2,3\}$ to  $\mathbb{N}$ is $\mathbb{N}\times\mathbb{N}\times\mathbb{N}=|\mathbb{N}|$, however, I don't know how should I work from the other direction.

答: The set of functions mapping $X\to Y$ is often denoted $Y^X$. It is known that $|Y^X| = |Y|^{|X|}$.


问:
General formula for error of an integration method
Let $h > 0$ be a real number. For any function $f$, let
$$M(f) = \sum_{i = 1}^{N}w_if(x_i),$$
where the weights $w_i$ and the nodes $x_i$ are predetermined real numbers (they may depend on $h$, but $h$ is fixed). The weights and nodes are assumed to be obtained from integrating a polynomial interpolating the points $(x_1, f(x_1)), \dots, (x_N, f(x_N))$.
I am interested in the error of the approximation of $\int_{-h}^{h}f(x)\,dx$, which is given by
$$E(f) = M(f) - \int_{-h}^{h}f(x)\,dx$$

My approach so far is:
Let $n$ be the smallest nonnegative integer for which $E(x^n) \neq 0$.
For any $x \in \mathbb{R}$, I can write
$$f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2} + \dots + \frac{f^{(n)}(c(x))}{n!}x^n,$$
where $c(x)$ is between $0$ and $x$.
Since $E$ is a linear function,
\begin{align}
E(f) &= f(0)E(1) + f'(0)E(x) + \dots + E\left(\frac{f^{(n)}(c(x))}{n!}x^n\right) \\
&= E\left(\frac{f^{(n)}(c(x))}{n!}x^n\right).
\end{align}
From other derivations I have read, it seems that the error always can be written as
$$E(f) = \frac{f^{(n)}(\xi)}{n!}E(x^n)$$
for some $\xi$ in the smallest open interval containing the $x_i$, $-h$, and $h$. I am wondering how to get from my formula to this one. That is, how to show
$$E\left(\frac{f^{(n)}(c(x))}{n!}x^n\right) = \frac{f^{(n)}(\xi)}{n!}E(x^n).$$
So I want to show that
$$\sum_{i = 1}^{N}w_i f^{(n)}(c(x_i))x_i^n - \int_{-h}^{h}f^{(n)}(c(x))x^n \,dx = f^{(n)}(\xi)\left(\sum_{i = 1}^{N}w_ix_i^n - \int_{-h}^{h}x^n\,dx \right).$$
I've tried using intermediate value theorem and mean value theorem, but it doesn't seem to work.

答:
I've found a counterexample using $$M(f) = 2hf(kh)$$ where $0 < k < 1$ is to be determined. We have that
$$E(x) = 2kh^2$$
is the first nonzero error. So the formula says that for any $f$ we should have
$$E(f) = 2kf'(c)h^2$$
for some $c \in (-h, h)$.
In the case of $f = x^2$ we get
$$E(f) = 2k^2h^3 - \frac{2}{3}h^3 = \left(2k^2 - \frac{2}{3}\right)h^3.$$
Therefore, the formula claims there exists $c \in (-h, h)$ such that
$$4kc = \left(2k^2 - \frac{2}{3}\right)h.$$
This means
$$c = \left(\frac{k}{2} - \frac{1}{6k}\right)h.$$
Taking $k = \frac{1}{100}$ gives
$$c = \left(\frac{1}{200} - \frac{50}{3}\right)h < -h$$
contradicting the statement.

One theorem that explains the general formula I was looking for is the Peano kernel theorem, which does the same taylor expansion as I did, but uses the integral form of the remainder instead and pulls the $E$ inside the integral.


问:
find the largest integer less than a number
I'm trying to figure out this problem for few hours now. please help.

Define $$\text{PCOM}(a,b,c) := \{ax + by + cz : x, y, z ≥ 0\}.$$

Given integers $t > c > b > a > 1,$ I need help to find the largest integer $w < t$ such that $w\notin\text{PCOM}(a,b,c).$

Example: $\text{PCOM}(3,5,7) =\{0,3,5,6,7,8,9,10,…\}.$  The largest integer less than $9$ that is not in $\text{PCOM}(3,5,7)$ is $4$.

答:
This is a particularly simple variant of what is known as the coin problem. Here's a dynamic programming solution that runs in time linear in $t$. We use an array $m[0\dots t]$ of booleans, which I'll refer to as either T or F. The value of $m[i]$ will represent whether or not $i$ can be represented as the sum of zero or more copies of $a, b, c$. We will also keep a variable $w$, representing the current largest value not in $\text{PCOM}(a, b, c)$.

First, we initialize $m[0] = \text{T}$ (since we can obviously get a sum of zero by not using any of the values) and for $0 < i < a$ we set $m[i]=\text{F}$. We also initialize $w=a-1$.

Having done that, the algorithm is

```<code>for i = a to t
   m[i] = m[i - a] OR m[i - b] OR m[i - c]
   if m[i] = F
      w = i
return w
</code>```

The only tweak is in the second line: if $i < b$ or $i < c$ we don't try to access the corresponding array element (since it would make no sense to do so here). There are some obvious improvements we can make in the algorithm, but I hope the idea is clear. If not, just ask.

Body:
This is a particularly simple variant of what is known as the coin problem. Here's a dynamic programming solution that runs in time linear in $t$. We use an array $m[0\dots t]$ of booleans, which I'll refer to as either T or F. The value of $m[i]$ will represent whether or not $i$ can be represented as the sum of zero or more copies of $a, b, c$. We will also keep a variable $w$, representing the current largest value not in $\text{PCOM}(a, b, c)$.

First, we initialize $m[0] = \text{T}$ (since we can obviously get a sum of zero by not using any of the values) and for $0 < i < a$ we set $m[i]=\text{F}$. We also initialize $w=a-1$.

Having done that, the algorithm is

for i = a to t
   m[i] = m[i - a] OR m[i - b] OR m[i - c]
   if m[i] = F
      w = i
return w

The only tweak is in the second line: if $i < b$ or $i < c$ we don't try to access the corresponding array element (since it would make no sense to do so here). There are some obvious improvements we can make in the algorithm, but I hope the idea is clear. If not, just ask.


问:
Video Lessons in Complex Analysis
Does anybody have some link for good video lessons of a complete course in Complex Analysis?

Grateful.

答:
Complex Analysis (2015)

by James Cook of Liberty University following Gamelin's Complex Analysis. His webpage also has a supplementary guide to Gamelin.

Title: Video Lessons in Complex Analysis


问:
Best way to denote mappings $\mathbb{Z}_m \to \mathbb{Z}_n$, and $\mathbb{Z}[t] \to \mathbb{Z}_m[t]$?
The mapping $\mathbb{Z} \to \mathbb{Z}_n$ defined by $a \mapsto a \pmod{n}$ is commonly denoted as such. Or, in the construction or interpretation of $\mathbb{Z}_n$ involving the congruence classes modulo $n$ of $\mathbb{Z}$, I've seen that same mapping denoted $a \mapsto [a]$, or $a\mapsto [a]_n$.

What would be the clearest way to denote a mapping $\mathbb{Z}_m \to \mathbb{Z}_n$ where $n \leq m$ and $n|m$ and such that $a \mapsto a$ where the LHS of that mapping is interpreted as $a \in \mathbb{Z}_m$ and the RHS is interpreted as $a \in \mathbb{Z}_n$?

Also, how would I denote the same kind of mapping $\mathbb{Z}[t] \to \mathbb{Z}_n[t]$ where I want $p(t) = a_0 + a_1t + \cdots + a_kt^k$ to map the polynomial $p(t) = b_0 + b_1t+\cdots + b_kt^k$ where $b_i = a_i \pmod{n}$ for each $0 \leq i \leq k$? i.e. that I want the same polynomial $p(t)$ but with its coefficients interpreted in $\mathbb{Z}_n$ as opposed to in $\mathbb{Z}$.

答:
I guess I would choose something along the line of $[a]_m \mapsto [a]_n$ for the first one, if I really wanted to specify that with symbols. Otherwise there is nothing wrong with describing the map with words, which is not uncommon at all. You are reducing mod $n$.

For the second one I would always use the standard notation

$$\overline{a_0} + \overline{a_1}t + \dotsc + \overline{a_k}t^k$$

as long as the integer $n$ is clear from the context or once again use words. You are also reducing mod $n$ in this case. If you have to vary between classes with respect to two integers $n$ and $m$ and you really do not want to describe the map with words, you could of course also use something like $[a_i]_n$ again, which already starts to look quite terrible.

Another option for both cases would be to say something like the obvious map or the canonical map from ... to ... This is also done by quite some authors.


问:
Decompose $\frac{1}{a \sin x +b}$ into sum of trigonometric functions
I am wondering if it is possible to write

$$f(x)=\frac{1}{a \sin x +b}$$

as a sum of functions that don't have trigonometric functions in the denominator? Or just pure trigonometric functions, without a constant added. So maybe something like this:

$$\frac{1}{a \sin x +b}=A\sin^3 x \cos^2x + B\sin x \cos ^4 x + C\frac{1}{\cos x}+ D\frac{\cos^2 x }{\sin x}+\dots $$

The reason I'm asking is that I need to integrate products of $f(x)$ with powers $\cos^n x$, and it would be really neat to get rid of the constant in the denominator, as integrals of the form $\int\frac{\cos^n x}{a \sin x +b}dx$ are extremely hard to solve, whereas integral tables contain solutions to $\int \sin^p x \cos^q x dx$ and similar integrals.

One potential starting point could be

$$\frac{1}{a \sin x +b}=\frac{a \sin x -b}{(a \sin x +b)(a \sin x -b)}=\frac{a \sin x -b}{(a^2 \sin^2 x -b^2)}=...$$
with the goal to eventually make the replacement $\sin^2 x +\cos^2 x =1$. Can such a method be formalized? Or is the decomposition I have in mind provable not possible? Thanks for your help!

答:
An idea for a solution just came to my mind, which I am going to post here hoping it might be useful for somebody in the future. Unfortunately, the condition doesn't work in my case, so that I might go with the Fourier series solution. But for $a\geq b$, we can write

$$\frac{1}{a \sin x +b}=\frac{1}{a}\frac{1}{\sin x +\frac{b}{a}}=\frac{1}{a}\frac{1}{\sin x +\sin\left(\arcsin\left(\frac{b}{a}\right)\right)}=\frac{1}{a}\frac{1}{2\sin\left(\frac{x+\arcsin\left(\frac{b}{a}\right)}{2}\right) \cos\left(\frac{x-\arcsin\left(\frac{b}{a}\right)}{2}\right)}.$$

Trigonometric expansion with Mathematica yields

$$\frac{1}{2\sin\left(\frac{x+\arcsin\left(\frac{b}{a}\right)}{2}\right) \cos\left(\frac{x-\arcsin\left(\frac{b}{a}\right)}{2}\right)}\\=\frac{1}{2} \sec \left(\frac{c}{2}\right) \sec (c) \sin \left(\frac{x}{2}\right) \sec \left(\frac{c}{2}-\frac{x}{2}\right)-\frac{1}{2} \csc \left(\frac{c}{2}\right) \sec (c) \sin \left(\frac{x}{2}\right) \csc \left(\frac{c}{2}+\frac{x}{2}\right)+\csc (c)$$

with $c=\arcsin\left(\frac{b}{a}\right)$.


问:
Find a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ that is strictly decreasing and has an image equal to $(-1,1)$.
Find a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ that is strictly decreasing and has an image equal to $(-1,1)$.

I tried creating asymptotes but that didn't work out. I am unsure of how to create a function that maps the reals to an interval.

One function I cam up with is $\frac{x}{(x+1)(x-1)}$ which has asymptotes at (-1,1) and the graph in between is strictly decreasing, but it doesn't work because the map also includes other values in $\mathbb{R}$.

Edit: Using hint, since $f:(-1,1)\rightarrow\mathbb{R}$ given by $f(x)=-\tan(\frac{\pi}{2}x)$ is strictly monotone, using the theorem below, we have that $f:\mathbb{R}\rightarrow(-1,1)$ given by $f^{-1}=-\frac{2arctan(x)}{\pi}$ is continuous. But $f:\mathbb{R}\rightarrow(-1,1)$ is the same as $f:\mathbb{R}\rightarrow\mathbb{R}$ because of the asymptotes of arctan. Clearly, it is also strictly decreasing. I don't think this is a sufficient proof but it is what I am thinking.

Theorem 3.29: Let I be an interval and suppose that the function $f : I \rightarrow\mathbb{R}$ is strictly monotone. Then the inverse function $f^{-1} : f(I)\rightarrow \mathbb{R}$ is continuous.

答:
HINT

\begin{align*}
f(x) = -\frac{2\arctan(x)}{\pi}
\end{align*}


问:
Number Theory : $11x \equiv44\pmod{64}$ how to solve?
how to solve this $11x\equiv 44\pmod{64}$ ?

attempt:

first of all need to find the inverst of $11x\equiv 44\pmod{64}$

by Euclid algorithm then:

$64=11\cdot 5 +9$

$11=1\cdot 9 +2$

$9=2\cdot4+1$

$1=9-2\cdot4$

what is the next step for the make the inverst I do not understand ?

Body:
As others have said, you can just eyeball it, since
$11 \times 4 = 44.$

The formal approach is as follows:

$1 = 9 - (2)4$

$= 9 - [11-9](4)$

$= (9)5 - (11)4$

$= [64 - (11)5]5 - (11)4$

$= [(64)5 - (11)25 - 11(4)$

$= (64)5 - 11(29).$

Therefore, $(11)29 \equiv -1 \pmod{64}.$

Therefore, $(11)35 \equiv (11)(64 - 29) \equiv 1 \pmod{64}.$

Therefore, $x \equiv (11 \times 35) \times x \equiv (44 \times 35) \equiv 4 \pmod{64}.$


问:
Comparing two box plots
Say, these are two boxplots for ratings for a movie (1 to 10) say Star Wars The Phantom Menace.
The first boxplot are ratings received by various experts for the same movie
The second boxplot are ratings received by the audience for the same movie again.

I see that the
First Range: 7-1=6
Second Range: 10-0=10

First max: 7
Second max: 10

Comparing the two box plots above, Is it fair to say that that there's more variation in the Second boxplot data?

Also, Is it fair to say that on an average the rating giving by experts is higher than that of the audience?

答:
Your charts are suggestive that the variation may be greater in the second. But this need not be the case.  Suppose the mark distribution was as follows, broadly consistent with the minimum - 1st quartile - median - 3rd quartile - maximum values in the box plots:

```<code>First:          Second:
Mark Number     Mark Number
 1    12         0      1
 4     1         5   2499
 5    13         6   1249
 6     1         7   1250
 7    25         10     1
</code>```

Then the first sets of marks has variance of about $5.65$ while the second has a variance of about $0.698$, i.e. the variation is much smaller in the second, contrary to the impression given by the boxplots. Other distributions could produce different variances.

That particular example has the mean of the second distribution higher than the mean of the first, but again it would be easy enough to to reverse this by changing a few of the numbers (e.g. in the first distribution if mark $1$ was allocated by $1$ and mark $4$ by $12$, or in the second distribution if mark $0$ was allocated by $1249$ and mark $5$ by $1251$).  The boxplots do not suggest to me which should have the higher mean.

Title: Comparing two box plots

Body:
Your charts are suggestive that the variation may be greater in the second. But this need not be the case.  Suppose the mark distribution was as follows, broadly consistent with the minimum - 1st quartile - median - 3rd quartile - maximum values in the box plots:

First:          Second:
Mark Number     Mark Number
 1    12         0      1
 4     1         5   2499
 5    13         6   1249
 6     1         7   1250
 7    25         10     1

Then the first sets of marks has variance of about $5.65$ while the second has a variance of about $0.698$, i.e. the variation is much smaller in the second, contrary to the impression given by the boxplots. Other distributions could produce different variances.

That particular example has the mean of the second distribution higher than the mean of the first, but again it would be easy enough to to reverse this by changing a few of the numbers (e.g. in the first distribution if mark $1$ was allocated by $1$ and mark $4$ by $12$, or in the second distribution if mark $0$ was allocated by $1249$ and mark $5$ by $1251$).  The boxplots do not suggest to me which should have the higher mean.


Body:
Find a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ that is strictly decreasing and has an image equal to $(-1,1)$.

I tried creating asymptotes but that didn't work out. I am unsure of how to create a function that maps the reals to an interval.

One function I cam up with is $\frac{x}{(x+1)(x-1)}$ which has asymptotes at (-1,1) and the graph in between is strictly decreasing, but it doesn't work because the map also includes other values in $\mathbb{R}$.

Edit: Using hint, since $f:(-1,1)\rightarrow\mathbb{R}$ given by $f(x)=-\tan(\frac{\pi}{2}x)$ is strictly monotone, using the theorem below, we have that $f:\mathbb{R}\rightarrow(-1,1)$ given by $f^{-1}=-\frac{2arctan(x)}{\pi}$ is continuous. But $f:\mathbb{R}\rightarrow(-1,1)$ is the same as $f:\mathbb{R}\rightarrow\mathbb{R}$ because of the asymptotes of arctan. Clearly, it is also strictly decreasing. I don't think this is a sufficient proof but it is what I am thinking.

Theorem 3.29: Let I be an interval and suppose that the function $f : I \rightarrow\mathbb{R}$ is strictly monotone. Then the inverse function $f^{-1} : f(I)\rightarrow \mathbb{R}$ is continuous.

Body: $f(x) = -\tanh x = \dfrac{e^x - e^{-x}}{e^x + e^{-x}}. \tag 1$


答: The other commenter's example was better but here's one using simple groups. Assume $G$ is a non-cyclic simple group and let $n$ be the order of $\sigma \in G$ and take the natural isomorphism $f: \langle \sigma \rangle \to \mathbb{Z}/n\mathbb{Z}$. Then if $f$ extended to a homomorphism $F: G \to \mathbb{Z}/n\mathbb{Z}$, it would be an isomorphism since its kernel would have to be trivial by the simplicity of $G$, contradicting the fact that $G$ was not cyclic.

Title: Let $G,K$ be groups with $H\le G$. It is not always possible to extend a homomorphism $\sigma:H\to K$ to $G$

Body:
This is Exercise 4.2 of Roman's, "Fundamentals of Group Theory: An Advanced Approach". According to this MSE search and this Approach0 search, it is new to MSE.

The Question:

Let $G$ and $K$ be groups with $H\le G$. Show that it is not always possible to extend a homomorphism $\sigma:H\to K$ to $G$.

Thoughts:

This looks, prima facie, like a job for the extreme cases, like $K$ (and hence $H$) being trivial, although I haven't found anything; the solution seems more elusive than that.

My second idea came in the form of supposing $G$ and $K$ are finite, then employing some trick - I don't know what yet - with the orders of the groups involved, like Lagrange's Theorem for the subgroup $H$ in $K$, aiming to get $\lvert \sigma'(G)\rvert\nmid \lvert K\rvert$ somehow, for some potential extension $\sigma':G\to K$. This seems feasible at first but, after a while, I have doubts.

Besides, I'm not entirely sure whether the extension is from $H$ to $G$ or from $K$ to $G$ (if you see what I mean). The word "extension" is defined only once prior to the exercise ibid. but it is in an entirely different context. For completeness, here is that definition:

Let $G$ be a group. We refer to subgroups $H$ and $K$ of $G$ for which $H\le K$ as an extension.

Please help :)


Title: Find a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ that is strictly decreasing and has an image equal to $(-1,1)$.

Body:
Sigmoid!

$$ f(x) = \frac{e^{-x}-1}{e^{-x}+1} $$


问:
Deriving a generating function from a pdf
i am trying to derive a generating function from a probability density function (PDF) on $x \geq 0$.

i think i'm supposed to apply the following formula to do this, where $f(x)$ is the PDF.

$\int_{0}^\infty e^{tx} \cdot f(x)dx$

the formula for my specific PDF is below.

$f(x) = \frac{e^{-x/3}}{3}$

so applying the formula from the beginning:

$\int_{0}^\infty e^{tx} \cdot \frac{e^{-x/3}}{3}dx$

i know that the answer is $\frac{1}{1-3t}$, but i am having a lot of trouble with the integration process. i have tried u substitution and integration by parts, but i just can't seem to crack it.

any advice for how to proceed? thank you in advance.

答:
Note that $e^{tx}\cdot e^{-x/3}=e^{x(t-1/3)}$.

Hence, $\int_{0}^\infty e^{tx} \cdot \frac{e^{-x/3}}{3}dx=\frac 13\int_{0}^\infty e^{x(t-1/3)}dx$. Now use $u$ substitution with $u=x(t-1/3)$. Treat $t$ as just a number.

Of course, you need $t<1/3$ for the integral to converge.


问:
Obtention and estimate error of trapezoid rule: Why was necessary to mention that $f''(\xi)$ is constant?
I'm considering the following integral

$$I=\int_a^b[f(a)+\Delta f(a)\alpha+\frac {f''(\xi)}{2}\alpha(\alpha-1)h^2]dx$$

where $a\le x\le b, h=b-a, \Delta f(a)=f(b)-f(a), \xi$ is a number between $a$ and $b$. If $\alpha=\frac{x-a}{h}$ then $dx=hd\alpha$. Thus

$$I=h\int_0^1[f(a)+\Delta f(a)\alpha+\frac {f''(\xi)}{2}\alpha(\alpha-1)h^2]d\alpha$$

If one assumes that for $h$ small, $f''(\xi)$ is approximately constant, then

$$I=h\left[\alpha f(a)+\frac{\alpha^2}{2}\Delta f(a)+\left(\frac{\alpha^3}{6}-\frac{\alpha^2}{4}\right)+f''(\xi)h^2\right]_0^1d\alpha$$

My question is, why is the author making this assumption of "If one assumes that for $h$ small, $f''(\xi)$ is approximately constant"? Wasn't already a constant?

Thank you in advance

答: The $\xi$ in the remainder term depends on the value of $\alpha$ (and $x$), so in principle it actually should be part of the integration, and is not a constant factor that can be pulled out of the integral. If however for all $\alpha$ (or $x$) in the relevant range, the quantity $f''(\xi)$ is bounded, then you can bound how large the integral of the last term might be.


问:
Continuous function with a restriction is Riemann integrable

Let $f:[0,1]\rightarrow \mathbb{R}$ be a bounded function whose restriction to $(0,1]$ is continuous.
Prove that $f$ is Riemann integrable.

There are plenty of definitions, worked examples and explanations on inclusive boundaries i.e. $[a,b]$.

But because of the exclusive boundary I don't think I could use $0$ as an infimum.
Does the slight difference in the boundary make $f$ a non-uniformly continuous function?
If so, how do I prove this differently so the argument still holds?

Any help will be greatly appreciated!

答:
Since $f$ is bounded there exist $m,M$ such that $m \leqslant f(x) \leqslant M$ for all $x\in [0,1]$.  For any $\epsilon > 0$, take a point $x_1 \in (0,1)$ such that $(M-m)x_1 < \epsilon/2$.

For partition $P: 0< x_1< x_2< \ldots < x_n = 1$, let $M_j =\sup_{x \in [x_{j-1},x_j]}f(x)$ and $m_j =\inf_{x \in [x_{j-1},x_j]}f(x)$.

We have $(M_1-m_1)x_1 \leqslant (M-m)x_1 < \epsilon/2$, and

$$U(f,P) - L(f,P) = (M_1-m_1)(x_1-0) +  \sum_{j=2}^n(M_j-m_j)(x_j - x_{j-1})\\ < \epsilon/2 + \sum_{j=2}^n(M_j-m_j)(x_j - x_{j-1})$$

I'll let you finish by showing that the partition points $x_2,\ldots, x_{n-1}$ can be chosen such that

$$\sum_{j=2}^n(M_j-m_j)(x_j - x_{j-1}) < \epsilon/2$$


问:
Solving a tensor equation from singularity condition
I apologize if this question has already been asked but I'm not sure what the best key-words are.

I have a tensor equation of the form: $(A + x\otimes b)c=0$.

Here, $A$ is a second-order tensor/matrix, and the rest are first-order tensors/vectors. $A, b$ are known but $x, c$ are unknown. I am interested in calculating a vector '$x$' such that $A + x\otimes b$ is singular. What would be an approach to solve such a problem?

Thanks!

答:
If $A$ is a square matrix, then it has a Jordan canonical form:
\begin{equation}
P^{-1}AP = J = \left(\begin{array}{ccc}
J_{i_1}(\lambda_1) & & \\
& \ddots & \\
& & J_{i_n}(\lambda_n)
\end{array}\right),
\end{equation}
where

$\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$

each $J_{i_m}(\lambda_m)$ is an $i_m\times i_m$ Jordan block.

For example, a $2\times 2$ Jordan block has the form
\begin{equation}
J_2(\lambda) = \left(\begin{array}{c}\lambda & 1\\0 & \lambda\end{array}\right).
\end{equation}
Let's suppose that $A$ is equal to a Jordan form. We address the other case afterward.

When dealing with concrete matrices, I prefer the notation $xb^{\mathsf{T}}$ over $x\otimes b$. If $x$ and $b$ are column vectors, then $xb^{\mathsf{T}}$ is a rank-1 matrix; each column is a scalar multiple of any other column in this matrix.

We assume that $A$ is in Jordan canonical form.

If the top entry of $b$ is not $0$, then we can engineer an $x$ so that the first column of $A - xb^{\mathsf{T}}$ is all zeros. Since $A$ is Jordan form, it has the form
\begin{equation}
A =
\left(\begin{array}{c}
\lambda_1 & \cdots\\
0 & \ddots \\
\vdots & \ddots\\
0 & \cdots
\end{array}\right),
\end{equation}
i.e. its first column has an eigenvalue in its top entry and zeros otherwise.

Consider
\begin{equation}
x = \left(\begin{array}{c}\lambda_1/b_1\\0\\\vdots\\0\end{array}\right).
\end{equation}
If we set $x$ to this, then
\begin{equation}
xb^{\mathsf{T}} =
\left(\begin{array}{c}
\lambda_1/b_1\\
0\\
\vdots\\
0
\end{array}\right)
\begin{array}{c}
\left(b_1, b_2,\ldots,b_N\right)\\
\\
\\
\\
\end{array}
=
\left(\begin{array}{ccc}
\lambda_1 & \lambda_1\cdot b_2/b_1 & \cdots\\
0 & 0 & \cdots\\
\vdots & \vdots & \ddots\\
0 & 0 & \cdots
\end{array}\right).
\end{equation}

With this engineered $xb^{\mathsf{T}}$, the first column  of $A - xb^{\mathsf{T}}$ is all zeros, and $A - xb^{\mathsf{T}}$ is singular.

Now suppose that $A$ is not in Jordan canonical form, but $J = P^{-1}AP$ is. If $A - xb^{\mathsf{T}}$ is singular, then $P^{-1}\left(A - xb^{\mathsf{T}}\right)P$ is, too.
\begin{eqnarray}
P^{-1}\left(A - xb^{\mathsf{T}}\right)P &=& P^{-1}AP - P^{-1}xb^{\mathsf{T}}P\\
&=& J - \left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}
\end{eqnarray}

We consider the same scheme as before. This time, however, we want $P^{-1}x$ to have the form
\begin{equation}
P^{-1}x =
\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right).
\end{equation}
To ensure that this happens, set
\begin{equation}
x =
P\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right),
\end{equation}
so that
\begin{equation}
P^{-1}x ~=~
P^{-1}P\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right)
~=~
\left(\begin{array}{c}
\lambda_1/\left(P^{\mathsf{T}}b\right)_1\\
0\\
0\\
\vdots\\
0\end{array}\right).
\end{equation}

With $x$ engineered this way, we have
\begin{equation}
\left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}
=
\left(\begin{array}{c}
\lambda_1/b_1\\
0\\
\vdots\\
0
\end{array}\right)
\begin{array}{c}
\left((P^{\mathsf{T}}b)_1, (P^{\mathsf{T}}b)_2,\ldots,(P^{\mathsf{T}}b)_N\right)\\
\\
\\
\\
\end{array}
=
\left(\begin{array}{ccc}
\lambda_1 & \lambda_1\cdot (P^{\mathsf{T}}b)_2/(P^{\mathsf{T}}b)_1 & \cdots\\
0 & 0 & \cdots\\
\vdots & \vdots & \ddots\\
0 & 0 & \cdots
\end{array}\right).
\end{equation}
This ensures that the first column of $J - \left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}$ is all zeros. Thus $J - \left(P^{-1}x\right)\left(P^{\mathsf{T}}b\right)^{\mathsf{T}}$ is singular, and $A - xb^{\mathsf{T}}$ is, too.


答:
Once you learned modular fraction you will probably never want to go back to Euclid's algorithm.  In your case if the RHS is not some nice number like 44, you want to find the inverse of 11 under modulo 64, then you apply inverse reciprocity
$$
\pmod{11}: 0 \equiv 1+64k \equiv -10-2k \implies k\equiv -5 \equiv 6
$$

Therefore
$$
\frac{1}{11}\equiv \frac{1+64\cdot 6}{11}\equiv \frac{1+(66-2)\cdot 6}{11} \equiv 6 \cdot 6-1 \equiv 35 \pmod{64}
$$

Title: Number Theory : $11x \equiv44\pmod{64}$ how to solve?

Body:
how to solve this $11x\equiv 44\pmod{64}$ ?

attempt:

first of all need to find the inverst of $11x\equiv 44\pmod{64}$

by Euclid algorithm then:

$64=11\cdot 5 +9$

$11=1\cdot 9 +2$

$9=2\cdot4+1$

$1=9-2\cdot4$

what is the next step for the make the inverst I do not understand ?


问:
From monoids to groups
I was looking at the case when you go from the monoid of natural numbers to the group of integers by means of a suitable equivalence relation. The key here was to find inverses for each natural and I was wondering how it could be generalized. I mean:

Given any monoid, when can I find inverses for your elements?

In other words, can I always go from a monoid to a group?
If not, when is it? When do not?

If you could tell me where I can find information about this (a book or paper) it would be very helpful. Thank you.

答:
Yes, you can always obtain a group from a monoid, through a process that is called group completion.

In general, the completion is a process very similar to the one you cited: going from $(\mathbb N,+)$ to $(\mathbb Z,+)$ is the standard, classic example; another one is going from the monoid $(\mathbb N^\ast,\cdot)$ to its group completion $(\mathbb Q_{>0},\cdot)$.

The "easy" case is when your monoid $(M,+)$ is commutative; its group completion, sometimes called Grothendieck group (for its ties with algebraic K-theory, if I'm not mistaken), can be defined as a quotient of $M\times M$ (with coordinate-wise operation, $(m,m')+(n,n'):=(m+n,m'+n')$) by the equivalence relation described as follows:
$$ (m,m')\sim(n,n') \iff \exists\;k\in M:\;m+n'+k=m'+n+k $$
Sometimes (i.e. when the cancellation law holds in $M$) the term $k$ is not necessary (you can just choose the identity of your monoid as $k$).
In your original example, you have that the group completion of $(\mathbb N,+)$ is made of equivalence classes of $(m,m')\in\mathbb N^2$, where $(m,m')\sim(n,n')\iff m+n'=m'+n$, hence you can see that "$(m,m')$ basically means $m-m'$".

Notice that this procedure is very reminiscent of the construction of rational numbers! In fact, as we mentioned before, if you take as $M$ the monoid of positive integers with multiplication $(\mathbb N^\ast,\cdot)$, you get as its group completion elements of $\mathbb N^*\times \mathbb N^*$ modulo the relation:
$$ (m,m')\sim(n,n') \iff m\cdot n'=m'\cdot n,$$
so $(m,m')$ is basically what we're used to denote as $\frac{m}{m'}$.

The most important application of this construction I know of is, as I mentioned early, K-groups in K-theory; for instance, real/complex vector bundles over a "nice" (e.g., compact Hausdorff) topological space, and finitely generated projective modules over a ring, up to isomorphisms, form a monoid $(M,\oplus)$ with the direct sum, and then $K_0$ is defined as the group completion of $M$.

Title: From monoids to groups


问:
Rewriting this Riemann sum as a definite integral
Can someone help me rewrite this Riemann sum as a definite integral?

$\displaystyle \lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n}\left(-7+\frac{14i}{n} \right)^9\sin\left(4+ \left(-7+\frac{14i}{n} \right)^8 \right)$

$\Delta x=\frac{1}{n}$ so this means $b-a=1$. If I rewrite the $\frac{14i}{n}$ as $14\frac{i}{n}$, then this means I could say that $a=0$ and $b=1$ so this means my definite integral is:

$$\int_{0}^{1}(-7+14x)^9\sin(4+(-7+14x)^8)dx$$ but I am not sure if this is correct or not.

If this IS correct, how would I do this integral with no integration technique other than u-substitution or manipulation? I can't seem to do it otherwise.

答:
Assuming partition points are of the form $x_i=a+i \frac{b-a}{n}$, it looks like $a=-7$ and $b-a=14$ so $b=7$. The $1/n$ that's pulled out of the sum is misleading. Rewrite it as $1/n=\frac{1}{14}\cdot \frac{14}{n}$. This leads to the integral
$$
\frac{1}{14}\int_{-7}^7 x^9 \sin(4+x^8)
$$
Hint: To evaluate the integral, take note of the bounds and types of functions.


答:
It's important to have so theory under you belt.

$11$ has a unique inverse inverse $\mod 64$ if and only if $\gcd(11,64)=1$ which it it does.

So solve $11x \equiv 44 \pmod {64}$ we multiply both sides by $11^{-1}$ (whatever that is) and get

$11^{-1}\cdot 11x \equiv 11^{-1}\cdot 44\equiv 11^{-1}\cdot 11\cdot 4\pmod {64}$ so

$x \equiv 4\pmod {64}$.

And that's it.  You don't need to figure out what $11^{-1}$ actually is (it is $35$ because $11\cdot 35\equiv 385=1+6\cdot 64$ so $35\cdot 11x\equiv 35\cdot 44\pmod {64}$ so $x\equiv 385x \equiv 1540 = 4 + 35\cdot 24\equiv 4 \pmod {64}$);  It's enough to know there is an inverse for $11$ and that it is unique.

If a number is invertible you can do division. And the will be okay if $11$ and $64$ are relatively prime.

=========

But note if $\gcd(11,64)\ne 1$ there is no inverse and this might have no solutions or multiple solutions.

$11x = 44 \pmod 66$ for example has $4$ as a solution but also $x\equiv 10,16,22,28,......$ as solutions.

And although $11x \equiv 43 \pmod {64}$ has a unique solution $x \equiv 35\cdot 43 \equiv 33 \pod {64}$  $11x \equiv 43 \pmod {66}$ doesn't have any solutions.


问:
Prove that $M_{n} - m_{n}$ has distribution $\beta (n - 1, 2)$
I have the following exercise, but I'm a bit lost, I'll share it with you:

Let $(U_{n})_{n\in\mathbb{N}}$ be a sequence of random variables i.i.d. with uniform distribution in the interval $[0, 1]$.
We define for each $n\in \mathbb{N}$ the random variables:
$$M_{n} = \max\{U_{1} ,...,U_{n}\}$$
$$m_{n} = \min\{U_{1} ,...,U_{n}\}$$

a) Prove that $M_{n} - m_{n}$ has distribution $\beta (n - 1, 2)$

b) Show that $M_{n} -m_{n}\ \underrightarrow{d}\ 1$

c) Calculate the limit in distribution of the sequence $(n (1 - M_{n} + m_{n})) \ n \in \mathbb{N}$. Is the limit a distribution known?

Now I have to:

$$P(\min\{U_{1} ,...,U_{n}\}>a)=P(U_{1}>a)P(U_{2}>a)\dotsi P(U_{n}>a)=(1-a)^n$$
$$\rightarrow F_{m_{n}}(a)=1-(1-a)^n \rightarrow f_{m_{n}}(a)=n(1-a)^{n-1}$$

$$P(\max\{U_{1} ,...,U_{n}\}\leq a)=P(U_{1} \leq a)P(U_{2} \leq a)\dotsi P(U_{n} \leq a)=a^n$$
$$\rightarrow f_{m_{n}}(a)=na^{n-1}$$

I don't know how to continue, could you help me?

答:
First, calculate the joint pdf of $m_n, M_n$. For any $0 \leq m \leq M \leq 1$, if you have $m < m_n, M_n \leq M$, we must have $m < U_i \leq M, \forall i = 1, \ldots, n$. Thus by independence
$$
P(m < m_n, M_n \leq M) = \prod_{i = 1}^n P(m \leq U_i \leq M) = (M - m)^n
$$ Also, as you already noted yourself, $P(M_n \leq M) = M^n$. So now, the complement with respect to the first component (i.e. keeping $M_n \leq M$ fixed) is:
$$
P(m_n \leq m, M_n \leq M) = P(M_n \leq M) - P(m < m_n, M_n \leq M) = M^n - (M - m)^n
$$ Taking the partial derivatives w.r.t. $m$ and then $M$ gives the density of $(m_n, M_n)$ over $0 \leq m \leq M \leq 1$:
$$
f_{m_n, M_n}(m, M) = n(n - 1)(M - m)^{n - 2},\ 0 \leq m \leq M \leq 1
$$ It is also not hard to see that other combinations of $m, M$ must yield $0$ for the density. So
$$
f_{m_n, M_n}(m, M) = \begin{cases} n(n - 1)(M - m)^{n - 2}\ &0 \leq m \leq M \leq 1 \\ 0 &\text{elsewhere} \end{cases}
$$ Now we do the transformation $T : (m, M) \mapsto (m, M - m) := (\mu, \delta)$. This maps the support $$\{(m, M) \in \mathbb{R}^2:0 \leq m \leq M \leq 1\}$$ of $(m_n, M_n)$ to the support $$\{(\mu, \delta) \in \mathbb{R}^2:0 \leq \mu \leq 1 - \delta,\ 0 \leq \delta \leq 1\}$$ of $(m_n, M_n -  m_n)$. Also, the inverse transformation is $T^{-1}: (\mu, \delta) \mapsto (\mu, \mu + \delta)$ with Jacobian $J(\mu, \delta) = \begin{bmatrix} 1 &0 \\ 1 &1 \end{bmatrix}$. Thus the transformed pdf is
\begin{gather*}
f_{m_n, M_n - m_n}(\mu, \delta) \\ = \begin{cases} f_{m_n, M_n}(T^{-1}(\mu, \delta))|J(\mu, \delta)| = n(n - 1)\delta^{n - 2}\ &0 \leq \mu \leq 1 - \delta,\ 0 \leq \delta \leq 1 \\ 0 &\text{elsewhere} \end{cases}
\end{gather*} So finally the (marginal) density of $M_n - m_n$ is:
$$
f_{M_n - m_n}(\delta) = \int_0^{1 - \delta}n(n - 1)\delta^{n - 2}\ d\mu = n(n - 1)(1 - \delta)\delta^{n - 2},\ \text{ for } 0 \leq \delta \leq 1
$$ and $0$  elsewhere. This is precisely the pdf of $\beta(n - 1, 2)$.


问:
How to show determinant of a specific matrix is nonnegative
How to show that
$$\det A= \det \begin{pmatrix}\cos\frac{\pi}{n}&-\frac{\cos\theta_1}{2}&0&0&\cdots&0&-\frac{\cos\theta_n}{2}
\\-\frac{\cos\theta_1}{2}&\cos\frac{\pi}{n}&-\frac{\cos\theta_2}{2}&0&0&\cdots&0\\
0&-\frac{\cos\theta_2}{2}&\cos\frac{\pi}{n}&-\frac{\cos\theta_3}{2}&0&\cdots&0\\
\vdots&\vdots&-\frac{\cos\theta_3}{2}&\ddots&\ddots&\vdots&\vdots\\\vdots&\vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\vdots&\vdots&\vdots&\ddots&\ddots&-\frac{\cos\theta_{n-1}}{2}\\
-\frac{\cos\theta_n}{2}&0&\cdots&0&0&-\frac{\cos\theta_{n-1}}{2}&\cos\frac{\pi}{n} \end{pmatrix} \geq 0,
$$
  where $\sum\limits_{i=1}^n\theta_i=\pi$, $0<\theta_i<\frac{\pi}{2}, i=1,\dots,n$, and $n\ge 3$? </p>

答:
(too long for a comment, but hopefully somebody more skilled than me can flesh this out as a full answer.)

The matrix in question is a symmetric, cyclic tridiagonal (a.k.a. periodic tridiagonal) matrix, and can be decomposed as

$$\mathbf A=\begin{pmatrix}\cos\frac{\pi}{n}&-\frac12\cos\theta_1&&\\-\frac12\cos\theta_1&\cos\frac{\pi}{n}&\ddots&\\&\ddots&\ddots&-\frac12\cos\theta_{n-1}\\&&-\frac12\cos\theta_{n-1}&\cos\frac{\pi}{n}\end{pmatrix}+\begin{pmatrix}1&0\\0&\vdots\\\vdots&0\\0&1\end{pmatrix}\cdot\begin{pmatrix}0&\cdots&0&-\frac12\cos\theta_n\\-\frac12\cos\theta_n&0&\cdots&0\end{pmatrix}$$

or in abbreviated form, $\mathbf A=\mathbf T+\mathbf U\mathbf V^T$. Using the Sherman-Morrison-Woodbury formula for determinants, we have

$$\det\mathbf A=(\det\mathbf T)\det(\mathbf I+\mathbf V^T\mathbf T^{-1}\mathbf U)$$

Now, a three-term recurrence can be set up to compute $\det\mathbf T$; letting $p_0=1$ and $p_k$ be the determinant of the $k\times k$ leading submatrix of $\mathbf T$, we can establish the recurrence

$$p_k=p_{k-1}\cos\frac{\pi}{n}-\frac{\cos^2 \theta_{k-1}}{4}p_{k-2}$$

and then $\det\mathbf T=p_n$. There is also an explicit formula for the first and last columns of the inverse of a tridiagonal matrix (the result of $\mathbf T^{-1}\mathbf U$) which I will have to look up (unless somebody beats me to it), and proving the nonnegativity of $\det\mathbf A$ boils down to proving that either $p_n$ and $\det(\mathbf I+\mathbf V^T\mathbf T^{-1}\mathbf U)$ have the same sign, or that one of the two is zero.


问:
Does the kernel need to be the full automorphism group of the induced subgraph?
Let $\Gamma$ be a simple graph. Suppose its automorphism group $G=\text{Aut}\Gamma$ is imprimitive on its vertex set $V$. Take a block system $\mathcal{B}$ of $V$. Let $B\in\mathcal B$, and let $K=G_{(\mathcal B)}$. If $K$ acts faithfully on the induced subgraph $[B]_\Gamma$, that is $K_{(B)}=1$, can we always say $K^B=\text{Aut}[B]_\Gamma$?

NB: $G_{(\mathcal B)}$ is the subgroup of $G$ which stabilises $\mathcal B$ pointwisely.

$G^\mathcal{B}:=G/G_{(\mathcal B)}$ is the group that acts on $\mathcal B$ faithfully.

The same way of definition applies to all other notations wherever it applies.

答:
The answer is "no"; I offer one counterexample.

I assume that imprimitive means vertex transitive and imprimitive (an imprimitive group
need not be transitive).

Take $X$ to be the icosahedron, a 5-regular arc-transitive
graph on 12 vertices with automorphism group $G$ of order 120.
Now let $Y$ be the line graph of the subdivision graph of the icosahedron, i.e.,
put a new vertex in the middle of each of the 60 edges of the icosahedron, then take the line graph of this subdivided graph.
Then $|V(Y)|=2|E(X)|=120$ and $H=\mathrm{Aut}(Y)$ is isomorphic to $G$ (by
theorems about automorphism groups of line graphs).
Since $G$ is arc transitive, $H$ is vertex transitive, in fact it acts regularly on the vertices of $Y$. Each vertex of $X$ gives rise to a clique
of size five in $Y$, and these cliques are blocks of imprimitivity for $H$. The subgroup of $H$ that fixes one of these cliques has order at most five, and so it does not induce the full automorphism group of the clique.

I am sure this construction will work with many other arc-transitive graphs.


答: Your first idea is not far from what you need. $f(x) = \frac{x}{x^2-1}$ maps $(-1,1)$ to $\mathbb{R}$. But you need its inverse: $g(x) = \frac{-2x}{1+\sqrt{1+4x^2}}$ is a continuous strictly decreasing function with image $(-1,1)$


问:
Show that there exists a sequence $c_n\rightarrow \infty$ such that $\frac{X_n}{c_n}\rightarrow 0$ almost surely
Let $(X_n)_n$ be a sequence of random variables. Show that there exists a sequence $c_n\rightarrow \infty$ such that $\frac{X_n}{c_n}\rightarrow 0$ almost surely. I am not sure how to start, may someone elaborate on the intuition to construct the $c_n's$? How would one sttart constructing them?

答:
Note that $X_n/c_n\to 0$ a.s. if $\mathsf{P}(|X_n/c_n|>\epsilon_n\text{ i.o.})=0$, where $\{\epsilon_n\}$ is a positive sequence converging to $0$. Then, given $\{\epsilon_n\}$, choose a sequence $\{c_n\}$ s.t.
$$
\sum_{n\ge 1}\mathsf{P}(|X_n|>c_n\epsilon_n)<\infty.
$$


问:
On derivatives of composite and inverse functions
Question

Suppose $g$ is the inverse of a one-to-one differentiable function $f$ and $H = g \circ g\ $. If $f(4) = 3\ $, $g(4) = 5\ $, $f'(4) = \frac 1 2\ $ and $f'(5) = 2\ $, find $H'(3)\ $.

My working

$\because H(x) = g^2(x)$

$\therefore H'(x) = 2g(x)g'(x)$

$\because g'(x) = \frac 1 {f'[f^{-1}(x)]}$

$\therefore H'(x) = \frac {2f^{-1}(x)} {f'[f^{-1}(x)]}$

$\because f(4) = 3$

$\therefore f^{-1}(3) = 4$

$\implies H'(3) = \frac {2(4)} {f'(4)} = \frac {2(4)} {\frac 1 2} = 16$

However, the answer given is $H'(3) = 1\ $. I am not sure where I have gone wrong (although I have a hunch it is probably in my derivation of $H'(x)$). Any help will be greatly appreciated :)

答:
You went wrong right at the beginning (assuming that you wrote the problem correctly).

$$H = g \circ g = g(g(x)) $$

Composition not multiplication. I'm assuming that you can get $g(3)= 4$ easily.

So
$$H' = g \circ g = g'(g(x))g'(x)$$

Using rule for derivate inverses.

$$g'(3) = \frac{1}{f'(g(3))} = \frac{1}{f'(4)} = 2$$
$$g'(4) = \frac{1}{f'(g(4))} = \frac{1}{f'(5)} = \frac{1}{2}$$


问:
The set of all functions mapping set $A$ to set $B$
Is $F$ as defined here the set of all functions from set $A$ to set $B$?

$F=\{f\in 2^{A\times B}:\forall x(x\in A\rightarrow\exists y (y\in B\wedge (x,y)\in f))\wedge \forall x,y_1,y_2 ((x,y_1)\in f\wedge (x,y_2)\in f\rightarrow y_1=y_2)\}$

If $A$ was non-empty, then $B$ would also have to be non-empty, but my construction in no way depended on either set being non-empty. How can this be? Have I made a mistake?

答: Assuming that you’re using the notation $2^S$ for the power set of $S$, then yes, you’ve correctly described the set of functions from $A$ to $B$. If $A\ne\varnothing$ and $B=\varnothing$, $A\times B=\varnothing$, so the only $f\in 2^{A\times B}$ is $\varnothing$, and it fails to satisfy the first clause of the definition and is therefore not actually in $F$. Thus, $F=\varnothing$ in this case, just as it should.</p>


问:
Show that the ring $\mathbb{Z}[i]/n\mathbb{Z}[i]$ is finite.
I'm pretty stuck with this problem. I hope you can help me.

Let $R=\mathbb{Z}[i] .$ Consider $n \in \mathbb{Z} \backslash\{0\}$ and the ideal $I=R n$ in $R$. Show that $a+b i \in I$ if and only if $n \mid a$ and $n \mid b .$ Show that $R / I$ is a finite ring.

I have already shown the first part, that n|a and n|b iff $a+b i \in I$.
But how do I show that the ring is finite?

I know that $R$ is a PID and thus that I is a principal ideal. So, $I=\{n(a+i b) \mid a, b \in \mathbb{Z}, \quad n \in \mathbb{N} /\{0\}\}$ must be generated by some element $z\in R$ ; $I=\langle z \rangle$. I think I'm supposed to use these results and that n|a and b. But I just can't seem to make any connections and where to go from here.

答: Hint: Consider the group homomorphism $$\begin{align} \mathbb Z[i] & \to (\mathbb Z/n\mathbb Z) \oplus (\mathbb Z/n\mathbb Z) \\ a+bi & \mapsto (a+n\mathbb Z, b+n\mathbb Z). \end{align}$$


答:
You need not resort to any stronger features of $\mathbb{Z}[\mathrm{i}]$ as a ring. Let us consider an arbitrary ring $(A, +, \cdot)$ whose additive subjacent group $(A, +)$ is finitely generated as a $\mathbb{Z}$-module. Then for any $n \in \mathbb{N}^{\times}\colon=\mathbb{N} \setminus \{0\}$ we have that $nA \leqslant_{\mathrm{b}} A$ is a bilateral (the English idiom would be two-sided, but I dislike it) ideal -- fact which is easy enough to ascertain -- and the quotient $A/nA$ is a finite ring.

To show this, let us use the abbreviation $\mathbb{Z}_r\colon=\mathbb{Z}/r\mathbb{Z}$ for any $r \in \mathbb{N}$ and let us also set $B\colon=A/nA$. Remark that $n \in \mathrm{Ann}_{\mathbb{Z}}(B)$, explicitly put $n$ is in the annihilator of the $\mathbb{Z}$-module $A/nA$. This means that this quotient abelian group can be naturally equipped with a $\mathbb{Z}_n$-module structure and it is furthermore easy to see that the lattices of $\mathbb{Z}$-submodules respectively $\mathbb{Z}_n$-modules of $B$ coincide.

Since $B$ is the quotient of a finitely generated $\mathbb{Z}$-module, it remains finitely generated over $\mathbb{Z}$ and therefore -- in light of the above observation regarding the lattices of submodules (with respect to the two distinct rings of operators) -- it is also finitely generated as a $\mathbb{Z}_n$-module. This means there exists a finite set $T \subseteq B$ such that $B$ can be expressed as a quotient (homomorphic image) of $\mathbb{Z}_n^{(T)}=\mathbb{Z}_n^T$, the free $\mathbb{Z}_n$-module on set $T$. As $\mathbb{Z}_n$ and $T$ are both finite, the free module $\mathbb{Z}_n^T$ is also finite and so therefore must also be its quotient $B$.

This general reasoning applies to your particular problem since indeed $\mathbb{Z}[\mathrm{i}]=\langle \{1, \mathrm{i}\}\rangle$ is finitely generated as a(n abelian) group.

P.S. The general argument above only establish finiteness and does little to assess that order of finiteness. In your particular case however, it is clear that the additive group $\mathbb{Z}[\mathrm{i}]$ is a free $\mathbb{Z}$-module of basis $\{1, \mathrm{i}\}$ and therefore isomorphic to $\mathbb{Z} \times \mathbb{Z}$ via the isomorphism:
$$\begin{align}
f \colon \mathbb{Z}\times\mathbb{Z} &\to \mathbb{Z}[\mathrm{i}]\\
f(r,s)&=r+s\mathrm{i},
\end{align}$$
isomorpism which has the property that $f[n(\mathbb{Z} \times \mathbb{Z})]=n\mathbb{Z}[\mathrm{i}]$. Bearing in mind that $n(\mathbb{Z} \times \mathbb{Z})=n\mathbb{Z} \times n\mathbb{Z}$, we thus infer that $f$ induces the following morphism of additive groups:
$$\mathbb{Z}[\mathrm{i}]/n\mathbb{Z}[\mathrm{i}] \approx (\mathbb{Z} \times \mathbb{Z})/(n\mathbb{Z} \times n\mathbb{Z}) \approx (\mathbb{Z}/n\mathbb{Z}) \times (\mathbb{Z}/n\mathbb{Z})=\mathbb{Z}_n \times \mathbb{Z}_n \quad (\mathbf{Gr}),$$
on grounds of which we can claim that $\left(\mathbb{Z}[\mathrm{i}]\colon n\mathbb{Z}[\mathrm{i}]\right)=\left|\mathbb{Z}[\mathrm{i}]/n\mathbb{Z}[\mathrm{i}]\right|=n^2$.

More generally, if ring $A$ had a free $\mathbb{Z}$-module for its subjacent additive group, then the subjacent additive group of the quotient ring $A/nA$ would also be a free $\mathbb{Z}_n$-module such that $\dim_{\mathbb{Z}}A=\dim_{\mathbb{Z}_n}A/nA$.

Title: Show that the ring $\mathbb{Z}[i]/n\mathbb{Z}[i]$ is finite.

Body:
I'm pretty stuck with this problem. I hope you can help me.

Let $R=\mathbb{Z}[i] .$ Consider $n \in \mathbb{Z} \backslash\{0\}$ and the ideal $I=R n$ in $R$. Show that $a+b i \in I$ if and only if $n \mid a$ and $n \mid b .$ Show that $R / I$ is a finite ring.

I have already shown the first part, that n|a and n|b iff $a+b i \in I$.
But how do I show that the ring is finite?

I know that $R$ is a PID and thus that I is a principal ideal. So, $I=\{n(a+i b) \mid a, b \in \mathbb{Z}, \quad n \in \mathbb{N} /\{0\}\}$ must be generated by some element $z\in R$ ; $I=\langle z \rangle$. I think I'm supposed to use these results and that n|a and b. But I just can't seem to make any connections and where to go from here.


问:
Flux through a circular loop
Question: Let, $F$=$r\over{||r||}^2$ = ${{xi + yj}}\over{x^2+y^2}$

Lets $\alpha$ be a simple closed curve in a plane that doesn't pass through origin. Show ${\int F\cdot nds }$ is equal to zero or 2pi according as the origin inside or outside $\alpha$

My work so far: I did the computation and found the flux using  ${\int F\cdot nds }$  to be 2$\pi$. I also used green's therom and found that div(F) is equal to zero.  I think these numbers are right, if you could double check though that would be nice. I understand that those two terms are different because div(F) is not defined at origin because you would be dividing by zero. At origin div(F) blows up and everywhere else its zero.  But I dont know how to approach the question that is asked. I dont know what my teachers means by "Show ${\int F\cdot nds }$ is equal to zero or 2pi....".  Is there more computation needed to answer this question? If so, what should I do next?

答:
Here's how I would prove the hard case:

Step 1: Show that $\int_\alpha F\cdot n\, ds=\int_{C_r} F\cdot n\, ds$,
where $C_r$ is a closed circle of radius $r$, chosen small enough so that the corresponding disk fits inside the closed region created by $\alpha$. You can show this fact that splitting up the region enclosed by $\alpha$ into two regions that avoid the origin (where a clockwise circle is created around). Divergence theorem applies.

Step 2: Solve with parameterization on the new integral.

Title: Flux through a circular loop


问:
Lagrange theorem help
As per Lagrange theorem it says I can get maximum or minimum under some constraint.

Example:

$f(x,y)=x^2+y^2+4$ under constraint $x+y=2$. I can use Lagrange theorem for this.

But I have a problem like $f(x,y)=x^2+y^2+4$  under constraint like $x$ should be minimum.

How can I solve this any theorem or solution?

答:
Use Lagrange multipliers with the constraint $\partial g/\partial x=0$. This will at least constrain $g$ to be extremal. So construct an auxillary function

$$F(x,y,\lambda)=f(x,y)+\lambda \frac{\partial g(x)}{\partial x}$$

and minimize with $\nabla F=0$

Title: Lagrange theorem help


问:
distance function between point and compact set

Let be $\left(X,\Vert \cdot \Vert \right)$ a normed space and $\emptyset \neq M\subseteq X$.

Consider the function:  $dist(\cdot,M):X\to \mathbb{R}$, where $dist(x,M)=\inf\limits_{y\in M}\Vert x-y\Vert$.

If $M$ is compact show that there exists a $y\in M$ for every $x \in X$ such that: $dist(x,M) = \Vert x-y\Vert$.

My approach:

Let $x\in X$ be arbitrary but fixed. We know that for every $\epsilon>0$ there exists a $y\in M$ with: $dist(x,M)+\epsilon>\Vert x-y\Vert$ and that $dist(x,M)$ is a lower bound for $\Vert x-y\Vert$ with $y\in M$. This allows me to construct a sequence $\left(y_n\right)_{n\in\mathbb{N}}$. Every sequence on a compact set contains a convergent subsequence. Let be $\left(y_{n_j}\right)_{j\in\mathbb{N}}$ a convergent subsequence. I define $\epsilon:=\frac{1}{j}$ and get: $0\leq\Vert x- y_{n_j}\Vert -dist(x,M) <\frac{1}{j}$. (I can omit the absolute value bars as $dist(x,M)$ is a lower boound and hence always smaller). Further, as $M$ is compact, the limit of each convergent sequence lies within $M$, so $\lim\limits_{j\to\infty} y_{n_j}=y \in M$. Applying the squeeze theorem, we can finally conclude $dist(x,M)= \Vert x-y\Vert$.

Is this o.k. or are there any gaps/flaws?

Any thoughts are welcome :)

答:
It seems that there is no flaw.

(1) Fact : A function $g(y)=\| x-y\|$ is continuous in a metric
space $(X,\|\ \|)$ i.e. When $\|y_n -y\|\rightarrow 0$, then
$g(y)=\lim_n\ g(y_n)$ (OP use this fact)

(2) Define $f(x)=\| x- M\|$ (For convenience, we use this notation)

Hence $f(x)= \inf_{y\in M}\ \{ g(x,y) \}$ where $g(x,y)=\|x-y\|$. Then prove that $f$ is 1-Lipschitz map.

Proof : By OP's posting, there is $x_M\in M$ s.t. $f(x)=\| x-x_M\|$.
Hence $$ f(y)=\|y-y_M\|\leq \|y-x\| + \|x-x_M\| = \|x-y\| + f(x)$$


答: Notice that the random vector $(m_n,M_n)$ is supported on the the region $R$ defined by $$R=\{(x,y)\in\mathbb{R}^2\big|x \leq y \leq 1,0\leq x\leq 1\}$$ Let $f_{m_n,M_n}$ denote the joint density of $(m_n,M_n)$, and fix $(x,y)\in R$. We have that $$f_{m_n,M_n}(x,y)=f_{M_n|m_n}(y|x)f_{m_n}(x)$$ You have successfully shown that $f_{m_n}(x)=n(1-x)^{n-1}$. Notice how $$F_{M_n|m_n}(y|x)=P(M_n \leq y\big|m_n=x)=\bigg(\frac{y-x}{1-x}\bigg)^{n-1}$$ Applying $\frac{\partial}{\partial y}$ yields $$f_{M_n|m_n}(y|x)=\frac{n-1}{1-x}\cdot \bigg(\frac{y-x}{1-x}\bigg)^{n-2}$$ Putting these pieces together provides us with our joint density $$f_{m_n,M_n} (x,y)=n(n-1)(y-x)^{n-2}$$ Can you take it from here?

Body:
I have the following exercise, but I'm a bit lost, I'll share it with you:

Let $(U_{n})_{n\in\mathbb{N}}$ be a sequence of random variables i.i.d. with uniform distribution in the interval $[0, 1]$.
We define for each $n\in \mathbb{N}$ the random variables:
$$M_{n} = \max\{U_{1} ,...,U_{n}\}$$
$$m_{n} = \min\{U_{1} ,...,U_{n}\}$$

a) Prove that $M_{n} - m_{n}$ has distribution $\beta (n - 1, 2)$

b) Show that $M_{n} -m_{n}\ \underrightarrow{d}\ 1$

c) Calculate the limit in distribution of the sequence $(n (1 - M_{n} + m_{n})) \ n \in \mathbb{N}$. Is the limit a distribution known?

Now I have to:

$$P(\min\{U_{1} ,...,U_{n}\}>a)=P(U_{1}>a)P(U_{2}>a)\dotsi P(U_{n}>a)=(1-a)^n$$
$$\rightarrow F_{m_{n}}(a)=1-(1-a)^n \rightarrow f_{m_{n}}(a)=n(1-a)^{n-1}$$

$$P(\max\{U_{1} ,...,U_{n}\}\leq a)=P(U_{1} \leq a)P(U_{2} \leq a)\dotsi P(U_{n} \leq a)=a^n$$
$$\rightarrow f_{m_{n}}(a)=na^{n-1}$$

I don't know how to continue, could you help me?


问:
Showing the tangent space of $SL(n,\mathbb{C})$

Showing the tangent space of $SL(n,\mathbb{C})=\{A\in GL(n,\mathbb{C}):\det(A)=1\}$ is $S=\{A\in GL(n,\mathbb{C}):\text{trace(A)}=0\}$

I have the definition of tangent space at a matrix $A$ is $T_A(M)=\{X\in M: \exists\gamma:[0,1]\to M,\text{is differentiable}, \gamma'(0)=X,\gamma(0)=A$

So if I take some matrix $B\in T_A(SL(n,C))$ I want to show it has $0$ trace , to show the tangent is contained in my set $S$

But I don't really have anything linking trace of a matrix to its determinant except for $\det(e^A)=e^{Tr(A)}$

答:
The map $X\mapsto \exp X$ is a diffeomorphism from a neigborhood of $0_n$ to a neighborhood of $I_n$.  Let $\gamma \colon [0,1[ \to SL(n,\mathbb{C})$, $\gamma(0)= I_n$. For some $\epsilon>0$ we can write $\gamma(t) = \exp \eta(t)$, with $\eta(\cdot)$ smooth, and $\eta(0) = O_n$.  Now $\gamma(t) \in SL(n,\mathbb{C})$ implies $Tr \,\eta(t) = 0$ for  all $t \in [0, \epsilon)$. We have $\gamma'(t) = \exp (\eta(t)) \cdot \eta'(t)$, and $\gamma'(0) = \eta'(0)$ has also trace $0$.

Therefore, the tangent space of $SL(n, \mathbb{C})$ at $I_n$ is $sl(n, \mathbb{C})$, the subspace of $n\times n$ complex matrices with trace $0$.

Note: Here we determined the tangent space of $SL(n,\mathbb{C})$ at $I_n$. At an arbitrary element $A\in SL(n, \mathbb{C})$ it will be $A\cdot sl(n, \mathbb{C})= sl(n,\mathbb{C})\cdot A$.


问:
Number Theory : $11x \equiv44\pmod{64}$ how to solve?
how to solve this $11x\equiv 44\pmod{64}$ ?

attempt:

first of all need to find the inverst of $11x\equiv 44\pmod{64}$

by Euclid algorithm then:

$64=11\cdot 5 +9$

$11=1\cdot 9 +2$

$9=2\cdot4+1$

$1=9-2\cdot4$

what is the next step for the make the inverst I do not understand ?

Body:
Your work (excluding 4.) means that $11x \equiv 1\pmod{64}$ has a solution.

$64=11\cdot 5 +9$

$11=9\cdot 1 +2$

$9=2\cdot4+1$

So

$\;1 = 9 - (2\cdot4) =$

$\quad\quad 9 - (11\cdot4 - 9\cdot4) =$

$\quad\quad64 - 11\cdot5 - (11\cdot4 - 9\cdot4) =$

$\quad\quad 64 - 11\cdot5 - \bigr(\,11\cdot4 - (64\cdot4 - 11\cdot20)\,\bigr) =$

$\quad\quad  64 + 11\cdot(-5) + 11\cdot(-4) + 4\cdot64 + 11\cdot(-20)=$

$\quad\quad  5\cdot64 + 11\cdot(-29)$

So $11 \cdot (-29) \equiv 11\cdot(35) \equiv 1 \pmod{64}$

Finally,

$\;11x\equiv 44\pmod{64} \implies x\equiv 35\cdot44 \equiv 4 \pmod{64}$


问:
Ratio of subsequences of divergent series
Suppose $a_n\in (0,\infty)$ and $\sum_{n\geq 1}a_n = \infty$. Can i find a subsequence of indices $\{n_k\}_{k\geq 1}$ such that
$$ \frac{\sum_{i=1}^{n_{k+1}} a_i}{\sum_{i=1}^{n_k}a_i} \rightarrow 1 \quad \wedge \quad \sum_{k\geq 1} \frac{1}{\sum_{i=1}^{n_k}a_i} < \infty$$

Thank you for the help.

Edit: For example if $a_n\leq 1$ for all $n$, then i can choose $n_k = \inf \{ n: \sum_{i=1}^{n} a_i \geq k^2\}$.

答:
Consider $a_n = 2^n$. Let $(n_k)$ be a subsequence indexing scheme i.e. $n_k < n_{k + 1}$ so that $n_k + 1 \leq n_{k + 1}$. Then because $\sum_{i = 1}^{N}a_i = \sum_{i = 1}^{N}2^i = 2^{N + 1} - 2$
$$
\frac{\sum_{i = 1}^{n_{k + 1}}a_i}{\sum_{i = 1}^{n_k}a_i} = \frac{2^{n_{k + 1} + 1} - 2}{2^{n_k + 1} - 2} \geq \frac{2^{n_k + 2} - 2}{2^{n_k + 1} - 2} = \frac{2 - \frac{2}{2^{n_k + 1}}}{1 - \frac{1}{2^{n_k + 1}}}
$$ So $\frac{\sum_{i = 1}^{n_{k + 1}}a_i}{\sum_{i = 1}^{n_k}a_i}$ is either divergent or converges to a number that is at least $2$. So for this sequence, the first condition is always violated for all possible subsequences.


问:
Proving Chromatic Number without 5-Color Theorem
We are currently studying planar graphs and I have come across a review exercise that is proving to be more trouble than it's worth. The problem is as follows:

Let G = (V;E) be a planar graph with at most 12 vertices. Without using the 4-colour theorem or the 5 colour theorem (Theorems 13.4.1 or 13.4.2 in the Discrete Math book), prove that G has chromatic number at most 5.

I've got a few tools at my disposal, mostly that of Euler's equation of $|V|+|F|=|E|+2$, and it's later derivation of $|E| \leq 3|V|-6$. From these two, I can prove the six color theorem, as we can say by Handshake Lemma there must be at least one vertex of degree 5 or less in order to not break the $|E| \leq 3|V|-6$ inequality. As, with 12 vertices max, that would mean there is a max of 30 edges. However, I keep coming back to the fact that it seems I'm going to have to basically pseudo-prove the 5 color theorem again. Can I utilize the amount of vertices to my advantage, as it bounds vertices, edges, and faces?

Any help would be appreciated.

答:
Hint: Show that
has a vertex with degree at most 4
.

First try contradiction.

If all vertices of G are of at least 5 degree, by handshake lemma and ||≤3||−6, we know that it requires |V|=12, |E|=30, and G to be a 5-regular graph.

Except this case, it's not possible for all vertices of G to have at least 5 degree, so for other cases, G has at least a vertex v with deg(v)  ≤ 4.

For the special case, use Brook's theorem.

For the more general case, prove by induction when n ≤ 12, remove v, which is similar to the proof of 5-colour theorem.


问:
Why it is easier to draw on some grids?

I would like to know is there any way to prove that drawing in a grid on the left would be more precise than in a grid on the right. For me, that make sense since we have more points on the left. But, is there a mathematical way to prove something like this?

答:
If you look at the middle two lines of your left grid, there is a point every $\frac 12$ unit while on the right there is only a point every unit.  Once you extend the grid enough beyond the boundary of an area, there will be about twice as many points inside the area with the grid on the left.  However, the blocks on the left have the same area of one square unit as the blocks on the right.

You have not defined what style of drawing you are doing or how you define precision.  Are you allowed to select line segments from each grid to make the drawing?  Do you fill in blocks to color an area?  Do you connect dots that are defined by the grid with lines?  In the third case, as the left grid has all the points of the right grid and more, you must be able to be more precise.  In the first and second, it probably depends on what picture you are trying to draw.  If it follows the grid lines on the right, the left will introduce jaggies for vertical lines.

Title: Why it is easier to draw on some grids?


问:
Groups generated by the subset of integer powers of group elements.
I am intrigued by the following question about finite groups. Given a group $G$ with elements $\{g_1, g_2, \cdots, g_n\}$, do there exist interesting sufficient conditions such that the group generated by the subset of some fixed integer power of the elements of the group, e.g., $S^m \equiv \{g_1^m, g_2^m, \cdots, g_n^m\}$ is a non-trivial proper subgroup $\langle S^m \rangle = G^\prime < G$?

There are many simple examples of this, e.g., among Abelian groups for any $m \,\lvert\, n$, or the symmetric group on $k$ elements for $m = 2$, etc., but I am looking for a more general principle.

答:
This happens whenever $G$ admits a quotient group $G/N$ of order $m$ (equivalently, a normal subgroup of index $m$), since every $m^{th}$ power is zero in this quotient by Lagrange's theorem and so the subgroup generated by the $m^{th}$ powers is contained in $N$. This explains both the abelian case and the case of the symmetric groups with $m = 2$.

Somewhat more generally, $G/H$ can have exponent $m$. This explains further cases such as the case that $G$ is a finite $p$-group and $m = p$; in this case we can take the quotient given by the Frattini subgroup.


答:
Consider , the ring homomorphism $f:\mathbb{Z}[i] \to \mathbb{Z}_{n}[i] $ defined by $f(a+ib)=(a+ib)(\text{mod n})$ with kernel $n\mathbb{Z}[i]$ .

Kernel $=\{a+ib : f(a+ib)=0 \}$
$=\{a+ib : (a+ib)\equiv 0 (\text{mod n})\}$
$=\{a+ib: a\equiv 0 (\text{mod n}),b\equiv 0 (\text{mod n})\} $

$=n\mathbb{Z}[i]$

Clearly, then , $\mathbb{Z}[i]/n\mathbb{Z}[i] \cong \mathbb{Z}_{n}[i] $

Clearly, $\mathbb{Z}_{n}[i]$ has finite number of elements.


问:
Does a non-fraction number have a reciprocal?
As far as I understand from this Wikipedia page, $\frac 73$ is a reciprocal of $\frac 37$;

But can a term "reciprocal" be applied to a non-fraction number?

For example, it would be correct to say:

reciprocal of $\frac 21$ is $\frac 12$

But would it also be correct to say:

reciprocal of 2 is $\frac 12$

?

答: Yes, every number have a reciprocal, except $0$. It's also called its multiplicative inverse.


问:
if $f$ is continuous at $(a,b)$ and differentiable at $(a,b)-\{c\}$ and $\lim_{x\to c}f'(x)=A$ then $f$ is differentiable at $c$ and $f'(c)=A$.
I'm having some issues with this proof. What I have so far is this: \

Let's divide the interval $(a,b)-\{c\}$ in $(a,c)\cup(c,b)$.

Because we know that the $\lim_{x\to c}f'(x)=A$ that means that, given $\epsilon>0$, we can find a $\delta>0$ such that $|f'(x)-A|<\epsilon$ whenever $|x-c|<\delta$.

We choose $x,y\in(c-\delta,c)$ such that $x<y$ and we consider the interval $[x,y]$. We know $f$ is differentiable in $(x,y)$ and by the MVT we can find a $z\in(x,y)$ such that:$$f'(z)=\frac{f(y)-f(x)}{y-x}$$
and because $x,y\in(c-\delta,c)$ that means $|f'(z)-A|<\epsilon$.

Let's do the same thing for the interval $(c,c+\delta)$ and find a $\bar{z}\in(c,c+\delta)$ such that $|f(\bar{z})-A|<\epsilon$.\

Here is where I'm stuck. And I'm also not sure it that approach is the correct one.

答:
Yes, continuity at $c$ is an essential requirement. Otherwise, you can make $f$ discontinuous at $c$, which would imply $f$ is not differentiable at $c$.
$$
f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}
$$
Since $f$ is continuous at $c$, $\lim_{x\to c}(f(x)-f(c))=\lim_{x\to c}(x-c)=0$, and you can apply L'Hospital's rule:
$$
\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c}\frac{f'(x)}{1}=A
$$


问:
Shortest path in weighted graph
Given a weighted graph $G(V,E)$ which edges could have negative values, and a source node $s$. There is no negative cycles in the graph and for each node $v$ there is a minimum weighted path from $s$ to it that uses at most $k$ edges.

Is there an algorithm that finds all minimum weighted paths in $k\cdot|E|$?

答: Construct a layered graph with $k+1$ layers $0,\dots,k$, where each edge $(i,j)\in E$ corresponds to an edge from $(i,\ell)$ to $(j,\ell+1)$.  Now use breadth-first or depth-first search from $s$.

Title: Shortest path in weighted graph


问:
Proof for Moore-Penrose inverse of transposed matrix

I would like to ask how to prove that Moore-Penrose inverse of $A^T$ is $(A^+)^T$. I know that I can do it by proving all 4 properties but I am stuck at proving that
$(A^+A)^T=A^+A$ and $(AA^+)^T=AA^+$
</p>

答:
I am stuck at proving that $(A^+A)^T=A^+A$ and $(AA^+)^T=AA^+$.

You cannot prove that because it isn't true in general (unless $A$ is real).

What you should prove is that both $A^T(A^+)^T$ and $(A^+)^TA^T$ are Hermitian, but this is obviously true because they are the transposes of $A^+A$ and $AA^+$, and the latter two products are Hermitian by the definition of $A^+$.


问:
Inequality of Sums (Cauchy-Schwarz?)
I have to show

$(2 \cdot n) \cdot 2 \cdot \sum_{i=1}^{n}t_i^2-\left ( 2 \cdot \sum_{i=1}^{n} t_i \right )^2>0$

Which is probably not advanced but I am extremly new to working with sums in linear algebra. This is regarding a problem showing that a point is a minimum in a function using ABC-criteria. I need to show the above property for the last step. I am not experienced AT ALL with this and would really love some help.

答:
This is just Cauchy Schwarz inequality, $ \sum a_i^2 \sum b_i^2 \geq (\sum a_i b_i)^2$

Set $ a_i = 2, b_i = t_i$, and we get that

$$ 4n \sum t_i^2 \geq (\sum 2 t_i)^2. $$

Title: Inequality of Sums (Cauchy-Schwarz?)