问:
If |a - b| < |a - c|, then |1 - $\frac{a}{b}$| < |1 - $\frac{a}{c}$|?
If |a - b| < |a - c|, then |1 - $\frac{a}{b}$| < |1 - $\frac{a}{c}$|

Is that true for all real values? If so, can someone point me to a proof?

(EDIT: I made a mistake in my context. When comparing "closeness", I should have put 'a' in the denominator. Then I'm pretty sure |a - b| < |a - c| implies |1 - $\frac{b}{a}$| < |1 - $\frac{c}{a}$|. However, I'm going to let the question remain, as it might be interesting to see what happens when a is put to the numerator.)

I used to think so but I may have come up with a counter-example. It's very confusing though so I want to first provide some background. Also, I don't know all the official terminology for these things, which is why the background might help.

The idea is how to measure the "closeness" of an approximation to some true value. There are two ways of doing it, absolute difference and relative ratio.

So if our true value is 2.5, and we want to test approximations of 2.25 and 2.75, we could say that 2.5 - 2.25 is 0.25 and 2.75 - 2.5 is also 0.25, so they are equally close.

But we could also try looking at the ratios. $\frac{2.5}{2.25}$ = 1.111..., and $\frac{2.5}{2.75}$ = 0.9090909... So this time, the second number is closer, when viewing it from the perspective of proportions.

Another concrete example. Our true value is 3, and our test values are 2.9 and 3.05. This time 3.05 is closer in both ways. In other words, the rule holds true:

|3 - 3.05| < |3 - 2.9|, then |1 - $\frac{3}{3.05}$| < |1 - $\frac{3}{2.9}$|

It seems impossible to ever violate that rule. If there is a smaller absolute distance between two numbers, then there should be a smaller percentage difference as well.

Now for my supposed counter-example.

a = $\sqrt{2}$, b = $\frac{7}{5}$, c = $\frac{10}{7}$

|a - b| = 0.014 213 562...

|a - c| = 0.014 357 866..., which implies b is closer.

|1 - $\frac{a}{b}$| = 0.010 152 545...

|1 - $\frac{a}{c}$| = 0.010 050 506..., which implies c is closer.

How is it possible? Does it have something to do with the coincidence that $\frac{b}{a}$ = $\frac{a}{c}$ due to square root "normalization"? By that I mean:

$\displaystyle \frac{\frac{7}{5}}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{7*\sqrt{2}}{5*2} = \frac{\sqrt{2}}{\frac{10}{7}}$

If so I can't fathom how. I feel like I might be missing something obvious.

Btw if someone can clarify what the official terminology is, that would be a big help. I might be able to search and find answers on my own if I just knew the proper names of these things.

Body:
Let $a=1$, $b=10^{-3}$, $c=2$

$$|1-10^{-3}|<|1-2|$$

but not

$$ |1-10^3|<|1-1/2|.$$


问:
Weak convergence of unit mass
I have found in my notes the statement bellow:

$\mu_n\Rightarrow\mu$ as $n\rightarrow \infty$, if and only if $x_n\rightarrow x$ as $n\rightarrow \infty$ where $\mu$ is the unit mass at $x$ and $\mu_n$ is a unit mass at $x_n$ for $n=1,2,...$.

I do not understand this statement clearly as why this holds.

Any help on this would be appreciated!

答:
I am going to assume $\mu_n$ and $\mu$ are probability measures on $\mathbb{R}$

The definition of $\mu_n \Rightarrow \mu$ is that $\int f \,d\mu_n \to \int f \,d\mu$ for all bounded continuous functions $f$.  If $\mu_n$ is a point mass at $x_n$ and $\mu$ is a point mass at $x$, then the above condition says that $f(x_n) \to f(x)$ for all bounded continuous functions $x$.  So you need to prove that $f(x_n) \to f(x)$ for all bounded continuous $f$ if and only if $x_n \to x$.

In one direction, if $x_n \to x$, then it simply follows from (one of) the definition(s) of continuity that $f(x_n) \to f(x)$ whenever $f$ is continuous.  The other direction is a bit trickier.  Suppose that $x_n$ does not converge to $x$.  Then there is a positive distance $\epsilon$ such that $|x_n - x| > \epsilon$ for infinitely many $n$.  Now let $f$ be a continuous function such that $f(x) = 1$ but $f(y) = 0$ whenever $|y-x| > \epsilon$ (I will leave it to you to figure out what such a function looks like).  Then $f(x) = 1$ but $f(x_n) = 0$ for infinitely many $n$, so $f(x_n)$ does not converge to $f(x)$.

ContentLicense: CC BY-SA 4.0

Title: Weak convergence of unit mass


问:
show that a function has a maximum
Here is the exercise :
Let f: R+ → R continues with f(0)=0=limx→+∞ f(x). Show that f admits a maximum on R+ = [0, +∞[.

My proof is the following :
As f(0)=0=limx→+∞ f(x): $\forall \epsilon > 0 \exists \alpha \forall x \geq \alpha : \mid f(x) \mid < \epsilon $. Then it is true for $\epsilon = M \in \mathbb{R} $ , so f is bounded on [$\alpha ; \infty$[ .
Then as f is continuous on [0;$\alpha$] it is bounded on this interval, lets say by N $\in \mathbb{R}$.
Then let X = max(M;N); then f is bounded by X on R+.
Hence f admits a supremum and because f is continuous on R+, the supremum is a maximum.

Is this reasoning correct?

答:
No, it is not correct. You have proved that $f$ is bounded above, but it doesn't follow from that that $f$ has a maximum. For instance, $x\mapsto-\frac1{x+1}$ is also bounded above on $\Bbb R^+$, but it has no maximum.

If you always have $f(x)\leqslant 0$, then $0$ is a maximum. Otherwise, take $x_0\in\Bbb R^+$ such that $f(x_0)>0$. There is some $\alpha>0$ such that $x\geqslant\alpha\implies f(x)<f(x_0)$. So, consider the maximum $M$ of the restriction of $f$ to $[0,\alpha]$ and then $M$ will be the maximum of $f$.

Title: show that a function has a maximum


答:
If $\left|c\right| \leq \left|b\right|$ we have

$$\frac1{\left|b\right|} \leq \frac1{\left|c\right|}$$

which can be combined with $\left|a - b\right| < \left|a - c\right|$ to derive $\left|1-\frac{a}{b}\right| < \left|1 - \frac{a}{c}\right|$

Title: If |a - b| < |a - c|, then |1 - $\frac{a}{b}$| < |1 - $\frac{a}{c}$|?

Body:
If |a - b| < |a - c|, then |1 - $\frac{a}{b}$| < |1 - $\frac{a}{c}$|

Is that true for all real values? If so, can someone point me to a proof?

(EDIT: I made a mistake in my context. When comparing "closeness", I should have put 'a' in the denominator. Then I'm pretty sure |a - b| < |a - c| implies |1 - $\frac{b}{a}$| < |1 - $\frac{c}{a}$|. However, I'm going to let the question remain, as it might be interesting to see what happens when a is put to the numerator.)

I used to think so but I may have come up with a counter-example. It's very confusing though so I want to first provide some background. Also, I don't know all the official terminology for these things, which is why the background might help.

The idea is how to measure the "closeness" of an approximation to some true value. There are two ways of doing it, absolute difference and relative ratio.

So if our true value is 2.5, and we want to test approximations of 2.25 and 2.75, we could say that 2.5 - 2.25 is 0.25 and 2.75 - 2.5 is also 0.25, so they are equally close.

But we could also try looking at the ratios. $\frac{2.5}{2.25}$ = 1.111..., and $\frac{2.5}{2.75}$ = 0.9090909... So this time, the second number is closer, when viewing it from the perspective of proportions.

Another concrete example. Our true value is 3, and our test values are 2.9 and 3.05. This time 3.05 is closer in both ways. In other words, the rule holds true:

|3 - 3.05| < |3 - 2.9|, then |1 - $\frac{3}{3.05}$| < |1 - $\frac{3}{2.9}$|

It seems impossible to ever violate that rule. If there is a smaller absolute distance between two numbers, then there should be a smaller percentage difference as well.

Now for my supposed counter-example.

a = $\sqrt{2}$, b = $\frac{7}{5}$, c = $\frac{10}{7}$

|a - b| = 0.014 213 562...

|a - c| = 0.014 357 866..., which implies b is closer.

|1 - $\frac{a}{b}$| = 0.010 152 545...

|1 - $\frac{a}{c}$| = 0.010 050 506..., which implies c is closer.

How is it possible? Does it have something to do with the coincidence that $\frac{b}{a}$ = $\frac{a}{c}$ due to square root "normalization"? By that I mean:

$\displaystyle \frac{\frac{7}{5}}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{7*\sqrt{2}}{5*2} = \frac{\sqrt{2}}{\frac{10}{7}}$

If so I can't fathom how. I feel like I might be missing something obvious.

Btw if someone can clarify what the official terminology is, that would be a big help. I might be able to search and find answers on my own if I just knew the proper names of these things.


答:
Hint: You are given that $ | a - b | < |a - c|$, and are asking if

$$ \frac { | a - b | } { |b| } < \frac{ |a-c|} { |c| }. $$

An obvious place to start would be if $ |b| < |c|$, then this statement might not follow.

This suggests we should look for a counterexample where $|b| < |c|$, so take a really small $b$ and a really large $c$.


问:
Uniform convergence and limit of an integral
Let $(E,\mathcal{A},\mu)$ be a measure space such that $\mu(E)<+\infty$.

Is it possible to find an example of a sequence $f_n$ converging uniformly to $f$ (with no assumption if they are non-negative, integrable) such that $\lim_n\int_Ef_nd\mu \neq \int_Efd\mu$? (maybe for $n, f_n,$ is non-negative and the integral can be $+\infty,$ for other $n,f_n$ takes values in $\mathbb{R}$ and is not integrable so we can't define the integral)

Any suggestions are appreciated.

答:
No.  If $f_n$ converges uniformly to $f$ on $E$, then
$$
\left| \int_E f_n \,d\mu - \int_E f \,d\mu \right| \leq \int_E |f_n - f| \,d\mu \leq \mu(E) \cdot \sup_{x \in E} |f_n(x) - f(x)|
$$
By definition of uniform convergence, the above sup tends to $0$ as $n \to \infty$.


问:
Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates.
Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates

I thought of using the second part of Sylow's Theorem that says; In particular, any two Sylow $p$-subgroups of $G$ are conjugate in $G$.

But I'm not sure if I have to prove that $H$ and $K$ are $p$-subgroups

答: Hint: try to prove that $N_G(H^g)=N_G(H)^g$ for any subgroup $H$ of $G$ and element $g \in G$.


问:
What are the following 'familiar' spaces these quotient spaces are homeomorphic to - homework assignment problem?
I'm trying to solve two topology assignment problems as in the images and I'm a bit lost how to solve them:

For Problem 1: I did a bit of search on the internet, but found that it's isomorphic to $\mathbb{R}$ as a group, but I found that it's not homeomorphic to $\mathbb{R}?$  So I'm having difficulty identifying $\mathbb{R}/\mathbb{Q}$ to any familiar space, as the assignment asks?

For problem 2b): I think it's topologically the half-open interval $[0, \infty),$ as the equivalence relation identifies all the points lying in any circle with center at the origin, i.e. $\forall r \ge 0,$ all $(x,y)$ so that $x^2+y^2=r^2$ are identified. So the quotient space should consist of all the classes $[(x,y)]$ with different values of $x^2 + y^2.$ These different values are just $[0, \infty),$ hence my answer. Is it correct?

For problem 2a): I feel it'd be just $\mathbb{R},$ by just using the above argument and noting that $y-x^3$ can take any real values. Is this correct?

答:
For the first problem consider any open interval $(a,b)$ in $\Bbb R$: it intersects every $\sim$-equivalence class (why?), so every non-empty open set in $\Bbb R$ intersects every $\sim$-equivalence class. Now let $q:\Bbb R\to\Bbb R/\!\sim$ be the quotient map, and suppose that $U$ is a non-empty open set in the quotient space. By definition $q^{-1}[U]$ is open in $\Bbb R$, and clearly it must be non-empty. It also must contain every $\sim$-class that it intersects, so what set must it be?

Your answers to the second problem are correct. The equivalence classes in (a) are the graphs of $y=x^3+a$ for $a\in\Bbb R$, and a set of these graphs is open in the quotient iff the corresponding set of constants $a$ is open in $\Bbb R$. And as you saw, the situation in (b) is similar.


问:
For $f(x)=x^n$, show $f'(x)=nx^{n-1}$ for rational number n
We have $x \in \Bbb R$ and $n\in\Bbb  Q$. I understand how to prove the statement when $n \in\Bbb R$, but does the condition $n \in\Bbb  Q$ make a difference?

答:
We wish to prove:

$$\frac{d}{dx}x^{n} = nx^{n-1}$$

For all $n\in\mathbb{Q}$. We have:

$$\frac{d}{dx}e^{\ln(x^{n})} = nx^{n-1}$$

$$\frac{d}{dx}e^{n\ln(x)} = nx^{n-1}$$

Chain rule:

$$\frac{d}{dx}(n\ln x)e^{n\ln(x)} = nx^{n-1}$$

$$\frac{n}{x}(e^{\ln(x)})^{n} = nx^{n-1}$$

$$\frac{n}{x}x^{n} = nx^{n-1}$$

$$nx^{n-1} = nx^{n-1}$$

This is true, and thus the original equation is true as well. $\blacksquare$

This proof requires knowing the derivatives of $e^{x}$ and $\ln(x)$. I believe the reason for $n\in \mathbb{Q}$ is that exponentiation has not been extended to irrational exponents yet.


问:
Dirac Delta Function "removing integral"
I am trying to proof the following  identity:
\begin{align}
 \delta(ax) = \frac{1}{|a|}\delta(x).
\end{align}
where $\delta$ represents the dirac delta function.
I found that
\begin{align}
  \int_{-\infty}^\infty  f(x) \delta(ax) \,\,dx = \frac{1}{|a|} \int_{-\infty}^\infty f(x)\delta(x)\,\,dx.
\end{align}
Consequently we get
\begin{align}
 \delta(ax) = \frac{1}{|a|}\delta(x).
\end{align}
by "removing" the integrals. Is this step actually valid?

答:
As pointed out by @GEdgar in the comments, equality here is in the sense of distribution, and integration of a Dirac delta against a function is not really a Riemann or Lebesgue integral. It's an abuse of notation. But what does all of that mean?

The space of distributions, often denoted $\mathcal{D}^{'}(\mathbb{R})$, is the space of all maps $f: \mathcal{C}_c ^\infty (\mathbb{R}) \to \mathbb{R}$ that are linear and continuous (whatever that means for now) where $\mathcal{C}_c ^\infty (\mathbb{R})$ is the space of smooth functions with compact support. For example, the dirac delta takes any function $\phi \in \mathcal{C}_c ^\infty (\mathbb{R})$ and maps it to its value at zero: $\phi(0) \in \mathbb{R}$. The action of dirac delta on $\phi$ is denoted: $\langle \delta, \phi \rangle$. It can be shown that such a map is linear and continuous, and hence, the dirac delta is a distribution. If a map $f$ is smooth enough, for example, a continuous function, then you can write: $\langle f, \phi \rangle = \int_{-\infty} ^{\infty} f \phi \; dx$ for any $\phi \in \mathcal{C^\infty _c(\mathbb{R})}$. However, this is not necessarily the case. With that in mind, let's move on to clarify your question.

Two distributions $f,g$ are said to be equal if $\forall \phi \in \mathcal{C^\infty _c(\mathbb{R})}$, we have:
$$ \langle f, \phi \rangle = \langle g, \phi \rangle
$$
In other words, both distributions induce the same effect on $\phi$. So, the equality: $\delta(ax) = \frac{1}{|a|} \delta(x)$ actually reads: the action of the distribution $\delta(ax)$ on an arbitrary fixed $\phi$ is equal to the action of the distribution $\frac{1}{|a|} \delta(x)$ on $\phi$. Formally; $\langle \delta(ax), \phi(x) \rangle = \langle \frac{1}{|a|}\delta(x), \phi(x) \rangle $. In order to formally prove this result we need to know what is the composition of a distribution.

The left hand-side of your equality is $\delta(ax)=\delta \circ g$ where $g(x)=ax$. Now that we know what a distribution is, we can move on and ask ourselves what the composition of the Dirac delta distribution with a function $g$ is. Is the composition of a distribution with a function $g$ a distribution? It turns out that the answer is yes if $g$ and its inverse are smooth. Which is true for our case (assuming $a \neq 0)$. This composition is defined as:
$$\begin{aligned}
\langle \delta \circ g, \phi \rangle &= \langle \delta, \phi \circ g^{-1} |\text{det } dg^{-1}| \rangle \\
& = \langle \delta, \phi (\frac{y}{a}) \frac{1}{|a|} \rangle \\
& = \frac{1}{|a|} \langle \delta, \phi (\frac{y}{a})  \rangle \\
& = \frac{1}{|a|} \phi(0) \\
& = \langle \frac{1}{|a|}\delta, \phi \rangle \\
\end{aligned}
$$

If we allow for an abuse of notation, the definition of a composition of a distribution with a smooth function is equivalent to changing the variabe of integration!
$$
\int_{-\infty}^{\infty} \delta(g(x)) \phi (x) \; dx = \int_{-\infty}^{\infty} \delta(y) \phi (\frac{y}{a}) \frac{1}{|a|}  \; dy
$$


问:
Find irreducible factor of the polynomial
Find all irreducible factors of the polynomial $x^{15}-1$ over $\Bbb{Q}$.

Clearly, this is a cyclotomic polynomial.

I can find all the irreducible factors but that is very calculative:

\begin{align}
\Phi_1(x)&=x-1\\
\Phi_3(x)&=\frac{x^3-1}{\Phi_1(x)}\\
\Phi_5(x)&=\frac{x^5-1}{\Phi_1(x)}\\
\Phi_{15}(x)&=\frac{x^{15}-1}{\Phi_1(x)\Phi_3(x)\Phi_5(x)}.
\end{align}
The last equation is very calculative. That's why I am looking for any less time consuming method.

Thanks!

Body:
You have $$\Phi_1(x) \Phi_5(x)= x^5-1$$

Therefore

$$\Phi_{15}(x)= \frac{x^{15}-1}{(x^5-1) \Phi_3(x)}= \frac{(x^5)^3-1}{(x^5-1)\Phi_3(x)}=\frac{x^{10}+x^5+1}{\Phi_3(x) }$$

One division left...


问:
Linear combinations of four-dimensional vectors
Let $v_1  = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$ $v_2  = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$ $v_3  =  \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}$ $v_4  =  \begin{pmatrix} 2 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ $\mathbb{R}^4$ vectors.

Show that every $v \in \mathbb{R}^{4\times1}$  can be written as vectors $(v_1,v_2,v_3,v_4)$ linear combination.

My attempt:

$\left[\begin{array}{cccc|l}
1 & 0 & 0 & 2  &v_1\\
1 & 1 & 0 & 0  & v_2\\
0 & 1 & 1 & 0  &v_3\\
0 & 0 & 1 & 1  &v_4\\
\end{array}
\right]$

Where do I go from here? Every input is appreciated.

答:
As $\det A=-1\ne 0$, given any vector $B=(b_1,b_2,b_3,b_4)$, we solve the linear system $AX=B$ with
$$A=\left(
\begin{array}{cccc}
 1 & 0 & 0 & 2 \\
 1 & 1 & 0 & 0 \\
 0 & 1 & 1 & 0 \\
 0 & 0 & 1 & 1 \\
\end{array}
\right);\;X=\left(
x_1,x_2,x_3,x_4
\right)$$
and thank to Cramer's theorem we get as unique solution  the components $X$ of $B$ in the base $\left(v_1,v_2,v_3,v_4\right)$.


问:
Tips finding the gcd of $(n^2-3n-1, n-4)$

I need tips on how to find the $gcd(n^2-3n-1, n-4)$ for any $n \in \mathbb{N}$.

I tried the two following methods (but got stuck):

Since I need to find the $gcd(n^2-3n-1, n-4)$, then there is a $q \in \mathbb{Z}$ which is the divisor of both $(n^3-3n-1)$ and $(n-4)$, giving each of them an answer of $q_1,q_2 ∈ \mathbb{Z}$ accordingly,

therefore, it is possible to write them as follow:

$n^2-3n-1 = q*q_1$

$n-4 = q*q_2$

Because $q,q_1,q_2 ∈ \mathbb{Z}$, multiplying them with each other produces another Integer (same with adding to them an Integer), so

$n^2-3n = q_3$ , $q_3 ∈ \mathbb{Z}$

$n = q_4$ , $q_4 ∈ \mathbb{Z}$

With this method, I got stuck here (didn't find a way how to continue).

Another method I've tried is:

According to the Euclidean Algorithm,

$n^2-3n-1 = (n-4)*q + r, q,r ∈ \mathbb{Z}$

Use $q=n$,

$n^2-3n-1 = (n-4)*n+r$ => $n-1=r$

$n^2-4n=(n-1)q+r$ (while $q=n$) => $-3n=r$

$n^2-n=-3n*n+r$ => $4n^2-n=4$

and so on...but also got stuck..

I'd appreciate any tips on how to continue with what I've tried or if I should try a different approach

答:
hint

Use this

$$b=aq+\color{red}{r}\implies \;gcd(a,b)=\;gcd(a,\color{red}{r})$$

with

$$b=n^2-3n-1=n(n-4)+n-1$$
$$=(n+1)(n-4)+\color{red}{3}$$

So,
$$gcd(n^2-3n-1,n-4)=gcd(n-4,3)$$
$$=gcd(n+2,3)=G$$
Thus

$$n\equiv 0\mod 3 \implies G=1$$
$$n\equiv 1 \mod 3\implies G=3$$
$$n\equiv 2\mod 3 \implies G=1$$


Title: Tips finding the gcd of $(n^2-3n-1, n-4)$

Body:
I need tips on how to find the $gcd(n^2-3n-1, n-4)$ for any $n \in \mathbb{N}$.

I tried the two following methods (but got stuck):

Since I need to find the $gcd(n^2-3n-1, n-4)$, then there is a $q \in \mathbb{Z}$ which is the divisor of both $(n^3-3n-1)$ and $(n-4)$, giving each of them an answer of $q_1,q_2 ∈ \mathbb{Z}$ accordingly,

therefore, it is possible to write them as follow:

$n^2-3n-1 = q*q_1$

$n-4 = q*q_2$

Because $q,q_1,q_2 ∈ \mathbb{Z}$, multiplying them with each other produces another Integer (same with adding to them an Integer), so

$n^2-3n = q_3$ , $q_3 ∈ \mathbb{Z}$

$n = q_4$ , $q_4 ∈ \mathbb{Z}$

With this method, I got stuck here (didn't find a way how to continue).

Another method I've tried is:

According to the Euclidean Algorithm,

$n^2-3n-1 = (n-4)*q + r, q,r ∈ \mathbb{Z}$

Use $q=n$,

$n^2-3n-1 = (n-4)*n+r$ => $n-1=r$

$n^2-4n=(n-1)q+r$ (while $q=n$) => $-3n=r$

$n^2-n=-3n*n+r$ => $4n^2-n=4$

and so on...but also got stuck..

I'd appreciate any tips on how to continue with what I've tried or if I should try a different approach

Body:
Suppose $d=(n^2-3n-1,n-4)$ so $$d|n^2-3n-1\\d|n-4$$
from $d|n-4 \to d|(n-4)(n+1)$ so
$$d|n^2-3n-4$$
and now $$d|A,d|B \to d|rA+sB \\d|(n^2-3n-1)-(n^2-3n-4) \to d|3\\
d=1,or,3$$


问:
Parallel transport equation of a covariant component
I am working through Paul Renteln's book "Manifolds, Tensors, and Forms" (As I am learning General Relativity). I have come across a derivation of a 'parallel transport equation':
$$\frac{d\gamma^i}{dt}\left(\frac{\partial Y^k}{\partial x^i}+\Gamma^k_{ij}Y^j\right)=0,$$

Definition of parallel transport:
(I have only included this so you know what the variables used are referring to)

Let $I\subseteq \mathbb{R}$ be an interval, and let $\gamma:I\rightarrow M$ be a parameterized curve in M with tangent vector field $X=\gamma_*(d/dt)$. Let $Y$ be any other vector field on $M$. We say that $Y$ is parallel transported along $\gamma$ if $\nabla_{X}Y=0.$

The nearly-full derivation is as follows:

Suppose that $X=X^i \partial_i$ and $Y=Y^j \partial_j$. Then:
$$\nabla_X Y=\nabla_X(Y^j\partial_j),$$
$$=X(Y^j)\partial_j+Y^j\nabla_X\partial_j,$$
$$=X(Y^j)\partial_j+Y^j\nabla_X\partial_j,$$
$$X(Y^j)\partial_j+Y^jX^i\nabla_{\partial_i}\partial_j,$$
$$=\left[X(Y^k)+\Gamma^k_{ij}X^i Y^j\right]\partial_k.$$
Hence, our condition for parallel transport $\nabla_X Y=0$, holds provided that:
$$X(Y^k)+\Gamma^k_{ij}X^iY^i=0.$$
Using the notion of integral curves we have:
$$X^i=X(x^i)=\gamma_*\left(\frac{d}{dt}\right)(x^i)=\frac{d}{dt}(x^i \circ \gamma)=\frac{d\gamma^i (t)}{dt},$$
So we write our previous expression as:
$$\frac{d}{dt}Y^k+\Gamma^k_{ij}\frac{d\gamma^i}{dt}Y^j=0,$$
or:
$$\frac{d\gamma^i}{dt}\left(\frac{\partial Y^k}{\partial x^i}+\Gamma^k_{ij}Y^j\right)=0,$$

 My Problem

My problem is regarding the first step (perhaps not directly related to parallel transport), when the following choices were made:
$$X=X^i \partial_i,$$
$$Y=Y^j \partial_j,$$
I am completely oblivious as to what this means/why they have been chosen? Would these be different if we were considering perhaps covariant components of a vector field $Y$? I am relatively new to this level of maths (and am a physicist) so am struggling to visualise what this means, I have also been given a question for homework regarding the parallel transport of the covariant component of a vector, so am wondering if I need to perhaps perform a different substitution for $Y$ and re-derive?

Any help is appreciated, Thanks! :)

答:
As per @Mark Wildon 's comment, the problem in question was not a choice, but an expression of the vector field $X$ as a linear combination of the standard basis of vector fields, thus for covariant components it would naturally look as such:
$$Y=Y_j\partial^j,$$


问:
Is the set of upper bounds squared either empty or a singleton?
Let $(P, \leq)$ be a partially ordered set, and let $A$ be a subset of $P$. The set $UB(A)$ is the set of upper bounds of $A$, where upper bound means greater than or equal to every element of $A$. Now, what happens if we form $UB(UB(A))$? I conjecture that it is either empty (if $P$ has no maximum), or it is a singleton (if $P$ does have a maximum). Is this correct?

答:
Your conjecture is not quite right, for two reasons.

First, you've forgotten the case when $UB(A) = \varnothing$. Then $UB(UB(A)) = P$, since every element of $P$ is an upper bound for the empty set.

Second, it's possible to get $UB(UB(A))$ to be a singleton even when $P$ has no greatest element - all you need is that $UB(A)$ has a greatest element, not all of $P$.

For an example demonstrating both behaviors, let $P = \{a,b\}$ with $a$ and $b$ incomparable. Then $UB(UB(\{a,b\})) = UB(\varnothing) = \{a,b\}$. And on the other hand, $UB(UB(\{a\}) = UB(\{a\}) = \{a\}$, even though $P$ has no greatest element.

To answer the question correctly, let's consider $UB(X)$ where $X\subseteq P$ is any upwards-closed set.

If $X$ is empty, then $UB(X) = P$.

If $X$ is non-empty, then $u$ is an upper bound for $X$ if and only if $u$ is a greatest element of $X$. Indeed, a greatest element of $X$ is just an upper bound for $X$ that happens to be an element of $X$. If $u$ is any upper bound for $X$, then there is some $x\in X$ with $x\leq u$ (since $X$ is nonempty), and hence $u\in X$ (since $X$ is upwards-closed), so $u$ is a greatest  element of $X$. Now greatest elements are unique if they exist, so: $$UB(X)  = \begin{cases} \varnothing & \text{ if $X$ has no greatest element}\\ \{g\} &  \text{ if $X$ has a greatest element $g$}.\end{cases}$$

Now for an arbitrary set $A\subseteq P$, just note that $UB(A)$ is upwards-closed and apply the above to get: $$UB(UB(A)) = \begin{cases} P & \text{ if $A$ has no  upper bound}\\ \varnothing & \text{ if $A$ has an upper bound but no greatest upper bound}\\ \{g\} &  \text{ if $A$ has a greatest upper bound $g$}.\end{cases}$$


答:
$\begin{vmatrix}1&0&0&2\\1&1&0&0\\0&1&1&0\\0&0&1&1\end{vmatrix}=\begin{vmatrix}1&0&0\\1&1&0\\0&1&1\end{vmatrix}-\begin{vmatrix}0&0&2\\1&1&0\\0&1&1\end{vmatrix}=1+\begin{vmatrix}0&2\\1&1\end{vmatrix}=-1\ne0$,

so the four vectors are linearly independent, so they span $\mathbb R^4$,

so any vector in $\mathbb R^4$ can be expressed as a linear combination of them.

Title: Linear combinations of four-dimensional vectors

Body:
Let $v_1  = \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$ $v_2  = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}$ $v_3  =  \begin{pmatrix} 0 \\ 0 \\ 1 \\ 1 \end{pmatrix}$ $v_4  =  \begin{pmatrix} 2 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ $\mathbb{R}^4$ vectors.

Show that every $v \in \mathbb{R}^{4\times1}$  can be written as vectors $(v_1,v_2,v_3,v_4)$ linear combination.

My attempt:

$\left[\begin{array}{cccc|l}
1 & 0 & 0 & 2  &v_1\\
1 & 1 & 0 & 0  & v_2\\
0 & 1 & 1 & 0  &v_3\\
0 & 0 & 1 & 1  &v_4\\
\end{array}
\right]$

Where do I go from here? Every input is appreciated.


问:
Erdős-Straus conjecture
The Erdős-Straus Conjecture (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:

$$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$

The basic approach to solving this problem outlined by Mordell [Ref1] is described below

By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that

a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution
$$\frac{4}{4t}=\frac{1}{t}$$

In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g.
$$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$

b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus

$$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$

will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.

Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$

The argument continues...

Mordell goes on to show in turn that the search can be reduced further to the cases when
$$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$
$$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$
$$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$
$$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$
$$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$

Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when
$$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$

Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n<1009$.

This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.

Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.

The question is:

Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?

[Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290.

答:
"Are there any other elementary approaches to solving this problem..." - this question is not asked properly, as this problem is not solved, only partially, as to my best knowledge (I mean I see this question is old, but it seems to also be true now). So if the question is about how to find residue classes other than Mordell's, but still not covering all primes thus not proving the conjecture, then I can show that the following $n$'s are all possible to be expanded into 3 terms $\forall p, x \in \mathbb{Z}^+$:

$n = (4p - 1)x - 1$

$n = (4p - 1)x - p$

The smallest prime number that is not covered by Mordell's residue classes is $4201$ that can be produced with the former, and is $1009$ with the latter.

You can see the identity here, in Part IV. - II. (little after the middle).

Update:

I attributed the above identities to myself, but Mordell would possibly kick me in the butt: he got all the residue classes from the condition $na + b + c = 4abcd$.

If I let $a = b = 1$, then $n = (4d - 1)c - 1$, and

if I let $a = d = 1$, then $n = (4c - 1)b - c$, so the above can be produced from his condition, they are just not part of the usual residue class analyses, I guess because they are hard to handle.

ContentLicense: CC BY-SA 3.0

Title: Erdős-Straus conjecture


答:
Since $H$ and $K$ are conjugate in $G$, there exists a $g\in G$ such that $H=gKg^{-1}$.

Consider $$N_G(K)=\{x\in G\mid xK=Kx\}.$$

Observe that

$$\begin{align}
gN_G(K)g^{-1}&=g\{x\in G\mid xK=Kx\} g^{-1}\\
&=\{gxg^{-1}\in G\mid xK=Kx\}\\
&=\{y=gxg^{-1}\in G\mid xKx^{-1}=K\}\\
&=\{y\in G\mid g^{-1}ygKg^{-1}y^{-1}g=K\}\\
&=\{y\in G\mid g^{-1}yHy^{-1}g=K\}\\
&=\{y\in G\mid yHy^{-1}=gKg^{-1}=H\}\\
&=\{y\in G\mid yH=Hy\}\\
&=N_G(H).
\end{align}$$

Hence they are conjugate in $G$.

Title: Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates.

Body:
Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates

I thought of using the second part of Sylow's Theorem that says; In particular, any two Sylow $p$-subgroups of $G$ are conjugate in $G$.

But I'm not sure if I have to prove that $H$ and $K$ are $p$-subgroups


问:
For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$
For $a,b,c>0$, such that: $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}=3$ prove that:$(a+b)^3+(b+c)^3+(a+c)^3\le24$

I couldn't solve this question and hence looked at the solution. The solution substitutes $x=\sqrt{a}$, $y=\sqrt{b}$, $z=\sqrt{c}$, hence $x^3+y^3+z^3=3$. We also have that $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3\le24$

which is also written as $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3\le \frac{8}{3}(x^3+y^3+z^3)^2$ which holds true, after expanding etc.

My question is, is there a simpler and more intuitive approach? And if not how was I supposed to think of the substitution used above?

答:
Here I will show the thoughts that could help you to do all the steps above.

The first substitution is unnecessary, but it simplifies the algebra, since you have $(\sqrt{a})^3+(\sqrt{b})^3+(\sqrt{c})^3=3$

Now, for the inequality we want to prove we have $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3$ on the left side. The general trick is to make inequality Homogeneous. It means that if you imagine $x,y,z$ to be lengths in meters, you will get $meters^6$ on the left side, which means that we want to get $meters^6$ on the right side as well. Since $(x^3+y^3+z^3)$ is $meters^3$, we want to square it to get $meters^6$.

The last step is just to find the coefficient before $(x^3+y^3+z^3)^2$. Here it helps to check the equality point $(x^3+y^3+z^3)^2=9$, but we want to have 24. Therefore we put $\frac{8}{3}$ as a coefficient and get $(x^2+y^2)^3+(y^2+z^2)^3+(z^2+x^2)^3 \leq \frac{8}{3}(x^3+y^3+z^3)^2$.


Body:
Bear in mind that $\gcd(a,b) = \gcd(a\pm kb, b)$ for all integers $k$.

So if $n^3 - 3n -1 = P(n)(n-4) + R(n)$ then $\gcd(n^2 -3n-1, n-4) = \gcd(n^2-3n-1 -P(n)(n-4), n-r) = \gcd(R(n), n-4)$.

And if we use synthetic division

$n^2   -3n -1 = n^2(n-4) +4n -3n -1=$

$n(n-4) +n-1=$

$n(n-4) +(n-4) +4-1=$

$n(n-4) + (n-4) + 3=$

$(n +1)(n-4) + 3$.

....

So $\gcd(n^2-3n-1,n-4)= \gcd((n+1)(n-4) + 3, n-4)=\gcd(3,n-4)=\gcd(3,n-4 +3)=\gcd(3,n-1)$.

So if $3|n-1$ or in other words if $n\equiv 1 \pmod 3$ then $\gcd(n^2-3n-1,n-4)=\gcd(3,n-1)=3$.

ANd if $3\not \mid n-1$ or in other words if $n\not \equiv 1$ or if $n\equiv 0, 2 \pmod 3$ and $n-1\equiv 2,1$ and $n-1$ and $3$ are relatively prime then $\gcd(n^2-3n-1,n-4)=\gcd(3,n-1)=1$.


问:
If $f'(x)$ is uniform continuous, then $\lim_{n \rightarrow\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)=f'(x)$.
Let $f\colon\mathbb{R}\rightarrow\mathbb{R}$ be continuous and differentiable, where $f'(x)$ is uniformly continuous. Then prove that the following is true for all $x \in \mathbb{R}$ $$\lim_{n \rightarrow\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)=f'(x).$$

My attempt:

Let $\varepsilon >0$ be arbitrary. Since $f'(x)$ is uniformly continuous, there exists $\delta$ such that if $|h| \leq \delta$, we have for any $x \in \mathbb{R}$ $$\frac{f(x+h)-f(x)}{h} \leq \varepsilon. $$

Consider the case where $h = \frac{1}{n}$. Then we can fix a $N \in \mathbb{N}$ such that $\frac{1}{N} \leq \delta$ and for any natural number $n \geq N$, we maintain the inequality above.

And now we can see that the equality is valid for any $n \geq N$: $$\lim_{n \rightarrow\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right)= \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} =f'(x).$$

that is, both limits converge since $f$ is differentiable and $f'(x)$ is uniformly continuous.

Note: I am not sure this is correct. I've read some hints which recommend using MVT, but I found no explanation on why one must use MVT nor how to use it.

答:
Lets define $n^{'}$ := $\frac{1}{n} $ Notice as $ n \to \infty$ we have $n^{'} \to 0 $

Then $\lim_{n \rightarrow\infty}n\left(f\left(x+\frac{1}{n}\right)-f(x)\right) = \lim_{n^{'} \to 0} \frac{f(x+n^{'})-f(x)}{n^{'}} = f'(x)$


问:
How to apply resolution principle on a product of sums
Following the book Discrete Mathematics and its Applications By Rosen, in the "foundations of logic and proofs" chapter, I came across this question

$\text{Use resolution principle to show that the compound proposition } (p \vee q) \wedge (\lnot p \vee q) \wedge (p \vee \lnot q) \wedge (\lnot p \vee \lnot q) \text{ is not satisfiable.}$

I can certainly prove this by using a truth table, but how can I apply resolution? To my knowledge to prove this via the Resolution principle, we must have clauses, which are disjunctions of literals. Each maxterm can be thought of as an individual clause, but how can I work with the whole premise at once?

答:
You have to separate the 'maxterms' otherwise it's meaningless to apply the resolution principle. You can apply an algorithm like this :

construct the set $S$ composed by the $4$ 'maxterm' literals :

$$S=\{ p \vee q \ ,\ \lnot p \vee q \ ,\ p \vee \lnot q \ ,\
\lnot p \vee \lnot q\}$$

So applying one time the resolution rule to the first two and last two literals you get $S=\{ q\ ,\ \neg q\}$ , applying again you get $S = \{\}$ so the proposition is not satisfiable.

The sense of this algorithm is that you infer $q$ by the first two literals but then you infer $\neg q$ by the last two literals, this means that, to have all the statement to be true, you need $q \land \neg q$, something not possible.


问:
Is there a rational function that meets all these criterias?
A zero at x = 3

A hole when x = 5

A vertical asymptote at x = -1

A horizontal asymptote at y = 3

A y-intercept at y = -2

What I have so far is $$ y=\frac{3(x-3)(x-5)}{(x+1)(x-5)} $$

I figured the y intercept right now is -9 but no matter how I transform it to try and get a y intercept of -2, it just messes up the whole equation. Is there a way to do a vertical shift without changing the entire equation? Thank you!

答:
In order for a rational function to have a zero at $x=3$, a hole at $x=5$, and a vertical asymptote at $x=-1$, it must be of the form

$$(x-3)(x-5)P(x)\over(x+1)(x-5)Q(x)$$

If there is a nonzero horizontal asymptote, $P$ and $Q$ must have the same degree. This still allows a lot of freedom to get a $y$-intercept of $-2$ and a horizontal asymptote of $y=3$. Can you take it from here? (If you're willing to allow additional zeroes and vertical asymptotes, you can let $P$ and $Q$ be linear; if you want to avoid them, you can take $P$ and $Q$ to be quadratics like $x^2+1$.)


问:
Is uncountable subset of separable space separable?
I have to prove that any uncountable $B\subseteq \mathbb{R}$, where $(\mathbb{R},\epsilon^1)$ is euclidean topology and topology on B is relative, is separable. And I know it's true because every subset of separable metric space is separable. </p>

<p>But what if we are given separable space $(X,\tau)$, $X$ uncountable, and $A \subseteq X$ uncountable subset with relative topology. Is $(A,\tau_A)$ separable and if it is, how to prove it? </p>

答: Consider the Niemytzki (or Moore) plane.  This space is separable (the family of points with both coordinates rational is dense), but the $x$-axis $A = \{ \langle x , 0 \rangle : x \in \mathbb{R} \}$ is an uncountable closed discrete subset (and so $A$ with the subspace topology is discrete, and is therefore not separable).</p>


问:
Convergence of $\sum_{n=1}^{\infty} \frac{(\cos(iz))^n}{n} $

Examine the series convergence : $$\sum_{n=1}^{\infty} \frac{(\cos(iz))^n}{n}$$

I transformed it into $$\sum_{n=1}^{\infty} \frac{(e^z+e^{-z})^n}{n2^n}$$ and I don't know what to do now.

答:
Let $$S(z)=\sum_{n\ge1}\frac{\cos(iz)^n}{n}.$$
Note that $$\frac1{1-x}=\sum_{n\ge0}x^n,\qquad |x|<1,$$
so
$$-\ln(1-x)=\sum_{n\ge1}\frac{x^n}{n},\qquad |x|\le 1, x\ne 1.$$
If $z\in\Bbb R$, $\cos(iz)=\cosh(z)$, and the series $S(z)$ diverges.

Let $z=a+bi$. We may see that
$$\cos(i(a+bi))=\cosh(a)\cos(b)+i\sinh(a)\sin(b).$$
Thus
$$\cos(iz)=re^{iT(z)},$$
where
$$r=\sqrt{\sinh^2(a)+\cos^2(b)},$$
and $$T(z)=T\in\left\{\arcsin\left(\frac{\sinh(a)\sin(b)}{r}\right),\quad \arccos\left(\frac{\cosh(a)\cos(b)}{r}\right),\quad \arctan\left(\tanh(a)\tan(b)\right)\right\}.$$
We know that regardless of the choice of $t$, $|e^{it}|=1$.

We can split the rest up into cases.

Case 1: $r< 1$ and $T(z)/\pi\not\in\Bbb Z$.

In this case we have that $|z|=|re^{iT}|=|r|<1$ so $S(z)$ converges.

Case 2: $r=1$ and $T(z)/\pi\not\in\Bbb Z$.

In this case $|z|=|re^{iT}|=|e^{iT}|=1$. But since $T/\pi$ is not an integer, $e^{iT}\ne 1$, so $S(z)$ converges to $-\ln(1-e^{iT(z)})$.

Case 3: $r>1$ and $T(z)/\pi\not\in\Bbb Z$.

This implies $|z|>1$ so $S(z)$ diverges.

Case 4: $r<1$ and $T(z)/\pi\in\Bbb Z$.

This gives $|z|=|r|<1$ so $S(z)$ converges.

Case 5: $r=1$ and $t(z)/\pi\in\Bbb Z$.

This gives $z=(-1)^{T/\pi}$. If $T/\pi$ is even, $z=1$ and $S(z)$ diverges. If $T/\pi$ is odd, $z=-1$ and $S(z)$ converges to $-\ln 2$.

Case 6: $r>1$ and $T/\pi\in\Bbb Z$.

This gives $|z|=|r|>1$ so $S(z)$ diverges.

To sum up (pun very intended), the series $S(z)$ converges iff $$\sqrt{\sinh^2(\Re(z))+\cos^2(\Im(z))}<1,$$ OR $$\sqrt{\sinh^2(\Re(z))+\cos^2(\Im(z))}=1\quad \text{AND}\quad \frac{1}{\pi}T(z)\not\in\Bbb Z,$$ OR $$\sqrt{\sinh^2(\Re(z))+\cos^2(\Im(z))}=1\quad \text{AND}\quad \frac{1}{\pi}T(z)\in\Bbb Z \text{  is odd}.$$


问:
Is this relation symmetric? $R={(1,1),(0,0)}\ $
R={(1,1),(0,0)}

A relation is symmetric if xRy and yRx but can it be x=y?

答:
A relation $ R $ defined at a set $ E $ is said to be symetric if, each time an element $ a $ is in relation with an element $ b$, then the element $ b $ is also in relation with $ a $.

$$(\forall a,b\in E)\; \Bigl(a\;R\; b\;\implies\;b\;R\;a\Bigr)$$

In your case, clearly, the relation $ R $ is symetric.

The relation is antisymetric if

$$(\forall a,b \in E)$$
$$\Bigl(a \;R\; b \;\wedge\;b\;R\;a\implies a=b\Bigr)$$

The negation is
$$(\exists a,b \in E)\ :$$
$$(\;a\;R\;b)\;\wedge\; (b\;R\;a)\;\wedge\;(a\ne b)$$

This is not the case. Your relation is also antisymetric.

It is transitive iff
$$(\forall a,b,c\in E)$$
$$(a\;R\;b)\wedge (b\;R\;c)\implies a\;R\;c$$

The negation is
$$(\exists a,b,c\in E)\;:$$
$$(a\;R\;b) \wedge (b\;R\;c)\;\wedge (a\;not\;R\;c)$$

this is not the case, so it is transitive.


问:
Showing that the segment joining $0$ to $z$ is perpendicular to the segment joining $0$ and $w$ iff $Re[z\bar{w}]=0$
I'm reading Beardon's Algebra and Geometry.

Suppose that $zw\neq0$. Show that the segment joining $0$ to $z$ is perpendicular to the segment joining $0$ to $w$ if and only if $Re[z\bar{w}]=0$.

From here I first expanded $z\bar{w}$:

$$i \left(a_2 b_1-a_1 b_2\right)+a_1 a_2+b_1 b_2$$

Then I obtained the real part of $z\bar{w}$:

$$a_1 a_2+b_1 b_2$$

From here, I was kinda stuck on how to proceed in the demonstration but then I decided to proceed with some of the most obvious examples for getting complex numbers that wouldfit perpendicularity: $(0+7i)$ and  $(7+0i)$ for which $0 \cdot 7+7\cdot0=0$ and we have one case of perpendicularity. This first case would be:

$$((a_1=0)\oplus (a_2=0))\wedge ((b_1=0)\oplus (b_2=0))$$

With this reasoning in mind, I thought that for all the other cases of perpendicularity, we must find any numbers that satisfy $a_1 a_2=n$ and $b_1 b_2=-n$.

Is my reasoning correct? Is there something I should add?

答: Two lines are perpendicular iff the scalar product of some vectors in their direction is zero. Apply this to the vectors $(a_1,b_1)$ and $(a_2,b_2)$.</p>


问:
why $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$ give contradiction?
I have some confusion in Baby Rudin book

My doubt is given below marked in red line and red circle

My attempt : Here we have already assume  that no point
of $K_1$ belongs to every $ K_{\alpha} $ that mean it implies already $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$

My doubts:  why  $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$ give contradiction ?

Edit:here $K_1$ is compact  and open cover are  $G_\alpha$ and so $K_1 \cap G_\alpha^c = \emptyset$ this implies  again $K_1 \cap K_\alpha=\emptyset$  ......why  its  contradicts?

答:
The key of the theorem is to get that precise contradiction, since you were assuming every finite subcollection of the compact sets $\{K_\alpha\}$ had nonempty intersection.

The way to get the contradiction is by assuming we don't have the result (reductio ad absurdum). Assuming that we don't have the result is assuming $\bigcap K_\alpha=\emptyset$.

If you fix one compact set $K_1$ then for every point $x\in K_1$ there is some $\alpha_x$ such that $x\notin K_{\alpha_x}$, since otherwise we would have $x\in K_\alpha\;\forall \alpha$, but that would mean $x\in\bigcap K_\alpha$, and we were assuming the contrary.

Then for every $x\in K_1$ there is some $\alpha_x$ such that $x\in G_{\alpha_x}:=K_{\alpha_x}^\mathsf c$. Every $G_\alpha$ is open, since compact subsets are closed in metric spaces. In consequence:

$\forall x\in K_1\;x\in G_{\alpha_x}\Rightarrow K_1\subset\bigcup_\limits{x\in K_1}G_{\alpha_x}$, so $\{G_{\alpha_x}\}_{x\in K_1}$ is an open cover of $K_1$. Since $K_1$ is compact, we must have a finite subcover $K_1\subset G_{\alpha_{x_1}}\cup\cdots\cup G_{\alpha_{x_n}}$.

Now, remember that $G_\alpha:= K_\alpha^\mathsf c$, so we actually have $K_1\subset K_{\alpha_{x_1}}^\mathsf c\cup\cdots\cup K_{\alpha_{x_n}}^\mathsf c$, which means

$K_{\alpha_{x_1}}\cap\cdots\cap K_{\alpha_{x_n}}=\big(K_{\alpha_{x_1}}^\mathsf c\cup\cdots\cup K_{\alpha_{x_n}}^\mathsf c\big)^\mathsf c\subset K_1^c$, which implies

$K_{\alpha_{x_1}}\cap\cdots\cap K_{\alpha_{x_n}}\cap K_1=\emptyset$

But $\{K_{\alpha_{x_1}},\cdots, K_{\alpha_{x_n}},K_1\}$ is a finite subcollection of the compact sets $\{K_\alpha\}$, so it should have nonempty intersection by hypothesis. We got our contradiction!

This must means that the thing we assumed is false, so the opposite is true: We have that $\bigcap K_\alpha\neq\emptyset$.


问:
Fibre sequences involving $BO(2)$, $BSO(3)$ and $BSU(2)$
In an answer to another question, it is remarked that there are fibre sequences
$$\mathbf{RP}^2 \to BO(2) \to BSO(3),$$
$$\mathbf{RP}^\infty \to BSU(2) \to BSO(3).$$

The second one is explained by another fact stated in this answer, namely: if we have an exact sequence $1 \to H \to G \to G/H \to 1$, then it induces a fibre sequence
$$ BH \to BG \to B(G/H).$$
I don't know how to obtain the first one though.

Does someone have a reference for either of these facts?

答:
You can directly verify that a short exact sequence of groups induces a fiber sequence of classifying spaces by using the long exact sequence in homotopy to compute say the homotopy groups of the fiber, and noting that classifying spaces are characterized by their homotopy groups.

Or you can write down models of classifying spaces and check.  Let $EG$ be a contractible space on which $G$ acts freely, etc. Then the map $BG \to B(G/H)$ is modeled by $$BG \simeq EG \times_G E(G/H) \simeq BH \times_{G/H} E(G/H) \to * \times_{G/H} E(G/H) \simeq B(G/H),$$ whose homotopy fiber is $BH$.

On the other hand, the sequence $$\mathbb{R}P^2 \to BO(2) \to BSO(3)$$ is induced by the group homomorphism $O(2) \hookrightarrow SO(3)$ sending $A$ to $A \oplus \det A$.  The fiber of $BO(2) \to BSO(3)$ is $SO(3)/O(2)$, which can be identified with $\mathbb{R}P^2$.  ($SO(3)$ acts transitively on lines in $\mathbb{R}^3$, and the stabilizer of a line is $O(2)$.)


问:
How to verify the order of DOPRI Runge-Kutta method?
I've written code in Fortran based on the RK8(7)-13 method by Dormand and Prince to solve the system $\mathbf{y}'=\mathbf{f}(t,\mathbf{y})$. The method is Runge-Kutta 8$^\text{th}$ order with an embedded 7$^\text{th}$ order method; I want to verify that the error on the output agrees with this. For a method of order $p$ and timestep $h$, we expect the local truncation error to be $\mathcal{O}(h^{p+1})$. Therefore, if I advance the program by one time step and calculate $\text{error} = |y_{\text{RK}}-y_{\text{exact}}|$ (say I know $y_{\text{exact}}$) and plot $\log(\text{error})$ vs. $\log(h)$ for several values of $h$, the slope of the line should be 9.

Here's the problem: it isn't. I've tried simple polynomials like $f(t,y) = 1$ and $f(t,y) = 2t$, but the slope of the line always seems to be between 1 and 2; this is also the case for $f(t,y) = y$. However, for $y'' + y = 0$ (which translates into the coupled system $\mathbf{f}(t,y_{1}, y_{2}) = (y_{2}, -y_{1})$), I found that the slope is almost exact 9 for larger values of $h$ - for small values the slope goes back to ~1. Furthermore, the magnitude of the error is $~10^{-20}$ for this small $h$. I was wondering if anyone had any ideas as to what the issue could be.

My ideas:

I've been misinformed as to what order means, and the method I've described is not representative of the order of the RK scheme. This is entirely possible, but the fact that one of the slopes was 9 seems too coincidental.

Something to do with precision. This could explain why the small $h$ values have unreliable errors since I'm "only" carrying 30 digits of precision, but this wouldn't explain the disagreement for the polynomial $f$.

Something specific to RK methods that I'm unaware of. I'm hoping it's this.

Programming error. I'm hoping it isn't this.

Mistake entering in the Butcher table. I'm really hoping it isn't this.

Side note: I've also tried plotting the global error as a function of $h$, but the results were even stranger, so I think the problem may be more fundamental.

答: If I remember correctly, in the special case where $f(t,y)$ is a polynomial in $t$ of degree $d$ and independent of $y$, every Runge-Kutta method of order $d+1$ gives the exact solution (which is a polynomial of degree $d+1$). So the Euler method solves $y' = 1$ exactly and the classical fourth-order RK-method solves $y' = t^3$ exactly (all up to round-off error). </p>


问:
Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates.
Let $H$ and $K$ be two conjugate subgroups in $G.$ Show that $N_G(H)$ and $N_G(K)$ are conjugates

I thought of using the second part of Sylow's Theorem that says; In particular, any two Sylow $p$-subgroups of $G$ are conjugate in $G$.

But I'm not sure if I have to prove that $H$ and $K$ are $p$-subgroups

答: If $H$ and $K$ are conjugate in $G$, then they belong to the same orbit of the action of $G$ by conjugation on the set of the subgroups of $G$, and thence their stabilizers are conjugate. But such stabilizers are precisely $N_G(H)$ and $N_G(K)$, respectively.


问:
On proving that a limit involving an integral and part of its integrand tends to zero.
Suppose $f$ is continuous, $F(x)=\int_{a}^{x}{f(t)dt}$ is bounded on $[a,b)$ and $\int_{a}^{b}{f(t)dt}$ converges as an improper integral. Suppose also that $g>0$, $g'$ is non-negative and locally integrable on $[a,b)$ and that $\lim_{x\to b^{-}}{g(x)}=\infty$

Prove:

$\lim_{x\to b^{-}}{\frac{1}{g(x)}\int_{a}^{x}{f(t)g(t)dt}}=0$

I´ve been trying prove this result that was part of an exercise in a real analysis PDF.
Can´t seem to get very far proving it so I am not sure if there is some theorem or lemma I´m not thinking of that will make it obvious or that I´ve been given insufficient information. I´ve tried integrating by parts which gives me: $\frac{1}{g(x)}\Big(F_0(t)g(t)\Big|_{a}^{x} -\int_{a}^{x}{F_0(t)g'(t)dt}\Big)$ and letting $F_0(x)$ be $\int_{x}^{b}{f(t)dt}$.
It seemed to help since: $\frac{1}{g(x)}\Big(F_0(t)g(t)\Big|_{a}^{x}\Big)\to 0$ as $x\to b^{-}$, but I can´t seem to get anywhere on the second integral.

I´m asking if anyone can help me prove it or explaing why it can´t be done.

答:
You are almost there. It remains to show that the growth of $g$ is offset by the vanishing of $F_0$ at $b$, so that
$$\Biggl\lvert\,\int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert$$
grows slower than $g$ near $b$. To exploit the vanishing of $F_0$, split the interval into two parts. One part where $\lvert F_0(t)\rvert$ may be large, but that stays at a safe distance from $b$, and the remaining part close to $b$ where $\lvert F_0(t)\rvert$ is small.

Thus, let $\varepsilon > 0$ be given, and pick $c \in (a,b)$ such that $\lvert F_0(t)\rvert \leqslant \varepsilon$ for $c \leqslant t \leqslant b$. For all $x \in (c,b)$ we have
\begin{align}
\Biggl\lvert\, \int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert &\leqslant \Biggl\lvert\, \int_a^c F_0(t)g'(t)\,dt\,\Biggr\rvert + \int_c^x \lvert F_0(t)\rvert\,g'(t)\,dt \\
&\leqslant \Biggl\lvert\, \int_a^c F_0(t)g'(t)\,dt\,\Biggr\rvert +  \varepsilon \bigl(g(x) - g(c)\bigr)\,.
\end{align}
From this it follows that
$$\limsup_{x \to b^-} \frac{1}{g(x)}\Biggl\lvert\, \int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert \leqslant \varepsilon \tag{$\ast$}$$
since $g(x) \to +\infty$ and the integral as well as $g(c)$ are constants (for every fixed $c$).

But $(\ast)$ holds for every $\varepsilon > 0$, hence
$$\limsup_{x \to b^-} \frac{1}{g(x)}\Biggl\lvert\, \int_a^x F_0(t)g'(t)\,dt\,\Biggr\rvert  \leqslant 0\,.$$
Since the expression on the left hand side is nonnegative, this means the limit exists and is $0$.


问:
Solve $\lim_{z\to i} (\frac{4z+i}{z+1}) = (\frac{5i}{1+i})$ using the formal definition of limit
I want to know if I answered this correctly:

$$|f(z) - L| = \left|\frac{4z+i}{z+1} - \frac{5i}{1+i}\right|\\
 = \left|\frac{4z+i+4zi-1-5zi-5i}{(z+1)(1+i)}\right| \\
=\left| \frac{4z-4i-zi-1}{(z+1)(1+i)}\right|\\
 = \left|\frac{4(z-i)-i(z-i)}{(z+1)(1+i)}\right|\\
=\left|\frac{(z-i)(4-i)}{(z+1)(1+i)} \right|$$

Then consider that $|z+1| > r$ and we can assume $\left|\frac{(z-i)(4-i)}{(z+1)(1+i)} \right|\leq \left|\frac{(z-i)(4-i)}{r(1+i)} \right|$. Thus $|z-i|< \frac{\varepsilon r(1+i)}{4-i}= \delta$ . This condition need to be satisfied together with $|z+1| > r$ and for that we can presume $$|z+1| = |z+1-i+i| = |i+1 +(z-i)| \\
\geq |1+i| - |z-i| > \sqrt2 - \delta > \sqrt2 -(\sqrt2 - r) = r.$$
Lastly, $\delta = \min\left\{\frac{{\varepsilon r(1+i)}}{4-i},\sqrt2 - r\right\}$

答: There is a problem there. Your answer contains a number $r$, but you don't tell us what that number is. Here's a suggestion: take $r=1$. Then take$$\delta=\min\left\{\varepsilon\frac{\sqrt2}{\sqrt{17}},\sqrt2-1\right\}.$$Those numbers $\sqrt2$ and $\sqrt{17}$ are $|1+i|$ and $|4-i|$ respectively.


问:
Let $U$ and $V$ be two $n\times n$ unitary matrices such that $UV=\lambda VU$.Then prove that $\lambda^n=1$
Let $U$ and $V$ be two $n\times n$ unitary matrices such that $UV=\lambda VU$.Then prove that $\lambda^n=1$

Totally stuck on it.Can I get some help?

答:
Hint: The determinant is invariant under base change, so how does the determinant of $U$, say, relate to its eigenvalues? Moreover the determinant is multiplicative, so what is $\det(UV)$, what is $\det(\lambda VU)$?

Edit: I just realised that the fact that they are unitary is a red hering of sorts. Multiplicativity of the determinant and non-singularity suffice.


问:
Suppose that $A,B$ are two $n\times n$ matrices such that $A,B$ commutes and $A,B$ are diagonalizable.
Suppose that $A,B$ are two $n\times n$ matrices such that $A,B$ commutes and $A,B$ are diagonalizable. Show that there is a polynomial $p(t)$ such that $B=p(A)$ where all $n$ eigenvalues of $A$ are distinct.

I know that $A,B$ are simultaneously diagonalizable. Then how can we proceed.

答:
This is not true. Consider $A=\operatorname{diag}(1,1,0)$ and $B=\operatorname{diag}(0,1,1)$. If $p(A)=B$, then
$$
\pmatrix{p(1)\\ &p(1)\\ &&p(0)}=p(A)=B=\pmatrix{0\\ &1\\ &&1}
$$
and hence $0=p(1)=1$, which is impossible. For the same reason, it is also impossible that $p(B)=A$.

Edit. However, $p$ exists if all eigenvalues of $A$ are distinct. Suppose $A$ and $B$ have been simultaneously diagonalised. It suffices to find a $p$ such that $p(\lambda_i(A))=\lambda_i(B)$, where $\lambda_i(\cdot)$ denotes the $i$-th diagonal entry in the (diagonal) eigenvalue matrix of $A$ or $B$.


问:
Is $(a+1)^n-a^n$ ever divisible by $n$?
If $n\geq 2$ is prime and $a\geq 1$ is an integer, then $(a+1)^n-a^n$ leaves a remainder of $1$ modulo $p$.

If $n$ is even or a multiple of $3$, then it follows ad-hoc that again $(a+1)^n-a^n$ is never divisible by $n$.

By simulations, the same should be true for all $n$. How would one prove this?

Prove that $(a+1)^n-a^n$ is never divisible by $n$.

答: Assume, for the sake of contradiction, that there exists $n$ and $a$ such that $n$ divides $(a+1)^n-a^n$. Choose $n$ so that the smallest prime factor $p$ of $n$ is minimal (the OP has noted that $p\ge5$), and write $n=mp$. Then we have $(a+1)^{mp} \equiv a^{mp}\pmod p$, or equivalently $(a+1)^{m} \equiv a^{m}\pmod p$ by Fermat's little theorem, or equivalently again $(a+1)^{\gcd(m,p-1)} \equiv a^{\gcd(m,p-1)}\pmod p$. This implies that $\gcd(m,p-1)>1$; however, $m$ is not divisible by any primes smaller than $p$ by assumption, which contradicts $\gcd(m,p-1)>1$ since all primes dividing $\gcd(m,p-1)$ are at most $p-1$.

Title: Is $(a+1)^n-a^n$ ever divisible by $n$?


问:
Induction hypothesis for inequality
The solution to a question I found online is in the image below, I'm confused on how did they go from $2^n + 2$ to$ 2^n + 2^n$. Where did they get the extra $2^n$ from? and then they got $2^{n+1}$ from $2^n + 2^n$. It's not multiplication so how did they get $2^{(n+1)}$ suddenly?

Prove that for all integers $n \geq 4,$ we have:

(a) $2 n+1<2^{n}$

Solution. We prove it by induction on $n .$ For $n=4,$ we have $2 \cdot 4+1=9<1$

Now, assume that $2 n+1<2^{n}$. Then,

$$
\begin{aligned}
2(n+1)+1 &=2 n+1+2 \\
&<2^{n}+2
\end{aligned} \quad \begin{array}{l}
\text { [by the IH }]
\end{array}
$$
$$
\begin{array}{ll}
<2^{n}+2^{n} & {\left[\text { as } n \geq 4,2^{n}>2\right]} \\
=2^{n+1}
\end{array}
$$

答:
You can as well prove the inductive step by using strong induction twice. Assume the hypothesis holds for $\frac{1}{2}n$, then we must prove it holds for $n$.
$$ 2\left(\frac{1}{2}n\right) + 1 < 2^{\frac{1}{2}n}$$
$$ n + 1 < 2^{\frac{1}{2}n}$$
Multiply both sides by $2$ so we get:
$$ 2n + 2 < 2^{\frac{1}{2}n} \cdot 2 $$
$$ 2n + 1 < 2n + 2 < 2^{\frac{1}{2}n + 1} $$
All we have to prove now is that $\frac{1}{2}n + 1 < n$ for $n >=4$. Again we prove this using strong induction.

To prove that $$\frac{1}{2}n + 1 < n$$
When $n=4$:
$$\frac{1}{2} \cdot 2 + 1 = 2 < 4$$
In the inductive step, assume the statement holds for $\frac{1}{2}n$, then prove it holds for $n$.

$$ \frac{1}{2}\left(\frac{1}{2}n\right) + 1 < \frac{1}{2}n $$
Now multiply both sides by $2$.
$$ 2 \cdot \frac{1}{2}\left(\frac{1}{2}n\right) + 2 \cdot 1 < 2 \cdot \frac{1}{2}n$$
$$ \frac{1}{2}n + 1 < \frac{1}{2}n + 2 < n $$
$$ \frac{1}{2}n + 1 < n $$
And we are done. (QED)

We used the same approach twice to prove both the original statement and the nested statement. A bit of a recursive proof really. I like to call this approach of repetitive induction Gilbert's Idea of Nested Strong induction (GINS). It might look long right now but remember it when you have to prove something a bit more complicated.

Title: Induction hypothesis for inequality


问:
Triangle inequality for infinite number of terms
We can prove that for any $n\in \mathbb{N}$ we have triangle inequality: $$|x_1+x_2+\cdots+x_n|\leqslant |x_1|+|x_2|+\cdots+|x_n|.$$

How to prove it for series i.e. $$\left|\sum \limits_{n=1}^{\infty}a_n\right|\leqslant \sum \limits_{n=1}^{\infty}|a_n|.$$
Can anyone help to me with this?

答:
I think in this case the easiest way is to show it via contradiction. Let's assume assume that $\sum\limits_{n=1}^{\infty}|a_n|$ and $\sum\limits_{n=1}^{\infty}a_n$ converge.

Now, we assume $\Big|\sum\limits_{n=1}^{\infty}a_n\Big|>\sum\limits_{n=1}^{\infty}|a_n|$ and therfore we define $\epsilon:=\Big|\sum\limits_{n=1}^{\infty}a_n\Big|-\sum\limits_{n=1}^{\infty}|a_n|>0$ . Then, by definition of convergence there exists an index $m_0$ such that for all $m>m_0$ we have $\Big|\sum\limits_{n=1}^{\infty}a_n-\sum\limits_{n=1}^{m}a_n\Big|<\frac{\epsilon}{2}\implies \Big|\sum\limits_{n=1}^{m}a_n\Big|>\sum\limits_{n=1}^{\infty}|a_n|\geq \sum\limits_{n=1}^{m}|a_n|$, which contradicts triangle inequality. Note that if $\sum\limits_{n=1}^{\infty}|a_n|=\infty$ then we get a contradiction immediately.


问:
How to find the inverse Laplace Transforms?
How do you find the inverse Laplace Transform of the following,

$$\frac{2 (s^2+4 s+5)^2+40}{(s^2+4 s+5)^2}$$

Separating them into complex coefficients is to long.

Body:
I have done it, you can see if it's useful to you, I.m not sure so if you look out the mistake please fix it..

$\begin{array}{l}
 F(s) = \frac{{2{{({s^2} + 4s + 5)}^2} + 4}}{{{{({s^2} + 4s + 5)}^2}}} = 2 + \frac{{40}}{{{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}^2}}} \\
 {L^{ - 1}}{\rm{[2]  =  }}2\delta (t) \\
 {\rm{\{ }}\int_0^\infty  {2.{e^{ - st}}\delta (t)dt = 2} {\rm{.}}{{\rm{e}}^{ - s0}}{\rm{ = 2\} }} \\
 G(s) = \frac{{40}}{{{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}^2}}}{\rm{  =  40}}{\rm{.}}\frac{1}{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}}.\frac{1}{{{\rm{[}}{{(s + 2)}^2} + 1{\rm{]}}}} = 40L\left[ {{e^{ - 2t}}.\sin t} \right].{L^{ - 1}}\left[ {{e^{ - 2t}}.\sin t} \right] \\
  =  > g(t) = {e^{ - 2t}}.\sin t*{e^{ - 2t}}.\sin t = \int_0^t {{e^{ - 2(t - u)}}.\sin (t - u).{e^{ - 2u}}.\sin u.du}  \\
  = {e^{ - 2t}}\int_0^t {(\sin t.\cos u - \cos t.\sin u)\sin u.du}  \\
  =  > g(t) = {e^{ - 2t}}\left[ {\sin t\underbrace {\int_0^t {\cos u.\sin u.du} }_{{I_1}} - \cos t\underbrace {\int_0^t {{{\sin }^2}udu} }_{{I_2}}} \right] \\
  =  > f(t) = 2\delta (t) + g(t) \\
 \end{array}$

You can evalue the integrals I1 and I2 yourself


答:
You have
$$
K_1\subseteq X\setminus(K_{\alpha_1}\cap K_{\alpha_2}\cap\dots\cap K_{\alpha_n})
$$
Now it's a general feature of sets that, if $A$ and $B$ are subsets of $C$, then $A\subseteq C\setminus B$ implies $A\cap B=\emptyset$, because if $a\in A$, then $a\in X\setminus B$ and therefore $a\notin B$.

The assumption on the family $(K_\alpha)$ is that the intersection of any finite set of members of the family is not empty, but $K_1,K_{\alpha_1},\dots,K_{\alpha_n}$ is a finite set of members of the family.

Maybe it is useful to see a different version of the proof. Fix a member $K$ of the family and the family of subsets of $K$ defined by $C_\alpha=K\cap K_\alpha$.

Here we're assuming that the given family is not empty, which also Rudin does (without mentioning it explicitly).

The family $(C_\alpha)$ is a family of closed subsets of $K$ (because in a metric space compact sets are closed) and $K$ is compact. If $\bigcap_{\alpha}C_\alpha=\emptyset$, then $(U_\alpha)$, where $U_\alpha=K\setminus C_\alpha$, is an open cover of $K$ and so it has a finite subcover:
$$
U_{\alpha_1}\cup U_{\alpha_2}\cup\dots U_{\alpha_n}=K
$$
that becomes
$$
C_{\alpha_1}\cap C_{\alpha_2}\cap\dots \cap C_{\alpha_n}=\emptyset
$$
But this means that
$$
(K\cap K_{\alpha_1})\cap (K\cap K_{\alpha_2})\cap\dots \cap (K\cap K_{\alpha_n})=\emptyset
$$
and therefore
$$
K\cap K_{\alpha_1}\cap K_{\alpha_2}\cap\dots \cap K_{\alpha_n}=\emptyset
$$
against the assumption.

Title: why $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$ give contradiction?

Body:
I have some confusion in Baby Rudin book

My doubt is given below marked in red line and red circle

My attempt : Here we have already assume  that no point
of $K_1$ belongs to every $ K_{\alpha} $ that mean it implies already $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$

My doubts:  why  $K_1 \cap K_{\alpha_1} \cap ...\cap K_{\alpha_n}=\emptyset$ give contradiction ?

Edit:here $K_1$ is compact  and open cover are  $G_\alpha$ and so $K_1 \cap G_\alpha^c = \emptyset$ this implies  again $K_1 \cap K_\alpha=\emptyset$  ......why  its  contradicts?


问:
How to prove $\int_{0}^{\infty} \frac{\sin x}{x(x^2+1)}dx=\frac{\pi}{2}(1-\frac{1}{e})$ by residue calculus?
How to prove $$I=\int_{0}^{\infty} \frac{\sin x}{x(x^2+1)}dx=\frac{\pi}{2}(1-\frac{1}{e})\, ?$$

Let $$f(z)=\frac{\sin z}{z(z^2+1)}=\frac{e^{iz}-e^{-iz}}{2i\,z(z^2+1)}.$$

Then I have no idea how to deal with $[0,\infty)$.

Can anyone please give me a hint?  Thanks in advance.

答: It helps to use $\tfrac{1}{x(x^2+1)}=\tfrac1x-\tfrac{x}{x^2+1}$ to reduce the problem to $I=\tfrac12\pi-J$ with$$J:=\int_0^\infty\tfrac{x\sin xdx}{x^2+1}=\tfrac12\Im\int_{\Bbb R}\tfrac{x\exp ixdx}{x^2+1}=\tfrac12\Im[2\pi i\lim_{x\to i}\tfrac{x\exp ix}{x+i}]=\tfrac{\pi}{2e}.$$


问:
The ideal sheaf of the base change of a closed immersion
Let $i:Z\to X$ be a closed immersion of schemes of quasi-coherent ideal sheaf $\mathcal{I}$ and $f:Y\to X$ be a morphism of schemes.
$\require{AMScd}$
\begin{CD}
W @>{j}>> Y\\
@V g V V @VV f V\\
Z @>{i}>> X
\end{CD}

Let $W$ be the fiber product. Then $j:W\to Y$ is a closed immersion by [https://stacks.math.columbia.edu/tag/01QR]. What is the ideal sheaf $\mathcal{J}$ corresponding to $j$?

Some examples (maybe I am wrong):

1.If $f$ is affine, $Y=\underline{\mathrm{Spec}}(\mathcal{A})$, $\mathcal{A}$ being a quasi-coherent $O_X$-algebra. Then $\mathcal{I}\mathcal{A}$ is an $\mathcal{A}$-module and $\tilde{\mathcal{IA}}=\mathcal{J}$.

2.If $Y=\underline{\mathrm{Proj}}(\mathcal{R})$, $\mathcal{R}$ being a quasi-coherent graded $O_X$-algebra. Then $\tilde{\mathcal{IR}}=\mathcal{J}$.

3.If $f$ is flat, then by EGA3, Ch.III, Proposition 1.4.15, $j_*O_W=f^*i_*O_Z$. Recall the SES $$0\to\mathcal{I}\to O_X\to i_*O_Z\to0.$$
As $f$ is flat, we have a SES $$0\to f^*I\to O_Y\to j_*O_W\to0.$$ So, $\mathcal{J}=f^*I$

What about the  general case?

答:
If we follow the chain of lemmas cited in the proof of the lemma you linked, we arrive at Stacks Project, Lemma 26.4.7 (tag 01HQ), which tells us that, in general, $\mathcal{J} = \operatorname{Im}(f^* \mathcal{I} \to \mathcal{O}_Y)$.

The key point (which implies the universal property of the fiber product for the closed subscheme corresponding to $\mathcal{J}$) is that a morphism $\varphi\colon W \to X$ factors through the closed immersion $i\colon Z \to X$ if and only if the map $\varphi^* \mathcal{I} \to \varphi^* \mathcal{O}_X = \mathcal{O}_W$ is zero. The forward implication is trivial. For the reverse implication, the topological part of this statement can be checked on stalks. The statement on sheaves follows from the adjunction between $f^*$ and $f_*$, which gives a correspondence between maps $\varphi^* \mathcal{I} \to \mathcal{O}_W$ and maps $\mathcal{I} \to \varphi_* \mathcal{O}_W$, so since these maps are zero, we get a map $\mathcal{O}_X/\mathcal{I} \to \varphi_* \mathcal{O}_W$ and thus a factorization $W \to Z \to X$. Details can be found in Stacks Project, Lemma 26.4.6 (tag 01HP).


问:
Analytical solution to coin-flip probability problem?
I recently came across the following probability problem:

You have a counter on the first square of a finite-length board

You flip a coin: if heads, you move the counter forwards 1 square; if tails, move forwards 2 squares

You keep flipping until you reach the end of the board

If $P(i)$ is the probability of landing on board square number $i$, for what value of $i>1$ is the value of $P(i)$ maximized?

Having manually calculated probabilities for $i=2,3,4,5,6,...$ is it clear (and intuitive) that $P(2)$ maximizes the quantity $P(i)$. Are there any nice ways (recursion, perhaps?) of demonstrating this analytically / algebraically?

答:
I will assume you start on square $0$.

Note that $P(1) = \frac{1}{2}$ and $P(2) = \frac{3}{4}$. Then, we have the following recurrence:

$$P(i) =\frac{P(i-1)}{2} + \frac{P(i-2)}{2}, P(1) = \frac{1}{2}, P(2) = \frac{3}{4}$$

This is because we have probability of $\frac{1}{2}$ of jumping forward $2$ spaces from square $i - 2$ and a probability of $\frac{1}{2}$ of jumping forward $1$ space from square $i - 1$.

Using this, we will now prove by induction that $P(2)$ must be maximal.

Assume $P(i), P(i-1) < P(2)$ for some $i\geq 4$, and then note:

$$P(i + 1) = \frac{P(i) + P(i - 1)}{2}\leq\max\big(P(i), P(i-1)\big) < 2$$

Then, since $P(3) = \frac{5}{8} < \frac{3}{4}$ and $P(4) = \frac{11}{16} < \frac{3}{4}$, we have that $P(i) < P(2)$ for all $i \geq 3$ and $P(1) < P(2)$, so $\boxed{P(2)\text{ is maximal.}}$


问:
Prove that composing two distribution functions results in another distribution function
If $F$ is a distribution function and $T: \Re \rightarrow \Re$ surjective and not decreasing, prove that $F \circ T$ is distribution function.

Can someone please give me a hint on how to prove this property of distribution functions?

I know which are the properties of function to be a distribution function, but I don't know how to apply those in the composition.

Thanks.

答:
Sketches (justifications will involve the given assumptions on $T$ and the fact that $F$ is a distribution function):

If $x \le y$ then $T(x) \le T(y)$ and thus $F(T(x)) \le F(T(y))$.

If $x \to \infty$, then $T(x) \to \infty$, so $F(T(x)) \to 1$.

If $x \to -\infty$, then $T(x) \to -\infty$, so $F(T(x)) \to 0$.

If $x \searrow x_0$, then $T(x) \searrow T(x_0)$ so $F(T(x)) \downarrow F(T(x_0))$.


问:
Partial derivatives of functions with a change in variable
Say if I have a function $x = x(u,v)$, and then two further functions $u= u(s,t)$ and $v = v(s,t)$, I was wondering how I would go about taking a partial derivative with respect to s.

答:
The rule is

$$\frac{\partial x}{\partial s} = \frac{\partial x}{\partial u}\frac{\partial u}{\partial s} + \frac{\partial x}{\partial v}\frac{\partial v}{\partial s}$$

You would add more terms of the same form if there were more variables like your $u$ and $v$. So you could write that if $x=x(u_1,u_2,\ldots,u_n)$, and each of the $u_i$ were functions $u_i = u_i(s_1,s_2,\ldots,s_m)$, then for each $j = 1,2,\ldots,m$,

$$\frac{\partial x}{\partial s_j} = \sum_{i=1}^{n}\frac{\partial x}{\partial u_i}\frac{\partial u_i}{\partial s_j}$$

Note that in this sum, $j$ is fixed (since your are looking for $\partial x/\partial s_j$) and the sum has one term for each of the $u_i$.

The $u_i$ are the "input slots" for $x$, each of which in turn depend on all of the $s_j$.


问:
Convex combination inequality
I have been doing some work which resulted in the following inequality, which I believe to be true, but am unsure how to prove. Let $\lambda_i, \mu_i \geq 0$, and $\sum_{i=1}^n\lambda_i=\sum_{i=1}^n\mu_i=1$. Is it true that
$$\sum_{i \neq j}\lambda_i\lambda_j \leq  2\sum_{i \neq j}\mu_i\lambda_j \;?$$

答:
Hint: Start with WTS

$$ \frac{ (\sum \lambda_i )^2 - \sum \lambda_i ^2 } { 2 } \leq ( \sum \lambda_i ) (\sum \mu_j) - \sum \lambda_i\mu_i. $$

Replace several terms by 1, then simplify again.

 WTS: $ 1 - \sum \lambda_i ^2  \leq 2  - 2 \sum \lambda_i\mu_i$

 WTS: $2 \sum \lambda_i\mu_i \leq 1 + \sum \lambda_i ^2$

 Since $ 2 \sum \lambda_i\mu_i \leq \sum ( \lambda_i ^2 + \mu_i^2)$ and $ \sum \mu_i^2 \leq ( \sum \mu_i)^2 =1$, the result follows.

In addition, equality holds iff $ \lambda_I = \mu_I = 1$ for some $I$. (in which case, both sides are 0).

Title: Convex combination inequality


问:
Improper integral $\int_{0}^{\infty}\tan^{-1}(\alpha x) \tan^{-1}(\beta x)/ x^2 dx, \alpha, \beta>0.$
I am facing difficulty in evaluation of the following improper integral:

$$\int_{0}^{\infty} \frac{\tan^{-1}(\alpha x) \tan^{-1}(\beta x)}{ x^2} dx,\ \ \alpha, \beta>0$$

I thought of applying Leibniz rule of differentiation under integral sign. But here we have two parameters $\alpha$ and $\beta$. I do not know how to apply Leibniz in such cases. I think I can first differentiate the given integral first w.r.t. first parameter $\alpha$ and denote that as $I'(\alpha)$ which is as follows:

$$I'(\alpha)= \int_{0}^{\infty} \frac{\tan^{-1}(\beta x)}{(1+\alpha^2 x^2) x }dx$$

Next I differentiate it w.r.t. $\beta$ and denote the result as $J(\beta)$ which is as follows:

$$J(\beta) =\int_{0}^{\infty} \frac{1}{(1+\alpha^2 x^2)(1+\beta^2 x^2)} dx.$$
This integral is quite easy to evaluate. But my question is that am I allowed to do all this. What are the other methods to evaluate this integral? Please suggest.

答:
Yes, you can apply Differentiation Under the Integral Sign as I show you bellow. Just a quick note, I'm applying a slightly modified version of Leibniz Rule, but in essence it's pretty much the same.

$$f(\alpha,\beta)=\int_{0}^{\infty}\frac{\arctan(\alpha x)\arctan(\beta x)}{x^2}dx$$

$$=\int_{0}^{\infty}\int_{0}^{\alpha}\int_{0}^{\beta}\frac{dydzdx}{(1+y^2x^2)(1+z^2x^2)}=\int_{0}^{\alpha}\int_{0}^{\beta}dydz\int_{0}^{\infty}\frac{dx}{(1+y^2x^2)(1+z^2x^2)}$$

$$=\int_{0}^{\alpha}\int_{0}^{\beta}\frac{dydz}{y^2-z^2}\int_{0}^{\infty}\left(\frac{y^2}{1+y^2x^2}-\frac{z^2}{1+z^2x^2}\right)dx$$

$$=\int_{0}^{\alpha}\int_{0}^{\beta}dydz\left[\frac{y\arctan(yx)-z\arctan(zx)}{(y-z)(y+z)}\right]^{\infty}_{0}=\frac{\pi}{2}\int_{0}^{\alpha}\int_{0}^{\beta}\frac{dydz}{y+z}$$

$$=\frac{\pi}{2}\int_{0}^{\alpha}dz[log(z+\beta)-log(z)]=\frac{\pi}{2}[(\beta+z)log(\beta+z)-z-zlog(z)+z]^{\alpha}_{0}$$

$$\boxed{f(\alpha,\beta)=\int_{0}^{\infty}\frac{\arctan(\alpha x)\arctan(\beta x)}{x^2}dx=\frac{\pi}{2}log\left(\frac{\left(\alpha+\beta\right)^{\alpha+\beta}}{\alpha^\alpha\beta^\beta}\right)}$$


问:
Dimension of polynomial ring
Suppose I have a polynomial ring $k[x_0,\cdots,x_n]$ and let $k[x_0,\cdots,x_n]_m$ denote the homogeneous part with degree $m$. How can I see that this $k[x_0,\cdots,x_n]_m$ has dimension $\binom{m+n}{m}$? Or more exactly, why does this give the Krull dimension?

答:
Hint:

If you denote $k_i$ the exponent of $xi$in such a monomial, it is the number of solutions in natural numbers of the equation
$$k_0+k_1+\dots+ k_n=m.$$

Title: Dimension of polynomial ring


问:
Probability of random line through a point intersecting a circle

Consider a circle of radius $r$ with center $C$ and let $p$ be a point outside the circle ( that is $\operatorname{dist}\left(p,C\right) > r$ ).

What is the probability that a random line through $p$ will intersect the circle $?$.

My intuition tells me it will be something like $r/\operatorname{dist}\left(p,C\right)$ but I would like a formal proof.

答: Assuming the "random line" is chosen with respect to a uniform measure on the angle it makes relative to some fixed orientation (e.g., the positive horizontal axis), then the probability is proportional to the angle measure subtended by the tangent lines drawn from the point $p$ to the circle.  This angle is a function of the radius $r$ and the distance $d > r$ between the circle's center and the point $p$, and is simply $$\theta = 2 \sin^{-1} \frac{r}{d}.$$  Therefore the desired probability is $$\frac{2}{\pi} \sin^{-1} \frac{r}{d}.$$


答:
A tree diagram is probably helpful here:

This shows which variables are functions of which other variables. From the diagram, you can use the chain rule to come up with

$$\frac{\partial x}{\partial s}=\frac{\partial x}{\partial u}\frac{\partial u}{\partial s}+\frac{\partial x}{\partial v}\frac{\partial v}{\partial s}$$

Title: Partial derivatives of functions with a change in variable

Body: Say if I have a function $x = x(u,v)$, and then two further functions $u= u(s,t)$ and $v = v(s,t)$, I was wondering how I would go about taking a partial derivative with respect to s.


问:
Contiguous Triangle Coloration
I somehow have this knack of imagining problems of my own, which I then myself find very hard to approach.  So there is no textbook to refer to, for the correct answer.

Here is a latest example of my ruminations, which goes into combinatorics

In how many ways can the triangle given below, consisting of 9 smaller triangles be colored, starting with 5 distinct colors, so that no two triangles that share an edge are of the same color?

I can sense that the problem can get more complicated as we venture into larger set of 16, 25, 36 and higher number of triangles in this arrangement having perfect square number of objects to fill. A generalization is most welcome, if it exists, but I would be glad even with a specific solution!!

答:
It’s actually a graph coloring problem: in how many ways can the $9$ vertices of the graph shown below be colored with $5$ colors so that adjacent vertices do not have the same color?

```<code>                       *
                       |
                       *
                      / \
                     *   *
                    /     \
               *---*---*---*---*
</code>```

The three vertices of degree $3$ are the key to counting such colorings.

Suppose first that all three of them are the same color; then each of the remaining six vertices can be any of the other $4$ colors. There are $5$ choices for the color of the vertices of degree $3$ and $4^6$ choices for the colors of the other six vertices, so for this case we have a total of $5\cdot4^6$ possible colorings.

Now suppose that the three vertices of degree $3$ get three different colors; there are $5\cdot4\cdot3=60$ ways to assign $3$ of the $5$ colors to those three vertices. There are $4$ choices of color for each of the vertices of degree $1$ and $3$ choices for each of the vertices of degree $2$, so for this case we have a total of $60\cdot 4^3\cdot 3^3$ colorings.

Finally, suppose that the three vertices of degree $3$ are colored with two different colors. There are $5$ ways to choose the color that appears twice, $3$ ways to assign it to two of these vertices, and $4$ ways to choose a color for the third of these vertices, so there are $60$ ways to color those three vertices with two colors. Each of the vertices of degree $1$ and the vertex between the two vertices of the same color can be colored with any of $4$ colors, and each of the remaining two vertices can be colored with any of $3$ colors, so for this case we have a total of $60\cdot4^4\cdot 3^2$ colorings.

The grand total is then $20\,480+103\,680+138\,350=262\,400$.

Title: Contiguous Triangle Coloration

Body:
It’s actually a graph coloring problem: in how many ways can the $9$ vertices of the graph shown below be colored with $5$ colors so that adjacent vertices do not have the same color?

                       *
                       |
                       *
                      / \
                     *   *
                    /     \
               *---*---*---*---*

The three vertices of degree $3$ are the key to counting such colorings.

Suppose first that all three of them are the same color; then each of the remaining six vertices can be any of the other $4$ colors. There are $5$ choices for the color of the vertices of degree $3$ and $4^6$ choices for the colors of the other six vertices, so for this case we have a total of $5\cdot4^6$ possible colorings.

Now suppose that the three vertices of degree $3$ get three different colors; there are $5\cdot4\cdot3=60$ ways to assign $3$ of the $5$ colors to those three vertices. There are $4$ choices of color for each of the vertices of degree $1$ and $3$ choices for each of the vertices of degree $2$, so for this case we have a total of $60\cdot 4^3\cdot 3^3$ colorings.

Finally, suppose that the three vertices of degree $3$ are colored with two different colors. There are $5$ ways to choose the color that appears twice, $3$ ways to assign it to two of these vertices, and $4$ ways to choose a color for the third of these vertices, so there are $60$ ways to color those three vertices with two colors. Each of the vertices of degree $1$ and the vertex between the two vertices of the same color can be colored with any of $4$ colors, and each of the remaining two vertices can be colored with any of $3$ colors, so for this case we have a total of $60\cdot4^4\cdot 3^2$ colorings.

The grand total is then $20\,480+103\,680+138\,350=262\,400$.


问:
How does the independent events rule prove independence?
We are taught to find if events A and B are independent it will satisfy the equation          P(A and B) = P(A)P(B), but how does this prove independence.

答:
Event A being independent of event B denotes that p(A) is unaffected by whether event B has occurred.

WLOG assume that $p(A) \neq 0 \neq p(B).$

Event A is independent of event B denotes that

$p(A) = p(A|B) \iff p(B) \times p(A) = p(B) \times p(A|B) = p(AB)$

by Bayes Theorem.


问:
Three numbers a, b and c are chosen at random, with replacement, from the numbers 1, 2, 3, ..., 99.

Three numbers a, b and c are chosen at random, with replacement, from the numbers 1, 2, 3, ..., 99.

The probability that $a^3+b^3+c^3-3abc$ is divisible by 3 is-

I can't really find a starting point to approach this question. Any help would be greatly appreciated.

答:
Fun question, lets break it down a little. Firstly I will introduce some modular arithmetic notation:

$x$ mod $y$ means the reminder when $x$ is divided by $y$.

For example: $25$ mod $7$ = $4$ as $25 = 3 \cdot 7 + \textbf{4} $

Another example: $19$ mod $3$ = $1$ as $19 = 6 \cdot 3 + \textbf{1} $

$\textbf{Property of Mods:}$

$x_1 + x_2 $ mod $y = x_1$ mod $y$ + $x_2$ mod $y$

Now onto mod $3$. There are $3$ seperate cases. Firstly $x$ mod $3$ can be $0 , 1 $ or $2$

$\textbf{Case 0} $

If $x$ mod $3 = 0$ then $x$ is a multiple of $3$ and can be written as $3k$ for some number $k$. And hence $x^3 = (3k)^3 = 27k^3 = 3y$. Where $y = 9k^3$. This is also a multiple of $3$ as $x^3 = 3y$

I.e if $x$ mod $3 = 0$ then $x^3$ mod $3 = 0$

$\textbf{Case 1} $

If $x$ mod $3 = 1$ then $x = 3k +1 $ for some integer k and hence $x^3 = (3k+1)^3 = 27k^3 + 27k^2 + 9k + 1  = 3(9k^3 +9k^2 + 3k) + 1$ and again if we let $y = 9k^3 +9k^2 + 3k$ then $x^3 = 3y + 1 $ and from our property of mods we have $x^3$ mod $3$ = $3y +1$ mod $3$ = $3y$ mod $3 + 1 $ mod $3$ = $ 0 + 1$ = 1. So to summarise: If $x$ mod $3 = 1$ then $x^3$ mod $3$ = $1$

$\textbf{Case 2} $

If $x$ mod $3 = 2$ then $x = 3k +2 $ for some integer k and hence $x^3 = (3k+2)^3 = 27k^3 + 54k^2 + 36k + 8  = 27k^3 + 54k^2 + 36k + 6 + 2 =  3(9k^3 + 18k^2 + 12k + 2) + 2$ and again if we let $y = 9k^3 + 18k^2 + 12k + 2$ then $x^3 = 3y + 2 $ and from our property of mods we have $x^3$ mod $3$ = $3y +2$ mod $3$ = $3y$ mod $3 + 2 $ mod $3$ = $ 0 + 2$ = 2. So to summarise: If $x$ mod $3 = 2$ then $x^3$ mod $3$ = $2$

Now this shows that $x$ mod $3$ = $x^3$ mod $3$ for all different choices of x.

Now onto your problem:

We want to find out what needs to happen for $a^3+b^3+c^3-3abc$ to be divisible by 3. This is the same as $a^3+b^3+c^3-3abc$ mod $3 = 0. $ Using our property of mods we have: $a^3+b^3+c^3-3abc$ mod $3$ =

$a^3$ mod $3$ + $b^3$ mod $3$ + $c^3$ mod $3$ - $3abc$ mod $3$.

Firstly as $3abc$ is a multiple of 3 we have $3abc$ mod $3$ = 0 so:

$a^3+b^3+c^3-3abc$ mod $3$  = $a^3$ mod $3$ + $b^3$ mod $3$ + $c^3$ mod $3$ But we know that $x^3$ mod $3$ =  $x$ mod $3$ so :

$a^3+b^3+c^3-3abc$ mod $3$ =  $a$ mod $3$ + $b$ mod $3$ + $c$ mod $3$ = $a + b + c $ mod $3$

Hence in order for $a^3+b^3+c^3-3abc$ to be divisible by 3 all we need is $a + b + c $  to be divisible by 3

If $a,b,c$ are all picked randomly from $1-99$ then a third of the numbers are exactly divisible by $3: 3,9,12,...,96,99.$ Another third are divisible with remainder 1: $1,4,7,...,94,97$ and the final third are divisible with reminder 2: $2,5,8,...,95,98$ and thus Probability( $a$ or $b$ or $c$ mod $3$ =$ 0 )$  = $\frac{1}{3} $and the Probability( $a$ or $b$ or $c$ mod $3$ =$ 1 )$  = $\frac{1}{3} $ and the Probability( $a$ or $b$ or $c$ mod $3$ =$ 2 )$  = $\frac{1}{3}.$

As $a + b + c$ mod $3$ = $a + b $ mod $3$ + $c$ mod $3$ and so in order to have $a + b + c$ mod $3$ = $0$ we can first pick $a$ and $b$ and not really care about what they are because there will always be one and exactly 1 value for $c$ mod $3$ that will ensure this value is $0$

i.e if  $a + b $ mod $3$ = $1$ then we just need a $c$ such that $c$ mod $3$ = $2$

or if $a + b $ mod $3$ = $0 $ then we just need a $c$ such that $c$ mod $3$ = $0 $

or if $a + b $ mod $3$ = $2 $ then we just need a $c$ such that $c$ mod $3$ = $1$

The probability of either of this required cases for $c$ is $\frac{1}{3} $

And thus the probability that $a^3+b^3+c^3-3abc$ is divisible by $3 = \frac{1}{3}$

Oskar :)


问:
Can you use modulus to make 0 > 2?
I wanted to create a rock-paper-scissors game that didn't use a lot of conditionals, and I was wondering if there were any mathematical way of representing the cycle of rock-paper-scissors. So Rock beats Paper beats Scissors beats Rock, or Rock > Paper > Scissors > Rock. If you assigned numbers to these, it wouldn't work: 0 > 1 > 2 > 0. But I was wondering if there were some way you could use modulus function to make this work? I understand that if you had a series of integers and modded them by 3 you would get a repeating cycle of 0, 1, 2, e.g. 60 % 3 = 0; 61 % 3 = 1; 62 % 3 = 2, 63 % 3 = 0, etc.

Do you think there could be any way to make this programmable using mods? I'm interested to know if there's an algorithm or some way to use this to create a cycle where 0>1>2>0. Just a warning, though, I don't study maths or know a lot about complex maths, so if you know of a way to do this, I'd really appreciate it if you tried not to overwhelm me with maths terms!

答:
Since there are only three options the simplest thing to do is an equality check rather than a less than sort of comparison, which is possibly what long tom meant.

Specifically, $x\equiv y+1$ modulo three if and only if $x$ beats $y$. Swap $x,y$ to check the other way round. If neither was true, it was a tie.

Title: Can you use modulus to make 0 > 2?


答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
There's not loss of generality if

I take the point $\ds{\color{red}{p}}$ as the origin of coordinates $\ds{\pars{~\mbox{i.e.}\ \pars{0,0}~}}$ and

the circle center $\ds{\color{red}{C}}$ on the positive-$\ds{\hat{x}}$-axis. Namely, the "circumference equation" is given by
$$
\pars{x - d}^{2} + y^{2} = r^{2}\,,\qquad d \equiv \on{dist}\pars{p,C} > r
$$

A "line equation", which is tangent to the circle,  is given by $\ds{y = \alpha x}$  where $\ds{\alpha}$ is determined by
$$
\pars{x - d}^{2} + \alpha^{2}\,x^{2} = r^{2} \implies
\pars{\alpha^{2} + 1}x^{2} - 2dx + d^{2} - r^{2} =0
$$

Since both values $\ds{x_{\pm}}$ of $\ds{x}$ must be equal ( by symmetry considerations ):
$$
\pars{-2d}^{2} - 4\pars{\alpha^{2} + 1}\pars{d^{2} - r^{2}} = 0 \implies
\alpha = \pm\,{r \over \root{d^{2} - r^{2}}}
$$
Both tangent lines to the circle span an angle =
$\ds{2\arctan\pars{r \over \root{d^{2} - r^{2}}}}$

Then, the answer is
\begin{align}
& \mbox{}
\\ &\
{2\arctan\pars{r/\root{d^{2} - r^{2}}} \over \pi} =
\bbx{{2 \over \pi}\,\arctan\pars{r \over \root{d^{2} - r^{2}}}} \\ &
\end{align}

Title: Probability of random line through a point intersecting a circle

Body:
Consider a circle of radius $r$ with center $C$ and let $p$ be a point outside the circle ( that is $\operatorname{dist}\left(p,C\right) > r$ ).

What is the probability that a random line through $p$ will intersect the circle $?$.

My intuition tells me it will be something like $r/\operatorname{dist}\left(p,C\right)$ but I would like a formal proof.


问:
How is the inequality $\|B_n(f)\|_{\infty} \leq \|f\|_{\infty}$ correct?
I am currently studying the sequence $(B_n(f))_{n=1}^{\infty}$ of Bernstein Polynomials $$(B_n(f))(x) := \sum_{k=0}^{n}f\left(\frac{k}{n}\right) {n \choose k}x^k(1-x)^{n-k},\quad \text{ where }\quad 0\leq x \leq 1,$$ and came across the following inequality in my analysis textbook: $$\|B_n(f)\|_{\infty} \leq \|f\|_{\infty} \tag{$*$}$$ I already showed that $|B_n(f)| \leq B_n(|f|)$ and $B_n(f) \geq 0$ whenever $f \geq 0$, however, how is the inequality $(*)$ correct? I feel like having shown the other two inequalities I should be able to deduce why $(*)$ is true $($as the author has not provided a proof$)$ but something is not clicking. Any suggestions are welcome.

Note: Showing $|B_n(f)| \leq B_n(|f|)$ and $B_n(f) \geq 0$ whenever $f \geq 0$ was an exercise in the textbook $-$ Showing $(*)$ is not as it was mentioned briefly in the book without proof.

答:
Note that $|B_n(f)(x)|\leq \|f\|_{\infty} \sum_{k=0}^n {n \choose k} x^k (1-x)^{n-k}$.

Now,
$$
1=1^n=(x+(1-x))^n=\sum_{k=0}^n {n\choose k} x^k(1-x)^{n-k},
$$
by the binomial identity.


问:
Application of Pythagorean Theorem in vectors in n-dimension space.
I'm attempting (and failing) at studying Real Analysis. I'm particularly stuck on this basic question about vectors, so if anyone could give me some hints with the algebra that be immensely appreciated. I'm also particularly bad with technology so I'll be posting a photo of my work typed into Microsoft Word.

Image of question and work performed

I have no idea how to progress past here. What are the algebraic steps for squaring a series? Is my attempt even remotely on the right path? Also note that the e vector represents the standard basis and not Euler's Number.

答: The $e_i$ of the standard basis are orthonormal. Since $\Vert x\Vert^2=\sum_{i=1}^jx_i^2$ while $\Vert y\Vert^2=x_{j+1}^2$, the result is $\sum_{i=1}^jx_i^2+x_{j+1}^2=\sum_{i=1}^{j+1}x_i^2$.


问:
Path on the torus

Prove that the following map is smooth and analyse its image as the real number $\alpha$ varies (how does it look like? Is it a submanifold?)

\begin{align*}
f_{\alpha}:\mathbb{R}&\to\mathbb{S}^1\times\mathbb{S}^1\\
t&\mapsto (e^{2\pi it}, e^{2\pi \alpha it})
\end{align*}

I've managed to prove that $f_{\alpha}$ is smooth and that $\text{Im}(f_{\alpha})$ is a closed curve on the torus $\Leftrightarrow \alpha\in\mathbb{Q}$. But I'm having trouble to formalize whether or not it is a submanifold of the torus. I actually don't know if it is true when the image is not a closed curve. How can I solve this?

答: Let $\alpha \in \mathbb{R}$ be irrational then we know the set $\{m+n\alpha|m,n \in \mathbb{Z} \}$ is dense in $ \mathbb{R}$. Let $x,y \in \mathbb{R}$ and for given $\delta >0 $ there exists $m,n\in \mathbb{Z}$ such that $|x \alpha-y+m+n\alpha|< \delta$. We have $||(e^{2 \pi ix},e^{2 \pi iy})-(e^{2 \pi i(x+n)},e^{2\pi i(x+n)\alpha} )||$=$|1-e^{2 \pi i(x \alpha-y +m+\alpha n)}| $. Since exponential function is continuous at zero for given any $\epsilon >0$ we can bound the quantity with that. Hence, it is dense.

Title: Path on the torus


问:
Is the direct image functor on sheaves of a point inclusion exact?
Given the inclusion of a point $i_x:\{x\}\to X$, is the associated direct image functor $(i_x)_*: Sh(\{x\}) \to Sh(X)$ an exact functor?

I know that in general this is not the case and the direct image functor is only left exact. I also know that if $\{x\}$ is a closed subset in X, then it should be an exact functor. So I was trying to construct a counterexample by looking at the inclusion of an open subset, but to be honest, I just lack the intuition for the subject to come up with something. Any hints?

答:
I actually think, that in case of single point it will always be exact. Let

$$ 0 \to \mathcal{F} \overset{\varphi}{\to} \mathcal{G} \overset{\psi}{\to} \mathcal{H} \to 0 $$

be exact, $\mathcal{F}, \mathcal{G}, \mathcal{H} \in Sh(\{x\})$. We only need to prove, that $(i_x)_*(\psi)$ is surjective. By definition, $ (i_x)_*\mathcal{F}(U) = \mathcal{F}(\{x\}) $, if $ x \in U $, otherwise it is just $ 0 $ (because $\mathcal{F}$ is a sheaf, so on empty set it gives 0). Same thing for $\mathcal{G}, \mathcal{H}$. Next, $ (i_x)_*\psi(U) = \psi(\{x\})$ (surjective, because $\psi(\{x\}) = \psi_x$ in case of one-pointed top. space), if $x \in U $, and trivial otherwise (also surjective, because between to trivial groups). So all $(i_x)_*\psi(U)$ are surjective, then $(i_x)_*\psi$ is surjective, and the sequence is exact.


问:
Understanding the construction of the smallest filtration making a random time a stopping time from Protter
I am struggling to understand a Lemma from Chapter 3 Section 5 of Protter's Stochastic Integration and Differential Equations.

Here, we take $\tau$ to be a random time on a complete probability space $(\Omega, \mathscr{F},P)$. Then set $\mathscr{F}_t^{00} = \sigma\{\tau \wedge t\}$, i.e. the smallest sigma algebra making $\tau \wedge t$ measurable. Then adjoin the null sets of $\mathscr{F}$ to get $\mathscr{F}_t^0$. Finally, we take $\mathscr{F}_t$ to be the right continuous version of $\mathscr{F}_u^0$. The Lemma states that the $\mathscr{F}_t$ so constructed forms the smallest filtration making $\tau$ a stopping time.

I am trying to understand why this is as they claim. So by definition, $\sigma \{ \tau \wedge t \}$ is generated by sets of the form $\{\tau \wedge t \ge r\}= \{\tau \ge r\} \cap [t,r] .$ From this, we can see that $\mathscr{F}_t^0$ is isomorphic to the Borel $\sigma$-algebra on $[0,t]$ together with the indivisible atom $(t,\infty)$. Hence indeed, we would have $\mathscr{F}_t^0 = \sigma\{ \tau \wedge u; u \le t\}$.

But how do we get the final identity for $E[Y| \mathscr{F}_t]$ from this? I cannot understand how we are able to replace the condition on $\mathscr{F}_t$ by $\tau$ here. It seems like we split $Y$ into $Y1_{\{t \ge \tau\}}+Y1_{\{\tau > t\}}$ but I cannot see why the conditional expectation of each given $\mathscr{F}_t$ are the expressions on the right. I would greatly appreciate some explanation on this.

答:
You know from your considerations that $\{\tau>t\}$ is an atom of $\mathcal{F}_t$. Since $\mathbb{E}(Y \mid \mathcal{F}_t)$ is $\mathcal{F}_t$-measurable, the random variable needs to be almost surely on the atom $\{\tau>t\}$, i.e. there is a constant $c \in \mathbb{R}$ such that

$$\mathbb{E}(Y \mid \mathcal{F}_t) =c \quad \text{a.s. on $\{\tau>t\}$}.$$

In order to determine $c$, note that, by the definition of the conditional expectation,

$$c \int_{\{\tau>t\}} \, d\mathbb{P} = \int_{\{\tau>t\}} \mathbb{E}(Y \mid \mathcal{F}_t) \, d\mathbb{P} = \int_{\{\tau>t\}} Y \, d\mathbb{P},$$

i.e.

$$c \,  \mathbb{P}(\tau>t) = \mathbb{E}(Y 1_{\{\tau>t\}}).$$ Dividing both sides by $\mathbb{P}(\tau>t)$ yields $c$, and we get

$$\mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau>t\}}= \frac{\mathbb{E}(Y 1_{\{\tau>t\}})}{\mathbb{P}(\tau>t)} 1_{\{\tau>t\}}. \tag{1}$$

It remains to determine the conditional expectation on $\{\tau \leq t\}$. To this end, note that

$$Z = \mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau \leq t\}} = \mathbb{E}(Y 1_{\{\tau \leq t\}} \mid \mathcal{F}_t),$$

is the $\mathcal{F}_t$-measurable random variable such that

$$\forall F \in \mathcal{F}_t\::\: \int_F Z \, d\mathbb{P} = \int_F (Y 1_{\{\tau \leq t\}}) \, d\mathbb{P}.  \tag{2} $$

Consequently, we are done if we can show that $Z=\mathbb{E}(Y \mid \tau) 1_{\{\tau \leq t\}}$ has the above properties. To this end, recall that by the factorization lemma $\mathbb{E}(Y \mid \tau) = h(\tau)$ for some measurable mapping $h:\mathbb{R}\to \mathbb{R}$. From

$$Z = h(\tau) 1_{\{\tau \leq t\}} = h(\tau \wedge t) 1_{\{\tau \leq t\}},$$

we see that $Z$ is $\mathcal{F}_t$-measurable (because $\{\tau \leq t\} \in \mathcal{F}_t$ and $\tau \wedge t$ is $\mathcal{F}_t$-measurable). It remains to check $(2)$. Note that $\mathcal{F}_t \subseteq \sigma(\tau)$. Hence, if $F \in \mathcal{F}_t$, then $F \cap \{\tau \leq t\} \in \mathcal{F}_t \subseteq \sigma(\tau)$, and so

$$\int_{F \cap \{\tau \leq t\}} \mathbb{E}(Y \mid \tau) \, d\mathbb{P} = \int_{F \cap \{\tau \leq t\}} Y \, d\mathbb{P}.$$

Equivalently,

$$\int_F Z \, d\mathbb{P} = \int_F 1_{\{\tau \leq t\}} Y \, d\mathbb{P},$$

which is exactly $(2)$. Combining both considerations, we conclude that

\begin{align*} \mathbb{E}(Y \mid \mathcal{F}_t) &=\mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau \leq t\}} + \mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau>t\}} \\ &= \mathbb{E}(Y \mid \tau) 1_{\{\tau \leq t\}} + \frac{\mathbb{E}(Y 1_{\{\tau>t\}})}{\mathbb{P}(\tau>t)} 1_{\{\tau>t\}}, \end{align*}

as claimed.


答:
By similarity, the probability for a point at distance $d$ of a circle of radius $r$ is indeed the same as for a point at distance $\dfrac dr$ of a unit circle, as only angles are involved. Hence the solution is indeed of the form

$$\phi\left(\frac rd\right).$$

As shown in the other answers, simple trigonometry gives the fraction of angles such that the line meets the circle.


问:
Do Boolean rings always have a unit element?

Let $(B, +, \cdot)$ be a non-trivial ring with the property that every $x \in B$ satisfies $x \cdot x = x$. How does one prove that such a ring $(B, +, \cdot)$ must have a unit element $1_B$? (Or, in case this is not true in general, what is a counterexample?)

BTW, I'm looking for an elementary proof, not requiring anything more than the definition of a ring, the definition of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

答: An example: the family of all finite subsets of a given infinite set.</p>


问:
What does the notation $\partial_{x,y} f$ mean?
For example, if I have a function $f \colon \mathbb{R}^4 \to \mathbb{R}^2$ such that $f(x,y,z,w)=(xyz+w^2,x-w-zy^2)$, what does it mean to find the following matrix:

$\partial_{p} f(p,q)$ where $p=(x,y)$ and $q=(z,w)$

My lecturer used this notation in his notes, but never really explained what it means or gave an example. Is it just supposed to be the Jacobian of $f$ but with the columns for $z$ and $w$ removed?

答:
Yes indeed this is just the notation for a 'part' of the Jacobian.

In general, if you have a function $f:\mathbb{R}^{d}\to\mathbb{R}^{k}$, then you define $\partial_{p}f$ with $p=(x_{1},\dots,x_{l})$ for some $l\in\{1,\dots,d\}$ as

$$\mathcal{J}_{f}(x)=:(\partial_{p}f(x),\partial_{q}f(x))$$

where $\mathcal{J}_{f}(x)$ denotes the Jacobian and $q=(x_{l+1},\dots,x_{d})$.

So in this general example $\partial_{p}f$ would be

$$\partial_{p}f(x):=\begin{pmatrix}\partial_{x_{1}}f_{1}(x) &\dots&\partial_{x_{l}}f_{1}(x)\\ \vdots&\ddots&\vdots \\\partial_{x_{1}}f_{k}(x) &\dots&\partial_{x_{l}}f_{k}(x)\end{pmatrix}$$


问:
Let $U$ and $V$ be two $n\times n$ unitary matrices such that $UV=\lambda VU$.Then prove that $\lambda^n=1$
Let $U$ and $V$ be two $n\times n$ unitary matrices such that $UV=\lambda VU$.Then prove that $\lambda^n=1$

Totally stuck on it.Can I get some help?

答:
Unitary is not needed, just non-singular. Eigenvalues are not needed either. Just use the multi-linearity and multiplicativity of the determinant:
$$
  \det(UV)=
  \det(\lambda VU)=\lambda^n\det(VU)=\lambda^n\det(V)\det(U)=\lambda^n\det(UV)
$$
so $\lambda^n=1$ since $\det(UV)\neq0$.</p>