问:
Trigonometry Airplane question. Finding bearing and distance.

A little background(if you don't care for my story, skip straight to the question): I've missed a few lectures from my teacher because I fell ill. Since I have no information to work with other than the bad notes a friend of mine has taken, I don't have much to work with. Now I'd love help with this problem because I've been at it for almost $3$ hours and nothing I try works. I don't need the answer at all, I have the answer, what I need is the process. I need to know how to get that answer.

**Now the question is-**An airplane travels due west for $1.5$ hours at $320$ mph. then it changes course to $S31°W$. Fine the airplane's distance from its point of departure and its bearing, after a  total flight time of $3.5$ hours.

What I know so far, sorry I wasn't as detailed as I should be. I have drawn it out, i know distance between from the starting point and when it changes direction being $480$. And the distance of the direction change to the end point which is $640$. I know the angle where the plane changes direction totals out to $121°$. I could be wrong with this information, but the point is I've taken many different approaches at this problem but have gotten no correct answers. I've taken it "outside of the box" and expanded the 480 line until i can drop it down to the end point and make the spot where those two lines meet a right angle, if that makes sense. I've used "sohcahtoa" and found out the lengths of that small right triangle, added it with the lengths of the large obtuse triangle to put it all together. The missing side is the hypotenuse and I used the Pythagorean theorem to find it, but the answer comes out wrong. I tried so many different values and processes, and nothing. For the missing angle I use the inverse of the tan function and got an answer for that, but it's also wrong. I'd appreciate any input, anything at all.

答: Draw the triangle. The first side is 480 miles long, the second is 640 miles long. You know the angle between the two sides. Then use the law of the cosines to figure out the length of the third side.</p>


问:
GMRES algorithm
Can you suggest me a reference (besides Wikipedia) where the GMRES (Generalized minimal residual method) algorithm is explained in full detail, in a nice and easy way to understand? A clearly written, easy to follow resource would be wonderful.

Thank you very much for your time!

答:
Read the book

Krylov solvers for linear algebraic system

Of the authors: Charles George Broyden and Maria Teresa Vespucci

This book treat all the algorithms that are the state of the art in linear system solution

ContentLicense: CC BY-SA 3.0

Title: GMRES algorithm

ContentLicense: CC BY-SA 4.0


问:
Prove/refute property about a constrained sequence of numbers
I need to prove or refute a property about a sequence of numbers.
Here is what is given to me:

Sequence ($a_1,a_2,...,a_k,a_{k+1},a_{k+2}$) containing $k+2$ numbers. Every number $0 < a_i \leq M, i=1,...,k+2$ for the same given constant $M$. Moreover, $\sum_{i=1}^{k+2} a_i = N$, for another given constant $N > 0$. The constants $N$ and $M$ are related by $kM \geq N, k > 1$.

Then, I need to prove or refute the following property: Can all consecutive pairs of numbers $a_i, a_{i+1}, i=1,...,k+1$ be defined such that $a_i + a_{i+1} > M$ ? Or does it lead to a violation of one of the given constraints?

I tried searching for something similar but in truth, I barely know what to search for. My tentative proofs do not really go anywhere meaningful, so I had to resort to more experienced math guys to help me with this. It has been a long time since I had to prove some property like this.

答:
If $a_i+a_{i+1}>M$ for all applicable $i$, then
$$ N=\sum_{i=1}^{k+2}a_i=\frac{a_1}2+\sum_{i=1}^{k+1}\frac{a_i+a_{i+1}}2+\frac{a_{k+2}}2>\frac{k+1}2M$$
and in fact
$$ N=\sum_{i=1}^{k+2}a_i=\sum_{j=1}^{\frac{k+2}2}(a_{2j-1}+a_{2j})>\frac{k+2}2M\qquad\text{if $k$ is even}.$$
Hence we certainly need the additional condition that
$$\tag1 \left\lceil\frac{k+1}2\right\rceil M<N.$$
In particular, $kM\ge N$ contradicts $(1)$ when $k=1$ or $k=2$. Hence we incidentally also need
$$\tag2 k\ge3 $$
(but as said, $(1)$ implies $(2)$ in the given context).
Finally, we also need
$$ \tag3 M>0$$
to allow for $0<a_i\le M$ in the first place.

On the other hand, assume we have $k\in\Bbb N$, $N,M\in \Bbb R$ such that $(1)$ and $(3)$ and $kM\ge N$. Then we can let
$$a_i=\frac N{k+2}. $$
By $(1)$ and $(3)$ and $N\le kM$, this makes
$$ 0<a_i<M.$$
We clearly have
$$ \sum a_i = (k+2)\cdot \frac N{k+2}=N$$
and by $(1)$,
$$ a_i+a_{i+1}=\frac{2N}{k+2}\ge\frac N{\left\lceil \frac{k+1}2\right\rceil}>M,$$
as desired.


问:
Yitang Zhang: Prime Gaps
Has anybody read Yitang Zhang's paper on prime gaps? Wired reports "$70$ million" at most, but I was wondering if the number was actually more specific.

*EDIT*$^1$:

Are there any experts here who can explain the proof? Is the outline in the annals the preprint or the full accepted paper?

答:
70 million is exactly what is mentioned in the abstract.

It is quite likely that this bound can be reduced; the author says so in the paper:

This result is, of course, not optimal. The condition $k_0 \ge 3.5 \times 10^6$ is also crude and there are certain ways to relax it. To replace the right side of (1.5) by a value as small as possible is an open problem that will not be discussed in this paper.

He seems to be holding his cards for the moment...

You can download a copy of the full accepted paper on the Annals page if your institution subscribes to the Annals.

Title: Yitang Zhang: Prime Gaps


问:
One element is given seperable and another purely inseperable then show that ...
Kindly help with the following question asked in my mid term exam which has been concluded.

If u $\in F$ is separable over K and c $\in F$ is purely inseparable over K, then
K(u,c) = K( u + c).

I am unable to get any idea on how this question will be solved. The irreducibility polynomial of u will be given by $x-u  \in K[x]$ and irreducibility polynomial of c will be given by $(x-c)^m \in F[x]$  for some m >1 , $m\in \mathbb{Z}$.

But I am unable to use it to prove K(u,c).  Let  I choose any polynomial in K(u+v) , so it will be given  by $ \frac{a_0 + a_1 (u+c) + a_2 (u+c)^2 +....}{ b_0 + b_1 (u+v) + b_2 (u+v)^2 ...}$ ,Now It is triavial to see that the polynomial also lies in K(u,v) .

If I choose a polynomial in K(u,v) then  I struggle to prove that it lies as K(u,v). I think here we have to use seperability but unable to use it.

So, Do u mind helping me? Also , I Hope my rest of argument is fine.

答:
We may assume $\operatorname{char}K=p>0$ (otherwise $c\in K$ and the statement is trivial). You already proved the inclusion $K(u+v)\subseteq K(u,c)$, for the other inclusion we only need to prove $u,c\in K(u+v)$. There is some $r\in\Bbb N$ with $c^{p^r}=d\in K$. Hence we have $(u+c)^{p^r}=u^{p^r}+d$, so $u^{p^r}\in K(u+c)$. Show

$K(u)$ is both separable and purely inseparable over $K(u^{p^r})$

$K(u)=K(u^{p^r})\subseteq K(u+c)$.

$u\in K(u+c)$

$c\in K(u+c)$

Conclude $K(u,c)\subseteq K(u+v)$ and then equality.


问:
Proving $1^k+2^k+ \dots + n^k = O(n^{k+1})$
I have to prove that : $1^k+2^k+ \dots + n^k = O(n^{k+1})$, where $k \in \mathbb{Z}^+$ . I followed a technique different from the book , and I want to check if my way is correct. I approach the problem using Induction.

Let $P(n)$ be the statement "$1^k+2^k+ \dots + n^k = \sum_{i = 1}^ni^k =  O(n^{k+1})$ ".

Induction basis: I will show that $P(1)$ is true. That equals: $1^k = O(1^{k+1})$. In other words, I have to prove that $\lim_{x\to \infty}\frac{1^x}{1^{x+1}} = L \geq 0$.

$\lim_{x\to \infty}\frac{1^x}{1^{x+1}} = \lim_{x\to \infty}\frac{1^x}{1^{x}1^1} = 1 \in \mathbb{R}$ so the Induction basis is proved.

Induction step: Let $P(l)$ be true, for some $l>1$. I will prove that $P(l) \rightarrow P(l+1)$.

As $P(l)$ is true, that means that $\sum_{i=1}^{l} = O(l^{k+1})$. Let's prove that $P(l+1)$ is true. I want to prove that $\sum_{i=1}^{l+1} = O((l+1)^{k+1})$ is true, so: $1^k + 2^k + \dots + l^k + (l+1)^k = O((l+1)^{k+1})$

I have to prove that: $\lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x+1}} = L \geq 0$

$
\begin{align}
\\
\lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x+1}}
 &= \lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^x \times (l+1)} \\
& = \frac{1}{l+1} \lim_{x \to \infty} \frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x}} \\
& = \frac{1}{l+1} [\lim_{x \to \infty} (\frac{1}{l+1})^x+ (\frac{2}{l+1})^x + \dots (\frac{l}{l+1})^x + 1 ]
\end{align}$ .

As $l > 1 \rightarrow l+1 > 2$ the above limit equals $0 + 0 + \dots + 0 + 1 = 1$. So $\frac{1}{1+l} \in \mathbb{R}$, and thus the induction step is proven, and the proof is concluded. $\blacksquare$

Is this way correct?

答:
Why not simply
$$\sum_{j=1}^nj^k\le \sum_{j=1}^nn^k =n\cdot n^k=n^{k+1}$$


问:
Tangent space of affine variety and set of $k$-morphisms

[Springer, Linear algebraic group, 4.1.9.(3)] Let $X$ be an affine algebraic variety. Let $k[\tau]=k[T] /\left(T^{2}\right)$ be the $k$-algebra of dual numbers ($\tau$ being the image of $T$). Show there is bijection of $T_{x} X$ onto the set of $k$-homomorphisms $\phi: k[X] \rightarrow k[\tau]$ such that $\phi(f)-f(x) \in k \tau$. (These are the '$k[\tau]$-valued points of $X$ lying over $x$'.)

$T_x X \cong (M_x/M_x^2)^*$, where $M_x\subset k[X]$ is the maximal ideal of functions vanishing in $x$.

Could you please help me with this problem? Thank you in advance!

答:
Given $\lambda \in (M_x/M_x^2)^*$, define $\phi(f) = f(x) + \lambda(f - f(x))\tau$ for all $f \in k[X]$.

Given $\phi \mathrel: k[X] \to k[\tau]$ satisfying your condition, define $\lambda(f) = \frac1\tau\phi(f)$ for all $f \in M_x$.


问:
Giving every solution of the equation $m^n=\overline{nAm}$ or proving there are no solutions to the equation
It should be explained that $\overline{nAm} = m + A\times10^{\mathrm{digit}(m)} + n\times 10^{\mathrm{digit}(m) + \mathrm{digit}(A+m)}$, where $\mathrm{digit}{(X)}$ is the number of digits of $X$.

My friends and I had been talking about a joke before we ran into a problem.

We know that $x^2=25$, so by deleting $2$ on both sides, we get $x = 5$, which is one of the solutions to the equation.

After that, we've been thinking about whether some other groups of solutions satisfy $m^n=\overline{nm}$ where $n,m\in\mathbf{N}_+$. We soon discovered that there are no solutions except $(m,n)=(5,2)$. We tried to make this question more interesting, so we decided to put some more numbers between $n$ and $m$. And therefore we tried to find solutions for $m^n=\overline{nAm}$, where $m, n\in\mathbf{N}_+$ and $A\in\mathbf{N}$.

We've looked around and tried multiple methods of this, but none of them seems to work. I would appreciate it if you give me your thoughts or solutions that point me the way.

答:
One way of looking at this problem:

You are looking for two numbers $m$ and $n$ such that, for $K$ the number of digits of $m$, we have:

$m^n\equiv m \pmod {10^K}$

$n=$ int$(m^n/10^L)$ for some $L\ge K$.

The first condition can be checked numerically fairly easily and then it seems a matter of chance whether or not the second condition holds.

Simple example

When $m=4$, all odd $n$ give a solution of the first condition. Then a quick use of a spreadsheet finds the solution
$$4^{17}=171798...4$$

This looks a suitable approach for a computer search.


问:
How to show that a valid rule is not derivable in intuitionistic propositional calculus.
First a word on notation, let the following be an inference rule that takes $\Gamma$ (a set of well-formed formulas) as premises and has $\psi$ (a well-formed formula) as a conclusion.

This is represented as

$$ \text{In:}\;\; \Gamma \\ \text{Out:}\;\; \psi $$

I'm using this two-line notation as a representation for an inference rule as a mathematical object, without ascribing any properties to it. By mentioning an inference rule in this way, I don't intend to make it part of the deductive system by fiat.

I want to show that the following rule is valid but not derivable in intuitionistic propositional calculus, since it is mentioned in passing in this paper, which I'm currently trying to read and make sense of.

$$ \text{In:} \;\; \lnot \alpha \to \beta \lor \gamma \\ \text{Out:} \;\;(\lnot\alpha \to\beta) \lor (\lnot\alpha\to\gamma) $$

Proving that it's valid seems fairly straightforward. I'm using the Heyting algebra semantics described here in an extremely naive way to show this. Let $A, B, G$ be subsets of $\mathbb{R}$ equipped with the standard topology. Let $E$ be defined as $\text{int}(A^c)$. Note that $E = \text{int}(E)$, i.e. $E$ is open. I'm pretty sure that showing that the conclusion is always a superset of the premise demonstrates the validity of the inference rule. The superset property seems like it would force the conclusion to be true when the premises are true, which is what we want out of an inference rule.

$$ \text{int}(E \cup B\cup G) \subset E \cup B \cup G $$
$$ \text{int}(E \cup B \cup G) \subset E \cup B \cup E \cup G $$
$$ \text{int}(E \cup B \cup G) \subset \text{int}(E) \cup B \cup \text{int}(E) \cup G $$
$$ \text{int}(E \cup B \cup G) \subset (\text{int}(E) \cup B) \cup (\text{int}(E) \cup G) $$

And now we unapply the interpretation, giving our rule as desired.

$$ \text{In:} \;\; \lnot \alpha \to \beta \lor \gamma \\ \text{Out:} \;\;(\lnot\alpha \to\beta) \lor (\lnot\alpha\to\gamma) $$

What's a good technique for showing that the rule is not derivable?

答:
Your translation to topological semantics is not quite right. $p\rightarrow q$ translates to $\text{int}(P^{c}\cup Q)$, not $\text{int}(P\cup Q)$. So what you wanted to show was $$\text{int}(\text{cl}(A)\cup B\cup G) \subseteq \text{int}(\text{cl}(A)\cup B) \cup \text{int}(\text{cl}(A)\cup G),$$ using the fact that $E^{c} = (\text{int}(A^c))^c = \text{cl}(A)$. Your proof doesn't work, since it relied on openness of $E$, and $\text{cl}(A)$ is not open in general.

In fact, this inclusion is not true in general. Take $A = \bigcup_{n\in \mathbb{N}^+} (\frac{1}{2n+1},\frac{1}{2n})$, $B = (-\infty,0)$, and $G = (0,\infty)$. Then $0\in \text{int}(\text{cl}(A)\cup B\cup G) = \mathbb{R}$, but $0\notin \text{int}(\text{cl}(A)\cup B) \cup \text{int}(\text{cl}(A)\cup G)$.

Ok, but what does this actually show? It shows that when $\alpha$, $\beta$, and $\gamma$ are propositional constants, $\lnot \alpha\rightarrow (\beta\lor \gamma)\not\models (\lnot \alpha\rightarrow \beta)\lor(\lnot \alpha\rightarrow \gamma)$. Equivalently, by completeness of IPL, $$\not\vdash (\lnot \alpha\rightarrow (\beta\lor \gamma))\rightarrow ((\lnot \alpha\rightarrow \beta)\lor(\lnot \alpha\rightarrow \gamma)).$$ That is, in the language of the paper you linked to, this rule is not derivable.

To show the rule is admissible means something different: Whenever we substitute sentences $\varphi_\alpha$, $\varphi_\beta$, and $\varphi_\gamma$ for $\alpha$, $\beta$, and $\gamma$, respectively, we have $$\text{If }\vdash \lnot \varphi_\alpha\rightarrow (\varphi_\beta\lor \varphi_\gamma),\text{ then }\vdash (\lnot \varphi_\alpha\rightarrow \varphi_\beta)\lor(\lnot \varphi_\alpha\rightarrow \varphi_\gamma).$$
Usually, to show that a rule is admissible, one has to do some induction on proofs. It seems that the original reference for admissibility of this rule is Theorem 3.1 in this paper (Ronald Harrop, Concerning Formulas of the Types $A\rightarrow B\lor C$, $A\rightarrow (Ex)B(x)$, J. Symbolic Logic Volume 25, Issue 1 (1960), 27-32; jstor.)


问:
Find the derivative of $f(x)=\text{sin}(x^2)\text{ln}(x)$.
I have the following task:

Find the derivative and state your answer in the simplest form: $$f(x) =\text{sin}(x^2)\text{ln}(x)$$

Here's my attempt:

\begin{align}
f'(x)&=\frac{d}{dx}\left(\sin \left(x^2\right)\right)\ln \left(x\right)+\frac{d}{dx}\left(\ln \left(x\right)\right)\sin \left(x^2\right) \\
&= \cos \left(x^2\right)\cdot \:2x\ln \left(x\right)+\frac{1}{x}\sin \left(x^2\right) \\
&= 2x\cos \left(x^2\right)\ln \left(x\right)+\frac{\sin \left(x^2\right)}{x}
\end{align}

I'm pretty sure that the answer is correct but I'm not sure if this is the simplest form. From a glance it does look like it can't be simplified anymore but obviously I could be wrong.

答:
You apply the product and chain rules correctly. There is not much to simplify further.

Usually, you can do simplification for algebraic expressions, but not much with transcendental functions.


问:
If $a+b+c=abc$ then $\sum\limits_{cyc}\frac{1}{7a+b}\leq\frac{\sqrt3}{8}$
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=abc$. Prove that:
$$\frac{1}{7a+b}+\frac{1}{7b+c}+\frac{1}{7c+a}\leq\frac{\sqrt3}{8}$$

I tried C-S:
$$\left(\sum_{cyc}\frac{1}{7a+b}\right)^2\leq\sum_{cyc}\frac{1}{(ka+mb+c)(7a+b)^2}\sum_{cyc}(ka+mb+c)=$$
$$=\sum_{cyc}\frac{(k+m+1)(a+b+c)}{(ka+mb+c)(7a+b)^2}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{k+m+1}{(ka+mb+c)(7a+b)^2}\leq\frac{3}{64abc},$$
but I did not find non-negative values of $k$ and $m$, for which the last inequality is true.

If we replace $7$ with $8$ so for $(a,b,c)||(28,1,5)$ this inequality would be wrong. Around this point the starting inequality is true, but we see that we can'not free use AM-GM because in AM-GM the equality occurs, when all variables are equal.

Thank you!

Body:
My second solution:

Remark: Using the Buffalo Way as well, but it is simpler.

Use @chenbai's substitutions $x=\frac{\sqrt{3}}{a}, y=\frac{\sqrt{3}}{b}, z=\frac{\sqrt{3}}{c}$.
We have $xy+yz+zx=3$. We need to prove that
$$\frac{xy}{x + 7y}+\frac{yz}{y + 7z}+\frac{zx}{z+7x}\leq\frac{3}{8}.$$

After homogenization, it suffices to prove that
$$\Big(\frac{xy}{x + 7y}+\frac{yz}{y + 7z}+\frac{zx}{z+7x}\Big)^2 \leq \frac{9}{64}\frac{xy+yz+zx}{3}$$
or (clearing the denominators)
$$f(x, y, z) \ge 0$$
where $f(x, y, z)$ is a homogeneous polynomial.

WLOG, assume $z = \min(x, y, z)$. Let $x = z + s, y = z + t$ for $s, t\ge 0$. We have
$$f(z + s, z + t, z) = q_6z^6 + q_5 z^5 + q_4 z^4 + q_3 z^3 + q_2 z^2 + q_1 z + q_0$$
where
\begin{align*}
q_6 &= 81920 s^2-81920 s t+81920 t^2, \\
q_5 &= 106496 s^3-107520 s^2 t+279552 s t^2+106496 t^3, \\
q_4 &= 49664 s^4-155648 s^3 t+312832 s^2 t^2+489472 s t^3+49664 t^4, \\
q_3 &= 18816 s^5-97904 s^4 t-39200 s^3 t^2+650272 s^2 t^3+202480 s t^4+18816 t^5,\\
q_2 &= 14112 s^5 t-135143 s^4 t^2+239198 s^3 t^3+244873 s^2 t^4+42336 s t^5, \\
q_1 &= 2646 s^5 t^2-23590 s^4 t^3+91322 s^3 t^4+30870 s^2 t^5, \\
q_0 &= 147 s^5 t^3-1078 s^4 t^4+7203 s^3 t^5.
\end{align*}
It is easy to prove that $q_5, q_4, q_3, q_1, q_0 \ge 0$. It suffices to prove that $4q_3 q_1 \ge q_2^2$ or
\begin{align}
&s^2 t^3\Big(1002582336 s^7-9318167153 s^6 t+34882209252 s^5 t^2-77849923910 s^4 t^3\\
&\quad +108085842468 s^3 t^4+72267463855 s^2 t^5+11141602752 s t^6+531062784 t^7\Big) \ge 0
\end{align}
which is true.

We are done.


问:
Proof of the Theorem of sign permanence
If I have a metric space with the Euclidean metric $(X,d)$ and a continuous function $f:X\to\mathbb{R}$, assumed that exists a point $x\in X$ such that $f(x)>0$ can I say that exists $\delta>0:f(y)>0,
 \forall(y)\in B(x,\delta)$, if yes why?

答:
Hint:

That follows from the continuity of $f$.

Prove by contradiction: assume that is not true and see what happens.

Notes.

If the conclusion if not true, then there exists (why? exercise!) a sequence $y_n\in B(x,\frac{1}{n})$ such that $f(y_n)\le 0$. Now think about the limit of $(y_n)$ and $(f(y_n))$.

Title: Proof of the Theorem of sign permanence


问:
How to find Laurent series?
I have a function $f(z)={z^2}/\sqrt{1+z^2}$. I need to expand it neighborhood of $z=\infty$ Here i made used $z^2+1=u^2$ then i had $f(u)=u-1/u$ but these substition seems useless to me. However, i cannot find any way to expand this function. Any suggestions?

答:
I expanded at $x=0$ the function
$$\frac{1}{ \sqrt{x^2+1}}=\sum _{n=0}^{\infty } \frac{x^{2 n} \left((-1)^n \Gamma \left(n+\frac{1}{2}\right)\right)}{\sqrt{\pi } n!}$$
Then I divided both sides by $x$
$$\frac{1}{ x\sqrt{x^2+1}}=\sum _{n=0}^{\infty } \frac{x^{2 n-1} \left((-1)^n \Gamma \left(n+\frac{1}{2}\right)\right)}{\sqrt{\pi } n!}$$
And finally  substituted $x=\frac{1}{z}$
$$\frac{z^2}{\sqrt{z^2+1}}=\sum _{n=0}^{\infty } \frac{z^{1-2 n} \left((-1)^n \Gamma \left(n+\frac{1}{2}\right)\right)}{\sqrt{\pi } n!}$$
Hope this helps

Title: How to find Laurent series?


问:
The probability of horse A winning a race is $\frac 13$ and that of B is $\frac 15$

The probability of horse A winning a race is $\frac 13$ and that of B is $\frac 15$. Find the probability that either of them will win

The two cases are where A wins and B loses and A loses and B wins

That gives
$$\frac {4}{15} + \frac{2}{15}$$
$$=0.4$$

But given ans is $0.533$

Where am I going wrong

答: The two events are mutually exclusive. For mutually exclusive events $A$ and $B$, the probability that $A$ or $B$ will occur is the sum of the probabilities $P[A]+P[B]$. Just add $\frac{1}{3}+\frac{1}{5}=0533..$


答:
Each term in the sum is $\le n^k$, and there are $n$ terms.  So clearly
$$1^k + 2^k + \ldots + n^k \le n \cdot n^k = n^{k+1}.$$

Your proof appears to correctly show that the sum grows as $n^{k+1}$ as $k\rightarrow\infty$ for each fixed value of $n$.  This isn’t what you were asked to show: you want to prove that the sum grows (no faster than) $n^{k+1}$ as $n\rightarrow\infty$ for each fixed $k$..... and in general you can’t conclude one of these facts from the other.  (For instance, which grows faster, $n^k$ or $k^n$?  The answer depends on which variable is fixed and which is becoming large.)

Title: Proving $1^k+2^k+ \dots + n^k = O(n^{k+1})$

Body:
I have to prove that : $1^k+2^k+ \dots + n^k = O(n^{k+1})$, where $k \in \mathbb{Z}^+$ . I followed a technique different from the book , and I want to check if my way is correct. I approach the problem using Induction.

Let $P(n)$ be the statement "$1^k+2^k+ \dots + n^k = \sum_{i = 1}^ni^k =  O(n^{k+1})$ ".

Induction basis: I will show that $P(1)$ is true. That equals: $1^k = O(1^{k+1})$. In other words, I have to prove that $\lim_{x\to \infty}\frac{1^x}{1^{x+1}} = L \geq 0$.

$\lim_{x\to \infty}\frac{1^x}{1^{x+1}} = \lim_{x\to \infty}\frac{1^x}{1^{x}1^1} = 1 \in \mathbb{R}$ so the Induction basis is proved.

Induction step: Let $P(l)$ be true, for some $l>1$. I will prove that $P(l) \rightarrow P(l+1)$.

As $P(l)$ is true, that means that $\sum_{i=1}^{l} = O(l^{k+1})$. Let's prove that $P(l+1)$ is true. I want to prove that $\sum_{i=1}^{l+1} = O((l+1)^{k+1})$ is true, so: $1^k + 2^k + \dots + l^k + (l+1)^k = O((l+1)^{k+1})$

I have to prove that: $\lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x+1}} = L \geq 0$

$
\begin{align}
\\
\lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x+1}}
 &= \lim_{x \to \infty}\frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^x \times (l+1)} \\
& = \frac{1}{l+1} \lim_{x \to \infty} \frac{ 1^x + 2^x + \dots + l^x + (l+1)^x  }{(l+1)^{x}} \\
& = \frac{1}{l+1} [\lim_{x \to \infty} (\frac{1}{l+1})^x+ (\frac{2}{l+1})^x + \dots (\frac{l}{l+1})^x + 1 ]
\end{align}$ .

As $l > 1 \rightarrow l+1 > 2$ the above limit equals $0 + 0 + \dots + 0 + 1 = 1$. So $\frac{1}{1+l} \in \mathbb{R}$, and thus the induction step is proven, and the proof is concluded. $\blacksquare$

Is this way correct?


问:
Show that $H$ is a normal subgroup of $G$.

Let $N$ be a normal subgroup of $G$. Let $H$ be the set of all elements $h$ of $G$ such
that $hn = nh$ for all $n \in N$. Show that $H$ is a normal subgroup of $G$.

My attempt: I've been trying to do this exercise and so far I've been able to show that $H$ is a normal subgroup of $N$ (the proof is similar to showing that the center is a subgroup), but that $H$ is a normal subgroup of $N$ does not always imply that $H$ be a normal subgroup of $G$. How could I show in this case that $H$ is indeed a normal subgroup of $G$?

答:
You need to start by showing $H\le G$.

I will use the two-step subgroup test.

Since for any $n\in N$, we have $en=n=ne$, so $e\in H$. Hence $H\neq\varnothing$.

By definition, since $H$ consists only of elements of $G$, we have $H\subseteq G$.

Let $h,k\in H$. Then $nh=hn$ and $nk=kn$, so $$k^{-1}n^{-1}=(nk)^{-1}=(kn)^{-1}=n^{-1}k^{-1},$$ but, multiplying by $n$ on the left and right, we have $nk^{-1}=k^{-1}n$, so $k^{-1}\in H$. Also,

$$\begin{align}
n(hk)&=(nh)k\\
&=(hn)k\\
&=h(nk)\\
&=h(kn)\\
&=(hk)n,
\end{align}$$

so $hk\in H$.

Hence $H\le G$.

For normality, for $n\in N$, $g\in G$, $h\in H$, we have

$$\begin{align}(ghg^{-1})n(ghg^{-1})^{-1}&=(ghg^{-1})n(gh^{-1}g^{-1})\\
&=g(h(g^{-1}ng)h^{-1})g^{-1}\\
&=g(g^{-1}ng)g^{-1}\\
&=(gg^{-1})n(gg^{-1})\\
&=n
\end{align}$$

by definition of $H$ and $g^{-1}ng\in N\unlhd G$. But then $(ghg^{-1})n=n(ghg^{-1})$. Hence $ghg^{-1}\in H$.

Hence $H\unlhd G$.


答:
So you have the probabilities:

$\frac{1}{3}$ A wins, $\frac{2}{3}$ A loses.

$\frac{1}{5}$ B wins, $\frac{4}{5}$ B loses.

But notice how, since these event are EXCLUSIVE (that means that $A$ winning implies $B$ not winning, to be said, being $\frac{1}{3}$ is equal to $P$[$A$ wins $\cap$ $B$ loses]), you don't have to multiply them, so you just sum them up, obtaining:
$$\frac{1}{3}+\frac{1}{5}=\boxed{\frac{8}{15} =0.5\bar{3}}$$

Body:
The probability of horse A winning a race is $\frac 13$ and that of B is $\frac 15$. Find the probability that either of them will win

The two cases are where A wins and B loses and A loses and B wins

That gives
$$\frac {4}{15} + \frac{2}{15}$$
$$=0.4$$

But given ans is $0.533$

Where am I going wrong


问:
What does Arg {Inf I(d)} means?
I am currently studying the phase-field method for fracture modeling. In an article by Miehe -"Thermodynamically consistent phase-field models of fracture: Variational principles and multi-field FE implementation", I came across on an identity that I don't understand basically it says:
$$
d = \mbox{Arg}\left\{\inf_{d \in W} I(d)\right\}.
$$
This expression came from the Euler type differential equation
$$
d(x) - l^2d''(x) = 0,
$$
which has the solution as an exponential function
$$
d(x)\ = \ \exp(-|x|/l)
$$
Variational principle of this differential equation is ,
$$
d = \mbox{Arg}\left\{\inf_{d \in W} I(d)\right\}.
$$
with $$ W = \{d\ |\ d(0)=1,\ d(\pm \infty)=0\}$$
where I(d) is a functional defined as an integral
$$
I(d) = \frac{1}{2}\int [d^2 + l^2d'^2] dV,
$$
and $l$ is a parameter.

Can anyone tell me what does that mean?

答: It returns the argument that makes the functional $I\equiv I(d)$ infimum such that the function $d$ belongs to the space of kinematically acceptable functions $W$, i.e., it satisfies the prescribed boundary conditions. It just explicitly emphasises that we are after the function (or argument) that makes the functional infimum and not the infimum value itself.

Title: What does Arg {Inf I(d)} means?


答: Because $f : X \to \Bbb{R}$ is continuous, if $A$ is any open subset of $\Bbb{R}$, the inverse image $f^{-1}[A]$ is an open subset of $X$. If $f(x) > 0$, put $\alpha = f(x)/2$ (or any positive real with $\alpha < f(x)$) and let $A$ be the open interval $(f(x) - \alpha, f(x) + \alpha)$. Then $Y = f^{-1}[A]$ is an open subset of $X$ such that $f(y) > 0$ for every $y \in Y$. Because $Y$ is an open subset of the metric space $X$ and $x \in Y$, $Y$ contains the open ball $B(x, \delta)$ for all sufficiently small $\delta > 0$.

Title: Proof of the Theorem of sign permanence

Body:
If I have a metric space with the Euclidean metric $(X,d)$ and a continuous function $f:X\to\mathbb{R}$, assumed that exists a point $x\in X$ such that $f(x)>0$ can I say that exists $\delta>0:f(y)>0,
 \forall(y)\in B(x,\delta)$, if yes why?


答:
By contradiction if $\forall\delta>0$ there exists $y\in B(x,\delta)$ such that $f(y)=0$ $\Longrightarrow$ then let $B_n:= B(x,1/n)$ so you can pick a $x_n \in B_n$ $\forall n\in \mathbb{N}$; you would have that $x_n \to x$ and, since $f$ is continuous $f(x_n) \to f(x)>0$, wich means $0\to f(x)>0$ which is impossible.

The fact that $x_n\to x$ implies $f(x_n) \to f(x)$ come from the fact that since $F$ is continuous, for all open neighborhod $U$ of $f(x)$, $x\in f^{-1}(U)$ is open in $X$ so it contains a $B_N$ for $N$ sufficiently large, so that $f(x_N)\in U$.

Title: Proof of the Theorem of sign permanence


答:
The claim $$\sum_{k=1}^n k^r=O(n^{r+1})\qquad(n\to\infty)\tag{1}$$ cannot be proven by induction with respect to $n$, because $n$ is a dummy variable which is burnt after the claim has been spoken out, because we have said $n\to\infty$ within the claim. That $(1)$ is true for $n:=5$ is absurd: Write $5$ for $n$ in the formula, and you obtain
$$1^r+2^r+3^r+4^r+5^r=O(5^{r+1})\qquad(5\to\infty)\ .$$
We all know from high school that
$$\eqalign{\sum_{k=1}^n k^0&=n=O(n^1),\cr
\sum_{k=1}^n k^1&={n(n+1)\over2}=O(n^2),\cr
\sum_{k=1}^n k^2&={n(n+1)(2n+1)\over6}=O(n^3)\ .\cr}$$
Following this one could maybe set up an induction proof with respect to the variable $r$ in $(1)$.


问:
Writing a mathematical statement: $n$ is a natural number bigger or equal to two
I want to know how to write: '$n$ is a natural number bigger or equal to two' in mathematical terms. My thoughts:

$$n\in\mathbb{N} \setminus \{0,1\}\tag1$$

$$n\in\mathbb{N}_{\ge2}\tag2$$

$$ n\in\mathbb{N}\space\wedge\space n\ge2\tag3$$

Which one is the best? Or am I wrong about this.

答: I think "$n\in\mathbb{N}$ and $n\ge2$" would be the best compromise, maybe also "$n\in\mathbb{N}$ such that $n\ge2$", which can be abbreviated as "$n\in\mathbb{N}$ s.t. $n\ge2$"


问:
Show that $L(f)=\sup\{L(P,f):P \in \mathcal P_{c}\}$,where $c\in \left(a,b\right)$
Assume $\mathcal P_{[a,b]}$ is the set of all partitions of $[a,b]$ and  $\mathcal P_{c}$ is the set of all partitions of $[a,b]$ containing $c$,where $c\in  \left(a,b\right)$,if $f:I \to \mathbb R$ is a bounded function over $I$,show that $$L(f)=\sup\{L(P,f):P \in \mathcal P_{c}\}$$

I know that

$$-M(b-a) \le L(P,f)  \le L(f) \le U(f) \le U(P,f)\le M(b-a)$$

Where $M$ is a real number such that $\forall x \in [a,b]:-M \le f(x) \le M$

And $$ U\left(P_{},f\right)=\sum_{i=0}^{n-1}M_{i}\Delta x_i$$

$$ L\left(P_{},f\right)=\sum_{i=0}^{n-1}m_{i}\Delta x_i$$

With $M_{i}=\sup \{f(x):x \in [x_{i},x_{i+1}]\}$ and $m_{i}=\inf \{f(x):x \in [x_{i},x_{i+1}]\}$,$$U(f)=\inf \{ U\left(P,f\right):P \in \mathcal P_{[a,b]} \}$$
$$L(f)=\sup \{ L\left(P,f\right):P \in \mathcal P_{[a,b]} \}$$

So I used the fact that $f$ is  bounded on $I$,but I don't know how to use $c$ to conclude the desired answer.

答:
The conclusion has nothing to do with the upper sum.

Hint.

Prove two inequalities:
$$L(f)=\sup\{L(P,f):P \in \mathcal P_{[a,b]}\}\le\sup\{L(P,f):P \in \mathcal P_{c}\}\tag{1}
$$
$$L(f)=\sup\{L(P,f):P \in \mathcal P_{[a,b]}\}\ge\sup\{L(P,f):P \in \mathcal P_{c}\}\tag{2}
$$

Note that (2) is trivial since $\mathcal{P}_c\subset \mathcal{P}_{[a,b]}$.

To show (1), note that for any partition $P$ of $[a,b]$ that is not in $\mathcal{P}_c$, you can always add a point $c$ to get a new partition $P'\in\mathcal{P}_c$. Now compare $L(P,f)$ and $L(P',f)$.


问:
Evaluating the matrix polynomial
I have a question that goes as follows:

Let $\alpha = \frac{\pi}{5}$ and $A=
    \begin{bmatrix}
    \cos \alpha & \sin \alpha \\
    -\sin \alpha & \cos \alpha \\    \end{bmatrix}
$
and let $B=A^4-A^3+A^2-A$

Then which of the following is true?
(i)B is singular (ii) B is non singular (iii) B is symmetric (iv) The determinant $|B| =1$

There may be multiple correct options.

Now here's what I did, I evaluated $A^2$ and $A^3$ and then it is pretty easy to conclude by induction that $$A^n = \begin{bmatrix}
    \cos n\alpha & \sin n\alpha \\
    -\sin n\alpha & \cos n\alpha \\    \end{bmatrix}$$

From this, I tried to find out $B$ by putting the values of $A^4, A^3, A^2, A$ and then applied the transformation formulae $\bigg($like $\sin C + \sin D=$ $2\sin\bigg( \frac{C+D}{2} \bigg)\cos\bigg( \frac{C-D}{2}$ $\bigg)$, $...$ etc$\bigg)$

By simplifying, and using the fact that $\alpha = \frac{\pi}{5}$ I got the matrix B as follows:

$$B = \begin{bmatrix}
    -4\cos \alpha \sin \frac{\alpha}{2} & 0 \\
   0 & -4\cos \alpha \sin \frac{\alpha}{2} \\    \end{bmatrix}$$

So, clearly, the matrix is symmetric, which makes (iii) true. But, how do I check whether the matrix is non-singular or not, and how do I check whether $|B|=1$? Using the manipulations that I performed, I seem to not be able to check those options, any help would be nice.

The answer is given as (ii), (iii), (iv) are true.

答:
One way to think of this is that you are essentially computing a geometric sum associated to $-A$.  As usual, we have that $(-A - 1)B = -A^5 + A = 1 + A$.  Since $-A - 1$ is non-singular (as you can compute by finding the determinant, by computing its eigenvalues, or just by explicitly writing down the inverse of a $2\times2$ matrix), we see that $B = -1$.  Sure enough, this is symmetric and has determinant $1$.

It may help in this case to think that we are essentially computing with complex numbers; the ring of complex numbers $\mathbb C$ embeds in the ring $\operatorname M_{2\times2}(\mathbb R)$ of $2\times2$ real matrices by $a + b i \mapsto \begin{pmatrix} a & b \\ -b & a \end{pmatrix}$ for all $a, b \in \mathbb R$.  In particular, you are really computing a sum inside $\mathbb C$ involving $z = \cos(\alpha) + i\sin(\alpha) = e^{i\alpha}$.  In this context, the determinant on $\operatorname M_{2\times2}(\mathbb R)$ pulls back to the squared-norm on $\mathbb C$, in the sense that
$$
\det \begin{pmatrix} a & b \\ -b & a \end{pmatrix} = a^2 + b^2 = \lvert a + b i\rvert^2
$$
for all $a, b \in \mathbb R$; and the transpose on $\operatorname M_{2\times 2}(\mathbb R)$ pulls back to complex conjugation on $\mathbb C$, in the sense that
$$
a + b i \mapsto \begin{pmatrix} a & b \\ -b & a \end{pmatrix}
\quad\text{and}\quad
\overline{a + b i} = a - b i \mapsto \begin{pmatrix} a & -b \\ b & a \end{pmatrix} = \begin{pmatrix} a & b \\ -b & a \end{pmatrix}^{\mathsf T}
$$
for all $a, b \in \mathbb R$ (and remember that $\overline{e^{i\theta}}$ equals $e^{-i\theta}$ for all $\theta \in \mathbb R$); so asking whether the alternating sum of matrices is symmetric, respectively has determinant $1$, is the same as asking whether the corresponding alternating sum of complex numbers is real, respectively lies on the unit circle centred at the origin.  Finally, since $\mathbb C$ is a field, every non-$0$ matrix in the image of $\mathbb C$ is invertible (with the inverse of the image of $z = a + b i$ being the image of $z^{-1} = \frac a{a^2 + b^2} - \frac b{a^2 + b^2}i$ for all $(a, b) \in \mathbb R^2 \setminus \{(0, 0)\}$); this is yet another way to see that $-A - I_2$ is non-singular.

Title: Evaluating the matrix polynomial


问:
Proof that the cartesian plane is an incidence geometry with only vector definition of a line.
I can get up to showing that it is an abstract geometry, but I cannot figure out how to show that for every two points, there is a unique line. The definition of a line in vector form is given
$L (AB) = \{X \in R \times R~ | ~X=A+t(B-A) ~\text{for some}~~ t \in \mathbb R\}$

$A,B,X$ are points.

I am trying to go about this by supposing A,B are in $L (CD)$ and $L (XY)$. From there I try to get $C=X$ and $D=Y$, but I can't do this. Any help or hint is appreciated.

The Cartesian Plane is the set of Points in $R \times R$ and the set of all lines $L (AB)$.

This is a problem in a textbook but I am not doing this for homework. I just find this subject really interesting so any help is appreciated.

答:
No, you can't get $C=X$ and $D=Y$. But you should be able to show that if $A,B\in L(CD)$ and $A,B\in L(XY)$, then $L(CD)\subset L(XY)$ (and so, symmetrically, $L(XY)\subset L(CD)$), whence the lines are equal. Here's a start:
\begin{align*}
A &= C + s(D-C) \quad\text{ for some } s\in\mathbb R\,; \\
B &= C + t(D-C) \quad\text{ for some } t\in\mathbb R\,; \\
A &= X + u(Y-X) \quad\text{ for some } u\in\mathbb R\,; \text{ and} \\
B &= X + v(Y-X) \quad\text{ for some } v\in\mathbb R\,.
\end{align*}
From this you should be able to show that $C\in L(XY)$ and likewise for $D$.</p>


问:
Matrices equation with one unknown matrix
Let us consider following equation:
$$
2X^2+2X=
\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right).
$$
I have to show that there is no real matrix which satisfy my equation.
I consider three diffrent way to do this.

Direct method. Let $$X=\left( \begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i
\end{array} \right).
$$ Then we can write euqation in equivalent form $X(X+I)=\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right)$, and we obtain system of nonlinear equations
$$
\left\{ \begin{array}{llll}
a^2+a+bd+cg=-\frac{1}{2} \\
ab+be+b+ch=\frac{5}{2}\\
ac+bf+ci+c=\frac{3}{2}\\
da+d+ed+fg=-1\\
db+e^2+e+fh=\frac{1}{2}\\
dc+ef+fi+f=1\\
ga+g+hd+ig=0\\
gb+he+h+ih=-2\\
gc+hf+i^2+i=-\frac{3}{2}
\end{array} \right.
$$
I was trying to deal with it, but it was impossible to solve for me (I can not use Mathematica, Matlab,etc)

Determinant method. From the fact, that $X(X+I)=\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right)$, we have that $\det X \det (X+I)=-\frac{11}{8}$, and then using direct method i was traing to write formula for $\det (X+I)$ in terms of $\det X$, but here also appear formulas,to difficult for me to interpret it.

My last idea to solve is to use formula for inverse matrix $3\times3$  ($X$ is invertible, which is consequence of the fact, that $\det X\neq 0$) and from relation $X+I=X^{-1}\frac{1}{2}\left( \begin{array}{ccc}
-1 & 5 & 3 \\
-2 & 1 & 2 \\
0 & -4 &  -3
\end{array} \right)$ we also obtain system of equations, which is much more complicated, that this previous one.

I would be grateful if you give me some hints.

答:
If there were a real matrix $X$ satisfying ...

Scalar-multiply the given equation with $2$ and add the identity matrix to get
$$(2X+\mathbb 1)^2 \;=\; \begin{pmatrix} -1& 10& 6\\
-4& 3& 4\\
0& -8& -5 \end{pmatrix}$$
Apply the determinant which produces $\,-25\,$ on the RHS, hence a contradiction because the resulting square on the LHS cannot get negative if $X$ is a real matrix.


问:
Transitive action of a lie group in a connected manifold
I am trying to see why is it that if we have a transitive action of a lie group $G\rightarrow M$ then if this action is transitive the connected component of the identity $G^0$ also acts transitively on $M$ and that for all $p\in M$, $G/G_0 \cong G_p/(G_p\cap G^0)$.

Since the action is transitive we know that $G/G_p\cong M$ and since the map $G\rightarrow G_p$ is a submersion we get that the map $G\rightarrow M, g\rightarrow g.p$ is open . I don't know how useful this is since we only know that $G^0$ is a closed set I don't think it has to be open since connected components don't necessarily need to be open, and also I am not sure if this map is even closed or if it is how I could try and prove it.

Any enlightment is appreciated. Thanks in advance.

答:
Manifolds (and thus also Lie groups) inherit the property of being locally connected from Euclidean space, and connected components of locally connected spaces are both closed and open. Hence $G_0$ is open in every Lie group.

Further, if we know that for every $q \in M$ the map $\phi_q : G \to M, g \mapsto g \cdot q$ is open, then the orbit $\phi_q(G_0) \subset M$ must be open.  Then, fixing a single $p \in M$, the union $\bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$ is open as a union of open sets. But note that $$M = \bigcup_{q \in M} \phi_q(G_0)=  \phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0),$$
because every $m \in M$ is equal to $e \cdot m \in \phi_m(G_0)$ if $e \in G_0$ is the identity.

But then we have written the connected set $M$ as a disjoint union of open sets $\phi_p(G_0) \cup \bigcup_{q \in M \setminus \phi_p(G_0) } \phi_q(G_0)$, and since the first set is nonempty, this must mean that the second one is empty, and hence $\phi_p(G_0) = M$.

(For academic integrity: Pieced this answer together from the comments to the very similar question here.)


答:
Writing mathematically an expression doesn't mean you have to use only symbols. Often, for clarity is better to combine symbols and words.

Also writing with only symbols tends to be more prone error.

It's better to say

"Any non-empty subset of natural numbers has a minimum"

or

"For any $A \subseteq \mathbb N$, if $A\ne\emptyset$ then there exists a minimum element $a\in A$"

that

$\forall A (A\subseteq \mathbb N \,\wedge A\ne\emptyset \implies \exists a(a\in A\, \wedge\forall b(b\in A \implies a\le b)))$

For the specific statement you are asking i would use something like "Let $n\in \mathbb N, n\ge 2$".

Actually, often $n$ denotes a natural number, so, unless the context could lead to confusions, i would only write "Let $n\ge 2$".

Title: Writing a mathematical statement: $n$ is a natural number bigger or equal to two

Body:
I want to know how to write: '$n$ is a natural number bigger or equal to two' in mathematical terms. My thoughts:

$$n\in\mathbb{N} \setminus \{0,1\}\tag1$$

$$n\in\mathbb{N}_{\ge2}\tag2$$

$$ n\in\mathbb{N}\space\wedge\space n\ge2\tag3$$

Which one is the best? Or am I wrong about this.


问:
Is this a square wave signal?
i have a decomposition of a square wave signal:

$$ y = \frac{4h}{\pi}(\sin(x) + \frac{1}{3}\sin(3x) + \frac{1}{5}\sin(5x) + ...) $$

I computed the fundamental wave and 2 harmonic waves:

$$ U_{r0} = 27.5e^{j90.8} $$
$$ U_{r1} = 35e^{j63} $$
$$ U_{r2} = 38 $$

Till here, it is correct.
Now i have to show the time function of this square wave and my solution is this one:

$$U_r(t) = 27\sin(628t+86.497) + 35\sin(628\cdot 3t+56) + 38.2\sin(628\cdot 5t)$$

But when i plot with Wolfram Alpha it does not look like a square wave.
Just too less harmonics or did i do something wrong?

答:
If I plot $\sin(x)+\sin(3x)/3+\sin(5x)/5$ I get a pretty nice square wave.  In your case the +86.497 and +56 introduce phase shifts, which will ruin the wave.  Your leading coefficients are also not in the ratio $1:\frac{1}{3}:\frac{1}{5}$ that they should be.  Presumably this came out of your calculation of $U_r0, U_r1, U_r2$, which you don't describe.  Note that $e^{90.8}$ is a very large number.  Where did that come from?

Added:  from your circuit, it appears the output voltage is taken from a divider.  So $U_o=\frac{R}{R+j\omega L+\frac{1}{j \omega C}}U_i=\frac{j \omega CR}{j \omega CR-\omega^2 CL+1}U_i$  The fact that the filter ratio depends upon the frequency means that a square wave in will not come out a square wave.

Title: Is this a square wave signal?


问:
What functions are solution to a homogeneous system of differential equations?
Given a vector $\vec{u} \in \mathbb{R}^n$. For what functions $\psi(t)$ can $\vec{x}(t) = \psi(t)\vec{u}$ be a solution of $\dot{\vec{x}} = A \vec{x}$ for some $n \times n$ matrix $A$?

I'm trying to prove that $\psi(t)$ has to be of the form $e^{\lambda t}$ for some constant $\lambda$, but I'm not sure about that and I do not know exactly how to prove that.

答:
We have
$$\dot \psi u= A \psi u$$

Suppose that $\psi(t_0)\neq0$ for some $t_0$. Then by continuity, in some neighbourhood

$$ A u = \frac{\dot\psi}\psi u$$

(and hence $\dot \psi/\psi$ is an eigenvalue of $A$ assuming $u\neq0$.) But the left-hand side is a constant; therefore so is the other, and the result follows.


问:
$1$-dimensional local Cohen--Macaulay ring whose normalisation is a module finite extension
Let $(R,\mathfrak m)$ be a local $1$-dimendlsional Cohen-Macaulay ring with total ring of fractions (https://en.m.wikipedia.org/wiki/Total_ring_of_fractions) $Q(R)$. Let $\bar R$ be the integral closure of $R$ inside $Q(R)$ . If $\bar R$ is a module finite extension of $R$, then is $R$ necessarily reduced?

答: I assume that $R$ is Noetherian. If $R$ is not reduced, let $z\neq 0$ be a nilpotent element and let $x$ a nonzero divisor in the maximal ideal. Then $z/x^n\in \overline{R}$ for all $n$. Take the suubmodule generated by all these elements. It is easy to see that this is not finitely generated.


问:
Laplace eigenvalues on unit disc intermediate step
If I try for a separable solution $u(r,\theta)=R(r)\Theta(\theta)$ to the equation $\Delta u=-ku$ where $k\geq0$ I run into some problems before getting to the Bessel functions part.

After separation the equation simplifies to $\frac{r^2}{R}R''+\frac{r}{R}R'+kr^2=-\frac{1}{\Theta}\Theta''$.

Now I would set both sides to be a constant $c$. But in every solution I have seen, they have set this constant to be $n^2$ (so that $c=n^2$ in my case) because $c\geq0$. But I don't understand how we can say $c\geq0$. Why is this true?

答: On the two-dimensional unit disk we have the parametrisation $x=r\cos\theta$ and $y=r\sin\theta$. Assuming $u$ is separable, we have $u=R\Theta$ where $\Theta$ is periodic. Setting the RHS to a constant $c$ gives $\Theta''+c\Theta=0$ so $\Theta$ is a linear combination of $\sin(\theta\sqrt c)$ and $\cos(\theta\sqrt c)$. If $c<0$ this forces $\Theta$ to be a linear combination of $i\sinh(\theta\sqrt{-c})$ and $\cosh(\theta\sqrt{-c})$ which violates periodicity. Thus $c\ge0$, and in particular, $\sqrt c$ must be an integer.


问:
Extreme value theorem and half-closed intervals
I undestand that the EVT states that if we have an interval like $[a, b]$ then if there is a continuous function $f$ defined over the interval then there are such $x_1, x_2\in [a, b]$ that $f(x_1) = \sup_{[a, b]} f$ and $f(x_2) = \inf_{[a, b]} f$ (in other words – the functions attains its extremes). So as I understand this fact implies that there can be no continuous (bijective?) function that'd transform eg. the interval [0, 1] into the $\mathbb R$ number line.

So my question is about sets like $[a, b)$. The set is not compact therefore the EVT does not apply yet it is somehow similar... Can we say that if we have a continuous $f:[a, b) \to \mathbb R$ then for some value(s) $f$ attains its maximum/minimum/both (but not none of those as with $(a, b)$)?

Hence there is no continuous bijection from eg. $[0, 1)$ to $\mathbb R$? If there is then how come? To me it seems that either we have the function bounded from either above or below by the $f(0)$ value and approaching $\pm\infty$ on the second end, or the function cannot be defined for every point on the interval if we want it to ,,produce'' both $+\infty$ and $-\infty$ in a continuous way (easier to draw than describe).

To generalise the examples: one can easily create a continuous bijection form $(a, b)$ to $\mathbb R$ whereas it is impossible with $[a, b]$ but how about those half-closed intervals? Does the EVT extend to those somehow?

Could someone clarify whether I understand this correctly?

答:
If $f:[a,b)\longrightarrow\mathbb{R}$ is continuous and injective then it's either monotonic increasing or decreasing (otherwise there would exist $x,y,z\in[a,b)$ such that $x<y<z$ and $f(x)<f(y)$ and $f(z)<f(y)$, which would lead by mean value theorem to $c\in(x,z)$ with $f(c)=\max\{f(x),f(z)\}\in(\min\{f(x),f(z)\},f(y))$).

If $f$ is monotonic increasing then $f(a)\leq f(x)$ for all $x\in[a,b)$, so $f([a,b))\neq \mathbb{R}$ and $f$ is not surjective, hence not bijective. Same for $f$ decreasing.

If you don task $f$ to be injective you can actually do that kind of map, for example, with the function: $$x\in(0,1]\longmapsto f(x):=\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}$$


问:
Quotient of a Lie algebra by a subalgebra - what is it?
The quotient $G/H$ of a group $G$ by its subgroup $H$ has a $G$-action - every transitive $G$-set is of this form.

However, the quotient space $\mathfrak g/\mathfrak h$ of a Lie algebra $\mathfrak g$ by its subalgebra $\mathfrak h$ is just a vector space. By analogy with the group case, I am trying to figure out whether it is still "special" in some way.

If $\mathfrak g$ is the Lie algebra of a Lie group or an algebraic group $G$ with some representation $V$, and $\mathfrak h$ is the Lie algebra of the stabilizer $H=G^v$ of some vector $v\in V$, then there is a way to identify the tangent space of the orbit $Gv$ at $v$ with the space $\mathfrak gv=\{gv\mid g\in\mathfrak g\}$, so that the surjective map $\mathfrak g\twoheadrightarrow\mathfrak gv$ sending $g$ to $gv$ has kernel $\mathfrak h$. Note that although $\mathfrak g$ acts on $V$, the subspace $\mathfrak gv$ is not in general closed under the $\mathfrak g$-action (and does not in general contain $v$).

So one possibility to relate spaces $\mathfrak g/\mathfrak h$ to quotients like $G/H$ would be to ask whether there exists a $\mathfrak g$-representation $V$ and a vector $v\in V$ such that $\mathfrak h=\{g\in\mathfrak g\mid gv=0\}$ and there is an isomorphism $\mathfrak g/\mathfrak h\cong\mathfrak gv$ compatible with the quotient maps $\mathfrak g\twoheadrightarrow\mathfrak g/\mathfrak h$ sending $g$ to $g+\mathfrak h$ and $\mathfrak g\twoheadrightarrow\mathfrak gv$ sending $g$ to $gv$.

Can this (or maybe something better) be always done?

答:
I believe the following holds: If $k$ is a field and $H \subseteq G$ is a closed subgroup of a linear algebraic group $G$, it follows the quotient $G/H$ is a smooth quasi projective scheme of finite type over $k$. We may embed $G/H \subseteq \mathbb{P}^d_k$ into projective $d$-space over $k$ for some integer $d \geq 1$ (it is not completely trivial to construct such an embedding).
As noted above: The tangent space $T_{\overline{e}}(G/H)$ of the scheme $G/H$ at the point $\overline{e}\in G/H$, where $e\in G$ is the multiplicative identity equals the quotient $Lie(G)/Lie(H)$ as left $H$ and  $Lie(H)$-module. There is an isomorphism of $H$-modules (and $Lie(H)$-modules)

$$ T_{\overline{e}}(G/H) \cong Lie(G)/Lie(H).$$

Question: "So one possibility to relate spaces $\mathfrak{g}/\mathfrak{h}$ to quotients like $G/H$ would be to ask whether there exists a $\mathfrak{g}$-representation $V$ and a vector $v \in V$ such that $\mathfrak{h}=\{g \in \mathfrak{g} : gv=0\}$ and there is an isomorphism $\mathfrak{g}/\mathfrak{h}≅\mathfrak{g}v$ compatible with the quotient maps $\mathfrak{g}\rightarrow \mathfrak{g}/\mathfrak{h}$ sending $g$ to $g+h$ and $\mathfrak{g}\rightarrow \mathfrak{g}v$ sending $g$ to $gv$. Can this (or maybe something better) be always done?"

Answer: "This" can be done and you may find it in the litterature. You should take a look at Borel's book "Linear algebraic groups" and Theorem 6.8, page 98. I believe this gives a construction of a quotient $G/H$ for any field $k$. Much work has been done on the construction of "quotients" in algebraic geometry, but Borel's book is a good place to start if your group is algebraic.

You are propably aware of the "stacks homepage" where more general "quotients" such as "algebraic spaces" and "algebraic stacks" are constructed.

Example 1. If $G$ is a semi simple algebraic group and $P\subseteq G$ a parabolic subgroup, it follows the quotient $\mathfrak{g}/\mathfrak{p}$ is the tangent space of the flag variety $G/P$ "at the identity". There is a canonical $\mathfrak{p}$-module structure

$$\rho: \mathfrak{p}\rightarrow \operatorname{End}(\mathfrak{g}/\mathfrak{p})$$

induced by the canonical action - the adjoint action - of $\mathfrak{p}$ on $\mathfrak{g}$.

Example 2. If $dim_k(V)=n+1$ and $l\in V$ is a line, you may let $P\subseteq SL(V)$ be the parabolic subgroup of elements fixing the line $l$. It follows there is an isomorphism

$$SL(V)/P \cong \mathbb{P}(V^*)\cong \mathbb{P}^n_k.$$

Hence the isomorphism

$$ T_{\overline{e}}(SL(V)/P) \cong T_{\overline{e}}(\mathbb{P}^n_k)$$

gives a canonical $\mathfrak{p}$-module structure on the tangent space of projective $n$-space at the "identity". If you  instead choose a vector subspace $W\subseteq V$ of dimension $m$ you get a similar result for the grassmannian. You get a geometric interpretation of these representations in terms of flag varieties and other quotient spaces.

Example 3. If $V:=k\{e_0,..,e_n\}$, $l:=(e_0)$, $V^*:=k\{x_0,..,x_n\}$ and $\mathcal{L}:=\mathcal{O}(d)$ for $d\geq 1$, it follows there is an isomorphism

$$ \phi:\Gamma(\mathbb{P}^n_k, \mathcal{L}) \cong Sym^d(V^*)$$

where $Sym^d(V^*)$ is the $d$'th symmetric product of the dual representation $V^*$ of $V$. The map $\phi$ is a map of $SL(V)$-modules. Hence the global sections of $\mathcal{L}$ is an (irreducible) $SL(V)$-module. Hence you may interpret some irreducible $SL(V)$-modules as global sections of invertible sheaves on the quotient $SL(V)/P$. This is the Borel-Weil-Bott construction.


问:
The particular integral of a 2nd order differential equation
Consider the following 2nd order differential equation, $$\frac{\text{d}^2 z}{\text{d} t^2} + \omega^2 z = C \text{e}^{\text{i}\omega t}$$where $z$ is complex, and $C$ and $\omega$ are unknown real constants.

Now that I am about to find the particular integral. I tried $z=A\text{e}^{\text{i}\omega t}$, so $\frac{\text{d}^2 z}{\text{d} t^2}=-A\omega^2 \text{e}^{\text{i}\omega t}$, and the differential equation becomes $-A\omega^2 \text{e}^{\text{i}\omega t} + A \omega^2 \text{e}^{\text{i}\omega t} = C \text{e}^{\text{i}\omega t}$, which implies $C=0$. That is obviously nonsensical.

[Edit: I am sorry. I mixed some $z$s and $t$s at first.]

So, could someone please explain where I am wrong or what alternative method(s) should I use?

答:
$$\frac{\text{d}^2 z}{\text{d} t^2} + \omega^2 z = C \text{e}^{\text{i}\omega t}$$
$${e}^{-i\omega t}z''\color{red} {- i\omega z'{e}^{-i\omega t}+iz'\omega {e}^{-i\omega t}}+\omega^2 z {e}^{-i\omega t} = C $$
$$({e}^{-i\omega t}z')'+(iz\omega {e}^{-i\omega t})'= C $$
$$({e}^{-i\omega t}z'+iz\omega {e}^{-i\omega t})'= C $$
$${e}^{-i\omega t}z'+iz\omega {e}^{-i\omega t}= Ct+C_2 $$
$$({e}^{i\omega t}z)'={e}^{2i\omega t}( Ct+C_2 )$$
Integrate.
$$z(t)=A \sin (\omega t)+B \cos (\omega t)+\dfrac {C}{2i \omega }te^{i \omega t}+\dfrac {C}{4\omega ^2}e^{i \omega t}$$


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

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

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

Another one would be something like

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

Yet another one would be

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

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

Edit: fixed typo in last example.

答:
This classic one:

Seven people went in a bar, and drank beer. Then, the waiter came for the bill which was 28 dollars. Then one of the seven people calculated the bill per person, and found it to be 13 dollars. Check out how:

They paid the bill. When the manager came back, he asked the waiter to which the waiter showed him how 13 added 7 times is 28. Check out how:


问:
Cauchy's minimum principle
At the moment I'm trying to understand a proof of the fundamental theorem of algebra. The author uses several facts known from analysis lectures. Under these facts is one I didn't know in the formulation. He states that every real and continuous function $f$ that is defined on a compact set $S$ has its minimum there.

Now my questions:

I suppose this does also hold for functions  $f: \mathbb{C} \rightarrow \mathbb{C}$. Is that right? Since he uses this for complex polynomial functions later on.

Do you recognize the fact in this statement I found here (Understanding the Minimum Principle in Complex Analysis)?

答:
Yes. In general, if you have a continuous function $f:X\to\mathbb{R}$ and $K$ a compact subset of a topological space $X$, then both $\min\{f(x):x\in K\}$ and $\max\{f(x):x\in K\}$ exist.

The linked question is more about the properties of analytic functions, in particular, the maximum principle.

Title: Cauchy's minimum principle


问:
Find all local extrema
I'm trying to solve an exercise that says:

Find all local extrema of the function:
$$g: \mathbb{R}^{2} \rightarrow \mathbb{R}, \quad g(x, y)=3 x^{4}-6 x^{2}-4 y^{3}+2 y^{2}+x y^{2}$$
I know I must solve the equation $\nabla g =(0,0)$. So I write:
$$12x^3-12x+y^2 = 0$$
$$-12y^2+4y+2xy=0$$
and try to solve them simultaneously. But I can't. Actually, I know a solution is (1,0). How can I find out the other solutions?

Title: Find all local extrema

Body:
Why not to use the trigonometric for three real roots ? The solutions would be
$$x_k=\frac{2 \sqrt{554689} \cos \left(\frac{1}{3} \left(2 \pi  k-\cos
   ^{-1}\left(-\frac{10909729}{554689 \sqrt{554689}}\right)\right)\right)-1}{1296}$$ with $k=0,1,2$.

Making the values rational to rather high accuracy
$$x_0=\frac{1789}{1808} \qquad x_1=\frac{78}{8345}\qquad x_2=-\frac{1737}{1735}$$
$$y_0=\frac{1849}{3711}\qquad y_1=\frac{2078}{6205}\qquad y_2=\frac{722}{4337}$$
$$g_0=-\frac{6317}{2296}\qquad g_1=\frac{133}{1783}\qquad g_2=-\frac{5177}{1731}$$


问:
A Question about Doctoral Theses in Mathematics
This is most definitely a soft question, which I'm sure may get some negative attention, and perhaps even be voted closed. However, I genuinely would like to generate answers on this matter as it concerns anyone that decides to get a doctorate in mathematics. (Since I believe this is the best forum to do so, here goes...)

Question: Does a doctoral thesis in mathematics have to contain more than an abstract, a proposition and proof of a new and exciting result?

I strongly believe that brevity is beauty to mathematicians and see no problem with a thesis that contains a lengthy abstract (~350 words) to generate enthusiasm and then jumps right into the heart of a novel and far-reaching result. The proof provided is extremely condensed in the spirit of Zagier, leaving almost all details that can be found elsewhere to be found elsewhere, in the works cited. If the result has merit, I see no reason why it can't be submitted or published in such a manner. To support my stance, I offer the following condensed theses:

David Rector, "An Unstable Adams Spectral Sequence", MIT (1966), 9 pgs.

Burt Totaro, "Milnor K-Theory is the simplest part of algebraic K-theory", Berkeley (1989), 12 pgs.

Herman Buvik, A New Proof of Torelli's Theorem, NYU (1962), 12 pgs.

Eva Kallin "A non-local function algebra" Berkeley (1963), 13 pgs.

Edmund Landau "Neuer Beweis der Gleichung $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} = 0$", Berlin University (1899), 14 pgs.

Barry Mazur, "On Embeddings of Spheres", Princeton (1959), 26 pgs.

John F. Nash "Non-Cooperative Games", Princeton (1950), 27 pgs.

Kevin Walker, "An Extension of Casson's Invariant to Rational Homology Spheres", Berkeley (1989). 29 pgs.

Although I'd appreciate to hear from anyone worth giving his two cents, I'm specifically eager to hear from those senior members of the community.

Thanks!

答:
In Littlewood's "Mathematician's Miscellany"
(fairly recently reissued as "Littlewood's Miscellany"),
this exact situation is discussed.
He says that, yes, even a two-sentence dissertation
could be acceptable.
He says that Picard's great theorem
(iirc) with a one-sentence proof is an example.

I $love$ this book and highly recommend it
to anyone with any sort of interest in mathematics.


问:
Finding a function which is not differentiable at $x_{0}$
Let $L \in \mathbb{R}$ and let $f$ be a function that is differentiable on a deleted neighborhood of $x_{0} \in \mathbb{R}$ such that $\lim_{x \to x_{0}}f'(x)=L$.

Find a function satisfying the above, and such that $f$ is not differentiable at $x_{0}$.

--

So I think that I do not completlely understand, when a function is indeed differentiable at $x_{0}$ and when it's not, and why in both cases I can still find its $f'$?

I will appreciate some explanation about that.

Moreover, I thought of $f(x)=x^x$ or $f(x)=\ln(x^x)$.

If I understand it correctly, than both my $f$'s does not differentiable at $x_{0}=0$, because:

$f'(0)=\lim_{x \to 0}\frac{x^x-0^0}{x-0}$ which is undefined?

or

$f'(0)=\lim_{x \to 0}\frac{\ln(x^x)-\ln(0^0)}{x-0}$ which is undefined?

Thanks a lot!

Body:
Take the function $f(x) = |x-x_0|$.

$\lim_{x\rightarrow x_0^-} \frac{|(x_0-h)+x_0|}{-h} = \frac{|h|}{-h}$ and

$\lim_{x\rightarrow x_0^+} \frac{|(x_0+h)+x_0|}{h} = \frac{|h|}{h}$.


问:
Prove a distance is a metric space
This is easy proof but i can't understand.

Let $X=\Bbb R^n$ and $d(x,y)=\max_{i\in\{1,n\}}|x_i-y_i|$

i) $d(x,y)\ge0$ and ii) $d(x,y)=d(y,x)$ are easy.
But iii) is $d(x,y)+d(y,z) \ge d(x,z)$ is so hard.

$d(x,y)+d(y,z) \ge d(x,z)$

<=> $\max_{i\in\{1,n\}}|x_i-y_i|+\max_{i\in\{1,n\}}|y_i-z_i| \ge \max_{i\in\{1,n\}}|x_i-z_i|$
We choose $j \in \{1,n\}$ so that $|x_j-y_j|$ is max, but we not sure $|y_j-z_j|$ and $|x_j-z_j|$ max too.

Please help me, thank you!

答:
Consider 3 vector $x,y,z$. $d(x,z)$ assumes its value at some $i_0 \in \{1,\ldots,n\}$ so that $d(x,z)=|x_{i_0}- z_{i_0}|$.

It's clear that $|x_{i_0}- z_{i_0}| \le |x_{i_0}- y_{i_0}| + |y_{i_0}- z_{i_0}|$ by the usual triangle inequality for $\Bbb R$ and also $|x_{i_0}- y_{i_0}|\le d(x,y)$ (a value at some index is $\le$ than the maximum over all coordinates) and also $|y_{i_0}- z_{i_0}| \le d(y,z)$.

So $$d(x,z)=|x_{i_0}- z_{i_0}| \le d(x,y) + d(y,z)$$

as required.

Title: Prove a distance is a metric space


问:
proof of Lehmann-Scheffe theorem
I'm reading proof of Lehmann-Scheffe theorem in Casella's Statistical Inference (2nd ed)

Theorem states that

Let $T$ be a complete sufficient statistic for a parameter $\theta$, and let $\phi(T)$be any estimator based only on T.

Then $\phi(T)$is the unique best unbiased estimator of its expected value

To prove $\phi(T)$ is best unbiased estimator, he use following lemma

Suppose $EW = \tau(\theta)$. Then $W$ is best unbiased estimator of $\tau(\theta)$ if and only if $W$ is uncorrelated with all unbiased estimator of $0$

Then he states proof of Lehmann-Scheffe theorem as follows:

Since $T$ is complete, there are no unbiased estimator of $0$ that are based on $T$

By Rao-Blackwell Theorem, It suffices to consider unbiased estimator of $0$ based on $T$

Hence, $\phi(T)$ is uncorrelated to all unbiased estimator of $0$ (as there is only one unbiased estimator of $0$, just $0$ itself.)

So by lemma, we got desired result

I cannot understand second step.

If we deal with to find or check best unbiased estimator, I think we can use Rao-Blackwell theorem since estimator based on $T$ always has smaller variance.

But in here, we want to check whether $\phi(T)$ is uncorrelated to all unbiased estimator of $0$, so I think it's not sufficient to check only estimator based on $T$.

How can I solve this problem?

(Indeed, I have other proof of Lehmann-Scheffe theorem. I just want to know how this proof make sense)

答:
Consider the possibility that there is a better unbiased estimator. The Rao-Backwell theorem says that the expected value of that unbiased estimator given $T$ has variance that is less than or equal to the original estimator. That means this improved estimator (which is a function of $T$) can be considered instead as the "better estimator". I other words, if I can show my estimator beats that one, then it also beats the original one.

Can you tell me where this proof is? The statement is Theorem 7.3.23, but I don't see the proof.

Your question is related to this one.

Title: proof of Lehmann-Scheffe theorem


问:
Tangent space of a product of algebraic group.
I need help with this problem from Shafarevich Basic Algebraic Geometry.

Let $G$ be an algebraic group and $\Psi:G\times G\rightarrow G$ the
  regular map defined by the group law. Let $T_{e}G$ and be
  $T_{e'}(G\times G)$ the tangent spaces to $G$ and $G\times G$ at their
  respective identity elements. Prove that $T_{e'}(G\times G)=T_{e}G\oplus T_{e}G$ and that $d_e \Psi:T_{e}G\oplus
 T_{e}G\rightarrow T_{e}G$ is given by addition of vectors.

I tried this way but I'm not acquainted with the tensor product. Let $T_{e'}(G\times G)=\left( \frac{M_{e'}}{M_{e'}^2} \right)^{*}$ where $M_{e'}=\{ f\in k[G\times G]\mid f(e')=0 \}$. But $k[G\times G]=k[G]\otimes_k k[G]$ so $M_{e'}=M_{e}\otimes M_{e}$.
Then
$$T_{e'}(G\times G)=\left( \frac{M_{e'}}{M_{e'}^2} \right)^{*} = \left(\frac{M_{e}}{M_{e}^2} \otimes \frac{M_{e}}{M_{e}^2}\right)^{*}=T_e G\oplus T_e G$$.

I'm not sure about the second and the third equalities. Could you help me, please? Thanks!

答:
Using Kahler differentials you may prove a more general lemma valid for arbitrary group schemes. There is a formula for the Kahler differentials of a tensor product: If $A,B$ are commutative unital $k$-algebras it follows there is a canonical isomorphism of $A\otimes_k B$-modules

$$F1.\text{       } \Omega^1_{A\otimes_k B/k}\cong \Omega^1_{A/k}\otimes_k B \oplus A\otimes_k \Omega^1_{B/k}.$$

This is Exercise II.8.3 in Hartshorne's book. Formula $F1$ relies on some basic properties of Kahler differentials that may be found in Hartshorne's book or in the Mastumura book "Commutative ring theory".
If $u: A\rightarrow A_1$and $v: B \rightarrow B_1$ are maps of commutative unital rings and $M$ is an $A$-module, $N$ a $B$-module there is the following formula:

$$M\otimes_k N\otimes_{A\otimes_k B} A_1\otimes_k B_1 \cong$$

$$ (M\otimes_A A_1)\otimes_k (N\otimes_B B_1).$$

If $A_1:=A/I, B_1:=B/J$ we get the formula

$$ M\otimes_k N \otimes_{A\otimes_k B} A/I\otimes_k B/J \cong$$

$$ (M/IM)\otimes_k (N/JN).$$

Hence

$$\Omega^1_{A\otimes_k B/k}\otimes_{A\otimes_k B}(A/I\otimes_k B/J) \cong$$

$$ A/I\otimes_k (\Omega^1_{B/k}\otimes_B B/J)\oplus (\Omega^1_{A/k}\otimes_A A/I)\otimes_k B/J.$$

If $I,J$ are ideals with $A/I\cong B/J\cong k$ we get

$$F2.\text{    }\Omega^1_{A\otimes_k B/k}\otimes_{A\otimes_k B}(A/I\otimes_k B/J) \cong$$

$$\Omega^1_{A/k}\otimes_A k \oplus \Omega^1_{B/k}\otimes_B k.$$

Example 1. If $G:=Spec(A)$ where $A$ is a finitely generated $k$-algebra with $k$ a field and $\mathfrak{m}$ is the maximal ideal corresponding to the identity element $e\in G$, it follows the residue field $\kappa(e)$ satisfies

$$\kappa(e) \cong A/\mathfrak{m} \cong k,$$

since $e$ is a $k$-rational point. Let $B=A$ and $I=J=\mathfrak{m}$ in F2. It follows  there is an isomorphism of $k$-vector spaces

$$\Omega^1_{A\otimes_k A/k}(e,e) \cong \Omega^1_{A/k}(e) \oplus \Omega^1_{A/k}(e).$$

Here $(e,e) \in G\times_k G$ is again a $k$-rational point. Dualizing we get an isomorphism

$$T_{(e,e)}(G\times_k G):= Hom_k(\Omega^1_{A\otimes_k A/k}(e,e) , k) \cong$$

$$ Hom_k(\Omega^1_{A/k}(e) \oplus \Omega^1_{A/k}(e),k)    \cong T_e(G) \oplus T_e(G).$$

Note that in F2 $k$ can be an arbitrary commutative unital ring.

Example 2. If $A:=\mathbb{Z}[x_{ij},t]/(det(x_{ij})t-1)$ and $G:=Spec(A):=GL(n,\mathbb{Z})$, it follows the unit element $I(n)$ is the $\mathbb{Z}$-valued point $\phi: A\rightarrow \mathbb{Z}$ of $G$ defined by $\phi(x_{ij})=0$ if $i\neq j$, $\phi(x_{ii})=1$ and $\phi(t)=1$. Let $\mathfrak{p}:=ker(\phi)$. It follows $I(n) \in G(\mathbb{Z})$ is a $\mathbb{Z}$-valued point (in the language of "representable functors"). F2 gives a formula for the "cotangent space" of the group scheme $G\times_{\mathbb{Z}} G$ at $(I(n),I(n))$ (or the non-closed point $(\mathfrak{p}, \mathfrak{p})$):

$$\Omega^1_{G\times_{\mathbb{Z}} G/\mathbb{Z}}(\mathfrak{p}, \mathfrak{p}) \cong $$

$$\Omega^1_{A/k}\otimes_A \kappa(\mathfrak{p}) \oplus \Omega^1_{A/k}\otimes_A \kappa(\mathfrak{p}).$$

At the level of tangent spaces we get

$$F3.\text{   }T_{(\mathfrak{p}, \mathfrak{p})}(G\times_{\mathbb{Z}} G)\cong T_{\mathfrak{p}}(G) \oplus T_{\mathfrak{p}}(G)$$

Note: The map $\phi:A \rightarrow \mathbb{Z}$ is a surjective map onto an integral domain, hence the kernel $\mathfrak{p}\subseteq A$ is a prime ideal - it is not a maximal ideal. Hence we cannot immediately define $GL(n,\mathbb{Z})$ unless we are willing to work with prime ideals.

You will find a discussion of the "language of representable functors" here:

Equivalence between two definitions of affine algebraic set.

This is a language used when studying more general group schemes such as $GL(n,\mathbb{Z})$ and $GL(n, \mathcal{O}_K)$ with $K$ a number field. You may defined the tangent space of any scheme $X$ at any point $\mathfrak{p}$, closed or non-closed. If you look in the litterature you will find that some authors speak of "distributions" when introducing tangent spaces (Jantzen's book "Representations of algebraic groups"). This is another approach that is much used.


问:
Three students are selected from a group of 10. What is the probability that George is selected if John is always one of the selected student.
Since John is always selected, that leaves picking 2 students out of 9. So

$$P(G) =\frac{\binom 91}{\binom 92} =\frac 14$$

But this is wrong. What’s the problem here?

答:
It is simply $\frac{2}{9}$ as there are two spots leaving John. So for George to be selected, he has to be one of the two getting selected. Here is another way to look at it -

$P = \frac{8}{9} \times \frac{1}{8} +  \frac{1}{9} \times \frac{8}{8} = \frac{2}{9}$


答: Take $f:\Bbb R\to\Bbb R$, with $$f(x)=\begin{cases}Lx~~~\text{ for }x<x_0\\Lx+1~\text{for }x\geq x_0\end{cases}$$

Title: Finding a function which is not differentiable at $x_{0}$

Body:
Let $L \in \mathbb{R}$ and let $f$ be a function that is differentiable on a deleted neighborhood of $x_{0} \in \mathbb{R}$ such that $\lim_{x \to x_{0}}f'(x)=L$.

Find a function satisfying the above, and such that $f$ is not differentiable at $x_{0}$.

--

So I think that I do not completlely understand, when a function is indeed differentiable at $x_{0}$ and when it's not, and why in both cases I can still find its $f'$?

I will appreciate some explanation about that.

Moreover, I thought of $f(x)=x^x$ or $f(x)=\ln(x^x)$.

If I understand it correctly, than both my $f$'s does not differentiable at $x_{0}=0$, because:

$f'(0)=\lim_{x \to 0}\frac{x^x-0^0}{x-0}$ which is undefined?

or

$f'(0)=\lim_{x \to 0}\frac{\ln(x^x)-\ln(0^0)}{x-0}$ which is undefined?

Thanks a lot!


问:
Challenging integration problem of an exponential function
While preparing a tutorial session for my students, I come across a very challenging integration, which is :$$ \int_{-\infty}^\infty e^{t^2 }\  \mathrm{d}t$$

I attempted many methods to solve it but with no use.
Your assistance will be greatly appreciated.

答: Assuming you are looking for the antiderivative of $e^{t^{2}}$: $$\int e^{t^2 }\  \mathrm{d}t = \frac{\sqrt{\pi}}{2}\operatorname{erfi}(x),$$where $\operatorname{erfi}(x)$ is the imaginary error function defined as $$\operatorname{erfi}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{t^{2}}\mathrm{d}t.$$  There is no way of expressing the antiderivative of $e^{t^{2}}$ in terms of elementary functions.


问:
Preimage of $A_{2k+1}$ in right multiplication action
Let $G$ be a finite group of order $n$ and let $\psi:G\to S_n$ be the homomorphism defined by the right multiplication action. I have to prove that:

$\text{(a)}$ Let $x\in G$ such that $o(x)=2$. Show that $\psi(x)\in A_n\iff4\mid n$.

$\textbf{My attempt:}$ Since $o(x)=2$, $o(\psi(x))|2\implies o(\psi(x))\in\{1,2\}$. If $o(\psi(x))=1$, the image of $x$ is the identity element and we have nothing to show. If $o(\psi(x))=2=\text{lcm}(type(\psi(x)))$ and we want that the condition $\psi(x)\in A_n$ is satisfied, the element $\psi(x)$ will be an element of order $2$ composed by the multiplication of an even number of $2$-cycles. The smallest $2$-cycle that satisfies this condition has to be in form  $(ab)(cd)$, so we need at least $4$ elements in $G$.

The operation that we do if we want $\psi(x)$ composed by a greater number of $2$-cycle is adding $2\cdot k$ $2$-cycles, for example:$$n=4\implies\psi(x)=(ab)(cd)$$$$n=8\implies\psi(x)=(ab)(cd)(ef)(gh)$$$$\vdots$$Conversely is really quick.

$\text{(b)}$ Assuming $|G|=2\cdot s$, where $s\in 2\mathbb Z+1$, show that $\psi^{-1}(A_n)\le G$ , concluding that $\exists K\le G$ such that $|G:K|=2$.

Here is my difficulty but I suppose that the subgroup $K$, for the fact that it's normal, is the kernel of the map.

$\text{(c)}$ Show that $K$ is characteristic in $G$.

$\textbf{My attempt:}$ This request follows easily from point $\text{(b)}$ because $K$ has index $2$ in $G\implies K\trianglelefteq G$ and

$$\begin{align}
|G:K|=2&\implies |K|=\dfrac{|G|}{2}=\dfrac{\not 2\cdot s}{\not2}=s\\
&\implies \gcd(|K|,|G:K|)=1\\
&\implies K\text{ is characteristic}.
\end{align}$$

Could you suggest to me a method to attack point $\text{(b)}$?

Thank for you help.

答:
The crucial idea for the entire answer, not just part (b), is that the action is right multiplication. Consequently, the kernel of $\psi$ is the identity element. (Note that  $\psi (g)$ maps each element $x$ to $xg$ and that $x$ only equals $xg$ if $g$ is the identity.)

Part (b)

Let $I$ be the image of $G$. Then $|I|=|G|$.

Now consider the subgroup $I\bigcap A_n$ of $S_n$. By the argument of part (a), $I$ is not contained in $A_n$. Then $|IA_n||I\bigcap A_n|=|I||A_n|$ and so $|I|=2|I\bigcap A_n|$.

Therefore $\psi ^{-1}(|I\bigcap A_n|)$ is a subgroup of $G$ of index $2$.


问:
Number of possible matrices using given conditions
I have a question, that goes as follows:

The number of all matrices $A=[a_{ij}], 1 \leq i,j \leq 4$ such that $a_{ij}= \pm1 $ and $\sum_{i=1}^4a_{ij}= \sum_{j=1}^4a_{ij}=0$ is?

So, I think this question means to say that the sum of all elements in a row, and all elements in a column is zero, and the possible value of the elements being only $+1,-1$. How do I form the combinations in this case?

I tried to find the number of solutions to the equation $a_{11}+a_{12}+a_{13}+a_{14}=0$ and then repeating the same process for every row and column but that didn't get me anywhere.
The answer says 90 and I cannot seem to be able to get there. I have seen another question asking the same, but the answer isn't very clear to me, and that question being 4 years old, I am posting again.

答:
The problem is same as placing two $1$s and two $0$s in each row and column.

Put two $1$s in a row. $\binom{4}{2}$ ways. Pick one of the columns in which a $1$ is placed and place another $1$. $\binom{3}{1}$ ways.

There is a $2 \times 2$ submatrix in which three $1$s are placed now. If fourth square also contains a $1$, entire matrix is determined. Thus one matrix is obtained.

$$
\begin{bmatrix}
1 & 0 & 1 & 0 \\
0 &   &  & \\
1 & & (1,0)& \\
0 & & &
\end{bmatrix}
$$

If the fourth square contains a zero, there are two ways to place a $1$ in that particular row (containing the zero in place of fourth $1$).

$$
\begin{bmatrix}
1 & 0 & 1 & 0 \\
0 &   &  & \\
1 & 1& 0& 0\\
0 & & &
\end{bmatrix}
$$

Now in the untouched column (fourth one above), there is only one way to put two $1$s. The remaining $2\times2$ submatrix can be filled in two ways by choosing a position for $1$.

$$
\begin{bmatrix}
1 & 0 & 1 & 0 \\
0 &   &  & 1 \\
1 & 1 & 0 & 0\\
0 &   & & 1
\end{bmatrix}
$$

Hence final answer $$\binom{4}{2}\cdot\binom{3}{1}\cdot(1+2\cdot2)=90$$


答:
In general, you have the following theorem.

Suppose $f:(a,b)\to\mathbb{R}$ is continuous and differentiable on $(a,b)\setminus\{c\}$ for some $c\in (a,b)$. If $\lim_{x\to c}f'(x)$ exists, then $f$ is differentiable at $x=c$ and
$$
f'(c)=\lim_{x\to c}f'(x).
$$

(This is a good exercise of the mean value theorem.)

So in order to find your desired example, $f$ must not be continuous at $x=c$. This is easy: just move it somewhere else. For example,
$$
f(x)=\begin{cases}
Lx&x\ne 0\\
1&x=0
\end{cases}.
$$

Notes.

If a function $f$ is not defined at $x=x_0$ or it is defined but not continuous at $x_0$, then you can immediate tell that $f$ is not differentiable at $x=x_0$.


问:
Discussion on the prolongation of the solution
I have the following

$$(PC)\begin{cases}y'=\dfrac{1}{x+1}y-\dfrac{4x}{x+1}y^2\\y(3)=\dfrac{2}{b} \end{cases}\text{ with }b\ne 0$$
Determine for which values of $b$, it exists at least one solution defined in $I=[0,4]$.

$\textbf{My attempt:}$ I noticed that the function $f(x,y):=\dfrac{1}{x+1}y-\dfrac{4x}{x+1}y^2$ is defined in $(-\infty,-1)\cup (-1,+\infty)$ and, since $3\in(-1,+\infty)$, I studied the Cauchy problem for $x>-1$.

The differential equation admits a unique local solution and a unique prolongation in its maximal domain.

We have a Bernoulli equation so I substituted $z(x)=\dfrac{1}{y(x)}$ and this substitution defines the new Cauchy problem:
$$(PC')\begin{cases}z'+\dfrac{1}{x+1}z=\dfrac{4x}{x+1}\\z(3)=\dfrac{b}{2} \end{cases}$$
and I got the solution $y(x)=\dfrac{x+1}{2x^2+2b-18}$, so I imposed $2x^2+2b-18\ne 0$ and $y(x)\ne 0$ for the substitution that I did$\implies|x|\ne\sqrt{9-b}$ and $x\ne -1$. Then I considered the solution in
$$(-\infty,-\sqrt{9-b})\cup (-\sqrt{9-b},\sqrt{9-b})\cup (\sqrt{9-b},+\infty)\bigcap (-1,+\infty).$$
If we want $I\subseteq I_{\text{SOL}}$ I think that it's correct to impose that $(-\sqrt{9-b},\sqrt{9-b})$ must contain $[0,4]\implies$ $$\begin{cases} -\sqrt{9-b}<0\text{ or}\\\sqrt{9-b}>4\end{cases}\implies\begin{cases}b<9\text{ or}\\b<-7\end{cases}.$$ For the other case I imposed that $[0,4]\subseteq (\sqrt{9-b},+\infty)$, so
$\sqrt{9-b}> 0$ which defines only the condition $b< 9$.

I always have difficulties in the study of the prolongation of the solution on a certain interval... :-(

答:
You found correctly that
$$
(1+x)z(x)=2(x^2+C)\implies 4·\frac{b}{2}=2·(9+C)
$$
The solution $y$ becomes singular where $z=\frac1y$ has a root. As $(x+1)$ has no root on $[0,4]$, it remains to consider the other side. If $C>0$ then there is no problem. For negative $C$ to avoid roots on the interval one needs $C<-16$. So either

$C=b-9>0\iff b>9$ or

$C=b-9<-16\iff b<-7$.


问:
Recommendations for good self-learning textbooks in group theory (or abstract algebra in general)
I am trying to self teach myself abstract algebra and am currently looking for some good self learning books for group theory. I have already done linear algebra 1 and 2 (vector spaces, spans, bases, finite dimensions, linear transformations and homo/isomorphisms and so on). I specifically want books of the same style as Terence Tao's "Real analysis I": a lot of exposition and explaining why a topic is needed in the first place. Thanks!</p>

答: I recommend Gallian's "Contemporary Abstract Algebra". It has plenty of exercises, half of which come with solutions.


答:
Let G be event that George is selected  and J be when John is selected .$$P(G|J)=\frac{n(G\cap J)}{n(J)}=\frac{\binom{8}{1}}{\binom{9}{2}}=2/9$$

Explanation: $n(J\cap G)=\binom{8}{1}$ as we are already having to select george and John and we are left to select one person from the rest $ 8$.

Title: Three students are selected from a group of 10. What is the probability that George is selected if John is always one of the selected student.

Body:
Since John is always selected, that leaves picking 2 students out of 9. So

$$P(G) =\frac{\binom 91}{\binom 92} =\frac 14$$

But this is wrong. What’s the problem here?


问:
Number of integers at most $x$ with exactly two distinct prime factors
I wish to find an asymptotic for the number of integers not exceeding $x$ with exactly two distinct prime factors. Here is a starting point:

Throughout $p$ and $q$ are primes. We are interested in $\frac{1}{2}\#\{(p,q,\alpha,\beta): p^{\alpha}q^{\beta} < x\}$. (The factor $\frac{1}{2}$ arises since e.g. for $x=10^4$ we get $2^{3}\cdot 5^4$ once from $p=2$, $q=5$, $\alpha = 3$, $\beta = 4$ and once from $p=5$, $q=2$, $\alpha = 4$, $\beta = 3$.) Another way to write it is $\frac{1}{2}\#\{(p,q,\alpha,\beta): q<x, p<(x/q^{\beta})^{1/\alpha} \} = \frac{1}{2}\sum_{q<x}\sum_{\alpha, \beta \geq 1}\sum_{p<(x/q^{\beta})^{1/\alpha}}1$. Taking care of the inner sum with the Prime Number Theorem, the count becomes
$$\frac{1}{2}\sum_{q<x}\sum_{\substack{\alpha, \beta \geq 1}} \bigg(\frac{x^{1/\alpha}}{q^{\beta/\alpha}\log \frac{x^{1/\alpha}}{q^{\beta/\alpha}}} + O\bigg(\frac{x^{1/\alpha}}{q^{\beta/\alpha}\log^2 \frac{x^{1/\alpha}}{q^{\beta/\alpha}}}\bigg)\bigg)$$

Any idea how to conviently proceed with the latter nasty sums (or perhaps another approach)? Any help appreciated!

答:
The asymptotic is
$$\frac{x\log \log x}{\log x}\,.$$
More generally, a result of Landau (1900) is that
$$\pi_k(x) \sim \rho_k(x) \sim \sigma_k(x) \sim \frac{x(\log \log x)^{k-1}}{(k-1)!\log x}\,,$$
where
\begin{align}
\pi_k(x) &= \# \{ n \leqslant x : \omega(n) = \Omega(n) = k\}\,,\\
\rho_k(x) &= \#\{ n \leqslant x : \omega(n) = k\}\,,\\
\sigma_k(x) &= \#\{ n \leqslant x : \Omega(n) = k\}\,,
\end{align}
and $\omega(n)$ is the number of distinct prime factors of $n$, while $\Omega(n)$ is the number of prime factors of $n$ counted with multiplicity,.

You are specifically asking for the asymptotics of $\rho_2$.

If we only want to show that, not Landau's result in full generality, the IMO easiest way is to obtain the asymptotics for $\pi_2(x)$, and then show that the numbers $p^{\alpha} q^{\beta}$ with $\alpha > 1$ or $\beta > 1$ are asymptotically negligible.

By symmetry, we can assume $\alpha > 1$, and we can ignore the condition $q \neq p$ (that can only make us overestimate the count, so if it still remains negligible, a fortiori the true count is negligible).

Given a prime $p$ and an exponent $\alpha > 1$, the count of numbers $p^{\alpha} q^{\beta} \leqslant x$, where $q$ is a prime (possibly equal to $p$) is the number of prime powers (including primes) not exceeding $x/p^{\alpha}$. Splitting into two parts according as $p^{\alpha} > \sqrt{x}$, for the part with $p^{\alpha} > \sqrt{x}$ we obtain the upper bound
$$\sum_{\substack{\sqrt{x} < p^{\alpha} \leqslant x/2 \\ \alpha \geqslant 2}} \sqrt{x} \leqslant \sqrt{x}\cdot \pi(\sqrt{x}) \leqslant C_1\cdot \frac{x}{\log x}$$
and for the part with $p^{\alpha} \leqslant \sqrt{x}$, since there are only $O(\sqrt{y})$ perfect powers not exceeding $y$ we obtain an upper bound of
$$C_2\cdot\sum_{\substack{p^{\alpha} \leqslant \sqrt{x} \\ \alpha \geqslant 2}} \pi\biggl(\frac{x}{p^{\alpha}}\biggr) \leqslant C_3\cdot \sum_{\substack{p^{\alpha} \leqslant \sqrt{x} \\ \alpha \geqslant 2}} \frac{x}{p^{\alpha}\log \frac{x}{p^{\alpha}}} \leqslant 2C_3\frac{x}{\log x} \sum_{\substack{p^{\alpha} \leqslant \sqrt{x} \\ \alpha \geqslant 2}} \frac{1}{p^{\alpha}} \leqslant C_4\frac{x}{\log x}\,.$$
Altogether,
$$\rho_2(x) - \pi_2(x) \in O\biggl(\frac{x}{\log x}\biggr)\,,$$
which is of smaller order than $\frac{x\log \log x}{\log x}$.

It remains to prove the assertion about $\pi_2$. It is easy to see
$$\pi_2(x) = \sum_{p \leqslant \sqrt{x}} \Biggl(\pi\biggl(\frac{x}{p}\biggr) - \pi(p)\Biggr)\,,$$
and
$$\sum_{p \leqslant \sqrt{x}} \pi(p) = \frac{\pi(\sqrt{x})\bigl(\pi(\sqrt{x}) + 1\bigr)}{2} \in O\biggl(\frac{x}{(\log x)^2}\biggr)\,,$$
whence we obtain
\begin{align}
\pi_2(x) &\sim \sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p}\biggr) \\
&\sim \int_2^{\sqrt{x}} \frac{\pi(x/u)}{\log u}\,du \\
&\sim \int_2^{\sqrt{x}} \frac{x}{u\log u(\log x - \log u)}\,du \\
&= x\int_{\log 2}^{\log \sqrt{x}} \frac{dv}{v(\log x - v)} \\
&= \frac{x}{\log x}\int_{\log 2}^{\log \sqrt{x}}\frac{1}{v} + \frac{1}{\log x - v}\,dv \\
&= \frac{x}{\log x} \int_{\log 2}^{\log \frac{x}{2}} \frac{dw}{w} \\
&= \frac{x}{\log x}\biggl(\log \log \frac{x}{2} - \log \log 2\biggr)\\
&\sim \frac{x\log \log x}{\log x}\,.
\end{align}
In that, the second asymptotic equality is admittedly not obvious, it is using a lemma by Landau. We can avoid using that, at the cost of a messier calculation, for example
\begin{align}
\sum_{p \leqslant \sqrt{x}} \pi\biggl(\frac{x}{p}\biggr) &\sim \sum_{p \leqslant \sqrt{x}} \frac{x}{p(\log x - \log p)} \\
&= \int_{2-\varepsilon}^{\sqrt{x}} \frac{x}{u(\log x - \log u)}\,d\pi(u) \\
&= \pi(\sqrt{x})\frac{\sqrt{x}}{\log \sqrt{x}} - \int_{2}^{\sqrt{x}} \pi(u) \frac{d}{du}\biggl(\frac{x}{u(\log x - \log u)}\biggr)\,du \\
&\sim -\int_{2}^{\sqrt{x}} \operatorname{Li}(u)\frac{d}{du}\biggl(\frac{x}{u(\log x - \log u)}\biggr)\,du \\
&= -\operatorname{Li}(\sqrt{x})\frac{\sqrt{x}}{\log \sqrt{x}}
+ \int_2^{\sqrt{x}} \frac{x}{u\log u(\log x - \log u)}\,du
\end{align}
using the asymptotics of $\pi(x)$ and Riemann-Stieltjes integrals. The asymptotic equalities here are fairly straightforward to establish.


问:
Convergence of $\int_{-\infty}^\infty \frac{1}{1+x^6}dx$
Okay, so I am asked to verify the convergence or divergence of the following improper integrals:

$$\int_{-\infty}^\infty \frac{1}{1+x^6}dx$$

and

$$\int_1^\infty \frac{x}{1-e^x}dx$$

Now, my first attempt was to use comparison criterion with $$\int \frac{1}{x^2}$$

and conclude that both of the improper integrals converge given that they are smaller than the general term $\frac{1}{x^2}$.

Is it the right path? Also, are the antiderivatives of the improper integrals given easy to find?

Thanks.

答:
If you exclude some irrelevant finite piece near the origin then $1/x^2$ easily bounds the first integral, yes.

For the second one it's easiest to just use something like $-e^{-x/2}$. The top is eventually smaller than any exponential and the bottom grows exponentially.

Neither has a simple antiderivative, and only the first one has an antiderivative expressible in terms of elementary functions.


问:
Is $\Bbb R$ closed under subtraction?
It is easy to see that:
$∀(x,y)∈ \Bbb R$

$(x+y) ∈ \Bbb R$

But for example $(-7)-(-10)≠(-10)-(-7)$

so if I understand correctly $\Bbb R$
is not closed under subtraction?

答: All you have shown is that subtraction is not commutative. Since $(\mathbb R,+)$ is an abelian group, it must be closed under subtraction when $a-b$ is defined as $a+(-b)$.

Title: Is $\Bbb R$ closed under subtraction?


答:
You are conflating commutativity and closure.

To assert closure, you'd need to assess

Is  $(-7)-(-10)\in\Bbb{R}?$

Which is true. However, you've already pointed out that $a-b \ne b-a.$

Title: Is $\Bbb R$ closed under subtraction?

Body:
It is easy to see that:
$∀(x,y)∈ \Bbb R$

$(x+y) ∈ \Bbb R$

But for example $(-7)-(-10)≠(-10)-(-7)$

so if I understand correctly $\Bbb R$
is not closed under subtraction?


问:
Intuition behind the Triple Product identity
The well-known identity:
$$
\textbf{c}\times(\textbf{a}\times\textbf{b})=(\textbf{c}\cdot \textbf{b})\textbf{a}-(\textbf{c}\cdot \textbf{a})\textbf{b}
$$
and its counterpart for a curl:
$$
\nabla \times(\nabla \times A)=\nabla(\nabla\cdot A)-\nabla ^2 A
$$
are extremely useful, but I can't wrap my head around the intuition (geometrical or otherwise).

How exactly can these results be interpreted?

答:
I know this answer is $6$ years late, but I just thought I'd share another perspective on the vector triple product. So we want to show that:$$\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$$
Since this is just for the sake of conveying intuition, begin by assuming that the vector $\vec{b}$ lies purely along the $x$-axis (i.e. $\vec{b}=b_{1}\hat{i}$). Then let $\vec{c}$ lie somewhere in the $xy$ plane (i.e. $\vec{c}=c_{1}\hat{i}+c_{2}\hat{j}$). Finally, let $\vec{a}$ lie anywhere (i.e. $\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$). Intuitively one can just rotate this whole set-up and have the same basic dynamics play out, so it's generalizable (but again, this is just for intuition purposes). Now we make a few simple but crucial realizations: the $\hat{i}$-component of $\vec{c}$ (i.e. $c_{1}$) actually has no influence on the cross product $\vec{a}\times(\vec{b}\times\vec{c})$, since the (absolute value of the) signed area of the parallelogram spanned by $\vec{b}$ and $\vec{c}$ (which represents $||\vec{b}\times\vec{c}||$) only depends on the length of the parallelogram's base (i.e. $||\vec{b}||=|b_{1}|$) and the parallelogram's height (i.e. $|c_{2}|$). Furthermore, you can intuit that the cross product's direction also doesn't depend on what $c_{1}$ is. A similar line of reasoning allows us to neglect the $\hat{k}$-component of $\vec{a}$ (i.e. $a_{3}$ has no effect on $\vec{a}\times(\vec{b}\times\vec{c})$).

Therefore, for the sake of intuition, let us set $c_{1}=a_{3}=0$. As a result, we now have $\vec{b}$ lying purely along the $x$-axis, $\vec{c}$ lying orthogonal to $\vec{b}$ along the $y$-axis, and $\vec{a}$ lying somewhere on the $xy$ plane. In fact, for the moment, let us go even further and pretend $\vec{b}=\hat{i}$ (i.e. $b_{1}=1$) and $\vec{c}=\hat{j}$ (i.e. $c_{2}=1$). We know, out of context, that any vector $\vec{a}$ in $\mathbb{R^2}$ can be represented as:$$\vec{a}=(\vec{a}\cdot{\hat{i}})\hat{i}+(\vec{a}\cdot{\hat{j}})\hat{j}$$ When it is the case then that $\vec{b}=\hat{i}$ and $\vec{c}=\hat{j}$, then we have $\vec{b}\times\vec{c}=\hat{k}$. Then, from basic cross product intuition, we see that the vector $\vec{a}\times(\vec{b}\times\vec{c})$ will be proportional to a $90$-degree clockwise rotation of vector $\vec{a}$ (when viewed looking at the $xy$ plane from the head of vector $\vec{b}\times{\vec{c}}$). Furthermore, the magnitude of this cross product vector is equal to the magnitude of vector $\vec{a}$ (think of the rectangle's area). In this case, because $\vec{b}\times{\vec{c}}=\hat{k}$, the appropriate linear transformation that accomplishes the $90$-degree clockwise rotation is: $$\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}$$
(where we have pretended that all our vectors are in $\mathbb{R^2}$ instead of $\mathbb{R^3}$, though a similar rotation matrix exists for the $\mathbb{R^3}$ case). Thus: \begin{align}\vec{a}\times(\vec{b}\times\vec{c}) &=\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}\begin{bmatrix} \vec{a}\cdot{\hat{i}}\\ \vec{a}\cdot{\hat{j}}\\ \end{bmatrix} \\
& = \begin{bmatrix}\vec{a}\cdot{\hat{j}}\\-\vec{a}\cdot{\hat{i}}\\ \end{bmatrix}\\&=(\vec{a}\cdot{\hat{j}})\hat{i}-(\vec{a}\cdot{\hat{i}})\hat{j}\\&=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\end{align}
Now let us generalize a bit so that instead we have $\vec{b}=b_{1}\hat{i}$ and $\vec{c}=c_{2}\hat{j}$. First, suppose that $b_{1}$ is positive. Then if $c_{2}>0$ as well, we have (by the exact same logic as above): $$\vec{a}=\biggr(\vec{a}\cdot{\frac{\vec{b}}{||\vec{b}||}}\biggr)\frac{\vec{b}}{||\vec{b}||}+\biggr(\vec{a}\cdot{\frac{\vec{c}}{||\vec{c}||}}\biggr)\frac{\vec{c}}{||\vec{c}||}$$
But this time, the magnitude of the cross product $||\vec{a}\times(\vec{b}\times\vec{c})||$ will be the product of all $3$ of their magnitudes (can you see why?). So we can rotate vector $\vec{a}$ by $90$-degrees clockwise and then scale it up by $||\vec{b}||||\vec{c}||$ after (or before, since it's linear) the rotation: \begin{align}\vec{a}\times(\vec{b}\times\vec{c})\\&=||\vec{b}||||\vec{c}||\begin{bmatrix}0&1\\-1&0\\\end{bmatrix}\begin{bmatrix}\vec{a}\cdot{\frac{\vec{b}}{||\vec{b}||}}\\ \vec{a}\cdot{\frac{\vec{c}}{||\vec{c}||}}\\ \end{bmatrix}\\&=||\vec{b}||(\vec{a}\cdot{\vec{c}})\hat{i}-||\vec{c}||({\vec{a}\cdot{\vec{b}}})\hat{j}\\&=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}\end{align}
I'll leave the rest of the cases as an exercise to the reader.