问:
Smullyan's Godel's Incompleteness Theorems Chapter 2, Exercise 8
The question asks to prove that for any two Arithmetic sets $A$ and $B$ there are sentences $X$ and $Y$ such that $X$ is true iff $A$ contains the Godel number of $Y$, and $Y$ is true iff $B$ contains the Godel number of $X$.

Since $A$ is arithmetic there is a predicate $F_A(v_1)$ that expresses $A$ and an predicate $H_A(v_1)$ that expresses $A^{\ast} := \left\{m \in \mathbb{N} : d(m)\in A\right\}$, where $d(m)$ is the diagonal of the Godel number $m$.  Then, if $h_A$ is the Godel number of $H_A$, then $H_A(h_A)$ is a sentences that is true iff the $A$ contains Godel number of $H_A(h_A)$.  I can't figure out how to cross-reference: finding $X$ that asserts the Godel number of $Y$ is in $A$ and finding $Y$ that asserts the Godel number of $X$ is in $B$.  Any help is appreciated.

答:
Here is Jeon's answer:

Let $A \subseteq \mathbb{N}$ and $B \subseteq \mathbb{N}$ be two Arithmetic sets.

Then there exist two formulas $H_A \left( v_1 \right)$ and $H_B \left( v_1 \right)$ such that $$ \left( \forall n \in \mathbb{N} \right) \left[ H_A \left( \overline{n} \right) \in \mathcal{T} \leftrightarrow n \in A \right] $$ and $$ \left( \forall n \in \mathbb{N} \right) \left[ H_B \left( \overline{n} \right) \in \mathcal{T} \leftrightarrow n \in B \right] $$ hold.

Denote $F \left[ \overline{x} , \overline{y} \right] : \equiv \forall v_1 \forall v_2 \left( v_1 = \overline{x} \supset \left( v_2 = \overline{y} \supset F \right) \right) $, where $F$ is any formula and $x$ and $y$ are any numbers.

Define $\mathrm{fst} : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ and $\mathrm{snd} : \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ by $$\mathrm{fst} \left( x , y \right) := g \left( E_x \left[ \overline{x} , \overline{y} \right] \right)$$ and $$\mathrm{snd} \left( x , y \right) := g \left( E_y \left[ \overline{x} , \overline{y} \right] \right)$$ respectively.

Since $\mathrm{fst}$ and $\mathrm{snd}$ are Arithmetic functions, there exist two formulas $\mathrm{Fst} \left( v_1, v_2, v_3 \right)$ and $\mathrm{Snd} \left( v_1, v_2, v_3 \right)$ such that $$\left( \forall x \in \mathbb{N} \right) \left( \forall y \in \mathbb{N} \right) \left( \forall z \in \mathbb{N} \right) \left[ \mathrm{Fst} \left( \overline{x} , \overline{y} , \overline{z} \right) \in \mathcal{T} \leftrightarrow \mathrm{fst} \left( x , y \right) = z \right]$$ and $$\left( \forall x \in \mathbb{N} \right) \left( \forall y \in \mathbb{N} \right) \left( \forall z \in \mathbb{N} \right) \left[ \mathrm{Snd} \left( \overline{x} , \overline{y} , \overline{z} \right) \in \mathcal{T} \leftrightarrow \mathrm{snd} \left( x , y \right) = z \right]$$ hold.

Define $$\begin{align*} F_1 \left( v_1 , v_2 \right) & : \equiv \forall v_3 \left( \mathrm{Snd} \left( v_1 , v_2 , v_3 \right) \supset H_A \left( v_3 \right) \right) , \\ F_2 \left( v_1 , v_2 \right) & : \equiv \forall v_3 \left( \mathrm{Fst} \left( v_1 , v_2 , v_3 \right) \supset H_B \left( v_3 \right) \right) . \end{align*}$$

Then $$F_1 \left[ \overline{g \left( F_1 \right)} , \overline{g \left( F_2 \right)} \right] \in \mathcal{T} \leftrightarrow H_A \left( \overline{g \left( F_2 \left[ \overline{g \left( F_1 \right)} , \overline{g \left( F_2 \right)} \right] \right)} \right) \in \mathcal{T}$$ and $$F_2 \left[ \overline{g \left( F_1 \right)} , \overline{g \left( F_2 \right)} \right] \in \mathcal{T} \leftrightarrow H_B \left( \overline{g \left( F_1 \left[ \overline{g \left( F_1 \right)} , \overline{g \left( F_2 \right)} \right] \right)} \right) \in \mathcal{T}$$ hold.

Now, take $X : \equiv F_1 \left[ \overline{g \left( F_1 \right)} , \overline{g \left( F_2 \right)} \right]$ and $Y : \equiv F_2 \left[ \overline{g \left( F_1 \right)} , \overline{g \left( F_2 \right)} \right]$.

Then $X \in \mathcal{T} \iff g \left( Y \right) \in A$ and $Y \in \mathcal{T} \iff g \left( X \right) \in B$.

And we are done.


问:
Show that solutions of $f'(x)=f(x+\frac{\pi}{2})$ have real roots.
Given that $f: \mathbb{R} \to \mathbb{R}$ is differentiable and $f'(x)=f(x+\frac{\pi}{2})$ for all $x\in \mathbb{R}$, prove that there exists $x_0$ such that $f(x_0)=0$

I thought that $f(x)=\cos(x)$ but I don't know how to prove it. Moreover is there a solution without using the $\cos(x)$?

答:
Let us consider the equation
$$
 f'(x) = f(x+a)
$$
with $a \ge 1$ instead. Assume that $f$ has no zeros. Without loss of generality we can assume that $f$ is strictly positive.

It follows that $f$ and $f'$ are strictly increasing, so that for $x < y$
$$ \tag {*}
 f(y) > f(x) + (y-x) f'(x) \, .
$$
In particular,  $\lim_{x \to \infty} f(x) = +\infty$.

On the other hand, using $(*)$ again,
$$
 f'(x) = f(x + a) > f(x) + a f'(x)
$$
implies
$$
 (a-1)f'(x) + f(x) < 0
$$

If $a=1$ then $f(x) < 0$ gives an immediate contradiction. If $a > 1$ then the last inequality shows that
$$
 e^{x/(a-1)} f(x)
$$
is decreasing, so that
$$
 f(x) \le f(0) e^{-x/(a-1)}  \, ,
$$
contradicting the fact that $\lim_{x \to \infty} f(x) = +\infty$.

Remark: If $0 < a \le 1/e$ then $e^{\lambda a} = \lambda$ has a real solution $\lambda$, and $f(x) = e^{\lambda x}$ satisfies the functional equation and has no zeros (compare How to solve differential equations of the form $f'(x) = f(x + a)$).

ContentLicense: CC BY-SA 4.0


问:
Is $x2^x = O(2^x)$?
Is $x2^x = O(2^x)$? Intuitively it seems like the answer is no. But if we write $x2^x=2^{\log x + x}$ then the answer (to me) is unclear. Thanks in advance!

答:
Use the definition of $O(.)$ with limits:

$\lim\limits_{x→∞} \frac{f(x)}{g(x)} \ne ∞ \implies f = O(g)$

Let $f(x) = x2^x$, $g(x)=2^x$

Then $\lim\limits_{x→∞} \frac{f(x)}{g(x)} = \lim\limits_{x→∞} \frac{x2^x}{2^x} = \lim\limits_{x→∞}x = ∞$ then we can't say $x2^x=O(2^x)$

To the contrary, if we let $f(x) = 2^x$, $g(x)=x2^x$,

$\lim\limits_{x→∞} \frac{f(x)}{g(x)} = \lim\limits_{x→∞} \frac{2^x}{x2^x} = \lim\limits_{x→∞}\frac{1}{x} = 0 \implies 2^x=o(x2^x)$

Title: Is $x2^x = O(2^x)$?


问:
Sigma notation formula $i=0$
I am trying to help a high school student to solve summation calculation problems. The $i$ value of most answers I found was based on $i=1$ like this:

Now, the $i$ value of our questions is $i=0$ (instead of $i=1$). Is there any ready formula available for us to solve these questions? And what if $i=2$ or other numbers? Thanks for your help!

答:
In general, you want to make sure that you start/stop the sum with the same terms. For example, this sum starts from 7 and ends at 20:
$$\sum_{i=1}^{14}(i+6).$$
If we want $i$ to start from $3$ (say), then we need to adjust the function inside the sum, and the upper limit, so that it still starts from 7 and ends at 20:
$$\sum_{i=3}^{16}(i+4).$$
In general, for any "shift" $D$, we have
$$\sum_{i=a}^b f(i) = \sum_{i=a+D}^{b+D}f(i-D) = \sum_{i=a-D}^{b-D}f(i+D).$$
In your specific case however, for 2, 3, and 4 you don't need to change anything, since starting from $i=0$ simply adds $0+$ in front. For number 1, you have
$$\sum_{i=1}^nc=\sum_{i=0}^{n-1}c=cn.$$
This is true for any $n$, so if you want you can put $n=N+1$ to get
$$\sum_{i=0}^N c=c(N+1).$$

Title: Sigma notation formula $i=0$


问:
Derivative of order 16 - is there a method to do so?
I have the following exercise:

Find the $16^{\text{th}}$ derivative of $y$, (i.e. $y^{(16)}$), for $y = \sin x$.

Is there any method to do so, or I simply have to differentiate the function $16$ times?

答:
Hint:

$y'=\cos x$

$y''= -\sin x=-y$

$y'''=- \cos x=-y'$

$y''''=\sin x=y$

Do you notice the recursion? $y^{4k}= \dots $?

ContentLicense: CC BY-SA 3.0


问:
Extending a field isomorphism to a field embedding of a bigger field
Let $E$ and $F$ be two fields and let $f:E \longrightarrow F$ be a field isomorphism. Let $\Omega$ be an algebraically closed field containing both $E$ and $F$. Is it possible to extend $f$ to a field embedding $\tilde{f}: \Omega \longrightarrow \Omega$?

I tried to apply Zorn's Lemma to the set

$$\{(M,f_M): E \subseteq M \subseteq \Omega \text{ is a field and } f_M : M \longrightarrow \Omega \text{ is a field embedding extending } f\}$$

with the following partial order relation

$$(M_1,f_{M_1}) \le (M_2,f_{M_2}) \Longleftrightarrow M_1 \subseteq M_2 \text{ and } f_{M_2} \text{ restricted to } M_1 \text{ equals } f_{M_1}.$$
The assumptions are satisfied but the maximal element could still have a domain strictly smaller than $\Omega$ and I don't know how (and if it is possible) to disprove that.

Is there a way to do it or is there a proof without using Zorn's lemma?

Many Thanks!

答: This doesn't seem to be true. Take $F=\Omega$ to be an algebraically closed field which is isomorphic to a proper subfield $E$ (i.e. has infinite transcendence degree over its prime field). Let $f:E\to F$ be such an isomorphism. Then it is not possible to extend $f$ to a homomorphism $\Omega\to\Omega$.


答:
Just use $i = 0$ in your image. Then for 2), 3), 4) you just add $0$ to your sum.
In 1) you add another c, so it changes the summation but it is easy to think about it. Try it yourself.

For $i \geq 2$ you can write it differently, e.g.
$$\sum\limits_{i=2}^n c = \sum\limits_{i=1}^n c - \sum\limits_{i=1}^1 c = cn - c = (n-1)c$$

Title: Sigma notation formula $i=0$

Body:
I am trying to help a high school student to solve summation calculation problems. The $i$ value of most answers I found was based on $i=1$ like this:

Now, the $i$ value of our questions is $i=0$ (instead of $i=1$). Is there any ready formula available for us to solve these questions? And what if $i=2$ or other numbers? Thanks for your help!


问:
I have a question about limit supremum and limit infimum.
Let $\{x_n\}_{n=1}^{\infty}$ and $\{y_n\}_{n=1}^{\infty}$ be sequences of real numbers. Does the following hold:

$$
\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n).
$$

This is what I have tried but I am not quite sure if it is correct.
$\text{Fix } K>1. \text{ Let }L=\inf_{1\le i \le k}y_i$. Now
$$
\sup_{1 \le i \le k}(x_i+y_i)\ge \sup_{1\le i \le k}(x_i+L)=L+\sup_{1\le i \le k}(x_i) =
\inf_{1\le i \le k}(y_i)+\sup_{1\le i \le k}(x_i).
$$
Now we take the limit as $k \rightarrow \infty$ and we have the desired result. Does this look OK?

答:
Let $~\limsup x_n=\overline x~$ and $~\liminf y_n=\underline y~,$ then $~\overline x~,~\underline y~\in \mathbb R.$

Let $~\epsilon>0~$ be given.

Then $~\exists~~k\in \mathbb N~$ such that
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x_n>\overline x-\epsilon~/~2~,~~\forall~~n\ge k~$ and
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~y_n>\underline y-\epsilon~/~2~,$ for infinitely many value of $~n~.$

Thus $~x_n+y_n>\overline x+\underline y-\epsilon~,$ for infinitely many value of $~n~.$

Therefore for given $~m\in\mathbb N~,~~\exists~~n_0\ge m~$ such that $$x_{n_0}+y_{n_0}>\overline x+\underline y-\epsilon$$
$$\implies \sup_{n\ge m}\left(x_n+y_n\right)\ge x_{n_0}+y_{n_0}>\overline x+\underline y-\epsilon~,$$ for each $~m\in \mathbb N~.$

Hence $$\overline x+\underline y-\epsilon\le\limsup\,(x_n+y_n)$$
Since $~\epsilon > 0~$ be arbitrary, $$\limsup x_n +\liminf y_n \le \limsup\,(x_n+y_n)~.$$


问:
Get numbers that have only 2,3 and 5 as prime factors
I am given an integer N. I have to find first N elements that are divisible by 2,3 or 5, but not by any other prime number.

```<code>N = 3
Results: 2,3,5
N = 5
Results: 2,3,5,6,8
</code>```

Mistake number = 55.. 55/5 = 11.. 11 is prime number.. so means that it divides by any other prime number and doesn't counts in..

I guess there is need of recursive function,but I cant imagine what would algorithm look like

I was sent here from stackoverflow.

They gave me advice.. 2^i*3^j*5^k , for i,j,k = 0,1,2... , but it won't give an ordered array, because 2*2 < 2*3 .. and so on

I am trying to get this work in C++

答:
I get similar type of question on competitive programming, although I'm beginner, my code is not much optimized but I think it will help.</p>
```<code>// 1 2 3 4 5 6 8 9 10 12 14 15 16 18 20 21......etc//
// 1 2 3 4 5 6 7 8 9  10 11 12 13 14 15 16......etc//

#include<iostream>
using namespace std;

int main()
{
  int n;
  int size = 1000;
  int arr[size];
  int count = 1;
  int str = 1;
  int temp = 0;

  cin>>n;
  temp = n;

  while(count <= temp)
  {
    //cout<<"looping count : "<<count<<endl;

    if(temp == 1)
    {
      arr[count] = str;
    }
    if(count == 1)
    {
      arr[count] = str;
    }
    else if(count > 1)
    {
      str += 1;

      if(str % 2 == 0 || str % 3 == 0 || str % 5 == 0)
      {
        arr[count] = str;
      }
      else
      {
        str += 1;
        arr[count] = str;
      }

    }

    count += 1;
  }

  for(count = 1; count <= temp; count++)
    cout<<"Array at "<<count<<" "<<arr[count]<<endl;

  cout<<"End str : "<<str<<endl;

  return 0;
}
</code>```

Body:
I get similar type of question on competitive programming, although I'm beginner, my code is not much optimized but I think it will help.

// 1 2 3 4 5 6 8 9 10 12 14 15 16 18 20 21......etc//
// 1 2 3 4 5 6 7 8 9  10 11 12 13 14 15 16......etc//

#include<iostream>
using namespace std;

int main()
{
  int n;
  int size = 1000;
  int arr[size];
  int count = 1;
  int str = 1;
  int temp = 0;

  cin>>n;
  temp = n;

  while(count <= temp)
  {
    //cout<<"looping count : "<<count<<endl;

    if(temp == 1)
    {
      arr[count] = str;
    }
    if(count == 1)
    {
      arr[count] = str;
    }
    else if(count > 1)
    {
      str += 1;

      if(str % 2 == 0 || str % 3 == 0 || str % 5 == 0)
      {
        arr[count] = str;
      }
      else
      {
        str += 1;
        arr[count] = str;
      }

    }

    count += 1;
  }

  for(count = 1; count <= temp; count++)
    cout<<"Array at "<<count<<" "<<arr[count]<<endl;

  cout<<"End str : "<<str<<endl;

  return 0;
}


问:
question about calculus $f(x)=e^x(x^2+x)$, derive $\frac{d^n\,f(x)}{dx^n}$
$f(x)=e^x(x^2+x)$, derive $\dfrac{d^n\,f(x)}{dx^n}$

may use Leibniz formula but i'm not sure:(

答:
It seems a lit bit hard but if you try to find the pattern from n=1 to n=2 ,
so $(e^x (x^2 + x))\prime$ = $e^x (x^2 + 3x +1)$ this is for n egale to 1 you cannot get the pattern , so for $n = 2\quad , (e^x(x^2 + x))\prime\prime = e^x (x^2 + 5x + 4 )$ , to deduce calculate for $n=3$  :
$(e^x(x^2+x))\prime\prime\prime$ = $e^x (x^2 + 7x + 9 )$ , the patern is $\dfrac{d^{n}}{dx^{n}}f(x) = e^x ( x^2 + (2n+1)x +n^2) ...$ If you want to prove the formula you could do the induction .
Hope my answer helped you!


问:
Prove that $2^{1/3}$ is irrational.
Proof. Assume by way of contradiction that $\sqrt[3] 2$ is rational. Then $\sqrt[3] 2$ can be written as $a/b$ where $ a,b\in\mathbb{N},b\neq0$. Then $(a^3/b^3 = 2) = (a^3 = 2b^3)$. Thus $2|a^3$ and by the lemma, $2|a$. Then $a=2x $ where $x\in\mathbb{Z} $. Substituting $a=2x$ into $a^3 = 2b^3$ we get $8x^3 = 2b^3$, simplifying we get $4x^3 = b^3$. Which implies $4|b^3$ and then by the lemma, $4|b$. This is a contradiction because if a is a multiple of $2$ and $b$ is multiple of $4$, then they are not in lowest terms. Hence $\sqrt[3] 2$ is irrational.

I think I am good up til the $2|a^3$ part. I don't know if I can say that, "by the lemma $2|a$". Any tips or help is much appreciated!

答:
The present problem may be resolved if we can show that for primes $p$ with
$p \mid a^n$, $n \ge 1$, then $p \mid a$.

We note that if

$p \not \mid a \tag 0$

then

$\gcd(p, a) = 1; \tag 1$

for the only divisors of $p$ are $p$ and $1$; and if $p \not \mid a$, we are left with $1$ as the sole common divisor of $p$ and $a$; now in light of (1) we have, via Bezout's identity,

$xp + ya = 1, \tag 2$

for some

$x, y \in \Bbb Z; \tag 3$

multiplying (2) by $a^{n - 1}$ yields

$xpa^{n - 1} + ya^n = a^{n - 1}, \tag 4$

which since $p \mid a^n$ yields

$p \mid a^{n - 1}; \tag 5$

we now repeat the argument, replacing $a^{n - 1}$ with $a^{n - 2}$; then from (2),

$xpa^{n - 2} + ya^{n - 1} = a^{n - 2}; \tag 6$

now, again by virtue of (4) we find

$p \mid a^{n - 2}; \tag 7$

continuing in this manner we reach

$p \mid a \tag 8$

after $n - 1$ repititions.  But this stands in contradiction to (0), which must thus be false; hence (8) binds.

Title: Prove that $2^{1/3}$ is irrational.


答:
Just by calculating $f',f''$ and $f'''$ by hand, you can see the following pattern:
$$\frac{\mathrm{d}^nf}{\mathrm{d}x^n}(x) = e^x(x^2+(2n+1)x+n^2).$$

Now, we can prove this by induction (the base steps are already done if you did the calculations). Assume $\frac{\mathrm{d}^nf}{\mathrm{d}x}(x) = e^x(x^2+(2n+1)x+n^2)$ for some $n\geq 1$, then $$\frac{\mathrm{d}^{n+1}f}{\mathrm{d}x^{n+1}}(x) = e^x(x^2+(2n+1)x+n^2)+e^x(2x+2n+1)$$

The latter can be simplified as $$e^x(x^2+(2(n+1)+1)x+n^2+2n+1)=e^x(x^2+(2(n+1)+1)x+(n+1)^2).$$ This completes the induction step.

Title: question about calculus $f(x)=e^x(x^2+x)$, derive $\frac{d^n\,f(x)}{dx^n}$

Body:
$f(x)=e^x(x^2+x)$, derive $\dfrac{d^n\,f(x)}{dx^n}$

may use Leibniz formula but i'm not sure:(


问:
How can I solve the curve integral of $\frac{cos(z)}{z^2}$ with the unit circle as curve?
I tried to make a partial fraction decomposition but that didn't work. Can someone show me how to do it with the residue theorem?</p>

答:
By Cauchy's integral formula:
$$\int_{\gamma}\frac{\cos(z)}{z^2}dz=\frac{2\pi i}{1!}\cos'(0)=0$$
where $\gamma$ is the unit circle curve.


问:
Finding the third vertex of a isosceles triangle given the vertices of the base and knowing that the medians are perpendicular
Full problem:
The points A(0, c) and B(c, 0) are the vetrices of the base of an isosceles triangle. In addition, the medians AD and BE are perpendicular to each other.
What are the coordinates of the third vertex?

Even if I figure out the lengths of all sides, medians included, how do I find the coordinates of the vertex?

答:
Extend $\overline{AD}$ to $\overline{AE}$ such that $AD = DE$.

Bisect $\overline{DE}$ at point $G$.

Draw line $\overleftrightarrow{BG}$

Intersect line $\overleftrightarrow{BG}$ and $\overleftrightarrow{CD}$ at point F.


问:
Given a recursion $a_{n+1}=\sqrt{a_n^2+a_n}$ with $a_1=1.$ Prove that $\lim{\left(a_n\right)}'=\frac12$ without using $a_n\sim\frac n2-\frac14\ln n$

Given a recursion $a_{n+ 1}= \sqrt{a_{n}^{2}+ a_{n}}$ with $a_{1}= 1.$ Prove that $\lim{\left ( a_{n} \right )}'= \frac{1}{2}$ without using the result I've got
$$a_{n}\sim\frac{n}{2}- \frac{1}{4}\ln n\,{\rm as}\,n\rightarrow\infty$$
Source: StackMath/@haidangel_ Estimations of some new recurrence sequences

of course I used L'Hopital rule to remake the problem, we may need other approaches. All your nice comments and solutions are welcome and appreciated. I need to the help, thanks a real lot !

答:
There is no derivative for an integer-valued variable $n$, but there's an obvious replacement: the difference. Since $a_{n+1}\ge a_n\ge1$, we must have $a_n\to\infty$ as $n\to\infty$ (otherwise, the monotone sequence would have a finite limit $a\ge1$ satisfying $a=\sqrt{a^2+a}$, that's impossible). Then,
$$a_{n+1}-a_n=\frac{a_n}{\sqrt{a^2_n+a_n}+a_n}=\frac1{\sqrt{1+\frac1{a_n}}+1}\to\frac12$$ as $n\to\infty$.


问:
Proof related to existence of root in an odd degree polynomials
Let n be odd and $a_0, a_1, \dots; a_{n-1}  \in \mathbb{R}$. Define $f : \mathbb{R} \to \mathbb{R}$ by
$f(x) = x_n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$.

How do I prove that if $x_1 \gt max\{1, 2 \sum_{i=0}^{n-1} |a_i|\}$, $f(x_1)>0$?

Progress: I was able to prove the following:

$|\frac{a_{n-1}}{x} + \dots + \frac{a_0}{x}| \lt \frac{1}{2}$ if $x=max$ $\{{1, 2 \sum_{i=0}^{n-1} |a_i|}\}$ and also

$|\frac{a_{n-1}}{x} + \dots + \frac{a_0}{x}| \lt \frac{1}{|x|} (|a_{n-1}| + \dots + |a_1| + |a_0|)$ if $x=max$ $\{{1, 2 \sum_{i=0}^{n-1} |a_i|}\}$


问:
Collecting math websites
I would like to know math websites that are useful for students, PhD students and researchers (useful in the sense most of the students or researchers—of a particular area—are using it). Maybe you can share which math websites you sometime use and why you use it.

Let me give my websites and why I use them:

MathOverflow and Stack Exchange: Very good to look up questions about how to do mathematics, what mathematics is, how it evolves, and of course to have an exchange with smart people, who can help you if you are stuck at something.

arXiv: I don't use it now, but I think that's the main website to read papers and to publish (research level)

The Stacks Project: It is very useful for me to look algebraic things ups with more explanation than in some lectures...

MathSciNet: To search for papers that may help you.

Number Theory Web: Since I want to become a number theorist, this site helps me to see which kind of things exist.

Do you have more websites that help you to understand maybe mathematics, or just to connect to other mathematicians?

Everything is welcome. Just post it as an answer and not a comment.

答:
I really enjoy the following schematic overview of various statistical, distributions, their relationships and properties. It's quite handy for giving students (and others!) a quick way of relating new distributions to distributions that they already know about.

Univariate Distribution Relationships

edit: typos

Title: Collecting math websites


问:
Interesting way to integrate a function with two variables
My question came up with this differential Equation:

$$(x + y)(\Bbb dx-\Bbb dy) = \Bbb dx+\Bbb dy$$

I've seen a lot of indian people in Youtube doing integration like this (i'll show you while i'm solving the ODE):

$$\begin{align}
\Bbb dx - \Bbb dy &= \frac{\Bbb dx + \Bbb dy}{x + y}\\
x - y &= \ln(x+y) + c
\end{align}$$

That implies:
$$\int \frac{\Bbb dx + \Bbb dy}{x + y} = \ln(x+y) + c$$

But how? How does it make sense? I'm trying to split the integral but i got this:

$$\begin{align}
\int \frac{\Bbb dx + \Bbb dy}{x + y} &=
\int \frac{\Bbb dx}{x + y} + \int \frac{\Bbb dy}{x + y}\\
&= \ln(x+y) + \ln(x+y)\\
&= 2\ln(x+y)
\end{align}$$

Honestly i still can't process this. But i've verified with an implicit differentation that the solution satisfies the ODE.

答:
It seems that the difficulty of understanding comes from a change of variables.
$$(x+y)(dx-dy)=dx+dy$$
So we will detail all steps of the change of variables :
$$\begin{cases}
X=x-y\\
Y=x+y
\end{cases} \quad\implies\quad \begin{cases}
dX=dx-dy \\
dY=dx+dy
\end{cases}$$
Puting them into the original equation :

$$Y\:dX=dY$$
This is an elementary separable ODE which solution is well known :
$$Y=c\:\exp(X)$$
$$x+y=c\:e^{x-y}$$
This is the solution expressed on the form of an implicit equation.

One cannot solve this implicit equation for $y(x)$ in terms of a finite number of elementary functions. One have either to expand into infinite series or to use a special function namely the Lambert W function.

For information the manner to get the explicit solution is :
$$(x+y)e^{x+y}=c\:e^{x-y}e^{x+y}=c\:e^{2x}$$
Let $z=x+y$ and $t=c\:e^{2x}$
$$ze^z=t\quad\implies\quad z=W(t)$$
https://mathworld.wolfram.com/LambertW-Function.html

$$x+y=W(c\:e^{2x})$$
$$y(x)=-x+W(c\:e^{2x})$$


问:
Hard Year 11 Irrational and Rational Number problem
Suppose that $a$ and $b$ are positive irrational numbers, where $a < b$. Choose any positive integer $n$ such that $1/n < b − a$, and let $p$ be the greatest integer such that $p/n < a$:

a) Prove that the rational number $\frac{p + 1}{n}$ lies between $a$ and $b$.

b) If $a = \frac{1}{\sqrt{1001}}$, and $b = \frac{1}{\sqrt{1000}}$, find the least possible value of $n$ and the corresponding value of $p$.

c) Hence use part a) to construct a rational number between $\frac{1}{\sqrt{1001}}$ and $\frac{1}{\sqrt{1000}}$.

I am currently not learned in mathematical induction so im not really sure how to solve this. was thinking if there was another way of solving this question.

This question was taken from the Cambridge Maths Extension 1 Year 11 Book. Would like an intuitive explanation that a high school student in year 11 like me can understand since even my teachers cant seem to solve this problem lol. Sorry if this is a hassle of your time.

答:
Hints: for part (a), you need to use that
\begin{align}
\frac{1}{n} &< b - a \\[4pt]
\frac{p}{n} &< a \, .
\end{align}

For part (b), we are told that
\begin{align}
b-a &= \frac{1}{\sqrt{1000}}-\frac{1}{\sqrt{1001}} \\[4pt]
&= \frac{\sqrt{1001}-\sqrt{1000}}{\sqrt{1000}\sqrt{1001}}
\end{align}
We need to find the smallest value of $n$ such that
$$
\frac{1}{n}<\frac{\sqrt{1001}-\sqrt{1000}}{\sqrt{1000}\sqrt{1001}}
$$
Rearrange this inequality to find the answer. Part (c) should be straightforward: use the values of $p$ and $n$ that you found in part (b) to get the answer.


问:
Minimum distance between parabolas
Is there a simple way to find the minimum distance between two parabolas?

For example, between y=-0.1(x-17)2 + 8.6 {7.726 < x < 19.134}
and y=-0.12(x-17.5)2 + 6.2 {10.313 < x < 18.829}

See graph of the two parabolas here (the exponential function simply determines the restricted domain for the parabolas)

Context: I have to ensure that there is a minimum of 1.8 units between these two parabolas, to ensure the "path" is at least 1.8 metres wide always.

Attempt: I thought about trying to find the closest distance visually then finding the normal and subbing in for x and y at that specific point, but that doesn't seem very scientific and I'm a bit stuck.

Thanks in advance

答:
In general, the shortest line between two curves is the one that is normal to both of them, unless this line is at the boundary of the curves.

Let $y_1(x)=a_1(x-b_1)^2+c_1$ and $y_2(x)=a_2(x-b_2)^2+c_2$.

Then the tangents at $x$ are $y_i'(x)=2a_i(x-b_i)$ and the normal lines are $$\mathbf{r}_i^N(x,t)=\begin{pmatrix}x\\y_i(x)\end{pmatrix}+t\begin{pmatrix}1\\-1/y_i'(x)\end{pmatrix}$$

The requirements on $x_i,t$ are that $$\mathbf{r}_1^N(x_1,t)=\begin{pmatrix}x_2\\y_2(x_2)\end{pmatrix},\quad -1/y_1'(x_1)=-1/y_2'(x_2)$$ These three equations, when solved, give $$x_1+t=x_2,\\ a_1(x_1-b_1)^2+c_1-\frac{t}{2a_1(x_1-b_1)}=a_2(x_2-b_2)^2+c_2,\\ a_1(x_1-b_1)=a_2(x_2-b_2)$$

Solving gives $$x_1=8.23107, x_2=10.1926, t=1.96149,\\
x_1=27.4119, x_2=26.1765, t=-1.23531,\\
x_1=15.3571, x_2=16.1309, t=0.77382$$

The first and third are within the given ranges; of these the distance of 2.26 is minimised by the first.

Title: Minimum distance between parabolas


问:
I need visual examples of topological concepts
I'm trying to understand basic concepts of topology, unfortunately I'm a very visual person, and as much documentation there is on how to come up with closed/open/clopen/etc. There are very few visual examples (using actual sets of integers, shapes, etc). So it's very hard for me to understand. Not everyone learns through text or mathematical definitions, so I figured this could help other people that have the same difficulties I have. Thanks for your help!

Can anyone get an example of a topology over a set.

I need one for a topology over a set, an open set, closed set, and clopen set. It needs to be something real. Like, say a set contained numbers { 1,2,3,4 }

It doesn't have to be that set, but it has to be a actual set of something (numbers, letters, geometric shapes), no labels (set X is the union of set Y and set Z), etc.

答: This picture is from Topology , by J. Munkres.


问:
First Course in Linear Algebra book suggestions?
I'm starting maths degree this September and hoping to do some reading on Linear Algebra before I start. What are some good introductory books?

Is there something that starts from the very beginning but also covers stuff in depth? (Is there a book that will be useful for both 1st year and 2nd year linear algebra modules?)

It would also be great if the book has lots of exercises (proof questions!) and solutions given as well.

Thanks in advance.

答:
Look at the two books:

Basic Linear Algebra by Blyth and Robertson

further Linear Algebra by Blyth and Robertson.

the second is the sequel to the first. These two books will teach you most of the stuff you will need to know. These books have a lot of exercises with solutions.

Also take a look at the classic by Hoffmann Kunze. It has lot of exercises without solutions.

The book by Axler "Linear Algebra Done Right" is decent but I will not suggest it if you are going to continue your studies towards modules as well. This book banishes determinants and polynomials in the study of linear operators. While this may be helpful over the complex field, this approach has limitations when you go to more general rings and modules.

There is another well written book "Finite Dimensional Vector spaces" by Halmos. I do not remember if it has sufficient exercises though.

For a little more advanced level, take a look at "Advanced Linear Algebra" by Steven Roman and "Matrix Analysis Vol. 1" by Horn and Johnson.


答:
You can notice that by derivation we will get $e^xP(x)$ with $P$ polynomial, also the degree of $P$ stays unchanged since $(e^x)'$ does not bring some extra $x$.

So we can set $\, f^{(n)}(x)=e^x(a_nx^2+b_nx+c_n)\, $ and try to find a recurrence relation for these sequences.

By derivation we get $\begin{cases}a_{n+1}=a_n&\quad a_0=1\\b_{n+1}=b_n+2a_n&\quad b_0=1\\c_{n+1}=c_n+b_n&\quad c_0=0\end{cases}$

Therefore $a_n$ is constant and $a_n=a_0=1$.

The sequence $b_n$ is telescopic and $$b_{n}=b_0+2\sum\limits_{k=0}^{n-1} a_k=b_{n}=1+2\sum\limits_{k=0}^{n-1} 1=2n+1$$

The sequence $c_n$ is also telescopic and $$c_n=c_0+\sum\limits_{k=0}^{n-1} b_k=0+2\sum\limits_{k=0}^{n-1} k+\sum\limits_{k=0}^{n-1} 1=2\times\frac{(n-1)n}2+n=n^2$$


问:
Escalator word problem
I've been trying to crack this, but I've just given up at this point. The problem sounds like this:

The escalator is always going down. When a boy runs up it, he counts 150 steps, and when he runs down it, he counts 30 steps. The escalator and the boy do not change their speed. How many steps would the boy count if the escalator were not to move?

I've tried many ways but i always seem to get that the discriminant is less than 0. Does anyone have an idea on how to solve this?

答:
Let $n$ be the number you are looking for, it is the "number of steps of the static escalator."

Assume that the boy runs $b$ steps per second (or any other time period of your choice) and the escalator moves down $e$ steps per second. Then the boy needs $\frac{150}b$ seconds to reach the top when running up. During this time, the
escalator moved down $150 \frac eb$ steps, so

$$150-150\frac eb = n.$$

Analogously,

$$30+30\frac eb = n.$$

This can be seen as a linear system of equations in the unknowns $\frac eb, n$. Your favorite way of solving linear systems of equations will then give the solution

$$\frac eb = \frac23, \quad n = 50,$$

so that the answer you are looking for is $n=50$.

Title: Escalator word problem


Body:
By Newton-Lebniz formula for $D^n[u(x) v(x)]$, we het
$$f(x)=(x^2+x)e^x \implies D^n[e^x(x^2+x)]= (D^n e^x) (x^2+x)+ {n\choose 1} (D^{n-1} e^x) D(x^2+x)+{n \choose 2} (D^{n-2} e^x) D^2(x^2+x)+0$$
$$f^{n}(x)=e^x(x^2+x)+n e^x (2x+1)+n(n-1)e^x=e^x[x^2+(2n+1)x+n^2].$$


问:
van der Waerden's original proof
I am looking for a book/site which has the English translation of the proof of van der Waerden's theorem as presented by van der Waerden himself. In other words is the translation of the paper:

van der Waerden, B. L. "Beweis einer Baudetschen Vermutung." Nieuw
  Arch. Wisk. 15, 212-216, 1927.

available anywhere?

Thanks.

答:
Digitized versions of the first and second series of the journal "Nieuw Archief voor Wiskunde" are available at Delpher, an open access site hosted by the Royal Library of the Netherlands (Koninklijke Bibliotheek).

The specific paper of B. L. van der Waerden[1] that you're looking for that appears in the second series of this journal can be accessed at the PURL: https://resolver.kb.nl/resolve?urn=MMKWG01:022157001:00228.
While this isn't a translation of the paper, having the original paper in hand is definitely an asset. Google Translate is decent enough that you will probably be able to understand most of the paper with its help.

Instead of a translation of the paper, you could also do as Andrés E. Caicedo suggests in the comments and take a look at the expository article written by van der Waerden himself in 1954 about how he arrived at the proof of Baudet's conjecture[2]. This article was translated from German into English and published in 1971[3], and was republished in 2009[4]. One of these versions might turn out to be more easily accessible. You will also find several of the reprints of this article indexed at zbMATH Open.

van der Waerden, B. L., Beweis einer Baudet'schen Vermutung. (German) Nieuw Arch. Wiskde. (2) 15, 212–216 (1928). JFM 53.0073.12

van der Waerden, B. L., Einfall und Überlegung in der Mathematik, 3. Mitteilung: Der Beweis der Vermutung von Baudet. (German) Elem. Math. 9, 49–56 (1954). MR0061079, ZBL 55.03802.

van der Waerden, B. L., How the proof of Baudet’s conjecture was found. Studies in pure mathematics: Papers in combinatorial theory, analysis, geometry, algebra, and the theory of numbers presented to Richard Rado on the occasion of his sixty-fifth birthday (Mirsky, L. (ed.)), Academic Press, London, 251–260 (1971). MR0270881, ZBL 241.05009.

Soifer, A., Ramsey theory before Ramsey: Van der Waerden tells the story of creation. In: The mathematical coloring book: Mathematics of coloring and the colorful life of its creators, New York, NY: Springer (ISBN 978-0-387-74640-1/hbk). pp. 309–319 (2009). MR2458293, ZBL 1221.05001.

Title: van der Waerden's original proof


问:
What if the sum of the elements of the inverse of a matrix is 0
If the sum of the elements of an inverse of a matrix $A$ is $0$, does it affect somehow the determinant or something else of the matrix $A$, where $A$ is a $3\times 3$ matrix?

I tried for a $2\times 2$ matrix, and I get that also the sum of the elements of the matrix A is also $0$, but I do not know if it applies for any  $n\times n$ matrix.

答:
This is not even true for a $2 \times 2$ matrix, take
$$A = \begin{pmatrix}
1 & 2 \\
2 & 3
\end{pmatrix}, \quad A^{-1} = \begin{pmatrix}
-3 & 2 \\
2 & -1
\end{pmatrix}$$


问:
If $4$ distinct integers $a, b, c, d$ are randomly selected from $n$ distinct integers, what is the probability that $a \lt b \lt c \lt d$?

a. If $i$ distinct integers $r_1, r_2, \ldots,r_i$, i $\leq n$, are randomly selected (one after other) from n distinct integers, what is the probability that $r_1 < r_2 < \cdots < r_i$?

b. Does the probability you got in part a of the question change if < is replaced with $\leq$?

I know that a question similar to part 'a' of this  question can be solved using the concept of Decision tree that orders the elements of the list, an application of which is to analyse the complexity of comparison-based sorting algorithm.

I came across a solution to part 'a' of the question that says the required probability is $C(n, i)/P(n,i)$, which is nothing but equal to the solution of the Decision tree based approach, which is $1/i!$.

What I don't understand is this - how $C(n,i)$ is equal to favourable number of outcomes and $P(n,i)$ is equal to total number of outcomes?

Figure: A decision tree for sorting three distinct elements.

Image taken from Discreet Mathematics and its Applications by Kenneth Rosen

答:
Another approach for solving $a)$ and $b)$ is the following:

$a)$ (I assume the numbers are selected without replacement):
Each sequence of $i$ numbers can be represented as a vector in $\{0,1\}^n$ with exactly $i$ coordinates being equal to $1$. For example, let $n=9$ and $i=4$, then $$1< 4< 5<8 \longleftrightarrow \{1,0,0,1,1,0,0,1,0\}\\ 2<3<7<9 \longleftrightarrow \{0,1,1,0,0,0,1,1,0\}\\ 5<6<7<8\longleftrightarrow \{0,0,0,0,1,1,1,1,0\}$$ This way, in order to create any feasible sequence we only have to choose the $i$ coordinates where the $1$s will be placed. This lead us to a probability of $$\dfrac{\dbinom{n}{i}}{n!/(n-i-1)!}.$$

$b)$ (I assume the numbers are selected with replacement):
In order to solve $b)$ all we have to do is to represent each sequence as a vector in $\{0,\ldots,n\}^i$ and count how many of these vectors have the sum of their coordinates equal to $i$. For example, let $n=9$ and $i=4$, then $$1\leq 2\leq 2\leq 6 \longleftrightarrow \{1,2, 0,0,0,1,0,0,0\}\\3\leq 3\leq 3\leq 3 \longleftrightarrow \{0,0,4,0,0,0,0,0,0\}\\ 1\leq 4\leq 8\leq 9 \longleftrightarrow \{1,0,0,1,0,0,0,1,1\}$$ By using stars and bars we get that the probability of having a favourable sequence is $$\dfrac{\dbinom{n+i-1}{i}}{n^i}.$$


问:
Wondering how these two sets are countably infinite
We have $A = \mathbb{R}$ and $B = \{x|x \in \mathbb{R} \land \exists y (y\in \mathbb{Z} \land |x-y| < \frac{1}{2})\}$

How is $A-B$ countably infinite?

I know the definition of set minus is $A \cap \overline{B}$, but I don't know how to translate $B$ and to see if this is a countable set.

答:
Observe:

$$A-B=\{z+\frac{1}{2}\mid z \in \Bbb Z\}$$ so it is in bijection with $\Bbb Z$.

If $x$ is a real and its fractional part is $< \frac12$ then $\lfloor x \rfloor$ witnesses that $x \in B$, so $x \notin A-B$. If the fractional part is $>\frac12$ then $\lceil x \rceil$ witnesses that $x \in B$ and also $x \notin A-B$ so the only reals in this difference are those with exactly fractional part $\frac12$, i.e. the set on the right hand side.


问:
Show that every perfect square can be written as the sum of the squares of two non integer rationals?
I was trying to prove the following statement:

Show that if $n$ is an integer, then $n^2$ can be written as $n^2=r^2+s^2$ where $r, s \in \mathbb{Q} \backslash \mathbb{Z}$, that is, $r$ and $s$ are non integer rationals.

My first idea was the most obvious one: to write $r=\frac{a}{b}$ and $s=\frac{c}{d}$ where $a, b, c, d \in \mathbb{Z}$ with $(a, b)=(c, d)=1$ and $b, d>1$.

I then tried a few things, but none of them lead to anything interesting. Could anyone help me? I'm looking for an elementary proof. I already know which integers can be expressed as the sum of two squares, and that primes of the form $4n+1$ have a unique representation as the sum of two squares, but I'm not sure how any of those would help me here.

答:
There are infinitely many primitive Pythagorean triples (i.e., where $a$, $b$ and $c$ are coprime), such as explained in Pythagorean triple. Choose any one where the $c$ in

$$a^2 + b^2 = c^2 \implies 1 = \frac{a^2}{c^2} + \frac{b^2}{c^2} \tag{1}\label{eq1A}$$

has $c \not\mid n$, e.g., $c \gt n$. Multiplying both sides of \eqref{eq1A} by $n^2$ gives

$$n^2 = \frac{(na)^2}{c^2} + \frac{(nb)^2}{c^2} \tag{2}\label{eq2A}$$

This shows we can use $r = \frac{na}{c}$ and $s = \frac{nb}{c}$ since both values are non-integral.

For example, since $(68, 285, 293)$ is a Pythagorean triple, for any $n \lt 293$ (actually, since $293$ is a prime number, we can do this for any $n$ which doesn't have a factor of $293$), we get

$$n^2 = \frac{(68n)^2}{293^2} + \frac{(285n)^2}{293^2} \tag{3}\label{eq3A}$$


问:
dimension over residue field
Sorry for my bad English.

Let $A$ be integral domain, and $k(\frak{p})$ be residue field of $\frak{p}\in $Spec $A$.

Let $M$ be $A$-module, and $f(\frak{p}):=$dim$_{k({\frak{p}})} M\otimes_A k(\frak{p})$.

Now, is for any $\frak{p}\in$ Spec$A$, $f({\frak{p}})\le f((0))$ true?

答:
I assume by 'for any' you mean 'for all' (otherwise the answer is obvious as we can just take $\mathfrak{p}=(0)$).

No; if $M$ is a torsion module we have $f((0))=0$ but if $M\ne0$ there are points $\mathfrak{p}$ with $f(\mathfrak{p})>0$. Take e.g. $\mathfrak{p}$ to be a maximal ideal with $\mathfrak{p}\supseteq\operatorname{Ann}(M)$, then $$f(\mathfrak{p})=\dim_{A/\mathfrak{p}}M\otimes_AA/\mathfrak{p}=\dim_{A/\mathfrak{p}}M/\mathfrak{p}M=\dim_{A/\mathfrak{p}}M>0$$
For a concrete example take $A=\Bbb Z$, $M=\Bbb Z/p\Bbb Z$ where $p$ is a prime number.

Title: dimension over residue field


问:
A question from Rudin RCA
I'm studying Rudin's RCA chapter 7 and get in trouble with exercise 8.

Let $V = (a, b)$ be a bounded segment in $R^1$. Choose segments $W_n \subset V$ in such a way that their union $W$ is dense in $V$ and the set $K = V - W$ has $m(K) > 0$. Choose continuous functions $\phi_n$ so that $\phi_n(x) = 0$ outside $W_n$, $0 < \phi_n(x) < 2^{-n}$ in $W_n$. Put $\phi = \sum\phi_n$ and define
$$
T(x) = \int_a^x\phi(t)dt\qquad(a<x<b)
$$
Prove the following statements:

(a) $T$ satisfies the hypotheses of that $T$ is one-to-one on $X$, and $T$ is differentiable at every point of $X$, with $X = V$.

(b) $T'$ is continuous, $T'(x) = 0$ on $K$, $m(T(K)) = 0$.

(c) If $E$ is a nonmeasurable subset of $K$ (see Theorem 2.22) and $A = T(E)$, then $\chi_A$ is Lebesgue measurable but $\chi_A \circ T$ is not.

(d) The functions $\phi_n$ can be so chosen that the resulting $T$ is an infinitely differentiable homeomorphism of $V$ onto some segment in $R^1$ and (c) still holds.

Can anyone point me out about the b $m(T(K)) = 0$ and $d$?
I would be very thankful for any answer.

答:
For (b), to prove that $m(T(K))=0$, one need the followings:

$T(K)$ is Lebesgue measurable (Why?)
(Hint: prove that $T(W)$ is $F_{\sigma}$ set, and theorem 2.10 could be helpful)

Construct a Lebesgue measurable function
$$f(T(y))=\chi_{T(K)}(T(y))=\begin{cases}1, & y\in K\\0, & y\in K^C\end{cases}$$ (Check why this holds. Hint: for $=1$, use the fact that $T$ is one-to-one; for $=0$, use the same fact and argue by contradiction)

By theorem 7.26 in the text, since $K\subset V\Rightarrow T(K)\subset T(V)$, we have
$$m(T(K))=\int_{T(V)}\chi_{T(K)}(x)dm(x)=\int_{V}\chi_{T(K)}(T(y))|J_T|dm(y)=\int_{V}\chi_{K}(y)|J_T(y)|dm(y)$$
$$=\int_{K}\chi_{K}(y)|J_T|dm(y)=\int_{K}0dm(y)=0$$
Note that this is a consequence of (iii).

For (d), choose $W_n=(\alpha_n,\beta_n)$ that are disjoint, construct
$$\varphi_n(x)=\begin{cases}2^{-n}\exp\left(\frac{(\frac{\beta_n-\alpha_n}{2})^2}{(x-\frac{\alpha_n+\beta_n}{2})^2-(\frac{\beta_n-\alpha_n}{2})^2}\right), &x\in(\alpha_n,\beta_n)\\0, &\text{otherwise}\end{cases}$$

Then check that $\varphi_n(x)$ satisfies all the conditions.

Hope this helps.

Title: A question from Rudin RCA


答:
Hint:

Minimize

$$(x'-x'')^2+((a'x^2+b'x+c')-(a''x''^2+b''x''+c''))^2.$$

The stationary points are given by

$$(x'-x'')+(2a'x'+b')((a'x'^2+b'x'+c')-(a''x''^2+b''x''+c''))=0,\\(x''-x')+(2a''x''+b'')((a'x'^2+b'x'+c')-(a''x''^2+b''x''+c''))=0.$$

By adding the two equations and as the parabolas do not intersect in the given domain, we have one linear equation

$$2a'x'+b'+2a''x''+b''=0.$$

We can eliminate $x''$ by drawing it from the latter and plugging it in one of the formers. This results in a cubic equation, having one or three real solutions.

But that's not all. As the curves are bounded, the minimum might occur at an endpoint. From the figure we can see that there are two candidate endpoints: the left endpoint of the lower parabola, and the right endpoint of the upper one.

Knowing $x'$ or $x''$, the stationarity equations are still cubic.

Title: Minimum distance between parabolas

Body:
Is there a simple way to find the minimum distance between two parabolas?

For example, between y=-0.1(x-17)2 + 8.6 {7.726 < x < 19.134}
and y=-0.12(x-17.5)2 + 6.2 {10.313 < x < 18.829}

See graph of the two parabolas here (the exponential function simply determines the restricted domain for the parabolas)

Context: I have to ensure that there is a minimum of 1.8 units between these two parabolas, to ensure the "path" is at least 1.8 metres wide always.

Attempt: I thought about trying to find the closest distance visually then finding the normal and subbing in for x and y at that specific point, but that doesn't seem very scientific and I'm a bit stuck.

Thanks in advance


答:
Fix $n \in \Bbb N$.

Simple fact: the following set
$$S'=\{ (r,s) \in \Bbb Q \times \Bbb Z : n^2= r^2+s^2 \} \cup \{ (r,s) \in \Bbb Z \times \Bbb Q : n^2= r^2+s^2\}$$
is the union of two finite sets, hence it is a finite set.

Less obvious fact: the set
$$S=\{ (r,s) \in \Bbb Q^2 : n^2= r^2+s^2 \}$$
is an infinite set. In particular $S \setminus S' \neq \emptyset$. This means that there exist $r,s \in \Bbb Q \setminus \Bbb Z$ such that
$$n^2=r^2+s^2$$

I leave to you to prove that $S'$ is a finite set. Let's see why $S$ is infinite.

$S$ is an infinite set because there exist infinitely many primitive pythagorean triples $(a,b,c)$. For any such Pythagorean triple you can construct $r,s$ by the map
$$(a,b,c) \mapsto \left( \frac{an}{c}, \frac{bn}{c} \right)$$
Indeed
$$\left( \frac{an}{c}\right)^2 + \left( \frac{bn}{c}\right)^2 = \frac{n^2(a^2+b^2)}{c^2}=n^2$$

Title: Show that every perfect square can be written as the sum of the squares of two non integer rationals?

Body:
I was trying to prove the following statement:

Show that if $n$ is an integer, then $n^2$ can be written as $n^2=r^2+s^2$ where $r, s \in \mathbb{Q} \backslash \mathbb{Z}$, that is, $r$ and $s$ are non integer rationals.

My first idea was the most obvious one: to write $r=\frac{a}{b}$ and $s=\frac{c}{d}$ where $a, b, c, d \in \mathbb{Z}$ with $(a, b)=(c, d)=1$ and $b, d>1$.

I then tried a few things, but none of them lead to anything interesting. Could anyone help me? I'm looking for an elementary proof. I already know which integers can be expressed as the sum of two squares, and that primes of the form $4n+1$ have a unique representation as the sum of two squares, but I'm not sure how any of those would help me here.


答:
By definition, $(p+1)/n\ge a$. However, $a$ is irrational, so equality cannot hold.

Next you want to show that $(p+1)/n<b$; you have
$$
\frac{p+1}{n}=\frac{p}{n}+\frac{1}{n}<a+(b-a)
$$
Note that it's never said (nor it is generally possible) that $p>0$, but this is not a problem.

For the second part, follow the rules: since
$$
\frac{1}{b-a}=\sqrt{1000\cdot1001}\bigl(\sqrt{1001}+\sqrt{1000}\,\bigr)
$$
you can use a calculator to show that $n=63293$. It turns out that $an\approx2000.5$, so $p=2000$.

I can't see how to avoid using a calculator.

Title: Hard Year 11 Irrational and Rational Number problem

Body:
Suppose that $a$ and $b$ are positive irrational numbers, where $a < b$. Choose any positive integer $n$ such that $1/n < b − a$, and let $p$ be the greatest integer such that $p/n < a$:

a) Prove that the rational number $\frac{p + 1}{n}$ lies between $a$ and $b$.

b) If $a = \frac{1}{\sqrt{1001}}$, and $b = \frac{1}{\sqrt{1000}}$, find the least possible value of $n$ and the corresponding value of $p$.

c) Hence use part a) to construct a rational number between $\frac{1}{\sqrt{1001}}$ and $\frac{1}{\sqrt{1000}}$.

I am currently not learned in mathematical induction so im not really sure how to solve this. was thinking if there was another way of solving this question.

This question was taken from the Cambridge Maths Extension 1 Year 11 Book. Would like an intuitive explanation that a high school student in year 11 like me can understand since even my teachers cant seem to solve this problem lol. Sorry if this is a hassle of your time.


问:
How to approximate integral resulting from physics problem?

Setup

I'm taking a physics class and need to calculate (or approximate) a particular integral. Let
$$\Delta(x):= 1-(1-x)x \frac{p^2}{m^2},$$
where $p^2:= p_\mu p_\nu\eta^{\mu\nu}$ with $\eta$ being the Minkowski metric. I'd like to show that approximately
$$\int_0^1 dx\, \ln\left({\Delta(x)}\right)\propto \sqrt{1-\frac{4m^2}{p^2}}\ln\left(\frac{2m^2-p^2+\sqrt{p^2(p^2-m^2)}}{2m^2}\right).$$

My attempt

My only idea was to Taylor the $\ln$. Assuming $p^2/m^2$ to be small we have
$$\ln(\Delta(x)) = \left(x^2-x\right)\frac{p^2}{m^2}+\frac{1}{2} (x-1) x \left(x-x^2\right)\left(\frac{p^2}{m^2}\right)^2 +O\left(\left({p^2}/{m^2}\right)^3\right).$$
Then
$$\begin{align*}
\int_0^1dx \ln\Delta(x)
&\approx \int_0^1 dx\, \left[\left(x^2-x\right)\frac{p^2}{m^2}+\frac{1}{2} (x-1) x \left(x-x^2\right)\left(\frac{p^2}{m^2}\right)^2\right]\\
&= -\frac{1}{6}\frac{p^2}{m^2} - \frac{1}{60}\left(\frac{p^2}{m^2}\right)^2.
\end{align*}$$
This seems completely off, or at least way too crude of an approximation.. Unfortunately, I don't really have much experience with integration so I don't really have any other smart ideas or tricks up my sleeve...

P.S. I have no idea how to name this question properly. If the title doesn't fit, please feel free to edit it.

答:
You can easily have the exact result.

Let
$$I=\int_0^1\log \big[1-k (1-x) x\big]\,dx$$
$$1-k (1-x) x=k x^2-k x+1=k(x-a)(x-b)$$
$$\log \big[1-k (1-x) x\big]=\log(k)+\log(x-a)+\log(x-b)$$ Using integration by parts
$$J=\int \log(x-c)\,dx=(x-c) \log (x-c)-x+ C$$
$$K(c)=\int_0^1 \log(x-c)\,dx=-(c-1) \log (1-c)+c \log (-c)-1$$

Now, use the above for
$$a=\frac{k+\sqrt{k-4} \sqrt{k}}{2 k} \qquad \text{and} \qquad b=\frac{k-\sqrt{k-4} \sqrt{k}}{2 k}$$

Edit

Rearranging all the logarithms leads to the simple result
$$\int_0^1  \ln\left({\Delta(x)}\right)\,dx=\frac{2 \sqrt{4 m^2-p^2} }{p}\tan ^{-1}\left(\frac{p}{\sqrt{4 m^2-p^2}}\right)-2$$ which is exact and looks simpler then the looked for approximation.

Expanded as a infinite series, this gives
$$\int_0^1  \ln\left({\Delta(x)}\right)\,dx=-\sum_{n=1}^\infty \frac{n!\,(n+1)!}{(2 n+3)!}\,\left(\frac{p^2}{m^2}\right)^n$$ In your post, you wrote the first and second terms.

This is quickly convergence since
$$a_n=\frac{n!\,(n+1)!}{(2 n+3)!}\implies \frac{a_{n+1}}{a_n}=\frac{n+1}{4 n+10}$$


问:
Euler's Beta function for positive integers derivation
I am aware that a proof for the Euler's Beta function for positive integers has already been asked, but I have tried to derive it myself without using the gamma function, and my result doesn't match. Any ideas?

My derivation (I used $n=sm+d$ and $t=\left(1-x\right)^m$):
\begin{align*}
B\left(m,n\right)&=\int_{0}^{1}{x^{m-1}\left(1-x\right)^{n-1}dx} \\
&=\frac{1}{m}\int_{0}^{1}{\left(1-x\right)^{n-1}dx^m} \\
&=\frac{1}{m}\int_{1}^{0}{\left(1-x\right)^{n-1}d\left(1-x\right)^m} \\
&=-\frac{1}{m}\int_{0}^{1}{\left(1-x\right)^{sm+d-1}d\left(1-x\right)^m} \\
&=-\frac{1}{m}\int_{1}^{0}{t^{\frac{sm+d-1}{m}}dt} \\
&=\frac{1}{m}\int_{0}^{1}{t^{\frac{sm+d-1}{m}}dt} \\
&=\frac{1}{m}\left(\frac{m}{sm+m+d}1^{\frac{sm+m+d-1}{m}}-\frac{m}{sm+m+d}0^{\frac{sm+m+d-1}{m}}\right) \\
&=\frac{1}{sm+m+d-1} \\
&=\frac{1}{m+n-1} \\
\end{align*}

Correct result:

$$B\left(m,n\right)=\frac{\left(m−1\right)!\cdot\left(n−1\right)!}{\left(m+n−1\right)!}$$

Body:
Ok, I've found a solution:

let $a_i=\int_{0}^{1}{\left(m+i-1\right)x^{m+i-2}\left(1-x\right)^{n-i}dx}$

let $u_i=\left(1-x\right)^{n-i},du_i=\left(i-n\right)\left(1-x\right)^{n-i-1}dx$

let $v_i=x^{m+i-1},dv_i=\left(m+i-1\right)x^{m+i-2}dx$

\begin{align*}
a_i&=\int_{0}^{1}{\left(m+i-1\right)x^{m+i-2}\left(1-x\right)^{n-i}dx}\\
&=\int_{0}^{1}{u_idv_i}\\
&=\left[u_iv_i\right]_0^1+\int_{0}^{1}{v_idu_i}\\
&=\left[\left(1-v_i^{\frac{1}{m+i-1}}\right)^{n-i}v_i\right]_0^1-\int_{0}^{1}{x^{m+i-1}\left(i-n\right)\left(1-x\right)^{n-i-1}dx}\\
&=0-\left(i-n\right)\int_{0}^{1}{x^{m+i-1}\left(1-x\right)^{n-i-1}dx}\\
&=\frac{n-i}{m+i}\int_{0}^{1}{\left(m+i\right)x^{m+i-1}\left(1-x\right)^{n-i-1}dx}\\
&=\frac{n-i}{m+i}a_{i+1}\\
&=a_{n-1}\prod_{j=1}^{n-2}{\frac{n-j}{m+j}}\\
&=\frac{m!\left(n-1\right)!}{\left(m+n-2\right)!}a_{n-1}
\end{align*}

\begin{align*}
B\left(m,n\right)&=\int_{0}^{1}{x^{m-1}\left(1-x\right)^{n-1}dx}\\
&=\frac{1}{m}\int_{0}^{1}{mx^{m-1}\left(1-x\right)^{n-1}dx}\\
&=\frac{1}{m}a_1\\
&=\frac{1}{m}\frac{m!\left(n-1\right)!}{\left(m+n-2\right)!}a_{n-1}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\int_{0}^{1}{\left(m+n-2\right)x^{m+n-3}\left(1-x\right)dx}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left[\int_{0}^{1}{\left(m+n-2\right)x^{m+n-3}dx}-\int_{0}^{1}{\left(m+n-2\right)x^{m+n-2}dx}\right]\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left(\left[x^{m+n-2}\right]_0^1-\left[\frac{m+n-2}{m+n-1}x^{m+n+1}\right]_0^1\right)\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left(1-\frac{m+n-2}{m+n-1}\right)\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}-\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\frac{m+n-2}{m+n-1}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-1\right)}{\left(m+n-2\right)!}-\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-2\right)}{\left(m+n-1\right)!}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-1-m-n+2\right)}{\left(m+n-2\right)!}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-1\right)!}
\end{align*}


问:
Show that the probability that first passage time is finite satisfies this BVP
I am doing some exercises in E-Li-Vanden-Eijnden's Applied Stochastic Analysis and I meet this problem:

(Exercise 3.19) Consider an irreducible Markov chain $\{X_n\}$ on a
finite state space $S$. Let $H\subset S$ and define the first passage
time $T_H=\inf\{n:X_n\in H\}$, $h_i=P_i(T_H<\infty)$. Prove that
$\boldsymbol{h}=\{h_i\}_{i\in S}$ (column vector) satisfies
$(\boldsymbol{I}-\boldsymbol{P})\cdot\boldsymbol{h}=0$ with boundary
condition $h_i=1$ for $i\in H$, where $\boldsymbol{P}$ is the
transition matrix, and $\boldsymbol{I}$ is the identity matrix.

I have been struggling for this problem for hours and do not have much progress. The only thing that I can do currently is, as suggested by the book, to define the Laplace matrix $\boldsymbol{L}=-(\boldsymbol{I}-\boldsymbol{P})$ by acting on any function
$$(\boldsymbol{L}f)(i)=\sum_{j\in S}p_{ij}(f(j)-f(i))$$,
where $p_{ij}$ are entries of $\boldsymbol{P}$.

According to this, I write
$$(\boldsymbol{I}-\boldsymbol{P})\cdot\boldsymbol{h}=(-\boldsymbol{L}h)(i)=\sum_{i,j\in S}p_{ij}(h(i)-h(j))$$

but starting from here, I get stuck. Can anyone provide any solutions regarding this? Thank you.

答:
You want to prove that the vector of hitting probabilities satisfies $h = Ph$, with the boundary condition $h_i=1$ for $i\in H$.

If the chain starts in $H$, you're guaranteed to hit it, so the boundary condition follows.

If you start outside $H$, you can condition on the first step (notation $P_i(\cdot) = P(\cdot | X_0 = i)$):

$$h_i=P_i(T_H < \infty) = \sum_{j\in S}P_i(T_H<\infty, X_1 = j)=\sum_{j\in S}P_i(T_H<\infty| X_1 = j)P(X_1=j) = \sum_{j\in S}h_jp_{ij}=\sum_{j\in S}p_{ij}h_j,$$

where we have used that $P_i(T_H<\infty| X_1 = j)=P_j(T_H<\infty)=h_j$ by the Markov property.

If you're interested, $h$ is in fact the minimal solution to this system (in the sense that each vector component is minimal).


问:
An explicit polynomial isomorphism between the real projective line and the unit circle.
Prove that the map $$w=\frac{\frac{1}{2}+it}{\frac{-1}{2}+it} , t\in \mathbb{R}$$ is a bijective map which maps the real axis $(-\infty,\infty)$ to the unit circle $|w|=1$

My try
$$w=\frac{\frac{1}{2}+it}{\frac{-1}{2}+it} $$
$$ |w|=\frac{\sqrt{\frac{1}{4}+t^2}}{\sqrt{\frac{1}{4}+t^2}}$$
$$ |w|=1$$
$$w=\frac{\frac{1}{2}+it}{\frac{-1}{2}+it}$$
$$t=\frac{(1+z)}{2i(z-1)} $$
So the map is onto.

For One-One,
$$Let, \ t,t'\in \mathbb{R}$$ be such that
$$\frac{\frac{1}{2}+it}{\frac{-1}{2}+it}=\frac{\frac{1}{2}+it'}{\frac{-1}{2}+it'}$$
Gives $t=t'$

答:
Here is an elementary explanation: If you define $z(t):=\frac{1}{2}+it$ you get

$$w(t):=\frac{z(t)}{-\overline{z}(t)}.$$

Question: "Prove that the map $w(t)$ is a bijective map which maps the real axis $(−∞,∞)$ to the unit circle $|w|=1$."

Answer: Hence if $w(t)=1$ it follows $z(t)=-\overline{z}(t)$.
If $z=a+ib$ and $z=-\overline{z}$ then $a=0$. Hence $w(t)=1$ is impossible and the map $w(t)$ you define is not bijective as claimed.

The map $w(t)$ is a map

$$w: \mathbb{R} \rightarrow S^1-\{1\}$$

where $S^1$ is the unit circle in $\mathbb{C}$.

It seems you get a well defined map

$$\phi: C:=\mathbb{P}^1_{\mathbb{R}} \rightarrow S^1$$

defined as follows:

$$\phi(v:u):= \frac{\frac{1}{2}v +iu}{-\frac{1}{2}v+iu}:=\frac{z(v,u)}{-\overline{z}(v,u)}.$$

Here $C$ denotes the real projective line. This is the set of non-zero vectors $(v,u)\in V:=\mathbb{R}^2$ modulo the well known equivalence relation: Two vectors $a,b\in V$ are equivalent iff there is a non-zero real number $\lambda$ with $\lambda a =b$.

You may check that if $\lambda \in \mathbb{R}-0$ it follows

$$\phi(\lambda v: \lambda u)=\phi(v:u)$$
hence the map $\phi$ seems to be well defined. When you restrict $\phi$ to the open set $D(v)\cong \mathbb{R}$ (the set of $(v:u)$ with $v\neq0$) you recover the map $w(t)$. Here you view the set of real numbers as a subset of the real projective line $C$ via the map

$$s: \mathbb{R} \rightarrow D(v) \subseteq C$$

defined by

$$s(t):=(1:t) \in D(v).$$

Hence you get an equality

$$ w(t)=\phi(s(t))$$

for all real numbers $t$. The inverse image $\phi^{-1}(1)$ is the set of
elements $(v:u)\in C$ with

$$ z(v,u)=-\overline{z}(v,u)$$

which is iff $v=0$. Hence $\phi^{-1}(1)\cong (0:a)\in C.$

Question: "Is the map a map from the real line to the unit circle? If no then how can one redefine the map so that it gives the unit circle?"

Answer: Hence $\phi^{-1}(1) \cong \{(0:1)\}:=\infty \in C$. Hence the map $\phi$ equals your map on $D(v)$ and $\phi(0:1)=1\in S^1$. Hence we have defined an isomorphism of algebraic varieties/real algebraic manifolds

$$\phi: \mathbb{P}^1_{\mathbb{R}} \cong S^1.$$

If you want to "redefine the map" map you must embed the real line as an open subset of the real projective line. Then the map $w(t)$ extends to an isomorphism of real algebraic manifolds between $C$ and $S^1$.

Example. Let $G:=\mathbb{Z}$ and define the left action

$$\sigma: G \times \mathbb{R} \rightarrow \mathbb{R}$$

by

$$\sigma(k,x):=x+ 2k\pi.$$

Define the map

$$w: \mathbb{R} \rightarrow S^1.$$

by

$$w(x):=cos(x)+isin(x):=e^{ix}\in S^1.$$

It follows $w(\sigma(k,x))=w(x+2k\pi )=w(x)$ hence you get a well defined map

$$\overline{w}: \mathbb{R}/G \cong S^1.$$

The map $\overline{w}$ is an "isomorphism": You have realized $S^1$ as the quotient of $\mathbb{R}$ by the group $G$. This is the map mentioned in the other post. It factors through the quotient $\mathbb{R}/G$.


问:
When is the diagonal a closed immersion with open image?
Let $A$ be a commutative unital $R$-algebra and consider the  multiplication map $m:A\otimes_R A\rightarrow A$ given by $m(a\otimes b) = ab$ with kernel $I$. The map $m$ is surjective, hence the image of $f: \operatorname{Spec}A\rightarrow \operatorname{Spec}A\times\operatorname{Spec}A$ is $V(I)$.

If we assume that $V(I)$ is also open, then is $f$ an open immersion?

As far as I know, an immersion with closed image is a closed immersion, but an immersion with open image maybe not an open immersion?

Thanks.

答:
It's helpful to remember the sheaf-theoretic conditions on immersions: an open immersion $f:Y\to X$ must have that the induced maps $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$ are isomorphisms for all $y\in Y$, while a closed immersion only implies that those maps are surjections. Assuming $f$ is a closed immersion and adding the condition $f(Y)$ open only implies that the kernel of  $\mathcal{O}_{X,f(y)}\to\mathcal{O}_{Y,y}$ is in the nilradical of $\mathcal{O}_{X,f(y)}$, which is not enough to guarantee that we have an open immersion unless $\mathcal{O}_{X,f(y)}$ is reduced for all $y\in Y$.

Here is an explicit example in your specific case. Let $R=k$ be a field and let $A=k[\epsilon]/(\epsilon^2)$. Then $A\otimes_R A\cong k[\epsilon_1,\epsilon_2]/(\epsilon_1^2,\epsilon_2^2)$, so both $\operatorname{Spec} A$ and $\operatorname{Spec} A\otimes_R A$ are single points, and if $\operatorname{Spec} m$ is an open immersion it must be an isomorphism. Hence $m$ should be an isomorphism, but one can immediately see that this is not the case as $1\otimes\epsilon-\epsilon\otimes1$ is in the kernel.


问:
Subfields of field of cyclotomic 5th root of 1
What are all the subfields of the cyclotomic 5th root of 1.

My attempt:The galois  group over $\Bbb{Q}$ is $(\Bbb{Z}/5\Bbb{Z})^*=\{1,a^1,a^2,a^3\}$.The only possible non trivial subgroup is the subgroup generated by $a^2$.So the only possible non trivial subfield is the fixed field of this subgroup.

答:
The Galois group shows that the only non-trivial subextension is quadratic, and algebraic number theory shows that the quadratic subfield of $\Bbb Q\left(\zeta_p\right)$ is $\Bbb Q\left(\sqrt{p^\ast}\right)$ where $p^* = (-1)^{\frac{p-1}2} p$ for any odd prime $p$, so in this case you get $\Bbb Q\left(\sqrt5\right)$.

It is left as an exercise to verify that indeed $\sqrt5 \in \Bbb Q\left(\zeta_5\right)$.


问:
The degree of an irreducible factor of a composite polynomial
Dummit and Foote Ch 13.2 Q17

Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$. Prove that every irreducible factor of the composite polynomial $f(g(x))$ has degree divisible by $n$.

Following theorem may be useful

Let $p(x)\in F[x]$ be an irreducible polynomial of degree $n$ over the field $F$ and let $K$ be the field $K=\frac{F[x]}{(p(x))}$. Then the degree of the extension is $n$.

Let $h_1,..., h_m$ be irreducible factors of $ f(g(x))$. By Chinese remainder theorem $$\frac{F[x]}{(f(g(x)))}\cong \frac{F[x]}{(h_1(x))} \times...\times \frac{F[x]}{(h_m(x))}$$. So the RHS talks about the degree of the irreducible factors. The LHS should incorporate the given information, i.e. $f$ is irreducible polynomial of degree $n$. Please give a hint on how to incorporate this given information in LHS.

Edit: If some other method works, please give a hint for that method.

Thanks.

答:
Let $p(x)$ be an irreducible factor of $f(g(x))$ and $\alpha$ a root of $p$ in some extension field of $F$. Then the degree of $F(\alpha)/F$ is $\deg p(x)$. Since $p(\alpha)=0$ and $p(x)\mid f(g(x))$ we have $f(g(\alpha))=0$, i.e. $g(\alpha)$ is a root of $f$, furthermore $g(\alpha)\in F(\alpha)$. Now consider the degrees in the field tower $$F(\alpha)/F(g(\alpha))/F$$
Since $f$ is irreducible we have as above $[F(g(\alpha)):F]=n$ and hence $[F(\alpha):F]=[F(\alpha):F(g(\alpha))]\cdot n$, i.e. $n\mid [F(\alpha):F]=\deg p(x)$.


问:
Given a recursion $\frac{a_{n+1}}{4a_{n+1}+n+1}=\frac{2(a_n+n)^2}{9n(4a_n+n)}$ with $a_1=1.$ Show that $a_n\sim\frac n2\,{\rm as}\,n\rightarrow\infty$

Given a recursion $\dfrac{a_{n+ 1}}{4a_{n+ 1}+ n+ 1}= \dfrac{2\left ( a_{n}+ n \right )^{2}}{9n\left ( 4a_{n}+ n \right )}$ with $a_{1}= 1.$ Prove that
$$a_{n}\sim\frac{n}{2}\,{\rm as}\,n\rightarrow\infty$$

I tried to solve the nonlinear asymptotic equation I posted before again and I made a few mistakes.. so I fixed them. I think the difficulty level is vicious. On the other hand_ Estimations of some new recurrence sequences, why the first expansions between both of them are similar ? I can't explain it, I need to the help, thank you so much.

答:
By regrouping $a_{n+1}$ terms we get:

$$a_{n+1}=-\frac{2(n+1)(a_n+n)^2}{-20\,n\,a_n-n^2+8{a_n}^2}$$

Since we want $a_n\sim \dfrac n2$ let set $a_n=\dfrac n2(1+u_n)$ and hope to transform the problem into proving $u_n\to 0$

After substitution we get $$u_{n+1}=\frac{3{u_n}^2}{9+6u_n-2{u_n}^2}$$

Let set $f(x)=\frac{3x^2}{9+6x-2x^2}$, then, $f(x)=x$ has three fixed points $x=-\frac 32,0,3$.

https://www.desmos.com/calculator/cskpxqtrmp

And effectively from the seed $u_1=\frac {2a_1}1-1=2-1=1$, it appears to actually converge to $0$ while decreasing.

You can show easily that the quadratic $D(x)=(9+6x-2x^2)>0$ on the interval $[0,1]$

$f'(x)=\dfrac{18x(x+3)}{D(x)^2}\ge 0$ when $x\ge 0$ so $f\nearrow$ on $\mathbb R^+$

In particular $f([0,1])\subset[f(0),f(1)]=[0,\frac 3{13}]\subset[0,1]$

It means that since $u_1=1\in[0,1]$ then by induction $\forall n,\ u_n\in[0,1]$

Let now study $f(x)-x=\dfrac{x(2x+3)(x-3)}{D(x)}<0$ for $x\in[0,1]$ therefore $u_{n+1}\le  u_n$

We can conclude that $u_n\searrow$ in compact $[0,1]$ so it has a limit, and since the only fixed point of $f$ in this interval is $0$ then $u_n\to 0$.


问:
Prove that $\sqrt[n]{|a_n|}≤q$ converges absolutely.
I was preparing some tasks for the learning group I am meeting today and found this one interesting, but I have no clue how to do it!
There's $0<q<1$ so that $\sqrt[n]{|a_n|}≤q$ converges absolutely.
How would the prove look like and what do I need to pay attention to? Is it correct that I could use the Comparison test for this one?
Oh and yeah, $a_n$ is a series, as far a I can see it.</p>

答:
This is a simple application of the comparison test.

$\lvert a_{n} \rvert \leq q^{n}$

and $q$ being in $(0,1)$, the series $\sum_{n=1}^{n=\infty}q^{n}$ converges. Hence

$ \sum_{n=1}^{\infty}\lvert a_{n}\rvert$ converges.


问:
Are all second order linear partial differential equations solvable in general?
Are all linear second order differential equations in both real and complex spaces solvable by numerical or analytical methods  when we are given just the equation provided that a solution exists somewhere in both real and complex spaces.Solution could be of the form of an implicit equation, parametric equation, infinite series, elementary functions, special functions, etc. Will at least one of the methods work in general for linear second order partial differential ntial equations in general?</p>

答:
The question is ambiguous because the meaning of "solvable" is not precise enough (in fact not defined). Do you mean :

Solvable on the form of an implicit equation ?

Solvable on the form of parametric equation ?

Solvable on the form of an infinite series ?

Solvable in terms of elementary functions ?

Solvable in terms of special functions ?

Etc.

Moreover an ODE can be not solvable yesterday and be solvable today  in terms of special function. For example this is the case of $y''(x)+\frac{1}{x}y'(x)+y(x)=0$ which was not solvable in terms of special functions before the Bessel functions where defined studied and standardized.

Many special functions were standardized in order to make solvable some differential equations  https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales .

Thus a today "non-solvable" ODE can become "solvable" tomorow not only because new methods of solving are likely to be invented but simply because new functions are likely to be defined and standardized.

The question is too wide for a definitive answer.


问:
Is the closure of $S^1-\{(0,1)\}$ compact?
My question is very simple : Is the closure of $S^1-\{(0,1)\}$ compact ? (where $S^1\subset \mathbb R^2$ is the unit circle). In somehow, I think it is, since it should be the circle, but in on other way, $S^1-\{(0,1)\}$ is homeomorphic to $\mathbb R$, so in somehow, if the closure of $S^1-\{(0,1)\}$ is compact, then so is the closure of $\mathbb R$ (which I guess it's not). So, at the end, I'm not so sure if the closure of $S^1-\{(0,1)\}$ is really compact. Any idea ?

答:
Remember the notion of closure depends on the surrounding space (i.e., the phrase "closure of $X$" is not a topological notion: it depends on more than just the topology of $X$, but also on how $X$ is embedded in some larger space).

So we can't talk about the closure of $S^1 \setminus \{(0, 1)\}$ or of $\Bbb{R}$ without knowing what surrounding space is intended. Here, in the original question, it is implicit that $S^1 \setminus \{(0, 1)\}$ is being thought of as a subspace of $\Bbb{R}^2$ in the standard way, which means the closure is the compact space $S^1$. When we talk about $\Bbb{R}$ we are usually thinking of it as a space in its own right (or perhaps as a subspace of $\Bbb{C}$) and, on that understanding, it is its own closure, so the closure is not compact.


问:
Matsubara sum arising from QFT and contour integral
In the lecture of E. Fradkin on quantum field theory, an example of Matsubara sum is performed using contour integration (see eq. 5.214 in the lecture). It reads
$$
\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2}
= \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \pi\cot(\pi z)\quad(\tau, x>0),
$$
where $C_{\pm}$ denotes the lines $z\pm i\epsilon$ followed counter clockwise (see this figure). I see that above equation can be deduced applying residue theorem to the finite box, obtained by truncating $C_{+}\cup C_{-}$, and taking a limit.

To evaluate the right hand side, the author deforms the contour into $C^{+}\cup C^{-}$ and calculate instead residues at $z=\pm ix$. I am curious about the mathematical justification of this process.

To me, equating $\oint_{C_{-}}(\text{integrand})$ with residue at $-ix$ seems illegitimate since $e^{iz\tau}$ blows up as $z\rightarrow -i\infty$. I don't get what 'deforming $C_{-}$ to $C^{-}$' even means.

One approach that I tried is the change of variable $z\rightarrow 1/z$, which amounts to passing to the other coordinate chart of Riemann sphere. It goes like
$$
(RHS)=\oint_{\Gamma} \frac{dz}{2\pi i} \frac{e^{i\tau/z}}{1+z^2x^2} \pi\cot(\pi/z),
$$
where $\Gamma$ is the image of $C_{+}\cup C_{-}$ followed counter clockwise. The problem here is twofold; first, $\Gamma$ passes through the essential singularity at $z=0$. Second, if $C_{+}\cup C_{-}$ is taken to be the limit of boxes $B_n$ and $\Gamma$ the limit of their images $B'_n$, each contour $B'_n$ encircles an infinite subset of the poles $\{\pm 1/n: n\in\mathbb{Z}\}$. I have no idea if I can apply the residue theorem and take a limit in this case.

edit

Following Svyatoslav's suggestion, I obtained (for $0<\tau<2\pi$)
$$
\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2}
= \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \frac{\pi e^{-i\pi z}}{\sin\pi z}
= \frac{\pi}{x} \frac{\cosh(\pi-\tau)x}{\sinh(\pi x)} \\
= \frac{\pi}{x} \left[ \cosh(\tau x) \coth(\pi x) - \sinh(\tau x) \right].
$$
However, it is claimed in (5.215) of the lecture that
$$
\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2}
\approx \frac{\pi}{2x}\coth(\pi x) e^{-|\tau|x} \quad \text{for small } \tau.
$$
Did the author make a mistake?

答:
This is @Svyatoslav's answer.

That the integrand blows up on the lower half-plane is correct. For a legitimate calculation, put $\pi e^{-i\pi z}$ instead of $\pi\cot(\pi z)$:

$$
\sum_{n=-\infty}^{\infty} \frac{e^{in\tau}}{n^2+x^2}
= \oint_{C_{+}\cup C_{-}} \frac{dz}{2\pi i} \frac{e^{iz\tau}}{z^2+x^2} \frac{\pi e^{-i\pi z}}{\sin\pi z}
= \frac{\pi}{x} \frac{\cosh(\pi-\tau)x}{\sinh(\pi x)} \\
= \frac{\pi}{x} \left[ \cosh(\tau x) \coth(\pi x) - \sinh(\tau x) \right].
$$

Here we assume, without loss of generality, $0\leq \tau<2\pi$. Note that as $t\rightarrow\infty,$

$$
e^{iz\tau}\frac{\pi e^{-i\pi z}}{\sin \pi z} \sim
\begin{cases}
e^{-\tau t},  &\quad z=it \\ e^{(\tau-2\pi)t}, &\quad z=-it
\end{cases}
$$

so we can close the contour around a large circle toward both the upper and lower half-planes.


问:
How do initial conditions apply to a general solution for a PDE
Given an equation, $$u_{xx} + u_{xt} -20u_{tt}= 0$$ it can be shown that there exists $\xi = t - 5x$ and a $\eta= t+4x$ such that

$$u_{\xi \eta} = 0 \space\space\space\space\space\space\space\space \text{ (first canonical form of a hyperbolic)}$$

Thus, we have

$$u(\xi,\eta) = F(\xi)+G(\eta)$$$$u(x,t) = F(t-5x)+G(t+4x)$$

Now lets say we have initial conditions of: $$u(x,0) = f(x) $$ $$ u_t(x,0) = g(x)$$ How do initial conditions change the solution? What exactly are we trying to solve for beyond this?

答:
We have a system from the Cauchy conditions:
$$
F(-5 x)+G(4 x) = f(x),\: F'(-5 x)+G'(4 x) = g(x), \: x \in \mathbb{R}.
$$
From the first equation we get $F(z) = f\left(-\frac{z}{5}\right)-G\left(-\frac{4 z}{5}\right)$. Now the solution has a form $u(t,x)=f\left(x-\frac{t}{5}\right)-G \left(4 x-\frac{4}{5}t\right)+G(t+4 x)$.

In order to define $G$ we should use the second Cauchy condition
$$
\partial_t u(x, 0) = -\frac{1}{5} f'(x)+\frac{9}{5} G'(4 x) = g(x).
$$
So, we have ODE, which defines the function $G$: $-\frac{1}{5}
   f'\left(\frac{x}{4}\right)+
   \frac{9}{5} G'(x) =g\left(\frac{x}{4}\right)$.
The solution of this ODE is $G(z) = C + \frac{1}{9} \int_0^z \left(5
   g\left(\frac{\tau}{4}\right)+f'\left(\frac{\tau}{4}\right)\right) d\tau$, $z \in \mathbb{R}$, $C$ is some constant.

So, the function
$$u(x, t) = \frac{1}{9} \left(-\int_0^{4
   x-\frac{4 t}{5}}
   \left(f'\left(\frac{\tau
   }{4}\right)+5
   g\left(\frac{\tau
   }{4}\right)\right) \, d\tau
   +\int_0^{t+4 x}
   \left(f'\left(\frac{\tau
   }{4}\right)+5
   g\left(\frac{\tau
   }{4}\right)\right) \, d\tau
   +9
   f\left(x-\frac{t}{5}\right)
   \right)$$
is the solution of your Cauchy problem.


答:
Let $w=e^{i\theta}$, with $\theta\in(0,2\pi)$. Then

$$e^{i\theta}=\frac{2it+1}{2it-1}$$ is equivalent to

$$t=-\frac i2\frac{e^{i\theta}+1}{e^{i\theta}-1}=\frac12\cot\dfrac\theta2.$$

In the given range, the cotangent is an invertible function.

Note that the circle is not complete, as $e^{i\theta}=1$ has no image. (And as stated, the question has a negative answer.)

Title: An explicit polynomial isomorphism between the real projective line and the unit circle.

Body:
Prove that the map $$w=\frac{\frac{1}{2}+it}{\frac{-1}{2}+it} , t\in \mathbb{R}$$ is a bijective map which maps the real axis $(-\infty,\infty)$ to the unit circle $|w|=1$

My try
$$w=\frac{\frac{1}{2}+it}{\frac{-1}{2}+it} $$
$$ |w|=\frac{\sqrt{\frac{1}{4}+t^2}}{\sqrt{\frac{1}{4}+t^2}}$$
$$ |w|=1$$
$$w=\frac{\frac{1}{2}+it}{\frac{-1}{2}+it}$$
$$t=\frac{(1+z)}{2i(z-1)} $$
So the map is onto.

For One-One,
$$Let, \ t,t'\in \mathbb{R}$$ be such that
$$\frac{\frac{1}{2}+it}{\frac{-1}{2}+it}=\frac{\frac{1}{2}+it'}{\frac{-1}{2}+it'}$$
Gives $t=t'$


问:
complex analysis -bounded but not holomorphic function
What are some examples of functions that are bounded but not holomorphic?

I know that some the not holomorphic functions are

Absolute value function, Piecewise functions (not analytic where the pieces meet)

答:
Well, you can take any bounded discontinuous function. For example,

$f(z) = 1$ if $z = 0$ and $f(z) = 0$ otherwise is bounded and discontinuous and hence not holomorphic.


答:
You are expecting too much about the closure operation. Take $S^1-\{(0,1)\}$, $\mathbb{R}$ e $(0,1)$. The closure of the first one is $S^1$, of the second one is $\mathbb{R}$ and the last is $[0,1]$. The closure are not homeomorphic.

There is a reason for that: the closure is a extrinsic property. Look at the definition: we start with a topological space $X$ and a subset $S$. The closure of $S$ at $X$ is the intersection of all closed sets of $X$ that contains $S$. It's clear from the definition that the closure of $S$ depends of the whole topological space, not the set on is own.

So, I will rewrite what I did write better:

The closure of $S^1-\{(0,1)\}$, as a subset of $\mathbb{R}^2$, is $S^1$

The closure of $(0,1)$, as a subset of $\mathbb{R}$, is $[0,1]$

The closure of $\mathbb{R}$, as a subset of $\mathbb{R}$, is $\mathbb{R}$

See? We are working with non homeomorphic topological spaces, that's why you can't say the closure is preserved by homeomorphism.

To make it more clear, look what happen when we consider $S^1-\{(0,1)\}$ as a subset of $S^1-\{(0,1)\}$: in this case, $S^1-\{(0,1)\}$ is already closed, and then its closure is $S^1-\{(0,1)\}$. The closure changes when we change the topological space.

So, answering your initial question:

The closure of $S^1-\{(0,1)\}$, as a subset of $\mathbb{R}^2$, is compact.

The closure of $S^1-\{(0,1)\}$, as a subset of $S^1-\{(0,1)\}$, isn't compact.

Title: Is the closure of $S^1-\{(0,1)\}$ compact?

Body: My question is very simple : Is the closure of $S^1-\{(0,1)\}$ compact ? (where $S^1\subset \mathbb R^2$ is the unit circle). In somehow, I think it is, since it should be the circle, but in on other way, $S^1-\{(0,1)\}$ is homeomorphic to $\mathbb R$, so in somehow, if the closure of $S^1-\{(0,1)\}$ is compact, then so is the closure of $\mathbb R$ (which I guess it's not). So, at the end, I'm not so sure if the closure of $S^1-\{(0,1)\}$ is really compact. Any idea ?


问:
How do I refactor this sailboat transform to solve for apparent wind direction for my sailboat plotting algorithm?
The core problem I would like assistance with is this:

How do I refactor this formula to transpose $a$?

$$\sin(a_0)\sin(a) \biggl(\frac{ \sin(\frac{a}{2})}{\sin(a_0 - a)}\biggr) ^2 = VW \cdot \eta$$

For more context, please read on.

I'm in a little over my head, and may be mixing up some terms, so please forgive my ignorance. I am attempting to write a route plotting algorithm that calculates a sailboat's $VMG$ (a measure of how optimal a sailboat's heading or direction of sail is) using live wind velocity and wind direction. The goal is to give the user a tool for optimizing pathfinding in the context of sailing. VMG is calculated using this formula:

$V$ = boat speed

$x$ = angle of wind relative to the front of the boat

$$VMG = V \cdot \cos(x)$$

In order to calculate boat speed from wind speed, I plan to use this formula, which is apparently widely accepted:

$VW$ = wind velocity

$a_0$ = true wind angle relative to the front of the boat

$a$ = apparent wind angle (i.e. wind generated by movement of the boat)

$$VB = VW \cdot \frac{\sin(a_0 - a)}{\sin(a)}$$

In order to solve this, I need to find $a$.

This article by Yoav Raz posits that a can be found by solving for it in his sailboat transform:

VW= wind velocity

$a_0$ = true wind angle relative to the front of the boat

$a$ = apparent wind angle (i.e. wind generated by movement of the boat)

$\eta$ = a complex measure of the boat's resistance to acceleration which for this purpose I will assume to be a low, constant number around 0.01)

$$\sin(a_0) * \sin(a) * \biggl(\frac{ \sin( \frac{a}{2})}{ sin(a_0 - a)}\biggr) ^2 = VW \cdot \eta$$

In practice, I will know the wind velocity and true wind angle, and as I said will be assuming $\eta$. Here is where the trouble starts: I am not very good at math anymore, and I find myself at a loss as to how to refactor this formula for my purposes. In order to use this formula in my algorithm, I need to refactor it into an expression that isolates $a$ given the other parameters( i.e. $a=f(x)$ ).

I've tried to refactor it by using trigonometric identities, which I think is the right path. The farthest I've gotten is this:

$$\sin(a_0) \sin(a)  \biggl( \frac{\sin(\frac{a}{2})}{\sin(a_0)\cos(a)-\cos(a_0)\sin(a)}\biggr)^2 = VW\cdot \eta$$

I know I'm kind of asking "please teach me trigonometry and calculus," so my apologies for that. I'm just hoping that even if I'm barking up the wrong tree that someone can help point me in the  right direction.

Edit:
It now occurs to me that the solution might be to simply iteratively attempt to solve this by trying inputs for a and attempting to solve for 1. The Raz article suggests doing this and starting with values slightly below a0. I could write a model to evaluate this expression with decreasing values until it arrives at a value a such that both sides resolve to 1.

Here are some more relevant resources to this problem space, if you would like to understand a bit more background around this topic:

The Physics of Sailing

Sailboat speed Vs. wind speed

Velocity Made Good

Polar Diagram

Apparent Wind Speed

答:
Another approach

Consider
$$f(a)=\sin(b) \, \sin(a) \, \biggl(\frac{ \sin( \frac{a}{2})}{ \sin(b - a)}\biggr) ^2 $$ Make again ${a=b-\sin ^{-1}(2 x)}$ to define $g(x)$.

Assuming that $x$ is small, expand $g(x)$ as a series around $x=0$ to get
$$g(x)=\frac{\sin ^2(b) (1-\cos (b))}{8 x^2}+O\left(\frac 1x\right)$$ and then an approximate solution
$$x_0=\sin (b)\sqrt{\frac{ (1-\cos (b))}{8 k}}$$

For the worked example, this gives $$x_0=\frac{1}{4} \sqrt{1-\frac{\sqrt{3}}{2}}\implies a=\frac{\pi }{6}-\sin ^{-1}\left(\frac{1}{2} \sqrt{1-\frac{\sqrt{3}}{2}}\right)\sim 0.339549$$ Now, Newton iterates will be
$$\left(
\begin{array}{cc}
n & a_n \\
 0 & 0.339549 \\
 1 & 0.393787 \\
 2 & 0.399457 \\
 3 & 0.399521
\end{array}
\right)$$


问:
how to change polar coordinate into cartesian coordinate using transformation matrix
I would like to change $(3,4,12)$ in $xyz$ coordinate to spherical coordinate using the following relation

It is from the this link. I do not understand the significance of this matrix (if not for coordinate transformation) or how it is derived. Also please check my previous question building transformation matrix from spherical to cartesian coordinate system. Please I need your insight on building my concept.

Thank you.

EDIT::

I understand that $ \left [ A_x \sin \theta\cos \phi \hspace{5 mm} A_y \sin \theta\sin\phi \hspace{5 mm} A_z\cos\theta\right ]$ gives $A_r$ but how is other coordinates $ (A_\theta, A_\phi)$ equal to their respective respective rows from Matrix multiplication?

答:
This is not the Matrix you're looking for. For a simple co-ordinate switch you can just use the relations:

$$\begin{align*}x &= \rho\sin\theta\cos\phi\\
y &= \rho\sin\theta\sin\phi \\
z &= \rho\cos\theta\end{align*}$$

And the inverse operations:

$$\begin{align*}\rho &= \sqrt{x^2 + y^2 + z^2}\\
\phi &= \arctan\dfrac yx\\
\theta &= \arctan\left(\frac{\sqrt{x^2 + y^2}}z\right)\end{align*}$$

However the matrix you've found is for mapping a vector between the co-ordinate systems. For example (using a textbook, Engineering Electromagnetics by Demarest. Example 2-6, p34)

Need to do an integration of $\int( r^3\cos\phi\sin\theta\cdot Ar) d\theta d\phi$

Where $Ar$ is a unit vector in the radial direction. The integral is over phi and theta but also dependent on phi and theta, therefore it's much easier to do this by switching back to cartesian coordinates by the relation:

$$Ar = \sin\theta\cos\phi\cdot Ax + \sin\theta\sin\phi\cdot Ay + \cos\theta\cdot Az$$

Once we substitute that straight in for Ar the integral looks longer but we've removed the dependence inside the integrand, so we can do the integration in a straight forward way.


问:
How to approach question with variable floor like this one
$$4x^2 - 3 \lfloor x \rfloor ^2 ≥ 9 $$ I don't quite understand how do we approach such questions? Is there an expansion what we need to similar to mod or is there a different method altogether?

答:
I don't think there's a standard type of question like this; it looks quite unfamiliar to me.

What I would do to approach this is consider $x=\lfloor x\rfloor+\{x\}$, where $\{x\}$ is the "fractional part". In other words, break $x$ into an integer $n$ and value $y$ with $0\leq y<1$. Then your inequality becomes
$$4(n+y)^2-3n^2\geq 9,$$
which simplifies to $n^2+8ny+y^2\geq 9$. If $n$ is sufficiently large or sufficiently negative then this will be true for all $y\in [0,1)$, and then for each of the few additional values of $n$ you can work out which values of $y$ (if any) in range make the equation true.


问:
Induction Hypothesis Should there be a plus sign

$$\sum_{i=0}^n2^i=2^{n+1}-1.$$

After going through the steps I come up with
$$2^{k+1}+1,$$
And I expected
$$2^{k+1}-1.$$

Therefore it is false.

Can someone check and see if this is correct?
I can't find a way to validate my answer.

答:
The second step of the induction requires that:

if we suppose $\sum_{i=0}^{n}2^i=2^{n+1}-1$ then it is true also for $n+1$? I mean it is true that $\sum_{i=0}^{n+1}2^i=2^{n+2}-1$? Yes, in fact by induction hypothesis:

$\sum_{i=0}^{n+1}2^i=\sum_{i=0}^{n}2^i+2^{n+1}=2^{n+1}-1+2^{n+1}=2(2^{n+1})-1=2^{n+2}-1$.


问:
Can't understand why (n-1)choose(-1) equals zero.
I have tried to wrap my head around the idea but I can't seem to get it, and google yields no coherent explanation. Would appreciate some clarification about this.</p>

答:
It's seems more like a matter of definition but if for example you want Pascal's rule to work also for negative numbers then:

$${n \choose {-1}}={n+1\choose 0}-{n\choose 0}=1-1=0$$

Also as @user10354138 pointed out in the comments $n \choose k$ denotes the number of subsets of size $k$ in a set of $n$ elements so if we want to extend this to negative numbers it would only make sense if it would be 0 since there are 0 subsets of size -1.


问:
lagrange multipliers fails
I am looking for a certain counter example. Assume a $C^1$ function $f$ is to be optimized with respect to a $C^1$ constraint $g=0$, and we have at a point $(x,y)$, the existence of a lagrange multiplier $\lambda$ with
$$
\begin{align}
&\nabla f(x,y)=\lambda\nabla g(x,y)\\
&\nabla g(x,y)\neq0\\
\end{align}
$$
But $f$ fails to have an extremum at this point with respect to $g=0$

thanks

答: Let $f(x,y):=x^3+y$ and $g(x,y):=y$. Then $\nabla f(0,0)=\nabla g(0,0)=(0,1)$, but $f(x,0)-f(0,0)=x^3$ assumes both signs in the immediate neighborhood of $(0,0)$.</p>

Title: lagrange multipliers fails


问:
lower semi-bounded imply symmetric
A quadratic form is a map $q: Q(q) \times Q(q) \rightarrow \mathbb{C}$, where $Q(q)$ is a dense linear subset of the Hilbert space $H$. If $q(\phi,\psi)=\overline{q(\psi,\phi)}$, then we say q is symmetric.
If $q(\phi,\phi)\geq -M||\phi||^2$ for some $M$, we say $q$ is semibounded.

If $q$ is semibounded, then it is automatically symmetric if $H$ is complex.

Anyone could give my a hint for the proof?

答:
I have made an attempt to solve this:
Let $\phi,\psi \in Q(q)$. We know that $q(\phi+\psi,\phi+\psi)\in \mathbb{R}$ and $q(\phi+i\psi,\phi+i\psi)\in \mathbb{R}$.
$$q(\phi+\psi,\phi+\psi)=q(\phi,\phi)+q(\psi,\psi)+q(\psi,\phi)+q(\phi,\psi)(1)$$
$$q(\phi+i\psi,\phi+i\psi)=q(\phi,\phi)+q(\psi,\psi)-iq(\psi,\phi)+iq(\phi,\psi) \;(2)$$

From (2), we know that $q(\psi,\phi)=q(\phi,\psi)$. From (1), $q(\psi,\phi)+q(\phi,\psi)$ is real. Thus, $q(\phi,\psi)$ and $q(\phi,\psi)$ are real. $q(\psi,\phi)=\overline{q(\phi,\psi)}$

Title: lower semi-bounded imply symmetric


问:
For which values of $m$ the line is tangent to the quadratic curve?
For which values of $m$, the line $y=2x-4$ is tangent to the curve $y=(m+3)x^2+mx ?$

We have a quadratic equation. the equation of slope of tangent line to it for specific $x$ can be find by $y'=(2m+6)x+m$. but when I compare it to $y=2x-4$, I can't find $m$.

答:
You are after a point $(a,b)\in\Bbb R^2$ such that:

$(a,b)$ belongs to the curve $y=(m+3)x^2+mx$;

$(a,b)$ belongs to the line $y=2x-4$;

the slope of $y$ at $(a,b)$ is equal to $2$.

This means that$$\left\{\begin{array}{l}b=(m+3)a^2+ma\\b=2a-4\\(2m+6)a+m=2\end{array}\right.$$There are only two solutions: $m=-2$ (in which case $a=2$ and $b=0$) and $m=22$ (in which case $a=-\frac25$ and $b=-\frac{24}5$).


问:
What are the finite subgroups of $SU_2(C)$?
Is there any reference which classifies the finite subgroups of $SU_2(C)$ up to conjugacy ?

What I know is that lifting the finite subgroups of $SO_3(R)$ by the map $SU_2(C) \rightarrow PSU_2(C)$ gives rise to the groups : cyclic groups of even order, dicyclic groups, binary tetrahedral/octahedral/icosahedral groups. But I dont know if they are the only one.

答:
The image of a finite subgroup of $\text{SU}(2)$ in $\text{SO}(3)$ is a finite subgroup of $\text{SO}(3)$; moreover, the kernel is either trivial or $\{ \pm 1 \}$. But $-1$ is the unique element of order $2$ in $\text{SU}(2)$, so any group of even order contains it.

I claim all the finite subgroups of odd order are cyclic. This follows because the inclusion $G \to \text{SU}(2)$ cannot define an irreducible representation of $G$ (since otherwise $2 | |G|$), hence it must break up into a direct sum of dual $1$-dimensional representations.

So once you know the finite subgroups of $\text{SO}(3)$, you already know the finite subgroups of $\text{SU}(2)$.


问:
Use the epsilon-delta definition to prove
Use $\epsilon$ - $\delta$ to prove that $\lim_{x\rightarrow0} \frac {\sqrt{4+x}-2} {x} = \frac{1}{4}$

Given $\epsilon > 0$, we want to find a $\delta$ such that if $|x-0| < \delta$, $\left|\frac{\sqrt{4+x}-2}{x} - \frac{1}{4}\right| < \epsilon$.
I multiplied by $\frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}$ to rationalize the numerator and combined the fractions to get $\left|\frac{2-\sqrt{4+x}}{4(\sqrt{4+x}+2)}\right|$, but I am not sure where to go from here to relate it to $\delta$.

答:
Extended hint:

$\sqrt{4+x}+2$ is continuous and defined at $0$ with output $4$, so for $\epsilon > 0$ arbitrary we can find a $\delta > 0$, such that
$$
\lvert x \rvert < \delta \implies \lvert (\sqrt{4+x} + 2) - 4 \rvert = \lvert 2 - \sqrt{4+x}\rvert < \epsilon.
$$
Also
$$
4\leq \lvert 4- \sqrt{4+x}-2\rvert + \lvert \sqrt{4+x}+2 \rvert < \lvert \sqrt{4+x} +2\rvert + \epsilon \implies \frac{1}{4 - \epsilon} > \frac{1}{\sqrt{4+x}+2}.
$$

Finish off from here.


问:
The normed linear space $l_p$ ,$1\leq p<\infty $ is separable.
The normed linear space $l_p$ ,$1\leq p<\infty $ is separable.

Proof:

The normed linear space $l_p$ ,$1\leq p<\infty $ is separable. The set $S$ of all sequence having finitely many non zero rationals hence $S$ is countable.

Now we claim that $S$ is dense subset of $X$.

Let $(x_n)_{n\geq 1}=x\in l^p,1\leq p<\infty,$ then $\sum_{i=1}^{\infty}|x_i|^p<\infty$. Thus this series is convergent hence tail of this series is also convergent i.e., for given $\epsilon>0$, there exist $n\in\Bbb{N}$ such that,
$\sum_{i=n}^{\infty}|x_i|^p<\epsilon^p/2.$
Since rationals are dense in $\Bbb{R}$, therefore for given $\epsilon>0$,
there exist rational coordinates $y_1,y_2,...,y_n$ such that,

$$|x_i-y_i|<(\frac{1}{2n})^\frac{1}{p}\epsilon.$$
Therefore,
$$\sum_{i=1}^{n-1}|x_i-y_i|^p<\frac{\epsilon^p}{2}.$$

Now $d(x,y)=||x-y||=(\sum_{i=1}^\infty|x_i-y_i|^p)^\frac{1}{p}=(\sum_{i=1}^{n-1}|x_i-y_i|^p+\sum_{i=n}^{\infty}|x_i|^p)^\frac{1}{p}<(\frac{\epsilon^p}{2}+\frac{\epsilon^p}{2})^\frac{1}{p}=\epsilon.$ Hence  any open ball containing $x$ contains some points of $S$, thus $S$ is dense in $l_p$ space . This completes the proof.

Thanks @Kavi Rama Murthy!

答:
The idea is correct but some changes are necessary in the presentaion:

You said:
let $y_1,y_2,...,y_n$ be non zero rational coordinates, Since rationals are dense in $\Bbb{R}$, therefore for given $\epsilon>0$,

$$|x_i-y_i|<(\frac{1}{2n})^{p}\epsilon.$$

You should say there exist rational numbers $y_1,y_2,...,y_n$ such that

$$|x_i-y_i|<(\frac{1}{2n})^{1/p}\epsilon.$$
and hence

$$\sum_{i=1}^{n-1}|x_i-y_i|^p<\frac{\epsilon^{p}}{2}.$$

There is confusion beteeen $n$ and $N$ . You could drop $N$ and use $n$ everywhere.


问:
Why can't we define the mapping cone as $M^\bullet\oplus L[1]^\bullet$?
Consider the short exact sequences of complexes on an abelian category

Let $\tilde{N}^\bullet$ be the direct sum $M^\bullet\oplus L[1]^\bullet$. I wonder why doesn't $\tilde{N}^\bullet$ has all the nice properties of the mapping cone $MC(\varphi^\bullet)$. (Of course, the only difference between them is the differential.)

I've verified that $\tilde{N}^\bullet$ is indeed a complex, that it has a natural morphism of complexes $\rho^\bullet:\tilde{N}^\bullet\to N^\bullet$ (given by composing the natural projection $\tilde{N}^\bullet\to M^\bullet$ with $\psi^\bullet$), and that it fits in the exact sequence

Another main feature of $MC(\varphi^\bullet)$ is that the morphism $MC(\varphi^\bullet)\to N^\bullet$ is a quasi-isomorphism. I think that this may be the problem here; perhaps $\rho^\bullet$ is not always a quasi-isomorphism.

I would love to understand why it may not be a quasi-isomorphism and, if possible, to find some concrete examples.

答:
Here is a concrete example : (by the way, this is a very common example in homological algebra, and you can very often check your assertions with it).

Consider the short exact sequence
$$ 0 \longrightarrow\mathbb{Z}\overset{\times 2}\longrightarrow\mathbb{Z}\longrightarrow\mathbb{Z/2Z}\longrightarrow 0$$
where every group is considered as a complex concentrated in degree 0.

Then the mapping cone of the map $\times 2$ is given by the complex $...\longrightarrow 0 \longrightarrow\mathbb{Z}\overset{\times 2}\longrightarrow\mathbb{Z}\longrightarrow 0\longrightarrow ...$ where the $\mathbb{Z}$ are concentrated in degree -1 and 0. Obviously the cohomology of this complex is $\mathbb{Z/2Z}$ in degree 0 and 0 otherwise so this is thus quasi-isomorphic to $\mathbb{Z/2Z}$.

On the other hand, the direct sum is the complex $...\longrightarrow 0 \longrightarrow\mathbb{Z}\overset{0}\longrightarrow\mathbb{Z}\longrightarrow 0\longrightarrow ...$ and its cohomology is given by $\mathbb{Z}$ in degree -1 and 0. So this is not quasi-isomorphic to $\mathbb{Z/2Z}$.


答:
$m$ can be found without using slope. The given line touches the curve at $(x,2x-4)$ so that
$$2x-4=(m+3)x^2+mx$$
is satisfied.  This is a quadratic in $x$,
$$(m+3)x^2+(m-2)x+4=0$$
and its discriminant is zero, since for a given $m$, the quadratic has a unique root (tangency point). Thus
$$D : (m-2)^2-4\cdot 4 \cdot (m+3)=0$$
$$\Rightarrow m^2-20m-44=0 \Rightarrow m=22,-2$$

Title: For which values of $m$ the line is tangent to the quadratic curve?

Body:
For which values of $m$, the line $y=2x-4$ is tangent to the curve $y=(m+3)x^2+mx ?$

We have a quadratic equation. the equation of slope of tangent line to it for specific $x$ can be find by $y'=(2m+6)x+m$. but when I compare it to $y=2x-4$, I can't find $m$.


问:
Formal proof of ($A$ and ($A$ implies $B$)) implies $B$?

$$(A ∧ (A → B)) ⇒ B$$

I know that this is a tautology, but apart from setting up a truth-table I dont know how I would go about a formal proof in discrete mathematics. thanks

答:
$(A \land (A\rightarrow B)) \rightarrow B $ is true when $A \land (A\rightarrow B)$ is false or $B$ is true. Therefore if it was false then $A \land (A\rightarrow B)$ would be true and $B$ would be false.

If $A \land (A\rightarrow B)$ is false then A is false and $A\rightarrow B$ is false.
But if $A\rightarrow B$ is false and $B$ is false then $A$ must be true, which is a contradiction.


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
From: Link

Try a proof by contraposition (are you familiar with this method of proof? A statement and its contrapositive are logically equivalent so proving the contrapositive of the original statement actually proves the original statement).

So what you want to do is assume that $n$ is not even (i.e., it is odd) and show that $n^2$ is not even (i.e., it is odd).

$n$ is odd means that you can write $n = 2k + 1$ for some integer $k$. Then

$$n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$

which is clearly odd. This completes the proof.

I hope this helps.  Please write back if you'd like to talk about this more.


问:
Evaluating B(3/4,5/4) without the Legendre duplication formula?
Suppose that one wanted to evaluate $B(3/4,5/4)$ where $B$ is the beta function.  One approach is as follows.

$$B(3/4,5/4)=\frac{\Gamma(3/4)\Gamma(5/4)}{\Gamma(3/4 + 5/4)}=\Gamma(3/4)\Gamma(5/4).$$

There is an identity, the Legendre duplication formula, $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$, and applying this with $z=3/4$ we get that our function equals $2^{-1/2}\sqrt{\pi}\Gamma(3/2)$.  Since $\Gamma(3/2)=(1/2)\Gamma(1/2)=\sqrt{\pi}/2$, we get a final answer of $\frac{\pi}{2\sqrt{2}}$.

Is there a more direct way to get this result? The integral formulas for the beta function all see to give horrible messes, e.g., $2\int_0^{\pi/2} \sin(\theta)^{1/2}\cos(\theta)^{3/2}\,\mathrm{d}\theta$, and I don't see a reasonable substitution to make for that.

答:
A way to solve the Beta-function without applying clever identities will be provided in this answer.

We want to evaluate
$$\operatorname{B}[3/4,5/4] = \int\limits_{0}^{1}t^{3/4-1}(1-t)^{5/4-1}\,\mathrm{d}t=\int\limits_{0}^{1}\frac{\sqrt[4]{1-t}}{\sqrt[4]{t}}\,\mathrm{d}t.$$
Substituting $u=\frac{\sqrt[4]{t}}{\sqrt[4]{1-t}}$ will give us
$$4\int\limits_{0}^{\infty}\frac{u^2}{(u^4+1)^2}.$$
From now on there are many ways to proceed. I believe you want to get to the $\arctan$ integral somehow. Splitting up the denomitor, Ostrogradski method, complex analysis are the first ones that come to my mind.

Edit: I would do it via complex analysis. You can add a semi-circle and show that the integral over it goes to zero. Then you have to calculate the residues.


问:
Selecting at least 4 questions of 12 and at least one of the first 3 questions must be taken?

There are 12 questions. Students must take at least 4 questions and at least one of the first 3 questions is taken too.

How many configurations are there?

My attempt

\begin{align}
\text{answer} &= \sum_{i=1}^3\left({3\choose i}\sum_{j=4-i}^9{9 \choose j}\right)\\
 &= 3415
\end{align}

Is it correct?

答:
Yes that is correct answer. Another way to approach it -

If there are $12$ questions, we have $2^{12}$ ways to select any number of questions including no selection $\big[{12 \choose 0} + {12 \choose 1} + ...+ {12 \choose 12}\big]$.

Out of that ${12 \choose 0} + {12 \choose 1} + {12 \choose 2} + {12 \choose 3} = 299$ are selections with less than $4$ questions.

Similarly if we do not select any of the first $3$ questions, there are $2^9$ ways to select questions and out of that ${9 \choose 0} + {9 \choose 1} + {9 \choose 2} + {9 \choose 3} = 130$ are selections with less than $4$ questions.

So the solution we are interested in $ = 2^{12} - 299 - (2^9 - 130) = 3415$.


问:
The smallest extension field of $\mathbb{F}_3$ containing all the zeros of $f(x)=x^{11}-1$
Firstly notice that $f(1)=0$ so $f(x)=(x+2)g(x)$ over $\mathbb{F}_3$ for some $g$. Moreover, $\text{ord}_{11}3 = 5$, ie $5$ is the smallest exponent $t$ such that $3^t=1 \mod 11$, so $5$ is the lcm of all the degrees of the factors of $f$, hence $f(x)=(x+2)g(x)h(x)$ with $\deg(g)=\deg(h)=5$.

Hence the splitting field of $f(x)$ over $\mathbb{F}_3$ is $L=\mathbb{F}_{3^5}$. Now, is $L$ the smallest extension of $\mathbb{F}_3$ containing all the zeros of $f(x)$? How to prove it?

答:
You are searching for the smallest field extension of $\Bbb F_3$ which contains a primitive $11$-th root of unity.

You're right that the smallest $t$ such that $11$ divides $3^t-1$ (which is the order of the multiplicative group of $\Bbb F_{3^t}$) is $t=5$.
Since the group is cyclic it contains all $11$-th roots of unity. That's it.

Add: We have $3^5-1= 11\cdot 22$ and so if $\omega$ is a primitive element of $\Bbb F_{3^5}$, then $\omega^{22}$ is a primitive $11$-th root of unity.


问:
complex analysis -bounded but not holomorphic function
What are some examples of functions that are bounded but not holomorphic?

I know that some the not holomorphic functions are

Absolute value function, Piecewise functions (not analytic where the pieces meet)

答: In fact, it is the case that almost all bounded functions are nonholomorphic, by which I mean that if you pick a random bounded function on some domain $D$, there is a $0\%$ chance it will be holomorphic.  Intuitively, one can simply assign all the elements of $D$ to arbitrary meaningless values (as long as the values are bounded) and the result will be discontinuous, and thus nonholomorphic.


答:
To go from another direction, you can construct a proof using the deduction theorem to transform the problem first into $\{A \wedge (A \to B)\} \vdash B$. Then this is proved by the following:

$A \wedge (A \to B)$ (hypothesis).

$A$ ($\wedge$-elimination on 1).

$A \to B$ ($\wedge$-elimination on 1).

$B$ (modus ponens on 2 and 3).

Title: Formal proof of ($A$ and ($A$ implies $B$)) implies $B$?

Body:
$$(A ∧ (A → B)) ⇒ B$$

I know that this is a tautology, but apart from setting up a truth-table I dont know how I would go about a formal proof in discrete mathematics. thanks