答:
Answer. $B$: Only if $k\ge n$.

First, we study the case $k=n$.

If $p(x)=c_0+c_1x+\cdots+c_k x^k$, and
$$
p(a_j)=b_j, \quad j=0,1,\ldots,k,
$$
then the above is a $(k+1)\times (k+1)$ linear system
$$
c_0+c_1 a_j+\cdots+c_n a_j^k=b_j, \quad j=0,1,\ldots,k,
$$
with unknowns the $c_j$'s. The determinant of the matrix $M$ the system is the Vandermonde one, hence non-vanishing, and thus the system possesses a unique solution
$$
c=(c_j)=M^{-1}b=M^{-1}(b_j).
$$
Hence, if the $b_j=p(a_j)$ all vanish, so does $p$.

Clearly, $\|p\|_V=\max\{|c_j|\}$ defines a norm in $V$.

Also,
$$
\|p\|_V=\|c\|\le \|M^{-1}\|\cdot\|b\|
$$
where
$$
\|c\|=\max_{j=1,\ldots,n}|c_j|, \quad \|A\|=\sup_{\|x\|=1}\|Ax\|.
$$
and since $Mb=c$
$$
\|b\|\le \|M\|\cdot\|c\|=\|p\|_V
$$
Thus $\|b\|=\max_{j=0,1,\ldots,k}\{|p(a_j)|\}$ is a norm equivalent to $\|p\|_V$. In particular, if
$\max_{j=0,1,\ldots,k}\{|p(a_j)|\}=0$, then $p\equiv =0$.

If $k>n$, then it is still a norm, since
$$
\max_{j=1,\ldots,n}\{|p(a_j)|\}=0\quad\Longrightarrow\quad
\max_{j=1,\ldots,k}\{|p(a_j)|\}=0\quad\Longrightarrow\quad p\equiv 0.
$$

If $k<n$, one can then $p(x)=(x-a_0)(x-a_1)\cdots(x-a_k)$ vanishes at the $a_j$'s and hence $\max_{j=1,\ldots,k}\{|p(a_j)|\}=0$, while it is a non-vanishing polynomial of degree $k+1\le n$.

Title: When is this norm defined?

Body:
This question was part of my linear algebra assignment and I was unable to solv eit. SO, I am asking you for help.

Consider the vector space V of real polynomials of degree less than or equal to n. Fix distinct real numbers $a_0 , …, a_k$ . For $p \in V$ define $max${${|p(a_j)| : 0\leq j \leq k}$} . It defines a  norm on V if:

A only if k<n

B only if $k\geq n$

C if $k+1 \leq n $

D if $k\geq n+1$

I tried thinking on which result / reasoning I should use but I am clueless on when the norm will not be defined .

Can you please help me through it?


问:
Why $J_k(0)^T$ ~ $J_k(0)$?
Why these two Jordan blocks are similar: $J_k(0)^T $ and $J_k(0)$ ?

答:
Because if $k_1+k_2+\dots + k_s=k$, all $k_i\geqslant 1$, $s>1$ then
$$
\left(\bigoplus_{i=1}^{s}J_{k_i}(0)\right)^{k-1}=O
$$
whereas
$$
J_k(0)^{k-1}\not=O.
$$

ContentLicense: CC BY-SA 4.0

Title: Why $J_k(0)^T$ ~ $J_k(0)$?


问:
Understanding the proof of every nonempy open subset $U\subset \mathbb{R}$ is an (at most countable) disjoint union of open intervals
Statement: every nonempy open subset $U\subset \mathbb{R}$ is an (at most countable) disjoint union of open intervals.

The proof came from the course notes (page 8) of a measure theory course at the link below

https://people.clas.ufl.edu/pascoej/files/6616notes01dec2017.pdf

Proof

First, verify that if $I$ and $J$ are intervals and $I\cap J\neq \emptyset$, then $I\cup J$ is an interval.

Given $x\in U$, let

$\alpha_x=\sup\{a:[x,a)\subset U\}$,

$\beta_x=\inf\{b:(b,x]\subset U\}$,

$I_x=(\beta_x,\alpha_x)$.

Verify that, for $x,y\in U$ either $I_x=I_y$ or $I_x\cap I_y=\emptyset$.

Indeed $x\sim y$ if $I_x=I_y$ is an equivalence relation on $U$. Hence, $U=\bigcup_{x\in U} I_x$  expresses $U$ as a disjoint union of nonempty intervals, say $U=\bigcup_{p\in P}I_p$ where $P$ is an index set and the $I_p$ are nonempty intervals. For each $q\in \mathbb{Q}\cap U$ there exists a unique $p_q$ such that $q\in I_{p_q}$. On the other hand, for each $p\in P$ there is a $q\in \mathbb{Q}\cap U$ such that $q\in I_p$. Thus, the mapping from $\mathbb{Q}\cap U$ to $P$ defined by $q\rightarrow p_q$ is onto. It follows that $P$ is at most countable.

Questions

I want to check my understanding of the two verifications that the author asked the reader to do. Here are my attempts.

Verify that if $I$ and $J$ are intervals and $I\cap J\neq \emptyset$, then $I\cup J$ is an interval.

Recall that, for a subset $I$ of the real numbers to be an interval it must satisfy that
$$ (\forall a,b \in I)(a<c<b\implies c\in I).$$
If $I\cap J\neq \emptyset$, then there exists $t,a_I,b_I,a_J,b_J$ such that
$$
a_I<t<b_I\text{ and }a_J<t<b_J
$$
where $a_I,b_I,t$ are all in $I$ and $a_J,b_J,t$ are all in $J$. We will keep such $t$ for later use.

For all $a,b\in I\cup J$, take any $y$ satisfying $a<y<b$, then either $y<t$ or $y>t$.
If $y<t$, then $y\in (a_I,t)$. So $y\in I$. If $y>t$, then $y\in (t,b_J)$. So $y\in J$. Therefore, $y\in I\cup J$ and $I\cup J$ is an interval.

Verify that, for $x,y\in U$ either $I_x=I_y$ or $I_x\cap I_y=\emptyset$.

I will prove that $I_x\cap I_y\neq\emptyset$, then $I_x=I_y$.

If $I_x\cap I_y\neq\emptyset$, then $I_x\cup I_y$ is an interval by we have just shown. Then
$$
I_x\cup I_y= \Big(\max(\beta_x,\beta_y),\  \min(\alpha_x,\alpha_y)\Big)
$$
By construction, $I_x=I_x\cup I_y =I_y$.

I only know about the basic topology and author clearly tried to avoid the notion of connected set. It would be great if someone could improve my proofs and help me with the terminology.  For example, the extended interval will mess up the use of max and min operators.

答:
You can simplify the first part as follows (there's no need to fix $a_I, b_I$ etc). Fix a $t\in I\cap J, a, b\in I\cup J, a < b$ and let $y\in (a, b)$. If $a, b\in I$ or $a, b\in J$, then $y\in I$ or $y\in J$ respectively by definition, so assume wlog $a\in I, b\in J$. Now, either $y\leq t\implies y\in (a, t]\subset I$ or $y\geq t\implies y\in [t, b)\subset J$, therefore $y\in I\cup J$.

It is fine to use $\max, \min$ with $-\infty<a<\infty\forall a\in\mathbb{R}$. But it is not entirely clear (at least the way you've written it) how $I_x = I_x\cup I_y = I_y$. I prefer the following. Without loss of generality, assume $x<y$, so that $[x, y]\subset U$ because $I_x\cup I_y$ is an interval contained in $U$.

Now, given $t\in I_x$, either $[x, t)\subset U$ or $(t, x]\subset U$ and in both cases (actually, there are three cases : $t<x, x<t<y, y<t$), it is clear that $[y, t)$ or $(t, y]$ is contained in $U$, hence $t\in I_y$. Repeat the argument to show that $I_y\subseteq I_x$.


问:
Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$
Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
 We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?

Thanks!

答:
For odd $n$ you can always write:
$$f_n(x) = \sum_{i=0}^n x^{2i} = (x^2+1)(x^{2n-2}+x^{2n-6}+x^{2n-10}+...)$$
For even $n$ this trick no longer works.</p>

ContentLicense: CC BY-SA 3.0


问:
Is it true that $ e^x-\sum_{k=0}^n \frac{x^k}{k!} \leq e^x\frac{x}{n}?$
Is it true that for $n\in\mathbb{N}$ and $x> 0$ we have
$$
e^x-\sum_{k=0}^n \frac{x^k}{k!} \leq e^x\frac{x}{n}?
$$

What I am getting is this:
Use Taylor's series with remainder. We know that
$$ e^\lambda = \sum_{k=0}^r \frac{\lambda^k}{k!} + \frac{e^{c\lambda}\lambda^{r+1}}{(r+1)!}, $$
for some $c \in [0,1]$. Therefore
$$ \frac{\lambda^{r+1}}{(r+1)!} \leq e^\lambda - \sum_{k=0}^r \frac{\lambda^k}{k!} \leq e^\lambda \frac{\lambda^{r+1}}{(r+1)!}. $$

We can get a different upper bound by comparison to a geometric series, when $\lambda < r+2$:
$$
\sum_{k=r+1}^\infty \frac{\lambda^k}{k!} \leq \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \left(\frac{\lambda}{r+2}\right)^t = \frac{\lambda^{r+1}}{(r+1)!} \frac{r+2}{r+2-\lambda}.
$$

答: The left side is $\sum\limits_{k=n+1}^{\infty} \frac  {x^{k}} {k!}$. Note that $k! >(k-1)! n$ for $k >n$. Hence an upper bound for LHS is $\frac  1 n\sum\limits_{k=n+1}^{\infty} \frac  {x^{k}} {(k-1)!}$ which is $\frac  x n\sum\limits_{k=n+1}^{\infty} \frac  {x^{k-1}} {(k-1)!} \leq \frac x n e^{x}$.


问:
inequality strong induction proof
I understand everything up to the induction hypothesis in this worksheet of mine. I don't understand why you were able to substitute in values of 3 for all the a's in the induction hypothesis. Can someone explain the answer to me by any chance.

here is the question: https://imgur.com/a/fdKMWY1

and here's the supposed answer I don't understand: https://imgur.com/a/QCtHDjr

答:
They did not substitute the $a_k$ for $3$. They used the statements $P(k)$, $P(k-1)$ and $P(k-2)$, those are:
$$
a_k < 3^k,\quad a_{k-1} < 3^{k-1},\quad a_{k-2} < 3^{k-2}.
$$
Hence, they conclude that
$$
a_{k+1} = \underbrace{a_k}_{<3^k} + \underbrace{a_{k-1}}_{<3^{k-1}} + \underbrace{a_{k-2}}_{<3^{k-2}} < 3^k + 3^{k-1} + 3^{k-2}.
$$

Title: inequality strong induction proof


问:
How to show $A$ is compact in $\Bbb{R}$ with standard topology?
Let $A = \{0\} \cup \{ \frac{1}{n}| n \in \Bbb{N}\} \subseteq \Bbb{R}$. I am struggling with showing whether $A$ is compact or not. I do not even know how to start.
Is $u=\{0\} \cup \left\{\left(\frac{1}{n+1}, \frac{1}{n-1}\right)| n \in \Bbb{N}\right\} \cap \{(1/2, 2)\}$ an open cover of A with no finite subcover? If so, what is the next step?

答: If you can use the Heine-Borel theorem is easy to see that the set of acumulation points $A' =\{0\} \subseteq A$ therefore $A$ is closed and observe that $\forall x \in A$ we have that $ |x - 1| \leq 1$ hence $A$ is bounded. So $A$ is a closed and bounded set, by Heine-Borel theorem we can conclude that $A$ is compact.


问:
Is $e^{\sigma B_t - \frac{\sigma^2}{2}t}$ a martingale?
I'm trying to prove that $M_t = e^{\sigma B_t - \frac{\sigma^2}{2}t}$ is a martingale (for $(B_t)_{t\geq 0}$ a standard brownian motion), but I cannot show that $E|M_t| <\infty$. Could you help me?

答:
If $X \sim N(\mu, \sigma^{2})$ the $Ee^{aX}=e^{a\mu}e^{a^{2}\sigma^{2}/2}$. [See https://www.statlect.com/probability-distributions/normal-distribution ]. Hence $EM_t=e^{-\sigma^{2}t /2} Ee^{\sigma B_t}=e^{-\sigma^{2}t /2}e^{\sigma^{2}t/2}=1$ for all $t$.

A similar calculation for conditional MGF given $\mathcal  F_s$ using independence of $B_t-B_s$ and $\mathcal F_s$ for $s <t$ proves the martingale property.


答:
There is a closed quotient form for the polynomial using the geometric series formula:

$$\sum_{k=0}^n x^{2k}=\frac{x^{2n+2}-1}{x^2-1}.$$

The theory of cyclotomic polynomials tells us how to reduce polynomials of the form $x^m-1$. In particular, the irreducible factors of $x^m-1$ are $\Phi_d(x)$ for each divisor $d\mid m$ (no repeated factors are present in this factorization). See references for $\Phi_n(x)$ for further details.

Thus,

$$\frac{x^{2n+2}-1}{x^2-1}=\prod_{\substack{d\mid (n+1) \\ d\ne1}}\Phi_d(x^2).$$

If $n+1$ is composite then the above factorization involves more than one cyclotomic polynomial hence $f_n(x)$ is reducible. On the other hand, if $n>1$ is odd we have

$$\frac{x^{2(n+1)}-1}{x^2-1}=\frac{x^n+1}{x+1}\frac{x^n-1}{x-1}.$$

Both quotients are in fact integer-coefficient polynomials:

$$\frac{x^n-1}{x-1}=x^{n-1}+\cdots+x+1, \quad \frac{x^n+1}{x+1}=\frac{(-x)^n-1}{(-x)-1}=(-x)^{n-1}+\cdots+(-x)+1.$$

Note the latter depended on $n$ being odd so that $(-x)^n=-x^n$.

The only exception is $n=1$, and in that particular case we know that $x^2+1$ is irreducible.

Title: Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$

Body:
Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed:
$$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$
Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + x^2+1$. Using Macaulay2 (powerful software package) I checked that:
$$f_4(x) = (x^4-x^3+x^2-x+1)(x^4+x^3+x^2+x+1)$$
 We get that
$$ f_5(x)=(x^2+1)(x^2-x+1)(x^2+x+1)(x^4-x^2+1)$$
The polynomial is reducible for $n=6, 7, 8, 9$ as far as I checked. I suspect that $f_n(x)$ is reducible over integers for all $n\ge 2$. Is this true?

Thanks!


问:
Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$
Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ is
$${\begin{vmatrix} x& py+q& rz+s\\
\alpha& p\beta+q& r\gamma+s\\
1& 1& 1\\
\end{vmatrix}} = 0$$

My Attempt:
The equation of line is
$x = py + q = rz + s$
$$\implies x = p (y + \frac {q}{p}) = r ( z + \frac {s}{r}) $$
$$\implies \frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$$
The equation of plane through this line is
$$a(x-0) + b(y+\frac {q}{p}) + c(z+\frac {s}{r}) = 0$$
where $a,b,c$ are the direction ratios of the line normal to the plane.
Now,
$$a\cdot 1 + b\cdot (y+\frac {q}{p}) + c\cdot (z + \frac {s}{r}) = 0$$
Also, the plane passes through the point $(\alpha, \beta, \gamma)$,
$$a(\alpha - 0) + b(\beta + \frac {q}{p}) + c(\gamma + \frac {s}{r}) = 0$$

How to proceed further from here?

答:
Hint:

$\frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$

One of the points on the line is $P \,$ $(0,-\frac{q}{p}, -\frac{s}{r}$).

Direction vector from this point to $Q \,$ $(\alpha, \beta, \gamma)$ is $\vec{PQ} \, (\alpha, \beta + \frac{q}{p}, \gamma + \frac{s}{r})$.

As the given line and $\vec{PQ}$ are in the plane we take a cross product of vectors $\, (\alpha, \beta + \frac{q}{p}, \gamma + \frac{s}{r})$ and $(1, \frac{1}{p}, \frac{1}{r}) \,$to find the normal vector to the plane say, $(a, b, c)$.

Then the equation of the plane is $a(x-\alpha) + b(y-\beta) + c(z-\gamma) = 0$


问:
Method of characterist PDE, transport equation with 3 variables
Let $u=u(t,x,y)$. How can I use the method of the characteristis to solve the following transport equation:

$$\frac{\partial{u}}{\partial{t}}-\frac{\sinh(t)}{x}\frac{\partial{u}}{\partial{x}}-\frac{\cosh(t)}{y}\frac{\partial{u}}{\partial{y}}=\tan(t) $$

where
$$ u(0,x,y)=x^2-y^2+1.$$

I don't know to solve this PDE because there is 3 variable. Anyone can shed light on how to solve this problem?

答:
$$\frac{\partial{u}}{\partial{t}}-\frac{\sinh(t)}{x}\frac{\partial{u}}{\partial{x}}-\frac{\cosh(t)}{y}\frac{\partial{u}}{\partial{y}}=\tan(t) $$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{dt}{1}=\frac{-x}{\sinh(t)}dx=\frac{-y}{\cosh(t)}dy=\frac{du}{\tan(t)}$$
First characteristic equation from solving $\frac{dt}{1}=\frac{-x}{\sinh(t)}dx$ :
$$x^2+2\cosh(t)=c_1$$
Second characteristic equation from solving $\frac{dt}{1}=\frac{-y}{\cosh(t)}dy$ :
$$y^2+2\sinh(t)=c_2$$
Third characteristic equation from solving $\frac{dt}{1}=\frac{du}{\tan(t)}$
$$u+\ln|\cos(t)|=c_3$$
General solution of the PDE expressed on the form of implicit equation $\Phi(c_1\:,\:c_2\:,\:c_3)=0$ with arbitrary function $\Phi$ of three variables :
$$\Phi\left(\left(x^2+2\cosh(t)\right)\:,\:\left(y^2+2\sinh(t)\right)\:,\:\left(u+\ln|\cos(t)|\right)\right)=0$$
Or equivalent $c_3=F(c_1\:,\:c_2)$ with arbitrary function $F$ of two variables :
$$ \left(u+\ln|\cos(t)|\right)=F\left(\left(x^2+2\cosh(t)\right)\:,\:\left(y^2+2\sinh(t)\right)\right)$$
General solution :
$$\boxed{u(t,x,y)=-\ln|\cos(t)|+F\left(\left(x^2+2\cosh(t)\right)\:,\:\left(y^2+2\sinh(t)\right)\right)}$$
INITIAL CONDITION : $u(0,x,y)=x^2-y^2+1$

$u(0,x,y)=-\ln|\cos(0)|+F\left(\left(x^2+2\cosh(0)\right)\:,\:\left(y^2+2\sinh(0)\right)\right)=F(x^2+2\:,\:y^2)$

$$u(0,x,y)=x^2-y^2+1=F(x^2+2\:,\:y^2)$$
Thus the function $F(X,Y)$ with $x^2=X-2$ and $y^2=Y$ must satisfy :
$$F(X,Y)=(X-2)-Y+1=X-Y-1$$
Now the function $F(X,Y)$ is known. We put it into the above general solution where $X=x^2+2\cosh(t)$ and $Y=y^2+2\sinh(t)$.

$$u(t,x,y)=-\ln|\cos(t)|+x^2+2\cosh(t)-y^2-2\sinh(t)-1$$
This is the particular solution which satisfies both the PDE and the initial condition.


问:
Divergence proof for $\frac{2^{\sqrt{n}}}{n^3}$ as $n\to\infty$
Can someone please check my proof on this divergence question (as I'm a bit rusty). I want to use a proof by contradiction using the definition of limit.

Prove $\displaystyle a_n=\frac{2^{\sqrt{n}}}{n^3}$ diverges.

Rewriting $\displaystyle a_n=\frac{e^{\sqrt{n}\log 2}}{n^3}$ and using $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$ then
$$a_n=\underbrace{\frac{1}{n^3}+\frac{\sqrt{n}\log 2}{n^31!}+\frac{n\log^22}{n^32!}+\frac{n^{3/2}\log^32}{n^33!}+\frac{n^2\log^42}{n^34!}+\frac{n^{5/2}\log^52}{n^35!}}_{\displaystyle\alpha}+\frac{n^3}{n^36!}+O(n^{7/2}),$$
hence
$$a_n=\alpha+\frac{1}{6!}+O(n^{7/2}),$$
where it's clear to me that $\alpha\to 0$ as $n\to\infty$. Now for a contradiction. Suppose $a_n$ converges to $L$ finite, then for all $\varepsilon>0$ we must have
$$|a_n-L|<\varepsilon\iff\left|\alpha+\frac{1}{24}+O(n^{7/2})-L\right|<\varepsilon.$$
But this can never be the case since $a_n$ is $O(n^{7/2})$ which is a contradiction. In other words, for all $M>0$ there exists $k\in\mathbb{N}$ such that $n>k$ implies $|a_n|>M$ (diverges).

If there are other ways to prove this I'd be interested too.

答:
What you did looks fine.

You can also note that the sequence $\left(\frac{2^n}{n^6}\right)_{n\in\Bbb N}$ is a subsequence of your sequence. But this subsequence diverges, since$$\frac{\frac{2^{n+1}}{(n+1)^6}}{\frac{2^n}{n^6}}=2\left(\frac n{n+1}\right)^6$$which is greater that $\frac32$ if $n$ is large enough. So, for some $N\in\Bbb N$ you have$$(\forall n\in\Bbb N):\frac{2^{N+n}}{(N+n)^6}\geqslant\left(\frac32\right)^n\frac{2^N}{N^6}$$and therefore the subsequence diverges indeed. So, your sequence diverges too.


答:
If you use your arguments based on the Taylor series and exploit that all terms are positive, you obtain
$$a_n > \dfrac{(\mathrm{log}(2))^7n^{\frac{7}{2}}}{n^3 7!} = \dfrac{(\mathrm{log}(2))^7}{7!}n^{\frac{1}{2}}\to\infty\;\text{as } n\to\infty.$$

Hence also $a_n\to\infty$.

Title: Divergence proof for $\frac{2^{\sqrt{n}}}{n^3}$ as $n\to\infty$

Body:
Can someone please check my proof on this divergence question (as I'm a bit rusty). I want to use a proof by contradiction using the definition of limit.

Prove $\displaystyle a_n=\frac{2^{\sqrt{n}}}{n^3}$ diverges.

Rewriting $\displaystyle a_n=\frac{e^{\sqrt{n}\log 2}}{n^3}$ and using $\displaystyle e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$ then
$$a_n=\underbrace{\frac{1}{n^3}+\frac{\sqrt{n}\log 2}{n^31!}+\frac{n\log^22}{n^32!}+\frac{n^{3/2}\log^32}{n^33!}+\frac{n^2\log^42}{n^34!}+\frac{n^{5/2}\log^52}{n^35!}}_{\displaystyle\alpha}+\frac{n^3}{n^36!}+O(n^{7/2}),$$
hence
$$a_n=\alpha+\frac{1}{6!}+O(n^{7/2}),$$
where it's clear to me that $\alpha\to 0$ as $n\to\infty$. Now for a contradiction. Suppose $a_n$ converges to $L$ finite, then for all $\varepsilon>0$ we must have
$$|a_n-L|<\varepsilon\iff\left|\alpha+\frac{1}{24}+O(n^{7/2})-L\right|<\varepsilon.$$
But this can never be the case since $a_n$ is $O(n^{7/2})$ which is a contradiction. In other words, for all $M>0$ there exists $k\in\mathbb{N}$ such that $n>k$ implies $|a_n|>M$ (diverges).

If there are other ways to prove this I'd be interested too.


答:
It seems correct.

Another way is to pass to a continuous variable and use L'Hôpital Rule. For every application of L'Hopital, you will get the same fraction but the denominator should be decresed by $1/2$.
$$\lim_{x\to+\infty}\frac{2^\sqrt{x}}{x^3} =\lim_{x\to+\infty}\frac{2^{\sqrt{x}} \frac{1}{2\sqrt x}\ln 2}{3x^2}=\frac{\ln2}{6}\,\lim_{x\to\infty}\frac{2^{\sqrt x}}{x^{2+1/2}} $$
So after 6 applications of L'Hopital, you'l, get
$$\lim_{x\to+\infty}\frac{2^\sqrt{x}}{x^3} = C \lim_{x\to\infty}2^{\sqrt{x}}=+\infty,$$ where $C$ is a positive constant that does not matter its concrete value.


Body: Note that  $f_n(x)(x^2-1) = x^{2n+2} - 1 = (x^{n+1}-1)(x^{n+1}+1)$. Now use unique factorization...


答: Using $\exp(x)=\sum_{k=0}^\infty\frac{x^k}{k!}$ as you did but now as $\exp(x)\geq\frac{x^k}{k!}$ for $x=\sqrt{n}\ln(2)\geq0$ and $k=8$ results in $a_n\geq\frac{n^4 \ln(2)^8}{8!n^3}=c n$ for some $c>0$. This implies the result.


答:
With L'Hospital we see that $\frac{\ln x}{\sqrt{x}} \to 0$ as $ x \to \infty.$ Hence there is $N \in \mathbb N$ such that

$$ 0 <  \frac{\ln n}{\sqrt{n}} \le \frac{\ln 2}{4}$$

for $n>N.$

But this means

$$\frac{2^{\sqrt{n}}}{n^3} \ge n$$

for $n>N.$


问:
How do I find the first term in an arithmetic sequence?
The last term of an arithmetic sequence is $63$, the common difference is $5$ and the sum is $426$. How do you find the first term?

That's all the information I have.

My attempt:

$63 = a + (n-1)(5)$

$63 = a + 5n - 5$

$68 = a + 5n$

I'm stuck here.

答:
You haven't yet used the fact that the sum is $426$. Using the formula for the sum of an AP gives
$$na+\frac{n(n-1)}{2}d=426.$$


答:
Let $g_n(x) = x^n + x^{n-1} + \ldots + x + 1$. Then $g_n(x) = \frac{x^{n+1}-1}{x-1}$. Your polynomial $f_n(x) = g_n(x^2)$. The factorization of $x^n-1$ over $\mathbb{Q}$ is known:
$$
x^n-1 = \prod_{d\mid n}{\Phi_d(x)},
$$
where $\Phi_d(x)$ is the $d$th cyclotomic polynomial. Thus, the factorization of $g_n(x)$ will lead to a factorization of $f_n(x) = g_n(x^2)$ in $\mathbb{Q}[x^2]$. It is then possible that this factorization may be refined further. So, the only chance of an irreducible $f_n(x)$ would be when $n = p-1$ for some prime: in this case, it is known that $g_{p-1}(x) = x^{p-1} + \ldots + x + 1$ is irreducible over $\mathbb{Q}$. However, for odd primes $p$, the splitting field $\mathbb{Q}(\zeta_p)$ is equal to $\mathbb{Q}(\zeta_{2p})$. Thus, $\zeta_p = \zeta_{2p}^2$, that is, $\zeta_{p}$ has a square root in the same field. So $\zeta_{2p}$ has a minimal polynomial of degree $p-1$ and it is a root of $f_{p-1}(x)$, which has degree $2(p-1)$. So, $f_{p-1}(x)$ must split into two irreducible factors.

Therefore, the answer is yes, $f_n(x)$ is always reducible when $n\geq 2$.


Title: How do I find the first term in an arithmetic sequence?

Body:
The last term of an arithmetic sequence is $63$, the common difference is $5$ and the sum is $426$. How do you find the first term?

That's all the information I have.

My attempt:

$63 = a + (n-1)(5)$

$63 = a + 5n - 5$

$68 = a + 5n$

I'm stuck here.

Body:
Hint

$$\sum _{k=0}^n (a+5 k)=\frac{1}{2} (n+1) (2 a+5 n)=426$$
Should be $a=8$


问:
If $\sum_{i=1}^{n} n a_n$ converge ,does $\sum_{i=1}^{n} a_n$ converge?
If $ a_n \geq 0 $, then is easy to get.
Cause $$\lim_{n \to \infty} \frac{a_n}{n a_n} = 0$$ , so $\sum_{i=1}^{n} n a_n$ converge, then $\sum_{i=1}^{n} a_n$ also converge.

When $ a_n $ not always positive, does $\sum_{i=1}^{n} a_n $ still converge? I guess $ \sum_{i=1}^{n} n a_n $ construct a not absolute convergence $ \sum_{i=1}^{n} n a_n $, then maybe $\sum_{i=1}^{n} a_n $ will diverge. Seem not so easy.

答: If the series $\sum_{n=1}^\infty na_n$ converges, then the series $\sum_{n=1}^\infty a_n$ converges too, by Abel's criterion:$$\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac1n(na_n),$$and the sequence $\left(\frac1n\right)_{n\in\Bbb N}$ is monotonic and its limit is $0$.


问:
Area under curve equals product of arc length and its projection
Find equation of a curve whose  area under it equals product of arc length  $L$ and its projection $(b-a)$ on the x-axis.

$$ A = L (b-a)$$

I was trying to establish a corollary to the Amazing Catenary property

答:
A constant map satisfies the equation. Let's prove that there is no other solution.

Suppose that $f$ is continuously differentiable. As mentioned, if $f$ is a solution (with adequate translation) it satisfies the functional equation:

$$ \int_0^x f(t) dt =x  \int_0^x \sqrt{ 1 + (f^ \prime(t))^2} dt \tag{1}.$$

The RHS, namely

$$R(x)= x  \int_0^x \sqrt{ 1 + (f^ \prime(t))^2} dt$$ remains invariant if we change $f$ into $g(x) = f(0)  - \int_0^x f^\prime(t) \ dt$ as $g^\prime(x) = - f^\prime(x)$.

However if for a point $x_0$, we have $f^ \prime(x_0) \neq 0$, the LHS of the equation $(1)$ will change locally around $x_0$ if $f$ is replaced with $g$. A contradiction. As this is independent of $x_0$, we get $f^\prime(x) = 0$ for all $x \in \mathbb R$ meaning that $f$ has to be constant.


问:
Why does the natural filtration induce a brownian motion
I am reading up on the brownian motion right now and there is sometthing claimed in my textbook that I don't understand.

Let $(B_t)_t$ be a Brownian motion with respect to some filter. Now define the natural filtration of $(B_t)_t$ as $\mathcal{F}_t = \sigma (B_s : 0 \leq s \leq t)$ for $t \geq 0$. Then $(B_t)_t$ is also a Brownian motion with respect to $(\mathcal{F}_t)_t$.

I think the only property that needs checking is that for $0 \leq s \leq t$ the increment $B_t-B_s$ is independent of $\mathcal{F}_s$. But why is this true? Since there is no proof I think this should be trivial, but I can't see how.

答:
The proof depends mainly on properties of multivariate normal distributions.

If $t_1 <t_2<...<t_k \leq s$ then $B_t-B_s,B_{t_1},B_{t_2},...,B_{t_k}$ have a multivariate normal distribution. Since then covariance of $B_t-B_s$ with the other variables are $0$ it follows that $B_t-B_s$ is independent of $\{B_{t_1},B_{t_2},...,B_{t_k}\}$ and this being true for all choices of $t_i$'s it follows that $B_t-B_s$ is independent of $\mathcal F_s$.


问:
How many 5-digit numbers such that the sum of their digits is a multiple of 4?
My textbook has the same problem, except for 5-digit numbers the sum of whose digits is a multiple of 5. The general approach is:

Select the first 4 digits. Number of possible ways of doing this = $9\times 10 \times 10 \times 10$ since the ten thousands-place digit cannot be zero. The sum of these 4 digits is $5k$, $5k+1$, $5k+2$, $5k+3$, or $5k+4$. If it's $5k$, the last digit may only be $0$ or $5$. Sum is $5k+1 \implies$ last digit is $4$ or $9$, and so on. In each case, there are only $2$ possible last digits.

Therefore the total number of such 5-digit numbers is $9\times 10 \times 10 \times 10 \times 2 = 18000$.

When we apply this approach to the "sum of digits is a multiple of 4" problem, we get,

Sum of first 4 digits is $4k \implies$ last digit is $0$, $4$, or $8$.

Sum of first 4 digits is $4k+1 \implies$ last digit is $3$ or $7$.

Sum of first 4 digits is $4k+2 \implies$ last digit is $2$ or $6$.

Sum of first 4 digits is $4k+3 \implies$ last digit is $1$, $5$ or $9$.

In two of these cases, we get three possible last digits and in the other two, we get two possible last digits. Our general approach clearly breaks down here. One way to go around this would be to ask ourselves, "how many ways can we select four digits such their sum is $4k$ or $4k+3$", and multiply that number by 3; repeat for $4k+2$ and $4k+1$ but multiply by 2. Add both these numbers to get the answer.

This seems too long. Is there a simpler way to do this?

答:
Denote by $a_i(n)$ $\>(0\leq i\leq3)$ the number of $n$-digit decimal strings having a digit sum leaving remainder $i$ modulo $4$, and collect the $a_i(n)$ into $a(n):=\bigl(a_0(n), a_1(n), a_2(n), a_3(n)\bigr)$, the latter considered as a column vector. We then have $$a(1)=(2,3,2,2)\ .$$ For the latter digits the multiplicities of the possible remainders are $(3,3,2,2)$, since the digit ${\tt 0}$ with remainder $0$ is allowed.

What we have written already allows to compute $$a_0(2)=3\cdot2+2\cdot3+2\cdot2+3\cdot2=22\ .$$
Indeed, when the first digit has remainder $0$ the second also has to have remainder $0$, and when the first digit has remainder $1$ the second has to have remainder $3$, etcetera. This idea, used for all possible remainders, leads to the recursion
$$ a(n+1)=A\ a(n)\qquad(n\geq1),\qquad A:=\left[\matrix{3&2&2&3\cr3&3&2&2\cr2&3&3&2\cr2&2&3&3\cr}\right]\ .$$
In this way we obtain
$$a(5)=A^4\ a(1)=\bigl(22500, 22498, 22500, 22502\bigr)\ ,$$
so that the number we are looking for is $22\,500$. In order to compute, e.g., $a(23)$ we better find a scheme for computing the higher powers of the "cyclic" matrix $A$.


问:
Probability that digits are at their proper places of $n$ possibilities
I can't solve this question although I have tried many times.

Question : Let $n>1$ , suppose that $n$ numbers are written down in random order.

${1,2,...,n}$

Number is said to be in the "right place" if for example 2 is located in the 2nd place.

1)What is the probability that 2 numbers will be in the right place ?

2)what is the probability that atleast 2 numbers will be in the right place?

My try (for the question number 2) :

all the probabilities are $n!$, $A$ is wanted probability, I chose $2Cn$ and the probabilities for the others is $(n-2)!$.

I got $\frac {A}{n!} = \frac{1}{2}$

答:
Number of arrangements where exactly two numbers are in the right places is a derangement of $(n-2)$ and the way to choose $2$ numbers that are in right places.

So, Number of favorable arrangements for $(1) = {n \choose 2} \times \, !(n-2)$

For second question, find the number of arrangements that have at most one number in the right place $( = \, !n + n \times \, !(n-1))$ and subtract from $n!$.

Now use these to find desired probability.


问:
An example which shows that $F(f) \subset \overline{Per(f)} \subset L(f) \subset \Omega(f)$ is restrict?
I have some questions on Dynamical Systems.

$(X,f)$ is a Dynamical System if $f:X \to X$ is a homeomorphism and $X$ is a compact space. Let's define
$$
\begin{align}
&Per(f):=\{ x \in X ; f^n(x)=x ,\text{ for some } n \in \mathbb{Z} \}\\
&L(f):=L_{+}(f) \bigcup L_{-}(f)
\end{align}
$$ where
$$
L_{+}(f)=\overline{\bigcup_{x\in X}\omega(x)}\quad\text{ and }\quad
L_{-}(f)=\overline{\bigcup_{x\in X}\alpha(x)},
$$ where

$\omega(x):=\{y : \exists (n_i), f^{n_i}(x) \to y \}$ and

$\alpha(x):=\{y : \exists (n_i), f^{ -n_i}(x) \to y \}$

and finally where $\Omega(f)$ is the set of non wandering points, i.e. all $x$ such that $\forall$ U open containing $x$ and $\forall$ $N>0$ there exists some $n>N$ such that $f^n(U) \cap U \ne \emptyset$ and
$$
F(f) := \{x \in X : f(x)=x\}
$$

1 - I was looking for an example which shows that $Per(f)$ could be closed. I think I found a simple example which shows that :
\begin{align}
&f : [0,1] \to [0,1]\\
&f(x)=x \quad , \forall x\in [0,1]
\end{align}
In this case the set $Per(f)$ is closed because we know the set of all fixed points is closed. am I right?
Now I'm looking for an example which shows that $Per(f)$ is not closed in general, I think If we find an example whose periodic points set is empty the question will be solved.

2- I already proved this;
\begin{align}
F(f) \subseteq \overline{Per(f)} \subseteq L(f) \subseteq \Omega(f)
\end{align}
Now I'm looking for an example which shows that the relation may be restrict what I mean is that;
\begin{align}
F(f) \subset \overline{Per(f)} \subset L(f) \subset \Omega(f)
\end{align}

3- The third question is also an example just like question $1,2$ I want an example which shows that;
\begin{align}
\overline{Per(f)} \neq \Omega(f)
\end{align}
For this we have to find an example which shows that;
\begin{align}
\Omega(f) \nsubseteq \overline{Per(f)}
\end{align}

I consider discrete dynamical systems. Can any one help?

答:
In this case the set $Per(f)$ is closed because we know the set of all fixed points is closed.

Yes (when $X$ is Hausdorff).

Examples such that:

1)) $F(f) \subset \overline{Per(f)}$. Let $X$ be a set $\{1,2,3\}$ endowed with the discrete topology and $f(1)=2$, $f(2)=3$, and $f(3)=1$. Then $F(f)=\varnothing$ but $Per(f)=X$.

2)) $\Omega(f) \nsubseteq\overline{Per(f)} \subset L(f)$. Let $X$ be the unit circle $\{z\in\Bbb C: |z|=1\}$ endowed with the topology inherited from $\Bbb C$, $\varphi$ be a real number such that $\varphi/\pi$ is irrational and $f:X\to X$, $x\mapsto xe^{\varphi i}$ be a rotation at the angle $\varphi$. It is easy to check that $Per(f)=\varnothing$ and $L(f)=\Omega(f)=X$.

3)) $L(f) \subset \Omega(f)$. I assume that $(n_i)$ in the definitions of $\alpha(x)$ and $\omega(x)$ is a strictly increasing sequence of natural numbers.

Let $X={\Bbb T}^{\Bbb T}$. By Tychonov Theorem, $X$ is a compact topological group with coordinate-wise multiplication.  Let  $g=(g_z)_{z\in\Bbb T}\in X$ be an element such that $g_z=z$ for each $z\in\Bbb T$. For each $x\in X$ put $f(x)=gx$.

We claim that $\Omega(f)=X$. Indeed, let $x$ be any element of $X$, $U$ be any neighborhood of $x$, and $N$ be any natural number.  Since $X$ is a topological group, there exists a neighborhood $V$ of the identity of $G$ such that $xVV{-1}\subset U$. A family $\{yV:y\in X\}$ is an open cover of a compact space $X$, so there exists a finite subset $F$ of $X$ such that $FV=X$. Therefore there exist an element $y\in F$ and natural numbers $n,m$ such that $g^n,g^m\in yV$ and $n>m+N$. Then $g^m\in g^nVV^{-1}$ and so $x\in x g^{n-m}VV^{-1}\subseteq  g^{n-m}U$.

We claim that sets $\alpha(x)$ and $\omega(x)$ are empty for each $x=(x_z)_{z\in\Bbb T}\in X$. Since the group $G$ is Abelian, $\omega(x)=\alpha(x^{-1})^{-1}$, it suffices to show that the set $\alpha(x)$ is empty.

Suppose to the contrary that there exists an increasing sequence $\{n_i\}$ of natural numbers such that a sequence $\{g^{n_i}x\}$ converges to a point $y=(y_z)_{z\in\Bbb T}\in X$. Let $U_0=\{z\in\Bbb T: |z-1|\le 1/\sqrt{2} \}$ be a neighborhood of the identity of the group $\Bbb T$. For each natural number $i$ put $$T_i=\{z\in\Bbb T: (n_j-n_k)z\in U_0^2\mbox{ for each }j,k>i\}.$$ Since the set $ U_0^2$ is closed in $\Bbb T$, the continuity of power on the group $\Bbb T$ implies that the set $T_i$ is closed for each natural number $i$. We claim that $\Bbb T=\bigcup_{i\in\Bbb N} T_i$. Indeed, let $z\in \Bbb T $ be any element. There exists a natural number $i$ such that $n_jzx_z\in y_zU_0$ for each $j>i_z$. Then $(n_i-n_k)z\in U_0^2$ for each $j,k>i$, so $z\in T_i$.

Since $\Bbb T=\bigcup_{i\in\Bbb N} T_i$ and $\Bbb T$ is a compact metrizable space, Baire theorem implies that there exists a natural number $i$ such that a set $T_i$ has non-empty interior. Therefore there exists an open arc $U\subseteq T_i$ of the circle $\Bbb T$. Let $\ell$ be the length of $U$. Since the sequence $\{n_i\}$ is increasing, there exists a number $j>i$ such that $n_j-n_{i+1}>2\pi/\ell$. But then $U_0^2\supseteq (n_j-n_{i+1})T_i\supseteq (n_j-n_{i+1}) U=\Bbb T$, a contradiction.

4)) $Per(f)$ is not closed. Let $X$ be the unit disk $\{z\in\Bbb C: |z|\le 1\}$ and $f:X\to X$, $x\mapsto xe^{|x|i}$. Then $Per(f)=\{x\in X: |x|/\pi\in\Bbb Q\}.$


答:
SAs per OP's work the line passes throught $A(0,-q/p,-s/r)$ its direction vector is $\vec L=(1,1/p,1/r)$. The plane passes through $B(u,v,w)$. Normal to the plane $\vec n=\vec{AB}\times \vec L=(n_1,n_2,n_3)$, then the equation of the plane is
$$n_1(x-u)+n_2(y-v)+n_3(z-w)=0.$$ This equation can also be written as
$$(\vec r-\vec B).(\vec{AB} \times \vec L)=0 \implies\begin{vmatrix} x-u & y-v & z -w  \\ 0-u & -q/p-v & -s/r-w \\ 1 & 1/p & 1/r \end{vmatrix}=0,$$ that gives the equation of the required plane.
Next, $C_2'\to p C_2, C_3' \to rC_3$ to get
$$\begin{vmatrix} x-u & py-pv & rx-rw \\ u & q+vp & s+rw \\ 1 & 1 & 1\end{vmatrix}=0$$
Next, Perform $R_1'\to R_1+R_2$ to get the required form:
$$\begin{vmatrix} x & py+q & rx+s \\ u & vp+q & rw+s \\ 1 & 1 & 1\end{vmatrix}=0$$

Body:
Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ is
$${\begin{vmatrix} x& py+q& rz+s\\
\alpha& p\beta+q& r\gamma+s\\
1& 1& 1\\
\end{vmatrix}} = 0$$

My Attempt:
The equation of line is
$x = py + q = rz + s$
$$\implies x = p (y + \frac {q}{p}) = r ( z + \frac {s}{r}) $$
$$\implies \frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$$
The equation of plane through this line is
$$a(x-0) + b(y+\frac {q}{p}) + c(z+\frac {s}{r}) = 0$$
where $a,b,c$ are the direction ratios of the line normal to the plane.
Now,
$$a\cdot 1 + b\cdot (y+\frac {q}{p}) + c\cdot (z + \frac {s}{r}) = 0$$
Also, the plane passes through the point $(\alpha, \beta, \gamma)$,
$$a(\alpha - 0) + b(\beta + \frac {q}{p}) + c(\gamma + \frac {s}{r}) = 0$$

How to proceed further from here?


问:
Solving integral by partial integration
Looking at the integral

$$\int\limits_1^{\infty} \dfrac{\sin(x)}{x} \ dx $$

Is it ok to write

$$\int\limits_1^{\infty} \frac{\sin(x)}{x} \ dx = \left[-\frac{\cos(x)}{\displaystyle x}\right]_1^{\infty} + \int\limits_1^{\infty} \dfrac{\cos(x)}{x^2} \ dx$$

and then saying that the existence is given since $ \left|\ \dfrac{\cos(x)}{x^2} \right|\ \leq \dfrac{1}{x^2}$

I've seen a proof on the existence of the above integral which is much more complicated than this, so I'm not sure if this can be correct.

答:
$$\int_1^{\infty } \frac{\sin (x)}{x} \, dx= \tag 1$$
$$=\left[\frac{-\cos x}{x}\right]_1^\infty-\int_1^{\infty } \frac{\cos x}{x^2}\,dx$$
$$\left[\frac{-\cos x}{x}\right]_1^\infty=\underset{x\to \infty }{\text{lim}}\left(-\frac{\cos x}{x}\right)+\frac{\cos 1}{1}=0+\cos 1=\cos 1$$
$$\int_1^{\infty } \frac{\cos x}{x^2}\,dx\le\int_1^{\infty } \frac{dx}{x^2}$$
Last integral converges, therefore the first integral $(1)$ converges

Title: Solving integral by partial integration


问:
A man throws 4 dice. What is the probability that same number shows up in at least two faces?
I saw a question  in internet , it seems very trivial.I tried to solve it in two different way but i obtained two different result.

I know that my second solution is correct but why is the former wrong ? What am i missing ?

Question: A man throws $4$ dice. What is the probability that same number shows up in at least two faces?

First Solution:

The probability of getting ONLY $2 $ dice showing the same face : $\frac{C(4,2).6.5.4}{6.6.6.6}=\frac{120}{216}$

The probability of getting ONLY $3 $ dice showing the same face : $\frac{C(4,3).6.5}{6.6.6.6}=\frac{20}{216}$

The probability of getting ONLY $4 $ dice showing the same face : $\frac{C(4,4).6}{6.6.6.6}=\frac{1}{216}$

$\therefore \frac{141}{216}=0.652$

Second Solution: All situations - all dice appears different faces

$1- \frac{6.5.4.3}{6.6.6.6}=0.722$

What am i missing ?

答:
Your $\frac{936}{1296}\approx 0.722$ is correct

Your first calculation is missing the case of two faces each appearing on two dice, which has probability $\frac{90}{1296}=\frac{15}{216}\approx 0.069$; this is the difference between your two answers


问:
Show that $\forall n \in \mathbb{Z}, P(n) \in \mathbb{Z}$ when $e^{2 i \pi P(n)} \underset{n \to +\infty}{\longrightarrow} 1$
I want to show that $\forall n \in \mathbb{Z}, P(n) \in \mathbb{Z}$, assuming I have both $e^{2 i \pi P(n)} \underset{n \to +\infty}{\longrightarrow} 1$ and $P \in \mathbb{R}[X]$.

My idea was to start doing it by mathematical induction, I write $P = \alpha X^N + Q$ where $\deg Q \leq N-1$. So, if I show that $\alpha \in \mathbb{Z}$, by mathematical induction all coefficients of $P$ will be in $\mathbb{Z}$ and then we'll have the desired result (initialisation is easy).

But I dont't find a way to exploit this track. Any help is welcome.

答:
Following @Kelenner's hint, if we do it by mathematical induction, we remark that $e^{2 i \pi P(n-1)} \underset{n \to +\infty}{\longrightarrow} 1$ then we also have that $e^{2 i \pi (P(n) -P(n-1))} \underset{n \to +\infty}{\longrightarrow} 1$. As $Q = P(X) - P(X-1)$ is of degree $< \deg P$, one has $Q(n) \in\mathbb{Z}$ for all $n \in \mathbb{Z}$.

As $P(n) - P(0)= \sum_{k = 1}^n P(k) - P(k-1) = \sum_{k = 1}^n Q(k)$, we have $\forall n \in \mathbb{Z}, P(n) - P(0) \in \mathbb{Z}$.

Then: $e^{2 i \pi P(0)} = e^{2 i \pi P(n)} e^{- 2 i \pi (P(n) - P(0))}\underset{n \to +\infty} {\longrightarrow} 1 \times 1 = 1$ then $P(0) \in \mathbb{Z}$.

Then, for all $n \in \mathbb{Z}$, $P(n) = P(0) + Q(1) + \dots + Q(n) \in \mathbb{Z}$.


问:
Bounding $\|A^n-B^n\|_F$ by $\|A-B\|_F$
Given
$$ \epsilon = \|A-B\|_F, $$
it is clear that
$$ \epsilon^n > \|(A-B)^n\|_F, $$
which follows from submultiplicativity.

I wonder if something related can be said about $\|A^n-B^n\|_F$?

答:
\begin{align}A^n-B^n&=A^n-A^{n-1}B+A^{n-1}B-A^{n-2}B^2+\cdots-B^n\\
&=A^{n-1}(A-B)+A^{n-2}(A-B)B+\cdots+(A-B)B^{n-1}
\end{align}
$$\therefore\ \|A^n-B^n\|\le(\|A^{n-1}\|+\|A^{n-2}\|\|B\|+\cdots+\|B^{n-1}\|)\|A-B\|\le\frac{\|A\|^n-\|B\|^n}{\|A\|-\|B\|}\|A-B\|$$

If the intent of the question is whether $\|A^n-B^n\|$ is small compared to $\|A-B\|$ then the answer is no. Take $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, then $$\|A-B\|_F=2,\qquad\|A^n-B^n\|_F=2n^2$$

Title: Bounding $\|A^n-B^n\|_F$ by $\|A-B\|_F$


问:
Borel-Cantelli Lemma
I have some difficulties understanding the following:

Let $(X_n)$ be a sequence of independent random variables s.t.

$P[X_n=1]=1−P[X_n=0]=\frac{1}{n}$

After using the Borell Cantelli lemma, I could show that:

$P(\limsup [X_n=0])=1$

$P(\limsup [X_n=1])=1$

Now using the fact that $\operatorname{[\limsup F_n]}^c  = [\liminf \operatorname{(F_n)}^c]$ I concluded that

$P(\liminf [X_n=1])=0$

$P(\liminf [X_n=0])=0$         as $P(A)+P(A^C)=1$.

This implies that as $X_n$ is either $0$ or $1$, $\liminf X_n$ is never constant with probability $1$, contradicting some of the implications of the Kolmogorov $0$-$1$ law.

Could someone please clarify this?
Much appreciated.

答:
Consider the events $A_n=[X_n=0]$. By definition, the event $\limsup A_n$ happens if and only if $A_n$ happens for infinitely many $n$, hence the general formula
$$
\limsup A_n=\bigcap\limits_n\bigcup\limits_{k\geqslant n}A_k.
$$
In our case, consider $N=\{n\mid X_n=0\}$, then
$$
\limsup A_n=[X_n=0\ \text{for infinitely many}\ n]=[N\ \text{is infinite}].
$$
Consider $Y=\liminf X_n$. This is a random variable $Y$ characterized by the fact that $Y=1$ if $X_n=1$ for every $n$ large enough, that is,
$$
[Y=1]=[\exists n,\ \forall k\geqslant n,\ X_k=1].
$$
and that $[Y=0]=[N\ \text{is infinite}]$.

Finally, for $\{0,1\}$-valued random variables $(X_n)$,
$$
\limsup[X_n=0]=[\liminf X_n=0].
$$
Likewise, again for $\{0,1\}$-valued random variables $(X_n)$,
$$
\liminf[X_n=0]=[(X_n)\ \text{converges and}\ \lim X_n=0]=[\limsup X_n=0],
$$
$$
\liminf[X_n=1]=[(X_n)\ \text{converges and}\ \lim X_n=1]=[\liminf X_n=1],
$$
and
$$
\limsup[X_n=1]=[X_n=1\ \text{for infinitely many}\ n]=[\limsup X_n=1].
$$

Title: Borel-Cantelli Lemma


问:
Suppose $A$ is an n by n matrix whose eigenvalues have a magnitude larger than 1. $P$ is a covariance matrix. Is it true that $APA' \geq P$?
Here, $A$ is an $n$ by $n$ matrix, whose eigenvalues all have a magnitude larger than 1. $P$ is a covariance matrix of dimension $n$. I guess it should be true that $APA' - P$ is positive semi-definite because $A$ is 'amplifying' the covariance $P$.

I guess to shows its correctness we may leverage on matrix decomposition such as $APA' - P = AQTQ'A' -QTQ'$. But I am not sure how to proceed especially how to incorporate the fact that the eigenvalues of $A$ all have magnititude larger than $1$.

答:
A counter example would be

$$
A =
\begin{bmatrix}
2 & 0 \\ 0 & -2
\end{bmatrix}, \quad
P =
\begin{bmatrix}
1 & 1 \\ 1 & 1
\end{bmatrix}, \quad
A\,P\,A'-P =
\begin{bmatrix}
3 & -5 \\ -5 & 3
\end{bmatrix},
$$

which has eigenvalues $-2$ and $8$. If you prefer a positive definite covariance matrix you could also use

$$
P =
\begin{bmatrix}
10 & 9 \\ 9 & 10
\end{bmatrix},
$$

which yields

$$
A\,P\,A'-P =
\begin{bmatrix}
30 & -45 \\ -45 & 30
\end{bmatrix},
$$

and has eigenvalues $-15$ and $75$.


问:
Can a theory be not satisfiable? Idk if I expressed myself right, can it be that a theory has no model?
According to lecture on FO logic:
A Σ-theory T is defined by a set of Σ-sentences.

According to another lecture on mathematical logic (predicate logic):
A theory is a satisfiable set T ⊆ FO (τ) of sentences, which is closed under |=, meaning it holds for all τ-sentences ψ with T |= ψ, that ψ ∈ T.

I am asking because of this true/false statement (*): There exists a signature Σ and a Σ-theory T such that no Σ-formulas are T-satisfiable.

A formula is T-satisfiable, if there exists a structure to satisfy both T and this formula.

The statement could be true if the theory is unsatisfiable, meaning it has no model. If the theory has a model that satisfies it, you could pick the formula to be one of the sentences of the theory.

答:
Can a theory be not satisfiable?

If you restrict the definition of theory to "satisfiable set" of sentences, the answer is NO.

But, in general, we have that a formula (set of formulas) is unsatisfiable iff it is inconsistent.

Thus, a single-axiom "theory", like e.g.$\{ \ \forall x (x \ne x) \ \}$ has no model.

Every inconsistent set of sentences is unsatisfiable, i.e. has no model.

The "standard" use of theory in logic is:

a set of sentences in a formal language.

See also David Marker, Model Theory: An Introduction (Springer, 2002), page 14:

Let $\mathcal L$ a language. An $\mathcal L$-theory $T$ is simply a set of $\mathcal L$-sentences.


问:
How can I count the number of colored combinations in a set of regions?
Let me first start out by saying as you might have guessed, this is a homework problem. Therefore I am not looking for an answer to the question. I am looking for help in how to analyze it (My textbook is horrid).

I have a grid of 9 squares surrounded by a circle.
I can use three colors to color each region with no region touching any other region with the same color (corners don't count). **`

I am asked to find the number of
  possible different colorings.

I drew a set of nodes with lines connecting neighboring regions, but I still have no idea how to figure out the number of combinations.

I also made a guess of three (which I know is wrong) since you can color the regions once with the three colors then map each color to a different one two times after the first coloring.

No blue is touching any other blue regions, no white is touching any other white regions, no red is touching any other red regions.

答: I've found that the simplest way to attack problems like this is really by straight enumeration.  Start by coloring the outer region, then see what colors that forces or eliminates; in this case, it means that none of the eight outer cells of your grid can share that color.  Now, pick another unforced cell (e.g., the top-left corner of the grid) and see how many ways that you can color it; in this case we have two separate colors that we can use, and whichever one we use then forces the colors on all of the outer cells of the grid, leaving us with just one un-forced color (the central square of the grid can be one of two colors), etc. While this is sort-of a graph coloring problem, it's really closer to a constraint satisfaction problem in disguise, and the methods used for solving a lot of the popular pen-and-paper puzzles like Sudoku/Kakuro/Picross/etc. are pretty closely related to this.</p>

ContentLicense: CC BY-SA 2.5


问:
Second degree linear derivate with exponential function, which general solution?
Suppose $y''-4y=xe^{2x}$.

Solution of the homogenous equation is $y_H(x) = C_{1} e^{2x} + C_{2} e^{-2}$, after solving the characteristic equation with the guess $y=e^{rx}$.

Now my instructor insists that the one solution is either

$y(x) = Ax^{2} e^{2x} + Bxe^{2x}$

or

$y(x) = Ax^{2}e^{2x} + Bxe^{2x}+Ce^{2x}$

...and then I get the solution $y(x) = y_{H}(x) +\frac{1}{8}x^{2}e^{2x}-\frac{1}{16}xe^{2x}$

but I cannot understand the choices. Why do just the forms work? I have tried many different forms there and get lost/wasted a lot time -- and my instructor just says that it is a "guess" and smiles. How can I find the one solution without guessing? For one solution, things such as $y(x)=e^{2x}$ or $y(x)=xe^{2x}$ work so I am a bit lost why the more complex form is "the solution" and why it provides all solutions.

For example, the case $y=e^{2x}$. We get $y'=2e^{2x}$ and $y''=4e^{2x}$ so

$y''-4y = 4e^{2x} - 4 e^{2x} = 0$

so $RHS = xe^{2x} = 0$ so $x=0$, a solution?! If not why? Ok it is naive, it does not contain all cases but how can I know that some form contains all?

答:
The "right" approach is highly context-dependent.  What follows is from the perspective of an introduction to differential equations attached to a first-year calculus course.

We describe an informal procedure, in which we make a plausible guess as to $y$, and test how well the guess works by calculating $y''-4y$ for this guess.  We then use the information obtained from the calculation of $y''-4y$ to improve our guess.  Of course more formal procedures are available, but it is nice to see how far one can get by playing a little.

A very reasonable "guess" for a particular solution is $y=xe^{2x}$.  Let's check whether it works.  If $y=xe^{2x}$, then $y'=2xe^{2x}+e^{2x}$, and $y''=4xe^{2x}+4e^{2x}$.  Then $$y''-4y=4e^{2x}$$

But $4e^{2x}$ is definitely not the same function as $xe^{2x}$.  And multiplying the guess $xe^{2x}$ by a constant $k$ will do nothing useful.

Note that things would have worked out nicely if the differential equation had, for example,  $y''-3y$ on the left instead of $y''-4y$.  But the $-4y$ is exactly the right thing to make the $xe^{2x}$ term disappear. When the person posing the problem chose $-4$, (s)he chose the only constant that would make our lives difficult. Nasty!

However, this gives us an idea.  If we try $y=x^2e^{2x}$, and calculate $y''-4y$, maybe the $x^2e^{2x}$ term will disappear, which is exactly what we want.  Calculate.  If $y=x^2e^{2x}$, then $y'=2x^2e^{2x}+ 2xe^{2x}$ and $y''=4x^2e^{2x}+8xe^{2x} +2e^{2x}$.
Thus
$$y''-4y=8xe^{2x} +2e^{2x}$$

Close!  We should clearly multiply our guess by $(1/8)$ to get the $xe^{2x}$ term right.  And we will not be quite there, since then there will still be an unwanted  $(1/4)e^{2x}$ term in $y''-4y$ to get rid of.

Look back on the work we did at the beginning with $y=xe^{2x}$.  Then $y''-4y$ turned out to be $4e^{2x}$.  So to get rid of the unwanted $(1/4)e^{2x}$, we can simply add
$(-1/16)xe^{2x}$ to the "guess" $(1/8)x^2e^{2x}$.

So we end up with the particular solution
$$y =\frac{x^2e^{2x}}{8}-\frac{xe^{2x}}{16}$$


问:
Infinitely many primes of the form $4n+3$
I've found at least 3 other posts$^*$ regarding this theorem, but the posts don't address the issues that I have.

Below is a proof that for infinitely many primes of the form $4n+3$, there's a few questions I have in the proof which I'll mark accordingly.

Proof: Suppose there were only finitely many primes $p_1,\dots, p_k$, which are of the form $4n+3$. Let $N = 4p_1\cdots p_k - 1$. This number is of the form $4n+3$ and is also not prime as it is larger than all the possible primes of the same form. Therefore, it is divisible by a prime $ \color{green}{ \text{(How did they get to this conclusion?)}}$. However, none of the $p_1,\dots, p_k$ divide $N$. So every prime which divides $N$ must be of the form $4n+1$ $ \color{green}{ \text{(Why must it be of this form?)}}$. But notice any two numbers of the form $4n+1$ form a product of the same form, which contradicts the definition of $N$. Contradiction. $\square$

Then as a follow-up question, the text asks "Why does a proof of this flavor fail for primes of the form $4n+1$? $ \color{green}{ \text{(This is my last question.)}}$

$^*$One involves congruences, which I haven't learned yet. The other is a solution-verification type question. The last one makes use of a lemma that is actually one of my questions, but wasn't a question in that post.

答:
Therefore, it is divisible by a prime (How did they get to this conclusion?).

All integers are divisible by some prime!

So every prime which divides N must be of the form 4n+1 (Why must it be of this form?).

Because we've assumed that $p_1, \dots, p_k$ are the only primes of the form 4n+3. If none of those divide N, and 2 doesn't divide N, then all its prime factors must be of the form 4n+1.

"Why does a proof of this flavor fail for primes of the form 4n+1? (This is my last question.)

Can you do this yourself now? (Do you understand how the contradiction works in the proof you have? What happens if you multiply together two numbers of the form 4n+3?)


问:
$K[[x]]$ has only one prime element - what is it?
$K[[x]]$ is the ring of formal power series over a field $K$. What can we say about the prime elements? In the exercise I should prove there exists only one prime element (to be precise one class of prime elements which are related to each other, i.e. if two elements $r,s$ are from this class than one can express the other one as a product with a unit $u,r=us$). I already know that the units are the elements where the constant part is not zero. That every ideal looks like $<x^n>$ and that for every nonzero element $f$ there exists a $x^n$ such that $f=ux^n$, where $u$ is a unit again.

How can I start?

答:
You have to look  for some element $p \ne 0$ which is not a unit and has the property that if $p \mid (f\cdot g)$, then $p \mid f$ or $p \mid g$. In this condition we may of course restrict to the case $f, g \ne 0$.

Certainly $p = x$ is the simplest non-zero power series which is no unit. Now assume $x \mid (f \cdot g)$ with $f, g \ne 0$. Then $f = ux^n, g = vx^m$ with units $u,v$. If $n > 0$, then $x \mid f$, and if $m > 0$, then $x \mid g$. Is it possible that $m = n = 0$? In that case the constant term of $f \cdot g$ would be non-zero so that $x$ could not divide $f \cdot g$.

Thus $x$ is a prime element.

Now let $p$ be an arbitrary prime element. It has the form $p = ux^n$ with a unit $u$ and $n > 0$ (note that $n = 0$ is impossible since then $p$ would be a unit). We conclude that $x^n$ is a prime element. But $x^n \mid (x \cdot x^{n-1})$ so that $x^n \mid x$ or $x^n \mid x^{n-1}$. Certainly $x^n \not\mid x^{n-1}$, thus $x^n \mid x$ which shows that $n = 1$.

Thus $p = ux$ which finishes the proof.


答: Your first solution is 141/216 and your second solution is 156/216. But your first solution doesn't include the case of "two pairs", which accounts for the missing 15/216.</p>

Title: A man throws 4 dice. What is the probability that same number shows up in at least two faces?

Body:
I saw a question  in internet , it seems very trivial.I tried to solve it in two different way but i obtained two different result.

I know that my second solution is correct but why is the former wrong ? What am i missing ?

Question: A man throws $4$ dice. What is the probability that same number shows up in at least two faces?

First Solution:

The probability of getting ONLY $2 $ dice showing the same face : $\frac{C(4,2).6.5.4}{6.6.6.6}=\frac{120}{216}$

The probability of getting ONLY $3 $ dice showing the same face : $\frac{C(4,3).6.5}{6.6.6.6}=\frac{20}{216}$

The probability of getting ONLY $4 $ dice showing the same face : $\frac{C(4,4).6}{6.6.6.6}=\frac{1}{216}$

$\therefore \frac{141}{216}=0.652$

Second Solution: All situations - all dice appears different faces

$1- \frac{6.5.4.3}{6.6.6.6}=0.722$

What am i missing ?


问:
Does there exists a sequence of polynomials and rational functions approximating an analytic function uniformly?
This question was asked in my complex analysis quiz and I was absolutely confused on which result to use.

Consider the function $f(z)=1/z$ on the annulus $A=[{z \in \mathbb{C} : 1/2 < |z|<2}]$.  Then

(a) Does there exists a sequence ${p_n(z)}$ of polynomials that approximate f(z) uniformly on compact subsets of A.

(b) DOes there exists a sequence ${r_n(z)}$ of rational functions , whose poles are contained in $\mathbb{C}/A$ and which approximate  f(z) uniformly on compact subsets of A.

Attempt: $1/z$ is analytic on $A$ so there will exist a sequence of analytic functions which converge uniformly to $1/z$ on compact subsets of $A$ but why should they be polynomials or rational functions specifically?

答:
To summarize the comments above :

(a)

Let $p_n(z)$ be polynomials uniformly approximating $f(z)$ on $A$. Consider the contour $\gamma = \{|z| = 1\}$.

By Cauchy's theorem, since polynomials are holomorphic everywhere in the interior of the circle $|z| = 2$, we get that $\int_{\gamma} p_n(z)dz = 0$ for all $n$.

However, it is well known that $\int_{\gamma} \frac 1z dz = 2 \pi i$.

Now, if $p_n(z) \to \frac 1z$ uniformly on $A$, in then particular we must have $\int_{\gamma} p_n(z)dz \to \int_{\gamma} \frac 1zdz$, which does not hold. Consequently, no sequence of polynomials can uniformly approximate $\frac 1z$ in this annulus (or for that matter, in any annulus centered at the origin by the same argument).

(b)

This is far more obvious : the function $f(z)$ has only one pole, which is outside $A$. Consequently, we can just take $r_n = f$ for all $n$.

The theorem that generalizes this kind of approximation, is Runge's theorem. It states :

Let $K \subset \mathbb C$ be compact and $f$ be a holomorphic on an open set containing $K$. Let $D$ be any set, which has at least one element from every bounded component of $\mathbb C \setminus K$. Then, we can find rational functions $r_n$ such that :

For all $n$, the poles of $r_n$ are contained in $D $.

$r_n \to f$ uniformly on $K$.

We also have the stronger Mergelyan's theorem, which removes the holomorphicity assumption on the boundary of $K$.

If $K$ is a subset of $\mathbb C$ such that $\mathbb C \setminus K$ is connected, then every continuous function $f:K \to \mathbb C$ with $f$ holomorphic on $K^{\circ}$ is approximable by polynomials uniformly on $K$.


问:
Solving a nonlinear least-square problem with a nonlinear equality constraint
Given three points $P_k \in \mathbb{R}^2$, $k \in \{0,1,2\}$, and two edges $P_0 P_1$ and $P_1 P_2$ connecting them, I'm trying to get my head around the following (representing a very much simplified version of a more involved problem):

$P_1$ is moved to a new position $P_1'$

$P_0$ and $P_2$ should be updated to $P_0'$ and $P_2'$ such that we have $\|P_1' - P_0'\| = \|P_1 - P_0\|$ and $\|P_2' - P_1'\| = \|P_2 - P_1\|$. In other words, the lengths of the edges connecting the three points should not change. This means that $P_0'$ and $P_2'$ can move on circles with respective radii $\|P_1 - P_0\|$ and $\|P_2 - P_1\|$ centred at $P_1'$

At the same time, $P_0$ and $P_2$ should move as little as possible, i.e. $\|P_0 - P_0'\|$ and $\|P_2 - P_2'\|$ should be minimal

A visualisation of the above might look as follows:

Now, this problem can be easily solved by hand: $P_0'$ and $P_2'$ must lie on the intersections of the edges $P_0 P_1'$ and $P_1' P_2$ with the circles mentioned above.

However, as the actual setting is more involved (e.g. there might be many more points and connected edges, or even a 2D network of them), I'd like to consider the problem in a (nonlinear) optimisation setting. As such, the first condition can be formulated as a (single) nonlinear equality constraint, i.e.
$$g(P_0', P_1', P_2') = (\|P_1 - P_0\| - \|P_1' - P_0'\|)^2 + (\|P_2 - P_1\| - \|P_2' - P_1'\|)^2 = 0.$$

In other words, the nonlinear equality constraint gives us a subset of $\mathbb{R}^4$ as feasible domain (though we only have two scalar degrees of freedom, the angles of the two edges). Furthermore, the second condition can be formulated as a nonlinear least-squares problem, i.e.
$$\min_{P_0', P_2'} f(P_0', P_2') = \frac12 \Big( \|P_0 - P_0'\|^2 + \|P_2 - P_2'\|^2 \Big).$$

For a general nonlinear least-squares problem without (nonlinear) equality constraint, I'd use a Gauss-Newton method to solve it numerically. With the constraint present, I'm not certain how to proceed.

[Edit]
One option might be to introduce a Lagrangian multiplier $\lambda$ and solve for $\nabla f - \lambda \nabla g = 0$, e.g. using a Newton-Lagrange or constrained Gauss-Newton approach as described in these lecture notes (Chapter 12). However, I have some doubts about the nonlinear equality constraint $g$ — we actually have an equality constraint for each individual edge (so two of them following the example), which results in multiple (two) Lagrangian multipliers. At the same time, summing these to form a single equality constraint as I did above looks valid (the individual terms are squared, so the only way for the constraint to vanish is for each term to vanish).

答:
Really, the applied approach can be modified.

Let

$r_0 = \|P_0P_1\|,\; r_2=\|P_1P_2\|\tag1$

$P_k\rightarrow b_k\in\mathbb C^3,\;P'_k\rightarrow z_k \in\mathbb C^3,\tag2$
then

$z_0(\varphi_0) = z_1 + r_0e^{i\varphi_0},\quad z_2(\varphi_2) = z_1 + r_2e^{i\varphi_2},\tag3$

$f(\varphi_0,\varphi_2) = \dfrac12(|z_0(\varphi_0)-b_0|^2+|z_2(\varphi_2)-b_2|^2).\tag4$
$ $

Proposed $2$-parametric model allows to avoid Lagrange multipliers method and looks the easiest for the given problem.


问:
If $R$ is noetherian then every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}.$
Here is the question I want to answer:

If $R$ is noetherian then every family $\mathcal{C}$ of ideals in $R$ has a maximal element in $\mathcal{C}$ relative to the partial order of set inclusion.

Here is a trial for the solution:

Is this trial correct or I need to use the axiom of choice somewhere

答:
You are using the axiom of choice exactly here:

If every chain is finite, then there is a maximal element.

This, as explained in the comments, is a weak version of Zorn's lemma, also known as Dependent Choice. The reason being that we have a partial order where every chain is finite, and therefore every chain has an upper bound: its largest element (because it's finite, and finite linear orders have a maximum).

And indeed, if Dependent Choice fails, then there is a partial order, a tree if you want to think about it, without maximal elements, where every chain is finite. That means that a recursive process cannot "go on forever", but nevertheless, it can always go "one more step".

This is counterintuitive, since we are used to Dependent Choice so much, that we don't even notice it. The key point is that recursion is defined with a function, so it's not "there is some ideal $A_2$ extending $A_1$", but rather a function that specifies which ideal that is. If you're only using the fact that "there is some extension", then you're subtly sneaking the axiom of choice into your proof in order to turn this into an actual recursive process.


问:
Does $A^{p^2}=I$ lead to $(A-I)^{p^2}=0$?
Suppose that $A$ is a $n\times n$ matrix with entries in a finite field $F_p$ where $p$ is prime, $1\leq n \leq p$. An exercise answer told me that $$A^{p^2}=I\Rightarrow (A-I)^{p^2}=0.$$

I don't know why this holds. I tried to expand the equation as
$$
0=A^{p^2}-I^{p^2}=\left( A-I \right) \left( A^{p^2-1}+A^{p^2-2}+\cdots +A+I \right),
$$
but still have no idea what info is given. Any help would be appreciated! Thank you in advance！

答:
The binomial coefficients $\binom{p}{k}$ are all multiples of $p$ for $0 <k < p$. Therefore
$$
(A-I)^{p^2}=((A-I)^p)^p=(A^p-I^p)^p=A^{p^2}-I=0
$$


问:
Help me to show that exist only one interpolation polynomial with such properties
Let $f:[a,b]\to \Bbb R$, $f \in C^2[a,b]$  and $x_0,x_1,x_2 \in [a,b]$ where $ x_0 \neq x_2$. Show that exists only one polynomial $p \in P_3$ for which  $$p(x_0)=f(x_0),\  p\prime(x_1)=f\prime(x_1),\ p\prime\prime(x_1)=f\prime\prime(x_1),\ p(x_2)=f(x_2)$$ Find form of polynomial and error.

答:
There are many ways to skin a cat.

I followed the first comment, which is equivalent as "show the following matrix is invertible"

$$
\begin{bmatrix}
x_0^3 & x_0^2 & x_0 & 1 \\
x_2^3 & x_2^2 & x_2 & 1 \\
3x_1^2 & 2x_1 & 1 & 0 \\
6x_1 & 2 & 0 & 0
\end{bmatrix}
$$

which leads to two obvious solutions. You can get the determinant, possibly using a computational tool, and immediately find that the determinant can factorize into $2(x_1-x_0)^3-2(x_1-x_2)^3$ which in turn factors into $2(x_2-x_0)(x_0^2+x_0x_2+x_2^2)$ which equals $0$ iff $x_2=x_0$.

Or you can try to convert the matrix using elementary operations into an upper triangular matrix. It's more obvious than how it writes out to be. First move the existing $0$s into lower left. Next see that if $x_1$ is zero, you get a trivial case. Next, you will rely on $x_2-x_0\neq0$ and $x_1(x_2-x_0)\neq0$ to get the upper triangular matrix with all non-zero diagonal terms, which demonstrates invertibility.

EDIT: I missed the question about the the form. Since the form is required, you should definitely follow the hint of the second comment.


答:
Question 1: Any integer is divisible by a prime (you actually don't even need the full strength of the unique factorization theorem; this is provable by induction).

Question 2: It's not divisible by any of the primes of the form $4n+3$ (namely, $p_1,\ldots,p_k$, which we're assuming is all of them), and it's not divisible by $2$ because it's odd. Thus, any prime dividing it must be of the form $4n+1$.

Question 3: Multiplying an even number of things of the form $4n+3$ together gives something of the form $4n+1$.

Title: Infinitely many primes of the form $4n+3$

Body:
I've found at least 3 other posts$^*$ regarding this theorem, but the posts don't address the issues that I have.

Below is a proof that for infinitely many primes of the form $4n+3$, there's a few questions I have in the proof which I'll mark accordingly.

Proof: Suppose there were only finitely many primes $p_1,\dots, p_k$, which are of the form $4n+3$. Let $N = 4p_1\cdots p_k - 1$. This number is of the form $4n+3$ and is also not prime as it is larger than all the possible primes of the same form. Therefore, it is divisible by a prime $ \color{green}{ \text{(How did they get to this conclusion?)}}$. However, none of the $p_1,\dots, p_k$ divide $N$. So every prime which divides $N$ must be of the form $4n+1$ $ \color{green}{ \text{(Why must it be of this form?)}}$. But notice any two numbers of the form $4n+1$ form a product of the same form, which contradicts the definition of $N$. Contradiction. $\square$

Then as a follow-up question, the text asks "Why does a proof of this flavor fail for primes of the form $4n+1$? $ \color{green}{ \text{(This is my last question.)}}$

$^*$One involves congruences, which I haven't learned yet. The other is a solution-verification type question. The last one makes use of a lemma that is actually one of my questions, but wasn't a question in that post.


问:
A problem related to physics, vector, integration
This problem is from my Homework. I have the idea of the solution. But I don't know if it is right or not.
I am quoting it from the question.
"Find the work done in moving a particle once around 3/4 th a circle C in the xy plane, If
the circle has center at the origin and radius 3 and the force field is given by,
$\vec{F} = (2x-y+z) \mathbf{i} + (x+y-z^2) \mathbf{j} + (3x-4y+4z) \mathbf{k}$

My idea is, if i calculate it for the whole circle (let it be x) and then do this thing

$a=x-\frac{3}{4}x$
then the value of a which I am getting is it the correct answer? Or, I am wrong?

答:
I think it is best to just solve this directly. Observe that the circle may be parametrized by

$$\mathbf{r}(\theta)=\cos(\theta) \mathbf{i} + \sin(\theta) \mathbf{j}, \quad \theta \in [0, \tfrac{3\pi}{2}]$$

and that the work is given by

$$\int_{C} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{3\pi/2} \mathbf{F}(\mathbf{r}(\theta)) \cdot \mathbf{r}'(\theta) \, d\theta.$$

When you do the dot product, you will get a trigonometric integral. I trust you to take it from there.


问:
Understanding the minimization of this "cost function"
I'm looking at this post, over on Stack Overflow. The relevant portion is as follows:

...first, you need a cost function... which would be something like

$err(x,y,z) = \sum_{i=1}^n\sqrt{[(x-xi)^2 + (y-yi)^2 + (z-zi)^2]} -
 di$

...where $x, y, z$ are coordinates of the current point in the
numerical solution and  $xi, yi, zi$ and $di$ are the coordinates and
distance towards the $i$th reference point. In order to solve this -
my advice is NOT to use Newton/Gauss or Newton methods. You need first
and second derivative of the aforementioned function - and those have
a finite discontinuation in some points in space - hence that is not a
smooth function and these methods won't work. What will work is direct
search family of algorithms for optimization of functions (finding
minimums and maximums. in our case - you need minimum of the
error/cost function).

Where I'm lost

I've seen cost functions only in the context of machine learning and, admittedly, to limited extent. Why is it that a 'cost function' works here, "intuitively"?

Moreover, it isn't obvious to me precisely what sort of algorithm OP is referring to ("direct search family of algorithms"), nor how that would be applied to a summation such as this in practice.

Could someone please elaborate, and offer a more in-depth explanation?

Many thanks.

答:
As @littleO commented, this is a very strange setting of the problem.

The distance between the source and the current point is
$$\tilde d_i=\sqrt{(x-x_i)^2 + (y-y_i)^2 + (z-z_i)^2}$$ while the measured value is $d_i$. So, if you have three points and no noise you can solve for $(x,y,z)$ the three equations $\tilde d_i=d_i$.

Otherwise, you could do it using as a "cost function"

$$\text{err}(x,y,z) = \sum_{i=1}^n\color{red}{\Bigg|}\sqrt{(x-x_i)^2 + (y-y_i)^2 + (z-z_i)^2}-d_i\color{red}{\Bigg|} $$ which is a totally different story (and, I would say, a very difficult problem from a numerical point of view.

If you are interested by a solution which would give you good estimates of $(x,y,z)$ to start with, let me know and I shall edit.


问:
Show that the ring of formal power series is a Euclidean domain.
Here is the question I want to solve part $(c)$ of it:

A commutative ring $R$ is local if it has a unique maximal ideal $\mathfrak{m}.$ In this case, we say $(R, \mathfrak{m})$ is a local ring. For example, if $R$ is a field, then $(R,(0))$ is a local ring, since the only proper ideal of a field is $(0).$

$(a)$ Let $(R, \mathfrak{m})$ be a local ring. Show that $R^* = R\setminus \mathfrak{m}.$

$(b)$ Show that, for a field $K,$ $R = K[[x]]$ is a local ring.

$(c)$ Show that $R = K[[x]]$ is a Euclidean domain.

I got hint to use that $(R, \mathfrak{m})$ be a local ring but I still do not know how to do that, could anyone help me in this proof please?

Euclidean Domain definition:

A euclidean domain $R$ is an integral domain together with a function $\sigma : R \rightarrow \mathbb N$  where $\sigma (r) = 0 \iff r= 0 $ (well ordered monoid) and such that, for each $a,b \in R,$ there exist $q,r \in R$ where $a = bq + r$ and $\sigma(r) < \sigma(b).$

答:
Hint:

Use as a Euclidean function $\sigma$ the order of a formal power series:
$$\sigma(a_0+a_1X+a_2X^2+\dots)=\min\{i\in\mathbf N\mid a_i\ne 0\}$$
and prove the existence of $q$ and $r$ by comparing the orders of the dividend $f$ and divisor $g$.

Some details:

Let $f,g\in K[[X]]$ and $\sigma(g)=r$.

If $\sigma(f)<\sigma(g)$, the solution is obvious: $q=0$, $\:r=f$.

If  $\sigma(f)\ge\sigma(g)$, I'll give  an example: suppose $\:f=X^3(X^2+2X^3+3X^4+4X^5+\dots)\:$ and  $g=X^3+X^4+X^5=X^3(1+X+X^2)$ (a polynomial is also a formal power series). Note that $u=1+X+X^2$ is a unit in $K[[X]]$, which is a discrete valuation ring. We deduce that
\begin{align}
f&=(X^2+2X^3+3X^4+4x^5+\dots)X^3\\
&=\underbrace{(X^2+2X^3+3X^4+4x^5+\dots)u^{-1}\!}_{\textstyle q}\;g,
\end{align}


问:
Remaining travel time calculation
I am a bit confused with a time calculation problem, given average speed <km/h> and remaining mileage  for travel.

I saw in a book, where they give average speed <km/h> of 120 and remaining mileage  of 20, they say that the remaining travel time is: 1 hour(s) and 40 minutes.

But when calculating this I get 16 minutes i.e. dividing 20 miles (about 32 km) to 120 km/h.

Why is it that in the book they have a remaining travel time of 1 hour and 40 minutes?

答:
There are different measurements of miles.

The US mile is 1 mile = 1.60934 km

The Swedish mile is 1 mile = 10 km

In the problem stated, Swedish mile is used. So, the procedure to calculate is correct, if it is for the US mile.

Title: Remaining travel time calculation


问:
An equivalence to quantifier elimination
So I am studying the theory of quantifer elimination and have come across the following equivalence. I am defining quantifier elimination as: a theory $T$ admits quantifier-elimination (QE) if every formula $\phi$ is equivalent to a quantifier-free one.

I then want to prove that $T$ admits QE if and only if for any quantifier-free formula $\phi(x,y_1,y_2,...,y_n)$ of $L$, there exists a quantifier-free formula $\psi(y_1,...,y_n)$ such that $$T \models  (\forall y_1,...,y_n)(\psi \iff(\exists x)\phi)$$.

I have proved the forward direction I think:

Suppose $T$ admits QE, and suppose $\phi(x,y_1,y_2,...,y_n)$ is a quantifier-free formula. Consider $\exists x\phi(x,y_1,y_2,...,y_n)$. Then as $T$ admits QE, $T \models  (\forall y_1,...,y_n)(\psi \iff(\exists x)\phi)$, where $\psi$ is quantifier free formula. I think this is simply by definitions. However, I do not know how to prove the converse, so any help would be hugely appreciated. Thank you.

答:
First note that for any formula $\phi(x)$ we have that $\forall x \phi(x)$ and $\neg \exists x \neg \phi(x)$ are equivalent.

Let $\zeta(y_1, \ldots, y_n)$ be any formula. We prove by induction on the structure of $\zeta$ that it is equivalent to a quantifier-free formula (modulo $T$). By the above we may assume that the only quantifiers that appear in $\zeta$ are existential quantifiers. The base case, where $\zeta$ is atomic, is trivial. The induction step for the logical connectives ($\wedge$, $\vee$, $\neg$, $\to$, $\bot$) is also trivial. So we are left with the case where $\zeta$ is of the form $\exists x \phi(x, y_1, \ldots, y_n)$. By the induction hypothesis we may assume that $\phi(x, y_1, \ldots, y_n)$ is equivalent (modulo $T$) to a quantifier-free $\phi'(x, y_1, \ldots, y_n)$. Then by assumption there is quantifier-free $\psi(y_1, \ldots, y_n)$ that is equivalent (modulo $T$) to $\exists x \phi'(x, y_1, \ldots, y_n)$. Putting things together we get
$$
T \models \forall y_1 \ldots y_n(\psi(y_1, \ldots, y_n) \leftrightarrow \zeta(y_1, \ldots, y_n)),
$$
and so we conclude that $\zeta$ is indeed equivalent to a quantifier-free formula (modulo $T$). This completes the induction and thus the proof.


问:
Changing the quality of the transient process in a nonlinear system (in Mathematica)
I urgently need advice and help.

I have a system of differential equations like this:

$\begin{cases} \frac{dx}{dt} == y[t] \cdot \alpha \cdot sin(\omega t) + \frac{d}{dt}(\alpha \cdot sin(\omega t))
   \\ \frac{dy}{dt} + h \cdot y(t) == \frac{d}{dt}(e^{-(x[t] - 2)^2}) \end{cases}$

Parameters: $\alpha = 0.3, h = 1, \omega = 2 \pi 0.5, x(0)=1/4, y(0)=0$

It corresponds to the following structural scheme:

The code that simulates such a system is shown below:
ClearAll["Global`*"]

```<code>pars = {\[Alpha]1 = 0.3, h1 = 1, \[Omega]1 = 2 Pi 0.5}

extr = Exp[-(x[t] - 2)^2]

sys =
 NDSolve[{x'[t] ==
    hpf1[t] \[Alpha]1 Sin[\[Omega]1 t] +
     D[\[Alpha]1 Sin[\[Omega]1 t], t],
   y'[t] + h1 y[t] == D[extr, t], x[0] == 1/4, y[0] == 0},
  x, {t, 0, 500}]
</code>```

The numerical solution is presented below:

```<code>Plot[{Evaluate[x[t] /. sys]}, {t, 0, 150}, PlotRange -> Full,
 PlotPoints -> 50]
</code>```

It can be seen that the transition process is a transition from the initial point to the final one with a certain character.

I need to change this character i.e. make the transition from one point to another exponentially.
Like this:

What are the ways to solve this problem? What to do, add a regulator or manipulate the system of differential equations?

Please help me!

答: The rate at which you converge is proportional to the gradient of your process, in your case $e^{-(x(t)-2)^2}$. However, for values of $x(t)$ far away from $2$ the gradient very quickly becomes tiny. For example if you use $x(0)=0$ the behaviour that you are trying to remove gets even more pronounced. In this case you could take the logarithm of the output of process before feeding it into the washout filter. However, this only works for processes for which you know that their outputs are always positive.


答:
Every number $n>1$ is divisible by some prime $p$ (which includes the case $n=p$). Assume otherwise and let $n$ be the smallest such number. As this $n$ is not prime, it has a nontrivivial divisor $d$ with $1<d<n$. By minimality of $n$, $d$ is divisible by some prime $p$. But then $p$ also divides $n$.

All numbers are of the form $4n$, $4n+1$, $4n+2$, or $4n+3$.
This is also true for primes $p$, but $p=4n$ is not possible and $p=2n$ only for $p=2$. Here, we have excluded $p=2$ as well as $p=4n+3$ by construction, which leaves only primes $p=4n+1$.

This proof fails for $p=4n+1$ because a number of the form $4n+1$ may well be the product of two numbers of the form $4n-1$. For example $3\cdot 7=21$. Therefore the step that at least one divisor must be of form $4n+1$ fails.


问:
Pullback of a Volume Form
Suppose we have a a diffeomorphism between two manifolds, $f: M \rightarrow N$ and a volume form $\Omega$ on $N$. Then is it true that $f^{*}(\Omega) = \Upsilon$ will always be a volume form on $M$?

My thinking is that for any arbitrary vector fields $(X_1, X_2,...,X_N) \in T_pM$ then

$\Upsilon(X_1, X_2,..., X_N) = f^{*}\Omega(X_1, X_2,..., X_N) = \Omega(df_x(X_1),df_x(X_2),...df_x(X_N))$

is always non vanishing given that $\Omega$ is a volume form on $N$? This would be because $\Omega$ is acting on the tangent vectors $df_x(X_i) \in T_{f(p)}N$, which is non-vanishing given that $\Omega$ is a volume form on N.

I'm new to differential geometry, so any mistakes with notation, or comments which aren't accurate, please do let me know!

答: Your agrument is basically coorect, the only observation you have not put in explicitly that that for a basis $\{X_i\}$ of $T_pM$ the vectors $df_x(X_i)$ form a basis of $T_{f(p)}N$, which is needed to conclude that the value of $\Omega$ on these tangent vectors is non-zero.

Title: Pullback of a Volume Form


问:
for every $v$ vector If $Av = Bv$ then $A=B$
I just can't grasp the idea, I mean I can give you billion vectors and matrices for which this
sentence is not correct.
take vector which build with multiple zeros and whatever different $A,B$ matrices you want
Can someone shed some light

答:
So After I racked my brain for a good day I will answer my own question for future readers:
The reason why you can not simply use vector of zeros or any other specific vector with any other specific matrices to disprove the sentence is the fact that If you give a specific example for A,B and v where Av=Bv but A!=B It does not necessarily mean Av=Bv for EVERY vector.

```<code>Let v = 0     A= 1 0    B= 1 0
        0        0 0      -1 0
</code>```

You simply can not guarantee Av and Bv are equal for all the other vectors

Body:
So After I racked my brain for a good day I will answer my own question for future readers:
The reason why you can not simply use vector of zeros or any other specific vector with any other specific matrices to disprove the sentence is the fact that If you give a specific example for A,B and v where Av=Bv but A!=B It does not necessarily mean Av=Bv for EVERY vector.

Let v = 0     A= 1 0    B= 1 0
        0        0 0      -1 0

You simply can not guarantee Av and Bv are equal for all the other vectors


问:
Interpreting relations - are they functions?
I have two descriptions of two different relations and I need to determine whether they are functions. However, for some reason I have a really hard time interpreting what's going on here (esp. in (1)), maybe someone can help:

(1) $\{(a,b) \in \mathbb{Z} \times \mathbb{Z} | 7 | (b-a), -a -3 \leq b \leq -a +3\}$

(2) $\{(a,b) \in \mathbb{Z} \times \mathbb{Z} | b = |a|\}$

I'd be grateful for any tips and pointers!

Cheers!

答:
Yes, both of them are functions. That's easy to see in the second case: that's simply the function $a\mapsto|a|$.

The first one is the function $f\colon\Bbb Z\longrightarrow\Bbb Z$ defined by$$f(a)=\text{only element of }\{a-3,a-2,\ldots,a+3\}\text{ which is a multiple of }7.$$Note that this makes sense, since $\{a-3,a-2,\ldots,a+3\}$ consists of $7$ consecutive integers, and therefore it has one and only one multiple of $7$. And, with this definition, $7\mid\bigl(f(a)-a\bigr)$ and $a-3\leqslant f(a)\leqslant a+3$.


问:
Approximate fixed point sequences
Let $(\ell_2,\|\cdot\|)$ be the usual Banach space of the real sequences with summable squared, and $B$ its closed unit ball. Define $T:B\longrightarrow B$ as

$$
T(x):=(\sqrt{1-\|x\|^{2}},x_{1},x_{2},\ldots),
$$
for all $x:=(x_{n})_{n\geq 1}\in B$. Then, it is easy to check that $T$ has not fixed points. However, I have read in some papers that the following holds:
$$
\inf\{\|x-T(x)\|:x\in B\}=0
$$

How can we prove this equality?

Many thanks in advance for your comments.

答:
I will denote the $m$th point in a sequence $x \in \ell^2$ by $x(m)$.

For each $m, n \in \Bbb{N}$, let
$$x_n(m) = \begin{cases} \dfrac{1}{\sqrt{n}} & \text{if } 1 \le m \le n \\ 0 & \text{if } m > n. \end{cases}$$
Then $\|x_n\| = 1$, hence $x_n \in B$. Further,
$$(T(x_n) - x_n)(m) = \begin{cases} -\dfrac{1}{\sqrt{n}} & \text{if } m = 1 \\ \dfrac{1}{\sqrt{n}} & \text{if } m = n + 1 \\ 0 & \text{otherwise.} \end{cases}$$
Hence,
$$\|T(x_n) - x_n\| = \frac{\sqrt{2}}{\sqrt{n}} \to 0.$$

Title: Approximate fixed point sequences


问:
Justifying a certain application of Lebesgue Dominated Convergence
In a proof I'm reading, I came across the following application of Lebesgue Dominated Convergence. We let $(X,d)$ be a compact metric space, $f\in C(X)$, $T:X\to X$ continuous, and $h\in L^1(X,\mathcal B,\mu)$, where $\mathcal B$ is the Borel $\sigma$-algebra corresponding to $d$, and $\mu$ is  some probability measure. Then it was established that $\mu$-a.e. $$\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}f(T^ix)h(x)=h(x)\int_Xf(y)d\mu(y),$$ and it is stated that by Dominated convergence we then have $$\lim_{n\to\infty}\frac1n\sum_{i=0}^{n-1}\int_Xf(T^ix)h(x)d\mu(x)=\int_Xh(x)d\mu(x)\int_Xf(y)d\mu(y).$$

To apply Dominated convergence, we need a.e. pointwise convergence, which is clear by the first display, and we need an integrable dominating function. Since $h\in L^1(X,\mathcal B,\mu)$, it seems that the first display also gives a dominating function, if we can show that the convergence $\frac1n\sum_{i=0}^{n-1}f(T^ix)\to\int fd\mu$ is uniform. Else, I don't see what the dominating function should look like. The problem is that we need to somehow bound $\frac1n\sum_{i=0}^{n-1}f(T^ix)$ a.e. by a constant, uniformly for all $n$.

Any help is much appreciated!

答:
Since $f$ is continuous and $X$ is compact, there exists $M$ such that $|f(x)| \leq M$ for every $x \in X$. In particular, for every $x \in X$, for all $n \in \mathbb{N}^*$,
$$\left|\frac1n\sum_{i=0}^{n-1}f(T^ix)\right| \leq \frac1n\sum_{i=0}^{n-1} M = M$$


问:
Difference in Robin's inequality assuming R.H.
If we assumed R.H. what is the best upper/lower bounds for $ E +\ln \theta(x) - \sum \limits_{p \leq x} \frac{\ln p}{p}$

And if one can also give the best upper/lower bounds for $ E +\ln x + (\frac{\theta(x)}{x}-1)- \sum \limits_{p \leq x} \frac{\ln p}{p}$

Where $E = -\gamma - \sum \limits_{P} \frac{\ln p}{p(p-1)} \approx -1.33258$

Any ref that gives partial answer (only lower bounds, or for one of the inequalities) will be helpful.

Thanks.

答: Note that $$\sum_{n\leq N}\frac{\Lambda\left(n\right)}{n}-\sum_{p\leq N}\frac{\log\left(p\right)}{p}\leq\sum_{p\geq2}\frac{\log\left(p\right)}{p\left(p-1\right)}$$ and, using the explicit formula $$\sum_{n\leq N}{}^{^{\prime}}\frac{\Lambda\left(n\right)}{n}=\log\left(N\right)-\gamma+\sum_{\rho}\frac{N^{-\rho}}{\rho}-\frac{1}{N}+\frac{1}{2}\log\left(\frac{N+1}{N-1}\right)$$ where the symbol $\prime$ indicates that if $N$ is a prime power then the term $n=N$ must be multiplied by $1/2$, we have $$E-\sum_{p\leq N}\frac{\log\left(p\right)}{p}-\log\left(N\right)\leq-\sum_{\rho}\frac{N^{-\rho}}{\rho}+\frac{1}{N}-\frac{1}{2}\log\left(\frac{N+1}{N-1}\right)$$ and so, if we assume RH, we obtain $$E-\sum_{p\leq N}\frac{\log\left(p\right)}{p}-\log\left(N\right)=O\left(N^{-1/2}\log^{2}\left(N\right)\right).$$ Now remains to observe that, since under RH we have $$\theta\left(N\right)=N+O\left(N^{1/2}\log^{2}\left(N\right)\right),$$ then $$\log\left(\theta\left(N\right)\right)-\log\left(N\right)=\log\left(1+O\left(N^{-1/2}\log^{2}\left(N\right)\right)\right)$$ and so $$E-\sum_{p\leq N}\frac{\log\left(p\right)}{p}+\log\left(\theta\left(N\right)\right)=O\left(N^{-1/2}\log^{2}\left(N\right)\right),$$ as indicated by Greg Martin.


问:
How do you graph $x + y + z = 1$ without using graphing devices?
How can I graph $x + y + z = 1$ without using graphing devices?

I equal $z = 0$ to find the graph on the xy plane. So I got a line, $y = 1-x$
But when I equal 0 for either the $x$ or the $y,$ I get $z = 1-y$ or $z = 1-x$ , and those are two different lines from different angles. Different graphing websites were telling me different answers...

Please don't show some crazy and complicated methods to graph this. I just want simple steps just as plugging $x,y,z$ as zeros and etc.

答:
Sketching 3D graphs can be super difficult to do by hand. When it comes to sketching simple planes, what I like to do is sketch the part of the plane that is only in a single octant (for example, in this case, we can consider only the first octant where $x,y,z \ge 0$). To do this, we find where the plane intersects each of the 3 axes.

For the $x$-intercept, set $y=z=0$ to obtain $x+0+0=1$, which yields the point $(1,0,0)$.

For the $y$-intercept, set $x=z=0$ to obtain $0+y+0=1$, which yields the point $(0,1,0)$.

For the $z$-intercept, set $x=y=0$ to obtain $0+0+z=1$, which yields the point $(0,0,1)$.

Finally, draw the given octant that contains these 3 intercepts and label each of these 3 points. Connect these points to form a triangle that represents the portion of the plane inside this octant.


问:
Minimise $|f''(x)|$ on an interval when you know the values of the function and the values of the derivative at the endpoints of the interval only.
Given $a, b,c \in \mathbb{R}$, let $A_{abc}$ be the set of twice-differentiable real functions $f:[0,1] \to \mathbb{R}$ such that $f(0) = 0$, $f(1) = a, f'(0) = b$ and $f'(1) = c.\ $ Find the infimum of {$\max_{x \in [0,1]}$$|f''(x)|: f \in A_{abc}$ }$\subset \mathbb{R}$ in terms of $a, b$ and $c$.

I'm not sure I've written that correctly: I'm sure I'll have to re-write that more clearly. Anyway, the title should be somewhat clear as to what I'm after.

Example 1:

$$$$
Example 2:

I think we should also add the constraint that $ f'(0) \geq 0$, because if we then reflect everything the functions in the x-axis, we get all the cases where $f'(0) < 0$, so we don't lose any generality by assuming $ f'(0) \geq 0$.

Anyway, I'd be very surprised if this wasn't some sort of duplicate as it's such a natural optimisation question: I just think I'm not using the right lingo.

答:
This answer is preliminary, but I’ll try to update it.

Given a twice-differentiable function $f$ from $[0,1]$ to $\Bbb R$ put $\|f\|=\sup_{x \in [0,1]} |f''(x)|$. If $f\in A_{abc}$ is any function then a function $g(x)=f(x)-ax$ belongs to $A_{0, b-a, c-a}$ and $\|g\|=\|f\|$, so it suffices to consider the case when $a=0$.

Given  $b$ and $c$, denote the required infimum by $I$.

Let $f\in A_{0bc}$ be any function. By Rolle’s theorem, there exists $y\in (0,1)$ such that $f’(y)=0$. By Lagrange’s theorem, there exist $x_1, x_2\in (0,1)$ such that $f’’(x_1)=\frac{f’(y)-f’(0)}{y-0}=-\frac by$ and  $f’’(x_2)=\frac{f’(1)-f’(y)}{1-y}=\frac c{1-y}$. This easily follows that $$I\ge \max\{|f’’(x_1)|, |f’’(x_2)|\}\ge |b|+|c|.$$

On the other hand, a polynomial $f(x)=(b+c)x^3-(2b+c)x^2+bx$ belongs to $A_{0bc}$ and $$I\le \|f\|= 2\max\{|2b+c|,|b+2c|\}.$$


答: $x+y+z=1$ forms a plane. This plane contains all three lines you've found (and lots more). But what you've done should be enough - if you know a few points on a plane (in this case, for example, you know (1,0,0), (0,1,0) and (0,0,1), and more), you can sketch the whole plane.</p>

Title: How do you graph $x + y + z = 1$ without using graphing devices?

Body:
How can I graph $x + y + z = 1$ without using graphing devices?

I equal $z = 0$ to find the graph on the xy plane. So I got a line, $y = 1-x$
But when I equal 0 for either the $x$ or the $y,$ I get $z = 1-y$ or $z = 1-x$ , and those are two different lines from different angles. Different graphing websites were telling me different answers...

Please don't show some crazy and complicated methods to graph this. I just want simple steps just as plugging $x,y,z$ as zeros and etc.


问:
Number of simple cycles(cycles with no repeating node) of odd length in a complete graph with 17 nodes.
I know that the number of simple cycles of length $k$ from $n$ nodes is $\frac{n!}{(n-k)!} \cdot\frac{1}{2k}$.

So i should add $$\frac{17!}{(17-3)!} \cdot\frac{1}{2\cdot3} + \frac{17!}{(17-5)!} \cdot \frac{1}{2\cdot5} + \cdots +\frac{17!}{(17-17)!} \cdot\frac{1}{2\cdot17}$$right?

But the problem is it is tedious to find this sum. Is there other way to solve this problem or is there a way to compute this sum faster?

答: $\frac{17!}{(17-3)!}$ can be simplified to $17 \cdot 16 \cdot 15$, and similar can be done for each other fraction. Collect common factors and simplify.


答: A matrix is just a linear transformation, if two transformations act exactly the same way on every possible input, they are the same transformation.</p>

Title: for every $v$ vector If $Av = Bv$ then $A=B$

Body:
I just can't grasp the idea, I mean I can give you billion vectors and matrices for which this
sentence is not correct.
take vector which build with multiple zeros and whatever different $A,B$ matrices you want
Can someone shed some light


问:
Find the interior angles of one polygon given a ratio with another
This question was asked in an exam of mathematics for which I am preparing and I was unable to solve it.

The ratio of number of sides of 2 regular polygons is 1:2. If each interior angle of 1st polygon is 120 then each interior angle of 2nd polygon is?

I can only think of that angle would be larger but I don't know any results which could be used( there is no specific syllabus for it in the exam).

This was asked in quiz (general concepts) section and I asked it on Puzzle stackexchange but It was closed there. So, I am asking for help here.

So, can you please help?

答:
Each exterior angle of a regular polygon is $\frac{360º}{n}$.

Imagine a point tracing the sides of the polygon. The exterior angle is the angle turned after each side is traced. Since the point ends up where it started, it completes one full revolution or $360º$. Therefore, the sum of all the exterior angles (of any polygon, not just regular ones) is $360º$, and since all $n$ angles are the same for a regular polygon, each angle is $\frac{360º}{n}$.

Since the exterior angle and the interior angle lie on a straight line, the interior angle and the exterior angle sum to $180º$. Therefore, each interior angle is $180º - \frac{360º}{n}$ (which is equivalent to $\frac{180º(n-2)}{n}$ using algebra).

The number of sides of the 1st polygon is given by $180º - \frac{360º}{n} = 120º \Rightarrow n =6$. Guess and check works fine as well.

Therefore, the 2nd polygon has $12$ sides. Now you just have to plug in $n = 12$ into the interior angle formula.


问:
Definiteness of quadratic form by completing the square
I want to check my thinking here.
We're asked to determine the definiteness of the matrix

$   \begin{pmatrix}
      3 & -1 & 1\\\ -1 & 1 & 2\\ 1 & 2 & 9
    \end{pmatrix}  $

We can write down the quadratic form as:

$3x_1^2  - 2 x_1 x_2 + x_2^2 + 4 x_2 x_3 + 6x_3^2 + 2 x_1 x_3 $

So by completing the square we arrive at

$(x_1 - x_2)^2 + (x_2 + 2x_3)^2 + (x_1 + x_3)^2  + x_1^2 - x_2^2 + x_3^2 $

But now it seems quite difficult to determine the definiteness, how would this usually be done?

答:
There is a quick way to check positive definiteness called Sylvester's criterion which states that a matrix is positive definite if and only if the upper-left hand matrices have positive determinant, i.e.

Applying it to your matrix:

$\begin{vmatrix} 3 \end{vmatrix} >0; \begin{vmatrix} 3 & -1 \\ -1 & 1\end{vmatrix} = 3 -1 = 2>0; \begin{vmatrix} 3 & -1 & 1 \\ -1 & 1 & 2 \\ 1 & 2 & 9 \end{vmatrix} = 15-11-3 = 1 > 0$

Therefore your matrix is positive definite.


问:
How to identify a specific type of lemniscate
I have the following equation:
\begin{equation}
f(x,y)\equiv -1296 x^4 \left(4 y^2+9\right)+9 x^2 \left(-128 y^4+864
   y^2+2187\right)-y^2 \left(8 y^2+81\right)^2=0
\end{equation}

Which represents a lemniscate (see picture).

I would like to know if this a well-known lemniscate (for example
Lemniscate of Gerono, Lemniscate of Bernoulli etc). I have tried to compare with the ones listed on wikipedia, but it seems that none fit this specific equation.

Is $f(x,y)$ a well-known lemniscate? or is there at least a a 'smarter' representation of $f(x,y)$?

edit: in polar coordinate the equation becomes:
\begin{equation}
-\frac{1}{4} r^6 (3 \sin (\phi )+2 \sin (3 \phi ))^2-\frac{81}{16} r^4 (2 \cos (2 \phi
   )+1)^2+\frac{6561}{256} r^2 (2 \cos (2 \phi )+1)
\end{equation}

答:
Due to the fact that variables $x$ and $y$ appear as squared in the equation of $f$, one can consider that this curve is issued from this one

$$\begin{equation}
g(x,y)\equiv -1296 x^2 \left(4 y+9\right)+9 x \left(-128 y^2+864
   y+2187\right)-y \left(8 y+81\right)^2=0
\end{equation}$$

Curves with equations $f(x,y)=0$ (blue) and its mother curve $g(x,y)=0$ (red).

Let us call $g$ the "mother function" of $f$.

Important: we will consider the mother function/curve only in the first quadrant $x,y \ge 0$.

In fact, we are faced to a general generation scheme.

When we apply to the curve of $g$ the (inverse) transformation

$$(x,y) \to (X=\sqrt{x},Y=\sqrt{y}) \tag{1}$$

we obtain the upper right quarter of the lemniscate situated in the first quadrant. The remaining parts of the curve are obtained by symmetries wrt to the axes.

The important point is to take a mother curve $g$ in the first quadrant:

crossing the $x$ axis in two points: $x=0$ and $x=x_1$.

having a derivable cartesian equation $y=\gamma(x)$ with an increasing then decreasing part, more precisely such that there exists $x_0$ with $\gamma'(x)>0$ if $x \in [0,x_0)$ and $\gamma'(x)<0$ if $x \in (x_0,x_1]$.

In fact this is enough for having an image curve by (1) with a vertical axis and a maximum due to the fact that the jacobian matrix of transformation (1):

$$J=\begin{pmatrix}\dfrac{1}{2\sqrt{x}}&0\\0&\dfrac{1}{2\sqrt{y}}\end{pmatrix}$$

is such that $J_{22}$ tends to $\infty$ if $y \to 0_+$ and $J_{11}$ tends to  $\infty$ if $x \to x_0$ (abscissa of the maximum of the mother curve).

Therefore, with these conditions, one can build many curves of type $g(x,y)=0$ yielding as many curves $f(x,y)=0$ having a lemniscate shape like this one.

Conclusion: For the reasons above, the given curve is a lemniscate among many others. I see no reason (in particular with its complicated coefficients) that it is amenable to one of the classical lemniscates (Bernoulli, Gerono, which are issued from second degree "mother" curves as shown on fig. 2 and 3).

The lemniscate of Gerono  ($x^4-x^2+y^2=0$) is issued from a parabola ($x^2 -x +y=0$) by the process described in the text (transformation (1) applied to its first quadrant part, followed by symmetries).

The lemniscate of Bernoulli $(x^2+y^2)^2=x^2-y^2$ is also issued from a (45° slanted) parabola with equation $(x+y)^2=x-y$ by the same process.

Remark: Lemniscates are still studied. See for example this nice 2014 article.


问:
How do I show that $A$ and $A^T$ are similar, given that the minimal polynomial of $A$ splits over $\mathbb{F}$?

Let $n\in\mathbb{N}$ and $A\in M_n(\mathbb{F})$. Assume that the minimal polynomial of the square matrix $A$ is defined as the minimal polynomial of the linear operator $x\mapsto Ax$ for all $x\in\mathbb{F}^n$, on $\mathbb{F}^n$. Now, assume that the minimal polynomial of $A$ splits over $\mathbb{F}$. Show that $A$ and $A^T$ are similar.

Two matrices $A$ and $B$ similar if there exists an invertible matrix $S$ such that $B=SAS^{-1}$. Finding such an $S$ for $B=A^T$ will solve the problem.

Since the minimal polynomial of $A$ splits over $\mathbb{F}$, let us denote it by $$m_A(t) = (t-\lambda_1)^{m_1}...(t-\lambda_k)^{m_k}$$ for some $k\in\mathbb{N}$.
We know that $m_A(A)$ corresponds to the zero linear operator, i.e. $m_A(A)x=0$, $\forall x\in \mathbb{F}^n$. I quickly noticed that if matrices $P_1,P_2,...P_q$ commute, then $(P_1P_2...P_k)^T = P_1P_2...P_k$. It may be useful to replace each $P_i$ with $(A-\lambda_i)^{m_i}$?

I'm not sure how to proceed from here. I'd appreciate any hints!

答:
You are overthinking it. If the minimal polynomial of $A$ splits over $\mathbb F$, $A$ and $A^T$ are similar to their Jordan forms over $\mathbb F$. However, as $\operatorname{rank}(\lambda I-A)^k=\operatorname{rank}(\lambda I-A^T)^k$ for each eigenvalue $\lambda$ of $A$ and each positive integer $k$, the two Jordan forms are the same (up to permutation of Jordan blocks). Hence $A$ and $A^T$ are similar.

Alternatively, that the minimal polynomial of $A$ splits over $\mathbb F$ implies that $A$ is similar to its Jordan form over $\mathbb F$. Since every Jordan block is similar to its own transpose via a reversal matrix (the matrix obtained by flipping the identity matrix left and right), $A$ is similar to $A^T$.