Title: what is $\lim_{k \to 0}\frac{(k+1)A-3kA+2A^3}{(k-A^2)^{\frac{3}{2}}}$ where $A=-\sqrt{2} \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$

Body:
let $k>0$ and define
$$A=-\sqrt{2} \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$$

and

$$\gamma(k)=\frac{(k+1)A-3kA+2A^3}{(k-A^2)^{\frac{3}{2}}}.$$
Where $\Gamma$ is the Gamma function.

What is

$$\lim_{k \to 0^{+}} \gamma(k)?$$

First try

So

$$\gamma(k)=\frac{A-2kA+2A^3}{(k-A^2)^{\frac{3}{2}}}=\frac{A^{3}}{A^{3}}
\frac{1/A^2-2k/A^2+2}{(k/A^2-1)^{\frac{3}{2}}}=\frac{1/A^2-2k/A^2+2}{(k/A^2-1)^{\frac{3}{2}}}$$

But I  stuck in $\lim_{k \to 0}k/A^2$?

Since $\Gamma(\frac{1}{2})=\sqrt{\pi}$ so $A\sim -\frac{\sqrt{2\pi}}{\Gamma(\frac{k}{2})}$ so

$$\gamma(k)=\frac{\frac{\Gamma^2(\frac{k}{2})}{2\pi}-2k\frac{\Gamma^2(\frac{k}{2})}{2\pi}+2}{(k\frac{\Gamma^2(\frac{k}{2})}{2\pi}-1)^{\frac{3}{2}}}=
\frac{
\frac{\Gamma^2(\frac{k}{2})(1-2k)+4\pi}{2\pi}
}{
\frac{
(k\Gamma^2(\frac{k}{2}) -2\pi)^{3/2}
}{(2\pi)^{3/2}}
}
$$

$$=\sqrt(2\pi)\frac{\Gamma^2(\frac{k}{2})(1-2k)+4\pi}{(k\Gamma^2(\frac{k}{2}) -2\pi)^{3/2}}$$

By using $\Gamma(z)=\frac1z - \gamma + o(1)$ (mentioned by @Brian Tung)

$$\Gamma^2(\frac{k}{2})=(\frac{2}{k} - \gamma + o(1))^2
=\frac{4}{k^2}+\gamma^2+o^2(1)-2\gamma \frac{2}{k} +2\frac{2}{k}o(1)-2\gamma o(1) $$

Second try

By knowing that  $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$  [(4.1 The complement formula)][2]

$$A\sim -\frac{\sqrt{2\pi}}{\Gamma(\frac{k}{2})} =-\sqrt{2\pi} \frac{\Gamma(1-\frac{k}{2})\sin(\pi \frac{k}{2})}{\pi} \sim -\sqrt{2\pi} \frac{\sin(\pi \frac{k}{2})}{\pi}=-\sqrt{\frac{2}{\pi}} \sin (\frac{k\pi}{2}) $$

so by Taylor expansion
$$A\sim -\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\} $$ and so

$$\gamma(k)=\frac{(1-2k)A+2A^3}{(k-A^2)^{\frac{3}{2}}}=
\frac{(1-2k)\left(
-\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\}
\right)+2
\left(
-\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\}
\right)^3}{\left(k-
\left(
-\sqrt{\frac{2}{\pi}} \left\{ (\frac{k\pi}{2}) -\frac{(\frac{k\pi}{2})^3}{3!} \right\}
\right)^2\right)^{\frac{3}{2}}}
$$

Body:
Composing Taylor series, we have
$$A=-\sqrt{2}\, \frac{\Gamma(\frac{k+1}{2})}{\Gamma(\frac{k}{2})}$$
$$A=-\sqrt{\frac{\pi }{2}} k+\sqrt{\frac{\pi }{2}} k^2 \log (2)+O\left(k^3\right)$$
$$A^2=\frac{\pi  k^2}{2}-\pi  k^3 \log (2)+O\left(k^4\right)$$
$$A^3=-\frac{\pi ^{3/2} k^3}{2 \sqrt{2}}+\frac{3 \pi ^{3/2} k^4 \log (2)}{2
   \sqrt{2}}+O\left(k^5\right)$$

So, (strangely written for me),
$$(k+1)A-3kA+2A^3=-\sqrt{\frac{\pi }{2}} k+\sqrt{\frac{\pi }{2}} k^2 (2+\log (2))+O\left(k^3\right)$$
$$\left(k-A^2\right)^{3/2}=k^{3/2}-\frac{3}{4} \pi  k^{5/2}+\frac{1}{2} k^{7/2} \left(\frac{3 \pi ^2}{16}+3
   \pi  \log (2)\right)+O\left(k^{9/2}\right)$$ Now, long division
$$\gamma(k)=-\sqrt{\frac{\pi}{2 k}}\left(1+\frac{1}{4} k (-8+3 \pi -\log (16))+O\left(k^2\right)\right)\sim -\sqrt{\frac{\pi}{2 k}}\to -\infty$$


问:
Bounding uniformly continuous functions

Show that if $f \colon \mathbb{R} \to \mathbb{R}$ is uniformly continuous, then there is an $L > 0$ such that $$|f(x)| \leq L \cdot (1 + |x|).$$

I'm quite stuck here. Any hints would be much appreciated.

答:
Suppose $f$ is uniformly continuous, and take $\varepsilon=1$. There exists some $\delta>0$ such that, for all $x,y\in\mathbb{R}$, $$|f(x)-f(y)| \leq 1$$ whenever $|x-y|\leq \delta$.

Fix any $x\geq0$ (the negative case is identical), and let $n = \left\lfloor\frac{x}{\delta}\right\rfloor$, so that $|x-n\delta| \leq \delta$. Then
$$\begin{align*}
|f(x)|&\leq |f(0)|+|f(x)-f(0)| \leq |f(0)|+|f(x)-f(n\delta)|+\sum_{k=0}^{n-1}|f((k+1)\delta)-f(k\delta)|\\
&\leq|f(0)| + 1+n
\end{align*}$$
But we know that $n\leq \frac{x}{\delta}$, so we get
$$\begin{align*}
|f(x)|&\leq|f(0)| + 1+ \frac{x}{\delta} \leq \max(1/\delta, |f(0)| + 1)\cdot (1+x)
\end{align*}$$
and we can just take $L=\max(1/\delta, |f(0)| + 1)$ to conclude.

ContentLicense: CC BY-SA 4.0

Title: Bounding uniformly continuous functions


问:
If $\sum_{k=1}^{\infty}kx^k$ = 30, compute x.
If $\sum_{k=1}^{\infty}kx^k$ = 30, compute x.

I don't think I can find $x$ because using $a/(1-r)$ is only one equation for two variables.

Looking it up on wolfram alpha, it says $x/(x-1)^2$ when $|x| < 1$. Can someone explain to me please?

答: $$\sum_{k=1}^{\infty}kx^k = 30 \implies \frac{x}{(x - 1)^2} = 30\quad\text{ for}\quad |x|<1\quad\implies\quad x=\frac{5}{6}$$


问:
$ \lim_{T_{0} \to +\infty} {\int_{\Gamma_{T_0}}df+\frac{\sin 4z}{z}dz} $
$$ \text{Here we get }f(x,y,z)=\cos y e^{\sin x-x^2-y^2},\text{try to find}\lim_{T_{0} \to +\infty} {\int_{\Gamma_{T_0}}df+\frac{\sin 4z}{z}dz} $$
$$\text{where }\Gamma_{T_0}:x(t)=te^{t^{2018}},y(t)=t^2 \sin (1+t^{2019}),z(t)=\phi (t),\phi\text{ is smooth here },\phi(0)=0,\phi'\geq 1,t\in[0,T_{0}] $$

Body:
$$ df=\cos ye^{\sin x-x^2-y^2}(\cos x-2x)dx-e^{\sin x-x^2-y^2}(\sin y+2y\cos y)dy+0dz $$
The original limit would be $$ \lim_{T_{0} \to +\infty} {\int_{0}^{T_0} (\cos ye^{\sin x-x^2-y^2}(\cos x-2x)\frac{dx}{dt}-e^{\sin x-x^2-y^2}(\sin y+2y\cos y)\frac{dy}{dt}+\frac{\sin 4\phi}{\phi}\frac{d\phi}{dt}})dt $$
$$ \lim_{T_{0} \to +\infty} {\int_{0}^{T_0} \{e^{\sin x-x^2-y^2}[(\cos x\cos y-2x\cos y)x'-(\sin y+2y\cos y)y']+\frac{\sin 4\phi}{\phi}\phi'}\}dt $$
$$ =\lim_{T_{0} \to +\infty} {\int_{0}^{T_0} \{e^{\sin x-x^2-y^2}[\cos y(\sin x-x^2-y^2)'+(\cos y)']+\frac{\sin 4\phi}{\phi}\phi'}\}dt $$
$$ =\lim_{T_{0} \to +\infty} {\int_{0}^{T_0} e^{\sin x-x^2-y^2}[\cos y(\sin x-x^2-y^2)'+(\cos y)']dt+\int_{0}^{+\infty}\frac{\sin 4\phi}{4\phi}}d4\phi $$
$$ =\lim_{T_{0} \to +\infty} {\int_{0}^{T_0} e^{\sin x-x^2-y^2}[\cos y(\sin x-x^2-y^2)'+(\cos y)']dt}+\frac{\pi}{2} $$
$$ =^{\text{integrate by parts}}\lim_{T_0 \to +\infty}\cos ye^{\sin x-x^2-y^2}\bigg|_{0}^{T_0}+\frac{\pi}{2} $$
$$ =\frac{\pi}{2}-1 $$


问:
Diagonalizable matrices in $M_{2\times 2}(\mathbb{F}_2)$
List all diagonalizable $2\times 2$ matrices over the a field $F$ consisting of two elements $0$ and $1$.

I want to try and do this using C++, but perhaps this isn't the place to ask. I have an idea as to how I'd do it.

答:
If $M$ is diagonalizable, we have $M=PDP^{-1}$ with $D$ diagonal. Since $D^2=D$, we get $M^2=M$. Conversely, if $M^2=M$, $M$ is annihilated by $X^2-X=X(X-1)$ which has simple roots and splits over $\mathbb{F}_2$. So $M$ is diagonalizable.

Hence the diagonalizable matrices are exactly the idempotents, i.e. $M$ such that $M^2=M$. This gives three possibilities for the minimal polynomial of $M$: $X$, $X-1$, and $X^2-X$. In the latter case, it must be equal to the characteristic polynomial as well.

We get the null and the identity matrix with the first two cases, and the matrices whose characteristic polynomial is $X^2-X$ in the last case. The latter correspond to the matrices $M$ with $\mbox{trace} \;M=1$ and $\det M=0$:
$$
M=\pmatrix{a&b\\ c&d}\qquad a+d=1\qquad ad-bc=0.
$$
This leaves a few cases to consider and this gives in the end eight diagonalizable matrices:
$$
\pmatrix{0& 0\\0&0}\;\pmatrix{1& 0\\0&1}\;\pmatrix{1& 0\\0&0}\;\pmatrix{1& 1\\0&0}\; \pmatrix{1& 0\\1&0}\;\pmatrix{0& 0\\0&1}\;\pmatrix{0& 0\\1&1}\;\pmatrix{0& 1\\0&1}.
$$
Conclusion: $50\%$ of the matrices are diagonalizable in $M_2(\mathbb{F}_2)$.

ContentLicense: CC BY-SA 3.0


问:
Why do we multiply in tree diagrams?
In probability, we always lay out the events through tree to see what depends on what. Then we were taught to "multiply" through that branch to get the probability of that event.

Why do we "multiply"? I've noticed we have the same sort of rule in calculus too (chain rule)

EDIT: I think my question was misinterpreted. I am asking just because we put a number (usually - if not almost fractional) beside a branch, why must we "multiply" through that number or that "thing" if I am being more abstract. Where "thing" could even mean a derivative operator $$\partial / \partial x$$

答:
This question has been bothering me for a while, most of the solutions state the analogies and examples so I am trying to provide a more generalized and mathematical approach.

Consider a generic tree (not a probability tree) as shown above, where value at root is equal to sum of all the values of its branches.

X is a parent of X1, which means its value of some fraction of X. X1 equals X times the ratio of X1 to sum of all its sibling nodes.

Y1 is child node of X1, so similiary it can be written as above.

Substituting the value of X1 in above equation gives represtation of Y1 in terms of X.

I would like to point out that traversing branches is not multiplication, it is division of root value in some fraction. however in case of probability, sum of all branches equlas to 1 and we do not explicitly show X1 + X2 + .... + Xn = 1 and ignore the denominator.

Consider the following example, it shows when x=1 why it is actually division but interpreted as multiplication in practical probability. If x is not 1,it still holds for a regular tree.

Title: Why do we multiply in tree diagrams?


问:
Double partial derivative proof
Let $z=f(x,y)$  where $x:=r \cos \theta$ and $y := r \sin \theta$.
I had to prove that $\frac {\partial ^{2}z}{\partial \ x^{2}} + \frac {\partial ^{2}z}{\partial \ x^{2}} = \frac {\partial ^{2}z}{\partial \ r^{2}} \ + \dots$ The whole equation was given.

My doubt:

I applied Jacobian and tried to solve this but I ended up in a loop where I was substituting the same things back and forth.

The solution provided said this.

$\frac {\partial ^{2}z}{\partial r^{2}} \ = \frac {\partial}{\partial r}(\frac {\partial z}{\partial r})  = \frac {\partial}{\partial r}(\cos \theta \frac {\partial z}{\partial x} + \sin \theta \frac {\partial z}{\partial y})$

$ = \cos \theta \ [ \cos \theta \frac {\partial}{\partial x} + \sin \theta \frac {\partial }{\partial y} ](\frac {\partial z}{\partial x}) \ + \sin \theta [ \cos \theta \frac {\partial }{\partial x} + \sin \theta \frac {\partial }{\partial y} ] (\frac {\partial z}{\partial y}).$

I do not understand the last step. What exactly is happening? Why do we use the partial derivative of $z$ with respect to $r$ (and skip the $z$ in the numerator) and multiply it with partial derivative of z with respect to $x$ and $y$? Since $x$ and $y$ are functions of $r$ and $\theta$ and since in $\frac {\partial z}{\partial x}$ we have kept $y$ constant are we accounting for that? I am really confused as this relevant theory wasn't taught in class and the professor hasn't been helpful to clear this part. Any insight is highly appreciated.

答:
It sounds like your question is really why
$$ \frac{\partial}{\partial r} = \cos(\theta) \frac{\partial}{\partial x} + \sin(\theta) \frac{\partial}{\partial y} $$
since this is what is being used in the last step.

The short answer: the chain rule

The long answer: $x$ and $y$ can be written as functions of $r$ and $\theta$, as you've written at the very top of your post. Then a function $f(x,y)$ can be thought of as $f\Big(x(r,\theta), \; y(r,\theta)\Big)$. Then the chain rule says that
$$ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial r} $$
You get the result by seeing that $\frac{\partial x}{\partial r} = \cos(\theta)$ and $\frac{\partial y}{\partial r} = \sin(\theta)$.

Title: Double partial derivative proof


问:
What is the sum of the squares of the roots of the equation $x^2 − 7[x] + 5 = 0?$ (Here $[x]$ denotes the greatest integer less than or equal to $x$)
I tried many different approaches to this question,

$α^2 = 7[α] - 5$ and

$β^2 = 7[β] - 5$

So combining both the equations we get,

$α^2 + β^2 = 7([α] + [β]) - 10$,

Which is the answer. However I am unable to simplify the LHS.

答:
$x^2–7[x]+5=0$

$\Rightarrow x^2=7[x]–5 \hspace{1cm}(1)$

$\Rightarrow x^2≤7x–5$ [Since $[x]≤x$]

$\Rightarrow x^2–7x+5≤0$

Therefore, $0.8≤x≤6.2$

i.e., $0≤[x]≤6$

Since for $[x]=0$, equation $(1)$ becomes, $x²=–5$, which is not possible.

Therefore, $1≤[x]≤6$.

In equation $(1)$

For $[x]=1, x^2=2 \Rightarrow x=\pm\sqrt{2}$

For $[x]=2, x²=9 \Rightarrow x=\pm\sqrt{3}$ [not possible because $x=\pm3$ and $[x]=2$ doesn’t make any sense]

For $[x]=3, x²=16 \Rightarrow x=\pm\sqrt{4}$ [not possible because $x=±4$ and $[x]=3$ doesn’t make any sense]

For $[x]=4, x²=23 \Rightarrow x=\pm\sqrt{23}$

For $[x]=5, x²=30 \Rightarrow x=\pm\sqrt{30}$

For $[x]=6, x²=37 \Rightarrow x=\pm\sqrt{37}$

Therefore, sum of all the roots of the given equation $=2+23+30+37=92$.

Hope it helps!!!!


问:
Help with deriving information from a generating function
Sequence of interest

I feel like I've got most of the way, I just need some help at the last step. I am considering an integer sequence which encodes information pertaining to certain weighted ordered rooted trees. I won't bore you with the details suffice to say I found a recurrence relation for the sequence and consequently a generating function. Given the arbitrary variables $(c_0, c_1, c_2, c_3)$, the recurrence relation is:

\begin{align}
a_0 &= c_0 \\
a_n &= c_1 a_{n-1} + c_2 \sum_{i=0}^{n-2}a_ia_{n-2-i} + c_3\sum_{i=0}^{n-3}\sum_{j=0}^{n-3-i}a_ia_ja_{n-3-i-j} \hspace{2em} \text{for $n \geq 1$}
\end{align}

and the generating function $A(x)$ satisfies:

$$A = c_0 + c_1 (Ax) + c_2 (Ax)^2 + c_3 (Ax)^3$$

I'd like to get an asymptotic formula for this sequence, based on the variables $(c_0,c_1,c_2,c_3)$.

Bonus points for an exact formula!

Simpler sequence

Given the case with no weightings where $c_0 = c_1 = c_2 = c_3 = 1$, we are just counting the number of ordered rooted trees with outdegree no greater than 3. I found this sequence on OEIS where it says

$$a_n = \frac{1}{n+1}\sum_{j=0}^{\left\lfloor n/2\right\rfloor} \binom{n+1}{n-2j}\binom{n+1}{j}$$

and $$a_n \sim \frac{cd^n}{n^{3/2}}$$

where

$$d = \frac{(6371+624\sqrt{78})^{2/3}+11(6371+624\sqrt{78})^{1/3}+217}{12(6371+624\sqrt{78})^{1/3}} \approx 3.610718613276\ldots$$

is the root of the equation $4d^3-11d^2-8d-16=0$ and

$$c = \sqrt{\frac{f}{\pi}} \approx 0.9102276936417\ldots$$

where

$$f = \frac{1}{9984}(9295 + (13(45085576939 - 795629568\sqrt{78}))^{1/3} + (13(45085576939 + 795629568\sqrt{78}))^{1/3})$$

is the root of the equation $3328f^3-9295f^2+1696f-128=0$.

I'm assuming these magical formulae were somehow derived from its generating function? But I'm afraid this is beyond my generatingfunctionology skills.

I didn't know if this was more suitable for math stackexchange or mathoverflow. Please feel free to redirect me.

Thanks for any help!

答:
Regarding how the formulas like the one on the OEIS can be found:

A sequence will always grow with exponential order $R^{-n}$, where $R$ is the absolute value of the singularity of $A$ closest to the origin. In the $c_0 = c_1 = c_2 = c_3 = 1$ case, Wolfram Alpha gives this for $A(x)$. Taking the root of the expression under the square root (as there are no other singularities), i.e. solving
$$16x^{12}+8x^{11}+11x^{10}-4x^9=0$$
yields a value which has a reciprocal with the same minimal polynomial as found for $d$ on OEIS. The $3/2$ exponent on $n$ in the asymptotic formula results from the fact the singularity comes from a square root.

A similar process can probably be done for the general case, although the results look fairly horrific very quickly. If the dominant singularity is a pole, the subexponential factor is a polynomial of degree one less than the order, instead of being $n^{-3/2}$.

Specifics are much better explained by this reference from a similar question.


问:
When is $S(m, k)=\sum_{n=1}^{\infty}\frac{1}{n(mn+k)}$ rational?

Let $k$ and $m$ be positive integers. Show that $$S(m, k)=\sum_{n=1}^{\infty}\frac{1}{n(mn+k)}$$ is rational if and only if $m$ divides $k$.

It is easy to see that when $m|k$ then the sum telescopes to a sum of finitely many rationals. But I am not able to prove the other way round. I tried integrals: $$\int_{0}^{1}\frac{1-x^t}{1-x}dx$$ for rational $t$ however all attempts failed. I tried to use feynman trick as $$f(\alpha)=\int_{0}^{1}e^{\alpha (1-x)}\frac{1-x^t}{1-x}dx$$ however the expression for $f’$ is not manageable (if I want to integrate it again). I also thought of another way of doing the trick by using Ramanujan’s master theorem (as I was getting $x^{something}\frac{\ln(x)}{1-x}$ but the integral was from 0 to 1).

Even if you tried to split the terms, convergence issues arise as $H_n$ diverges. How do I do this?

答:
Hint

$$S(m, k)=\sum_{n=1}^{\infty}\frac{1}{n(mn+k)}=\frac{1}{k}H_{\frac{k}{m}}$$ should be sufficient.

Edit

With regard to the integral approach

$$\int\frac{1-x^t}{1-x}dx=-\frac{x^{t+1} }{t+1}\, _2F_1(1,t+1;t+2;x)-\log (1-x)$$ where appears the gaussian hypergeometric function. Assuming $t >0$
$$\int_0^1\frac{1-x^t}{1-x}dx=\psi (t+1)+\gamma=H_t$$


问:
Intuition on non-unique ideal factorization if not integrally closed
I'm aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.

It's clear that $\mathbb{Z}[\sqrt{-3}]$ is not a Dedekind domain, as it is not integrally closed.  I'm having difficulty demonstrating directly that $\mathbb{Z}[\sqrt{-3}]$ does not admit unique factorization of ideals.

Obviously we only need one example of an ideal that doesn't have a unique factorization into prime ideals, but a good answer would provide either a process or some intuition in finding such an example.

答:
In the ring ${\mathbf Z}[\sqrt{-3}]$, the ideal $P = (2,1+\sqrt{-3})$ is prime since it has index 2 in the ring. Note $P^2 = (4,2+2\sqrt{-3},-2 + 2\sqrt{-3}) = (4,2+2\sqrt{-3}) = (2)(2,1+\sqrt{-3}) = (2)P$, where $(2)$ is the principal ideal generated by 2 in ${\mathbf Z}[\sqrt{-3}]$. If there were unique factorization of ideals into products of prime ideals in ${\mathbf Z}[\sqrt{-3}]$ then the equation $P^2 = (2)P$ would imply $P = (2)$, which is false since $1+\sqrt{-3}$ is in $P$ but it is not in $(2)$.

In fact we have $P^2 \subset (2) \subset P$ and this can be used to prove the ideal (2) does not even admit a factorization into prime ideals, as follows. If $Q$ is a prime ideal factor of (2) then $(2) \subset Q$, so $PP = P^2 \subset Q$, which implies $P \subset Q$ (from $Q$ being prime), so $Q = P$ since $P$ is a maximal ideal in ${\mathbf Z}[\sqrt{-3}]$. If (2) is a product of prime ideals then it must be a power of $P$, and $P^n \subset P^2$ for $n \geq 2$, so the strict inclusions $P^2 \subset (2) \subset P$ imply (2) is not $P^n$ for any $n \geq 0$.

The "intuition" that the ideal $P = (2,1+\sqrt{-3})$ is the key ideal to look at here is that $P$ is the conductor ideal of the order ${\mathbf Z}[\sqrt{-3}]$.  The problems with unique factorization of ideals in an order are in some sense encoded in the conductor ideal of the order. So you want to learn what a conductor ideal is and look at it in several examples.  For example, the ideals $I$ in ${\mathbf Z}[\sqrt{-3}]$ which are relatively prime to $P$ (meaning $I + P$ is the unit ideal (1)) do admit unique factorization into products of prime ideals relatively prime to $P$. That illustrates why problems with unique factorization of ideals in ${\mathbf Z}[\sqrt{-3}]$ are closely tied to the ideal $P$.

If you look at the ideal notation $P' = (2,1+\sqrt{-3})$ in the larger ring ${\mathbf Z}[(1+\sqrt{-3})/2]$, which we know has unique factorization of ideals, then we don't run into any problem like above because $P' = 2(1,(1+\sqrt{-3})/2) = (2)$ in ${\mathbf Z}[(1+\sqrt{-3})/2]$, so $P'$ is actually a principal ideal and the "paradoxical" equation $PP = (2)P$ in ${\mathbf Z}[\sqrt{-3}]$ corresponds in ${\mathbf Z}[(1+\sqrt{-3})/2]$ to the dumb equation $P'P' = P'P'$. (The ideal $P'$ in ${\mathbf Z}[(1+\sqrt{-3})/2]$ is prime since the quotient ring mod $P'$ is a field of size 4: ${\mathbf Z}[(1+\sqrt{-3})/2]$ is isom. as a ring to ${\mathbf Z}[x]/(x^2+x+1)$, so ${\mathbf Z}[(1+\sqrt{-3})/2]/P' = {\mathbf Z}[(1+\sqrt{-3})/2]/(2)$ is isom. to $({\mathbf Z}/2{\mathbf Z})[x]/(x^2+x+1)$, which is a field of size 4.)


答:
An asymptotic follows from Theorem VI.6 ("Singular Inversion") in Flajolet and Sedgewick's Analytic Combinatorics (freely legally available online at the link). The theorem concerns generating functions which solve equations of the form $Y(z) = z \phi(Y(z))$, where $\phi$ is a polynomial or more generally a power series; these count trees of various sorts. Setting $Y(z) = z A(z)$ lets us put your problem in this framework, with $\phi(u) = c_0 + c_1 u + c_2 u^2 + c_3 u^3$. We need $\phi(u)$ to satisfy three sets of conditions (equations (35) and (36) on pages 402 and 403, together with a third condition detailed on page 404):

$\phi(0) \neq 0, [u^n] \phi(u) \ge 0, \phi(u) \neq \phi_0 + \phi_1 u$,
and

within the open disc of convergence of $\phi$, there is a (necessarily unique) positive real $\tau$ such that $\phi(\tau) - \tau \phi'(\tau) = 0$ (the characteristic equation).

the exponents of the nonzero terms of $\phi(u)$ are not all divisible by some $p \ge 2$ ($\phi(u)$ is aperiodic).

Then the radius of convergence of $Y(z)$ is $\rho = \frac{\tau}{\phi(\tau)} = \frac{1}{\phi'(\tau)}$. For example, when $c_i = 1$ the characteristic equation is

$$(\tau^3 + \tau^2 + \tau + 1) - \tau (3 \tau^2 + 2 \tau + 1) = - 2 \tau^3 - \tau^2 + 1 = 0$$

which has a unique positive real root $\tau \approx 0.657$ (an exact value can be written down using the cubic formula) and which gives $\rho \approx 0.277$.

Theorem: With the above hypotheses, $y_n = [z^n] Y(z)$ has an symptotic expansion beginning

$$y_n \sim \sqrt{ \frac{\phi(\tau)}{2 \phi''(\tau)} } \frac{\rho^{-n}}{\sqrt{\pi n^3}}.$$

Again with $c_i = 1$ we have $\rho^{-1} \approx 3.61 \dots$ so this is the $d$ which appears in the result you cited from the OEIS, and the constant factor is

$$\sqrt{ \frac{\tau^3 + \tau^2 + \tau + 1}{(12 \tau + 4)\pi} } \approx 0.252$$

but since $Y(z) = z A(z)$ our indexing is off by $1$ and $y_n = a_{n-1}$. So to get $a_n = y_{n+1}$ we multiply by $\rho^{-1}$ and get a constant factor $\approx 0.910$ which matches up with the result you cited from the OEIS again. Using the cubic formula and a bunch of algebraic manipulations it would be possible to match the exact results using radicals but it doesn't sound fun.

All of these computations simplify substantially if $c_3 = 0$, since then all of the cubic equations become quadratic. In this case we can write down an explicit closed form for $Y(z)$ using the quadratic formula and analyze that closed form using the binomial theorem to get asymptotics. This case is an elaboration on that one; the asymptotics come from a square-root singularity although it's harder to tell that that's what's happening.

The first formula you give can be obtained from Lagrange inversion, which gives that if $\frac{Y(z)}{\phi(Y(z))} = z$ then $Y$ (which is the compositional inverse of $\frac{w}{\phi(w)}$) has coefficients given by

$$[z^n] Y(z) = \frac{1}{n} [w^{n-1}] \phi(w)^n.$$

When $c_i = 1$ this gives

$$\begin{eqnarray*} a_n &=& \frac{1}{n+1} [w^n] (1 + w + w^2 + w^3)^{n+1} \\
 &=& \frac{1}{n+1} [w^n] (1 + w)^{n+1} (1 + w^2)^{n+1} \\
 &=& \frac{1}{n+1} [w^n] \left( \sum_{i=0}^{n+1} {n+1 \choose i} w^i \right) \left( \sum_{j=0}^{n+1} {n+1 \choose j} w^{2j} \right) \\
 &=& \frac{1}{n+1} \left( \sum_{i+2j = n} {n+1 \choose i} {n+1 \choose j} \right) \end{eqnarray*}$$

which is equivalent to the desired result. A nicer special case is $c_0 = c_2 = 1, c_1 = c_3 = 0$, or $\phi(w) = 1 + w^2$, which corresponds to binary trees and gives $a_{2n+1} = 0$ and

$$a_{2n} = \frac{1}{2n+1} [w^{2n}] (1 + w^2)^{2n+1} = \frac{1}{2n+1} {2n+1 \choose n} = \frac{1}{n+1} {2n \choose n}$$

which is the Catalan numbers. These have a similar asymptotic expansion because their generating function also has a square-root singularity, but the base of the exponent is $4$. It would be good to understand this case as a warmup exercise if you haven't already done so.

Title: Help with deriving information from a generating function

Body:
Sequence of interest

I feel like I've got most of the way, I just need some help at the last step. I am considering an integer sequence which encodes information pertaining to certain weighted ordered rooted trees. I won't bore you with the details suffice to say I found a recurrence relation for the sequence and consequently a generating function. Given the arbitrary variables $(c_0, c_1, c_2, c_3)$, the recurrence relation is:

\begin{align}
a_0 &= c_0 \\
a_n &= c_1 a_{n-1} + c_2 \sum_{i=0}^{n-2}a_ia_{n-2-i} + c_3\sum_{i=0}^{n-3}\sum_{j=0}^{n-3-i}a_ia_ja_{n-3-i-j} \hspace{2em} \text{for $n \geq 1$}
\end{align}

and the generating function $A(x)$ satisfies:

$$A = c_0 + c_1 (Ax) + c_2 (Ax)^2 + c_3 (Ax)^3$$

I'd like to get an asymptotic formula for this sequence, based on the variables $(c_0,c_1,c_2,c_3)$.

Bonus points for an exact formula!

Simpler sequence

Given the case with no weightings where $c_0 = c_1 = c_2 = c_3 = 1$, we are just counting the number of ordered rooted trees with outdegree no greater than 3. I found this sequence on OEIS where it says

$$a_n = \frac{1}{n+1}\sum_{j=0}^{\left\lfloor n/2\right\rfloor} \binom{n+1}{n-2j}\binom{n+1}{j}$$

and $$a_n \sim \frac{cd^n}{n^{3/2}}$$

where

$$d = \frac{(6371+624\sqrt{78})^{2/3}+11(6371+624\sqrt{78})^{1/3}+217}{12(6371+624\sqrt{78})^{1/3}} \approx 3.610718613276\ldots$$

is the root of the equation $4d^3-11d^2-8d-16=0$ and

$$c = \sqrt{\frac{f}{\pi}} \approx 0.9102276936417\ldots$$

where

$$f = \frac{1}{9984}(9295 + (13(45085576939 - 795629568\sqrt{78}))^{1/3} + (13(45085576939 + 795629568\sqrt{78}))^{1/3})$$

is the root of the equation $3328f^3-9295f^2+1696f-128=0$.

I'm assuming these magical formulae were somehow derived from its generating function? But I'm afraid this is beyond my generatingfunctionology skills.

I didn't know if this was more suitable for math stackexchange or mathoverflow. Please feel free to redirect me.

Thanks for any help!


问:
For all topologies of R, are the open sets of R union of open intervals?
My textbook gave the following proposition which followed the section on Euclidean topology:

"A subset S of R is open if and only if it is a union of open intervals"

but did not state whether this proposition is true only for the Euclidean topology or any topology on R. There are topologies on R that can be only closed intervals, but since infinite union of open interval can be closed intervals, I think the proposition is true for any topology on R right?

答: In the topology consisting of the three open sets $\emptyset$, $\Bbb R$ and $\{0\}$, the open set $\{0\}$ is not a union of open intervals.


问:
Prove $f(x)$ is constant.
Suppose $ f: \mathbb{R} \mapsto \mathbb{R}$ be continous. For any interval $[a,b]$, there exists $x_0\in (a,b)$ such that either $f(x_0)=\max\limits_{a\le x \le b} f(x)$ or $f(x_0)=\min\limits_{a\le x \le b} f(x)$. Prove $f(x)$ is constant.

We may consider apply Reductio ad Absurdum. Suppose $f(x)$ is not constant, then there exist $\alpha<\beta$ such that $f(\alpha)\neq f(\beta).$ But how to reduce the contradiction?

答:
Suppose $f$ is continuous and not constant, then by the conditions given, for each $(a,b)$ there exists a non empty set $\mathfrak M_{(a,b)}=\{M_1,M_2, ... \}$ such that $a<M_1<M_2<...<b$, with $$f(M_k)=M=\max_{x\in [a,b]} f(x)$$

In other words, $f$ attains it's maximum inside the set $(a,b)$ exactly $\#\mathfrak M_{(a,b)}$ times (which may be infinite) and at each point inside that set.

We then observe that $x\in[a,M_1)\implies f(x)<f(M_1)$, since we assumed that $M_1$ was the smallest point inside the set where $f$ attains its maximum in the set. This directly implies that $$\max_{x\in[a,M_1]}f(x)=f(M_1)$$ but $M_1\notin(a,M_1)$, which violates the condition that it's maximum is attained in the interior of the set.

Arguing similarly for the minimum, we may declare $\mathfrak m_{(a,b)}=(m_1, m_2,...)$ the set of minimums on $(a,b)$ with $a<m_1<m_2<...<b$ and $$f(m_k)=m=\min_{x\in[a,b]} f(x)$$

Again: $$x\in[a,m_1)\implies f(x)>f(m_1)\implies \min_{x\in[a,m_1]}f(x)=f(m_1)$$
with $m_1\notin (a,m_1)$ violating the same condition.

However, we have that $m_1<M_1$ or $m_1>M_1$. Suppose that $m_1<M_1$, then it is not guaranteed that $f(y)=\max_{x\in[a,m_1]}f(x)\implies y\notin(a,m_1)$, which would satisfy the condition on that range. We can argue similarly for $m_1>M_1$.

Thusly, we must combine the two: noticing in the case of $m_1<M_1$ that $$x\in(m_1, M_1)\implies m_1<f(x)<M_1\implies\min_{x\in[m_1,M_1]}f(x)=f(m_1), \max_{x\in[m_1,M_1]} f(x)=f(M_1)$$ neither of which are in the set $(m_1,M_1)$. and arguing  similarly for $m_1>M_1$

Title: Prove $f(x)$ is constant.


问:
Why is local cohomology $R$-linear functor?
Let $ R = k[x_1, \ldots, x_n] $ and $\mathfrak{m} = (x_1, \ldots, x_n)$. Let $N$ be an $R$-module, then the $\mathfrak{m}$-torsion functor is defined as $\Gamma_{\mathfrak{m}}(M) = \{ m \in M \text{ s.t.} \mathfrak{m}^nr = 0 \text{ for some } n \in \mathbb{N} \}$.

Now, in Brodmann and Sharps Local cohomology 1.2.1, it says:

Since $\Gamma_{\mathfrak{m}}(-)$ is covariant and $R$-linear, it is automatic that each local cohomology functor $H_{\mathfrak{m}}^{i}(-), i \geq 0$ is covariant and $R$-linear.

Why is this true?
It is clear for me, why $\Gamma_{\mathfrak{m}}(-)$ is a $R$-linear, left-exact functor.
I also know that for a short exact sequence of $R$-modules $$0 \to A \overset{f} \to B \to C \to 0,$$ there is a long exact sequence of cohomology modules s.t. we get $$H_{\mathfrak{m}}^{i}(A) \overset{H_{\mathfrak{m}}^{i}(f)} \to H_{\mathfrak{m}}^{i}(B) \to H_{\mathfrak{m}}^{i}(C).$$
So it makes sense that $H_{\mathfrak{m}}^{i}(-)$ is a functor for all $i \geq 0$. But why does it have to be $R$-linear?

While the map $f$ is from $A$ to $B$, the map $H_{\mathfrak{m}}^{i}(f)$ is from $\ker(\Gamma_{\mathfrak{m}}(d_A^{i}))/Im(\Gamma_{\mathfrak{m}}(d_A^{i-1}))$ to $\ker(\Gamma_{\mathfrak{m}}(d_B^{i}))/Im(\Gamma_{\mathfrak{m}}(d_B^{i-1}))$, where $d_A^{i}$ are the maps in an injective resolution of $A$ and $d_B^{i}$ in an injective resolution of $B$ respectively. Since the choice of injective resolution of $A$ and $B$ doesn't matter, how do I know so much about the map $H_{\mathfrak{m}}^{i}(f)$?

Clearly, $ H_{\mathfrak{m}}^{i}(f)$ depends on $f$, but how is it defined, resp. how does one see that is $R$-linear?

(My goal is to show that if $f$ is multiplication by $x \in R$, then $H_{\mathfrak{m}}^{i}(f)$ is still multiplication by $x \in R$.)

答:
First of all, "$H^i_{\mathfrak{m}}$ is $R$-linear" does not mean "$H^i_{\mathfrak{m}}(f)$ is $R$-linear for all $f$". Instead, it means $$H^i_{\mathfrak{m}}(\alpha f + g) = \alpha H^i_{\mathfrak{m}}(f) + H^i_{\mathfrak{m}}(g)$$ for all $f,g : A \to B$ and all $\alpha \in R$. However, the statement "$H^i_{\mathfrak{m}}(f)$ is $R$-linear for all $f$" is also true (and necessary in order for us to interpret the RHS of the above equation)! Once we unravel the definition of $H^i_{\mathfrak{m}}(f)$, this will be easy to prove.

So, it's not exactly true that the choice of injective resolution doesn't matter. Different injective resolutions for $A$ may actually yield different sets for $H^i_{\mathfrak{m}}(A)$. However, all the possible versions of $H^i_{\mathfrak{m}}(A)$ are isomorphic, and these isomorphisms are compatible with the maps $H^i_{\mathfrak{m}}(f)$. As a result of all this, we can compute cohomology groups (and maps between them) using any injective resolution we want (and just remember that the result is only well-defined up to isomorphism).

Well, how is $H^i_{\mathfrak{m}}(f)$ defined? First, we choose injective resolutions for $A$ and $B$:
$\require{AMScd}$
\begin{CD}
0 @>>> A @>>> A_0 @>>> A_1 @>>> \cdots\\
@. @VfVV \\
0 @>>> B @>>> B_0 @>>> B_1 @>>> \cdots
\end{CD}
Now, we fill this in to a morphism of complexes:
\begin{CD}
0 @>>> A @>>> A_0 @>d^0_A>> A_1 @>d^1_A>> \cdots\\
@. @VfVV @Vf_0VV @Vf_1VV \\
0 @>>> B @>>> B_0 @>d^0_B>> B_1 @>d^1_B>> \cdots
\end{CD}

Important exercise: how do we know this can be done? Hint: build the maps inductively, and use the definition of "injective module".

Next, we apply $\Gamma_{\mathfrak{m}}$ to the deleted resolutions:
\begin{CD}
0 @>>> \Gamma_{\mathfrak{m}}(A_0) @>\Gamma_{\mathfrak{m}}(d^0_A)>> \Gamma_{\mathfrak{m}}(A_1) @>\Gamma_{\mathfrak{m}}(d^1_A)>> \cdots\\
@. @V\Gamma_{\mathfrak{m}}(f_0)VV @V\Gamma_{\mathfrak{m}}(f_1)VV \\
0 @>>> \Gamma_{\mathfrak{m}}(B_0) @>\Gamma_{\mathfrak{m}}(d^0_B)>> \Gamma_{\mathfrak{m}}(B_1) @>\Gamma_{\mathfrak{m}}(d^1_B)>> \cdots
\end{CD}

Finally, we can state the definition of $H^i_{\mathfrak{m}}(f)$:
$$H^i_{\mathfrak{m}}(f)([x]) = [\Gamma_{\mathfrak{m}}(f_i)(x)]$$
for $x \in \ker(\Gamma_{\mathfrak{m}}(d^i_A))$, where $[\bullet]$ means "the equivalence class (aka coset) of $\bullet$".

Important exercise: This is a well-defined function, i.e. $[\Gamma_{\mathfrak{m}}(f_i)(x)] = [\Gamma_{\mathfrak{m}}(f_i)(y)]$ whenever $[x] = [y]$ for $x, y \in \ker(\Gamma_{\mathfrak{m}}(d^i_A))$.

Important note: This construction of $H^i_{\mathfrak{m}}(f)$ relied on three arbitrary choices:

An injective resolution of $A$

An injective resolution of $B$

An extension of $f$ to a morphism of complexes

We've already discussed how the choices of injective resolution for $A$ and $B$ don't end up mattering, up to isomorphism. Moreover, it turns out that choosing a different morphism of complexes would result in exactly the same map $H^i_{\mathfrak{m}}(f)$ (not just up to isomorphism)! Thus, $H^i_{\mathfrak{m}}(f)$ is well-defined.

Once we have this definition, we can check that $H^i_{\mathfrak{m}}(f)$ is always $R$-linear! We let $a, a' \in H^i_{\mathfrak{m}}(A)$ and $\lambda \in R$ be arbitrary. Then $a = [x], a' = [x']$ for some $x, x' \in \ker(\Gamma_{\mathfrak{m}}(d^i_A))$. Now
\begin{align*}
H^i_{\mathfrak{m}}(f)(\lambda a + a') &= H^i_{\mathfrak{m}}(f)(\lambda [x] + [x'])\\
&= H^i_{\mathfrak{m}}(f)([\lambda x + x']) & \text{(definition of quotient module)}\\
&= [\Gamma_{\mathfrak{m}}(f_i)(\lambda x + x')] & \text{(definition of } H^i_{\mathfrak{m}}(f)\text{)}\\
&= [\lambda \Gamma_{\mathfrak{m}}(f_i)(x) + \Gamma_{\mathfrak{m}}(f_i)(x')] & \text{(} \Gamma_{\mathfrak{m}}(f_i) \text{ is } R\text{-linear)}\\
& = \lambda [\Gamma_{\mathfrak{m}}(f_i)(x)] + [\Gamma_{\mathfrak{m}}(f_i)(x')] & \text{(definition of quotient module)}\\
&= \lambda H^i_{\mathfrak{m}}(f)([x]) + H^i_{\mathfrak{m}}(f)([x']) & \text{(definition of } H^i_{\mathfrak{m}}(f)\text{)}\\
&= \lambda H^i_{\mathfrak{m}}(f)(a) + H^i_{\mathfrak{m}}(f)(a')
\end{align*}
and we're done!

I want to emphasize that this is where the definition of $H^i_{\mathfrak{m}}$ as a functor comes from (not the long exact sequence)! Indeed, the map $H^i_{\mathfrak{m}}(f)$ in the long exact sequence doesn't make sense to write down before we've defined $H^i_{\mathfrak{m}}(f)$ (i.e. before we've made $H^i_{\mathfrak{m}}$ into a functor).

We still haven't shown that $H^i_{\mathfrak{m}}$ is an $R$-linear functor, but I believe you can do that from here! Definitely begin by proving that $\Gamma_m$ is $R$-linear. Let me know if you get stuck!


问:
How is the boundary of the clopen set [0,1) empty?
I don't get why the boundary of a clopen set is empty. If you take $A = [0,1)$ in $\mathbb R$, then isn't the closure of this the smallest closed super set that contains $A$ which is $[0,1]$.  Isn't the interior, $(0,1)$?  So the boundary would be the closure minus the interior, aka, ${0}$ and ${1}$?

I'm confused. Thanks for all of your great and quick replies!! No wonder I was so confused, I was completely misunderstanding the definition of clopen. I've learned a lot from all of you!

答:
You are correct that the boundary of the set $A=[0,1)$, considered as a subset of $\mathbb{R}$, is the set $\{0,1\}$, by exactly the reasoning you described.

However, treating $A$ as a topological space in its own right (i.e., given the subspace topology,  a.k.a. the relative topology), the boundary of $A$ is empty.  The same is true of any topological space $X$: the boundary of $X$, as a subset of itself, is empty.

Why? Well, the closure of $X$ in $X$ is the smallest closed subset of $X$ that contains $X$. The only subset $X$ whatsoever that contains $X$ is $X$ itself; and $X$ is closed as a subset of $X$ (since it is the complement of the open set $\varnothing$), so the closure of $X$ as a subset of $X$ is just $X$ again. Similarly, because $X$ is already open as a subset of itself, the interior of $X$ is also $X$. Thus $$\partial X=\overline{X}\setminus\mathrm{int}(X)=X\setminus X=\varnothing.$$

In short, the issue is that "boundary" is a term that depends on two pieces of information: the set we are taking the boundary of, and the "ambient" space that we are working in. The same is true of "closure" and "interior"; there is no such thing as the closure or the interior of a set. It is always relative to some "ambient" set where everything takes place (which may be left implicit, if it is assumed that it will be clear from context).


答:
Suppose $f(\alpha)<f(\beta)$ for contradiction.

Let $A:=\{x:f(x)\le f(\alpha), x\le \beta\}$. It is non-empty since $\alpha\in A$ and bounded above by $\beta$, so it has a supremum $y=\sup A$. By continuity, $f(y)\le f(\alpha)$.

Similarly consider $B:=\{x:f(x)\ge f(\beta), y\le x\}$. It is also non-empty and bounded below by $y$, so it has an infimum $z=\inf B$. Again, $f(z)\ge f(\beta)$

Now consider the interval $[y,z]$ which has a point $w$ that is either a max or min inside the interval $(y,z)$. This would be a contradiction as $w\in A$ or $w\in B$. Hence $y=z$. But this then contradicts that $f$ is continuous.

Title: Prove $f(x)$ is constant.

Body:
Suppose $ f: \mathbb{R} \mapsto \mathbb{R}$ be continous. For any interval $[a,b]$, there exists $x_0\in (a,b)$ such that either $f(x_0)=\max\limits_{a\le x \le b} f(x)$ or $f(x_0)=\min\limits_{a\le x \le b} f(x)$. Prove $f(x)$ is constant.

We may consider apply Reductio ad Absurdum. Suppose $f(x)$ is not constant, then there exist $\alpha<\beta$ such that $f(\alpha)\neq f(\beta).$ But how to reduce the contradiction?


问:
Magic dice question.
A pair of dice are magic. When they are rolled together, then half the time, they behave like a normal pair of dice, but the other half of the time they miraculously roll a double ( equally likely to be any of the six possible doubles).  What is the expected value of the product of the two scores when the pair of dice are rolled?</p>

答:
Think of the sample space as $(M,n,m)$ where $M=0$ if the dice behave normally and $M=1$ otherwise. $n,m \in \{1,...,6\}$. $M, n,m $ are independent and uniformly distributed as appropriate. Hence $P[(M,n,m) ] = {1 \over 2 \cdot 6^2}$.

Define $f(M,n,m) = \begin{cases} nm,& M=0 \\
n^2, & M=1\end{cases}$.

Then $E[f] = {1 \over 2 \cdot 6^2} (\sum_n \sum_m nm + 6 \sum_n n^2) = {1 \over 2 \cdot 6^2} ((\sum_n n)^2 + 6 \sum_n n^2) = {329 \over 24} \approx 13.708$.

Title: Magic dice question.


问:
How to prove $2a^4+2b^4 \geq c^2(2a^2+2b^2-c^2)$ for any positive $a$, $b$ and $c$?
How to prove $2a^4+2b^4 \geq c^2(2a^2+2b^2-c^2)$ for $a,b,c>0$?

Body: $$2a^4+2b^4 \geqslant c^2(2a^2+2b^2-c^2) \iff (a^2-b^2)^2+(a^2+b^2-c^2)^2\geqslant 0$$


问:
Showing linear independence of a set
Let $A$ be a $3\times 3$ matrix with characteristic polynomial $p(x)=-x^3$ and suppose $w\in \Bbb R^3$ is a vector such that $A^2w\neq0$. I am supposed to show that $B=\{w,Aw,A^2w\}$ is a basis for $\Bbb R^3$.

To show that this set is linearly independent, $A^3w$ must be equal to $0$ but I cannot see why.

I am so sorry for this very easy question.

I appreciate any help

答:
I have found a proof that does not use Cayley-Hamilton theorem.

The roots of $p(x) = -x^3$ are precisely the eigenvalues of $A$, so $A$ has only one eigenvalue: zero. We will prove that $B=\{w,Aw,A^2w\}$ is a basis for $\mathbb{R}^3$ by first showing that $Aw$ is not a multiple of $w$ and then that $A^2w$ is not a linear combination of $w$ and $Aw$.

If $Aw$ was a multiple of $w$ then $w$ would be an eigenvector of $A$ and $Aw$ would be zero, contradicting assumption. Thus, $w$ and $Aw$ are linearly independent.

Let $V = \mathrm{span}(w, Aw)$. For a proof by contradiction, suppose that $A^2w$ is a linear combination of $w$ and $Aw$. In this case, $Av \in V$ for every $v \in V$. Thus, we can consider the restriction $B$ of $A$ to $V$. Every eigenvalue of $B$ is an eigenvalue of $A$. Therefore, the only eigenvalue of $B$ is zero. Let $aw + bAw$ be the eigenvector of $B$ associated with the eigenvalue. Then

$$
aAw + bA^2w = 0.\tag1
$$

Now, $Aw \ne 0$ and $A^2w \ne 0$ so $a \ne 0$ and $b \ne 0$. Therefore, we can write $(1)$ as

$$
A^2w = -\frac{a}{b}Aw,
$$

but this means that $\frac{a}{b} \ne 0$ is an eigenvalue of $A$ contradicting the fact that its characteristic polynomial is $p(x)=-x^3$. Thus, $A^2w$ is not a linear combination of $w$ and $Aw$.

Title: Showing linear independence of a set


问:
Mathematics Engineering: How do you prove the power rule?
Consider the real-valued function  $f(x)=x^r$ where $r$ is a real number.

(1) For what values of $x$ and $r$, is this function differentiable?

(2) How do you prove the power rule, when $f(x)$ is differentiable?

It would be nice if the proof is accessible to a 12th grade student.

EDIT: Perhaps, I have not been very clear about my question.  I am not satisfied with the "textbook proofs" I have seen.  Please established the existence (of the derivative) first before proving the power rule.  That is, please consider first question first.  This is the reason why I call the question a mathematical engineering question.  (If we can we bring high level math down to Grade 12 level, without loosing precision, then I call it good mathematical engineering.)

答:
Prove power rule from first principle via binomial theorem and taking leading order term, now for negative exponents, we can use a trick. Consider:

$$ x^k \cdot x^{-k} = 1$$

The above identity holds for all $ x \in R -{0}$, differentiate it:

$$ kx^{k-1} x^{-k} + x^k \frac{d}{dx} x^{-k} =0$$

$$  \frac{d}{dx} x^{-k} =  -\frac{k}{x^{k+1}}$$

So, without any generalized binomial theorem, a student who knows binomial theorem + product rule would be able to do this :^)

It is definitely differentiable for all $r \in \mathbb{R}$ and all x in R except for case when $r$ is negative ; then x can't be zero


问:
Finding the Rank of power of a matrix
Let $A$ be $6 \times 6$ matrix with $A^5 = I$, Find the rank($A^2$).

The only thing I can think about this question is
$A^5 = I$ implies $x^5 - 1 =0$ is annihilating polynomial. Can I conclude or continue solving the problem with that idea ?

How can we solve these type of problems. What are the results I have to know. Is there a general idea for solving them ??

答:
We have

$$A^2A^3=A^3A^2=A^5=I.$$

Thus, $A^2$ is inverible. Conclusion ?

Title: Finding the Rank of power of a matrix


答:
Let $ I = [a, b] $ be any closed interval and assume, without loss of generality, that $ f $ attains its maximum value $ M $ in the interior; the case where it attains its minimum value instead is similar. Suppose $ f(a) < f(b) $ without loss of generality, I'll obtain a contradiction from this assumption.

Since $ f $ is continuous, $ [a, b] \cap f^{-1}(\{ M \}) $ is a closed and bounded set, therefore it has minimum and maximum values $ x_{\textrm{min}}, \, x_{\textrm{max}} $. If $ x_{\textrm{min}} = a $ then $ f(a) = M $ and this contradicts $ f(a) < f(b) $, so we can assume $ x_{\textrm{min}} > a $.

Now, since $ x_{\textrm{min}} $ is the minimum value for which $ f $ attains its maximum value $ M $, it can't attain the value $ M $ anywhere in the interval $ (a, x_{\textrm{min}}) $. Therefore it must instead attain its minimum value $ m $ on the interval $ [a, x_{\textrm{min}}] $ somewhere in the interior. Similar to the above, take the maximum value such that $ f $ attains its minimum $ m $ there, call it $ y_{\textrm{max}} $. If $ y_{\textrm{max}} = x_{\textrm{min}} $ then $ m = M $ and therefore $ f(a) = m = M $ once more, which yields the same contradiction as above with $ f(a) < f(b) $. Therefore we must have $ y_{\textrm{max}} < x_{\textrm{min}} $, in which case $ f $ attains neither its maximum nor its minimum on $ [y_{\textrm{max}}, x_{\textrm{min}}] $ in the interior, yielding a contradiction.

We conclude that $ f(a) < f(b) $ is impossible, and we can similarly show that $ f(a) > f(b) $ is impossible, so we must have $ f(a) = f(b) $. Since $ a, b $ were arbitrary, it follows that $ f $ is constant.

Title: Prove $f(x)$ is constant.


问:
For the discrete topology, why must the singletons be in every basis?
I know the set of all singletons are a basis for a discrete topology, but if we had another basis, why must it also include all the singletons?</p>

答: Given a topological space $(X,\tau)$, a basis for $(X,\tau)$ is a collection $\mathcal B \subseteq \tau$ such that for each $U \in \tau$ and each $x \in U$ there exists $B \in \mathcal B$ with $x \in B \subseteq U$. So, if $\tau = \mathcal P(X)$, for each $x \in X$ there exists $B \in \mathcal B$ with $x \in B \subseteq \{x\}$. It follows that $B = \{x\}$ is in $\mathcal B$, as required.


问:
Is there an error-correcting code for bits expressed in 3-D?
As an outsider looking into the field of coding-theory, I am wondering if there are examples of an error-correcting code where the bits are expressed in 3-D space?

Though I am not sure this exists, I am imagining a "lattice" extension of a 2-D barcode generated with the Reed-Solomon polyonmial. In 3-D, the row (x), column (y) and height (z) are used as information to encode the barcode. Would someone be able to point me to some examples of these codes? I am particularly interested in how this added dimension will change error-correction and message capacity.

答:
Not really an answer, but I started thinking, then typing, and then thinking again. Posting whatever popped into my mind :-(

The Reed-Solomon codes used in 2D barcode really correct byte errors. Or perhaps more precisely, symbol errors. IIRC the QuickResponse code has 8-bit symbols (=bytes)., but Reed-Solomon codes exist for each possible symbol size, 9-bit, 10-bit, whatever. Their math is simple enough. Should symbol errors be relevant to your application, it is trivial to construct a Reed-Solomon code to deal with them.

The way I see it the challenges in your idea are technical. The standard of Quick Response codes specifies a handful of formats, and those squary parts are designed in a way that makes it possible for example to identify the orientation of the barcode in relation to the scanner as well as to deduce the choice of format.

Any 3D-barcode would likely need something similar.

But the details depend on the type of printing/scanning available to both whoever produces such 3D-barcode labels and whoever tries to interpret them.

If asked to design an error/orientation coding scheme for such a system, I would need to know at least:

What kind of range of information content are we supposed to handle with this system? Would a larger 2D-barcode do just as well?

What can go wrong with the interpreter? Orientation? Distance? Identifying a 3D-orientation probably needs something other than those squares. But is that a problem we want to solve?

What kind of errors are possible? Are some types of error more likely than others (probably want to concentrate on those)?

How is the data encoded into the 3D-barcode (at the level of bits)?

Those are largely engineering problems. But we need at least a rough idea of these before we can begin the design of the error correction scheme. A few general remarks.

If it turns out that errors in interpreting "nearby" bits correlate with each other (often happens in 2D-barcodes because a small part of the barcode may be damaged), then (possibly only then) we should probably think in terms of symbol errors. Basically treating a small lump of adjacent illegible bits as a single error event. This. Is a tell-tale sign to go with Reed-Solomon.

On the other hand, it may turn out that a more typical error event involves isolated bits only. Say, the least significant bit of the height of an individual part of the 3D-barcode is often misinterpreted, but in a haphazarz way, all across the 3D-barcode. If this is a prevalent error, then we need either a different kind of an error-correcting code altogether. Or, possibly we need to restructure the encoding of data to the barcode in a different way.

With all that off my chest, I try to say something about your $xyz$. If your Reed-Solomon code uses $m$-bit symbols, we need a way to encode them in the barcode. A part of it would be just a numbers game.

If we decide to use $12$-bit symbols, and the scanner is able to reliably distinguish $8$ different heights, we could lump together a 2x2 square of adjacent locations and build a symbol of $12$ bits out of those. If there are only $4$ levels of heights, that 2x2 square would produce only $8$-bit symbols. Because RS-codes deal with symbol errors, this kind of encoding could deal with the type of error events, where damage to the barcode is typically limited to small areas.

But if a typical damage to the 3D-barcode amounts to the top layer being scrubbed away, the simple scheme of the previous paragraph is mostly useless. After all, with it the data was in the height information. Hence scrubbing would disturb several symbols. This is a much more complicated error event. For starters the errors are asymmetrical. The height may get reduced by scrubbing, but it much less likely that the height of a barcode would become higher. A totally different kind of scheme is required. We could try and use erasure recovery properties of RS-codes, building symbols differently, but I don't see that working either. I have next to no knowledge about coping with asymmetrical errors, so cannot even suggest a scheme.


问:
Formalizing composition of sequences
The following first equality is intuitively evident:
$$
\lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n
= \lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n
=e^{x+y}
$$
how can I give a formal argument?

Informally it is something like
$$
\lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{\lim_{n\to \infty}\left(x+y+\frac{xy}{n}\right)}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n
$$
Is there a "classical" property of sequences that backs this argument? Something like limit of composition, but here we speak of sequence only:
$$
(b_n),\quad b_n \to b\quad\text{ for }n\to\infty \implies
\lim_{n\to \infty} \left(1+\frac{b_n}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{b}{n}\right)^n
$$

EDIT As Qiaochu Yuan is important to define what I have already defined.
This computation is at the beginning of "Real Analysis", lets say in the chapter of sequences. Later, in the notes of complex analysis, I develop formal series with derivation and everything is the standard way (I think).

By this question I tried another approach, very early (1) define
$$
e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n
$$
(2) derive the product as depicted in this question
(3) derive the derivation of the $e^x$.

Thus I have (a) no logarithm, (b) no derivation of $e^x$ yet.

答:
The desired statement is true. It follows from taking logarithms and writing

$$\begin{eqnarray*} \log \left( 1 + \frac{b_n}{n} \right)^n &=& n \log \left( 1 + \frac{b_n}{n} \right) \\
 &=& n \left( \frac{b_n}{n} + O \left( \frac{b_n^2}{n^2} \right) \right) \\
 &=& b_n + O \left( \frac{b_n^2}{n} \right). \end{eqnarray*}$$

Taking the limit as $n \to \infty$ gives that the limit of the logarithms is $\lim_{n \to \infty} \left( b_n + O \left( \frac{b_n^2}{n} \right) \right) = b$ (using that $b_n$ is convergent and hence bounded, so the $O \left( \frac{b_n^2}{n} \right)$ term vanishes in the limit).

Depending on how you've defined exponentials and logarithms and proven their properties this argument may be circular, since it depends on the first two terms of the Taylor series of the logarithm, and one of several ways to compute the Taylor series of the logarithm is to use $\log xy = \log x + \log y$ which is more or less equivalent to $e^{x+y} = e^x e^y$.

There are alternatives for proving that $e^{x + y} = e^x e^y$ which avoid having to prove this; for example, taking for granted that $\frac{d}{dx} e^x = e^x$ and $e^0 = 1$ (which together uniquely determine $e^x$ by the existence and uniqueness theorems for ODEs), we have first

$$\frac{d}{dx} (e^x e^{-x}) = e^x e^{-x} - e^x e^{-x} = 0$$

so $e^x e^{-x}$ is a constant, and plugging in $x = 0$ gives $e^x e^{-x} = 1$. Next,

$$\frac{d}{dx} \left( e^{x+y} e^{-x} e^{-y} \right) = e^{x+y} e^{-x} e^{-y} - e^{x+y} e^{-x} e^{-y} = 0$$

so $e^{x+y} e^{-x} e^{-y}$ is a constant in $x$, and plugging in $x = 0$ gives $e^y e^{-y} = 1$ and hence $e^{x+y} = e^x e^y$.

Title: Formalizing composition of sequences


问:
The composition of a homotopy and a quotient map
I feel like I have some fallacies, please help me clear them. About something that seems to be very basic.

Say we have a homotopy $h:[0,1]\times [0,1]\to [0,1]$ with $h(x,t)=tx$. Then there is this quotient map $q:[0,1]\to S^1$ that identifies $0$ and $1$. Hence compositing them gives a continuous map $qh:[0,1]\times[0,1]\to S^1$ that is a homotopy from a loop in $S^1$ to the constant map, so does it follow that the loop $f(t)=e^{2\pi it}$ is nullhomotopic? Or which part is wrong?

答: In general, you could say your $f$ is homotopic to the constant map on “$[0,1]$,” but the loop $S^1$ is not. To show a loop is homotopic to another, your domain should be $S^1,$ not an interval.


问:
If a function of two variables has a unique critical point, which is a local maximum, is it a global maximum?

$f(x,y)$ has partial derivatives in all $\mathbb R^2$ and a unique critical point at $(x_0,y_0)$ (local maximum). Is it a global maximum?

I know that in compact sets, it isn't enough to say that if a point is the only maximum inside the set, then it's a global maximum, because in the frontier of the set it could happen that the function has a maximum greater than the inside one. But for the entire $\mathbb R^2$ there is no frontier, therefore can I admit that this unique point is a point of global maximum?

答:
Loose description of the geometry: imagine a flat plane and then you put a lone hill with a peak on it. Now tilt the plane a little. Now you still have one peak, but hopefully you can also see that you have introduced a saddle point. Imagine sliding the saddle point location off to infinity to make it effectively no longer there.

Basic description of the construction I saw somewhere once: you can have something whose cross sections in one direction are cubic, but whose cross-sections in the other direction is a sum of exponentials in just the right way for one of the cubic cross-sectional peaks to meet the peak of one of the exponential sum cross-sections. This can happen with no other critical points, and of course the cubic component alone is unbounded in either direction (so there is no global max or min). The direction that the exponentials taper off in aligns with the idea of the saddle point being slid away to infinity.

Consider $f(x,y)=x^3+axe^{my}+be^{ny}$ for some constants $a,b,m,n$. We have:
$$\begin{align}
\frac{\partial}{\partial x}f(x,y)&=3x^2+ae^{my}&\frac{\partial}{\partial y}f(x,y)&=amxe^{my}+bne^{ny}
\end{align}$$

To find a critical point, the $\frac{\partial}{\partial x}$ shows us $a$ must be negative. We are just looking for one example, so we pick some numbers for the constants that make the algebra that follows simpler. With $a=-3$, $b=1$, $m=1$, $n=3$:

$$\begin{align}
\frac{\partial}{\partial x}f(x,y)&=3x^2-3e^{y}&\frac{\partial}{\partial y}f(x,y)&=-3xe^{y}+3e^{3y}\\
&=3(x^2-e^y)&&=3e^y(e^{2y}-x)
\end{align}$$

So at a critical point, $x=e^{2y}$ according to the latter relation.

Then using the $\frac{\partial}{\partial x}$ relation:
$$e^{4y}-e^y=0\implies e^{3y}=1\implies y=0$$
which in turn says that $x=1$. So the only critical point is at $(1,0)$. Is it a local min, max, or neither?

$$\begin{align}
f_{xx}(1,0)&=6&f_{xy}(1,0)&=-3&f_{yy}=6
\end{align}$$

So the Hessian determinant is $6^2-(-3)^2=27>0$, so this critical point is either a local max or a local min. OK, it turns out to be a local min, since for example $f_{xx}(1,0)$ is positive. But $f$ has no global min, since you could fix $y$ and let $x\to-\infty$ and get unbounded large negative output.

Take this $f$ and negate it to get the direct counterexample. Here is an image of $-f$ from GeoGebra. Note how there would be a saddle point, except it has been slid off the map (to infinity) to the left.


问:
Proof that $\operatorname{rank}(\mathbf A\mid\mathbf B) = \max(\operatorname{rank}(\mathbf A), \operatorname{rank}(\mathbf B))$
By $(\mathbf A \mid \mathbf B)$, I mean the matrix obtained by adjoining $\mathbf A$ to $\mathbf B$, e.g., if $\mathbf A = \left(\begin{smallmatrix}1&2\\3&4\end{smallmatrix}\right)$ and $\mathbf B = \left(\begin{smallmatrix}6\\7\end{smallmatrix}\right)$, then
$$(\mathbf A \mid \mathbf B) = \begin{pmatrix}
1&2&6\\
3&4&7
\end{pmatrix}.$$

I have tried to prove it but I don't know how to proceed. Similarly there is the result
$$\operatorname{rank}\left(\frac{\mathbf A}{\mathbf B}\right) = \operatorname{rank}(\mathbf A)+\operatorname{rank}(\mathbf B)$$
which I suspect is proved using a similar technique, but I'm unsure of how to go about it.

答:
The comments from @Sebastion Schulz show that your conjectures don't stand up.

You don't need any of this to prove
$$
\renewcommand\arraystretch{1.2}\operatorname{rank}\left[\left(\begin{array}{c|c}\mathbf A & \mathbf O \\\hline\mathbf B & \mathbf I     \end{array}\right)\left(\begin{array}{c|c}\mathbf I & \mathbf C \\    \hline \mathbf O & \mathbf D \end{array}\right)\right] = \operatorname{rank}(\mathbf A) + \operatorname{rank}(\mathbf D)
$$
when $\mathbf A$ is invertible.

Pre-multiplying by $\left(\begin{array}{c|c}\mathbf A & \mathbf O \\\hline\mathbf B & \mathbf I     \end{array}\right)$ in these circumstances is just making some invertible row operations and this does not change the rank of $\left(\begin{array}{c|c}\mathbf I & \mathbf C \\    \hline \mathbf O & \mathbf D \end{array}\right)$.

By elementary column operations we can reduce $\left(\begin{array}{c|c}\mathbf I & \mathbf C \\    \hline \mathbf O & \mathbf D \end{array}\right)$ to $\left(\begin{array}{c|c}\mathbf I & \mathbf O \\    \hline \mathbf O & \mathbf D \end{array}\right)$, whose rank is $\operatorname{rank}(\mathbf I) + \operatorname{rank}(\mathbf D)=\operatorname{rank}(\mathbf A) + \operatorname{rank}(\mathbf D)$.


答:
You can use the basic fact that for any sequence $c_n \stackrel{n\to\infty}{\longrightarrow}0$ $(c_n \neq 0)$ you have

$$\left(1+c_n\right)^{\frac 1{c_n}}\stackrel{n\to\infty}{\longrightarrow}e$$

So with $c_n =\frac{b_n}n$ and $b_n \stackrel{n\to\infty}{\longrightarrow}b$, you get

\begin{eqnarray*} \left(1+c_n\right)^n
& = & \left(\left(1+c_n\right)^\frac 1{c_n}\right)^{c_n\cdot n}\\
& = & \left(\left(1+c_n\right)^\frac 1{c_n}\right)^{b_n} \\
& \stackrel{n\to\infty}{\longrightarrow} & e^b
\end{eqnarray*}

Here the continuity of $f(x,y) = x^y$ in both variables $x$ and $y$ is used which can easily be established considering,for example, first the logarithm of this function.

Hence, you have

$$\lim_{n\to\infty} \left(1+\frac{b_n}n\right)^n=e^b = \lim_{n\to\infty} \left(1+\frac{b}n\right)^n$$

Title: Formalizing composition of sequences

Body:
The following first equality is intuitively evident:
$$
\lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n
= \lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n
=e^{x+y}
$$
how can I give a formal argument?

Informally it is something like
$$
\lim_{n\to \infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{\lim_{n\to \infty}\left(x+y+\frac{xy}{n}\right)}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{x+y}{n}\right)^n
$$
Is there a "classical" property of sequences that backs this argument? Something like limit of composition, but here we speak of sequence only:
$$
(b_n),\quad b_n \to b\quad\text{ for }n\to\infty \implies
\lim_{n\to \infty} \left(1+\frac{b_n}{n}\right)^n
=\lim_{n\to \infty} \left(1+\frac{b}{n}\right)^n
$$

EDIT As Qiaochu Yuan is important to define what I have already defined.
This computation is at the beginning of "Real Analysis", lets say in the chapter of sequences. Later, in the notes of complex analysis, I develop formal series with derivation and everything is the standard way (I think).

By this question I tried another approach, very early (1) define
$$
e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n
$$
(2) derive the product as depicted in this question
(3) derive the derivation of the $e^x$.

Thus I have (a) no logarithm, (b) no derivation of $e^x$ yet.


问:
Finding infimum and proving continuity of a function
Let $f : [a,b] \to\mathbb R$ and for every $x \in [a,b]$ there exists $y \in [a,b]$ such that $|f(y)| < \frac12|f(x)|$.

Find $\inf\{|f (x)| : x \in [a, b]\}$, and show that $f$ is not continuous on $[a, b]$.

I am fairly new to this type of calculus so I am not being able to start an approach regarding this, I think infimum will be a function out of the range of $f$ as according to the definition there will always be another term less than half of it.

答:
Part One.

Let $I=\{|f(x)|: x\in [a,b]\}.$ Let $D=\inf I.$

Then $D=0.$ Proof by contradiction: Suppose $D>0.$ By def'n of $D$ we have $$(i).\quad I\subset [D,\infty).$$ Now if $E>D$ then $$(ii).\quad I\cap [D,E]\ne \emptyset.$$ Otherwise, by $(i)$ we would have $I\subset [E,\infty),$ implying $ D=\inf I\ge \inf \,E>D.$

So in $(ii)$ take $E=3D/2.$  Then there exists $x\in [a,b]$ with $|f(x)|\in [D,E]=[D,3D/2].$ But then there exists $y\in [a,b]$ with $|f(y)|<|f(x)|/2\le 3D/4<D,$ contrary to $(i).$ This is the desired contradiction that follows from assuming $D>0$.

Part two.

$0=D=\inf I\not \in I.$ Otherwise for some $x\in [a,b]$ we would have $|f(x)|=0$ but then there would exist $y\in [a,b]$ with $|f(y)|<|f(x)|/2=0.$


答:
Any path $[0, 1] \to X$ is a nullhomotopic map, no matter if its image has self-intersections. So
$$
[0,1] \to S^1 : \;  t \mapsto e^{2\pi it}
$$
is nullhomotopic.

But loop is a map of the form $S^1 \to X$. The loop
$$
S^1 \to S^1 : \;  t \mapsto t
$$
is not nullhomotopic.

Title: The composition of a homotopy and a quotient map

Body:
I feel like I have some fallacies, please help me clear them. About something that seems to be very basic.

Say we have a homotopy $h:[0,1]\times [0,1]\to [0,1]$ with $h(x,t)=tx$. Then there is this quotient map $q:[0,1]\to S^1$ that identifies $0$ and $1$. Hence compositing them gives a continuous map $qh:[0,1]\times[0,1]\to S^1$ that is a homotopy from a loop in $S^1$ to the constant map, so does it follow that the loop $f(t)=e^{2\pi it}$ is nullhomotopic? Or which part is wrong?


答:
I think you're mistaking a clopen set, which is both open and closed, with an interval (set) that is half open, and half closed. The two are not the same. Indeed, in this case, the "half-open/half-closed" interval $[0, 1)$ is not clopen in $\mathbb R$: it is neither open nor closed.

Your reasoning about the boundary of $[0, 1)$ is spot-on: it is indeed the set of two elements: $\{0, 1\}$. Now that we know that $[0, 1)$ is not clopen, then, we have no contradiction. The boundary of a clopen set is indeed empty, as you know, but then, since $[0, 1)$ is not clopen, there's not conflict.

The link above (to the entry Wikipedia entry "clopen set") will help clarify what it means to be a clopen set, and you'll find some nice examples to help illustrate such sets.

Title: How is the boundary of the clopen set [0,1) empty?

Body:
I don't get why the boundary of a clopen set is empty. If you take $A = [0,1)$ in $\mathbb R$, then isn't the closure of this the smallest closed super set that contains $A$ which is $[0,1]$.  Isn't the interior, $(0,1)$?  So the boundary would be the closure minus the interior, aka, ${0}$ and ${1}$?

I'm confused. Thanks for all of your great and quick replies!! No wonder I was so confused, I was completely misunderstanding the definition of clopen. I've learned a lot from all of you!


问:
If a $3\times 4$ matrix has $(2,3,1,0)$ spanning its null space, find the $RREF$ of the matrix.
I know that the rank of the matrix must be $3$, since there is exactly one free variable (since exactly one vector spans the null space).

But I am not sure how to reconstruct the exact $RREF$ that would have this vector spanning its null space. All I know is that the last column would be a free variable, and that the other three columns would be linearly independent, since again there is only one variable in the basis of the null space.

I actually would not even think that it is possible to get the $RREF$ with just this information. We can know that $2\times$column$1$ $+ 3 \times $column$2$ $+$column$3$ equals $0$ for all three rows, but how does that uniquely determine a matrix?

答: Observe that the null space and row space of the matrix are orthogonal complements. So $v=[x,y,z,w]^T$ is a vector in the row-space of the matrix iff$$\langle[2,3,1,0]^T,v\rangle=0\iff 2x+3y+z=0\\v=[x,y,z,w]^T=x[1,0,-2,0]^T+y[0,1,-3,0]^T+w[0,0,0,1]^T$$We want to select three linearly independent row vectors so one possible option is$$\begin{bmatrix}1&0&-2&0\\0&1&-3&0\\0&0&0&1\end{bmatrix}$$Note that we cannot determine the matrix uniquely. We may change the order of the row vectors or take some other linearly independent vectors in the row-space and row-reduce the matrix to obtain another valid choice.


问:
Digamma function Identities
I encountered following expression

$$\psi ^{(0)}\left(z+\frac{1}{4}\right)-\psi ^{(0)}\left(z-\frac{1}{4}\right)$$

Searching in different resources I managed to find these identities involving the digamma function $\psi^{(0)}(x)=\frac{\mathrm{d}}{\mathrm{d}x}\ln\Gamma(x)=\frac{1}{\Gamma (x)}\frac{\mathrm{d} \Gamma (x)}{\mathrm{d} x}$.

$\psi ^{(0)}(z+1)-\psi ^{(0)}(z-1)=\frac{1}{z}+\frac{1}{z-1}$

$\psi ^{(0)}\left(z+\frac{1}{2}\right)-\psi ^{(0)}\left(z-\frac{1}{2}\right)=\frac{1}{z-\frac{1}{2}}$

$\psi ^{(0)}\left(z+\frac{1}{4}\right)-\psi ^{(0)}\left(z-\frac{1}{4}\right)=?$

Is there any similar expression for the third case? Or are there any simplification of any other form?

答:
I am not sure that we can simplify more than you did.

However, making the problem more general, we can write
$$\psi ^{(0)}\left(z+\frac{1}{n}\right)-\psi ^{(0)}\left(z-\frac{1}{n}\right)=2\sum_{k=1}^\infty \frac{\psi ^{(2 k-1)}(z) }{n^{(2 k-1)}\,(2 k-1)!}$$

If you make $n=4$ and use only $4$ terms in the summation, you have an almost perfect match.

Title: Digamma function Identities


答:
The easiest way to calculate the 36 cross sums (1/2) the problem is to note that

$$(1 + 2 + 3 + 4 + 5 + 6) \times (1 + 2 + 3 + 4 + 5 + 6) = 441.$$

Therefore, 1/2 of the time, the expected product is

$$\frac{441}{36}.$$

$$1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 91.$$

Therefore, 1/2 of the time, the expected product is

$$\frac{91}{6} = \frac{546}{36}.$$

$$\left(\frac{1}{2}\right) \times \left(\frac{441}{36} + \frac{546}{36}\right)
~=~ \frac{987}{72} ~\approx~ 13.708.$$

Title: Magic dice question.

Body: A pair of dice are magic. When they are rolled together, then half the time, they behave like a normal pair of dice, but the other half of the time they miraculously roll a double ( equally likely to be any of the six possible doubles).  What is the expected value of the product of the two scores when the pair of dice are rolled?</p>


问:
Find the number of integral ordered pairs $(a, b)$ such that $a^2 + b^2 + ab = 1$
I simplified it to $(|a+b|+1)(|a+b|-1) = ab$ because I thought that i could find the no. of ordered pairs by finding the number of divisors of RHS, but the RHS isn't a constant so now I'm stuck. Please help.

Body:
$$a^2 + b^2 + ab = 1\implies b = \frac{\pm\sqrt{4 - 3 a^2} - a}{2}
\quad\text{for}\quad -10\le a \le 10$$

$$\implies (a,b)= (-1,1)\quad  (-1,0)\quad (0,1)\quad  (0,-1)\quad (1,0)\quad  (1,-1)\quad $$


答:
The half-open interval $[0,1)$ is not clopen. A set is clopen if it is both closed and open, and $[0,1)$ is neither closed nor open.

If a set is clopen, then it is open so it is equal to its own interior; and it is closed so it is equal to its closure. The difference between a set and itself is empty.


问:
How can I show that $f$ is bilinear and onto?
Let $\{e_1,e_2,\cdots,e_n\}$ and $\{f_1,f_2, \cdots, f_m \}$ be the standard ordered bases for $\Bbb R^m$ and $\Bbb R^n$ respectively and let $\{E_{11},E_{12},\cdots,E_{mn} \}$ be the standard basis for $\Bbb R^{mn}$ appearing in some order. Consider the map $f : \Bbb R^m \times \Bbb R^n \longrightarrow \Bbb R^{mn}$ defined by $$\sum\limits_{i,j} a_{ij} (e_i,f_j) \longmapsto \sum\limits_{i,j} a_{ij} E_{ij},\ a_{ij} \in \Bbb R.$$

Is this map bilinear? Is $f$ onto? Our instructor suggests that this map is bilinear and onto. Suppose I take $(e_1+e_2,f_1) \in \Bbb R^n \times \Bbb R^m.$ Then how can I show that $f ((e_1+e_2,f_1) = E_{11} + E_{21}\ $? I can't see how it is the case. Can anybody please help me in this regard?

Thanks for your time.

EDIT $:$ Every element of $\Bbb R^n \times \Bbb R^m$ can be written as $\left (\sum\limits_{i=1}^{n} a_i e_i, \sum\limits_{j=1}^{m} b_j f_j \right ).$ So if $f$ acts bilinearly on such elements then we should have $$f \left ( \left (\sum\limits_{i=1}^{n} a_i e_i, \sum\limits_{j=1}^{m} b_j f_j \right ) \right ) = \sum\limits_{i,j} a_i b_j  E_{ij}.$$

Now the question is whether $f$ is onto or not?

RE-EDIT $:$ I don't think $f$ thus defined is onto. For instance let us take $m=n=2.$ Let $\{e_1,e_2\}$ be the standard basis of $\Bbb R^2.$ Let $E_{11} = (1,0,0,0), E_{12} = (0,1,0,0), E_{21} = (0,0,1,0)$ and $E_{22} = (0,0,0,1).$ Now any element of $\Bbb R^2 \times \Bbb R^2$ is of the form $((x,y),(z,w)),$ $x,y,z,w \in \Bbb R.$ Then by the definition of $f$ it follows that $$f((x,y),(z,w)) = (xz,xw,yz,yw).$$ This is clearly not onto since there does not exist any element in $\Bbb R^2 \times \Bbb R^2$ which maps to $(1,2,3,4) \in \Bbb R^4$ under $f.$

Can anybody please check my argument above? Thanks.

If my argument above is fine then the bilinear map $\otimes : \Bbb R^2 \times \Bbb R^2 \longrightarrow \Bbb R^2 \otimes \Bbb R^2 \cong \Bbb R^4$ fails to become onto as well.

答:
The equality $(e_1+e_2,f_1)=(e_1,0)+(e_2,f_1)$ should solve your problems. This is true due to the algebraic properties of $\mathbb{R}^n\times\mathbb{R}^m$.

There does seem to be something wrong with the map $f$ as it is written down in your post. There are no $E_{ij}$ basis vectors of $\mathbb{R}^{nm}$. You have taken the standard basis $\{E_1,\ldots,E_{nm}\}$. So a good map would be $$\sum_{i\leq n}a_i(e_i,0)+\sum_{j\leq m}b_j(0,f_j)\mapsto \sum_{i\leq n} a_i E_i+\sum_{j\leq m} b_jE_{j+n}. $$


问:
A similar (little) Fermat's Theorem result and repunit multiple
I got this exercise in arithmetic class (I'm a french student but let me translate the problem)
 In this thread I only talk about questions from question 2) on the paper.

Let n and p be 2 integers greater or equal than 2 such that $gcd(n,p)=1$. We
want to show that there is $ k \in \mathbb{Z} $ such that $n^k \equiv 1\mod  p  $.
(I find this closed to De Fermat's little theorem)

Q1) Find the smallest k for n=4 and p=7

Q2) If p is prime say something about the existence of k

Q3) p is no more prime Let $r_k$ be the rest of the division
algorithm of $n^k$ by $p$. Show that there is two positive and
different integers i and j such that $r_i =r_j$

Q4) We suppose that $i<j$ . Show that $n^{j-i} \equiv 1 \mod p$

Let n be an odd number that doesn't end by 5.

Q5) Show that $gcd(9n;10)=1$

Q6) Show that there is a multiple of n which
is a repunit number (base 10)

Q7) With n=101 find a multiple of n
which is a repunit (give your approach)

With the pigeonhole principle I answered Q3. For Q4 I calculate $n^{j-i} -1 $ but I don't find any whole numbers times p which give me  $n^{j-i} -1 $.

(I don't know if it's the same symbol for english but '^' stands for the gcd)

For Q5) there is only 4 cases :

$n \equiv 1 \mod 10$

$n \equiv 3 \mod 10$

$n \equiv 7 \mod 10$

$n \equiv 9 \mod 10$

From that that's easy to show that $gcd(9n,10)=1.$

For the Q6) we got the same hypothesis from last proof now we have (9n)^10=1. So it exists an integer l such that $(9n)^l \equiv 1 \mod 10$.
Then there is k in $\mathbb{Z}$ such that $$ (9n)^l = 10k+1$$
I know that for $n \geq 2 $ we have $9U_n = 10^n -1$ with Un the nth repunit number in base 10.
But maybe this is obvious but I don't manage to show that there is a multiple of n which is a repunit number (always base 10).

So if you can help me with this problem, please.

(For Q7 with a smalll python programm I've got 101*11 is a repunit number)

EDIT 1 I changed the french notation to be more comprehensible

答:
For Q4,
\begin{align*}
n^i n^{j-i} = n^j \equiv n^i &\mod p\\
\implies n^i(n^{j-i}-1) \equiv 0 &\mod p
\end{align*}
using the fact that $\operatorname{gcd}(n,p)=1$ you should be able to finish from there.

For Q6, applying the result the other way (taking $10$ to the power instead of $9n$) yields
$$10^k -1 \equiv 0 \mod 9n,$$ which bears a closer resemblance to the formula you want to apply. Does this help?


问:
I get $ \int fg' \, dx = 2fg + \int fg'\,dx + C $ while performing integration by parts. What did I do wrong?
I am reaching a paradox by using integration by parts. I must be going wrong but unable to figure where I went wrong. Can you help me understand what I did wrong here?

We have two differentiable functions $ f(x) $ and $ g(x) $. Now we start with $ \int f(x) g'(x) dx $ and proceed like this:

$$
\int fg' \, dx = fg + \int f'g \, dx + C = fg + fg + \int fg'\,dx + C = 2fg + \int fg'\,dx + C.
$$

What went wrong here?

Update:

As told in the first comment to this post I have made a very silly error. Here is the fix now:

$$
\int fg' \, dx = fg - \int f'g \, dx + C = fg - fg + \int fg'\,dx + C = \int fg'\,dx + C.
$$

So there is no issue.


问:
Galois Group of a Polynomial Problem help
Can someone help me find the Galois group of the following polynomial:

$f = x^5 + x^4 + x^3 + x, f \in \Bbb Z_2[X]$

Don't know how to use LaTeX here, so forgive me about that.

答:
You can factor that polynomial as

$$f(x) = x(1 + x^2 + x^3 + x ^4) = x(x+1)(1+x+x^3)$$

$(1+x+x^3)$ is irreducible because it has no roots (and if it was reducible one of its factors would have degree $1$).

Let $\alpha$ be a root of $1 + x + x^3$. Then, $[\mathbb{F}_2 (\alpha): \mathbb{F}_2] = 3$. Let $L$ be the splitting field of $f$, which is also the splitting of $1 + x + x^3$.

Now observe the following:

$$(\alpha^2)^3 + \alpha^2 + 1 = (1 + \alpha + \alpha^3)^2 = 0$$

Since $1 + x + x^3$ is the minimal polynomial of $\alpha$; $\alpha^2 \neq \alpha$. Hence, we get $2$ roots of $1 + x + x^3$ in $\mathbb{F}_2 (\alpha)$, so all three of them must be there.

The Galois group has the same order as the degree of the splitting field, so it has order $3$, which means it is $C_3$.


问:
Symmetrizability of conservation laws via mathematical entropy
Let
$$
\frac{\partial \boldsymbol{u}}{\partial t}
+
\sum_{j=1}^d \frac{\partial}{\partial x_j} \boldsymbol{f}_j(\boldsymbol{u})
=
\boldsymbol{0},
$$
be a system of conservation laws and let us assume that we are able to find an additional conservation law
$$
\frac{\partial U(\boldsymbol{u})}{\partial t}
+
\sum_{j=1}^d \frac{\partial}{\partial x_j} F_j(\boldsymbol{u})
=
0,
$$
where $U$ is a strictly convex function. (This is common for many physical system.) Can we now say that the function $U$ is a strictly convex mathematical entropy and conclude that the original system of conservation laws is symmetrizable (and thus locally well-posed)?

In Godlewski, Raviart: Numerical Approximation of Hyperbolic Systems of Conservation Laws (1996) the authors say that a convex function $U$ is a mathematical entropy if there exist functions $F_j$ (entropy fluxes) such that
$$
U'(\boldsymbol{u})\boldsymbol{f}_j'(\boldsymbol{u})
=
F_j'(\boldsymbol{u}).
$$
If this condition holds then, of course, the additional conservation law is satisfied. However, this does not work the other way around. And yet, in Examples 3.1 and 3.2 the authors only check whether the additional conservation law, and not the condition above, holds. Could someone clear this up for me? Is it always sufficient to find an additional conservation law to conclude symmetrizability?

答:
Let us follow the paper by Friedrichs and Lax [1]. Multiplying scalarly the system of conservation laws by the gradient $U'(\boldsymbol u) = \partial U/\partial \boldsymbol u$ of the entropy gives
$$
U'(\boldsymbol u) \frac{\partial \boldsymbol u}{\partial t} + \sum_{j=1}^d U'(\boldsymbol u) \frac{\partial}{\partial x_j} {\boldsymbol f}_j(\boldsymbol u) = \boldsymbol 0
$$
and the chain rule gives
$$
\frac{\partial U(\boldsymbol u)}{\partial t} + \sum_{j=1}^d \frac{\partial {F}_j(\boldsymbol u)}{\partial x_j} = \boldsymbol 0
$$
where $F'_j = U' {\boldsymbol f}'_j$. Now, if we multiply the system of conservation laws by the hessian matrix $U''(\boldsymbol u)$ of the entropy, we find $$
U''(\boldsymbol u) \frac{\partial \boldsymbol u}{\partial t} + \sum_{j=1}^d U''(\boldsymbol u) {\boldsymbol f}'_j(\boldsymbol u) \frac{\partial \boldsymbol u}{\partial x_j} = \boldsymbol 0 .
$$
By virtue of convexity, the matrix $U''$ is symmetric positive definite. Moreover, the matrices $U'' {\boldsymbol f}'_j$ are symmetric. Therefore, the above system is symmetric hyperbolic, and the initial system is symmetrizable.

Friedrichs, K. O., & Lax, P. D. (1971). Systems of conservation equations with a convex extension. Proceedings of the National Academy of Sciences, 68(8), 1686-1688. doi:10.1073/pnas.68.8.1686


问:
How is the boundary of the clopen set [0,1) empty?
I don't get why the boundary of a clopen set is empty. If you take $A = [0,1)$ in $\mathbb R$, then isn't the closure of this the smallest closed super set that contains $A$ which is $[0,1]$.  Isn't the interior, $(0,1)$?  So the boundary would be the closure minus the interior, aka, ${0}$ and ${1}$?

I'm confused. Thanks for all of your great and quick replies!! No wonder I was so confused, I was completely misunderstanding the definition of clopen. I've learned a lot from all of you!

答:
$[0,1)$ is not clopen as a subset of $\mathbb{R}$.

A clopen set is by definition a set that is both open and closed. Hence, the closure as well as the interior would be equal to the set itself, leaving an empty boundary.

As you already remarked, the interior of $[0,1)$ is $(0,1)\neq[0,1)$, hence $[0,1)$ is not open.
Furthermore, the closure of $[0,1)$ is $[0,1]\neq[0,1)$, hence $[0,1)$ is not closed either. Thus $[0,1)$ is certainly not clopen.


问:
Arc length of ellipse in polar coordinates
I have the equation of an ellipse given in Cartesian coordinates as
$\left(\frac{x}{0.6}\right)^2+\left(\frac{y}{3}\right)^2=1$ .

I need the equation for its arc length in terms of $\theta$, where $\theta=0$ corresponds to the point on the ellipse intersecting the positive x-axis, and so on.

So converting to polar coordinates with the substitutions
\begin{Bmatrix}
x=r\cos\theta\\y=r\sin\theta
\end{Bmatrix}
gives $r(\theta)=\frac{1.8}{\sqrt{9\cos^2\theta+0.36\sin^2\theta}}$ , as is illustrated here on Desmos: https://www.desmos.com/calculator/h27qsdnotm.

Then substituting into the arc length formula for polar equations:
$L=\int\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} \ d\theta$
gives the rather ugly integral $L=\int\sqrt{\frac{1.8^2}{9\cos^2\theta+0.36\sin^2\theta}+\frac{1944^2\sin^2\theta\cos^2\theta}{(216\cos^2\theta+9)^3}}$ . The online https://www.integral-calculator.com/ was unable to determine an antiderivative, so my question is whether or not the above expression can be expressed in terms of elementary functions, or if there are other methods of finding the equation of the arc length of an ellipse in polar coordinates with respect to $\theta$.

答:
It is no good idea to convert to polar coordinates, it just makes things more difficult: the polar equation of an ellipse is not simple.

Stick to the parametric equation and use

$$\int\sqrt{\dot x^2+\dot y^2}\,d\theta=\int\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}\,d\theta.$$

It turns out that this integral does not have a closed form and a special function was introduced to represent it: the... elliptic integral !

https://www.wolframalpha.com/input/?i=integrate+sqrt%28a%5E2+cos%5E2+theta%2Bb%5E2+sin%5E2+theta%29


问:
sum of $9$ consecutive natural numbers and $10$ consecutive natural numbers are equal

Find the smallest natural number which can be expressed as sum of $9$
consecutive natural numbers and also $10$ consecutive natural numbers.

I considered these sequences of natural numbers: $x+1,\cdots,x+9\quad\text{and}\quad y+1,y+2,\cdots,y+10\quad$ therfore :

$$9x+45=10y+55=a\quad\text{where}\quad x,y,a\in\mathbb{N}$$

But I don't know how to find minimum of $a$.

答:
Note that $9x+45 = 9(x+5)$ is a multiple of $9$ at least as large as $54$, and $10y + 55 = 5(2y+11)$ is a multiple of $5$ but not $10$ (why?) and at least as large as $55$.

The smallest number fitting both these criteria must be the multiple of $9$ ending with $5$. Since $9$ and $5$ are co-prime, these are just odd multiples of $45$ (so they don't end with a $0$). Then we just have $45 \times 3 = 135$ as the smallest number fitting the criteria.

Indeed, $135 = 11 + 12 + ... +18+19$ and $135 = 9+10+... + 17+18$.


答:
Let those numbers be $$x+1,x+2,...,x+9$$ and $$y,y+1,...,y+9$$ Then, $$a = 9x + 45 = 10y + 45 \\ 9x = 10y \implies x = \frac{10y}9$$
Since $\gcd(9,10)=1$, it follows that $9$ divides $y$. Minimum such natural $y$ is $9$. Hence $$a=10\cdot 9 + 45 = 135$$

Title: sum of $9$ consecutive natural numbers and $10$ consecutive natural numbers are equal

Body:
Find the smallest natural number which can be expressed as sum of $9$
consecutive natural numbers and also $10$ consecutive natural numbers.

I considered these sequences of natural numbers: $x+1,\cdots,x+9\quad\text{and}\quad y+1,y+2,\cdots,y+10\quad$ therfore :

$$9x+45=10y+55=a\quad\text{where}\quad x,y,a\in\mathbb{N}$$

But I don't know how to find minimum of $a$.


答:
$9x=10y+10\implies y+1\equiv0\mod 9$ so $y=9k-1,k\in\Bbb Z^+$. The smallest such $y$ is $9-1=8$ for which we get $x=10$ and $a=135$.

Note that if we take $0\in\Bbb N$, then the smallest permissible $y$ is $9(0)-1=-1$ for which we get $x=0$ and $a=45$.


问:
What's the Lie algebra associated to positive-defined Hermitian matrices?
I'm studying polar decomposition $GL(n,\Bbb C)\cong U(n)\times Herm^+(n)$, where $Herm^+(n)$ are positive-defined Hermitian matrices. I was trying to understand what happens to the associated Lie algebras, and I was wondering what $Herm^+(n)$ Lie algebra would look like.

What I tried:
I know that considering $F:SL(n,\Bbb R)\to SL(n,\Bbb R):A\mapsto A^\top A$, since $SO(n)=F^{-1}(I_n)$, we can prove that $\mathfrak{so}(n)$ is the space of skew-symmetric matrices by calculating $$\mathfrak{so}(n)=\ker (dF_{I_n})=\left\{A\in\Bbb R^{n\times n} : \left. \frac d{dt}\right\vert_{t=0}F(I_n+tA)=0\right\}.$$
I wondered if I could do something similar for $Herm(n)$, defining $G:\Bbb C^{n\times n}\to\Bbb C^{n\times n}:A\mapsto A^*-A$ and $Herm(n)=G^{-1}(O_n)$, but I then realised that being $Herm(n)$ a linear subspace of $\Bbb C^{n\times n}$ I would just find $Herm(n)$ again. Is this a correct answer? Is $Herm(n)$ itself its own associated Lie algebra (with the commutator as Lie bracket)?

And what about positive-defined ones? What useful characterization may I take advantage of?

Thanks in advance!

答:
Hermitian matrices (positive-definite or otherwise) aren't closed under multiplication so they aren't a Lie group and don't have a Lie algebra. They also aren't closed under commutator bracket so they aren't themselves a Lie algebra either. However, they are $i$ times the skew-Hermitian matrices $\mathfrak{u}(n)$, which do form a Lie algebra (this requires that we ignore positive-definiteness), and we do have

$$\mathfrak{gl}_n(\mathbb{C}) \cong \mathfrak{u}(n) \oplus i \mathfrak{u}(n).$$

In other words, every matrix $M$ is uniquely the sum of a skew-Hermitian matrix $\frac{M - M^{\dagger}}{2}$ and a Hermitian matrix $\frac{M + M^{\dagger}}{2}$. This is the Lie algebra incarnation of polar decomposition; despite not being a Lie algebra, there is still an exponential map from Hermitian matrices to positive-definite Hermitian matrices.


问:
Prove $f(x)$ is constant.
Suppose $ f: \mathbb{R} \mapsto \mathbb{R}$ be continous. For any interval $[a,b]$, there exists $x_0\in (a,b)$ such that either $f(x_0)=\max\limits_{a\le x \le b} f(x)$ or $f(x_0)=\min\limits_{a\le x \le b} f(x)$. Prove $f(x)$ is constant.

We may consider apply Reductio ad Absurdum. Suppose $f(x)$ is not constant, then there exist $\alpha<\beta$ such that $f(\alpha)\neq f(\beta).$ But how to reduce the contradiction?

答:
Not sure of my approach, this is very equivalent to @Ege Erdil answer. Let us consider the interval $[\alpha, \alpha + \epsilon]$ and suppose $f$ non constant, in the way $f(\alpha) < f(\alpha + \epsilon)$.

According to the given condition, $\exists x_0 \in (\alpha, \alpha + \epsilon) : f(x_0) = \max_{x \in [\alpha, \alpha + \epsilon]} f(x) \vee f(x_0) = \min_{x \in [\alpha, \alpha + \epsilon]} f(x)$.

As $f$ is continuous, we can take the limit of this relation where $\epsilon \to 0$:

$$\lim_{\epsilon \to 0} \max_{x \in [\alpha, \alpha + \epsilon]} f(x) = f(\alpha + \epsilon)$$

$$\lim_{\epsilon \to 0} \min_{x \in [\alpha, \alpha + \epsilon]} f(x) = f(\alpha)$$

We see that $\alpha + \epsilon \not\in (\alpha, \alpha + \epsilon)$, and $\alpha \not\in (\alpha, \alpha + \epsilon)$, hence contradicting the given condition. The same reasoning can be done for $f(\alpha) > f(\alpha + \epsilon)$. It comes that $f$ must be constant.

Title: Prove $f(x)$ is constant.


答:
I guess we’re not including zero as a natural number here or the answer would be very simple (45).

So the sum of a sequence of consecutive natural numbers is average times count. We know the count and the average will be an integer for odd-length sequences and half more than an integer for even length sequences - we can bring that back to an integer by doubling the average to an odd number and halving the count, but the result for a 10-length sequence will be odd

Thus we know that our answer is divisible by $9$ and by $5$. Discarding $45$ the next is $90$, which is not odd so we move on to $135$. This implies that the average is $15$ for the length-9 sequence and $13.5$ for the length-10. So
$$ 11+12+13+14+15+16+17+18+19 = 8+9+10+11+12+13+14+15+16+17 =135$$


问:
How to find the unknown values in this Numerical Integration type?
Given the following type of numerical integration:

$$I(f)=\int_0^1 f(x) \, dx  \approx  \frac 12 f(x_{0}) +c_1 f(x_1) $$

a) Find the values of: the coefficient $c_1$ and points $x_0$ and $x_1$ so that the above formula numerical integration to be as accurate as possible.

b)Find the error of the formula of the numerical integration in (a).

c)What is the degree of accuracy of the numerical integration formula in (a)?

答:
We need to interpret "as accurate as possible." One common interpretation is that it is dead on for $f(x)$ identically equal to $1$, for $f(x)=x$, for $f(x)=x^2$, and so on for as long as possible.

If the formula is to give the right answer for $f(x)=1$ (and hence for $f(x)$ any constant function) we need $c_1=\frac{1}{2}$.

For the formula to give the right answer for $f(x)=x$, we then need
$\frac{1}{2}(x_0+x_1)=\int_0^1 x\,dx=\frac{1}{2}$. So we need $x_0+x_1=1$.

For the formula to give the right answer for $f(x)=x^2$, we  need
$\frac{1}{2}(x_0^2+x_1^2)=\int_0^1 x^2\,dx=\frac{1}{3}$. So we need $x_0^2+x_1^2=\frac{2}{3}$.

Solve the equations $x_0+x_1=1$, $x_0^2+x_1^2=\frac{2}{3}$  for $x_0$ and $x_1$. We get $x_0=\frac{1}{2}\left(1-\frac{\sqrt{3}}{3}\right)$ and $x_1=\frac{1}{2}\left(1+\frac{\sqrt{3}}{3}\right)$.

Remarks: $1.$ This little problem connects with important mathematics. For detail, look for Gaussian quadrature. It turns out that by choosing suitable not uniformly distributed points, and suitable weights, one can produce numerical integration procedures that are extremely efficient. This is particularly important in situations where function evaluation is "expensive."

$2.$ We did not try to get the right answer for $x^3$, since obviously $x_0$ and $x_1$ were determined when we dealt with $x^2$. However, we get a nice bonus! It turns out the formula is exact for $x^3$. This is not hard to show, all we need to do is to verify that $\frac{1}{2}(x_0^3+x_1^3)=\frac{1}{4}$. That is a short calculation.

So the formula gives dead on right answers for $1$, $x$, $x^2$, and $x^3$. Thus, by linearity the formula  is dead on for all polynomials of degree $\le 3$.


问:
Equivalent condition of the following group presentation to give a group of order $mn$; the other side
Let $G$ be a group (not necessary abelian) generated by $a, b$ that has a group presentation: $$\langle a, b \mid a^m = 1, b^n = 1, ba = a^rb\rangle. $$In my question Equivalent condition of the following group presentation to give a group of order $mn$ , I was able to understand that if $G$ is of order $mn$ then $r^n = 1 \pmod{m}$. I thought that I have proven the opposite side (if $r^n = 1 \pmod{m}$, then $G$ is of order $mn$), but my proof was erroneous. Especially I'm having a difficulty proving that all $a^xb^y$s are distinct for distinct $(x,y)$s, so that $G$ will have $mn$ distinct elements(I know that all elements in $G$ can be written in the form of $a^xb^y$, because of $ba = a^rb$ and Prove that the following equality holds in a group). How can I prove this?

答:
$\renewcommand{\phi}{\varphi}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$You can do it by bounding the group from below.

Construct the semidirect product $S$ of a cyclic group $\Span{\alpha}$ of order $m$ by a group $\Span{\beta}$ of order $n$, with $\beta$ acting on $\Span{\alpha}$ by raising to the $r$-th power. This is a group of order $m n$.

Let $G$ be your group. Now von Dyck's theorem yields that there is a unique homomorphism
\begin{align}
\phi :\ &G \to S
\end{align}
such that $\phi(a) = \alpha$ and $\phi(b) = \beta$, and $\phi$ is clearly surjective. Since the $\alpha^{x} \beta^{y}$ are distinct, for $0 \le x < m$ and $0 \le y < n$, so are the corresponding $a^{x} b^{y}$.


问:
$S^{n + m}$ is not homeomorphic to $M \times N$
Let $M$ and $N$ be two oriented manifolds such that $m = \dim M$ and $n = \dim N$. Using the de Rham cohomology I have to show that $S^{n + m}$ is not homeomorphic to $M \times N$. I have no idea how to proceed, but maybe the next statement could be helpful: if $M$ is an oriented and compact manifold such that $m = \dim M$, then $\dim H^m(M) \geq 1$. I think I have to use just geometry and topology statements, not analysis theorems like Fubini.

答: I think an idea would be to use the Kunneth Formula applied to the cohomology groups of $H^k(M\times N)$ and the fact that we know the cohomology groups of the sphere,i.e they are trivial except for the $0-$degree and the top degree, and that this are homeomorphic invariant to get a contradiction.


问:
Cubic diophantine equation in two variables
I am trying to solve the Diophantine equation $xy^2 + 2xy + x - 243y = 0$.

I simplified it to $x(y^2 +2y +1) = 243y$ but I am stuck on what to do now.  Any help would be appreciated.

Body:
$$xy^2 + 2xy + x - 243y =(x)y^2+(2x-343)y+x=0 \implies x=0,y=0$$
$$y = \frac{(243-2x) \pm 9 \sqrt{3} \sqrt{243 - 4 x}}{(2 x)}
\quad\text{for}\quad -1500\le x \le 1500\\
\implies (x,y)=(-486,-2)\quad (-108,-4)\quad (-30,-10)\quad
(24,8)\quad (54,2)\quad $$


答: I assume that $M,N$ are manifolds of dimensions $m,n>0$ (this part is obviously necessary, otherwise we would get trivial counterexamples as $S^n\simeq \{0\}\times S^n)$. If $M\times N\cong S^{n+m}$ then $M,N$ are compact and orientable (why?). Now notice that $\dim H^m(M)>0$ and $\dim H^0(N)>0$, so by the Künneth formula we get $\dim H^m(M\times N)\geq \dim H^m(M)\otimes_{\mathbb{R}}H^0(N)>0$, but what is $\dim H^m(S^{n+m})$?

Body: Let $M$ and $N$ be two oriented manifolds such that $m = \dim M$ and $n = \dim N$. Using the de Rham cohomology I have to show that $S^{n + m}$ is not homeomorphic to $M \times N$. I have no idea how to proceed, but maybe the next statement could be helpful: if $M$ is an oriented and compact manifold such that $m = \dim M$, then $\dim H^m(M) \geq 1$. I think I have to use just geometry and topology statements, not analysis theorems like Fubini.


问:
Is it possible for a function to not have a decreasing and increasing interval? If so, how do we show it mathematically?
Is it possible for a function to not have a maxima or a minima? (S.t. I can't find the decreasing and increasing interval.) If so, how do we show it mathematically?

I was practicing and found these two functions.

$a. f(x) = x+\sqrt{x^2-1} $ and $b. f(x) = \frac{x^2}{x^2+4} $

$a. f'(x) = 1 + \frac{x}{\sqrt{x^2-1}} = 0$

However, I can't find the extrema.

I tried to multiply with $\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}$ and obtain
$x\sqrt{x^2-1}=-x^2+1$

But I still can't find the extrema.
If I only paid attention to $\frac{1}{\sqrt{x^2-1}}$, I'll obtain $x=-1$ or $x=1$.

Thus, $f(-1)=-1$ as the minimum value and $f(1)=1$ as the maximum value.

But I think this is wrong.

$b. f'(x) = \frac{2x(x^2+4)-x^2(2x)}{(x^2+4)^2}=0$

$x=0$

$---0+++$

So I know that $0$ will give us the minimum value. But what about the maxima?

$f''(x)=\frac{8(-3x^2+4)}{(x^2+4)^3}$

$f''(0)=\frac{1}{2} >0$ So 0 will really gives the minimum.

So, the function decreases when x<0 and increases when x>0.

答:
For the first problem, $1 + \frac{x}{\sqrt{x^2-1}} = 0$ gives $$\frac{x}{\sqrt{x^2-1}}=-1\\x=-\sqrt{x^2-1}\\x^2=x^2-1$$ which has no solution, so the derivative never vanishes.  That means that the function has no extremum at a point of differentiability, but we still have to consider other points.

The natural domain of the function is $|x|\geq1$ and the function fails to be differentiable at $x=\pm1$.  The function takes its absolute minimum value of $-1$ at $x=-1$, which you can show by proving that $f$ decreases when $x<-1$.  (Obviously, $f(x)>0$ for $x\geq1$.)


答:
a. As you wrote, $f'(x)=1+\frac x{\sqrt{x^2-1}}$. It turns out that $f'(x)<0$ if $x<-1$ and that $f'(x)>0$ if $x>1$. So, $f$ is strictly decreasing on $(-\infty,-1]$ and strictly increasing on $[1,\infty)$. So, since $f(-1)=-1$ and since $f(1)=1$, the minimum of $f$ is $-1$ and it has no maximum.

$b.$ Note that$$f(x)=1-\frac4{x^2+4}.$$So, since $x^2+4$ is strictly increasing on $[0,\infty)$ and strictly decreasing on $(-\infty,0]$, $f$ is strictly decreasing on$[0,\infty)$ and strictly increasing on $(-\infty,0]$. Therefore, its minimum is $0(=f(0))$ and it has no maximum.

Title: Is it possible for a function to not have a decreasing and increasing interval? If so, how do we show it mathematically?

Body:
Is it possible for a function to not have a maxima or a minima? (S.t. I can't find the decreasing and increasing interval.) If so, how do we show it mathematically?

I was practicing and found these two functions.

$a. f(x) = x+\sqrt{x^2-1} $ and $b. f(x) = \frac{x^2}{x^2+4} $

$a. f'(x) = 1 + \frac{x}{\sqrt{x^2-1}} = 0$

However, I can't find the extrema.

I tried to multiply with $\frac{\sqrt{x^2-1}}{\sqrt{x^2-1}}$ and obtain
$x\sqrt{x^2-1}=-x^2+1$

But I still can't find the extrema.
If I only paid attention to $\frac{1}{\sqrt{x^2-1}}$, I'll obtain $x=-1$ or $x=1$.

Thus, $f(-1)=-1$ as the minimum value and $f(1)=1$ as the maximum value.

But I think this is wrong.

$b. f'(x) = \frac{2x(x^2+4)-x^2(2x)}{(x^2+4)^2}=0$

$x=0$

$---0+++$

So I know that $0$ will give us the minimum value. But what about the maxima?

$f''(x)=\frac{8(-3x^2+4)}{(x^2+4)^3}$

$f''(0)=\frac{1}{2} >0$ So 0 will really gives the minimum.

So, the function decreases when x<0 and increases when x>0.


问:
How can I derive the Legendre function of first kind in terms of the hypergeometric function?
I was reading in Wikipedia about Legendre's differential equation. I was particularly interested in the simple case of the equation given by
$$
\left(1-x^2\right)y'' -2xy' + \lambda(\lambda+1)y = 0 \tag{1}
$$
In the article, it is stated that one of the two linearly independent solutions to $(1)$ is given by
$$
P_{\lambda}(x) =  \,_2F_1 \left(-\lambda, \lambda+1; 1; \frac{1-x}{2}\right) \tag{2}
$$
where $\,_2F_1$ represents the hypergeometric function. Since by definition, the hypergeometric function ${}_2F_1(a,b;c;t)$ is a solution to the hypergeometric differential equation given by
$$
t(1-t) \ \ddot{y} + \left[c-(a+b+1)t \right] \dot{y} - ab\,y = 0 \tag{3}
$$
if $(2)$ is a solution to $(1)$ then I should be able to re-write $(1)$ to look like $(3)$ with the corresponding parameters. With this in mind, taking
\begin{align*}
a &= - \lambda\\
b &= \lambda +1\\
c &= 1\\
t & =  \frac{1-x}{2}
\end{align*}
From where we also get that
$$
\dot{y} = \frac{dy}{dt} =  \frac{dy}{dx} \cdot  \underbrace{\frac{dx}{dt}}_{\color{blue}{-2}} = -2y'
$$
which also tells us that
$$
\ddot{y} = -2 y''
$$
Substituting this into $(3)$ gives
\begin{align}
\frac{1-x}{2}\left(1-\frac{1-x}{2}\right)\ddot{y} + &\left[1-\left(-\lambda+(\lambda +1)+1\right)\frac{1-x}{2} \right] \dot{y} - (-\lambda)(\lambda +1)y = 0\\
\implies  \frac{1}{4}\left(1-x^2 \right)\left(-2 y''\right) +& \left[1-2\frac{1-x}{2} \right] \left(-2 y'\right) +\lambda(\lambda +1)y = 0\\
\implies \left(1-x^2 \right)y'' &+ 4x y' -2 \lambda(\lambda +1)y = 0 \tag{4}
\end{align}
and although $(4)$ looks really similar to $(1)$, they're not the same equation.

Could anyone tell me where did I make a mistake? Or alternatively, does anyone know another simple way to show that the Legendre function can be written in terms of the hypergeometric function? Thank you!

答:
In the proposed derivation, it is correctly establish that $\dot y=-2y'$ but we have also $\ddot y=4y''$ (and not $\ddot y=-2y''$ as stated).
With this correction, the ODE are identical.


答:
Suppose that there exist $a$ and $b$  such that $f(a) \neq f(b)$. Let $y:=\max \{x \in [a,b] | f(x) =f(a)\}$ and $z:=\min \{x\in [y,b]| f(x)=f(b)\}$, then $y<z$.

Note that if $x \in (y,z)$ then $f(x) \neq f(a) \neq f(b)$ .

But $f$ attains an extremal value in $(y,z)$, WLOG assume that its the minimum. Then there exist a $d$ such that  $f(d)\leq\min \{f(a),f(b)\}<\max \{f(a),f(b)\}$ so by the IVT there exist a $c\in (y,z)$ such that $f(c)=\min \{f(a),f(b)\}$ a contradiction.

Title: Prove $f(x)$ is constant.