问:
Arctan and Log Integration Problem
I have been trying to evaluate this integral: $\displaystyle \int_0^{\infty} \frac{\arctan^2 x \log^2 (1+x^2)}{x^2}\,dx$, but have failed despite all my attemps. I tried using the trig substitution $x = tan(a)$, and through a series of steps, was able to simplify the integral to $I(b)$ = $4\displaystyle \int_0^{\pi/2} \frac{(a\log(\cos(a))^b}{(\sin(a))^b}\,da$. I then tried to differentiate under the integral sign but failed. What should I do now?

Note: b = 2 in the latter integral.

Title: Arctan and Log Integration Problem

Body:
Let's start integrating by parts:
$$\int_0^\infty\frac{\arctan^2\left(x\right)\log^2\left(1+x^2\right)}{x^2}dx=$$
$$\underbrace{-\left[\frac{\arctan^2\left(x\right)\log^2\left(1+x^2\right)}{x}\right]^\infty_0}_{0}+\underbrace{4\int_0^\infty\frac{\arctan^2\left(x\right)\log\left(1+x^2\right)}{1+x^2}dx}_{I_1}+\underbrace{2\int_0^\infty\frac{\arctan\left(x\right)\log^2\left(1+x^2\right)}{x(1+x^2)}dx}_{I_2}$$

To solve $I_1$, let $x\rightarrow \frac{y}{\sqrt{1-y^2}}$. Then, let's rewrite $\arctan(x)$ as $\arcsin\left(\frac{x}{\sqrt{1+x^2}}\right)$ and apply the series expansion of $\arcsin^2(x)$:

$$I_1=-4\int_0^1\frac{\arcsin^2\left(y\right)\log\left(1-y^2\right)}{\sqrt{1-y^2}}dy=-2\sum_{n=1}^{\infty}\frac{\left(2^nn!\right)^2}{n^2(2n)!}\int_0^1\frac{y^{2n}\log\left(1-y^2\right)}{\sqrt{1-y^2}}dy$$

Bearing in mind that:
$$\frac{\partial}{\partial b}\int_0^1t^{a-1}\left(1-t\right)^{b-1}dt=\frac{\partial}{\partial b}\mathfrak{B}(a,b)$$
$$\int_0^1t^{a-1}\left(1-t\right)^{b-1}\log\left(1-t\right)dt=\mathfrak{B}(a,b)\left[\mathfrak{\psi^{(0)}}(b)-\mathfrak{\psi^{(0)}}(a+b)\right]=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+b\right)}\left[\mathfrak{\psi^{(0)}}(b)-\mathfrak{\psi^{(0)}}(a+b)\right]$$
and
$$\Gamma\left(z\right)\Gamma\left(z+\frac{1}{2}\right)=2^{1-2z}\sqrt{\pi}\ \Gamma\left(2z\right)$$

$$I_1=-\sum_{n=1}^{\infty}\frac{\left(2^nn!\right)^2}{n^2(2n)!}\frac{\Gamma\left(n+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(n+1\right)}\left[\mathfrak{\psi^{(0)}}\left(\frac{1}{2}\right)-\mathfrak{\psi^{(0)}}\left(n+1\right)\right]=-\pi\sum_{n=1}^{\infty}\frac{\mathfrak{\psi^{(0)}}\left(\frac{1}{2}\right)-\mathfrak{\psi^{(0)}}\left(n+1\right)}{n^2}$$

$$=-\pi\sum_{n=1}^{\infty}\frac{-2\log(2)-\gamma+\gamma-H_n}{n^2}=2\pi\zeta(2)\log(2)+\pi\sum_{n=1}^{\infty}\int_0^1\frac{dx}{1-x}\frac{1-x^n}{n^2}$$

$$=2\pi\zeta(2)\log(2)+\pi\int_0^1\frac{\zeta(2)-Li_2(x)}{1-x}dx\\=2\pi\zeta(2)\log(2)-\pi\underbrace{\left[\left(\zeta(2)-Li_2(x)\right)\log\left(1-x\right)\right]^1_0}_{0}+\pi\underbrace{\int_{0}^{1}\frac{\log^2\left(1-x\right)}{x}dx}_{2\zeta\left(3\right)}$$
$$\boxed{I_1=2\pi\zeta\left(2\right)\log(2)+2\pi\zeta\left(3\right)}$$

To solve $I_2$, let $x\rightarrow {\frac{1}{x}}$:
$$I_2=\underbrace{\int_{0}^{\infty}\frac{2x\arctan\left(\frac{1}{x}\right)\left(\log^2{\left(1+x^2\right)-4\log(x)\log(1+x^2)}\right)}{1+x^2}dx}_{A}+\underbrace{8\int_{0}^{\infty}{\frac{x\arctan\left(\frac{1}{x}\right)\log^2{\left(x\right)}}{1+x^2}dx}}_{B}$$

$$A=\underbrace{\left[\frac{\arctan\left(\frac{1}{x}\right)\log^3\left(1+x^2\right)}{3}-2\arctan\left(\frac{1}{x}\right)\log^2\left(1+x^2\right)\log(x)\right]_{0}^{\infty}}_{0}+\\\frac{1}{3}\int_{0}^{\infty}\frac{\log^3{\left(1+x^2\right)}}{1+x^2}dx-2\int_{0}^{\infty}{\frac{\log^2{\left(1+x^2\right)}\log{\left(x\right)}}{1+x^2}dx}+2\int_{0}^{\infty}{\frac{\arctan\left(\frac{1}{x}\right)\log^2{\left(1+x^2\right)}}{x}dx}$$

Let $x\rightarrow \sqrt\frac{y}{1-y}$:

$$A=\int_{0}^{1}\frac{\frac{1}{3}\log^3{\left(1-y\right)}-\frac{1}{2}\log^2{\left(1-y\right)}\log{\left(y\right)}}{\sqrt{y\left(1-y\right)}}dy+\int_{0}^{1}{\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{y(1-y)}dy}$$
Notice that:

$$\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{y(1-y)}dy=\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{y}dy+\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{1-y}dy\\=\frac{\pi}{2}\underbrace{\int_{0}^{1}\frac{\log^2{\left(1-y\right)}}{y}dy}_{2\zeta(3)}-\underbrace{\int_{0}^{1}\frac{\arcsin(\sqrt{y})\log^2{\left(1-y\right)}}{y}dy}_{y\rightarrow \frac{x^2}{1+x^2}}+\int_{0}^{1}\frac{\arccos(\sqrt{y})\log^2{\left(1-y\right)}}{1-y}dy\\=\pi\zeta(3)-\underbrace{2\int_0^\infty\frac{\arctan\left(x\right)\log^2\left(1+x^2\right)}{x(1+x^2)}dx}_{I_2}-\underbrace{\left[\frac{\arccos(\sqrt{y})\log^3\left(1-y\right)}{3}\right]^1_0}_{0}-\frac{1}{6}\int_{0}^{1}\frac{\log^3{\left(1-y\right)}}{\sqrt{y(1-y)}}dy$$

Therefore
$$A=\pi\zeta(3)-I_2+\frac{1}{6}\lim_{a\rightarrow 1/2\\ b\rightarrow 1/2}\frac{\partial^3}{\partial^3a}\mathfrak{B}(a,b)-\frac{1}{2}\lim_{a\rightarrow 1/2\\ b\rightarrow 1/2}\frac{\partial^3}{\partial^2a\partial b}\mathfrak{B}(a,b)$$

$$A=\pi\zeta(3)-I_2+\left(-2\pi\zeta(3)-\frac{4}{3}\log^3(2)-2\pi\zeta(2)\log(2)\right)-\left(\pi\zeta(3)-4\pi\log^3(2)\right)$$

$$\boxed{A=-2\pi\zeta(3)-2\pi\zeta(2)\log(2)+\frac{8}{3}\log^3(2)-I_2}$$

$$B=8\int_{0}^{1}dy\int_{0}^{\infty}{\frac{x^2\log^2{\left(x\right)}}{\left(1+x^2\right)\left(1+y^2x^2\right)}dx}=8\int_{0}^{1}\frac{dy}{1-y^2}\int_{0}^{\infty}\left(\frac{\log^2{\left(x\right)}}{1+x^2}-\frac{y^2\log^2{\left(x\right)}}{y^2+x^2}\right)dx$$

$$=8\int_{0}^{1}\frac{\frac{3\pi}{4}\zeta\left(2\right)\left(1-y\right)-\frac{\pi}{2}ylog^2\left(y\right)}{1-y^2}dy\\\boxed{B=6\pi\zeta\left(2\right)\log{\left(2\right)}-\pi\zeta\left(3\right)}$$

$$I_2=-2\pi\zeta(3)-2\pi\zeta(2)\log(2)+\frac{8}{3}\log^3(2)-I_2+6\pi\zeta\left(2\right)\log{\left(2\right)}-\pi\zeta\left(3\right)\\
\boxed{I_2=-\frac{3\pi}{2}\zeta\left(3\right)+2\pi\zeta\left(2\right)\log{\left(2\right)}+\frac{4\pi}{3}\log^3{\left(2\right)}}$$

Hence:

$$\boxed{\int_0^\infty\frac{\arctan^2\left(x\right)\log^2\left(1+x^2\right)}{x^2}dx=\frac{\pi}{2}\zeta\left(3\right)+4\pi\zeta\left(2\right)\log{\left(2\right)}+\frac{4\pi}{3}\log^3{\left(2\right)}}$$


问:
What is the formula for series like $-a1, -a2, +a3, +a4, -a5, -a6...$
Just wonder how to come up with the "formula" for any series like:

$-a1, -a2, +a3, +a4, -a5, -a6...$

or maybe:
$-, -, -, +, +, +$

or any of these variations with sequences of same signs but alternating.

Sorry for the poorly formulated question.

答:
To get the sequence $-1, -1, +1, +1, -1, -1, ...$ you can put
$$s(n) = \begin{cases} -1,\textit{if }\ n = 4k \textit{ or }\ 4k + 1\\
+1,\textit{if  }\ n = 4k + 2 \textit{ or }\ 4k + 3,\end{cases}$$
or you can define it inductively:
$$s(0) = s(1) = -1,\\
s(n) = \begin{cases} -s(n-1),\textit{if }\ s(n-1) = s(n-2)\\
\phantom{-}s(n-1),\textit{otherwise}\end{cases}.$$

ContentLicense: CC BY-SA 4.0


问:
Identifying isomorphic graphs
I'm having trouble solving this exercise I found about isomorphic graphs.

It reads

Among the following graphs, which pairs are isomorphic? For each pair that are isomorphic, describe an isomorphism (bijection of vertices preserving adjacency) between them. For each pair that are not isomorphic, give a property preserved under isomorphism that one graph has but the other graph does not.

I've been referencing the vertices and counting the cycles but I don't know if that's all there is to it. I know that a, b and d have vertices that line up well, but c's don't. On the other hand a, b and c have cycles for 1 of 5 but d's is 4. Any help is appreciated.

答:
The graph $G_2$ is the only one containing a cycle of length $3$, so it isn't isomorphic to any of the other graphs.

Among $G_1$, $G_3$ and $G_4$ only $G_4$ contains a cycle of length $4$.

The remaining two graphs $G_1$ and $G_3$ are in fact isomorphic. One possible isomorphism $G_1\to G_3$ is given by
$$
1\mapsto 1,\
2\mapsto 2,\
3\mapsto 3,\
4\mapsto 8,\
5\mapsto 9,\\
6\mapsto 10,\
7\mapsto 4,\
8\mapsto 5,\
9\mapsto 6,\
10\mapsto 7.
$$

I found this by considering $G_1$ as the two $5$-cycles $1,2,3,4,5$ and $6,7,8,9,10$ connected by $5$ edges. In the same way $G_3$ can be described as two $5$-cycles $1,2,3,8,9$ (top half) and $10,4,5,6,7$ (bottom half) connected by $5$ edges. Just check that the $5$ connecting edges are identified under the above map.

Title: Identifying isomorphic graphs


问:
Is the inverse function smooth?
Imagine that we have a function $Inv$ that maps $A \rightarrow A^{-1}$, where A is an invertible square matrix. now my questions is: how do i see that this function is arbitrarily often differentiable? </p>

答: Hint: The inverse is the adjugate matrix divided by the determinant, hence its entries are ratios of two polynomials (i.e. a rational function) in the entries of $A$. So, you are essentially asking why a rational function is infinitely differentiable (provided that the denominator is nonzero).

ContentLicense: CC BY-SA 3.0

Title: Is the inverse function smooth?


问:
When is a set open in the topology of pointwise convergence?
If $E$ is a set, then the topology $\rho(E)$ generated by $$p_x(f):=|f(x)|\;\;\;\text{for }f:E\to\mathbb R$$ for $x\in E$ is called the topology of pointwise convergence on $\mathbb R^E$.

Are we able to show that if $\Gamma\subseteq C(E,\tau)$ is $\tau$-equicontinuous, then$^1$ $\left.\rho(E)\right|_\Gamma=\left.\kappa(E,\tau)\right|_\Gamma$? If not, what do we need to assume to show that?

Moreover, I would like to know how exactly we can show that $$\kappa(E,\tau)\subseteq\left.\rho(E)\right|_{C(E,\:\tau)}\tag1.$$

Is there a useful characterization of $\rho(E)$-openness? For example, if $\tau$ is a topology on $E$, is $C(E,\tau)\in\rho(E)$? And given an arbitrary subset $\Gamma\subseteq C(E,\tau)$, how can we show that $\Gamma\in\rho(E)$?

I know that generally, if $X$ is a vector space and $\sigma$ is the topology generated by a family $P$ of seminorms on $X$, then $$\mathcal B_P:=\left\{\varepsilon\bigcap_{p\in F}U_p:F\subseteq P\text{ is finite and }\varepsilon>0\right\}$$ is an analytic basis for $\sigma$, where $$U_p:=\{x\in X:p(x)<1\}\;\;\;\text{for }p\in P.$$

This clearly gives an abstract characterization of being $\rho(E)$-open.

$^1$ If $(X,\sigma)$ is a topological space and $B\subseteq X$, then $\left.\sigma\right|_B:=\{O\cap B:O\in\sigma\}$ denotes the subspace topology on $B$.

答:
In quite concrete terms $O \subseteq F \subseteq \Bbb R^E$ is open in the topology of pointwise convergence (relative to $F$) iff

For every $f \in O$ there exist finitely many $x_1,\ldots, x_n \in E$ and $r>0$ such that $$\{g \in F\mid \forall 1 \le i \le n: |f(x_i) - g(x_i)| < r \}\subseteq O$$

This is quite standard. Note that it’s just the product-topology on $\Bbb R^E$ and its subsets in mild disguise.


问:
Composition of equivalence relations and quotient set

Let $R$ be an equivalence relation on non empty set $X$. Prove that $X/(X/R\circ R \circ \dots R) = R$

Since $R$ is an equivalence relation we have $$R\circ R \circ \dots R = R$$So we should prove that $X/(X/R) = R$ but I don't know the meaning of $X/(X/R)$. According to the definition $$X/R=\{[x]_R:x\in X\}$$ So what's the meaning of $X/(X/R)$?

Edit: I found that this problem belongs to Set Theory With Applications. Here are some definitions which relate to this problem:

Definition $1$: Let $X$ be a nonempty set. By a partition $\mathcal{P}$ of $X$ we mean a set of nonempty subsets of $X$ such that

a) If $A,B \in \mathcal{P}$ and $A\not = B$ then $A \cap B = \emptyset  $

b) $\bigcup_{C \in \mathcal{P}}C = X$

Definition $2$: Let $R$ be an equivalence relation on a nonempty set $X$. For each $x \in X$, we define $$x/R = \{ y\in X | yRx \}$$ and $$X/R = \{ x/R | x \in X \}$$

Definition $3$: Let $\mathcal{P}$ be a partition of a nonempty set $X$. We define a relation $X/\mathcal{P}$ on $X$ by $x(X/\mathcal{P})y$ if and only if there exists a set $A \in \mathcal{P}$ such that $x,y \in A$.

答:
The exercise is part of showing that, given a non-empty set $X$, there is a one-to-one correspondence between

equivalence relations on $X$, and

partitions of $X$.

So start with an equivalence relation $R$ on $X$. Show first that
$$
X/R
=
\{ x/R : x \in X  \}
$$
is a partition of $X$. You are done, because $y, z \in X$ belong to an element $x/R$ of $X/R$ iff by definition of $x/R$ you have $y R x$ and $z R x$ iff $y R z$.


问:
Show that $cd[(c+d)^2-3] > 0$ where $c > 0$ and $d > 0$
Show that $cd[(c+d)^2-3] > 0$ where $c > 0$ and $d > 0$

Can anyone help me with how to proceed?

答: This is false. Dividing both sides by $cd$ (this is allowed since they're both positive), $$(c+d)^2 -3>0$$ $$\therefore(c+d)^2>3,$$ for which there are many counterexamples. If $c,d \in \mathbb{Z}$, then try using the inequality above and using induction.


问:
limit of $e^z$ at $\infty$
What's the limit of $e^z$ as $z$ approaches infinity?

I am given that the answer is "There is no such limit."

Is this correct, and if so, am I correct to demonstrate this by showing that as $y$ tends to infinity along the $y$-axis, the magnitude of $e^z$ remains $1$, i.e. it doesn't have infinite magnitude, thus it cannot be tending to infinity? And does this mean $e^z$ has an essential singularity at infinity?

答: If you take two sequences, e.g. $z_n:=n$, and $w_n:=2\pi ni$, then $e^{z_n}\to+\infty$ and $e^{w_n}\equiv 1\to 1$, so $\not\exists\displaystyle\lim_{|z|\to\infty}e^z$.</p>

Title: limit of $e^z$ at $\infty$


问:
Finding maximum area of a pentagon
Out of all the pentagons in the shape of a rectangle overlapped by an isosceles triangle, with perimeter P fixed, determine the dimensions of the one with maximum area.

答: Your value of lambda in terms of P does not agree with what I came up with.  I got 3/(4*(3+sqrt(3))P - but yours appears to work, so I am not sure what I did wrong.</p>

Title: Finding maximum area of a pentagon


问:
The precise general expression for the sequence generated by my code
From the following code, I would like to extract the general formula in terms of $n$

```<code>   public static int second_loop(int n ){
        int count = 0;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                for(int k=i; k<=2*j - 1; k++)
                    count++;
        return count;
    }
</code>```

I manage to get the following approximation: 0.5 * n^2 * (n + 1).

Comparing the values that I got from the approximation above with the count (i.e., expected value) from the code, I got the following:

```<code>       expected value | approximation
N=1 |         1       |    1
N=2 |         6       |    6
N=3 |        19       |    18
N=4 |        43       |    40
N=5 |        82       |    75
</code>```

However, the approximation is wrong. Someone has tried to help me answer it but it’s still wrong.

My question is how can I get the precise expression?

I am not working or study math at the moment so this question might be super simple to solve, but unfortunately not for me.

答:
Correcting the typos in my comment:

Your sum is $$\sum_{i=1}^n \sum_{j=1}^n \sum_{k=i}^{2j-1} 1  = \sum_{j=1}^n \sum_{i=1}^{\max(2j, n)} (2j-i)$$ Splitting the $j$ sum into parts where $2j\leq n$ and $2j>n$, using the formulas for $\sum j$ and $\sum j^2$, and conditioning on whether $n$ is odd or even, the sum is $(28m^3 + 9m^2 - m)/6$ if $n=2m$ (values 6, 43, 139, ...) and $(m + 1) (28 m^2  + 23 m + 6)/6$ if $n=2m+1$ (values 1, 19, 82, ...).

For example, if $n=2m+1$ then the sum is $\sum_{j=1}^m \sum_{i=1}^{2j} (2j-i) + \sum_{j=m+1}^{2m+1} \sum_{i=1}^n (2j-i)$. The sum from $i=1$ to $2j$ of $(2j-i)$ is $j(2j-1)$ and the sum from $i=1$ to $n$ of $(2j-i)$ is $2jn-n(n+1)/2$, so we get

$$ \sum_{j=1}^m j(2j-1) + \sum_{j=m+1}^{2m+1}(2jn-n(n+1)/2)$$

Using the sum-of-squares formula, this is

$$ m(m+1)(2m+1)/3 - m(m+1)/2 + 2(2m+1)(m+1)(3m+2)/2 - (m+1)^2(2m+1) $$

which simplifies to $(m+1)(28m^2+23m+6)/6$.


答:
The first inequality to be proved is $\sup_{0\leqslant x\leqslant 1}|f(x)|\leqslant\int_0^1(|f(x)|+|f'(x)|)\,\mathrm dx$, or
$$\sup_{0\leqslant x\leqslant 1}|f(x)|-\int_0^1|f(x)|\,\mathrm dx\leqslant\int_0^1|f'(x)|\,\mathrm dx.$$
This statement is essentially unchanged if $f$ is replaced by $-f$, and we do this just in the case when $\sup_{0\leqslant x\leqslant 1}|f(x)|= \sup_{0\leqslant x\leqslant 1}-f(x)$; otherwise, when
$$\sup_{0\leqslant x\leqslant 1}|f(x)|= \sup_{0\leqslant x\leqslant 1}f(x),$$ we keep the original $f$. Thus, either way, we may assume the above equality. Since $\int_0^1f(x)\,\mathrm dx\leqslant\int_0^1|f(x)|\,\mathrm dx$, it is therefore enough to show that
$$\sup_{0\leqslant x\leqslant 1}f(x)-\int_0^1f(x)\,\mathrm dx\leqslant\int_0^1|f'(x)|\,\mathrm dx.$$
Now observe that each side of this inequality is unchanged if the value of $f$ is shifted  by a constant, and so we reduce $f$ by its average value over the range of integration. Hence it is sufficient to prove that
$$\sup_{0\leqslant x\leqslant 1}f(x)\leqslant\int_0^1|f'(x)|\,\mathrm dx,$$ where $f$ satisfies the condition
$$\int_0^1f(x)\,\mathrm dx=0.$$
$\qquad$From this last condition, the function $f$, by its continuity, is either identically zero—a trivial case, which we will ignore—or takes both positive and negative values. Consequently there is a point $b$ in the range at which $f$ is positive and maximal:
$$\sup_{0\leqslant x\leqslant 1}f(x)=f(b).$$ Again by the continuity of $f$, there is an interval containing $b$ with an end point  $a$ such that $f(a)=0$ and $f(x)>0$ for either (A) $a<x<b$ or (B) $b<x<a$. In case A, because $\int_0^1f'(x)\,\mathrm dx\leqslant\int_0^1|f'(x)|\,\mathrm dx$, it follows that
$$\int_0^1|f'(x)|\,\mathrm dx\geqslant\int_a^b|f'(x)|\,\mathrm dx\geqslant\int_a^bf'(x)\,\mathrm dx=f(b)-f(a)=f(b),$$ which is the required result. In case B, the result follows similarly by interchanging $a$ and $b$ and replacing $f'$ by $-f'$.$$***$$

To show the second inequality, we proceed initially as in the first paragraph above, replacing $\sup_{0\leqslant x\leqslant 1}|f(x)|$ by $f(\frac12)$ and $|f'(x)|$ by $\frac12|f'(x)|$, and conclude that it is sufficient to prove the following inequality:
$$2f(\tfrac12)\leqslant\int_0^1|f'(x)|\,\mathrm dx,$$where (as before) $f$ satisfies the condition
$$\int_0^1f(x)\,\mathrm dx=0.$$ Now, arguing as above by continuity of $f$, there are $a$ and $b$, with $0\leqslant a<\frac12<b\leqslant1$, such that $f(a)=f(b)=0$. Then
$$\int_0^1|f'(x)|\,\mathrm dx\geqslant\int_a^{1/2}|f'(x)|\,\mathrm dx+\int_{1/2}^b|f'(x)|\,\mathrm dx$$
$$\geqslant\int_a^{1/2}f'(x)\,\mathrm dx-\int_{1/2}^bf'(x)\,\mathrm dx$$
$$=f(\tfrac12)-f(a)-f(b)+f(\tfrac12)$$
$$=2f(\tfrac12),$$ as required.

Title: Prove $\sup_{0\le x\le 1}|f(x)|\le\int_0^1(|f(t)|+|f'(t)|)dt$

Body:
Let $f\in C^1([0,1])$. Prove the following:

$$\sup_{0\le x\le 1}|f(x)|\le\int_0^1(|f(t)|+|f'(t)|)dt$$

and

$$|f(1/2)|\le\int_0^1(|f(t)|+\frac12|f'(t)|)dt$$

Note that $e^{-x}(e^xf(x))'=f(x)+f'(x)$. So $\int_0^1(|f(t)|+|f'(t)|)dt\ge|\int_0^1(f(t)+f'(t))dt|\ge\int_0^1(e^tf(t))'dt=ef(1)-f(0)$


问:
Prove $0\le a_n \le a_{n+1} \le 1\ $ for all $n$, if $ a_{n+1}=\sqrt[3]{\frac{1+a_n}{2}}$ and $0\le a_1\le 1$
I know that $0\le a_1\le 1$ and $ a_{n+1}=\sqrt[3]{\frac{1+a_n}{2}}$ for all $ n\ge 1$

Prove that $0\le a_n\le 1\ $ for all $n$, and $a_{n+1}\ge a_n$ for all $n$.

I think I shall assume the opposite and show that this leads to a false conclusion.

I don't know where to start and would appreciate any help I can get.

答: Hint: $f(x) = \sqrt[3]{\frac{1+x}{2}}$ is an increasing function.


问:
Graph theory : meaning of "bottleneck"
Hall's theorem states: Let G be a bipartite graph with vertex partition L;R. There is matching in G that covers L iff no subset of L is a bottleneck.
what is the meaning of bottleneck?</p>

答:
A set S is called bottleneck if
|S| > |N(S)|, where N(S) is the neighbors of S.
(a vertex v is said to be neighbor to u if (v, u) is an edge in the graph).

In other words, a set S is bottleneck if it has elements more than neighbors (as Berci commented.)

Title: Graph theory : meaning of "bottleneck"


问:
Math question efficiency
A solar collector has 1000 Btu/min of radiant energy available on a clear sunny day. The collector can transfer 450 Btu/min to a storage tank. What is the efficiency of the system?

I used n= energy-output/energy-input *100

I plugged it all in $\;\dfrac{450}{1000}*\times 100\%$.

I got $45\%$, however on my answer choices it says the answer is $48\%$?

Exact Question. Maybe I used the wrong formula?

A solar collector has 1000 Btu/min of radiant energy available on a clear sunny day. The collector can transfer 450 Btu/min to a storage tank. What is the efficiency of the system?

A.  17%

  B.  25%

  C.  37%

  D.  48%

答:
You used the correct formula, and your calculation is the correct answer: $45\%$.

There must be a misprint/typo in the solution manual/text.

Title: Math question efficiency


答: Maybe you should try to prove it by induction ! You have $0\le a_1\le1$ and you can prove that $a_1\le a_2$. Now that you have the base case, you can assume $0\le a_n\le1$ and $a_{n-1}\le a_{n}$ and prove that $0\le a_{n+1}\le1$ and $a_n\le a_{n+1}$. (Use the hint given by one of the answers above)

Title: Prove $0\le a_n \le a_{n+1} \le 1\ $ for all $n$, if $ a_{n+1}=\sqrt[3]{\frac{1+a_n}{2}}$ and $0\le a_1\le 1$

Body:
I know that $0\le a_1\le 1$ and $ a_{n+1}=\sqrt[3]{\frac{1+a_n}{2}}$ for all $ n\ge 1$

Prove that $0\le a_n\le 1\ $ for all $n$, and $a_{n+1}\ge a_n$ for all $n$.

I think I shall assume the opposite and show that this leads to a false conclusion.

I don't know where to start and would appreciate any help I can get.


答:
The recurrence is $$a_{n+1}=\sqrt[3]{\frac{1+a_n}{2}}$$
Let's go step by step. First, we'll prove that $$0\leq a_n\leq 1$$ for all $n$.

We are given that $0\leq a_1\leq 1$. Now suppose $0\leq a_k\leq 1$ for some $k\in\mathbb{N}$. This gives $$ \left(\frac{1}{2}\right)^{\frac{1}{3}} \leq a_{k+1} = \left( \frac{a_{k} + 1}{2} \right)^{\frac{1}{3}} \leq 1$$ which means $0\leq a_{k+1} \leq 1$. By the principle of weak induction, we have $0\leq a_n\leq 1$ for all $n$.

To show $a_{n+1} \geq a_n$ for all $n\in\mathbb{N}$, we outline a proof by contradiction. Suppose $a_{n+1} < a_n$ for some $n\in\mathbb{N}$. Using the recurrence in the problem, this would mean $$2a_n^3 - a_n - 1 > 0 \implies (a_n - 1)\left(2\left(a_n + \frac{1}{2}\right)^2 + \frac{1}{2} \right) > 0 \implies a_n > 1$$ which is a contradiction, since we have already shown that $0\leq a_n\leq 1$ for all $n\in\mathbb{N}$.
So we have $a_{n+1} \geq a_n$ for all $n\in\mathbb{N}$ as desired.


答:
Hints

I'd strongly suggest you try mathematical induction.

You'll probably find the following facts useful:

i. If $\ 0\le x\le1\ $ then $\ 0\le\frac{1+x}{2}\le1 $.

ii. If $\ 0\le y\le 1\ $ then $\ 0\le\sqrt[3]{y}\le1\ $.


问:
Is a cumulative distribution function defined for all real $x$?
Suppose that $X$ is a continuous random variable whose probability density function is given by

$$
f(x) = \left\{
        \begin{array}{ll}
            x & \quad 0\leqq x \leqq 1 \\ -x+2 & \quad 1\leqq x\leqq 2 \\ 0 & \quad \text{otherwise}
        \end{array}
    \right.
$$

I found the CDF by integration, obtaining

$$
F(x) = \left\{
        \begin{array}{ll}
            \frac{1}{2}x^2 & \quad 0\leqq x \leqq 1 \\ -\frac{1}{2}x^2+2x-1 & \quad 1\leqq x\leqq 2
        \end{array}
    \right.
$$

which definitely seems correct,

as it is nondecreasing and continuous, with $F(0)=0$ and $F(2)=1$.

My question is, is the CDF defined for $x<0$ and $x>2$? It would make sense for me if the CDF was the constant function $y=1$ for $x>2$ as the CDF specifies the probability that our random variable is less than or equal to $x$, and $P(X\leqq 3)=F(3)$ is definitely equal to $1$. Similarly, would the CDF be $y=0$ for $x<0$. In other words, my question is, is the CDF defined for all real $x \in (-\infty,\infty)$, much like the PDF?

答:
Yes. Your intuition is correct. Indeed the correct CDF is:

$$
F(x) = \left\{
        \begin{array}{ll}
            0 & \quad x \leq 0 \\
\frac{1}{2}x^2 & \quad 0< x \leq 1 \\ -\frac{1}{2}x^2+2x-1 & \quad 1< x\leq 2\\
1&\quad 2\leq x
        \end{array}
    \right.
$$


问:
Limit of Itô-Integral
We consider the following Itô-Integral $$Z_t := \int_0^t \exp(-\mu s) W_s ds$$ for $\mu\geq0$. I wonder if I could calculate the limit as $t\to\infty$ in some convergence, but how should I start? Some ideas?

Edit: Is it possible to get convergence almost surely to some random variable? Or in $L^2$?

答:
Apply Ito's lemma to $-\frac{1}{\mu}\exp(-\mu t)W_t$. You get

$$d\left(-\frac{1}{\mu}\exp(-\mu s)W_s\right)=-\frac{1}{\mu}\exp(-\mu s)dW_t+\exp(-\mu t)W_sds$$

Taking integrals and setting $W_0=0$

$$-\frac{1}{\mu}\exp(-\mu t)W_t=\int_0^t-\frac{1}{\mu}\exp(-\mu t)dW_s+\int_0^t\exp(-\mu s)W_sds$$

Rearranging

$$\int_0^t\exp(-\mu s)W_sds=\int_0^t\frac{1}{\mu}\exp(-\mu s)dW_s-\frac{1}{\mu}\exp(-\mu t)W_t$$

The 2nd expression converges to zero a.s. because $W_t/t$ tends to zero a.s.. The 1st term is a normal random variable with mean 0 and variance $\int_0^t \frac{1}{\mu^2}\exp(-2\mu s)ds$, which converges to $\frac{1}{2\mu^3}$. Hence $\int_0^t\exp(-\mu s)W_sds$ converges a.s. to $\int_0^t\frac{1}{\mu}\exp(-\mu s)dW_s$. It also converges in distribution to a $N(0,\frac{1}{2\mu^3})$ random variable.

Title: Limit of Itô-Integral


问:
The proof that L is not regular language
Let $L=\{x$| the number of $0$ in $x$ is not equal to $1$} is the language on the alphabet {$0,1$}, prove that $L$ is not RL(regular language)

I am having trouble with this. It seems that pumping lemma is not useful in this question.

答:
Hint: Use the relation of Nerode.
Let $L$ be a language over the alphabet $\Sigma$, (here $\Sigma=\{0,1\}$).

Define the relation $\equiv_L$ as

$u\equiv_L v :\Leftrightarrow (\forall w\in\Sigma^*: uw\in L\Leftrightarrow vw\in L).$

This is an equivalence relation and so you can form the equivalence classes.
If the index (=number of equivalence classes) is finite, then $L$ is regular.

Here the number of equivalence classes is infinite. For instance, $1$, $11$, $111$ and so on lie in different classes, e.g. $1\not\equiv_L 11$ since for $w=0$, we have $10\not\in L$ but $110\in L$.


Body:
$a_n \ge 0$ is trivial.

$2a_{n+1}^3-2=a_n-1 \implies 2(1-a_{n+1})(a_{n+1}^2+a_{n+1}+1)=1-a_n$

Therefore $a_n \le 1 \implies a_{n+1}\le 1$

Finally $1-a_n \ge 2(1-a_{n+1})\cdot 1 \ge 1-a_{n+1} \implies a_{n+1} \ge a_n$.


问:
Usage of generating functions in the given enumeration problem
There are $48$ flags in a ship that should be sent to the other ships,the flags are white,blue,black,red and for each one of the colors there are $12$ flags.
We choose $12$ of the flags.

How any messages can be sent assuming an odd number of black flags and an even number of blue flags have been used?

Consider the following exponential generating function:

$$\sum_{k=0}^{n}P(n,k)\frac{x^k}{k!}\tag{I}$$

Where $P(n,k)=k!\binom{n}{k}=k!C(n,k)$ is the number of ways to distribute $k$ distinct objects into $n$
distinct bins such that no bin can contains more than one object.

And so $$[x^k/k!]\sum_{k=0}^{n}P(n,k)\frac{x^k}{k!}=P(n,k)$$

I want to use the same approach to answer the question:
We can consider the coefficient of $x^{12}/(12)!$ in $f(x)$ where:

$$f(x)=\sum_{k=0}^{48}P(n,k)\frac{x^{2k}}{(2k)!}\sum_{k=0}^{48}P(n,k)\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k=0}^{48}P(n,k)\frac{x^{k}}{k!}\right]^2$$

And I "guess" since we only interested in the coefficients so the convergence is not what we should be worry about,and so I think the answer is given by :

$$[x^{12}/(12)!]\sum_{k \ge 0}^{}P(n,k)\frac{x^{2k}}{(2k)!}\sum_{k \ge 0}^{}P(n,k)\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k \ge 0}^{}P(n,k)\frac{x^{k}}{k!}\right]^2$$

Since $P(n,k)=k!C(n,k)[0 \le k \le n]$,where $[]$ is Iverson bracket and so the terms with $k > n$ will be equal to $0$ (ignoring the convergence ).

But the book I'm working with says that we should be looking for the coefficient of $x^{12}/{(12)!}$ in :
$$ \sum_{k \ge 0}^{}\frac{x^{2k}}{(2k)!}\sum_{k \ge 0}\frac{x^{2k+1}}{(2k+1)!}\left[\sum_{k \ge 0}\frac{x^{k}}{k!}\right]^2$$

But I don't understand why that is,because it looks to me a conflict with $\text{(I)}$

答:
Here is what I think is a correct approach.

First, it will simplify the generating function if we disregard the fact that there are only $12$ flags of each type.  Since we only need to send a message of length $12$, we may regard the number of flags of each type as unlimited.

The EGF for the number of sequences of length $n$ of white flags is then
$$1 + x + \frac{1}{2!} x^2 + \frac{1}{3!} x^3 + \dots = e^x$$
and the same for red flags.

The EGF for the number of sequences of length $n$ of an even number of blue flags is
$$1 + \frac{1}{2!} x^2 + \frac{1}{4!} x^4 + \dots = \frac{e^x + e^{-x}}{2}$$

The EGF for the number of sequences of length $n$ of an odd number of black flags is
$$x + \frac{1}{3!} x^3 + \frac{1}{5!} x^5 + \dots = \frac{e^x - e^{-x}}{2}$$

A sequence of white, blue, black or red flags is the "star product" of the four sequences above, so its EGF is
$$f(x) = e^x \cdot e^x \cdot \frac{e^x + e^{-x}}{2} \cdot \frac{e^x - e^{-x}}{2}$$

The answer to the original problem is given by $12! [x^{12}]f(x)$, a computation which we leave to the reader.


问:
If $m$ and $n$ are reversed numbers (like $123$ and $321$) and $m * n = 1446921630$ , find $(m+n)$.

If $m$ and $n$ are reversed numbers (like $123$ and $321$) and $m * n = 1446921630$ , find $(m+n)$.

What I Tried: I found $1446921630 = 2 * 3^5 * 5 * 7 * 11^2 * 19 * 37$ , but that did not really give me useful information.

A little information I got is that $m$ and $n$ each will have a factor of $11$ , and at-least $1$ factor of $3$, as both will be divisible by $3$ and $11$ .

I could have assumed the numbers to be of the form $10x + y$ or something, but I can't as I don't know how many digits both $m$ and $n$ will have, and that will more like Trial and Error.

Another thing is that among $m$ and $n$, one will be even and one will be odd. The odd number will start with an even digit which is not $0$, and the even number cannot end with $0$. Also $5$ should divide the odd number and it will end with $5$. That is all I could conclude.

Can anyone help me?

答:
Observe that $mn \approx 1.44 \times 10^9$, $5 \mid mn$ and $4 \not\mid mn$, so one of the two numbers is odd ending with the digit $5$ and another number is even but not divisible by $4$.  WLOG, assume $5 \mid m$ and $2 \mid n$.  Make an estimate $n \approx 5 \times 10^4$, so that $m \approx 2.88 \times 10^4$.  This drives us to the guess that the first two digits of $m$ are $2$ and $8$.  We now have $28 ?? 5$.

Use the criteria that $9$ and $11$ divides both $m$ and $n$ to exhaust all possible candidates.  We focus on these two criteria first since these criteria are invariant upon reversion of digits.  By trial, it's easy to find that $m = 28215$, so $n = 51282$, and thus $m + n = 79497$.


问:
How to find the last page of a book when some numbers are skipped?
The problem is as follows:

Julia enumerated consecutevely the $204$ pages of her diary, starting
from $1$, excluding those numbers where the digits $1$ and $7$ appear
together in any order. For example, the numbers $17$ and $117$ did not
appear in her journal, however $107$ appears in her diary. Using this
information, what was the number you wrote on the last page of your
journal?

The alternatives are as follows:

$\begin{array}{cc}
1.&\textrm{219 pages} \\
2.&\textrm{199 pages}\\
3.&\textrm{200 pages} \\
4.&\textrm{218 pages} \\
\end{array}$

I'm confused how to assess this problem. So far what I have found is that in he given conditions it seems that the only numbers not allowed between 1 to 204 would be:

$\textrm{11, 17, 71, 77, 117, 117}$

these account for six digits.

Since the number of pages in her diary is $204$ then this would mean that this number would be displaced to the right by six units asuming that those pages are excluded and she would write the inmediate number following, so when she reaches $10$, the next page would be $12$.

Then this would mean that is $204+6=210$. But this doesn't appear in the alternatives. Did I made some mistake or something?. Can someone help me here please?.

答:
Count by cases based on the number of digits:

$\bullet$ Only such two-digit numbers are $17$ and $71$.

$\bullet$ For three-digit numbers, consider those containing $17$. These are  $117, 170,171,172,...179$.

That gives a total of $\mathbf{13}$ such numbers, and so the last page would be numbered $204+13= 217$, had $217$ been permissible. Instead, it would be assigned the next number, i.e. $\color{blue}{218}$.


问:
Applying Zorn's lemma to the set of submodules
Let $A$ be a ring, and $B$ is finitely generated faithful $A$-module.

Let $Γ$＝｛N｜N is submodule of B and B/N is faithful A-mod｝

I want to show $Γ$ has maximal element with Zorn's Lem.

I only need to show that every chain of $Γ$ has upper bound・・・①

But I do not know how to show ①.

According to my text book, ① is obvious because

$N∈Γ$⇔｛$ab1,・・・,abn$｝⇒$a＝0$ for all $a∈A$ ・・・②

（$B$ is finitely generatedly module over $A$,so we denote $B$＝$Ab1＋・・・＋Abn$）

But I don't know why this is obvious.
I think ② is obvious but ②⇒① is not obvious.
Thank you for your help.

答:
Let $\mathcal{C}= \{N_i: i \in I\}$ be a chain in $\Gamma$. Put $$N:= \bigcup_{i \in I} N_i$$
Since we have a chain, $N$ is a submodule of $B$. Moreover, $B/N$ is faithful. To see this, assume $x \in A$ satisfies $xB \subseteq N$. We must show that $x = 0$.

With your notation, let $B$ be generated as an $A$-module by $b_1, \dots, b_n$. Then because we have a chain, there is $i \in I$ with $xb_j \in N_i$ for all $j =1, \dots, n$ and consequently $x B \subseteq N_i$. Since $B/N_i$ is faithful, necessarily $x= 0$.

Hence, $N$ is an upper bound for the chain $\mathcal{C}$.


问:
Cubes as the sum of odd integers
It is well known that

$1^3=1$

$2^3=3+5$

$3^3=7+9+11$

$4^3=13+15+17+19$

$5^3=21+23+25+27+29$

and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)

We will consider the array
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
&&&&\ddots
\end{matrix}
\end{align*}
and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$.
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
\vdots \\
A_{(i-1)1}&...&A_{(i-1)(i-1)}\\
A_{i1}&...&A_{ij} &...&A_{ii}
\end{matrix}
\end{align*}
It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration.
\begin{align*}
A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\\
&=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\\
&=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\\
&...\\
&=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\\
&=(i-1)i-1.
\end{align*}
Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.

Let me know if there is any clarification necessary!

答:
The OP's claim can be summarized by saying that $n^3$ can be expressed as
the sum of $n$ consecutive odd numbers starting with $n(n+1) - (2n - 1)$ and ending with $n(n+1) - 1$.

This is easy to show by a direct proof, using a result for $n^2$ that
is well-known and easy to prove by induction.

Take the series of odd numbers in reverse order, from largest to
smallest.  Then we are saying that:
\begin{eqnarray}
  n^3 & = & \big[(n^2 + n - 1) + (n^2 + n - 3) + \ldots + (n^2 + n - (2n - 1))\big] \\
      & = & n (n^2 + n) - [1 + 3 + \ldots + 2n - 1]
\end{eqnarray}
This in turn gives:

\begin{equation}
  n^2 = 1 + 3 + \ldots + 2n - 1
\end{equation}

which is an exercise in elementary proof by induction.

Title: Cubes as the sum of odd integers


答:
(I'll use, in Definition 3 the notation $R_{\mathcal P}$ in place of $X/\mathcal P$, which I find ambiguous because there are two uses of the slash.)

Suppose $x R y$.

Then $x,y \in x/R$.

Since $X/R$ is a partition of $X$, you apply Definition 3 and (with $x/R$ in place of $A$) conclude that $x R_{X/R} y$.

For the converse, if $x R_{X/R} y$, again by Definition 3, there exists $A \in X/R$ such that $x,y \in A$.

Now, if $A \in X/R$ means that $A$ is an equivalence class of $R$;
so if $x,y \in A$ then $x R y$.

So you see, it''s only a matter of proving that $X/R$ is a partition of $X$ (which follows almost immediately from the definitions) and that the relation defined in Definition 3 is, indeed, an equivalence relation (which is also easy).

Essentially this result tells you that equivalence relations and partitions over a set are the same, under a different guise, since you can start with an equivalence  relation $R$, get the related partition $X/R$ and the equivalence related with that partition $R_{X/R}$ is the same equivalence you started with.

A related result is that if you start with a partition $\mathcal P$, then $\mathcal P = X/R_{\mathcal P}$.

Title: Composition of equivalence relations and quotient set

Body:
Let $R$ be an equivalence relation on non empty set $X$. Prove that $X/(X/R\circ R \circ \dots R) = R$

Since $R$ is an equivalence relation we have $$R\circ R \circ \dots R = R$$So we should prove that $X/(X/R) = R$ but I don't know the meaning of $X/(X/R)$. According to the definition $$X/R=\{[x]_R:x\in X\}$$ So what's the meaning of $X/(X/R)$?

Edit: I found that this problem belongs to Set Theory With Applications. Here are some definitions which relate to this problem:

Definition $1$: Let $X$ be a nonempty set. By a partition $\mathcal{P}$ of $X$ we mean a set of nonempty subsets of $X$ such that

a) If $A,B \in \mathcal{P}$ and $A\not = B$ then $A \cap B = \emptyset  $

b) $\bigcup_{C \in \mathcal{P}}C = X$

Definition $2$: Let $R$ be an equivalence relation on a nonempty set $X$. For each $x \in X$, we define $$x/R = \{ y\in X | yRx \}$$ and $$X/R = \{ x/R | x \in X \}$$

Definition $3$: Let $\mathcal{P}$ be a partition of a nonempty set $X$. We define a relation $X/\mathcal{P}$ on $X$ by $x(X/\mathcal{P})y$ if and only if there exists a set $A \in \mathcal{P}$ such that $x,y \in A$.


问:
Vector Identity with index notation
I'm trying to proving the vector identity given by
$$(\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot(\mathbf{B}\times \mathbf{D}))\mathbf{C}-(\mathbf{A}\cdot(\mathbf{B}\times \mathbf{C}))\mathbf{D}$$

I started from left hand side
$$((\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D}))_i=\epsilon_{ijk}(\mathbf{A}\times \mathbf{B})_j (\mathbf{C}\times \mathbf{D})_k$$
$$=\epsilon_{ijk}\epsilon_{jpq }\epsilon_{klm}A_pB_qC_lD_m$$
I don't idea what to do now. I tried using product property but that's just messed everything.

Please help me though this.

答:
$$
\begin{aligned}
 \left[ \left( \mathbf{A}\times \mathbf{B} \right) \times \left( \mathbf{C}\times \mathbf{D} \right) \right] ^k&=\epsilon _{ij}^{k}\left( \epsilon _{mn}^{i}A^mB^n \right) \left( \epsilon _{pq}^{j}C^pD^q \right)\\
 &=\epsilon ^{ijk}\left( \epsilon _{mni}A^mB^n \right) \left( \epsilon _{pqj}C^pD^q \right)\\
 &=\delta _{pqj}^{ijk}\left( \epsilon _{mni}A^mB^n \right) \left( C^pD^q \right)\\
 &=\delta _{pqj}^{kij}\left( \epsilon _{mni}A^mB^n \right) \left( C^pD^q \right)\\
 &=\delta _{pq}^{ki}\left( \epsilon _{mni}A^mB^n \right) \left( C^pD^q \right)\\
 &=2\left( \epsilon _{mni}A^mB^n \right) \left( C^{k}D^{i} \right)\\
 &=\left( \epsilon _{mni}A^mB^n \right) \left( C^kD^i \right) -\left( \epsilon _{mni}A^mB^n \right) \left( C^iD^k \right)\\
 &=\left( \epsilon _{mni}A^mB^nD^i \right) C^k-\left( \epsilon _{mni}A^mB^nC^i \right) D^k\\
 &=A^m\left( \epsilon _{nim}B^nD^i \right) C^k-A^m\left( \epsilon _{nim}B^nC^i \right) D^k\\
 &=\left( \mathbf{A}\cdot \left( \mathbf{B}\times \mathbf{D} \right) \right) C^k-\left( \mathbf{A}\cdot \left( \mathbf{B}\times \mathbf{C} \right) \right) D^k\\
\end{aligned}
$$
As desired.

Title: Vector Identity with index notation


问:
Completing the square with simple polys
I am suppose to rewrite $x^2 + x + 1$ by completing the square. I don't really know what that means but I know that if I add 3 at the end of this I get </p>

<p>$$(x + 2) (x - 1) - 3$$ this is the same as the original now but the answer isn't right. What is wrong with what I did? It seems a lot cleaner than the answer.</p>

答:
When I hear completing the square, I think of a method to solve a quadratic equation such as

$$x^2+x+1=0$$

The idea is to make the left side a square and take the square root of both sides.  So what constant do we add to both sides?  Since

$$(x+a)^2=x^2+2ax+a^2$$

We take half of our $x$ term to find our value for $a$.

$$2a=1,a=\frac12$$
$$(x+\frac12)^2=x^2+x+\frac14$$

So to complete the square, we subtract $\frac34$ from both sides to get

$$x^2+x+\frac14=(x+\frac12)^2=-\frac34$$

In this case, the problem would have no real roots.  My guess is the form your teacher wants is

$$(x+\frac12)^2+\frac34$$

Title: Completing the square with simple polys


问:
Neighbours of white squares on a chess board and Hall's theorem.
Consider an $m \times n$ chess board with $mn$ even, $m,n \geq 2$, and one black and one white square removed. Label the white squares $1,\dots, \ell$ where $l = mn/2 - 1$, and for each $i \in \{1,\dots,\ell\}$, let $A_i$ be the set of (black) neighbours of white square $i$. How can I prove that for any $S \subseteq \{i,\dots, \ell\}$,
$$ \left|\bigcup_{i \in S}A_i \right| \geq |S|,$$
i.e., for any set of $k$ white squares, there are at least $k$ black squares neighbouring at least one white square in this set. It seems intuitively obvious, but I'd like to understand how to prove this rigorously (so that I can use Hall's theorem to prove that there exists a domino tiling for such a board).

答:
For integers $m,n\ge2$ with $mn$ even, let $P(m,n)$ denote the statement: on an $m\times n$ chessboard, any $k$ white squares border at least $k+1$ black squares, provided $0\lt k\lt mn/2$.

The statement $P(m,n)$ implies that an $m\times n$ chessboard will still satisfy Hall's condition (for matching white squares to adjacent black squares) after one black and one white square is removed.

Theorem. If $m,n\in\mathbb N$, $m,n\ge2$, and $mn$ is even, then $P(m,n)$ holds.

The theorem follows directly from the fact that the chessboard graph (whose vertices are the squares, two squares being adjacent if they share a side) has a Hamiltonian circuit. Alternatively, it can be proved by induction using the following lemmas. The easy details are left to the reader.

Lemma 1. $P(2,2)$ and $P(2,3)$ hold.

Lemma 2. Suppose $m,n\in\mathbb N$, $m,n\ge2$, $mn$ even. If $P(m,n)$ holds then $P(n,m)$ holds.

Lemma 3. Suppose $m,n_1,n_2\in\mathbb N$; $m,n_1,n_2\ge2$; and $mn_1$ and $mn_2$ are even. If $P(m,n_1)$ and $P(m,n_2)$ hold, then $P(m,n_1+n_2)$ holds.


问:
Interesting convex function inequality
Let $f:(0,+\infty)\to \mathbb R$ a convex function. Prove that:

$$20\int^3_0 f(x)\,dx + 10\int^6_0 f(x)\,dx \ge 12 \int^5_0 f(x)\,dx + 15 \int^4_0 f(x)\,dx$$
I tried to use the Newton-Leibniz formula but I did not get to any relevant point.

答:
Hint : Apply the inequality $$f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$$
with $x=3u$, $y=6u$, $t_1=1/3$ and also $t_2=2/3$. Then add, and integrate from $0$ to $1$, change the variables, and to finish multiply by $60$.

Title: Interesting convex function inequality


Body:
If $m$ and $n$ are reversed numbers (like $123$ and $321$) and $m * n = 1446921630$ , find $(m+n)$.

What I Tried: I found $1446921630 = 2 * 3^5 * 5 * 7 * 11^2 * 19 * 37$ , but that did not really give me useful information.

A little information I got is that $m$ and $n$ each will have a factor of $11$ , and at-least $1$ factor of $3$, as both will be divisible by $3$ and $11$ .

I could have assumed the numbers to be of the form $10x + y$ or something, but I can't as I don't know how many digits both $m$ and $n$ will have, and that will more like Trial and Error.

Another thing is that among $m$ and $n$, one will be even and one will be odd. The odd number will start with an even digit which is not $0$, and the even number cannot end with $0$. Also $5$ should divide the odd number and it will end with $5$. That is all I could conclude.

Can anyone help me?

Body:
Just another approach...

$1$st, estimate the number of digits in $m,n$:
$$
\sqrt{1446921630} \approx 38038,
$$
so we expect that $m,n$ are of the form
$$
m = 10^4 a+10^3b+10^2c+10d+e, \\
n = 10^4 e+10^3d+10^2c+10b+a, \\
$$
where $a,b,c,d,e$ are just digits ($a,e \ne 0$).

$2$nd,
$$
m n = 10^8ae+10(\ldots) + ae,
$$
so:

 a) $ae\le 14$;

 b) $ae$ has last digit $0$.
From here we have only one pair for $a\le e$:
$$a=2, \;e=5.$$

$3$rd,
now $mn$ has the form
$$
mn = 10\cdot 10^8 + 10^7(2d+5b) + 10^2(...) + 10(2d+5b) + 2\cdot  5.
$$
and we have estimation:
$$
mn > 10\cdot 10^8+10^7(2d+5b).
$$
$$
mn < (2\cdot 10^4 + 10^3(b+1))(5\cdot 10^4+10^3(d+1)) \le 10\cdot 10^8 + 10^7(2d+5b)+10^6\cdot 100.
$$
from here we conclude:

 a) $34.69 <2d+5b < 44.69$

 b) last  digit of $2d+5b$ is $2$.

Definitely $2d+5b = 42$.
Therefore, only $2$ possible cases work for $b,d$:

 $b=6, d=6$;

 $b=8, d=1$.

$4$th: so we have $2$ candidates:

a) $m = 26?65$, $n = 56?62$,

b) $m = 28?15$, $n = 51?82$.

case a) is too big: even if $?=0$, we'll have $mn>14.6\times 10^8$.

case b) leads to $m=28215, n=51282$.


答:
The first and third Levi-Civita symbols contract to $\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$, so the component is$$\color{blue}{\epsilon_{pqj}}(A_pB_qC_iD_j-A_pB_qC_jD_i)=(A\cdot B\times D)C_i-(A\cdot B\times C)D_i,$$where the blue factor uses $\epsilon_{jpq}=\epsilon_{pqj}$.

A proof with less calculation uses the fact that, since $C\times D$ is orthogonal to the plane $C,\,D$ span, $(A\times B)\times(C\times D)$ is a linear combination of $C,\,D$ that changes sign when they're exchanged, and also when $A,\,B$ are. This, together with multilinearity, gives the desired result up to a proportionality constant, which can be fixed with an example.

Title: Vector Identity with index notation

Body:
I'm trying to proving the vector identity given by
$$(\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D})=(\mathbf{A}\cdot(\mathbf{B}\times \mathbf{D}))\mathbf{C}-(\mathbf{A}\cdot(\mathbf{B}\times \mathbf{C}))\mathbf{D}$$

I started from left hand side
$$((\mathbf{A}\times \mathbf{B})\times (\mathbf{C}\times \mathbf{D}))_i=\epsilon_{ijk}(\mathbf{A}\times \mathbf{B})_j (\mathbf{C}\times \mathbf{D})_k$$
$$=\epsilon_{ijk}\epsilon_{jpq }\epsilon_{klm}A_pB_qC_lD_m$$
I don't idea what to do now. I tried using product property but that's just messed everything.

Please help me though this.


问:
A limit on binomial coefficients
Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.

What I can do is just use Stolz formula. But I could not proceed.

答:
The limit is $\frac{1}{2}$.
We have
$$\begin{eqnarray*}
\sum_{k=0}^n \log {n\choose k}
&=& \sum_{k=0}^n \log n! - \sum_{k=0}^n \log k! - \sum_{k=0}^n \log (n-k)! \\
&=& (n+1)\log n! - 2\sum_{k=1}^n \log k!.
\end{eqnarray*}$$
But
$$\begin{eqnarray*}
\sum_{k=1}^n \log k!
&=& \sum_{k=1}^n \sum_{j=1}^k \log j \\
&=& \sum_{j=1}^n \sum_{k=j}^n \log j \\
&=& \sum_{j=1}^n (n-j+1)\log j \\
&=& (n+1)\sum_{j=1}^n \log j - \sum_{j=1}^n j\log j \\
&=& (n+1)\log n! - n^2 \frac{1}{n}\sum_{j=1}^n \frac{j}{n}\log \frac{j}{n} - \sum_{j=1}^n j\log n \\
&=& (n+1)\log n! - n^2\int_0^1 dx\ x\log x - \frac{n(n+1)}{2}\log n + O(n\log n) \\
&=& (n+1)\log n! +\frac{n^2}{4} - \frac{n(n+1)}{2}\log n + O(n\log n).
\end{eqnarray*}$$
(The error estimate above can probably be tightened.)

Using Stirling's approximation we find
$$\begin{eqnarray*}
\frac{1}{n^2}\sum_{k=0}^n \log {n\choose k}
&=& -\frac{n+1}{n^2}(\log n! - n\log n) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& -\frac{n+1}{n^2}(-n + O(\log n)) - \frac{1}{2} + O\left(\frac{\log n}{n}\right) \\
&=& \frac{1}{2} + O\left(\frac{\log n}{n}\right).
\end{eqnarray*}$$

Title: A limit on binomial coefficients


问:
Python Reed-Solomon encoding in a GF(2^8)
I'm working on a school project and I have to encode a message using Reed-Solomon encoding procedure and Python.
I couldn't wrap my head around a way to divide my generator polynomial $g(x)$ with my information polynomial $I(x)x^{n-k}$, and I stumbled upon this implementation (https://rextester.com/ZMBYT68318) :

```<code>def RSEncode(self, argMesg, errSize):

        # prepare the generator polynomial
        polyGen = self._rsGenPoly(errSize)

        # prepare the output buffer
        outBuffer = (len(argMesg) + errSize)
        outBuffer = [0] * outBuffer

        # initialise the output buffer
        for mesgPos in range(0, len(argMesg)):
            mesgChar = argMesg[mesgPos]
            outBuffer[mesgPos] = ord(mesgChar)

        # begin encoding
        for mesgPos in range(0, len(argMesg)):
            mesgChar = outBuffer[mesgPos]
            if (mesgChar != 0):
                for polyPos in range(0, len(polyGen)):
                    tempValu = self.__gfMult(polyGen[polyPos], mesgChar)
                    outBuffer[mesgPos + polyPos] ^= tempValu

        # finalise the output buffer
        for mesgPos in range(0, len(argMesg)):
            mesgChar = argMesg[mesgPos]
            outBuffer[mesgPos] = ord(mesgChar)

        # return the output buffer
        return (outBuffer)

</code>```

What I can't understand is why a multiplication is working when I kept reading that I had to divide $I(x)x^{n-k}$, my information offset by $n-k$ parity bits, with $g(x)$ my generator polynomial so that my final encoded message would be
$$
E(x) = I(x)x^{n-k} + r(x)
$$
$r(x)$ being the remainder of my euclidean division.
I know that in a Galois field, division is multiplication by the inverse modulo p, but it doesn't seem the be the case here, am I missing something ? (Obviously yes !).

Additionnaly, if you have a clue on how to divide two polynomials, I'm definitely good with any advices you may give !

Thanks in advance for the help !

答:
I finally figured that encoding was actually an implementation of a Synthetic division :
https://en.wikipedia.org/wiki/Synthetic_division

Being new on this website, I also wonder why my post was downvoted without any comment mentioning why, so if anyone can help me understand the reason so I can be a better user of the website, I'm down :) !

Bye !


问:
(Nice) functions whose Taylor-Maclaurin expansion have alternating signs after every two/three terms?
Seeing some users wonder about closed form of series of real numbers with signs alternating after pairs/triples (eg, $1+2-3-4+5+6-7-8+\ldots$ and $1+2+3-4-5-6+7+\ldots$) I got curious whether

do there exist (sufficiently nice?) functions whose Taylor Maclaurin expansion $\ldots$ has signs alternating after every two/three or more terms?

I remembered that $f^{(n)}(\sin x)$ does change sign in our desired manner $${f^{(0)}(\sin x)=\sin x,f^{(1)}(\sin x)=\cos x,f^{(2)}(\sin x)=-\sin x,}$$ $${f^{(3)}(\sin x)=-\cos x,f^{(4)}(\sin x)=\sin x,\ldots}$$

so expanding about some small positive $h$

$$\sin x = \sin h + \dfrac{\cos h}{1!}(x-h) - \dfrac{\sin h}{2!}(x-h)^2 - \dfrac{\cos h}{3!}(x-h)^3 + \dfrac{\sin h}{4!}(x-h)^4 + \ldots $$

which is not as interesting a formula as formula of $\sin x$ about zero.

Thus my question :

do there exist (sufficiently nice?) functions whose Taylor Maclaurin expansion about zero has signs alternating after every two/three or more terms?

Here nice probably means some simple combination of known continuous functions e.g., trig, exponential, hyperbolic etc.

I'm not sure of its uses. Just curiosity. Thank you!

答:
This is possible, and here is a general way how you can do it. The result will be expressed via only the exponential function, even though the exponent may be a complex number. You may want to rephrase the result in terms of real numbers (using Euler's formula), in which case $\sin$ and $\cos$ may pop out.

Suppose you want the values of the higher derivatives to be $f^{(n)}(0)=a_n$ where $(a_n)_{n\ge 0}$ is an arbitrary real (or even complex) sequence, periodic with period $N$ (i.e. $a_{n+N}=a_n$ for all $n\ge 0$).

Let $\varepsilon$ be the primitive $N$th root of unity, i.e. $\varepsilon=e^{\frac{2\pi i}{N}}$. Look at the following matrix:

$$\begin{bmatrix}1&1&1&\cdots&1\\1&\varepsilon&\varepsilon^2&\cdots&\varepsilon^{N-1}\\1&\varepsilon^2&\varepsilon^4&\cdots&\varepsilon^{2(N-1)}\\\vdots&\vdots&\vdots&&\vdots\\1&\varepsilon^{N-1}&\varepsilon^{2(N-1)}&\cdots&\varepsilon^{(N-1)^2}\end{bmatrix}$$

This matrix is a regular matrix (its determinant is nonzero), because it is a Vandermonde matrix and $1, \varepsilon, \varepsilon^2, \ldots,\varepsilon^{N-1}$ are all different.

Thus, the columns of the matrix span $\mathbb C^N$ and so you can find $\alpha_0, \alpha_1,\ldots,\alpha_{N-1}$ such that $(a_0,a_1,\ldots,a_{N-1})^T=\sum_{k=0}^{N-1}\alpha_k\cdot(1,\varepsilon^k,\varepsilon^{2k},\ldots,\varepsilon^{k(N-1)})^T$.

Now, for a fixed $k=0,1,2,\ldots,N-1$, note that the function $f_k(x)=e^{\varepsilon^k x}$ has as an $n$th derivative $f_k^{(n)}(x)=\varepsilon^{nk}e^{\varepsilon^k x}$, so $f_k^{(n)}(0)=\varepsilon^{nk}$. In other words, in the Maclaurin series you will have the coefficients precisely equal to $1, \varepsilon^k,\varepsilon^{2k},\ldots$, which will be periodic with period $N$.

Now it is the matter of a simple check that the function $f$ built up as:

$$f(x)=\sum_{k=0}^{N-1}\alpha_k f_k(x)=\sum_{k=0}^{N-1}\alpha_k e^{\varepsilon^k x}$$

(using those same coefficients $\alpha_0,\ldots,\alpha_{N-1}$ we got before) is the function you want, and it will have exactly the desired coefficients in its Maclaurin series, due to linearity of those coefficients.

Example: How to get back to $\cos$: you want the coefficients to be $1,0,-1,0,1,0,-1,0,\ldots$, i.e. the period to be $N=4$. Take $\varepsilon=i$ (the $4$th root of unity) and express $(1,0,-1,0)^T$ as a linear combination of $(1,1,1,1)^T, (1,i,-1,i)^T, (1,-1,1,-1)^T$ and $(1,-i,-1,i)^T$. The coefficients will end up being $0,\frac{1}{2}, 0, \frac{1}{2}$, respectively. Thus, the function will end up being:

$$\frac{1}{2}e^{ix}+\frac{1}{2}e^{-ix}$$

for which Euler's formula says that it is the same as $\cos x$.


问:
Proving that $\mathbb{N}$ is well-ordered without appealing to induction.

The above proof is from an introductory paper on ordinals. I'm confused by the proof since it seems to assume that every subset of $\mathbb{N}$ has a smallest element, which is something the paper doesn't prove. I went online looking for proofs that $\mathbb{N}$ is well-ordered, but all the ones I've found use induction in one way or another, so I wonder how one could prove that $\mathbb{N}$ is well-ordered without appealing to induction.

In case it is relevant, the paper defines $\mathbb{N}$ by means of the Axiom of Infinity:

答:
You can't really do that.

For a partial order $(P,<)$ we say that $P$ has "capacity for induction"1 if whenever $A\subseteq P$ is such that all the minimal elements are in $A$, and whenever $a\in P$ is such that $\{x\in A\mid x<a\}\subseteq A$, then $a\in A$, then $A=P$.

Theorem. $(P,<)$ has capacity for induction if and only if for every non-empty $A\subseteq P$, there is a minimal element in $A$ (relative to the restricted order).

In other words, capacity for induction is equivalent to being well-founded.

Corollary. If a $P$ is a linear order, then it has the capacity for induction if and only if it is well-ordered.

This means that we can't prove that $\Bbb N$ has the capacity for induction without using the fact that it is well-ordered. Luckily we are not working with the Peano axioms, we are working in $\sf ZF$, where we can prove the fact that our set $\Bbb N$ is in fact well-ordered without appealing to induction principles internal to $\Bbb N$.

Finally, as a small side note, the paper you're reading should state that $\Bbb N$ is not just "a set" with these properties, but in fact the smallest such set.

This is an ad-hoc term, but the definition is not ad-hoc.


问:
Given a right triangle whose side lengths are all integer multiples of 8, how many units are in the smallest possible perimeter of such a triangle?

Given a right triangle whose side lengths are all integer multiples of 8, how many units are in the smallest possible perimeter of such a triangle?

Does this mean the side lengths and not the hypotenuse or does it mean all three sides?

答: The smallest right triangle with integer lengths is the $3 - 4 - 5$ right triangle. Since none of those lengths are multiples of 8, we must scale each side up by a factor of 8 to create a right triangle whose side lengths are all integer multiples of 8. This triangle has perimeter $3\cdot 8 + 4 \cdot 8 + 5 \cdot 8 = (3 + 4 + 5 )\cdot 8 = 12\cdot 8 = \boxed{96}$ units.


答:
An easier if less interesting example than @StinkingBishop's:

If you want $n$ pluses followed by $n$ minuses, use e.g. $\sum_{j=0}^{n-1}x^j\cdot(1-x^n)f(x^{2n})$, where $f$ has positive coefficients for every power, such a with $f(x)=\exp x$ (I take this example to get an infinite radius of convergence). I've mentioned the cases $n\in\{2,\,3\}$ in a comment.

Title: (Nice) functions whose Taylor-Maclaurin expansion have alternating signs after every two/three terms?

Body:
Seeing some users wonder about closed form of series of real numbers with signs alternating after pairs/triples (eg, $1+2-3-4+5+6-7-8+\ldots$ and $1+2+3-4-5-6+7+\ldots$) I got curious whether

do there exist (sufficiently nice?) functions whose Taylor Maclaurin expansion $\ldots$ has signs alternating after every two/three or more terms?

I remembered that $f^{(n)}(\sin x)$ does change sign in our desired manner $${f^{(0)}(\sin x)=\sin x,f^{(1)}(\sin x)=\cos x,f^{(2)}(\sin x)=-\sin x,}$$ $${f^{(3)}(\sin x)=-\cos x,f^{(4)}(\sin x)=\sin x,\ldots}$$

so expanding about some small positive $h$

$$\sin x = \sin h + \dfrac{\cos h}{1!}(x-h) - \dfrac{\sin h}{2!}(x-h)^2 - \dfrac{\cos h}{3!}(x-h)^3 + \dfrac{\sin h}{4!}(x-h)^4 + \ldots $$

which is not as interesting a formula as formula of $\sin x$ about zero.

Thus my question :

do there exist (sufficiently nice?) functions whose Taylor Maclaurin expansion about zero has signs alternating after every two/three or more terms?

Here nice probably means some simple combination of known continuous functions e.g., trig, exponential, hyperbolic etc.

I'm not sure of its uses. Just curiosity. Thank you!


问:
Find the distance of a vertex from the incentre of a triangle.
$F$ is the incentre of $\Delta ABC$. $AF$ is joined to meet $BC$ at $D$ and the circumcircle of $\Delta ABC$ at $E$. If $ED = 4cm$ and $DF = 3cm$, find $8AI$.

MY WORK :-
Regarding this question I could think of only 2 formulas :
${AF\over FD} = {AC+AB\over BC}$ and $AB*AC = BD*DC + AD^2$ however none of them are helping in finding the solution. I feel as if the solution is revolving around these formulas, but still I am not able to reach to a fruitful end.

答:
We use Fact 5 that $E$ is center of circle passing through $B,F,C,J$ where is $J$ is $A$-excenter. Thus $$EF=7=EB=EC=EJ$$

Now by power of point of $D$ wrt these two circles

$$AD\cdot DE =BD\cdot CD= DF\cdot DJ$$

$$\Rightarrow (AF+3)\cdot 4 = 3\cdot 11$$

$$\Rightarrow 8AF = 42$$


问:
Algebra problem with isomorphism
Let $G=(-1,1)$ and the binary operation  such that
$x*y=\frac{x\sqrt{1-y^2}+y\sqrt{1-x^2}}{\sqrt{1-(xy)^2+2xy\sqrt{(1-x^2)(1-y^2)}}}$.
Prove that (G,*) it is a isomorph group with the group (R,+).
I tried to find the bijective function $f:G\to R$ such that  $f(x*y)=f(x)+f(y)$   but I could not.

答:
Let $x=\sin\alpha$, $y=\sin\beta$, so $\alpha,\beta\in(-\pi/2,\pi/2)$. (always do such substitution when there is a $\sqrt{1-x^2}$)

Then $x*y = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sqrt{1 - \sin^2\alpha\sin^2\beta + 2\sin\alpha\sin\beta\cos\alpha\cos\beta}} = \frac{\sin(\alpha+\beta)}{\sqrt{\cos^2\alpha\cos^2\beta + \sin^2(\alpha+\beta)}} = \frac{t}{\sqrt{1+t^2}} = \tanh\mathrm{asinh}\,t = g(t) $ where $t = \frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta} = \tan\alpha+\tan\beta = \frac{x}{\sqrt{1-x^2}} + \frac{y}{\sqrt{1-y^2}} = f(x)+f(y) $.

(I have replaced $1$ in the denominator by $(\sin^2\alpha+\cos^2\alpha)(\sin^2\beta+\cos^2\beta)$ because we need something symmetric with respect to $\alpha\leftrightarrow\beta$, and we need to get rid of that complicated term with the factor $2$; we did it by including it into $\sin^2(\alpha+\beta)$.)

So $x*y=g(f(x)+f(y))$, and since in fact $g(f(x))=x$, $f$ and $g$ indeed give the desired isomorphism.

Title: Algebra problem with isomorphism


问:
How to conclude that $ a \leq \mathrm{sup}_{n} a_n = \mathrm{sup} \{a_n : n \in \mathbb{N}\}$
This is part of the exercise 5 of chapter 1 in Real Analysis by N. L. Carothers. The whole problem is the following:

Suppose that $a_n \leq b$ for all $n$, and that a = $\lim_{n\to\infty}a_n$ exists. Show that $a \leq b$. Conclude that $a \leq \mathrm{sup}_{n} a_n = \mathrm{sup} \{a_n : n \in \mathbb{N}\}$.

So far what I have done is that I have used the definition of a limit to conclude that $\left| a_n - a \right| < \epsilon \leftrightarrow a_n + \epsilon > a > a_n - \epsilon \leftrightarrow b + \epsilon \geq a_n + \epsilon > a > a_n - \epsilon \leftrightarrow a_n - \epsilon < a < b + \epsilon$. Since $\epsilon$ can be made arbitrarily small, we conclude that $a < b$.

So I am not sure how to show the missing equality, and how to use the first part of the proof in the $\mathrm{sup}_n$ part.

答:
Your conclusion that $a < b$ is not right,
for example
$\forall n\ \left(1 - \frac{1}{n}\right) < 1$, but $\lim_\limits{n\to\infty}\ \left(1 - \frac{1}{n}\right) = 1 \not < 1$

So you can conclude only that $a\le b$.

(formally, you show that $\forall \epsilon > 0\ a < b + \epsilon$
 hence $a$ is lower bound of $\{b + \epsilon\ |\ \epsilon > 0\}$
and by definition of infinum
  $a \le \inf{\{b + \epsilon\ |\ \epsilon > 0\}} = b$)

Second part:
by definition of supremum
$\forall n\ a_n \le \mathrm{sup} \{a_n : n \in \mathbb{N}\}$

You can take $b = \mathrm{sup} \{a_n : n \in \mathbb{N}\}$, and use first part.


问:
Does $L=K(\theta)$ implies that $\hat{\mathscr{O}_L}=\hat{\mathscr{O}_K}(\theta)$?
Let $A$ be a Dedekind ring, $K=\operatorname{Frac}A$, $L/K$ a finite separable extension and $B$ the integral closure of $A$ in $L$. (Perhaps we should suppose that $K$ is a global field.)

Each non-zero prime ideal $\mathfrak{p}\in \operatorname{Spec}A$ determines an absolute value $|\cdot|_{\mathfrak{p}}$ of $K$ and each prime ideal $\mathfrak{P}\in\operatorname{Spec}B$ over $\mathfrak{p}$ determines an absolute value $|\cdot|_{\mathfrak{P}}$ of $L$ which extends $|\cdot|_{\mathfrak{p}}$. We denote $\hat{K}$ and $\hat{L}$ the completions of $K$ and $L$ with respect to their respective absolute values.

We also denote $\hat{A}$ the valuation ring of $\hat{K}$ (which coincides with the completion of $A$ and the closure of $A$ in $\hat{K}$) and $\hat{B}$ the valuation ring of $\hat{L}$ (which coincides with the completion of $B$, the closure of $B$ in $\hat{L}$ and the integral closure of $\hat{A}$ in $\hat{L}$).

If $L=K(\theta)$ for some $\theta\in L$, I understand that $\hat{L}=\hat{K}(\theta)$. But I also know that $\hat{B}=\hat{A}[\alpha]$ for some $\alpha$. I wonder if we can take $\alpha=\theta$. If not, how can we calculate $\hat{B}$? I'm having trouble calculating it even in the simplest cases (for example when $A=\mathbb{Z}$ and $L=\mathbb{Q}(i)$).

答:
$L=K(\theta)$ an finite extension of number field.

A completion $L_Q$ above $K_P$ gives an embedding $h:K(\theta)\to \overline{K_P}$, it maps $\theta$ to a root $h(\theta)$ of $f\in K[x]$ the minimal polynomial of $\theta$.

$h(L)=K(h(\theta))$ is dense in $K_P(h(\theta))$ and the latter is complete so $L_Q=K_P(h(\theta))$.

Then we look at $O_{L_Q}$ the closure of $O_L$ in $L_Q$. The main theorem is that it is a DVR, with uniformizer $(\pi_Q)$. Take some $b_Q\in O_L$ such that $O_L/(\pi_Q)=O_L/Q=(O_K/P)[b_Q+P]$. Let $S_Q$ be some representatives in $O_K[b_Q]$ of the residue field $O_L/Q$. By definition of a complete DVR $$O_{L_Q} = \{ \sum_{n\ge 0} s_n \pi_Q^n, s_n\in S_Q\}= \sum_{j=0}^{f(Q/P)-1} b_Q^j \sum_{m=0}^{e(Q/P)-1} \pi_Q^m O_{K_P}=O_{K_P}[b_Q,\pi_Q]$$
It works the same way when replacing $b_Q$ by $b_Q^p+\pi_Q$ and $\pi_Q$ by $(b_Q^p+\pi_Q)^{N(Q)-1}-1$ (prove that the latter is an uniformizer if $p$ isn't an uniformizer) thus $$O_{L_Q} = O_{K_P}[b_Q^p+\pi_Q]$$


问:
Mean curvature of a level set
On page 256 http://zakuski.utsa.edu/~jagy/papers/Michigan_1991.pdf there is a formula for the mean curvature of a level set of a function. Im interested in the case n=3. How do you prove this formula? I can prove it for the case of a graph, so i thought i would use the implicit function theorem to make it a graph locally, but i dont see how I will get the original function in the final answer. Alternatively, I'm aware this equation is just the divergence of the normal vector, but how do you show that H can be defined this way? I seem to need to refer to a chart, but i can't seem to find one that gives a nice answer. Any help would be much appreciated

Tom

答:
As I recall, what I did originally was just to rotate everything. Put another way, if your level surface is a graph over the xy plane, $z=f(x,y),$ in such a way that $f(0,0)=0$ and $\nabla f(0,0) = (0,0),$ then the mean curvature at the origin is just the Laplacian $\Delta f(0,0),$ although I always liked to divide by 2, or $n$ in $\mathbb R^{n+1}.$ As soon as you need to rotate in order to get the level surface into that position, instead of just the Laplacian you get a mixture of products of first and second partials because of the rotation step. I think i say this in the paper, take the formula for a level set and just plug in $g(x,y,z) = f(x,y) - z$ and see what you get.

What is your background in Riemannian geometry?

Title: Mean curvature of a level set


答:
You can reduce your identity to the well-known vector identity given by

$$
\mathbf{X}\times(\mathbf{C}\times\mathbf{D})=(\mathbf{D}\cdot\mathbf{X})\,\mathbf{C}-(\mathbf{C}\cdot\mathbf{X})\,\mathbf{D}.
$$

Now just set $\mathbf{X}=\mathbf{A}\times\mathbf{B}$ and use the following property of the triple product

$$
\mathbf{A}\cdot(\mathbf{B}\times\mathbf{C})=\det(\mathbf{A},\mathbf{B},\mathbf{C})=\det(\mathbf{C},\mathbf{A},\mathbf{B})=\mathbf{C}\cdot(\mathbf{A}\times\mathbf{B}),
$$

that is, you can cyclically permute the vectors in the determinant.

Title: Vector Identity with index notation


问:
Inverting Triangular Matrices
A $d × d$ triangular matrix $L$ with non-zero diagonal entries can be expressed in the form $(\Delta + A)$, where $Δ$ is an invertible diagonal matrix and $A$ is a strictly triangular matrix. Show how to compute the inverse of $L$ using only diagonal matrix inversions and matrix multiplicatons/additions. Note that strictly triangular matrices of size $d × d$ are always nilpotent and satisfy $A^d = 0$.

It is mentioned a page before that $(I+A)^{-1}=I-A+A^2-A^3+A^4+\dots$ which I see is obtained via Taylor Expansions, using this hint, my attempt:

$(\Delta+A)^{-1}=\\\Delta^{-1}\Delta(A+\Delta)^{-1}=\\\Delta^{-1}[(A+\Delta)\Delta^{-1}]^{-1}=\\\Delta^{-1}(I+A\Delta^{-1})^{-1}$

Expanding:

$\Delta^{-1}[I-A\Delta^{-1}+(A\Delta^{-1})^2-(A\Delta^{-1})^3+\cdots]$

And since $A$ is a nilpotent matrix, any $A^d=0$, therefore:

$(\Delta+A)^{-1}=\Delta^{-1}(I-A\Delta^{-1})$

Although something in my mathematical gut tells me this is ridiculous.

SOURCE: Linear Algebra and Optimization for Machine Learning: A Textbook (Charu C. Aggarwal); page 18 problem 1.2.11

答:
You have that $A^d=0$ only for $d$ the dimension of the triangular matrix. So what you actually have is that $A^k=0$ for $k\ge d$. (This can be seen easily for
$A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$)

Furthermore, in your taylor expansion, you have $\left(A\Delta^{-1}\right)^k=\underbrace{A\Delta^{-1}A\Delta^{-1}\ldots A\Delta^{-1}}_{k\text{ times}}\neq A^k(\Delta^{-1})^k$ since matrix multiplication is not commutative. However since $A\Delta^{-1}$ is also a strictly triangular matrix of size $d\times d$ we still have $\left(A\Delta^{-1}\right)^k=0$ for $k\ge d$.

Therefore

$$
(\Delta+A)^{-1}=\Delta^{-1}\sum_{k=0}^{d-1} \left(-A\Delta^{-1}\right)^k=\Delta^{-1}\left(I-\left(A\Delta^{-1}\right)+\ldots\left(-A\Delta^{-1}\right)^{d-1}\right)
$$

Which satisfies the requirements (diagonal matrix inversion and matrix multiplication and addition). And a little sanity check never hurts : we indeed get that the inverse of $\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$ is $I-\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}+\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}$

Title: Inverting Triangular Matrices


问:
Given $P(3i) = P(1 − i) = 5$, find $P(2)$
I'm given that $P(x)$ be a monic 4th degree polynomial with real coefficients and that $P(3i) = P(1 − i) = 5$. I'm supposed to compute P(2), which is 31, but I need to show the work for it.

If $3i$ and $1-i$ were roots, I would know that $-3i$ and $1+i$ were also roots, but I'm not sure how you get there. And I tried $P(3i)-5 = P(1 − i)-5 = 0$ (shifting down 5 units such that $Q(x) = P(x)-5$ and $3i$ and $1-i$ were roots of $Q(x)$ but that didn't seem to go anywhere. Can someone point me in the right direction?

答:
Since the polynomial has real coeficients $-3i$ and $1+i$ also gives value $5$ so \begin{align}P(x) &= (x-3i)(x+3i)(x-1+i)(x-1-i)+5\\
&= (x^2+9)((x-1)^2+1)+5\\
\end{align}


问:
Expectation of a log random variable
I have an equation that requires taking the natural log of a random variable. When trying to figure out how to evaluate the expression, I came across this paper:

Y. W. Teh, D. Newman and M. Welling (2006), A Collapsed Variational Bayesian Inference Algorithm for Latent Dirichlet Allocation, NIPS 2006, 1353–1360.

They use a second order Taylor expansion around $_0=\mathbb{E}[x]$ to approximate $\mathbb{E}[\log(x)]$ to obtain

$$\mathbb{E}[\log(x)]\approx\log(\mathbb{E}[x])−\frac{\mathbb{V}[x]}{2\mathbb{E}[x]^2}$$

I can't seem to derive that equation and was wondering if someone might be able to help show that it is true? Also, does that result hold if instead we wanted to evaluate $\mathbb{E}[\log(-x)]$ where $x\in(-\infty,0)$?

答:
A second order Taylor expansion around $\mathbb{E}(x)$ gives

$$log(x)\approx log(\mathbb{E}(x))+\frac{1}{\mathbb{E}(x)}\times(x-\mathbb{E}(x))-\frac{1}{2}\frac{1}{\mathbb{E}(x)^2}(x-\mathbb{E}(x))^2$$

Take expectations

$$\mathbb{E}(log(x))\approx  \mathbb{E}(log(\mathbb{E}(x)))+\frac{1}{\mathbb{E}(x)}\times(\mathbb{E}(x)-\mathbb{E}(x))-\frac{1}{2}\frac{\mathbb{V}ar(x)}{\mathbb{E}(x)^2}$$

because the 2nd term vanishes and the expectation of a constant is the constant:

$$\mathbb{E}(log(x))\approx  log(\mathbb{E}(x))-\frac{1}{2}\frac{\mathbb{V}ar(x)}{\mathbb{E}(x)^2}$$

Title: Expectation of a log random variable


问:
Sequence of real random variables.
Let $X_{n}$ be a sequence of real random variables over a probabilistic space $(\Omega,\mathcal{A},\mathbb{P})$, $\{x_{n}\}_{n}$ be a sequence of real numbers such that $x_{n}\searrow_{n} x$ and such that $\mathbb{P}(X_{n}\leq x_{n})\geq 1/2$ for all $n\in\mathbb{N}$. I want to prove that
$$\limsup_{n}\mathbb{P}(X_{n}\leq x)\geq \cfrac{1}{2} - \epsilon$$

for an arbitrary $\epsilon > 0$.

We have that, for all $\delta > 0$, there exists $N\in\mathbb{N}$ such that $x_{n} - x \leq \delta$ for all $n\geq N$. Thus, for this delta, we have that

$$\limsup_{n}\mathbb{P}(X_{n} \leq x)\geq\limsup_{n}\mathbb{P}(X_{n} \leq x_{n} - \delta)$$

In this point i don't know how to proof that. It seems to be easy.

Thanks in advance.

答: Suppose $\ x_n>x\ $, for all $\ n\ $ and $\ X_n=\frac{x+x_n}{2}\ $ with probability $1$.  That is, the distribution of $\ X_n\ $ has a single atom of weight $1$ at $\ X_n=\frac{x+x_n}{2}\ $. Then $\ \mathbb{P}\big(X_n\le x_n\big)=1\ge\frac{1}{2}\ $ for all $\ n\in\mathbb{N}\ $, but $\ \mathbb{P}\big(X_n\le x\big)=0\ $ for all $\ n\in\mathbb{N}\ $, so $\ \limsup_\limits{n}\,\mathbb{P}\big(X_n\le x\big)=\lim_\limits{n} \mathbb{P}\big(X_n\le x\big)=0\ $.

Title: Sequence of real random variables.


问:
Find the Maclaurin series of $\ln{(x+\sqrt{x^2+1})}$ with accuracy up to $o(x^{2n})$
I need to find the Maclaurin series of $\ln{(x+\sqrt{x^2+1})}$ with accuracy up to $o(x^{2n})$ and present the series as a sum of members with the same formula. Since we haven't studied integrals yet, I'd prefer suggestions without them. For the Maclaurin series, we studied the formula $f(x)=f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^{2}}{2!}+...+\frac{f^{(n)}(0)x^{n}}{n!}+o(x^{n})$. From this formula, we derived $\ln{(1+x)}=\sum_{k=1}^{n}{\frac{(-1)^{k-1}x^{k}}{k}}+o(x^{n})$. So, trying to use one of these, I thought of two approaches:

$\ln{(x(1+\frac{\sqrt{x^2+1}}{x}))}=\ln{x}+\ln{(1+\frac{\sqrt{x^2+1}}{x})}$ but then that root fraction is not suitable for the $\ln$ formula

the derivative of $\ln{(x+\sqrt{x^2+1})}$ is $(1+x^{2})^{-\frac{1}{2}}$. But then I will have to find the n-th derivative for the basic formula, which I wouldn't be able to. How should I procede?

答:
Not a full answer but a few steps you will encounter on the way to one. There is a pattern if you keep differentiating, and keep rearranging to get the following landmarks, which will keep it all manageable.

Show that, (as you have already done in your method 2)
$$\sqrt{1+x^2}f'(x)=1$$

then (using the product rule on the line above) that,
$$(1+x^2)f''(x)+xf'(x)=0$$
then (using the product rule on the line above),
$$(1+x^2)f'''(x)+3xf''(x)+f'(x)=0$$
and so on, as required...

You can then extract the values of $f(0), f'(0),f''(0), f'''(0),...$ and use them in the formula that you tell me you have studied.

This method, although long, avoids integration, as you requested.


答:
You don't want to use integration, but I don't think you can avoid it entirely.  First note that

$$f(x) = \ln(x+\sqrt{x^2+1}) = \mbox{arcsinh } x, $$

and as you note

$$f'(x) = (1+x^2)^{-1/2}.$$

The generalized binomial theorem

https://en.wikipedia.org/wiki/Binomial_theorem#Newton's_generalized_binomial_theorem

gives you

$$f'(x) = \sum_{n=0}^{\infty} \frac{ \left(-\frac{1}{2}\right)_n}{n!}x^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n(2n)!}{2^{2n}(n!)^2}x^{2n}.$$

Since you know how to differentiate polynomials, then it's just barely "integration" to un-differentiate this series and get your answer.  Realize that any constant could be added to your answer and still differentiate to the above.  So you can plug in $x=0$ at the end and work out that constant.  (It'll be $0$.)

Title: Find the Maclaurin series of $\ln{(x+\sqrt{x^2+1})}$ with accuracy up to $o(x^{2n})$

Body:
I need to find the Maclaurin series of $\ln{(x+\sqrt{x^2+1})}$ with accuracy up to $o(x^{2n})$ and present the series as a sum of members with the same formula. Since we haven't studied integrals yet, I'd prefer suggestions without them. For the Maclaurin series, we studied the formula $f(x)=f(0)+\frac{f'(0)x}{1!}+\frac{f''(0)x^{2}}{2!}+...+\frac{f^{(n)}(0)x^{n}}{n!}+o(x^{n})$. From this formula, we derived $\ln{(1+x)}=\sum_{k=1}^{n}{\frac{(-1)^{k-1}x^{k}}{k}}+o(x^{n})$. So, trying to use one of these, I thought of two approaches:

$\ln{(x(1+\frac{\sqrt{x^2+1}}{x}))}=\ln{x}+\ln{(1+\frac{\sqrt{x^2+1}}{x})}$ but then that root fraction is not suitable for the $\ln$ formula

the derivative of $\ln{(x+\sqrt{x^2+1})}$ is $(1+x^{2})^{-\frac{1}{2}}$. But then I will have to find the n-th derivative for the basic formula, which I wouldn't be able to. How should I procede?


问:
Integral of a vector field
I'm trying to evaluate the following integral:

$ \int_C(y+\sin x) dx +(z^2+\cos y)dy+(x^3)dz$

Where $C$ is the curve: $c(t) = (\sin t, \cos t, \sin 2t) $. Note that
  $C$ lies on the surface $z= 2xy$.

Question:
Could I just use the parameterization of the surface S $\phi(u,v)=(u, v,2uv) $ with $-1\leq  u\leq 1 $ and $-1\leq v \leq 1$ and use Stokes' Theorem to evaluate $ \int_S \nabla \times  F\cdot dS$ which should be equal to the required integral?

My only trepidation about the above method is that $c(t)$ may not necessarily be on the boundary of $S$, which would be required to use Stokes' Theorem.

答: Your instinct to use Stokes's Theorem is a good one. But you need to use the region in that surface whose boundary is $C$, as you understood. Think polar coordinates, since you've got a good start with the parametrization of the curve $C$. Try using $\phi(r,t) = (r\sin t, r\cos t, r^2\sin 2t)$. (Note that this is a parametrization of a region in your surface!) I'll let you figure out the $r$ and $t$ bounds.</p>

Title: Integral of a vector field


问:
Composition of equivalence relations and quotient set

Let $R$ be an equivalence relation on non empty set $X$. Prove that $X/(X/R\circ R \circ \dots R) = R$

Since $R$ is an equivalence relation we have $$R\circ R \circ \dots R = R$$So we should prove that $X/(X/R) = R$ but I don't know the meaning of $X/(X/R)$. According to the definition $$X/R=\{[x]_R:x\in X\}$$ So what's the meaning of $X/(X/R)$?

Edit: I found that this problem belongs to Set Theory With Applications. Here are some definitions which relate to this problem:

Definition $1$: Let $X$ be a nonempty set. By a partition $\mathcal{P}$ of $X$ we mean a set of nonempty subsets of $X$ such that

a) If $A,B \in \mathcal{P}$ and $A\not = B$ then $A \cap B = \emptyset  $

b) $\bigcup_{C \in \mathcal{P}}C = X$

Definition $2$: Let $R$ be an equivalence relation on a nonempty set $X$. For each $x \in X$, we define $$x/R = \{ y\in X | yRx \}$$ and $$X/R = \{ x/R | x \in X \}$$

Definition $3$: Let $\mathcal{P}$ be a partition of a nonempty set $X$. We define a relation $X/\mathcal{P}$ on $X$ by $x(X/\mathcal{P})y$ if and only if there exists a set $A \in \mathcal{P}$ such that $x,y \in A$.

答:
By definition 2, we have $X/R := \{[z]_R: z\in X\}$ where $R$ is a relation over $X$ and $[z]_R=\{y\in X:\langle y,z\rangle\in R\}$

First we note that $X/R$, being a set of $R$-equivalence classes, indeed partitions $X$.

Using the symbol $/\!\!/$ to avoid confusion, by definition 3, we have $X/\!\!/\mathcal P:=\{\langle x,y\rangle\in X^2:\exists A\in\mathcal P~.(x\in A\land y\in A)\}$ where $\mathcal P$ partitions $X$.

So let us inspect $X/\!\!/(X/R)$.

It is a relation over $X$, whose pairs of elements are both in the same part from $(X/R)$.  Since these parts are the $R$-equivalence classes, that entails that $X/\!\!/(X/R)$ is exactly the relation $R$.

$$\begin{align}X/\!\!/(X/R)&=\{\langle x,y\rangle\in X^2:\exists A\in (X/R)~.(x\in A\land y\in A)\}\\&= \{\langle x,y\rangle\in X^2:\exists A\in \{[z]_R:z\in X\}~.(x\in A\land y\in A)\}\\&~~\vdots\\&=\{\langle x,y\rangle\in X^2:\langle x,y\rangle\in R\}\\&=R\end{align}$$


问:
Showing that $R$/~ is compact
This is question from general topology.

Question : Let $a$~$b$ iff $a-b$ is an integer. Show that $R$/~ is compact.

But, my attempt : $R$/~ is homoemorphic to $(0, 1]$ . (It’s easily shown)

However, $(0, 1]$ is not closed w.r.t standard metric topology on $R$. Then by heine-borel, it should not be compact...

What did I do wrong?

答:
But, my attempt : $R$/~ is homoemorphic to $(0, 1]$ . (It’s easily shown)

This is were you are wrong: $\mathbb{R}/\sim$ is actually not homeomorphic to $(0,1]$.

Consider $S^1=\big\{z\in\mathbb{C}\ \big|\ \lVert z\rVert=1\big\}$, i.e. the standard sphere and let

$$f:S^1\to \mathbb{R}/\sim$$
$$f(z)=[arg(z)/2\pi]_\sim$$

where $arg$ stands for the complex argument. I leave as an exercise that this is a homeomorphism.

Title: Showing that $R$/~ is compact


问:
How to sum numerator and denominator of a fraction?
I want to do sum over this.  Can apply the summation to top and bottom separately?

$$\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$
$$=\frac{\sum\limits_{i=1}^{n}-a(x_i-\mu)^2}{\sum\limits_{i=1}^{n}x_i}$$

Is this correct?

Where can I find the rules to summations of a division, product, or addition?

Edit:

I want to solve for y,
$$\frac{5}{y}=\sum\limits_{i=1}^{n} \frac{-a(x_i-\mu)^2}{x_i}$$
$$\frac{y}{5}=\sum\limits_{i=1}^{n} \frac{x_i}{-a(x_i-\mu)^2}$$

Is this correct?

答:
No. For example $ \frac{3 + 2}{1 + 1} = \frac{5}{2} \neq 5 = \frac{3}{1} + \frac{2}{1} $. There are no basic rules for division or product. Of cource, $ \sum a_n + b_n = \sum a_n + \sum b_n $ which follows from definition.

One trick to use with products though is that $ \ln \prod a_n = \sum \ln a_n $.


答:
Consider the following two continuous maps:
$$i: [0,1] \to \Bbb R$$
the inclusion map $i(x)=x$, and the projection
$$\pi : \Bbb R \to \Bbb R / \sim$$

You can check that the composition $\pi \circ i$ is continuous and surjective.

Thus $\Bbb R / \sim$ is the coninuous image of $[0,1]$. Since $[0,1]$ is compact, so is $\Bbb R / \sim$.

Note that $\pi (i(0)) = \pi (i (1))$, so this is not a homeomorphism.

Title: Showing that $R$/~ is compact

Body:
This is question from general topology.

Question : Let $a$~$b$ iff $a-b$ is an integer. Show that $R$/~ is compact.

But, my attempt : $R$/~ is homoemorphic to $(0, 1]$ . (It’s easily shown)

However, $(0, 1]$ is not closed w.r.t standard metric topology on $R$. Then by heine-borel, it should not be compact...

What did I do wrong?


问:
Is this correct arguing?
Is this correct to conclude this way?

Premise $\neg p$

Premise $(\neg p \vee q) \to r$

Therefore $\neg p \wedge (\neg p \vee q) \to r$

Therefore $\neg p \to r$

答:
First, I would say you really need parentheses on line 3, i.e.

Therefore $\neg p \wedge \color{red}((\neg p \vee q) \to r\color{red})$

Second, yes, all inferences are valid: line 3 follows from lines 1 and 2, and line 4 follows from line 3. So, in that sense it is 'correct', as user2661923 points out in their answer.

However, and third, ... few people would regard this as 'good reasoning', let alone a 'proper proof'. The whole idea of a proof is that you break it down to small steps that are intuitively obvious. In this case, few people would find the step from line 3 to 4 'intuitively obvious'. Indeed, you might as well have skipped line 3 altogether. This is why proof systems typically provide you with certain rules that reflect inferences that are 'intuitively obvious'. See GrahamKemp's answer for more details on that.

Title: Is this correct arguing?


问:
The article "a" before the adjective "unique"
When we are talking about a universal property, we say that there exists "a" unique morphism that satisfies blah blah blah. However, I, a non-native speaker of English, feel it should be "the" unique morphism since it is uniquely determined. I agree that it is natural to say "a" unique morphism up to isomorphisms*, but I am not sure why we say "a" unique morphism even when it is strictly unique. Is there any good mathematical explanation for this? (One attempt is that since the existence is not guaranteed, "the" sounds somehow awkward to native English speakers.) Any linguistic intuition will also be appreciated.

*Another question: I often see the notation "unique up to isomorphism", but why can we omit the article "a" in this case despite the fact that there can be multiple choices of isomorphisms?

答:
This is an interesting question as much about English language usage as mathematics.

First,

there exists the unique morphism

simply sounds wrong. That's reason enough to stick with the conventional "a". You don't want your readers or listeners to stumble over unfamiliar usage.

I can venture a reason. When you say "the" you are  assuming existence, which is what your sentence is trying to assert. The proof usually runs along the lines "here is one, and in fact it's the only one". Using "a unique" somehow suggests the two separate steps.

As for

unique up to isomorphism

The same unfamiliarity holds. You could say

you could say

unique up to an isomorphism

but it wouldn't sound exactly right.

Grammatically, I read "up to isomorphism" as an adverb phrase modifying the adjective "unique" which in turn modifies the noun that names the essentially unique thing you are writing about.


问:
Matrix exponent $A^n$ in modular arithmetic $\mathbb{Z}/10\mathbb{Z}$
Let $A=\begin{bmatrix}1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}, A_{ij} \in \mathbb{Z}/{10\mathbb{Z}}$.

Determine how many different $A^n$ there are and which of them are equal to $A$.

My attempt: Simple multiplication doesn't yield any results, so I assume I'm supposed to apply some algebraic property of $\mathbb{Z}/10\mathbb{Z}$ but I'm not sure which one would be helpful in this case. Maybe something about it being a cyclic group.  Any hints would be helpful!

Edit:

The matrix $B = A^2 = \begin{bmatrix}0 & 6 & 2 \\ 6 & 1 & 6 \\ 2 & 6 &0 \end{bmatrix}$ is symmetric and it seems that $B^5 = B$, is there anything that would justify this and maybe enable discovery without direct calculation?

答:
This is a very interesting question.

EDIT 06.01.2021

Introducing what I call the equivalent 2x2-matrix appoach simplifies and unifies the task of finding $q(d)$ for all dimensions. The task now boils down to finding the order of an exponential congruence generalized to the 2x2 matrix.

EDIT 01.01.21

Explicit formulas for the mod(m) of $\alpha$ and $\beta$.

EDIT 30.12.20

Pursuing the second approach, in the case of even dimension $d$ we find a simple exponential congruence $\beta^k \equiv 1 \text{ mod(m)}$ ($\beta$ and $m$ are known integer functions of $d$) for which we need to find the smallest positive solution of the exponent $k$.

The general solution for this restricted case is still open (to me). Help from experts in number theory is greatly appreciated.

The more complicated case of odd $d$ has still to be done.

EDIT 27.12.20

Extension to two approaches to solve the generalized problem:

Brute force matrix multiplication (as in the original post)

Employing the characteristic polynom of the matrix and the Cayley-Hamilton theorem to derive a linear recurrence of the Fibonacci type

Generalized problem

Let $a(d)$ be a square matrix of dimension $d$ formed by consecutively filling in the numbers 1 through d^2 in lexicographic order.

The elements are hence

$$a_{ik}=i+d(k-1)\tag{1}$$

Now form the set $s(d) = (a, a^2, a^3, ..., a^q)$ of consecutive powers of the matrix $a$ taken modulo $m=d^2+1$, which we designate by $[..]$. The power $q$ is defined by the maximum value of $q$ for which all elements in the set $s(q)$ are different.

Task: find $q(d)$

Solution

0) Equivalent 2x2 matrix

This approach is a compact formulation of the results of my earlier studies documented before in this answer. It is valid for all dimensions $d$ and is based on the following theorem 1 which I provide here for the moment without proof.

Theorem 1

Consider the matrix

$$A(d) = \begin{bmatrix}\alpha(d) & 1  \\ \beta(d)&0 \end{bmatrix}\tag{0.1}$$

where

$$\alpha(d) = \frac{1}{2} d(d^2+1)\tag{0.2a}$$
$$\beta(d) = \frac{1}{12}d^3(d^2-1)\tag{0.2b}$$

and let

$$m=1+d^2\tag{0.3}$$

Then the number $q(d)$ of different powers of the original matrix $a$ of the OP defined in $(1)$ is given for $d$ even by the order $k$, and for $d$ odd by $k+1$ in the exponential congruence

$$A^{k+1} \equiv A \text{ (mod m)}\tag{0.4}$$

where the action of the module function on a matrix is defined as the action on its elements. Generalizing slightly the terminology of exponential congruences the order of a matrix $A$ modulo $m$ is defined as the smallest number $k>0$ for which $(0.4)$ holds.

1) Brute force calculation of the powers of the matrix of the OP

Let us start small.

For $d=2$ the first 9 powers of

$$a = \begin{bmatrix}1 & 2  \\ 3&4 \end{bmatrix}$$

mod 5 are

$$
\begin{array}{cc}
a = \begin{bmatrix}1 & 2  \\ 3&4 \end{bmatrix},
[a^2] = \begin{bmatrix}2 & 0  \\ 0&2 \end{bmatrix},
[a^3] = \begin{bmatrix}2 & 4  \\ 1&3 \end{bmatrix},
[a^4] =  \begin{bmatrix}4 & 0  \\ 0&4 \end{bmatrix},\\
[a^5] =  \begin{bmatrix}4 & 3  \\ 2&1 \end{bmatrix},
[a^6] = \begin{bmatrix}3 & 0  \\ 0&3 \end{bmatrix},
[a^7] = \begin{bmatrix}3 & 1  \\ 4&2 \end{bmatrix},\\
[a^8] = \begin{bmatrix}1 & 0  \\ 0&1 \end{bmatrix},
[a^9] = \begin{bmatrix}1 & 2  \\ 3&4 \end{bmatrix}
\end{array}
\tag{2}$$

We see that $[a^9] = a$ so that $q(2)=8$.

For $d=3$, the case of the OP, we have

$$
\begin{array}{ccc}
a= \begin{bmatrix}1 & 2&3  \\ 4&5&6\\7&8&9 \end{bmatrix},
[a^2] = \begin{bmatrix}0 & 6&2  \\ 6&1&6\\2&6&0 \end{bmatrix},
[a^3] = \begin{bmatrix}8&6&4  \\ 2&5&8\\6&4&2 \end{bmatrix},\\
[a^4] = \begin{bmatrix}0&8&6  \\ 8&3&8\\6&8&0 \end{bmatrix},
[a^5] = \begin{bmatrix}4&8&2  \\ 6&5&4\\8&2&6 \end{bmatrix},
[a^6] = \begin{bmatrix}0&4&8\\9&4&9\\8&4&0 \end{bmatrix},\\
[a^7] = \begin{bmatrix}2&4&6\\8&5&2\\4&6&8\end{bmatrix},
[a^8] = \begin{bmatrix}0&2&4\\2&7&2\\4&2&0\end{bmatrix},
[a^9] = \begin{bmatrix}6&2&8\\4&5&6\\2&8&4\end{bmatrix},\\
[a^{10}] = \begin{bmatrix}0&6&2\\6&1&6\\2&6&0\end{bmatrix}
\end{array}
\tag{3}$$

Interestingly, for $d=3$, the case of the OP, there is no power $k\gt 1$ for which $[a^k] = a$ but we have $[a^{10}] = [a^2]$, hence $q(3)=9$.

Now we let mathematica work out the first 20 values of $q(d)$:

$$q(d) =\\ \{1,8,9,32,7,2,41,24,81,40,\\31,56,33,56,225,512,57,120,61,800\}\tag{4}$$

2) Employing the characteristic polynomial of the matrix and reducig the problem to a linear recurrence relation of second order

The characteristic polynomial of a matrix $A$ is defined by

$$p(A,\lambda) = \left| \lambda  I - A\right|\tag{5}$$

and the characteristic equation $p(A,\lambda)=0$ gives the eigenvalues $\lambda$ of $A$ as roots.

For our matrix $a$ defined by $(1)$ the list of the first ten polynomials for $d=1..10$ in the format $\{d,p\}$ is

$$
\begin{array}{c}
 \{1,\lambda -1\} \\
 \left\{2,\lambda ^2-5 \lambda -2\right\} \\
 \left\{3,\lambda ^3-15 \lambda ^2-18 \lambda \right\} \\
 \left\{4,\lambda ^4-34 \lambda ^3-80 \lambda ^2\right\} \\
 \left\{5,\lambda ^5-65 \lambda ^4-250 \lambda ^3\right\} \\
 \left\{6,\lambda ^6-111 \lambda ^5-630 \lambda ^4\right\} \\
 \left\{7,\lambda ^7-175 \lambda ^6-1372 \lambda ^5\right\} \\
 \left\{8,\lambda ^8-260 \lambda ^7-2688 \lambda ^6\right\} \\
 \left\{9,\lambda ^9-369 \lambda ^8-4860 \lambda ^7\right\} \\
 \left\{10,\lambda ^{10}-505 \lambda ^9-8250 \lambda ^8\right\} \\
\end{array}
\tag{6}$$

The general form of $p(a,d)$ can be written as

$$p(a,d) =  \lambda ^d-\alpha ( d) \lambda ^{d-1}-\beta ( d) \lambda ^{d-2}\tag{7}$$

where

$$\alpha (d)=\frac{1}{2} d \left(d^2+1\right)\tag{8a}$$

$$\beta (d)=\frac{1}{12} d^3 \left(d^2-1\right)\tag{8b}$$

Now the theorem of Cayley and Hamilton states that a matrix obeys its own characteristic equation.

For example from $(6$) we get for $d=2$ the equation

$$a^2 = 5a + 2\tag{9}$$

where we have dropped writing down the unit matrix $I$.

This equation allows to write higher powers of $a$ in terms of a linear combination of lower powers. Indeed, if we also take into account that all terms should be taken mod $d^2+1=5$ for $d=2$, the elements of the set $s$ are can be calculated as follows

$$a, \\a^2 \to 5a+2\equiv 2 \text{ (mod 5)}, \\a^3=5a^2+2a \equiv_{5} 2a \\a^4 = 2a^2\to 2\times 2= 4\\a^5 = 4a\\a^6 \equiv_{5} 4a^2\to 4 \times 2 = 8\equiv_{5} 3\\a^7=3a\\ a^8 = 3a^2  \to 3 \times 2 = 6 \equiv_{5} 1 $$

Hence

$$s(d=2)  = \{a, 2, 2a, 4, 4a, 3, 3a, 1\}$$

which has $8$ different elements, hence $q(2)=8$.

The generalization is not difficult. We have to employ $(7)$ to get the general power of $a$.

We are going to do this by solving the corresponding difference equation

$$c_{k} = \alpha c_{k-1} + \beta c_{k-2}\tag{10}$$

with the initial conditions

$$c_1 = a, c_2 = a^2\tag{11}$$

$(10)$ is a linear difference equation of the Fibonacci type which can be easily solved by standard methods.

The result is with two constants $u$ and $v$

$$c(k) = u \; . 2^{-k} \left(\alpha -\sqrt{\alpha ^2+4 \beta }\right)^k\\+v\;. 2^{-k} \left(\alpha +\sqrt{\alpha^2+4 \beta }\right)^k\tag{12} $$

Modular arithmetic

We pause here for a moment to study the modular arithmetic of the function $\alpha(d)$ and $\beta(d)$, i.e. their remainder mod $m=(1+d^2)$.

For $\alpha$ we have

$$\alpha(d \text{ even})\equiv 0\tag{13a}$$

$$\alpha(d \text{ odd})\equiv_{m} \frac{1}{2}(1+d^2)\tag{13b}$$

For $\beta$ things are more complicated.

The first few terms for $\beta(d)$ mod($m$) are

$$ \beta(d) |_{d=1}^{d=20} \equiv_{m} \\ \{0, 2, 8, 12, 16, 1, 22, 23, 22, 69, 12, 2, 158, 68, 172, 174, 172, 3, 154, 137\}\tag{14a}$$

We can analyse the structure of this sequence by plotting it.

For even d we get

and for odd d

We see that there are "hidden" sub-sequences.
Specifically, three and six sub-sequences for even and odd d, respectively.

Closer inspection shows that we have explicit formulas for the sequences as follows

$\beta(d \text{ even))}\equiv_{m}
\left\{
 \begin{array}{ll}
        \frac{d}{6} & \;\; d \equiv_{6} 0\\
        12 \left(\frac{d-2}{6}\right)^2+\frac{9 (d-2)}{6}+2 & \;\; d \equiv_{6} 2\\
  24 \left(\frac{d-4}{6}\right)^2+\frac{33 (d-4)}{6}+12 &  \;\; d \equiv_{6} 4\\
 \end{array}
\right. \tag{14b}$

$\beta(d \text{ odd))}\equiv_{m}
\left\{
 \begin{array}{ll}
        132 \left(\frac{d-1}{12}\right)^2+\frac{24 (d-1)}{12}+2 & \;\; d \equiv_{12} 1\\
       108 \left(\frac{d-3}{12}\right)^2+\frac{56 (d-3)}{12}+8 & \;\; d \equiv_{12} 3\\
  84 \left(\frac{d+7}{12}\right)^2-\frac{96 (d+7)}{12}+28 &  \;\; d \equiv_{12} 5\\
60 \left(\frac{d+5}{12}\right)^2-\frac{48 (d+5)}{12}+10 &  \;\; d \equiv_{12} 7\\
36 \left(\frac{d+3}{12}\right)^2-\frac{16 (d+3)}{12}+2 &  \;\; d \equiv_{12} 9\\
12 \left(\frac{d+1}{12}\right)^2 &  \;\; d \equiv_{12} 11\\
\end{array}
\right. \tag{14c}$

Now that we have explicit formulas for the mod(m) of $\alpha$ and $\beta$ we can dive deeper.

Let us start with the case of even $d$ where $\alpha\equiv_{m}=0$ and the characteristic equation is simply

$$a^3 = \beta(d) a\tag{15}$$

(to do: the next steps have to be revised in view of $(14b)$ and $(14c)$).

Explicitly the first 10 equations are in the format {d, m, equn} are

$$
\begin{array}{c}
 \left\{2,5,a^3\equiv 2 a\right\} \\
 \left\{4,17,a^3\equiv 12 a\right\} \\
 \left\{6,37,a^3\equiv a\right\} \\
 \left\{8,65,a^3\equiv 23 a\right\} \\
 \left\{10,101,a^3\equiv 69 a\right\} \\
 \left\{12,145,a^3\equiv 2 a\right\} \\
 \left\{14,197,a^3\equiv 68 a\right\} \\
 \left\{16,257,a^3\equiv 174 a\right\} \\
 \left\{18,325,a^3\equiv 3 a\right\} \\
 \left\{20,401,a^3\equiv 137 a\right\} \\
\end{array}
\tag{16}$$

All sets $s$ start with $\{a,a^2,a^3\}$. Now $a^3$ has to be replaced from the corresponding by $\beta(d)  a$, the next element is $\beta a^2$, and so on.

Hence

$$s = \{a,a^2, \beta a, \beta a^2, \beta^2 a, \beta^2 a^2, ...,\beta^k a, \beta^k a^2 \}\tag{17} $$

The last two elements of s are a repetition of the first two if

$$\beta(d \text{ even})^k \equiv 1 \text{  mod}(m)\tag{18a}$$

Remember that $m=1+d^2$ and that $\beta$ is given by $(8b)$.

Let $k_{min}$ be the smallest positive value which solves $(18a)$. The number $k_{min}$ is called the order of $\beta$ modulo $m$. Then the number of elements of the set s is given by

$$q(d) = |s| = 2 k_{min}\tag{18b}$$

Our task is hence reduced for even $d$ to the problem of finding the order of $\beta(d)$ modulo $1+d^2$.

Concerning its general solution we have to look up the theory of exponential congruences. Nice examples are Solution to exponential congruence and https://www.youtube.com/watch?v=oCgWwB465p8.

However, an upper limit of $k$ can be found using the little Fermat theorem in the formuation of Euler which states that if $m$ and $\beta$ have no common divisors (which is the case in our examples) then

$$k = k_{\phi} = \phi(m)\tag{19}$$

solves $(18a)$. Here $\phi(m)$ is Euler's totient function. Unfortunately, in most cases $k_{\phi}$ is not the smallest solution of $(18a)$.

Indeed, for d=8 we have m=65, $\beta=23$, $\phi(65) = 48$ but we have already $23^{12}\equiv 1 \text{ mod 65}$

It is helpful to notice that the order of $\beta$ modulo $m$ must be a divisor of $\phi(m)$. Hence we have reduced the possible number of orders. In the case of prime $m$ we have $\phi(m) = m-1=d^2$.

An even more effective procedure to find the order is described here https://math.stackexchange.com/a/165728/198592.

Discussion

For the specific problem of the OP which has $d=3$ the answer is 9. There are 9 different matrices.

Two values appear twice ($q(12)=q(14)=56$.

The case $d=6$ is remarkable since here already $[a^3] = a$

I have not found a general formula for $q(d)$. This might be difficult since the sequence is not contained in https://oeis.org/.

Matrix multiplication and taking the module

We can verify that for two matrices $a$ and $b$ with integer elements we have the relation

$$[ a.b ] = [[a].[b]]\tag{5}$$

Hence the two operations can be interchanged.

Group properties?

The set $s$ does not form a group for the simple reason that the matrix $a$ is singular and hence has no inverse.

The even powers have an inverse but in general have non integer elements, and also do not generate a unit element.

Rather, for each $d$, the set $s$ is an algebra with a multiplication rule defined by the characteristic equation.


答: In the context of universal properties a unique morphism may be different depending on the objects involved, so there can be many unique morphisms. For example in the category of rings $\mathbb{Z}$ is inital but the morphisms to $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ are each unique, but distinct. This makes it more reasonable to say that there exist a unique morphism since the unique morphism would imply there was only one morphism.

Title: The article "a" before the adjective "unique"

Body:
When we are talking about a universal property, we say that there exists "a" unique morphism that satisfies blah blah blah. However, I, a non-native speaker of English, feel it should be "the" unique morphism since it is uniquely determined. I agree that it is natural to say "a" unique morphism up to isomorphisms*, but I am not sure why we say "a" unique morphism even when it is strictly unique. Is there any good mathematical explanation for this? (One attempt is that since the existence is not guaranteed, "the" sounds somehow awkward to native English speakers.) Any linguistic intuition will also be appreciated.

*Another question: I often see the notation "unique up to isomorphism", but why can we omit the article "a" in this case despite the fact that there can be multiple choices of isomorphisms?