问:
Unifying the connections between the trigonometric and hyperbolic functions
There are many, many connections between the trigonometric and hyperbolic functions, some of which are listed here. It is probably too optimistic to expect that a single insight could explain all of these connections, but is there a holistic way of seeing the parallels between $\sin$ and $\sinh$, $\cos$ and $\cosh$? Can all of these seemingly disparate connections be shown to be essentially the same, or at least very similar?

Geometric connections

Sine and cosine parameterise the unit circle $x^2+y^2=1$, just as hyperbolic sine and cosine parameterise the 'unit hyperbola' $x^2-y^2=1$. Both circles and hyperbolas are conic sections.

The sector of the circle connecting the points $(0,0)$, $(1,0)$, and $(\cos t,\sin t)$ has an area of $t/2$. The region of the hyperbola connecting the points $(0,0)$, $(1,0)$, and $(\cosh t,\sinh t)$ has an area of $t/2$. This can even be used to define the hyperbolic functions geometrically, and many authors do the same with the trigonometric functions.

Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts.

The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$. The compound angle formulae are almost identical to their hyperbolic counterparts, save for a pesky minus sign:
\begin{align}
\sin(x+y) &= \sin(x)\cos(y)+\cos(x)\sin(y) \\
\sinh(x+y) &= \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \\[4pt]
\cos(x+y) &= \cos(x)\cos(y) \color{red}{-} \sin(x)\sin(y) \\
\cosh(x+y) &= \cosh(x)\cosh(y) \color{blue}{+} \sinh(x)\sinh(y) \, .
\end{align}

In general, given a trigonometric function, it is possible to write down the corresponding hyperbolic identity using Osborn's rule: replace every occurrence of $\cos$ with $\cosh$; replace every occurrence of $\sin$ by $\sinh$; but negate the product of two $\sinh$ terms.

Analytic connections

$\sin$ is the unique solution to the initial value problem
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 1 \\
f(0) &= 0 \, ,
\end{align}
and the corresponding initial value problem for $\sinh$ is the same, except $f''(x) = \color{blue}{+}f(x)$.

Likewise, the initial value problem for $\cos$ is
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 0 \\
f(0) &= 1 \, ,
\end{align}
and again we see a mysterious sign change for $\cosh$: $f''(x) = \color{blue}{+}f(x)$.

It follows that the higher-order derivatives of $\sin$ and $\sinh$ form periodic sequences.

If we solve the initial value problems shown above, we obtain the exponential forms of all $4$ functions:
\begin{align}
\sin x &= \frac{e^{\color{green}{i}x}-e^{-\color{green}{i}x}}{2\color{green}{i}} \quad{} \cos x = \frac{e^{\color{\green}{i}x}+ e^{-\color{green}{i}x}}{2} \\[3pt]
\sinh x &= \frac{e^{x}-e^{-x}}{2} \quad{} \cosh x = \frac{e^x + e^{-x}}{2} \, .
\end{align}

All $4$ functions are analytic, and their Taylor series bear a striking resemblance to each other:
\begin{align}
\sin x &= x \color{red}{-} \frac{x^3}{3!} + \frac{x^5}{5!} \color{red}{-} \frac{x^7}{7!} + \ldots \\[4pt]
\sinh x &= x \color{blue}{+} \frac{x^3}{3!} + \frac{x^5}{5!} \color{blue}{+} \frac{x^7}{7!} + \ldots \\[4pt]
\cos x &= 1 \color{red}{-} \frac{x^2}{2!} + \frac{x^4}{4!} \color{red}{-} \frac{x^6}{6!} + \ldots \\[4pt]
\cosh x &= 1 \color{blue}{+} \frac{x^2}{2!} + \frac{x^4}{4!} \color{blue}{+} \frac{x^6}{6!} + \ldots
\end{align}

And Euler's formula
$$
e^{ix} = \cos x + i \sin x
$$
is replaced by the underwhelming
$$
e^x = \cosh x + \sinh x \, .
$$

答:
The trigonometric functions are related to the imaginary exponential $e^{iv}=\cos v+i\sin v$.

The hyperbolic functions are related to the real exponential $e^{u}=\cosh u+\sinh u$.

They both are special cases of the complex exponential $e^z=e^{u+iv}$.

The associated conics are obtained by

$$1=e^{iv}e^{-iv}=(\cos v+i\sin v)(\cos v-i\sin v)=\cos^2u-i^2\sin^2u,$$

$$1=e^{v}e^{-v}=(\cosh v+\sinh v)(\cosh v-\sinh v)=\cosh^2u-\sinh^2u.$$

A unifying ODE is

$$z''''=z$$ giving the characteristic polynomial $$\omega^4=1,\\\omega=\pm1,\pm i$$

and the solution

$$z=a\cos t+b\sin t+c\cosh t+d\sinh t.$$

Just think

$$\text{real}\leftrightarrow\text{imaginary}.$$


问:
A differentiation and double differential function proof

This is a question which I received from my friend as he was not able to solve it. Even I and my professor too failed to solve this question. Pls someone help me with this question.

I have been able to proved it till here-

g(0) 9

答:
Suppose $f(x)$ meets the conditions of the problem. Note that since $|f(0)| \le 1$, $2\sqrt 2 \le |f'(0)| \le 3$. Now $-f(x), f(-x),$ and $-f(-x)$ also meet all the conditions. At least one of the four functions will have both derivative and second derivative $\ge 0$ at $x = 0$. Let $h(x)$ be the one with both $h'(0) \ge 0, h''(0) \ge 0$.

So $2\sqrt 2 \le h'(0) \le 3$ and $h''(0) \ge 0$. If $h'''(0) < 0$, we are done. Otherwise, Let $a = \sup \{b \ge 0 \mid h''' \ge 0 \text{ on } [0,b]\}$, so $h''' \ge 0$ on $[0,a)$.

$h''$ is increasing on this interval, and so $h''(x) \ge h''(0) \ge 0$ everywhere. This in turn means that $h'$ is increasing on the interval as well. So $h'(x) \ge h'(0) \ge 2\sqrt 2$. Since $h'$ is continuous, it is integrable and $$h(a) - h(0) = \int_0^a h'(x)\,dx \ge 2\sqrt 2 a$$
But $|h(a) - h(0)| \le |h(a)| + |h(0)| \le 2$. So
$$2 \ge 2\sqrt 2a\\\frac 1{\sqrt 2} \ge a$$

Because $h'(a) \ge 2\sqrt 2$ and continuous, there is some neighborhood of $a$ where $h' > 0$. But by the choice of $a$, there are points $c$ in that neighborhood where $h'''(c) < 0$, and therefore $h'(c)h'''(c) < 0$.

From this it follows that either $f'(c)f'''(c) < 0$ or $f'(-c)f'''(-c) < 0$, and further the point $c$ or $-c$ is in $(-b, b)$ for any $b > \frac 1{\sqrt 2}$, for which $b = 3$ certainly qualifies.

ContentLicense: CC BY-SA 4.0


问:
Can the following expression be related to the multinomial formula?
The following formula
$$(n,p,q,r\ \text{odd})\quad \sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \times \begin{cases} 1 & \text{if}\ p< q <r\\
 \frac{1}{2} & \text{if exactly two indices are equal}\\
 \frac{1}{3!}& \text{if}\ p=q=r= \frac{n}{3} \end{cases}$$
reminds me very much of the "multinomial formula"
$$ \left( x_1 + x_2 + \cdots + x_d \right)^n = \sum_{\left|\boldsymbol{\alpha}\right|=n} \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}}\, \mathbf{x}^{\boldsymbol{\alpha}}
$$
where $\boldsymbol{\alpha}\in \mathbb{N}^d,\ \left|\boldsymbol{\alpha}\right| = \alpha_1 + \alpha_2 + \cdots + \alpha_d $  and
$$ \begin{split}
 \genfrac{(}{)}{0pt}{0}{\left|\boldsymbol{\alpha}\right|}{\boldsymbol{\alpha}} & = \genfrac{(}{)}{0pt}{0}{n}{\alpha_1} \genfrac{(}{)}{0pt}{0}{n-\alpha_1}{\alpha_2}\, \cdots \, \genfrac{(}{)}{0pt}{0}{n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1}}{\alpha_d}= \frac{n!}{\alpha_1 !\, (n-\alpha_1)!} \frac{(n-\alpha_1)!}{\alpha_2 !\, (n-\alpha_1-\alpha_2)!}\, \cdots\, \frac{(n-\alpha_1 -\alpha_2 - \cdots - \alpha_{d-1})!}{\alpha_d !\, 0 !} \\
 & = \frac{n!}{\alpha_1 !\, \alpha_2 ! \, \cdots \, \alpha_d!}
 \end{split}
$$
is the number of times the facteur $x^{\boldsymbol{\alpha}}:= x_1^{\alpha_1} x_2^{\alpha_2} \cdots x_d^{\alpha_d}$ appears in the expansion of $\left( x_1 + x_2 + \cdots + x_d \right)^n$.

In our case $d=3,\ \boldsymbol{\alpha}= (p,q,r)$.

There could be other possibilities but I'm thinking of taking $(x_1,x_2,x_3)= \left(1, \frac{1}{2}, \frac{1}{3} \right)$ then I need to find a "simple" function of $\boldsymbol{\alpha}$, $f:\mathbb{N}^3 \to \mathbb{N}^3 $  which would take value $(1,0,0)$ if $p<q<r$ (i.e. if $\boldsymbol{\alpha}$ view as a function $1\mapsto \alpha_1=p,\ 2\mapsto \alpha_2=q,\ 3\mapsto \alpha_3=r$ is injective), take value $(0,1,0)$ if exactly two indices are equal (or if $\boldsymbol{\alpha}$ view as a the previous function takes only two values) and $(0,1,1)$ if $p=q=r$.

If we find such a function then the first sum can be rewritten
$$\sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \mathbf{x}^{f(\boldsymbol{\alpha})}$$

The final step, if possible would be to relate $\mathbf{x}^{f(\boldsymbol{\alpha})}$ to $\mathbf{x}^{\boldsymbol{\alpha}}$ by integration or derivation or maybe something different.

答:
Yes, they are related. The coefficients in the case bracket account for the fact that you require $p\leqslant q \leqslant r$, whereas the usual formula for the multinomial theorem makes no such assumption about $\alpha_1$, $\alpha_2$ and $\alpha_3$.

If $p$, $q$ and $r$ are pairwise distinct, the coefficient $\binom{n}{p,q,r}$ would appear $3! = 6$ times in the multinomial formula, corresponding to the ways of assigning the values of $p$, $q$ and $r$ to $\alpha_1$, $\alpha_2$ and $\alpha_3$.

Similarly, if $p=q\neq r$, the coefficient $\binom{n}{p,q,r}$ would appear $3$ times in the multinomial formula, corresponding to the ways of assigning the values of $p$, $q$ and $r$ to $\alpha_1$, $\alpha_2$ and $\alpha_3$.

Finally, if they are all the same, the coefficient appears only once, as there is a signle way to assign the values.

So, aside from the restriction that $r$ be odd, $3!$ times your expression would be another representation of the expression you get from the multinomial theorem for $(1+1+1)^n$.
Put another way, your expression equals

$$
\frac1{3!}\sum_{\genfrac{}{}{0pt}{1}{p+q+r =n}{\max\{p,q,r\} \text{ is odd}}}
\binom{n}{p,q,r}
$$


问:
Technical detail in Rudin's proof of Cartan's uniqueness theorem
In Rudin's "Function theory in the Unit ball of $\mathbb{C}^n$", page 23, the following theorem, due to Cartan, is stated and proven:

Suppose that $\Omega$ is a bounded connected domain in $\mathbb{C}^n$, $F:\Omega\rightarrow\Omega$ is holomorphic, $F(p)=p$ and $F'(p)=Id$ for some $p\in\Omega$. Then $F(z)=z$ for all $z\in\Omega$.

Now, the proof is very simple and clear, but there is one technical detail that baffles me. First of all, one can obviously assume $p=0$, so that there is a ball $B_1\subseteq\Omega$ of radius $r_1$ in which $F$ has a homogeneous expansion
$$F(z)=z+\sum_{s=2}^{\infty}F_s(z)$$
where every $F_s$ is a map from $\mathbb{C}^n$ to $\mathbb{C}^n$ whose components are homogeneous polynomials of degree s.
Now, we want to prove by induction on $m$ that every $F_m$ is zero. To do so, we assume the property to be true for $s<m$ and, for  $k\in\mathbb{N}$, we consider the $k$-th iterate $F^k$ of $F$.
This is where Rudin loses me: the proof continues saying that $F^k$ has homogeneous expansion
$$F^k(z)=z+kF_m(z)+\dots$$
in B_1.
I cannot see why $B_1$ suits for all $k$. It is true that the series of the $k$-th iterate converges on a sufficiently small ball surrounding the origin, but I do not understand how the radius could be independent from $k$. My point is that the image of $B_1$ is not necessarily contained in $B_1$, and a priori we know that the series expansion for $F$ is valid only on this ball.
I have found some ways to get around the problem (e.g. restricting the radius, or observing that $F$ is a biholomorphism near the origin which fixes the origin) but none of them seems to work: a fundamental part of the proof is showing that this expansion works for every $k$ in the same ball, since we want $F_m$ to be identically zero on an open set.

Is there something (possibly quite trivial) that I am missing?

答: It doesn't matter that $F(B_1)$ may not be a subset of $B_1.$ If you have $F_k(z) = z+kF_m(z)+\cdots$ for $|z|$ sufficiently small, then this must hold in $B_1.$ That's because $F_k$ is holomorphic on $\Omega,$ hence on $B_1,$ hence has a homogenous expansion in $B_1.$ The coefficients for that expansion depend only the behavior of $F_k$ near $0,$ which is given by $z+kF_m(z)+\cdots.$


问:
Empty theory is a sequence of isolated points
I'm working my way through Henson and van den Dries Model Theory lecture notes doing odd-numbered exercises. Yet I am stuck on the exercise 4.17:

Show that if $T$ is the empty theory in the language $L$ of equality, then the space $S_0(T)$ consists of a sequence of points $(T_n | n \geq 1)$ that are isolated, together with a point $T_\infty$ to which this sequence converges.

Nowhere in the notes do they define what an "empty theory" is (I am only aware of an empty language). How would such a thing even look like? What are the consequences of it? In particular, wouldn't it imply everything?

Then the "isolated points" are also not defined anywhere. How could I possibly start proving the statement? It seems more of a topology question to me than model theory...

Any help or hints would be appreciated!

答:
Let me fill in a few details of Noah's excellent answer. The empty theory $T$ in the language $\mathcal{L}=\{=\}$ is just that – the theory that contains no $\mathcal{L}$-sentences. Any set is a model of $T$. You ask if $T$ implies everything; this is not the case. Indeed, the sentences derivable from $T$ are precisely the $\mathcal{L}$-sentences that hold in every set. These are also called the tautologies of the predicate calculus. For instance, the following is a theorem of $T$: $$\forall v\forall w(v=w\vee v\neq w),$$ since it is true in every set (ie, true in every model of $T$). On the other hand, consider the $\mathcal{L}$-sentence $\phi:\equiv\exists v\exists w(v\neq w)$. $\phi$ is not a theorem of $T$, since there are models of $T$ in which it does not hold. For example, consider a model $\{\star\}$ consisting of a single element. Note that $\neg\phi$ is not a theorem of $T$ either, since $\phi$ holds in the model $\{\star,\bullet\}$ consisting of $2$ elements. So, in other words, both $T\cup\{\phi\}$ and $T\cup\{\neg\phi\}$ are consistent. Hopefully this clarifies the question a bit.

Now, recall that $S_0(T)$ is the set of all complete (and consistent) $\mathcal{L}$-theories that contain $T$. Since every $\mathcal{L}$-theory contains $T$ (because the empty set is a subset of any set), this means $S_0(T)$ is just the set of all complete $\mathcal{L}$-theories. Topologically, a basis of clopen sets for $S_0(T)$ is parametrized by the set of all $\mathcal{L}$-sentences: for any $\mathcal{L}$-sentence $\phi$, we define $[\phi]\subseteq S_0(T)$ to be the set of all elements of $S_0(T)$ that contain $\phi$. So, if $\phi\equiv\forall v\forall w(v=w\wedge v\neq w)$, then $[\phi]=S_0(T)$, since $\phi$ holds in any model of $T$. On the other hand, $[\neg\phi]=\emptyset$. (Why?)

For a less vacuous example, consider the sentence $\phi\equiv\exists v\exists w(v\neq w)$. Then both $[\phi]$ and $[\neg\phi]$ are non-empty. Indeed, if $M$ is a model of $\phi$ (say $M=\{\star,\bullet\}$), then the set of all $\mathcal{L}$-sentences that hold in $M$, denoted $\operatorname{Th}_\mathcal{L} M$, is a complete and consistent $\mathcal{L}$-theory containing $\phi$. So $\operatorname{Th}_\mathcal{L}M\in [\phi]$. Likewise, if $N$ is a model of $\neg\phi$ (say $N=\{\star\}$), then $\operatorname{Th}_\mathcal{L} N\in[\neg\phi]$.

So, this is the topology we're working with. Recall that a point $x$ in an arbitrary topological space $X$ is "isolated" if $\{x\}$ is an open subset of $X$. In particular, a complete $\mathcal{L}$-theory $S$ is isolated in $S_0(T)$ if $\{S\}$ is open. As an exercise, show that this holds if and only if there exists an $\mathcal{L}$-sentence $\phi$ such that $[\phi]=\{S\}$. Furthermore, can you show that, for such $\phi$, we have $T\models\phi\to\psi$ for every $\psi\in S$? We then say that $S$ is isolated by $\phi$. As a final hint, to complement Noah's answer, consider the sentence $$\phi_n\equiv\exists v_1\dots\exists v_n\bigwedge_{i\neq j}v_i\neq v_j\wedge\forall w\bigvee_{i=1}^n w=v_i.$$ Up to isomorphism, what does a model of $\phi_n$ look like? (Can there be two non-isomorphic models of $\phi_n$?) In particular, what can you conclude about the open set $[\phi_n]\subseteq S_0(T)$? (Remember that two isomorphic $\mathcal{L}$-structures $M\cong N$ are elementarily equivalent: $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$.) I'll leave my comments at this; hopefully you can now work out the details yourself.

Edit: Okay, here's a complete solution to the question. At each stage, try to read a small bit at a time and work out the rest for yourself! For each $n\in\mathbb{N}$, let $\phi_n$ be the sentence given above, which says that there exist precisely (and no more than) $n$ elements. Now, because $\mathcal{L}$ consists of only equality, any function is an $\mathcal{L}$-embedding, and a function between $\mathcal{L}$-structures is an isomorphism if and only if it is a bijection. In particular, suppose we have models $M\models\phi_n$ and $N\models\phi_n$ for some $n\in\mathbb{N}$. Then $|M|=|N|=n$ (why?) and so we can find a bijection – ie an $\mathcal{L}$-isomorphism – $M\to N$. In particular, since isomorphic structures are elementarily equivalent, this means $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$, and so (since $M$ and $N$ were arbitrary) this shows that any two models of the sentence $\phi_n$ have the same $\mathcal{L}$-theory. Denoting this theory by $T_n$, we then have $[\phi_n]=\{T_n\}$, and so in particular $T_n$ is isolated.

Now, there are $\mathcal{L}$-structures in which $\phi_n$ does not hold for any $n\in\mathbb{N}$ – just consider any infinite set. To deal with these models, first fix any infinite cardinal $\kappa\geqslant\aleph_0$, and suppose that $M$ and $N$ are two models of $T$ of size $\kappa$. (Ie, two sets of size $\kappa$.) Then there exists a bijection $M\to N$, which is an $\mathcal{L}$-isomorphism by the comments above, and so $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$. Thus any two models of cardinality $\kappa$ have the same complete $\mathcal{L}$-theory, for any $\kappa\geqslant\aleph_0$. In fact, this means that any two infinite models have the same complete $\mathcal{L}$-theory! Indeed, suppose $|M|=\kappa$ and $|N|=\lambda$ for $\kappa,\lambda\geqslant\aleph_0$. Without loss of generality assume $\kappa\geqslant\lambda$. Since $\mathcal{L}$ is finite, the Löwenheim-Skolem theorem tells us that there exists an elementary substructure $M'\preccurlyeq M$ with $|M'|=\lambda$. Then $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}M'$, and, by the argument above, since $|M'|=\lambda=|N|$, we also have $\operatorname{Th}_\mathcal{L}M'=\operatorname{Th}_\mathcal{L}N$, so $\operatorname{Th}_\mathcal{L}M=\operatorname{Th}_\mathcal{L}N$ as desired. Thus any two infinite models of $T$ have the same complete $\mathcal{L}$-theory; call this theory $T_\infty$.

Now, we first claim that $S_0(T)=\{T_n\}_{n\in\mathbb{N}}\cup\{T_\infty\}$. Indeed, let $M$ be any model of $T$ (ie any set). If $M$ is finite, then $\operatorname{Th}_\mathcal{L}M=T_{|M|}$, and if $M$ is infinite, then $\operatorname{Th}_\mathcal{L}M=T_\infty$, by the arguments above. In particular, every complete $\mathcal{L}$-theory appears in $\{T_n\}_{n\in\mathbb{N}}\cup\{T_\infty\}$, as desired. By the argument above we've shown that each $T_n$ is isolated, by the formula $\phi_n$, so it remains only to show that the sequence $(T_n)_{n\in\mathbb{N}}$ converges to $T_\infty$. This amounts to the following statement: for any $\mathcal{L}$-sentence $\phi$ with $T_\infty\in[\phi]$, there exists some $k\in\mathbb{N}$ such that $T_n\in[\phi]$ for any $n\geqslant k$.

To see this, suppose $\phi$ is an $\mathcal{L}$-sentence such that $T_n\notin[\phi]$ for arbitrarily large values of $n$. This means that, for every $k\in\mathbb{N}$, we can find $n\geqslant k$ such that $\phi$ does not hold in any $\mathcal{L}$-structure of size $k$. By the compactness theorem, this means that the following family of sentences is consistent: $$\Sigma=\{\exists v_1\dots\exists v_k\bigwedge_{i\neq j}v_i\neq v_j\}_{k\in\mathbb{N}}\cup\{\neg\phi\}.$$ Indeed, for any finite subset $\Delta\subset\Sigma$, there will be a largest value of $k$ such that the sentence $\exists v_1\dots\exists v_k\bigwedge_{i\neq j}v_i\neq v_j$ appears in $\Delta$. We will then have $M\models\Delta$ for any set $M$ such that $k\leqslant|M|\in\mathbb{N}$ and $T_{|M|}\notin[\phi]$, meaning that $\Delta$ is consistent. So, by compactness, $\Sigma$ is consistent, and hence we can find a model $M\models\Sigma$. Then $M$ is infinite (why?), so $\operatorname{Th}_\mathcal{L}M=T_\infty$ (why?), but $M\models\neg\phi$, whence $T_\infty\notin[\phi]$. This proves the desired result, and so we are done.


问:
An integral involving Fresnel integrals $\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$
I need to calculate the following integral:
$$\int_0^\infty \left(\left(2\ S(x)-1\right)^2+\left(2\ C(x)-1\right)^2\right)^2 x\ \mathrm dx,$$
where
$$S(x)=\int_0^x\sin\frac{\pi z^2}{2}\mathrm dz,$$
$$C(x)=\int_0^x\cos\frac{\pi z^2}{2}\mathrm dz$$
are the Fresnel integrals.

Numerical integration gives an approximate result $0.31311841522422385...$ that is close to $\frac{16\log2-8}{\pi^2}$, so it might be the answer.

答:
Step 1. Reduction of the integral

Let $I$ denote the integral in question. With the change of variable $v = \frac{\pi x^2}{2}$, we have

$$ I = \frac{1}{\pi} \int_{0}^{\infty} \left\{ (1 - 2 C(x) )^{2} + (1 - 2S(x))^{2} \right\}^{2} \, dv $$

where $x = \sqrt{2v / \pi}$ is understood as a function of $v$. By noting that

$$ 1-2 S(x) = \sqrt{\frac{2}{\pi}} \int_{v}^{\infty} \frac{\sin u}{\sqrt{u}} \, du \quad \text{and} \quad 1-2 C(x) = \sqrt{\frac{2}{\pi}} \int_{v}^{\infty} \frac{\cos u}{\sqrt{u}} \, du, $$

we can write $I$ as

$$ I = \frac{4}{\pi^{3}} \int_{0}^{\infty} \left| A(v) \right|^{4} \, dv \tag{1} $$

where $A(v)$ denotes the function defined by

$$ A(v) = \int_{v}^{\infty} \frac{e^{iu}}{\sqrt{u}} \, du. $$

Step 2. Simplification of $\left| A(v) \right|^2$.

Now we want to simplify $\left| A(v) \right|^2$. To this end, we note that for $\Re u > 0$,

$$ \frac{1}{\sqrt{u}}
= \frac{1}{\Gamma\left(\frac{1}{2}\right)} \frac{\Gamma\left(\frac{1}{2}\right)}{u^{1/2}}
= \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-ux}}{\sqrt{x}} \, dx
= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} e^{-ux^{2}} \, dx \tag{2} $$

Using this identity,

\begin{align*}
A(v)
&= \frac{2}{\sqrt{\pi}} \int_{v}^{\infty} e^{iu} \int_{0}^{\infty} e^{-u x^2} \, dx du
 = \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \int_{v}^{\infty} e^{-(x^2-i)u} \, du dx \\
&= \frac{2 e^{iv}}{\sqrt{\pi}} \int_{0}^{\infty} e^{-v x^2} \int_{0}^{\infty} e^{-(x^2-i)u} \, du dx
 = \frac{2 e^{iv}}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-v x^2}}{x^2-i} \, dx.
\end{align*}

Thus by the polar coordinate change $(x, y) \mapsto (r, \theta)$ followed by the substitutions $r^2 = s$ and $\tan \theta = t$, we obtain

\begin{align*}
\left| A(v) \right|^2
&= A(v) \overline{A(v)}
 = \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-v (x^2+y^2)}}{(x^2-i)(y^2 + i)} \, dxdy \\
&= \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{\frac{\pi}{2}}  \frac{r e^{-v r^2}}{(r^2 \cos^{2}\theta-i)(r^2 \sin^{2}\theta + i)} \, d\theta dr \\
&= \frac{2}{\pi} \int_{0}^{\infty} \int_{0}^{\frac{\pi}{2}}  \frac{e^{-v s}}{(s \cos^{2}\theta-i)(s \sin^{2}\theta + i)} \, d\theta ds \\
&= \frac{2}{\pi} \int_{0}^{\infty} \frac{e^{-v s}}{s} \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{s \cos^{2}\theta-i} + \frac{1}{s \sin^{2}\theta + i} \right) \, d\theta ds \\
&= \frac{2}{\pi} \int_{0}^{\infty} \frac{e^{-v s}}{s} \int_{0}^{\infty} \left( \frac{1}{s -i(t^2 + 1)} + \frac{1}{s t^2 + i (t^2 + 1)} \right) \, dt ds.
\end{align*}

Evaluation of the inner integral is easy, and we obtain

\begin{align*}
\left| A(v) \right|^2
&= 2 \int_{0}^{\infty} \frac{e^{-v s}}{s} \Re \left( \frac{i}{\sqrt{1 + is}} \right) \, ds.
\end{align*}

Applying $(2)$ again, we find that

\begin{align*}
\left| A(v) \right|^2
&= 2 \int_{0}^{\infty} \frac{e^{-v s}}{s} \Re \left( \frac{i}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-(1+is)u}}{\sqrt{u}} \, du \right) \, ds \\
&= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-v s}}{s} \int_{0}^{\infty} \frac{e^{-u} \sin (su)}{\sqrt{u}} \, du\, ds \\
&= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \int_{0}^{\infty}  \frac{\sin (su)}{s} \, e^{-v s} \, ds\, du \\
&= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \arctan \left( \frac{u}{v} \right) \, du \\
&= \frac{4\sqrt{v}}{\sqrt{\pi}} \int_{0}^{\infty} e^{-vx^{2}} \arctan (x^2) \, dx \qquad (u = vx^2) \tag{3}
\end{align*}

Here, we exploited the identity

$$ \int_{0}^{\infty} \frac{\sin x}{x} e^{-sx} \, dx = \arctan \left(\frac{1}{s}\right), $$

which can be proved by differentiating both sides with respect to $s$.

Step 3. Evaluation of $I$.

Plugging $(3)$ to $(1)$ and applying the polar coordinate change, $I$ reduces to

\begin{align*}
I
&= \frac{64}{\pi^{4}} \int_{0}^{\infty} \int_{0}^{\infty} \int_{0}^{\infty} v e^{-v(x^{2}+y^{2})} \arctan (x^2) \arctan (x^2) \, dx dy dv \\
&= \frac{64}{\pi^{4}} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\arctan (x^2) \arctan (x^2)}{(x^2 + y^2)^2} \, dx dy \\
&= \frac{64}{\pi^{4}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{\arctan (r^2 \cos^2 \theta) \arctan (r^2 \sin^2 \theta)}{r^3} \, dr d\theta \\
&= \frac{32}{\pi^{4}} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{\arctan (s \cos^2 \theta) \arctan (s \sin^2 \theta)}{s^2} \, ds d\theta. \qquad (s = r^2) \tag{4}
\end{align*}

Now let us denote

$$ J(u, v) = \int_{0}^{\infty} \frac{\arctan (us) \arctan (vs)}{s^2} \, ds. $$

Then a simple calculation shows that

$$ \frac{\partial^{2} J}{\partial u \partial v} J(u, v) = \int_{0}^{\infty} \frac{ds}{(u^2 s^2 + 1)(v^2 s^2 + 1)} = \frac{\pi}{2(u+v)}. $$

Indeed, both the contour integration method or the partial fraction decomposition method work here. Integrating, we have

$$ J(u, v) = \frac{\pi}{2} \left\{ (u+v) \log(u+v) - u \log u - v \log v \right\}. $$

Plugging this to $(4)$, it follows that

\begin{align*}
I
&= -\frac{64}{\pi^{3}} \int_{0}^{\frac{\pi}{2}} \sin^2 \theta \log \sin \theta \, d\theta
 = -\frac{16}{\pi^{3}} \frac{\partial \beta}{\partial z}\left( \frac{3}{2}, \frac{1}{2} \right)
\end{align*}

where $\beta(z, w)$ is the beta function, satisfying the following beta function identity

$$ \beta (z, w) = 2 \int_{0}^{\infty} \sin^{2z-1}\theta \cos^{2w-1} \theta \, d\theta = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}. $$

Therefore we have

\begin{align*}
I
&= \frac{16}{\pi^{3}} \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(2)} \left\{ \psi_{0} (2) - \psi_{0} \left(\tfrac{3}{2} \right) \right\}
 = \frac{8}{\pi^2} \int_{0}^{1} \frac{\sqrt{x} - x}{1 - x} \, dx
 = \frac{8 (2 \log 2 - 1)}{\pi^2},
\end{align*}

where $\psi_0 (z) = \dfrac{\Gamma'(z)}{\Gamma(z)}$ is the digamma function, satisfying the following identity

$$ \psi_{0}(z+1) = -\gamma + \int_{0}^{1} \frac{1 - x^{z}}{1 - x} \, dx. $$

ContentLicense: CC BY-SA 3.0


问:
what value of K does the system have a unique solution
$\begin{cases}x_1 + kx_2 - x_3 = 2\\2x_1 - x_2 + kx_3 = 5\\x_1 + 10x_2 -6x_3= 1\\
\end{cases}$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\2&-2&K&5\\1&10&-6&1\end{array}\right]$

I've been trying echelon form where i took $R_2 = R_2 - 2R_1$ and $R_3 = R_3-R_1$
and reduced it

So I have $\left[\begin{array}{ccc|c}1&K&-1&2\\0&-1-2K&K+2&1\\0&10-K&-5&-1\end{array}\right]$

But now I am not sure how i could remove $10-K$ with $-1-2K$ any help would be appreciated

答:
Without solving you the problem, you can ALWAYS proceed with the following way if you don't spot any easy factorisation.

Suppose that, in your matrix, you have two values $a_{i,j}$ and $a_{i+1,j}$ you want to reduce in that way. You can ALWAYS exploit the property of linearity, and proceed with the substitution $R_{i+1} = a_{i,j}R_{i+1} - a_{i+1,j}R_{i}$, providing that $a_{i,j} \neq 0$.

In your case, proceed with the substitution: $R_3 = (-1-2K)R_3 - (10-K)R_2$, providing (i.e. check at the end if the solution allows it) that $-1-2K \neq 0$.


问:
find the remainder for $\color{purple}{TETRATION}$ such that $( {^{2021}2021}(\mod13))$
It is given that $ {^{2021}2021}$ is a tetration and I want to find the remainder when it is divided by $13$.

Briefly , I am looking for $ {^{2021}2021}\pmod{13}$

My work :

$\color{purple}{Firstly:}$ I found $2021\equiv 6\pmod{13}$

$\color{blue}{Secondly:}$ I thought that $6^{\color{red}{a}}\equiv 6^{\color{red}{b}}\pmod{13}$ should be satisfied.

$\color{orange}{Thirdly:}$ $a\equiv b\pmod{12}$ by Euler's phi function.Then , i said that $a\equiv 11$ and $b\equiv -1$

$\color{red}{Fourthly:}$ I said that $6^{11^m}\equiv 6^{(-1)^m}\pmod{13}$ and there left $2020$ superscript for tetration. Therefore , I said that it must be equal to $6^{(-1)^{2020}}\equiv6\pmod{13}$

$\therefore$ Answer is $6$

Is my solution correct , if not , can you share any trick or shortcut for finding remainders when we encounter tetrations..

$\color{red}{NOTE:}$  WHAT IS TETRATION $\rightarrow $  https://en.wikipedia.org/wiki/Tetration

$\color{red}{NOTE-2:}$ I would rather colorful answer  for the sake of @terasalisbon :)

$\color{GREEN}{THANKS...}$

Body:
$^{2021}2021 = 2021^{^{2020}2021}\pmod{13}\equiv 2021^{^{2020}2021\pmod {12}}\pmod {13}\equiv 2021^{2021^{^{2019}2021}\pmod{12}}$

$6^{5^{^{2019}2021}\pmod{12}}\pmod{13}$

Now the order of $5^2 \equiv 1 \pmod 12$ so $5^{odd} \equiv 5 \pmod{12}$ and $5^{even}\equiv 1\pmod {12}$ and $^{2019}2021$ is odd.

So $6^{5^{^{2019}2021}\pmod{12}}\pmod {13}\equiv 6^5 \equiv 6^2*6^2*6\equiv (-3)^2*6\equiv 45\equiv 2 \pmod {13}$.


问:
Number of points on elliptic curves over finite fields
I noticed that for elliptic curves of the form

$y^2≡x^3+a\pmod p$

sometimes the number of points is always $p+1$ for any choice of $a$. This seems to be the case for all $p ≡ 5 \pmod 6$.

Moreover, when this does not happen, i.e., for $p ≡ 1 \pmod 6$, it looks like there are exactly zero curves of such form where the number of points is $p+1$.

Can someone point me towards the right direction as to why this happens?

答:
The case $p \equiv 5 \pmod 6$ was already solved here, so I'll just handle the other one.

Denote $S_a = \{ (x,y) | y^2=x^3+a \}$. Then $\bigcup S_a = \mathbb{F}_p^2$. We want to prove $|S_a| \neq p$ for non-zero $a$ (with our point at infinity this would mean that number of points on the curve is $ \neq p+1$).

Firstly, look at $S_0$ (this case actually won't be needed, but I think it's good to show it as well). $y^2=x^3$ either if $(x,y)=(0,0)$ (that's one point) or if $y^2=t$ and $x^3=t$ for some non-zero $t$. Since $\mathbb{F}_p^{\times} \sim \mathbb{Z}_{p-1}$, $t$ is a sixth power, so we have $\frac{p-1}{6}$ choices for $t$, two solutions for $y^2=t$ and three solutions for $x^3=t$. Therefore $|S_0|=1+\frac{p-1}{6} \cdot 2 \cdot 3 = p$.

Now we will just investigate cases:

$a$ is not a quadratic residue

Then for any $y$, $y^2-a$ is non-zero. So for any particular $y$, equation $y^2-a=x^3$ has either zero or three solutions. Therefore $|S_a|$ is divisible by $3$, so it can't be equal to $p$.

$a$ is not a cubic residue

Then for any $x$, $x^3+a$ is non-zero, so $y^2=x^3+a$ regarded as an equation for $y$ has zero or two solutions. Thus $|S_a|$ is even, so it's not $p$.

$a$ is both quadratic and cubic residue

Well, $|S_a|$ is just

$$ \sum_x \#\{y:y^2=x^3+a\} $$

Every term in this sum is equal to two if $x^3+a$ is quadratic residue, zero if it's quadratic non-residue, and one if it's zero. Consider $b$ that is a quadratic non-residue. Then

$$ \#\{y:b^3y^2=x^3+a\} + \#\{y:y^2=x^3+a\} = 2 $$

for any $x$, so

$$ 2p = \sum_x \#\{y:b^3y^2=x^3+a\} + \#\{y:y^2=x^3+a\} = \sum_x \#\{y:b^3y^2=x^3+a\} + |S_a|$$

But $\sum_x \#\{y:b^3y^2=x^3+a\}$ is the same as $\sum_x \#\{y:y^2=x^3+ab^{-3}\} = |S_{ab^{-3}}|$: and now $ab^{-3}$ is a quadratic non-residue, so we already know that it's not equal to $p$. Therefore we also have $|S_a| \neq p$.


问:
Definition of a limit when there are no cones?
A very pedantic question: If a category $\mathcal{C}$ admits no cones, then does it contains all limits vacuously? Or does the definition of a limit imply the existence of at least one limit?

答:
You should distinguish more carefully between a diagram which admits no cones and a category which admits no cones. As the other answer explains, it is only the empty category that admits no cones over any diagram, so this property is not very interesting.

Let’s consider the other property, then. For instance the identity functor of the discrete category on two objects admits no cone, and this implies it admits no limit, since a limit cone is in particular a cone. This is one of the three ways a diagram can lack a limit: it can have no cones at all, or it can no cone through which every other cone factors (this is called a weak limit cone) or it can have no weak limit cone, factorizations through which are always unique.


问:
Computing turning number of closed curve
I'm trying to compute the turning number of a closed plane curve, which should be a really straight forward computation. But my result is not a whole number, i.e. the total curvature is not a multiple of $2\pi$, which is impossible so I'm making a mistake somewhere. I know from plotting the curve that the turning number should be $-2$. I've been looking for hours but cannot seem to find it. Maybe someone can spot the mistake... Here is my work:

Let $\beta(t) = ((1+2\cos(t))\sin(t), (1+2\cos(t))\cos(t))$ be a regular closed plane curve. To compute the turning number $n_{\beta}$ I want to use the following:

$$n_{\beta} = \frac{1}{2\pi}\int_{0}^{L}K_{\beta}(t)dt\, ,$$

where $L$ is the period of $\beta$ and $K_{\beta}(t)$ is the curvature. I can compute the curvature like this:

$$K_{\beta} = \frac{1}{||\dot{\beta}(t)||^3}\cdot\det({\dot{\beta}(t), \ddot{\beta}(t)})$$

Computations:

$$\dot{\beta}(t) = (\cos t + 2\cos 2t, -\sin t -2\sin 2t)$$
$$\ddot{\beta}(t) = (-\sin t - 4 \sin 2t, -\cos t - 4 \cos 2t)$$
$$||\dot{\beta}(t))|| = (4\cos t + 5)^{1/2}$$
$$\det({\dot{\beta}(t),\ddot{\beta}(t)}) = -6 \cos t - 9$$

This is giving me the curvature

$$K_{\beta}(t) = \frac{-6 \cos t -9}{(4 \cos t + 5)^{3/2}}$$

At this point I integrated the curvature with the help of Wolframalpha, which did it numerically but I'm pretty sure it's $-2\pi-1$, when from the picture it should be $-4\pi$. Oddly enough, if I integrate the curvature without the power of $\frac{3}{2}$ in the denominator i actually get the correct result. Can anyone help me spot my mistake? Thanks a lot!

Edit: Here's a link to see the graph: https://www.desmos.com/calculator/vcxbpj2ios

答: Is this your curve, given by your equations?


问:
Proof of Cartan's Magic formula
I'm trying to prove Cartan's magic formula, $\mathcal{L}_{v}(X) = i_{v}(dX)+d(i_{v}X)$ for X a p-form.

My attempt so far:

We can choose a basis (t,$x^{i}$) such that $v$ = $d/dt$. In this basis

$\mathcal{L}_{v}(X) = \frac{\partial }{\partial t}(X_{\mu_{1}...\mu_{p}})$, $v^{\alpha} = \delta^{\alpha}_{0}$ and set $\frac{\partial }{\partial t} = \partial_{0}$

Then I evaluate:

$i_{v}(dX) = i_{v}((p+1)\partial_{[\nu_{1}}X_{\mu_{1}...\mu_{p}]}) = (p+1)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}...\mu_{p}]}$

and:

$ d(i_{v}X) = d(v^{\alpha}X_{\alpha\mu_{2}...\mu_{p}}) = (p)\partial_{[\mu_{1}}v^{\alpha}X_{\alpha\mu_{2}...\mu_{p}]} = (p)v^{\alpha}\partial_{[\mu_{1}}X_{\alpha\mu_{2}...\mu_{p}]}$ (as $v^{\alpha}$ is constant)

Putting things together:

$i_{v}(dX)+d(i_{v}X) = (p+1)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}...\mu_{p}]}+(p)v^{\alpha}\partial_{[\mu_{1}}X_{\alpha\mu_{2}...\mu_{p}]} = (p+1)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}...\mu_{p}]}-(p)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}\mu_{2}...\mu_{p}]} = v^{\alpha}\partial_{[\alpha}X_{\mu_{1}\mu_{2}...\mu_{p}]}$

Using $v^{\alpha} = \delta^{\alpha}_{0}$ we get $i_{v}(dX)+d(i_{v}X) =\partial_{[0}X_{\mu_{1}\mu_{2}...\mu_{p}]}$

I think this expression however is incorrect, as I don't see how $\partial_{[0}X_{\mu_{1}\mu_{2}...\mu_{p}]} = \partial_{0}X_{[\mu_{1}\mu_{2}...\mu_{p}]} = \partial_{0}X_{\mu_{1}\mu_{2}...\mu_{p}} =\mathcal{L}_{v}(X)$

Is this the correct approach to take or am I missing something obvious? Thank you very much!

答:
I think taking a slightly more abstract route might be clearer in this context. Consider the following definition:

Let $M$ be a smooth manifold and $r\in\Bbb{Z}$. By a superderivation of degree $r$ acting on differential forms on $M$, we shall mean a collection of $\Bbb{R}$-linear maps $\theta^{(k)}:\Omega^k(M)\to\Omega^{k+r}(M)$ (though we usually don't put the index $k$, which is easily inferred from context) such that the "Leibniz rule" is satisfied:

For every $\alpha\in \Omega^k(M), \beta\in\Omega^l(M)$, we have $\theta(\alpha \wedge \beta)= [\theta(\alpha)]\wedge \beta + (-1)^{kr} \alpha \wedge[\theta(\beta)]$.

Here, $\Omega^k(M)$ is defined to be the space of all differential forms on $M$ of degree $k$ (and by convention we put $\Omega^k(M)=\{0\}$ if $k<0$). There are two nice features about such superderivations:

They are local: if two forms $\alpha$ and $\beta$ agree on an open set $U\subset M$, then $\theta(\alpha)$ and $\theta(\beta)$ also agree on $U$.

They are completely determined by their action on smooth functions and $1$-forms; i.e if you know $\theta(f)$ and $\theta(dg)$ for every $f,g\in \Omega^0(M)=C^{\infty}(M,\Bbb{R})$ then you know $\theta(\omega)$ for every $k$-form $\omega$.

To prove the first bullet point, its a standard bump function argument. For the second assertion, write $\omega$ as a linear combination $\sum\omega_I dx^I$, where $I=i_1,\dots, i_k$ is an increasing multi-index; then the Leibniz rule and $\Bbb{R}$-linearity will completely determine the outcome.

Finally, one last thing to note:

If $\theta_1$ and $\theta_2$ are superderivations of degree $r_1$ and $r_2$, then we define their "supercommutator" $[\theta_1,\theta_2]:= \theta_1\circ \theta_2 -(-1)^{r_1r_2}\theta_2\circ \theta_1$. This is a super-derivation of degree $r_1+r_2$.

The proof is a direct calculation, and the funny sign makes everything work out in the end.

For your actual question, note that the exterior derivative $d$ is a superderivation of degree $+1$, while the interior product $\iota_v$ is a superderivation of degree $-1$. Hence, we can consider their supercommutator
\begin{align}
\theta_v := [d,\iota_v] = d\circ\iota_v - (-1)^{(1)(-1)}\iota_v\circ d = d\circ \iota_v + \iota_v\circ d
\end{align}
This is now a super-derivation of degree $0$ (i.e it maps $k$-forms to $k$-forms, and satisfies the product rule with no minus signs).

It is also standard to check (starting with the flow definition) that the Lie derivative $L_v$ is also a super-derivation of degree $0$. So, in order to check $L_v = \theta_v$, all you have to do is verify that for all $f,g\in C^{\infty}(M)$, we have $L_v(f)=\theta_v(f)$ and $L_v(dg)= \theta_v(dg)$. But now this should be quite simple.

So, really the point of this answer was to show you that once you set up appropriate definitions, there are actually very few things you have to check: just check equality on functions and exact $1$-forms; because of the Leibniz rule this will then imply equality on all $k$-forms. So in a sense this is like a nice "uniqueness in extension" type of argument.

Title: Proof of Cartan's Magic formula


问:
How do get combination sequence formula?
What would be a closed-form formula that would determine the ith value of the sequence

1, 3, 11, 43, 171...

where each value is one minus the product of the previous value and 4?

Thanks!

答:
The OEIS gives the closed form formula of this sequence to be
$$
\frac{2^{(2n+1)} + 1}{3}.
$$</p>


问:
Reason to apply the Koszul sign rule everywhere in graded contexts
I'm copy-pasting this question I asked in MO that received no answer.

The Koszul sign rule is a sign rule that arises from graded commutative algebras. For instance, let $\bigwedge(x_1,\dots, x_n)$ be the free graded commutative algebra generated by $n$ elements of respective degrees $\lvert x_i\rvert$. Then, the sign $\varepsilon(\sigma)$ of a permutation $\sigma$ on $(x_1,\dotsc, x_n)$ is given by
$$x_1\wedge\dotsb\wedge x_n=\varepsilon(\sigma)x_{\sigma(1)}\wedge\dotsb\wedge x_{\sigma(n)},$$
which comes from the fact that in a graded commutative algebra one has by definition $a\wedge b = (-1)^{\lvert a\rvert\lvert b\rvert}b\wedge a$.

There is also an antisymmetric Koszul sign rule which arises from graded anticommutative algebras and it's just the previous sign times the sign of the permutation. Both signs are used for instance in Lada and Markl - Symmetric brace algebras.

However, I've been seeing the Koszul sign rule used in any graded context and even for operations that are not products in some algebra. For example, from Roitzheim and Whitehouse - Uniqueness of $A_\infty$-structures and Hochschild cohomology, given graded maps of graded algebras $f,g:A\to B$, if we want to evaluate $f\otimes g$ in an element $x\otimes y$, apparently we need to apply the sign rule to get
$$(f\otimes g)(x\otimes y)=(-1)^{\lvert x\rvert\lvert g\rvert}f(x)\otimes g(y),$$
but I see no mathematical reason to do that, it just seems to be a convention.

A more complex example of application of the Koszul sign rule is in the definition of brace algebra (also in the Lada and Markl paper).

I could give many more examples. In some of them I can understand the reason. For instance, the differential of a tensor product of complexes $C$ and $D$ cannot simply be $d_C\otimes 1_D+ 1_C\otimes d_D$ (it can be defined this way if we use the sign rule when we apply it to elements, but in any case it needs the sign). But maps in general need not be differentials. In other cases, the signs appear in nature and one use this sign rule to justify them, as in $A_{\infty}$-algebras, but this feels too artificial for me and doesn't really explain why we should use that sign rule.

So, in the end, every time there is a sequence $(x_1,\dotsc, x_n)$ of graded objects of any kind and not necessarily all of them of the same kind (elements, maps, operations, …), and related in any way (they can be multiplied, or applied, or whatever), we use the Koszul sign rule to permute the sequence.

To me all of this seems more philosophical than mathematical, and as I said it feels to be just a convention. But, is there some general mathematical reason to use the sign rule in such an extensive way? And if it's just a convention, why should we use it? From my experience, it gets very messy when it comes to applying the sign rule to larger formulas, and in the end everything is just a $+$ or $-$ sign, so I see no advantage.

答:
First of all I should say that I am not familiar with those cohomology theories and high-tech gadgets you mentioned in your problem statement. But recently I learn some graded algebra stuff, and I would like to write what I know about this sign rule. Given two graded abelian groups $A=\{A_m\}_{m\in M}, B=\{B_m\}_{m\in M}$ where $(M, +, 0)$ is a monoid, the graded tensor product is $$A\otimes B=\left\{\bigoplus_{i+j=m}A_i\otimes B_j\right\}_{m\in M}.$$ Most widely-used examples are $M=\mathbb{N}, \mathbb{Z}, \mathbb{Z}_2.$ In theses cases, the graded brading is an isomorphism $$A\otimes B\xrightarrow{\cong}B\otimes A$$ given index-wise by $A_i\otimes B_j\xrightarrow{(-1)^{i+j}}B_j\otimes A_i,$ which is exactly the Koszul sign rule. This rule appears in

cup product operation on cocycles :  $\alpha^p\smile\beta^q=(-1)^{pq}(\beta^q\smile\alpha^p)$

wedge product of differential forms: $\alpha\wedge\beta=(-1)^{\text{deg}(\alpha)\text{deg}(\beta)}\beta\wedge\alpha$

super geometry (including Grassmann algebra), etc.

As far as I understand, and as the second example suggests, the reason for this sign change is keeping track of "orientation" for computations. On the other hand, this is necessary as otherwise we cannot obtain some important isomorphisms like

Künneth formula $H^*(M\times N)\cong H^*(M)\otimes H^*(N)$

uniqueness of determinants

Also, again as the second example suggests, the Koszul sign convention is related with sign of permutations.


问:
Question about spectral radius of a positive matrix
I am learning the Perron-Frobenius theorem from some lecture notes. Let $X \in \mathbb{R}^{n}_{++}$ be a square matrix with each element being strictly positive. The theorem says that the spectral radius of matrix $X$, denoted by $\rho(X)$, is an eigenvalue of X, and all other eigenvalues have a strictly smaller absolute value. But why does this implies that
$$
\rho(X) = \sup_{x: ||x||_2 \leq 1} ||Xx||_2 = \sup_{x: ||x||_2 \leq 1} \sqrt{x^T X^T X x}?
$$
If $X$ is further symmetric, I can see it is true. Any idea why this is true in general?

*The same question has also been asked and answered here:
Is spectral radius = operator norm for a positive valued matrix?

答:
The statement that you have made is not true. For example, consider the matrix
$$
X = \pmatrix{10 & 100\\1 & 10}.
$$
Its eigenvalues are $0,20$, which means that $\rho(X) = 20$. On the other hand, we find that $\|X\|_2 = 101 > \rho(X)$.


问:
Does a non-measurable set always contain a bounded non-measurable subset?
This question might be easy but I lack the proper intuition of non-measurable sets.

The only example of a non-measurable set I can think of is the Vitali set.

答:
I will write down explicitly what is meant in @Teresa Lisbon's comment.

Let $S$ be an unbounded non-measurable set.
For $n\ge 1$ call $$S_n=S \cap [-n; n]$$
Argue by contradiction that all the $S_n$ are measurable. Then their union is measurable.
However
$$\bigcup_{n=1}^{\infty} S_n = \bigcup_{n=1}^{\infty} S \cap [-n, n] = S \cap \bigcup_{n=1}^{\infty} [-n, n] = S \cap \Bbb R = S$$
Hence $S$ is measaurable, contradicting our assumptions.

This means that some $S_n$ is not measurable. And clearly $S_n \subseteq S$ is bounded.


问:
Finding my mistake in a "proof" that there is no non-trivial knot
I am working on a presentation about the complexity of determining whether a knot is trivial, and while reading about some topological stuff I somehow managed to "show" that there are no non-trivial knots. Obviously, there has to be something wrong with my "proof", but I can't see where I did something wrong, or maybe which of the things I read about are inaccurate.

Here are the definitions and propositions I've picked up from various papers.

Definition 1: An embedding $N\hookrightarrow M$ is proper, if $\partial N=N\cap\partial M$. And a properly embedded hypersurface $S\subset M$ is essential in $M$, if it cannot be homotoped into $\partial M$ while keeping $\partial S$ fixed in place, and if the fundamental group $\pi_1(S)$ of $S$ injects into $\pi_1(M)$.

Proposition 2: Let $K$ be a tame knot embedded in $\mathbb S^3$, and let $T_K\subseteq\mathbb S^3$ be an open tubular neighborhood of $K$. The knot $K$ is trivial, if and only if the knot complement $M_K:=\mathbb S^3\setminus T_K$ of $K$ contains an essential disk (whose boundary certainly is a longitude of $\overline{T_K}$).

(as far as I know this is a corollary from the fact that a knot embedded in $\mathbb R^3$ is trivial, if and only if it has a spanning disk, that is, a disk embedded in $\mathbb R^3$ whose boundary is the knot traversed once.)

Proposition 3: The complement $T'$ of the interior of a solid torus $T$ in $\mathbb S^3$ is homeomorphic to a solid torus, the longitudes of $T$ are the meridians of $T'$ and the meridians of $T$ are the longitudes of $T'$.

Proposition 4: Every solid torus $T$ contains an essential disk, and every such disk has a meridian of $T$ as boundary.

And here is my corollary from those propositions.

My corollary: Every tame knot is trivial. Proof. Let $K\subset\mathbb S^3$ be any tame knot. Since $K$ is an embedding of the circle $\mathbb S^1$ into $\mathbb S^3$, any open tubular neighborhood $T_K\subseteq\mathbb S^3$ of $K$ is homeomorphic to $\mathbb S^1\times(D^2)^\circ$, i.e. the interior of a solid torus. By proposition 3 the knot complement $M_K=\mathbb S^3\setminus T_K$ is homeomorphic to a solid torus which, by proposition 4, contains an essential disk (with a meridian of $M_K$ or a longitude of $\overline{T_K}$ as boundary). Hence, $K$ is trivial by proposition 1.

Can you see where I've made a mistake or which propositions are inaccurate?

答:
Proposition 3 is false, or at least phrased in a misleading way: the complement of a knotted solid torus is S³ is certainly not a solid torus: this happens if and only if the solid torus is unknotted, essentially because of your argument.

Actually, more is true: a theorem of Gordon and Luecke says that the homeomorphism type of the complement of a knotted solid torus determines the type of the knot.

What's true is that the complement of a standard, revolution solid torus in S³ (which certainly is unknotted) is a solid torus: this doesn't apply to knoted solid tori.


问:
Average of Two Quadratic Forms
I'm trying to show the following: $$\frac{1}{2}(x^{T}Ax +y^{T}Ay) \leq x^{T}Ay$$ with A symmetric and $||x||_{2}, ||y||_{2} \leq 1$. I'm not sure if it's true or not and was wondering if anyone had any insight into this? Thanks a ton!

答:
For any symmetric matrix $A$, we have
$$
\frac 12 (x - y)^TA(x - y) = \frac{x^TAx + y^TAy}{2} - x^TAy.
$$
With that, we can conclude that your inequality will hold for all $x,y$ (with norm less than $1$) if and only if $A$ is negative semidefinite. If $A$ is positive semidefinite, then the opposite inequality will hold. In other cases, neither inequality holds for all $x,y$ (with norm less than $1$).

Title: Average of Two Quadratic Forms


问:
How to get sign right for parametric area for circle
For finding the area of a circle of radius $r$ using $x=r \cos(t)$ and $y=r \sin(t)$ with $t$ from $0$ to $2 \pi$, it seems that the formula usually given for parametric area, which is $\int_a^b y(t)x'(t)dt$, gives a negative result and all the examples I could find use the formula of $\int_a^b x(t)y'(t)dt$

From what I've read, for the area formula to work, you'd need a closed curve but the sign depends on the direction? From what I read on brilliant (the above paragraph, as well as the circle example), my negative sign is due to the using the first formula and not the second... however, if I use my formula with $\int_{2\pi}^{0} r \sin(x) \cdot -1 r \sin(x) \ dx$ then it works out well and I get $\pi r^2$ with the positive sign

Then reading this answer to a user with a similar issue, the suggestion is due to the sign of the product of the terms inside more than anything.

Which is the right formula to use, and is there a good way to predict which one would give the correct sign?

答:
Both formulas are correct; care needs to taken though in determining the integration limits of the parametrized variable.

In the case of $\int y(t)x'(t)dt=\int y (x)dx$, the integration is over the range $x\in [-r,r]$, which corresponds to $ [\pi,0]$ under the parameterization $x=r\cos t$ (i.e. $-r= r\cos\pi$ and $r=r\cos 0$) and the area integral for the circle is given by

\begin{align}
Area & =\int_{-r}^r [y-(-y)]\>dx = 2\int_{-r}^r y\>dx\\
&= 2 \int_\pi^0 y(t)x’(t) \>dt\\
&= 2 r^2 \int_\pi^0(-\sin^2t)\>dt\\
&= \pi r^2
\end{align}


问:
$f \in O(\log(n))$?
How can I find the complexity of

```<code>i = 1;
while( i <n)
{
 i = 2*i;
}
</code>```

using summation?

I have $f(n) = \sum_{i=1, i=2*i}^{n-1}1$, but I am not sure how to solve that. I think $f \in O(\log(n))$

答:
In your while-loop $i$ increases exponentially.

On the $k^{th}$ iteration, you have $i= 2^{k}$ and will end when $2^k\ge n$.

So, the loop will exit as soon as $k\ge \log_2(n)$.

Title: $f \in O(\log(n))$?


问:
Determinant of a block matrix with entries on the diagonal, right above the diagonal and corner
I am currently trying to take the determinant of a block matrix on the form
$$\left[\begin{matrix} I & K_1 & 0 & 0 &  \cdots & 0&0
\\ 0 & I & K_2 & 0 & \cdots & 0 & 0
\\\vdots &\vdots&\vdots&\ddots &&\vdots&\vdots
\\0&0&0&0&\cdots&I&K_{N-1}
\\K_N&0&0&0&\cdots&0&I\end{matrix}\right]$$
where $I$ is an $M \times M$ identity matrix, and $K_i$ are almost tridiagonal $M\times M$ matrices, but which also have entries in the top right and bottom left corners.
$$K_i=\left[\begin{matrix} \eta_{i,1} & \eta_{i,2}Q_{i,1} & 0 & 0 &  \cdots & 0&\eta_{i,M}Q_{i,M}^*
\\ \eta_{i,1}Q_{i,1}^* & \eta_{i,2} & \eta_{i,3}Q_{i,2} & 0 & \cdots & 0 & 0
\\ 0&\eta_{i,2}Q_{i,2}^* & \eta_{i,3} & \eta_{i,4}Q_{i,3} & \cdots & 0 & 0
\\\vdots &\vdots&\vdots&\ddots&&\vdots&\vdots
\\0&0&0&0&\cdots&\eta_{i,M-1}&\eta_{i,M}Q_{i,M-1}
\\\eta_{i,1}Q_{i,M}&0&0&0&\cdots&\eta_{i,M-1}Q_{i,M-1}^*&\eta_{i,M}\end{matrix}\right]$$
Here, $Q_{i,j}$ are complex variables, and $Q_{i,j}^*$ denotes complex conjugate. Also, $\eta_{i,j}$ are arbitrary real variables. Is there in general any way to find the determinant of such a matrix?

答:
For computing the determinant of your matrix, $\left[\begin{matrix} I & K_1 & 0 & 0 &  \cdots & 0&0
\\ 0 & I & K_2 & 0 & \cdots & 0 & 0
\\\vdots &\vdots&\vdots&\ddots &&\vdots&\vdots
\\0&0&0&0&\cdots&I&K_{N-1}
\\K_N&0&0&0&\cdots&0&I\end{matrix}\right]$, an equivalent form can be constructed by subtracting $K_N \times$ the first row from the last row, which gives $$\left[\begin{matrix} I & K_1 & 0 & 0 &  \cdots & 0&0
\\ 0 & I & K_2 & 0 & \cdots & 0 & 0
\\\vdots &\vdots&\vdots&\ddots &&\vdots&\vdots
\\0&0&0&0&\cdots&I&K_{N-1}
\\0 &-K_N K_1&0&0&\cdots&0&I\end{matrix}\right]$$ In here, subtract $-K_N K_1 \times$  the second row from the last row, which gives $$\left[\begin{matrix} I & K_1 & 0 & 0 &  \cdots & 0&0
\\ 0 & I & K_2 & 0 & \cdots & 0 & 0
\\\vdots &\vdots&\vdots&\ddots &&\vdots&\vdots
\\0&0&0&0&\cdots&I&K_{N-1}
\\0 &0&K_N K_1 K_2&0&\cdots&0&I\end{matrix}\right]$$ Continuing like this leads to
$$\left[\begin{matrix} I & K_1 & 0 & 0 &  \cdots & 0&0
\\ 0 & I & K_2 & 0 & \cdots & 0 & 0
\\\vdots &\vdots&\vdots&\ddots &&\vdots&\vdots
\\0&0&0&0&\cdots&I&K_{N-1}
\\0 &0&0&0&\cdots&0&I +(-1)^{N-1}K_N\prod_{j=1}^{N-1} K_j\end{matrix}\right]$$ and the sought determinant of this strict triangular matrix is $\det( I + (-1)^{N-1}  K_N\prod_{j=1}^{N-1} K_j)$. This can be evaluated with the formula
$\det(I+A) = \sum_{k=0}^M \operatorname{Tr}(\Lambda^{k} A)$, see here, also for details how to technically proceed. Since $A = (-1)^{N-1}K_N\prod_{j=1}^{N-1} K_j$, I wonder if this can be achieved elegantly.

For the individual determinants of tridiagonal matrices with corners, usually done with transfer matrices, you can refer to the treatment e.g. in here: http://pcteserver.mi.infn.it/~molinari/RMT/RM_tridiagonal.pdf


问:
continuous extension at x=0

The function $f(x) = (\arctan \left(\frac 1x\right))^2$ has a continuous extension at $x=0$ because $\displaystyle \lim_{x \to 0} f(x)= \dfrac{\pi^2}4$.

I solved it using squeeze theorem.

I am trying to use the same method for $\arctan$ because: $\dfrac{-\pi}2 < \arcsin < \dfrac{\pi}2$ . Does it work?

答:
The arcsinus function is only defined for $ x\in [-1,1] $.

So, to the function $$x\mapsto \arcsin(\frac 1x)$$

is defined if
$$0<\frac 1x \le 1$$
or
$$-1\le \frac 1x <0$$

which is equivalent to
$$x\ge 1 \text{ or } x\le -1$$

this tells us, that the function $$x\mapsto (\arcsin(\frac 1x))^2$$

is not defined near $ 0 $, so, we cannot speak about
$$\lim_{x\to 0}(\arcsin(\frac 1x))^2$$

Title: continuous extension at x=0


问:
Specifically, what is Pi besides a number allowing you to calculate the circumference of a circle?
What is Pi? Why is Pi the only number able to calculate the circumference of a circle? Also, how was Pi discovered?</p>

答:
Usually, Pi ($\pi$) is defined to the number that is the ratio of the circumference of a circle to its diameters. That is, if you measure a circle perfectly and find the circumference  equals $C$ and the diameter equals $d$, then you will have $$\pi = \frac{C}{d}.$$

Now, there are many other good numbers out there for determining the circumference of a circle. For example, the number $\tau$ is the ratio of the circumference of a circle, $C$, to its radius, $r$. That is $$\tau = \frac{C}{r}.$$ However, since the diameter is twice the radius ($d = 2r$) this all comes down to the same idea and we have $$\pi = \frac{C}{d} = \frac{C}{2r}.$$ Therefore we have $\tau = 2\pi$.

The value of $\pi$ (or at least approximations) were most likely discovered through measuring the circumference and diameter very precisely. As these measurements were carried out, someone probably noticed a pattern and that the ratios were very close. He or she then began to measure more precisely and noticed that the ratio was in fact constant.


问:
How many times do you have to download a 100,000 word document if a random 5% of the document is deleted every time you download it?
If there is an online document with 100,000 completely unique words and every time you download it, 5% of the words are randomly deleted, how many times do you have to download it before you get all 100,000 words?</p>

答:
Take a look at the first word. There is a 5% it is deleted the first time. The probability it is deleted on both the first and second download is $0.05^2$.  The probability it is deleted on all downloads after $N$ downloads of the document is $0.05^N$. That is never equal to 0 no matter how big $N$ is. Thus, you cannot ever be certain that even the first word of the document is downloaded, let alone all the words in the document.

There is a formula in the answer to this question.

I implemented it two different ways in Mathematica. The first way is literally from the formula as written. The second way is an attempt to make it more numerically stable. I assume $n$ is large and$m$ is a fraction (between 0 and 1) of $n$ so that $m=f \times n$. Then, I take the log of the terms with the binomial coefficients because those numbers are huge in this case. Do the calculations on the log-scale and then exponentiate back to get the correct term. I check that both give almost the same answer in the case of that question where $m$ and $n$ are relatively small ($n=100$ and $m=10$, which means $f=0.1$). The first function won't even run with large numbers like this problem, i.e. $n=100000$.

```<code>P1[t_, n_, m_] :=
 Sum[(-1)^j Binomial[n, j] (Binomial[n - j, m]/Binomial[n, m])^t, {j,
   0, n}]

P2[t_, n_, f_] := N[1 + Sum[(-1)^j
     Exp[(j t Log[1 - f] + j Log[n] - LogGamma[1 + j]) + (-j + f j +
        j^2 - f j^2 - f j t + f j^2 t)/(2 (-1 + f) n)], {j, 1, n}]]

</code>```

Next, I found that for your question
P2[8,100000,0.95] is 0.999996

P2[5,100000,0.95] is 0.969233

P2[4,100000,0.95] is 0.535181

P2[3,100000,0.95] is 0.00000356

That is, you will almost never download the whole document after 3 download attempts, there is a 53.5% of success after 4 download attempts, 96.9% chance after 5 download attempts.

Intuition: on average, each download captures 95% of the the missing words. So, after two downloads, you have only about 250 missing words. After 3 downloads, about 12 missing words. The probability that you capture all those 12 words in the fourth download is $0.95^{12} \approx 0.54$.  If you don't capture all of them, you will most likely only have 1 or 2 missing now and you are practically guaranteed to catch them in the next one or two downloads.

Body:
Take a look at the first word. There is a 5% it is deleted the first time. The probability it is deleted on both the first and second download is $0.05^2$.  The probability it is deleted on all downloads after $N$ downloads of the document is $0.05^N$. That is never equal to 0 no matter how big $N$ is. Thus, you cannot ever be certain that even the first word of the document is downloaded, let alone all the words in the document.

There is a formula in the answer to this question.

I implemented it two different ways in Mathematica. The first way is literally from the formula as written. The second way is an attempt to make it more numerically stable. I assume $n$ is large and$m$ is a fraction (between 0 and 1) of $n$ so that $m=f \times n$. Then, I take the log of the terms with the binomial coefficients because those numbers are huge in this case. Do the calculations on the log-scale and then exponentiate back to get the correct term. I check that both give almost the same answer in the case of that question where $m$ and $n$ are relatively small ($n=100$ and $m=10$, which means $f=0.1$). The first function won't even run with large numbers like this problem, i.e. $n=100000$.

P1[t_, n_, m_] :=
 Sum[(-1)^j Binomial[n, j] (Binomial[n - j, m]/Binomial[n, m])^t, {j,
   0, n}]

P2[t_, n_, f_] := N[1 + Sum[(-1)^j
     Exp[(j t Log[1 - f] + j Log[n] - LogGamma[1 + j]) + (-j + f j +
        j^2 - f j^2 - f j t + f j^2 t)/(2 (-1 + f) n)], {j, 1, n}]]

Next, I found that for your question
P2[8,100000,0.95] is 0.999996

P2[5,100000,0.95] is 0.969233

P2[4,100000,0.95] is 0.535181

P2[3,100000,0.95] is 0.00000356

That is, you will almost never download the whole document after 3 download attempts, there is a 53.5% of success after 4 download attempts, 96.9% chance after 5 download attempts.

Intuition: on average, each download captures 95% of the the missing words. So, after two downloads, you have only about 250 missing words. After 3 downloads, about 12 missing words. The probability that you capture all those 12 words in the fourth download is $0.95^{12} \approx 0.54$.  If you don't capture all of them, you will most likely only have 1 or 2 missing now and you are practically guaranteed to catch them in the next one or two downloads.


问:
How to show that these integrals don't converge
I'm trying to solve these two improper integrals $$\int_{1}^{\infty} \dfrac{x+4}{2x^2 +x -3}\, dx$$ and $$\int_{-\infty}^{-1} \dfrac{x+4}{2x^2 +x -3}\, dx = 0$$
I evaluated the indefinite integral and then I tried to solve the definite integral using substitution, but it didn't really work out for me. I checked out WolframAlpha, but there's no step-by-step solution for these integrals. Could someone please show me how to solve them step-by-step, as after hours of attempts I keep failing to do so.

答:
You need to do partial fraction decomposition:
$$\frac{x+4}{2x^2+x-3}=\frac{x+4}{(2x+3)(x-1)}=\frac A{2x+3}+\frac B{x-1}$$
Then $$A(x-1)+B(2x+3)=x+4$$or $$A+2B=1\\-A+3B=4$$
Add the last two equations, and you get $5B=5$ or $B=1$, and therefore $A=-1$.
Then your integral becomes much simpler. Can you take it from here?


问:
Zero rows in echelon matrix
In the MIT lecture notes (pg. 1) for Linear Algebra, it states;

The third row is zero because row 3 was a linear combination of rows 1 and 2; it was eliminated.

Here, $A = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 8 & 10 \end{bmatrix}$ and $U = \begin{bmatrix} 1 & 2 & 2 & 2 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix}$.

Does this mean  that if the row of the echelon matrix is all zeros, the corresponding row in A can be represented as a linear combination of the other rows in A?

And if the above statements is true, does this also apply to reduced row echelon matrix?

答:
(1) Yes, if a row of the echelon matrix is all zeros, then that row (but remember to consider possible partial pivoting) is a linear combination of other rows in the original matrix. Generally, the coefficients such that forms this linear combination can be found solving the following linear system for $a$ and $b$:
$$\begin{bmatrix}1 & 2 \\ 2 & 4 \\ 2 & 6 \\ 2 & 8 \end{bmatrix}\cdot\begin{bmatrix}a \\ b \end{bmatrix}=\begin{bmatrix}3 \\ 6 \\ 8 \\ 10\end{bmatrix}$$

(2) Yes, since it applies to echelon forms it also applies to row reduced echelon forms.

Title: Zero rows in echelon matrix


问:
Finding the length of the curve defined by $x(t)=t^2-2t$ and $y(t)=t^3-4t$ for $0\leq t\leq2$
So I was given the following prompt:

What is the length of the curve defined by the parametric equations $x(t)=t^2-2t$ and $y(t)=t^3-4t$ for $0\leq t\leq2$?

I understand the setup for the length formula, but I'm a bit confused about what the integral would look like for this equation.

Any help would be appreciated!

答:
what you have is essentially:
$$L=\int dL$$
where we know that:
$$dL=\sqrt{(dx)^2+(dy)^2}=\sqrt{\frac{(dx)^2+(dy)^2}{dt^2}}dt=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$
which is where the formula comes from, now work out the derivatives:
$$y=t^3-4t$$
$$\frac{dy}{dt}=3t^2-4\Rightarrow y'^2=(3t^2-4)^2$$
$$x=t^2-2t$$
$$\frac{dx}{dt}=2t-2=2(t-1)\Rightarrow x'^2=4(t-1)^2$$
now just combine the two:
$$(y')^2+(x')^2=(3t^2-4)^2+4(t-1)^2=9t^4-20t^2-8t+20$$
now remember we are square-rooting then integrating wrt $t$ so:
$$I=\int_0^2\sqrt{9t^4-20t^2-8t+20}\,dt$$


问:
Why is Peano arithmetic undecidable?
I read that Presburger arithmetic is decidable while Peano arithmetic is undecidable. Peano arithmetic extends Presburger arithmetic just with the addition of the multiplication operator. Can someone please give me the 'intuitive' idea behind this?

Or probably a formula in Peano arithmetic that cannot be proved. Does it have something to do with the self reference paradox in Gödel's Incompleteness Theorem?

答:
@hardmath:
"It's a good point, that undecidability of a formal first-order theory implies incompleteness but not the converse" - that is simply not true.
The theorem (that is actually an easy consequence of Post's theorem) we have here is:

For any FO-theory T, if T is recursively axiomatizable and complete, then it is decidable.

This does not mean that an undecidable theory is necessarily incomplete - it can be complete, but lack  recursive axiomatization. That is e.g. the case of ${Th(N, +, \times, 0, 1)}$. Moreover, there are theories which are both recursive and decidable, but incomplete, e.g. $T=\{0=0\}$.

Title: Why is Peano arithmetic undecidable?


问:
Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.

Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$

The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:

\begin{align}
\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx &= \int \left(\:x^4+3x^2+4x+3+\frac{18}{x+1} \right )dx \\
&= \int \:x^4dx+\int \:3x^2dx+\int \:4xdx+\int \:3dx+\int \frac{18}{x+1}dx \\
&= \frac{x^5}{5}+x^3+2x^2+3x+18\ln \left|x+1\right|+ c, c \in \mathbb{R}
\end{align}

The only issue I had is that the polynomial long division took quite some time. Is there another way to do this that is less time consuming? The reason I ask this is that, this kind of question can come in an exam where time is of the essence so anything that I can do to speed up the process will benefit me greatly.

答:
If you do the substitution $x+1=t\iff x=t-1$ you only need do a (long) multiplication:

$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx=
\int \frac{\left((t-1)^2+t+2\right)\left((t-1)^3+7\right)}{t}dt=
$$
$$=\int \frac{\left(t^2-t+3\right)\left(t^3-3t^2+3t+6\right)}{t}dt=\int \frac{t^5-4 t^4+9 t^3-6 t^2+3 t+18}{t}dt=$$
$$=\int\left( t^4-4 t^3+9 t^2-6 t+3 +\frac{18}{t}\right)dt
$$
And now it's simple.


答:
As written by Andrei

$$f(x)=\frac{x+4}{2x^2+x-3}$$
$$=\frac{x+4}{(x-1)(2x+3)}$$
$$\sim \frac{5}{5(x-1)} \;\; (x\to 1)$$

but

$$\int_1^2\frac{dx}{x-1} \text{ is divergent}$$
thus

$$\int_1^2f(x)dx \text{ is divergent }$$
and so does $\int_1^{+\infty}f(x)dx$.

Title: How to show that these integrals don't converge

Body:
I'm trying to solve these two improper integrals $$\int_{1}^{\infty} \dfrac{x+4}{2x^2 +x -3}\, dx$$ and $$\int_{-\infty}^{-1} \dfrac{x+4}{2x^2 +x -3}\, dx = 0$$
I evaluated the indefinite integral and then I tried to solve the definite integral using substitution, but it didn't really work out for me. I checked out WolframAlpha, but there's no step-by-step solution for these integrals. Could someone please show me how to solve them step-by-step, as after hours of attempts I keep failing to do so.


答:
Multiplying by 8 and expanding, we WTS

$$ 16 a^3   + 7 b^3  + 7 c^3 + 6 a b^2  + 6 a c^2 \geq 12 a^2 b + 12 a^2c + 12 abc + 3b^2 c + 3 bc^2. $$

This follows by taking AM-GM creatively with terms on the left, to get terms on the right.

Give it a try. If you're stuck, explain what you've tried and why it doesn't seem to work.

As an example, one of the inequalities we can use is:

 $ 6 a^3 + 6 ab^2 \geq 12 a^2 b $

Title: Proof then $a^3+b^3+c^3-3abc\geq \left(\displaystyle\frac{b+c}{2}-a\right)^3$

Body:
If $a,  b,  c$ are non-negative numbers, prove  then

$a^3+b^3+c^3-3abc\geq \left(\displaystyle\frac{b+c}{2}-a\right)^3$

$8a^3+8b^3+8c^3-24abc\geq\displaystyle(b^3+3b^2c+3bc^2+c^3-6a(b^2+2bc+c^2)+12a^2(b+c)-8a^3)$

$16a^3+7b^3+7c^3\geq 3b^2c+3bc^2-6ab^2+12abc-6ac^2+12a^2b+12a^2c$

$16a^3+7b^3+7c^3+6ab^2+6ac^2\geq 3b^2c+3bc^2+12abc+12a^2b+12a^2c$

$AM-GM$

$\frac{6a^3+6ab^2}{2}\geq 6\sqrt{a^4b^2}$

$6a^3+6ab^2\geq 12a^2b...(1)$

$\frac{6a^3+6ac^2}{2}\geq 6\sqrt{a^4c^2}$

$6a^3+6ac^2\geq 12a^2c...(2)$

$\frac{4a^3+4b^3+4c^3}{3}\geq 4^3\sqrt{a^3b^3c^3}$

$4a^3+4b^3+4c^3\geq 12abc...(3)$

$\frac{b^3+b^3+c^3}{3}\geq ^3\sqrt{b^6c^3}$

$2b^3+c^3\geq 3b^2c...(4)$

$\frac{b^3+c^3+c^3}{3}\geq ^3\sqrt{b^3c^6}$

$b^3+2c^3\geq 3bc^2...(5)$

$(1)+(2)+(3)+(4)+(5)$, we get

$16a^3+7b^3+7c^3+6ab^2+6ac^2\geq 3b^2c+3bc^2+12abc+12a^2b+12a^2c$

Proved.


问:
Show that $f(x) = 2x^2+4$ is reducible over $\mathbb{C}$
I know that this question has already been asked here before, but there's just one part that I didn't get. By definition a polynomial is irreducible whenever we can write $f(x) = g(x)h(x)$ and $g(x)$ or $h(x)$ is a unit. In our case we have

$f(x) = 2(x^2+2)$, but $2$ is a unit over $\mathbb{C}$, right? so why is $f(x)$ reducible? Am I using the right definition?

Body:
Hint

$$x^2+2=x^2-(i\sqrt{2})^2$$


问:
How to show the $f. (g .h)$ is a reparametrization of $(f .g). h ?$
Now days Im reading Allen Hatcher book

I have some confusion in Allen Hatcher book

My confusion is given below in red line

Author didn't explain in detail

My doubt :How to show the $f. (g .h)$ is a reparametrization
of $(f .g). h ?$

答:
You just have to write the explicit definitions of the two compositions of paths and $\varphi$:

$(f\cdot (g\cdot h))(t)$ is:

$f(2t)$ if $t\in[0,\frac{1}{2}]$

$g(4t-2)$ if $t\in[\frac{1}{2},\frac{3}{4}]$

$h(4t-3)$ if $t\in[\frac{3}{4},1]$.

$((f\cdot g)\cdot h)(t)$ is:

$f(4t)$ if $t\in[0,\frac{1}{4}]$

$g(4t-1)$ if $t\in[\frac{1}{4},\frac{1}{2}]$

$h(2t-1)$ if $t\in[\frac{1}{2},1]$.

$\varphi(t)$ is:

$\frac{t}{2}$ if $t\in[0,\frac{1}{2}]$

$t-\frac{1}{4}$ if $t\in[\frac{1}{2},\frac{3}{4}]$

$2t-1$ if $t\in[\frac{3}{4},1]$.

Now, using the explicit definitions, you just have to check case by case (3 cases: $t\in[0,\frac{1}{2}]$,$t\in[\frac{1}{2},\frac{3}{4}]$ and $t\in[\frac{3}{4},1]$ ) that $((f\cdot g)\cdot h)(\varphi(t))=(f\cdot (g\cdot h))(t)$.


问:
To find the area of the region outside $r = 3 + 4 \sin(θ)$ and inside $r = 5 + 2 \sin(θ)$ from $0 \leq θ \leq π/2$
The answer I got is wrong. I started by finding the derivative and then substituting my values for theta, not sure if that is right.</p>

答:
Please see the graph between $0$ and $\frac{\pi}{2}$. You need to simply consider the lower limit and upper limit of the radius between two curves.

The integral is simply

$\displaystyle \int_0^{\pi/2}\int_{3+4\sin\theta}^{5+2\sin\theta} r \ dr \ d\theta$


问:
Order linear such that extends a partial order
Let $\le$ a partial order on a set $A$. Prove that there exists a linear order $\le ^*$ on $A$ such that $a\le b$ implies $a\le ^*b$.

I considered $\mathcal{F}$ the family of all $\le _0 \subseteq A\times A$ such that $\le\subseteq\le _0$ and $\le _0$ is a partial order on $A$.

Then, using Zorn's lemma I concluded that $\mathcal{F}$ has a maximal element $\le ^*$. I proved that this is a partial order on $A$.

Now, I suppose this is not a linear order, then there exists $a,b\in A$ such that $(a,b)\not\in\le ^*$ and $(b,a)\not\in\le ^*$.

Where is the contradiction here? First, I considered $\le _0^* : = \le ^*\cup \{ (a,b)\}$ and I wanted to prove that this is a partial order, but I can't prove the transitivity.

Any hint? Thanks.

答: Hint: You can't just arbitrarily include $(a,b)$.  Set $M(a)=\{x:x\le a\}$, $M(b)=\{x:x\le b\}$, $N(a)=\{x:x\ge a\}$, $N(b)=\{x:x\ge b\}$.  Look at these sets, and choose $(a,b)$ or $(b,a)$ based on them.</p>


问:
Is $\log u\in W^{1,p}_{loc}(\Omega)$ if $u\in W^{1,p}_{loc}(\Omega)~~u>0$??
I trying to show that if  $u\in W^{1,p}_{loc}(\Omega)~~u>0$ then $\log u\in W^{1,p}_{loc}(\Omega)$.

My attempt

Let $D\subset\subset\Omega$ and $\varphi\in C^{\infty}_0(D)$
$$\int_{D}\frac{\partial \log u}{\partial x_i}\varphi dx=\int_{D}\frac{\partial u}{\partial x_i}\frac{1}{u}\varphi dx=\int_{D}\frac{\partial u}{\partial x_i}\frac{\varphi}{u} dx~~(1)$$

Since $u\in W^{1,p}_{loc}(\Omega)$ then $\exists~~g_1\ldots g_n\in L^{p}(D)$ such that $\forall \varphi\in C^\infty_0(D)$
$$\int_{D}\frac{\partial u}{\partial x_i}\varphi dx=-\int_{D}g_i\frac{\partial \varphi}{\partial x_i}dx$$

In (1) this yields

$$\int_{D}\frac{\partial \log u}{\partial x_i}\varphi dx=-\int_{D}g_i\frac{\partial \frac{\varphi}{u}}{\partial x_i}dx$$

i got stuck here, any remark or hint would be helpful. Thank you

答:
The assertion is untrue. Consider $\Omega = (-1, 1)$ and $u(x) := \lvert x \rvert$. Clearly $u>0$ (in an a.e. sense) and $u \in L^1(\Omega)$. Since $u'(x) = \frac{x}{\lvert x \rvert}$ a.e. in a weak sense (this is can easily be computed using integration by parts), we have, because of $\log\circ u$ being $C^1$ in $\Omega \setminus \lbrace 0 \rbrace$, that $\log(u(x))' = \frac{x}{\lvert x \rvert^2}$ is the only possible candidate for the weak derivative. But $\log(u(x))'$ does not live in $L^1_{loc}(\Omega)$:
$$
\int_{-\frac{1}{2}}^\frac{1}{2} \frac{\lvert x \rvert}{\lvert x \rvert^2}~\mathrm{d}x = 2 \int^{\frac{1}{2}}_0 \frac{1}{x}~\mathrm{d}x = 4 - 2\lim_{x \downarrow 0} \log(x) = \infty
$$

Quick sidenote: If $u>0$ and $u \in W^{1, p}(\Omega)$ and $f \in C^1([0, \infty))$ with $f'$ bounded, then $f(u) \in W^{1, p}(\Omega)$. About the $loc$ case I am not quite sure.


问:
Quadratic substituted into itself : wrong?
Let $x^2=x+1$, the solution is the golden ratio phi :$x=\frac{1\pm\sqrt{5}}{2}$.

Now I tried $x=x^2-1$ substituted in the equation itself : $$(x^2-1)^2=x^2\Rightarrow x^4-3x^2+1=0$$
$$\Rightarrow x=\pm\sqrt{\frac{3\pm\sqrt{5}}{2}}$$

Now we get 4 roots because one step is not iff giving back $x^2-1=\pm x$.

However we can recover the numerical values of the first equation solutions.

What happens now if I use $(x^2-1)^2=x+1\Rightarrow x(x^3-2x-1)=0$ ?

Could the numerical values be recovered in this case ? I do not think, so what step is used that is not reversible, only $x^2=x+1$ was used ?

答:
When you square $M = W$ to $M^2 = W^2$ you add in the extraneous solutions of $M = -W$.

So $x = (x^2 -1)$ has two solutions.  And $x^2 =(x^2 -1)^2$ will have $4$ solutions: the two solutions to $x=x^2-1$ as well as two new and extraneous solutions to $x =-(x^2-1)$.

$x = \sqrt{\frac {3 \pm \sqrt 5}2} = \frac {1 \pm \sqrt 5}2$ are the two solutions to $x = x^2-1$.

$x = -\sqrt{\frac {3 \pm \sqrt 5}2} = -\frac {1 \pm \sqrt 5}2$ are the two solutions to $x = -(x^2 -1)$.

Notice that $( \frac {1 \pm \sqrt 5}2)^2=\frac {3 \pm \sqrt 5}2$ so $ \sqrt{\frac {3 \pm \sqrt 5}2} = \frac {1 \pm \sqrt 5}2$


问:
Limsup of average of a function less than limsup of function
Let $f:[0,\infty)\to\mathbb{R}$ be continuous. I would like to show the following:
\begin{equation}
  \limsup_{t\to\infty}\frac{1}{t}\int_{0}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau\leq
  \limsup_{t\to\infty}\,\lvert f(t)\rvert.
\end{equation}
In this post, an answer is provided in terms of sequences. Does the same ''strategy'' work here as well?

Following these arguments, we have for $0< s\leq t$,
\begin{align}
  \frac{1}{t}\int_{0}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau&=
  \frac{1}{t}\int_{0}^{s}\lvert f(\tau)\rvert\,\mathrm{d}\tau+
  \frac{1}{t}\int_{s}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau\\
  &\leq\frac{1}{t}\int_{0}^{s}\lvert f(\tau)\rvert\,\mathrm{d}\tau+
  \frac{t-s}{t}\sup_{\tau\in[s,t]}\lvert f(\tau)\rvert\\
  &\leq\frac{1}{t}\int_{0}^{s}\lvert f(\tau)\rvert\,\mathrm{d}\tau+
  \frac{t-s}{t}\sup_{\tau\geq s}\,\lvert f(\tau)\rvert.
\end{align}
Taking the upper limit as $t\to\infty$, we get
\begin{equation}
  \limsup_{t\to\infty}\frac{1}{t}\int_{0}^{t}\lvert f(\tau)\rvert\,\mathrm{d}\tau\leq
  \sup_{\tau\geq s}\,\lvert f(t)\rvert.
\end{equation}
Do I now simply pass to the limit as $s\to\infty$ or is there any ''trap'' I have not thought over?

答:
There is no trap. You proved that for all $s>0$ the inequality
$$\limsup_{t \to \infty} \frac{1}{t} \int_0^t |f( \tau)| \ \mathrm d \tau \le \sup_{\tau \ge s} |f( \tau)|$$
holds. In particular, taking the infimum of RHS you get
$$\limsup_{t \to \infty} \frac{1}{t} \int_0^t |f( \tau)| \ \mathrm d \tau \le \inf_{s>0} \left( \sup_{\tau \ge s} |f( \tau)| \right)$$
Which is exactly the inequality you needed:
$$\limsup_{t \to \infty} \frac{1}{t} \int_0^t |f( \tau)| \ \mathrm d \tau \le \limsup_{s \to \infty} |f( s)| $$


答:
Since $1$ is a pole, let first simplify by making the change $u=x-1$

$$\int_1^\infty \frac{x+4}{2x^2+x-3}\,dx=\int_0^\infty\frac {u+5}{u(2u+5)}\,dx=\int_0^\infty\frac{du}u-\int_0^\infty\frac{du}{2u+5}$$

The first integral has an issue in $0$ (i.e. $\ln(u)\to-\infty$), the second one is continuous there (the second integral is so congergent in $0$) so the issue in $0$ cannot be removed by summing and the overall integral is divergent.

But even in the case the integral would have started at $x=2$, we still have an issue at infinity:

Indeed for $u\gg 1$ we have $\begin{cases}u+5>u\\2u+5<3u\end{cases}\implies \dfrac{u+5}{u(2u+5)}>\dfrac 1{3u}$

Which is divergent at infinity (i.e. $\ln(u)\to+\infty$), so the integral is divergent again.

For the last part of the exercise, you similarly have a pole in $-\frac 32$ and the same issue at minus infinity.


问:
Integral with delta Dirac power
Is it possible to calculate the integral:
$$J=\int_{-\infty}^{+\infty}f(x)\delta(x-x_0)^kdx$$
wih $k\in\mathbb{R}$?
I know that in the Colombeau algebra the distribution $\delta(x)^2$ is defined. What happens if the Delta function is raised to a real number different from $2$? Thanks in advance.</p>

答:
For $k$ being an integer:

You certainly know that

$\int f(x) \delta (x-x_0) dx = f(x_0)$

This is true regardless of what $f(x)$ is, even when $f(x)$ itself contains a $\delta$-function.

So your integral gives the highly singular result:

$\int f(x)\delta^{k-1}(x-x_0) \delta(x-x_0) dx = f(x_0)\delta^{k-1}(x_0-x_0)$

But you asked about all reals.  Don't know what to tell you for non-integer $k$.

Title: Integral with delta Dirac power


问:
Slutsky's theorem and the asymptotic normality of MLE
I understand that $$\left[I(\theta_0)\right]^{1/2}(\hat{\theta}-\theta_0)\overset{d}{\rightarrow}N(0,1),$$
where $I(\theta_0)$ is the Fisher information at $\theta_0$, $\hat{\theta}$ is a MLE of $\theta$.

My note says $I(\theta_0)$ can be replaced by $I(\hat{\theta})$ justified by Slutsky's theorem. Here is my proof:
$$I(\hat{\theta})^{1/2}(\hat{\theta}-\theta_0)=I(\hat{\theta})^{1/2}I(\theta_0)^{-1/2}I(\theta_0)^{1/2}(\hat{\theta}-\theta_0)\overset{d}{\rightarrow}N(0,1),$$
since $I(\hat{\theta})^{1/2}I(\theta_0)^{-1/2}\overset{p}{\rightarrow}1$, we have $$\left[I(\hat{\theta})\right]^{1/2}(\hat{\theta}-\theta_0)\overset{d}{\rightarrow}N(0,1)$$

Is this correct? If not, what part is wrong?

In this proof, I used $I(\hat{\theta})^{1/2}I(\theta_0)^{-1/2}\overset{p}{\rightarrow}1$. This hold when $\hat{\theta}\overset{p}{\rightarrow}\theta_0\implies I(\hat{\theta})^{1/2}\overset{p}{\rightarrow}I(\theta_0)^{1/2}$. Is $I(\theta)$ always continuous for it to hold, or the continuity of $I(\theta)$ is an assumption we need to make?

答: Your proof is correct. Also, $I(\theta)$ is not always continuous, so you have to assume that it is continuous (only needed to be continuous close to $\theta$) in order to use Slutzky's Theorem.


问:
Looking for analytical proof that this function given as a power series is constant.
Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$

Is there an analytical way to prove that $P_t(x)$ is a continuous function on $x\in(0,1)$ and:

$P_t(x)=1$ for all $x:\ 0\le 1-x\le\frac1t$,

$P_t(x)<1$ for all $x:\ \frac1t< 1-x\le1$.

I proved this for trivial cases $t=1,2$ but did not find a way to deal with general $t$.

The behavior of the function for $t=1,2,3,4$ is demonstrated below:

答:
This contribution is a development of the answer given by Joshua and aims mainly at clarification of the remaining questions. It will be assumed that $t>1$.

As already indicated in the mentioned answer it is advantageous to replace $x$ with $1-x$ and consider the function
$$
P_t(x)=(1-x)^tF_t\left(z_t(x)\right)\tag1
$$
with
$$
F_t(z)=\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n z^n.\tag2
$$
and
$$
z_t(x)=x(1-x)^{t-1}.\tag3
$$

First of all let us investigate the behavior of $z_t(x)$. While $x$ runs from $0$ to $1$ the function first monotonically increases from $0$ to $z^*_t$ and then monotonically decreases back to $0$, the maximum $z^*_t$ being attained at $x=\frac1t$. Thus any value $z:\ 0\le z<z^*_t$ is attained exactly by two values of $x$: $0\le x_1<\frac1t$ and $\frac1t<x_2\le 1$. On the domain $x\in[0,\frac1t]$ the function can be inverted and we shall refer to the inverse function (defined on the domain $z\in[0,z^*_t]$) as $z^{-1}_t(z)$.

For convenience we introduce the function
$$
X_t(x)=z^{-1}_t(z_t(x)).
$$
Observe:
$$
X_t(x)\begin{cases}
=x,& 0\le x\le \frac1t;\\
<x,& \frac1t<x\le 1.
\end{cases}\tag4
$$

The next figure shows the function $z_t(x)$ and indicates the action of the function $X_t(x)$ for $x>\frac1t$.

Let us now investigate the convergence of the series $F_t(z)$ in the neighborhood of $z=0$.

The ratio of the series coefficients reads:
$$\begin{align}
\frac{a_{n}}{a_{n-1}}&=\frac n{n+1}\frac{(nt+t)!}{n!(nt+t-n)!}\frac{(n-1)!(nt-n+1)!}{(nt)!}z\\
&=\frac 1{n+1}\frac{(nt+t)!}{(nt)!}\frac{(n(t-1)+1)!}{(n(t-1)+t)!}z\\
&=\frac {nt+1}{n+1}\frac{nt+2}{n(t-1)+2}\cdot\frac{nt+3}{n(t-1)+3}\cdots\frac{nt+t}{n(t-1)+t}
z,\tag5
\end{align}$$
so that
$$
\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=t\left(\frac t{t-1}\right)^{t-1}z.
$$

Thus the series $F_t(z)$ converges if
$$
|z|<\tilde z_t=\frac1t\left(1-\frac1t\right)^{t-1}.\tag6
$$
Observe that $\tilde z_t$ is precisely the maximum value $z^*_t$ of the function $z_t(x)$.

Substituting in (4) $z=z^*_t$ and expanding in powers of $\frac1n$ one obtains
$$
\frac{a_{n}}{a_{n-1}}=1-\frac32\frac1n
+O\left(\frac1{n^2}\right),
$$
so that by Raabe test:
$$
\lim_{n\to\infty}n\left(\frac{a_{n-1}}{a_{n}}-1\right)=\frac32>1
$$
the series converges also for $z=z^*_t$.

After the absolute convergence is established one can demonstrate applying the Rothe-Hagen identity:
$$\forall x: 0\le x\le\frac1t:\ F_t\left(z_t(x)\right)=\frac1{(1-x)^t}.\tag7
$$
The proof is given in the answer of Joshua and need not be repeated.

Using (7) one obtains:
$$\begin{align}
P_t(x)&=(1-x)^tF_t(z_t(x))\\
&=\frac{(1-x)^t}{(1-X_t(x))^t}
\underbrace{\left[(1-X_t(x))^t F_t(z_t(X_t(x)))\right]}_{=1}\\
&=\left(\frac{1-x}{1-X_t(x)}\right)^t.\tag8
\end{align}$$

In view of the property (4) one concludes that indeed:
$$
P_t(x)\begin{cases}
=1,& 0\le x\le \frac1t;\\
<1,& \frac1t<x\le 1.
\end{cases}\tag9
$$


问:
Solving $u_x+2u_y=1$, $u(0,y)=e^{-y^{2}}$

Solving $u_x+2u_y=1$, $u(0,y)=e^{-y^{2}}$ for $x\in \mathbb{R},y\in\mathbb{R}$

So I tried to use characteristics method, with $\frac{dx}{1}=\frac{dy}{2}=\frac{du}{1}$

Which gives me $\frac{dy}{dx}=2$, $\frac{du}{dx}=\frac{1}{2}$

Which I solved to get $y=2x+c_1$, $u=\frac{1}{2}x+c_2$

Then taking $G(c_1)=c_2$ I get that $u=\frac{1}{2}x+G(y-2x)$

And using the initial conditions gives $G(y)=e^{-y^{2}}$

So my solution is $u=\frac{1}{2x}+e^{-(y-2x)^{2}}$

答: Write the PDE as $(u_x,u_y,-1)\cdot (1,2,1) = 0$.  What is the flow of the constant vector field $V(x,y,u) = (1,2,1)$? Clearly it is $$(x(t),y(t),u(t)) = \Phi(t, (x_0,y_0,u_0)) = (x_0+t, y_0+2t, u_0+t).$$However, setting $x_0 = 0$, we have $u_0 = e^{-y_0^2}$. So the problem boils down to "eliminating" $t$ in $$x(t) = t, \qquad y(t) = y_0 +2t, \qquad u(t) = e^{-y_0^2}+t$$and solving for $u(t)$ in terms of $x(t)$ and $y(t)$ (then effectively dropping $t$ everywhere). We immediately have that $$u(x,y) = e^{-(y-2x)^2}+x.$$Let's see that this works: for $x=0$, we have $u(0,y) = e^{-y^2}$ as required. And also $$u_x(x,y) = 4(y-2x)e^{-(y-2x)^2} + 1\qquad\mbox{and}\qquad u_y(x,y) = -2(y-2x)e^{-(y-2x)^2}. $$Then obviously $u_x+2u_y=1$, so we are done.


答:
Horner's method for division:

$$(x^2+x+3)(x^3+7)=x^5+x^4+3x^3+7x^2+7x+21$$

\begin{array}{*{7}{r}}
& 1 & 1  & 3 & 7 & 7 & 21 \\
+ & \downarrow & -1 & 0 & -3 & -4 & -3 \\
\hline
\times -1 & \color{red}1 &  \color{red}0 &  \color{red}3 &  \color{red}4 &  \color{red}3 & \color{cyan}{18}
\end{array}
so the quotient is $\;x^4+3x^2+4x+3$ and the remainder is $\;18$.

Title: Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.

Body:
Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$

The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:

\begin{align}
\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx &= \int \left(\:x^4+3x^2+4x+3+\frac{18}{x+1} \right )dx \\
&= \int \:x^4dx+\int \:3x^2dx+\int \:4xdx+\int \:3dx+\int \frac{18}{x+1}dx \\
&= \frac{x^5}{5}+x^3+2x^2+3x+18\ln \left|x+1\right|+ c, c \in \mathbb{R}
\end{align}

The only issue I had is that the polynomial long division took quite some time. Is there another way to do this that is less time consuming? The reason I ask this is that, this kind of question can come in an exam where time is of the essence so anything that I can do to speed up the process will benefit me greatly.


问:
Why is $\operatorname{rank}(A^* A)=\operatorname{rank}(A) \Leftrightarrow$ $(A^* Ax=0 \Leftrightarrow Ax=0)$?
Let $A \in M_{m\times n}(F)$ and $x \in F^n$.

$A^*$ is the adjoint of $A$.

Why is $\operatorname{rank}(A^* A)=\operatorname{rank}(A)$ equivalent to $A^* Ax=0$ if and only if $Ax=0$?

答:
If $\operatorname{rank}(A^*A)=\operatorname{rank}(A)$ it means $\operatorname{rank}(A)=n-\dim(\operatorname{null}(A))$ and $\operatorname{rank}(A^*A)=n-\dim(\operatorname{null}(A^*A))$ so   $\dim( \operatorname{null}(A))=\dim(\operatorname{null}(A^*A))$ if exist $x$ such that $Ax\ne 0$ and $A^*Ax=0$ then $x\in \operatorname{null}(A^*A)$but but $x$ not belong to $\operatorname{null}(A)$ and since $\operatorname{null}(A)\subset \operatorname{rank}(A^*A)$ so   $$\dim( \operatorname{null}(A))\ne \dim(\operatorname{null}(A^*A))$$contradiction!

on the other hand :
we know $\operatorname{null}(A)\subset \operatorname{null}(A^*A)$ in general

and since $A^*Ax=0$ if and only if $Ax=0$ it means $\operatorname{null}(A^*A)\subset \operatorname{null}(A)$ and so $\operatorname{null}(A^*A)=\operatorname{null}(A)$ and

So :
$\operatorname{rank}(A^*A)=\operatorname{rank}(A)$


问:
Evaluate $\int_{0}^\infty \frac{\sin(x)}{\sqrt{x^2+1}} dx$
The first substitution that came to my mind was
$$
\begin{cases}
x = \tan(\theta) \\
dx = \sec^2\theta d\theta \\
\textrm{Bounds:} \, \, 0 \to \, \pi/2
\end{cases}
$$
We now have:
$$
\int_0^{\frac{\pi}{2}}\sin(\tan(\theta))\cos^4\theta \, d \theta
$$
But that really doesn't seem to help. Any advice is greatly appreciated.

答: Changing the integration variable to $t$ for ease of comparison with Eq. 11.5.7 in @metamorphy's DLMF link, $\nu=0,\,x=1$ gives$$\int_0^\infty(1+t^2)^{-1/2}\sin t\mathrm dt=\frac{\pi}{2}(I_0(1)-L_0(1)),$$with $I_0$ a modified Bessel function of the first kind and $L_0$ a modified Struve function.


问:
Is $F(2)-F(-2)$ the same thing as $f(2)-f(-2)$?
 I thought that they were the same thing and that I would just use the graph numbers but the answer is not $4$ so apparently I'm wrong.

Based on the picture, we have $f$ to be the linear function on $[-2,-1]$ given by $f(-2)=-2$ and $f(-1)=0$. And $f$ to be the linear function on $[-1,2]$ given by $f(-1)=0$ and $f(2)=2$.

答:
Hint: Long way... Notice that $f$ is a linear function on $[-2,-1]$ and $[-1,2]$ so you can explicitly find it. Then you can antiderivate (integrate) $f$ on the desired interval.

So you have $$f(x)=\left\lbrace\begin{array}{rcl}2x+2 &\mathrm{if}& -2\leq x\leq -1 \\ \frac{2}{3}x+\frac{2}{3} &\mathrm{if}& -1\leq x\leq 2 \end{array}\right. $$

Then you can integrate (antiderivate) $f$ and evaluate as needed.

You could use the following to compute: $$F(2)-F(-2) = \int_{-2}^2 f(x)dx = \int_{-2}^{-1} 2x+2\ dx + \int_{-1}^{2} \frac{2}{3}x+\frac{2}{3} dx$$


问:
Non-periodic paths of length 4 in a graph and the Lovasz Local Lemma
Let $G$ be a graph with maximum degree $\Delta$ and consider colouring its edges, each with one of $k$ colours. A path $v_0v_1v_2v_3v_4$ in a $G$ is $\textit{periodically coloured}$ if $v_0v_1$ and $v_2v_3$ have the same colour and $v_1v_2$ and $v_3v_4$ have the same colour.

a) Prove that $G$ has a colouring with no periodically coloured paths (of length $4$) and at most $\lceil 5\Delta^{3/2} \rceil$ colours.

b) Give a sequence of graphs $G_n$ to show that even if $\Delta$ is fixed and $k > 100\Delta^{3/2}$, the probability that a random $k$-edge-colouring of $G_n$ has no periodically coloured path of length $4$ can be arbitrarily small.

In a) I can give a proof but with $5$ replaced by a larger constant. Consider applying Lovasz Local Lemma (in its symmetric form - see e.g. Theorem 2.3 in https://theory.stanford.edu/~jvondrak/MATH233A-2018/Math233-lec02.pdf) on the events $A_i =$ 'the $i$-th path is periodically coloured'. Clearly $\mathbb{P}(A_i) = \frac{k^2}{k^4} = \frac{1}{k^2}$ and, for any $i$, $d+1 \leq $ 'number of paths of length $4$ which have at least one edge in common with the $i$-th path of length $4$'. (In this number we include the path of $A_i$ itself and that's why we can have $+1$ on the left-hand side.)

Now here's how I estimate the latter number. If the first path is $v_0v_1v_2v_3v_4$, then for the second path there are $4$ choices for which edge of the first path to be common and $4$ choices for what its position in the second path be. After that, there are at most $\Delta^3$ choices for the other three edges from the second path.

So the Local Lemma shall be applicable if $\frac{e}{k^2}16\Delta^3 \leq 1$, but this requires $k\geq \lceil 4e^{1/2}\Delta^{3/2} \rceil$ and unfortunately, $4e^{1/2} \sim 6.59 > 5$. So is there a way to improve the above counting in order to take care of this multiplicative constant issue? Or I somehow need to use the general form of the Local Lemma (but with what real numbers)?

Update: Thanks to Misha Lavrov for the simple example for b).

Any help appreciated!

答:
We cannot significantly improve the degree in the dependency graph.

To see this, consider a path $(v_0, v_1, v_2, v_3, v_4)$ in an infinite $\Delta$-regular tree. Starting from each vertex $v_i$, there are at least $\Delta-2$ edges off the path, at least $(\Delta-2)(\Delta-1)$ paths of length $2$, and at least $(\Delta-2)(\Delta-1)^2$ paths of length $3$; if we ignore lower-order terms, these are basically $\Delta, \Delta^2, \Delta^3$ paths that share no edges with this path.

Now, we can get approximately $16 \Delta^3$ distinct paths that overlap with this one as follows:

Pick $i \in \{0,1,2,3\}$ and $j \in \{0,1,2,3\}$.

Take the union of a length-$j$ path from $v_i$, the edge $v_i v_{i+1}$, and a length-$(3-j)$ path from $v_{i+1}$; there are roughly $\Delta^j \cdot \Delta^{3-j}=\Delta^3$ choices, and remain so if we make sure not to share any other edges with the path.

Even in a finite graph, we see the same thing if we take a sufficiently large $\Delta$-regular tree, for almost all paths.

We should also not expect to benefit from the asymmetric local lemma. That's useful when we have events of legitimately distinct types, but that's not what's happening here.

It is possible that we can fix the problem in some other way; maybe we can choose the colors by a different random method. It might help to start with a proper $(\Delta+1)$-edge-coloring, and then randomly determine the color of an edge based on its color in the preliminary coloring somehow.

I think it is more likely that there is a mistake in the constant in the result you are trying to prove.


问:
Solving AM-GM without expanding
Prove that $(p^2+p+1)(q^2+q+1)(r^2+r+1)(s^2+s+1) ≥ 81pqrs$ for all $p, q, r, s ≥ 0$. I am not sure how to solve this problem. I know that this is supposed to be an AM-GM question, so I tried expanding it. It didn't take long for me to realize that expanding is probably not the way to go to solve this problem. I also tried taking the fourth root of both sides, but that didn't solve anything either. How can I prove this? Any help would be appreciated.

答:
I have the original problem in my math book, "The Art of Problem Solving: Intermediate Algebra", and it says that all $p,q,r,s$ are positive, which is important to note to solve this problem.

Apply AM-GM to the terms $p^2,p,1$. We get $\frac{p^2+p+1}3 \ge \sqrt[3]{p^2*p*1}$. Simplifying and rearranging we get $p^2+p+1 \ge 3p$. Now apply the same logic to each of the other expressions to get $q^2+q+1 \ge 3q$, $r^2+r+1 \ge 3r$, and $s^2+s+1 \ge 3s$. Since all of $p,q,r,s$ are positive, then all of these expressions are positive, too, so we can safely multiply all the inequalities together to get $(p^2+p+1)(q^2+q+1)(r^2+r+1)(s^2+s+1) \ge 81pqrs$ as desired.

Title: Solving AM-GM without expanding


问:
Die Rolling Question about probability.
So if I had a six sided die and rolled it six times what is the probability that I get a six, and why? I am only a sophomore in highschool so please don't talk about anything too advanced unless it is necessary to answer the question. Thanks!</p>

答:
Rolling once the probability of rolling a 6 is 1/6 and the probability of not rolling a 6 is 5/6.  If you roll it twice the probability of rolling no sixes is 5/6 x 5/6 or, 25/36.  So, the probability of rolling at least one 6 is 1 - 25/36 or 11/36.

IF you roll it 6 times, the probability of getting at least one 6 is 1 - ((5/6) ^ 6) or about 66.5%.

If you are looking for the probability of rolling exactly one 6 that is equal to 1/6 x ((5/6) ^ 5) * 6, or about 40.2%.  The 1/6 is for rolling a 6; the (5/6)^5 is for not rolling a 6 5 times, the 6 is to account for the all the possible positions where the 6 comes up in the first through sixth rolls.

Title: Die Rolling Question about probability.


答:
Let $A$ be the event "we got six in one of the rolls". Now remember that $P(A)=1-P(A^c)$. Lets calculate $P(A^c)$ because it is much simpler (where the $c$ stands for complement, i.e the event did no occur).

Now, if we want to calculate the probability that six was not the result of any of the experiments, the probability for that is:

$P(A^c)= (\frac {5}{6})^6$.

Now we use the complement formula and we get $$P(A)=1-P(A^c)=1- (\frac {5}{6})^6=0.665$$

Title: Die Rolling Question about probability.

Body: So if I had a six sided die and rolled it six times what is the probability that I get a six, and why? I am only a sophomore in highschool so please don't talk about anything too advanced unless it is necessary to answer the question. Thanks!</p>


问:
Showing Equivalence between a Proposition and the Axiom of Choice
I am trying to establish the equivalence between the following statement and the Axiom of Choice:

Suppose that $R$ is a relation between two non-empty sets $A$ and $B$. There is a function $F$ with the same domain as $R$ such that $F \subseteq R$.

I have established the direction from the Axiom of Choice to this statement. My proof goes as follows.

For each set $x$, we may define the following set:
\begin{equation*}
A_{x} = \left\{y \in B: \langle x,y \rangle \in R\right\}.
\end{equation*}
Specifically, if $x \in \operatorname{dom}\left(R\right)$, then $A_{x} \neq \emptyset$.
Further, we may define the following set:
\begin{equation*}
S = \left\{A_{x}: x \in \operatorname{dom}\left(R\right)\right\}.
\end{equation*}
It is obvious that $S$ is a set of non-empty sets and $\bigcup{S} \subseteq B$. According to the Axiom of Choice, there is a function $h: S \to \bigcup{S}$ such that
\begin{equation*}
\left(\forall T \in S\right)\left(h\left(T\right) \in T\right).
\end{equation*}
We may define a function $F: \operatorname{dom}\left(A\right) \to B$ such that
\begin{equation*}
F\left(x\right) = h\left(A_{x}\right),\ x \in \operatorname{dom}\left(R\right).
\end{equation*}
It is obvious that $F \subseteq R$.

I am wondering how to establish the reverse direction. I understand that there are already several equivalent forms of the Axiom of Choice. For instance, the following statement is the axiom used in my proof above:

Suppose that $\mathscr{F}$ is a family of non-empty sets. There is a function $h: \mathscr{F} \to \bigcup{\mathscr{F}}$ such that for each $A \in \mathscr{F}$, $h\left(A\right) \in A$.

We also have the following versions:

Suppose that $\mathscr{F}$ is a family of disjoint non-empty sets. There is a function $h: \mathscr{F} \to \bigcup{\mathscr{F}}$ such that for each $A \in \mathscr{F}$, $h\left(A\right) \in A$.

Suppose that $M$ is a non-empty set. Then there is a function $h: \mathscr{P}\left(M\right)\setminus \left\{\emptyset\right\} \to M$ such that for all non-empty subsets $A$ of $M$, $h\left(A\right) \in A$.

Can anyone help with completing the proof?

答: Assume $M$ is a non-empty set, and apply the condition to the relation $$R = \{(A,B) \in (\mathscr P(M) \setminus \{\emptyset\}) \times M : B \in A\}.$$


问:
Evaluate $\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$.

Evaluate:
$$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx$$

The only thing I can think of doing here is long division to simplify the integral down and see if I can work with some easier sections. Here's my attempt:

\begin{align}
\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx &= \int \left(\:x^4+3x^2+4x+3+\frac{18}{x+1} \right )dx \\
&= \int \:x^4dx+\int \:3x^2dx+\int \:4xdx+\int \:3dx+\int \frac{18}{x+1}dx \\
&= \frac{x^5}{5}+x^3+2x^2+3x+18\ln \left|x+1\right|+ c, c \in \mathbb{R}
\end{align}

The only issue I had is that the polynomial long division took quite some time. Is there another way to do this that is less time consuming? The reason I ask this is that, this kind of question can come in an exam where time is of the essence so anything that I can do to speed up the process will benefit me greatly.

答:
As substitution by parts goes nowhere It simply do what you did but skip the step of multiplying out the polynomial $ \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}=\frac {(x(x+1) + 3)(x^3+7)}{x+1}= x(x^3 +7) + \frac 3{x+1}(x^3 + 7)=$

$x(x^3 +7) +\frac 3{x+1}(x^2(x+1) -x(x+1) + x+1 + 6)=x(x^3 +7) + 3(x^2-x+1) + \frac {18}{x+1} $

Saves one or two tablets of ibuprofen.