问:
How to get better at calculus?
I'm still more or less at the beginning of my journey to become a mathematician, but I already recognized that there is a huge gap relating to my skills between solving problems in linear algebra and solving problems in calculus. I really like calculus, but solving problems in linear algebra seems way more easy to me then solving problems in calculus, and that bothers me - a lot. Most of the problems in calculus look like this:

"Show the identity of ..."

"Show that the limit of ... is ..."

"Solve ..."

"Show inductive that ..."

All of these tasks have something in common: They include a lot of playing with equations. Linear algebra, on the other hand, requires a (how to say it in English?) "structural approach" or "structural thinking". Its more about the relations between different objects then about writing down $15$ different steps until you showed that the left side is indeed the same as the right side of an equation.

But since it would be way too easy to simply accept the fact that I'm not that good at calculus, I want to get better at it. But where do I have to start? Does anyone have useful hints for me?

Edit:

Since this was asked in the comments, I'll give an example of a task I would have my problems with. In this case, it's simply about showing that the limit of the $p$-norm is equal to the right side.

$$\displaystyle\lim_{p\to\infty}\lVert \textbf{x}\rVert_{p}=\lVert \textbf{x}\rVert_{\infty}$$

答:
I had to do a crash course in Calculus. I was doing some stuff and honestly, it was the trig that was messing me up the whole time. I don't know $e$ and I don't pretend to. But what I did instead was sort of miraculous.
$$y = | \text{baseline} - \text{Y-axis} |$$
$$t = \text{length of a to b} $$
$$x=2$$
$$m = \frac{sec(x)}{2}$$
$$n = | y - t |$$
$$P = (m*2) + (n*2)$$
$$L_n = \frac{P}{2} + \frac{n}{2}$$

If you add the $L_n$ answers together, you get an integrand of the entire curve.

If you average them using the $V = L_0 + L_1 + L_2 + ...$ and dividing by the largest integer, $G = n-1$ then you'll have the integral of the equation: $VG$

Finally, if you the differential of a Gaussian equation, you do the first algorithm I made, and replace $x$ with the answer to it. The derivative will be divided into the first answer. This is the differential. And obviously you can keep going in derivatives.

Title: How to get better at calculus?


问:
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$?
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.

答:
For $|x| \le 1$ the Taylor series
$$
 \cos(x) = 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
$$
is an alternating series with terms that decrease in absolute value. It follows that for these $x$
$$
 \cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le  1 - \frac{x^2}{2!}+ \frac{x^2}{4!} = 1 - \frac{11}{24} x^2
$$
and therefore
$$
 1 - \cos(x) \ge \frac{11}{24} x^2  \, .
$$
That is an even better estimate since $\frac{11}{24}  > \frac 13$.

The same approach can be used to show that $1 - \cos(x) \ge \frac{x^2}{3}$ holds on the larger interval $[-2, 2]$:
$$
 \cos(x) \le 1 - \frac{x^2}{2!}+ \frac{x^4}{4!} \le 1 - \frac{x^2}{2!}+ \frac{4 x^2}{4!} = 1 - \frac 13 x^2
$$
because the $x^{2n}/({2n})!$ terms decrease in absolute value for $n \ge 1$.

ContentLicense: CC BY-SA 4.0


问:
Solving Recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n\geq 1, a_0=2$?
How can I solve recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n \geq1 $ and $a_0 = 2$?

Body:
Put
$$a_n=5.3^n.b_n$$

then

$$5.3^n.b_n=3.5.3^{n-1}.b_{n-1}+5.3^n$$

thus

$$b_n=b_{n-1}+1$$

So,
$$b_n=n+2,\; \; and \; \; a_n=5.3^n(n+2)$$


问:
when $C$ is a maximal chain
let $R$ be partial order on $A$.

suppose $C \subseteq A$ is a chain, then $C$ is maximal iff for every $a\in A \setminus C$ there is a $c\in C$ s.t $a$ and $c$ are $(a,c)\notin R \\and \\(c,a)\notin R$

the only thing that I know that $C$ is maximal chain that if you add an elemt to it it's any more :(

any solution or hint would be great and thank in advance

答:
A slightly easier-to-parse version of the statement of the problem is the following:

Let $R$ be a partial order on the set $A$, and suppose that $C\subseteq A$ is a chain in the poset $P =(A,R)$. Show that $C$ is maximal iff for every element $a\in A$ that is not an element of the chain (equivalently, $a\in\overline{C}$), there exists an element $c$ of the chain $C$ such that $a$ and $c$ are incomparable.

Suppose that $C$ is a maximal chain in $P$. Fix $a\in \overline{C}$. Then, for each element $c\in{C}$, $a$ and $c$ are either comparable or incomparable. Suppose that for every element $c\in{C}$, we have that $a$ and $c$ are comparable. Then, the set $C\cup\{a\}$ is a chain, contradicting the maximality of $C$. Thus, for some $c\in C$, we must have that $a$ and $c$ are incomparable. Since $a$ was chosen arbitrarily, we are done.

Now, suppose that for every $a\in \overline{C}$, there exists $c\in C$ such that $a$ and $c$ are incomparable. We have to show that the chain $C$ is maximal. Assume to the contrary. Then, there exists an element $a\in\overline{C}$ such that $a$ and $c$ are comparable for every element $c$ of the chain, contradiction. Thus, $C$ is maximal.

The proof is complete. $\square$

Title: when $C$ is a maximal chain


问:
Maclaurin expansion for sech$(x)$
I am a bit unsure where I have gone wrong in working this out.

Sech$(x)=2/(e^x+e^{-x}).$

Maclaurin expansions:

$e^x = 1+ x +  x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 -  x^4/24;$

so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$  (I think this is where I have gone wrong.)

The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).

How would I work this out?

答:
You are wrong exactly where you thought to be. The reason is that the Taylor expansion of a function $f$ takes into account the derivatives of the function itself, hence writing what you did, means that the derivative of $\frac2{e^x+e^{-x}}$ is $\left[(e^x)'+(e^{-x})'\right]^{-1}$ which is clearly false.

The most straightforward way to compute the Taylor expansion near a point $x_0$ is to compute the derivatives of that function at $x_0$:
$$
f(x)=\sum_{k=0}^{+\infty}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k
$$

Title: Maclaurin expansion for sech$(x)$


问:
Prove $L(b) = \ln(b)$
While thinking about the Taylor expansion of $b^x$, to take the derivative of $b^x$ we need to find:
$$L(b) = \lim_{h \to 0} \frac{b^h-1}{h}$$

It can be shown numerically that $L(2) \approx 0.7 $ and $L(3) \approx 1.1 $ thus by the Intermediate Value Theorem there must be a $ \{ c \in \mathbb{R} \mid 2<c< 3\}$ such that $L(c) = 1 $. Of course, we call that value Euler's number or e.

I have heard it asserted that $L(b) = \ln(b)$ but am unsure how to derive that result?

Title: Prove $L(b) = \ln(b)$

Body:
Hint:

$$L(b)=\frac{d}{dx}\left(b^x\right)\mid_{x=0}$$


问:
natural induced split exact sequence from $ 0\to \Bbb Z\xrightarrow{p}\Bbb Z\to \Bbb Z_p\to 0$

Multiplication by the prime $p:\Bbb Z\to \Bbb Z$ fits in a short exact sequence
$$ 0\to \Bbb Z\xrightarrow{p}\Bbb Z\to \Bbb Z_p\to 0$$
Use this to derive the natural split exact sequence
$$0\to H_n(X)/pH_n(X)\to H_n(X;\Bbb Z_p)\to\text{ker}\{p:H_{n-1}(X)\to H_{n-1}(X)\}\to 0$$

This is exercise 4.5.1 of Bredon's topology and geometry textbook. $0\to \Bbb Z\xrightarrow{p}\Bbb Z\to \Bbb Z_p\to 0$ induces $0\to \Delta_{*}(X)\otimes\Bbb Z\to \Delta_{*}(X)\otimes\Bbb Z\to\Delta_*(X)\otimes \Bbb Z_p\to 0$ so a long exact sequence $\cdots\to H_n(X)\to H_n(X)\to H_n(X;\Bbb Z_p)\to H_{n-1}(X)\to H_{n-1}(X)\to\cdots$. Then by first isomorphism theorem, $H_n(X)/pH_n(X)\simeq H_n(X;\Bbb Z_p)$. Hence, it suffices to show $H_n(X;\Bbb Z_p)\to H_{n-1}(X)$ is injective. How can I?

Edit: My naive approach is that if $\partial_*$ be a connecting homomorphism between $H_n(X;\Bbb Z_p)$ and $H_{n-1}(X)$ and let $\Delta_{*}(X)\otimes\Bbb Z\xrightarrow{i} \Delta_{*}(X)\otimes\Bbb Z\xrightarrow{j}\Delta_*(X)\otimes \Bbb Z_p$ so that $H_n(X)\xrightarrow{i_*}H_n(X)\xrightarrow{j_*} H_n(X;\Bbb Z_p)$, then for $[[c]]\in H_n(X;\Bbb Z_p)$, if $\partial_*([[c]]) = [[i^{-1}\circ (\partial\otimes 1)\circ j^{-1}(c)]] = 0 \iff c=0$ as $i$ maps $\Bbb Z\to p\Bbb Z$. Is this valid?

答:
The point is that your long exact sequence actually takes the form

$\cdots\to H_n(X)\xrightarrow{p} H_n(X)\to H_n(X;\Bbb Z_p)\to H_{n-1}(X)\xrightarrow{p} H_{n-1}(X)\to\cdots$

So $H_n(X;\Bbb Z_p)$ surjects onto its image in $H_{n-1}(X)$, which by exactness is equal to the displayed kernel.

The first isomorphism theorem does not give $H_n(X)/pH_n(X) \cong H_n(X; \Bbb Z_p)$, but rather $H_n(X)/pH_n(X) \cong \operatorname{im}(H_n(X) \to H_n(X;\Bbb Z_p))$, and this image of course injects into $H_n(X;\Bbb Z_p)$ and is also the kernel of the map from $H_n(X;\Bbb Z_p)$ to $H_{n-1}(X)$.


问:
Can $f[g(x)]$ and $g[f(x)]$ be used to prove that a function is onto?
I am very confused as to whether we can use  $f[g(x)]$ and $g[f(x)]$ to show that a function is onto.

Let's say I have a function :

$y = f(x) = x + 1$
from $\mathbb R \to \mathbb R$

Is it possible that if I find $f[g(x)]$ and $g[f(x)]$ and they give one $y$ and one $x$ (i.e., $f[g(x)] = y$ and $g[f(x)] = x$)
then my function will be onto?

Sometimes this method did worked but other times even though I got one $y$ and one $x$, the function turned out to be not onto.

Kindly explain.
Thanks

答:
If $f\circ g$ is "onto" (surjective), then $f$ must be onto.

Proof: Let $f:X\to Y, g:Z\to X$, and suppose $f\circ g$ is onto. Let $y\in Y$ be an arbitrary element. As $f\circ g$ is onto, then there is $z\in Z$ such that $(f\circ g)(z)=f(g(z))=y$. But then $f(x)=y$ if you set $x=g(z)$. In other words, for every $y\in Y$ you can find $x\in X$ such that $f(x)=y$. This proves that $f$ is onto.


Title: Solving Recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n\geq 1, a_0=2$?

Body: How can I solve recurrence relation $a_n = 3a_{n-1} + 5 \cdot 3^n$ for $n \geq1 $ and $a_0 = 2$?

Body:
Let $ n\in\mathbb{N} $, we have for any $ k\leq n-1 $ the following : $$ a_{n-k}=3a_{n-k-1}+5\times 3^{n-k} $$

Thus : \begin{aligned}3^{k}a_{n-k}&=3^{k+1}a_{n-k-1}+5\times 3^{n}\\ \Longrightarrow \sum_{k=0}^{n-1}{\left(3^{k}a_{n-k}-3^{k+1}a_{n-k-1}\right)}&=5n\times 3^{n}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{n}-3^{n}a_{0}&=5n\times 3^{n}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{n}&=3^{n}\left(5n+2\right)\end{aligned}

Since $ n $ is arbitrary, the result stands for all $ n\in\mathbb{N} $.

Hope that helps.


答:
The question finds an immediate answer by noting that $b^h=e^{\ln b^h}=e^{h\ln b}$.
Therefore
$$L(b) = \lim_{h\to 0}\frac{b^h-1}{h}=\lim_{h\to 0}\frac{e^{h\ln b}-1}{h} = \ln b\lim_{t\to 0}\frac{e^t-1}{t}=\ln b.$$
I have multiplied and divided by "$\ln b$" the fraction and substituted $h\ln b = t$ to get the result.

Title: Prove $L(b) = \ln(b)$

Body:
While thinking about the Taylor expansion of $b^x$, to take the derivative of $b^x$ we need to find:
$$L(b) = \lim_{h \to 0} \frac{b^h-1}{h}$$

It can be shown numerically that $L(2) \approx 0.7 $ and $L(3) \approx 1.1 $ thus by the Intermediate Value Theorem there must be a $ \{ c \in \mathbb{R} \mid 2<c< 3\}$ such that $L(c) = 1 $. Of course, we call that value Euler's number or e.

I have heard it asserted that $L(b) = \ln(b)$ but am unsure how to derive that result?


答:
When you have a first-order recurrence like this, you can often simply ‘unwind’ it:

$$\begin{align*}
a_n&=3a_{n-1}+5\cdot3^n\\
&=3(3a_{n-2}+5\cdot3^{n-1})+5\cdot3^n\\
&=3^2a_{n-2}+2\cdot5\cdot3^n)\\
&=3^2(3a_{n-3}+5\cdot3^{n-2})+2\cdot5\cdot3^n\\
&=\color{red}{3^3a_{n-3}+3\cdot5\cdot3^n}\\
&\;\;\vdots\\
&=\color{blue}{3^ka_{n-k}+k\cdot5\cdot3^n}\\
&\;\;\vdots\\
&=3^na_0+n\cdot5\cdot3^n\\
&=3^n(2+5n)
\end{align*}$$

By the time you get to the red line, the pattern should be pretty clear; I’ve expressed it in general terms in the blue line. Then you can jump to the end and simplify to get a closed form for $a_n$. The step from the red to the blue line isn’t rigorous, so properly speaking you should prove by induction on $n$ that the formula $a_n=3^n(5n+2)$ really does satisfy the recurrence with the right initial condition, though in many cases, like this one, the pattern really is very clear.


问:
Basic question about how to form left adjoint of forgetful functor $U: R\text-\mathrm{Mod}\text-S \to R\text-\mathrm{Mod}$ as tensor product with $S$?
My algebra is a bit rusty, and I'm having some trouble forming a certain tensor product of modules over a non-commutative ring. Not sure if there is a typo, or if I'm forgetting something obvious. On page 87 of Mac Lane's Categories for the Working Mathematician (CWM) he offers a table of examples of left adjoints to forgetful functors. One of row of the table offers this example:

$U: R\text-\mathrm{Mod}\text-S \rightarrow R\text-\mathrm{Mod}$

taking $(R,S)$-bimodules to left $R$-modules, with the left adjoint given by:

$F:A\rightarrow A\otimes S$

taking left $R$-modules to $(R,S)$-bimodules. Presumably $S$ is intended to be a ring containing $R$, so that $S$ can serve as an $(R,S)$-bimodule here. But wouldn't $A$ is need to be a right $R$ module, so that we can even form the tensor product $A\otimes_R S$? Even if that were the case, wouldn't $A\otimes_R S$ then be merely a right $S$ module, when what is required is actually an $R,S$ bi-module? I'm apparently missing something, because I can't get the "types" correct.

答:
The tensor product should be over $\mathbb{Z}$. This doesn't require any relationship between $R$ and $S$ to be specified in advance.

If $R$ and $S$ are algebras over a common commutative ring $k$ and "$(R, S)$-bimodule" means "compatible $k$-module structures" then the tensor product should be over $k$.


问:
Counting dimensions for submanifolds
I am trying to find the conditions under which a complex circle manifold intersects with a linear subspace. I have a linear subspace of dimension T-K (defined as the null subspace of a $K \times T$ complex matrix) (i.e., a subspace of the space of complex vectors of size $T$, $\mathbb{C}^T$). I have control over $T$ but $K$ is fixed. The complex circle manifold is defined as $\mathcal{C} =\{\mathbf{x} \in \mathbb{C}^T|\; |x_1| = |x_2|=\cdots = |x_T|=1\}$. It has a dimension of $T/2$ as far as I see it (complex dimensions). Both the subspace and the complex circle manifolds are submanifolds of the Euclidean space. I am confused with the concept of dimensions counting (i.e., saying that $(T-K) + (T/2) > T$ would guarantee that the intersection will be non-empty) and I am not sure if it fits here. The reason for my confusion is that if I count the dimensions of two hyperspheres of different radii they will, of course, exceed $T$ dimensions, yet obviously, there is no intersection between them.

Please accept my apology as I am not a math expert and I might be missing something clear.

答:
Suppose $K = 1$.  Consider the $1 \times T$ matrix
$$A = \begin{bmatrix} 0 & 0 & \cdots & 0 & 1 \end{bmatrix}\!.$$
The nullspace of $A$ is the linear subspace
$$L = \{\mathbf{z} = (z_1, \ldots, z_T) \in \mathbb{C}^T \colon z_T = 0\}.$$
Note that $L$ has complex dimension $T - 1$.  In other words, $L$ has the largest possible complex dimension inside of $\mathbb{C}^T$ (without being $\mathbb{C}^T$ itself).  Nevertheless, $L$ does not intersect the torus
$$\mathcal{C} = \{\mathbf{z} \in \mathbb{C}^T \colon |z_1| = 1, \ldots, |z_T| = 1\},$$
because no point of $L$ has the property that $|z_T| = 1$.

Title: Counting dimensions for submanifolds


答:
One approach is to define $e$ as the unique number $b$ for which
$$
\lim_{h \to 0}\frac{b^h-1}{h}=1 \, .
$$
It then directly follows that $e^x$ is its own derivative. Then, since $b^h=e^{h\log(b)}$, it can be shown that the above limit is equal to $\log(b)$ in general. (See Gauge_name's answer.)

Title: Prove $L(b) = \ln(b)$


问:
Equation of two variables with no solutions in a given countable set
Let $(x_{n})$ be a sequence of non-negative real numbers diverging to infinity. Can we find a real number $r\in \mathbb{R}$ such that $x_{n}-x_{m}\neq r$ for all $n,m\in \mathbb{N}$?

答: $\mathbb N\times\mathbb N$ is countable, so certainly there are countably many differences $x_n-x_m$. However, there are uncountably many $r\in\mathbb R$.


答:
I guess the question was rather : can we find a sequence $(x_n)$ of positive numbers that diverge to $+\infty$ such that for all $r\in \mathbb{R}^+$, there exists $n,m$ such that $u_n -u_m = r$.

The answer is no.

In fact, $\{u_n - u_m; n,m\in \mathbb{N}\}$ is countable, while $\mathbb{R}$ is not.

Title: Equation of two variables with no solutions in a given countable set

Body: Let $(x_{n})$ be a sequence of non-negative real numbers diverging to infinity. Can we find a real number $r\in \mathbb{R}$ such that $x_{n}-x_{m}\neq r$ for all $n,m\in \mathbb{N}$?


问:
Determining the joint density function -- what's wrong with my solution?
Let $T_1$ be the time between a car accident and the claim reported to the insurance company. Let $T_2$ be the time between the report of the claim and payment of the claim. The joint density function of $T_1$ and $T_2$, $f(t_1, t_2)$,  is constant over the region $0<t_1, t_2<6, t_1+t_2<10$ and zero otherwise. Determine $f(t_1, t_2)$.

My attempt: Let $f(t_1, t_2) = k$. Then,
$$\int_0^6 \int_0^{10-y} k \; dx\; dy = \int_0^6 k(10-y) \; dy = 1 \implies k\left[60-\frac{35}{2}\right] = 1$$

which ultimately gives $k = \frac{1}{42}$, however, the correct answer is $k = \frac{1}{34}$. Where have I gone wrong?

答:
Blue line: the area on which you should have integrated.

Orange line: the area on which you actually integrated.


答:
The problem with your method is that your integral is incorrect. Your limits correspond to the region $0 < t_1< 10, 0<t_2<6, t_1 + t_2 < 10$.

Here's an alternative approach. Let $R$ denote the region in question. First of all, note that
$$
\iint_R k\,dx\,dy = k\cdot \iint_R dx\,dy = 1,
$$
where $R$ denotes the region of interest. We can rearrange the above to find that
that $k = 1/A$, where $A = \iint_R 1\,dx\,dy$ is the area of $R$. To compute the area, subtract the area of a triangle from the area of a square. In particular, define
$$
R_1 = \{x,y : 0 < x < 6, 0 < y < 6\},\\
R_2 = \{x,y: 0 < x < 6, 0 < y < 6 \text{ and } x + y \geq 10\}.
$$
Verify that $R = R_1 \setminus R_2$, which is to say that $R$ contains the points of $R_1$ not included in $R_2$. $R_1$ is a square with area $6 \times 6 = 36$, and $R_2$ is a triangle with area $\frac 12 2 \times 2 = 2$. Thus, the area of $R$ is $36 - 2 = 34$.

Thus, we arrive at the correct answer $k = 1/A = 1/34$.

Title: Determining the joint density function -- what's wrong with my solution?

Body:
Let $T_1$ be the time between a car accident and the claim reported to the insurance company. Let $T_2$ be the time between the report of the claim and payment of the claim. The joint density function of $T_1$ and $T_2$, $f(t_1, t_2)$,  is constant over the region $0<t_1, t_2<6, t_1+t_2<10$ and zero otherwise. Determine $f(t_1, t_2)$.

My attempt: Let $f(t_1, t_2) = k$. Then,
$$\int_0^6 \int_0^{10-y} k \; dx\; dy = \int_0^6 k(10-y) \; dy = 1 \implies k\left[60-\frac{35}{2}\right] = 1$$

which ultimately gives $k = \frac{1}{42}$, however, the correct answer is $k = \frac{1}{34}$. Where have I gone wrong?


问:
linear transformation- cant continue the proof
as part of studying for an exam i encountered a proof question:

Let vector spaces $U,V$ over a field $F$ and Linear maps $T,S:U\rightarrow V$ be given.

Prove that there exist bases $B_1$,$B_2$ of $U$ and $C_1,C_2$ of $V$ such that $[T]$ from $B_1$ to $C_1$ = $[S]$ from $B_2$ to $C_2$  if and only if $rank(T)=rank(S)$.
Here the rank of a linear map is defined as the dimension of its image: $rank(T)=dim(Im(T))$.

I know that in order to prove this i need to show both ways.

for the first way, i know that if $[T]=[S]$ than it doesnt matter in in what basis the matrixes are, the ranks of the matrixes will be equal, therefore $rank(T)=rank(S)$.

for the other way, i know that $rank(T)=rank(S)$ $\Rightarrow$ $dim(Im(T))=dim(Im(S))$ $\Rightarrow$ from the rank-nullity theorom i will know that $dim(ker(T))=dim(ker(S))$.

from the point that the dimensions of the kernels are equal i dont know how to continue the proof. i would like someone to continue the proof for me. and i dont know if the first way is true as well so if someine can tell me if it is true

答:
For such $T,S:U\to V$, if $\text{rank}(T)=\text{rank}(S)=n$, choose two ordered bases of $Im(T)$ and $Im(S)$, denoted as $\mathcal{B}_1=\{b_0,..,b_{n-1}\}$ and $\mathcal{B}_2=\{b_0',...,b_{n-1}'\}$. Choose $c_i$ and $c_i'$ s.t. $T(c_i)=b_i$ and $T(c_i')=b_i'$. I claim that $\{c_i\}_{i\leq n-1}$ and $\{c_i'\}_{i\leq n-1}$ are both independent sets (why?). Extend them to be basis of $U$ if they are not and make sure new $c_i$ and $c_i'$ satisfy $T(c_i)=T(c_i')=0$ when you extend it. Denote these two basis as $\mathcal{C}_1$ and $\mathcal{C}_2$ respectively. Then, the matrix representation of $T$ under $\mathcal{B}_1$ and $\mathcal{C}_1$ and $S$ under $\mathcal{B}_2$ and $\mathcal{C}_2$ are the same, because the two linear maps have the same rank (that is a matrix with 1 in the first $\text{rank}(T)=\text{rank}(S)$ many entries of diagonal and all other entries are zero).

For the other direction, note that basis change cannot alter the dimension of a matrix representation, so if $T$ and $S$ have different ranks, the matrix representation cannot be the same.


问:
definition of cycle theoretic fibre
I am studying the definition of Chow variety on Kollar's Rational Curves on Algebraic Varieties, and I am having some trouble in understanding Definition 3.9.

Here we have a proper morphism of schemes $g_i:U_i\rightarrow W$ where $U$ is irreducible and $W$ is reduced. There is an open subset $W_i\subset g_i(U_i)\subset W$ such that $g_i$ is flat over it and of relative dimension $d$. Now he takes $T$ to be the spectrum of a DVR and a morphism $h:T\rightarrow W$ which sends the closed point $T_0\mapsto w\in W$ and the generic point to a point in $W_i$. Now let $h^\ast U_i$ be the pullback of $U_i$ via $h$ and define $J\subset\mathcal{O}_{h^\ast U_i}$ to be the ideal given by functions whose support is contained in the special fiber of $h^\ast U_i$ over $T$.

Set  $f:(U_i)'_T=\text{Spec}(\mathcal{O}_{h^\ast U_i}/J)\rightarrow T$: it is a flat map by construction so the central fibre $Z_0$ has pure dimension $d$. Then one names the cycle theoretic fibre of $g_i$ at $w$ along $h$ to be the cycle $[Z_0]\in Z_d(g_i^{-1}(w)\times_w T_0)$.

What I think I am not understanding properly is how to picture the scheme $(U_i)'_T$. Moreover, and I think that this is the important question, I don't understand what is the point behind the construction of $(U_i)'_T$. I mean, why can't we just take the special fibre of $h^\ast U_i\rightarrow T$ and define its fundamental cycle to be the cycle theoretic fibre of $g_i$ at $w$ along $h$?

答:
(Five years past the point of this being helpful, but I found this while trying to answer the same question myself and thought I'd leave the answer I eventually came up with:)

The intuition for $(U_i)_T'$, and the difference between your suggested construction and the construction given in Kollár, is in how it treats components of $h^* U_i$ lying over the special fiber of $T$. This can be seen in a simple example, where we take $T=W=\mathrm{Spec}\ k[[t]]$ and $U=\mathrm{Spec}\ k[[t][x]/(tx)$ (this isn't really proper, but you can compactify it to be proper and everything else works the same). Since $T=W$, $h^* U$ is just $U$. Note that $U\to W$ is not flat, since the generic fiber is $0$-dimensional and the special fiber is an $\mathbb A^1$. The ideal $J$ here is $(x)$ (since where $t\neq 0$, $x=xtt^{-1}=0$), and thus $U_T'=\mathrm{Spec} k[[t]]$. That is, $U'_T$ is the closed subset of $U$ obtained by discarding the vertical fiber lying over the closed point of $T$, so now $U'_T\to W$ is flat (in fact, an isomorphism).

So, in this case, the resulting special fiber of $U'_T$ over the closed point of $T$ is just a point, while the special fiber of $U\to T$ over the closed point is a curve. Away from this fiber, both families are $0$-dimensional. The cycle-theoretic fiber "knows" that it should be a limit of $0$-dimensional cycles (thus $0$-dimensional), while the scheme-theoretic fiber can jump dimensions arbitrarily.

ContentLicense: CC BY-SA 3.0

Title: definition of cycle theoretic fibre


问:
Interpretation of probability distributions from a frequentist perspective
I was reading Kline’s (2011) textbook on structural equation modeling when I found something extremely interesting: Kline explained that, under a frequentist perspective, probability is based on the expected relative frequency over a large number of trials.  In other words, there isn't a probability associated with whether or not a particular guess is correct.

I'm just a bit confused then on the purpose and interpretation of the negative binomial distribution.  If, given a frequentist perspective, probability can't be associated with singular events, what does the negative binomial distribution mean? Doesn't it describe the the trial on which the rth success occurs in repeated independent Bernoulli trials?

答: Say you have a set of repeated Bernoulli trials until you get to the $r$th success. Think of that set of Bernoulli trials as one negative binomial experiment. The frequentist interpretation says that if you run that experiment over and over, as the number of experiments gets larger and larger, the distribution of the number of Bernoulli trials required for $r$ successes converges to the negative binomial distribution.


问:
Topology - Dunce Cap Homotopy Equivalent to $S^2$
So I'm trying to find two spaces with isomorphic homology groups but where the spaces aren't homotopy equivalent.

From my work so far, taking the Dunce Cap as a triangle with the edges identified as $aa(a^{-1})$ if that makes sense, the homology group would be $\Bbb Z$ for $n=0,2$ and $0$ elsewhere.

However how do I prove that it isn't homotopy equivalent to $S^2$? I'm relatively confident it isn't as it's homotopy equivalent to $D^2$?

答: No, for the dunce cap, $H_2=0$, and it is contractible!


问:
Quasi-finite morphism into affine scheme is affine?
Let $f: X\rightarrow\operatorname{Spec}A$ be a quasi-finite morphism of schemes, then is $f$ an affine morphism? If this is wrong, what if $A$ is a field?

Thanks for any answers.

答: No. Consider the punctured plane mapping in to the plane: $\Bbb A^2_k\setminus \{(0,0)\}\to \Bbb A^2_k$. This is quasi-finite (every fiber is one or zero points) but isn't affine because an affine morphism with affine target must have affine source.


问:
solve $\int_{c}(1-z)^2dz$; given that $c$ is the contour defined by$x=3\cos\theta$ and $y=3+3\sin\theta$; from $\theta=0$ to $\theta=\pi/2$
As the title says I need to solve this integral, but this is the first problem I have to solve of this type and the example given by our professor is almost entirely different, and he is yet to answers my messages asking about that...

So if you can give me some insight on how can I solve this problem, I would appreciate very much.

答:
Here we can define the curve C as a set of z $\in \mathbb C$ . Note that C is not a closed curve, and so the Cauchy-Goursat theorem you mentioned in some comments is not relevant.

Let $C := \{z \in \mathbb C| z = (3cos(\theta))+(3+3sin(\theta))i , \theta \in [0,\pi/2]\}$.
Then, by definition of line integrals, we have that if C is parameterized by $(x,y)=(x(t),y(t)), z=x+yi$, the following holds: $$\int_{C} f(z)dz = \int_a^b f(z(t))\cdot z'(t)dt$$ for $t \in [a,b]$.

This is a basic property of line integrals, and I'll leave the calculations up to you.


问:
Factorization algorithm for a semiprime of 29-digits
I'm looking for the fastest prime factorization algorithm for a semiprime of 29-digits.
There are many factorization algorithms there, so as far as I know (my background is not maths:)the key factor is the length of the number to be factorized. In this case, it has only 29 digits.
Many thanks</p>

答:
For a 29 digit semiprime number, elliptic curve factorization will suffice.

For larger semiprime numbers (around 92 digits and beyond) of a general form, the general number field sieve is the fastest factorization algorithm available.

As a practical matter, just download a program called YAFU, probably from SourceForge, for whatever platform you have. YAFU will factor semiprime numbers for you intelligently, starting with elliptic curve, and switching over to GNFS if needed. YAFU relies on GGNFS and ECM, so download those too as compiled binaries, and supply the paths to the folders the executables are in, in the yafu.ini configuration file, and you should be good to go. The configuration is straightforward, there is a ggnfs_dir and a ecm_path setting you have to set. Also choose how many threads to dedicate to the process. YAFU factors a 100 digit semiprime number (like the RSA-100 challenge number) in about an hour on my machine.

To run YAFU, simply run the yafu-Win32.exe or whatever on the command line and then type:

factor(YOUR_NUMBER_HERE)

So, to factor the RSA-100 number, you would type:

factor(1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139)

For a number of your size, 29 digits, I imagine it will take a few seconds.

If you are interested in learning the GNFS, first start by reading the Wikipedia on it to get a general overlay of the algorithm. When I was trying to learn it, I read a lot of C code, but the problem with that code is it was meant for performance, not readability. Anyways, I ended up writing an implementation of the GNFS in C#. The goal, beyond learning the GNFS, was to make the code readable, not performance, so people could read it to learn the algorithm, like I wish I had when I was learning. If you are interested, my code is on my GitHub. I also have uploaded a number of whitepapers that I read and referenced in order to help me write the application. The address is https://github.com/AdamWhiteHat/GNFS


问:
Verify that the vector field $\vec{V}(x,y,z)$ = $\frac{y\vec{i}-x\vec{j}}{x^2+y^2}$ is irrotational when $(x,y) \neq (0,0)$
By definition a vector field is irrotational when $curl(\vec{V}) = 0$.

If we take $(x,y) \neq (0,0)$, say $(x,y) = (1,1)$, Then $\vec{V}(x,y,z) = \frac{yi}{2} - \frac{x}{2}j +0k$

Then, $\Delta \times \vec{V}$

$= (\vec V_{z,y} - \vec V_{y, z})\vec i +  (\vec V_{x, z} - \vec V_{z, x})\vec j +  (\vec V_{y, x} - \vec V_{x, y})\vec k = -4\vec k \ne 0; \tag 3$

Therefore it is not irrotational. But it's supposed to be! So how can I show that it is?

答:
Of course you can compute ${\rm curl}({\bf v})$ right away, and obtain $\equiv{\bf 0}$. In fact your field
$${\bf v}(x,y,z):=\left({y\over x^2+y^2}, {-x\over x^2+y^2},0\right)$$
is the gradient field of the "function"
$$f(x,y,z):=-\,{\rm polar\ angle\ of}\,(x,y)\ .$$
This "polar angle" is not globally defined on ${\mathbb R}^2\setminus\{{\bf 0}\}$, but its gradient is.


问:
What is a thing that is homomorphized?
This is a purely syntax question. If $A$ is a structure and $\pi$ is a homomorphism sending $A$ to something.

We say that $\pi$ is a homomorphism of $A$.

Similarly we might want to say $A$ is a ____ of $\pi$.

What is that word? I found myself writing "$A$ is a structure for which $\pi$ is a homomorphism" but that really was verbose and convoluted.

What I had considered:

$A$ is a domain, but this doesn't carry the importance that $\pi$ is a homomorphism of $A$, it only says that $\pi$ is a function which maps $A$ to something.

答:
I'm not aware of any context where we have a fixed map $\pi$ but consider different structures on the domain of $\pi$ with respect to which $\pi$ remains a homomorphism, and this is the only sort of context I can think of where "$A$ is the domain of $\pi$" would seem inappropriate to me. So I don't think such a term exists.

(Contrast this with "$\pi$ is a homomorphism from $A$" - this language is useful since we do often consider multiple homomorphisms out of the same structure.)

EDIT: you clarify that this is in fact the context you're in. I stand by my claim that this context is rare enough that there probably isn't a standard term in the literature. That said, the term which fits the bill most cleanly for me is "compatible," and I would suggest the phrase "$A$ is a compatible domain for $\pi$," or - if you only ever talk about varying domains and not codomains - "$A$ is compatible with $\pi$." But of course unless I'm wrong and there is after all a standard term this should be defined explicitly in your text.

Alternatively, a phrase like "$A$ is a domain object for $\pi$" has a categorical flavor which may be desired (this is motivated by Ty Jensen's comment above). Personally I think this sort of thing is clunkier, but I could see it being preferred.

Title: What is a thing that is homomorphized?


答:
This is false in general, even if you assume that $f$ is of finite presentation. For example, take $f$ to be an open embedding of a non-affine scheme into an affine scheme.

If $A$ is a field then things are better. Namely, by [1, Tag06RT] if you assume that $f$ is of finite type, then $X$ is finite and discrete as a topological space. It’s then easy to see that if $X=\{x_1,\ldots,x_n\}$ then $X=\mathrm{Spec}\left(\prod\mathcal{O}(\{x_i\})\right)$.

[1] https://stacks.math.columbia.edu/

Title: Quasi-finite morphism into affine scheme is affine?

Body:
Let $f: X\rightarrow\operatorname{Spec}A$ be a quasi-finite morphism of schemes, then is $f$ an affine morphism? If this is wrong, what if $A$ is a field?

Thanks for any answers.


问:
Cloture of forward orbits of ternary dilation and Cantor set
I wonder if anyone can give me sensible proof.

The Cantor set $C$ is defined by
\begin{align*}
  C_0&=[0,1],\\
  C_1&=[0,\tfrac13]\cup [\tfrac23,1],\\
  C_2&=[0,\tfrac19]\cup [\tfrac29,\tfrac13]\cup [\tfrac23,\tfrac79]\cup [\tfrac89,1],\\[-6pt]
  &\ \,\vdots\\[-12pt]
  C&=\bigcap_{n=0}^\infty C_n.
\end{align*}

Observe that
$$C=\left\{ x\in [0,1] \quad  | \quad   x=\sum_{k=1}^\infty\frac{2\epsilon_k}{3^k} \quad \text{where}
 \quad \epsilon_k \in \{0,1\}^{\mathbb N}\right\}.$$

Consider the ternary dilation $x\longmapsto T(x)=3x \quad \text{mod} \, \, 1 $.

Question:  Show that there exists $x_0 \in  [0 ; 1)$ with the closure of its forward orbit $\{T^n(x_0) : n \in \mathbb N \}$ are identical to $C$.

\begin{align*}
T(x)=& 3x - \lfloor 3x \rfloor \\
T(T(x))= & T(3x - \lfloor 3x \rfloor) \\
= & 3(3x - \lfloor 3x \rfloor) - \lfloor 3(3x - \lfloor 3x \rfloor)\rfloor \\
 = & 3^2 x - \lfloor 3^2 x \rfloor\\
\vdots \\
T^n(x) = & 3^n x - \lfloor 3^n x \rfloor.
\end{align*}

So, by induction,
\begin{align*}
\overline{\{T^n(x_0)\}}=&  \overline{\bigcup_{k=1}^{n}\{3^n - \lfloor 3^n x\rfloor}\} \\
& = \bigcup_{k=1}^{n}\{\overline{3^n - \lfloor 3^n x\rfloor}\}.
\end{align*}

Another way consists to do some calculation from
\begin{align*}
x=& \sum_{k=1}^\infty\frac{2\epsilon_k}{3^k} & \\
=& 2 \epsilon_1 + \sum_{k=}^\infty\frac{2\epsilon_{k+1}}{3^k} & \\
=& \sum_{k=1}^\infty\frac{2\epsilon_k}{3^k} & \text{mod} \, \, 1
\end{align*}
Thus $T^n(x)= \sum_{k=1}^\infty\frac{2\epsilon_{k + n}}{3^k}$, hence $|T^n(x)- x| = \sum_{k=1}^\infty\frac{2(\epsilon_{k+n} - \epsilon_k)}{3^k}\leq \sum_{k=1}^\infty\frac{2}{3^k}=1$
So it isn't clear that the claim is true.

Any help is welcome.

答:
Consider the set $W$ of all finite words in the alphabet $\{0,1\}$.  This is a countable set, so we may write $W=\{w_n: n\in \mathbb N\}$.

Now consider the infinite word $w$ obtained by concatenating the $w_n$ one after another.

If the digits of $w$ are used in the ternary expansion of $x$, then the orbit of $x$ under $T$ will be dense.

Notice that the collected works of Shakespeare, if encoded in a binary fashion, will appear as part of $w$, and so will all of the questions and answers in MSE.

EDIT: Let me try to be a bit more specific regarding the choice of the point with  a dense orbit.

Take the universal word $w$ constructed above as the concatenation of all finite
words $w_n$, and write
$$
  w=\alpha _1\alpha _2\alpha _3\ldots ,
  $$
with each $\alpha _i$ in $\{0, 1\}$.
Defining
$$
  x_0=\sum_{k=1}^\infty\frac{2\alpha _k}{3^k},
  $$
I claim that $\{T^n(x_0):n\in {\bf N}\}$ is dense in the Cantor set $C$.  To see this, let $x$ be any point in $C$, and write
$$
  x=\sum_{k=1}^\infty\frac{2\epsilon _k}{3^k},
  $$
with each $\epsilon _i$ in $\{0, 1\}$.  Given any $\varepsilon >0$ let us therefore find some $n$ such that $|T^n(x_0)-x|<\varepsilon $.

We first  choose $N$ such that the truncated sum
$$
  x'=\sum_{k=1}^N\frac{2\epsilon _k}{3^k}
  $$
satisfies $|x-x'|<\varepsilon /2$.  By increasing $N$, if necessary, we may suppose that $3^{-N}<\varepsilon /2$.

Next, consider  the finite word
$\epsilon _1\epsilon _2\epsilon _3\ldots \epsilon _N$.  Since all finite words appear in the set $W$ defined above, and since $w$ is the concatenetion of
all words in $W$,  there must be some $n$ such that
$$
  \alpha _{n+1}=\epsilon _1, \quad   \alpha _{n+2}=\epsilon _2, \quad \cdots ,  \quad \alpha _{n+N}=\epsilon _N.
  $$
As noted by the OP, to apply $T$ to any element in the Cantor set has the same effect as shifting its  ternary
digits.  Therefore
$$
  T^n(x_0) = \frac{2\alpha _{n+1}}{3} + \frac{2\alpha _{n+2}}{3^2} + \cdots  + \frac{2\alpha _{n+N}}{3^N} + \sum_{k=N+1}^\infty \frac{2\epsilon _{n+k}}{3^k} =
$$$$ =
  x' + \sum_{k=N+1}^\infty \frac{2\epsilon _{n+k}}{3^k}.
  $$
This implies that
$|T^n(x_0) -x'|<3^{-N}$, so
$$
  |T^n(x_0) -x|\leq    |T^n(x_0) -x'| + |x'-x| < \varepsilon /2+\varepsilon /2=\varepsilon .
  $$


问:
Ring is Noetherian if it admits a faithful finitely generated module with ACC on submodules generated by ideals

I have a (commutative unitary) ring $A$ and a faithful finitely generated module $M$ over $A$ which satisfies ACC on submodules of the form $IM$, where $I\subset A$ is an ideal. I want to prove that $A$ is Noetherian.

I have already proved that, because $M$ is finitely generated, if $M$ is a Noetherian $A$-module, then $A/Ann(M)$ is a Noetherian ring. So, given my hypothesis, to prove that $A$ is Noetherian as a ring it suffices to prove that $M$ is Noetherian as $A$-module.

I've been given a sketch of proof, but I'm stuck with filling in the details. Here's the sketch:

Proceed by contradiction and assume $M$ is not Noetherian. Consider the set $\{IM \mid I\subset A \text{ is an ideal and }M/IM \text{ is not Noetherian}\}$ and reduce to the case where $M$ is not Noetherian, but for every ideal $I\neq 0$, $M/IM$ is Noetherian. Then consider the set $\Gamma = \{N\subset M \mid M/N \text{ is a faithful } A\text{-module}\}$, find a maximal element in $\Gamma$ and conclude.

Here's my attempt at using the hints I've been given: using ACC on submodules of the form $IM$ I found a maximal submodule $JM$ such that $M/JM$  is not Noetherian, so I thought substituting $M$ with $M/JM$, but $M/JM$ is clearly not faithful so I don't think that's the right approach.

Then I tried to assume the reduction to just $M$ not Noetherian had been made and tackled the set $\Gamma$. I managed to find a maximal element (using Zorn's lemma), but I'm clueless on how to use that maximal element to conclude.

I don't need the solution straight away, just some help to understand the hints better would be very much appreciated.

(I hope I explained everything clearly cause english is not my first language.)

答:
Edit (12/31): This result turns out to have a bit of history. It was published in 1973, though there were other formulations around prior to that [1]. One finds that this result has been reproduced in Matsumura's textbook, Commutative Ring Theory [2]. The method of proof is essentially the same as what I had answered prior to looking it up, with the first step proceeding in basically the same way. The second step is done a little differently, so I provide both in this answer. Since the approaches are similar in content, I made the answer community wiki.

First step: Let $IM$ be maximal such that $M/IM$ is not Noetherian as an $A$-module. Let $I_0=\operatorname{Ann}(M/IM)$. Then, $I_0M=IM$ and $M/I_0M$ is a faithful $A/I_0$-module. Thus, replacing $M$ with $M/I_0M$ and $A$ with $A/I_0$, we may assume $M/JM$ is Noetherian for all nonzero ideals $J$ of $A$, but $M$ is not Noetherian. We will arrive at a contradiction by showing that this $A$ is Noetherian (and thus this $M$ is Noetherian).

Second step: (The approach in the references) By Zorn's Lemma, we get a maximal submodule $N\subset M$ such that $M/N$ is a faithful $A$-module. We may assume $M/N$ is not Noetherian. So, replacing $M$ with $M/N$, we have

(a) $M$ is not Noetherian

(b) If $I$ is a nonzero ideal of $A$, then $M/IM$ is Noetherian

(c) $M$ is faithful, and for all nonzero submodules $N\subset M$, the $A$-module $M/N$ is not faithful.

Let $N$ be any submodule of $M$. We will prove $N$ is finitely generated. If $N=0$, then that is the case. Suppose $N\ne 0$. By statement (c), there exists $0\ne a\in A$ such that $aM\subset N$. Because $N/aM$ is a submodule of $M/aM$, which is Noetherian, $N/aM$ is finitely generated. Furthermore, if $x_1,\dots,x_n$ generate $M$, then $ax_1,\dots,ax_n$ generate $aM$. So, $aM$ is finitely generated. Therefore, $N$ is finitely generated. It follows that $M$ is Noetherian, violating statement (a). This is the desired contradiction.

(A different approach) Any ascending chain of submodules $M_i$ of $M$ such that $M_i\neq M_{i+1}$ for all $i$ has the property that $M/M_i$ is faithful for all $i$ (otherwise, $M_i\supset IM$ eventually and we know $M/IM$ is Noetherian, contradiction). Thus, if $N$ is a maximal submodule of $M$ such that $M/N$ is a faithful $A$-module, then $M/N$ is Noetherian. Hence, $A$ is Noetherian. This is a contradiction.

Having arrived at a contradiction, we conclude that our original $M$ is Noetherian, as wanted.

References:

[1] E. Formanek. "Faithful Noetherian Modules". Proc. AMS. 1973. 41(2): 381-3.

[2] H. Matsumura. Commutative Ring Theory. 2006. p. 18 (Theorem 3.6).


问:
How to solve these?

Inverse Trigonometric Functions

They are incomplete and I don't know how to complete them.
Who can help me?

1st

$$
\int\frac 1{ x \sqrt{x^{6} - 4}}dx
$$
I tried with:
$$u = x^3 $$
$$du= 3x^2dx$$
but this is not completed,

2nd

oops, is not $$x^2$$ is only "x"
$$
\int \frac 1{ x \sqrt{x-1}}dx
$$
$$u =  \sqrt x $$
$$du= \frac 1{\sqrt x} dx$$
where is the sqrt(x) on du?

3

$$
\int \frac {e^x}{ \sqrt{4-e^x}}dx
$$
I think in this
$$u=e^x $$
$$du = e^xdx$$
I have a bad English syntax, but I know read English++ Ok..

答:
Hint

$\int\frac {dx}{ x \sqrt{x^{6} - 4}}$, let $x^6=4\sec^2 u$.

$\int \frac {dx}{ x \sqrt{x-1}}$, let $x=\sec^2 u$.

$\int \frac {e^x }{ \sqrt{4-e^x}}dx$, let $e^x=4\cos^2 u$.

Keep in mind these trigonometric identities: $\cos^2u + \sin^2u = 1$ and $1 + \tan^2 u = \sec^2 u$.

For $(1)$, observe that in the denominator, you have $\sqrt{x^6-4}$. Now look at the trigonometric identities, you observe that $\tan^2 u = \sec^2 u - 1$. Which is pretty similar to $x^6 - 4$. Number $(2)$ is similar.

For number $(3)$, observe that in the denominator, you have $\sqrt{4-e^x}$. Now look at the trigonometric identities, you observe that $\sin^2u = 1 - \cos^2u$, which is pretty similar to $4-e^x$.

Title: How to solve these?


问:
Maclaurin expansion for sech$(x)$
I am a bit unsure where I have gone wrong in working this out.

Sech$(x)=2/(e^x+e^{-x}).$

Maclaurin expansions:

$e^x = 1+ x +  x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 -  x^4/24;$

so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$  (I think this is where I have gone wrong.)

The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).

How would I work this out?

答:
The simplest way t get it is to write $\;\operatorname{sech} x=\frac1{\cosh x}$ and to use the MacLaurin expansion  of $\cosh x$ at order $4$:
$$\cosh x=1+\frac{x^2}2+\frac{x^4}{24}+o(x^4)$$
You obtain the expansion of its reciprocal  dividing  $1$ by the MacLaurin expansion of $\cosh x$ along increasing powers, up to order 4, truncating every term at order $4$ in this process:
$$\begin{array}[t]{r|}
1\phantom{{}-\frac{x^2}2-\frac{x^4}{24}} \\
-1-\frac{x^2}2-\frac{x^4}{24}\\\hline
-\frac{x^2}2-\frac{x^4}{24} \\
\frac{x^2}2+\frac{x^4}{4} \\ \hline
\frac{5x^4}{24}\\ -\frac{5x^4}{24}\\\hline 0
\end{array}
\begin{array}[t]{l}
1+\frac{x^2}2+\frac{x^4}{24} \\\hline
\color{red}{1 -\frac{x^2}2+\frac{5x^4}{24}}
\end{array}$$

Title: Maclaurin expansion for sech$(x)$


答: A contractible space has no higher homology groups, so your example doesn't work. A working example would be $X=\mathbb CP^2$ and $Y=S^2\vee S^4$. They have the same homology, but are not homotopy equivalent. One way of seeing this, is looking at the cohomology.</p>

Title: Topology - Dunce Cap Homotopy Equivalent to $S^2$

Body:
So I'm trying to find two spaces with isomorphic homology groups but where the spaces aren't homotopy equivalent.

From my work so far, taking the Dunce Cap as a triangle with the edges identified as $aa(a^{-1})$ if that makes sense, the homology group would be $\Bbb Z$ for $n=0,2$ and $0$ elsewhere.

However how do I prove that it isn't homotopy equivalent to $S^2$? I'm relatively confident it isn't as it's homotopy equivalent to $D^2$?


问:
Why is $\epsilon = \dfrac{\alpha}{2}$ in this lemma of $\int_0^1f(t)\,dt = 0$ if and only if $f(t)=0$?

Lemma: Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous with $f(t) \ge 0$ for all $t \in [0,1].$ Then $\int_0^1f(t)\,dt = 0$ if and only if $f(t)=0$ for all $t \in [0,1]$.

For the non-trivial part, the author did by contradiction:

If $f$ is not identically $0$, since $f \geq 0$, there exists $t_0$ such that $f(t_0) > 0$. Let $\alpha = f(t_0)$ and $\epsilon = \dfrac{\alpha}{2}$. For this value of $\epsilon$, by continuity of $f$ at $t_0$, there is a $\delta$ such that $f(t) \in \left(\dfrac{\alpha}{2},\dfrac{3\alpha}{2}\right)$ for $t \in(t_0-\delta,t_0+\delta)$

which implies
$$\int_0^1f(t)\,dt \ge \int_{t_0-\delta}^{t_0+\delta} f(t)\,dt \ge \int_{t_0-\delta}^{t_0+\delta} \dfrac{\alpha}{2}\,dt = \alpha\delta > 0$$

My point is, why did author chose $\epsilon = \dfrac{\alpha}{2}$?

What I usually do is to backtrace $\epsilon$. So, I assumed $\epsilon = k $.

By the definition of continuity, for a given $\epsilon, $

$$ |f(t) - f(t_0)| < \epsilon $$
$$ f(t) \in \left(f(t_0)-\epsilon \, ,f(t_0)+\epsilon\right) $$
$$ f(t) \in \left(\alpha-k \, ,\alpha+k\right) $$

There exists $\delta$ such that, for $t \in(t_0-\delta,t_0+\delta)$

$$\int_0^1f(t)\,dt \ge \int_{t_0-\delta}^{t_0+\delta} f(t)\,dt \ge \int_{t_0-\delta}^{t_0+\delta} k\,dt = 2k\delta > 0$$

I can get the same result by just having $\epsilon$ as it is. Then why did the author chose $\epsilon = \dfrac{\alpha}{2}$?

Am I missing something?

答:
No, if you choose $$\epsilon=k$$

You will get

$$\int_0^1f(t)dt\ge \int_{t_0-\delta}^{t_0+\delta}f(t)dt$$
$$\ge \int_{t_0-\delta}^{t_0+\delta}(\alpha-k)dt$$

$$=2\delta(\alpha-k)$$
You can, if you want, take $\epsilon=\frac{\alpha}{3} $


问:
What does mean $df/dg$ (derivative of a function with relation to another)?
I have on a textbook a system of two ODEs as follow

$$\dfrac{df}{dx}=af(x)g(x)\\
\dfrac{dg}{dx}=-bg(x)$$

So, the author studies $df/dg$ writing $\dfrac{df}{dg}=-\dfrac{a}{b}f(x)$.

I understand the arithmetic, but I am not sure of mathemathical/geometric meaning.

Thank you in advance for light and maybe some lemmas or theorems of reference that it is valuable (are when is valuable).

答:
This notation is often difficult to make sense of. What, for instance, would it mean to differentiate $\sin(x)$ with respect to $\cos(x)$? To begin with, we must write $\sin(x)$ in terms of $\cos(x)$:
$$
\sin(x)=\sqrt{1-\cos(x)^2} \, ,
$$
ignoring the $\pm$ signs that should be on the square root for now. This means that
$$
\frac{d\sin(x)}{d\cos(x)}=\frac{d\sqrt{1-\cos(x)^2}}{d\cos(x)} \, .
$$
But this is the same as
$$
\frac{d}{du}\left(\sqrt{1-u^2}\right)\Biggr|_{u=\cos(x)} \, .
$$
So every time you see
$$
\frac{df(x)}{dg(x)} \, ,
$$
note that $f(x)=h(g(x))$ for some function $h$, and write
$$
\frac{dh(g(x))}{dg(x)}=\frac{dh(u)}{du}\Biggr|_{u=g(x)} \, .
$$
The $u$ is a dummy variable that can then be replaced by whatever function you are 'differentiating with respect to'. You can use $g(x)$ as a dummy variable, but it often leads to confusion, and is considered by many people to be poor style.

To give another example, let's consider the chain rule,
$$
\frac{dy}{dx}=\frac{dy}{du} \cdot \frac{du}{dx}
$$
Say $y=\sin(u)$, where $u=x^2$. Then, the chain rule appears to say
$$
\frac{d}{dx}(\sin(x^2))=\frac{d}{dx^2}\left(\sin(x^2)\right) \cdot \frac{d}{dx}(2x) \, .
$$
It seems difficult to parse the first term on the RHS. The easiest way to do this is to write it instead as
$$
\frac{d}{du}(\sin(u)) \Biggr|_{u=x^2} \, .
$$
However, we can also treat $x^2$ just as we would any other variable, and make sense of the term that way, too.


问:
2 dimensional symmetric random walk
Let $\left\{\left(X_{n}, Y_{n}\right)\right\}_{n=0}^{\infty}$ be a 2 -dimensional symmetric random walk. Namely, this is a Markov chain where $\left(X_{n+1}, Y_{n+1}\right)$ takes one of the following 4 values with equal probability:
$$\left(X_{n}+1, Y_{n}\right),\left(X_{n}-1,Y_{n}\right),\left(X_{n}, Y_{n}+1\right),\left(X_{n}, Y_{n}-1\right)$$
Suppose that $X_{0}=Y_{0}=0$.

Define $T:=\inf \left\{n \geq 0: \max \left(\left|X_{n}\right|,\left|Y_{n}\right|\right)=3\right\} .$
I want to find the value of $\mathbb{E}[T]$ and $\mathbb{P}\left(X_{T}=3, Y_{T}=0\right)$.
Thanks for your help!

答: The answer above is entirely correct. Am just here with some simple code to give you confidence that $\mathbb{E}[T] = \frac{135}{13} $

Title: 2 dimensional symmetric random walk


问:
Rational Expression Question. (Word problem)
Joe got a mark of $\dfrac{44}{50}$ on one test and $\dfrac{32}{x}$ on another test. If the average mark on the two tests was 80%, what value was the second test out of?

My revised attempt:

Still confused, is this correct? Seems odd that the answer I am getting is a repeating decimal:

$$\dfrac 12 \left(\dfrac{44}{50}+\dfrac{32}{x}\right)=.8$$

$$\dfrac{44}{50}+\dfrac{32}{x}=\dfrac{.8}{.5}$$

$$\dfrac{32}{x}=\bigg(\dfrac{8}{5}-\dfrac{44}{50}\bigg)x$$

$$32=\dfrac{8}{5}x-\dfrac{44}{50}x$$

$$50\bigg(32=\dfrac{8}{5}x-\dfrac{44}{50}x\bigg)$$

$$1600=80x-44x$$

$$\dfrac{1600}{36} = \dfrac{36x}{36}$$

$$\boxed{x=44.4}$$

答:
There are unfortunately two possible interpretations of "average mark." Some teachers would think of the average mark as $\frac{44+32}{50+x}$. Essentially they would be computing a weighted average. In that case, solving $\frac{44+32}{50+x}=0.8$, we get $x=45$. </p>

<p>It is quite possible that this is the intended solution, for the alternative $\frac{1}{2}\left(\frac{44}{50}+\frac{32}{x}\right)=0.8$ does not yield a nice $x$. </p>

Body:
There are unfortunately two possible interpretations of "average mark." Some teachers would think of the average mark as $\frac{44+32}{50+x}$. Essentially they would be computing a weighted average. In that case, solving $\frac{44+32}{50+x}=0.8$, we get $x=45$.

It is quite possible that this is the intended solution, for the alternative $\frac{1}{2}\left(\frac{44}{50}+\frac{32}{x}\right)=0.8$ does not yield a nice $x$.


问:
Given three equivalent statements, prove equality between two sets of variables
What technique should I use to solve the following problem?

Would I utilize the division algorithm?

Let $ m, n, r, s ∈ \mathbf{Z}$.

If $m^2 + n^2 = r^2 + s^2 = mr + ns$,

prove that $m = r$ and $n = s$.

Body:
We have
$$
(m-r)^2+(n-s)^2=m^2-2mr+r^2+n^2-2ns+s^2=0.
$$

Hence $m-r=0$ and $n-s=0$.


问:
Solve $ \int\frac{1}{\sin(x)-\cos(2x)}dx $ ...
Weierstrass substitution :
$$\tan(\frac{x}{2})=t$$  $$\sin(x)=(\frac{2t}{1+t^2})$$  $$\cos(x)=(\frac{1-t^2}{1+t^2})$$ $$dx=(\frac{2\,dt}{1+t^2})$$

Than : $$\cos(2x)=\cos^2(x)-\sin^2(x)$$

P.s I tried using these substitutions, but I couldn't get so far .
Need a bit help to solve this problem, if it's possible by using these substitutions.

Thank you in advance :)

Body: Since $\cos2x=2\cos^2x-1=\frac{1-6t^2+t^4}{(1+t^2)^2}$, $\sin x-\cos 2x=\frac{-1+2t+6t^2+2t^3-t^4}{(1+t^2)^2}$, so your integral is$$\int\frac{2(1+t^2)}{-1+2t+6t^2+2t^3-t^4}dt.$$Sincce $-1+2t+6t^2+2t^3-t^4=-(t^2-4t+1)(t+1)^2$ (the repeated root is easily guessed with the rational root theorem), we seek a partial fraction decomposition:$$\begin{align}\frac{-2(1+t^2)}{(t^2-4t+1)(t+1)^2}&=\frac{At+B}{t^2-4t+1}+\frac{Ct+D}{t^2-4t+1},\\-2(1+t^2)&=(At+B)(t+1)^2+(Ct+D)(t^2-4t+1)\\&=(A+C)t^3+(2A+B-4C+D)t^2\\&+(A+2B+C-4D)t+B+D.\end{align}$$Since $A+C=0,\,2A+B-4C+D=-2,\,A+2B+C-4D=0,\,B+D=-2$, the solution is$$\int\frac{-2(1+t^2)dt}{(t^2-4t+1)(t+1)^2}=\int\left(\frac{-4/3}{t^2-4t+1}+\frac{-2/3}{(t+1)^2}\right)dt,$$which you can finish.


答:
$\vec V=\dfrac{y}{x^2+y^2}{\hat{\imath}}-\dfrac{x}{x^2+y^2}{\hat{\jmath}}+0{\hat{k}}, (x,y)\ne (0,0)$.

\begin{align}
\vec\nabla \times\vec V&=\begin{vmatrix}
{\hat{\imath}} & {\hat{\jmath}} & {\hat{k}} \\
\dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
\dfrac{y}{x^2+y^2} & -\dfrac{x}{x^2+y^2} & 0
\end{vmatrix}\\
&=0{\hat{\imath}}+0{\hat{\jmath}}+\{-\dfrac{1}{x^2+y^2}+\dfrac{2x^2}{(x^2+y^2)^2}-\dfrac{1}{x^2+y^2}+\dfrac{2y^2}{(x^2+y^2)^2}\}{\hat{k}}\\
&=\vec 0
\end{align}
So $\vec V$ is irrotational.

Title: Verify that the vector field $\vec{V}(x,y,z)$ = $\frac{y\vec{i}-x\vec{j}}{x^2+y^2}$ is irrotational when $(x,y) \neq (0,0)$

Body:
By definition a vector field is irrotational when $curl(\vec{V}) = 0$.

If we take $(x,y) \neq (0,0)$, say $(x,y) = (1,1)$, Then $\vec{V}(x,y,z) = \frac{yi}{2} - \frac{x}{2}j +0k$

Then, $\Delta \times \vec{V}$

$= (\vec V_{z,y} - \vec V_{y, z})\vec i +  (\vec V_{x, z} - \vec V_{z, x})\vec j +  (\vec V_{y, x} - \vec V_{x, y})\vec k = -4\vec k \ne 0; \tag 3$

Therefore it is not irrotational. But it's supposed to be! So how can I show that it is?


答:
Since in a neighbourhood of the origin $\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have $\operatorname{sech}(x)=1-\frac{x^2}{2}+Ax^4+O(x^6)$
and since
$$ \left(1+\frac{x^2}{2}+\frac{x^4}{24}\right)\left(1-\frac{x^2}{2}+Ax^4\right)=1+\left(A-\frac{5}{24}\right)x^4+O(x^6) $$
we must have $A=\frac{5}{24}$. This unique non-trivial coefficient can also be found by considering that

$$\cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{\pi^2(2n+1)^2}\right)\tag{1} $$
$$\operatorname{sech}(x)=\prod_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^m 4^m x^{2m}}{\pi^{2m}(2n+1)^{2m}}\tag{2} $$
$$\operatorname{sech}(\sqrt{x})=\prod_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^m 4^m x^{m}}{\pi^{2m}(2n+1)^{2m}}\tag{3} $$

$$ [x^2]\operatorname{sech}(\sqrt{x})=\frac{16}{\pi^4}\sum_{a>b\geq 0}\frac{1}{(2a+1)^2(2b+1)^2}+\frac{16}{\pi^4}\sum_{n\geq 0}\frac{1}{(2n+1)^4}\tag{4} $$

and the RHS of $(4)$ just depends on the values of $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$.

Title: Maclaurin expansion for sech$(x)$

Body:
I am a bit unsure where I have gone wrong in working this out.

Sech$(x)=2/(e^x+e^{-x}).$

Maclaurin expansions:

$e^x = 1+ x +  x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 -  x^4/24;$

so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$  (I think this is where I have gone wrong.)

The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).

How would I work this out?


Title: Solve $ \int\frac{1}{\sin(x)-\cos(2x)}dx $ ...

Body:
Weierstrass substitution :
$$\tan(\frac{x}{2})=t$$  $$\sin(x)=(\frac{2t}{1+t^2})$$  $$\cos(x)=(\frac{1-t^2}{1+t^2})$$ $$dx=(\frac{2\,dt}{1+t^2})$$

Than : $$\cos(2x)=\cos^2(x)-\sin^2(x)$$

P.s I tried using these substitutions, but I couldn't get so far .
Need a bit help to solve this problem, if it's possible by using these substitutions.

Thank you in advance :)

Body:
hint

$$\sin(x)-\cos(2x)=$$
$$\sin(x)-1+2\sin^2(x)=$$
$$(\sin(x)+1)(2\sin(x)-1)$$

and

$$\frac{1}{\sin(x)-\cos(2x)}=$$
$$\frac 13\Bigl(\frac{-1}{\sin(x)+1}+\frac{2}{2\sin(x)-1}\Bigr)$$


问:
From discrete dynamical system to continuous one
To me it seems that is not always the case that a discrete dynamical system of the form
$$ x_{k+1} = f(x_k) $$
can be seen as the discretization of a continuous one like
$$ \dot{x}(t) = f(x), $$
while we can always go in the opposite direction.

My question is: is there some keyword I can search in the literature to understand better this connection or do you have any reference to suggest?

I would like to understand if there are some sufficient conditions guaranteeing the possibility of doing this transition, I don't even need a constructive way to build this continuous version, "just" existence would already be a great thing to understand.

To be clear, I would appreciate even results on this sort of "continuous prolongation" based on increasing the dimension of the phase space, i.e. for example for a discrete system of $\mathbb{R}^2$ if there is a dimension $d\geq 2$ such that on it there is a well defined continuous system whose discretization at certain time instants correspond to the discrete one. This question should be easier since there is not the constraint in the dimension.

I know it is quite a broad question, but I need at least some terminology to look for.

Up to now I have read about embeddability of homeomorphisms into flows and this seems to answer to my question in some cases, so I just wanted to know if the embeddability into flows is the right thing to study or if there are other perspectives.

P.S.
About getting an approximation of them I am finding very interesting this question Links between difference and differential equations? at the moment.

答:
As you allude to in your question, there is indeed a construction which inputs a discrete dynamical system in the form of a self-diffeomorphism $f : M \to M$ of a smooth manifold $M$, and outputs a continuous dynamical system called the suspension flow of $f$. This suspension flow is defined on a manifold $M_f$ of one dimension higher than $f$, namely the quotient of $M \times \mathbb R$ under the equivalence relation generated by $(x,t) \sim (f(x),t-1)$. This suspension flow is also generated in the usual fashion by a vector field on $M_f$, namely the quotient vector field of the $\frac{d}{dt}$ vector field on $M \times \mathbb R$ equipped with coordinates $(x,t)$, $x \in M$, $t \in \mathbb R$. Thus, $f$ can be realized as the time $1$ map of the suspension flow, restricted to the image of $M$ embedded in $M_f$.

There are further generalizations of this construction beyond the realm of self-diffeomorphisms.

In the case of a self-homeomorphism $f : M \to M$ where $M$ is just a topological space, the construction of the suspension flow on $M_f$ works exactly as stated, producing an action of $\mathbb R$ on $M_f$. However, $M_f$ is not a smooth manifold and so its rather tricky to express the suspension flow as being generated by a vector field.

There is a still further generalization to any continuous self-map $f : M \to M$ where $M$ is again a general topological space. There is again a "suspension flow" construction, defined on the quotient of $M \times [0,\infty)$ by the equivalence relation $(x,t) \sim (f(x),t-1)$ for $x \in M$, $t \in [1,\infty)$. The reason for restricting to positive time is that $f$ itself can only be iterated with positive exponents; there is no inverse map $f^{-1}$.


问:
Union and Intersection, Problem with {a}
In some book (Deiser, Mengenlehre, 3. Aufl.) about set theory the author defines the general union and intersection:

$\cap$M = {x|for all a $\in$ M also x $\in$ a}

$\cup$M = {x|there exists some a $\in$ M with x $\in$ a}.

He then brings the example: $\cap${a} = $\cup${a} = a. If you just apply the definition it does not make sense to me. Do you see that by just applying the definition? What'd make sense is this: $\cap${{a}} = $\cup${{a}} = a. Am I right or do I make a mistake?

答:
Yes, it follows from the definition.

$\bigcap\{a\}$ is by definition the set of all $x$ such that $x\in y$ for every member $y$ of the set $\{a\}$. The only member of the set $\{a\}$ is $a$, so by definition $x\in\bigcap\{a\}$ if and only if $x\in a$. The sets $\bigcap\{a\}$ and $a$ therefore have the same members and hence are the same set.

Similarly, $\bigcup\{a\}$ is by definition the set of all $x$ such that $x$ belongs to at least one member of $\{a\}$. The only member of $\{a\}$ is $a$, so if $x$ belongs to $\bigcup\{a\}$, then $x$ must belong to $a$. Conversely, if $x\in a$, then certainly $x$ belongs to at least one member of $\{a\}$, so $x\in\bigcup\{a\}$. Thus, $\bigcup\{a\}\subseteq a$ and $a\subseteq\bigcup\{a\}$, so $\bigcup\{a\}=a$.


答:
For $\int \frac{dx}{ x \sqrt{x^2-1}}$
let $x = \cosh(y)$,
so $\sqrt{x^2-1} = \sinh(y)$
and $dx = \sinh(y) dy$.

The integral becomes
$\int \frac{\sinh(y) dy}{\cosh(y)\sinh(y)}
=\int \frac{ dy}{\cosh(y)}
=\int \frac{2 dy}{e^y + e^{-y}}
=\int \frac{2 e^y \,dy}{e^{2y}+1}
=\int \frac{2 dz}{z^2+1}
$
with $z = e^y$.
This last is standard, so just substitute back.

Title: How to solve these?

Body:
Inverse Trigonometric Functions

They are incomplete and I don't know how to complete them.
Who can help me?

1st

$$
\int\frac 1{ x \sqrt{x^{6} - 4}}dx
$$
I tried with:
$$u = x^3 $$
$$du= 3x^2dx$$
but this is not completed,

2nd

oops, is not $$x^2$$ is only "x"
$$
\int \frac 1{ x \sqrt{x-1}}dx
$$
$$u =  \sqrt x $$
$$du= \frac 1{\sqrt x} dx$$
where is the sqrt(x) on du?

3

$$
\int \frac {e^x}{ \sqrt{4-e^x}}dx
$$
I think in this
$$u=e^x $$
$$du = e^xdx$$
I have a bad English syntax, but I know read English++ Ok..


答:
Add $m^2 + n^2$ to $r^2 + s^2$ noting that each of the two sums equals $mr + ns$

$$
m^2 + n^2 + r^2 + s^2 = 2mr + 2ns \\
m^2 - 2mr + r^2 + n^2 - 2ns + s^2 =  0 \\
(m - r)^2 + (n - s)^2 = 0
$$

which for $m, n, r, s$ integer holds if and only if $m = r$ and $n = s$.

Title: Given three equivalent statements, prove equality between two sets of variables

Body:
What technique should I use to solve the following problem?

Would I utilize the division algorithm?

Let $ m, n, r, s ∈ \mathbf{Z}$.

If $m^2 + n^2 = r^2 + s^2 = mr + ns$,

prove that $m = r$ and $n = s$.


答:
Hint

$$\sin x-\cos2x=2s^2+s-1=(s+1)(2s-1)$$ where $s=\sin x$

Now use https://mathworld.wolfram.com/PartialFractionDecomposition.html for $$\dfrac1{2s^2+s-1}=\dfrac a{s+1}+\dfrac b{2s-1}$$

Now $1+\sin x=1+\cos(\pi/2-x)=\dfrac2{\sec^2(\pi/4-x/2)}$

For the other integral, use https://en.m.wikipedia.org/wiki/Weierstrass_substitution#The_substitution


问:
Extension field equality
Suppose you have a subfield $F$ of a field $E$. If we want to prove that two extension fields are equal, e.g. $F(a) = F(b,c,d)$ for $a,b,c,d \in E$, why is it sufficient to prove that $a \in F(b,c,d)$ and $b,c,d \in F(a)$?

答:
Per definition, $F(b,c,d)$ is the smallest subfield of $E$ containing $F$ and the elements $b,c,d$. If $F(a)$ happens to contain $b,c,d,$ then it is a subfield containing $F,b,c$ and $d$.

Since it is trivial to show that $F(b,c,d)$ is the intersection of all subfields of $E$ containing $F,b,c,d,$
it follows that $F(b,c,d) \subset F(a)$.

The other inclusion follows similary.

Title: Extension field equality


问:
finite subsets proof

If $A$ is an infinite set then $\mathcal{F}(A),$ the set of finite subsets of $A$ has the same cardinality as $A$. Is the following proof of the above fact correct?

Note that $|A_k|\leq |A^k|$ for any $k\geq 0,$ where $A_k$ is the set of subsets of $A$ of cardinality $k$  since the map $f : A_k\to A^k, f(\{a_1,\cdots, a_k\}) = (a_1,\cdots, a_k)$ is injective (indeed for two different subsets, at least one element differs, so that element will be mapped to a different tuple by $f$). Then $\mathcal{F}(A) = \cup_{k\geq 0} A_k$, which is a disjoint union as $A_i \cap A_j = \emptyset$ for $i\neq j$ (the cardinality of a finite set is unique). So $|\mathcal{F}(A)| = \sum_{k\geq 0} |A_k| \leq \sum_{k\geq 0}|A^k| = \sum_{k\geq 1}|A^k| = \sum_{k\geq 1} |A| = \aleph_0 |A|$. Also, $g : A\to A_1, g(a) = \{a\}$ is clearly a bijection, so $|A| = |A_1| \leq |\mathcal{F}(A)|$ as $A_1\subseteq \mathcal{F}(A)$ and if $B\subseteq A$ then there is clearly an injective (identity) function from $B$ to $A.$ So $|\mathcal{F}(A)| = |A|$ by transitivity (for infinite sets this also applies).

答:
Small issue: the function $f : A_k \to A^k$ you wrote down is not well-defined! For example, should $f(\{2,3\}) = f(\{3,2\})$ be $(2,3)$ or $(3,2)$? You gave no way to decide. You could try to fix this in a few ways:

Put a total order on $A$ (if you assume the axiom of choice, this is always possible), and then define $f$ so that it lists the elements in ascending order.

Put a total order on every element of $A_k$ (if you assume the axiom of choice, this is always possible), and then define $f$ so that it lists the elements in ascending order.

Redesign the argument: instead build a surjection $A^k \to A_k$.

To me, approach #3 feels most natural -- there is an "obvious map" $(a_1, \dots, a_k) \mapsto \{a_1, \dots, a_k\}$ that feels like it should be a surjection. The only issue is that $\lvert \{a_1, \dots, a_k\} \rvert$ need not equal $k$! So we have the wrong codomain: instead, this is a surjection $A^k \to A_{\leq k}$. This shows $\lvert A_{\leq k} \rvert \leq \lvert A^k \rvert$, and we also have $\lvert A_k \rvert \leq \lvert A_{\leq k} \rvert$, so transitively $\lvert A_k \rvert \leq \lvert A^k \rvert$.

Another small issue: you definitely showed $\lvert \mathcal{F}(A) \rvert \leq \aleph_0 \lvert A \rvert$, and that $\lvert A \rvert \leq \lvert \mathcal{F}(A) \rvert$. For your proof to be complete, you should say somewhere that $\aleph_0 \lvert A \rvert = \lvert A \rvert$.

Edit: as a side note, I should also mention that it's necessary to assume the axiom of choice here: choice is equivalent to the claim that $a^2 = a$ for all infinite cardinals $a$, and if we had $\lvert A \rvert^2 > \lvert A \rvert$ here the claim would be false.

Title: finite subsets proof


问:
Multi-input function surjectivity and injectivity
How does surjectivity and injectivity apply to multi-input (multivariable) function?
Let say we have a function from R x R to R:

f(x,y) = x + y + 6

If we assume that f(x,y) = 9, so:

9 = x + y + 6

And thus:

3 = x + y

So does it mean that the function is not injective for pairs x,y (for example) f(1.5, 1.5) and f(0, 3) as both are giving the same output equal 3? Or is it not working that way?

And in terms of surjectivity, if x, y ∈ R and f(x,y) = z , function will be surjective if the output will simply be z ∈ R?

Best regards

答: The function is surjective if for every element of the codomain, say $z$, there exists an element of the domain whose image is $z$. In this case, $f$ is surjective since for each $z \in \mathbb R$ we have $f(z,-6) = z$. The injectivity part is fine!


问:
How do I simplify this sum?
What technique should I use for simplifying

$$\sum_{k=1}^\infty \frac{k-1}{k(k+1)(k+2)} ?$$

答:
Let $$g(k)=\dfrac{k-1}{k(k+1)(k+2)}=f(k+1)-f(k)$$ where $f(m)=\dfrac{am+b}{m(m+1)}$ where $a,b$ are arbitrary constants

so that using  https://en.m.wikipedia.org/wiki/Telescoping_series,

the partial sum
$$\sum_{k=1}^ng(k)=g(n+1)-g(1)$$

Compare the coefficients of $k^2,k$ and the constants to find $a,b$

Finally set $n\to\infty$

Title: How do I simplify this sum?


答:
It is better to have the half-angle substitution after the partial decomposition and on $t= \frac\pi2 -x $ instead, as seen below

\begin{align}
 \int\frac{1}{\sin x-\cos 2x}dx
 & =-\int \frac{1}{\cos t+2\cos^2t -1}dt\\
& =\int \frac{1}{(1-2\cos t)(1+\cos t )}dt\\
&= \frac13 \int \left(\frac1{1+\cos t}+ \frac1{\frac12- \cos t}\right)dt\\
&= \frac13 \int \left(\frac12\sec^2\frac t2 -\frac{\frac23 \sec^2\frac t2}{\frac13-\tan^2\frac t2}\right)dt\\
&= \frac13 \tan\frac t2 -\frac4{3\sqrt3} \tanh^{-1}(\sqrt3\tan \frac t2)+C
\end{align}


问:
Writing a 2nd order PDE as a system of equations
I want to turn this 2nd order equation into a system of first order equations but I am unsure about whether I can get rid of the $u$ or not

$$u_{xy}-u_x+u_y+10u u_{xx}$$

To write this as a system of equations so I can determine whether its semi-linear, quasilinear or nonlinear I thought that

$\xi\equiv u_x$

$\eta \equiv u_y$

$$\xi_y-\xi+\eta+10u \xi_x=0$$

$$\xi_y=\eta_x$$

Form a system of 2 equations.

But $u$ is still in it? Does this say something about whether it is quasi-linear or not? Is $u\equiv u(x,y)$ still? If it is then the equation is in the form

$\xi(x,y)_y+f(x,y)\xi(x,y)_x+g(\xi,\eta)=0$

I cannot tell what classification these equations have.

答:
As far as I can tell, "quasilinear" means "linear in the highest-order derivatives".
Here the highest-order derivatives are $u_{xy}$ and $u_{xx}$, and the equation is indeed linear in those. The fact that the coefficient of $u_{xx}$ involves $u$ is no problem for quasilinear.  On the other hand, since this coefficient does involve $u$ rather than depending only on the independent variables $x,y$, your equation is not semi-linear.</p>


问:
Why do books titled "Abstract Algebra" mostly deal with groups/rings/fields?
As a computer science graduate who had only a basic course in abstract algebra, I want to study some abstract algebra in my free time. I've been looking through some books on the topic, and most seem to 'only' cover groups, rings and fields. Why is this the case? It seems to me you'd want to study simpler  structures like semigroups, too. Especially looking at Wikipedia, there seems to be a huge zoo of different kinds of semigroups.</p>

答: Any theorem on groups is simultaneously a theorem of monoids, semigroups and magmas, since each layer is defined as a refinement of the previous layer. Once you finish a book/course in abstact algebra and move on to applying it in your field, there is a plethora of resources for using the theory of semigroups and monoids in computer science, so you'll be set.</p>

ContentLicense: CC BY-SA 2.5


问:
Long run percentage of customers who wait for a bus less than x units of time if customers arrive according to a homogenous Poisson process?
Assume that customers arrive to a bus stop according to a homogenous Poisson process with rate $\alpha$ and that the arrival process of buses is an independent renewal process with interarrival distribution $F$. What is the long run percentage of customers who wait for a bus less than $x$ units of time?

I tried to solve this problem by considering a renewal reward process. In the case that $x$ is very large so that it imposes no restriction we may proceed as follows to solve a similar problem: we say that a new cycle of length $T$ begins each time a bus arrives. Further, we assume customers pay us money at a rate of $1$ per time unit while they wait for a bus. Then our "reward rate" at any time corresponds to the number of people who are waiting at that time. So, by the renewal reward theorem we have that: average number waiting = $E[R]/E[T]$, where $R$ is the total reward during a cycle. In that case, $E[R] = 1/2E[NT] = 1/2\alpha E[T^2]$, where $N$ is the number of arrival in a cycle, which follows from the definition of a Poisson process and since the points of the Poisson process are uniformly distributed in a cycle conditional on the total number of points in that cycle.

But I don't know how to define the reward with the restriction imposed by $x$. I tried the following conditional expectation: $E[R|T,N]=N x1_{T>x}/2+NT1_{T\leq x}/2$, but got stuck. Can someone please help me?

答:
I ended up writing my comment as a full solution.  Main ideas:

1.) Poisson process has zero probability of arrival at a particular time -- i.e. there is zero probability of a 'collision'

2.) Define a renewal reward process, where a new epoch begins each time a bus arrives, and a reward of 1 for each customer that waited less than $x$ in time.  The long-run probability / percentage of customers with a sufficiently short wait ($\lt x$) is given by

$p=\lim_{t\to \infty}\frac{R(t)}{N(t)}$

where $N(t)$ of course is the r.v. for the Poisson counting process

3.)  The Poisson process is memoryless, which is something we repeatedly exploit, e.g. allows the above renewal reward process to properly work, and the associated renewal function is quite simple $E[N(t)]=m(t)=\frac{t}{\mu}$, and so on.

The calculations

$p=\lim_{t\to \infty}\frac{R(t)}{N(t)}=\lim_{t\to \infty}\frac{\frac{R(t)}{t}}{\frac{N(t)}{t}}= \lim_{t\to \infty}\frac{R(t)}{t}\cdot \lim_{t\to \infty}\frac{t}{N(t)}$

i.)  $\lim_{t\to \infty}\frac{t}{N(t)}=\mu$ by SLLN

ii.) $\lim_{t\to \infty}\frac{R(t)}{t}=\frac{E[R_1]}{E[T_1]}$ where $R_1$ and $T_1$ are rewards and time associated with first epoch.  Using conditioning

$E\big[R_1\big]= E\big[E[R_1\vert T_1=t]\big]$ and in particular

if $t\in(0,x)$ then $E[R_1\vert T_1=t]=m(t)=\frac{t}{\mu}$

if $t\geq x$ then $E[R_1\vert T_1=t]=m(x)=\frac{x}{\mu}$

These both hold by memorylessness. In the latter case this reads: if the epoch length t is at least $x$ long what is the expected number of Poisson arrivals in the interval $(t-x,t)$, but by memorylessness this is the same as the expected number of Poisson arrivals in the interval $(0,x)$ and this is given by $m(x)=\frac{x}{\mu}$.  For the former case it reads, what is the expected number of Poisson arrivals between the beginning and ending of the epoch, i.e. in the interval $(0,t)$ which is given by $m(t)=\frac{t}{\mu}$

Thus

$E\big[R_1\big]= E\big[E[R_1\vert T_1=t]\big]=\big(\int_{0}^x \frac{t}{\mu} dF(t)\big)+ \big(\int_{x}^\infty \frac{x}{\mu} dF(t)\big) = \frac{1}{\mu}\Big(\int_{0}^x t dF(t)\big)+ \big(\int_{x}^\infty x dF(t)\big)\Big)$

$\implies E\big[R_1\big]=\frac{1}{\mu}\cdot E\big[\min(T_1,x)\big]$

putting this all together

$p=\lim_{t\to \infty}\frac{R(t)}{t}\cdot \lim_{t\to \infty}\frac{t}{N(t)}=\Big(\frac{1}{\mu}\cdot E\big[\min(T_1,x)\big]\cdot\frac{1}{E[T_1]}\big)\cdot \big(\mu\big)= \frac{E[\min(T_1,x)]}{E[T_1]}$

$= \frac{\int_0^x Pr(T_1\gt t)dt}{E[T_1]}$


问:
Balls in a box, volume off smaller balls an bigger balls, will it change with same weight?
If I have $10$ pounds of stainless steel balls of $2$ inch, then together they have a (pack)volume. Will the (pack)volume change if I switch to $10$ pounds of stainless steel $4 $ inch balls?

(I think it will change because of the air between the balls is more with the $4$ inch balls, so it will be a bigger box I need, but not sure).

答: It depends how you pack. Let's assume that you have a very large number of balls, so that you can ignore the effects of the balls on the edge. If you pack them the same way, say simple cubic, then yes, you have bigger air gaps, but you have fewer of them. In fact, from geometrical considerations only, you can show that the same arrangement has the same fraction of filled and void. Look up packing fraction. See this link for example.</p>


问:
Identifying the differentials of the cellular chain complex of a (fat) geometric realization
Let $X_\bullet$ be a simplicial set, $\|X_\bullet\|$ its fat geometric realization and $|X_\bullet|$ its geometric realization, see here for definitions.

The fat geometric realization $\|X_\bullet\|$ has a natural CW complex structure with $n$ cells in bijection with $X_n$ and attaching maps done in accordance with face maps. One thus expects:

Plausible fact 1. The Moore complex $CX_\bullet$ is naturally isomorphic with the cellular homology complex of this CW complex structure on $\|X_\bullet\|$.

And indeed one can find this fact stated in Dupont's Curvature and Characteristic Classes at the beginning of the proof of Proposition 5.15:

The geometric realization $|X_\bullet|$ also has a CW complex structure with $n$ cells in bijection with the nondegenerate simplices of $X_n$ (Proposition 2.3 in Goerss-Jardine) and with attaching maps done in accordance with face maps. One similarly expects that the quotient complex $CX_\bullet/DX_\bullet$, where $DX_\bullet$ is the subcomplex generated by all degenerate simplices, is the cellular homology complex of the natural CW complex structure on $|X_\bullet|$. Also, since the normalized chain complex $NX_\bullet$ is naturally isomorphic to $CX_\bullet/DX_\bullet$ one expects

Plausible fact 2. The normalized Moore complex $NX_\bullet$ is naturally isomorphic to the cellular homology complex of this CW complex structure on $|X|$.

And indeed one finds this statement as Proposition 13.8 in P. May's Classifying Spaces and Fibrations

Both points above are clearly true of the underlying graded abelian groups. The sources I cite further claim that this is true of the differentials involved. This seems utterly believable since in both cases the attaching maps of are "done in accordance with face maps" and it seems attaching is done "without multiplicities" in the sense that all attaching maps either "collapse something" and should have degree $0$ or are affine homeomorphisms onto their image and should have degree $\pm1$. Neither fact is given a full proof, however.

Question. Is there a nice write-up somewhere of these facts? I.e. of the computations identifying the differentials of the cellular homology complexes of $\|X_\bullet\|$ and $|X_\bullet|$ with those of the Moore complex and the normalized Moore complex respectively? Alternatively if it's not too much work could someone spell it out?

答:
The semi-simplicial case. This is a partial answer that should contain most of what is needed for a clean proof. The proof for the normalized case should be very similar, although slightly complicated by the fact that some simplices will be "modded out". I'd still like a clean reference where the details are worked out! A good reference to a textbook/paper that treats either or both of the differentials will get the bounty.

A fact that ought to be true

Let $X_\bullet$ be a semi-simplicial set, i.e. a contravariant functor $\Delta_\text{inj}\to\mathrm{Set}$. Let $\Delta^\bullet$ be the usual co-semi-simplicial topological space with $\Delta^\bullet[n]=\Delta^n$ the standard $n$-simplex and the $d^i:[p]\to[q]$ realized by the usual affine maps. I'll take for granted that

Fact.
$$
\|X_\bullet\|
=
\int^n\Delta^n\times X_n
=
\mathrm{coeq}\bigg(
\bigsqcup_{\varphi:[p]\hookrightarrow[q]} \Delta^p\times X_q\rightrightarrows\bigsqcup_{[r]} \Delta^r\times X_r
\bigg)
$$
is a CW complex with the filtration given by the
$$
\|X_\bullet\|^{(n)}
=
\mathrm{coeq}\bigg(
\bigsqcup_{\substack{\varphi:[p]\hookrightarrow[q]\\q\leq n}} \Delta^p\times X_q\rightrightarrows\bigsqcup_{r\leq n} \Delta^r\times X_r
\bigg)
$$
and that the topology works out perfectly. In particular that there are pushout diagrams
$$
\begin{array}{ccc}
\displaystyle\bigsqcup_{X_n} \partial\Delta^n & \hookrightarrow & \displaystyle\bigsqcup_{X_n} \Delta^n \\
\downarrow & &\downarrow\\
\|X_\bullet\|^{(n-1)} & \hookrightarrow & \|X_\bullet\|^{(n)}
\end{array}
$$
for the canonically defined maps
$$
\bigsqcup_{\alpha\in X_n} \phi_{\alpha}:\bigsqcup_{X_n} \Delta^n\longrightarrow\|X_\bullet\|^{(n)}
$$
That arise in the coequalizers above, i.e.:
$$
\bigsqcup_{\alpha\in X_n} \phi_{\alpha}:\bigsqcup_{X_n} \Delta^n
\hookrightarrow
\bigsqcup_{r\leq n} \Delta^r\times X_r
\xrightarrow{~\mathrm{coeq}~}
\|X_\bullet\|^{(n)}
$$
Also $\|X_\bullet\|$ has the weak topology relative to this filtration.

Setup

I guess to make the claim about the differentials work it is enough to slightly modify the cellular chain complex of $\|X_\bullet\|$. I.e. rather than use maps $D^{n}\to \|X_\bullet\|^{(n)}$ which restrict to maps $\partial D^{n}\to \|X_\bullet\|^{(n-1)}$, where $D^n$ is the euclidean radius 1 ball in $\Bbb{R}^n$, we use the maps $\Delta^{n}\to\|X_\bullet\|^{(n)}$ implicit in the coequalizer. In particular this gives us the following picture

The cyan $\color{cyan}{\sim}$ are isomorphisms in homology by general properties of homology. We endow every relative homology group $H_p(\Delta^p,\partial \Delta^p)$ with the basis element $[\mathrm{id}_{\Delta^p}]$. This provides us with an isomorphism of abelian groups
$$
\begin{array}{ccccc}
\displaystyle
\big(CX_\bullet\big)_n
& \simeq &
\displaystyle
\bigoplus_{X_{n}}H_n(\Delta^n,\partial\Delta_n)
&
\displaystyle\overset{H(\bigsqcup_{\alpha\in X_n}\phi_\alpha)}\simeq &
\displaystyle
H_n(\|X_\bullet\|^{(n)}, \|X_\bullet\|^{(n-1)})
\\
\end{array}
$$
Under the connecting homomorphism this basis vector becomes the basis vector $\partial[\mathrm{id}_{\Delta^p}]=\sum_{i=0}^p(-1)^i[\mathrm{id}_{\Delta^p}\circ d^i]=\sum_{i=0}^p(-1)^i[d^i]$ using the structure maps of the co-semi-simplicial set $\Delta^\bullet$.

Now to conclude it will be enough to note that for every $(n+1)$-simplex $\alpha\in X_{n+1}$ and its associated attaching map $\phi_\alpha:\Delta^{n+1}\to\|X_\bullet\|^{(n+1)}$ the two paths below are the same:

Using the commutativity of the first diagram one can alternatively show that the orange and purple paths below define the same map:

Computation

Let $\alpha_0\in X_{n+1}$ be an $n+1$-simplex. The orange path does the following:

Compute the Moore differential of the basis element $\alpha_0\in (CX_\bullet)_{n+1}$, i.e.
$$\alpha_0\mapsto\sum_{i=0}^{n+1}(-1)^id_i(\alpha_0).$$

Send it to the following sum
$$\sum_{i=0}^{n+1}(-1)^id_i(\alpha_0)\mapsto\sum_{i=0}^{n+1}(-1)^i [id_{\Delta^n}]_{d_i(\alpha_0)}$$ where $[id_{\Delta^n}]_{d_i(\alpha_0)}$ designates the previously highlighted basis vector of the $d_i(\alpha_0)$-th copy of $H_n(\Delta^n,\partial\Delta_n)$ in $\bigoplus_{X_n}H_n(\Delta^n,\partial\Delta^n)$.

Compute its image under the map induced in homology $\displaystyle \Big(\bigsqcup_{\beta\in X_n}\phi_\beta\Big)_*$ of the attaching maps in dimension $n$:
$$
\begin{array}{rcl}
\displaystyle
\sum_{i=0}^{n+1}(-1)^i [id_{\Delta^n}]_{d_i(\alpha_0)}
& \mapsto &
\displaystyle
\sum_{i=0}^{n+1}(-1)^i \Big(\bigsqcup_{\beta\in X_n}\phi_\beta\Big)_*[id_{\Delta^n}]_{d_i(\alpha_0)}
\\
& &
\displaystyle
\quad
=
\sum_{i=0}^n(-1)^i
\Big[\phi_{d_i\alpha_0}:\Delta^n
\to
X_n
\Big]
\\
& &
\displaystyle
\quad
=
\sum_{i=0}^n(-1)^i
\Big[\phi_{d_i\alpha_0}
\Big]
\end{array}
$$

On the other hand, the purple path does the following:

send $\alpha_0\in X_{n+1}$ to the element $[\mathrm{id}_{\Delta^{n+1}}]\in H_{n+1}(\Delta^{n+1},\partial\Delta^{n+1})$ embedded at the $\alpha_0$-th summand in $\bigoplus_{X_{n+1}}H_{n+1}(\Delta^{n+1},\partial\Delta^{n+1})$:
$$
\alpha_0\mapsto
[\mathrm{id}_{\Delta^{n+1}}]\in H_{n+1}(\Delta^{n+1},\partial\Delta^{n+1})
\overset{\iota_{\alpha_0}}\hookrightarrow
\bigoplus_{X_{n+1}}H_{n+1}(\Delta^{n+1},\partial\Delta^{n+1})
$$
let's call it $[\mathrm{id}_{\Delta^{n+1}}]_{\alpha_0}$

Apply the connecting homomorphism to $[\mathrm{id}_{\Delta^{n+1}}]_{\alpha_0}$, i.e.
$$
\begin{array}{rcl}
[\mathrm{id}_{\Delta^{n+1}}]_{\alpha_0}
& \mapsto &
\sum_{i=0}^n(-1)^i
[d^i:\Delta^{n}\hookrightarrow \Delta^{n+1}]_{\alpha_0}
\\
& &\quad =
\sum_{i=0}^n(-1)^i
[d^i]_{\alpha_0}
\end{array}
$$

Apply the map induced in homology $\Big(\bigsqcup_{\alpha\in X_{n+1}}\phi_\alpha\Big)_*$:
$$
\begin{array}{rcl}
\displaystyle\sum_{i=0}^n(-1)^i
[d^i]_{\alpha_0}
& \mapsto &
\displaystyle
\sum_{i=0}^n(-1)^i
\Big(\bigsqcup_{\alpha\in X_{n+1}}\phi_\alpha\Big)_*
[d^i]_{\alpha_0}
\\
& &
\displaystyle
\quad
=
\sum_{i=0}^n(-1)^i
\Big[\Delta^n
\xrightarrow{d^i}
\Delta^{n+1}
\xrightarrow{\phi_{\alpha_0}}
X_{n+1}
\Big]
\\
& &
\displaystyle
\quad
=
\sum_{i=0}^n(-1)^i
\Big[\phi_{\alpha_0}\circ d^i
\Big]
\end{array}
$$

Conclusion

Thus, it would be sufficient if all the the attaching maps $\phi_\alpha$, $\alpha\in X_{n+1}$ satisfy, for all $\alpha\in X_{n+1}$ and all $i\in[n+1]$,
$$\phi_\alpha\circ d^i=\phi_{d_i\alpha}$$
Which I believe will be the case by virtue of the Fact above and the way the maps $\phi_\alpha$, $\alpha\in X_n$ are defined.


问:
How can I deduce $\cos\pi z=\prod_{n=0}^{\infty}(1-4z^2/(2n+1)^2)$?
Using the infinite product of $\sin(\pi z)$, one can find the Hadamard product for $e^z-1$:

$$e^z-1  =2ie^{z/2}\sin(-iz/2)= 2i e^{z/2} (-iz/2) \prod_n \left(1+\frac{z^2}{4\pi n^2}\right)\\= e^{z/2} z \prod_n \left(1+\frac{z^2}{4\pi n^2}\right).$$

I don't see a way to find the product for $\cos\pi z$. A naive attempt is letting $\{a_n\}\subset{\Bbb C}$ be all the zeros of $\cos(\pi z)$ and showing the possible convergence of
$$
\prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right)
$$

Is there an alternative way to find the Hadamard product in the title for $\cos\pi z$?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\prod_{n = 0}^{m}
\bracks{1 - 4z^{2}/\pars{2n + 1}^{2}}
\right\vert_{\,m\ \in\ \mathbb{N}_{\,\geq 1}}} =
\prod_{n = 0}^{m}
{\pars{n + 1/2}^{2} - z^{2} \over \pars{n + 1/2}^{2}}
\\[5mm] = &\
\on{f}_{m}\pars{z}\on{f}_{m}\pars{-z}
\end{align}
where
\begin{align}
\on{f}_{m}\pars{z} & \equiv \prod_{n = 0}^{m}
{n + 1/2 - z \over n + 1/2} =
{\pars{1/2 - z}^{\overline{m + 1}} \over
\pars{1/2}^{\overline{m + 1}}}
\\[5mm] & =
{\pars{1/2 - z + m}!\,/\,\Gamma\pars{1/2 - z} \over
\pars{1/2 + m}!\,/\,\Gamma\pars{1/2}}
\\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\,
{\root{\pi} \over \Gamma\pars{1/2 - z}}\ \times
\\[2mm] &\
{\root{2\pi}\pars{1/2 - z + m}^{1 - z + m}\,\,\,
\expo{-1/2 + z - m} \over
\root{2\pi}\pars{1/2 + m}^{1 + m}\,\,
\expo{-1/2 - m}}
\\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\,
{\root{\pi} \over \Gamma\pars{1/2 - z}}\ \times
\\[2mm] &\
{m^{1 - z + m}\,\,\,\,\bracks{1 + \pars{1/2 - z}/m}^{\,m}
 \over
m^{m + 1}\,\,\,\bracks{1 + \pars{1/2}/m}^{\,m}}\expo{z}
\\[5mm] & \stackrel{{\rm as}\ m\ \to\ \infty}{\sim}\,\,\,
{\root{\pi} \over \Gamma\pars{1/2 - z}}\,m^{-z}
\end{align}
Then,
\begin{align}
&\bbox[5px,#ffd]{\left.\prod_{n = 0}^{m}
\bracks{1 - 4z^{2}/\pars{2n + 1}^{2}}
\right\vert_{\,m\ \in\ \mathbb{N}_{\,\geq 1}}}
\\[5mm] = &\
\lim_{m \to \infty}\braces{\bracks{{\root{\pi} \over \Gamma\pars{1/2 - z}}\,m^{-z}}\bracks{{\root{\pi} \over \Gamma\pars{1/2 + z}}\,m^{z}}}
\\[5mm] = &\
{\pi \over \pi/\sin\pars{\pi\bracks{1/2 + z}}} =
\bbx{\cos\pars{\pi z}} \\ &
\end{align}


问:
Find $\mathbb{P}[\exists n>0 : X_n=j \mid X_0=i]$ in Markov Process
Let $\left\{B_n\right\}_{n=0}^{\infty}$ be an i.i.d. process, such that, for every $n\in\mathbb{N}\cup{0}$:

$$\mathbb{P}[B_n=-1]=p \\ \mathbb{P}[B_n=+1]=q$$

Such that $p+q=1$. Let $\left\{X_n\right\}_{n=0}^\infty\subset\mathbb{S}$ be a markov chain, where $\mathbb{S}=\left\{i\right\}_{i=0}^K=\left\{0,1,...,K\right\}$ for some $K\in\mathbb{N}$, such that:

$$X_n=X_{n-1}+B_{n}\qquad0<X_{n-1}<K \\ X_n=X_{n-1}\qquad\qquad \ \ X_{n-1}\in\left\{0,K\right\}$$

I drew the diagram of the process in order to see things more clearly:

Define the probabilty $\rho_{ij}$ for every $i,j\in\mathbb{S}\setminus\left\{0,K\right\}$ such that:

$$\rho_{ij}\triangleq\mathbb{P}[\exists n>0 : X_n=j \mid X_0=i]$$

(In other words, $\rho_{ij}$ is the probability that the chain would somewhen visit the state {$j$}, given that the initial state is {$i$}. It is clear that if the initial state is $X_0=0$, then it is certain that the chain would visit the state {$0$} again, but never any other state. In other words, $\rho_{0k}=\delta[k]$ where $\delta$ is the delta of kronecker.)

It is given that $p=q=1/2$. Find $\rho_{ij}$ for every $i,j\in\mathbb{S}\setminus\left\{0,K\right\}$.

I feel like I've tried everything. It seems that doing this straightforward is impossible, since the total number of paths that must be considered is too large. I understand that the expression would be different for the three cases $i=j,i>j$ and $i<j$. Finding $\rho_{ii}$ shouldn't be a problem if I knew $\rho_{ij}$ for $i>j$ and $i<j$, since:

$$\rho_{ii}=p\rho_{i-1,i}+q\rho_{i+1,i}=[\rho_{i-1,i}+\rho_{i+1,i}]/2$$

And obviously $i>i-1$ & $i<i+1$.

Thank you very much!

答:
A suggestion:
Merge states $1$ and $2$, and also $K-1$ and $K$. Find the transition matrix $P_{(K-2)\times (K-2)}$ for the new $(K-2)$-state Markov chain, which the elements of $P^n$ gives you the transition probabilities for states $3$ to $K-2$. A conditional probability would yield the desired result for the inner states of the two merged states.


问:
How come $\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} \binom{n+1}{2k} \frac{k!}{2}$ is so close to $\frac{n^n}{n!}$ for $x \leq 6$?
I just recently was puzzled because of this puzzle I posted on Puzzling SE:

Number sequence (yet again?)

I'll skip all the details, but @hexomino's answer was very close to mine as in the following:

$$\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} \binom{n+1}{2k} \frac{k!}{2}\approx \frac{n^n}{n!}$$

I wonder why? They're totally unrelated, and one of them has a ugly $n^n$ term in it, but how come they're very similar for $n \leq 6$?

The left side has terms to the nearest tenth:

1, 2, 4.5, 10.5, 26, 67

Right side's terms to the nearest tenth:

1, 2, 4.5, 10.7, 26, 64.8

How come they're so similar? And does the similarity continue for $n > 6$ (I haven't checked yet)?

答: It's just a coincidence for small $n$. By the time  $n=19$, the left-hand side is larger by a factor of $17$.


问:
Specific ten digit number
I'm trying to solve this little problem. So far no luck. Could anyone help? Thanks in advance :)

What is the ten digit number such that the i-th digit is the number of i's in the number ( 0<= i <= 9) (the 0th position is number of zeros in the number, the 1st position is number of ones in the number,…)?

Title: Specific ten digit number

Body:
Here is one solution:
$$6210001000$$
$$(0123456789)$$


问:
How to solve these?

Inverse Trigonometric Functions

They are incomplete and I don't know how to complete them.
Who can help me?

1st

$$
\int\frac 1{ x \sqrt{x^{6} - 4}}dx
$$
I tried with:
$$u = x^3 $$
$$du= 3x^2dx$$
but this is not completed,

2nd

oops, is not $$x^2$$ is only "x"
$$
\int \frac 1{ x \sqrt{x-1}}dx
$$
$$u =  \sqrt x $$
$$du= \frac 1{\sqrt x} dx$$
where is the sqrt(x) on du?

3

$$
\int \frac {e^x}{ \sqrt{4-e^x}}dx
$$
I think in this
$$u=e^x $$
$$du = e^xdx$$
I have a bad English syntax, but I know read English++ Ok..

答:
For $\int\frac 1{ x \sqrt{x^{6} - 4}}dx$,
you didn't go far enough.

Write it as
$\int\frac{x^2}{ x^3 \sqrt{x^{6} - 4}}dx
$.
Make your substitution of $y = x^3$,
so $dy = 3 x^2 dx$,
and we get
$\int \frac{dy}{3 y \sqrt{y^2-4}}$
and this is essentially the same as
$\int \frac 1{ x \sqrt{x^2-1}}dx$
with $4$ instead of $1$
(but write $4$ as $2^2$).

Title: How to solve these?


问:
Conjecture regarding the sum of prime factors of $x!$ and $(x-1)!$?
I think using dodgy means I can show the following:

Let $\lambda(x)$ be the sum of  primes in $x$. For example:

$$ \lambda(2) = 2$$
$$ \lambda(4) = 2 + 2 = 4$$
$$ \lambda(6) = 3 + 2 = 5$$

Then for large $n$:

$$ \frac{\lambda (n!)}{\lambda ((n-1)!)} \sim \frac{\ln (n!)}{\ln (n-1)!}$$

Or more precisely substracting $1$:

$$ \frac{\lambda (n)}{\lambda ((n-1)!)} \sim \frac{\ln (n)}{\ln (n-1)!}$$

Or even better:

$$ \frac{\lambda (n)}{\lambda ((n-1)!)} - \frac{\ln (n)}{\ln (n-1)!} = o(1)$$

Can someone prove or disprove this conjecture?

For example:

$$ \frac{\lambda(8!)}{\lambda(7!)} = \frac{31}{26} \approx 1.23$$

And

$$ \frac{\ln(8!)}{\ln(7!)} \approx 1.244 $$

Substracting $1$:

$$ .23  \approx .24$$

答:
The question does not make sense, why ?

First $ \frac{\lambda((n+1)!)}{\lambda(n!)} \sim \frac{\ln((n+1)!}{\ln n!} \implies \frac{\lambda((n+1)!)}{\lambda(n!)}\sim 1 $ but this is not that impressive, it only means that for every $\epsilon>0$ we have that $ (1-\epsilon)^n < \lambda(n!) < (1+\epsilon)^n$ for large enough $n$ (so basically $ \lambda(n!)$ grows sub-exponential), for instance i could easily said that :$$ \frac{\lambda((n+1)!)}{\lambda(n!)} \sim \frac{(n+1)^3}{n^3}$$

Second $ \frac{\lambda(n+1)}{\lambda(n!)} \sim \frac{\ln(n+1)}{\ln n!}$ is simply wrong take $n$ such that $ n+1$ is prime number and so $ \lambda(n+1) = n+1$  and $ \lambda (n!) \sim \frac{\pi^2 n^2}{12 \ln n}$ and $\ln n! \sim n \ln n$ and so what you are saying is that $ \frac{12 \ln n}{\pi^2 n} \sim \frac{1}{n}$ which obviously is not correct.

Third This part tells you nothing simply because $ \frac{\lambda(n+1)}{\lambda(n!)} = o(1)$ so i could easily said that : $$ \frac{\lambda(n+1)}{\lambda(n!)} - \frac{1}{\ln \ln n} = o(1)$$

Also the function $\lambda(n)$ is very chaotic as most of the arithmetic function in number theory because they are related to the prime numbers and those guys are the definition of chaotic in small intervals( for instance when $n+1$ is highly composite number $\frac{\lambda(n+1)}{\lambda(n!)}$ will approach $0$ waaay faster than if $n+1$ was prime number).

In these cases we find the Average\Sum of such function to have beautiful results such as $ \lambda (n!) \sim \frac{\pi^2 n^2}{12 \ln n}$.

Hope its helpfull.


问:
Find a basis of the generated subspace
Consider the vectors
$$
v_1 =\begin{pmatrix}
1\\ 0\\ 0\\ 1
\end{pmatrix}, \quad v_2=\begin{pmatrix}
\sqrt{2}\\ 1\\ 2\\0
\end{pmatrix}, \quad v_3 =\begin{pmatrix}
\sqrt{2}+2\\ 1\\ 2\\ 2
\end{pmatrix}
$$
and let $S$ be the subspace generated by $v_1, v_2, v_3$. Find the dimension of $S$ and a basis for $S$.

${\bf My \ solution:}$ the dimension of $S$ is clearly $2$ since since $v_1, v_2, v_3$ are linearly dependent (or, better, the rank of the matrix whose columns are the three vectors is $2$). In order to find a basis I use the echelon form. It is
$$
\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&2 &2\\
1&0&2
\end{pmatrix}\xrightarrow{}\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&0&0\\
0&0&0
\end{pmatrix}
$$
which means that a basis for $S$ is given by the vectors $v_1$ and $v_2$.

Is my solution correct or am I missing something?

Thank you in advance!

${\bf EDIT:}$ My steps for row echelon form:
[
\begin{split}
&\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&2 &2\\
1&0&2
\end{pmatrix}\xrightarrow{R_1-R_3}\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&2&2\\
0&\sqrt{2} &\sqrt{2}
\end{pmatrix}\xrightarrow[R_3\leftrightarrow R_4]{R_3-2R_2}\\[5pt] &\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&\sqrt{2}&\sqrt{2}\\
0&0&0
\end{pmatrix}\xrightarrow{R_3-\sqrt{2}R_2}\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&0&0\\
0&0&0
\end{pmatrix}
\end{split}
]

答:
It's correct, however an overkill.

You observed correctly that $v_3=2v_1+v_2$, so they are linearly dependent, but clearly $v_1,v_2$ are not parallel, hence independent, so the dimension they span is indeed $2$.

But then any $2$ independent elements of the span form a basis, so just take $v_1$ and $v_2$.

Title: Find a basis of the generated subspace


问:
Liouville's formula
I have some questions concerning a proof of Liouville's formula:

$$W'(t)=\text{tr}(A) W(t)$$ where $W$ is the Wronskian of the homogenous ODE.

If the vectors in the columns of the fundamental matrix $X(t)$ are given by $ x_1,...,x_n$. Then it is said that $$W'(t)=\sum_{i=1}^{n}det(x_1,...,A(t)x_i,...,x_n)$$ where $W'$ is the time-derivative of the Wronskian of the given fundamental matrix. Until now, I am totally okay with that. Then they say: Let $X$ be the identity matrix as the fundamental matrix. And there comes my first question? Isn't the time derivative of a matrix with constant numbers always zero?

But my real problem comes when they try to trace the other cases back to this example by saying that at each time and each fundamental matrix $T(t)$, there is a matrix $C$, such that $$det(T)'(t)=(det(X) det(C))'(t)=\text{tr}(A)det(X)(t) det(C)=\text{tr}(A)det(T)
$$ How so? I mean would this not imply that $X$ (being the identity) is always a fundamental matrix of the differential equation with fundamental matrix $T(t)$? Cause I don't know how can possibly argue that the second equality sign is right, where comes this tr$(A)$ into play?

答:
You are confused with the terminology.

Consider the initial value problem

$$ \tag 1
u'=A u, \quad u(0)=  u_0.
$$

The matrix analogue is

$$ \tag 2
\Psi'=A u, \quad \Psi(0)=  \Psi_0,
$$
where $\Psi: \mathbb{R} \rightarrow \mathbb{C}^{n \times n}$. There is a unique solution $\Psi$, given $\Psi_0$, the columns of being solutions of $ (1) $.

Liouville's formula is
$$
det \Psi(x)= det \Psi(0) \exp \left( \int_0^x Tr A \right).
$$

Here det $\Psi$ is called the Wronskian of $\Psi$.

If, $\Psi_0=I$, then $\Psi $ is called the canonical fundamental matrix and we write $\Psi=\Phi$. Then, we have
$$ \Phi'=A \Phi, \quad \Phi(0)= I. $$

Any other solution matrix of $ (2) $ can be written as
$$
\Psi(x)=\Phi(x)\Psi_0.
$$

$Tr A$ comes into play in the Proof of Liouville's formula.

Title: Liouville's formula