答:
I think your real difficulty is that we don't say "$f$ is integrable," but rather "$f$ is integrable over $D$" where $D$ is some domain.

Your definition works fine if the domain is an interval that doesn't contain $0$.

Title: Is $\frac{1}{x}$ integrable?

Body: I am taking a course on theory of integrals and measure, where I have learned a very fundamental definition of integral. One of the things I've learned is that $f$ is integrable iff $\int|f|<\infty$. Then, according to this definition, $f(x)=\frac{1}{x}$ wouldn't be integrable, since the integral of $|f(x)|=|\frac{1}{x}|$ diverges. However, I can calculate $\int_a^b\frac{1}{x}dx$ for any $a, b \in (-\infty, +\infty)$ and this isn't problematic at all. How do I deal with this?


问:
Probability density function of $W = X + Y$ where $X \sim \mathrm{Unif}[0,1]$ and $Y \sim \mathrm{Exp}(\lambda)$ are independent random variables?
This is the question that I was given:

And this was the provided solution:

I can't seem to make sense of it - firstly, what exactly is the question asking you to do? How did they know to divide the range of $w$ in that way? And why did they use those particular limits for the integrals?

答:
The point is that both random values $X$ and $Y$ yield only positive values. Thus, $X+Y = w$ requires that $X$ yields a value $\geq w$, because otherwise the sum will surely be larger than $w$.

Note though that the convolution formula deals with that correctly - you just have to be carefull when computing the integral. Let $$
  f_W(w) = \int_0^1 f_X(x)f_Y(w-x) \,dx  \text{.}
$$
unconditionally. Then watch what happens if you evaluate $f_W(\omega)$ with $\omega < 1$. Once $x$ reaches $\omega$, $\omega-x$ becomes zero, and negative afterwards. You thus evaluate $f_Y$ for negative arguments, where its simply zero. But if you aren't carefull, you might have simply replaced $f_Y(w-x)$ with $\lambda e^{-\lambda(w-x)}$, and that function isn't zero for $x > w$!

So all that happens here is that your densities are defined piecewise, i.e. for different ranges of arguments they're represented by different algebraic expressions. When computing the integral, you have to take that into account.

ContentLicense: CC BY-SA 3.0


问:
(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $

$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor  + \ldots + \lfloor \log_2 n \rfloor  = 1994 $$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

The first few terms of this series shows me that summation $\lfloor \log_2 n \rfloor$ for $n=1$ to $n=10$ give $2^{n +1}$.

答:
As noted, the sequence goes like

$$ 0,\underset{2}{\underbrace{1,1}},\underset{4}{\underbrace{2,2,2,2}},\underset{8}{\underbrace{3,3,3,3,3,3,3,3}},4,4,\ldots$$

i.e, every natural number $k$ occurs $2^k$ times.

So desired is $$ \sum k\cdot 2^k =1994$$

It's quick enough to attack directly:

$$ 1\cdot2 + 2\cdot4 + 3\cdot8 + 4\cdot16 + 5\cdot 32 + 6\cdot 64 + 7\cdot 128 = 1538$$

Next up is $8$ repeating $x$ times till $1994$ $$1538 + 8\cdot x = 1994$$

$$\Rightarrow x=57 $$

Last term of our sequence can be found by counting the number of repeating units: $$n = (1+2+4+\ldots+128) + 57 = \boxed{312}$$

ContentLicense: CC BY-SA 4.0


问:
Solve $\sin x = 1 - x$
How would you be solve sin x = 1 - x, without drawing the graph and manually measuring the point of intersection?</p>

答:
As @Gerben answered, building the Taylor series around $x=\frac \pi 6$ gives
$$y=x+\sin(x)-1=\frac{\pi -3}{6} +\frac{2+\sqrt{3}}{2} x+\sum_{n=2}^\infty \frac{\sqrt{3} \sin \left(\frac{\pi  n}{2}\right)+\cos \left(\frac{\pi n}{2}\right)}{2 n!} \left(x-\frac{\pi }{6}\right)^n$$
Truncated to $O\left(\left(x-\frac{\pi }{6}\right)^6\right)$ for example and using series reversion, then
$$x=\frac{\pi }{6}+t+\left(1-\frac{\sqrt{3}}{2}\right)
   t^2+\left(3-\frac{5}{\sqrt{3}}\right) t^3+\left(\frac{67}{6}-\frac{77}{4
   \sqrt{3}}\right) t^4+\left(\frac{232}{5}-\frac{803}{10 \sqrt{3}}\right)
   t^5+O\left(t^{6}\right)$$ where $t=\frac{y+\frac{3-\pi }{6}}{1+\frac{\sqrt{3}}{2}}$.

Making $y=0$ that is to say $t=-\frac{\left(2-\sqrt{3}\right) (\pi -3)}{3} $, this gives
$x=0.51097342938847$ while the exact solution is
$x=0.51097342938857$.

Title: Solve $\sin x = 1 - x$


问:
The Funtion Complex - Product Adjunction of simplicial sets and diagram chasing
I run into the following claim in the book Simplicial Homotopy Theory (in the proof of Proposition 5.2.). Given $i:K \rightarrow L$ inclusion of simplicial sets and $p:X\rightarrow Y$ fibration.

Someone explained to me that I can use the exponential law to get the identification, but I don't get his idea, for it involves careful interchanging lim/colim etc. Is there any written down proof for this fact? Moreover, I wish to know that if there is general way to understand the interplay between adjunction and commutative diagram so that I might guess from the first diagram what kind of equivalent diagram should I look for.

答:
Yes, there is a written proof of this very long and boring exercise: https://arxiv.org/abs/1902.06074

The general statement is dubbed "Theorem 2.6", and although the obvious application is to model structures, it has absolutely nothing to do with algebraic topology, just old plain category theory.


答: Denote the sum by $S_{n}$. Note that for any $k\in\mathbb{N}$ there are $2^k$ positive integers $x$ for which $[\log_{2}(x)]=k$, and those are $x=2^{k},2^{k}+1,\ldots,2^{k+1}-1$. Thus we have $$S_{2^{k}-1}=0 + (1+1) + (2+2+2+2+) + \cdots + \bigl((k-1)+(k-1)+\cdots + (k-1)\bigr)$$ where there number of $(k-1)$ terms is $2^{k-1}$.  It follows that $$S_{2^{k}-1} = (k-2)2^{k}+2$$ Putting $k=8$ we see that $S_{255}=1538<1994$ and putting $k=9$ we see that $S_{511}=3586>1994$. Thus it's clear that our $n$ should satisfy $2^{8}-1<n<2^{9}-1$. Now we have $$1994=S_{n}=S_{255}+(n-255)8=8n-502$$ which gives $n=312$.

Title: (AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $

Body:
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor  + \ldots + \lfloor \log_2 n \rfloor  = 1994 $$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

The first few terms of this series shows me that summation $\lfloor \log_2 n \rfloor$ for $n=1$ to $n=10$ give $2^{n +1}$.


Body:
We'll find a maximal $m$ for which $$\sum_{k=1}^mk2^k\leq1994.$$
Indeed, $$\sum_{k=1}^mk2^k=2\sum_{k=1}^mk2^{k-1}=2\left(\sum_{k=1}^mx^{k}\right)'_{x=2}=2\left(\frac{x(x^m-1)}{x-1}\right)'_{x=2}=$$
$$=2\cdot\frac{(m+1)2^m-1-2^{m+1}+2)}{(1-1)^2}=(m+1)2^{m+1}-2^{m+2}+2.$$
Id est, $$(m+1)2^{m+1}-2^{m+2}+2\leq1994,$$ which gives $m=7$.

Now, $$\frac{1994-((7+1)2^{7+1}-2^{7+2}+2)}{8}=57$$ and we obtain:
$$n=1+2^1+...+2^7+57=2^8-1+57=312.$$


问:
A question on convergence of a sequence of functions
Let $f_n : (a,b) \to \mathbb{R}$ be functions that have finite number of maxima and minima, for $n = 1,2,3...$. Let D be a countable dense subset of $(a,b)$. If sequence $\{f_n\}$ converges to $f$ such that the convergence is non-uniform at the points $x \in D$ and uniform at the points $x \in (a,b)\setminus D$, then does it imply that $f(x) = 0 \forall x \in (a,b)\setminus D$ ?

EDIT : The question didn't come out as what i expected. So I am posting a new one. sorry for the inconvinience.

答: Certainly not.  Perhaps you might want to further quantify what you mean by "non-uniform".  Note that the sequence of functions $f_i(x) = c$ where $c$ is a fixed non-zero number converges pointwise (as well as uniformly, of course), and is a counterexample.</p>


问:
Find the formula of the sequence using generating functions
I have the following sequence given:

$$\sum_{k=1}^{n} (-1)^{k}k^{2}$$

How can I do this? The sequence goes like this:

$$-1 + 4 - 9 + 16 - 25 + 36 - ...$$

So it doesn't have any variables inside. It looks like a geometric sequence for me, so the simple known formula would do the whole thing, but I actually doubt it's that simple.

So there are some generating functions that are quite similar, like:

$$\sum (-1)^{n}x^{n} = x - x^{2} + x^{3} - x^{4} + ...$$

But, uhm, well...what's next? The lack of x variable seems a bit strange to me.

Body:
The generating function is
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^kk^2x^n
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^kk^2x^n\\
&=\sum_{k=0}^\infty(-1)^kk^2\frac{x^k}{1-x}\\
&=\frac1{1-x}\sum_{k=0}^\infty(-1)^kk^2x^k\tag{1}
\end{align}
$$
Starting with $\frac1{1+x}$ and taking derivatives, we get
$$
\begin{align}
\frac1{1+x}&=\sum_{k=0}^\infty(-1)^kx^k\\
\frac1{(1+x)^2}&=\sum_{k=0}^\infty(-1)^k(k+1)x^k\\
\frac2{(1+x)^3}&=\sum_{k=0}^\infty(-1)^k(k+2)(k+1)x^k
\end{align}\tag{2}
$$
Since $k^2=(k+2)(k+1)-3(k+1)+1$, we have
$$
\begin{align}
\sum_{k=0}^\infty(-1)^kk^2x^k
&=\frac2{(1+x)^3}-\frac3{(1+x)^2}+\frac1{1+x}\\
&=\frac{(x-1)x}{(1+x)^3}\tag{3}
\end{align}
$$
Thus, the generating function we want is
$$
\begin{align}
\frac1{1-x}\sum_{k=0}^\infty(-1)^kk^2x^k
&=\frac1{1-x}\frac{(x-1)x}{(1+x)^3}\\
&=-\frac{x}{(1+x)^3}\\
&=-\sum_{n=0}^\infty\binom{-3}{n-1}x^n\\
&=\sum_{n=0}^\infty(-1)^n\binom{n+1}{n-1}x^n\\
&=\sum_{n=0}^\infty(-1)^n\binom{n+1}{2}x^n\tag{4}
\end{align}
$$
Therefore,
$$
\sum_{k=0}^n(-1)^kk^2=(-1)^n\binom{n+1}{2}\tag{5}
$$


问:
Find the function that minimises the following integral
How can I find the function $f(t)$, $t\geq0$ continuous and everywhere differentiable that minimises the following integral

$$L = \int_0^\infty f(t)^2 + f'(t)^2dt$$

such that $f(0)=x$.

Intuitively I am asking what is the function $f$ that starts at $x$ and wants to approach $0$ the fastest but keeping the derivative smaller. My intuition tells me that somehow the function $f$ should be a decreasing exponential. But how to prove it? how does one tackle this problem?

Here is what I tried to do:
I tried to think of a similar discrete problem: find the sequence $h_n$ that minimises

$$L = \sum_{i=1}^\infty h_i^2 + (h_{i-1}-h_i)^2$$ such that $h_0=x$

Some observations:
If $x=0$ then trivially $h_i=0$ $\forall i$

On the other hand, if $h_i=0$ $\forall i$, then $L = x^2$. (here the function would suddenly drop to $0$)

If instead we had gradually decreased it with one step: $h_1=\frac{x}{2}$ and $0$ otherwise, then $L=\frac{x^2}{4}+\frac{x^2}{4}+\frac{x^2}{4}=\frac{3x^2}{4}$ which is lower than the previous example, showing that it is better to decrease it gradually...

Then my intuition tells me that if $h_i = \alpha h_{i - 1}$ for $\alpha<1$, then $h_i=x\alpha^i$ and $h_{i-1}-h_i = x\alpha^{i-1}(1-\alpha)$.

For this case, this means that $L = \sum_{i=1}^\infty (x\alpha^i)^2 + (x\alpha^{i-1}(1-\alpha))^2=x^2\sum_{i=1}^\infty (\alpha^2)^i+x^2(1-\alpha)^2\sum_{i=1}^\infty (\alpha^2)^{i-1}$

$$L=x^2\frac{\alpha^2}{1-\alpha^2}+x^2(1-\alpha)^2\frac{1}{1-\alpha^2}=\frac{x^2}{1-\alpha^2}(2\alpha^2-2\alpha + 1)$$

Now if we try to find $\alpha$ that minimises this expression we can take its derivative

$\frac{dL}{d\alpha}=x^2\frac{(4\alpha-2)(1-\alpha^2)-(2\alpha^2-2\alpha + 1)(-2\alpha)}{(1-\alpha^2)^2}=0 \Leftrightarrow (4\alpha-2)(1-\alpha^2)-(2\alpha^2-2\alpha + 1)(-2\alpha)=0 \Leftrightarrow 4\alpha-4\alpha^3-2+2\alpha^2+4\alpha^3-4\alpha^2+2\alpha=0\Leftrightarrow -2\alpha^2+6\alpha-2=0$

$\alpha=1+\frac{1-\sqrt{5}}{2}$ which seems to be related to the golden ratio

答:
This is the sort of problem that the calculus of variations was invented to solve.  According to that formalism, $f$ will extremize the functional (in the sense that small "perturbations" of $f$ will not change the value of $L$) if it satisfies the Euler-Lagrange equation
$$
\frac{d}{dt} \left( \frac{\partial Q}{\partial f'} \right) = \frac{\partial Q}{\partial f},
$$
where $Q$ is defined to be the integrand $(f')^2 + f^2$.
Doing these derivatives, we obtain
$$
f'' = f.
$$

In other words, $f = A e^t + B e^{-t}$ for some values of $A$ and $B$.

To carry this further, we note that if $A \neq 0$, $L \to \infty$, and this cannot be a minimum of $L$.  So we must have $f(t) = B e^{-t}$.  Requiring that $f(0) = x$ then sets $B=x$, with the result that
$$
f(t) = x e^{-t}.
$$

This function will be the minimum if a minimum exists;  and the value of $L$ in this case can be shown to be $L = x^2$.


问:
Is $\operatorname{SL}_2(\mathcal O_K)$ dense in $\operatorname{SL}_2(\mathbb R)$?
Let $K$ be a real quadratic number field of discriminant $D>0$ with $\mathcal O_K$ being its ring of integers.

There are two embeddings $K \hookrightarrow \mathbb R$ and using them $\mathcal O_K \hookrightarrow \mathbb R^2$ can be reagarded as discrete subgroup of $\mathbb R^2$. But when you regard only one embedding $\mathcal O_K \hookrightarrow \mathbb R$ is a dense subgroup of $\mathbb R$.

Now let us consider $\operatorname{SL}_2(\mathcal O_K)$. Embedded into $\operatorname{SL}_2(\mathbb R)^2$ this is still discrete. But if you regard only one embedding $\operatorname{SL}_2(\mathcal O_K) \hookrightarrow \operatorname{SL}_2(\mathbb R)$ this is not discrete anymore. For example you have

$$\left\{\left(\begin{matrix}
1 & \lambda\\
0 & 1\\
\end{matrix}\right) \in \operatorname{SL}_2(\mathcal O_K) : \lambda \in \mathcal O_K\right\}
$$
whose closure is
$$\left\{\left(\begin{matrix}
1 & x\\
0 & 1\\
\end{matrix}\right) \in \operatorname{SL}_2(\mathbb R) : x \in \mathbb R\right\}.
$$
On the other hand I'm not sure if for all $x \in \mathbb R$ the matrix $\operatorname{diag}(x,x^{-1})$ is part of the closure of $\operatorname{SL}_2(\mathcal O_K)$.

So my question is: What is the closure of $\operatorname{SL}_2(\mathcal O_K)$ in $\operatorname{SL}_2(\mathbb R)$? Is $\operatorname{SL}_2(\mathcal O_K)$ maybe dense in $\operatorname{SL}_2(\mathbb R)$?

答:
It is dense because $\pmatrix{d&0\\0&d^{-1}}$ is in the closure : take $a,b\in O_K$ such that $a- d,b- d^{-1}$ are very small, $u\in (0,1)$ is the fundamental unit such that $O_K^\times=\pm u^\Bbb{Z}$,  $|ab-1|\in [u^{2n+2},u^{2n}]$,
then $$\pmatrix{a&(ab-1)u^{-n}\\ u^n &b}\in SL_2(O_K)$$

Next $\pmatrix{0&1\\-1&0},\pmatrix{1&*\\0&1},\pmatrix{1&0\\*&1}$ are in the closure which must be $$SL_2(\Bbb{R})=\pmatrix{1&0\\*&1}\pmatrix{*&0\\0&*}\pmatrix{1&*\\0&1}\cup \pmatrix{0&1\\-1&0}\pmatrix{*&0\\0&*}\pmatrix{1&*\\0&1}$$


问:
When do projective modules give generators?
I am reading arXiv:math/0111066 [math.RA] and am stuck understanding the proof of Lemma 1.4.  The only part I don't follow is the claim:

If $B$ is a projective (right) module over a ring $R$ which is simple, then $B$ is a generator for $\operatorname{Mod}-R$.

I don't see how to leverage the hypothesis that $R$ is simple.  $B$ is the direct summand of a free module (and free modules generate) but why does that show that $B$ generates (and not the complement of $B$?)

Update: One possible reference is Lang's Algebra section XVII, 7.  Note that $B$ becomes a left $R'(B)=\operatorname{End}_R(B)$ module in a natural way.  Morita's Theorem says that $B$ is a generator if and only if the natural map $R\rightarrow \operatorname{End}_{R'}(B)$ is onto ($B$ is "balanced") and $B$ is finitely generated projective over $R'(B)$.  However, I have no idea how to verify this...

答:
It is easy to see that $R$ itsef is a generator.

I will provide you with an epimorphism $P \twoheadrightarrow R$, notice that this is the same of providing you with a non-zero map, because $R$ is simple. Such a "cover" of a generator shows that also $P$ is a generator.

Being projective, $P$ is a split submodule of $R^{(\Gamma)}$, the free module generated by some set $\Gamma$. Now, consider all the projections $\pi_i: R^{(\Gamma)} \to R$, composing with the inclusion $P \hookrightarrow R^{(\Gamma)} \to R$, there must be at least one $i$ such that the projection is non-zero, because $P$ is a nontrivial module.


问:
Limit with variable: non-defined expression
I have a given limit that depends on a variable $a$:

$$\lim_{x \rightarrow \infty} \left (\frac{e^{ax}}{1 - ax} \right)$$

I understand cases for $a < 0 \implies \lim = 0$ and $a > 0 \implies \lim = -\infty$. However, for the case $a = 0$, the expression $ax$ which is basically $0\cdot \infty$ in undefined. I somehow know, that the result will be $\lim (\frac{e^0}{1}) = 1$ but I am not sure how to justify that $0\cdot\infty$ is $0$ in this case.

Thanks for any ideas or an explanation!

答:
If we choose the value $a=0$ we evaluate the expression $\frac{e^{ax}}{1 - ax}$ before passing to the limit so your result would be
$$\lim_{x\to\infty}\left(\left[\frac{e^{ax}}{1 - ax}\right]_{a=0}\right)=\lim_{x\to\infty}\frac{e^0}{1-0}=1$$</p>


问:
Are there any simple way to show rigorously that the largest hyper-rectangle inscribed in a hyper-ellipsoid is the one we expect?
Let $S_n=\{x\in \mathbb R^n: \sum_{j=1}^n (x_j/a_j)^2\leq1\}$ be a filled ellipsoid, and $C_n=\{x\in \mathbb R^n: 0\leq x_j \leq 1,\forall j\}$. Let $$\mathcal T= \{U\in M_n(\mathbb R): \text{eigenvalues of $U$ are orthogonal, and }\\UC_n+ y\subseteq S_n \text{ for some } y\in\mathbb R^n \},$$
and
$$
T=\{|\det U|: U\in \mathcal T\},
$$
i.e. $T$ is the set of volumes of a (hyper-)rectangular parallelpiped inside the ellipsoid.

How can I show rigorously that the maximal volume is given by
$$
\sup T= \frac{2^n}{n^{n/2}} \prod_{1\leq j \leq n} a_j,
$$
and that this can only be achieved when $U$ has eigenvalues $\lambda_j= a_j n^{-1/2}$?

All the results above are obvious if we only consider parallelpipeds with sides parallel to some coordinate axis. But when I also take other parallelpiped into consideration, the problem seem to get pretty messy. Is there any neat way to formulate this and prove it?

答:
You can view a hyper-ellipsoid as a hyper-sphere stretched by $a_k$ along axis $x_k$. For a hyper-sphere the inscribed hyper-rectangle of maximum volume is of course a hyper-cube. Hence the maximum volume hyper-rectangle inscribed in a hyper-ellipsoid is nothing but a stretched hypercube, because stretching preserves the ratio of volumes.

A non axis-aligned hyper-rectangle cannot have maximum volume, because if you apply the inverse stretching (the one carrying the hyper-ellipsoid to a hyper-sphere) it doesn't become a hyper-cube.


问:
Constructive vs computable real numbers
I find it confusing that all of the following statements are true :

The computable real numbers are countable. $-\hspace{-3pt}-$ Alan Turing, "On Computable Numbers, with an Application to the Entscheidungsproblem"

In constructive analysis, the real numbers are uncountable. $-\hspace{-3pt}-$ Errett Bishop, Foundations of Constructive Analysis

"every mathematical statement [in constructive analysis] ultimately expresses the fact that if we perform certain computations within the set of positive integers, we shall get certain results" $-\hspace{-3pt}-$ Ibid.

Perhaps I am misunderstanding something.

I suppose I really have two questions. In constructive analysis :

Why isn't every real number computable?

How is it possible to construct an uncountable set?

答:
A real number $x$ is computable if there exists a program $P(n)$ that on input $n$ outputs the decimal expansion of $x$ up to the $n$th digit. For example, $\pi$ is one such  computable number - you can approximate $\pi$ as close as you want with a computer.

From this we can deduce $1$. Each computer program can be specified by a unique integer (it's binary representation), and hence there are a countable number of such programs, and therefore at most a countable number of computable real numbers.

However as $2$ says, the real numbers are uncountable. You can find a classical diagonalization proof of this fact here.

So if the real numbers are uncountable but computable numbers are countable then there must exist uncountably many real numbers which are not computable. Examples of uncomputable real numbers are rather exotic - they do not usually appear in natural problems.

However that does not mean they are sparse - indeed since every interval of the real line contains uncountable many numbers it must contain uncountable many uncomputable numbers. One example of an uncomputable number is Chaitin's constant.

The third statement you cite is the most sophisticated. Essentially, uncomputable numbers are not operational - you cannot use them in ordinary calculations. So for all practical purposes you could just work with the set of computable real numbers - and that is what constructive analysts do. Computable real numbers are algebraically closed - every reasonable operation you do with them will result in another computable numbers. That is the essence of the statement.

Regarding your question on how to build uncountable sets - if a set $S$ is countably infinite, then the set of all subsets of $S$ is uncountable. This is Cantor's theorem.

Title: Constructive vs computable real numbers


问:
On Jacobson's proof of the Smith normal form in a PID

theorem：If$ A$ ∈$ M_{m×n}(D)$, D is PID, then$ A$ is equivalent to a matrix
which has the "diagonal" form:
\begin{pmatrix}
d_1       & 0           & \cdots    & 0          &\cdots  & 0\\
 1          & d_2       &  \cdots  & 0          & \cdots  & 0\\
\vdots & \vdots   & \ddots  & \vdots & \cdots & \vdots \\
0           & 0          &\cdots  & d_r        & \cdots & 0 \\
\vdots & \vdots &\cdots & \vdots  & \ddots  & \vdots \\
0           & 0          &\cdots  &0            & \cdots
& 0 \\
\end{pmatrix}
It is belongs to $ M_{m×n}(D)$.

I can understand the proof about D is ED.this thought is similar with linear algebra. When D is PID,i can't understand The following step：

let $f(a)$ be a number of  prime factors of $a $.$A$=$(a_{ij})_{m×n}$,we assume $f(a_{11})$ is minimal number(otherwies take elementary transformation).if $a_{11}$ not  division $a_{1k}$, we assume k=2. We want to eliminate the elements in the first row and the first column of the matrix expect $a_{11}$  let $d$=gdc($a_{11}$, $a_{12}$) then we have $xa_{11}$+$ya_{12}$=$d$
Consider a matrix B：
\begin{pmatrix}
x & s & \cdots &0 \\
y & t  & \cdots &0 \\
\vdots & \vdots & \ddots & \vdots \\
0           & 0          &  \cdots & 1          \\
\end{pmatrix}
s = $a_{12}d^{-1}$    t =$a_{11}d^{-1}$. we can see this matrix is invertible matrix over$ F$ . F is quotient field of $D$.  Multiplying $A$ on the right by $B$, we can get a new matrix which frits row is $( d,0,a_{13},\cdots,a_{1n}) $we can Continue to do this will get the matrix we want.But $B$ maybe not invertible matrix over D.

This theorem comes from jacobson Basic algebra page 182~184.

答:
Let's examine more closely the generalization from Euclidean domains to PIDs in Jacobson's proof of the Smith normal form algorithm.

In the Euclidean domain case he is essentially using invertible (elementary) linear transformations whose effect on a row (or column) is to view it as the gcd of its entries, then apply the Euclidean algorithm to reduce the row to one whose only nonzero entry is the gcd of all initial entries $\,a_i.\,$ It's the same as  the below method to compute a gcd $\,(a_1, a_2,\ldots, a_n)\,$ of a "row" of $n$ naturals not all $\,0$.

Let $\,a_i\neq 0\,$ be the least entry. If $\,a_i\,$ divides all entries then we are done. Else $\,a_i\nmid a_j\,$ so replace $\,a_j\,$ by $\,a_j' := a_j\bmod a_i\neq 0\,$ which is $< a_i,\,$ so now the row has smaller least element, so by induction we can reduce it to a row with same gcd whose least element divides all others. Finally use the least entry (= gcd) to mod out all the others to $\,0,\,$ then swap the least entry  into the first index.

Generally a PID has no "mod" (remainder) so we can't use that method to  compute gcds. But it does have linear (Bezout) gcds, so we can use them to achieve  the descent to the gcd in a single-step, i.e. instead of using $\,(a,\,b) = (a,\,b\bmod a)\,$ we can do $\,(a,\,\,b) = (d,\, 0)\,$ for $\,d = \gcd(a,b),\,$ This too is an invertible linear transformation since, by Bezout $\,d = a u + b v\,$ for some $\,u,v,\,$ thus

$\qquad\qquad\qquad (a\ \ \,b) \begin{pmatrix} 1 & \!\!\!-q\\ 0 & 1\end{pmatrix} =\, (a\ \ \, \overbrace{b\!-\!aq}^{ b\bmod a})\quad \text{[Euclidean descent step]}$

$\qquad\qquad\qquad (a\ \ \,b) \begin{pmatrix} u &\!\! -b/d \\v & a/d\end{pmatrix} =\, (d\ \ \,0)\ \ \ \quad\text{[PID descent step]}$

Both matrices are unimodular (have $\det = 1)\,$ so they preserve the gcd $\,(a,b)\,$.

So the above row gcd algorithm also works if we replace the mod steps by this step of doing the gcds a pair at a time - which exploits the fact that gcd is associative, so $\,(a,b,c) = ((a,b),c),\,$ hence e.g. in row form  $\,(a,b,c) = (d,0,c) = (d,c,0) = (e,0,0),\,$ for $\, e=(d,c)$.

One "size" metric that the Bezout reduction step decreases is the number of prime factors, since if $\,a\nmid b\,$ then the gcd $\,(a,b)\,$ divides $\,a\,$ properly (else $\,a\mid (a,b)\Rightarrow\,a\mid b),\,$ so, by unique factorization, the gcd arises by deleting at least one prime factor from $\,a,\,$ so it has fewer prime factors than $\,a\,$ (counting multiplicity).  Said equivalently, the algorithm terminates because it produces a chain of proper factors $\,\ldots a_3\mid a_2\mid a_1\,$ and such chains cannot be infinite (i.e. "divides" is well-founded in a UFD), because each proper divisor has fewer prime factors. See also the Dedekind-Hasse PID generalization of the Euclidean division algorithm .

Jacobson's proof proceeds essentially as above, except we need to extend those $\,2\times 2\,$ matrices to $\,n\times n\,$ matrices with $\,\det = 1\, $ by appending $1$'s on the diagonal and $0$'s off-diagonal.


问:
Definite integral of function multiplied by its dependent variable
It is well known that $$ \int_a^b f(x) dx = (b-a)\bar{f} $$
where $\bar{f}$ is the average value of $f(x)$ in $[a,b]$

I was wondering if there is a similar approach for $$ \int_a^b xf(x) dx $$
involving the average value. I am pretty sure that i have seen something about this in the past but I can't find anything right now since I don't even know how to search for it

Thank you all in advance

答:
The first formula you wrote is usually taken to be the definition of "average value."

The second integral is closely related to a similar quantity known as the "first moment of $f$" (you've written down "the first moment around $0$").

If you think of $f(x)$ as the amount of mass at position $x$ from the center of a lever-arm (which is centered at $x = 0$), then the integral represents the torque about the pivot of the lever.

Because of the appearance of the integral, a natural way to evaluate it is to use integration by parts (at least when $f$ has an obvious antiderivative).

Anyhow, if you search for "moment integrals", you'll find a great deal more information.


问:
Presentation of subgroup
Let $G$ be the group with presentation
$$G = \langle \sigma_1 ,\sigma _2 , \sigma _3\, |\, \sigma _1^2=\sigma _2^2=\sigma _3^2=(\sigma _1\sigma _2)^p =(\sigma _2\sigma _3)^q =(\sigma _3\sigma _1)^r =1\rangle .$$
I want to have a presentation for the subgroup $H$ generated by the elements $\tau _1=\sigma _1\sigma _2,\, \tau _2=\sigma _2\sigma _3,\ \tau _3=\sigma _3\sigma _1$.

Is it enough to conjecture the presentation
$$ H=\langle \tau _1,\tau _2 ,\tau _3\, |\, \tau _1^p=\tau _2^q=\tau _3^r=\tau _1\tau _2\tau _3 =1 \rangle$$
and argue that the relations in the presentation of $G$ clearly imply the ones conjectured for $H$ and that the relations of $H$ imply the relations in the presentation of $G$.

答:
No, this is not sufficient, as there may be "hidden" relations. In your example, $H$ does indeed have the given presentation, and it is presumably possible to show this "by hand" using properties of the group $G$, but it is not "clearly" true.

Alternatively, you can use the Reidemeister-Schreier method. This is a process to find a presentation of a given subgroup of a group, and it works nicely here because $H$ has finite index in $G$. There are two approaches to the method: the classical combinatorial/algebraic version as covered in Section 2.3 of the book Combinatorial group theory by Magnus, Karrass and Solitar, or in Section II.4 of Lyndon and Schupp's book of the same name, and the topological version using coverings of presentation complexes (see here). Personally, I favour the topological version, as I view it as a "visualisation" of the algebraic version.

The process can also be done on a computer (see here), although the computer will want $p,q,r$ to be fixed.

Title: Presentation of subgroup


答:
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor  + \ldots + \lfloor \log_2 n \rfloor  = 1994 $$

Let $f(k)=\lfloor\log_2k\rfloor$. Since $\log$ is increasing we know that

$f(2^0)=f(1)=0$

$f(2^1)=f(2)=f(3)=1$

$f(2^2)=f(4)=f(5)=f(6)=f(7)=2$

$f(2^3)=f(8)=f(9)=\cdots=f(15)=3$

$f(2^4)=f(16)=f(17)=\cdots=f(31)=4$

etc.

So there are 2 ones, 4 twos, 8 threes, 16 fours, 32 fives, 64 sixes, 128 sevens, 256 eights, etc. We can multiply these to form the sequence $(2,8,24,64,160,384,896,2048)$. Summing the first 7 terms of this sequence gives $1538$, so we're missing $1994-1538=456$. Since $456/8=57<2048$ then our $n$ should be $2+4+8+16+32+64+128+57=311$, but we must also account for the $f(1)=0$ term, so in fact $n=312$.


问:
Historical definition of a group
Wikipedia states that van Dyck (1882) was the first to give the definition of a group in the modern way.  Before this, what were some of the original axioms or conditions for groups?  I mean, how were groups originally defined/restricted?  I understand groups were originally studied as groups of permutations, but that is all I know.

Thanks.

答:
The Wikipedia page looks pretty complete to me.

Like all other main concepts in mathematics, it took time for a precise definition of a group to arise; it was never axiomatized from the start.

See also the book A History of Abstract Algebra and the articles
The Foundation Period in the History of Group Theory and The Evolution of Group Theory: A Brief Survey.

Title: Historical definition of a group


答:
No, there is no connection. Specifically, there is no function of $\overline{f}$, $a$, and $b$ that reliably produces the given integral. To see this, we just need to find two functions $f$ and $g$, over a common interval $[a, b]$, with the same average value $\overline{f} = \overline{g}$.

Let $f(x) = 1$ and $g(x) = 2x$, over the interval $[0, 1]$. Then $\overline{f} = \overline{g} = 1$, but
$$\int_0^1 xf(x)\, \mathrm{d}x = \int_0^1 x\, \mathrm{d}x = \frac{1}{2}$$
and
$$\int_0^1 xg(x) \, \mathrm{d}x = \int_0^1 2x^2 \, \mathrm{d}x = \frac{2}{3}.$$

Generally speaking, the two integrals have very little to do with each other. Even the difficulty of computing each integral tends to be incomparable. The classic example is $\int e^{x^2} \, \mathrm{d}x$, which cannot be expressed with elementary functions. But, if we multiply by $x$, then we get the very tractable integral
$$\int xe^{x^2} \, \mathrm{d}x = \frac{1}{2}e^{x^2} + C.$$
But, multiplying $xe^{x^2}$ by $x$ again, we get an integral as intractable as the first:
$$\int x^2e^{x^2} \, \mathrm{d}x = \int x\left(xe^{x^2}\right) \, \mathrm{d}x = x\cdot \frac{1}{2}e^{x^2} - \frac{1}{2}\int e^{x^2} \, \mathrm{d}x,$$
hence we can compute $\int x^2e^{x^2} \, \mathrm{d}x$ if and only if we can compute $\int e^{x^2} \, \mathrm{d}x$ (which we can't, easily).

Title: Definite integral of function multiplied by its dependent variable

Body:
It is well known that $$ \int_a^b f(x) dx = (b-a)\bar{f} $$
where $\bar{f}$ is the average value of $f(x)$ in $[a,b]$

I was wondering if there is a similar approach for $$ \int_a^b xf(x) dx $$
involving the average value. I am pretty sure that i have seen something about this in the past but I can't find anything right now since I don't even know how to search for it

Thank you all in advance


问:
Showing that $c_n=\cos\frac\pi{2^n}$ satisfies the recurrence $c_{n+1} = \sqrt{\frac{1+c_n}2}$

From the identity $\cos(\alpha) = \sqrt{\frac{1+\cos(2\alpha)}{2}}$ with $0 \leq \alpha \leq \frac{\pi}{2}$, show that the sequence $c_n = \cos(\frac{\pi}{2^n})$ satisfies the recurrence relation $$c_{n+1} = \sqrt{\frac{1+c_n}{2}}$$

What I did was:

Say that $\alpha_n = \frac{\pi}{2^n}$. We first must prove $0 \leq \alpha_n \leq \frac{\pi}{2}$. By induction:

Base case is trivial: $0 \leq \frac{\pi}{2^1} \leq \frac{\pi}{2}$.

For $p(n) \Rightarrow p(n+1)$:
$0 \leq \frac{\pi}{2^n} \leq \frac{\pi}{2} \iff 0 \leq \frac{\pi}{2^n}\frac{1}{2} \leq \frac{\pi}{2}\frac{1}{2} \iff 0 \leq \frac{\pi}{2^{n+1}} \leq \frac{\pi}{4} < \frac{\pi}{2}$

So it is true that $\{a_n\}^\infty_{n=1} \in [0, \frac{\pi}{2}]$ and we can write $c_n = \cos(a_n)$.

Therefore:

$$c_1 = \cos(\alpha_1) = \cos(\frac{\pi}{2}) = 0$$

$$c_n = \cos(\alpha) \iff 1+c_n = 1+\cos(\alpha)\iff \frac{1+c_n}{2} = \frac{1+\cos(\alpha)}{2}$$
$$\iff \sqrt{\frac{1+c_n}{2}} = \sqrt{\frac{1+\cos(\alpha)}{2}}$$

But $\sqrt{\frac{1+c_n}{2}}$ would be the following term for $c_n$, meaning $c_{n+1} = \sqrt{\frac{1+c_n}{2}}$.

This was my atempt to prove it but I can't tell if it actually proves the recurrence. How could I prove recurrence relations in sequences like this?

答: You're doing much more work than you have to. As @Surb notes, you just need to verify $\cos\tfrac{\pi}{2^{n+1}}=\sqrt{\tfrac{1+\cos\tfrac{\pi}{2^n}}{2}}$, which is the case $\alpha=\tfrac{\pi}{2^{n+1}}$.


答:
1.Because there is at least one real number that is not computable.

2.
Step 1 : 0.0 , 0.1

Step 2 : 0.00 , 0.01 , 0.10 , 0.11

Step 3 : 0.000 , 0.001 , 0.010 , 0.100 , 0.011 , 0.101 , 0.110  , 0.111

step n : 0.(all combinations 1,0 of length n)

continue above indefinitely in binary to get all values in the interval 0,1.

Title: Constructive vs computable real numbers

Body:
I find it confusing that all of the following statements are true :

The computable real numbers are countable. $-\hspace{-3pt}-$ Alan Turing, "On Computable Numbers, with an Application to the Entscheidungsproblem"

In constructive analysis, the real numbers are uncountable. $-\hspace{-3pt}-$ Errett Bishop, Foundations of Constructive Analysis

"every mathematical statement [in constructive analysis] ultimately expresses the fact that if we perform certain computations within the set of positive integers, we shall get certain results" $-\hspace{-3pt}-$ Ibid.

Perhaps I am misunderstanding something.

I suppose I really have two questions. In constructive analysis :

Why isn't every real number computable?

How is it possible to construct an uncountable set?


答:
Let $\alpha = \frac{\pi}{2^{n+1}}$

$c_{n+1} = \cos\frac{\pi}{2^{n+1}} = \cos\alpha$ and $c_n = \cos\frac{\pi}{2^n} =  \cos\frac{2\pi}{2^{n+1}} = \cos(2\alpha)$

Now, $\cos(\alpha) = \sqrt{\dfrac{1+\cos(2\alpha)}{2}} \Rightarrow c_{n+1} = \sqrt{\dfrac{1+c_n}{2}}$

Title: Showing that $c_n=\cos\frac\pi{2^n}$ satisfies the recurrence $c_{n+1} = \sqrt{\frac{1+c_n}2}$

Body:
From the identity $\cos(\alpha) = \sqrt{\frac{1+\cos(2\alpha)}{2}}$ with $0 \leq \alpha \leq \frac{\pi}{2}$, show that the sequence $c_n = \cos(\frac{\pi}{2^n})$ satisfies the recurrence relation $$c_{n+1} = \sqrt{\frac{1+c_n}{2}}$$

What I did was:

Say that $\alpha_n = \frac{\pi}{2^n}$. We first must prove $0 \leq \alpha_n \leq \frac{\pi}{2}$. By induction:

Base case is trivial: $0 \leq \frac{\pi}{2^1} \leq \frac{\pi}{2}$.

For $p(n) \Rightarrow p(n+1)$:
$0 \leq \frac{\pi}{2^n} \leq \frac{\pi}{2} \iff 0 \leq \frac{\pi}{2^n}\frac{1}{2} \leq \frac{\pi}{2}\frac{1}{2} \iff 0 \leq \frac{\pi}{2^{n+1}} \leq \frac{\pi}{4} < \frac{\pi}{2}$

So it is true that $\{a_n\}^\infty_{n=1} \in [0, \frac{\pi}{2}]$ and we can write $c_n = \cos(a_n)$.

Therefore:

$$c_1 = \cos(\alpha_1) = \cos(\frac{\pi}{2}) = 0$$

$$c_n = \cos(\alpha) \iff 1+c_n = 1+\cos(\alpha)\iff \frac{1+c_n}{2} = \frac{1+\cos(\alpha)}{2}$$
$$\iff \sqrt{\frac{1+c_n}{2}} = \sqrt{\frac{1+\cos(\alpha)}{2}}$$

But $\sqrt{\frac{1+c_n}{2}}$ would be the following term for $c_n$, meaning $c_{n+1} = \sqrt{\frac{1+c_n}{2}}$.

This was my atempt to prove it but I can't tell if it actually proves the recurrence. How could I prove recurrence relations in sequences like this?


问:
Linearly independent subset of the domain in a linear transformation implies linear independence of subset of codomain
Let $T : R^n \rightarrow  R^m$ be a linear transformation and $S = \{\vec{v_1}, \vec{v_2}, · · · , \vec{v_k}\}$ be a subset of $R^n$. Prove that if the set $\{T(\vec{v_1}), T(\vec{v_2}), · · · , T(\vec{v_k})\}$ is a linearly independent subset of $R^m$, then $S$ is a
linearly independent subset of $R^n$

My attempt:

Suppose set $\{T(\vec{v_1}),T(\vec{v_2}),...,T(\vec{v_k})\}$ is linearly independent.

$T(\vec{v_1}) = A\vec{v_1}$, where $A$ is the standard matrix of the transformation. Realize that the set is a set of columns of $AV$ where $A\in\mathcal{M}_{mxn}, V\in\mathcal{M}_{nxk}$. Thus the set of columns of $AV$ is linearly independent. To prove that $S$ is linearly independent, $V$ has to be invertible. But I can't prove that, since I don't have any information regarding matrix $A$. Or is there any other way?

答:
\begin{align*}
\alpha_1v_1+...+\alpha _k v_k=0&\implies T(\alpha _1v_1+...+\alpha _kv_k)=0\\
&\implies \alpha _1T(v_1)+...+\alpha _kT(v_k)=0\\
&\implies \alpha _1=...=\alpha _k=0.
\end{align*}

I let you justify each step properly.


问:
Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?

Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?

If you replace $\sec^2$ with $\tan^2 + 1$, it should be $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$, right?

However, it seems that it is not multiplication here : $\sec^2 (\tan^{-1} (\frac{x}{2}))$, when the $1$ becomes independent of the $\tan^2$ somehow.

Could someone explain why?

答:
The correct statement is $\sec^2(y)=1+\tan^2(y)$.

Now let $y=\tan^{-1}\left(\dfrac x2\right)$.

I hope this helps you see why.


问:
How to calculate $\sin(37°)$ with a Taylor approximation?
How to calculate

$$\sin(37°)$$

with a Taylor approximation accurate to 3 decimal digits?

I know it is not a difficult question, but I have no answers of my book and so far I have only determined the Taylor approach of the umpteenth order and not approached or calculated anything in any sense.

Body:
Hint:

$ \sin (a+x) = \sin(a) + x \cos(a) + O(x^2) $

$ a = 36^\circ = 2 \pi /10 $

$ x = 1^\circ = 2 \pi /360 \approx 0.0175 $

$ \sin(a) = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$, $\cos(a)=\frac{1+\sqrt{5}}{4}$ (see Wikipedia)


问:
Groups and automorphisms of group-1
Why valid results don't understand for groups $V$ seen as automorphisms of group

$G$ is also valid when it is considered those groups as subgrupos of $G$.

For instance, the following result is true:

(to see An Introduction the Theory of Finite Groups,Kurzweil and Bernd Stellmacher)

(A) Let be $V\leq \operatorname{Aut}\left( G\right),\ N\vartriangleleft G$ and $V$-invariant.
Consider the semidirect product of $V$ with $G$. Let
$C_{G}(V)=\{g\in G:g^{v}=g\}$. For finite groups the following applies:

If $(|V|,|G|)=1$, then $C_{G/N}(V)=C_{G}(V)N/N$.

Assuming (A), because (B) it is true??

(B) Let be $K$ a subgroup of $G$,  $N\vartriangleleft G$ and $K$ normalizer $N$.
   Let $C_{G}(K)=\{g\in G:g^{k}=g\}$. Thus,

If $(|K|,|G|)=1$, then  $C_{G/N}(K)=C_{G}(K)N/N$.

"Excuse!

English is not my origin language."

答:
Once you have (A), note that in (B) $C_{G}(K)$ is the same thing as $C_{G}(V)$, where $V$ is the image of $K$ in $\operatorname{Aut}(G)$. In fact an element $k \in K$ is mapped onto the inner automorphism $x \mapsto x^{k}$.

A similar argument holds for $C_{G/N}(K)$.

Title: Groups and automorphisms of group-1


Title: How to calculate $\sin(37°)$ with a Taylor approximation?

Body:
How to calculate

$$\sin(37°)$$

with a Taylor approximation accurate to 3 decimal digits?

I know it is not a difficult question, but I have no answers of my book and so far I have only determined the Taylor approach of the umpteenth order and not approached or calculated anything in any sense.

Body:
Expand around $\pi/5$
$$\sin x = \sin(\pi/5)+\cos(\pi/5) \left(x-\frac{\pi }{5}\right)-\frac{1}{2} \sin(\pi/5) \left(x-\frac{\pi }{5}\right)^2+\\+\frac{1}{6} \cos(\pi/5) \left(x-\frac{\pi }{5}\right)^3+\frac{1}{24} \sin(\pi/5)\left(x-\frac{\pi }{5}\right)^4+O\left(x^5\right)$$
You get $0.601$ plugging $x=37/180 \pi$


问:
What is the significance of using Warped Products?
I have recently started studying differential geometry. I came across this idea of Warped Product Manifolds and I was wondering how is it different from regular Cartesian products of manifolds? What special properties or special intuition this gives compared to regular cartesian product of manifolds? Any insights are welcome. Thank you in advance!</p>

答:
Given two riemannian manifolds $(M,g_M)$ and $(N,g_N)$, there is one natural way to construct a product manifold $M\times N$ but plenty natural ways to construct product riemannian manifolds $(M\times N,g)$. The most simple way is to define $g = g_M \oplus g_N$, but it is a very restricted way as in this construction, geodesics are the curves whose projection on $M$ and $N$ are geodesics. Wraped product is made by choosing a positive function $f$ on $M$ and saying that $g_{m,n} = {g_M}_m \oplus f(m){g_N}_n$: in this construction, moving along $M$ modify the metric in the $N$ component.

The more important example of a natural wraped product is while choosing a polar decomposition in an euclidean space. The diffeomorphism
\begin{align}
\varphi : \mathbb{R}_+^* \times \mathbb{S}^n & \longrightarrow \mathbb{R}^{n+1}\setminus\{0\} \\
(r,\theta) & \longmapsto r\theta
\end{align}
allows to identify $\mathbb{R}^{n+1}\setminus\{0\}$ as a product of two manifolds but to have an isometry, you have to put on the product manifold the metric $\varphi^*g_{\text{eucl}}$. If $\overset{\circ}{g}$ denotes the standard metric of the sphere of radius $1$, one can show that in the product manifold $\mathbb{R}^*_+\times\mathbb{S}^n$, the metric $\varphi^*g_{\text{eucl}}$ has the form
\begin{align}
\mathrm{d}r^2 \oplus r^2 \overset{\circ}{g}
\end{align}
and thus, we can identify the euclidean space with this wraped product: wraped product come naturally. It reflects that at $r$ fixed, you are not on a sphere with radius $1$ but with radius $r$ (whose metric is $r^2$ times those of the sphere of radius 1)

You can generalize this construction, in a general riemannian manifold $(M,g)$. With some curvature assumptions, there exists some submanifold $Y$ such that there is a natural trivial bundle $Y\times \mathbb{R}^k$ which is isometric to $(M,g)$ while given a wraped product metric, etc.


问:
The name for a factor that determines the contribution of an individual within a team
Some time ago I read an article on Wikipedia about a number/factor that describes the contribution factor of an individual within a group and I can't find the correct name for this number/figure anymore and I hope someone knows what I mean and can help out.

If I remember correctly, the general idea is that you consider for example 5 individual contributors A, B, C, D, E and let them solve tasks in teams of e.g. 3 with different permutations and measure the output of the team. For example team A,B,D achieves an output of 12. Team A,D,E accomplishes an output of 10 and so on. The goal is to assign every individual a "contribution factor" that basically measures how well the team performs. There is a special name/number for this that I forgot and it's not just solving linear equations.

Googling is hopeless, does someone here know what I mean?

答: It sounds like you are looking for the Shapley Value.</p>


问:
Limit points / cluster points from $a:$ $\mathbb {N} \rightarrow \mathbb {Q}$
So i have the bijection $a:$ $\mathbb {N} \rightarrow \mathbb {Q}$.

I was asked to find the limit points / cluster points of the sequence $a_n = a(n)$.

I know that this bijection exists since $\mathbb {N}$ and $\mathbb {Q}$ are both countable. After reading up in my script, I know how they constructed it using Cantor's diagonal argument. This is where I'm stuck though, since I just read that apparently this sequence has every real number as a limit point of the sequence. How can this be proven?

答:
The sequence is given by $$a_{\frac{q(q-1)}{2}+p}=\frac{p}{q},\quad p\leq q.$$

If $\lambda \in (0,1]\cap \mathbb Q$, then $\lambda =\frac{p}{q}$ for some $p,q\in\mathbb N^*$, $p\leq q$. Taking $$m_k=\frac{kq(kq-1)}{2}+kp,$$
then $x_{m_k}=\lambda $.

If $\lambda =0$, the argument is similar.

If $\lambda \in (0,1)\setminus \mathbb Q$, by density of $\mathbb Q$ in $\mathbb R$, there are $(p_k)$, $(q_k)$ where $p_k\leq q_k$ and $$\lim_{k\to \infty }\frac{p_k}{q_k}=\lambda .$$
Set $m_k=\frac{kq_k(kq_k-1)}{2}+kp_k$, and then $(x_{m_k})$ is a subsequence that converges to $\lambda $.


问:
Is there any redundancy here?
Let $\Phi$ be the class of all functions $\phi:[1, \infty) \rightarrow[0, \infty)$ satisfying the following properties:

$\phi$ is lower semi-continuous;

$\phi(1)=0$;

for each sequence $\left\{t_{n}\right\} \subseteq(1, \infty), \lim _{n \rightarrow \infty} \phi\left(t_{n}\right)=0$ if and only if $\lim _{n \rightarrow \infty} t_{n}=1$.

Is there any redundancy between those conditions? I mean, doesn't $1$ and $3$ implies $2$?

答:
Yes. Indeed by (3) there exists $\lim_{x\to 1}\phi(x)$, hence
$$
0 \overset{(hp)}{\le} \phi(1) \overset{(1)}{\le} \liminf_{x\to 1}\phi(x)\overset{(above)}{=}\lim_{x\to 1}\phi(x) \overset{(3)}{=}0.
$$

Title: Is there any redundancy here?


答:
The easiest approach may be to use the Taylor series $\sin x=x-{1\over6}x^3+{1\over120}x^5-\cdots$, since $37^\circ=37\pi/180\approx0.646\lt1$:

$$\sin(37^\circ)=\sin\left(37\pi\over180\right)\approx\left(37\pi\over180\right)-{1\over6}\left(37\pi\over180\right)^3+{1\over120}\left(37\pi\over180\right)^5$$

(noting that the next term in the alternating sum is considerably less than $1/5040\approx0.0002$). I wouldn't want to complete the decimal calculation by hand, but it's relatively straightforward with a calculator, even if you have to use an approximation like $\pi\approx3.1416$ on a pocket calculator that lacks a button for $\pi$ and only does arithmetic.


问:
Get Transformation Matrix from Points
I have built a little C# application that allows visualization of perpective transformations with a matrix, in 2D XYW space. Now I would like to be able to calculate the matrix from the four corners of the transformed square. Here is a screenshot to help you understand what I am working with:

The idea is to allow the user to move the corners of the square and update the matrix. The corners are currently located at (1,1), (1,-1), (-1,-1), (-1,1). Is there an algorithm that will calculate a 3x3 matrix given four 2D points?

If I understand this correctly, every matrix corresponds to a set of four points, and every set of four points corresponds to one or more equivalent matrices ('equivalent' meaning 'producing identical transformations').

I searched for an algorithm to do this, but didn't have much luck.

I figured out that I could do it by creating eight equations, one for each variable in the four points, and then setting one of the matrix values to one, and solving for the other eight with algebra. However, the equations grow much too complicated to do this all successfully on pencil and paper.

This is the process I used to try to get it working.

So this is the basic matrix transformation formula.

$\begin{pmatrix}a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}\begin{pmatrix}x\\
y\\
z
\end{pmatrix}=\begin{pmatrix}ax+by+cz\\
dx+ey+fz\\
gx+hy+iz
\end{pmatrix}$

The resulting point is then converted from homogeneous to Euclidean coordinates.

$\begin{pmatrix}x\\
y\\
z
\end{pmatrix}$=$\begin{pmatrix}\frac{x}{z}\\
\frac{y}{z}
\end{pmatrix}$

So given a collection of points, we transform them like this.

$\begin{pmatrix}a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}\begin{pmatrix}x_{n}\\
y_{n}\\
z_{n}
\end{pmatrix}=\begin{pmatrix}x'_{n}\\
y'_{n}
\end{pmatrix}$

These are the formulas used for the transformation.

$x'_{n}=$$\frac{ax_{n}+by_{n}+cz_{n}}{gx_{n}+hy_{n}+iz_{n}}$

$y'_{n}=$$\frac{dx_{n}+ey_{n}+fz_{n}}{gx_{n}+hy_{n}+iz_{n}}$

We then define four base points, which are the corners of our square.

$x'_{0}=1$

$y'_{0}=1$

$z'_{0}=1$

$x'_{1}=1$

$y'_{1}=-1$

$z'_{1}=1$

$x'_{2}=-1$

$y'_{2}=-1$

$z'_{2}=1$

$x'_{3}=-1$

$y'_{3}=1$

$z'_{3}=1$

This gives us the following system of equations, in the form that lets us determine the transformed points from the matrix.

$x'_{0}=$$\frac{a+b+c}{g+h+i}$

$y'_{0}=$$\frac{d+e+f}{g+h+i}$

$x'_{1}=$$\frac{a-b+c}{g-h+i}$

$y'_{1}=$$\frac{d-e+f}{g-h+i}$

$x'_{2}=$$\frac{-a-b+c}{-g-h+i}$

$y'_{2}=$$\frac{-d-e+f}{-g-h+i}$

$x'_{3}=$$\frac{-a+b+c}{-g+h+i}$

$y'_{3}=$$\frac{-d+e+f}{-g+h+i}$

Now we want to reverse the transformation, and find the matrix that produces the above points. Since we have 9 unknowns and 8 equations, we need to add another equation.

$i=1$

Now all that is left is to solve the system equations to find the formulas for the matrix values. I'm not patient enough nor good enough with algebra to do this myself, so I used an online calculator to solve the system of equations. The formulas it gave almost worked, but had some glitches with y-coordinates.

I think this can be narrowed down to 2 questions:

Are the above calculations wrong, or does the online calculator have a bug?

Is there an easier algorithm for this?

答:
I'll use $f(x, y)$ to represent the 2D point $(x, y)$ multiplied by an assumed 3x3 transformation matrix:

$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}
:=
\begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}
\begin{pmatrix}
x\\
y\\
1
\end{pmatrix}$

$f(x, y) :=
\begin{pmatrix}
\frac{x'}{z'}\\
\frac{y'}{z'}
\end{pmatrix}
=
\begin{pmatrix}
\frac{ax + by + c}{gx + hy + i}\\
\frac{dx + ey + f}{gx + hy + i}
\end{pmatrix}$

Given the transformation matrix, the corners of a square centered on $(0, 0)$ can be transformed:

$f(1, 1) = (x_1, y_1)\\
f(1, -1) = (x_2, y_2)\\
f(-1, -1) = (x_3, y_3)\\
f(-1, 1) = (x_4, y_4)$

Forgive me for using slightly different naming and notation than the original question. It was a long time ago.

The problem is to find a transformation matrix (values for $a$ through $i$) given these 8 output variables. I used the following equations to compute a solution.

First, some temporary variables to simplify the process:

$j = x_1 - x_2 - x_3 + x_4\\
k = -x_1 - x_2 + x_3 + x_4\\
l = -x_1 + x_2 - x_3 + x_4\\
m = y_1 - y_2 - y_3 + y_4\\
n = -y_1 - y_2 + y_3 + y_4\\
o = -y_1 + y_2 - y_3 + y_4$

Then values for the matrix elements, calculated in reverse order:

$i = 1\\
h = \frac{(jo - ml)i}{mk - jn}\\
g = \frac{kh + li}{j}\\
f = \frac{y_1(g + h + i) + y_3(-g - h + i)}{2}\\
e = \frac{y_1(g + h + i) - y_2(g - h + i)}{2}\\
d = y_1(g + h + i) - f - e\\
c = \frac{x_1(g + h + i) + x_3(-g - h + i)}{2}\\
b = \frac{x_1(g + h + i) - x_2(g - h + i)}{2}\\
a = x_1(g + h + i) - c - b$

I have a proof of concept implementation: https://jsfiddle.net/kendfrey/oxm5L6q0/5/

For the unhealthily curious, my working out was done in Lean: https://gist.github.com/kendfrey/28c49ebc1c28e543ca6322e167fb8fd8

Notes:

Since there are 9 unknown variables and 8 equations, I'm free to choose a value for $i$.

This works in the general case but has some degenerate cases where division by zero occurs.

OwnerDisplayName: Kendall Frey

Title: Get Transformation Matrix from Points


Body:
For $\alpha=\frac{37\cdot\pi}{180}=0.65477$ we take 3 terms in the Maclaurin series:
$$
\sum_{k=0}^2 \frac{(-1)^k 0.64577^{1 + 2 k}}{(1 + 2 k)!}=0.64577-\frac{0.65477^3}{6}+\frac{0.65477^5}{120}=0.601823
$$
This is as close as $10^{-5}$ to the exact value.


问:
Should a high school introductory calculus class teach $\varepsilon$-$\delta$ proofs?
It seems to me that most high school students are comfortable with the intuitive notion of a limit ("as $x$ gets arbitrarily close to $c$, $f(x)$ gets arbitrarily close to $L$") and gain little insight from learning the $\varepsilon$-$\delta$ definition. Is the added rigor of the $\varepsilon$-$\delta$ definition worth teaching at the high school level?</p>

答: As I self-teaching student, I believe that the epsilon-delta definition is very useful. The intuitive definition worked great for me, but the rigor behind the epsilon-delta definition is vital. I may be a bit biased as I am planning to pursue a math-related profession in the future, but still think all students should know the definition. Not being able to use the epsilon-delta definition is like understanding the process behind addition, but not really being able to do the procedure (I have a younger sibling hence the analogy). Yes, the epsilon-delta definition took me around a week to learn how to apply and use, but it was worth it in the end. Also, you can PROVE limits with epsilon-delta.</p>


问:
Submanifolds are preserved under diffeomorphisms
I've been reading books about differential manifolds. There is a remark in the book that says

Remark: Submanifolds are preserved under diffeomorphism, i.e If $A$ is a submanifold of $M$ and $\tau : M \rightarrow N$ is a diffeomorphism, then $\tau(A)$ is a submanifold of $N$.

Can anyone enlighten me why is this true or is there post before regarding to this fact.

Any help would be much appreciated!

答:
Recall that being a submanifold is a local property: if $M^n$ is a manifold and $Y \subset M$, then $Y$ is a submanifold of $M^n$ of dimension $m$ if for every point $p\in Y$, there exists an open subset $U$ in $M$ with $p\in U$ and a chart $\varphi: U \to \mathbb{R}^n$ such that $\varphi\left(Y\cap U \right) = V^m\cap \varphi(U)$ where $V^m$ is a linear subspace of $\mathbb{R}^n$, of dimension $m$. This says that there exists local charts such that in those charts, $Y$ is an a linear subspace of the chart.

Now, if $f : M \to N$ is a diffeomorphism, and if $Y \subset M$ is a submanifold, take a chart $\varphi : U \subset M \to \mathbb{R}^n$ that is used in the definition for $Y$ to be a submanifold. Let $\tilde{Y} = f(Y) \subset N$,$\tilde{U} = f(U) \subset N$ and define $\tilde{\varphi} : \tilde{U} \to \mathbb{R}^n$ to be $\tilde{\varphi}(p) = \varphi(f^{-1}(p))$. It is clear that $\tilde{\varphi}$ is a chart in which we can read that $\tilde{Y}$ is a submanifold! We have then shown that $\tilde{Y}$ is a submanifold of $N$. So every submanifold of $M$ gives birth to a submanifold of $N$. Conversly, as $f^{-1}: N \to M$ is also a diffeomorphism, every submanifold of $N$ gives birth to a submanifold of $M$, and there is a correspondance between the set of submanifolds of $M$ and the set of submanifolds of $N$.

In fact, diffeomorphic manifolds are "the same", but drawn differently. A diffeomorphism is a dictionnary between them. Thus, every geometric property of one is translated into a geometric property of the other, and bijectively.


答:
Using Taylor's poloynomial with degree $p$ around $a=\frac{\pi}{6}$, the error is bounded by
$$
|R_p| =\left| \dfrac{f^{(p+1)}(\xi)}{(p+1)!}\left(\frac{37 \pi}{180}-\frac{\pi}{6}\right)^{p+1}\right|\leq \frac{1}{(p+1)!} \left(\frac{7 \pi}{180}\right)^{p+1}
$$
so, just select a convenient $p$ and use the corresponding Taylor polynomial to get the approximation.

For $p=3$ you get na error bound of $ 0.928304 \times 10^{-5}$ and for $p=4$ na error bound of $0.226828\times 10^{-6}$


问:
Is $\frac{1}{x}$ integrable?
I am taking a course on theory of integrals and measure, where I have learned a very fundamental definition of integral. One of the things I've learned is that $f$ is integrable iff $\int|f|<\infty$. Then, according to this definition, $f(x)=\frac{1}{x}$ wouldn't be integrable, since the integral of $|f(x)|=|\frac{1}{x}|$ diverges. However, I can calculate $\int_a^b\frac{1}{x}dx$ for any $a, b \in (-\infty, +\infty)$ and this isn't problematic at all. How do I deal with this?

答:
I think the OP is referring to the fact that, if a function $f$ is the derivative of another function $g$, then
$$
  \int_a^bf(x)\, dx = g(b)-g(a).
  \tag{1}
  $$
Since $1/x$ is the derivative of $\ln(|x|)$,  we would then have  that
$$
  \int_a^b{1\over x}\, dx = \ln(|b|) - \ln(|a|).
  \tag{2}
  $$

The error in this argument is that for (1) to hold, one needs that $g'(x)=f(x)$ for EVERY $x$ in $[a,b]$.  Since
$$
  {d\over dx}\ln(|x|) = {1\over x}
  $$
is only true if $x\neq 0$, one may not allow 0 to be in the interval  $[a,b]$ for (2) to hold!

Title: Is $\frac{1}{x}$ integrable?


问:
Prove that if polynomial's $f(x)=x^6+ax^3+bx^2+cx+d$ , all roots are real, then $a=b=c=d=0$.
$$f(x)=x^6+ax^3+bx^2+cx+d$$

I have some trouble proving it and all I know is that I should be using Vieta's formulas. Can anyone give me a hint?

答:
You have $f(x)=0$

Let $\alpha_1, \alpha_2 \dots, \alpha_6$ be the roots of the equation.

$\sum_1^6 \alpha_k=\dfrac{0}{1}$ or simply $0$. (Why?)

$\sum_{1\le i \lt j \le 6} \alpha_i \alpha_j=0$

$(\sum_1^6 \alpha_k)^2-2\sum_{1\le i \lt j \le 6} \alpha_i \alpha_j= \sum_1^6\alpha_i^2=0 \implies \alpha_1= \cdots=\alpha_6=0$

Can you continue from here?


问:
Problem with four equal-sized Circles
I think the following is true but haven't managed to prove it yet.

Consider a circle with center $O$ and radius $r$. Choose three points $A,B,C$ on or inside the circle such that all sides of $\triangle ABC$ have length greater than $r$. Show that the three circles of radius $r$ centered at $A,B,C$ cannot intersect outside $C(O,r)$.

答:
Take the point $O=(0,0); A=(r,0); B=(0,r); C=(-r,0)$. Now we have that (length) $L_{ab}=L_{bc}=\sqrt{2}r$ and $L_{ac}=2r$ which are obviously $>r$. Write the equation for circle A as $\left(x-r\right)^{2}+y^{2}=r^{2}$ and circle B $x^{2}+\left(y-r\right)^{2}=r^{2}$ and the intersection are the solution to the equation $\left(x-r\right)^{2}+y^{2}=x^{2}+\left(y-r\right)^{2}$. By simple calculations you get a line $y=x$. Substitute in one of the prevous equation and you get (for $y$) $\left(y-r\right)^{2}+y^{2}=r^{2}\rightarrow y\left(y-r\right)=0$ whose solutions are $y=0; y=r$. But wait, we said before that the intersection must be in the line $y=x$ so the points of intersection must be $(0,0)$ and $(r,r)$, where the last one is obviously outside the circle centered in $O=(0,0)$.

The case limit for $A=(r,0); B=\left(r\cos\theta,r\sin\theta\right)$ and $C$ somewhere else is when $L_{ab}=r$. Since $L_{oa}=L_{ob}=L_{ab}=r$, we have an equilateral triangle △OAB, where $\theta$ must be $\pi/3$. A bit of calculations and we have that the intersections are $(0,0)$ and $\left(\frac{3}{2}r,\frac{\sqrt{3}}{2}r\right)$ and the line is $y=\frac{x}{\sqrt{3}}$.

Title: Problem with four equal-sized Circles


问:
Comparing Two Statements of the Rank Theorem
I don't think this a duplicate, even though a similar question appears here.

Let $m\geq n$ and let $F:\mathbb R^n\to \mathbb R^m$ be a $\mathcal C'$ mapping s.t. rank $F'(x)=r\leq n$ for all $x\in E\subseteq \mathbb R^n.\ $Fix $a\in E$ and set $A=F'(a)$ and let $P$ be a projection in $\mathbb R^m$ onto the $Y_1=$range of $A$. Set $Y_2=kerP.$

Then the claim is that there are open sets $U\subseteq E$ and $V$ in $\mathbb R^n$ s.t $a\in U$; and there is a  bijective $\mathcal C'$ map $H:V\to U$ s.t $\tag1 F(Hx)=Ax+\varphi (Ax)$

where $\varphi$ is a $\mathcal C'$ function mapping $A(V)$ into $U$.

This is,of course, a somewhat abbreviated version of the statement that appears in blue Rudin. The proof is not hard, but it seems rather abstruse and uniformative, compared to the version I first learned, which follows my comments, and question here.

$H$ seems to be simply a change of coordinates (diffeomorphism),but I can't find an easy geometric interpretation of $\varphi$. It is easy to show that $P$ restricted to $F(U)$ is a bijection onto $A(V)$ but this is immediate from the other version below.

In fact, it seems a lot easier to understand the idea in the following, different (?) version of the Rank Theorem:

Suppose $U, V$ are open in $\mathbb R^n,\mathbb R^m$,resp. and let $F:U\to V$ be a $\mathcal C'$ map s.t. $F'(x)$ has rank $r$ for all $x\in U$.

Then, there exist open sets $U_1,V_1\in \mathbb R^n,\mathbb R^m$, resp. and  diffeomorphsims $\varphi:U_1\to U_0$ and $\psi:V_1\to V_0$ s.t. for fixed $a\in U$, $\varphi (a)=0;\ \psi (f(p))=0$ and

$\tag2 \psi\circ f\circ \varphi ^{-1} (x_1,x_2,\cdots ,x_r,x_{r+1},\cdots x_n)=(x_1,\cdots ,x_r,0,0,\cdots ,0)$.

This formulation is simpler and more intuitive. Are the two versions equivalent? If not, how is Rudin's "better" than the other?

答:
The formulation of the rank theorem in Rudin is indeed quite confusing. However it is not quite so bad once the interpretation of the theorem is fleshed out.

Rank Theorem (Rudin): Suppose $m, n, r$ are nonnegative integers, $m > r$, $n > r$, $\mathbf{F}$ is a $C^1$-mapping of an open set $E\subseteq \mathbb{R}^n$ into $\mathbb{R}^m$, and $\mathbf{F}'(\mathbf{x})$ has rank $r$ for every $x\in E$. Fix $\mathbf{a} \in E$, put $A = \mathbf{F}'(\mathbf{a})$, let $Y_1$ be the range of $A$, and let $P$ be a projection in $\mathbb{R}^m$ whose range is $Y_1$. Let $Y_2$ be the null space of $P$. Then there are open sets $U$ and $V$ in $\mathbb{R}^n$, with $\mathbf{a} \in U$, $U \subseteq E$, and there is a
$1$-$1$ $C^1$-mapping $\mathbf{H}$ of $V$ onto $U$ (whose inverse is also of class $C^1$) such that
$$\mathbf{F}(\mathbf{H}(\mathbf{x})) = A\mathbf{x} + \varphi(A\mathbf{x})\qquad (\mathbf{x}\in V)$$
where $\varphi$ is a $C^1$-mapping of the open set $A(V)\subseteq Y_1$ into $Y_2$.

Let's contrast this with some other standard textbook formulations of the rank theorem, this one from Lee's Introduction to Smooth Manifolds, for example.

Rank Theorem (Lee): Suppose $M$ and $N$ are smooth manifolds of dimensions $m$ and $n$, respectively, and $F:N\rightarrow M$ is a smooth map with constant
rank $r$. For each $p \in M$ there exist smooth charts $(U,\psi)$ for $N$ centered $p$ and
$(V,\phi)$ for $M$ centered at $F(p)$ such that $F(U)\subseteq V$, in which $F$ has a coordinate representation of the form
$$\hat{F}(x^1,\cdots x^r,x^{r+1},\cdots, x^n) = (x^1,\cdots,x^r,0,\cdots ,0),$$
where $\psi^{-1}\circ F \circ \phi = \hat{F}: \mathbb{R}^n \rightarrow \mathbb{R}^m$.

To make the comparison more transparent, let me disregard the $C^1$ condition in Rudin's version of the theorem and just assume everything is smooth if necessary. We also don't need the full manifold formulation of Lee's version, so we will just take $M = \mathbb{R}^m$ and $N = \mathbb{R}^n$.

The key content of the rank theorem is that the local image of a rank $r$ map is an $r$-dimensional manifold. The difference between the two versions is that Lee's rank theorem fully flattens out the resulting manifold by mapping it to the linear subspace $(x^1,\cdots x^r,0,\cdots 0)$. He accomplishes this by transforming both the domain and the codomain.

On the other hand, Rudin never touches the codomain. He instead chooses coordinates on the domain so that the resulting image submanifold is given as an orthogonal decomposition (more precisely, the decomposition will be orthogonal if $P$ is chosen to be an orthogonal projection, otherwise it will be oblique) into the tangent plane at a point $\mathbf{a}$ (the $A\mathbf{x}$ term), plus deviations orthogonal to this tangent plane (the $\varphi(A\mathbf{x})$ term). Therefore you can view Rudin's version as sort of intermediate, where the image submanifold of $\mathbf{F}$ is straightened out partially along the tangent plane, but not quite all the way.

It follows that to go from Rudin's version to Lee's version, all we need to do is to fully "straighten out" this manifold. The map that does this is $f(\mathbf{x}) = \mathbf{x} - \varphi(P\mathbf{x})$. Intuitively, this map removes the deivations orthogonal to the tangent plane. Then we have
$$f(\mathbf{F}(\mathbf{H}(\mathbf{x})) = f(A\mathbf{x} + \varphi(A\mathbf{x})) = A\mathbf{x} + \varphi(A\mathbf{x}) - \varphi(A\mathbf{x}) = A\mathbf{x},$$
where we use the fact that $PA\mathbf{x}=A\mathbf{x}$ and $P\varphi(A\mathbf{x}) = \mathbf{0}$. Note that $f$ is in fact a smooth diffeomorphism, with inverse $f^{-1}(\mathbf{x}) = \mathbf{x} + \varphi(P\mathbf{x})$. Since $A$ is of rank $r$, there exists invertible matrices $Q_1$ and $Q_2$ such that $Q_1AQ_2 = I_r$, where $I_r$ is the matrix which is zero everywhere except the first $r$ diagonal entries, which are $1$. Let $Q_1\mathbf{y}=\mathbf{x}$. Then
$$Q_1f(\mathbf{F}(\mathbf{H}(Q_2\mathbf{y})) = I_r\mathbf{y}.$$
Therefore we can take the coordinate charts $\psi$ and $\phi$ in Lee's formulation to be $\phi = \mathbf{H}\circ Q_2$ and $\psi^{-1} = Q_1\circ f$.

Since we have the relations between the coordinate charts in Lee's version and the various maps in Rudin's version, we can easily go between the two formulations. These relations also makes precise the missing steps it takes to fully "straighten out" the image manifold.


问:
Is the condition $\phi(ab) = \phi(b)\phi(a)$ basically equivalent to being a homomorphism?
Let $\phi$ be a map from one group to another. Is the condition $\phi(ab) = \phi(b)\phi(a)$ basically equivalent “up to isomorphism” (or whatever) to the usual condition $\phi(ab) = \phi(a)\phi(b)$ of being a homomorphism? Is the correspondence basically given by “do things right to left instead of left to right”?

I’m asking because I was reading Herstein’s proof of Cayley’s Theorem in Topics in Algebra, where he considers the map $\tau_g: G \to G$ defined by $\tau_g(x) = xg$, which is right-multiplication by $g$, and then shows that $\tau_{gh} = \tau_h \circ \tau_g$ (here my notation differs from Herstein; by $\tau_h \circ \tau_g$ I mean the map defined by $(\tau_h \circ \tau_g)(x) = \tau_h(\tau_g(x))$). But I thought from the “usual definition”, being a homomorphism should show that $\tau_{gh} = \tau_g \circ \tau_h$. Then when I look up a proof of Cayley’s Theorem (e.g. in Wikipedia), it considers left-multiplication by $g$ instead (that is, we define $\tau_g(x) = gx$), in which case we indeed have $\tau_{gh} = \tau_g \circ \tau_h$.

So this made me think that it shouldn’t matter, for all practical/“structural” purposes, whether our condition for homomorphisms was $\phi(ab) = \phi(a)\phi(b)$ or $\phi(ab) = \phi(b)\phi(a)$. Am I correct in thinking this? Is there a rigorous justification of this?

Edit: To clarify what I mean: Herstein’s definition of function composition is $(\sigma \circ \tau)(x) = \tau(\sigma(x))$. By this definition, we needed to consider right-multiplication by $g$ in the map $\tau_g$ in our proof of Cayley’s Theorem. But by the “usual” definition of function composition as $(\sigma \circ \tau)(x) = \sigma(\tau(x))$, we would need to consider left-multiplication by $g$. Does it “just happen” that these two ways of going about things work out the same? Why should Herstein's requirement of homomorphism, which is that $\tau_{gh}(x) = \tau_h(\tau_g(x))$ for all $x$, not affect things -- and basically we just had to do it "from the right" instead of "from the left"? Does this principle hold more generally?

答:
Suppose we have $D_4=\langle a^4=b^2=1, \ bab=a^{-1}\rangle$.

I take $f:D_4 \to D_4$ with $f(x)=x$.

If your definition was right (or at least promising) then $f$ would be a homomorphism.

But $ab=f(ab)=f(b)f(a)=ba$ which is not true.

Hence here, the identity map is not a homomorphism in your sense.


问:
The difference of cubed diffrence of Wiener processes
Problem

I need to calculate the limit of $X_n$ in mean squared when $n$ tends to infinity. $X_n$ is given as

$$X_n = \sum_{i=1}^n (W_{ti/n} - W_{t(i-1)/n})^3$$

where $W$ is Wiener process.

My attempt

If we prove that $E(X_n) = 0$ and $Var(X_n) \to 0$, we are done.
However, I get stuck at the very beginning when I try to calculate the expectation.

Issues

If we denote $W_{ti/n}$ as $Y_i$, I struggle evaluating $E(Y_i^2Y_{i-1})$ after rising the difference to cube. Even though I showed that $\sum_{i=1}^n Y_i \text{ and } \sum_{i=1}^n Y_{i-1}$ are almost the same with the exception of $Y_n$ and $Y_0$, I don't know how to get that $0$.

答:
Use the fact that $Y_i-Y_{i-1}$ is independent of $Y_{i-1}$ and $\mathsf{E}Y_i^k=0$ for odd $k$, i.e.,
\begin{align}
\mathsf{E}Y_i^2Y_{i-1}&=\mathsf{E}[(Y_i-Y_{i-1})^2Y_{i-1}]+2\mathsf{E}[(Y_i-Y_{i-1})Y_{i-1}^2]+\mathsf{E}Y_{i-1}^3 \\
&=\mathsf{E}[(Y_i-Y_{i-1})^2]\mathsf{E}Y_{i-1}+2\mathsf{E}[Y_i-Y_{i-1}]\mathsf{E}Y_{i-1}^2+\mathsf{E}Y_{i-1}^3=0.
\end{align}

Alternatively, notice that $Z_i:=Y_i-Y_{i-1}\sim N(0,t/n)$ so that $\mathsf{E}Z_i^k=0$ for odd $k$, which implies that $\mathsf{E}X_n=\sum_{i=1}^n\mathsf{E}Z_i^3=0$. Next, since the increments of $(W_t)$ are independent,
$$
\operatorname{Var}(X_n)=\sum_{i=1}^n \mathsf{E}Z_i^6=5!!\cdot\sum_{i=1}^n \left(\frac{t}{n}\right)^3=\frac{15 t^3}{n^2}.
$$


答:
I agree with the other answers that are already provided.  However, I think that there is a possible source of confusion that should be resolved.

In Trigonometry/Analytical Geometry, the domain of the sine and cosine functions are angles, which have a unit of measurement, the degree.  So $37^{\circ}$ is a dimensioned number in the same way that 1 foot is a dimensioned number.

In Calculus/Real Analysis, this (necessarily) changes.  The domain  of the sine and cosine functions are dimensionless real numbers (i.e. $\pi/4$ rather than $45^\circ$).  The reason for this alteration in the sine and cosine functions is to facilitate (for examples)

Using the Taylor series of the sine and cosine functions to attack problems

Using the ArcTan function to attack a problem like $\int_0^1 \frac{dx}{1 + x^2}.$

Some of the confusion centers around ambiguity in the connotation of the term radian.  That is, does $(\pi/4)$ radians represent a dimensioned number, with 1 radian representing a unit of measurement of an angle similar to what 1 degree represents?  This connotation is often useful in allowing the Trigonometry/Analytical Geometry student to transition into Calculus/Real Analysis
and immediately begin solving Calculus problems without requiring that the student first acquire a deep understanding of the theory.

The alternative connotation of 1 radian, is that it is used to represent the  dimensionless proportion of a specific arc length of a portion of the unit circle.  This proportion (i.e. ratio) is taken against the arc length of 1 complete revolution of the unit circle (i.e. the circumference of the unit circle).  The circumference of the unit circle is $2\pi.$

Therefore, the term $(\pi/4)$ radians for example, can be interpreted in two totally distinct ways:

equivalent to $45^\circ$.

representing the arc length of 1/8-th of a complete revolution around the unit circle.  This interpretation preserves the idea that $(\pi/4)$ radians (for example) is a dimensionless number.

Often, when a transitioning student uses the term radians, they are intending the first connotation above.  When someone experienced in Calculus/Real Analysis uses the term radians, they are intending the second connotation above.

For an illustration of the sine and cosine functions having as their domain, dimensionless numbers, see What is the physical meaning of sine, cosine and tangent of an obtuse angle?.


答:
The identity
$$\sec^2\theta\equiv1+\tan^2\theta$$
holds for all values of $\theta$, not just some values- that's why it's an identity, as opposed to an equation. Replacing $\theta$ by $\tan^{-1}(\frac{x}{2})$ yields the correct identity:
$$\sec^2(\tan^{-1}(\frac{x}{2}))\equiv1+\tan^2(\tan^{-1}(\frac{x}{2}))\equiv1+\left(\tan(\tan^{-1}(\frac{x}{2}))\right)^2\equiv1+\frac{x^2}{4}$$
Edit
Thinking that $\sec^2\theta$ means $\sec^2\times\theta$ is meaningless- $\sec^2$ or indeed any other trigonometrical function has no meaning unless it has an argument, ie the value the that is an input into the trigonometical function (in your case $\tan^{-1}\frac{x}{2}$).

Body:
Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?

If you replace $\sec^2$ with $\tan^2 + 1$, it should be $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$, right?

However, it seems that it is not multiplication here : $\sec^2 (\tan^{-1} (\frac{x}{2}))$, when the $1$ becomes independent of the $\tan^2$ somehow.

Could someone explain why?


问:
How this system transferred into this state variables
Take a look at this transfer function:
$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 2s + 1}
$$

The author of the book I'm reading says, the transfer function can be represented as

$$
\frac{Y(s)}{U(s)} = \frac{1}{s^2 + 2s + 1} = \frac{1}{(s+1)} \frac{1}{(s+1)}
$$

The block diagram is

Then the author states the following equations

$$
\begin{align}
\dot{x}_1 &= x_2 - x_1 \\
\dot{x}_2 &= -x_2 + u \\
y &= x_1
\end{align}
$$
Notice: the last equation is given. I'm aware of the fact that state variables representation is not unique but I failed to understand this method. How did he come up with this representation?

答:
The direct translation of the transfer relation is $(D+1)^2y=u$, with $D$ as the derivation operator. Now name the intermediate result $z=(D+1)y$ with consequently $(D+1)z=u$, then one can read this as a first order system
\begin{align}
y'&=-y+z\\
z'&=-z+u
\end{align}
which are the claimed equations.

The state here is $(y,z)$. If you want the systematic variables $\dot {\bf x}=f({\bf x},{\bf u})$, $y=g({\bf x},{\bf u})$, then you need to assign ${\bf x}=(x_1,x_2)$ to this state in some order, so that then $y$ only occurs as a function of the state ${\bf x}$


问:
Does there exist a bijection that is not eventually equal to the identity, but has a finite number of "crosses"?
Consider a  bijection $f: \mathbb{N} \to \mathbb{N}$:

We define the (possible infinite) number of "crosses" or "intersections" by the following:

For every pair $i,j$ with $i<j, $ define:

$$
\alpha_{ij}=\begin{cases}
0&\ \text{if} \ f(j) > f(i)\\[8pt]
1&\ \text{if} \ f(j) < f(i)\\[8pt]
\end{cases}
$$

Then the number of (possibly infinite) intersections is $\ s(f) = \displaystyle \sum_{i=1}^\infty \left( \sum_{j=i+1}^\infty\alpha_{ij} \right)$.

(s depends only on the bijection $f$).

Also, define a bijection eventually equal to the identity to be one in which $ \exists N$ such that $f(n) = n \quad \forall n \geq N$.

My conjecture is that a bijection $f: \mathbb{N} \to \mathbb{N}$ is eventually equal to the identity $ \iff s$ is finite, but I don't know how to prove this formally.

Note that my conjecture would be false with bijections $f: \mathbb{Z} \to \mathbb{Z}$, for example:

The pigeonhole principle came to mind, but I'm not sure if it's useful or necessary for a proof.

答:
The following is more precise details of the sketch in my comment.

Suppose $s$ is finite. Then this means that eventually there are no crosses. In other words there is some $n_0$ such that $f$ is increasing on $[n_0,\infty)$. Now let $n_1>n_0$ be such that for all $y\leq n_0$ there is some $x\leq n_1$ such that $f(x)=y$. We claim that $f$ is the identity after $n_1$.

So fix $x>n_1$ and suppose $f(x)\neq x$.

Case 1. $f(x)<x$. Then there must be some $y<x$ such that $f(y)\geq x$, since otherwise the  interval $[1,x]$ is mapped to $[1,x-1]$. Note that $y>n_0$ since $f$ maps $[1,n_0]$ into $[1,n_1]$ and $n_1<x\leq f(y)$. So $n_0<y<x$ and $f(x)<x\leq f(y)$.  This contradicts the assumption that $f$ is increasing after $n_0$.

Case 2. $f(x)>x$. By similar reasoning, this forces $x>n_0$ and there must be $y>x$ such that $f(y)\leq x$. Contradiction.


答: You don't need to know how $a$ is defined. Let $a : \Bbb{N} \to \Bbb{Q}$ be any bijection. Given any real number $x$ and $\epsilon > 0$, there are infinitely many rational numbers $q \neq x$ with $|x - q| < \epsilon$ and as each such $q$ is $a_k$ for some $k$, $x$ is a limit point of the sequence $a_1, a_2, \ldots$. You can describe a subsequence that converges to $x$ explicitly, by defining $n_k$ to be the smallest $n$ such that $a_n \neq x$ and $|x - a_n| < \frac{1}{k}$ for $k = 1, 2, \ldots$. Then $\lim_{k \to \infty}a_{n_k} = x$.

Title: Limit points / cluster points from $a:$ $\mathbb {N} \rightarrow \mathbb {Q}$

Body:
So i have the bijection $a:$ $\mathbb {N} \rightarrow \mathbb {Q}$.

I was asked to find the limit points / cluster points of the sequence $a_n = a(n)$.

I know that this bijection exists since $\mathbb {N}$ and $\mathbb {Q}$ are both countable. After reading up in my script, I know how they constructed it using Cantor's diagonal argument. This is where I'm stuck though, since I just read that apparently this sequence has every real number as a limit point of the sequence. How can this be proven?


问:
What is the solution of $\cos(x)=x$?
There is an unique solution with $x$ being approximately $0.739085$. But is there also a closed-form solution?

答:
Not a closed form but a way to generate very good approximations of the Dottie number.

Since the solution is close to $\frac \pi 4$, we have for
$$y=x-\cos(x)$$ $$y=\frac{\pi -2 \sqrt{2}}{4} +\left(1+\frac{1}{\sqrt{2}}\right) \left(x-\frac{\pi }{4}\right)+\frac{1}{\sqrt{2}}\sum_{n=2}^\infty\frac{\sin \left(\frac{\pi  n}{2}\right)-\cos \left(\frac{\pi  n}{2}\right)}{n!}\left(x-\frac{\pi }{4}\right)^n$$ Truncating to some order $O\left(\left(x-\frac{\pi }{4}\right)^n\right)$ and using series reversion, we should get things like
$$x=\frac \pi 4+\frac{32 \left(11482+8119 \sqrt{2}\right) t}{\left(2+\sqrt{2}\right)^{11}}-\frac{16
   \left(4756+3363 \sqrt{2}\right) t^2}{\left(2+\sqrt{2}\right)^{11}}+\frac{32
   \left(5333+3771 \sqrt{2}\right) t^3}{3
   \left(2+\sqrt{2}\right)^{11}}+O\left(t^4\right)$$ where $t=\frac{1}{4} \left(\sqrt{2}-2\right) \left(4 y+\pi -2 \sqrt{2}\right)$. Making $y=0$ and using this very truncated series would give
$$x \sim 0.739085133238$$ to be compared to the exact
$0.739085133215$

Playing with the $n$ of $O\left(\left(x-\frac{\pi }{4}\right)^n\right)$, we could get the following results
$$\left(
\begin{array}{cc}
n & x_{(n)} \\
 1 &  \color{red}{0.739}536133515238 \\
 2 &  \color{red}{0.739}100520482138 \\
 3 &  \color{red}{0.739085}585917040 \\
 4 &  \color{red}{0.7390851}49503943 \\
 5 &  \color{red}{0.739085133}811963 \\
 6 &  \color{red}{0.7390851332}38222 \\
 7 &  \color{red}{0.73908513321}6073 \\
 8 &  \color{red}{0.7390851332151}98 \\
 9 &  \color{red}{0.73908513321516}2 \\
 10 & \color{red}{0.739085133215161}
\end{array}
\right)$$

Edit

For the fun of it, using the $\large 1,400$ years old approximation
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$ solving the cubic equation
$$x^3+4 x^2+\pi ^2 x-\pi ^2=0$$ gives as an approximation
$$x\sim -\frac{2}{3} \left(2+\sqrt{3 \pi ^2-16} \sinh \left(\frac{1}{3} \sinh
   ^{-1}\left(\frac{128-63 \pi ^2}{2 \left(3 \pi
   ^2-16\right)^{3/2}}\right)\right)\right)=0.738305$$ that is to say a relative error of $0.1$%.

Title: What is the solution of $\cos(x)=x$?


问:
Inverse of a symmetric positive diagonally dominant, doubly stochastic matrix
I am trying to prove that the inverse of a $\mathbf{positive}, \mathbf{symmetric}, \mathbf{diagonally-dominant}, \mathbf{doubly-stochastic}$ matrix and having the following $\mathbf{property (3)}$ -   has positive diagonal elements and negative non-diagonal elements.

I'll describe the below definitions.

1)Diagonally dominant matrix - $|a_{ii}| > \sum_{i \neq j}|a_{ij}| \quad \forall i$

2) Doubly stochastic matrix - sum of entries in any row or column is unity.  i.e., $\quad \sum_{i} a_{ij} = \sum_{j} a_{ij} = 1 \quad \forall i,j$

3) Let $b_1, b_2,..,b_{N-1}$ be a decreasing sequence of positive numbers. The non-diagonal elements of the matrix satisy:  $$\qquad a_{ij} = b_{|i-j|} \qquad \forall i \neq j$$

Need to show that the inverse of such a matrix $\mathbf{A}$ will have positive diagonal and negative non-diagonal elements.

I have come till proving that the row sum and column sum of $\mathbf{A}^{-1}$ is 1. But not able to prove further.

答:
That was an intriguing conjecture, but unfortunately it is not always true.

Consider a 5x5 matrix $A$ of the form you described with $b_1=0.15$, $b_2=b_3=0.05$ and $b_4=0.0125$.

The inverse of this matrix has several strictly positive off-diagonal elements ($a^{-1}_{15}=a^{-1}_{51}\approx0.0189$ and $a^{-1}_{24}=a^{-1}_{42}\approx0.004$).


问:
What is the probability that there is an error in both blocks?
A computer program consists of two blocks written independently by two different programmers.   The  first  block  has  an  error  with  probability  $0.2$,  the  second  block  has  an  error  with probability $0.3$.  If the program returns an error, what is the probability that there is an error in both blocks?

答:
Hint:

$\Pr(X\mid Y) = \dfrac{\Pr(X\cap Y)}{\Pr(Y)}$

and in particular letting $X=A\cap B$ and $Y= A\cup B$ and noting that $A\cap B$ is a subset of $A\cup B$ you have

$$\Pr(A\cap B\mid A\cup B) = \dfrac{\Pr(A\cap B)}{\Pr(A\cup B)}$$


答:
Consider this figure:

So your answer is $\frac{0.06}{.2 + .3 - .06}$ which is simply the intersection over the union, as shown by JMoravitz.

Title: What is the probability that there is an error in both blocks?

Body: A computer program consists of two blocks written independently by two different programmers.   The  first  block  has  an  error  with  probability  $0.2$,  the  second  block  has  an  error  with probability $0.3$.  If the program returns an error, what is the probability that there is an error in both blocks?


问:
Find the function that minimises the following integral
How can I find the function $f(t)$, $t\geq0$ continuous and everywhere differentiable that minimises the following integral

$$L = \int_0^\infty f(t)^2 + f'(t)^2dt$$

such that $f(0)=x$.

Intuitively I am asking what is the function $f$ that starts at $x$ and wants to approach $0$ the fastest but keeping the derivative smaller. My intuition tells me that somehow the function $f$ should be a decreasing exponential. But how to prove it? how does one tackle this problem?

Here is what I tried to do:
I tried to think of a similar discrete problem: find the sequence $h_n$ that minimises

$$L = \sum_{i=1}^\infty h_i^2 + (h_{i-1}-h_i)^2$$ such that $h_0=x$

Some observations:
If $x=0$ then trivially $h_i=0$ $\forall i$

On the other hand, if $h_i=0$ $\forall i$, then $L = x^2$. (here the function would suddenly drop to $0$)

If instead we had gradually decreased it with one step: $h_1=\frac{x}{2}$ and $0$ otherwise, then $L=\frac{x^2}{4}+\frac{x^2}{4}+\frac{x^2}{4}=\frac{3x^2}{4}$ which is lower than the previous example, showing that it is better to decrease it gradually...

Then my intuition tells me that if $h_i = \alpha h_{i - 1}$ for $\alpha<1$, then $h_i=x\alpha^i$ and $h_{i-1}-h_i = x\alpha^{i-1}(1-\alpha)$.

For this case, this means that $L = \sum_{i=1}^\infty (x\alpha^i)^2 + (x\alpha^{i-1}(1-\alpha))^2=x^2\sum_{i=1}^\infty (\alpha^2)^i+x^2(1-\alpha)^2\sum_{i=1}^\infty (\alpha^2)^{i-1}$

$$L=x^2\frac{\alpha^2}{1-\alpha^2}+x^2(1-\alpha)^2\frac{1}{1-\alpha^2}=\frac{x^2}{1-\alpha^2}(2\alpha^2-2\alpha + 1)$$

Now if we try to find $\alpha$ that minimises this expression we can take its derivative

$\frac{dL}{d\alpha}=x^2\frac{(4\alpha-2)(1-\alpha^2)-(2\alpha^2-2\alpha + 1)(-2\alpha)}{(1-\alpha^2)^2}=0 \Leftrightarrow (4\alpha-2)(1-\alpha^2)-(2\alpha^2-2\alpha + 1)(-2\alpha)=0 \Leftrightarrow 4\alpha-4\alpha^3-2+2\alpha^2+4\alpha^3-4\alpha^2+2\alpha=0\Leftrightarrow -2\alpha^2+6\alpha-2=0$

$\alpha=1+\frac{1-\sqrt{5}}{2}$ which seems to be related to the golden ratio

答:
We assume that $x>0$. The other case is similar.

Claim 1: It suffices to assume that $f(t) <x$ for all $t>0$.

Proof of claim: Let be $f$ be a continuously differentiable function so that $f(t_1) >x$ for some $t_1 >0$. We show that this $f$ cannot attain the minimum value.

Since the integral $\int_0^\infty f^2$ has to be finite, there must be $t_2 > t_1$ so that $f(t_2) <x$. By continuity, there is $t_0 \in (t_1, t_2)$ so that $f(t_0) = x$. Then $g(t) = f(t+ t_0)$ satisfies $g(0) = x$ and
$$\int_0^\infty (g^2 + g'^2 ) = \int_{t_0}^\infty (f^2 + f'^2) < \int_{0}^\infty (f^2 + f'^2).$$
So $f$ cannot attain the minimum.

Claim 2: it suffices to assume $f(t) > -x$ for all $t>0$

Proof of claim: similar as in the proof of claim 1: find $t_0$ so that $f(t_0) = -x$ and consider $g(x) = -f(t+t_0)$.

In particular, it suffices to consider only those $f$ that is bounded.

Let $h = f - xe^{-t}$. Then $h(0) = 0$. Also

\begin{align}
\int_0^\infty f^2 + (f')^2 &= \int_0^\infty (h^2 +2xh e^{-t} + x^2 e^{-2t}) + ((h')^2 -2xh' e^{-t} + x^2 e^{-2t})\\
&= \int_0^\infty (h^2 + (h')^2) + 2x \int_0^\infty (h-h') e^{-t} + 2x^2\int_0^\infty e^{-2t} \\
&= \int_0^\infty (h^2 + (h')^2) - 2x \int_0^\infty (he^{-t})'  + x^2.\\
&=x^2 + 2xhe^{-t}\bigg|_{t=0}^\infty + \int_0^\infty (h^2 + (h')^2), \\
&=x^2 + \int_0^\infty (h^2 + (h')^2),
\end{align}
where in the last equality we used $h(0) = 0$ and $he^{-t}$ tends to $0$ as $t\to \infty$ since $f$ is bounded. Thus the problem is transformed into this:

Find the minimum of
$$x^2 + \int_0^\infty (h^2 + h'^2)$$
along all continuously differentiable function $h$ with $h(0) = 0$.

Since for this question it is obvious that $h(t) = 0$ is the (global) minimum, $f = xe^{-t}$ is the global minimum for your problem.


问:
What is the coefficient of $\sqrt{2}$ in $f(\sqrt{2})$ where $f(x) = 3 - x + 2x^2 - 5x^3 +4x^4?$

What is the coefficient of $\sqrt2$ in $f(\sqrt2)$ where $f(x) = 3 - x + 2x^2 - 5x^3 +4x^4?$

Hi
I'm studying polynomial rings.
I don't get what the question is asking me to do.
Should I just plug in $\sqrt2 $ in $f(x)$?
Isn't the definition of coefficient the number in front of a variable? How can there be a coefficient of a number, i.e. $\sqrt2$?

Am I missing something..

答: Well, when you compute $f(\sqrt{2})$ you'll obtain a result of the form $a + b \sqrt{2}$, where $a,b$ are integers. You are just being asked to say what $b$ is.


问:
Proof $(A-\{a\}) \sim (B - \{b\}) $
I must prove the following:

"let $A$ and $B$ two set, with $A \neq \emptyset$ and $B \neq \emptyset$ and $A \sim B$, and $a \in A$ and $b \in B$, then $(A-\{a\}) \sim (B - \{b\})$

The symbol $\sim$ is used for "equipotency"!!

Thanks in advance!

答:
Let $f:A\to B$ a bijection. If $f(a)=b$, we may of course just restrict $f$ to $A\smallsetminus\{a\}$. Otherwise, there is some unique $c\in A\smallsetminus\{a\}$ and some unique $d\in B\smallsetminus\{b\}$ such that $f(a)=d$ and $f(c)=b.$ (Why?) How can we use these facts, and the function $f,$ to construct the bijection we're looking for?

In general, we don't need the casewise approach that I was hinting at, above. If $f:A\to B$ is a bijection and $a\in A,b\in B$, then we know there is a unique $c\in A$ and a unique $d\in B$ such that $f(c)=b$ and $f(a)=d.$ (Why?) You should be able to show that $c=a$ if and only if $b=d.$

Now, given such an $f,$ let's construct a function $g:A\to A$ defined by $$g(x)=\begin{cases}c & \text{if }x=a,\\a & \text{if }x=c,\\x & \text{if }x\in A\smallsetminus\{a,c\}.\end{cases}$$ Now, what can you say about $f\circ g:A\to B,$ given what you know about $f$ and how we defined $g$? What about the restriction of $f\circ g$ to $A\smallsetminus\{a\}$?

Title: Proof $(A-\{a\}) \sim (B - \{b\}) $


问:
What happens if the greater limit becomes smaller when you change them for u-substitution?
If I am doing u-substitution on a definite integral, and my upper limit is $b$ and my lower limit is $a$, and I change the limits to match the u, what happens if my $b$ turns out to be lower than my $a$ (so the now b is smaller than a)?

Does my new b still go at the top, or must I maintain the order?

For example, if $u = 1 + cosx$, and $b = \frac{\pi}{2}$, and $a = \frac{\pi}{3}$, when I change them, I get $b = 1$, and $a = 1.5$.

Now $a$ is greater than $b$.

So which goes at the top of the integral sign (which is the upper limit)?

答: $ \int_b^a u(x) dx= - \int_a^b u(x) dx.$


问:
Are there other kinds of bump functions than $e^\frac1{x^2-1}$?
I've only seen the bump function $e^\frac1{x^2-1}$ so far.  Where could I find examples of functions $C^∞$ on $\mathbb{R}$ that are zero everywhere except on $(-1,1)$?

Are there others that do not involve the exponential function?  Are there any with a closed form integral?  Is there a preferred function?

答:
Juan Arias de Reyna (2017) shows the existence of a bump function with the characteristics you mention. The paper is in the arXiv (1702.05442 [math.CA]). Specifically, as he says, it is a smooth function $\varphi$ such that

(a) $\mathrm{supp}(\varphi) = [-1,1]$,

(b) $\varphi(t) > 0$ for any $t \in (-1, 1)$,

(c) $\varphi(0) = 1$,

(d) and there is a constant $k > 0$ such that for any $t \in \mathbb{R}$

$$\varphi'(t) = k(\varphi(2t+1) - \varphi(2t-1))$$

(e) later is shown that must be $k=2$.

That is, a function that is non-null around zero, is 1 at zero, and satisfies that funny condition. That condition, as a matter of fact, comes from the intuition that such a bump function's derivative looks like two transformed copies of the original function.

Though there's no neat formula for the function, the author gives four different expressions for it.

As an integral of a power series (involving its Fourier transform);

as a limit to a sequence of functions defined by convolution;

as “a limit of a sequence of step functions”;

as the measure of a subset of $[0,1]^{\mathbb{N}}$ parametrized by its argument – also, this way of looking at the function lets you interpret it as a probability, or a measure in itself (see the paper for details).

This are all quite involved, but the proposed function has the benefit of being entirely constructive, as opposed to the usual ones. Bear in mind, splitting the line based on inequalities is usually non-constructive (though in this case they might be).


问:
$u$ is harmonic and $v = \phi(u)$. Prove $v$ is subharmonic.
Problem Statement:
$\phi:\mathbb{R}\rightarrow \mathbb{R}$ be smooth and convex. Assume $u$ is harmonic and $v = \phi(u)$. Prove $v$ is subharmonic.

My attempt at a following solution is below:

We hope to show that $\Delta \phi(u) \geq 0$. We have:
\begin{equation}
        \Delta\phi(u) = \sum_{i=1}^n\phi'(u)\frac{\partial^2 u}{\partial x_i^2} = \phi'(u)\Delta u = 0
    \end{equation}

So we can atually conclude that $\phi(u)$ is harmonic. This this correct? It seems too simple to be true.

Any feedback would be greatly appreciated!

答:
Since $\phi$ is convex, one has $\phi''(s)\ge0$. In fact,
\begin{eqnarray}
\Delta\phi(u) &=& \sum_{i=1}^n\frac{\partial}{\partial x_i}\frac{\partial \phi(u)}{\partial x_i} \\
&=& \sum_{i=1}^n\frac{\partial}{\partial x_i}\phi'(u)\frac{\partial u}{\partial x_i} \\
&=& \sum_{i=1}^n\bigg[\frac{\partial \phi'(u)}{\partial x_i}\frac{\partial u}{\partial x_i} +\phi'(u)\frac{\partial^2 u}{\partial x_i^2} \bigg]\\
&=& \sum_{i=1}^n\bigg[\phi''(u)\left(\frac{\partial u}{\partial x_i}\right)^2 +\phi'(u)\frac{\partial^2 u}{\partial x_i^2} \bigg]\\
&=& \phi''(u)\sum_{i=1}^n\left(\frac{\partial u}{\partial x_i}\right)^2 +\phi'(u)\sum_{i=1}^n\frac{\partial^2 u}{\partial x_i^2}\\
&=&\phi''(u)\sum_{i=1}^n\left(\frac{\partial u}{\partial x_i}\right)^2 \\
&\ge&0
\end{eqnarray}
which means that $\phi(u)$ is subharmonic.


问:
Looking for an equation that roughly generates the given graph
I would like to precursor this with the fact that I am in no way a mathematician. Anyways, I am looking help finding an equation, series of equations, or any information on a good starting point for generating a graph roughly like the one in the picture below. Also, I'd like to be able to rotate it around 0,0. Any help is greatly appreciated.

Example of what I would roughly like to generate:

答:
The solution by @robert timmer-arends is good.

Here is a "a posteriori" possible full justification.

In fact one can relate the curve obtained to a general family, those of hypotrochoids, here given by:

$$z=5e^{\color{red}{1}it}-ae^{\color{red}{3}it} \ \ \iff \ \ \begin{cases}x&=&5\cos(t)-a\cos(3t)\\y&=&5\sin(t)-a\sin(3t) \end{cases}\tag{1}$$

represented on the left of the figure for values of $a$ ranging from $0$ to $5$. For the ratio $\color{red}{3/1}$, see remark below.

These curves can in a second step be "skewed" using a linear transform. Here (see the right part of the figure corresponding to $a=5$), we have taken:

$$\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}\ \ 1&-\tfrac12\\-1& \ \ \ 1 \end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\tag{2}$$

Plugging (1) into (2), no surprise that we find expressions close to those of Robert.

Remark: The ratio $\color{red}{3/1}$ in (1) is compulsory for having $3-1=2$ loops. Had we taken $z=5e^{\color{red}{1}it}-ae^{\color{red}{13}it}$, we would have generated $13-1=12$ loops.


问:
differentiating a complex function $f=z\bar{z}$ w.r.t. $z$
If $z$ is a complex number, what is the derivative $df/dz$, where $f=z\bar{z}$? The straight differentiation gives me $df/dz=\bar{z}+z(d\bar{z}/dz)$. Here,

$d\bar{z}/dz = \lim_{\Delta z\rightarrow 0} \frac{\overline{z+\Delta z}-\bar{z}}{\Delta z}=\ldots = \lim_{\Delta z\rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}$

If I take $\Delta z$ as real, then this last limit is one and $df/dz=z + \bar{z} = 2\,{\rm Re}(z)$. If I take $\Delta z$ as imaginary, this limit is -1 and then $df/dz = \bar{z}-z=-2\,{\rm Im}(z)$. It looks as if the function $f$ is not differentiable, am I correct?

Just to show you what I need. I have a function $F=\int_V \left[A\phi^2({\bf r}) d{\bf r}+\ldots\right]\,dV$ that I need to transform into the Fourier space (there are extra terms that don't allow me to stay in real space). By the Parseval's identity, the function above is $F=\int_\Omega \left[A|\tilde\phi({\bf k})|^2+\ldots\right]\, d{\bf k}$. Now, I need to differentiate the integrand w.r.t. $\tilde\phi({\bf k})$ which leads to the problem above. Is there any trick that would help here or I am wrong completely?

答: As was pointed out, this function is nowhere analytic. However, is it differentiable at $0$ and this is the only point where it is so. It's one of those strange-case functions that's differentiable at isolated points.


问:
How does adding a primitive root of unity to a number field affect the ring of integers?
We know that if $\xi$ is a primitive $n^\text{th}$-root of unity, then the ring of integers $\mathcal{O}_{\mathbb{Q}(\xi)}$ of $\mathbb{Q}(\xi)$ is $\mathbb{Z}[\xi]$.

Can we generalise this result to say much about the ring of integers $\mathcal{O}_{K(\xi)}$ of $K(\xi)$, where $K / \mathbb{Q}$ is some finite algebraic extension?

Is it the case that $\mathcal{O}_{K(\xi)} = \mathcal{O}_{K}[\xi]$?

If this is not generally true, do we have a characterisation of circumstances where this may hold?

Failing that, do we have an alternate description of $\mathcal{O}_{K(\xi)}$ in terms of $\mathcal{O}_{K}$?

I would appreciate any comments, or even just a reference for these kinds of results.

答:
By a result in Hilbert's report, the ring of integers in a compositum KL of normal extensions is the obvious one generated by the rings of integers in $K$ and $L$ if the discriminants of $K$ and $L$ are coprime.

If the discriminants are not coprime, the result is fals in general. In particular, the ring of integers in ${\mathbb Q}(\sqrt{3},\sqrt{-1})$ is not ${\mathbb Z}[\sqrt{3}][\sqrt{-1}]$.


问:
Volume with spherical polar coordinates
Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.

I convert to spherical polar coordinates.

$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$

$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$

I can't figure out how to take $\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$, which makes me think I made a mistake somewhere.

EDIT: Thanks for all the answers.

Now I understand how the limits of $\theta ,r,z$ works.

I don't fully understand where the function "disappear".

$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$

Why isn't it then:

$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$

答:
This is much easier to solve in cylindrical coordinates. $$x=r\cos\theta\\y=r\sin\theta\\z=h$$
Then the limits for $r$ are $0$ and $1$, the limits for $\theta$ are from $-\frac\pi4$ to $\frac{3\pi}4$, and the limits for $h$ are $0$ and $4-r^2$.
With these, $$V=\int_{-\frac\pi4}^{\frac{3\pi}4}d\theta\int_0^1dr\cdot r\int_0^{\sqrt{4-r^2}}dh$$

Note see comment below. Since $y>0$, the lower limit for $\theta$ is $0$, not $-\pi/4$

Title: Volume with spherical polar coordinates


问:
Elaboration of rules for Gomory’s cutting plane algorithm.
The Gomory’s cutting plane algorithm is as follows:

I’m also looking at the theorem:

Theorem 2 (Gomory). Suppose Gomory's algorithm is implemented by:

using the lexicographic simplex algorithm for LP solving and

deriving Gomory cut from the fractional variable with the smallest index.

Then the algorithm will terminate in finite numbers of iterations.

Now in the example, I have the following optimal LP tableau:

\begin{array} {|r|r|}\hline \text{Basic vars.} & x_1 & x_2 & x_3 & x_4 & x_5 &  \\ \hline x_3 & 0 & 0 & 1 & -1/3 & 1/3 & 7/3 \\ \hline x_1 & 1 & 0 & 0 & 1/3 & -1/3 & 5/3 \\ \hline x_2 & 0 & 1 & 0 & 1/3 & 2/3 & 20/3 \\ \hline  & 0 & 0 & 0 & -1 & -1 & -15 \\ \hline  \end{array}

The example says that I should generate a Gomory’s cut from row $1$ of the tableau, i.e. the one with $x_3$ as the basic variable, as it has the smallest fractional index. But amongst the three basic variables $x_1,x_2,x_3$ that are fractional, shouldn’t $x_1$ be the one with the smallest fractional index? Am I misunderstanding something?

答: I'm working on this too so I wouldn't trust my answer completely. However, I think you should actually derive a Gomory cut from the fractional variable with the largest index.


问:
How to find the position vector given $3$ colinear points?
$P$, $Q$ and $R$ are $3$ collinear points in space, with position vectors $p$, $q$ and $r$ respectively. Point $Q$ lies between $P$ and $R$ such that the distance from $P$ to $Q$ is twice the distance from $Q$ to $R$.

Which of the following is true?

How can I compute this? is $PQ$ = $2QR$?

答:
Sketch the situation on paper as if it were in 2D. Now $ \vec r$ = $\vec{p} + \vec{PR} \:$ and $\vec{PR} = \vec{pq} + \frac{1}{2} \cdot \vec{pq}$. By writing $\vec{pq} = q - p$, you obtain: $\vec{r} = \frac{3}{2} \vec{q} - \frac{1}{2}\vec{p}$.

(Recall that $\vec{pq} = \vec q - \vec p$).

Maybe the following image clears things up: