问:
Integrate: $\int\limits_0^{2\pi } {\exp \left( {ar\cos ({\psi _a} - \theta ) + b{r^2}\cos ({\psi _b} - \theta )....} \right)} d\theta $?
I am trying to find the following integral in the close form:

$\int\limits_0^{2\pi } {\exp \left( {ar\cos ({\psi _a} - \theta ) + b{r^2}\cos ({\psi _b} - \theta ) + c{r^3}\cos ({\psi _c} - \theta )} \right)} d\theta $

Where, $a,b,c,r$ is non-negative real number.

I was trying to use the form of zero-order modified Bessel identity that is,
Assuming $a > 0$,
$$ I_0(x) = \frac{1}{\pi}\int_0^\pi e^{x\cos\theta}d\theta $$
Where $I_0(x)$ is a modified Bessel function of the first kind.

My approach:

Converting ${ar\cos ({\psi _a} - \theta ) + b{r^2}\cos ({\psi _b} - \theta ) + c{r^3}\cos ({\psi _c} - \theta )}=Kcos({\psi _K}-\theta)$. Then use the identity mentioned above. But I am not sure if it is possible to convert ${ar\cos ({\psi _a} - \theta ) + b{r^2}\cos ({\psi _b} - \theta ) + c{r^3}\cos ({\psi _c} - \theta )}=Kcos({\psi _K}-\theta)$?

Is there any other approach that I might take? Any suggestion would help.

答:
I shall try to make it completely general.

Changing notations, the agument of the exponential function is
$$S=\sum_{n=1}^p a_n r^n \cos (\psi_n-\theta) $$ Expanding the cosines, we have
$$S=\Big[\sum_{n=1}^p a_n r^n \cos (\psi_n)\Big]\cos (\theta)+\Big[\sum_{n=1}^p a_n r^n \sin (\psi_n)\Big]\sin (\theta)$$ that is to say
$$S=A_c \,\cos (\theta)+A_s \,\sin (\theta)$$ and then the problem is
$$\int_0^{2\pi} \exp\Big[A_c \,\cos (\theta)+A_s \,\sin (\theta)\Big]\,d\theta=2 \pi  I_0\left(\sqrt{A_c^2+A_s^2}\right)$$


问:
Cyclic inequality with a+b+c=3
Let $a;b;c$ be non-negative real numbers such that $a+b+c=3$
Prove:
$(a+b)(b+c)(c+a) \geq (a+bc)(b+ca)(c+ab)$

Can someone prove that problem without using $pqr$ or $uvw$ method? Thanks so much.

答:
Using the AM-GM inequality, we have
$$(a + bc)(b + ca) \leqslant \left[\frac{(a + bc)+(b + ca)}{2}\right]^2 = \frac{(a + b)^2(c + 1)^2}{4} .$$
Therefore
$$(a+bc)(b+ca)(c+ab) \leqslant \frac{\displaystyle \prod (a+b) \prod (a+1)}{8}.$$
It remains to prove that
$$(a+1)(b+1)(c+1) \leqslant 8.$$
Which is true because
$$(a+1)(b+1)(c+1) \leqslant \left(\frac{a+1+b+1+c+1}{3}\right)^3=8.$$
Done.

ContentLicense: CC BY-SA 4.0

Title: Cyclic inequality with a+b+c=3


问:
How to Use Baire's Theorem to Show a set is not Nowhere Dense?
Given a sequence of closed, successively inclusive sets,in the complete metric space, R, the real numbers.  I wish to use Baire's Catagory Theorem (BCT) to show that at least one of them is not nowhere dense (nd) .
A corollary to BCT, I understand, states that a sequence of nd sets in R has an nd union. If the union were not nd, then at least one of the sequence of sets would not be nd.

So,  if I can show the closure of the union to have a non-empty interior, then one of the closed sets would not be nd, by the corollary.
I believe the union is closed, because the closed sets grow inclusively, but I have not shown this to be true.

Further, if I can show the union has a non-empty interior, then, with the corollary, job done!   Am I on the right track?

答:
There is no guarantee that the union of an increasing nest of closed sets is closed: consider the sets $\left[\frac1n,1\right]$ for $n\in\Bbb Z^+$, for instance.

You also don’t want to look at the closure of the union. If $\Bbb Q=\{q_n:n\in\Bbb N\}$ is an enumeration of the rationals, and we set $F_n=\{q_k\in\Bbb Q:k\le n\}$ for $n\in\Bbb N$, then each $F_n$ is a nowhere dense closed set, but $\operatorname{cl}\bigcup_{n\in\Bbb N}F_n=\Bbb R$, which certainly has non-empty interior. You need to look just at the union of your nest. In this example that union is $\Bbb Q$, which does indeed have empty interior, as ensured by the Baire category theorem.

If you can show that the union of your closed sets has non-empty interior, then you can indeed conclude that on of the closed sets has non-empty interior.


问:
Proving $ \sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + 4} \leq \frac{3}{3 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})}$
Problem proposed for JBMO practice in symmetrical inequalities (Chebyshev, rearrangement):

For every positive real numbers $a, b, c$, for which $a + b + c = 3$ we have:
$$\sum_{cyc} \frac{1}{a^6 + b^6 + 3c^3 + 4} \leq \frac{3}{3 + 2(\sqrt{ab} + \sqrt{bc} + \sqrt{ca})}$$

My attempt:

Step 1

AM-GM states that $$a^6 + b^6 \geq 2a^3b^3$$

Therefore, $$ LHS \leq \sum_{cyc} \frac{1}{2a^3b^3 + 2c^3 + 2 + c^3 + 2}$$

Step 2

AM-GM states that $2a^3b^3 + 2c^3 + 2 \geq 6abc$ and that $c^3 + 1 + 1 \geq 3c$

This means that $$ LHS \leq \sum_{cyc} \frac{1}{6abc + 3a} = \frac{1}{3} \sum_{cyc} \frac{1}{2abc + a}$$

Step 3

Rearrangement inequality states that $$\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \leq a + b + c = 3$$
This means: $$ RHS \geq \frac{1}{3}$$

Remaining of the proof and further ideas

We should prove $$ \sum_{cyc} \frac{1}{2abc + a} \leq 1$$

I should probably apply Chebyshev but I could get no good result after applying its variants...

Thanks in advance!

Body:
By AM-GM, we have $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \le a + b + c = 3$. Thus,
$\mathrm{RHS} \ge \frac{1}{3}$.

By Chebyshev's sum inequality, we have
\begin{align}
a^3 + b^3 + c^3 - (a+b+c) &= (a-1)(a^2 + a) + (b-1)(b^2+b) + (c-1)(c^2+c)\\
&\ge \frac{1}{3}(a-1 + b-1 + c-1)(a^2+a + b^2+b + c^2 + c)\\
& = 0.
\end{align}
Thus, we have $a^3 + b^3 + c^3 \ge a + b + c = 3$. Thus, we have
\begin{align}
a^6 + b^6 + 3c^3 + 4 &= (a^6 + 1) + (b^6 + 1) + 3c^3 + 2\\
&\ge 2a^3 + 2b^3 + 3c^3 + 2\\
&= 2(a^3 + b^3 + c^3) + c^3 + 2\\
&\ge 8 + c^3.
\end{align}
Thus, $\mathrm{LHS} \le \sum_{\mathrm{cyc}} \frac{1}{8 + c^3}$.

It suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{1}{8 + c^3} \le \frac{1}{3}$$
or (tangent line method)
$$\sum_{\mathrm{cyc}} \left(\frac{1}{8 + c^3} - \frac{1}{9} + \frac{1}{27}(c-1) \right) \le 0$$
or
$$\sum_{\mathrm{cyc}} \frac{(c^2-2c-5)(c-1)^2}{27(8 + c^3)} \le 0$$
which is true since $x^2-2x-5 < 0$ for all $0 \le x \le 3$.


问:
True or false: $\lim_{n\to\infty}\int_0^1\frac{\sin(2n\pi x)}{\ln x}\,dx=\frac\pi2$
Numerical result shows that

$$
\lim_{n\to\infty}\int_0^1\frac{\sin(2n\pi x)}{\ln x}\,dx=\frac\pi2,\qquad
\lim_{n\to\infty}\int_0^1\frac{\sin\left((2n+1)\pi x\right)}{\ln x}\,dx=-\frac\pi2.
$$

I want to know if it's true or my computer is just lying to me.

What I've tried:

Something like here and here. Maybe it can work but I don't know how to deal with the limit. It just ruins everything.

Residue theorem for
$$f:z\mapsto\frac{\exp(i\cdot2n\pi z)}{\log z}$$
using a curve around the $1/4$ unit disk without center and outer edge: $(\epsilon_1, 0)\to(1-\epsilon_2, 0)\to(0, 1-\epsilon_2)\to(0,\epsilon_1)\to(\epsilon_1, 0)$.

It didn't work. Maybe using directly $z\mapsto \sin (2n\pi z)/\log z$ can lead to something but I didn't try.

Substitution using $t = 2nx$, then dividing the integration interval into $2n$ parts and trying to get a Riemann sum. I failed.

Any other ideas?

答:
Both results follow from the well known Dirichlet integral $\int_0^{\infty}\frac{\sin x}{x}=\frac{\pi}{2}$

We first notice that $g(x)=\frac{1}{\log x}+\frac{1}{1-x}$ is continuous on $[0,1]$ (where we define by continuity $g(0)=1, g(1)=1/2$ since $1-x+\log x=-(1-x)^2/2+O((1-x)^3), \log x=-(1-x)(1+O(1-x))$ near $1^-$)

Hence by the Riemann-Lebesgue lemma:

$\lim_{m\to\infty}\int_0^1\sin(m\pi x)g(x)dx=0$ so the required limits are $-\lim_{n\to\infty}\int_0^1\frac{\sin(2n\pi x)}{1-x}dx$ and  $-\lim_{n\to\infty}\int_0^1\frac{\sin((2n+1)\pi x)}{1-x}dx$ respectively once we show that those exists.

Changing variables $x \to 1-x$ and using $\sin(2n\pi (1-x))=-\sin(2n\pi x)$ the first integral becomes

$\int_0^1\frac{\sin(2n\pi x)}{x}dx=\int_0^{2n\pi}\frac{\sin y}{y}dy$ and the limit follows from the Dirichlet integral as noted.

Obviously $\sin((2n+1)\pi (1-x))=\sin((2n+1)\pi x)$ so the second integral preserves the minus sign and the limit is then $-\pi/2$


问:
What are good ways of understanding a permutation group from a set of generators?
I'm trying to understand the structure of a Rubik's Cube-style puzzle I'm playing with; I have an expression of the solutions as the permutation group generated by four elements of $S_{16}$, each a simple cycle. My first inclination is to check all words of a given length (I think I can check all 8-element words easily, and maybe all 9- or 10-element words with a little effort) to see which ones move as few pieces as possible and then try and use that to understand more of the structure (for instance, I already know that there are elements of odd sign, so it's not a subgroup of $A_{16}$); above and beyond that, though, what tricks and techniques should I be aware of?  In particular, are there any good ways of finding useful/meaningful expressions of my group as some sort of wreath product, or of understanding its quotient in $S_{16}$ (assuming it's even normal!)?  I'm presuming that understanding $S_{16}/G$ will provide a sense of the constraints that any valid position has to satisfy...

EDIT: For anyone who's curious enough and has better tools to play with it than I do, the four generators are the cycles: $\alpha = (1\quad 2\quad 3\quad 7\quad 11\quad 10\quad 9\quad 5)$ , $\beta = (2\quad 3\quad 4\quad 8\quad 12\quad 11\quad 10\quad 6)$, $\gamma = (5\quad 6\quad 7\quad 11\quad 15\quad 14\quad 13\quad 9)$, and $\delta = (6\quad 7\quad 8\quad 12\quad 16\quad 13\quad 14\quad 10)$.  I wouldn't be surprised to learn it's all of $S_{16}$, but that still leaves a lot of obviously-open questions from a puzzle perspective (minimal-length words to swap two elements and the like), and any good suggestions on ways of finding particular elements of the group (where to look for interesting commutators and conjugates, etc.) would be welcome; I have a little background from doing some basic Rubik's-Cube stuff (for instance, my first look was at the element $(\alpha^2\beta^2)^2$ ), but not much.

ContentLicense: CC BY-SA 3.0

Body:
Using Maple's group package:

with(group):
 a:= [[1,2,3,7,11,10,9,5]]:
 b:= [[2,3,4,8,12,11,10,6]]:
 c:= [[5,6,7,11,15,14,13,9]]:
 d:= [[6,7,8,12,16,13,14,10]]:
 G:= permgroup(16, {a,b,c,d}):
 grouporder(G);

  20922789888000

Since this is 16!, the four cycles do generate all of $S_{16}$.


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

Body:
The answer is: YES, we can!

$$
\begin{align}
(\sqrt{1001}+\sqrt{999})^2&=2000+2\sqrt{1001\times 999}
\\
&=2000+2\sqrt{(1000+1)(1000-1)}
\\
&=2000+2\sqrt{1000^2-1}
\end{align}
$$
\begin{align}
\text{and that: }(2\sqrt{1000})^2&=4000
\\
&=2000+2\sqrt{1000^2}
\end{align}

Since $2000+2\sqrt{1000^2-1}<2000+2\sqrt{1000^2}$

$\therefore \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$


问:
Is Laplace transform of piecewise continuous injective?
Let $(\mathcal{L}f)(s)$ be the Laplace transform of a piecewise continuous function $f(t)$ defined for $t\geq 0$. If $(\mathcal{L}f)(s)=0 $ for all $s\in\mathbb{R^+}$ does this imply that $f(t)= 0$ for all $t\geq 0$ ?

答:
You didn't assume anything on the convergence of the integral or integrability of $f$.

If $$F(s_0)=\lim_{T\to\infty} \int_0^T f(t)e^{-s_0 t}dt$$ converges for some $s_0$ then
$$g(T)=\int_0^T f(t)e^{-s_0 t}dt$$ is bounded and for all $\Re(s)>\Re(s_0)$ (integration by parts)
$$F(s)=\lim_{T\to\infty} g(T) e^{-(s-s_0)T}+\int_0^T g(t)(s-s_0)e^{-(s-s_0)t}dt=\int_0^\infty g(t)(s-s_0)e^{-(s-s_0)t}dt$$
converges absolutely.

Thus $F$ is analytic for $\Re(s)>\Re(s_0)$.

If $f$ is not identically zero then $g(a) \ne 0$ for some $a>0$ thus for $n$ and $b$ large enough $$2^{-n}\sum_{k=0}^n {n\choose k} \frac{F(s_0+1-i(2k-n)/b)}{1-i(2k-n)/b} e^{i(2k-n)a/b}=\int_0^\infty g(t) \cos(t/b-a/b)^n e^{-t}dt \ne 0$$
so $F$ is a non-zero analytic function whence it can't vanish on a real segment.


问:
Question about Bishop's "constructive analysis" for people familiar with bishop style constructive analysis only
It has been a while since i read Bishop's book "constructive analysis", recently I dug it out of my book shelve and started to read. I came around this observation on the top of page 85.

"A subset Y of $\mathbb{R}$ can be bounded as a
metric space but not bounded as a subset of $\mathbb{R}$."

I have tried to figure out how this can be and came up with nothing so far. Can anybody give me a hint?

PS: As far as I understand Bishop is referring to the metric induced by the absolute value.

PPS: The definition provided:
A metric space $(X, p)$ is called bounded if there exists a real number
$C > 0$, called a bound for $(X, p)$, such that $p(x, y) \leq C \forall x,y \in X$.
A subset $Y$ of a non void metric space $X$ is bounded if, for all
(equivalently, some) $x$ in $X$, the set $Y\cup\{x\}$ with the induced metric p* is a bounded metric space.

答: Let $Y=\{n\in\mathbb N:n\text{ is the first counterexample to the Goldbach conjecture}\}$.  We don't know whether $Y$ has an element, but if it has one then it has only that one. That is, for any two elements of $Y$, the distance between them is $0$. So (assuming that Bishop's definition of "bounded metric space" is reasonable, i.e., that it's what I would use) I conclude that $Y$ is bounded as a metric space (under any metric). But to be able to say that $Y$ bounded as a subset of $\mathbb R$, we would need (again assuming a reasonable definition) to exhibit real numbers $c$ and $r$ (the center and radius of a metric ball) and prove that all elements of $Y$ are are within $r$ of $c$. And we don't know how to do that.  [You might think we could be sneaky and take $c$ to be the unique element of $Y$ if $Y$ has an element and $0$ otherwise, because then any positive $r$ would work. But that "definition" of $c$ isn't constructive; we don't know which of the two cases holds. Or, as a constructivist would say, we don't know that $Y$ is inhabited or empty.]


问:
I need to solve $12x-30y+24z=18$ equation as a diophantine equation where $x,y,z$ is whole number
I know i can divide with $6$ so I get $2x-5y+4z=3$, I replaced $2x-5y$ with $u$ so I got $u+4z=3$. I see that $u=3$ and $z=0$ is a solution but I got stuck. How can I continue? Is there a way to solve this using extended euclidean algorithm?

Thank you for your help.

答:
Experiments in a spreadsheet show that $x$ $y$ $z$ combinations only add up to a given value if we increment $y$ by $4$ and $z$ by $5$ for a given value of $x$. Further experimentation shows offsets of $1$ and $2$, respectively to make the result zero.

There are infinite solutions starting with $x\in\mathbb{Z}$ and
$$y = 4 n + 2 x + 1,\qquad z = 5 n + 2 x + 2\qquad n \in Z$$
Here are samples with $x\in\{-2,-1,0,1,2,3\}$

For $n=-1$
\begin{equation}
\cdots\quad
(-2,-7,-7)\quad
(-1,-5,-5)\quad
(0,-3,-3)\quad
(1,-1,-1)\quad
(2,1,1)\quad
(3,3,3)\quad\cdots
\end{equation}
For $n=0$
\begin{equation}
\cdots\quad
(-2,-3,-2)\quad
(-1,-1,0)\quad
(0,1,2)\quad
(1,3,4)\quad
(2,5,6)\quad
(3,7,8)\quad\cdots
\end{equation}

For $n=1$
\begin{equation}
\cdots\quad
(-2,1,3)\quad
(-1,3,5)\quad
(0,5,7)\quad
(1,7,9)\quad
(2,9,11)\quad
(3,11,13)\quad\cdots
\end{equation}
for $n=2$
\begin{equation}
\cdots\quad
(-2,5,8)\quad
(-1,7,10)\quad
(0,9,12)\quad
(1,11,14)\quad
(2,13,16)\quad
(3,15,18)\quad\cdots
\end{equation}
For $n=3$
\begin{equation}
\cdots\quad
(-2,9,13)\quad
(-1,11,15)\quad
(0,13,17)\quad
(1,15,19)\quad
(2,17,21)\quad
(3,19,23)\quad\cdots
\end{equation}
and so on.


问:
Prove that this function is integrable and equal to zero
If $f:[a,b] \to R$ is a bounded function such that $f(x)=0$ except for $x \in \{c_1,c_2,...,c_n\} \subset [a,b]$. Prove that $f$ is integrable in $[a,b]$ and that $\int_a^b f=0$.

I think this can be proved by Riemann criterion or by it's corollary. By finding the upper and lower sums, and since there are only certain points where the function isn't equal to zero its upper and lower sums must be 0.

But the only problem is that I don't know what partition to take, any thoughts?

Is there a way to determine what partition to take in this kind of problems, or it's purely intuition/arbitrary?

答: It doesn't matter. As long as the longest length in the partition is less than $\min\{c_{k+1}-c_k\}$, each point will be in a single partition. So, in the Riemann sum, only $n$ intervals will give nonzero values, and you can make these contributions as small as you want by making the partition finer.


问:
Which topics of mathematics should I study?
I'm a first year econometrics student with a great interest in mathematics. I very much enjoy my study, but still I am interested to learn about more topics in mathematics which are not part of my study. Some of the main topics which I am already familar with or which will soon be covered in my study are: calculus, linear algebra, optimization, statistics/probability/combinatorics.

Which topics in mathematics would you advise me to study still (that are not part of this list)? I have a general interest in mathematics, so any advice on interesting and or essential topics in mathematics that are worth studying is appreciated.

If possible, could you also give me advice on books/references which I should study from?

Thanks in advance.

Edit: To be more specific, I am looking for topics on which the general consensus is that they are essential to know of for any mathematician or very interesting to study.

答:
Real and complex variables, fractal geometry, complex dynamics, partial differential equations, and numerical analysis.  Definitely not abstract algebra.

Of course, you might not be me.

Edit

To be crystal clear - I'm joking.  I'd go as far to say that anyone who seriously advises you against studying abstract algebra is a bad person.  Of course, the language of abstract algebra suffuses so much of modern mathematics that an analyst can scarcely do without it.  The point is that Dedalus is correct, you must follow your passion.

As a professor, students frequently ask me for advice on courses, majors, careers, and graduate school.  Often, students state that they might want to go to graduate school because it looks like my job is pretty chill.  That's an insufficient reason - you've got to have the passion to make it through a Ph.D.

That's my 2 cents, anyway.


问:
What is the solution on the interval (-3, 3) for this ODE?
I took Differential Equations a few years ago and am reading back through my old book in a little more detail, especially with regard to intervals of validity for solutions. I found one of the example problems which finds a solution only for the intervals (-$\infty$, -3) and (3, $\infty$) using the standard integrating factor method. One of the section problems is to go back and find the solution on the interval (-3, 3). This is the equation: $$(x^2 - 9)\frac{dy}{dx}+xy=0$$
I've banged my head against the wall for a little bit before I remembered to check the Existence and Uniqueness theorem for IVP's to see if a solution exists on the interval: $$f(x,y)=-\frac{x}{x^2-9}y$$ and $$\frac{\partial f}{\partial y}=-\frac{x}{x^2-9}$$
Both of which are defined everywhere but x=-3 and x=3 so I believe a unique solution must exist on all of the intervals I mentioned above. I follow the books solution fine for the example but in this case I don't see any way to move forward on (-3,3) without bringing in complex numbers which I'm not very confident with. Further plugging the equation into Symbolab gives a plot with no solution on this interval. Usually I'm missing something simple but could someone enlighten me as to what it is?

答:
I imagine the book solves the equation for the intervals $(-\infty,-3)$ and $(3,\infty)$ by setting up

$$ y' + \frac{x}{x^2-9}y = 0 $$

Then multiplying both sides of the equation by the integrating factor $\sqrt{x^2-9}$. This is where the original domain restriction appears. But we will not use this integrating factor! Integrating factors are not unique, and using different integrating factors may give us solutions for different intervals. Let's find the integrating factor we need for $(-3,3)$. The general form of the integrating factor of this ODE is

$$ \exp\biggl({\int\frac{x}{x^2-9}}\biggr) = \exp\biggl(\frac{\ln\lvert x^2-9\rvert}{2}+C\biggr) = C\cdot\sqrt{\lvert x^2-9\rvert}$$

Where $C$ is nonzero. Since $\lvert -a \rvert = \lvert a \rvert$, a valid integrating factor is

$$\sqrt{9-x^2}$$ You will notice this is defined on the required interval $(-3,3)$. From here, apply the integrating factor method as normal to reach your desired solution.

This is one of my first answers on this site, so I would appreciate any formatting and/or etiquette tips on how to improve my responses. I hope this was helpful!


问:
How did $e^x$ get equal to $e^{2x}$?
$$  \begin{vmatrix}
e^x \, \cos(2x) & e^x \, \sin(2x) \\
e^x \, \cos(2x)- 2 e^x \, \sin(2x) & e^x \, \sin(2x)+ 2 e^x \, \cos(2x) \\
\end{vmatrix}  $$

I am finding Wronskian examples to work with and in this particular. I'm getting a different answer which is $2e^{2x}$. However, the example used $2e^x$. I rechecked it and it still gives me the same answer.  I don't know where i went wrong somewhere I know this is basic for some but if  $2e^{2x}$ is correct then is there an explanation for this? Can $e^x$ be used in place of $e^{2x}$? Thank you.

答:
The Wronskian determinant is

$e^x\cos(2x)(e^x\sin(2x)+2e^x\cos(2x))-e^x\sin(2x)(e^x\cos(2x)-2e^x\sin(2x)$

$=2e^{2x}\cos^2(2x)+2e^{2x}\sin^2(2x)=2e^{2x}$.

Note that $e^xe^x=e^{x+x}=e^{2x}$.

Title: How did $e^x$ get equal to $e^{2x}$?


问:
How to understand this hermitian vector bundle?
I am reading Huybrechts's complex geometry. In page 166, he gives the definition of hermitian structure.

Let $E$ be a complex vector bundle over a real manifold $M$. An hermitian structure $h$ on $E\to M$ is an hermitian scalar product $h_{x}$ on each fibre $E(x)$ which depend smoothly on $x$. This pair $(E,h)$ is called an hermitian vector bundle.

This definition is very natural. However, then he gives an example which make me confused.

Let $L$ be a holomorphic line bundle and let $s_{1},s_{2},...,s_{k}$ be global sections generating $L$ every where. Picking a trivialization $\phi: L_{U}\cong U\times \mathbb{C}$ around a point $x$. Then one defines an hermitian structure on $L$ by

$$h(t)=\frac{|\phi(t)|^{2}}{\sum |\phi(s_{i})|^2}.$$

where $t$ is a point in the fiber $L(x)$.

I am very confused. We know $L$ is a line bundle, so $L(x)$ is an complex one dimensional vector space. So if we want to find a hermitian metric on $L(x)$, it is enough to find a positive number for it. Thus we should construct a funtion $h:=h(x)$ which is a smooth function on $U$. I do not know why he uses $h(t)$. $h(t)$ is a function on the fiber $L(x)$, not the function on the manifold.

One explantion is he may want to refer $h(x)=\frac{|\phi(t)|^{2}}{\sum |\phi(s_{i})|^2},$ where $t$ is a point in $L(x)$. This is also strange, since then this hermitian metric will depond on the choice of $t$ and hence it is not natural.

Could you tell me how to explain this?

答:
Let $t$ is an arbitrary section of $L$ representing the germ.  Now you use the local trivialization $\phi$ to define $h(t)\colon U\to\mathbb{R}$ by
$$
\frac{\lvert\phi_2(t(\cdot))\rvert^2}{\sum_i\lvert\phi_2(s_i(\cdot))\rvert^2}
$$
where $\phi_2$ is the $\mathbb{C}$-coordinate of $\phi\colon L\vert_U\to U\times\mathbb{C}$ (at the appropriate point of $U$).


问:
If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$
If $a, b, c, d>0$, such that $a+b+c=1$, prove that: $$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$

I tried solving it as follows:
$$a^3+b^3+c^3=3abc+1-3(ab+bc+ac).$$

From Schur we have that:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c).$$

Hence $1+9abc\ge 4(ab+bc+ac)$.

This is as far as I got. I do not know how to introduce the $\min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ to my inequalities. Could you please explain to me how to solve it?

Body:
WLOG, assume $c = \min(a, b, c)$.

Let $P(x) = a^3 + b^3 + c^3 + abc x$ and $Q(x) = 2(\frac{a+b}{2})^3 + c^3 + (\frac{a+b}{2})^2 c x$ where $x \ge 0$.
We have
\begin{align}
P(x) - Q(x) &= a^3 + b^3 - 2(\tfrac{a+b}{2})^3
- ((\tfrac{a+b}{2})^2 - ab)c x \\
&= \frac{3}{4}(a+b)(a-b)^2 - \frac{1}{4}(a-b)^2 cx\\
&= \frac{1}{4}(a-b)^2(3a + 3b - cx)\\
&\ge \frac{1}{4}(a-b)^2(3a + 3b - \tfrac{a+b}{2} x)\\
&= \frac{1}{8}(a-b)^2(a+b)(6 - x). \tag{1}
\end{align}
Also, since $a + b + c = 1$, we have $Q(x) = \frac{1}{4}(x+3)c^3 + \frac{1}{4}(-2x+3)c^2 + \frac{1}{4}(x-3)c + \frac{1}{4}$.

Now, we split into two cases:

If $d > \frac{15}{4}$, by (1), we have
\begin{align}
a^3 + b^3 + c^3 + abcd &\ge a^3 + b^3 + c^3 + \frac{15}{4} abc\\
&= P(\tfrac{15}{4}) \\
&\ge Q(\tfrac{15}{4}) \\
&= \frac{1}{4}(\tfrac{15}{4} +3)c^3 + \frac{1}{4}(-2\cdot \tfrac{15}{4} + 3)c^2 + \frac{1}{4}(\tfrac{15}{4}-3)c + \frac{1}{4}\\
&= \frac{3}{16}c(3c-1)^2 + \frac{1}{4}\\
&\ge \frac{1}{4}\\
&= \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).
\end{align}

If $d\le \frac{15}{4}$, by (1), we have
\begin{align}
a^3 + b^3 + c^3 + abcd &= P(d) \\
&\ge Q(d)\\
&= \frac{1}{4}(d+3)c^3 + \frac{1}{4}(-2d+3)c^2 + \frac{1}{4}(d-3)c + \frac{1}{4}\\
&= \frac{1}{9} + \frac{1}{27}d + \frac{1}{108}(3c-1)^2(3cd + 9c - 4d + 15)\\
&\ge \frac{1}{9} + \frac{1}{27}d\\
&= \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).
\end{align}

We are done.


问:
Do we ever study "mixed" categories?
Consider a category whose object class includes the class of all topological spaces and the class of all topological groups. Furthermore, let the hom-sets between any two objects be the usual hom-sets in the category of topological spaces, unless both objects are topological groups, in which case the hom-set between them will be the usual hom-set in the category of topological groups.

Are "mixed" categories such as this ever studied?

答:
Yes:

if categories $\Bbb A$ and $\Bbb B$ are disjointly contained in a (bigger) category $\Bbb H$ as full subcategories, and suppose that $\Bbb H$ has no other objects, then I call $\Bbb H$ a bridge between $\Bbb A$ and $\Bbb B$, see my paper.

As Matt E. answered, your particular example is not a bridge, because we allow both directions $\Bbb A\not\to\Bbb B$ and $\Bbb B\not\to\Bbb A$. However, if we keep only one direction, say $\Bbb A\not\to\Bbb B\ $ (letting the homsets $\hom(b,a):=\emptyset$), we do get a category.

Such directed bridges from $\Bbb A$ to $\Bbb B$ straightly correspond to profunctors, i.e. functors of the form $\Bbb A^{op}\times\Bbb B\to\Bbb{Set}\ $ (just take the restriction of the hom functor). In the paper I show that bridges are just the so called Morita contexts connecting two profunctors.

A pair of adjoint functors $F\dashv G$ determines one single profunctor (up to isomorphism), namely $F_*:=\hom(F-,-)$ which is the same as $G_*:=\hom(-,G-)$.

This is the case for a profunctor (as category) iff $\Bbb A$ is coreflective and $\Bbb B$ is reflective subcategory therein.

The two profunctors corresponding to the two directions in your example arise from the forget forgetful functor $\Bbb{TopGr}\to\Bbb{Top}$.

Title: Do we ever study "mixed" categories?


问:
Why it is important to write a function as sum of even and odd functions?
For the function $f(x)$ we can write it as sum of even and odd functions:

$$f(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{\text{Even}}+\underbrace{\frac{f(x)-f(-x)}{2}}_{\text{Odd}}$$

My question is why it is important for us to write a function as sum of these two even and odd functions? Is there any application of that?

答:
The answer by KCd mentions in passing what I'll talk about, but I'll elaborate on it: the short answer is Fourier analysis.

Splitting a function into odd and even components is an extremely useful problem-solving technique when working with the Fourier transform, and the associated Fourier series. A function that is purely even or purely odd is easier to find the Fourier transform/series of.

That may seem like a niche topic, but Fourier analysis is one of the most powerful and widely used mathematical techniques. You cannot go far into any STEM field without encountering it, and so making Fourier analysis easier is more significant than you might think.

There is a wealth of knowledge on the internet about what Fourier analysis is and how it works, so I won't reiterate it here. I've found this YouTube video as a good introduction to the topic.


问:
Does $\{\mathbf{x} \mid \mathbf{x} \text{ is an isolated point of }S\text{ and }f(\mathbf{x})\ne 0\}$ have measure zero in $\mathbb{R}^n$?
I am reading "Analysis on Manifolds" by James R. Munkres.

Let $f:A\to Y$.

In this book, Munkres defines $\lim_{x\to x_0} f(x)$ only for $x_0$ which is a limit point of $A$.

So, if $x_0$ is an isolated point of $A$, we cannot compute $\lim_{x\to x_0} f(x)$.

There is the following theorem in this book:

Theorem 13.5(on p.109)

Let $S$ be a bounded set in $\mathbb{R}^n$; let $f:S\to\mathbb{R}$ be a bounded continuous function. Let $E$ be the set of points $\mathbf{x_0}$ of $\text{Bd}\,S$ for which the condition $$\lim_{\mathbf{x}\to\mathbf{x_0}} f(\mathbf{x})=0$$ fails to hold. If $E$ has measure zero, then $f$ is integrable over $S$.

And its proof is the following:

My question is here:

If $\mathbf{x_0}$ is an isolated point of $S$ and $f(\mathbf{x_0})\ne 0$, then $f_S$ is not continuous at $x_0$.

I think Munkres need to show that $\{\mathbf{x} \mid \mathbf{x} \text{ is an isolated point of }S\text{ and }f(\mathbf{x})\ne 0\}$ has measure zero in $\mathbb{R}^n$ to use the following theorem:

Let $Q$ be a rectangle in $\mathbb{R}^n$; let $f:Q\to\mathbb{R}$ be a bounded function. Let $D$ be the set of points of $Q$ at which $f$ fails to be countinuous. Then $\int_Q f$ exists if and only if $D$ has measure zero in $\mathbb{R}^n$.

答: There can only be a countable number of isolated points for any set in $\mathbb R^{n}$. And any countable set has measure $0$.


问:
Finding a point on the line
Suppose I have a line ax+by+c=0 and some points (x1, y1), (x2, y2), (x3, y3), ....(xn, yn). I wanna have a point on the line such that the total distance from every point to this point on the line is minimum. So is there some formulae or tweak or something that can be of help?</p>

答:
Let your point on the line is (x', y') where y' = -(ax'+c)/b.

Then find the distance to all the other points using pythagoras therom.

Ex: sqrtOf((xi-x')^2 + (yi-y')^2).

Get the summation as f(x) = sqrtOf((x0-x')^2 + (y0-y')^2) + sqrtOf((x1-x')^2 + (y1-y')^2) + ...

Find the first derivative f'(x). Then solve f'(x) = 0.

Title: Finding a point on the line


问:
Lebesgue density application
I have been working on this problem but I have not been able to figure out how to solve it.

For every $x\in (0,1)$ and $\epsilon>0$ there exists $0<r<\epsilon$
such that $\int_{x-r}^{x+r}f(x)dx\geq 2r$. Show $f\geq 1$ for almost
every $x\in [0,1]$.

This is my attempt: Assume $f<1$ for almost every x, and define $E=\{x\in [0,1]:f<1\}$. By our assumption, we must have that the measure of E is close to 1; i.e. $m(E)<1-\delta$ for some $delta>0$. Let $x_0\in E$ be a point of densitiy. Then, there exists $\epsilon >0$ such that $\frac{m(E\cap (x_0-\epsilon, x_0+\epsilon)}{2\epsilon}<2/3$ (by the Lebesgue density Thm). Then, $m(E\cap (x_0-\epsilon, x_0+\epsilon)<\frac{4}{3}\epsilon$. Also, by assumption we have that $\int_{x_0-\epsilon}^{x_0+\epsilon}f(x)dx\geq 2\epsilon$.

Here is all I have. I do not know how to use $f<1$, nor how to finish the problem. Any help will be very much appreciate it.

答: Lebesgue's differentiation Theorem tells you that almost all points are Lebesgue points. If $x$ is  a Lebesgue point then $\frac 1 {2r} \int_r^{r} f(x) d x\to  f(x)dx$ as $r \to 0$.  Hence $f(x) \geq 1$ for every Lebesgue point $f$.  (Take $\epsilon =\frac 1n$ to get $0<r_n <\frac 1 n$ such that $\int_{x_n-r_n}^{x_n+r_n} f(x)dx \geq 2r_n$. Divide by $2r_n$ and take the limit).

Title: Lebesgue density application


问:
Weird geometry question ($AG^3=CE^2\times AB$)
$ABC$ is a triangle with $AB = 2AC$ and $E$ is the midpoint of $AB$. The point $F$ lies on the line $EC$ and the point $G$ lies on the line $BC$ such that $A, F, G$ are collinear and $FG = AC$. Show that $AG^3=CE^2\times AB$

答:
The claim is true. However, in order to have a $F \ne E$ so that $FG = 1$, one need to look for a $F$ outside the line segment $CE$ (more precisely, on the ray start at $C$ pointing away from $E$).

For simplicity, we will only consider the case $AC = AE = EB = FG = 1$.

For any point $P$ in the plane, let $z_P = (u_P,v_P,w_P)$ be its barycentric coordinates with respect to $\triangle ABC$. i.e $u_P, v_P, w_P$ are the three real numbers such that:

$$\vec{P} = u_P \vec{A} + v_P \vec{B} + w_P\vec{C}\quad\text{ and } \quad u_P + v_P + w_P = 1$$

Since $E$ is mid-point of $AB$, we have $z_E = (\frac12,\frac12,0)$.

Since $F$ lies on $CE$, $$z_F = (-s,-s,1+2s)$$ for some number $s$. In order to met the condition $FG = 1$ for a $F \ne E$, we need $u_F$ to be negative which is equivalent to $s$ to be positive!

Since $G$ lies on $BC$, $u_G = 0$. Together with the fact $G$ lies on $AF$,
we find $$z_G = \left(0,-\frac{s}{1+s},\frac{1+2s}{1+s}\right)$$

Notice $\displaystyle\;\frac{u_F - u_A}{u_G - u_A} = \frac{-s-1}{-1} = s+1$. We have
$$AF = (1+s)AG \implies FG = AF-AG = sAG$$

If $FG = 1$, we will have $$AG = \frac1s\quad\text{ and }\quad AF = 1 + \frac1s$$

To proceed, we need another formula to compute $AF$.

Let $\alpha = \frac12\angle BAC$, $\beta = CE = 2\sin\alpha$ and $\delta = AG = \frac1s$.

Since $AC = AE$, if we let $M$ be the mid-point of $EC$, $\triangle AMF$ is a right angled triangle. It is easy to see
$FM : CM = s + \frac14 : \frac14$, this implies

$$AF^2 = AM^2 + MF^2 = \cos^2\alpha + \sin^2\alpha(4s+1)^2
= 1 + 2\beta^2 s(2s+1)
$$

Combine with $AF = FG + AG = 1 + \delta$ and $s = \frac1{\delta}$, we obtain

$$(1+\delta)^2 = AF^2 = 1 + 2\beta^2 \frac{\delta+2}{\delta^2}
\quad\implies\quad \delta^3 = 2\beta^2$$

The last equality is precisely what we want to show:

$$AG^3 = \delta^3 = 2\beta^2 = AB(CE)^2$$


问:
Fixed point iteration VS Bisection
I need to answer a multiple choice question that goes like this -

Does fixed point iteration method always faster than bisection method?"

Yes

No

Now, I know that fixed point is faster, I have seen the math behind the convergence

But I wonder if there's like a specific case where I would pick bisection over fixed point ?
How would you answer that question cause I feel like I didn't get enough information or the question isn't well phrased.

答:
Fixed point iteration is not always faster than bisection. Both methods generally observe linear convergence. The rates of convergence are $|f'(x)|$ for fixed-point iteration and $1/2$ for bisection, assuming continuously differentiable functions in one dimension.

It's easy to construct examples where fixed-point iteration will converge much slower than bisection (sublinear convergence). For example, the iterations $x_{n+1}=\sin(x_n)$ are well-known to converge very slowly, much slower than using bisection on $f(x)=x-\sin(x)$.

The main advantage of bisection is the fact that it is guaranteed to give linear convergence. The main drawback is also that it is guaranteed to only give linear convergence, whereas other methods can converge faster in some situations.

Bisection is also only applicable in some scenarios (the function must be defined over an ordered domain (e.g. not $\mathbb R^2$) and known to attain two different signs (e.g. not $x^2$)). It also does not give an analytic estimate of the solution, so it's less useful from a theoretical standpoint.

Title: Fixed point iteration VS Bisection


答:
You'll just need to get some universities syllabi an pick some topics to study. The syllabi do have a lot of subjects but they also provide you an order in which they should be studied that will restrain a little of your choices. You won't be lost with a diversity of fields of study because you'll have to cover some fundamentals first.

Cambrige and Oxford have nice materials for guiding your study - It'll also be useful in the case that you know some of the first subjects, you'll be able to pick more advanced stuff. The folowing resources are going to be very useful:

How to Become a Pure Mathematician

This website has book recomendations and also information on what order it should be studied.

All the Mathematics You Missed: But Need to Know for Graduate School

This book comments on the importance of some math fields for the mathematical study.

Cambridge Syllabus

Oxford Syllabus

I've found both syllabi enlightening and informative, with them you're going to have something similar to the site I mentioned before: fields of study, the order in which they should be studied, some intruction on what you should be able to do after covering the topics and a little about the importance of them.

There are also some all-in-one books and book collections that you should look:

Mathematics, It's Content Method and Meaning

Fundamentals of Mathematics

The World of Mathematics

Mathematics Form and Function

What it Mathematics?

Princeton Companion to Mathematics

For the end, as a personal suggestion: Don't get afraid, just get the books and start reading, when the things start to become dark you can use the torches of our fellow members to lighten your path! Good Luck!

Title: Which topics of mathematics should I study?

Body:
I'm a first year econometrics student with a great interest in mathematics. I very much enjoy my study, but still I am interested to learn about more topics in mathematics which are not part of my study. Some of the main topics which I am already familar with or which will soon be covered in my study are: calculus, linear algebra, optimization, statistics/probability/combinatorics.

Which topics in mathematics would you advise me to study still (that are not part of this list)? I have a general interest in mathematics, so any advice on interesting and or essential topics in mathematics that are worth studying is appreciated.

If possible, could you also give me advice on books/references which I should study from?

Thanks in advance.

Edit: To be more specific, I am looking for topics on which the general consensus is that they are essential to know of for any mathematician or very interesting to study.


问:
How is $\mathbb C(x)^*$ a free group?
My professor gave an example of a free group as the field of rationals over one variable  $\mathbb C(x)^*$  but I do not know how. could anyone explain this to me, please?

答:
You mean a free abelian group, a free group is completely different.

$\Bbb{C} (x)^\times/\Bbb{C}^\times$ (isomorphic to the quotients of monic polynomials) is a free abelian group, generated by the $x-a,a\in \Bbb{C}$.

$\Bbb{C} (x)^\times$ is not a free abelian group as $(-1)^2= 1$.  Also for any fixed $z\in \Bbb{C}^\times$, there is a $z^{1/n}$ for all $n$.

Title: How is $\mathbb C(x)^*$ a free group?


问:
Structure constant in Weyl basis
$$(e_{ij})_{lk} = \delta_{il}\delta{jk}$$
Which satisfy the following commutation relations
$$[e_{ij},e_{kl}] = \delta_{jk}e_{il} - \delta_{il}e_{kj}$$
So that*
$$c_{sm,kr}^{ij} = \delta^{i}_{s}\delta_{mk}\delta^{j}_{r} - \delta_{k}^{i}\delta_{rs}\delta^{j}_{m}$$

Hello. I am having trouble to understand the step * (since the both equations before it is just definitions). Particularly, i am having trouble to understand how to get this equality, and why the structure constant has this notation

答:
By definition, $c_{sm,kr}^{ij}$ is the coefficient of $e_{ij}$ in the expansion of $[e_{sm},e_{kr}]$.

According to your commutation relations, substituting indices as necessary,

$$ [e_{sm},e_{kr}]=\delta_{mk}e_{sr}-\delta_{sr} e_{km}. $$

Thus the coefficient of $e_{ij}$ is $\delta_{mk}$ if $ij=sr$ and $-\delta_{sr}$ if $ij=km$, and $0$ otherwise.

Thus we can write it as a combination $\square\delta_{mk}+\square(-\delta_{sr})$ where the first square is $1$ if $ij=sr$ and $0$ otherwise and the second square is $1$ if $ij=km$ and $0$ otherwise.

Thus, the squares are $\delta^i_s\delta^j_r$ and $\delta^i_k\delta^j_m$ and we may conclude

$$ c_{sm,kr}^{ij} = \delta^i_s\delta^j_r \delta_{mk}-\delta^i_k\delta^j_m\delta_{sr}. $$

Title: Structure constant in Weyl basis


问:
How to solve permutations
Vanessa, Samantha, Hannah, and Jordyn are all part of a 12 member volleyball team. Four members of the team are randomly selected to fill the water jug for the day. What is the probability that all four of them will be chosen?</p>

答:
Number of ways to select 4 people out of the 12 in the volleyball team - $^{12}C_{4}$

Out of these combinations, there is only 1 which has all 4 of the required people, Vanessa, Samantha, Hannah, and Jordyn.

So the probability becomes $\frac{1}{^{12}C_{4}}$

Also, do not tag a question by geometry if it isn't about geometry at all. Thanks

Title: How to solve permutations


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

Body:
$$\frac{1}{\sqrt{1000}+\sqrt{1001}}<\frac{1}{\sqrt{1000}+\sqrt{999}}$$

$$\implies \sqrt{1001}-\sqrt{1000}<\sqrt{1000}-\sqrt{999}$$

$$\implies \sqrt{1001}+\sqrt{999}<2\sqrt{1000}$$


问:
Mind too small for quadratic equation problem
I was struggling for a long time with the following question, eventually unable to find it. I'm really frustrated with it because the answer is not difficult per se but to me requires a complete shift in approaching the problem, and I am just unable to make such a leap by myself. So, is it a hard question or am I just stupid?

Here goes: Two pipes together can fill a reservoir in 6 hr 40 min. Find the time each alone will take to fill the reservoir if one of the pipes can fill it in 3 hr less time than the other.

Edit: The solution is in my textbook so not quite asking for that. Once you formulate the quadratic equation it's straightforward but for me this step really takes a leap of the imagination to find it and I wonder if I'm alone in this. Can't really post the answer because then it'll be obvious.

So basically I'm asking if anyone can formulate the equation and tell me if they found it difficult.

Thanks for the willingness to answer already!

答:
Here is the logic:-

Suppose that pipe-1 can fill the tank in x hours.

Then, in 1 hour's time, ($\dfrac {1}{x}$) of tank have been filled.

Similarly, for pipe-2, in 1 hour's time, ($\dfrac {1}{x - 3}$) of the tank have been filled.

If they worked together, in 1 hour's time, $(\dfrac {1}{x} + \dfrac {1}{x - 3})$ of the tank have been filled.

Then,  $(\dfrac {1}{x} + \dfrac {1}{x - 3}) = ...$


问:
Basis of Complex vector space, complex conjugate
I am not sure whether if this statement is correct,

For each basis $\beta =\left \{ \vec x_{1} , \vec x_{2},..., \vec x_{n}  \right \}$ of $\mathbb{C}^n$, the set $\alpha=\left \{ \overline{\vec x_{1}} , \overline{\vec x_{2}},...,  \overline{\vec x_{n}}  \right \}$  is a basis of $\mathbb{C}^n$.

Here is my thought,
I think the statement is false, the complex conjugate of a vector in $\mathbb{C}^n$ is different from itself. If all the entries of a vector are real, then the complex conjugate is just itself, it does not change, so $\left \{ \vec x_{1} , \vec x_{2},..., \vec x_{n}  \right \}=\left \{ \overline{\vec x_{1}} , \overline{\vec x_{2}},...,  \overline{\vec x_{n}}  \right \}$ but this only works for the real vectors.

I am really not sure if $\left \{ \overline{\vec x_{1}} , \overline{\vec x_{2}},...,  \overline{\vec x_{n}}  \right \}$ is also a basis for $\mathbb{C}^n$ if $\left \{ \vec x_{1} , \vec x_{2},..., \vec x_{n}  \right \}$ is a basis for $\mathbb{C}^n$, and is this true for each basis?
And how do you prove it if it is true or disprove it?
Maybe somehow related to linear independence?

I have tried a few example, and it seems that this statement is true?
Thank you for your help!

答: The statement is true. I will change the notation slightly by using capital letrres for vectors. Suppose $\{X_1,X_2,..,X_n\}$ is a basis. If $\sum c_k\overline {X_k}=0%$ then we can take conjugate on both sides to get $\sum \overline {c_k} X_k=0$.  This implies $\overline {c_k} =0$ for each $k$ and hence $c_k=0$ for eachy $k$. Hence $\{ \overline  {X_1}, \overline  {X_2},..,\overline  {X_n}\}$ is linearly independent Since the sapce $\mathbb  C^{n}$ is $n$ dimensional it is a basis.


问:
Lebesgue measurable implies caratheodory measurable
The definition of Lebesgue measure is

$E$ is Lebesgue measurable if $\forall \epsilon > 0,$ there exists an open set $O$ such that $E \subset O$ and $m_*(O-E) < \epsilon$.

Caratheorody measurable is

$E$ is Caratheodory measurable, if $m_*(A) = m_*(A \cap E) + m_*(A \cap E^c)$ for all $A \in \Omega$.

I want to show that the top definition implies the bottom definition. Since $m_*(A) \le m_*(A \cap E) + m_*(A \cap E^c)$ is trivial by subadditivity, we only need to show $m_*(A) \ge m_*(A \cap E) + m_*(A \cap E^c)$. In order to use the Lebesgue measurability, let $E$ be Lebesgue measurable, and let $O$ be some open set such that $A \subset O$. Then
\begin{align*}
 m_*(A \cap E) + m_*(A \cap E^c) \le m_*(O \cap E) + m_*(O \cap E^c)
\end{align*}
But I'm not sure where I go from there. I want to use $\epsilon$ somewhere but I'm not sure how. Any hint would be much appreciated.

答:
Attention: We are going to use the terminology used in the question, including $m_*$ to denote Lebesgue outer measure:

$E$ is Lebesgue measurable if $\forall \epsilon > 0$, there exists an open set $O$ such that $E \subset O$ and $m_*(O-E) < \epsilon$.

and

$E$ is Caratheodory measurable, if $m_*(A) = m_*(A \cap E) + m_*(A \cap E^c)$ for all $A \subseteq \Omega$.

Let us prove that if $E$ is Lebesgue measurable, then $E$ is Caratheodory measurable.

Suppose $E$ is is Lebesgue measurable. Given any $\epsilon >0$, let $O_\epsilon$ be an open set such that  $E \subset O_\epsilon$ and $m_*(O_\epsilon-E) < \epsilon$.

Given any $A \subseteq \Omega$, we have, taking any  any $\epsilon >0$
\begin{align*}
\mu_*(A) &\leqslant \mu_*(A \cap E) + \mu(A\cap E^c) \\
& \leqslant \mu_*(A \cap O_\epsilon ) + \mu_*((A\cap O_\epsilon ^c) \cup (A\cap O_\epsilon \cap E^c))\\
&  \leqslant \mu_*(A \cap O_\epsilon ) + \mu_*(A\cap O_\epsilon ^c) +\mu_*(A\cap O_\epsilon \cap E^c)\\
& \leqslant \mu_*(A \cap O_\epsilon ) + \mu_*(A\cap O_\epsilon ^c) +\mu_*(O_\epsilon \cap E^c)\\
&  \leqslant \mu_*(A) + \epsilon
\end{align*}
where in the last step we used the fact that open sets are Caratheodory measurable.

So we have proved that, for any  $\epsilon >0$,
$$
\mu_*(A) \leqslant \mu_*(A \cap E) + \mu(A\cap E^c) \leqslant \mu_*(A) + \epsilon
$$
So
$$
\mu_*(A) = \mu_*(A \cap E) + \mu(A\cap E^c)
$$
So $E$ is Caratheodory measurable.


问:
What is the correct way to simplify this derivative? (Derivative divided by another derivative.)
I am trying to simply some terms involving derivatives for a project I'm working on.

Here's a basic rule I believe to be true for dividing derivatives:

$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$$

Therefore we can simply the following as:

$$\frac{(\frac{y_x}{||r_x||})_x}{y_{xx}} =\frac{\frac{y_x}{||r_x||}}{y_{x}}=\frac{1}{||r_x||}$$

I think this is correct.

But I am having trouble simplifying a similar equation in a way that performs as expected:

$$\frac{(\frac{1+y_x}{||r_x||})_x}{y_{xx}} = \frac{(\frac{y_x+1}{||r_x||})_x}{(y_x+1)_x} =\frac{(\frac{y_x+1}{||r_x||})}{(y_x+1)} = \frac{1}{||r_x||}$$

I thought it was reasonable to say $y_{xx} = (y_x+1)_x$ because the added constant 1 will be removed by the second derivative.

Is this correct? Is it then true:

$$\frac{(\frac{y_x}{||r_x||})_x}{y_{xx}} = \frac{(\frac{1+y_x}{||r_x||})_x}{y_{xx}}$$

I feel like this is not possible and it's causing problems. If what I did was wrong, why was it wrong?

Thanks.

答:
I don't know why you are avoiding applying the quotient rule...

The relation you are (actually) using is the chain rule,
$$  \frac{\mathrm{d} y}{\mathrm{d} t} = \frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} x}{\mathrm{d} t}  \text{.}  $$

So, from your first example, we should expect
$\displaystyle   \left(\frac{y_x}{||r_x||} \right)_x = \frac{1}{||r_x||} y_{xx}  \text{.}  $
Of course, we can just evaluate the left-hand side:
\begin{align*}  \left(\frac{y_x}{||r_x||} \right)_x &= \frac{||r_x|| y_{xx} - y_x ||r_x||_x}{||r_x||^2}  \\
&= \frac{1}{||r_x||} y_{xx} - \frac{y_x ||r_x||_x}{||r_x||^2}  \text{.}
\end{align*}
Therefore, a condition that your first example is correct is $\displaystyle y_x ||r_x||_x = 0$.

Your second example has a similar difficulty.


问:
Is there some really powerful summation method?
I know that Borel summation gives you a value if the coefficients are bounded by $n!C^n$.

 Is there a more powerful summation method (with nice properties comparable to Borel) that sums series with coefficients bounded by $(kn)!$ or even $(n^k)!$ ?

答:
In many situations “the most powerful possible generalization” of Borel’s method is applicable.  It was introduced in (zillions of) Écalle’s papers (for example, it is discussed in Chapter 3 of his “NATO” paper).

(Unfortunately, as Écalle’s papers go, they all seem to be “popular expositions” of a [non-existent — or existing in a Platonic sense only] fundamental giga-mega-treatise.  This is not surprising: in its entirety, his work seems to be based on half-a-hundred of very deep non-trivial and intertwined ideas.  It is hard to imagine how to put the tangle of these ideas on paper.)

If all you are interested in is summation, Balser’s books may be a better source.  He untangled a small chunk of these ideas: see the notion of “multisummability”.  (I learned it from the book Formal Power Series and Linear Systems of Meromorphic Ordinary Differential Equations; I can see other people recommending From Divergent Power Series to Analytic Functions: Theory and Application of Multisummable Power Series as well.


问:
Writing Linear Complex Equations from two points
I'm just wondering how I write, suppose a line that goes through origin and point $z=a+bi$ into equation form?

答:
Hint

Pretend that you are working in $\mathbb{R^2}$ rather than $\mathbb{C}.$

One form of the general equation for a line is $y = mx + b.$

From the information given, you have enough to solve for $m$ and $b$.


答:
Suppose that one pipe takes $t$ hours to fill the reservoir, and the other pipe takes $t-3$ hours.

Then in $t(t-3)$ hours, pipe 1 has filled $t-3$ reservoirs, and pipe 2 has filled $t$ reservoirs. Together, they fill $2t - 3$ reservoirs in $t(t-3)$ hours, or $1$ reservoir in $\frac{t(t-3)}{2t-3}$ hours.

Converting $6$ hours $40$ minutes into hours, we get:

$$\frac{t(t-3)}{2t-3} = \frac{20}{3}$$
$$\Rightarrow 3t(t-3) = 20(2t-3)$$
$$\Rightarrow 3t^2 - 9t = 40t - 60 \implies 3t^2 - 49t + 60 = 0$$
$$\Rightarrow (3t - 4)(t - 15) = 0 \implies t = 15$$

so working alone, one pipe fills it in $15$ hours, and the other fills it in $12$.

The smaller solution is discarded because firstly, it would make $t-3$ negative (which does not make sense), and working alone should result in a longer time than working together.

As a sanity check, you can repeat this process with $t = 15$ (or using any other method), and doing this gives you $\frac{20}{3}$ hours to fill up the reservoir.

Title: Mind too small for quadratic equation problem

Body:
I was struggling for a long time with the following question, eventually unable to find it. I'm really frustrated with it because the answer is not difficult per se but to me requires a complete shift in approaching the problem, and I am just unable to make such a leap by myself. So, is it a hard question or am I just stupid?

Here goes: Two pipes together can fill a reservoir in 6 hr 40 min. Find the time each alone will take to fill the reservoir if one of the pipes can fill it in 3 hr less time than the other.

Edit: The solution is in my textbook so not quite asking for that. Once you formulate the quadratic equation it's straightforward but for me this step really takes a leap of the imagination to find it and I wonder if I'm alone in this. Can't really post the answer because then it'll be obvious.

So basically I'm asking if anyone can formulate the equation and tell me if they found it difficult.

Thanks for the willingness to answer already!


问:
Does folded cube have a name?
Folding a square to produce a torus has an analogy with "folding" a cube to produce what? Is there a name for the resulting 4D structure?</p>

答: Such an object is also called a genus-3 handlebody.</p>

Title: Does folded cube have a name?


问:
Parameterization of a triangle in $\mathbb R^3$ (edge and surface)
Let $S$ be a triangle with vertices at A=$(1,2,0)$, B=$(2,3,2)$, and C=$(0,0,4)$. Find a parametrization of the triangle and its contour.

I have set up the parametric equations for each edge

$\begin{pmatrix} 2 \\ 3 \\ 2\end{pmatrix}$-$\begin{pmatrix} 1 \\ 2 \\ 0\end{pmatrix}$=$\begin{pmatrix} 1 \\ 1 \\ 2\end{pmatrix}$

first edge: $r_1(t)=\begin{pmatrix} 1 \\ 2 \\ 0\end{pmatrix}  + t\cdot \begin{pmatrix} 1 \\ 1 \\ 2\end{pmatrix}=\begin{pmatrix} 1+t \\ 2+t \\ 0+2t\end{pmatrix}$

likewise for the others

$r_2(t)=\begin{pmatrix} 2-2t \\ 3-3t \\ 2+2t\end{pmatrix}$

$r_3(t)=\begin{pmatrix} 0+t \\ 0+2t \\ 0-4t\end{pmatrix}$

for each $t \in [0,1]$

What can I do now? How to put them together, and how to get the parametrization of the whole surface and not only the edges?

答:
What you have done is all you need to do for the edges. You could parametrize so you could go from $0 \leq t \leq 1, 2 \leq t \leq 2, 2 \leq t \leq 3$.

$r_1(t)= (1, 2, 0) + (1, 1, 2)t = (1 + t, 2 + t, 2t) \,$ for $( 0 \leq t \leq 1)$

$r_2(t)= (2, 3, 2) + (-2, -3, 2) (t-1) = (4-2t, 6-3t, 2t) \,$ for $( 1 \leq t \leq 2)$

$r_3(t)= (0, 0, 4) + (1, 2, -4) (t-2) = (-2 + t, -4 + 2t, 12-4t) \,$ for $( 2 \leq t \leq 3)$

Now if you have to parametrize the surface, find the equation of the plane it is on. First take two directional vectors you found above and do a cross product to find the normal vector to the plane-

$(-2, -3, 2) \times (1, 1, 2) = (-8, 6, 1)$

Take any point on the surface say $(0, 0, 4)$ then we have,

$ -8(x-0) + 6(y-0) + (z-4) = 0$

Parametrization of the surface of the plane the given triangle is on,

$ \,z = 4 + 8x - 6y$

EDIT:

On the parametrization that you asked,

If you take parametrization of two of your edges, $r_1(s) = (1 + s, 2 + s, 2s) \,(0 \leq s \leq 1)$

$r_2(t) = (2 - 2t, 3 - 3t, 2 + 2t) \,(0 \leq t \leq 1)$

You can combine these two to parametrize your triangle.

$r(s, t) = (1 + s - 2t, 2 + s - 3t, 2s + 2t) \, \, (0 \leq s, t \leq 1)$

If you take cross product $r_t \times r_s$, you will get the normal vector $(-8, 6, 1)$ which is what I used for other parametrization.


答: You can tell without calculation if you can visualize the graph of the square-root function; specifically, you need to know that the graph is concave (i.e., it opens downward).  Imagine the part of the graph of $y=\sqrt x$ where $x$ ranges from $999$ to $1001$.  $\sqrt{1000}$ is the $y$-coordinate of the point on the graph directly above the midpoint, $1000$, of that interval.  $\frac12(\sqrt{999}+\sqrt{1001})$ is the average of the $y$-coordinates at the ends of this segment of the graph, so it's the $y$-coordinate of the point directly above $x=1000$ on the chord of the graph joining those two ends.  The concavity of the graph shows that the chord lies below the graph.  So $\frac12(\sqrt{999}+\sqrt{1001})<\sqrt{1000}$.  Multiply by $2$ to get the numbers in your question.

Title: Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$

Body:
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?


问:
How many "words" of length n is it possible to create from {a,b,c,d} without any "a"s after a "b"?
I think this is a hard question and I didn't come up with a solution. Could you please give me a hint or approach?

For instance, "bdca" is not ok, but "aabcd" is ok.

It is somehow similar to this question, but with some differences.

Thanks in advance.

答:
Hint

$\sum_{k=1}^n [f(k) \times g(k)].$

$f(k)$ is the # of possible sequences of length $k$, where the very first occurrence of the letter $b$ is on the $k$-th letter.

Note that this categorization facilitates mutually exclusive groupings.

$g(k)$ is the # of possible sequences of length $(n-k)$ where the letter $a$ is excluded from this sequence.

Two items need special handling:

What happens if the first $b$ in the sequence is on the last (i.e. $n$-th) letter?

What happens if the sequence does not contain any $b$'s.


问:
Find the altitude of the triangle given the equation of sides

The sides of a triangle are on the lines $2x+3y+4=0$, $ \ \ x-y+3=0$, and $5x+4y-20=0$. Find the equations of the altitudes of the triangle.

Should I find the vertices first? Or is there a direct way? Actually, I tried finding the vertices, using the substitution method but I find it hard to turn it into an equation.

答:
Plan: For the given equation of line as $$ax + by + c = 0 \tag{1}$$ you can find the general equation of perpendicular lines as $$bx - ay + d = 0 \tag{2}$$
for some $d\in \mathbb R$. Now, you just need to find the values of $d$.

In order to find $d$, find the vertex point (the intersection of the other two sides than $(1)$) and plug that into $(2)$ (because the perpendicular line should pass through that vertex). Since $a$ and $b$ are known values you can derive the value of $d$ from there.

Solution: Take the third equation, for example. $$5x+4y-20=0 \tag{1}$$
Then, the general equation of perpendicular lines is $$4x-5y+d = 0 \tag{2}$$
Now, we will find the intersection of the other two lines than $(1)$.
$$\begin{cases}
2x + 3y + 4 = 0 \\
x - y + 3 = 0\end{cases} \implies (x,y) = \left(-\frac{13}{5} , \frac{2}{5}\right)$$
Now plug this into $(2)$ to get
$$4\left(-\frac{13}{5}\right) - 5 \left(\frac{2}{5}\right) + d = 0 \\ d=  \frac{62}{5}$$
Thus, the equation of altitude to the side which is given by $(1)$ is $$\boxed{4x-5y + \frac{62}{5} = 0}$$ You can find the other two altitudes with the same method.


问:
Summation of a series with 2 different methods gives 2 different answers
The objective here is to find the value of $S$, where $S$ is given by,
$$S = 1-{1\over2}+{1\over3}-{1\over4}+...$$
I did this using two methods, but both the methods give different answers.

Method 1:

Let
$$S' = 1+{1\over2}+{1\over3}+{1\over4}+...$$
Now,

$1+{1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+...$

$\;\;\;\;-1\;\;\;\;\;\;\;-{1\over2}\;\;\;\;\;\;\;-{1\over3}\;\;\;...$

$\frac{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}{}$

$1-{1\over2}+{1\over3}-{1\over4}+{1\over5}-{1\over6}+...$

So,
$$S = S'-S' = 0$$

Method 2:

Using the formula for infinite Geometric Progression, we can say
$$ 1 + x + x^2 + x^3 + ... = \frac{1}{1-x}$$
Integrating both sides with respect to $x$ from $0$ to $t$ we get,
$$[x+{x^2\over2}+{x^3\over3}+{x^4\over4}+...]^t_0 = [-ln(1-x)]^t_0$$
$$t+{t^2\over2}+{t^3\over3}+{t^4\over4}+... = -ln(1-t)$$
Putting $t = -1$,
$$-1+{1^2\over2}-{1^3\over3}+{1^4\over4}-... = -ln(2)$$
$$1-{1\over2}+{1\over3}-{1\over4}+... = ln(2)$$
So,
$$S=ln(2)$$
After some searching I found that the 2nd method gives the correct answer. But why is the 1st method wrong?

答:
I want to add a (minor?) point to the other answers.  You considered ∑aₙ and ∑(-1)ⁿaₙ in the case aₙ=n⁻¹.  What happens when one replaces this with aₙ=nˢ?

Answer: For any s except s=‒1 your argument “would work”.

(Quotes are needed since I assume a particular way of regularization of divergent series for s≥‒1.  However, I never saw any other way meaningfully used in the case of power-of-n series.)

Indeed, for s<‒1 your approach works due to the convergence of ∑aₙ.  For s>‒1, both series may be regularized by Dirichlet method; the key observation is that Dirichlet regularization is preserved by dilution (which replaces a₁+a₂+… by 0+a₁+0+a₂+0+…).  (Note that the alternating series is also Chesaro/Abel/Borel summable — to the same answer as by Dirichlet.  The Chesaro summability is almost completely elementary.)

However, for s=‒1 the Dirichlet regularization gives ∞ (due to a pole of ζ(z) when z=1).  So in your calculation you get ∞‒∞, and you see that it is not necessarily 0.

Essentially, what you demonstrated is that when one gets ∞ with an “advanced” summation method, one should be as careful as with the naive summation method.


答:
The answer is

$$3^n+ \sum_{k=1}^n 3^{k-1}\cdot 3^{n-k}=n3^{n-1}+3^n.$$

There are $3^n$ possible words with no occurrence of $b$.  Assume the first $b$ occurs at position $k$.  Then each of the first $k-1$ positions has three possible choices ($a, c, \text{ or } d$), and each of the last $n-k$ positions has three possible choices ($b, c, \text{ or } d$).

Title: How many "words" of length n is it possible to create from {a,b,c,d} without any "a"s after a "b"?

Body:
I think this is a hard question and I didn't come up with a solution. Could you please give me a hint or approach?

For instance, "bdca" is not ok, but "aabcd" is ok.

It is somehow similar to this question, but with some differences.

Thanks in advance.


问:
example of a flat but not faithfully flat ring extension
I am learning commutative algebra and there is a definition about faithfully flat modules or ring extensions. I can't think of an example of a flat but not faithfully flat ring extension or module. Can someone help me with it? thanks</p>

答: Take $f: A \rightarrow B$ to be a ring homomorphism such that the corresponding morphism of affine schemes $\operatorname{Spec}B \rightarrow \operatorname{Spec}A$ is not surjective, but only flat. There is an easy way to do this: Remember that localizing a ring $R$ in a multiplicative subset $S$ gives a flat ring homomorphism $R \rightarrow S^{-1}R$. However, this ring homomorphism is faithfully flat iff $S^{-1}R=R$.


问:
Among 26 natural numbers from $\{1,2,3,...,100\}$ there are at least $2$ integers whose difference is atmost $3$
The heading says the question. Is it true that we can find 2 integers with their difference atmost 3 ?

I have taken pairs with difference $3$ and after that kind of lost.

Any help will be appreciated. Thank you.

答:
hint pigeonhole

$100 \div 4$

..

Body:
hint pigeonhole

$100 \div 4$

..


答: Hint: Among $3$ natural numbers from $\{1,2,3,4,5,6,7,8\}$ there are at least $2$ integers whose difference is at most $3$. Do you believe this? Why?

Title: Among 26 natural numbers from $\{1,2,3,...,100\}$ there are at least $2$ integers whose difference is atmost $3$

Body:
The heading says the question. Is it true that we can find 2 integers with their difference atmost 3 ?

I have taken pairs with difference $3$ and after that kind of lost.

Any help will be appreciated. Thank you.


问:
Asymptotic expansion of the inverse of $x\mapsto x+x^\phi$ near zero
Consider a continuous real-valued monotone increasing function $f:\mathbb R^+\to\mathbb R^+$ satisfying $f\!\left(x+x^\phi\right)=x,$ where $\phi=\frac{1+\sqrt5}2$ is the golden ratio. Here is a plot of the function $f(z)$:

I need to find an asymptotic expansion of $f(z)$ for $z\to0^+$ in terms of powers of $z$. By manually balancing coefficients, I was able to find a few first terms:
$$f(z)=z-z^\phi\color{red}+\phi\,z^{2\,\phi-1}+\mathcal O\left(z^{3\,\phi-2}\right), \quad z\to0^+.$$
How can I find more terms and, ideally, a general formula for all terms of this series?

Update: A wrong sign in formula is corrected (red).

答:
I notice that factoring $z$ out yields $f^{-1}(z)=z\big(1-z^{1/\varphi}\color{Red}{+}\varphi z^{2/\varphi}+\mathcal{O}(z^{3/\varphi})\big)$. Thus, let's write

$$ f^{-1}(z)=z\sum_{k=0}^{\infty} a_k z^{k/\varphi} $$

for some coefficients $a_k$ with $a_0=1$. Substitute $z=x+x^{\varphi}=x(1+x^{1/\varphi})$ to get

$$ x=x(1+x^{1/\varphi})\sum_{k=0}^{\infty} a_k(1+x^{1/\varphi})^{k/\varphi}x^{k/\varphi}. $$

Divide by $x(1+x^{1/\varphi})$, then expand Newton-Binomial series:

$$ \frac{1}{1+x^{1/\varphi}} = \sum_{k=0}^{\infty} a_kx^{k/\varphi}(1+x^{1/\varphi})^{k/\varphi} $$

$$ \begin{array}{ll} \displaystyle \sum_{n=0}^\infty (-1)^nx^{n/\varphi} & \displaystyle = \sum_{k=0}^{\infty} a_k x^{k/\varphi}\sum_{j=0}^{\infty} \binom{k/\varphi}{j}x^{j/\varphi} \\ & \displaystyle =\sum_{n=0}^\infty x^{n/\varphi} \left[\sum_{k=0}^n \binom{k/\varphi}{n-k} a_k\right]. \end{array} $$

Equating coefficients yields a recurrence for $a_n$:

$$ a_n=(-1)^n-\sum_{k=1}^{n-1} \binom{k/\varphi}{n-k}a_k. $$

This also shows there's a sign error in your quadratic term.


问:
Forward difference of Bernoulli polynomials
I am working on a research project on Bernoulli polynomials, which are defined as
$$B_n(x) = \sum_{j} \binom{n}{j}B_j x^{n-j}$$
where $B_j$ are the Bernoulli numbers. There is also the "generating function"
$$\frac{ze^{zx}}{e^z-1} = \sum_n B_n(x)\frac{z^n}{n!}$$
Many papers I have found cite the following property and use it extensively in their proofs:
$$\Delta B_n(x) = n x^{n-1}$$
where $\Delta$ is the forward-difference operator from discrete calculus. However, this fact is always stated without proof, and I am wondering what the justification is.

答:
One way is to use the generating function:
\begin{align}
  \sum_n \Delta B_n(x) \frac{z^n}{n!}
    &= \sum_n(B_n(x+1) - B_n(x)) \frac{z^n}{n!} \\
    &= \frac{z e^{z(x+1)}}{e^z-1} - \frac{z e^{zx}}{e^z-1} \\
    &= z e^{zx} \\
    &= \sum_m x^m \frac{z^{m+1}}{m!} \\
    &= \sum_n x^{n-1} \frac{z^n}{(n-1)!}
\end{align}
Then compare coefficients.


问:
Prove Closed Set
From Stephen Abbott Understanding Analysis book, exercise 3.2.2 there is a set $A$ :
$$A=\left\{(-1)^n + \frac{2}{n}:n=1,2,3,...\right\}$$
I know that this is not an open set because no neighborhood is contained
in $A$, but then how do I properly prove that $A$ is a closed set based on the definition in the book : "A set $F\subseteq\Bbb R$ is closed if it contains its limit points"?

答:
A simple way to prove this from your definition is to first identify the limit points and then to verify that $A$ contains them.

First, note that $(-1)^n + \frac{2}{n}$ for $n = 1, 2, 3, \dots$ splits into two convergent subseries:

$$
\lim_{k\to\infty} (-1)^{2k} + \frac{2}{2k} = 1
$$

and

$$
\lim_{k\to\infty} (-1)^{2k+1} + \frac{2}{2k+1} = -1.
$$

Therefore, it has exactly two limit points: $-1$ and $1$.

Finally, $1 \in A$, but $-1 \notin A$, so $A$ does not contain all of its limit points. Therefore, $A$ is not closed.

Title: Prove Closed Set


问:
Comparing $\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$
Without the use of a calculator, how can we tell which of these are larger (higher in numerical value)?

$$\sqrt{1001}+\sqrt{999}\ , \ 2\sqrt{1000}$$

Using the calculator I can see that the first one is 63.2455453 and the second one is 63.2455532, but can we tell without touching our calculators?

Body: $\sqrt{1001}+\sqrt{1000}>\sqrt{1000}+\sqrt{999} \\ \implies \dfrac{1}{\sqrt{1001}-\sqrt{1000}}>\dfrac{1}{\sqrt{1000}-\sqrt{999}}\\ \implies\sqrt{1000}-\sqrt{999}>\sqrt{1001}-\sqrt{1000} \\ \implies 2\sqrt{1000}>\sqrt{999}+\sqrt{1001}$


问:
Proving or disproving a claim about limits

I was asked to prove or disprove this claim. I've tried to find a counter example but I've got a feeling now that the statement is correct.
My guess is they want us to prove it using the squeeze rule, but how would I approach this?
I can see straight away that $0 < f(x) \le1$, but how does that help me? should I perhaps try to use the contrapositive or proof by contradiction?

答:
Hints

Does $f(x)$ have to be a positive function?

Can $f(x)$ be positive for $x < 0$?

Can $f(x)$ have two specifications, one for positive $x$ and one for negative $x$?

Does $g(x)$ have to be a positive function?

Can you define a function $f(x)$ such that $xf(x)$ goes towards $-\infty$ as $x$ approaches $0$ from above?


答:
Transition matrix with two states, the string has an $b$ or it doesn't.  State 0 will be no bs and state 1 will be some bs.  $M_{i, j}$ is "how many ways can we transist from a string of type $i$ to a string of type $j$" (by appending characters) :

$$M = \begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}$$

For example, $M_{1, 0} = 0$ there is no way to go from a string with some bs to a string with no bs.  $M_{1, 1} = 3$ there are 3 ways to go from a string with some bs to a string with some bs (by appending b, c, or d).

Initial state is $V=[1, 0]$ since the empty string is one string which does not contain a $b$.  Final state can be either containing or not containing a string so it is $F=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$

$$C = VM^nF$$ counts the number of ways to go from the initial state (empty string) to the final state (any string reachable).  $M$ is a jordan normal form matrix so a closed form for it's exponent is well known:

$$C = [1, 0]\begin{bmatrix} 3^n & n \cdot 3^{n-1} \\ 0 & 3^n \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = 3^n + n3^{n-1}$$


答:
Suppose , not then there exist $a_{1}\leq a_{2}\leq ...\leq a_{26}$ such that $a_{j+1}-a_{j}\geq 4$ for $j=1,2,...,25$
Then $a_{26}-a_{1}=\sum_{j=1}^{25} a_{j+1}-a_{j}\geq 100$
But we know $a_{26}-a_{1}\leq 99$ since it's belongs to that set $\{1,2,..,100\}$.This possible contradiction leads to the solution.


问:
Why is it that for $n<60$, the simple groups are precisely the cyclic groups $\mathbb{Z}_N$ for prime $n$?
I was reading this Wikipedia article and came across this

For groups of order $n<60$, the simple groups are precisely the cyclic groups $\mathbb{Z}_n$ for prime $n$.

I was wondering, why is this true for $n<60$? What is so special about this value?

答:
Note that 60 is the least non square free product of 3 powers of distinct primes(= $2^2\bullet3\bullet5$), all non-negative integers less than 60 is one of the following cases and any group of this order is either abelian or not simple or both:

p-group : Either cyclic (abelian) or not-simple (By the 1st Sylow)

$p^a q^b$ : Burnside's $p^a q^b$ theorem states they are not simple.

pqr : Assume some group of order pqr is simple and p<q<r then calculate the smallest numbers can be $n_p, n_q, n_r$, then count the smallest number of elements of Sylow p,q,r-subgroups reveals that the sum is exceed pqr which is a contradiction.

0 : $<e>$, which is abelian and not simple since everything is equal to e, and has no proper normal subgroup.


问:
Proving a group is a normal subgroup from its order
Suppose $G$ and $H$ are groups such that $H$ is a subgroup of $G$, and $o(G)=120$ and $o(H)=24$, and there exists a in $G-H$ such that $aH=Ha$.
Prove that $H$ is a normal subgroup of $G$.

I've been stuck on this for days now. have tried Lagrange theorem, indexes, and many other methods with no luck. please help.

答:
Consider the normalizer $N_G(H)$. We can say that $H\subseteq N_G(H)\subseteq G$.

By Lagrange's theorem, $|H|$ divides $|N_G(H)|$ which also divides $|G|$.

Since $[G:H]=5$ is prime, there are no intermediate divisors between $|H|$ and $|G|$, so we conclude that $|N_G(H)|$ must be one of these, so $N_G(H)$ is one of $H$ or $G$. Can you finish?


答:
I assume $a \notin H$.

If $a^2 \in H \cup aH$ then $H \cup aH$ is a subgroup of $G$. By Lagrange theorem, this is impossible. Similarly, we see that $a^3 \notin H \cup aH \cup a^2H$ and $a^4 \notin H \cup aH \cup a^2H \cup a^3H$. Thus, we have five disjoint cosets of $H$ in $G$: $H$, $aH$, $a^2H$, $a^3H$, $a^4H$. Their union is all of $G$.

Also, $a^kH = Ha^k$ for every $k=0, 1, 2, 3, 4$. Thus, $H$ is normal.

Title: Proving a group is a normal subgroup from its order

Body:
Suppose $G$ and $H$ are groups such that $H$ is a subgroup of $G$, and $o(G)=120$ and $o(H)=24$, and there exists a in $G-H$ such that $aH=Ha$.
Prove that $H$ is a normal subgroup of $G$.

I've been stuck on this for days now. have tried Lagrange theorem, indexes, and many other methods with no luck. please help.


问:
How solve this probability problem?
We have $n$ beds labeled as $B_1, B_2,.., B_n$ and we have $n$ person labeled as $P_1, P_2, ..., P_n$. We assign each bed to one person. What is the probability of which person and bed subscripts don't equal? For instance, we can't assign $B_1$ to $P_1$.

If $n=2$, the solution is $1\over 2$

If $n=3$, the solution is $1\over 3$

If $n=4$, the solution is $3\over 8$

I just wrote all possibilities for $n=2,3,4$.

What about $n$ bed and people?

答: This a classic derangement problem. Check it out here</p>

Title: How solve this probability problem?


问:
Minimal Polynomials Annihilating an Abelian Torsion-Free Group
Let $A$ be an abelian torsion-free group. Let $\theta \in\operatorname{Aut}A$. Assume that $\theta$ has a finite period in $\operatorname{Aut} A$, say $n$. Obviously $\theta^n-1$ annihilates $A$ (i.e. $A^{\theta^n-1}=\{0\}$, where $\theta^n-1$ is now the obvious endomorphism of $A$).
Thus, there is a polynomial of minimal degree with integral coefficients annihilating $A$, say $g(\theta)$. Can we assume that it is also a monic polynomial?

I would like to say that, for instance, $g(\theta)\mid \theta^n-1$.

答: If you've studied injective envelopes or tensor products this can be handled in a smooth way. Let $E(A) = A \otimes \mathbb{Q}$ be the injective envelope of $A$. Then $\phi: A \to A$ can be considered $\phi:A \to E(A)$ and since $A \leq E(A)$ and $E(A)$ is injective, there is an extension $\tilde \phi : E(A) \to E(A)$. Since $E(A)$ is torsion-free the extension turns out to be unique and still an automorphism. Since $E(A)$ is a vector space over $\mathbb{Q}$, you can use all your old minimal polynomial ideas. This works over commutative domains (and Ore domains), but be careful when you use Gauss's lemma if the domain is not a UFD. :-)</p>


问:
Weak derivatives of function defined over $\Bbb C$

Compute $D u$ in the sense of distributions, where
$u =(x+iy)^{-1}$ is a function defined in $\mathbb{C}$ and
$D=\frac{\partial }{\partial x}+i\frac{\partial }{\partial y}$.  (Hint:  use the integration by parts
formula)

The Solution to the question is $2\pi\delta$

I don't know how to solve it Please help me.

答:
Show that $D\frac1{x+iy}$ is a radial distribution, with $\Phi(z)=\int_0^1 \phi(e^{2i\pi t}z)dt=\varphi(|z|^2)$ we get

$$\langle D\frac1{x+iy},\phi \rangle=
\langle D\frac1{x+iy},\Phi \rangle=-\langle \frac1{x+iy}, D\Phi\rangle$$ $$=-\int_0^\infty \frac1{re^{it}} \int_0^{2\pi}  D\Phi(re^{it}) dt rdr=-\int_0^\infty \frac1{re^{it}}  \int_0^{2\pi} \varphi'(r^2) 2re^{it} dt rdr$$ $$=-2\pi \int_0^\infty \varphi'(r^2)2rdr=2\pi \varphi(0)=2\pi \phi(0)$$


问:
Proving basic norm inner product inequality using first principles
Say I have an inner product space $(\mathbb{V}, \langle.,.\rangle)$. Now I want to prove the inequality (without using Cauchy Schwarz):

$$\|u\|^2 + \|v\|^2 \geq 2 |\langle u,v \rangle|$$.

So I started using the following property of inner products:

$$\langle v-u,v-u\rangle \geq 0$$ which will simplify to:

$$\langle v,v\rangle + \langle u,u\rangle \geq \langle v,u\rangle + \langle u,v\rangle$$

and now I am stuck. Because $\mathbb{V}$ can be a complex vector space I don't know how to move forward. How to get the required statement?

答:
Hint:  $\|u-cv\|^{2}\geq 0$ for any complex number $c$. Put $c=\frac {\langle v, u \rangle} {\|v\|^{2}}$ and simplify using the fact that $\|a+b\|^{2}=\|a\|^{2}+\|b\|^{2}+2 \Re \langle a, b \rangle$.

You get $| \langle u, v \rangle | \leq \|u\|\|v\|$ so $2| \langle u, v \rangle | \leq 2\|u\|\|v\| \leq \|u\|^{2}+\|v\|^{2}$.


问:
Why the null space of quotient map is $U$?
I am reading the textbook Linear Algebra Done Right Chapter 3 section E on Products and Quotients of Vectors Spaces.

It tried to prove the dimension of a quotient space is equal to $\text{dim }V/U = \text{dim }V -\text{dim }U$.

Before that it defines the quotient map $\pi$ as follow:

Suppose $U$ is a subspace of $V$. The quotient map $\pi$ is the linear map $\pi:V \to V/U$ defined by
  $$\pi(v) = v+U$$
  for $v \in V$.

I can understand that the range of $\pi$ is $v+U$ which is $V/U$ according to the definition of $V/U$.

But I don't understand why the null space of $\pi$ is $U$. The book said it is due to the proof like this following:

Suppose $U$ is a subspace of $V$ and $v,w \in V$. Then the following are equivalent:
  $$v-w \in U$$
  $$v+U=w+U$$
  $$(v+U) \cap (w+U) \neq \emptyset$$

答:
Here's another way of thinking about it. Suppose that $\nu' \in$ $null\ \pi$. Then, we can see that:

$\pi(\nu') = \nu' + U = 0 + U$

*Because $0 + U$ is the additive identity of $V/U$.

And according to the additional proof that you provided, we see that $\nu'- 0 = \nu' \in U$.

So we see that whenever we map $\nu' \in U$ through $\pi$, we will always get $0 + U = U$ because the affine set of $\{\nu' + u : \nu',u \in U\} = \{0 + (\nu' + u)\}$. Since taking any vector from $U$ will map to the null space of $\pi$, we can see now that $null \ \pi = U$.


问:
How can I find all the solutions of $\sin^5x+\cos^3x=1$

Find all the solutions of $$\sin^5x+\cos^3x=1$$

Trial:$x=0$ is a solution of this equation. How can I find other solutions (if any). Please help.

答:
Hint: $ \sin^5 x\leq \sin^2 x$ and $ \cos ^3 x \leq \cos^2 x $.

Hint: Pythagorean Identity for trigonometric functions.


问:
Showing the boundary of a cylinder has Lebesgue-measure zero
Let $\partial C=\{(x,y,z) \in \Bbb{R^3}\enspace\vert  \enspace x^2+y^2 = r, \enspace0\leq z\leq h\}.$ I'm trying to show $\partial C$ has Lebesgue-measure zero in $\Bbb{R^3}$.

Let $f: I \to \Bbb{R^2}$, s.t $f=(r\cos x, r\sin x)$, where $I=[0,2\pi]$. Now $f$ is Riemann-integrable and its graph surface $G_f = B(0,r)\subset\Bbb{R^2}.$
Therefore $m_n(G_f)=0$. Then $\partial C=[0,h] \times G_f$. This cartesian product has Lebesgue measure zero, since $[0,h]$ is bounded in $\Bbb{R^2}$. Therefore $m_n(\partial C)=0$. Is this right?

答:
Note that $\partial C = \{(x,y)| x^2+y^2 = r \} \times [0,h]$, so
$m_3 (\partial C) = h \cdot m_2 \{(x,y)| x^2+y^2 = r \}$. Hence it suffices to
show that the measure of the circle $rS^1$ is zero.

Note that for fixed $x$ the indicator $y \mapsto 1_{r S^1} ((x,y))$ is zero except for at most 2 points. Hence $\int 1_{S^1} = \int_{x \in \mathbb{R}} \int_{y \in \mathbb{R}} 1_{r S^1} ((x,y)) dy dx = \int_{x \in \mathbb{R}} 0y dx = 0$


问:
Function from empty set to empty set
On page 2 of Dummit & Foote's Abstract Algebra (3rd ed.), Proposition 1/(4) states:

(Let $f:A \to B$.) If $A$ and $B$ are finite sets with the same number of elements (i.e., $\lvert A \rvert = \lvert B \rvert$), then $f:A \to B$ is bijective if and only if $f$ is injective if and only if $f$ is surjective.

Firstly, may I confirm that empty set is treated as finite set?

If so, let $A = B = \emptyset$, then we have $\lvert A \rvert = \lvert B \rvert = 0$.

Secondly, may I ask if we could talk about function from empty set to empty set?

If so, to me the definition of such function goes like "for each $x\in A$, there exists a unique $b\in B$ being assigned with x". And this should be vacuously true. So the function is well-defined.

Now, the statement "for each $a,b\in A$, if $a\neq b$, then $f(a)\neq f(b)$" is vacuously true, so the function is injective. Meanwhile, the statement "for each $b\in B$, there exists $a\in A$ such that $f(a)=b$" is vacuously true, so the function is surjective and thus bijective.

Finally, may I ask if this completes the proof for the $\emptyset$ case of Proposition 1/(4)?

答:
Yes, the empty function, from the empty set to the empty set, is vacuously bijective, and your proof is fine.

In fact, any function with the empty set as domain is vacuously injective, but if the codomain is non-empty it won't be surjective.

And if the domain is non-empty, then there are no functions with the empty set as codomain.

Title: Function from empty set to empty set


问:
The probabilty of the a 6-letter word is a palindrome
a cat is walking on the keyboard. Given that the cat walked just on letters (with an equal chance for each letter A-Z) and wrote a 6-letter word, what is the probability that this word is a palindrome.</p>

答:
Let XYZ be the first 3 letters.

The probability that the fourth letter is Z is $1/26$. The probability that the fifth letter is Y is $1/26$ and the probability of the sixth letter being X is $1/26$ as well.

So the probability of your word being a palindrome is $1/26^3$


问:
The letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

Dr. Lizardo writes down a sequence of $n$ letters, where each letter is A, B, or C, and the letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

What I have tried so far

The last letter is predefined in terms of "A or not" based on the first $n-1$ letters. If there are an odd number of As so far, the last one has to be A, otherwise, B or C. Therefore, there are $3^{n-1}$ ways to write the first $n-1$ letters. However, the last letter can either have 1 choice or 2 choices.

I am thinking maybe we need to do $S_n = 3^{n-1} + S_{n-1}$. However, I need a closed form, not a recurrance. How can I do this?

Thank you!

Note: $S_n = 3^{n-1} + S_{n-1}$ seems to be working for the first few values of $n$ (I tested through $n=4$), so finding a closed form of that recurrance would work as well.

答:
Setting $S_0 = 1$ (the empty sequence definitely has an even number of 'A's) and then defining $S_n = 3^{n-1} + S_{n-1}$ for $n \geq 1$ gives
$$S_n = S_0 + \sum_{k=0}^{n=1}3^k = 1 + \frac{1-3^n}{1-3} = \frac{3^n+1}{2}$$

where we have used the formula for the finite geometric series
$$\sum_{k=0}^{n-1}ar^k = \frac{a(1-r^n)}{1-r}$$

Another way to see this is to use generating functions:
$$P(x,y,z) = (x+y+z)^n = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n}} S(i,j,k)x^iy^jz^k$$
is the generating function for the number of sequences $S(i,j,k)$ with $i$ 'A's, $j$ 'B's, and $k$ 'C's.  We want to find $$\sum_{\substack{i,j,k \geq 0 \\ i+j+k=n \\ 2|i}} S(i,j,k)$$

To impose the assumption that $i$ is even, we take the even part with respect to $x$:
$$\frac{P(x,y,z) + P(-x,y,z)}{2} = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n \\ 2|i}} S(i,j,k)x^iy^jz^k$$

and then we just evaluate at $x=y=z=1$:
$$\frac{(1+1+1)^n + (-1+1+1)^n}{2} = \sum_{\substack{i,j,k \geq 0 \\ i+j+k=n \\ 2|i}} S(i,j,k)$$

which we note simplifies once again to $\frac{3^n+1}{2}$


问:
Strictly invariant sets of the rotation transformation on a discrete space.
Fix an integer $n>0$. Consider the space $X=\{a_0,a_1,...,a_{n-1}\}$ with transformation $T:X\to X$ defined by $T(a_i)=a_{i+1(\text{ mod n})}$.

What are the strictly invariant sets of this space? (i.e the $A\subset X$ with $T^{-1}A=A$.)

I think the only strictly invariant sets are $X$ and $\emptyset$, but I am having some trouble proving this claim.

答: Let $A$ be a non-empty invariant set. If $a_j\in A$, then $a_j\in T^{-1}(A)\cap T(A)$, hence $a_{j+1\mod n}$ and $a_{j-1\mod n}$ also are in $A$. This proves that $A=X$.</p>


答:
Good analysis, but a closed form requires that you get down in the mud.

Note : possible exception is if someone can provide a more elegant computation than my kludgy approach.

Let $c = \lfloor \frac{n}{2}\rfloor.$

That is $c$ is the largest integer such that $2c \leq n$.

Then, you want

$$\sum_{k=0}^c f(k)$$,

where $f(k)$ is the exact # of ways of having $2k$ A's in the sequence.

There are $\binom{n}{2k}$ ways of distributing the $2k$ A's.

There are $2^{(n-2k)}$ ways of choosing B or C for the $(n - 2k)$ other slots.

Therefore,

$$f(k) = \binom{n}{2k} \times 2^{(n-2k)}.$$

It is unclear to me whether the summation can be expressed in a simpler fashion.

Title: The letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

Body:
Dr. Lizardo writes down a sequence of $n$ letters, where each letter is A, B, or C, and the letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?

What I have tried so far

The last letter is predefined in terms of "A or not" based on the first $n-1$ letters. If there are an odd number of As so far, the last one has to be A, otherwise, B or C. Therefore, there are $3^{n-1}$ ways to write the first $n-1$ letters. However, the last letter can either have 1 choice or 2 choices.

I am thinking maybe we need to do $S_n = 3^{n-1} + S_{n-1}$. However, I need a closed form, not a recurrance. How can I do this?

Thank you!

Note: $S_n = 3^{n-1} + S_{n-1}$ seems to be working for the first few values of $n$ (I tested through $n=4$), so finding a closed form of that recurrance would work as well.


答: Let $c$ be a complex number of unit modulus such that $\langle u,cv\rangle=|\langle u,v\rangle|$. The inequality in question then follows from $\langle u-cv,u-cv\rangle\ge0$.

Title: Proving basic norm inner product inequality using first principles

Body:
Say I have an inner product space $(\mathbb{V}, \langle.,.\rangle)$. Now I want to prove the inequality (without using Cauchy Schwarz):

$$\|u\|^2 + \|v\|^2 \geq 2 |\langle u,v \rangle|$$.

So I started using the following property of inner products:

$$\langle v-u,v-u\rangle \geq 0$$ which will simplify to:

$$\langle v,v\rangle + \langle u,u\rangle \geq \langle v,u\rangle + \langle u,v\rangle$$

and now I am stuck. Because $\mathbb{V}$ can be a complex vector space I don't know how to move forward. How to get the required statement?


问:
Is a sequence of all the same numbers monotonic?
I'm wondering based on the definition of monotonicity:

A sequence where $a_n\geq a_{n+1}$ for all $n\in\mathbb{N}$ is monotonic.

So given that the sequence $a_n = 3$ is all the same numbers and is neither increasing or decreasing, is it monotonic?

答:
Yes, a constant sequence (a number repeated indefinitely) is inceed monotonic: it is both monotonic non-decreasing, and monotonic non-increasing.

Hence, one can require that a sequence be strictly monotonic increasing or strictly monotonic decreasing. Under such a restriction, a constant sequence is neither strictly increasing nor strictly decreasing monotonically.


问:
Finding Intersection given probabilities
If I have events $A$ and $B$ where $P(A) = 0.20$ and $P(B) = 0.13$. Can we find the $P (A \cap B)$ and show that events are independent?

答:
The independence of two events $A$ and $B$ can be characterized by their joint probability. Two events $A$ and $B$ are independent if and only if their joint probability of $A\cap B $ equals the product of their individual probabilities.

$P(A\cap B)=P(A).P(B)$ implies $P(A \lvert B)= \frac{P(A\cap B)}{P(B)}=P(A)$


问:
Picard theorem in complex analysis - Example
Let $f:\mathbb{C}\rightarrow \mathbb{C}$ be a non-constant entire function. Is it true that for all $a,b > 0$, there exists $s \in \mathbb{C}$ such that $|s|>a$ and $|f(s)|>b$.

I have picard's theorem in mind, but it only says about $f(s)$ , i am not sure how to prove the part $|s|>a$.

Any ideas how to prove or disprove ?
Thanks

答:
It is true. Otherwise there would be numbers $a, b > 0$ such that $|f(z)| \le b$ for all $z$ with $|z| > a$. That would imply that $f$ is bounded
by
$$
 M = \max(b, \, \max_{|z| \le a } |f(z)| )
$$
in contradiction to Liouville's theorem (and also contradicting Picard's little theorem).


问:
Forward difference of Bernoulli polynomials
I am working on a research project on Bernoulli polynomials, which are defined as
$$B_n(x) = \sum_{j} \binom{n}{j}B_j x^{n-j}$$
where $B_j$ are the Bernoulli numbers. There is also the "generating function"
$$\frac{ze^{zx}}{e^z-1} = \sum_n B_n(x)\frac{z^n}{n!}$$
Many papers I have found cite the following property and use it extensively in their proofs:
$$\Delta B_n(x) = n x^{n-1}$$
where $\Delta$ is the forward-difference operator from discrete calculus. However, this fact is always stated without proof, and I am wondering what the justification is.

答:
You can think of the space of polynomials as the vector space generated by the monomials $1$, $x$, $x^2$, $x^3$, $\ldots$. The Bernoulli polynomials are a different basis for the same vector space and can be defined by the property that the linear map $L$ that transforms each monomial into the corresponding Bernoulli polynomial,
$$
Lx^n=B_n(x),
$$
transforms the forward difference of a polynomial into the derivative of that polynomial:
$$
L\Delta f(x)=\frac{d}{dx}f(x).
$$
(Writing down the system of linear equations relating the forward difference of $f$ to the derivative of $f$ and then solving that system is one way to obtain $L$ and hence the Bernoulli polynomials. Your summation formula for the Bernoulli polynomials and the recurrence for the Bernoulli numbers, $\sum_{k=0}^n\binom{n+1}{k}B_k=\delta_{n,0}$, naturally emerge from the solution.)

Now the property you want to prove has $L$ and $\Delta$ in the opposite order: you want to prove
$$
\Delta B_n(x)=\Delta L x^n=\frac{d}{dx} x^n
$$
or, more generally, $\Delta L f(x)=\frac{d}{dx}f(x)$.

In fact, the three operators $L$, $\Delta$, and $\frac{d}{dx}$, which all act on the space of polynomials, are mutually commuting,
$$
\frac{d}{dx}\Delta f(x)=\Delta\frac{d}{dx}f(x),\quad \frac{d}{dx}Lf(x)=L\frac{d}{dx}f(x),\quad \Delta Lf(x)=L\Delta f(x).
$$
The first of these follows from the chain rule; the second can be checked by verifying from your definition that $B'_n(x)=nB_{n-1}(x)$, which is the rule in the case $f(x)=x^n$, and using linearity of $L$ and $\frac{d}{dx}$. The third one, which is the one you want to prove, then follows from the second: apply $L\Delta f(x)=\frac{d}{dx}f(x)$ to the polynomial $Lf(x)$ to obtain $L\Delta Lf(x)=\frac{d}{dx}Lf(x)=L\frac{d}{dx}f(x)$. Since $L$ is an invertible operator, it follows that $\Delta Lf(x)=\frac{d}{dx}f(x)$.

By the way, the operators $\frac{d}{dx}$ and $\Delta$ are not invertible, a fact that amounts to the observation that the antiderivative and the anti-forward-difference are not uniquely defined.

Added: This expands on the parenthetical remark in the first paragraph. Let $f(x)=\sum_{i=0}^n f_ix^i$ be a polynomial of degree $n$ and let $f'(x)=\sum_{i=0}^{n-1}g_ix^i$, $\Delta f(x)=\sum_{i=0}^{n-1}h_ix^i$. Comparing the first and last lines of
\begin{align}
\sum_{i=0}^{n-1}h_i x^i &= \Delta f(x)\\
&=f(x+1)-f(x)\\
&=\sum_{i=0}^nf_i((x+1)^i-x^i)\\
&=\sum_{i=1}^nf_i\sum_{j=0}^{i-1}\binom{i}{j}x^j\\
&=\sum_{i=1}^n if_i\sum_{j=0}^{i-1}\frac{1}{i}\binom{i}{j}x^j\\
&=\sum_{i=0}^{n-1} g_i\sum_{j=0}^{i}\frac{1}{i+1}\binom{i+1}{j}x^j
\end{align}
gives a system of linear equations relating the $h_i$ to the $g_i$. The coefficient matrix is upper triangular, so its inverse is found by straightforward iteration. Carrying out the inversion gives
$$
\sum_{i=0}^{n-1}g_ix^i=\sum_{i=0}^{n-1}h_iB_i(x).
$$
This is where the prescription "replace every power $x^i$ in $\Delta f(x)$ with $B_i(x)$ to obtain $f'(x)$" comes from.
Taking this linear algebra problem as the definition of the Bernoulli polynomials allows a great many things to be easily explained, in my view.


答: Yes, every constant sequence is monotone, in fact simultaneously monotone non-decreasing and monotone non-increasing. </p>

Title: Is a sequence of all the same numbers monotonic?

Body:
I'm wondering based on the definition of monotonicity:

A sequence where $a_n\geq a_{n+1}$ for all $n\in\mathbb{N}$ is monotonic.

So given that the sequence $a_n = 3$ is all the same numbers and is neither increasing or decreasing, is it monotonic?


答:
This is a bit more of a linear algebra proof which is probably not what you are seeking but I'd still like to share it anyways. Define two sequences $a_n$ being the number of length-$n$ sequences with even number of As and $b_n$ being the number of length-$n$ sequences with odd number of As. Clearly $a_0 = 1$ and $b_0 = 0$

Notice that the if you have an length-$(n-1)$ sequences with even number of As: there are two ways to add one character (either B or C) so that the new length-$n$ sequence has even As and there is one way to add one character (A) so that the new length-$n$ sequence has odd As.

Similarly if you have an length-$(n-1)$ sequences with odd number of As: there is one way to add one character (A) so that the new length-$n$ sequence has even As and there are two ways to add one character (B or C) so that the new length-$n$ sequence has odd As.

From here we can define a recurrence relation in terms of matrices
$$ \left( \begin{matrix} a_{n} \\ b_{n} \end{matrix} \right) =
\left( \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right)
\left( \begin{matrix} a_{n-1} \\ b_{n-1} \end{matrix} \right) =
\left( \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right)^n
\left( \begin{matrix} a_{0} \\ b_{0} \end{matrix} \right)$$

If we diagonalize this matrix we get
$$ \left( \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right) =
\left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)
\left( \begin{matrix} 1 & 0 \\ 0 & 3 \end{matrix} \right)
\frac{1}{2} \left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)$$

$$ \left( \begin{matrix} a_{n} \\ b_{n} \end{matrix} \right) =
\left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)
\left( \begin{matrix} 1 & 0 \\ 0 & 3^n \end{matrix} \right)
\frac{1}{2} \left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)
\left( \begin{matrix} 1 \\ 0 \end{matrix} \right) =
\frac{1}{2} \left( \begin{matrix} 3^n+1 \\ 3^n-1 \end{matrix} \right)$$

thus $a_n = \frac{3^n+1}{2}$


问:
Expressing a vector as a sum of eigenvectors
Let $A$ be a positive definite matrix and $x$ be arbitrary vector. Show that $x$ can be written as $x = \sum y_i$, where each $y_i$ is an eigenvector of $A$ corresponding to eigenvalue $\lambda_i$ and $\lambda_i$ are all different.

I would be glad for any help with this problem.

答: A real positive-definite matrix is necessarily symmetric, thus diagonalizable, thus possesses a basis of eigenvectors, so any vector can be written as $x=\sum y_i$ with $y_i$ eigenvectors. Summing all $y_i$ with the same eigenvalue is still an eigenvector with this same eigenvalue, so we can make the $y_i$ in the sum to have different eigenvalues.


问:
derivative $f(z)=\sqrt [n] {g(z)}$ and $f(z)=\log(g(z))$
How I can compute the derivative at a point of function $f(z)=\sqrt [n] {g(z)}$ and $f(z)=\log(g(z))$ if I know $f(z_0)$ for a fixed $z_0$ and if I have a suitable plane cut?

Is it true that we can not care about $\log(g(z))$ because every brunch has the same derivative and we can say that $(\log (g(z))’(a)= \frac {g’(z)}{g(z)}(a)$?

答:
It is true that for $f(z) = \log(g(z))$ one has
$$
 f'(z) = \frac{g'(z)}{g(z)}
$$
independently of the branch. That reflects the fact that different local branches of the logarithm differ by a constant.

That is not the case for $f(z)=\sqrt [n] {g(z)}$ where different local branches differ by a constant factor. Taking the logarithmic derivative of $(f(z))^n = g(z)$ gives
$$
 f'(z) = \frac 1n f(z) \frac{g'(z)}{g(z)}
$$
so that $f'(z)$ depends on the branch chosen for $f(z)$.


答:
Knowing only probabilities $P(A)=0.20$ and $P(B)=0.13$ it is impossible to unambiguously define probability for $P(A \cap B)$, which can have any value from $[0,0.13]$, so is not possible to determine independence.

Also let me point out that for the given probabilities for $A,B$, if the intersection is empty $A \cap B = \emptyset$, then $A,B$ are, of course, dependent.

Title: Finding Intersection given probabilities

Body: If I have events $A$ and $B$ where $P(A) = 0.20$ and $P(B) = 0.13$. Can we find the $P (A \cap B)$ and show that events are independent?


问:
If a model predicts that a loan will default, what is the probability that it will actually default?
The loan default at a certain bank is $1\%$. the bank has developed a model to try to predict which loans will default. The model's predicted positive rate is $3\%$; $99\%$ of the loans that default are predicted as such by the model. If a model predicts that a loan will default, what is the probability that it will actually default?

答:
This is a simple application of Bayes formula

$$P(\text{default} | \text{predicted default})=\frac{P(\text{predicted default} | \text{default})P(\text{default})}{P(\text{predicted default})}$$

$$=\frac{0.99\times 0.01}{0.03}=0.33$$


问:
There are at least two primes with power 1 in $k!$ for $k>11$
My question is: need to prove that $\forall \ k \geq 11, \exists$ at least 2 prime numbers with power of 1, in k! factorization. I know I missed something, and that is why I'm looking for a hint. Thanks for help!

*** I can use that $\forall \ k > 5$ there exist at least $2$ prime numbers in the next interval: $(k,2k)$.

答:
For a fixed $k$, consider the prime numbers in the interval $(\lceil \frac{k}{2} \rceil, 2 \lceil \frac{k}{2} \rceil)$ (observe that $k/2 > 5$). According to the result you said you are allowed to assume, there are at least two such prime numbers.

Note that if $p$ is in that interval, then $2p > n$ so $p$ is the only multiple of $p$ in the interval $[1,k]$. Then, $p$ appears only once in the factorization of $k!$. This is because to count the power of $p$ in $k!$ we only need to look at multiples of $p$ (since $p$ is prime, those are the only ones that contribute a power of $p$ to $k!$). But the only multiple is $p$, so the power of $p$ is $1$ in the factorization.

Since this is true for at least two different prime numbers, you get your result.

EDIT: You don't need to distinguish between $k$ even or odd. $\lceil \frac{k}{2} \rceil$ is the integer $\geq k$ that is closest to it, so you have $\lceil \frac{k}{2} \rceil > \frac{k}{2}$. Hence, $k+1 \geq 2 \lceil \frac{k}{2} \rceil > k$ and you can apply the reasoning above regardless of the parity of $k$.