问:
Inequalities proofs in Spivak's Calculus
From Chapter 1, page 9 of Spivak's Calculus, 3rd edition:

In fact, it is convenient to consider the collection of all positive numbers, denoted by $P$, as the basic concept, and state all properties in terms of $P$:

(P10) (Trichotomy law) For every number $a$, one and only one of the following holds:
(i) $a=0$, (ii) $a$ is in the collection $P$, (iii) $-a$ is in the collection $P$.

(P11) (Closure under addition) If $a$ and $b$ are in $P$, then $a+b$ is in $P$.

(P12) (Closure under multiplication) If $a$ and $b$ are in $P$, then $a\cdot b$ is in $P$.

This is question 8 on page 15:

Although all the basic properties of inequalities were stated in terms of the collection $P$ of all positive numbers, and < was defined in terms of $P$, this procedure can be reversed. Suppose that P10-P12 are replaced by

(P'10) For any numbers $a$ and $b$ one, and only one, of the following holds: (i) $a=b$, (ii) $a<b$, (iii) $b<a$.

(P'11) For any numbers $a$, $b$, and $c$, if $a<b$ and $b<c$, then $a<c$.

(P'12) For any numbers $a$, $b$, and $c$, if $a<b$, then $a+c<b+c$.

(P'13) For any numbers $a$, $b$, and $c$, if $a<b$ and $0<c$, then $ac<bc$.

Show that P10-P12 can then be deduced as theorems.

And here's the solution as in the manual:

Two applications of P'12 show that if $a<b$ and $c<d$, then $a+c<b+c<b+d$, so $a+c<b+d$ by P'11. In particular, if $0<b$ and $0<d$, then $0<b+d$, which proves P11. It follows, in addition, that if $a<0$, then $-a>0$; for if $-a<0$ were true, then $0=a+(-a)<0$, contradicting P'10. Consequently, any number $a$ satisfies precisely one of the conditions $a=0$, $a>0$, $a<0$, the last being equivalent to $-a>0$. This proves P10. Finally, P'13 shows that if $0<a$ and $0<c$, then $0<ac$, which proves P12.

My question is, why doesn't P10 immediately follow from P'10, by substituting $0$ for $b$ ?

答:
Because then what we would have would be:

For any number $a$, one, and only one, of the following holds: (i) $a=0$, (ii) $a<0$, (iii) $0<a$.

It is not obvious that you can jump from this to:

For any number $a$, one, and only one, of the following holds: (i) $a=0$, (ii) $0<-a$, (iii) $0<a$.

You will need some connection between $<$ and the arithmetic operations in $\Bbb R$.


问:
Convergence/Divergence of summation
Is it possible to determine if the following sum converges as $k$ tends to infinity? Wolfram Alpha gave a version in terms of the incomplete gamma function however I didn't see how that could help me.
$$\sum_{n=0}^{k}\frac{k!\left(-1\right)^{n}}{x^{n}\left(k-n\right)!}$$

答:
Call your sum $f_k(x)$.  The exponential generating function of this is

$$ \eqalign{g(x,t) &= \sum_{k=0}^\infty f_k(x) t^k/k!\cr &= \sum_{n=0}^\infty \sum_{k=n}^\infty \frac{(-1)^n}{x^n (k-n)!} t^k \cr
&= \sum_{n=0}^\infty \frac{(-1)^n t^n e^t}{x^n} \cr
&= \frac{x e^t}{t+x} }$$
converging if $|t| < |x|$.  This has a pole at $t = -x$.  But if $f_k(x)$ converged
as $k \to \infty$, $g(x,t)$ would converge for all $t$, with no pole.  So it must diverge, in fact it must be unbounded.

ContentLicense: CC BY-SA 4.0

Title: Convergence/Divergence of summation


问:
$\int _{B_\epsilon (0)}\frac{1}{|y|^{n-2}}dy=nw_n\int^\epsilon_0\frac{r^{n-1}}{r^{n-2}}dr$
$\int _{B_\epsilon (0)}\frac{1}{|y|^{n-2}}dy=nw_n\int^\epsilon_0\frac{r^{n-1}}{r^{n-2}}dr$
where $w_n$ is a volume of unit sphere of $n$ dimension, and $y\in\mathbb{R}^n$.

This is an elementary change of variable calculus, but could anybody help me show how this equality holds?

答:
Please be careful with the terms as they both tend to infinity! But the proof you were looking is given by an application of "Cavalieri’s principle":

$\int \limits_{B_{\varepsilon}}\frac{1}{|y|^{n-2}}dy=\int \limits_0^{\varepsilon} \int \limits_{\partial B_{r}}\frac{1}{|\xi|^{n-2}}d\xi dr =\int \limits_0^{\varepsilon}\frac{1}{r^{n-2}}\cdot vol(\partial B_r)dr=n\cdot \omega_n\int \limits_0^{\varepsilon} \frac{r^{n-1}}{r^{n-2}}dr$

where $vol(\partial B_r)=n\cdot \omega_n r^{n-1}.$


问:
An inconsistency between the nabla operator in differential geometry and classical mechanics
I've recently finished a course in differential geometry, and I was having a look at some of my old classical mechanics notes. I found the following written in my notes:

Under the influence of a conservative potential energy $U$, a particle of mass $m$ and position $\underline{\mathrm{r}}$ obeys the equation
$$m\underline{\ddot{\mathrm{r}}}=-\nabla U\tag{1}$$

While the motivation and meaning of this equation is very clear, there is a slight problem. In my differential geometry course, I learned that the $\nabla$ operator, also known as the covariant or semicolon derivative, is defined to act on an $(r,s)$ tensor, in components, as

$$(\nabla\mathbf{T})^{i_1..i_r}_{j_1..j_s~k}=\\ \partial_kT^{i_1..i_r}_{j_1..j_s}\\+\Gamma^{i_1}_{k~l}T^{l~i_2..i_r}_{j_1..j_s}+\dots+\Gamma^{i_r}_{k~l}T^{i_1..i_{r-1}~l}_{j_1..j_s}\\-\Gamma^l_{k~j_1} T^{i_1..i_r}_{l~j_2..j_s}-\dots -\Gamma^l_{k~j_s}T^{i_1..i_r}_{j_1..j_{s-1}~l}$$

In particular the covariant derivative of a scalar $\phi$ should be
$$(\nabla\phi)_i=\partial_i\phi$$
This is a covector. However, going back to (1),
$$m\underline{\ddot{\mathrm{r}}}=-\nabla U$$
The left hand side is a vector! Position is a vector, so its time derivative should be a vector - right??

So what's the deal here? Is equation (1) wrong? It's used quite a lot in classical mechanics, with correct results too. In (1) should I instead write $\nabla^{\mathrm{T}}U$ to indicate the transpose? Or should I just ditch the symbol altogether and write
$$m\underline{\ddot{\mathrm{r}}}=-\sum_{i}\underline{\mathrm{e}_i}\partial_iU$$
Which, although correct, is rather inelegant? I would appreciate some advice here.

--

ADDENDUM

This other user seems to define the nabla operator contravariantly. Is this standard outside of differential geometry and topology?

答:
Right, they are different nablas. $\nabla U$ here is actually referring to the gradient, which by definition is $\text{grad}(U):= g^{\sharp}(dU)=g^{\sharp}(\nabla U)$; i.e the vector field associated to the $1$-form $dU=\nabla U$ (here this $\nabla U$ refers to the one from differential geometry; it is a $(0,1)$ tensor field/covector field/one-form).

And if you want to write $F=ma$ in truly differential geometric notation, you'd write (letting $\gamma$ denote the trajectory of the particle)
\begin{align}
m\nabla_{\dot{\gamma}}\dot{\gamma} &= -\text{grad}(U)=-g^{\sharp}(dU)=-g^{\sharp}(\nabla U).
\end{align}
"mass times acceleration is force = minus gradient of potential energy"

As an aside:

In terms of a chart, this expression yields
\begin{align}
m\left(\ddot{x}^i + \Gamma^i_{jk}\dot{x}^j\dot{x}^k\right) &= -g^{ij}\dfrac{\partial U}{\partial x^j}
\end{align}
(everything evaluated at appropriate places, and of course I really mean $x\circ \gamma$). If you use cartesian coordinates in $\Bbb{R}^3$ this reduces to the simple $m\ddot{x}=-\frac{\partial U}{\partial x}$ likewise for $y$ and $z$.


答:
note the orthogonality

$-1\cdot \text{trace}\big(BA\big)= \text{trace}\big(B^TA\big)= \text{trace}\big(B^TA\big)^T= \text{trace}\big(A^TB\big)= \text{trace}\big(AB\big)= \text{trace}\big(BA\big)$

$\implies  \text{trace}\big(BA\big)= 0$

using this we can compute the squared Frobenius norm of A+B

$\Big \Vert A+B\Big \Vert_F^2= \text{trace}\big(A^TA\big) +\text{trace}\big(B^TA\big)+\text{trace}\big(A^TB\big)+\text{trace}\big(B^TB\big) $

$= \text{trace}\big(A^2\big) +0+0 -\text{trace}\big(B^2\big) $

$=0 $

$\implies A+B = \mathbf 0\implies A = -B $

so $A$ is both symmetric and skew symmetric, i.e. $A=A^T=-A^T \implies A=\mathbf 0=-B = B$

Title: Let $A$ is a real symmetric matrix, $B$ is a real antisymmetric matrix, $A^2 = B^2$, prove $A = B = 0$.

Body:
Let $\boldsymbol{A}$ is a real symmetric matrix, $\boldsymbol{B}$ is a real antisymmetric matrix, $\boldsymbol{A}^2 = \boldsymbol{B}^2$, prove $\boldsymbol{A} = \boldsymbol{B} = \boldsymbol{0}$.

I tried the second-order matrix.
Let $\boldsymbol{A} = \begin{bmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{bmatrix}$,
$\boldsymbol{B} = \begin{bmatrix} b_{11} & b_{12} \\ {-b_{12}} & b_{22} \end{bmatrix}$, $a_{ij}, b_{ij} \in \mathbb{R}$.

\begin{align}
\boldsymbol{A}^2 &= \begin{bmatrix} {a_{11}^2 + a_{12}^2} & {a_{11}a_{12} + a_{12}a_{22}} \\ {a_{11}a_{12} + a_{12}a_{22}} & {a_{12}^2 + a_{22}^2} \end{bmatrix} \\
\boldsymbol{B}^2 &= \begin{bmatrix} {b_{11}^2 - b_{12}^2} & {b_{11}b_{12} + b_{12}b_{22}} \\ {-b_{11}b_{12} - b_{12}b_{22}} & {-b_{12}^2 + b_{22}^2} \end{bmatrix}
\end{align}

Because $\boldsymbol{A}^2 = \boldsymbol{B}^2$, I got
\begin{align}
a_{11}^2 + a_{12}^2 &= b_{11}^2 - b_{12}^2 \\
a_{11}a_{12} + a_{12}a_{22} &= b_{11}b_{12} + b_{12}b_{22} \\
a_{11}a_{12} + a_{12}a_{22} &= -b_{11}b_{12} - b_{12}b_{22} \\
a_{12}^2 + b_{22}^2 &= -b_{12}^2 + b_{22}^2
\end{align}

Hence,
\begin{align}
a_{11}a_{12}+a_{12}a_{22}=b_{11}b_{12}+b_{12}b_{22}=0
\end{align}

I lost my momentum here.


问:
Inverse function of a polynomial
What is the inverse function of $f(x) = x^5 + 2x^3 + x - 1?$ I have no idea how to find the inverse of a polynomial, so I would greatly appreciate it if someone could show me the steps to solving this problem. Thank you in advance!</p>

答:
You need to solve the equation

$$x^5 + 2x^3 + x - 1=y$$ for $x$. Unfortunately, such quintic equations are known to have no closed-form solution in general, and this one does not escape the rule.

Anyway, there is a little backdoor, as a quintic can be (after painful computation) reduced to the form known as Bring Quintic Form

$$x^5-x-a=0.$$

Under this particular form, the solutions of $x$ in terms of $a$ are called the Bring radicals of $a$. So if you accept this special univariate function in your toolbox, then you can invert the quintic polynomials.

The cases of linear, quadratic, cubic and quartic polynomials can be solved with the usual functions, with increasing difficulty.

ContentLicense: CC BY-SA 3.0

Title: Inverse function of a polynomial


问:
Limit ($x \to \infty$) of integral of positive and continuous function (in $\mathbb{R}$ and centered at $0$) always finite?
For all functions $f(x)$ that are positive and continuous everywhere in the real numbers. If $\lim\nolimits_{x\to \infty}f(x)=0$, is $\lim\nolimits_{x\to \infty}F(x)$ convergent for all possible $f(x)$? $F(x)$ denotes the indefinite integral of $f(x)$ centered at $0$.

I first thought so until I found a possible counterexample:
$$
f(x) =\begin{cases} \frac{1}{x-2}+2 & \text{if }\; x<1,\\ \frac{1}{x} & \text{if }\; x\geq 1.\end{cases}
$$

Edit: I would also like to know / possibly see a proof if my counterexample is actually valid or not.

答:
Here's an example where everything is even fully differentiable on the real line.  Define $f$ on $\mathbb{R}$ by
$$
f(x) = \begin{cases} -x+2, & x \leq 1\\
\frac{1}{x}, & x > 1.
\end{cases}
$$
This function is continuous everywhere (in fact, differentiable) and has $\lim_{x\to +\infty} f(x)=0$.    However, the antiderivative
$$F(x) = \begin{cases} -\frac{x^2}{2}+2x-\frac{3}{2}, & x \leq 1\\
\ln x, & x > 1.
\end{cases}
$$
will have $\lim_{x \to +\infty} F(x) = +\infty$.

Edit: with your updated question, yes, your counterexample is right.


问:
studying for clep calculus exam inverse functions and derivates
I was taking a practice clep calculus exam I found online and I do not understand how the correct answer was derived.

$f(x) = x^3 + x$.  and $h(x)$ is an inverse function of.  find $h'(2)$???  I know that $f(h(x)) = x$ and if $f'(h(x))h'(x) = 1 $, $h'(x) = 1/f'(h(x))$.  My understanding is that if $f(2) = 10$ then $h(10) = 2$ because they are inverses.  Could somebody help me with my misunderstanding?  I think the correct answer was $1/4$.

答:
Sure your understanding is correct but you’re evaluating at the wrong point.

You’re interested in $h’(2)$, so try to evaluate the right hand side $1/f’(h(x))$ at $2$. You need $h(2)$. What value, when plugged into $f$, gives $2$?

More rigorously you should prove that $f$ is invertible, depending on how strict your exam is. But this is easy, since $f$ is strictly increasing and continuous.


问:
Ways for checking the correctness of DE
How can I check the correctness of the solution of the DE?

say the question is,

$$ \frac{dy}{dx} = \frac{-8x+3y-31}{-3x+y-11}$$

I solve it using the method of exact DE, and my solution is,

$$4x^2- 3xy+31x+\frac{y^{2}}{2}-11y=c, c\in \mathbb{R}$$

and the other example,

$$\frac{dy}{dx}=\frac{y}{x}+\frac{9}{2}xe^{\frac{-2y}{x}}+\frac{9}{2}xe^{\frac{2y}{x}}$$
and the potential solution is,
$$y = \frac{-x}{2}\ln|\tan(9x+c)|$$

How can I check the answer like this question?

答:
$$\frac{dy}{dx}=\frac{y}{x}+\frac{9}{2}xe^{\frac{-2y}{x}}+\frac{9}{2}xe^{\frac{2y}{x}}$$
$$\left ( \dfrac yx \right)'=\frac{9}{2}e^{\frac{-2y}{x}}+\frac{9}{2}e^{\frac{2y}{x}}$$
$$w'=9(e^w+e^{-w})$$
Where $w=2\dfrac yx$
$$\int \dfrac {de^w}{e^{2w}+1}=9x+C$$
$$\arctan e^w=9x+C$$
$$w=\ln  |\tan (9x+C)|$$
$$y=\dfrac x2 \ln  |\tan (9x+C)|$$
looks like you made a sign mistake. If you need to check your answer just differentiate $y$.
$$y'= \frac{1}{2}\ln|\tan(9x+c)+ \frac{x}{2}\dfrac {\tan (9x+C)'}{\tan(9x+c)}$$
$$y'= \frac{y}{x}+ \frac{x}{2}\dfrac {\tan (9x+C)'}{\tan(9x+c)}$$
Note that:
$$\tan (9x+C)=e^{2y/x}$$

For the first differential equation:
$$4x^2- 3xy+31x+\frac{y^{2}}{2}-11y=c, c\in \mathbb{R}$$
Differentiate:
$$8x- 3y-3xy'+31+yy'-11y'=0$$
$$y'(-3x+y-11)=-8x+3y-31$$
$$y'=\dfrac {-8x+3y-31} {-3x+y-11}$$

Title: Ways for checking the correctness of DE


答: A counterexample is $f(x) = \frac{1}{\sqrt{x^2+1}}$

Title: Limit ($x \to \infty$) of integral of positive and continuous function (in $\mathbb{R}$ and centered at $0$) always finite?

Body:
For all functions $f(x)$ that are positive and continuous everywhere in the real numbers. If $\lim\nolimits_{x\to \infty}f(x)=0$, is $\lim\nolimits_{x\to \infty}F(x)$ convergent for all possible $f(x)$? $F(x)$ denotes the indefinite integral of $f(x)$ centered at $0$.

I first thought so until I found a possible counterexample:
$$
f(x) =\begin{cases} \frac{1}{x-2}+2 & \text{if }\; x<1,\\ \frac{1}{x} & \text{if }\; x\geq 1.\end{cases}
$$

Edit: I would also like to know / possibly see a proof if my counterexample is actually valid or not.


问:
How to show that $x, y$ $\in$ $\mathbb{R}$ exist such that $f(x) < 0 < f(y)$ using a Epsilon-Delta proof?
For a small part of a school assignment (real Analysis) I have to show that $x, y$ $\in$ $\mathbb{R}$ exist such that $f(x) < 0 < f(y)$

Let $f$: $\mathbb{R}$ $\longrightarrow$ $\mathbb{R}$ differentiable in $0$ and assume that $f(0)$ = $0$ and $f'(0) = 1$

I already have something like this:

$$\lim\limits_{x\to0}\frac{f(x)-f(0)}{x-0} = f'(0)2$$
in other words:

$$\lim\limits_{x\to0}\frac{f(x)-0}{x} = \lim\limits_{x\to0}\frac{f(x)}{x} = 1$$

en then I would think:
$$\left| \frac{f(x)}{x} - 1 \right|....$$

and then I need to prove this with an Epsilon-Delta proof, but I got stuck.

答:
By definition of limit, with $\varepsilon = \frac12$, there is a $\delta>0$ such that
$$
-\frac12<\frac{f(x)}x - 1< \frac12\\
\frac12 < \frac{f(x)}x < \frac32
$$
for all $x$ with $0<|x|<\delta$. Among these $x$, pick one that is negative and one that is positive.


答:
Let $\epsilon>0$ with $\epsilon<1$.

By definition of the limit you showed existing and converging to $1$, there exists $\delta>0$ such that if $0<|x|<\delta$, then
$$
1-\epsilon<\frac{f(x)}{x}<1+\epsilon.
$$
When $x$ is positive ... and when $x$ is negative ...

Title: How to show that $x, y$ $\in$ $\mathbb{R}$ exist such that $f(x) < 0 < f(y)$ using a Epsilon-Delta proof?

Body:
For a small part of a school assignment (real Analysis) I have to show that $x, y$ $\in$ $\mathbb{R}$ exist such that $f(x) < 0 < f(y)$

Let $f$: $\mathbb{R}$ $\longrightarrow$ $\mathbb{R}$ differentiable in $0$ and assume that $f(0)$ = $0$ and $f'(0) = 1$

I already have something like this:

$$\lim\limits_{x\to0}\frac{f(x)-f(0)}{x-0} = f'(0)2$$
in other words:

$$\lim\limits_{x\to0}\frac{f(x)-0}{x} = \lim\limits_{x\to0}\frac{f(x)}{x} = 1$$

en then I would think:
$$\left| \frac{f(x)}{x} - 1 \right|....$$

and then I need to prove this with an Epsilon-Delta proof, but I got stuck.


问:
Is there a closed form for $\int_0^\infty \frac{\sin(x)}{x^2+a^2} dx$?
As the title says, is there a closed form for
$$
\int_0^\infty \frac{\sin(x)}{x^2+a^2} dx \,?
$$

The one with $\cos(x)$ instead of $\sin(x)$ can be calculated via a simple application of the Residue theorem, but the computation uses the fact that the function is even, so the same trick does not work for this one.

Clarification In case this was unclear, I do not care about the method used to calculate this integral.

答:
First, we enforce the substitution $x\mapsto ax$.  Then, we use the convolution theorem of the Fourier Transform to write

$$\begin{align}
I(a)&\equiv \int_0^\infty \frac{\sin(x)}{x^2+a^2}\,dx\\\\
&=\frac1a\int_0^\infty \frac{\sin(ax)}{x^2+1}\,dx\\\\
&=\frac1a \int_{-\infty}^\infty \frac{\sin(ax)}{x^2+1}\,H(x)\,dx\\\\
&=\frac1a\text{Im}\left(\int_{-\infty}^\infty \frac{H(x)}{x^2+1}e^{iax}\,dx\right)\\\\
&=\frac1{2\pi a} \text{Im}\left(\left(\pi e^{-|a|}\right)*\left(\pi \delta(a)+\text{PV}\left(\frac ia\right)\right)\right)\\\\
&=\frac1{2a}\text{PV}\left(\int_{-\infty}^\infty \frac{e^{-|a-a'|}}{a'}\,da'\right)\\\\
&=\frac{e^{-a}\text{Ei}(a)-e^{a}\text{Ei}(-a)}{2a}
\end{align}$$

where $\text{Ei}(a)=\int_{-\infty}^a \frac{e^{x}}{x}\,dt$ is the exponential integral, interpreted in the Cauchy Principal Value sense for $a>0$.


问:
Time at which there have been more people in a session
Let's say I have a document with all data of a Zoom (or whatever platform for online classes and conferences) session that gives me the following for each person in the session:

The time at which they have joined it.

How much time they have spent in the session until leaving.

My question is: given only this information, how can I find (if I can) the time/interval of hours at which there have been more people in the session?

I would really appreciate any kind of help, thank you!

答:
You know the times when people entered and left.  Sort each of these lists in increasing order.  The number of people in the session at time $t$ is the difference between the number of entering times $\le t$ and the number of leaving times $< t$.  Compute this for all the entering times and see which
makes this largest.  The interval you want is from that entering time until the next leaving time.  Of course there may be ties, so there could be several such intervals.


问:
Extending a function beyond the completion/closure of its domain
In  analysis there are certain theorems that tell under which conditions you can continuously extend a continuous functions to the closure/completion of its domain (which actually give the same set, since when talking about completions we have to be (at least) in a metric space).

The theorems I have in mind are these three:

Every uniformly continuous mapping between a metric space and a complete metric can be extended uniformly continuous to a mapping between the completion of the domain and the range.

Every continuous function on a dense subset can be continuously extend to the closure of that subspace (which is by definition of "dense", the whole space).

The theorem from here.

What I'm looking for is a collection of theorems that tell me under which circumstances one can extend a  continuous function beyond the completion or the closure of it's domain.

The only theorem of this kind that I know of is the famous Tietze-Urysohn extension theorem (which applies to function whose domain is already a closed set).

I'd be also happy with a reference, as long as the theorems listed there are quickly accessible (i.e. don't require reading through a thicket of abstract/very specialized definitions before getting to the theorems - the more "concrete" the theorems are (as in $\mathbb{R}^n$, or a metric space), the better.

答:
Extensions problems come in two basic shapes:

unique extension from a non-closed set to its closure

non-unique extension from a closed set to a larger set

So it makes sense that many extension theorems fall into one or the other class. Sometimes you just have to use two of them.

But extension theorems for Lipschitz functions, from the simple McShane-Whitney theorem to the more sophisticated Kirszbraun's theorem and its generalizations, do not care whether the domain is closed or not.

The recent two-volume book Methods of geometric analysis in extension and trace problems by Brudnyi and Brudnyi looks like the reference you want. It has separate chapters for results for $\mathbb R^n$ and for general metric spaces.


问:
Differential of defining Equation System (from David Mumfords Complex Algebraic Geometry)
In David Mumford's book Algebraic Geometry I, Complex Projective Varieties
the proof of Bezout's Theorem (5.16 page 80) contains terminology
I not understand. Mumford speaks at pages 83 & 84 about
'differentials of equation system (that defines a variety)'. I know what differentials of
regular functions are, but I don't know what kind of thing is a
'differential of an equation system ' in classical complex algebraic geometry.
We assume:

Let $X^r,Y^s \subset \mathbb{P}^n$ complex algebraic varieties of dimensions
$r, s$ with $r+s=n$.
Let $L \subset \mathbb{P}^n$ a linear subspace with $L \cap X =  \emptyset$.
After a reasonable choice of cordinates we can assume
$L= V(X_0,...,X_r)$ where $X_i \in \mathbb{C}[X_0,...,X_n]$.

Then the rational map

$$ \sigma:(\mathbb{P}^n - L) \times \mathbb{C} \to \mathbb{P}^n - L$$

def by

$$ \sigma(x_0,...,x_n, t) = (x_0,...,x_r,t x_{r+1},..., t x_n)$$

restricts to regular map $\text{res } \sigma: X \times \mathbb{C} \to \mathbb{P}^n$.
Define the algebraic set

$$ \mathcal{X} = \{(x,y,t) \ \vert \ \sigma_t x=y \} \subset
X \times Y \times \mathbb{C} $$

and consider the restriction of the $3$rd projection
$\pi: \mathcal{X} \to \mathbb{C} $. Locally, $\mathcal{X}$ is defined
by $n$ equations:

$$ \tag{EQ} X_i (\sigma_t x) = X_i (y), \ 1 \ge i \ge n     $$

$$    X_1,...,X_n \text{ affine coord in } \mathbb{P}^n      $$

Fix a $(x,y, t) \in \mathcal{X}$.

My question:  Mumford uses terminology in this context
that I don't understand. On pages 83 & 84 he talked about 'differentials
of the equations (EQ) defining $\mathcal{X}$ restricted to tangent space $T_{(x,y,t),X \times Y, \{t\}} $, then ...'

I'm not familar with this terminology. What is a 'differential of
an equation system' like (EQ) in this example?

Note, that we can regard $X_i$ as functions $\mathbb{A}^n \to \mathbb{C}$
from affine parts of $\mathbb{P}^n$, therefore compositions like
$X_i \circ \sigma_t$ can be also regarded as locally defined functions
$X \to \mathbb{C}$ and we can associate to differentaible functions a differential in usual way.

But here Mumford associates a differential to equations, instead to
functions and I not understand it's meaning and construction. Is there a standard method in classical algebraic geometry to calculate a 'differential'
of a equation system? In simple words: what is it?

答: When talking about algebraic varieties it is common to speak interchangeably between "equations" that define a variety and the corresponding regular functions which vanish on the variety.  So, when referring to the differential of an equation $f(x_1,\dots,x_n)=g(x_1,\dots,x_n)$, that would usually just mean the differential of $f(x_1,\dots,x_n)-g(x_1,\dots,x_n)$, the function whose vanishing is equivalent to that equation.


问:
There are $450$ students in our class. I have to choose the students who are ranked in the $80$th percentile or above.
There are $450$ students in our class. I have to choose the students who are ranked in the $80$th percentile or above.

Which measure of average would be the most meaningful – mean, median, mid-range, or mode? Do we at all need any measure of average?

can anyone please help me ?

答:
First, you need to decide what kind of score you will
use to represent a student's accomplishments in a class.
That score could be based on the mean of three exams,
the median of five exams, the highest score among four
exams, or whatever the instructor thinks is the best
measure of student performance.

Then you will have $n = 450$ scores. Sort then from
smallest to largest. Roughly speaking, the 80th percentile will lie at
the the $360$th of the sorted scores. The top $90$
scores will be at or above the 80th percentile.

If scores are integers, then it is possible to have
several sorted scores tied at the 89th percentile. If
that happened, you would have to decide whether to
include all or none of the students with these tied
scores. So if you know in advance you will be interested
in percentiles, it is best not to round scores to be integers.

Example in R without rounding:

```<code>set.seed(223)
x = 100*rbeta(450, 7, 3)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  17.45   59.79   70.72   68.75   79.44   96.30
quantile(x, .8)
     80%
82.22081
</code>```

Here is a stripchart of all 458 scores with a vertical red line at the 80th percentile:

```<code>stripchart(x, pch="|")
abline(v = 82.22, col="red")
</code>```

The plot is somewhat congested around the 80th percentile,
so let's take a detailed look at the actual values in
the vicinity of the 90th percentile. We begin by
sorting the 450 scores from smallest to largest.
Then we print scores in positions 355 to 365.

```<code>x.s = sort(x)
x.s[355:365]
[1] 81.84596 81.98908 82.17073 82.17978 82.21221 82.22057
[7] 82.22177 82.23840 82.32506 82.38705 82.59776
</code>```

The 80th percentile  $82.22081$ is between scores $82.22057$ and $82.221777.$ There are no ties, so there is no ambiguity about drawing the line between the lower 80% and the upper 20% of the sample. [Caution: Various statistical computer programs and textbooks have slightly different
ways of giving an exact value at the 'borderline' between scores of rank 360 and 361, but that usually makes no practical difference in interpretation.]

With rounding to integers:

Call the rounded scores y and the sorted rounded scores y,s:

```<code>y.r = round(x.s)
summary(y.s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  17.00   60.00   71.00   68.77   79.00   96.00
quantile(y.s, .80)
80%
 82

stripchart(y.s, ylim=c(-.2,20), meth="stack", offset=1/6, pch="-")
 abline(v=82, col="red")
</code>```

Now, the 80th percentile is at rounded score $82.$ However,
several students have rounded score $82.$ To be precise, there are 14 scores tied at $82.$

```<code>sum(y.s == 82)
[1] 14

y.s[350:365]
[1] 81 82 82 82 82 82 82 82 82 82 82 82 82 82 82 83
</code>```

So we have a choice to include or exclude score $82$
among the "top 20%". Inclusion leads to 80.89% below
and exclusion leads to 77.78% below.

```<code>mean(y.s < 82);  mean(y.s <= 82)
[1] 0.7777778
[1] 0.8088889
</code>```

In the grand scheme of things the difference may not matter. But if the "top 20%" get A's and those just
below get B's, it will matter to the 14 students tied
at the borderline. So it is a good idea to realize
there there may be ties at the borderline and to
have a clear policy how to handle them.

Body:
First, you need to decide what kind of score you will
use to represent a student's accomplishments in a class.
That score could be based on the mean of three exams,
the median of five exams, the highest score among four
exams, or whatever the instructor thinks is the best
measure of student performance.

Then you will have $n = 450$ scores. Sort then from
smallest to largest. Roughly speaking, the 80th percentile will lie at
the the $360$th of the sorted scores. The top $90$
scores will be at or above the 80th percentile.

If scores are integers, then it is possible to have
several sorted scores tied at the 89th percentile. If
that happened, you would have to decide whether to
include all or none of the students with these tied
scores. So if you know in advance you will be interested
in percentiles, it is best not to round scores to be integers.

Example in R without rounding:

set.seed(223)
x = 100*rbeta(450, 7, 3)
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  17.45   59.79   70.72   68.75   79.44   96.30
quantile(x, .8)
     80%
82.22081

Here is a stripchart of all 458 scores with a vertical red line at the 80th percentile:

stripchart(x, pch="|")
abline(v = 82.22, col="red")

The plot is somewhat congested around the 80th percentile,
so let's take a detailed look at the actual values in
the vicinity of the 90th percentile. We begin by
sorting the 450 scores from smallest to largest.
Then we print scores in positions 355 to 365.

x.s = sort(x)
x.s[355:365]
[1] 81.84596 81.98908 82.17073 82.17978 82.21221 82.22057
[7] 82.22177 82.23840 82.32506 82.38705 82.59776

The 80th percentile  $82.22081$ is between scores $82.22057$ and $82.221777.$ There are no ties, so there is no ambiguity about drawing the line between the lower 80% and the upper 20% of the sample. [Caution: Various statistical computer programs and textbooks have slightly different
ways of giving an exact value at the 'borderline' between scores of rank 360 and 361, but that usually makes no practical difference in interpretation.]

With rounding to integers:

Call the rounded scores y and the sorted rounded scores y,s:

y.r = round(x.s)
summary(y.s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  17.00   60.00   71.00   68.77   79.00   96.00
quantile(y.s, .80)
80%
 82

stripchart(y.s, ylim=c(-.2,20), meth="stack", offset=1/6, pch="-")
 abline(v=82, col="red")

Now, the 80th percentile is at rounded score $82.$ However,
several students have rounded score $82.$ To be precise, there are 14 scores tied at $82.$

sum(y.s == 82)
[1] 14

y.s[350:365]
[1] 81 82 82 82 82 82 82 82 82 82 82 82 82 82 82 83

So we have a choice to include or exclude score $82$
among the "top 20%". Inclusion leads to 80.89% below
and exclusion leads to 77.78% below.

mean(y.s < 82);  mean(y.s <= 82)
[1] 0.7777778
[1] 0.8088889

In the grand scheme of things the difference may not matter. But if the "top 20%" get A's and those just
below get B's, it will matter to the 14 students tied
at the borderline. So it is a good idea to realize
there there may be ties at the borderline and to
have a clear policy how to handle them.


问:
Does identifying countably many points prevent a quotient space from being second countable?
There is a well-known counterexample to the claim that quotient spaces of second countable spaces are second countable. It is $\mathbb{R} / \sim$ where
$$
x \sim y \iff x = y \text{ or } x, y \in \mathbb{Z}.
$$

I want to know if there is a generalization to this. Specifically, is it true that any quotient space of an uncountable second countable space within which we identify countably many points will fail to be second countable? Follow up question: is there a nice necessary and sufficient condition for quotient spaces of second countable spaces being second countable? Maybe if the quotient map is open? I know that's sufficient but I don't know if it's necessary.

答: No, you an identify $[0,1]$ to a point and the result is homeomorphic to $\Bbb R$ again, and there we even identify uncountably many points. So it has to be decided on  a case by case basis. If $X$ is compact, Hausdorff, second countable and we identify a closed set to a point, then $X{/}\sim$ is again second countable. This even holds for such an $X$ that is non-compact (but $T_2$ and $C_2$) and where we identify a compact set to a point. (In that case the quotient map is perfect). If the quotient map is closed and all classes are compact, we have the same general theorem we can use: that $C_2$ is preserved by perfect maps..


问:
Let's say you have a graph defined by $f(t) = 2t + 8 $ if $t \le 3$ $ f(t) = 2 $ if $t \gt 3 $ what is the area?
So, I have two functions which let me reiterate

$ f(t) = -2t + 8$

if $0 \leq $ t $ \leq 3 $

$f(t) =  2 $ if $t \gt 3$

Now, what this looks like is a downward slope from y = 8 and x = 0 that stops reducing at about t = 3 and every f(t) becomes 2.

When I solve for the Area of 2 or A(2) I do the following calculations :

$\frac{1}{2}(2)(6)$ The height at t = 2 is exactly 6, so my answer should be 6. I look at the back of the book and find that the answer is 12.
I have no idea why the answer is 12.

I then look at A(5) which should be the area of the first piece of the graph. (Which for arguments sake, we're going to say is 12) and then add that to the area of the base and the height since the remainder of the graph is a rectangle we do not need to do the base times height formula.

So, it becomes an interval of [3,6] which is 6-3 = 3 times the height which is 2.
This should just be 3*2 based on my understanding, which is 6. So, 12 + 6 is just 18.

I turn to the back of the book and find 21. I'm kind of lost here, please help me.

答:
This may help:

An area is defined between two limit values of $x$.

The area for any range is simply given by high-school calculus:

$$A = \int\limits_{x = x_l}^{x_u} f(x)\ dx$$


问:
How to prove that $S^1 \times S^3$ identifying antipodal points is again $S^1 \times S^3$.
I am studying the compactification of Minkowski spacetime as the space of generators $\mathcal{PN}$ of the null cone $\mathcal{N}$ of a six dimensional real manifold of signature 2. Concretely I am following the book "An introduction to twistor theory" by Huggett and Tod
(see the image). Topologically the say that $\mathcal{PN}$ is $S^1\times S^3$ identifying antipodal points (because each generator cuts the 5-sphere twice), which is again $S^1\times S^3$. Does anybody know why $S^1\times S^3/\sim$ is again $S^1\times S^3$?

And can this be extended to $S^1\times S^n$ in general?

Thank you.

答: $S^1$ and $S^3$ can be identified with the unit spheres in $\mathbb C$ and $\mathbb C^2$ respectively.  Then the map $$(z, w) \mapsto (z^2, z w)$$ where $z \in \mathbb C$, $w \in \mathbb C^2$ and $\lvert z \rvert = \lVert w \rVert = 1$ identifies exactly the antipodal points.


问:
Does $X\times Y$ have countable chain condition?
Let $X$ and $Y$ have countable chain condition. Does $X\times Y$ have countable chain condition?

Thanks for your help.

答:
Consistently not necessarily.  A consistent counterexample, under, for example, the assumption $\mathbf{V} = \mathbf{L}$ or just $\diamondsuit$, would be a Souslin line, which has the ccc, but whose square does not.

However, also consistently yes.  In particular, Martin's Axiom, $\sf{MA}$ (or even the fragment $\sf{MA} (\aleph_1)$), implies that any product of ccc spaces is ccc.


问:
Uniquess of a matrix in terms of eigenvalues and eigenspaces.
This question is going to sound weird but I really cannot get my head around it.
Suppose an unknown matrix has known eigenvalues and known eigenspaces.
Then, is the matrix necessarily unique?

I mean, if $A$ is 3X3 unknown matrix and there are three distinct eigenvalues and eigenspaces, then we can find the matrix by the usual analysis $PDP^{-1}$. Is this matrix necessarily unique?

答:
For an $n \times n$ matrix with $n$ distinct eigenvalues, yes.  Of course I'm assuming  that "known" includes knowing which eigenspace goes with which eigenvalue.

If there are fewer than $n$ distinct eigenvalues, and the eigenspaces don't span the whole space, it might not be unique.  Thus both
$\pmatrix{0 & 1\cr 0 & 0\cr}$ and $\pmatrix{0 & 2\cr 0 & 0\cr}$ have the same eigenvalues ($0$) and eigenspaces (span of $\pmatrix{1\cr 0}$).


问:
Are there infinitely many pairs of primes $(p,q)$ such that $p^2+1=2q$?
For example, $(3,5), (5,13)$ or $(11,61)$ satisfies this condition, but I have no idea how to prove or disprove that such pairs exist infinitely many. Is there any prior researches about it?

答:
OEIS/A048161 says

A048161: Primes $p$ such that $q = (p^2 + 1)/2$ is also a prime.

Primes which are a leg of an integral right triangle whose hypotenuse is also prime.

It is conjectured that there are an infinite number of such triangles.

So, no one knows whether infinitely many such pairs exist.


答:
In this context it is a bit easier to think about linear transformations (i.e. maps from a vector space to itself) than about matrices, even if for each linear transformation there is a unique matrix and vice versa (after we fix a basis of our vector space).

The reason that the linear transformation perspective is easier here is that we have a clear way of describing for a linear transformation what it means to be unique: if for every vector in the vector space we know where it is mapped, then apparently our knowledge uniquely determines the transformation. If there are some elements of which we have no idea where the transformation maps them, then there might be multiple options, or even none.

Now for your question. The answer is:

If the eigenspaces together span the space (equivalently: if the total space is the direct sum of the eigenspaces), then the linear transformation is uniquely determined by these eigenspaces and corresponding eigenvalues and otherwise not.

Proof:
Let the eigenspaces be $E_1, E_2, \ldots E_k$ and the corresponding eigenvalues be $\lambda_1, \ldots, \lambda_k$. Let $V$ be the space on which the transformation acts and let $T \colon V \to V$ be the name of the transformation.

When indeed $V = E_1 \oplus \ldots \oplus E_k$ then every vector $v \in V$ can be decomposed as $v = v_1 + \ldots + v_k$ with $v_i \in V_i$ and it is easy to see that $Tv = \lambda_1 v_1 + \ldots + \lambda_k v_k$ and so we know what $Tv$ is and hence $T$ is determined uniquely.

When $E_1 \oplus \ldots \oplus E_k$ is strictly smaller than $V$ then we can pick $v \in V$ outside this space and we have no information whatsoever about what $Tv$ is, so $T$ is not determined uniquely. This happens for instance when the Jordan form of $T$ has non-trivial blocks.

Title: Uniquess of a matrix in terms of eigenvalues and eigenspaces.

Body:
This question is going to sound weird but I really cannot get my head around it.
Suppose an unknown matrix has known eigenvalues and known eigenspaces.
Then, is the matrix necessarily unique?

I mean, if $A$ is 3X3 unknown matrix and there are three distinct eigenvalues and eigenspaces, then we can find the matrix by the usual analysis $PDP^{-1}$. Is this matrix necessarily unique?


问:
Unsure if $f(x)\to \pm \infty$ for $x$ tending to the endpoints when it is strictly decreasing between them
Suppose that a continious function $f$ is strictly decreasing on an interval $(a,b)$. When I wrote in a text, "Since $f$ is decreasing on this interval, we have that $f(x)\to \infty$ as $x\to a^{+}$ and $f(x)\to -\infty$ as $x\to b^{-}$" my teacher says "Reason not completely clear" to this part. Could anyone explain what I'm missing here? I could have asked my teacher, but this comment was over one year ago.

答: Strictly decreasing does not necessarily mean $f(x)$ goes to $\pm \infty$ at either endpoint.  Consider $f(x) = -x$.  Then $f$ is strictly decreasing on every interval $(a,b)$, but so long as $a,b$ are finite we do not have that $f(x)\to\infty$ as $x\to a^{+}$ or $f(x)\to -\infty$ as $x\to b^{-}$ on any such interval.


问:
Collecting math websites
I would like to know math websites that are useful for students, PhD students and researchers (useful in the sense most of the students or researchers—of a particular area—are using it). Maybe you can share which math websites you sometime use and why you use it.

Let me give my websites and why I use them:

MathOverflow and Stack Exchange: Very good to look up questions about how to do mathematics, what mathematics is, how it evolves, and of course to have an exchange with smart people, who can help you if you are stuck at something.

arXiv: I don't use it now, but I think that's the main website to read papers and to publish (research level)

The Stacks Project: It is very useful for me to look algebraic things ups with more explanation than in some lectures...

MathSciNet: To search for papers that may help you.

Number Theory Web: Since I want to become a number theorist, this site helps me to see which kind of things exist.

Do you have more websites that help you to understand maybe mathematics, or just to connect to other mathematicians?

Everything is welcome. Just post it as an answer and not a comment.

答:
On /r/math, there is A Compilation of Useful, Free, Online Math Resources. It is somewhat geared towards students in scope, but references many tools used by research mathematicians as well (see especially the comment on subject-specific resources). It contains most of the links already in this thread, and several not yet listed here - a selection of the latter group of links:

Open Problem Garden

Proof Wiki

Librarian is a Chrome extension that provides easy access to references, BibTeX, and comments for all papers on the arXiv.

The Inverse Symbolic Calculator takes in a constant and tries to find a closed form for it.

Approach0 is a mathematical search engine, particularly useful for cases where you would like to search for a specific formula that e.g. may use different variable names.

Groupprops, the group properties wiki

Database of Local Fields

The Knot Atlas

Title: Collecting math websites


问:
Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.
In a triangle $ABC$ the point $D$ is on $BC$ such that $\overline{BD}=\frac{2}{3}\cdot \overline{BC}$ and the point  $E$ is on $AB$ such that $\overline{AE}=\frac{3}{5}\cdot \overline{AB}$.

The lines $AD$ and $CE$ intersect in $S$.

Write $AS$ as a multiple of $AD$ and $CS$ as a multiple of $CE$.

We have a triangle as follows, right?

But how can we get an information for $AS$ and $CS$ ? Could you give me a hint? Do we maybe use vector operations?

$$$$

EDIT :

Using vectors we have $$\vec{AS}+\vec{SD}=\vec{AD} \Rightarrow \vec{AS}=\vec{AD}-\vec{SD} \Rightarrow \vec{AS}=\vec{AD}+\vec{DS}$$ Does this help us ? How can we continue?

答:
Hint:

Applying Ceva's theorem we have:

$$\frac{AE}{EB}\cdot\frac{BD}{DC}\cdot\frac{CG}{AG}=1$$

Puting values we obtain:

$AG=3 CG$

As can be seen in figure, M is midpoint of BC(you have to prove it) and in triangle ABD; AM, DE and BC intersect in point F. Also we have:

$MD=\frac 23 BC-\frac 12 BC=\frac 16 BC$

Applying Ceva's theorem on this triangle we have:

$$\frac{AE}{EB}\cdot\frac{BM}{MD}\cdot \frac{DS}{AS}=\frac 32\cdot \frac {BC/2}{BC/6}\cdot \frac{DS}{AS}=1$$

Which gives $AS=4.5 DS$ , or $AS=\frac 9{11}AD$

Similarly you can find other ratio.


问:
Finding the result of $1/2*1/3...*1/1000$
Let the binary operation $x*y=(x+y)/(1+xy)$ where $x,y\in (-1,1)$
Find $(1/2)*(1/3)*...*(1/1000)$

I found that $1/x*1/y=(x+y)/(1+xy)$ but I can not find a rule for $n$ such numbers and apply induction to calculate that sum.

Could you please tell me what would be another way to solve the problem?

答: HINT: To solve this problem, compute the product $(x_0)(x_1)...(x_n)$ of finitely many terms using your binary operation. Find the pattern, and prove what you see; It is a pretty symmetric result. After this, computing the final value should not be too hard. If you want a complete answer, then ask for one in the comments after you have tried it yourself.


答:
This is covered by the generalized Bunyakovsky conjecture (aka generalized Dickson conjecture), for the polynomials $f_1(x) = 2x+1$ and $f_2(x) = 2 x^2 + 2 x + 1$.
However, the only case where this conjecture is known to be true is for just one polynomial of degree $1$ (Dirichlet's theorem).

Title: Are there infinitely many pairs of primes $(p,q)$ such that $p^2+1=2q$?

Body: For example, $(3,5), (5,13)$ or $(11,61)$ satisfies this condition, but I have no idea how to prove or disprove that such pairs exist infinitely many. Is there any prior researches about it?


问:
Experimental Analysis of an Algorithm - How to prove that the graph is $O(n\log n)$?
This question is probably stupid, but I've been trying to figure this out for hours and I still couldn't find anything about it. Probably I'm just too lost.

So basically, I'm analysing an algorithm by doing an asymptotic and experimental analysis. The asymptotic analysis went well and I concluded that my algorithm is $O(n\log n)$. The problem is the experimental analysis. Firstly, I did some tests to get the time taken for each input. For example:

$n = 1, t = 0 | n = 2, t = 2 | n = 4, t = 8 | n = 8, t = 24 | n = 16, t = 64 | n = 32, t  = 160 (...)$

If I do the graph with this example, I can see that it is an O(n log n)/linearithmetic graph, but how do I prove it? Do I have to calculate the order of growth? If so, how do I do it?

Thanks for the help!

答:
You can try to see (e.g., by plotting) if the function $\frac{T(n)}{n}$  has a logarithmic growth (to reduce the number of hidden constants, instead of checking $T(n)$ itself).

For example, you can try to use non-linear least square curve fitting (e.g., with R) and fit a logarithm function (of the form Const + A * log(n), e.g.,) and compare the growths. The following R code shows an example of the same on randomly generated time

```<code>n <- 1:100
T <- 5*n*(log(10*n))+2+10*runif(1) # randomly generated time complexity T(n)

nlmod <- nls(T/n ~  Const + A * log(n), start = list(A = 1, Const=1), trace=TRUE)
summary(nlmod)
# Formula: T/n ~ Const + A * log(n)

# Parameters:
#  Estimate Std. Error t value Pr(>|t|)
#  A      4.14962    0.07108   58.38   <2e-16 ***
#  Const 15.05845    0.26675   56.45   <2e-16 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

# Residual standard error: 0.6563 on 98 degrees of freedom

# Number of iterations to convergence: 1
# Achieved convergence tolerance: 3.671e-09

plot(n,T/n, main = "nls(*), data, true function and fit, n=100", pch=19)
lines(n, predict(nlmod), col = 2)
lines(n, predict(nlmod,n), col = 2, lwd=2)
legend("bottomright", legend=c("T(n)/n", "fitted"),
   col=c("black", "red"), pch=c(19,NA), lty=c(NA,1), lwd=2)
</code>```

You can also compute the distance (e.g., MSE, KL-divergence etc.) between the fitted logarithmic curve and the actual $\frac{T(n)}{n}$, to ensure that it's sufficiently small for large enough $n$.

Body:
You can try to see (e.g., by plotting) if the function $\frac{T(n)}{n}$  has a logarithmic growth (to reduce the number of hidden constants, instead of checking $T(n)$ itself).

For example, you can try to use non-linear least square curve fitting (e.g., with R) and fit a logarithm function (of the form Const + A * log(n), e.g.,) and compare the growths. The following R code shows an example of the same on randomly generated time

n <- 1:100
T <- 5*n*(log(10*n))+2+10*runif(1) # randomly generated time complexity T(n)

nlmod <- nls(T/n ~  Const + A * log(n), start = list(A = 1, Const=1), trace=TRUE)
summary(nlmod)
# Formula: T/n ~ Const + A * log(n)

# Parameters:
#  Estimate Std. Error t value Pr(>|t|)
#  A      4.14962    0.07108   58.38   <2e-16 ***
#  Const 15.05845    0.26675   56.45   <2e-16 ***
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

# Residual standard error: 0.6563 on 98 degrees of freedom

# Number of iterations to convergence: 1
# Achieved convergence tolerance: 3.671e-09

plot(n,T/n, main = "nls(*), data, true function and fit, n=100", pch=19)
lines(n, predict(nlmod), col = 2)
lines(n, predict(nlmod,n), col = 2, lwd=2)
legend("bottomright", legend=c("T(n)/n", "fitted"),
   col=c("black", "red"), pch=c(19,NA), lty=c(NA,1), lwd=2)

You can also compute the distance (e.g., MSE, KL-divergence etc.) between the fitted logarithmic curve and the actual $\frac{T(n)}{n}$, to ensure that it's sufficiently small for large enough $n$.


答: (https://artofproblemsolving.com) is a very good math site. It has artofporblemsolving books and it also has a lot of math games to practice your math skills. Alcumus is the best game on there (https://artofproblemsolving.com/alcumus)</p>

Title: Collecting math websites

Body:
I would like to know math websites that are useful for students, PhD students and researchers (useful in the sense most of the students or researchers—of a particular area—are using it). Maybe you can share which math websites you sometime use and why you use it.

Let me give my websites and why I use them:

MathOverflow and Stack Exchange: Very good to look up questions about how to do mathematics, what mathematics is, how it evolves, and of course to have an exchange with smart people, who can help you if you are stuck at something.

arXiv: I don't use it now, but I think that's the main website to read papers and to publish (research level)

The Stacks Project: It is very useful for me to look algebraic things ups with more explanation than in some lectures...

MathSciNet: To search for papers that may help you.

Number Theory Web: Since I want to become a number theorist, this site helps me to see which kind of things exist.

Do you have more websites that help you to understand maybe mathematics, or just to connect to other mathematicians?

Everything is welcome. Just post it as an answer and not a comment.


问:
inequality of $(ax+by+cz)^n$
The Cauchy Schwarz inequality says
$$
(ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2).
$$

I found that there is a kind of analogous inequality for $(ax+by+cz)^n$
$$
(x+y+z)^n \leq 3^n(x^n+y^n +z^n).
$$
if I remember it correctly.

How to prove this inequality?

答: The factor can be improved to $3^{n-1}$. For odd $n$ you have to assume probably that $x,y,z \geq 0$. Then this is Jensen’s inequality for the monomial function $x \mapsto x^n$: $$\left(\frac{x+y+z}3\right)^n \leq \frac{x^n + y^n + z^n}3.$$

Title: inequality of $(ax+by+cz)^n$


问:
Stochastic Processes, practice problem taken from One Thousand Exercises in Probability
I am struggling with probability and specifically markov chains. I find making them requires much creativity and insight. I took this problem from Grimmett's One Thousand Exercises in Probability and I am having difficulty setting up a markov chain and finding the transition probabilities for this example problem.

The question is:

A sequence $(W_n)_{ n>0}$ of random words is recursively made like this:

$W_0$ is a word, with length $N$ and the letters are taken from the set $\{a,b\}$

Suppose that $W_n$ is a word. Then we draw $N$ letters at random, one at a time with replacement from $W_n$. Then $W_{n+1}$ is the word of length $N$ obtained by recording the letters drawn.

Let $X_n$ be the number of letter $a$'s within $W_n$

Argue $(X_n)_{ n>0}$ is a Markov chain and find the transition probabilities

Show $X$ is martingale wrt to filtration $(F_n)_{ n\ge 0}$ generated by $X$ itself.

Use the Optional Stopping theorem to compute $P_k (V_n < V_0)$ for $0 <k < N$

There are more parts to this problem but the main confusion is on 1,2 in how I can find the transition probabilities as I am lost.

Any help or tips would be greatly appreciated !

答:
Note that if $(U_i)$ is an iid sequence, and $X_0$ is independent of $(U_i)$, then the recursively defined $X_{n+1} := f(X_n, U_n)$ is always a Markov chain with respect to its natural filtration, for any measurable $f$. (Check this yourself in the simple case of a countable state space.) In a discrete setting like we have here, measurability is typically obvious, so you can think of this as saying "any recursive thing where you get the next element by doing something to the previous one with an independent source of randomness will be Markov". The fact that $f$ does not depend on $n$ and the $U_n$ have the same distribution is what gives you that it is a (homogeneous) Markov (chain).

In your case, you get the next word $W_{n+1}$ from the previous one $W_n$ by taking an independent samples $(U^{(n)})$ with replacement of (1,2,..., $N$) and then using them as indices into $W_{n}$, so in your case $f(x, (u_1,\ldots, u_N)) = (x[u_1], ..., x[u_N])$.

For transition probabilities, you view $W_n$ as a bag of $A$ $a$'s and $B$ $b$'s with $A+B=N$. For each index, you have an $A/N$ chance of getting $a$ and $B/N$ chance of getting $b$. So the chance to go from $W_n$ to a fixed sequence $(x_1, \ldots, x_N)$ with $x_i \in \{a, b\}$ is $(A/N)^{A'} (B/N)^{B'}$ where $A',B'$ are the number of $a$'s or $b$'s in $(x_1,\ldots, x_N)$.

I'll stop here and let you think about it. In the future, if you have a question with many parts, ask them in separate questions as you progress.


问:
Finding the accumulation points of $A=[0,1]$
I'm having some trouble finding the set of all accumulation points of $A=[0,1]$. I already prove that $0$ and $1$ are accumulation points, but I can't find a way to prove that any $p \in(0,1)$ is also an accumulation point.

If anyone could help me, it would be great.

Thank you very much!

答:
Every point of $[0,1]$ is an accumulation point. To see this, let $x\in[0,1]$. If $\varepsilon>0$, do you understand why the intersection $\big((x-\varepsilon,x+\varepsilon)\setminus\{x\}\big)\cap[0,1]$ is non-empty? If so, you understand that $x$ is an accumulation point of $[0,1]$.

On the other hand, any point outside $[0,1]$ is not an accumulation point. Indeed, if $x\not\in[0,1]$, then we can find $\delta>0$ so that $(x-\delta,x+\delta)\cap[0,1]=\emptyset$.


答: Consider any neighbourhood of $p$ of the form $(p-a,p+b),a,b>0$. If $p-a\ge0$, then $\frac{p+(p-a)}2\in A\cap(p-a,p+b)$. Otherwise if $p+b\le1$, then $\frac{p+(p+b)}2\in A\cap(p-a,p+b)$. If the above is not true, then $A\subset(p-a,p+b)$ and $\frac p2\in A\cap(p-a,p+b)=A$.

Title: Finding the accumulation points of $A=[0,1]$

Body:
I'm having some trouble finding the set of all accumulation points of $A=[0,1]$. I already prove that $0$ and $1$ are accumulation points, but I can't find a way to prove that any $p \in(0,1)$ is also an accumulation point.

If anyone could help me, it would be great.

Thank you very much!


问:
Recurrence Relation with Square Root
Well, I was doing a problem on recurrence relation , where there was given a an recurrence relation and we had to find $a_{n}$ or simplify the recurrence.

The recurrence relation was
$$\begin{align} a_{n} &= \frac{1}{16}(1 + 4a_{n+1} + \sqrt{1+24a_{n+1}} ) \\ a_{0} &= 1 \end{align}$$

What I tried was I reached the equation
$$4a_{n} - a_{n+1} = \frac{1+\sqrt{1+24a_{n+1}}}{4} $$
Using formula for quadratic roots and putting $$4a_{n} - a_{n+1} = x = \frac{1+\sqrt{1+24a_{n+1}}}{4}$$
$$\implies Ax^2 + Bx + C = 0$$
$$ A = 2, B = -1, C = -3a_{n+1}$$
And then putting the value of $x$ I tried to solve it but nothing fruitful was obtained.

Edit: I made a mistake in the recurrence relation and now $n$ has been swapped by $n+1$

Title: Recurrence Relation with Square Root

Body:
Hint:

Denote $b_n=\sqrt{1+24a_{n}}$. Then you get
$$
4b_n^2-4=b_{n+1}^2+6b_{n+1}+5,
$$
whence
$$
2b_n=\pm (b_{n+1}+3).
$$


问:
Is n over itself, over itself ... recursively equal to one, or square-root of n?
Suppose that $x = \frac{n}{x} $ where n a constant real number and we wish to solve for x. Solving analytically, $x^2 = n \rightarrow x = \sqrt{n}$. But, suppose we plugged in the value of x recursively so that $x = \frac{n}{\frac{n}{\frac{n}{...}}}$. Does this mean that $x = \frac{n}{\frac{n}{\frac{n}{...}}} = \sqrt{n}$? Intuitivly, I would expect it to be 1.

答: Usually we define infinite nesting as the $k\to\infty$ limit of the result with $k$ items. In this case the sequence is $x_0=n,\,x_{k+1}=n/x_k$, a sequence that alternates bewteen $n$ and $1$, and for $n\ne1$ doesn't converge.


问:
Let $A\le B$ and $B\le C$. What are the possibilities of ${\rm ord}(B)$ if ${\rm ord}(A)=60$ and ${\rm ord}(C)=4200?$
I had a nice problem in my abstract algebra homework, and when I wanted to go deeper and generalize it, I got a little bit stuck.

Let $A$ be a subgroup in $B$ and $B$ be a subgroup in $C$. Is there any equation that can connect ${\rm ord}(A)$, ${\rm ord}(B)$, ${\rm ord}(C)$?

I would appreciate if you can point me in the right direction.

答: By Lagrange's Theorem, ${\rm ord}(A)=60\mid {\rm ord}(B)$ and $ {\rm ord}(B)\mid 4200={\rm ord}(C)$.


答:
The number of $a$'s in $W_n$ only depends on the numbers of $a$'s in $W_{n-1}$. To see this, note that there are $X_{n-1}$ $\ a$'s in $W_{n-1}$, so to generate $W_n$ we're drawing $N$ elements with replacement from a word with $X_{n-1}$ $\ a$'s and $N- X_{n-1}$ $\ b$'s. So the conditional probability distribution of $X_n$ is, after doing some counting :
$$\begin{align}\mathbb P (X_n = x | X_{n-1},\ldots,X_0) &= \mathbb P (X_n = x | X_{n-1}) \\
&= {N\choose x}\left(\frac{X_{n-1}}{N}\right)^x \left(\frac{N-X_{n-1}}{N}\right)^{N-x} \\
&= {N\choose x}\left(\frac{X_{n-1}}{N}\right)^x \left(1-\frac{X_{n-1}}{N}\right)^{N-x}\end{align} $$
You just need to be a bit cautious about the states $X_{n-1}\in\{0,N\}$, which are absorbing states, I'll let you take care of it.

To show that $X$ is a martingale, you have to show that $\mathbb E[X_n|X_{n-1}] = X_{n-1}$. To do so, you can notice that the conditional distribution of $X_n|X_{n-1}$ is, as can be seen from question 1), a binomial distribution of parameters $(N,\frac{X_{n-1}}{N})$. So you can directly compute its mean and be able to conclude that it is indeed a martingale with respect to the filtration $(F_n)_{n\ge 0}$. (You also have to treat the case $X_{n-1}\in\{0,N\}$ separately).

What is $V$ ?

Title: Stochastic Processes, practice problem taken from One Thousand Exercises in Probability

Body:
I am struggling with probability and specifically markov chains. I find making them requires much creativity and insight. I took this problem from Grimmett's One Thousand Exercises in Probability and I am having difficulty setting up a markov chain and finding the transition probabilities for this example problem.

The question is:

A sequence $(W_n)_{ n>0}$ of random words is recursively made like this:

$W_0$ is a word, with length $N$ and the letters are taken from the set $\{a,b\}$

Suppose that $W_n$ is a word. Then we draw $N$ letters at random, one at a time with replacement from $W_n$. Then $W_{n+1}$ is the word of length $N$ obtained by recording the letters drawn.

Let $X_n$ be the number of letter $a$'s within $W_n$

Argue $(X_n)_{ n>0}$ is a Markov chain and find the transition probabilities

Show $X$ is martingale wrt to filtration $(F_n)_{ n\ge 0}$ generated by $X$ itself.

Use the Optional Stopping theorem to compute $P_k (V_n < V_0)$ for $0 <k < N$

There are more parts to this problem but the main confusion is on 1,2 in how I can find the transition probabilities as I am lost.

Any help or tips would be greatly appreciated !


问:
Orthonormal Matrix and Transpose.
If Q is an orthogonal matrix that has orthonormal vectors as columns, I can total understand the following result
$$Q^TQ=I$$
but $QQ^T=I$ only if Q is a square matrix and not if Q is a rectangular matrix.
I cannot understand why the second result is true. When I'm doing $QQ^T=I$ I'm not taking dot product of orthonormal vectors and yet it yields as identity matrix for a square matrix. How is this possible?

Edit: Orthonormal changed to Orthogonal

答:
Suppose that $Q \in \mathbb{R}^{n \times n}$ has orthonormal columns $q_1,\dots,q_n \in \mathbb{R}^{n \times 1}$. Then the map $L_Q : \mathbb{R}^{n \times 1} \to \mathbb{R}^{n \times 1}$ given by $L_Q(x) = Qx$ for all $x \in \mathbb{R}^{n \times 1}$ is bijective since it takes the standard basis of $\mathbb{R}^{n \times 1}$ to the basis $q_1,\dots,q_n$ (why $n$ orthogonal vectors in $\mathbb{R}^{n \times 1}$ form a basis?). Then there exists $P \in \mathbb{R}^{n \times n}$ such that $(L_Q)^{-1} = L_P$, which means that $QP=I=PQ$. Moreover, since $Q^\textsf T Q = I$, it follows that
$$P = IP = (Q^\textsf T Q)P = Q^\textsf T (QP) = Q^\textsf TI = Q^\textsf T.$$

Title: Orthonormal Matrix and Transpose.


问:
Binomial distribution with fractional number of successes
I have a stochastic variable with binomial distribution $T\sim B(n,\frac{1}{4})$

I need to estimate $n : P(0.30 < \frac{T}{n} < 0.70) > 0.95$

I have tried to resolve this problem with

$P(0.30 < \frac{T}{n} < 0.70) = P(T <0.70n) - P(T < 0.30n)$

but I don't know how can I proceed because I don't know the $\operatorname{Img}(T)$

答:
No such $n$ exists.  For sufficiently "large" $n$, we can use a normal approximation to the binomial and show that the function $$R(n) = \Phi\left(\frac{9\sqrt{n}}{5\sqrt{3}}\right) - \Phi\left(\frac{\sqrt{n}}{5\sqrt{3}}\right)$$ where $\Phi(z) = \Pr[Z \le z]$ is the standard normal CDF, has derivative $$r(n) = \frac{e^{-27n/50}(9-e^{8n/15})}{10\sqrt{6\pi n}}$$ with critical point at $n = \frac{15}{4} \log 3 \approx 4.1198$.  This is a global maximum on $n > 0$, but $$R(4.1198) \approx 0.389891 < 0.95,$$ with very little hope of the exact binomial probability getting that large.  But we make a table anyway:

$$\begin{array}{c|c|c|c}
n & \lceil 0.3n \rceil & \lfloor 0.7n \rfloor & \Pr[0.3n < T < 0.7n] \\
\hline
 1 & 1 & 0 & 0 \\
 2 & 1 & 1 & \frac{3}{8} \\
 3 & 1 & 2 & \frac{9}{16} \\
 4 & 2 & 2 & \frac{27}{128} \\
 5 & 2 & 3 & \frac{45}{128} \\
 6 & 2 & 4 & \frac{945}{2048} \\
 7 & 3 & 4 & \frac{945}{4096} \\
 8 & 3 & 5 & \frac{10395}{32768} \\
\end{array}$$

The maximum probability occurs at $n = 3$, but $9/16 < 0.95$.


问:
Prove that $|\sin x|\le|x|$ without using the continuity property
Without using any property about continuity of sine function, Prove that $|\sin x|\le|x|$.
First come to my mind is that using taylor series to approximate the sine function, where we know that $\sin x=\sum_{i=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$, but i am not quite see how to prove the inequality.</p>

答:
For $0\leq\phi\leq{\pi\over2}$ consider the two points $(\cos \phi,\pm\sin \phi)$ on the unit circle. Their distance is $2\sin \phi$, while the length of the circular arc between them is $2\phi$. Therefore we have
$$2\sin\phi\leq 2\phi\qquad(0\leq\phi\leq{\pi\over2})\ .$$
Putting $\phi:=|x|$ we obtain
$$|\sin x|\leq |x|\qquad(0\leq|x|\leq{\pi\over2})\ ,$$
and for $|x|\geq{\pi\over2}$ one obviously has $|x|\geq{\pi\over2}>1\geq|\sin x|$.</p>


问:
What is the value of the $x$, when $\sqrt{x}=-1$?
I tried to solve this problem, but check this out:
$$\sqrt{x} = -1$$
$$x^\frac{1}{2}=-1$$
$$\frac{1}{2}\ln x = \ln (-1) = \pi i$$
$$\ln x = 2 \pi i$$
$$x = e^{2\pi i}$$

And if we plug this into the first equation:
$$\sqrt{e^{2\pi i}} = \sqrt{(-1)^2} = \sqrt{1} = 1$$

This makes a paradox and there is actually no solution to the equation $\sqrt{x} = -1$ !

What is the value of the $x$ which satisfies the first equation? If there isn't, why can't we make an another imaginary number like $i$, where $i = \sqrt{-1}$?

答:
$\sqrt{x}$ is defined as the non-negative square root aka the principle square root. This is due to the fact that the each number has two square roots. For example, $4$ has $2$ and $-2$. However, for $\sqrt{x}$ to be a function, it must map every $x$-value to only one $y$, thus only one of the $2$ possible values of the square root may be chosen to have $\sqrt{x}$ be a function. So, the positive value is chosen by convention. This thus means that it is impossible for $\sqrt{x}$ to be negative by definition.

This problem is similar to why $|x|$ cannot be negative. The $|x|$ is defined as the distance to $0$, and distances are non-negative, thus $|x|$ cannot be negative by definition.


问:
When is the product of symmetric matrices symmetric?
Suppose I have two symmetric matrices $A, B\in\mathbb{R}^{n\times n}$. What qualifiers do I need to ensure that $B^{-1}AB$ is also symmetric? Is $B$ being diagonal enough?

答: Since $B^{-1}AB=B^{-1T}A^TB^T=(BAB^{-1})^T$, $B^{-1}AB-(B^{-1}AB)^T=(BAB^{-1}-B^{-1}AB)^T$, so the desired condition is equivalent to $BAB^{-1}=B^{-1}AB$ i.e. $[A,\,B^2]=O$, which occurs if $A,\,B^2$ are simultaneously diagonalizable. Clearly, however, not all diagonal $B$ satisfy $[A,\,B^2]=O$, even if $A$ is symmetric.


问:
How to denote the BNF syntax of propositional calculus?
tldr: how to design a BNF grammar for propositional calculus acknowledging the order of precedence between different logical connectives and disallowing stuff like $p \lor q \land r$?

Hi,

lately I’ve been thinking about the (E)BNF notation for syntax for different areas and I’ve thought about the propositional calculus. How to define a BNF syntax for it? I know that for simple arithmetic expressions (even excluding binary minus and division but those are analoguous to + and *) we could use something like that

```<code><expression> ::= <term> + <expression> | <term>
<term> ::= - <term> | <factor> * <term>
<factor> ::= (<expression>) | <number>
</code>```

and one could think that logical connectives sort of resemble those operators above but there are some differences right? Eg. when I googled for a BNF syntax for logic I found some that are very simple like here and completely ignore the precedence order.

So as I was taught the precedence order is:

negation

conjunction and disjunction (they are equally important)

implication and equivalence

When compared to the syntax designed for arithmetic above we see that the unary minus is exactly like negation but addition doesn't map to disjunction
So what I always thought was that a formula like $p \wedge q \vee r$ is illegal since it is not clear whether it is $(p \wedge q) \vee r$ or $p \wedge (q \vee r)$ (yet another reason for choosing polish for logic) but I do not know how to define a syntax that would make these not show up in the production process. I've seen BNFs for logic online which simply ignored that or assumed that $\wedge$ takes precedence over $\vee$ for which there is no real reason in my opinion since both are distributive (contrary to multiplication and addition as in the first example out of which only one is).

So how to define such a syntax using simple BNF rules?

edit: ok so clearly I wasn't precise enough and that's my bad. So the model for propositional calculus I was taught and which I believe is quite logical is to apply the order of precedence as listed above. About associativity... I haven't given it much thought to be frank and personally I'd bracket stuff like $p \to q \to r$ since it has two interpretations that are not equivalent and I don't really know which one would be the obvious one.
But for other connectives the parentheses are not necessary since we have tautologies like $((p\lor q) \lor r) \leftrightarrow (p \lor (q \lor r))$ (the same follows for conjunction and equivalence) hence we see that we don't need to add any parentheses there. But as I said I am not so sure about this one.

But there is some confusion below about the order of precedence. So as I've said above I consider disjunction having a higher order of precedence than implication etc. So a formula like $\lnot p \lor q \to r$ is to be interpreted as $((\lnot p) \lor q) \to r$.

The whole issue I am having is with the disjunction and conjunction mix where I don't consider $p \lor q \land r$ to be a well formed formula.

答:
The crucial point is to require external parentheses for every binary connective, so that priority between binary connectives is unambiguous.

Formally, given a (usually countably infinite) set of atomic formulas, the BNF syntax of formulas of propositional calculus is the following:

\begin{align*}
\langle \text{formula} \rangle &::= \langle \text{atomic} \rangle \mid \lnot \langle \text{formula} \rangle \mid (\langle \text{formula} \rangle \land \langle \text{formula} \rangle)
\\
&\quad \mid (\langle \text{formula} \rangle \lor \langle \text{formula} \rangle) \mid (\langle \text{formula} \rangle \to \langle \text{formula} \rangle)
\end{align*}

Note that this means that, given some atoms $p$, $q$ and $r$, the strings $p \land q$ and $p \lor q \land r$ and $p\lor (q \land r)$ are not formulas. On the contrary, the strings $(p \land q)$ and $((p \lor q) \land r)$ and $(p\lor (q \land r))$ are formulas.


问:
Summing the trigonometric series $\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)....$
This is a problem of Loney's "Plane trigonometry, part 2"

$$\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)...=\sum_{n=0}^{+\infty}\frac{1}{2^n}\sin[(n+1)a]$$

Here is what I've got so far:

Let $S=\sin(a)+\frac{1}{2}\sin(2a)+\frac{1}{2^2}\sin(3a)+\frac{1}{2^3}\sin(4a)_....\tag{1}$

$C=1+\cos(a)+\frac{1}{2}\cos(2a)+\frac{1}{2^2}\cos(3a)+\frac{1}{2^3}\cos(4a)...\tag{2}$

$C+iS=1+[\cos(a)+i\sin(a)]+\frac{1}{2}[\cos(2a)+i\sin(2a)]+\frac{1}{2^2}[\cos(3a)+i\sin(3a)]+...\tag{3}$

$=1+e^{ia}+\frac{1}{2}(e^{ia})^2+\frac{1}{2^2}(e^{ia})^3+...\tag{4}$

I am stuck here. I am trying to find the general formula for $(4)$, $(4)$ resembles a geometric series, but not quite. I have looked into $-\log(1-x)$ with $x=e^{ia}$, but this doesn't work.

If I let $e^{ia}=x$, then I have a power series

$$1+x+\frac{1}{2}x^2+\frac{1}{2^2}x^3+...$$

But I can't come up with the general formula.

The answer at the back of the book says the sum of this series is $\frac{4\sin(a)}{5-4\cos(a)}$.

My guess with this problem is that you will arrive at a formula that is the sum of $C$ and $S$, then by equating the real and imaginary part, you will be able to arrive at the sum of this series.

答: $$\sum_{k=0}^\infty z^k = \frac{1}{1-z}, \quad |z| < 1$$ implies with the choice $z = e^{ia}/2$ which does have magnitude less than $1$, that $$\sum_{k=0}^\infty \left(\frac{e^{ia}}{2}\right)^k = \frac{1}{1-e^{ia}/2}.$$  Now split this into real and imaginary components; i.e. $$\frac{2}{2 - e^{ia}} = \frac{2}{(2 - \cos a) - i \sin a} = \frac{2((2 - \cos a) + i \sin a)}{(2 - \cos a)^2 + \sin^2 a},$$ and the rest is simple.  The missing factor of $2$ returns when you adjust the summation accordingly.


问:
What's the probability of a set of only three digits appearing in a random 9 digit set?
I'd like to know the method for correctly calculating the probability of a random sequence of $9$ numbers only containing $3$ unique, different numbers. For the purpose of this question: there are 10 numbers $0,1,2,3,4,5,6,7,8,9$

i.e. the probability of this: $123123123$ - for all unique combinations of $3$ digits (e.g $071$ in $071717717$)

My initial instinct is $(1/3)^9$ - is this correct?

答:
And there is the bull at red cape brute force method ... here using Mathematica:

```<code>SantasLittleDigitCounter[m_] := (j = 0;
 Do[ If[Length[Union[IntegerDigits[i, 10, m]]] == 3, j++], {i, 0, 10^m - 1}]; j)
</code>```

Then:

```<code>SantasLittleDigitCounter[9]
</code>```

2178000

which matches exactly the super theoretical solution derived above by Mario Carneiro.

A few minor comments:

Ran this while having dinner.

A neater/better approach in Mathematica is usually to generate all the numbers we want to test in one go (say lis = Range[$0$, $10^9-1$]), and then test them all in one go, using faster functional approaches such as Map rather than the more old-fashioned procedural Do loop-de-loop ... but because the number of terms we have to test is so large (a billion), the memory requirements of generating them all in one go is a bit messy ... so went back to just using good old Do. I suppose one could parallel-optimise this, if desired.

The original poster has noted that numbers like 21 count as a 3-digit hit, because it is represented as the 9 digit {0,0,0,0,0,0,0,2,1}. This is the reason for using the form IntegerDigits[i, 10, m] ... which pads integer i (in base 10) with zeroes to fill $m=9$ slots.

Body:
And there is the bull at red cape brute force method ... here using Mathematica:

SantasLittleDigitCounter[m_] := (j = 0;
 Do[ If[Length[Union[IntegerDigits[i, 10, m]]] == 3, j++], {i, 0, 10^m - 1}]; j)

Then:

SantasLittleDigitCounter[9]

2178000

which matches exactly the super theoretical solution derived above by Mario Carneiro.

A few minor comments:

Ran this while having dinner.

A neater/better approach in Mathematica is usually to generate all the numbers we want to test in one go (say lis = Range[$0$, $10^9-1$]), and then test them all in one go, using faster functional approaches such as Map rather than the more old-fashioned procedural Do loop-de-loop ... but because the number of terms we have to test is so large (a billion), the memory requirements of generating them all in one go is a bit messy ... so went back to just using good old Do. I suppose one could parallel-optimise this, if desired.

The original poster has noted that numbers like 21 count as a 3-digit hit, because it is represented as the 9 digit {0,0,0,0,0,0,0,2,1}. This is the reason for using the form IntegerDigits[i, 10, m] ... which pads integer i (in base 10) with zeroes to fill $m=9$ slots.


问:
$\lim_{z\to\infty} \frac{4z^2}{(z-1)^2} = 4$
So, I'm having trouble performing the $\varepsilon$ - $\delta$ for this proof.

I made the change from $z$ to $\frac{1}{z}$ so the limit becomes $\lim_{z \to 0} \frac{4}{(1-z)^2} = 4$.

Next, I applied the definition for the limit: $ \left|  \frac{4}{(1-z)^2} - 4  \right| < \varepsilon$ when $\left| z \right| < \delta$. Then, after performing some math I get to $ \left|  \frac{-z^2 +2z}{(1-z)^2}\right| < \frac{|z||z-2|}{|z-1|} < \varepsilon $.

Finally I choose $\delta<5$ that makes $ z-2 < 3$ and $ z-1 < 4$ and $ \delta $ is min{$5, \frac{4 \varepsilon}{3} $}.

I don't know if that is right or if there is a better way to do it.

Edit: This problem is from "Complex variables and applications" Brown & Churchill

答:
When you get to the step of setting an upper bound on $\delta$ it needs to be chosen so that  $\left| \frac {z-2}{z-1}\right|$ is bounded.

$\delta \le 4$ will not do the trick!  We can't let the denominator equal zero.

Choose $\delta \le \frac 12$ (we could choose any value less than 1, but this is easy to work with)

In wich case $\left |\frac {z-2}{z-1}\right|< 5$

$\delta = \min (\frac \epsilon 5, \frac 12)$


问:
How to prove that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is not continuous in $(0, 0)$ using an Epsilon-Delta proof?
In an assignment for the course Real-Analysis I need to proof that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is not continuous in $(0, 0)$ using an Epsilon-Delta proof.

However, me and my fellow students have been trying to figure it out for some time now, but we can't...

答:
Hint: In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$ and

$f(x, y)=\frac{2xy}{x^2 + y^2}=\frac{2r^2\cos\theta\sin\theta}{r^2}=2\cos\theta\sin\theta$.

When $r \to 0$, the limit does not exist. Why?


答:
Edit: My first post only showed that $f$ cannot be continuous by declaring $f(0,0) = 0$. Now I adjusted the proof to show that $f$ is not continuous at $(0,0)$.

Let us suppose for the sake of contradiction that $f$ is continuous at $(0,0)$ and $\lim_{(x,y)\rightarrow(0,0)}f(x,y) = L$. Note that $f(x,x) = \frac{2x^2}{x^2+x^2} = 1$ and $f(2x,x) = \frac{4x^2}{4x^2+x^2} = \frac{4}{5}$. Now let $\epsilon = \frac{1}{20}$ and let $\delta > 0$ be such that $|f(x,y)-L| < \epsilon$ whenever $(x,y)$ is within $\delta$ units of $(0,0)$. Since the points $(\frac{\delta}{4},\frac{\delta}{4})$ and $(\frac{\delta}{2},\frac{\delta}{4})$ are within $\delta$ units of $(0,0)$, we see that $\frac{1}{20} > |f(\frac{\delta}{4},\frac{\delta}{4}-L| = |1-L|$ and $\frac{1}{20} > |f(\frac{\delta}{2},\frac{\delta}{4})-L| = |\frac{4}{5}-L|$. It follows that $\frac{1}{10} = \frac{1}{20}+\frac{1}{20} > |1-L|+|L-\frac{4}{5}| \ge |1-\frac{4}{5}| = \frac{1}{5}$, which yields the desired contradiction.

Title: How to prove that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is not continuous in $(0, 0)$ using an Epsilon-Delta proof?

Body:
In an assignment for the course Real-Analysis I need to proof that $f(x, y)=\frac{2xy}{x^2 + y^2}$ is not continuous in $(0, 0)$ using an Epsilon-Delta proof.

However, me and my fellow students have been trying to figure it out for some time now, but we can't...


答:
Suppose that the limit exists and it is equal to $L$.

For any $\varepsilon>0$, we can find $\delta>0$ so that $|(x,y)|<\delta$ implies that $|f(x,y)-L|<\varepsilon$.

Fix $\varepsilon>0$, but arbitrary. Find such $\delta>0$, so for $|x|<\delta$ and $y=0$ we have that $|(x,y)|<\delta$, thus $|f(x,y)-L|<\varepsilon$, but since $y=0$ we have that $f(x,y)=0$, so $|L|<\varepsilon$. Likewise, for $|x|<\sqrt{\frac{1}{2}\delta}$ we have that $|(x,x)|<\delta$, so $|f(x,x)-L|<\varepsilon$, but $f(x,x)=2x^2/(x^2+x^2)=1$, so $|1-L|<\varepsilon$.

So we have concluded that $|L|<\varepsilon$ and $|1-L|<\varepsilon$ and this is true for any $\varepsilon>0$. Is this possible for $\varepsilon=1/10$?


问:
How can we draw a line parallel $l_1$ and $l_2 $ to that passes of $p$

How can we draw a line parallel $l_1$ and $l_2 $ to that  passes of $p$ ?

Please help me.

Thank you.$\Large\color{red}{☺}$

答:
As you called that segments , lines, I think you are working in Non-Euclidean geometry (Hyperbolic) and you are using the Klein's model. In this model, especially when you consider $\mathbb R^2$, The points are represented by the usual points in the interior of a circle (unit disk) and lines are represented by the chords, straight line segments with endpoints on the boundary of the disk.

Here, as you drew two lines and regarding the place of the point $P$. we cannot draw a chord as problem desired. I made some lines for you:

Note that, as @newzad remarked correctly, if you don't indicate the certain geometry , then your claim, homework or your problem will be absolutely wrong.


问:
How to find a "simple" fraction between two other fractions?
If we have two fractions $a = { a_1 \over a_2} $ and $c = {c_1 \over c_2}$ with $a<c$,

how to find the fraction

$b = { b_1 \over b_2 }$ ,  $a < b < c$

for which some measure of simpleness like

$b_1^2+b_2^2$

is minimal, without trying every possible fraction?

答:
Here's a simple and efficient algorithm, expressed explicitly in Python. It resembles the Euclidean algorithm for computing greatest common divisors, and it can be derived from the descriptions in other answers in terms of continued fractions and/or the Stern-Brocot tree, but it's also straightforward to give a proof of correctness that doesn't use those concepts directly. We'll give the code first, and then sketch that proof of correctness.

Here's the code:

```<code>def simplest_in_interval(r, s, u, v):
    """
    Find simplest fraction in an open subinterval of the positive reals.

    Given nonnegative integers r, s, u and v with r*v < s*u, return a pair
    x, y of relatively prime positive integers such that x/y is the
    simplest fraction in the open interval (r/s, u/v) (in the case v > 0),
    or the simplest fraction in the interval (r/s, ∞) (in the case v = 0).
    """
    a, b, c, d = 1, 0, 0, 1
    while s <= r or u <= v:
        q = r // s
        r, u, b, d = r - q * s, u - q * v, b + q * a, d + q * c
        q = v // u
        s, v, a, c = s - q * r, v - q * u, a + q * b, c + q * d
    return a + b, c + d
</code>```

A note for non-Python-speakers: the // operator in Python performs floor division; that is, a // b gives the value of $\lfloor a/b\rfloor$.

Here's an example of the above function in action: finding the simplest fraction in the interval (314/100, 315/100). (Note that there's no requirement for the input fractions to be given in lowest terms.)

```<code>>>> simplest_in_interval(314, 100, 315, 100)
(22, 7)
</code>```

Sketch of proof of correctness.

Write $J$ for the originally-specified open interval $(r/s, u/v)$. As the algorithm progresses, we maintain eight separate integer-valued variables: $a$, $b$, $c$, $d$, $r$, $s$, $u$ and $v$. Now at every stage of the algorithm, the following loop invariants hold:

$b$, $c$, $r$ and $v$ are nonnegative

$a$, $d$, $s$ and $u$ are positive

$rv < su$ (in fact, $su - rv$ stays constant), so $r$, $s$, $u$ and $v$ determine an open subinterval $K = (r/s, u/v)$ of the positive real line (where as a special case we allow $v=0$ to represent a right endpoint of $\infty$)

$ad - bc = 1$, and $a$, $b$, $c$ and $d$ describe a bilinear transformation $T: x \mapsto (ax + b) / (cx + d)$

$T$ maps $K$ bijectively (and monotonically) to the original interval $J$, so establishes an order-preserving bijection between the rationals in $K$ and those in $J$

The above invariants can be established in the usual way, by checking that they're  true just before the while loop is entered for the first time (at which point $K = J$ and $T$ is the identity map), and then by checking that if they're true at the start of any given iteration of the loop body, then they're also true at the end of the loop body for that iteration.

When the while loop exits, we must have $r < s$ and $v < u$, or in other words, $r/s < 1 < u/v$. So at that point the interval $K$ contains $1$. Now every fraction in the original interval $J$ has the form $(ap + bq) / (cp + dq)$ for some $p/q$ in $K$, and assuming that $p/q$ is written in lowest terms, it follows from the fact that $ad-bc = 1$ that $(ap + bq) / (cp + dq)$ is also written in lowest terms. Since $a$, $b$, $c$ and $d$ are all nonnegative, the simplest possible fraction arises when $p$ and $q$ are minimal - that is, when $p = q = 1$, and that simplest fraction is $(a + b) / (c + d)$.

There's one final thing we need to know, and that's that the while loop does exit - that is, that the while loop condition eventually becomes false, and that we don't simply enter an infinite loop. For this, it's enough to observe that on every iteration of the loop, either $r$ or $v$ decreases, and neither ever increases. Since both $r$ and $v$ stay nonnegative, we can't continue those decreases indefinitely, so the loop must terminate after finitely many iterations. This completes the sketch of the proof.

Body:
Here's a simple and efficient algorithm, expressed explicitly in Python. It resembles the Euclidean algorithm for computing greatest common divisors, and it can be derived from the descriptions in other answers in terms of continued fractions and/or the Stern-Brocot tree, but it's also straightforward to give a proof of correctness that doesn't use those concepts directly. We'll give the code first, and then sketch that proof of correctness.

Here's the code:

def simplest_in_interval(r, s, u, v):
    """
    Find simplest fraction in an open subinterval of the positive reals.

    Given nonnegative integers r, s, u and v with r*v < s*u, return a pair
    x, y of relatively prime positive integers such that x/y is the
    simplest fraction in the open interval (r/s, u/v) (in the case v > 0),
    or the simplest fraction in the interval (r/s, ∞) (in the case v = 0).
    """
    a, b, c, d = 1, 0, 0, 1
    while s <= r or u <= v:
        q = r // s
        r, u, b, d = r - q * s, u - q * v, b + q * a, d + q * c
        q = v // u
        s, v, a, c = s - q * r, v - q * u, a + q * b, c + q * d
    return a + b, c + d

A note for non-Python-speakers: the // operator in Python performs floor division; that is, a // b gives the value of $\lfloor a/b\rfloor$.

Here's an example of the above function in action: finding the simplest fraction in the interval (314/100, 315/100). (Note that there's no requirement for the input fractions to be given in lowest terms.)

>>> simplest_in_interval(314, 100, 315, 100)
(22, 7)

Sketch of proof of correctness.

Write $J$ for the originally-specified open interval $(r/s, u/v)$. As the algorithm progresses, we maintain eight separate integer-valued variables: $a$, $b$, $c$, $d$, $r$, $s$, $u$ and $v$. Now at every stage of the algorithm, the following loop invariants hold:

$b$, $c$, $r$ and $v$ are nonnegative

$a$, $d$, $s$ and $u$ are positive

$rv < su$ (in fact, $su - rv$ stays constant), so $r$, $s$, $u$ and $v$ determine an open subinterval $K = (r/s, u/v)$ of the positive real line (where as a special case we allow $v=0$ to represent a right endpoint of $\infty$)

$ad - bc = 1$, and $a$, $b$, $c$ and $d$ describe a bilinear transformation $T: x \mapsto (ax + b) / (cx + d)$

$T$ maps $K$ bijectively (and monotonically) to the original interval $J$, so establishes an order-preserving bijection between the rationals in $K$ and those in $J$

The above invariants can be established in the usual way, by checking that they're  true just before the while loop is entered for the first time (at which point $K = J$ and $T$ is the identity map), and then by checking that if they're true at the start of any given iteration of the loop body, then they're also true at the end of the loop body for that iteration.

When the while loop exits, we must have $r < s$ and $v < u$, or in other words, $r/s < 1 < u/v$. So at that point the interval $K$ contains $1$. Now every fraction in the original interval $J$ has the form $(ap + bq) / (cp + dq)$ for some $p/q$ in $K$, and assuming that $p/q$ is written in lowest terms, it follows from the fact that $ad-bc = 1$ that $(ap + bq) / (cp + dq)$ is also written in lowest terms. Since $a$, $b$, $c$ and $d$ are all nonnegative, the simplest possible fraction arises when $p$ and $q$ are minimal - that is, when $p = q = 1$, and that simplest fraction is $(a + b) / (c + d)$.

There's one final thing we need to know, and that's that the while loop does exit - that is, that the while loop condition eventually becomes false, and that we don't simply enter an infinite loop. For this, it's enough to observe that on every iteration of the loop, either $r$ or $v$ decreases, and neither ever increases. Since both $r$ and $v$ stay nonnegative, we can't continue those decreases indefinitely, so the loop must terminate after finitely many iterations. This completes the sketch of the proof.


答:
Alternative approach:

Suppose that it is given that $(x,y)$ approaches $(0,0)$ along the line $y = kx$, where $k$ is a fixed positive real number.  Then, you can compute the (non-zero) limit in terms of $k$.  That is, the limit will vary, as $k$ varies.

Therefore, since $(x,y)$ can approach $(0,0)$ along any line $y = kx$, where (for example) $k \in \mathbb{R^+}$, and since the limit will vary as $k$ varies, it is impossible for there to be a single limit that would pertain, regardless of the value of $k$.

That is, it is impossible for there to be a single limit that would pertain, regardless of the line $(y = kx)$ along which $(x,y)$ approached $(0,0)$.


答:
In any neighborhood of $(0,0),$ regardless of the size of delta, there will be a point $(x,x)$ such that $f(x,x) = 1$ and there will be a point $(x,-x)$ such that $f(x,-x) = -1$

For $\epsilon = 1$ and any $\delta > 0$ and any value of $L,$ there is a point $(x,y)$ with $d((x,y),(0,0))<\delta$ and $|f(x,y) - L| > \epsilon.$


问:
How to write the equation of an off-center circle in terms of r (parameterization)
I've looked around online for a while and I can't seem to find a question that properly captures what I'm talking about (or answers that really help me).

For a homework problem, I have to find the area bounded by $x^{2} + y^{2} = 4$ and $(x-1)^{2} + (y-1)^2 = 1$ using double integrals in polar form. I can easily parameterize the first equation as just $r = 2$ using simple algebra.

However, for the second equation, I get stuck quickly. I can first expand the equation to get $x^{2} -2x + 1 + y^{2} -2y + 1 = 1$. I can simplify to get $x^2 + y^2 + 1 = 2(x + y)$. From here I can substitute the polar equivalents of x and y, to get

$r^2 + 1 = 2\big(rcos(\theta) + rsin(\theta)\big)$

From here I have no idea what to do. The math to get here seems simple enough, yet it's somehow wrong.

Once I have the second equation in terms of $r$, I'd imagine I could set the two equations equal to each other, and find at what angles they are equal (giving me the range of $\theta$ over which I need to integrate), then I'd have the second integral for $r$ to be over the bounds of the two equations. But then, I don't know what function I'd be integrating over that specific range.

Thanks for any help. I really am at a loss as to where to turn

答:
In fact this polar equation is in two parts ($r_1$ and $r_2$: see equations (2) and (3) below) which is not surprising when you consider figure 1.

Write your equation under the form:

$$r^2 - 2r\underbrace{(\cos(\theta) + \sin(\theta))}_{f(\theta)}+1=0 \tag{0}$$

and consider it as a quadratic equation, or even simpler as factorisable under the form:

$$[r-f(\theta)]^2+\underbrace{1-f(\theta)^2}_{-2 \sin(\theta)\cos(\theta)}=0$$

Consequently

$$[r-f(\theta)]^2=2 \sin(\theta)\cos(\theta)=\sin(2\theta)$$

which means:

$$r-f(\theta)=\pm \sqrt{\sin(2\theta)}\tag{1}$$

under the condition that $\theta$ belongs to the first quadrat:

$$0\le \theta \le \pi/2$$

in order that $\sin(2\theta) \ge 0$ (needed for taking the square root in (1)).

The previous expressions gives rise to 2 polar equations:

$$r_1=\cos(\theta) + \sin(\theta)-\sin(2\theta) \ \ \ \ \text{(blue arc below)}\tag{2}$$

$$r_2=\cos(\theta) + \sin(\theta)+\sin(2\theta) \ \ \ \ \text{(red arc below)}\tag{3}$$

Fig. 1: the two circles.

Important remark: you need only the blue arc of the second circle.

Another remark: If you want to consider (0) as a quadratic equation, you will find the same $r_1$ and $r_2$ as its roots using discriminant:

$$\Delta=4((\cos(\theta) + \sin(\theta))^2-1)=4(2\cos(\theta)\sin(\theta))=4 \sin(2\theta) $$


问:
Radical equation $x- 5\sqrt{x} + 4 = 0$
Solve: Radical equation $x- 5\sqrt{x} + 4 = 0$

I think I understand what I had did wrong. Instead of squaring a product of a number I was squaring the equation.

Title: Radical equation $x- 5\sqrt{x} + 4 = 0$

Body:
$$x-5\sqrt{x}+4=0$$
$$x+4=5\sqrt{x}$$
$$x^2+16+8x=25x$$
$$x^2-17x+16=0$$
$$x_{1,2}=\frac{17\pm\sqrt{17^2-16\cdot 4}}{2}$$
$$x_1=16\qquad;\qquad x_2=1$$


问:
Quantifier elimination of valued fields.
I have been reading the quantifier elimination theorem for valued fields, Theorem 3.26 on page 35 of these lecture notes by Lou van den Dries: http://homepages.math.uic.edu/~freitag/valfields.pdf

I do not quite understand why we need the $|E|^+$-saturation in the proof.

I mean I know that we need it to use the QE-test but I do not understand why it does not reduce the generality of our proof.
And how we can prove that the ACVF is saturated? Is it because the types are in the form of polynomials?

答:
Let me try to address your questions one by one.

I do not quite understand why we need the $|E|^+$-saturation in the proof.

There are several places in the proof where saturation is explicitly used. For example, in Case 1, van den Dries writes "By the saturation assumption on $(F,B)$, we can find $y\in B$ such that $\bar{y}\notin k_{B\cap iK}$." Here, this amounts to writing down a partial type in the variable $y$ which expresses the condition "$y\in B$ and $\bar{y}\notin k_{B\cap iK}$" and checking that it is consistent (by compactness) and uses at most $|E|$-many parameters. Then $|E|^+$-saturation of $(F,B)$ implies that the partial type is realized in $(F,B)$.

I know that we need it to use the QE-test but I do not understand why it does not reduce the generality of our proof.

Why would it reduce the generality? The QE test says that if we check some semantic condition on models of $T$, then we can conclude that $T$ has QE. The semantic condition involves an arbitrary model $M\models T$ and an $|M|^+$-saturated model $N\models T$. To understand why we may assume that $N$ is $|M|^+$-saturated, you need to read the proof of the QE test! For example,  see Theorem 7.14 in these lecture notes on model theory by van den Dries and Henson.

And how we can prove that the ACVF is saturated?

ACVF is a theory. Saturation is a property of a model. So "ACVF is saturated" makes no sense.

It is a standard theorem that sufficiently saturated models exist: For any cardinal $\kappa$, any first-order theory $T$, and any model $M\models T$, $M$ has an elementary extension which is $\kappa$-saturated. In particular, every algebraically closed valued field embeds (elementarily) in an algebraically closed valued field which is as saturated as you need it to be.

But note that in the statement of the QE test, you are handed an arbitrary model $M$ and an $|M|^+$-saturated model $N$. You don't need to construct $N$ or prove that it is $|M|^+$-saturated. You just need to extend homomorphisms from substructures of $M$ to $N$.

Is it because the types are in the form of polynomials?

It's not clear what it would mean to "types to be in the form of polynomials". But whatever you mean by that, it's not true in ACVF - formulas also include information about the valuation, not just polynomial equations. Quantifier elimination will allow you to reduce understanding the type of a tuple to understanding equations and valuation relations between polynomials in the tuple (since QE says every formula is equivalent to Boolean combination of atomic formulas, and every atomic formula is equivalent to an equality or a valuation relation between polynomials), but you can't assume this in the proof of QE.


问:
Limit with a set proof
Let $x_n = n^{-1}$ and let $X \subset \mathbb{N}$ so that $\sum_{n \in X} x_n$ is convergent. Define $p(m) = 2^m$ and let $Y = \{x \in X | p(m) \leq x < p(m+1)\}$. Prove that $\liminf_{m\to\infty} \frac{cardinal(Y)}{p(m)}$ converges to $0$.

This is what I have so far:

Assume $\sum_{n \in X} x_n$ converges. Then $\lim_{n\to\infty} x_n = 0$.By definition of the limit, for all $\epsilon > 0$, there is a $N \in \mathbb{N}$ such that $|x_n| = |n^{-1}| = |\frac{1}{n}| < \epsilon$ for $n \geq N$

We need that $\liminf_{m\to\infty} \frac{cardinal(Y)}{p(m)} = 0$. We require an $N_2 \in \mathbb{N}$ so that $|\frac{cardinal(Y)}{p(m)}| = \frac{cardinal(Y)}{p(m)} < \epsilon_2$ for $n \geq N_2$.

I have no idea how to solve this problem particularly because it spans and requires arguments from two topics: sets and limits. Any assistance is appreciated.

Edit: I tried using the comparison test as an argument but to no avail.

答:
Since the set $X$ is a subset of the naturals, and the sum converges absolutely, you can assume that the sequence $x_n$ can be rearranged to be monotonically decreasing.

The sequence $p(m)$ partitions $X$, so you can bound the sum in terms of contributions from each block:

$$\sum_{n}x_n\geq \sum_m \frac{|Y(m)|}{p(m+1)}.$$

Suppose otherwise that the liminf is a positive number. Then there's a positive number $c>0$ and a sequence $m_k$ such that $|Y(m_k)|/p(m_k)\geq c$ for all $k>0$. So the above sum can be bounded:

$$\sum_{n}x_n\geq \sum_{k} \frac{cp(m_k)}{p(m_k+1)}= \sum_{k} c/2=\infty $$

Title: Limit with a set proof


问:
Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$
I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits_{0}^{1} \frac{\log(1-u^2+u^4)}{1+u^2}\mathrm{d}u.$$But now I am stuck again.

答:
Svyatoslav showed that $$\int_{0}^{1} \frac{\log(1-x+x^{2})}{\sqrt{x}(1+x)} \, \mathrm dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx- 4 G.$$

The following is a slightly different way to show that $$\int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx = 2 \pi \log (3).  $$

First notice that $$ \begin{align} \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx &= \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx+ \int_{-\infty}^{\infty} \frac{\log(1-\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx \\ &= \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx + \int_{\infty}^{-\infty} \frac{\log(1+\sqrt{3}u + u^{2})}{1+u^{2}} (- \mathrm du) \\ &= 2 \int_{-\infty}^{\infty} \frac{\log(1+\sqrt{3}x+x^{2})}{1+x^{2}} \, \mathrm dx \end{align}$$

In a previous answer I used  contour integration to show that $$I(a,b,\theta) = \int_{-\infty}^{\infty} \frac{\log \left(a^{2}+2ax \cos \theta + x^{2}\right)}{x^{2}+b^{2}} \, \mathrm dx = \frac{\pi}{b} \, \log \left(a^{2}+2ab \sin \theta +b^{2} \right), $$ where $a, b >0$ and $0 <  \theta < \pi$.

Therefore, $$ \int_{-\infty}^{\infty} \frac{\log(1-x^{2}+x^{4})}{1+x^{2}} \, \mathrm dx = 2  I \left( 1,1, \frac{\pi}{6} \right) = 2 \pi \log(3).$$


问:
Genus of a curve
My question is:

How to show that for all $g>0$ there exists an algebraic curve of genus $g$?
The intuition of this is simple, but I don't know how to show it.

答:
The Wikipedia article Hyperelliptic curve states:

In algebraic geometry, a hyperelliptic curve is an algebraic curve of genus $g > 1$, given by an equation of the form
$$    y^{2}+h(x)y=f(x) $$
where $f(x)$ is a polynomial of degree $n = 2g + 1 > 4$ or $n = 2g + 2 > 4$ with $n$ distinct roots, and $h(x)$ is a polynomial of degree $< g + 2$ (if the characteristic of the ground field is not $2$, one can take $h(x) = 0$).

The article goes on to refer to the Riemann-Hurwitz formula
which can be used to give the proof you want.

Title: Genus of a curve


问:
Is the packing density of an ellipse the same as that of a circle?
It is well-known that the densest packing of circles in the plane is the close hexagonal packing, with a density of $\frac{\pi\sqrt{3}}6\approx0.9069$:

By applying an affine transformation, we obtain a packing of ellipses with the same density:

However, not every ellipse packing arises from such a transformation, as we can rotate the ellipses at different angles. So it does not follow that the ellipse must have the same packing density.

Are there any ellipses which have a higher packing density than the circle?

I realize this question may be very difficult to answer in the affirmative, given the subtlety and difficulty of most packing problems, but I am curious to hear of bounds on the problem (possibly as a function of the aspect ratio) and expert opinions on the likely answer, if nothing else.

It is worth noting that in the three-dimensional case, the packing density of an ellipsoid can be higher than that of the sphere, with some ellipsoids having packing densities of at least $0.7585\ldots$ compared to the sphere's $\frac{\pi}{\sqrt{18}}\approx0.7405$ (see Wolfram MathWorld, J. Wills).

答:
Some further investigation reveals that ellipses do indeed have the same packing density as circles, though I hadn't seen it explicitly stated anywhere before so it seems worth leaving this question up for future readers.

In L. Fejes Tóth's 1949 paper Some packing and covering theorems, it is noted that a consequence of Theorem $1$ is that every centrally symmetric convex set has a packing density equal to its translational or lattice packing density, from which the desired result follows: we can assume that an optimal ellipse packing uses only translations, then apply an affine transformation to yield a circle packing of the same density, which could be improved were it any lower than $\pi\sqrt{3}/6$. (He mentions in footnote 8 that this result may have been independently given in an October 1949 lecture of K. Mahler.)

However, Tóth notes that there may still be non-lattice packings that attain the same density, so one can still ask whether there exist nontrivial deviations from the hexagonal packing with an optimal $\pi\sqrt{3}/6$ density. (I do not know whether this is the case, though I would be surprised to find it was true.)