问:
Does trace commutes with Laplacian?
Suppose $(M,g)$ be a Riemannian manifold and $T$ a $(0,2)-$tensor. Is it true that trace commutes with Laplacian? i.e.
$$\mathsf{tr\ }\Delta T=\Delta \mathsf{tr\ }  T?$$

In my point of view it is true because $\Delta T_{kl}=g^{ij}\nabla_i\nabla_j T_{kl}$  and we know that any trace commutes with covariant derivative (if I am not mistaken in this case/ as Prof. Lee response to this post show the trace should be independent of covariant derivative indices that I think is here.).

So if $\mathsf{tr\ } T=0$ then $\mathsf{tr\ }\Delta  T=0$? e.g. if $\omega$ is a two form then $\mathsf{tr\ }\Delta  \omega=0$?

答:
Due to this answer of Professor Lee, because the second trace is independent from covariant derivative indices so $$\mathsf{tr\ }\Delta T=\mathsf{tr_{12}\ }(\mathsf{tr_{34}\ }\nabla\nabla T)=\mathsf{tr_{34}\ }(\mathsf{tr_{12} }\nabla\nabla T)=\mathsf{tr_{34}\ }(\nabla\mathsf{tr_{12} }\nabla T)=\mathsf{tr_{34}\ }(\nabla\nabla\mathsf{tr_{12} } T)=\Delta \mathsf{tr\ }  T$$
So trace commutes with Laplacian. and if $\mathsf{tr\ }  T=0$ then $\mathsf{tr\ }\Delta  T=0$. like Laplacian of differential 2-forms.

Title: Does trace commutes with Laplacian?


问:
If Anna goes from point A to point B, each step can only move up or move right. How many method(s) is / are there?(reference the grid below)
If Anna goes from point A to point B, each step can only move up or move right. How many method(s) is / are there?(reference the grid below)

I’ve just recently learned permutations and combinations. Therefore, I understood how to answer problems regarding grids but only the type of questions with definite number of columns and rows. As shown in the picture above, the grids are different in size. I’m confused on how to solve this. I’d be happy if anyone could help.. Thank you.

答:
Simpler such problems ask for the number of paths from $A$ to $B$ "in a $(7\times11)$-grid". Here the numbers $7$ and $11$ give the full information about the grid in question, and there is then also a simple answer in terms of a formula containing these two numbers.

But your grid is much more complicated. To describe the grid you needed an explicit figure, or the full contents of a $01$-matrix of size $5\times6$. A "formula" for the number of paths would have to operate with the full contents of this arbitrary matrix.

Instead you can think about an algorithm that solves this kind of problem for arbitrary input figures. Such an algorithm could begin with writing $1$ at $A$ and then writing recursively the sum of the "entering numbers" at each "ready" lattice point, until you are at $B$, as in Pascal's triangle.

ContentLicense: CC BY-SA 4.0


问:
Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A\times B$ is closed in $X\times Y$.
Show that if $A$ is closed in $X$ and $B$ is closed in $Y$, then $A\times B$ is closed in $X\times Y$.</p>

答:
Let $ \left\{a_n \right\}_{n=0}^{\infty}\subseteq A, \left\{b_n\right\}\subseteq B$, be series, which converge respectively to the limits $a,b$. Since A is closed, then $ \lim_{n\rightarrow \infty }a_n=a\in A$, and similarly, since B is closed $ \lim_{n\rightarrow \infty}b_n=b \in B$. Therefore, if we define $ x_n=(a_n,b_n)$, then $ \lim_{n\rightarrow \infty}x_n=(\lim_{n\rightarrow \infty}a_n, \lim_{n\rightarrow \infty}b_n)=(a,b)\in A\times B$, and therefore, $A\times B$ is closed in $X\times Y$.

Let's note that this is the direction if this is a metric space. Otherwise, we have to show that $ (X\times Y)/(A\times B)$ is open, by using the fact that $X/A, Y/B$ are open.

ContentLicense: CC BY-SA 3.0


问:
$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$
Let $a,b,c>0$ and $a^2+b^2+c^2=3$, prove

$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$$

This inequality looks simple but I do not know how to solve it. The straightforward method is to bash the inequality with brutal computation method and assume $x = \min \{ x,y,z \}$, $y = x+a$, $z = x + b$, etc. I wonder if we can prove it using classical inequality.

答:
By Cauchy-Schwarz Inequality we have

$\begin{align}
~~~~~~\sum_\text{cyc}\left(a+3b\right)\sum_\text{cyc}\dfrac{a^{2}+3b^{2}}{a+3b}\ge\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\cdot
\end{align}$

So it suffices to show that

$\begin{align}
~~~~~~&\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\ge3\sum_\text{cyc}\left(a+3b\right)\iff\\&\iff\sqrt{a^{2}+3b^{2}}+\sqrt{b^{2}+3c^{2}}+\sqrt{c^{2}+3a^{2}}\ge\sqrt{12\left(a+b+c\right)}\cdot
\end{align}$

With $a^{2}+b^{2}+c^{2}=3$, the last inequality is a "known unsolved inequality" (and it's true).

See:

$\sum\limits_{cyc} \sqrt{a^2+3b^2}\geq \sqrt{12(a+b+c)}$

Prove that $\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$ if $(a+b+c)^2(a^2+b^2+c^2)=27$

a^2+b^2+c^2=3

Inequality 88


问:
Prove that the natural numbers are either the result of the subtraction of two elements in a particular set or exist as elements in the set
Prove that in a set  made of  natural numbers  which every element  follows  ${a_n}$<2n  and  ${a_n}$<${a_{n+1}}$,
every natural number is either found in the set or the result of the subtraction of two elements of this set.
obviously the first element of the sequence will be one but the other elements have other possibilities for example the set could be 1,3,4,7,9,  or it could be 1,3,5,7,9,... how do we go on to prove this?

答:
You can do it by contradiction.

Let's suppose some number $n$ does not occur in the sequence. Then, the numbers $a_1,\dots,a_n$ are all distinct and lie in $\{1,\dots,n-1,n+1,\dots,2n-1\}$.

That set can be partitioned into $n-1$ pairs of numbers with distance $n$. By the pigeonhole principle, in one of the pairs both numbers must be one of the $a_1,\dots,a_n$.


问:
A "Cantor-Schroder-Bernstein" theorem for partially-ordered-sets
If A and B are partially-ordered-sets, such that there are injective order-preserving maps from A to B and from B to A, is there necessarily an order-preserving bijection between A and B ?</p>

答: Here's an example with scattered countable linear orderings: a set of order type $\omega^*\omega$ and a set of order type $1+\omega^*\omega$.


问:
Composition of polynomials and Zariski-topology
I am learning algebraic geometry and have a rather vague question about the the correspondence between polynomial ideals and closed sets in the Zariski-topology in the context of composition of polynomials.

Let $\mathbb{K}$ be an algebraically closed field, polynomials $f,g_1,...,g_n\in\mathbb{K}[x_1,...,x_n]$ and $\overline{\tau}_{\mathbb{A}_\mathbb{K}^n}$ be the set of (Zariski-)closed sets (e.g. $Z(f)\in\overline{\tau}_{\mathbb{A}_\mathbb{K}^n}$ is the zero locus or algebraic set of the ideal $(f)$...). As we have basic identities like $Z(g_1\cdot g_2)=Z(g_1)\cup Z(g_2)$ and $Z(g_1,g_2)=Z(g_1)\cap Z(g_2)$ I wonder if there are also identities of the kind
\begin{align}
Z(f(g_1,...,g_n))=\theta(Z(f),Z(g_1),...,Z(g_n))
\end{align}
for some $n+1$-ary function $\theta:\left(\overline{\tau}_{\mathbb{A}_\mathbb{K}^n}\right)^{n+1}\to\overline{\tau}_{\mathbb{A}_\mathbb{K}^n}$. I did never encounter this situation in my algebraic geometry literature. Is there any material on this?

答:
$g_1, \cdots, g_n$ together define a morphism $g: \Bbb K^n \to \Bbb K^n$.

$$\begin{array}{cl}
& Z(f(g_1, \cdots, g_n)) \\
=& \{ x \in \Bbb K^n \mid f(g_1(x), \cdots, g_n(x)) = 0 \} \\
=& \{ x \in \Bbb K^n \mid (g_1(x), \cdots, g_n(x)) \in Z(f) \} \\
=& \{ x \in \Bbb K^n \mid g(x) \in Z(f) \} \\
=& g^{-1}(Z(f))
\end{array}$$

I am afraid there might be no expression strictly of the form you want; this is the best I can do.


问:
How does a constant magnitude in a system of ODEs affect the convergence of a numerical method?
Consider the following ODE system to be solved numerically:

\begin{align}
 \dot{x}_1 &= -x_1
\\
 \dot{x}_2 &= -2x_2 + \beta x_4
\\
 \dot{x}_3 &= 2x_2
\\
 \dot{x}_4 &= x_1 - \beta x_4
\end{align}
When I solved the system numerically, I found that increasing the value of $\beta$ increases the number of iterations until convergence in some solvers (e.g. Matlab ode113), but the number of iterations stays relatively constant in other solvers (e.g. Matlab ode15s), and in fact it does it in fewer iterations.

My question: how does the magnitude of $\beta$ in this system affect the rate of convergence of a numerical method? Is there any specific relation that clearly shows that?

ode15s

ode113

答:
Your system converges to a fixed point with $x_1^*=x_3^*=x_4^*=0$ and $x_2^*=c_1+c_2+c_3+c_4$ where $x(0)=c$. That last follows from the fact that the sum over the derivatives cancels all the terms on the right side, so that the sum over the components is constant.

The non-stiff solverode113 uses Adams-Bashford-Moulton, but in a PECE implementation which makes this an explicit method. As such it has a bounded stability region for any order that is used internally. Assume that the order stabilizes. Then the step size controller will increase the step size $h$ until that boundary is reached and surpassed for $\lambda h$ where $\lambda=-\beta$ is the dominant eigenvalue of the Jacobian. Remember that the stability region is where the step propagator is contracting (non-expanding) if the flow around the exact solution is contracting (non-expanding).

At this point, outside the stability region, the distance from the fixed point will grow, and with it the step error, so that the step size gets regulated down. Good step size controllers will minimize the resulting oscillation. The end result still is that for the most part of the integration interval you get $\beta h=const.$ where the constant is some value typically in the interval $[2,3]$. The number of evaluations over some interval $[0,T]$ will then be about $N=T/h$ which is proportional to $\beta$.

Note that this behavior is largely independent of the target error tolerance, the equilibrium step size will not change, only the equilibrium distance from the stationary point will change.

In the stiff solver ode15s the BDF methods that are truly implemented as implicit methods do not have (all) such restricted stability regions. Thus the error estimator is less wrong and keeps the step size in the region that is optimal for the error tolerance. As the components that have reached their asymptotic value fall below the absolute error tolerance, they do not influence the step size controller. Thus the step size dynamic is largely oriented to approximate the $e^{-t}$ and $e^{-2t}$ components correctly, which is independent of $\beta$ and gives thus the almost constant step counts.


问:
Probability of sequences of dice rolls
I would really appreciate help with a calculation of probabilities.

Let's say I have a die with six sides. I roll the die 7 times and get the following sequence:
5, 1, 2, 2, 4, 5, 1

Given some random other sequence (e.g. 6, 1, 3, 2, 5, 5, 5), what is the probability that there are exactly two numbers in the same position? Exactly three numbers in the same position? And so on.

For example, with the above sequences there are three numbers in the same position: 1 in second position, 2 in fourth position and 5 in sixth position.

答:
Hint:

Without Loss of Generality, the first sequence is

1, 1, 1, 1, 1, 1, 1.

Title: Probability of sequences of dice rolls


问:
Why $4b^3p=2a^2+qb$ has no solutions for integers $a,b$ and two odd primes $p<q$?
Empirically I do not find any integer solutions for $4b^3p=2a^2+qb$, where $a$ and $b$ are non-zero integers and $2<p<q$ are two odd primes.

If it is provable, why this equation does not provide a solution?

答:
If $b=0$, we quickly find $a=0$, which indeed gives a solution.

Write $b=uv^2$ where $u$ is square-free.
Then
from $$2a^2=uv^2\cdot{(4u^2v^4p-q)} $$
we conclude that $v\mid a$ and write $a=vw$:
$$2w^2=u\cdot\underbrace{(4u^2v^4p-q)}_{\text{odd}} $$
We conclude $u=2t$ and so
$$w^2=t\cdot{(16t^2v^4p-q)} $$
It seems promising to find solutions with $w=1$ and $t=\pm1$, which amounts to looking for primes $p$ and integers $v$ such that one of $(2v)^4p\pm1$ is prime.
Even with these additional restrictions we find lots of solutions $(p,q,a,b)$, such as
$$ (3,47,1,2)^1, (5,79,1,2), (7, 113, 1, -2), \ldots, (41, 409999, 5, 50),\ldots,(97, 1553, 1, -2),\ldots, (983, 15727, 1, 2),
(991, 64946177, 8, -128),
(997, 1292113, 3, -18),\ldots (1000003, 38416115249, 7, -98),\ldots, (1000000007, 256000001791, 2, 8),\ldots$$

$^1$ see answer by cr001.


问:
Coloring dots in a circle with no two consecutive dots being the same color
I ran into this question, it is not homework. :)

I have a simple circle with $n$ dots, $n\geqslant 3$. the dots are numbered from $1\ldots n$.

Each dot needs to be coloured red, blue or green. No dot can be couloured the same as his neighbors. How many options are there to organize the colours?

Hint: we need to use recurrence relation.

答:
$$A_{n+2} = A_{n+1}+2*A_{n}$$

for a coloring of a sequence of dots $d_1,d_2,...,d_n$, it is a valid coloring iff one of two mutually exclusive conditions is met.

(1) $d_1\ne d_n\ne d_{n-1}\ne d_1$ and $d_1,d_2,...,d_{n-1}$ is a valid coloring.

(2) $d_1 = d_{n-1}\ne d_{n-2}$ and $d_n\ne d_1$ and $d_1,d_2,...,d_{n-2}$ is a valid coloring.

in the first case, each unique choice of $d_1,d_2,...,d_{n-1}$ yields a unique choice for $d_n$, and in the second case, each choice of $d_1,d_2,...,d_{n-2}$ yields a unique choice for $d_{n-1}$ and two possibilities for $d_n$. these possibilities do not overlap, so we simply add them.


问:
Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. How do I prove that $A'$ is closed?
I was asked this simple question a short while ago.

Let $A\subseteq\mathbb{R}^k$, and let $A'=\{x\in\mathbb{R}^k \mid x\in\partial(A\setminus\{x\})\}$. Prove that $A'$ is closed.

I think I was able to explain it, but I am not sure what I claim is true. (The following is not a formal solution because I am not sure the reasoning is correct) My explanation goes as follows:

Let $x\in A'$. Say $x$ is in the interior of $A$, then $x\notin\partial(A\setminus\{x\})$ because there exists a ball $B(x;r)\subseteq A$ by definition. By a smilar logic if x is in the exterior of $A$, I claim that $x\notin\partial(A\setminus\{x\})$. Thus $x\in\partial A$ (that is $A'\subseteq\partial A$). From here I am quite sure it is not too hard to explain that $A'$ is closed.

I doubt my reasoning, but I am not sure what have I missed.
Care to shed some light over the matter?

答:
I believe you get a better insight if you forget about $\mathbb{R}^k$ and balls. Let's work in a (Hausdorff) topological space $X$, which your case is a specialization of.

What are the points $x$ such that $x\in\partial(A\setminus\{x\})$?

First of all they don't need to belong to $A$. For instance, with $A=(0,1)\subseteq\mathbb{R}$, $0$ belongs to the set $A'$.

Let's see: $x\in\partial(A\setminus\{x\})$ means that every neighborhood of $x$ intersects both $A\setminus\{x\}$ and its complement. However, $x$ surely belongs to the complement of $A\setminus\{x\}$, so the condition just reads
$$
\textit{every neighborhood of $x$ intersects $A\setminus\{x\}$}
$$
which is the standard definition of limit point.

Now it's easier, isn't it? You need to show that if $y$ has the property that each of its open neighborhoods intersects $A'$, then $y\in A'$.

Take such an element $y$ and an open neighborhood $V$ of $y$. Then there exists $x\in A'$ such that $x\in V$. Since the space $X$ is Hausdorff, there exists a neighborhood $U$ of $x$ such that $U\subseteq V$ and $y\notin U$.

Since $U$ is a neighborhood of $x$, there exists a point $z\in A$ (different from $x$, but it's irrelevant) such that $z\in U$. Now necessarily $z\ne y$, but $z\in V$ and we're done.


问:
Let $p$ be a prime number. Find all solutions of the congruence $x^2\equiv x\pmod p$
I used Fermat's Little Theorem to get

$x^2 \equiv x\pmod 2$

So $2k = x^2-x$ for some integer $k$

$= 2k = x(x-1)$, note that this is a product of two consecutive numbers, so we can rewrite it as $2l$ for some integer $l$

$ = 2k = 2l$

$= k = l$

I don't think this tells me anything meaningful. I think my wrong step is when I start to toy around with the product of consecutive integers, but I am not sure how else to proceed.

答:
$x^2\equiv x\pmod {p}$ simply means that $p|x^2-x$, or $p|x(x-1)$.

Now since $(x,x-1)=1$, either $p|x$ or $p|x-1$, thus giving $x=kp$ or $x=kp+1$ as the only solutions $(k\in \mathbb Z)$.


问:
Calculating the integral of $\int_0^\infty \int _{-\infty}^\infty \frac{x e^{-x}}{0.2(\sin(1.5x)+y)^2+0.1}dy dx$
I was wondering if this has a name (I was taking the Gamma distribution and tried to create a 2D with "wiggly-tail" version of it). And also if there is a way to solve it? I used numerical methods to estimate that the integral (i.e. normalizing-constant) should be around $20, 21$. Wolfram alpha puts it at $5\sqrt 2 \pi\approx 22.2144$.

$$\int_0^\infty \int _{-\infty}^\infty \frac{x e^{-x}}{0.2(\sin(1.5x)+y)^2+0.1}dy dx$$

答:
In the integral over $y$ you can treat $x$ as a parameter and introduce a new integration variable $u=y + \sin(1.5x)$, $du=dy$ to get an integral
$$ \int_0^\infty \Big(\int_{-\infty}^\infty \frac{xe^{-x}}{0.2 u^2+0.1} du\Big) dx = \Big( \int_0^\infty xe^{-x} dx\Big)\Big(\int_{-\infty}^\infty \frac{1}{0.2 u^2+0.1} du\Big)$$
Do you know how to calculate this?


答:
$$\int\frac{e^{-x} x}{\frac{1}{5} \left(\sin \left(\frac{3
   x}{2}\right)+y\right)^2+\frac{1}{10}}\,dy=5 \sqrt{2} e^{-x} x \tan ^{-1}\left(\sqrt{2} \left(\sin \left(\frac{3
   x}{2}\right)+y\right)\right)$$ Assuming $x \geq 0$, then
$$\int_{-\infty}^{+\infty}\frac{e^{-x} x}{\frac{1}{5} \left(\sin \left(\frac{3
   x}{2}\right)+y\right)^2+\frac{1}{10}}\,dy=5 \pi\sqrt{2}\,x\,   e^{-x} $$ One integration by parts to get
$$5 \pi\sqrt{2}\,\int_0^p x\,   e^{-x}\,dx=5 \pi\sqrt{2}   \left(1- (p+1)e^{-p}\right) \to 5\pi \sqrt{2} $$

Title: Calculating the integral of $\int_0^\infty \int _{-\infty}^\infty \frac{x e^{-x}}{0.2(\sin(1.5x)+y)^2+0.1}dy dx$

Body:
I was wondering if this has a name (I was taking the Gamma distribution and tried to create a 2D with "wiggly-tail" version of it). And also if there is a way to solve it? I used numerical methods to estimate that the integral (i.e. normalizing-constant) should be around $20, 21$. Wolfram alpha puts it at $5\sqrt 2 \pi\approx 22.2144$.

$$\int_0^\infty \int _{-\infty}^\infty \frac{x e^{-x}}{0.2(\sin(1.5x)+y)^2+0.1}dy dx$$


问:
Find the area of the shaded region within the smaller equilateral triangle

WLOG, let the side length of one of the smaller equilateral triangles by $1$. So the overall area of one of the smaller equilateral triangles is $\frac{\sqrt{3}}{4}.$ To find the area of the shaded region (within the smaller triangle), I toyed with the idea of coord-bashing (I know, a bad method, but I didn't know what else to do), but quickly dismissed it. It would take way too much time.

I am at a loss for what to do, can someone please provide me with a hint (not solution) that will help me on my way?

答:
EDIT: replacing hint with a solution after consent from OP

WLOG, sides of each small equilateral triangle is $1$.

As $UV$ and $XY$ are parallel to $BC$, $AM$ divides all of them in the same ratio. From that, we get

$XT = 4 \cdot \frac{3}{5} = \frac{12}{5} \implies ST = \displaystyle \frac{2}{5}$. Similarly we find $PQ = \displaystyle \frac{1}{5}$.

We also observe that $\angle QPR = \angle RST = 60^0$. So, $\triangle PRQ \sim \triangle SRT$.

Altitude from $R$ to $OP$ is $1/3$ times altitude from $S$ to $OP$.

So, Area $\triangle PQR = \displaystyle \frac{1}{15} \cdot$ Area $\triangle OPS$

$\therefore \ $ shaded part is $\frac{14}{15}$ times the dotted area.


问:
Confusion in proof for writing Poincares series in terms of Kloosterman sum
I am currently self reading " Spectral theory of Riemann zeta function" by Yoichi Motohashi. The example is in first chapter and of poincare series. I am not able to justify from 1.1.5 to the next line. Actually I don't know what does mean by classifying summands according to h (mod l). Any help/Hint would be appreciated. I am stuck here since yesterday.

答:
Since the summary has very often the expression $h/l$ , it is sensible to ask: how can I divide this fraction in integral part and fractional part? We can also hope, since some functions over there are periodic (e.g. $\exp( i\cdot) $) that things with the same fractional part will contribute in a similar fashion.

Hence we set $h = l n + h'$, with $h' = 1, \ldots, l $ and $n= - \infty, \ldots, +\infty$. The new $h'$ is the "classification $\pmod{l}$ " of the summand relative to $h$. Recall that $h$ has to be comprime with $l$, so that $ 1= (ln+h', l) = (h', l) $: the classification index has to be coprime with $l$. If $g(l, h) $ is your summand, you can then write

$$\sum_{l=1}^{\infty}\sum_{h=-\infty, (h, l) =1}^{+\infty} g(l, h) = \sum_{l=1}^{\infty} \sum_{h'=1}^{l} \sum_{n=-\infty}^{+\infty} g(l, nl+h') $$


问:
Why relative velocity must be perpendicular to velocity of one of the bodies for closest approach
I came across this question that asked the same, but I couldn't understand the reasoning. I feel it wasn't answered clearly, and the OP also sought clarification.

The answer stated,

From a physics perspective -- (here we go again, answering physics questions on a math site! ;) ) -- it may make more sense if you go into the inertial reference frame where the ship being approached is stationary. This involves subtracting the (vector) velocity of the ship being approached from the velocities of both ships. [...] one ship is stationary, its velocity is zero, so it stays at the same place all the time. It's at a point. (The red point.) The other ship moves along some straight line path past that point. The minimum distance from the point to the line is a perpendicular.

But this doesn't explain why $V_{ba} \perp V_{b}$. In fact, I think OP tried to ask the same.

But the question I'm asking John is marked in red above. The two lines in your diagram are perpendicular, the solid blue line represents the velocity vector of the patrol boat B as you rightly said. But this must mean that the dashed blue line is a relative velocity, (as well as being the shortest distance). Can you explain if the dashed line represents the velocity of B relative to S?

This is what I understand so far: (1-3 are covered in the linked question's accepted answer)

Assume $V_{a}$ is velocity of first body and $V_{b}$ the velocity of the second

$V_{ba} = V_{b} - V_{a}$ can be taken as object A is stationary and object B moves with $V_{ba}$

For distance of closest approach, it's as simple as dropping a perpendicular from A to $V_{ba}$

This part, I don't get (and unexplained in the linked question). For closest approach to happen, $V_{ba} \cdot V_{b} = 0$.

Why must $V_{ba} \perp V_{b}$?

Given $V_{ba}$ and $V_{b}$ are in different reference frames, how do I derive the relation that their dot product must be zero?

答:
We have two boats $A$ and $B$ in the plane  with
$$A(t)=a+t u,\qquad B(t)=b+t v\qquad(-\infty<t<\infty)\ ,$$
whereby it is assumed that the velocities $u$ and $v$ are linearly independent. We are interested in the vector
$$d(t):=B(t)-A(t)=(b-a)+t(v-u)\ .$$
Since $\bigl|d(t)\bigr|\to\infty$ when $|t|\to\infty$ there is a certain $t_*$ for which $|d(t)|$ is minimal, and this $t_*$ can be found by putting
$${d\over dt}|d(t)|^2={d\over dt}\bigl(d(t)\cdot d(t)\bigr)=2 d(t)\cdot d'(t)$$
to zero. As $d'(t)=v-u$ it follows that
$$d(t_*)\cdot (v-u)=0\ .$$
This means that at the time $t_*$ of "nearest contact" the relative speed of the two boats is orthogonal to the vector $d(t_*)$ connecting the two boats.


答: This problem can be solved using dynamic programming, let's define the function $n(P_1, P_2) =$ number of ways to go from $P_1$ to $P_2$. Then we need to find $n(A,B)$, which can be computed by solving smaller subproblems (in bottom-up fashion, from left to right and from bottom to top) and combining the results of the subproblems, as shown in the below figure

Body:
If Anna goes from point A to point B, each step can only move up or move right. How many method(s) is / are there?(reference the grid below)

I’ve just recently learned permutations and combinations. Therefore, I understood how to answer problems regarding grids but only the type of questions with definite number of columns and rows. As shown in the picture above, the grids are different in size. I’m confused on how to solve this. I’d be happy if anyone could help.. Thank you.


问:
Could someone explain the underlined part from Introduction to Stochastic Programming book (2nd Edt.) of JR Birge -F Louveaux?

Questions:

Q1- Why only finitely many different such hyperplanes?

Q2- How does this complete the proof?

答:
It seems that the proof tries to establish that $K_2(\xi)$ has a finite number of defining linear inequalities, therefore, is a polyhedron. However, it's not clear to me that the argument is entirely correct.

First, it may not be possible to find an $x$ and $\xi$ s.t. $x$ is not in $K_2(\xi)$. For example, if $W(\omega)=[-1, 1]$ for all $\omega$ and $y\in\mathbb{R}^2$, then $\text{pos}W(\omega)=\{t\,|\,W(\omega)y=t, y=[y_1,y_2]^T\ge 0\}=\{t\, |\, t = y_1-y_2, \, y_1\ge 0, y_2\ge 0\}=\mathbb{R}$. So no such $x$ and $\xi$ exists in the assumption in the beginning of the proof. Of course, if no such $x$ exists, then $K_2(\xi)$ is the entire space, so it's a polyhedron. However, the proof does not seem to consider this possibility.

Second, there could be infinitely many separating hyperplanes that pass through the origin and separate a point from a convex cone. It doesn't seem to make sense to claim there are only finitely many such hyperplanes, if only separation argument is invoked.

Claim a) directly follows from the fact: ''the projection of a polyhedron is a polyhedron.'' Because $K_2(\xi)$ is the projection of the following polyhedron $\{(x,y) \,|\, y\ge 0, W(\omega)y=h(\omega)-T(\omega)x\}$ to the space of $x$. This fact can be proved by the Fourier-Motzkin elimination (see e.g. Projection of a polyhedron is a polyhedron).


问:
Prove that for $W$ is closed vector subspace of $V$, $V/W\cong W^\perp$ as well as $V^*/W^\perp\cong W^*$ by a different method
This is a well-known result in functional analysis, in terms of dual of quotient spaces and annihilators of subspaces. Let me formulate the problem first and a new attempt to prove it:

Let $W$ closed vector subspace of $V$, $V/W$ the quotient space, and $W^\perp$ the annihilator of $W$.

(1) Show that the map $S$ taking an element $f\in W^\perp$ into the linear functional $S(f)$ defined by
$$S(f)([x])=f(x),\forall[x]\in V/M$$
is well-defined, maps $W^\perp$ into $(V/W)^*$ and induces an isometry between $W^\perp$ and $(V/W)^*$.

(2) Show that the map $T$ taking an element $[f]\in V^*/W^\perp$to the linear functional $T([f])$ defined by $$T([f])(x)=f(x),\forall[f]\in V^*/W^\perp\text{ and }x\in W$$
is well-defined, maps $V^*/W^\perp$ into $W^*$ and induces an isometry between $ V^*/W^\perp$ and $W^*$.

Many texts, like in Brezis', directly proves the existence of isometric isomorphisms in both cases. That is, prove both $S$ and $T$ are bijective, linear, and are isometries. After reading the proofs in Brezis' book, I'm thinking of constructing inverse maps of $S$ and $T$, another way to prove both results and here are my thoughts so far:

For (1), I consider $g\in (V/W)^*$ and define the map $T(g)=g\circ Q$, where $Q:V\to V/W$. I wish to show that $T:(V/W)^*\to W^\perp$ (this is not the same as $T$ in (2)), continuous, $T$ is inverse of $S$ with both norm equal to $1$.

For (2), to construct the inverse map, I'm thinking of using Hahn-Banach theorem to extend an element $g\in W^*$ to an element $f\in V^*$ with the same norm. Then show that $[f]=[\tilde{f}]$ for any two such extensions, then the map $g\to [f]$ is well-defined.

Up to now, I don't know how to continue with these forumulations. I can prove that $S$ and $T$ are well-defined, but having problem showing the rest using this method. Can anyone help me finish the argument? Thank you.

答:
I figured this out:

Take $v_1-W,v_2-W\in V/W$, this implies $v_1-v_2\in W$. Since $f\in W^\perp$, $f(v_1-v_2)=0$, $f(v_1)=f(v_2)$ by linearity. Hence $S(f)$ is well-defined.

By the hint, we show that $T$ maps $V/W$ into $W^\perp$, and is linear, bijective, and is an isometry: It is clear that $T$ is linear. For $x\in V$,
$$T(g)(x)|=|g(x-W)|\leq\|g\|\|x-W\|\leq\|g\|\|x\|$$

Hence $T(g)\in V^*$. If we take $x\in W$, $T(g)(x)=g(W)=0$, so $T(g)\in W^\perp$. So $T$ maps $(V/W)^*$ into $W^\perp$. By the above argument, it is clear that $T$ is linear and injective. To show that it is surjective, let $h\in W^\perp$ and define the map $g:V/W\to\mathbb{C};g(x-W)=h(x),\forall x\in V$. $g$ is well-defined, since $W\subset\ker h$. $g$ is actually linear, and $\forall x\in V$ and $\forall y\in W$,
$$|g(x-W)|=|h(x)|=|h(x-y)|\leq\|h\|\|x-y\|$$

which implies $g\in(V/W)^*$. Hence, $T(g)=h$, $T$ is surjective.

And recall that in Exercise 2, we prove that $\|Q\|_{op}=1$, so we have $\|T(g)\|=\|g\circ Q\|\leq\|g\|$. To prove the converse, let $\{x_n-W\}\subset V/W$ be a sequence s.t. $\|x_n-W\|<1$ and $|g(x_n-W)|\to\|g\|$. Then, $\forall n$, $\exists y_n\in W$ s.t. $\|x_n-y_n\|<1$. Hence,
$$\|T(g)\|\ge\|T(g)(x_n-y_n)\|=|g(x_n-y_n)|\to\|g\|$$

Hence, $\|T(g)\|=\|g\|$. Then we have shown the above three results, and it remains to check that $S$ is the inverse of $T$.
$$\forall[x]\in V/W,T\circ S(f)([x])=T(f)(x)=f([x])\Rightarrow T\circ S=\text{id}_{W^\perp}$$
$$\forall x\in V,S\circ T(g)(x)=S(g)([x])=g(x)\Rightarrow S\circ T=\text{id}_{(V/W)^*}$$

Since the inverse of an isometry is again an isometry, we conclude that $S$ is an isometry.

To show $T$ is well-defined, take $[f],[f']\in V^*/W^\perp$, then $\forall x\in W$, $[f-f'](x)=0$, which implies $f-f'\in W^\perp$. $0=(f-f')(x)\Leftrightarrow f(x)=f'(x)\Leftrightarrow T[f](x)=T[f'](x)$.

Consider the map $S:W^*\to V^*/W^\perp;S(w^*)=v^*-W^\perp$. First, we show it is well defined, independent of choices of Hahn-Banach extension. Suppose $w^*$ has Hahn-Banach extensions $v_1^*$ and $v_2^*$, then $w^*=v_1^*|_W=v_2^*|_W$. This implies $(v_1^*-v_2^*)|_W=0\Rightarrow v_1^*-v_2^*\in W^\perp\Rightarrow v_1^*-W^\perp=v_2^*-W^\perp$.

Obviously, $S$ is linear. $S$ is injective: We can rewrite $S(v^*|_W)=v^*-W^\perp$, and $v^*|_W=0$ if and only if $v^*\in W^\perp$. By analytic Hahn-Banach theorem, $\forall w^*\in W^*$, $\exists v^*\in V^*$ s.t. $v^*|_W=w^*$, which implies $S$ is surjective. Furthermore, the theorem implies that $\|v^*\|=\|w^*\|$. Use this result to prove that $S$ is an isometry: $\|S(w^*)\|\leq\|v^*\|=\|w^*\|$. To prove the converse, as noticed when proving injectivity, $w^*=(v^*-W^\perp)|_W$, hence $\| w^{*}\|\leq\| v^*-w^{*'}\|,\forall w^{*'}\in W^\perp$. Take infimum over all $w^{*'}\in W^\perp$, $\|w^*\|\leq\|v^*-W^\perp\|=\|S(w^*)\|$. Hence $\|S(w^*)\|=\|w^*\|$.

It remains to check that $T$ is the inverse of $S$:
$$\forall w^*\in W, T\circ S(w^*)(x)=T([v^*])(x)=v^*(x)=w^*(x)\Rightarrow T\circ S=\text{id}_{W^*}$$

where in the last equality is consequence of Hahn-Banach extension.
$$\forall[f]\in V^*/W^\perp, S\circ T([f])(x)=S(f)(x)=[f](x)\Rightarrow S\circ T=\text{id}_{V^*/W^\perp}$$

Note that in the last equality, there is actually another extension $[\tilde{f}]$, as given in the hint. But we have shown that $S$ is well-defined, so it is independent of choices of extensions. Since the inverse of an isometry is again an isometry, we conclude that $T$ is an isometry.


问:
Prove the orthogonality relation of Chebyshev polynomials of the first kind
The Chebyshev polynomials of the first kind are obtained from the recurrence relation
$$\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=x\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x)~.\end{aligned}$$
Prove that:
$$\int _{-1}^{1}T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}={\begin{cases}0&~{\text{ if }}~n\neq m~,\\\\\pi &~{\text{ if }}~n=m=0~,\\\\{\frac {\pi }{2}}&~{\text{ if }}~n=m\neq 0~.\end{cases}}$$
I tried to prove it using $x = \cos \theta$ and using the defining identity $T_n(\cos \theta) = \cos n\theta$, but I couldn't... I arrived to this integral, which for me is hard to solve:
$$2\int_0^{\pi/2}\frac{\cos n\theta\, \cos m\theta}{\sin \theta}$$

答:
$n=m=0 \Rightarrow \int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\int \frac{dx}{\sqrt{1-x^2}}=arcsin x$

$\int \frac{dx}{\sqrt{1-x^2}}=arcsin x \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\pi$

$x=cos \theta $ and $n=m\neq 0 \Rightarrow \int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=-\int\cos^2n\theta d\theta$

$-\int\cos^2n\theta d\theta=-(\frac \theta2+\frac{\sin 2n\theta}{4n}) \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\frac\pi 2 $

let $m\neq n$

$x=cos\theta \Rightarrow dx=-sin\theta d\theta$

$x=cos\theta \Rightarrow T_m(x)=\cos m\theta$

$\int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=-\int \cos m\theta \cos n\theta{d\theta}$

$\int \cos m\theta \cos n\theta{d\theta}=\frac{\sin(m-n)\theta}{2(m-n)}+\frac{\sin(m+n)\theta}{2(m+n)} \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=0$


问:
Can we think of grothendieck topologies (in ZFC + $U$)?
(Here a set means an element of a universe $U$.)

We can define a cite as an ordered pair ($C$, $T$) which is that $C$ is the category and $T$ is a grothendieck topology. Usually, we define a category and a grothendieck topology as a set, so is a cite. However, I don't know cites is a set. That is because for me it is unclear that grothendieck topologies that can be given to a category is a set(as is well known, a collection of sets is not always a set.) Can we think of grothendieck topologies?

答:
As Asaf mentions in the comments, you mean site not cite.

Let us fix some small category $\mathcal{C}$. Then a Grothendieck topology consists for each object $X$ of a collection of sieves on $X$. A sieve is a collection of arrows into $X$. Of course, all of these need to satisfy certain axioms, but that is irrelevant for determining whether or not things are small (i.e. sets).

Since $\mathcal{C}$ is small $\operatorname{Hom}(A, X)$ is a set for all objects $A$ and $X$. So the collection of all arrows into $X$,
$$
\operatorname{Hom}(-, X) = \bigcup_{A \in \mathcal{C}} \operatorname{Hom}(A, X),
$$
is a set because there is only a set of objects in $\mathcal{C}$. So a sieve on $X$ is just some element of the powerset $\mathcal{P}(\operatorname{Hom}(-, X))$, which is a set. The Grothendieck topology has for every $X$ a collection of sieves on $X$, which is just an element of $\mathcal{P}(\mathcal{P}(\operatorname{Hom}(-, X)))$. We need such a collection of sieves for every object $X$, so a Grothendieck topology is an element of
$$
G = \prod_{X \in \mathcal{C}} \mathcal{P}(\mathcal{P}(\operatorname{Hom}(-, X))),
$$
which is again a set. Not every element in this set will be a Grothendieck topology (we included way too many things). So we can take
$$
G' = \{ g \in G : g \text{ is a Grothendieck topology on } \mathcal{C} \}.
$$
Then $G'$ is the collection of all Grothendieck topologies on $\mathcal{C}$ and by construction $G'$ is a set.


问:
Formulating the axiom of choice with a new function symbol without also getting global choice
Is there a way to formulate the axiom of choice by adding a single "choice-like" function symbol to set theory without also adding the axiom of global choice?

Bourbaki's $\tau$ or variations thereof can be defined as taking sets, sets or proper classes, or propositions with a designated free variable as arguments. If $\tau$ takes a set as an argument and returns a set, then it is a unary function in the theory of sets.

I'm pretty sure that adding any version of $\tau$, even one that is restricted to taking sets as arguments, gives you a theory with the axiom of global choice. In this Wikipedia article, $\tau$ is described as picking an arbitrary element from every nonempty set; proper classes and propositions are not mentioned. I'm assuming that the various flavors of $\tau$ are basically equivalent.

Here's my attempt to add a function symbol that sends every set to the graph of its choice function. I'm pretty sure that doing so gives you the axiom of global choice.

Since doing the obvious thing didn't work, I'm wondering whether what I'm trying to do is even possible.

I've seen formulations of the axiom of choice that use indexed families of sets of choice functions.

Here's the choice function formulation given on Wikipedia expressed using slightly different syntax. For every nonempty set $X$, there exists a choice function $f$.

$$ \forall X \supsetneq \emptyset \mathop.\;\; \exists f : X \to \cup X \mathop.\;\; \forall A \in X \mathop. \;\; f(A) \in A $$

Choice functions are unary functions and as such have a natural encoding as sets.

$$ f \;\text{as a set} \;\; \text{is equal to} \;\; \{ (x, f(x)) \mathop| x \in X \} $$

It, seems like, in principle, we could add a new function symbol $\sigma$ that just sends every nonempty set to the graph of a choice function. $\sigma$ is supposed to be analogous to a Skolem function for the axiom of choice. I'm pretty sure that this step strengthens the axiom because the newly minted $\sigma$ can be used to define $\tau$.

$$ \sigma(\emptyset) \;\; \text{is arbitrary} $$
$$ \sigma(X) = \;\; \text{$f$ as a set where $f$ is the choice function for $X$} $$

$\sigma$ can now be treated as a function symbol in a theory of sets. It isn't expressible as a function because its domain and codomain are proper classes, but it can be a function symbol in the same way that $\in$ is a relation symbol.

We can define Bourbaki's $\tau$ in terms of $\sigma$ in the following way.

$$ \tau(X) \;\;\text{can be defined as}\;\; \sigma({X})(X) $$

答:
No. Once you add a symbol, and you require that it chooses an element from each set, you've effectively added global choice.

The reason why global choice is stronger than the axiom of choice is that for each family of sets there exists a choice function, but we cannot amalgamate these choice functions over all sets uniformly (unless, of course, we can choose one for each set, but then we already appeal to global choice).

Now, the problem with adding a symbol to the language is that it is interpreted in the structure. So it is now a fixed choice function which we can always refer to. And that's just global choice.

If it's any consolation, every model of ZFC can be extended to a model of ZFC+Global choice without adding sets, if we instead add a new symbol for the choice function. This can be done via class forcing.


答:
Hint:

We have a reference sequence to match. In how many ways, can there be a match in exactly two positions? There are $\displaystyle {7 \choose 2}$ ways to choose two positions.  Each position matches with a probability of $\displaystyle \frac{1}{6}$ and is different with a probability of $\displaystyle \frac{5}{6}$.

Title: Probability of sequences of dice rolls

Body:
I would really appreciate help with a calculation of probabilities.

Let's say I have a die with six sides. I roll the die 7 times and get the following sequence:
5, 1, 2, 2, 4, 5, 1

Given some random other sequence (e.g. 6, 1, 3, 2, 5, 5, 5), what is the probability that there are exactly two numbers in the same position? Exactly three numbers in the same position? And so on.

For example, with the above sequences there are three numbers in the same position: 1 in second position, 2 in fourth position and 5 in sixth position.


问:
$\lim_{x\to 0^{+}} \frac{\cos (\sinh x)- \cosh(\sin x)}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4}$
I have to solve the limit
$$\lim_{x\to 0^{+}} \frac{\cos (\sinh x)- \cosh(\sin x)}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4}$$

Applying Taylor's series.

$$\cos(\sinh x)=1- \frac{x^2}{2}+ o(x^2)$$

$$\cosh(\sin x)=1+ \frac{x^2}{2}+ o(x^2)$$

but I don't know how to treat the denominator because the derivatives are not defined in $0$.
Can someone please give me some hints?

答:
You can treat the denominator also as follows by writing

$$\frac{\cos (\sinh x)- \cosh(\sin x)}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4} = \frac{x^2}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4}\cdot \frac{\cos (\sinh x)- \cosh(\sin x)}{x^2}$$

Now, you can handle the first fraction by substituting $x=t^2$
\begin{eqnarray*}\frac{x^2}{(e^{\sqrt{x}}-5^{\sqrt{x}})^4}
& \stackrel{x=t^2}{=} & \frac{t^4}{(e^{t}-5^{t})^4} \\
& = & \left(\frac{t}{e^{t}-5^{t}}\right)^4 \\
& \stackrel{t\to 0}{\longrightarrow} & \frac 1{(1-\ln 5)^4}
\end{eqnarray*}

You can handle the second fraction by your already found Taylor series to get

$$\lim_{x\to 0+}\frac{\cos (\sinh x)- \cosh(\sin x)}{x^2}=-1$$


问:
Application of the principle of conservation of angular momentum and the principle of conservation of energy
A satellite is launched in a direction parallel to the surface of the earth with a velocity of $36900km/hr$ from an altitude of $500km$. Determine

(a) the maximum altitude reached by the satellite.

(b) the maximum allowable error in the direction of launching if the satellite is to go into orbit and come no closer than $200km$ to the surface of the earth.

I know that I have to apply the principle of conservation of angular momentum

$r_1 m v_1 \sin \phi_1 = r_2 m v_2 \sin \phi_2$

where
$r =$ distance of the satellite from the center of the earth
$m =$ mass of the satellite
$v =$ velocity of the satellite
$\phi =$ the angle the velocity makes with the radius vector

and the principle of conservation of energy

$T_1 + V_1 = T_2 + V_2$

Where $T =$ kinetic energy and $V =$ potential energy.

I know I have to use the two principles but I don't know how to use them. May someone please help me with how to use them or offer an alternative solution to the question.

答:
I figured out part (a)

The altitude of the satellite is maximum and minimum for $\phi = 90^\circ$ because the satellite is revolving around the planet

thus from the conservation of angular momentum

$r_1 v_1 = r_2 v_2$

$r_2 = \frac{r_1 v_1}{v_2}$ --- (i)

Also from the conservation of energy

$T_1 + V_1 = T_2 + V_2$  --- (ii)

Where
$T = \frac{1}{2}m v^2$ and $V = \frac{GMm}{r^2}$ where $G$ is the gravitational constant and $M$ is the mass of the earth

but $GM = g R^2$ where $R$ is the radius of the earth

So combining equation (i) and (ii)

$r_2 = \frac{v_1^2 r_1^2}{2g R^2 - v_1^2 r_1}$


问:
What is the visual intuition behind solving equations with linear algebra ? (row picture vs column picture)
I'm visiting this course on linear algebra, where the teacher propose that you can see a set of equations in row or column picture.

The first way, the usual way, is to think of it as rows:

```<code>2x-y = 0
-x+2y = 3
</code>```

You can plot a line on a graph for the first equation, which is all the possible answers for this equation. Then plot a second line for the second equation. Where lines meet is the point where the answer is valid for both equation, giving us a solution.

The second way of thinking about it is via the column picture

```<code>x [2 -1]' + y [-1 2]' = [0 3]'
</code>```

You then can draw the first vector, the second vector and find how much of them arrive at the answer vector.

However I can't wrap my head around why that works. The main thing that is bugging me is that he made a vector of the two x's and another with the two y's and when drawing those the first vector (with two x's), he draws it on a x, y plane, with x and y components. I don't see how the x suddenly is y, and how that relates to the first picture (row picture).

答: In the second case, you need to give other names to the coordinates in the plane where you're drawing, say $(u,v)$ instead of $(x,y)$. That is, you're drawing the vectors $(u,v)=(2,1)$, $(u,v)=(-1,2)$ and $(u,v)=(0,3)$ in the $uv$-plane (feel free to write them as column vectors if you like, but I was too lazy to do that here), and then you ask how many steps you need to go along the first vector ($x$ steps) and along the second vector ($y$ steps) in order to obtain the third vector.


问:
Surprisingly elementary and direct proofs
What are some examples of theorems, whose first proof was quite hard and sophisticated, perhaps using some other deep theorems of some theory, before years later surprisingly a quite elementary, direct, perhaps even short proof has been found?

A related question is MO/24913, which deals with hard theorems whose proofs were simplified by the development of more sophisticated theories. But I would like to see examples where this wasn't necessary, but rather the theory turned out to be superfluous as for the proof of the theorem. I expect that this didn't happen so often. [Ok after reading all the answers, it obviously happened all the time!]

答:
In general proofs that are very intricate and laborious (like brute force for example) give way to more elegant proofs throughout the ages. However, these proofs rely on an arsenal of concepts much more sophisticated than their predecessors. But a language is more sophisticated conceptual offset by an economy in brute force.

I think the most beautiful example that satisfies your question is the story of the proofs of Sylow's Theorem. See in this paper a elegant proof for exemple.

The Sylow theorems have been proven for the first time by the Norwegian mathematician Ludwig Sylow in 1872. See wikipedia for references.

"Classical proofs" using the Conjugacy class equation and insights in counting are proofs very laborious. A proof with Conjugacy class equation can be seen in the excelent book Abstract Algebra: An Introduction by Thomas W. Hungerford.

In outher execelent Hungerford's book we see a simple, elegant and sophisticated  proof ( see p. 93)  using the concept of group action ( see p. 88) .


问:
Example of a map from the natural numbers to the positive rational numbers less than 1
I want an example of a bijective function $f:\mathbb{N} \to \{n \in \mathbb{Q}:0<n<1\}$

There is already a question about an example of a map between $\mathbb{N}$ and $\mathbb{Q}$.

I am not really a layperson but you can approximate me as a layperson. So I cannot derive an example for this.

MOTIVATION:

a few days I was thinking about some crazy functions and roots of those function. The weierstrass factorization theorem also says that any continuous function can be written as the product involving its roots.

Yesterday I saw a video on the product formula of $\sin(x)$.

And I started wandering that can a function have roots which are all rational numbers or maybe roots which all positive rational numbers less than 1.

To get started at all I have to start here

$$\prod_{n=1}^\infty\left(x-a_nQ_n \right)$$

Where $Q_n$ is an example of a bijective function $f:\mathbb{N} \to \{n \in \mathbb{Q}:0<n<1\}$

答:
Although such functions exist, it is not easy (maybe not possible) to get a nice pretty formula for one. You are not going to get anything as simple as the function in lisyarus's comment:

$$f(n) = \frac{1}{n + 2}$$

A common way to demonstrate the bijection is to enumerate the rationals in a systematic way e.g.

$$ \frac{1}{2},  \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{2}{4}, \frac{3}{2}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, . . .$$

So, I list all of the $\frac{n}{2}$ rationals (all one of them) then all of the  $\frac{n}{3}$, then  $\frac{n}{4}$, etc. I map $1$ to the first in my list, $2$ to the second,etc. With some effort, you might be able to develop a formula for this.  You could easily calculate what $1000000$ mapped to without calculating all of the previous terms.  It could be coded simply and efficiently in most programming languages.

However, there is a problem.  The function is clearly surjective but it is not injective as, even in my short sample above, $1$ maps to $\frac{1}{2}$ and $5$ maps to $\frac{2}{4}$ which is the same.  So, we need to eliminate these duplicates.  This is clearly possible but it is much harder to jump to $1000000$ without calculating all previous terms (it is beyond my skills in number theory).

A key point here is that to a modern mathematician, a function does not necessarily have a nice pretty formula. In some cases, we might even prove that a function exists but have no way to calculate it.

It is quite common that it is easy to find an injection from a set $A$ to another set $B$ so that intuitively $A$ is smaller than or the same size as $B$ but also you can find an injection the other way.  It could be that neither function is surjective so intuitively $A$ is strictly smaller than $B$ and $B$ is strictly smaller than $A$.  You need to learn to accept this with infinite sets.  In fact, it is characteristic of them.  The map: $n \rightarrow n + 1$ naively proves that $\mathbb{N}$ is smaller than itself.  Now, back to the case of an injection in both directions.  There is a very useful theorem: Schröder–Bernstein theorem which proves that, in this case, a bijection exists. So, once you have found an injection in both directions you may invoke this theorem and stop without calculating an explicit bijection.

A similar result applies to the case of a surjection in both directions but this requires the Axiom of Choice and hence is not so basic.


问:
What happens to the argument of a trigonometric function when multiplied by a constant?
so I was training my derivation skills when I stumbled upon deriving trigonometric functions,, and during one step of the chain rule I wit the following:

$$ -35\cos(-15x) \cdot (-15) = 525\cos(15x)$$

and at this point I am confused, why does the sign of the argument of $\cos$ change all of a sudden? :( why would it affect its argument too?

thank you very much in advance for the help...

答: Since $\cos y$ is an even function, $\cos(-15x)=\cos(15x)$ is an optional sign change for the argument.


问:
Geometry of Set of States of Random Process
Consider the following random process in the unit circle. Starting at $(0,0)$, move up, down, left, right a distance $1/2$ with equal probability.

At the next step there are three possibilities: stay where you are with probability $1/2$, go a distance $1/4$ in a direction orthogonal to the previous step with probability $1/4$ (*).

Keep iterating, at step $k$ stay where you are with probability $1/2$, go a distance $1/2^k$ in a direction orthogonal to the previous step with probability $1/4$.

Can we describe the set of points this process can visit?

It seems to have a fractal nature, and I think the closest it gets to the circle is $(2/3,1/3)$.

(*) In context this probability might not be $1/4$ and there are some absorbing states. The idea is to encode products of non commuting orthogonal projections $p, 1-p, q, 1-q$ as points in the unit circle. $p$ would be up, $1-p$ down and similar for $q$ and $1-q$.

答:
Here is a FragM script that renders the fractal (and unit circle) using distance estimation:

```<code>#version 330 core

#include "MathUtils.frag"
#include "Progressive2D.frag"

uniform int Iterations; slider[0,10,100]

vec3 color(vec2 q)
{
  q *= 1.05;
  float dq = length(vec4(dFdx(q), dFdy(q)));
  vec2 p = q;
  p *= 2.0;
  float dp = length(vec4(dFdx(p), dFdy(p)));
  p = abs(p);
  if (p.x > p.y)
  {
    p.xy = p.yx;
  }
  p.y -= 1.0; // first step has no stay-put option
  vec2 dir = vec2(1.0, 0.0);
  float dist = 0.5;
  for (int i = 0; i < Iterations; ++i)
  {
    float d = dot(p, dir) / dist;
    if (d < -2.0/3.0)
    {
      p += dist * dir;
      dir.xy = dir.yx;
    }
    if (d > 2.0/3.0)
    {
      p -= dist * dir;
      dir.xy = dir.yx;
    }
    dist *= 0.5;
  }
  return vec3(tanh(clamp(min(length(p) / dp, abs(length(q) - 1.0) / dq), 0.0, 4.0)));
}
</code>```

Here is what it looks like, within the unit circle:

There is some ambiguity about "orthogonal to previous step", perhaps, I chose to interpret it as "orthogonal to the last step that actually moved some distance".

Body:
Here is a FragM script that renders the fractal (and unit circle) using distance estimation:

#version 330 core

#include "MathUtils.frag"
#include "Progressive2D.frag"

uniform int Iterations; slider[0,10,100]

vec3 color(vec2 q)
{
  q *= 1.05;
  float dq = length(vec4(dFdx(q), dFdy(q)));
  vec2 p = q;
  p *= 2.0;
  float dp = length(vec4(dFdx(p), dFdy(p)));
  p = abs(p);
  if (p.x > p.y)
  {
    p.xy = p.yx;
  }
  p.y -= 1.0; // first step has no stay-put option
  vec2 dir = vec2(1.0, 0.0);
  float dist = 0.5;
  for (int i = 0; i < Iterations; ++i)
  {
    float d = dot(p, dir) / dist;
    if (d < -2.0/3.0)
    {
      p += dist * dir;
      dir.xy = dir.yx;
    }
    if (d > 2.0/3.0)
    {
      p -= dist * dir;
      dir.xy = dir.yx;
    }
    dist *= 0.5;
  }
  return vec3(tanh(clamp(min(length(p) / dp, abs(length(q) - 1.0) / dq), 0.0, 4.0)));
}

Here is what it looks like, within the unit circle:

There is some ambiguity about "orthogonal to previous step", perhaps, I chose to interpret it as "orthogonal to the last step that actually moved some distance".


答:
The $525$ simply comes from $(-35)(-15)$ (two negatives multiply to give a positive).

The argument changes sign because $\cos (-x) = \cos x$. You can justify this in a number of ways, the easiest being to look at the signs of the trigonometric ratios in the four quadrants. Cosine is positive in the first and fourth quadrant, and the fourth quadrant can be thought of as angles in the range $[-\frac{\pi}2,0)$ (the "mirror" of positive acute angles). Another way is to look at the graph of the cosine function and observe that it's an even function. Still another way is to study the behaviour of the Maclaurin series for cosine. And so on.

Title: What happens to the argument of a trigonometric function when multiplied by a constant?

Body:
so I was training my derivation skills when I stumbled upon deriving trigonometric functions,, and during one step of the chain rule I wit the following:

$$ -35\cos(-15x) \cdot (-15) = 525\cos(15x)$$

and at this point I am confused, why does the sign of the argument of $\cos$ change all of a sudden? :( why would it affect its argument too?

thank you very much in advance for the help...


问:
How to solve an integral over a distribution function
To find the reservation wage $w^*$ in a search labour model it is possible to derive the following expression

$ w^* = b + \frac{\lambda}{\rho + \eta} \int_{w^*} (w-w^*)dG(w), \hspace{15mm} (1)$

where $b=-1 , \lambda=0.2, \rho=0.005 \hspace{1mm}$  and  $\hspace{1mm} \eta=0.02$. Further, $G(w)\sim LN(1,1)$, that is a log-normal distribution with $\mu=1$ and $\sigma=1$. The pdf of such distribution is $\frac{1}{x \sigma \sqrt{2 \pi}} exp\Big(-\frac{(ln(x)-\mu)^2}{2 \sigma^2}\Big)$.

In this example the solution is (apparently) $w^* = 7.439$ with a graphical representation. It should not be possible to find an analytical solution. I have two main questions

(A) $\textbf{What does it mean to integrate over a distribution?}$

I do not need a full theoretical explanation, but more of a "how to handle this". I have found some answers that suggest inserting the pdf of the distribution in the place of x (which in this example is w) and integrate over x. I.e.

$ w^* = b + \frac{\lambda}{\rho + \eta} \int_{w^*} (\frac{1}{w \sigma \sqrt{2 \pi}} exp\Big(-\frac{(ln(w)-\mu)^2}{2 \sigma^2}\Big)-w^*)dw,$

However I cannot confirm this.

(B) $\textbf{How to handle the semi-definite integral?}$

Combined it is generally a question of

(=) $\textbf{How to solve for $w^*$ in equation (1)?}$.

Any comments or helpful suggestions is very much appreciated! It is my first time posting here, so bear that in mind.

Best regards

答:
As I wrote in a comment, I interpreted the integral as $$\displaystyle\int^\infty_{w^*} (w-w^*)dG(w)=\int^\infty_{w^*} (w-w^*)G'(w)\,dw=\int^\infty_{w^*} w\,G'(w)\,dw-w^*\,\int^\infty_{w^*} G'(w)\,dw,$$ where
$$G'(w)=\frac{1}{w\,\sigma \sqrt{2 \pi}} \exp\left(-\frac{(\ln w-\mu)^2}{2 \sigma^2}\right).$$ The rightmost integral can be written in terms of the complementary cdf of standard normal distribution (https://en.wikipedia.org/wiki/Q-function) as $\displaystyle Q\left(\frac{\ln w^*-\mu}{\sigma}\right)$, the other one can be rewritten, using $w=\exp(\ln w)$ and
$$\ln w-\frac{(\ln w-\mu)^2}{2\sigma^2}=\mu+\frac{\sigma^2}2-\frac{(\ln w-\mu-\sigma^2)^2}{2\sigma^2},$$ as $\displaystyle \exp\left(\mu+\frac{\sigma^2}2\right)\,Q\left(\frac{\ln w^*-\mu-\sigma^2}{\sigma}\right)$. Then, the equation becomes
$$w^*=-b+c\,\exp\left(\mu+\frac{\sigma^2}2\right)\,Q\left(\frac{\ln w^*-\mu-\sigma^2}{\sigma}\right)-c\,w^*\,Q\left(\frac{\ln w^*-\mu}{\sigma}\right),$$ where $\displaystyle c=\frac{\lambda}{\rho + \eta}$.

The equation can be solved only numerically. With your numerical values, $c=8$, and in Pari/GP, I'd write

```<code>solve(x=1,10,-x-1+4*exp(1.5)*erfc((log(x)-2)/sqrt(2))-4*x*erfc((log(x)-1)/sqrt(2)))
</code>```

(since $\displaystyle Q(x)=\frac12\,\operatorname{Erfc}\left(\frac{x}{\sqrt{2}}\right)$) with result $7.4594961135988564221204777767475134525$.

Body:
As I wrote in a comment, I interpreted the integral as $$\displaystyle\int^\infty_{w^*} (w-w^*)dG(w)=\int^\infty_{w^*} (w-w^*)G'(w)\,dw=\int^\infty_{w^*} w\,G'(w)\,dw-w^*\,\int^\infty_{w^*} G'(w)\,dw,$$ where
$$G'(w)=\frac{1}{w\,\sigma \sqrt{2 \pi}} \exp\left(-\frac{(\ln w-\mu)^2}{2 \sigma^2}\right).$$ The rightmost integral can be written in terms of the complementary cdf of standard normal distribution (https://en.wikipedia.org/wiki/Q-function) as $\displaystyle Q\left(\frac{\ln w^*-\mu}{\sigma}\right)$, the other one can be rewritten, using $w=\exp(\ln w)$ and
$$\ln w-\frac{(\ln w-\mu)^2}{2\sigma^2}=\mu+\frac{\sigma^2}2-\frac{(\ln w-\mu-\sigma^2)^2}{2\sigma^2},$$ as $\displaystyle \exp\left(\mu+\frac{\sigma^2}2\right)\,Q\left(\frac{\ln w^*-\mu-\sigma^2}{\sigma}\right)$. Then, the equation becomes
$$w^*=-b+c\,\exp\left(\mu+\frac{\sigma^2}2\right)\,Q\left(\frac{\ln w^*-\mu-\sigma^2}{\sigma}\right)-c\,w^*\,Q\left(\frac{\ln w^*-\mu}{\sigma}\right),$$ where $\displaystyle c=\frac{\lambda}{\rho + \eta}$.

The equation can be solved only numerically. With your numerical values, $c=8$, and in Pari/GP, I'd write

solve(x=1,10,-x-1+4*exp(1.5)*erfc((log(x)-2)/sqrt(2))-4*x*erfc((log(x)-1)/sqrt(2)))

(since $\displaystyle Q(x)=\frac12\,\operatorname{Erfc}\left(\frac{x}{\sqrt{2}}\right)$) with result $7.4594961135988564221204777767475134525$.


问:
Square Root and Winding Number
Let $f\in C(S^{1},S^{1})$ such that $w(f)=0$ where $w$ denotes winding number. Can we conclude that there exists a function $g\in C(S^{1},S^{1})$ such that $f=g^{2}$?</p>

答: Yes. Parameterize the domain by $\varphi:[0,1]\to S^1$, $t\mapsto e^{2\pi it}$, and consider $\hat{f}\circ \varphi$ a lift of $f\circ \varphi$ to the universal covering space such that $\hat{f}\circ \varphi(0)=0$. Since the function has winding number zero, $\hat{f}\circ\varphi(0)=\hat{f}\circ\varphi(1)$. So we may take $\varphi\circ \frac{1}{2}\hat{f}\circ\varphi$ as $g$ such that $g^2=f$.</p>

Title: Square Root and Winding Number


问:
Why is proving the Riemann Hypothesis so hard?
The Riemann Hypothesis is considered by many to be the most important unsolved problem in pure mathematics.

Several attempts have been made in the last 150 years (here some of them are reported).

RH is the only problem that has been listed in both Hilbert's 23 Problems and the Millennium Problems by the Clay Institute; and yet it remains unsolved, seemingly resisting all attacks, and quickly becoming a piece of mathematical folklore as an "impossible" problem.

For instance, it is reported that Hilbert himself declared:

If I were to awaken after having slept for a thousand years, my first question would be:

"Has the Riemann hypothesis been proven?"

Implications and applications of this possible result have already been addressed many times on this website (reporting here some of them):

What does proving the Riemann Hypothesis accomplish?

What is so interesting about the zeroes of the Riemann $\zeta$ function?

Why do mathematicians care so much about zeta functions?

What is the link between Primes and zeroes of Riemann zeta function?

What I'm asking is: why is a conjecture on the zeroes of a specific complex function so hard to prove (or disprove)?

What are the main obstacles and obstructions to this problem's solution?

Edit: although the question has already been addressed before on Math.SE (here and, in some way, here), no answers have admittedly been given, and I personally find that the comments this question received in the last hours (for which I am grateful) addressed the problem much more clearly than the comments in the questions above.

I believe there's room for improvement, but I apologize if the question is too broad or violates the guidelines in some other way.

答:
This is a bit an opinion based question and answer.

The RH is about $\log\zeta(s),\frac1{\zeta(s)},\frac{\zeta'(s)}{\zeta(s)}$, not $\zeta(s)$.

On the $\zeta(s)$ side we can easily exploit that it is the Dirichlet series of the integers.

Surprisingly (or not?) on the $\log\zeta(s),\frac1{\zeta(s)},\frac{\zeta'(s)}{\zeta(s)}$ side we can't, and complicated structures appear, eg. the primes.

The same kind of structures appear for many other Dirichlet series (the Dirichlet L-functions, more generally the Selberg class) and the RH is (more or less) assumed to hold for all of them.

This set of Dirichlet series with a RH is discrete/isolated: you can't change slightly the coefficients without loosing one of the key properties (analytic continuation, functional equation, Euler product, growth of the coefficients).

So we need a setting where all those key properties are present: an arithmetical-analytical-algebraic setting. It is hard, by definition.

In practice most elementary approaches to the RH fail because they apply the same way to $(G(\chi_5)/5^{1/2})^{-1/2}L(s,\chi_5)+\overline{(G(\chi_5)/5^{1/2})}^{-1/2}L(s,\overline{\chi_5})$ lacking only the Euler product, and having a bunch of zeros off the critical line.


问:
How many numbers to describe a subspace of a vector space?
What is the minimum number of numbers to describe a generic subspace of dimension $n$ of a vector space of dimension $d$?

Here are some attempts.
For example, consider $d=4$.

To describe a point, I need zero numbers.

To describe a one-dimensional subspace, I need at most $4-1 = 3$ numbers (e.g. a unit vector
living on the line describing the one-dimensional subspace). The constraint of unit normalization reduces the number of numbers required by one.

To describe a two-dimensional subspace, I need at most $(4-1) + (4-1) - 1 =5$ numbers. The two $(4-1)$ refer to two orthogonal unit vectors forming a basis for the two-dimensional space, and the last $-1$ is because the unit vectors are orthogonal. I'm unsure if this is the minimum number of numbers needed to describe a two-dimensional subspace of a four-dimensional space.

To describe a three-dimensional subspace, I believe I only need to describe a unit vector perpendicular to that subspace. This leaves us with $4-1 = 3$, the same as the one-dimensional case.

To describe the whole space, I need zero numbers.

More generally, building off the idea of creating an orthogonal basis for the subspace of interest (or building an orthogonal basis for the subspace orthogonal to the subspace of interest), I expect that the result will be $$f(n,d) = dn - \frac{1}{2}n(n+1) \text{ for } n<\frac{d}{2}$$ and $f(d-n, d)$ otherwise. However, I'm left wondering if I can make do with fewer numbers.

答:
If the $n$-dimensional subspaces of a $d$-dimensional vector space $V$ would form a vector space, then what you are looking for would be its dimension.

Even though that is not the case, if $V$ is a real vector space the space $n$-dimensional subspaces does carry the structure of a smooth manifold, which is called the Grassmannian $\operatorname{Gr}(n, V)$.

Its dimension is given by $n(d-n)$.

For example, $\operatorname{Gr}(2,\mathbb{R}^4)$ has dimension $2(4-2) = 4$ (in your description with $5$ numbers, you have one "superfluous" degree of freedom which is the choice of the orthogonal basis).

$n$
dimension of $\operatorname{Gr}(n, \mathbb{R}^4)$

$0$
$0(4-0) = 0$

$1$
$1(4-1) = 3$

$2$
$2(4-2) = 4$

$3$
$3(4-3) = 3$

$4$
$4(4-4) = 0$

See also this related question for some geometric explanation:

What is the dimension of this Grassmannian?


问:
Prime solutions of an equation
I am trying to find out if there are any prime solutions to the equation $24p + 1 = q^2$ except $(2,7)$, $(5,11)$ and $(7,13)$. I have no idea how to approach this. I've run some brute force calculations and nothing turns up, but of course that's not a proof.

答:
We have $24p+1=q^2$. This gives us $$24p=q^2-1\space  \text{or} \space 24p=(q-1)(q+1)$$
Observe that this means $p|(q-1)(q+1)$. Since the difference between $q-1$ and $q+1$ is $2$,either $p|(q-1)$ or $p|(q+1)$ (assuming $p>2$)

This gives us two cases:

Case I: $p|(q-1)$

This gives us $\exists k\in \mathbb Z:kp=q-1$, thus $q+1=kp+2$

So We have $$24p=kp(kp+2)\space \text{or}\space 24=k(kp+2)$$
We can now simply check all the factors of $24$.

this gives us $(5,11)$ as the only solution for this equation in form of $(p,q)$ at $k=2$

Case II: $p|(q+1)$

This gives us $\exists k\in \mathbb Z:kp=q+1$, thus $q-1=kp-2$

So We have $$24p=kp(kp-2)\space \text{or}\space 24=k(kp-2)$$
We can now simply check all the factors of $24$.

this gives us $(7,13)$ as the only solution for this equation in form of $(p,q)$ at $k=2$

Adding in the case when $p=2$, thus giving another solution $(2,7)$.

$\therefore (2,7),\space (5,11),\space (7,13)$ are the only solutions for the given equation.

Title: Prime solutions of an equation


答:
$(q-1)(q+1) = q^2 - 1 = 24p.$

Factors of $(24)$ are the set $S = \{1,2,3,4,6,8,12,24\}.$

Assume that $p$ is relatively prime to $(24)$.

Since $p$ is a prime, $p > 2 \implies$ at least one of $(q-1), (q+1)$ is relatively prime to $p$ which implies that at least one of $(q-1)(q+1)$ must be in $S$.

Therefore, therefore $q$ can not be greater than $25$.

Alternatively, if $p$ is not relatively prime to $(24)$, then either $p=2$ or $p=3$.  These cases can be checked manually.

Title: Prime solutions of an equation

Body: I am trying to find out if there are any prime solutions to the equation $24p + 1 = q^2$ except $(2,7)$, $(5,11)$ and $(7,13)$. I have no idea how to approach this. I've run some brute force calculations and nothing turns up, but of course that's not a proof.


问:
Generalizing the apothem characterization of area for regular polygons, to convex regular polyhedra (and more?).
Recall: Given a convex, regular $n$-gon, of side length $s$, and apothem $a$, we may calculate its area $A$ as

$$A = \frac 1 2a n s$$

or, if we let $P$ be the perimeter (i.e. $P=ns$), then we have the formula we (or at least I) was typically given in school:

$$A = \frac 12 aP$$

To the uninitiated, the apothem is the line segment which is the radius of the inscribed circle of the regular polygon. It goes from the center of the polygon (in terms of this circle) to the midpoint of any side.

Question/Goal: Is there a way to generalize this formula to three dimensions, somehow relating the volume and surface area, to the length of some segment (from the center of some sphere to the middle of a face)? Up to some constant, of course. And presumably as a result, we can only consider convex, regular polyhedra and polytopes.

So, what I seek is some formula like

$$V = kaS$$

where $S$ denotes the surface area, and $k$ is some constant.

How would we prove this formula?

What about four dimensions and higher, if possible? Can it be done there as well?

Attempts & Research: (three dimensions)

So my first guess was to check the regular three-dimensional solids -- namely the convex ones, the Platonic solids.

Looking into things a bit more, the term I want (in 3D at least) to generalize the apothem is the "radius of the insphere" or "inradius" -- the insphere is a sphere whose faces are tangent to every face of the polyhedron. You can thus think of a 2D insphere giving you the length of the apothem for polygons (if we take the usual convention that a $2$-sphere is a circle).

Anyhow. Suppose we have a cube of edge length $s$. There should be no difficulty in seeing that the radius of the insphere is $s/2$, the volume is $s^3$, and the surface area is $6s^2$. Thus, we see that

$$V = kaS \implies s^3 = k \frac s 2 6s^2 \implies k = \frac 1 3$$

This hints at a possible formula for 3D, $V = aS/3$, but it needs to be checked.

Luckily, the platonic solids page on Wikipedia has a list of inradii among other things. (We'll still call it $a$ to match our convention.) For the tetrahedron, we have $S = s^2 \sqrt 3$, $a = s/2\sqrt 6$, and $V = s^3/6\sqrt 2$. Then we try our formula:

$$kaS = \frac{1}{3} \cdot \frac{s}{2\sqrt 6} \cdot s^2 \sqrt 3 = \frac{s^3}{6 \sqrt 2} =V$$

So we might be on the right track. We can check the others too for the necessary constant $k$. We use the Wikipedia notations of

$$\varphi := \frac{1+\sqrt{5}}{2} = 2 \cos \frac \pi 5 \qquad \xi := 2 \sin \frac \pi 5 = \sqrt{ \frac{5 - \sqrt 5}{2}}$$

as necessary.

Octahedron: $\displaystyle k = \frac{V}{aS} = \frac{s^3 \sqrt{2}}{3} \cdot \frac{\sqrt 6}{s} \cdot \frac{1}{2\sqrt 3 s^2} = \frac{\sqrt{12}}{6 \sqrt{3}} = \frac 1 3$

Dodecahedron: $\displaystyle k = \frac{V}{aS} =  \frac{5}{2} \frac{\varphi^3}{\xi^2} s^3 \cdot \frac{2 \xi}{\varphi^2 s} \cdot \frac{\sqrt{3-\varphi}}{15 \varphi s^2} = \frac{\sqrt{3-\varphi}}{3 \xi} = \frac 1 3$ (final equality per WA)

Icosahedron: $\displaystyle k = \frac{V}{aS} = \frac{5 \varphi^2 s^3}{6} \cdot \frac{2 \sqrt 3}{\varphi^2 s} \cdot \frac{1}{5 \sqrt 3 s^2} = \frac 1 3$

Thus, in each case of the platonic solids, the formula

$$V = \frac 1 3 a S$$

holds up.

Attempts & Research: (four dimensions)

So far, we've seen two formulas:

$$A = \frac 1 2 a P \qquad V = \frac 1 3 a S$$

I conjecture, then: the $n$-dimensional volume $V_n$ is related to the $n$-dimensional surface area $S_n$ and the polytope's inradius $a$, by the formula

$$V_n = \frac 1 n a S_n$$

So it's natural to look at the four dimensional case as well. Indeed, we see that it works, as follows, going through the list of regular polytopes on Wikipedia:

$5$-cell (post about volume): $\displaystyle V_4 = \frac{1}{4} \cdot 1 \cdot V_3$

$8$-cell (tesseract, some info from here): $\displaystyle V_4 = \frac 1 4 \cdot \frac{1}{2} s \cdot 8s^4 = s^4$, as desired

$16$-cell (some info from here): $\displaystyle V_4 = \frac1 4 \cdot \frac{\sqrt{2}}{4} s \cdot 16 \frac{\sqrt{2}}{12} s^3 = \frac{1}{6} s^4$, as desired

$24$-cell (some info from here): $\displaystyle V_4 = \frac 1 4 \cdot \frac{1}{\sqrt 2} s \cdot 24 \frac{\sqrt 2}{3} s^3 = 2s^4$, as desired

$120$-cell (some info from here): $\displaystyle V_4 = \frac 1 4 \cdot \frac{7 + 3 \sqrt 5}{4} s \cdot 120 \frac{15 + 7 \sqrt 5}{4} s^3 = \frac{15}{4} (105 + 47 \sqrt 5)s^4$, as desired (WA verified)

$600$-cell (some info from here): $\displaystyle V_4 = \frac 1 4 \cdot \sqrt{ \frac{9 + 4 \sqrt 5}{8} } s \cdot 600 \frac{\sqrt 2}{12} s^3 = \frac{25}{4} (2 + \sqrt 5) s^4$ (WA verified)

Thus the conjecture holds for the regular four-dimensional convex polytopes.

Question (albeit reiterated somewhat):

The question remains, though: does this hold true in all dimensions? That is, for a convex, regular, $n$-dimensional polytope, with

$n$-dimensional hypervolume $V_n$

inradius $a$

$n$-dimensional hyper-surface-area $V_{n-1}$ (or "surface volume" I guess? Not sure what the term is)

do we have

$$V_n = \frac 1 n a V_{n-1}?$$

If so, how do we prove this? Moreover, what (if any) is the underlying reason as to why this holds? (As in, how should one be able to "expect" this answer, from a motivational or intuitive aspect, if that makes any sense.)

答:
This relation between volume and surface area holds true for all circumscribed polytopes, that is, polytopes for which there exists a sphere (the in-sphere) that touches every facet (e.g. the regular polytopes).
If the polytope $P\subset\Bbb R^n$ is $n$-dimensional, its in-sphere has radius $r$, the volume is $V_n$ and the surface area is $V_{n-1}$, then indeed

$$V_n = \frac1n r V_{n-1}.$$

Proof.

Suppose that $P$ has facets $f_1,...,f_m$ and $f_i$ has $(n-1)$-dimensional volume $v_i^{n-1}$. Let $p_i$ be the pyramid with base face $f_i$ and apex at the center of the in-sphere. Since $P$ is circumscribed, each pyramid has then height $r$. The $n$-dimensional volume of this pyramid is known to be

$$v_i^n = \frac 1n r v_i^{n-1}.$$

The pyramids define a decomposition of the volume of $P$, as well as of its surface, and thus the volume $V_n$ is the sum of the $v_i^n$, and the surface area $V_{n-1}$ is the sum of the $v_i^{n-1}$.
The desired relation follows.


问:
$y=e^ {-x}\sqrt{x+2}$. what is the volume of solid formed when shaded region is rotated completely about the x-axis?
$e^ {-x}\sqrt{x+2}$. what is the volume of solid formed when shaded region is rotated completely about the x-axis?
I am doing the chapter further calculus in the sub chapter integration by parts. I know how to do regular questions but not this one  I have never encountered a question like this where the solid shaded region is completely rotated about the x-axis- Any help is much appreciated . Thank you.

答:
The volume of rotation about the $x$ axis is given by the following formula: $$\pi\int_{a}^{b}\left(f\left(x\right)\right)^2\,\mathrm{d}x\text{,}$$ where, in this case, $f\left(x\right)=e^{-x}\sqrt{x+2}$, $b=0$ and $a=-2$ (the $x$-intercept of $f\left(x\right)$). $a$ is found by setting $y$ (in the original equation) equal to $0$: $$0=e^{-x}\sqrt{x+2}$$ Either $e^{-x}=0$ or $\sqrt{x+2}=0$. The former will only $\to{0}$ if $x\to\infty$. However, the latter will equal $0$ when $x=-2$ (since $\sqrt{-2+2}=0$). This means the lower bound $a$ is equal to $-2$.

Substituting these values gives the equation: $$\pi\int_{-2}^{0}\left(e^{-x}\sqrt{x+2}\right)^2\,\mathrm{d}x=\pi\int_{-2}^{0}e^{-2x}\left(x+2\right)\,\mathrm{d}x$$ Then, using the integration by parts formula: $$\int{uv^\prime}=uv-\int{vu^\prime}$$ and substituting $u=x+2$ and $v^\prime=e^{-2x}$, the integral $\int_{-2}^{0}e^{-2x}\left(x+2\right)\,\mathrm{d}x$ can be solved: $$u=x+2\implies{u^\prime}=1$$ $$v^\prime=e^{-2x}\implies{v}=-\frac{e^{-2x}}{2}$$ $$\therefore\int{e^{-2x}}\left(x+2\right)\,\mathrm{d}x=-\frac{\left(x+2\right)e^{-2x}}{2}-\int-\frac{e^{-2x}}{2}\,\mathrm{d}x$$ Solving for $\int-\frac{e^{-2x}}{2}\,\mathrm{d}x$ using $u$ substitution: $$u=-2x\implies\frac{\mathrm{d}u}{\mathrm{d}x}=-2\implies\mathrm{d}x=-\frac{\mathrm{d}u}{2}$$ $$\therefore\int-\frac{e^{-2x}}{2}\,\mathrm{d}x=\int\left(-\frac{1}{2}\times-\frac{1}{2}\right)e^u\,\mathrm{d}u=\frac{1}{4}\int{e^u}\,\mathrm{d}u=\frac{e^u}{4}=\frac{e^{-2x}}{4}$$ Substituting $\frac{e^{-2x}}{4}$ back into the equation gives: $$\int{e^{-2x}}\left(x+2\right)\,\mathrm{d}x=-\frac{\left(x+2\right)e^{-2x}}{2}-\frac{e^{-2x}}{4}=-\frac{2\left(x+2\right)e^{-2x}}{4}-\frac{e^{-2x}}{4}$$ $$=-\frac{\left(2x+4\right)e^{-2x}}{4}-\frac{e^{-2x}}{4}=-\frac{\left(2x+4\right)e^{-2x}-e^{-2x}}{4}=-\frac{\left(2x+5\right)e^{-2x}}{4}$$ Finally, multiply the integral by $\pi$: $$\pi\int_{-2}^{0}e^{-2x}\left(x+2\right)\,\mathrm{d}x=\pi\left(\left.-\frac{\left(2x+5\right)e^{-2x}}{4}\right\vert_{-2}^{0}\right)$$ Evaluating this with the upper and lower bounds and multiplying the result by $\pi$, as shown above, will yield the final answer of $38.95429594394258$, or approximately $39$.


问:
Subset sum problem is NP-complete?
If I know correctly, subset sum problem is NP-complete. Here you have an array of n integers and you are given a target sum t, you have to return the numbers from the array which can sum up to the target (if possible).

But can't this problem be solved in polynomial time by dynamic programming method where we construct a table n X t and take cases like say last number is surely included in output and then the target becomes t- a[n]. In other case, the last number is not included, then the target remains same t but array becomes of size n-1. Hence this way we keep reducing size of problem.

If this approach is correct, isn't the complexity of this n * t, which is polynomial? and so if this belongs to P and also NP-complete (from what I hear) then P=NP.

Surely, I am missing something here. Where is the loophole in this reasoning?

Thanks,

答:
If you express the inputs in unary you get a different running time than if you express them in a higher base (binary, most commonly).

So the question is, for subset sum, what base is appropriate?  In computer science we normally default to the following:

If the input is a list or collection, we express its size as the number of items

If the input is an integer, we express its size as the number of bits (binary digits)

The intuition here is that we want to take the more "compact" representation.

So for subset sum, we have a list of size $n$ and a target integer of value $t$.  Therefore it's common to express the input size as $n$ and $t=2^k$ where $k$ is the number of bits needed to express $t$.  So the running time is $O(n 2^k)$ which is exponential in $k$.

But one could also say that $t$ is given in unary.  Now the size of $t$ is $t$, and the running time is $O(n t)$, which is polynomial in $n$ and $t$.

In reductions involving subset sum (and other related problems like partition, 3-partition, etc) we must use a non-unary representation if we want to use it as an NP-Hard problem to reduce from.

Title: Subset sum problem is NP-complete?


问:
Calculating the intersection points of two circles, of which the centers are located on the x axis.
I'm trying to solve the problem mentioned in the title, and the answer detailed here provides exactly what I was looking for: https://stackoverflow.com/a/3349134/4777480...except for the fact that the author skips several steps, leaving me confused as to how he ended up with the results he obtained!

For starters, there's this line:

a = (r02 - r12 + d2 ) / (2 d)

Thankfully, a reply to the answer does clarify this one and I did verify it using the steps they detailed.

But leaves me with the last two steps...

How did we end up with this result?

P2 = P0 + a ( P1 - P0 ) / d

And how did that give us the intersection coordinates? (I think in the first line it should be 'a' instead of 'h' but I'm not sure, can someone clarify?)

x3 = x2 +- h ( y1 - y0 ) / d

y3 = y2 -+ h ( x1 - x0 ) / d

A step-by-step explanation would be greatly appreciated, I want to understand the full process so I can implement it correctly in my code.

Body:
Let the centers be $(x_0,0)$ and $(x_1,0)$, and the radii $r_0,r_1$.

We need the solutions of the system

$$\begin{cases}(x-x_0)^2+y^2=r_0^2,\\(x-x_1)^2+y^2=r_1^2.\end{cases}$$

By subtraction,

$$(2x-x_1-x_0)(x_0-x_1)=r_1^2-r_0^2$$

and

$$2x=x_1+x_0-\frac{r_1^2-r_0^2}{x_1-x_0}.$$ $y$ follows.


问:
Determine all positive integers $r$ such that there exist at least two pairs of positive integers $(m, n)$ satisfying the equation $2^m = n! + r$
Determine all positive integers $r$ such that there exist at least two pairs of positive integers $(m, n)$ satisfying the equation $$2^m = n! + r.$$

I have found that for $r=2$, we have $m=2,n=2$ and $m=3,n=3$ and for $r=8$ we have $m=5,n=4$ and $m=7,n=5$

Further, $r$ can not be odd. How we can prove in general. Please give some idea.

答:
If we have two distinct solutions , we have $$2^a-b!=2^c-d!=r$$

hence $$2^a-2^c=b!-d!$$ hence $$2^c(2^{a-c}-1)=d!\cdot (\frac{b!}{d!}-1)$$

WLOG $a>c$ and $b>d$

Because of $2^c>d!$ , $\frac{b!}{d!}-1$ must be even , hence $\frac{b!}{d!}$ must be odd.
It is easy to show that this is only possible , if $b=d+1$ and $b$ is odd. Hence the right side
reduced to $(d+1)!-d!=d\cdot d!$ , so we get $$2^c(2^{a-c}-1)=d\cdot d!$$ with even $d$

Because of $2^c>d!$ , $2^{a-c}-1$ must be smaller than $d$ , but contain every odd prime smaller than $d$ as a prime factor. This shows $d\le 4$ (if $d\ge 5$ even , the largest prime smaller than $d$ is at least $\frac{d}{2}$ because of Bertrands postulate , hence $2^{a-c}-1$ must be at least $3\cdot \frac{d}{2}>d$ , since the largest prime below $d$ is also at least $5$) , hence $d=2$ or $d=4$ giving the two posted solutions. Those are the only solutions.


问:
Use the $\epsilon$-$\delta$ definition of a limit to prove $\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$
$\lim_{x\rightarrow 3} \frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6}=-1$

You should start by writing $\frac{x+6}{x^{4}-4x^{3}+x^{2}+x+6} + 1$ in the way $\left ( x-3 \right )g\left ( x \right )$

(a) Determine $g(x)$

(b) Could we choose $\delta = \min\lbrace 1, \frac{\varepsilon }{n} \rbrace$ for some $n$? Why?

(c) If we choose $\delta = \min\lbrace \frac{1}{4}, \frac{\varepsilon }{m} \rbrace$ , what is the smallest integer $m$ that we could use?

I managed to answer the part (a): By making the sum of the fraction with +1, and factoring the numerator, I get the following:

$\frac{\left ( x-3 \right )\left ( x^{3} - x^{2} - 2x - 4 \right )}{x^{4} - 4x^{3} + x^{2} + x + 6}$

where

$g\left ( x \right ) = \frac{\left ( x^{3} - x^{2} - 2x - 4 \right )}{x^{4} - 4x^{3} + x^{2} + x + 6}$

But my problem is when developing the part (b).

In the part (b) I am using the formal limit definition, but I am not managing to reach a result that allows me to find the answer.

I need help to solve this exercise, please

答:
This is a bit complicated question. You need to find a suitable bound for the rational function $g(x) $ for values of $x$ in a certain neighborhood of $3$. In formal terms we need two positive numbers $h, K$ such that $|g(x) |<K$ whenever $|x-3|<h$.

This clearly requires an analysis of numerator and denominator of $g$. Note that the function can not be bounded if the denominator of $g$ vanishes. Since it does not vanish at $3$ we have a guarantee by continuity of polynomials that the denominator will not vanish in some neighborhood of $3$.

Finding such a neighborhood (ie finding $h$ described above) is possible using the limit of denominator $q(x) =x^4-4x^3+x^2+x+6$ as $x\to 3$. But before we do that let's observe that $q(3)<0,q(4)>0$ and hence $q$ vanishes somewhere between $3,4$ and thus $h$ must be less than $1$.

Thus for part b) the answer is no. You can't expect a $\delta$ like $\min(1,\dots)$ to work. You should observe that $h$ described above forms the first argument of $\min$ in expression for $\delta$.

Part c) is about showing that we can take $h=1/4$. Let us now prove that $q(x)\to - 9$ as $x\to 3$. Clearly $$|q(x) +9|=|x-3||x^3-x^2-2x+5|$$ and $$|x^3-x^2-2x+5|=|x(x-2)(x+1)+5|$$ If $|x-3|<1/2$ then $5/2<x<7/2$ and hence $$|x(x-2)(x+1)+5|<(7/2)(3/2)(9/2)+5<32$$ and thus for any $\epsilon>0$ we have $$|q(x) +9|<\epsilon$$ if $|x-3|<\min(1/2,\epsilon /32)$. Using $\epsilon =8$ we can see that $$|q(x) +9|<8$$ if $|x-3|<1/4$. Thus $|q(x) |>1$ if $|x-3|<1/4$.

Next we deal with numerator $p(x) =x^3-x^2-2x-4$ of $g(x) $. Clearly if $|x-3|<1/4$ then $2.5<x<3.5$ and $$|p(x) |=|x(x-2)(x+1)-4|<28$$ Thus it follows that if $|x-3|<1/4$ then $$|g(x) |=|p(x) |/|q(x) |<28$$
It should now be clear that $$|(x-3)g(x)|<28|x-3|$$ and this is less than $\epsilon$ if $|x-3|<\min(1/4,\epsilon/28)$. This proves the desired limit in question.

It does not however make sense to find the smallest positive integer which can replace $28$ above. Finding such numbers becomes an algebraic problem related to solution of inequalities and is really not in the spirit of epsilon-delta definition of limit.


问:
Does $4b^3pq=2a^2+b$ has a solution for non-zero integers $a,b$ and odd primes $p<q$?
Just recently, I had an equation on my hands that I thought had no solutions:
Why $4b^3p=2a^2+qb$ has no solutions for integers $a,b$ and two odd primes $p<q$?

I was on a wrong path and now have an equation $4b^3pq=2a^2+b$ that really do not brings me any non-zero integer solution $a,b$ where $2<p<q$ are odd primes. What I see is that $b$ must be even and when I substitute $b=2t$ I have to solve $0=16t^3pq-t-a^2$ for $t$ (quadratic equation) and analyze the discriminant?

答:
The correct resulting equation is of substituting $b=2t$ is:
$$16t^3pq=a^2+t$$

We transform our equation to:
$$(16t^2pq-1)t = a^2$$

Let $\ell$ be any prime factor of $t$, and let $\ell^n$ be the highest power of $\ell$ that divides $\color{red}{a}$. Then $\ell^{2n}$ is the highest power of $\ell$ that divides $a^2$; now $\ell$ does not divide $16t^2pq - 1$ at all, so $\ell^{2n}$ must also be the highest power of $\ell$ that divides $t$.

Since $\ell$ is arbitrary, we conclude that $t$ is in fact a square, say $t = g^2$. Then our equation becomes:
$$(16g^4pq-1)g^2 = a^2$$

which gives:
$$16g^4pq-1 = \left(\frac ag\right)^2$$

which is impossible, since the squares mod $16$ are $0, 1, 4, 9$.


答:
Assume $r$ is such that there are two solutions $(m,n)$ and $(m',n')$ to the equation (all in positive integers). Wlog $n'\le n$.
If  $n=n'$, then also $m=m'$, hence  $1\le n'<n$, making $n!$ even. As already mentioned by the OP, it follows that $r$ is even.

Eliminate $r$:
$$ 2^m-2^{m'}=n!-n'!$$
As $2^m$, $2^{m'}$, $n!$ are even, so is $n'!$, i.e., we also have $n'\ge 2$ and from $n'!+r\ge 2!+2=4$, we have $m>m'\ge 2$.
Factor a little:
$$ 2^{m'}(2^{m-m'}-1) = n'!\left(\prod_{k=n'+1}^nk-1\right)$$
As $2^{m'}>n'!$, the expression in parentheses must be even, hence the product therein must be odd. This is only possible if $n=n'+1$ and $n$ is odd.
So the above simplifies to
$$\tag12^{m'}(2^{m-m'}-1) = (n-1)!\cdot (n-1)=(n-2)!\cdot (n-1)^2$$
Note that everything is determined by $n$ alone, i.e., not only $$n'=n-1,$$ but also
$$\tag2m'=v_2((n-1)!) + v_2(n-1),$$ $$r=2^{m'}-n'!,$$ and ultimately $$m=\log_2(n!+r),$$ which we hope to be an integer.
Manually checking small $n\in\{1,3,5\}$, we find the two solutions already given by the OP.
So henceforth assume $n>5$.

From $(2)$ and $2^{m'}>(n-1)!$, we find
$$n-1\ge 2^{v_2(n-1)}>\frac{(n-1)!}{2^{v_2((n-1)!)}}\ge \prod_{k=1\atop\text{odd}}^{n-1}\ge (n-2)(n-4), $$
contradiction.
We conclude that there are no further solutions.

Title: Determine all positive integers $r$ such that there exist at least two pairs of positive integers $(m, n)$ satisfying the equation $2^m = n! + r$

Body:
Determine all positive integers $r$ such that there exist at least two pairs of positive integers $(m, n)$ satisfying the equation $$2^m = n! + r.$$

I have found that for $r=2$, we have $m=2,n=2$ and $m=3,n=3$ and for $r=8$ we have $m=5,n=4$ and $m=7,n=5$

Further, $r$ can not be odd. How we can prove in general. Please give some idea.


问:
$\lVert A\lVert<1$ implies $1-A$ invertible only true in complete spaces?
It is a well known fact that

if in a Banach space $X$ a bounded linear operator $A:X\to X$ satisfies $\lVert A\lVert<1$, then $1-A$ has a bounded inverse.

I was wondering wether completeness is actually a necessary condition. I think the following argument is corect and doesn't use completeness:

Clearly for any $N\in\mathbb{N}$ $$\left(\sum_{k=0}^N A^k\right)(1-A)=1-A^{N+1}.$$
Since $\lVert A^{N+1}-0\lVert\leq\lVert A\lVert^{N+1}\rightarrow 0$ the limit of the right hand side exists and equals $1$. Hence also $$\lim_{N\rightarrow\infty}\left(\sum_{k=0}^N A^k\right)(1-A)=1$$ and therefore $(1-A)^{-1}=\lim_{N\rightarrow\infty} \sum_{k=0}^N A^k$. To see that the inverse is bounded I note that by continuity of the norm $$\lVert(1-A)^{-1}\lVert=\lim_{N\rightarrow\infty} \left\lVert \sum_{k=0}^NA^k\right\lVert\leq\sum_{k=0}^N \lVert{A}\lVert^k=\frac{1}{1-\lVert A\lVert}<\infty.$$

答: Without some completeness hypothesis, the infinite sum expressing the inverse may not converge in the given space.</p>


问:
Relationship between conditional expected values
For a random variable $X \ge 0$ and a parameter $t \ge 0$ consider $E[X | X \ge t]$ and $E[X| X < t]$.

Is there anything one could derive about the relationship between $E[X]$, $E[X | X \ge t]$ and $E[X| X < t]$?

Edit: I came up with $E[X] = p(X \ge t)* E[X| X \ge t] + p(X < t)* E[X| X < t]$  which feels painfully obvious... Is there anything else?

答:
Hint

$$\mathbb E[X\mid X\in A]:=\frac{\mathbb E[X\boldsymbol 1_{\{X\in A\}}]}{\mathbb P\{X\in A\}}.$$


问:
Finitely generated group with subexponential growth and surjection onto $\mathbb{Z}$ has finitely generated kernel.
I am trying to solve exercise 6.E.20 from the book of Clara Löh on Geometric Group Theory.

Let $G$ be a finitely
generated group with subexponential growth that admits a surjective homomorphism $\pi:G\rightarrow\mathbb{Z}$.  Show that then the kernel of $\pi$ is also finitely
generated.

I am still missing the part where you use the fact that $G$ has subexponential growth.

Take $T$ a finite generating set of $G$ and take $g\in G$ such that $\pi(g)= 1$. For any $t\in T$, if $\pi(t)=n$, we can write $t=tg^{-n}\cdot g^n$. Since $\pi(tg^{-n})=0$, we can find finite $S\subset\operatorname{ker}\pi$ such that $S\cup \{g\}$ generates $G$.

Now take $x=t_1t_2\ldots t_k\in\operatorname{ker}\pi$ with $t_i\in T$. This can be rewritten as $$x= {t_1g^{-n_1}}g^{n_1}{t_2g^{-n_2}}g^{-n_1}g^{n_1+n_2}{t_3g^{-n_3}}g^{-n_1-n_2}\cdots
        g^{n_1+\cdots+n_{k-1}}{t_kg^{-n_k}}g^{-n_1-\cdots -n_{k-1}}{g^{n_1+\cdots+n_k}}$$
$$={t_1g^{-n_1}}\cdot g^{n_1}{t_2g^{-n_2}}g^{-n_1}\cdot g^{n_1+n_2}{t_3g^{-n_3}}g^{-n_1-n_2}\cdots
        g^{n_1+\cdots+n_{k-1}}{t_kg^{-n_k}}g^{-n_1-\cdots -n_{k-1}}$$
Here, $n_i = \pi(t_i)$ and the final factor vanishes since $n_1+n_2+\cdots+n_k=0$ because $x\in\operatorname{ker}\pi$. We have shown that every element in the kernel is generated by the set $\{g^nsg^{-n}\mid s\in S\}$. Now Löh claims that this $n$ must be bounded and hence that the kernel is finitely generated.
I feel like this is a consequence of the subexponential growth but I cannot make this rigorous.

Subexponential growth means that for a generating set $T$, the growth function does not quasi-dominate an exponential.

The same question has been asked here, but it gives another proof method which is less direct.

答:
Let $K=\ker \pi$ and, as in your post,  let $G = \langle S, g \rangle$, with $S$ a finite subset of $K$. Then $K = \langle g^i S g^{-i} : i \in {\mathbb Z} \rangle$.

For $N \ge 0$, let $K_N = \langle g^i S g^{-i}  : 0 \le i \le N \rangle$, and for $N \le 0$ let $K_N = \langle g^{-i} S g^i : 0 \le i \le -N \rangle$.

So we have two ascending chains of subgroups of $K$:  $K_0 \le K_1 \le K_2 \le \cdots$ and $K_0 \le K_{-1} \le K_{-2} \le \cdots $.

If both of these chains stabilize after finitely many steps, then $K$ is generated by a finite number of the $K_i$ and hence $K$ is finitely generated, and we are done. So we can assume that one of them does not stabilize, and by swapping $g$ and $g^{-1}$ if necessary, we can assume that $K_0 \le K_1 \le K_2 \le \cdots$ does not stabilize.

Suppose that, for some $s \in S$ and $N_s  > 0$, we have $$g^{N_s} s g^{-N_s} \in \langle s,gsg^{-1},\ldots,g^{N_s-1}sg^{-(N_s-1)} \rangle.$$ Then the same condition holds for all $N \ge N_s$.

If such an $N_s$ existed for all $s \in S$, then the condition would hold for all $N \ge \max(N_s)$ and all $s \in S$. But then we would have $K_n \le K_{n-1}$ for all $n \ge \max(N_s)$, and the chain $K_0 \le K_1 \le K_2 \le \cdots$ would stabilize, contrary to assumption.

So there exists $s \in S$, such that $g^n s g^{-n} \notin \langle s, gsg^{-1},\ldots, g^{n-1}sg^{-(n-1)} \rangle$ for all $n > 0$.

We claim that the subsemigroup of $G$ generated by $gs$ and $g^2$ is free, which implies that $G$ has exponential growth.

If not then let $w_1$ and $w_2$ be distinct  (positive) words in $g^2$ and $gs$ of smallest total length such that $w_1 =_G w_2$. Then one of $w_1,w_2$ - say $w_1$ - must end in $gs$ and the other in $g^2$, since otherwise we could cancel the final letters and get shorter equal words.

Now, since $\pi(w_1) = \pi(w_2)$, they must both contain the same number of occurrences of $g^2$ and of $gs$ - suppose there is a total of $n$ occurrences of $g$ in both words.

Then, when we rewrite $w_1$ and $w_2$ in $G$ to collect the powers of $g$ to the right (as you have done in your post), we get $w_1' g^{n} = w_2'g^n$, where $w_1'$ and $w_2'$ are words in conjugates $g^ksg^{-k}$ of $s$ for $k \ge 0$.

Since $w_1$ ends in $gs$, and $w_2$ ends in $g^2$, the largest such $k$ occurring in $w_1'$ will be $g^nsg^{-n}$ at the end of $w'$, but the largest $k$ in $w_2$ will be less than $n$. So we get $g^n s g^{-n} \in \langle s, gsg^{-1},\ldots, g^{n-1}sg^{-(n-1)} \rangle$, contrary to assumption.


问:
Is this definition of a function correct?
So im trying to teach myself some logic and how quantifier scope and bound works and I saw the following defintion of a function which left me scratching my head.

$$\forall a\in A \exists b\in B:((a,b)\in R\land \forall b'\in B:(a,b')\in R\implies b=b').$$

We can see from the defintion that the variables $a$ and $b$ are bound in the entire expression by the brackets however my confusion arises for the variable $b'$ clearly it is bounded in the set $B$ however for the implication no brackets are used to bound $b'$ in the implication so basically my question is should we instead write

$$\forall a\in A \exists b\in B:((a,b)\in R\land \forall b'\in B:((a,b')\in R\implies b=b')).$$

With the addition of the extra brackets we have bounded the $b'$ in the implication as it falls within the quantifiers scope is this correct unnecessary or wrong?.

Thanks in advance.

答: I'd say that your version is actually better. In the context of the definition of functions, where the author is establishing the uniqueness of $R(a)$ using Russell's definite descriptions, one should understand the last quantifier to range over the whole following conditional, but as written it could be interpreted as ranging over the antecedent only, and leaving the consequent with a free variable.


答:
For showing a limit on solutions, consider the formlation of $24p=q^2-1 = (q+1)(q-1)$

Then both $(q+1)$ and $(q-1)$ must be even, and one will be divisible by $2p$. However this gives the remaining factor as no more than $12$, so $2p \leq 12+2$ and $p$ can be no bigger than $14/2 = 7$.

Once this is established, the full set of solutions you found can be obtained by evaluating  $p\in \{2,3,5,7\}$.

Title: Prime solutions of an equation


问:
How do I show that the map $u \to \int u^2$ is weakly continuous
I am new-ish to the material.
How do I show that the map $u \to \int u^2$ is weakly continuous in the space is $H^1_0(\Omega)$ using the fact that $H^1_{0}(\Omega)$ is compactly embedded in $L^2(\Omega)?$

答:
Let $(u_k)_{n \in \mathbb{N}} \subseteq H_0^1(\Omega)$ that converges weakly to $u \in H_0^1(\Omega)$ and let $(u_{n_k})_{k \in \mathbb{N}}$ be an arbitrary subsequence. Because of the compact embedding $H_0^1(\Omega) \hookrightarrow \hookrightarrow L^2(\Omega)$ there is a subsequence $(u_{j_{n_k}})_{k \in \mathbb{N}}$ that strongly converges in $L^2(\Omega)$. This limit has to be $u$, since strong limits are weak limits and those are unique. But this immediately implies
$$
\int_{\Omega} u_{j_{n_k}}^2~\mathrm{d}x = \lVert u_{j_{n_k}}\rVert_{L^2(\Omega)}^2 \overset{k \rightarrow \infty}{\longrightarrow} \lVert u \rVert_{L^2(\Omega)}^2
$$
in $\mathbb{R}$. Because of the following Lemma we have $\displaystyle \int_\Omega u_k^2~\mathrm{d}x \overset{k \rightarrow \infty}{\longrightarrow} \int_\Omega u^2~\mathrm{d}x$, i.e. weak continuity:

Lemma: Let $(X, \lVert \cdot \rVert_X)$ be a normed space and $(x_k)_{k \in \mathbb{N}} \subseteq X$ a sequence. If there is some $x \in X$ such that every subsequence of $(x_k)_{k \in \mathbb{N}}$ has itself a subsequence that converges to $x$, then $(x_k)_{k \in \mathbb{N}}$ converges to $x$:

Proof: Assume this were not the case, i.e. $(x_k)_{k \in \mathbb{N}}$ does not converge to $x$. Then. by definition, there is $\varepsilon > 0$ and a subsequence $(x_{n_k})_{k \in \mathbb{N}}$ such that
$$
\lVert x_{n_k} - x\rVert_X \geq \varepsilon.
$$
for all $k \in \mathbb{N}$. But then $(x_{n_k})_{k \in \mathbb{N}}$ can not have a convergent subsequence anymore.


问:
Coloring dots in a circle with no two consecutive dots being the same color
I ran into this question, it is not homework. :)

I have a simple circle with $n$ dots, $n\geqslant 3$. the dots are numbered from $1\ldots n$.

Each dot needs to be coloured red, blue or green. No dot can be couloured the same as his neighbors. How many options are there to organize the colours?

Hint: we need to use recurrence relation.

答:
Let $a_n$ be the number of ways of coloring a ring of $n$ dots. Now consider a ring of $n+1$ dots that have been properly colored. There are two possibilities: either dots $1$ and $n$ have the same color, or they do not.

If they do not have the same color, then removing dot $n+1$ leaves one of the $a_n$ properly colored rings of $n$ dots. Conversely, if we insert a dot $n+1$ into a properly colored ring of $n$ dots, we must give it the third color in order to have a properly colored ring of $n+1$ dots. Thus, there are $a_n$ properly colored rings of $n+1$ dots in which dots $1$ and $n$ have different colors.

If dots $1$ and $n$ have the same color, imagine removing dot $n+1$ and merging dots $1$ and $n$ into a single dot; you now have a properly colored ring of $n-1$ dots. Conversely, if you start with a properly colored ring of $n-1$ dots, you can split dot $1$ into two dots of the same color, one adjacent to dot $2$ and one adjacent to dot $n-1$; call the former dot $1$ and the latter dot $n$. Now insert dot $n+1$ between the new dots $1$ and $n$; this time you have two choices for its color, so there are $2a_{n-1}$ properly colored rings of $n+1$ dots in which dots $1$ and $n$ have the same color.

The recurrence is therefore $a_{n+1}=a_n+2a_{n-1}$, and I’ll leave it to you to work out the initial conditions.