问:
Piecewise continuous differentiable
if I have a piecewise continuously differentiable function. How do I see that on each open interval, where the derivative is continuous, there is a continous extension on the larger closed interval?</p>

答:
This is true when $f$ satisfies the condition: the lateral limits exist. And false in other cases.

Let $f:[a,b]\to\mathbb{R}$ be a piecewise continuously differentiable function. Then there is a partition $P=\{x_i\}_{i=1}^n$ of $[0,1]$ (i.e. $a=x_0<x_1<\ldots<x_n=b$) such that each $I_i=(x_{i-1},x_i)$ is a maximal interval where $f$ is continuously differentiable.

We want to extend some $g_i=f|{I_i}$ to the whole interval $\overline{I_i}=[x_{i-1},x_i]$. Define $g_i':\overline{I_i}\to\mathbb{R}$ as $g_i'|{I_i}=g$ and

$$g_i'(x_{i-1})=\lim_{x\to x_{i-1}^-}g(x),\quad g_i'(x_i)=\lim_{x\to x_i^+}g(x).$$

exists since lateral limit exists.

Is easy to see that $g'$ is a continuous extension of $g$. Note that you can't always extend $f$ in the whole $[a,b]$ since $\lim_{x\to x_i^-}g(x)$ is not necessary the same value as $\lim_{x\to x_{i-1}^+}g(x)$, so here we have $g'_i(x_{i})\neq g'_{i+1}(x_i)$.

If $f$ is not bounded this is false, note that $f(x)=\tan{x}$ where $\tan$ is definied and $f(x)=0$ where $tan$ is not definied is a piecewise differentiable function but if $P$ is a partition that maximices intervals where $f$ is continuously differentiable, $\lim_{x\to x_i^-}g(x)$ and $\lim_{x\to x_i^+}g(x)$ doesn't exists ($g$ defined as before).

If $f$ is discontinuous in some point of a maximal differentiable partition but bounded, this is also false, consider the function $f(x)=\sin{\frac{1}{x}}$ defined on $[-1,1]\setminus\{0\}$ and $f(0)=0$ then $x_0=-1$, $x_1=0$, $x_2=1$. Then  $\lim_{x\to 0^-}g(x)$ and $\lim_{x\to 0^+}g(x)$ doesn't exists ($g$ defined as before). In other cases it is easy to give examples.

Title: Piecewise continuous differentiable


问:
Integral solutions to the diophantine equation $y^2=x^3+2017$.
Case 1: $y^2=0 \mod{3}$
\begin{align*}
& y^2 = 3b \quad (b=3k^2) \implies x^3=3a +2  \\
& 3b = 3a + 2 + 2017 \implies b-a = 673 \\
& k^2 = \frac{a-2}{3} + 225 \\
\end{align*}
Then I just took $a =2$ since 225 is already a perfect square. Thus I got the solution $(x,y) = (2,\pm 45)$. This is the only solution I was able to get.

I tried to do the second case, i.e, $y^2 \equiv 1 \mod 3$ as I did Case 1 but did not have any luck. Here are my proceedings.

Let $y^2 = 3b + 1$ and $x^3 = 3a$ where $a = 9k^3$
Then doing the same thing above we get,
\begin{equation*}
k^3 = \frac{b+3}{9} - 75
\end{equation*}

Now I am stuck, since I cannot find any solutions to the above equation.

答:
Comment:

You can try this method:

Take floor of $2017^{\frac 13}=12$

You can check 12 and less to see it gives a solution:

$2017=12^3+17^2=y^2-x^3$

Which gives $y=\pm 17$ and $x=-12$

Among other numbers bellow 12 only 2 gives integer for y:

$2017=45^2-2^3=y^2-x^3$

Which give what you found; $x=2$ and $y=\pm 45$

This method is for small numbers.

ContentLicense: CC BY-SA 4.0


问:
Determinant factorisation using specific method

Consider the determinant $$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix}$$

I know that $x+3$ and $x+4$ are factors of this determinant. I wish to find the final factor (which is $(x+5)$) specifically by considering the coefficients of $x^2$ and $x^3$. How can I do this? Using this particular method?

答:
Method $1$

If you already know that $(x+5)$ is a factor of the determinant,
put $x = -5$ in the determinant

$$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix}$$

and check if it vanishes.

$$\begin{vmatrix}-1&2&3\\-1&-4&-3\\0&2&2\end{vmatrix} = \begin{vmatrix}0&6&6\\-1&-4&-3\\0&2&2\end{vmatrix} = \begin{vmatrix}0&0&0\\-1&-4&-3\\0&2&2\end{vmatrix}=0$$

Method $2$

Expand the determinant $$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix} = x^3 + 12x^2 + 47x + 60$$

and consider the equation $$x^3 + 12x^2 + 47x + 60 = 0$$

Sum of the three roots of the above equation is $-12$

If two roots are known to be $-3$ and $-4$, the third root must be $-12 - (-3-4) = -5$


问:
Why am I getting wrong answer for this Q
There are 12 books on a shelf. In how many ways can we select at least one of them .

So , I wrote the answer as 12C1. This means that selecting 1 book out of 12 books.

But it is $2^{12 }- 1$. Please help in this.

答: Note that at least 1 means that you can chose any value $\ge 1$, so, it's just $$\sum_{n=1}^{12}\binom{12}{n}$$ which can be written as $$\left(\sum_{n=0}^{12}\binom{12}{n}\right)-\binom{12}{0}$$ which would be $2^{12}-1$


答:
"At least one of them" means that we can select $1,2,3,4,\cdots,12$ books as opposed to just 1. So the number of ways can be found by $$\sum_{k=1}^{12}\binom{12}{k}$$
Consider the binomial theorem: $$(1+r)^{12}=\sum_{k=0}^{12}\binom{12}{k}r^k$$
In the case of $r=1$, we have $$2^{12}=\sum_{k=0}^{12}\binom{12}{k}$$
So the required number of ways is $$\sum_{k=1}^{12}\binom{12}{k}=\sum_{k=0}^{12}\binom{12}{k}-\binom{12}{0}=2^{12}-1$$

Note: if you are not familiar with sigma notation, note that $$\sum_{k=0}^na_k=a_0+a_1+\cdots +a_n$$

Title: Why am I getting wrong answer for this Q

Body:
There are 12 books on a shelf. In how many ways can we select at least one of them .

So , I wrote the answer as 12C1. This means that selecting 1 book out of 12 books.

But it is $2^{12 }- 1$. Please help in this.


问:
Why is 9 involved in this problem instead of 3?
So, lets say it costs $1.25 per square yard to paint a wall. The wall is 18x25 feet.

How much will it cost to paint the wall.

First, I don't understand what is better. Converting 18x25 feet into yards which is (54x75) or converting down the cost from yard to feet which is aprox $0.416.

But lets say that it's better to convert feet to yards in this situation.

X is the cost.

X=((54*75)/9)*1.25

But again. There are 3 feet in a yard, and it looks like that 9 is squared. But I fail to understand why it's that way.

Can anyone explain? Thanks!

答:
The wall is not 18 x 25 feet. The wall is 18 feet x 25 feet, which is 18 x 25 square feet.

Converting 18 x 25 square feet into square yards does not give 54 x 75 square yards. Yards are bigger than feet. 18 feet, that's not 54 yards, it's 6 yards. 18 x 25 square feet give 6 x (25/3) square yards.

So the cost is 6 x (25/3) x $1.25, whatever that comes to.

ContentLicense: CC BY-SA 3.0


Title: Telescoping sum of powers

Body:
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
 \vdots &=& &\vdots & \\
 3^3-2^3 &= &3(3^2) &-3(3) &+1\\
 2^3-1^3 &= &3(2^2) &-3(2) &+1\\
 1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\
n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n
\end{array}
$$

Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$

 How do this transformation was done?

Thanks!

Body:
An idea: since

$$A^3-B^3=(A-B)(A^2+AB+B^2)$$

we get then

$$(n-1)^3-(n-2)^3=\left[(n-1)-(n-2)\right]\left[(n-1)^2+(n-1)(n-2)+(n-2)^2\right]\ldots\ldots$$


问:
Name the function $f(x,y) = f(x,0) + f(0,y)$
I have a function $f(x,y)$ where $x \in R$, $y \in R$ and $f \mapsto R$.

What would you call this function when I add these properties to $f$:

$f(x+r,y) = f(x,y) + f(r,0)$

$f(x,y+r) = f(x,y) + f(0,r)$

for any $r \in R$

答:
I found the name that I can use.  The concept called linear map is what I was looking for.

To use the term, I am happy to include the homogeneity condition to my list

$f(cx,cy) = cf(x,y)$ for any $x$, $y$ and $c \in R$

(Thanks for guiding me in the right direction)


答:
Suppose the determinant is $f(\lambda)=(x+4)(x+3)(x-\lambda)$. (No coefficient in front as is clear from the main diagonal).

Substitute a value of $x$ different from $-3,-4$ into the determinant and solve for it. (Easiest would be expanding along $C_1$ or $R_3$.)

Set that numerical value equal to $f(\lambda)$ which is a linear equation in $\lambda$. You should obtain $\lambda=-5$.

Title: Determinant factorisation using specific method

Body:
Consider the determinant $$\begin{vmatrix}x+4&2&3\\-1&x+1&-3\\0&2&x+7\end{vmatrix}$$

I know that $x+3$ and $x+4$ are factors of this determinant. I wish to find the final factor (which is $(x+5)$) specifically by considering the coefficients of $x^2$ and $x^3$. How can I do this? Using this particular method?


问:
Asymptotic expansion for the following integral
I was trying to find the asymptotic expansion for
$$\int_0^1 \sqrt{t(1-t)}(t+a)^{-x} \; \mathrm{d}t,$$ for $a>0$ as $x \rightarrow \infty$. I have already tried re-writing $$(t+a)^{-x}=\exp(- x\log(t+a)).$$ However, by using Laplace's method for asymptotic expansions I get that $f(t) =  \sqrt{t(1-t)}$ vanishes everywhere.

Any ideas are appreciated/welcome. Thank you.

答:
There ia an antiderivative which involves the Appell hypergeometric function of two variables.

Using the bounds, this leads to
$$\int_0^1 \sqrt{t(1-t)}(t+a)^{-x} \,dt=\frac{\pi}{8} \,  (a+1)^{-x} \,\, _2F_1\left(\frac{3}{2},x;3;\frac{1}{a+1}\right)$$ where appears the gaussian hypergeometric function. This is the exact solution.

I have been unable to find the asymptotics of the gaussian hypergeometric function when its second argument tends to infinity.

Since, in their answers, @gary and @Vajra  proposed simpler approaches, I give up.


答:
The minimum of the phase function occurs at the endpoint $t=0$. We have
$$
\log (t + a) = \log a + \frac{t}{a} + \mathcal{O}(t^2 )
$$
and
$$
\sqrt {t(1 - t)}  = \sqrt t  - \frac{1}{2}t^{3/2}  + \mathcal{O}(t^{5/2} )
$$
as $t\to 0+$. Thus, by Laplace's method (http://dlmf.nist.gov/2.3.iii), the leading order asymptotics is
$$
e^{ - x\log a} \Gamma \left( {\frac{3}{2}} \right)\frac{{a^{3/2} }}{{x^{3/2} }} = \frac{{\sqrt \pi  }}{2}\frac{{a^{3/2 - x} }}{{x^{3/2} }}.
$$

Title: Asymptotic expansion for the following integral

Body:
I was trying to find the asymptotic expansion for
$$\int_0^1 \sqrt{t(1-t)}(t+a)^{-x} \; \mathrm{d}t,$$ for $a>0$ as $x \rightarrow \infty$. I have already tried re-writing $$(t+a)^{-x}=\exp(- x\log(t+a)).$$ However, by using Laplace's method for asymptotic expansions I get that $f(t) =  \sqrt{t(1-t)}$ vanishes everywhere.

Any ideas are appreciated/welcome. Thank you.


问:
Can I get a verification with a Line Integral I= $\int_\Gamma x(a+y)(ydx+xdy)$?
$O=(0,0),~ A=(3,0),~ B=(0,3)$

$I= \int_\Gamma x(a+y)(ydx+xdy)$

$I= 36$, $a \in \mathbb {R}$

The first portion of the line is $OA$, then $AB$ and finally $BO$

AB is characterized by $\{(x,y) \in \mathbb {R^2}$, $x^2 + y^2 = 9$, $x \ge 0$, $y \ge 0\}$

And I need to find a. I applied Green formula and after some calculations I got to $a=\dfrac {16}{\pi}$ which I am  not sure if it is right. Sorry for format or bad english, it is not my first language.

Link to how I solved it, last integral was put into wolfram https://i.imgur.com/oV5L4lk.jpg

EDIT: I might have gotten a better answer as both $x$ and $y$ are positive and my $AB$ might be actually a quarter of a circle so I use the constraints $0$ and $\dfrac{\pi}{2}$ for my integral with polar coordinates and it results in $a = 4$

答:
You made a mistake in your integral. See below. Also your diagram shows $AB$ as a straight line instead of circular arc.

$a \displaystyle \int_0^{3} \bigg[\int_0^{\sqrt{9-y^2}} x \ dx \bigg] \ dy = a \displaystyle \int_0^{3} \bigg[\frac{x^2}{2} \bigg]_0^{\sqrt{9-y^2}} \ dy$

$ = \frac{a}{2} \displaystyle \int_0^{3} (9-y^2) \ dy = 9a$

So $9a = 36 \implies a = 4$.

You could also set up your integral in polar coordinates as,

$a \displaystyle \int_0^{\pi/2} \int_0^3 r^2 \cos \theta \ dr \ d\theta = 9a$


问:
$\text{rank}(I+AB) = \text{rank}(I+BA)$
I want to prove $\text{rank}(I+AB) = \text{rank}(I+BA)$ without using property of for invertible matrix $A$,  $\text{rank}(AB) = \text{rank}(BA)= \text{rank}(B)$. [If $A$ is invertible, $1+AB = A(A^{-1} +B)$ and since $A$ is invertible, $\text{rank}(A(A^{-1}+B)) = \text{rank}((A^{-1}+B)A) = \text{rank}(1+BA)$.

First of all, I am not sure this holds without the condition of invertibility of $A, B$.

My trial is as follows :

First I want to prove $N(I+AB) = N(I+BA)$.  [Here $N$ denote the null space, and the rank is the dimension of null space]

Let $v \in N(I+AB)$, then $(I+AB) v = 0$ implies $v = - ABv$ so $Bv= - BABv$ which means $Bv \in N(I+BA)$.
Let $n = \text{rank}(I+AB)$ that means $v \in \{v_1, \cdots, v_n\}$ and I stuck at this moment...

Are there any straightforward or easy ways to prove this?

答:
It's not difficult to verify that if $\lambda \neq 0$ is an eigenvalue of $AB$ corresponding to $v \neq 0$, then $Bv$ is an eigenvector for $BA$ corresponding to $\lambda$. If we have $ABv = \lambda v$, then since $\lambda \neq 0$, we have $\lambda v \neq 0$, and hence $Bv \neq 0$. Then
$$BA(Bv) = B(ABv) = B(\lambda v) = \lambda Bv.$$
We can actually prove something stronger: if $E_1$ and $E_2$ are the eigenspaces of $AB$ and $BA$ respectively, with respect to eigenvalue $\lambda \neq 0$, then
$$f : E_1 \to E_2 : v \mapsto Bv$$
is an isomorphism between $E_1$ and $E_2$, with inverse
$$g : E_2 \to E_1 : v \mapsto \lambda^{-1}Av.$$
We have shown $f$ has a valid domain and codomain. By swapping the roles of $A$ and $B$, the same can be said about $\lambda g$, and hence $g$. It is clear they are linear, so we just need to show they are inverses.

We have, for any $v \in E_1$ (i.e. eigenvector of $AB$),
$$(g \circ f)(v) = \lambda^{-1} ABv = \lambda^{-1}\lambda v = v.$$
Similarly, if $v \in E_2$, then
$$(f \circ g)(v) = B(\lambda^{-1}Av) = \lambda^{-1}BAv = v.$$
Therefore, $f$ is an isomorphism, and $E_1$ and $E_2$ have the same dimension.

How does this help us? Suppose $A$ and $B$ are $n \times n$ matrices. Note that $N(I + AB)$ is the eigenspace of $AB$ corresponding to $\lambda = -1 \neq 0$, and similarly for $N(I + BA)$. If we denote them by $E_1$ and $E_2$ respectively, then by the rank-nullity theorem,
$$\operatorname{rank}(I + AB) = n - \operatorname{null}(I + AB) = n - \dim E_1 = n - \dim E_2 = \operatorname{rank}(I + BA).$$

Title: $\text{rank}(I+AB) = \text{rank}(I+BA)$


问:
How do you go from the first step where the summation of $(1-p)^{k-1}$ transform to the fraction
$$\sum_{k=1}^n p(1-p)^{k-1}=p \frac{1-(1-p)^n}{1-(1-p)}$$

I don't know how to simplify the summation. Please help me. Thank you so much.

答:
Hint

For all $a\neq 1$, one can prove that $$\sum_{k=0}^na^k=\frac{1-a^{n+1}}{1-a},$$
for all $n\in\mathbb N$.


问:
Show an inequality using scalar product of vectors
How can I prove that:

$$0<r^2<(r^2+(wl)^2)((1-w^2lc)^2+(wrc)^2)\tag1$$

$\forall c>0$ and all the other variables are bigger than zero using the scalar product of the vectors $A=(r,wl)$ and $B=(1-w^2lc,wrc)$?

I do not know how to get started on this problem and I do not see how I can use the  scalar product of vectors to prove this inequality.

答: For $r\ne0$, $r^2>0$ is trivial. The other inequality is $(A\cdot B)^2<(A\cdot A)(B\cdot B)$, which is Cauchy–Schwarz for non-parallel vectors (otherwise we could only claim $\le$). In particular, check that $r(1-w^2lc)+wl(wrc)=r$. But $A$ could be parallel to $B$, so $<$ is in general incorrect.


问:
Directed graphs, chains, paths...
First of all - what's a directed graph? Is this just some normal graph but with no loops or edges? I don't really understand definitions found in the web...

What is a path? I mean, as far as I know, it is a directed chain, where the chain is a "path between two vertices, connected anyhow". Ok, I understand what the chain is, but what is the DIRECTED chain?

答:
Let's use the simplest possible definitions.

A graph is a set $(V,E)$ of vertices and edges between those vertices, where an edge is an unordered pair $e=(v_1,v_2)$ of vertices. Thus $(v_1,v_2)=(v_2,v_1)$.

A directed graph is a set $(V,E)$ of vertices and directed edges between those vertices, where a directed edge is an ordered pair $\tilde e=(v_1,v_2)_\mathrm d$ of vertices. I'm using a subscript $\mathrm d$ to indicate the pair is ordered. Thus $(v_1,v_2)_\mathrm d\neq(v_2,v_1)_\mathrm d$.

In these definitions, there is no such thing as a 'partially directed graph' with some edges of each type. If you want a directed graph with some pair of vertices linked in both directions, you just include both $(v_1,v_2)_\mathrm d \;\&\; (v_2,v_1)_\mathrm d$.

An (undirected) path is any sequence of edges $(e_1,e_2,\cdots,e_n)$ connecting vertices continuously, which is invariant under reversal, so $(e_1,e_2,\cdots,e_n) = (e_n,\cdots,e_2,e_1)$. A directed path is the same without the reversal property. Generally you think about directed paths if and only if you're thinking about directed graphs.

A simple path is a path with no repeated vertices.

Title: Directed graphs, chains, paths...


问:
Calculate $\sum_{k=1}^{\infty}\frac{|\sin(k x)|}{1+k^2}$
In https://math.stackexchange.com/a/4015346/198592 it was shown that the sum

$$s(x) = \sum_{k=1}^{\infty}\frac{\sin(k x)}{1+k^2}$$

is exactly expressible in terms of the hypergeometric function.

I wonder what happens íf we replace the sine by its absolute value, i.e. ask for the properties of the function $a(x)$ defined by the sum

$$a(x) = \sum_{k=1}^{\infty}\frac{|\sin(k x)|}{1+k^2}$$

Questions

(1) can you find a closed expression for $a(x)$ if $x$ is a rational multiple of $\pi$?

(2) is there a closed expression for $a(x)$?

Here is a plot of the two functions $a(x)$ and $s(x)$

答:
Some first results

In order not to spoil the fun of those who wish to find solutions be themselves I provide here only the answers to some question (1) without deriving the results.

$$a(\frac{\pi}{2}) = \frac{1}{4} i \left(H_{-\frac{1}{2}-\frac{i}{2}}-H_{-\frac{1}{2}+\frac{i}{2}}\right)\simeq 0.72033 \tag{1a}$$

$$a(\frac{\pi}{2}) =\frac{1}{4} \pi  \tanh \left(\frac{\pi }{2}\right)\tag{1b}$$

Here $H_z$ is the harmonic number of the (complex) argument $z$

$$\begin{align}
a(\frac{\pi}{3})=\frac{i}{4 \sqrt{3}} \left(-H_{-\frac{1}{3}+\frac{i}{3}}-H_{-\frac{2}{3}+\frac{i}{3}}+H_{-\frac{1}{3}-\frac{i}{3}}+H_{-\frac{2}{3}-\frac{i}{3}}\right)\simeq 0.784626
\end{align}\tag{2a}$$

$$a(\frac{\pi}{3})=\frac{\pi  \sinh \left(\frac{2 \pi }{3}\right)}{\sqrt{3} \left(1+2 \cosh \left(\frac{2 \pi }{3}\right)\right)}\tag{2b}$$

These results show an interesting relation of $s(x)$ to the harmonic numbers and - less obviously - to explicit functions via contour integrals and residues.

General solution to question (1)

With a nudge from @NN2's solution I have found, using Cauchy's theorem and residues, the following simple finite sum for the case of a rational argument

$$a(x=\pi \;\frac{p}{q}) = a_{Q}(p,q): = \frac{\pi}{2 q}\sum _{j=1}^{q-1} | \sin \left(\pi \frac{ p }{q} j\right)| \frac{ \sinh \left(\frac{2 \pi }{q}\right)}{ \cosh \left(\frac{2 \pi }{q}\right)-\cos \left(\frac{2 \pi  j}{q}\right)}\tag{s.1}$$

This formula is the full answer to problem part (1) of the OP, and thus generalizes the special results of the initial post in the second form of the equations (i.e. $(1.b)$ and $(2.b)$ ).

Derivation

We have to consider the sum

$$a_Q(p,q) = \sum_{k=1}^{\infty} \frac{| \sin\left(\pi (\frac{p}{q}) k\right)|}{1+k^2}\tag{s.2}$$

In order to simplfy the sum, we start with the observation that the numerator of the summmand has period $q$ in $k$ to split the summation over $k$ into chunks of length $q$ writing

$$k= m q + j\tag{s.3a}$$

with the index ranges

$$m=0,1,...\tag{s.3b}$$
$$j=1,2,...,q-1\tag{s.3c}$$

notice that the $j$ needs to run only up to $q-1$ because the $\sin$-factor vanishes.

Hence our sum becomes the double sum

$$a_{Q}(p,q) = \sum_{j=1}^{q-1} \sum_{m=0}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) j)|}{1+(m q + j)^2}\tag{s.4}$$

Let us now calculate the $m$-sum (designating it following @NN2)

$$u(q,j) = \sum_{m=0}^{\infty}\frac{1}{1+(m q + j)^2}\tag{s.5}$$

Partial fraction decomposition, shifting the summation index $m\to n-1$ and extracting a factor $\frac{1}{q}$ leads to

$$\begin{align}
u(q,j) &=
\sum_{m=0}^{\infty}\frac{1}{2 i} \left(\frac{1}{-i+m q + j}-\frac{1}{+i+m q + j}\right)
\\
&=\frac{1}{2 i q} \sum_{n=1}^{\infty}
\left(\frac{1}{n-1 +\frac{-i+j}{q}}-\frac{1}{n-1 +\frac{ i+j}{q}}\right)
\\
&=\frac{1}{2 i q} \sum_{n=1}^{\infty}
\left(\left(\frac{1}{n}-\frac{1}{n-1 +\frac{+i+j}{q}}\right)-\left(\frac{1}{n}-\frac{1}{n-1 +\frac{-i+j}{q}}\right)\right)
\end{align}
\tag{s.6}$$

Here, in order to obtain two summands with convergent sums, we have added and subtracted a term $\frac{1}{n}$.

Using the definition of the harmonic number for a general argument $z$

$$H_{z}=H(z) = \sum_{m=1}^{\infty}(\frac{1}{m}-\frac{1}{m+z})\tag{s.7}$$

we can write

$$\begin{align}
u(q,j)
&=\frac{1}{2 i q} \left(H(\frac{+i+j}{q}-1)-H(\frac{-i+j}{q}-1)\right)\\
&=\frac{1}{q}\Im(H(\frac{+i+j}{q}-1))
\end{align}
\tag{s.8}$$

We can also easily find an integral representation of $u$ as follows. Writing the summand as

$$\begin{align}
\frac{1}{1+(m q+j)^2}&=\Im \left(\frac{-1}{i+m q + j}\right)=-\Im(\int_{0}^{\infty}e^{-(i+m q + j)}\,dt )\\
&=\int_{0}^{\infty}\sin{t}\;e^{-(m q + j)}\,dt
\end{align}
\tag{s.9}$$

and performing the sum over $m$ gives finally

$$u(q,j) =\int_0^{\infty } \frac{e^{-j\; t} \sin (t)}{1-e^{-q\; t}} \, dt \tag{s.10}$$

A similar formula was first given in the answer of @NN2 (https://math.stackexchange.com/a/4024768/198592).

Summarizing up to this point, we have obtained for the infinite sum $a_{Q}(p,q)$ a finite sum over $(q-1)$ summands which however contain the more awkward harmonic functions of a complex argument, or, equivalently, a non elementary integral.

Luckily, formula $(s.1)$ shows that there is a similar sum containing more elementary functions.

The key to its derivation is extending the one-sided sum over $m$ into a two-sided sum. It turns out that this is possible if we work under the $j$-sum and take into account the specific numerator.

Derivation A)

Consider, under the sum over $j$ from $1$ to $q-1$,

$$\begin{align}
&\sum_{m=-\infty}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) j)|}{1+(m q + j)^2}\\
&= \sum_{m=-\infty}^{-1}\frac{|\sin(\pi (\frac{p}{q}) j)|}{1+(m q + j)^2}+\sum_{m=0}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) j)|}{1+(m q + j)^2}
\end{align}
\tag{s.11} $$

and the first sum can be transformed as follows

$$\begin{align}
&\sum_{m=-\infty}^{-1}\frac{|\sin(\pi (\frac{p}{q}) j)|}{1+(m q + j)^2}=\sum_{m=1}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) j)|}{1+(-m q + j)^2}\\
&=\sum_{m=1}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) j)|}{1+(m q - j)^2}
\overset{j\to q-i}=\sum_{m=1}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) (q-i))|}{1+(m q - q+i)^2}\\
&=\sum_{m=1}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) i)|}{1+((m-1) q - +i)^2}=\sum_{m=0}^{\infty}\frac{|\sin(\pi (\frac{p}{q}) i)|}{1+(m q  +i)^2}
\end{align}
\tag{s.12}$$

Hence we can write (with a factor $\frac{1}{2}$)

$$a_{Q}(p,q) = \frac{1}{2} \sum_{j=1}^{q-1} |\sin(\pi (\frac{p}{q}) j)| \sum_{m=-\infty}^{\infty}\frac{1}{1+(m q + j)^2}\tag{s.13}$$

Derivation B)

Let

$$\begin{align}
S_1 = \sum_{j=1}^{q-1} | \sin(\pi \frac{p}{q} j) |  \sum_{m \ge 0} \frac{1}{1+(q m+j)^2 }=\frac{1}{2}(S_1+S_2)
\end{align}
\tag{s.14}$$

where $S_2$ has the same values as $S_1$ and differs only in the direction of the summation index $j$.

We proceed with $S_2$ as follows

$$\begin{align}
S_2 &= \sum_{j=q-1}^{1} | \sin(\pi \frac{p}{q} j)|  \sum_{m \ge 0} \frac{1}{1+(q m+j)^2 }\\
&\overset{j\to q-i}=
\sum_{i=1}^{q-1} | \sin(\pi \frac{p}{q}(q- i)) | \sum_{m \ge 0} \frac{1}{1+(q m+q-i)^2 }\\
&=\sum_{i=1}^{q-1} \left|\sin(\pi \frac{p}{q} i) \right| \sum_{m \ge 0} \frac{1}{1+(q(m+1)-i)^2 }\\
&=\sum_{i=1}^{q-1} \left|\sin(\pi \frac{p}{q} i) \right| \sum_{m \ge 0} \frac{1}{1+(-q(m+1)+i)^2 }\\
&\overset{m+1\to n}=\sum_{i=1}^{q-1} \left| \sin(\pi \frac{p}{q} i) \right| \sum_{n=-1}^{-\infty}  \frac{1}{1+(q n+i)^2 }
\end{align}\tag{s.15}
$$

hence from $(s.14)$ follows

$$S_1 =\frac{1}{2}
\sum_{j=1}^{q-1}  \left| \sin(\pi \frac{p}{q} j) \right| \left( \sum_{m =-1}^{-\infty} + \sum_{m =0}^{\infty} \right) \frac{1}{1+(q m+j)^2}\tag{s.16}$$

hence the two sums add up to a double-sided sum:

$$\begin{align}
a_{Q}(p,q) = S_1=\frac{1}{2}\sum_{j=1}^{q-1} | \sin(\pi \frac{p}{q} j)|  \sum_{m =-\infty}^{\infty} \frac{1}{1+(q m+j)^2 }
\end{align}\tag{s.17}$$

Now the double-sided sum

$$v(q,j) = \sum_{m=-\infty}^{\infty}\frac{1}{1+(m q + j)^2}\tag{s.18}$$

can be calculated by evaluating the contour integral

$$\int_C (\pi \cot(\pi m) \frac{1}{q} \frac{1}{1+(m q +j)^2}\, dm\tag{s.19}$$

over a circle of radius $R$ concentric around the origin in the complex $m$-plane. In the limit of $R \to \infty$ the integral vanishes so that $v$ is given by the sum of the residues at the two poles of $\frac{1}{1+(m q +j)^2}$ which are at $m = \frac { \pm i - j} {q}$

This leads for $1 \le j \le q-1$ to

$$v(q,j) = \frac{\pi  \left(\coth \left(\frac{\pi +i \pi  j}{q}\right)+\coth \left(\frac{\pi -i \pi  j}{q}\right)\right)}{2 q}\tag{s.20}$$

This can be simplified to

$$v(q,j) = \frac{\pi  \sinh \left(\frac{2 \pi }{q}\right)}{q \left(\cosh \left(\frac{2 \pi }{q}\right)-\cos \left(\frac{2 \pi  j}{q}\right)\right)}\tag{s.21}$$

which completes the derivation of $(s.1)$.

Afterwards I noticed that Mathematica can do the two-sided sum immediately and afterwards can be used to complete the simplification.

Remark: the fact that the values of the two sums for $a_{Q}$ (i.e. using either $u$ or $v$) with the "weight factor" $|\sin(\pi \frac{p}{q}j)|$ are the same might lead at first sight to the conjecture that $v$ is the solution of the integral $(s.10)$ for $u$. But this is not the case: $u$ and $v$ are different.

Plot

Here is a discrete plot of $a_{Q}(p,q)$

The graph is gratifying as is comes very close to the origial graph for real $x$ presented in the OP.

Notice that $a_Q$ can not serve as the general solution for real $x$. We compare the two graphs in the discussion.

Discussion

§d.1

It is interesting to compare the graphs for the function $a_Q(x\in \mathbb{Q})$ if taken at discrete values (as it is intended) and for continuous values $a_Q(x\in \mathbb{R})$. What emerges looks to me like the stalactites hanging of the ceiling of a cave

The function $a_Q(x)$ with $x\in \mathbb{R}$ creates the "ceiling", and its minima (downwards peaks) are the values at the discrete argument $x\in \mathbb{Q}$.

Notice as the main result that the function $a_{Q}(x \in \mathbb{R})$ is different from the original function $a(x)$ almost everywhere (i.e. except for a countable number of points).

§d.2

Type of curve of $a(x)$

For the sake of definiteness we consider the partial sum

$$a(x,n) = \sum_{k=1}^{n} \frac{|\sin(x k)|}{(1+k^2)}$$

For fixed $n$ this is a continuous function of $x$.

The first and second derivatives are calculated from the corresponding expression for $|\sin(k x)|$, i.e.

$$\frac{d}{dx} |\sin(k x)| = \frac{d}{dx} \sqrt{\sin(k x)^2}=\frac{k \sin (k x) \cos (k x)}{\sqrt{\sin ^2(k x)}}\tag{d.1a}$$

$$\left(\frac{d}{dx}\right)^2 |\sin(k x)| = -k^2 \sqrt{\sin ^2(k x)}=-k^2 |\sin(k x)|\tag{d.1b}$$

The first derivative is discontinuous at the points where the sin vanishes, i.e. at $x = \pi \frac{m}{k}$, i.e. it has the value $-k$ for $x$ coming from below and $+k$ for $x$ coming from above. Hence $|\sin(k x)|$ and has a jump of size $2k$.

The second derivative of $|\sin(k x)|$ is continuous.

This reflects in the corresponding expressions for the partial sum which are show in the next picture

Summing up we find that the first derivative of $a(x)$ is discontinuous in all $x$ which are rational multiples of $\pi$. Without mentioning it, these were exactly the points for which $a$ had to be calculated in the OP.  Hence the graph of $a'(x)$ is a sawtooth at any scale, probably even self-similar, we could call it a "rough" curve.

Casual question (1): what is the length of the graph of $a'(x)$?

Casual question (ii): what is the first derivative of $a(x)$ at $x=\frac{1}{e}$?

§d.3

I have tried to understand the sum over $j$ in the expression $(s.1)$ in the limit $q\to\infty$ as a Riemann integral, but with no avail.

Replacement of $|\sin(x)|$

I don't know if this leads to something useful but we can write

$$\begin{align}|\sin(x)| =
\sin(x)\times \text{sgn}(\sin(x)) \\
=\sin (x)  \; \Im\left(\frac{4}{\pi}\text{arctanh}\left(e^{i x}\right)\right)\\
= \sin (x) \; \Re\left(\frac{4}{\pi }\text{arctan}\left(e^{i \left(x-\frac{\pi }{2}\right)}\right)\right)
\end{align}
\tag{3} $$

here

$$\begin{align}
\text{arctanh}(z)  = \frac{1}{2} \left(\log(1+z)-\log(1-z) \right)\\
\text{arctan}(z)  = \frac{1}{2 i} \left(\log(1+i z)-\log(1-i z) \right)\end{align}\tag{4}$$


问:
what is the derivative of $\sec^2x$
I'm using the quotient rule and so I rewrite $\sec^2x$ as $\frac{1}{\cos^2x}$

Then set $u = 1,$  $u'= 0$
then $v = \cos^2x$ and $v'= -2\sin x \cos x$

I then get $\frac{2\sin x \cos x}{\cos^4x}$

but when I simplify that I don't get $2\sec^2x\tan x$ which is the correct answer?
I'm not really sure what I'm doing wrong

答: $$\frac{2\sin x \cos x}{\cos^4x}=\frac{2(\sin x)}{(\cos x)[\cos^2x]}=2\tan x \sec^2 x$$

Title: what is the derivative of $\sec^2x$


问:
measure defined in terms of an integral
Here's a problem from a midterm I took 2 decades ago!  Let $A$ be a Borel measurable set on the real line and $\lambda$ be the Lebesgue measure. Define a measure $\mu(A)=\int_A f d\lambda$ where $f(x)=1/x^2$ for $x\ne 0$ and $f(0)=\infty$.

Is $\mu$ finite? $\sigma$-finite?

Now I know it's not finite, since if $A$ is any interval containing $0$ then the integral is infinite.  I'm not sure about $\sigma$-finite though. Any interval that does not contain $0$ will have finite measure, and there's a countable number of these to cover everything but $\{0\}$, so it seems to me that the answer depends on the following....

What is $\mu(\{0\})$?

答:
$\mu$ is $\sigma$-finite.  $$
\mathbb R = \{0\} \cup \bigcup_{n=2}^\infty [1/n,n]
\cup \bigcup_{n=2}^\infty [-n,-1/n]
$$
and all those sets have finite measure.

Title: measure defined in terms of an integral


答:
In the development of the determinant by Sarrus' rule, no term is a quadratic polynomial, so only

$$(x+4)(x+1)(x+7)$$ yields cubic and quadratic terms, namely $$x^3+12x^2.$$

Hence by Vieta, the third root must be $$-12+3+4.$$


Title: what is the derivative of $\sec^2x$

Body:
I'm using the quotient rule and so I rewrite $\sec^2x$ as $\frac{1}{\cos^2x}$

Then set $u = 1,$  $u'= 0$
then $v = \cos^2x$ and $v'= -2\sin x \cos x$

I then get $\frac{2\sin x \cos x}{\cos^4x}$

but when I simplify that I don't get $2\sec^2x\tan x$ which is the correct answer?
I'm not really sure what I'm doing wrong

Body: $$\frac{2 \sin(x)\cos(x)}{\cos^2(x)\cdot \cos^2(x)}=2 \sec^2 (x) \tan(x)$$


问:
Evaluating a summation of inverse squares over odd indices
$$ \sum_{n=1}^\infty \frac{1}{n^2}  = \frac{\pi^2}{6}$$

I want to evaluate this sum when $n$ takes only odd values.

答:
Note that
$$\sum_{n \text{ is even}} \dfrac1{n^2} = \sum_{k=1}^{\infty} \dfrac1{(2k)^2} = \dfrac14 \sum_{k=1}^{\infty} \dfrac1{k^2} = \dfrac{\zeta(2)}4$$
Also,
$$\sum_{k=1}^{\infty} \dfrac1{k^2} = \sum_{k \text{ is odd}} \dfrac1{k^2} + \sum_{k \text{ is even}} \dfrac1{k^2}$$
Hence,
$$\sum_{k \text{ is odd}} \dfrac1{k^2} = \dfrac34 \zeta(2)$$</p>


答:
You simplified wrong. In $$\frac{2 \sin x \cos x}{\cos^4x}$$  start by cancelling $\cos x$ from top and bottom. Now its

$$\frac{2 \sin x }{\cos^3x}.$$

That's the same as your answer.

Title: what is the derivative of $\sec^2x$


Title: what is the derivative of $\sec^2x$

Body: $$\frac{2\sin(x)\cos(x)}{\cos^4(x)} = 2\cdot\frac{\sin(x)}{\cos( x)}\cdot\frac{1}{\cos^2(x)} = 2\tan(x)\sec^2(x)$$


问:
What is the limit of $f(x)=\frac{\sqrt{(x-2) \left(x^2+2 x-8\right)}}{x^2-4}$ when $x\to 0$?
I tried to find out the value of limit but when I took $x_-\rightarrow0$ then a result is obtained again when I took $x_+\rightarrow0$ another result was obtained. Can anyone please explain why is this happening?

Is there anything wrong with the question or I was doing it wrong?

I am quite weak at determining limits

Edit: The problem stated above has been solved. The question was wrong instead of $x \to 0$ it should be $x \to 2$ (this will satisfy the question).

答:
I'm sure you meant to evaluate the limit as $x \to 2$.

Factorize, and remember that $\sqrt {y^2}$ is not just $y$; it is $|y|$. This is crucial.

What's the behavior of the function $\frac {|y|}{y}$ near zero?

Solution:

 $$ \frac {\sqrt{(x-2)(x^2 +2x -8)}}{x^2 -4}=\frac {\sqrt{(x-2)(x+4)(x-2)}}{(x-2)(x+2)}= \frac {\sqrt {(x-2)^2} \sqrt {x+4} }{(x-2)(x+2)} = \frac {|x-2|}{x-2}  \frac {\sqrt {x+4}}{x+2} $$

 Now as $x \to 2$ the function $\frac {\sqrt {x+4}}{x+2} \to \frac {\sqrt {6}}{4}$, however the other half $\frac {|x-2|}{x-2}$ (aka $sign(x-2)$) is equal to $1$ as we approach $2$ from the right, and is equal to $-1$ as we approach $2$ from the left. Therefore the solution will be $-\frac {\sqrt {6}}{4}$ as $x \to 2^{\mathbf -}$ and $\frac {\sqrt {6}}{4}$ as $x \to 2^{\mathbf +}$.


问:
Is the realm of sets belonging to weakly reachable from below stages of the cumulative heirarchy a model of ZF?
I want to know how high is the universe of sets in ZFC as seen from below.

We say that a stage $V_\alpha$ of the cumulative hierarchy is reachable from below if and only if $\alpha$ is isomorphic to an ordering on a set that belongs to a prior stage. Formally:

$\text{reachable} (V_\alpha) \iff \exists S \in V_\alpha \exists f \exists \mathcal R \\ [ f: \alpha \hookrightarrow S  \land \forall \beta,\gamma \in \alpha \ (\beta < \gamma \Leftrightarrow f(\beta) \ \mathcal R \ f(\gamma) )]$

Now we define hierarchical gap as a nonempty stage of the hierarchy that is not reachable from below.

Define: $\text{weakly-reachable} (V_\alpha) \iff |\{ V_{\beta <\alpha}| V_\beta \text { is hierarchical gap } \}| < |V_\alpha|$

A stage is hereditarily weakly-reachable from below if it and every stage prior to it is weakly-reachable from below.

Does the realm of all sets beloning to hereditarily weakly-reachable from below stages constitute a model of ZF? Or ZF can prove the existence of a non weakly-reachable stage from below?

If we redefine weak reachability as:

$\text{weakly reachable} (V_\alpha) \iff |\{ V_{\beta <\alpha}| V_\beta \text { is hierarchical gap } \}| < |\alpha| $

Can ZF prove existence of a non weakly reachable stage?

答:
$\mathsf{ZF}$ proves that there are lots of non weakly reachable stages, even according to the stronger definition. In fact, $\mathsf{ZF}$ proves that the class of indices of such stages is a club.

The key point is that the indices of levels of $V$ which are not reachable from below form a club (this is easiest to see if we use a simpler definition of reachability: $V_\alpha$ is reachable from below iff $\alpha$ is in bijection with some $S\in V_\alpha$). To see this, think about iterating the function $$\xi\mapsto\sup\{\aleph^*(V_\zeta): \zeta<\xi\}.$$

Now $\mathsf{ZF}$ proves$^*$ that, given any club $C$, there is some ordinal $\theta\in C$ such that $\theta\cap C$ has ordertype $\theta$; this is a standard exercise. Applying this to the club of indices of non-reachable-from-below levels of $V$ gives the desired result for the second sense of weak reachability; for the first sense, just consider the intersection of $C$ with the club of ordinals satisfying $\vert\alpha\vert=\vert V_\alpha\vert$.

And in fact this really shows that $\mathsf{ZF}$ proves that there is a club of such ordinals.

$^*$Strictly speaking, $\mathsf{ZF}$ proves a theorem scheme: for each formula $\varphi$, $\mathsf{ZF}$ proves "If $\varphi$ defines a club of ordinals, then there is some ordinal $\theta$ such that $\varphi(\theta)$ and the set of ordinals $<\theta$ satisfying $\varphi$ has ordertype $\theta$."


答:
Firstly,lets get rid of p,which appears everywhere:

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

Now,lets denote $a=1-p$,where $a$ is an integer.

Now denote with $S=\sum_{k=1}^{n}a^{k-1}$

Lets calculate the product between $a$ and $S$:

$aS=\sum_{k=1}^{n}a^{k}$

$aS-S=\sum_{k=1}^{n}a^{k}-S$

Now lets subtract $S$ from the sum:

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

$(a-1)S=a^n-1$

$S=\frac{a^n-1}{a-1}$

$S=\frac{1-a^n}{1-a}$

Now all that we have to do is to replace a with $p-1$

$S=\frac{1-(1-p)^n}{1-(1-p)}$

$pS=p\frac{1-(1-p)^n}{1-(1-p)}$

$p\frac{1-(1-p)^n}{1-(1-p)}=pS=p\sum_{k=1}^{n}(1-p)^{k-1}=\sum_{k=1}^{n}p(1-p)^{k-1}$

Hope this helps.If you have any questions,feel free to ask me.

Title: How do you go from the first step where the summation of $(1-p)^{k-1}$ transform to the fraction

Body:
$$\sum_{k=1}^n p(1-p)^{k-1}=p \frac{1-(1-p)^n}{1-(1-p)}$$

I don't know how to simplify the summation. Please help me. Thank you so much.


问:
Meaning of Euler Reflection Formula
A book that I am reading says that Euler reflection formula $$\Gamma (x)\Gamma (1-x)=\frac{\pi}{\sin \pi x}$$, in a sense, shows that $1/\Gamma(x)$ is half of the sine function.

Does anyone understand what the book is talking about?

答:
I think the book meant analytic structure. The right side has simple poles at $x\in\mathbb{Z}$, $\Gamma(x)$ at $x\in\mathbb{Z}_{\le 0}$ and $\Gamma(1-x)$ at $x\in\mathbb{Z}_{>0}$. So to prove the identity, it suffices to check coincidence of the residues and behaviour at infinity.

One may also think about comparing product representations:
\begin{align}
\frac{1}{\Gamma(x)}&=x\prod_{n=1}^{\infty}\frac{1+\frac{x}{n}}{(1+\frac{1}{n})^x},\\
\sin\pi x &= \pi x \prod_{n=1}^{\infty}\left(1+\frac{x}{n}\right)\left(1-\frac{x}{n}\right).
\end{align}

This picture has further generalizations - for example, $q$-deformed gamma function is in the same sense a "half" of the Jacobi theta function.

Title: Meaning of Euler Reflection Formula


问:
Is $\inf \{P(|X| > \varepsilon) : E[X] = 0, \operatorname{Var}(X) = 1\} = 0$ for all $\varepsilon > 0$?
If $\varepsilon > 0$, is it true that the infimum of $\{P(|X| > \varepsilon) : E[X] = 0, \operatorname{Var} (X) = 1\}$ is zero? That is, for every $a > 0$, can we always find a random variable $X$ with zero mean and unit variance that satisfies $P(|X| > \varepsilon) < a$?

My first attempt to prove this statement was by using the Chebyshev's inequality: if $E[X] = 0$ and $\operatorname{Var} (X) = 1$, then $E[X^2] = \operatorname{Var} (X) + E[X]^2 = 1$, so we have
$$P(|X| > \varepsilon) \leq \frac{1}{\varepsilon^2}.$$
However, this inequality seems to be unhelpful, especially if $\varepsilon \leq 1$. Moreover, the right hand side does not even depend on $X$ or any other variable, so I cannot make the upper bound of $P(|X| > \varepsilon)$ to approach zero in this inequality.

Next, I tried to construct a sequence $\{X_n\}_{n\in\mathbb{N}}$ of random variables based on the uniform and U quadratic distributions. The idea is to reduce the support of $X_n$ to be less than $[-\varepsilon, \varepsilon]$ as $n \to \infty$, so that $P(|X_n| > \varepsilon) = 0$ eventually. More precisely, each $X_n$ has a probability density function defined by
$$f_{X_n} = \frac{2n+1}{2a_n^{2n+1}}x^{2n} \quad (x \in [-a_n,a_n]),$$
where $a_n = \sqrt{1+\frac{2}{2n+1}}$, and zero otherwise (this formulation ensures that $E[X_n] = 0$ and $\operatorname{Var}(X_n) = 1$ for all $n$). Unfortunately, $a_n$ converges to $1$ in this case, and $P(|X_n| > \varepsilon) \neq 0$ for $\varepsilon < 1$.

After these attempts, I began to doubt the validity of this proposition. After all, wouldn't a random variable with variance of $1$ surely "spread over" to the value of $1$ if it is centered at $0$? This is only my intuition and I have yet to prove this. But if that's the case, then $P(|X| > \varepsilon)$ surely can't approach $0$ for $0 < \varepsilon < 1$.

Any help on proving or disproving the statement in my question is appreciated.

答:
You can use a random variable $X$ that only takes two possible values (one with large and one with small absolute value):

Let $a>0$ and set $p=\frac{1}{1+a^2} \in (0,1)$. Now define a random variable $X$ such that
$$\begin{align*}P(X=a) &= p \\
P(X=-\frac{1}{a}) &= 1-p
\end{align*}$$

Then
$$E[X] = ap+(-\frac{1}{a})(1-p) = \frac{a}{1+a^2} - \frac{1}{a}\cdot\frac{a^2}{1+a^2} = 0 $$
and
$$\operatorname{Var}(X) = a^2 p + (-\frac{1}{a})^2 (1-p) =
\frac{a^2}{1+a^2} + \frac{1}{a^2}\cdot\frac{a^2}{1+a^2} = 1$$
so $X$ has zero mean and unit variance.

Now let $a \to \infty$ so that $p \to 0$ and $P(|X|>\varepsilon) \to 0$ for any fixed $\varepsilon > 0$.

Note that you cannot have $P(|X|>\varepsilon) = 0$ exactly, when $\varepsilon < 1$, because if $X$ only takes values in $[-\varepsilon, \varepsilon]$ and $E[X] = 0$, then
$$\operatorname{Var}(X) = E[X^2] \leq \varepsilon^2 < 1$$
and you cannot have unit variance.


问:
the distribution of registered events in Poisson process
Events, occurring according to a Poisson process with rate $\lambda$, are registered by a counter. However, each time an event is registered the counter becomes inoperative for the next  $b$ units of time and does not register any new events that might occur during that interval. Let  $R(t)$ denote the number of events that occur by time $t$ and are registered.

For $t\ge (n-1)b$, find $\mathbb{P}(R(t)\ge n)$

I found a comment from another post, saying that the distribution of $R(t)$ is the identical to $N(t-(n-1)b)$. I don't see the equivalence between these two distributions. Any help would be appreciated!

答:
By the strong Markov property and time homogeneity of $N$ (alternatively, independent and identically distributed interarrival times$^\star$), the jump times $(T_n)_{n\in\mathbb{N}}$ of $R$ are given by
\begin{align*}
T_1 &= S_1&& \\
T_2 &= T_1 + b + S_2&& \\
T_3 &= T_2 + b + S_3&& \\
&\,\,\,\vdots&&
\end{align*}
where $(S_n)_{n\in\mathbb{N}}$ are independent and identically distributed with $S_1 \sim \text{Exp}(\lambda)$.

Note that $T_n = S_1 + \sum_{i=2}^n (S_i + b)$ for $n\geq2$. Now, with $n\in\mathbb{N}$ and $t\geq0$,
\begin{align*}
\mathbb{P}(R(t) \geq n)
&=
\mathbb{P}\!\left(T_n \leq t\right) \\
&=
\mathbb{P}\!\left(\sum_{i=1}^n S_i \leq t - (n-1)b\right)\!.
\end{align*}
Let $(\tau_n)_{n\in\mathbb{N}}$ denote the jump times of $N$. Since $(S_n)_{n\in\mathbb{N}}$ follow the same distribution as the interarrival times $\tau_{n} - \tau_{n-1}$ of $N$ (under the convention $\tau_0 := 0$), we find that
\begin{align*}
\mathbb{P}\!\left(\sum_{i=1}^n S_i \leq t - (n-1)b\right)
=
\mathbb{P}(\tau_n \leq t - (n-1)b)
=
\mathbb{P}(N(t-(n-1)b)\geq n)
\end{align*}
for $t\geq (n-1)b$. Collecting results, we find that
\begin{align*}
\mathbb{P}(R(t) \geq n) = \mathbb{P}(N(t-(n-1)b) \geq n)
\end{align*}
for $n\in\mathbb{N}$ and $t\geq (n-1)b$.

$^\star$Let me know if additional details or explanations are needed; heuristically, you may think of the process starting anew at times $T_n + b$.


问:
Finding no of ways to select identical objects
If there are 10 notes of 10 dollars . Then no of ways to take 1 note is 10.I.e $10C_1.$

but I think it should be 1 since all notes are identical

$\frac{10C_1}{10}$

Why am I wrong ?

答:
I think you are right. The binomial coefficient is used when the objects are distinct. Here you have just one object {10}, e.g. 1 choose 1 is 1.

Equally, you are doing 10 choose 0 as there is no choice to be made. Or, if you have {10,10,...,10}, and do 10 choose 1, you must realize everything is the same and divide what you get by 10, again leading to 1.

Equivalently imagine you choose the first number out of a rearrangement of 10 objects as your method of choosing the number. The answer is the binomial coefficient 10 choose 10 slots for the 10 objects. They are identical so after you’ve chosen the slots they all fall in place. Still, answer is 1.


问:
Approach to test convergence of $\int_1^\infty \frac{\sqrt{x+1}}{x^2} dx$
The following improper integral converges:
$\int_1^\infty \frac{\sqrt{x+1}}{x^2} dx$

This integral CAN be evaluated, however is there a simpler way to test for convergence/divergence without evaluating it? I tried to use the comparison test, but could not come up with a suitable function.

答: The $x\approx1$ region has a finite integrand; for $x\gg1$, the integrand is asymptotic to $x^{-3/2}$.


答: Since$$\lim_{x\to\infty}\frac{\frac{\sqrt{x+1}}{x^2}}{x^{-3/2}}=1$$and since $\displaystyle\int_1^\infty x^{-3/2}\,\mathrm dx$ converges, then so does your integral.

Title: Approach to test convergence of $\int_1^\infty \frac{\sqrt{x+1}}{x^2} dx$

Body:
The following improper integral converges:
$\int_1^\infty \frac{\sqrt{x+1}}{x^2} dx$

This integral CAN be evaluated, however is there a simpler way to test for convergence/divergence without evaluating it? I tried to use the comparison test, but could not come up with a suitable function.


答:
Comparison test as follows

$$\int_1^\infty \frac{\sqrt{x+1}}{x^2} dx<
\int_1^\infty \frac{\sqrt{2x}}{x^2} dx=\sqrt2
 \int_1^\infty \frac{1}{x^{3/2}} dx=2\sqrt2
$$


问:
isometry group of double torus
Consider the double-torus, i.e., the orientable genus-$2$ surface. By embedding it into $\mathbb{R}^3$, we can get an induced metric from the usual Euclidean metric.

What's the isometry group of this double-torus? How does one think of it?

答:
Generically, the isometry group will be trivial.

In any case, the isometry group will be finite (unlike the genus zero or one case where a surface of revolution is possible).  See, for example, the book Finite groups of mapping classes of surfaces by Zieschang.

The fact that your surface is embedded in three-space should give a further non-trivial restriction on the allowable isometry groups.  Points realising the diameter will be permuted by any isometry.  This should prevent the surface from turning "inside out".

Title: isometry group of double torus


问:
Show that the sequence defined as $x_{n+1}=\frac{1}{1+x_n}$ converge
Show that the sequence $(x_n)$ defined as $x_{n+1}=\frac{1}{1+x_n},x_0=0$ converge. My attempt is to show that $(x_n)$ is Cauchy so i would like to have a feedback on my proof,please.

First of all we show by induction that $2>x_n\ge0$. The previous inequality is true for $n=0$ so we can suppose that it works up to a certain $n$. We want to proof right now that it holds fot $n+1$ and so show that $2>x_{n+1}\ge0$:

$2>x_{n+1}\ge0 \iff 2>\frac{1}{1+x_n}\ge0$.

It is trivial that $x_{n+1}\ge0$ by induction hypothesis (because $x_n\ge0$). Moreover, as $2>x_n\ge0$, we have that

$x_{n+1}=\frac{1}{1+x_n}\le1<2$. So we can conclude that $2>x_n\ge0 \ \forall n\ge0$

Then, $\forall p\ge2$ we have:

$|x_p-x_{p-1}|=\Big|\frac{x_{p-2}-x_{p-1}}{(1+x_{p-1})(1+x_{p-2})}\Big|\le 1 \cdot|x_{p-2}-x_{p-1}|=...=1^{p-1}|x_0-x_1|=1^{p-1}$

We show now that $(x_n)$ is Cauchy. So, by définition of Cauchy sequence, we have to show that $\forall \epsilon>0 \ \exists N \ \forall m,n>N$: $|x_m-x_n|<\epsilon$. But,

$|x_m-x_n|\le|x_m-x_{m-1}|+|x_{m-1}+x_{m-2}|+...+|x_{n+1}-x_n|\le(m-n-1)$

So $(x_n)$ is Cauchy

I am not reall sure if this proof holds, because i think that in case where $m=n-1$ it doesn't work,or...?

答:
Let $x_{1}>0$ and $\displaystyle x_{n+1}=\frac{1}{1+x_{n}}$. Notice that $x_{n}>0$ (by induction) and that :
$$
|x_{n+2}-x_{n+1}|=\left|\frac{1}{1+x_{n+1}}-\frac{1}{1+x_{n}}\right|=\frac{|x_{n}-x_{n+1}|}{(1+x_{n+1})(1+x_{n})}\leq\frac{4}{9}|x_{n}-x_{n+1}|
$$
Hence, $x_{n}$ is a contraction and hence it is Cauchy, and so converges by completeness of $(\mathbb{R},|.|)$. Let $L$ be its limit. We have that $\displaystyle L=\frac{1}{1+L}$ that is :
\begin{align*}
0&=L^{2}+L-1\\&=(L+0.5)^{2}-0.25-1\\&=\left(L-\frac{-1+\sqrt{5}}{2}\right)\left(L-\frac{-1-\sqrt{5}}{2}\right)
\end{align*}
We obtain that either $\displaystyle L=\frac{-1+\sqrt{5}}{2}$, or $\displaystyle L=\frac{-1-\sqrt{5}}{2}$. However, since $L\geq0$ and since $x_{n}>0$, we conclude that :
$$
\lim_{n\to\infty} x_{n}=\frac{-1+\sqrt{5}}{2}
$$
where $\displaystyle\varphi:=\frac{-1+\sqrt{5}}{2}$ is the golden ratio.

As $\text{Theo Bendit}$ mentioned, What's wrong with your proof is that it  fails to make $\left|x_{m}-x_{n}\right|$ small.


问:
Combinatorics on seats and circular tables
How many ways are there to arrange 6 people around a circular table with 7 seats? (Two seatings are considered the same if one is a rotation of the other.)?

I know that if it was just 6 seats it would be $\frac{6!}{6}$ but with 7 seats??? I'm not too sure about this one

答:
The empty seat 'fixes' the arrangement and so you simply have $6!$ possibilities.

An alternative answer

Let the invisible man join the dinner party. Then there are $\frac{7!}{7}$ possibilities.


问:
Lebesgue integration and countable partitions
The book "A Primer of Lebesgue Integration" by H.S. Bear defines Lebesgue integration through lower and upper sums $L(f,P) = \sum m_i\mu(E_i)$ and $U(f,P)=\sum M_i\mu(E_i)$ where infinite countable partitions are allowed.

The typical definition of Lebesgue integration that one encounters involves the supremum of simple functions. These simple functions are like lower sums. However, they are finite linear combinations and not infinite linear combinations.

When approaching Lebesgue integration through upper and lower sums why is it necessary to consider infinite countable partitions when simply functions seem to deal with only finite partitions?

答: After thinking about my own question I think the example $f=1/x^2$ highlights the problem with using simple functions for upper sum approximations. There is no problem in using simple functions from below to approximate $\int_1^\infty 1/x^2dx$ from below. But a simple function $\phi\ge f$ approximating from above will necessarily have infinite integral because $\phi$ will not be able to take on arbitrarily small values to well approximate $1/x^2$ for large $x$.</p>


问:
Sequence/List Formula
Consider the following sequence $\{1,1,2,1,2,3,4,1,2,3,4,5,6,7,8,1,...,16,1,...,32,1...\}$; it is the list of integers $1$ to $2^0$, then $1$ to $2^1$, then $1$ to $2^2$, etc. Let $f(n)$ return the $n$th entry in the sequence.

What is a closed form for $f(n)$? Currently I've just been building the sequence as a list and using the $n$th entry... but an analytical expression would be quite helpful. So far what I have is:
\begin{equation}
f(n)=j\quad \text{ if }\quad n=2^{m+\lceil \log_2 j\rceil}+j-1
\end{equation}
where $m$ is a positive integer or $0$. It's a little small but we're taking the ceiling of $\log_2 j$ in the above equation. This question has been bothering me for a while because it seems like I should've gotten this immediately... but I couldn't do it!

答:
Analytical, With the final value of in this case $2^{d-1}$ at $f(2^d-1)$

$$f(n)=\sum_{d=1}^{\infty}\bigg( \bigg( \sum_{m=1}^{d-1}2^{d-m-1}\sum_{j=2^{d-1}-2^{d-m}}^{2^{d-1}-2^{d-m-1}} \frac{1}{2^d}\sum_{k=0}^{2^d-1} e^{\frac{2 i \pi k (n+j)}{2^d}}\bigg) + \sum_{k=0}^{2^d-1}e^{\frac{2 i \pi k (n+2^{d-1})}{2^d}}\bigg)$$

For practical purpose you can stop after 2^d>=n ofcourse.

How to create repeated integers see also:
Explicit sum formula for repeated integers

Title: Sequence/List Formula


答:
You can simply consider the emty seat as the $7-th$ person which is seating at the table, so the answer is $\frac{7!}{7}=6!$

Note:You can do this because there is one only emty seat.If there would be two,or more, this trick won't work anymore,because then,you would have more persons that are identical.

Lets solve,for example,if there are 6 people and 10 seats.
The first person  has to choose between all 10 seats.The second one,gets to choose between the remaining 9 seats so the answer is $\frac{10*9*8*7*6*5}{10}$.
It is divided by 10 because we count every seating 10 times, because if one seating is obtained from the rotation of other,it is considered the same and we can rotate a seating in 10 total ways

Title: Combinatorics on seats and circular tables

Body:
How many ways are there to arrange 6 people around a circular table with 7 seats? (Two seatings are considered the same if one is a rotation of the other.)?

I know that if it was just 6 seats it would be $\frac{6!}{6}$ but with 7 seats??? I'm not too sure about this one


问:
Proving if a function is continuous and not one-one then it has many such points.
Let $g$ be a continuous function on an interval $A$ and let $F$ be the set of points where $g$ fails to be one-to-one; that is $$F = \{x \in  A : f(x)=f(y) \text{ for some $y \neq x$ and $y \in A$} \}$$
Show that if $F$ is non-empty then it is uncountable.

An attempt at the solution.

Since F is non-empty there exists some $c_1,c_2$ $\in A$ such that $c_1\neq c_2 $ and $g(c_1)=g(c_2)$.
Choose a point $z$ between $c_1$ and $c_2$ with $g(z) \neq g(c_1)$
Now $\forall$ $L$ satisfying $g(z)<L<g(c_1)$ or $g(z)>L>g(c_1)$ by $IVT$ there will be a point $K_1$ $\in (z,c_1)$ satisfying $g(K_1)=L$. Notice that $g(z)<L<g(c_2)$ or $g(z)>L>g(c_2)$ is also true for the same value of $L$ which by $IVT$ leads to the conclusion that there is also another point $K_2 \in (z,c_2)$  satisfying $g(K_2)=L$. Now how can i prove that there are uncountable number of points? Am i correct upto here?

答:
You are almost correct, and almost there. If you say: let $z \in [c_1, c_2]$ s.t. $g(z) \neq g(c_1)$, you assume that such a point exists. You should remark that you can assume this without loss of generality, because if this is not true, then $g(z) = g(c_1)$ always holds so that $[c_1, c_2] \subset F$ and we are done.

You can then assume without loss of generality that $g(z) > g(c_1)$ (the other case is identical). You are right that for any $L$ strictly between $g(z)$ and $g(c_1)$, you can deduce by the IVT that there exist points strictly to the left of $z$ and strictly to the right of $z$ that are mapped to the same point. Since you can do this for any $L$, this actually implies that $[c_1, c_2] \backslash \{z\} \subset F$, so you are also done.


问:
Can Infinitely Many Zero Vectors be Summed?
Let $V$ be an infinite dimensional vector space $V$ with a basis $\{e_\alpha\}_{\alpha \in A}$ and $\{E_\alpha\}_{\alpha \in A}$ a corresponding set of subspaces (where $E_\alpha = span(e_\alpha)$).

Then V has a representation as the direct sum, $V = \oplus_{\alpha \in A} E_\alpha$.

I.e., each $v \in V$ is a finite sum of vectors $v_\alpha \in V_\alpha$

I have seen this also expressed as each $v \in V$ = $\Sigma_{\alpha \in A} v_\alpha$ where only finitely many of the $v_\alpha \ne 0$.

Lets assume that V has no norm or inner product so there is no possibility of defining an infinite sum by convergence. Is it valid to say that the infinite number of zero vectors sum to zero ?

Besides idle curiosity, the question gains relevance when considering the inclusions $\{i_\alpha\}_{\alpha \in A}$ and projections $\{p_\alpha\}_{\alpha \in A}$ associated with the direct sum.
I would like to say that $v = \Sigma_{\alpha \in A} i_\alpha \circ p_\alpha (v)$.

There is a well-answered question Why is $\infty \cdot 0$ not clearly equal to $0$? which looks at summing infinitely many zeroes. This generally seems to to be answered as it depending on context: I don't see this context mentioned, hence this post.

答:
It is valid to say whatever you want, as long as you define what you mean by it.  In this case, there is a common convention to define an infinite sum $\sum_{\alpha\in A} x_\alpha$ as long as $x_\alpha=0$ for all but finitely many $\alpha$, even in the absence of anything like a topology as would usually be needed to define infinite sums.  This convention is convenient so that you can refer to such sums without constantly needing to restrict the index set to different finite subsets.

(If you like, you could also consider this as a limit of finite partial sums with respect to the discrete topology--the finite partial sums are eventually constant since only finitely many of the terms are nonzero.)


问:
How do you show that a recurrence relation holds?
There's a problem that goes as follows:

Let $a_n$ be the number of binary strings of length $n$ in which there are no consecutive 0s. Show that $a_1 = 1, a_2 = 3$ and that $a_n = a_{n-1} + a_{n-2}$

I am not sure how exactly am I supposed to "show" the final bit. Proofs by induction involve the closed formula for $a_n$.

Is there a formal, general proof to this or do I have to show using words and possibly figures?

答:
In several cases in recurrence relationships, the recurrence is due to the nature of the objects which allow them to be put into disjoint classes, each of which has as number of elements the number of elements given by a previous number of the recurrence. Not all recurrences have such a straight forward interpretation, nor all objects to count can be put into a nice looking recurence, tho.

In the case at hand, if $S$ is the set of all these binary sequences there is a natural splitting into two subsets: $S_0$, $S_1$ are those sequences that start with $0$ and $1$, respectively. Clearly $S = S_0 \cup S_1$, and this union is a disjoint union, so that
$a_n = |S| = |S_0| + |S_1|$.

Now notice how if you already put a $1$ at the start, then you can put whatever valid combination (i.e. not two consecutive 0's allowed) after it and generate one of the sequences from $S$. So $S_1$ has as number of elements just $a_{n - 1}$.

On the other hand, this doesn't work immediately if you start with a $0$, since you cannot put immediately another $0$, since you would have $00...$ which is not valid, so by force, you put a $1$ to begin with $01...$ and then you have $n - 2$ slots to fill now in any way, as long as there are not two $0's$. Hence, $S_0$ has as number of elements $a_{n - 2}$.

You conclude from this $a_n = a_{n - 1} + a_{n - 2}$.


问:
Possible tricks to use for solving linear equation involving binomial coefficients
Consider these two linear equation:

$$ 2a \binom{18}{j-1} - 4b \binom{18}{j} = \binom{18}{j-2}$$
True for $j=16,17$, find $a$ and $b$

This came up as part of a problem I was doing.. expanding everything out and doing it really ugly. Are there any tricks I could use to solve the system fast considering they are involving binomial coefficients? One thing I've considered doing is adding both equations and pascal's identity but not sure if that actually makes anything easier.

答:
$$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{n+1-r}{r}$$
Thus, if you divide both side of the equation by $\binom{18}{j-1}$ , then you can simplify it and that will decrease your calculations.


问:
Implicit differentiation: $x^2y - 2x^3 - y^3 + 1 = 0$
Hi I'm stuck on one problem in my study guide. It's the only one with an = 0 at the end of it.

Differentiate:

$x^2y - 2x^3 - y^3 + 1 = 0$

答:
If you don't feel comfortable with messing around with $dx$ and $dy$ like DonAntonio does in his answer, you can also view this simply as an application of the chain rule. Say you have an equation $$
  F(x,y) = 0 \text{.}
$$

If you let $x$ be the independent $y = y(x)$ be the dependent variable, this becomes $$
  F(x, y(x)) = 0
$$
Applying chain rule yields $$\begin{aligned}
  &&F_x + F_yy' &= 0 \\
  \Rightarrow
  &&y' &= -\frac{F_x}{F_y} \quad\text{if $F_y \neq 0$}
\end{aligned}$$
where $F_x$, $F_y$ are the partial derivatives of $F$ with respect to $x$,$y$.

For your problem $F(x,y) = x^2y−2x^3−y^3+1=0$, you get $$\begin{aligned}
  &&F_x &= 2xy - 6x^2 \\
  &&F_y &= x^2 - 3y^2 \\
  \Rightarrow
  &&y' &= -\frac{F_x}{F_y} = \frac{6x^2 - 2xy}{x^2 - 3y^2}
\end{aligned}$$


问:
Math Help on number theory Modular Arithmetic Systems
Can Someone please help me with this question it is coming up in my AOPS Alcumus and I don't know how to do it can someone show me please

Suppose that $x$ is an integer that satisfies the following congruences:\begin{align*}
3+x &\equiv 2^2 \pmod{3^3} \\
5+x &\equiv 3^2 \pmod{5^3} \\
7+x &\equiv 5^2 \pmod{7^3}
\end{align*}What is the remainder when $x$ is divided by $105$?

答:
HINT

Notice that $105=3\times 5 \times 7$. We know that

$\begin{align*}
x &\equiv 2^2\equiv 1 \pmod{3} \\
x &\equiv 3^2\equiv 4 \pmod{5} \\
x &\equiv 5^2\equiv 4 \pmod{7}
\end{align*}$

Can you complete it now using the Chinese Remainder Theorem?

https://en.wikipedia.org/wiki/Chinese_remainder_theorem

 Notice that $x \equiv1 \equiv 4\pmod{3}$ and so $x$ is $4$ modulo  $3,5$ and $7$. Hence $x=105k+4$ and the remainder is $4$.


问:
Solving $n\log{n}=10^9$
The logarithm here is of base $2$. I tried to solve the following equation to find out $n$, where $n \in \mathbb{R}$,

$$n\log{n}=10^9$$

Edited: My naive approach was to raise both sides to exponential $2$,

$$2^{n\log{n}}=2^{10^9}$$
$$2^{\log{n^n}}=2^{10^9}$$
$$n^n=2^{10^9}$$

I got stuck here. I saw another solution suggesting to use Lambert $W$ function,

$$n=\frac{10^9\times \log{2}}{W(10^9\times \log{2})}$$

Question 1: Can you please tell me if it's possible to solve it without Lambert function as I did? But I got stuck at $n^n=2^{10^9}$, so how to proceed from here?

Question 2: How the solution that used Lambert $W$ function get to that factor $n=\frac{10^9\times \log{2}}{W(10^9\times \log{2})}$?

答:
Trying to avoid Lambert, here is a strategy to get close to the solution.

You may know $2^{10}=1024\approx10^3$. From this it seems $2^{30}\approx10^9$. So lets look at $$x\log x=2^{30}$$ Note this $x$ will be larger than the original $n$ we are looking for. If $x=2^k$, then $$k2^k=2^{30}$$ $$\log k+k=30$$ We know $\log16+16=20$ and $\log32+32=37$ So in the solution, $\log k$ is between 4 and 5. We pretend linearity to take $\log k\approx4+\frac{10}{17}$. Then $$k\approx25+\frac{7}{17}$$ $$n\approx x\approx2^{25+7/17}$$

A calculator check gives $$\left(2^{25+7/17}\right)\log\left(2^{25+7/17}\right)\approx 1.134\cdot10^9$$ From here, putzing around with the exponent can show $$\left(2^{25+1/4}\right)\log\left(2^{25+1/4}\right)\approx 1.0075\cdot10^9$$

Title: Solving $n\log{n}=10^9$


问:
How to prove the two dimensional distribution is Delta function?
Let $f(t,x)=\frac{Y(t)}{a(\pi t)^{\frac{1}{2}}}e^{-\frac{x^2}{4t}},$ $Y(t)$ is Heaviside function. Prove that $$\partial_tf(t,x)-\partial^2_xf(t,x)=\delta_{(0,0)}$$ in distribution $\mathcal{D}'(R^2).$ I am familar with one dimensional,but I have some problem to compute the two dimensional distribution. It would be nice if you can give me some trick. Do I need use tensor product?

答:
Hint $f(t, x)\in L^1_{\text{loc}}({\mathbb R}^2)\subset {\cal D}'({\mathbb R}^2)$ because $f\in L^1(I_t\times {\mathbb R_x})$ for every bounded interval $I$. Let $\phi\in {\cal D}({\mathbb R}^2)$, one has
\begin{equation}
\langle\partial_t f - \partial^2_x f, \phi\rangle = \int_{\mathbb R^2}f(t, x)(-\partial_t\phi(t, x) - \partial^2_x\phi(t, x)) d t d x
\end{equation}
\begin{equation}
= \lim_{\epsilon\to 0^+}\int_{[\epsilon,+\infty)\times {\mathbb R_x}}f(t, x)(-\partial_t\phi(t, x) - \partial^2_x\phi(t, x)) d t d x
\end{equation}
Cut the integral in two integrals, integrate by parts in order to remove the derivatives on $\phi$.

Use the derivatives
\begin{equation}
\begin{array}{rcl}{\partial }_{t} \left({t}^{{-1}/2} {e}^{{-\frac{{x}^{2}}{4 t}}}\right)&=&\left(\frac{{x}^{2}}{4} {t}^{{-5}/2}-\frac{1}{2} {t}^{{-3}/2}\right) {e}^{{-\frac{{x}^{2}}{4 t}}}\\
{\partial }_{x} \left({t}^{{-1}/2} {e}^{{-\frac{{x}^{2}}{4 t}}}\right)&=&-\frac{x}{2} {t}^{{-3}/2} {e}^{{-\frac{{x}^{2}}{4 t}}}\\
{\partial }_{x}^{2} \left({t}^{{-1}/2} {e}^{{-\frac{{x}^{2}}{4 t}}}\right)&=&\left(\frac{{x}^{2}}{4} {t}^{{-5}/2}-\frac{1}{2} {t}^{{-3}/2}\right) {e}^{{-\frac{{x}^{2}}{4 t}}}
\end{array}
\end{equation}


答:
For $j=16,17$ and using $\binom{18}{r}=\binom{18}{18-r}$, we have
$$2a\binom{18}{3}-4b\binom{18}{2}=\binom{18}{4}$$
$$2a\binom{18}{2}-4b\binom{18}{1}=\binom{18}{3}$$

Writing out each binomial coefficient and cancelling common terms (eg, $18\cdot 17$ cancels off in first one), we have
$$16a-6b=30 \tag{A}$$
$$51a-12b=136 \tag{B}$$

On $B-2A$ is obtained,
$$a=4 , \quad b =17/3$$

Title: Possible tricks to use for solving linear equation involving binomial coefficients

Body:
Consider these two linear equation:

$$ 2a \binom{18}{j-1} - 4b \binom{18}{j} = \binom{18}{j-2}$$
True for $j=16,17$, find $a$ and $b$

This came up as part of a problem I was doing.. expanding everything out and doing it really ugly. Are there any tricks I could use to solve the system fast considering they are involving binomial coefficients? One thing I've considered doing is adding both equations and pascal's identity but not sure if that actually makes anything easier.


问:
Solving $n! = 10^9$
What is the largest $n\in \mathbb{R}$ for which one can solve within one second a problem using an algorithm that requires $f (n)$ bit operations, where each bit operation is carried out in $10^−9$ seconds, with $f (n) = n!$?

The number of bits carried out in $1$ second is $\frac{1}{10^{-9}}=10^9$.

We can estimate that using the fact that,

$$12! \lt 10^9 \lt 13!$$

Question 1: Is there a more systematic approach to solve it? I tried to replace $ n!$ with its upper bound $n^n$, but I am not sure if this is correct as the solution for $n!$ might be less than the solution of $n!$ upper bound $n^n$, or the upper bound solution of $n^n$ is acceptable answer for $n!$ based on question requirement?

$$n^n = 10^9$$

答:
So first, if this were a "real-world" problem. Then the answer is 12 because $n$ should be an integer. (E.g. you can have a graph with 12 nodes, but not with 12.5 nodes.)

If we're trying to figure out for which real number $n$ do we have $n! = 10^9$ then we need to understand what that question even means because factorials are only defined for whole numbers. The most common extension of the factorial function to the reals (and complex numbers) is the Gamma function and we can use this (and a computer) to get an answer. The Gamma function is related to the factorial function by $\Gamma(n + 1) = n!$ when $n$ is a whole number.

So the question we want to ask our computer is: solve $\Gamma(n + 1) = 10^9$. But probably we should modify this to: solve $\log \Gamma(n + 1) = 9 \log 10$ since computers have good algorithms for computing $\log \Gamma(x)$ and $10^9$ is quite large so it might be harder to work with numbers at that scale.

If we do this, we get the answer $n \approx 12.2900691486\ldots$.

https://www.wolframalpha.com/input/?i=solve+loggamma%28n+%2B+1%29+%3D+9+log%2810%29

Addendum: so this method also gives a good way to compute the largest integer with $n! \le 10^9$ by rounding down. It's advantage is that if you have a computer algebra system with existing methods, you can program this in 1 line. E.g. in Mathematica, you could write

```<code>Floor[n] /. NSolve[{LogGamma[n + 1] == 9 Log[10] && n >= 0}, Reals]
</code>```

But now let's say we don't have methods already for computing log-gamma or numeric solvers. Then probably the next best thing is to program (or work by hand) some sort of binary search:

compute $10!, 20!, 30!, \ldots$ etc. until you have two values with $a! < 10^9 < b!$. In our case that would be $10! < 10^9 < 20!$. You can use a larger gap than $10$ if the number is really-really-really big.

then just repeatedly divide the interval in two: $10! < 10^9 < 15!$

$10! < 10^9 < 13!$

$11! < 10^9 < 13!$

$12! < 10^9 < 13!$

Binary search is generally quite fast. Here we have to compute only 6 factorials: 10, 20, 15, 13, 11, 12. If we did a linear search: 1!, 2!, 3!, ... it would take twice as long.

Title: Solving $n! = 10^9$

Body:
So first, if this were a "real-world" problem. Then the answer is 12 because $n$ should be an integer. (E.g. you can have a graph with 12 nodes, but not with 12.5 nodes.)

If we're trying to figure out for which real number $n$ do we have $n! = 10^9$ then we need to understand what that question even means because factorials are only defined for whole numbers. The most common extension of the factorial function to the reals (and complex numbers) is the Gamma function and we can use this (and a computer) to get an answer. The Gamma function is related to the factorial function by $\Gamma(n + 1) = n!$ when $n$ is a whole number.

So the question we want to ask our computer is: solve $\Gamma(n + 1) = 10^9$. But probably we should modify this to: solve $\log \Gamma(n + 1) = 9 \log 10$ since computers have good algorithms for computing $\log \Gamma(x)$ and $10^9$ is quite large so it might be harder to work with numbers at that scale.

If we do this, we get the answer $n \approx 12.2900691486\ldots$.

https://www.wolframalpha.com/input/?i=solve+loggamma%28n+%2B+1%29+%3D+9+log%2810%29

Addendum: so this method also gives a good way to compute the largest integer with $n! \le 10^9$ by rounding down. It's advantage is that if you have a computer algebra system with existing methods, you can program this in 1 line. E.g. in Mathematica, you could write

Floor[n] /. NSolve[{LogGamma[n + 1] == 9 Log[10] && n >= 0}, Reals]

But now let's say we don't have methods already for computing log-gamma or numeric solvers. Then probably the next best thing is to program (or work by hand) some sort of binary search:

compute $10!, 20!, 30!, \ldots$ etc. until you have two values with $a! < 10^9 < b!$. In our case that would be $10! < 10^9 < 20!$. You can use a larger gap than $10$ if the number is really-really-really big.

then just repeatedly divide the interval in two: $10! < 10^9 < 15!$

$10! < 10^9 < 13!$

$11! < 10^9 < 13!$

$12! < 10^9 < 13!$

Binary search is generally quite fast. Here we have to compute only 6 factorials: 10, 20, 15, 13, 11, 12. If we did a linear search: 1!, 2!, 3!, ... it would take twice as long.


问:
support vs state space
In one of my examples, I saw that the state space for a binomial distribution is $({0,1,...,n})$. I then thought that for $X_1, X_2,...,X_n$ iid RVs, where $f_\theta(x_i)=\theta x_i^{\theta -1}$ for $x \in (0,1)$, and $0$ otherwise, the state space is $(0,1)^n$. However, it was said to be $\Bbb R^n$, meaning other values outside $(0,1)$ are also counted in the state space (which would make sense, since the state space is the set of values assumed by our RV), and that $(0,1)^n$ is rather the support. But, following this logic, how comes the state space for the binomial distribution isn't $\Bbb N$, where for all $N>n,$ the probability that the RV is equal to that value is zero? Or ss it because the binomial distribution isn't defined for $n+1$ onwards?

答:
I don't know a formal definition of "state space" for distributions that are not discrete, but for discrete distributions, in some cases the support is bigger than the state space. The support is always a closed set.

The support of the distribution taking values in $(0,1)^n$ that you describe is the closed set $[0,1]^n.$

Here is the definition: The support of a probability distribution taking values in $\mathbb R^n$ is the set of points whose every open neighborhood has positive probability.

Suppose a discrete probability distribution assigns positive probability to each of $1,\,1/2,\, 1/3,\,1/4,\,\ldots\,.$ Then $0$ is a member of the support since every open neighborhood of $0$ has positive probability.

Now suppose
\begin{align}
& \Pr(X=x)>0 \text{ for } x = 1,2,3,\ldots \text{ and } \sum_{x=1}^\infty \Pr(X=x) = 1 \\[8pt]
\text{and } & \Pr(Y=y\mid X=x) = \frac 1 {x-1} \text{ for } y = 1,\ldots,x-1
\end{align}
and consider the discrete probability distribution of the random variable $Y/X \in(0,1).$ The state space in $\mathbb Q\cap[0,1]$ and the support is the larger set $[0,1].$

Title: support vs state space


问:
Ring Homomorpism
Consider the polynomial ring $R[X]$ and the ring $M_2(R)$ of $2\times 2$ matrices with entries in R. We define a map $t: R[X] \to M_2(R)$ by

$$t(f) = \begin{pmatrix}f(0)&  f'(0)\\0 &  f(0)\end{pmatrix}$$

where $f'$ denotes usual derivative of $f$.

Show that t is a homomorphism. Determine the kernel and the image of t. Is ker(t) a principal ideal.

I have shown that t is a homomorphism, however am stuck on the rest of the question. Thank you for any help in advance.

答:
Hint: $ $ in $\,S = R[x]/(x^2)\,$ the linear map $\, g\to fg = (a+bx)g\,$ satisfies

$$\begin{eqnarray} (a+bx) 1 &=\,& a + bx \\ (a+bx) x &=\,& 0 + ax\end{eqnarray}$$

hence its matrix in the basis $\, \langle1,x\rangle\,$ is

$$ \left[\begin{array}{cc} a & b \\ 0 & a\end{array}\right] = \left[\begin{array}{cc} f(0) &\!\! f'(0)\\ 0 &\!\! f(0)\end{array}\right] $$

Title: Ring Homomorpism


问:
What is the difference between Quaternions and Bicomplex Numbers?
So, I know Quaternions are basically 4 dimensional Complex numbers, and the dimensions can double forever to Octonions, Sedinions, etc. I recently heard about bicomplex numbers, which are also sort of 4 dimensional complex numbers. I looked into it a bit, and I found that bicomplex numbers are a commutative version of quaternions. But I wanted to know how exactly that works. How is the multiplication defined? Also can that be extended to 8-Dimensional biquaternions, 16-D bioctonions, etc? And how are those defined exactly? I know with regular hypercomplex numbers, you lose properties as you go up in dimensions, so does the same thing happen with the bi- ones? If so does it just lose the property that it would normally lose one dimension lower? Or do they not lose properties at all? Any answers would be greatly appreciated.</p>

答:
Bicomplex numbers, or tessarines is a commutative algebra as other people said.

It has zero divisors. And I would say, being commutative is a lot more important than disadvantages from zero divisors!

Moreover, zero divisors are not a disadvantage at all, because it allows bicomplex numbers to generalize not only the concept of complex numbers, but also hyperbolic (split-complex) numbers. They are included in bicomplex numbers and have zero divisors.

There is nothing essentially bad regarding zero divisors: this means that we have some elements by which we cannot divide, but we anyway had one such real number: zero.

Unlike what the other answer said, of course there is norm (modulus) in bicomplex numbers.

They can be represented as real matrices of certain structure with usual matrix operations, so one can experiment with them in any computer algebra system that allows to apply analytic functions to matrices.

Images source:https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.527.356&rep=rep1&type=pdf


问:
Prove the existence of $n\in\mathbb{N}$ such that $0<1/n<\epsilon$
I'm seeking to prove that for any $\epsilon>0$, there exists an $n\in\mathbb{N}$ such that
$$
  0<\frac{1}{n}<\epsilon.
$$

I think that—in some form—I'm going to need to use the fact that for any $x\in\mathbb{R}$, we can find $a\in\mathbb{N}$ such that $a\geq x$ (i.e. the Archimedean property).

Or perhaps I could construct a set with an infimum of zero and somehow use that to generate an element between zero and $\epsilon$.

What's the most sensible approach here?

答:
Consider $n$ equal to $\lceil\frac{1}{\epsilon} \rceil$ (that means the smallest natural number that is greater or equal to $\frac{1}{\epsilon}$). Then you can easily see that
$$0<\frac{1}{n} = \frac{1}{\lceil\frac{1}{\epsilon}\rceil} <\frac{1}{\frac{1}{\epsilon}}=\epsilon,$$
so for any $\epsilon\in\mathbb{R}$, you can just pick $n\in\mathbb{N}$ that way and the condition is verified (for that $n$ and for any other natural number grater that $n$).


答:
Applying the identity : $$\frac{\binom{n}{r}}{\binom{n}{r-1}}=\frac{n+1-r}{r}$$

On

$$ 2a \binom{18}{j-1} - 4b \binom{18}{j} = \binom{18}{j-2}$$

First divide:

$$ 2a  - 4b \frac{\binom{18}{j}}{\binom{18}{j-1}} = \frac{\binom{18}{j-2}}{\binom{18}{j-1}}$$

Apply the Id

$$2a - 4b (\frac{19-j}{j} )= (\frac{j-2}{19-j})$$

This is much better because, the solution's of the equation directly fall out of differencing the above expression over j=16 and j=17


答:
The lambert $W(M)$ is by definition the unique real solution of $x$ if $xe^x =M$.

If we have to solve $w^w = K$ and $w> 0$ we can use this to solve

$$\ln w^w = \ln K$$

$$w\ln w = \ln K$$

$$e^{\ln w} \ln w = \ln K$$

$$\ln w = W(\ln K)$$

$$w = e^{W(\ln K)}$$

We are almost done but $e^{W(****)}$ is considered not simplified enough.

Note by definition that $W(M) e^{W(M)} = M$ we will always have $e^{W(M)} = \frac M{W(M)}$.  And that is considered simplified enough.

$$w = \frac {\ln K}{W(\ln K)}$$

.....

So a general solution:

$w^w =K$ has solution $w = \frac {\ln K}{W(\ln K)}$.

Now your problem is:

$$n\log_2 n = 10^9$$

$$2^{n\log_2 n} = 2^{10^9}$$

$$n^n = 2^{10^9}$$

So $n = \frac {\ln (2^{10^9})}{W(\ln (2^{10^9}))} = \frac {10^9\ln 2}{W(10^9\ln 2)}$

Title: Solving $n\log{n}=10^9$

Body:
The logarithm here is of base $2$. I tried to solve the following equation to find out $n$, where $n \in \mathbb{R}$,

$$n\log{n}=10^9$$

Edited: My naive approach was to raise both sides to exponential $2$,

$$2^{n\log{n}}=2^{10^9}$$
$$2^{\log{n^n}}=2^{10^9}$$
$$n^n=2^{10^9}$$

I got stuck here. I saw another solution suggesting to use Lambert $W$ function,

$$n=\frac{10^9\times \log{2}}{W(10^9\times \log{2})}$$

Question 1: Can you please tell me if it's possible to solve it without Lambert function as I did? But I got stuck at $n^n=2^{10^9}$, so how to proceed from here?

Question 2: How the solution that used Lambert $W$ function get to that factor $n=\frac{10^9\times \log{2}}{W(10^9\times \log{2})}$?


问:
Why the radial delta function is not equal to the product of two 1D delta functions
The first proof that comes to my mind is that the units of left side and right side does not match. Delta functions take an input and spit out an output of units = 1/[units of input].

What would be the rigorous proof of this?

Statement to be proven: (also given below):

$$\delta \left ( r \right )\space \space  \neq \space \space \delta \left ( x \right )\delta \left ( y \right )$$

答:
Delta Function in Two Dimensions

The problem with this question is that the radial delta function is not $\delta(r)$, where $\delta$ is the usual $1$-dimensional delta function.

An approximation of $\delta(x)$ is $ne^{-\pi n^2x^2}$, so
$$
\begin{align}
\delta(x)\,\delta(y)
&=ne^{-\pi n^2x^2}ne^{-\pi n^2y^2}\tag{1a}\\
&=n^2e^{-\pi n^2r^2}\tag{1b}
\end{align}
$$
Now, let's see what the integral of $n^2e^{-\pi n^2r^2}$ is (in polar coodinates)
$$
\begin{align}
\int_0^{2\pi}\int_0^\infty n^2e^{-\pi n^2r^2}\,r\,\mathrm{d}r\,\mathrm{d}\theta
&=\int_0^{2\pi}\int_0^\infty\frac1{2\pi}\,e^{-u}\,\mathrm{d}u\,\mathrm{d}\theta\tag{2a}\\
&=1\tag{2b}
\end{align}
$$
Explanation:

$\text{(2a)}$: $u=\pi n^2r^2$

$\text{(2b)}$: evaluate the integrals

So, it seems that $n^2e^{-\pi n^2r^2}$ is an approximation of the radial delta function. Taking the limit says that $\delta(x)\,\delta(y)$ is the radial delta function.

The Radial Delta Function is not $\boldsymbol{\delta(r)}$

However, if we use $\delta(r)$ in place of $\delta(x)\,\delta(y)$, we get
$$
\begin{align}
\int_0^{2\pi}\int_0^\infty ne^{-\pi n^2r^2}\,r\,\mathrm{d}r\,\mathrm{d}\theta
&=\int_0^{2\pi}\int_0^\infty\frac1{2\pi n}\,e^{-u}\,\mathrm{d}u\,\mathrm{d}\theta\tag{2a}\\
&=\frac1n\tag{2b}
\end{align}
$$
which tends to $0$.

The Radial Dirac Delta Function
$$
\begin{align}
\int_0^{2\pi}\int_0^\infty\overbrace{\frac{n}{\pi r}e^{-\pi n^2r^2}}^{\large\frac1{\pi r}\delta(r)}\,r\,\mathrm{d}r\,\mathrm{d}\theta
&=\int_0^{2\pi}\int_0^\infty \frac1{\pi}e^{-\pi r^2}\,\mathrm{d}r\,\mathrm{d}\theta\tag{3a}\\
&=1\tag{3b}
\end{align}
$$
Thus, the radial delta function is $\frac1{\pi r}\delta(r)$.


问:
Let K1 and K2 be parallel lines with line K3 passing through both of them. Points A, H are not located on these lines.
Let $K_1$ and $K_2$ be parallel lines with line $K_3$ passing through both of them. Points $A$ and $H$ are not located on these lines. Points $C$ and $F$ are on line $K_3$. Draw $AG$, $AH$ and $AF$. Point $E$ is on line $AF$ such that $EC$ is perpendicular to $DC$. Find the measurement of the angle at point $H$ that is labeled as $x$.

I noticed that if we extend $CE$ and $AG$ so that they intersect at a point ($x$), $AXDC$ will be a cyclic quadrilateral. Same for $CH$ and $AG$.

答:
We have for $\triangle ECD$, by exterior angle sum,
$$\angle EDH = \angle CED + \angle ECD = 90+2\theta$$

So $\angle ADH = \angle EDH = 90+2\theta$.

For $\triangle ADH$, by exterior angle sum,
$$x=\angle DAH + \angle ADH = 90+2\theta+\theta=90+3\theta$$


问:
Explanation of inference rules with quantifiers
I'm referring to the inference rules listed at this page of Wikipedia (https://en.m.wikipedia.org/wiki/Sequent_calculus), under the section of $LK$ system. There are four rules for the quantifiers; however we have restrictions only for $\forall R$ and $\exists L$. The motivations for $\forall R$ are clear, because in order to conclude that $\forall x A(x) \Rightarrow B$, one must prove that $A(t) \Rightarrow B$ for a generic $t$ (i.e. not a variable used before). However why one can't say the same for the rule $\forall L$?

Plus, in the case of $\exists L$ I'm not sure to understand why we have the restrictions, and anyway I don't understand why we have it only in one case. Eventually, can anyone tell me what book (or material in general) I could use to clarify these things?

答:
To make this question self-contained, these are the rules in question:

$$\forall L: \frac{\Gamma, A[t/x] \vdash \Delta}{\Gamma, \forall x A \vdash \Delta}$$

$$\forall R: \frac{\Gamma \vdash A[y/x], \Delta}{\Gamma \vdash \forall x A, \Delta}$$

In the case of $\forall R$, you are trying to conclude $\forall x A$.  Therefore your assumption must be $A[y/x]$ where $y$ is a new variable, so that $y$ is a generic element, not a variable you have used before which may have some additional assumptions on it. Therefore there are restrictions necessary for $\forall R$.

However, in the case of $\forall L$, you start with $A[t/x] \Rightarrow \Delta$ and are trying to conclude $\forall x A \Rightarrow \Delta$. The $\forall x A$ appears as an assumption. Therefore no restriction is necessary, because you already know that $A$ holds for all $x$.  Therefore you can substitute any $t$ for $x$, and apply the original $A[t/x] \Rightarrow \Delta$ to conclude $\Delta$.

The situation for $\exists$ is similar.


答:
You have almost said it yourself if you accept the Archimedian property: Let $\epsilon > 0$. Then there exists $n \in \mathbb{N}$ such that

$$n > \frac{1}{\epsilon} \implies \epsilon > \frac{1}{n}.$$

Title: Prove the existence of $n\in\mathbb{N}$ such that $0<1/n<\epsilon$

Body:
I'm seeking to prove that for any $\epsilon>0$, there exists an $n\in\mathbb{N}$ such that
$$
  0<\frac{1}{n}<\epsilon.
$$

I think that—in some form—I'm going to need to use the fact that for any $x\in\mathbb{R}$, we can find $a\in\mathbb{N}$ such that $a\geq x$ (i.e. the Archimedean property).

Or perhaps I could construct a set with an infimum of zero and somehow use that to generate an element between zero and $\epsilon$.

What's the most sensible approach here?


问:
Lyapunov function for a non linear 3 dimensional system
How it is possible to find a Lyapunov function for the following system?
\begin{cases}
  \dot {x_1}=x_2+x_3 & \\
         \dot {x_2}=-\sin x_1-x_3 & \\
 \dot {x_3}=-\sin x_1+x_2 &
\end{cases}
I have tried with $V(x)=\frac{1}{2}x^TPx+\int_0^{x_1} \sin{(s)} ds$ but I can't find out the right elements of P in order to obtain $\dot{V}<0$. Can you help me?

答:
The Lyapunov function (and also the first integral) of the system is
$$
V(x_1,x_2,x_3)= 1-\cos x_1+\frac12 x_2^2+\frac12 x_3^2.
$$
Indeed,
$$
\dot V(x_1,x_2,x_3)= \sin x_1\cdot x_1+x_2\dot x_2+x_3\dot x_3
$$
$$
= \sin x_1 \cdot (x_2+x_3)+x_2(-\sin x_1-x_3)+x_3(-\sin x_1+x_2)=0
$$
Since $V$ is positive definite and $\dot V$ is nonpositive, the origin is stable. It is not asymptotically stable because $V$ remains constant on the solutions of the system (and doesn't tend to zero).


问:
Help me with this this system of equations
Help me with this  system of equations
$$a+b = 3 -c$$
$$\frac{1}{a}+\frac{1}{b}= \frac{5}{12}-\frac{1}{c}$$
$$ a^3+b^3 = 45 -c^3$$</p>

答:
We have

$$\begin{align}
a+b+c & = 3 \\
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} &= \frac{5}{12} \\
a^3 + b^3 + c^3 & = 45 \\
\end{align} $$

The second equation tells us that $\frac{ab+bc+ca}{abc} = \frac{5}{12} $, or that $12(ab+bc+ca) = 5(abc)$.

Using $(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a) = 3(3-a)(3-b)(3-c)$, we get that
$27-45 = 3[27-9(a+b+c) + 3(ab+bc+ca) - abc]$, or that

$$-6 = 3(ab+bc+ca) - abc.$$

Hence, we get that $ab+bc+ca = -10 $, $abc = -24 $. by solving the simultaneous equations.

Hence, using Vieta, we get that $a, b, c$ are (permutation of) roots to the cubic

$$x^3 - 3x^2 - 10 x + 24 = 0. $$

This cubic is $(x-2)(x-4)(x+3) = 0$, and thus we have $\{a,b,c\} = \{2, 4, -3\} $.


答:
Note: that $0 < \frac 1n < \epsilon \iff n > \frac 1\epsilon > 0$

So you have to show:

If $\epsilon > 0$ then $\frac 1{\epsilon}$ exists (that's easy; $\mathbb R$ is a field so multiplicative inverses exist).

If $\epsilon > 0$ then $\frac 1{\epsilon} > 0$.  (You should have proven that already and can take that as a given without even acknowledging it is in question.)[The proof is If $\frac 1\epsilon \le 0$ and $\epsilon > 0$ we would have $1 =\frac 1\epsilon \cdot \epsilon \le 0\cdot \epsilon = 0$; a contradiction].

For any $M = \frac 1{\epsilon}\in \mathbb R$, there exists a natural number $n: n> M$.

And 3: (which says nothing more or less than "The Natural numbers are not bounded above") is the real heart of the issue.

Depending on the level of the book you may or may not be able to take that for granted.

But it is equivalent to the Archimedian principal.  $1 > 0$ so for any $M =\frac 1{\epsilon} \in \mathbb R$ there is an $n\in \mathbb N$ so that $n*1 > M$.

.....

To sum up:

$\epsilon > 0$.

So $\frac 1{\epsilon} > 0$.

So there exists an $n\in \mathbb N$ so that $n > \frac 1\epsilon$.

So $0 < \frac 1n < \epsilon$.

That's all you have to do.


答: Since$$\iint f(r,\,\theta)\delta(r)\delta(\theta)drd\theta=f(O)=\iint f(x,\,y)\delta(x)\delta(y)\underbrace{dxdy}_{rdrd\theta},$$we have the dimensionally homogeneous result $\delta(r)\delta(\theta)=r\delta(x)\delta(y)$. @MarkViola's comment has noted a more general finding we ca get by writing $f(p)$ for some point $p$ as two different double integrals.

Title: Why the radial delta function is not equal to the product of two 1D delta functions

Body:
The first proof that comes to my mind is that the units of left side and right side does not match. Delta functions take an input and spit out an output of units = 1/[units of input].

What would be the rigorous proof of this?

Statement to be proven: (also given below):

$$\delta \left ( r \right )\space \space  \neq \space \space \delta \left ( x \right )\delta \left ( y \right )$$


问:
Yet another Collatz generalization and conjecture
Observe that Collatz (the $\frac{3n+1}{2}$ version) can be written as: $$n_{i+1}:= \begin{cases}n_i+\left\lceil\frac{n_i}{2}\right\rceil\quad\text{ if odd }n_i \\
n_i-\left\lceil\frac{n_i}{2}\right\rceil\quad\text{ if even }n_i
\end{cases}$$

Looking at it this way gives the following pleasingly simple table for the difference between between consecutive terms, where $\Delta n := n_{i+1}-n_i$:

$$
\begin{array}{|l|l|} \hline
n_i&\Delta n\\ \hline
1&+1 \\ \hline
2&-1 \\ \hline
3&+2 \\ \hline
4&-2 \\ \hline
5&+3 \\ \hline
\ldots&\ldots\\ \hline
\end{array}
$$

I apologize for not having worked out a nice short-form version of this, but consider a generalization of the above to give a construction like:

$$
\begin{array}{|l|l|} \hline
n_i&\Delta n\\ \hline
1&+1 \\ \hline
2&+2 \\ \hline
3&-1 \\ \hline
4&-2 \\ \hline
5&+3 \\ \hline
6&+4 \\ \hline
7&-3 \\ \hline
8&-4 \\ \hline
9&+5 \\ \hline
\ldots&\ldots\\ \hline
\end{array}
$$

In fact, let $k$ be the number determining that cyclic count, so that $k=1$ for the standard Collatz in the first table, and $k=2$ in the second table, and $k=3$ would give $\{1,2,3,-1,-2,-3,4,5,6,-4\ldots\}$. And by that definition, let $f(n,k)$ return the corresponding $\Delta n$ using $k$ as described.

Then the next term for regular Collatz could be found as $C(n):=n+f(n,1)$. More generally, let's define $$C_k(n):=n+f(n,k),$$ so e.g. by examining that second table, we see that $C_2(8)=8-4=4$ and $C_2(9)=14$.

This construction seems like it might have some interesting properties. In particular, I conjecture that

$\forall k,n\in\mathbb N : C_k(n)$ will eventually reach $k$, assuming Collatz itself holds.

Furthermore, if you take any such sequence and number it starting with $S_0=n$, it seems we have

$$\gcd(S_i,k) \leq \gcd(S_{i+1},k).$$

This means that once you have $k|S_j$ for some $j$, all future terms can be divided through by $k$ to give a standard Collatz sequence, but this is definitely not the case before reaching $S_j$.

I'm interested to see if anyone can find a counterexample or simple reason why this should or should not hold. Any reasonably constructive feedback would be acceptable as an answer, and of course if this has been investigated before, by all means let me know.

Sample Mathematica code if anyone wants it:

```<code>Clear[f, c];
SetAttributes[{f, c}, Listable];
f[n_, k_] := With[{m = Mod[n, 2 k, 1]}, If[m <= k, 1, -1] (Floor[(n - 1)/(2 k)] (k) + Mod[n, k, 1])]
c[n_, k_] := NestWhileList[# + f[#, k] &, n, # != k &, 1, 1000]

c[27, Range[6]] // Column
</code>```

will give you sequences for $C_k(27)$ for $k \leq 6$.

Body:
Let $n\in\mathbb{Z^+}$ and define $$f(n)=(-1)^{n+1}\cdot\lceil\frac{n}{2}\rceil+n$$
Example for
$n=27$: $\{41,62,31,47,71,107,161,242,121,182,91,137,206,103,155,233,350,175,263,395,593,890,445,668,334,167,251,377,566,283,425,638,319,479,719,1079,1619,2429
3644,1822,911,1367,2051,3077,4616,2308,1154,577,866,433
650,325,488,244,122,61,92,46,23,35,53,80,40,20,10,5,8,4,2,1,2,1,..\}$

We can use modulus instead of $(-1)^n$, then we have to round down (using the floor function) instead of up:

$$f(n)=(n \bmod 2 + 1)n-\lfloor\frac{n}{2}\rfloor$$

The difference is that we subtract from the double of $n$ when $n$ is odd.


问:
Probability of a specific binary string sub-sequence occurring
I've been wracking my brains for a while to try to come up with a non-brute-force solution for this problem.

If you have have a random binary sequence of length $N$, what is the probability that some sub-sequence $X$ is present? My first thought was that it only depends on $N$ and the length of $X$, but from experimentation it's clear that it depends on the contents of $X$.

It seems similar to the problem where you want to know the probability of at least $A$ heads and $B$ tails when flipping a coin $C$ times, but I'm not sure how to then impose the restriction of a specific sequence (Not that I know how to solve that problem either).

I've computed a few results using brute force:

$N=5$ , $X = "111"$, $P(N,X) = 0.25$

$N=5$, $X = "101"$, $P(N,X) = 0.34375$

$N=5$, $X = "100"$, $P(N,X) = 0.375$

答:
Here is a way to get at least a recurrence relation:

Let us look at the string "100". We want to find out how many strings of length $n$ do not contain it.

Clearly such a string can only end in ...1, ...10, or ...000.

Now define:

$a_n$: Number of bitstrings of length $n$ that do not contain "100" and end in ...1

$b_n$: Number of bitstrings of length $n$ that do not contain "100" and end in ...10

$c_n$: Number of bitstrings of length $n$ that do not contain "100" and end in ...000

Then we have the recurrence relations
$$\begin{align*}
a_{n+1} &= a_n + b_n + c_n \\
b_{n+1} &= a_n \\
c_{n+1} &= c_n
\end{align*}$$

(because, for example, a string of length $n+1$ that counts towards $a_{n+1}$ can be created by appending "1" to any string belonging to $a_n$, $b_n$ or $c_n$; a string that counts towards $b_n$ can only be created by appending "0" to a string belonging to $a_n$)

Starting with $a_3 = 4$, $b_3 = 2$ and $c_3 = 1$ we can iteratively calculate the values $a_n$, $b_n$ and $c_n$.

The number of binary sequences of length $n$ that do contain "100" is then given by $2^n - (a_n+b_n+c_n)$.


问:
Show that $1+\sum_{k=1}^\infty\left\{1-\left(k+\frac12\right)\log\left(\frac{k+1}{k}\right)\right\}=\frac12\times\log(2\pi)$
I tried to prove that the constant in Stirling's formula is $\sqrt {2\pi}$. Then I used the Euler-Maclaurin formula, so finally this is what I need to prove. Wikipedia gives a hint that we need to use Wallis' product, but I couldn't prove it. Please help.

答:
Let's define two sequences, $ \left(a_{n}\right)_{n\geq 0} , \left(b_{n}\right)_{n>0}\in\mathbb{R}^{\mathbb{N}} $ as follows : $$ \left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{n^{n}\mathrm{e}^{-n}\sqrt{n}}{n!}\\ \left(\forall n\in\mathbb{N}^{*}\right),\ b_{n}=\ln{\left(\frac{a_{n+1}}{a_{n}}\right)} $$

Let $ n\in\mathbb{N}^{*} $, we have $ \left(\forall k\in\mathbb{N}^{*}\right) $, the following : \begin{aligned} b_{k}&=\ln{\left(\left(\frac{k+1}{k}\right)^{k+\frac{1}{2}}\mathrm{e}^{-1}\right)}\\ &=-1+\left(k+\frac{1}{2}\right)\ln{\left(1+\frac{1}{k}\right)} \end{aligned}

Thus : \begin{aligned} 1+\sum_{k=1}^{n-1}{\left(1-\left(k+\frac{1}{2}\right)\ln{\left(1+\frac{1}{k}\right)}\right)}&=1+\sum_{k=1}^{n-1}{\left(\ln{\left(a_{k+1}\right)}-\ln{\left(a_{k}\right)}\right)}\\ &=1+\ln{\left(a_{n}\right)}-\ln{\left(a_{1}\right)}\\ &=\ln{\left(a_{n}\right)} \end{aligned}

Since : $$ 1-\left(n+\frac{1}{2}\right)\ln{\left(1+\frac{1}{n}\right)}=\underset{\overset{n\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{n^{2}}\right) $$

Then the series $ \sum\limits_{n\geq 1}{\left(1-\left(n+\frac{1}{2}\right)\ln{\left(1+\frac{1}{n}\right)}\right)} $ converges, which means that $ \ln{\left(a_{n}\right)}\underset{n\to +\infty}{\longrightarrow}\ell_{0}\in\mathbb{R} $, hence, denoting $ \ell=\mathrm{e}^{\ell_{0}} $, we have : $$ n!\underset{n\to +\infty}{\sim}\ell\sqrt{n}\left(\frac{n}{\mathrm{e}}\right)^{n} \ \ \ \ \fbox{$\begin{array}{rcl}1\end{array}$}$$

Now we just need to find what $ \ell $ is, to do so we'll be using the fact that : $$ \frac{}{}\frac{2}{\left(2n+1\right)\pi}\left(\frac{2^{2n}\left(n!\right)^{2}}{\left(2n\right)!}\right)^{2}=\frac{W_{2n+1}}{W_{2n}}\underset{n\to +\infty}{\longrightarrow}1  \ \ \ \ \fbox{$\begin{array}{rcl}2\end{array}$}$$

Were $ W_{n} $, for $ n\in\mathbb{N} $, stands for Wallis integral : $$ \left(\forall n\in\mathbb{N}\right),\ W_{n}=\int_{0}^{\frac{\pi}{2}}{\sin^{n}{x}\,\mathrm{d}x} $$

I'm not gonna detail the second result, simply because it's easy to prove, also it is an exercise that, I think, has been solved many times in this website.

Using our first result, we can write that : $$ \frac{2}{\left(2n+1\right)\pi}\left(\frac{2^{2n}\left(n!\right)^{2}}{\left(2n\right)!}\right)^{2}\underset{n\to +\infty}{\sim}\frac{2^{4n}}{n\pi}\times\frac{\ell^{4}n^{4n+2}\mathrm{e}^{-4n}}{\ell^{2}\left(2n\right)^{4n+1}\mathrm{e}^{-4n}}\underset{n\to +\infty}{\sim}\frac{\ell^{2}}{2\pi} $$

Thus $ \ell = \sqrt{2\pi} $, which means $ \ell_{0} $, which is the sum of our series, would then be equal to $ \frac{1}{2}\ln{\left(2\pi\right)} $.


问:
Sequence/List Formula
Consider the following sequence $\{1,1,2,1,2,3,4,1,2,3,4,5,6,7,8,1,...,16,1,...,32,1...\}$; it is the list of integers $1$ to $2^0$, then $1$ to $2^1$, then $1$ to $2^2$, etc. Let $f(n)$ return the $n$th entry in the sequence.

What is a closed form for $f(n)$? Currently I've just been building the sequence as a list and using the $n$th entry... but an analytical expression would be quite helpful. So far what I have is:
\begin{equation}
f(n)=j\quad \text{ if }\quad n=2^{m+\lceil \log_2 j\rceil}+j-1
\end{equation}
where $m$ is a positive integer or $0$. It's a little small but we're taking the ceiling of $\log_2 j$ in the above equation. This question has been bothering me for a while because it seems like I should've gotten this immediately... but I couldn't do it!

答:
In case you need this for a purely coding purpose, perhaps you would benefit from re-defining your function $f(n)$ as "the sum of digits in the binary representation of $n$"; more technically,

$$ f(n) = \sum_i^{\infty} (n\ \mathrm{mod}\ 2^{i+1})\ \mathrm{div}\ 2^i, $$

or

$$ f(n) = \sum_i^{\lfloor n \rfloor} (n\ \mathrm{mod}\ 2^{i+1})\ \mathrm{div}\ 2^i. $$

Here $a\ \mathrm{div}\ b$ is the integer quotient of $a$ and $b$, i.e. $b \cdot \lfloor a/b \rfloor$, and $a\ \mathrm{mod}\ b$ is the remainder on division of $a$ by $b$, i.e. $(a - b \cdot \lfloor a/b \rfloor)$.

Title: Sequence/List Formula


问:
Efficient way of polynomial multiplication with dependency in Bipartite graph form
We use a Bipartite graph: $G=(U,V,E)$, with vertices: $U $ and $V$ to represent some task dependency.

Every $U$ is a unique task, and every $V$ is a degree one polynomial, e.g. $(1+v_ix)$.

Our goal is for all the tasks, get their complete polynomial product coefficient.

For $u_1$, the job is to compute $\prod_{(u_1, v_i) \in E} (1+v_ix) = \sum_i g_ix^i$, define $G_1 = (g_0, g_1, \cdots)$ .

And we assume the connection degree of every task node $u$ is bounded by $C$.

Is there any efficient way to obtain all $G_i$ for every $u_i \in U$ ?

答:
"Of course" there is always the idea of treating each task independently. This yields a cost of $\Theta(\sum_i \delta(u_i) \log^2(\delta))$ multiplications of $v_i$, where $\delta$ is the degree. You just do your polynomial expansion for each task. A naïve approach for the multiplications yields a $\Theta(\sum_i \delta(u_i)^2)$, but with fast Fourier transform algorithms you get the said bound. Globally, this cost is $O(n^2 \log^2(n))$. EDIT : It's even $O(|E| \log^2(\delta_{\max})) < O(|E| \log^2(|V|))$.

BUT you could try to find a way to reuse already existing multiplications. Typically, if all the tasks are the same, you could compute only one, and copy it.
This is an interesting idea, but I don't think it's possible to do it in less than $O(m \log^2(n))$ in most cases.

Indeed, even if you had a perfect and instant oracle/algorithm for minimizing the number of multiplications, I guess in most (random) cases I don't think you'll win a significant margin. Unless two tasks have exactly the same dependency set, in the end you'll have to do at least one long multiplication for each task ; and that long multiplication will probably cost you $\delta \log(\delta)$ (apart from marginal cases where you use a polynomial of size about the same as), and at least $\delta$. So, in any case, you couldn't go below a $O(m \log(n))$ for dense graphs, or $O(m)$ if you're very lucky and have a wonderful algorithm.