问:
Sum of binomial coefficients so that the sum equals ${n\choose n/2}$
I am trying to find $m$ (either exact or order of $m$ in terms of $n$) such that:

$$\sum_{k=0}^m\binom{n}{k} = {n\choose \frac{n}{2}}$$

I was thinking of applying Stirling's approximation for factorials for large $n$, so $\displaystyle\binom{n}{\frac{n}{2}}\sim\frac{2^n}{\sqrt{n}}$. But looking at $\displaystyle{n\choose k}=\frac{n!}{k!(n-k)!}$ and that the sum over $k$ starts from $k=0$, I cannot use Stirling's approximation for $k!$ for small $k$.

I was also looking at sum of binomial coefficients on Wiki which gives an upper bound for the sum and so am wondering if getting an upper bound is the best we can do or if we can find $m$ precisely?

答:
This is an interesting problem from a numerical point of view.

Transposed in the real domain, you are looking for $m$ such that

$$\color{blue}{\frac{\, _2F_1(1,m-n+1;m+2;-1)}{\Gamma (m+2) \,\,\Gamma (n-m)} =\frac{2^n}{\Gamma (n+1)}-\frac{1}{\Big[\Gamma \left(\frac{n}{2}+1\right)\Big]^2}}$$ where $m$ and $n$ are real numbers.

This equation is not difficult to solve using Newton method with $m_0=\frac n 2$. This starting point is justified by the left part of the trivial double inequality
$$\binom{n}{m} \leq\sum_{k=0}^m\binom{n}{k}\leq (m+1)\binom{n}{m}$$ which means that we already know that $m \leq \frac n 2$. I did not find any simple way to use the right part of the above inequality (this is no more true : have a look at the $\color{red}{\text{ update}}$) .

For example, for $n=10$, the iterates are
$$\left(
\begin{array}{cc}
k & m_k \\
 0 & 5.000000000 \\
 1 & 3.419647982 \\
 2 & 3.407971414 \\
 3 & 3.407943361
\end{array}
\right)$$

Below are some results (I let you rounding the results the way you want).
$$\left(
\begin{array}{cc}
n & m \\
 10 & 3.40794 \\
 20 & 7.41879 \\
 30 & 11.5964 \\
 40 & 15.8702 \\
 50 & 20.2093 \\
 60 & 24.5969 \\
 70 & 29.0227 \\
 80 & 33.4793 \\
 90 & 37.9619 \\
 100 & 42.4665 \\
 110 & 46.9903 \\
 120 & 51.5309 \\
 130 & 56.0864 \\
 140 & 60.6554 \\
 150 & 65.2365 \\
 160 & 69.8287 \\
 170 & 74.4310 \\
 180 & 79.0426 \\
 190 & 83.6628 \\
 200 & 88.2910 \\
 210 & 92.9267 \\
 220 & 97.5694 \\
 230 & 102.219 \\
 240 & 106.874 \\
 250 & 111.535 \\
 260 & 116.202 \\
 270 & 120.874 \\
 280 & 125.550 \\
 290 & 130.232 \\
 300 & 134.918
\end{array}
\right)$$

This looks to be very close to linearity. Using these numbers, a quick and dirty linear regression for $m=a +b \,n$ leads to $R^2=0.999957$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & -2.6778 & 0.2028 & \{-3.0938,-2.2618\} \\
 b & +0.4561 & 0.0011 & \{+0.4538,+0.4585\} \\
\end{array}$$

Using this empirical model for $n=400$, it gives $m=179.775$ while the solution is $181.986$.

As I wrote in comments, this also works for non integer values of $n$. For $n=123.456$, $m=53.1037$.

Update

I managed to use
$$\sum_{k=0}^m\binom{n}{k}\leq (m+1)\binom{n}{m}$$ defining the function
$$f(m)=(m+1)\binom{n}{m}- {n\choose \frac{n}{2}}$$ which was expanded as a series to
$O\left(\left(m-\frac{n}{2}\right)^3\right)$. Solving the quadratic, the approximate solution is given by
$$m=\frac n 2-\frac{n}{1+\sqrt{n} \sqrt{(n+2) \psi ^{(1)}\left(\frac{n}{2}\right)-\frac{3
   n+8}{n^2}}}$$ which is a much better starting point as shown below
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{solution} \\
 50 & 20.5142 & 20.2093 \\
 100 & 43.4413 & 42.4665 \\
 150 & 66.8507 & 65.2365 \\
 200 & 90.5099 & 88.2910 \\
 250 & 114.329 & 111.535 \\
 300 & 138.261 & 134.918
\end{array}
\right)$$

The asymptotics of the approximation is
$$m=\frac n2 \left(1-\sqrt{\frac 2 n}+\frac 1 n+O\left(\frac{1}{n^{3/2}}\right)\right)$$

Update

I found later this question; @user940 gave a very interesting asymptotic approximation. Adapted to your problem, we look for the solution $m$ of the equation
$$2^{n-1} \left(1-\text{erf}\left(\frac{n-2 m}{\sqrt{2n}
   }\right)\right)=\binom{n}{\frac{n}{2}}$$ that is to say
$$\text{erf}\left(\frac{n-2 m}{\sqrt{2n} }\right)=1-\frac{2\, \Gamma \left(\frac{n+1}{2}\right)}{\sqrt{\pi } \,\Gamma \left(\frac{n}{2}+1\right)}$$ This can be inversed using approximations of the error function (have a look here).

This would give
$$\left(
\begin{array}{cc}
 50  & 20.7060 \\
 100 & 42.9608 \\
 150 & 65.7299 \\
 200 & 88.7840 \\
 250 & 112.028 \\
 300 & 135.410
\end{array}
\right)$$ which is significantly better for large values of $n$.

Concerning the asymptotics of $n$, using
$$\text{erf}(x)=1+e^{-x^2} \left(-\frac{1}{\sqrt{\pi }
   x}+O\left(\frac{1}{x^3}\right)\right)$$ we have
$$m=\frac n 2-\frac {\sqrt n } 2 \sqrt{W(t)}\qquad \text{where} \qquad t=\frac 12\left(\frac{\Gamma \left(\frac{n+2}{2}\right)}{\Gamma \left(\frac{n+1}{2}\right)} \right)^2$$ $W(.)$ being Lambert function. So, as expected earlier, a logarithmic contribution in the asymptotics of $m$.


问:
injective holomorphic functions
If I have a bounded, connected, open subset of the complex plane, and a function that is holomorphic on it, continuous on its closure, and injective on its boundary, is my function necessarily injective?

It seems it is not true for arbitrary connected regions. Is it true for simply connected regions?

答:
No. For example, $f(z)=z+1/z$, your domain given by $r<|z|<R$ with $r<1<R$, $rR\neq1$, then the boundary circles are mapped injectively to two confocal ellipses, but $f(i)=f(-i)=0$.

edit: If the region is simply-connected and bounded by a Jordan curve then $f$ must be injective: The image (under $f$) of the boundary is a Jordan curve, hence the winding number of the image curve around any point is either $1$ (if the point is inside) or $0$ (if outside). By the argument principle, the number of preimages of any point is the winding number, hence $f$ is injective.

ContentLicense: CC BY-SA 3.0

Title: injective holomorphic functions


问:
Increasing subsequence with given limit
Given a limit ordinal $\alpha$ that is not a cardinal, we can set $\kappa=|\alpha|$ and find a one-to-one sequence $\langle \alpha_\nu\mid\nu<\kappa\rangle$  with range $\alpha$. We want to rigorously construct an increasing subsequence with limit $\alpha$.

This question has been asked before, but I am having difficulties to show that the corresponding subsequence has limit $\alpha$.

I believe one should use the fact that $\alpha$ is a limit ordinal, but I don't see how to apply it.

答:
Here is the argument in more detail.

Let $\eta_0=0$. Suppose that $\gamma<\kappa$, and that $\eta_\xi$ has been defined for $\xi<\gamma$ so that $\langle\alpha_{\eta_\xi}:\xi<\gamma\rangle$ is strictly increasing, and if

$$A_\xi=\{\nu<\kappa:\alpha_{\eta_\zeta}<\alpha_\nu\text{ for all }\zeta<\xi\}$$

for each $\xi<\gamma$, then

$$\eta_\xi=\min\{\nu\in A_\xi:\eta_\zeta<\nu\text{ for all }\zeta<\xi\}\,.\tag{1}$$

If $\sup_{\xi<\gamma}\alpha_{\eta_\xi}=\alpha$, we’re done. Otherwise, let

$$A_\gamma=\{\nu<\kappa:\alpha_{\eta_\xi}<\alpha_\nu\text{ for all }\xi<\gamma\}\ne\varnothing\,,$$

and let

$$A_\gamma'=\{\nu\in A_\gamma:\eta_\xi<\nu\text{ for all }\xi<\gamma\}\,.$$

If $A_\gamma'=\varnothing$, then for each $\nu\in A_\gamma$ there is a $\xi<\gamma$ such that $\nu<\eta_\xi$; this violates the minimality of $\eta_\xi$ in $(1)$, so $A_\gamma'\ne\varnothing$, and we set $\eta_\gamma=\min A_\gamma'$. Then $\langle\alpha_{\eta_\xi}:\xi\le\gamma\rangle$ is strictly increasing, and $(1)$ is satisfied for $\xi\le\gamma$, so the recursive construction goes through until at some stage $\gamma$ we have $\sup_{\xi<\gamma}\alpha_{\eta_\xi}=\alpha$, and at that point $\langle\alpha_{\eta_\xi}:\xi<\gamma\rangle$ is the desired subsequence.

ContentLicense: CC BY-SA 4.0

Title: Increasing subsequence with given limit


问:
A question in a solved example related to differentiability of an analytic function
This question was part of an assignment and its solution was given but I am having trouble understanding a conclusion.

So, I am asking for help here.

Question is: Let f: $\mathbb{C} \to \mathbb{C}$ be a holomorphic function and let u be it's real part and v be it's imaginary part.Then , for x,y $\in \mathbb{C}$ , $|f'(x+iy)|^2$ is equal to?

So, a line in solution is $|f'(z)|^2 = {u_x}^2+ {v_x}^2$ . why is it true ?

why it shouldn't be $|f'(z)|^2 = {u_x}^2 +{u_y}^2 +{v_x}^2 +{v_y}^2$ ? as in this equation I differentiated u wrt to y and v wrt to x as u and v are both functions of x and y?

Can you please give reason for that

答: $f'(z)=\lim_{h \to 0} \frac {f(z+h)-f(z)} h$. If you take the limit through real values if $h$ you get $f(z)=u_x+iv_x$ so $|f'(z)|^{2}=u_x^{2}+v_x^{2}$.


问:
Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?

I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.

Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?

Thank you in advance.

Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.

答: I know Octave and Freemat, both claim similarity to Matlab but I don't have first hand experience on them. This article  compares Matlab, Octave, Scilab and Freemat. The source is this MO question answer of which may be of some help.</p>

Title: Free online mathematical software


问:
A problem related to $2x^3-3x^2-x+\frac{3}{2}=0$

let $f(x)=2x^3-3x^2-x+\frac{3}{2}$.Then prove that
$$\int_{1/8}^{7/8} f(f(x)) \text d x\neq \frac{3}{4}$$

Factorising$$f(x)=2(x-1.5)(x-\frac{1}{\sqrt{2}})(x+\frac{1}{\sqrt{2}})$$
$$f(f(x))=2(f(x)-1.5)(f^2(x)-0.5)$$
But i don't see anything nice from this.
Maybe if i could prove that $$f(f(x+\frac{1}{2}))=-f(f(x-\frac{1}{2}))$$ then the integral would turn out to be zero.

also i dont think we have to find the exact value of integral for second part if we could just set up an inequality or prove that the integral is negative we are done!

Please note that i am intersted in a proof without actually finding $f(f(x))$

答:
You have the right idea about exploiting some sort of symmetry in $f(f(x))$, but the integral doesn't evaluate zero.

Anyway, for all $y \in \mathbb{R}$, we have $$f(1-y) = 2(1-y)^3-3(1-y)^2-(1-y)+\dfrac{3}{2} = -2y^3+3y^2+y-\dfrac{1}{2} = 1-f(y).$$

Hence, for all $x \in \mathbb{R}$, we have$$f(f(1-x)) = f(1-f(x)) = 1-f(f(x)),$$ where we have used the above property for $y = x$, and then for $y = f(x)$.

Now, let $I = \displaystyle\int_{1/8}^{7/8}f(f(x))\,dx$. Replace $x$ with $1-x$ to get $I= \displaystyle\int_{1/8}^{7/8}f(f(1-x))\,dx$.

By adding these expressions for $I$ together, we get $$2I = \int_{1/8}^{7/8}f(f(x))+f(f(1-x))\,dx = \int_{1/8}^{7/8}1\,dx = \dfrac{3}{4},$$ and thus, $I = \dfrac{3}{8} \neq \dfrac{3}{4}$.


问:
Show $f(x,y)=\frac{x^3\sin(x-y)}{x^3-y^3}$ for $x\neq y$, $f(x,y)=0$ for $x=y$ is continuous at $(0,0)$.
I know that once we bound $f$ for $x\neq y$ we are done because $f=0$ for $x=y$. I got that for $x\neq y$, $|f(x,y)|\le \left|\frac{x^3}{x^2+xy+y^2}\right|$ because $\left|\frac{\sin(a)}{a}\right|\leq 1$ for $a\neq 0$. But I don't know how to continue.

答:
so following @Kelenner's hint, $x^2+xy+y^2 = \frac{x^2+y^2}{2} +\frac{(x+y)^2}{2} \geq \frac{x^2+y^2}{2}$, in particular it must be positive, and $|f(x,y)|\leq |\frac{x^3}{x^2+xy+y^2}| \leq |\frac{2x^3}{x^2+y^2}| \leq |\frac{2x^3}{x^2}| =2|x|$, which tends to $0$ as $(x,y)\rightarrow(0,0)$.

We can also say that $|f(x,y)|\leq |\frac{x^3}{x^2+xy+y^2}| \leq |\frac{2x^3}{x^2+y^2}| \leq \frac{2x^2(x^2+y^2)^{\frac{1}{2}}}{x^2+y^2} \leq 2x^2(x^2+y^2)^{\frac{1}{2}}$ which also tends to $0$.


问:
Find $f:f(xf(x)+f(y))=f(x)^2+y$
Find $f:f(xf(x)+f(y))=f(x)^2+y$

Domain and co-domain is real numbers

I did the following:

Let $s=f(0)$

Then $f(f(y))=s^2+y$ so $f$ is surjective

Also, $f(x)=f(y)\implies f(xf(x)+f(y))=f(xf(x)+f(x))\implies x=y$ so $f$ is injective

So, $f$ is bijective. Letting $f(x)=0, y=0$

we get $f(f(0))=0 \implies s=f(0)=0$

In fact $f(f(x))=x$ for ALL $x$

Letting $x=1,y=0$ we get

$f(f(1))=f(1)^2\implies 1=f(1)^2 \implies f(1)=1$ or  $f(1)=-1$

If $f(1)=1$ we let $x=1$ to get

$f(y+1)=f(y)+1$

By induction this leads to $f(x)=x$ for all integers

If $f(1)=-1$ we let $x=1$ to get

$f(y-1)=f(y)+1$

By induction we get $f(x)=-x$ for all integers

How to extend the domain over real numbers I don't know, any help would be appreciated

答: Let $x=f(u)$, by using $f(x)=f(f(u))=u$ one gets now $f(u f(u)+f(y))=u^2+y$. By replacing $u$ with $x$, it follows that for all $x$, $f(x)^2 = x^2$, hence for all $x$, $f(x) = \pm x$. It remains to see if the sign depends on $x$.

Title: Find $f:f(xf(x)+f(y))=f(x)^2+y$


问:
How do I show $y=\ \frac{1}{2}-\frac{1}{2^{x}+1\ }$ is odd?
Of course I know that $f(-x)=-f(x)$, but I get confused as to how $$\frac{1}{2}-\frac{2^{x}}{2^{x}+1\ }=-\frac{1}{2}+\frac{1}{1+2^{x}}$$ In the steps of algebra.

答:
There's really no trick:

$\frac{1}{2}-\frac{1}{2^{-x}+1\ }= \frac{1}{2^{x}+1\ }- \frac{1}{2} \Longleftrightarrow$

$1 = \frac{1}{2^{-x}+1\ }+ \frac{1}{2^{x}+1\ }$ and multiplying by the product of the denominators  we get $\Longleftrightarrow$

$1= \frac{2^{x}+1 + 2^{-x}+1}{(2^{x}+1)(2^{-x}+1)}=\frac{2^{x}+1 + 2^{-x}+1}{2^{x}+1 + 2^{-x}+1}$. Since these steps are reversible the initial equality is true.


问:
Quadratic form positive semidefinite if limits in every direction are nonnegative?
Let
$$q(x_1,\ldots,x_n) = \sum_{i,j=1}^{n}a_{ij}{x_i}{x_j}, \quad a_{ij}\in \mathbb{R}.$$
be a quadratic form with real coefficients.
Suppose that the limit is nonnegative in every direction. That is, for any unit vector $u$,
$$ \lim_{t \rightarrow \infty} q(tu) \geq 0.$$
Is $q$ necessarily positive semidefinite?
If the condition is strengthened to
$$ \lim_{t \rightarrow \infty} q(tu) > 0,$$
is it necessary positive definite?</p>

答:
Hint: if the coefficients are constant, then you don't need the limit. Bilinearity allows to evacuate $t^2>0$, so all you get is </p>

<p>$\forall u\in\mathbb R^n\, q(u) = \langle Au,u\rangle\ge 0$ where $A$ is a square symetric matrix.</p>

Body:
Hint: if the coefficients are constant, then you don't need the limit. Bilinearity allows to evacuate $t^2>0$, so all you get is

$\forall u\in\mathbb R^n\, q(u) = \langle Au,u\rangle\ge 0$ where $A$ is a square symetric matrix.


问:
If one system has a feasible solution then other does not have
If we have 2 linear equality and inequatlity systems:

$(1)$ $A x < = b $ $ x >= 0$.

$(2)$ $b^T y < 0, A^T y > = 0, y >= 0$

where A $in$ $R^mxn$ and c $in$ $R^n$ are given and x $in$ $R^n$ and y $in$ $R^m$ are decision variables.
Prove that exactly one of the two systems (1) and (2) has a feasible solution. Which means that 1 has a feasible solution iff (2) does not have a feasible solution.

I have troubles doing the forward part of the proof and understanding it geometrically.

I imagine the forward part of the proof to be something like if there exists, $x>=0$ that satisfies the constraint of the primal problem and is thus feasible primal problem.. I don't know how to proceed further. Any guidance would be much appreciated.

答:
Assume $(1)$ is feasible.  Suppose on the contrary that $(2)$ is feasible as well.

We have $Ax \le b$ and  $y \ge 0$.

Hence $y^TAx\le y^Tb$.

Since $y^TA \ge 0$ and $x \ge 0$, hence $y^TAx \ge 0$ but this contradicts $y^Tb < 0$.

Edit:

For the other direction, if the primal is infeasible, then the dual is either unbounded or infeasible. However, since $y=0$ is a feasible solution to $(D)$, we conclude that it is unbounded. Hence we can find a $y$ that satisfy $(2)$.


问:
Solving a hard limit by approximating the sine function
I think I'm supposed to solve this limit using approximations for the behavior when $x$ is close to $0$, but I don't know how to apply the logic for the aproximation.

$$\lim_{x \rightarrow 0} (\frac{1}{sin^2(x)} - \frac{1}{x^2})$$

In the context of this exercise the textbook mentions the following method for approximating:
$$f(x) = \frac{1}{1+x+x^2}$$
Since $\lim_{x \rightarrow 0} f(x) = 1$, then $lim_{x \rightarrow 0} [f(x) - 1] =0$; but $f(x)-1 = -x\frac{x+1}{1+x+x^2}$ so, as $x \rightarrow 0$, $f(x) \approx 1-x$. Also for $x$ near $0$ $f(x) - 1 + x = x^3\frac{1}{1+x+x^2}$ so $f(x) \approx 1-x + x^2$.

I think I grasp the logic for it, but I can't apply this to solving limits like the one with $\frac{1}{sin^2(x)}$.

Note: at this point we haven't seen Taylor Polinomials yet, just this method for approximation near $0$.

Body:
$$y= \frac{1}{\sin^2(x)} - \frac{1}{x^2}=\frac{x^2-\sin^2(x)}{x^2\,\sin^2(x)}$$ Now, using Taylor series
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$
$$\sin^2(x)=x^2-\frac{x^4}{3}+O\left(x^6\right)$$
$$y=\frac{-\frac{x^4}{3}+O\left(x^6\right) } {x^2\left(x^2-\frac{x^4}{3}+O\left(x^6\right) \right) }=\frac{1}{3}+O\left(x^2\right)$$


问:
Let f ∈ S3 be a permutation which is not the identity. Prove that there exists g ∈ S3 such that fg ≠ gf.
Question: Let f ∈ S3 be a permutation which is not the identity. Prove that there exists g ∈ S3 such that fg ≠ gf.

My attempt:

First list out all the possible permutations of f (which there are 5 of them as it is not identity); then list out all possible permutations of g (which there are 6 of them).
Then I try each set of multiplication of fg and gf to check they are not equal in order to prove the hypothesis.

However, I do feel like I'm using the wrong method to prove this hypothesis as each permutation of f could multiple with 6 different permutations of g; It would take forever to complete every sets of multiplication by this method. May I ask is there an easier way to solve this question? I feel like I'm missing something out...

答:
Actually show that $(1\, 2)$ doesn't commute with any member of $S_3$ except identity and itself. Then you prove indirectly that for every non-identity element there exist another element with which it doesn't commute. Can you see why?

Suppose $f \in S_3$ and $f\neq e$ and $f\neq (1\, 2)$ then take $g=(1\, 2)$.

If $f = (1\, 2)$ then take any other non-identity element as $g$ as you have shown $(1\, 2)$ doesn't commute with any member of $S_3$ except identity and itself so you are done.

EDITED ON 21-11-2020: Another answer which is a proof.

Theorem: $\forall n\geq 3, Z(S_n)={e}$, where $Z(G)$ is center of $G$ and $e$ is identity.

That means identity is the only element that commutes with every elemnt of $S_n\forall n\geq 3$.(From now on whenever I say $S_n$, just assume $n\geq 3$).

In other words for every non-identity element $f\in S_n$ there exists another non-identity element $g\in S_n$ s.t. $fg\neq gf$.

Proof:-

First observe that to prove $fg=gf$, we need to show that $fg(x)=gf(x)\forall x\in \{1,2,3,\cdots ,n\}$.

But to prove $fg\neq gf$, we need to show that $\exists x\in \{1,2,3,\cdots ,n\}\text{s.t.} fg(x)\neq gf(x)$.

Let $a,b,c\in \{1,2,3,\cdots ,n\}$.

Let an arbitrary $f\neq e\in S_n$ s.t. $f(a)=b$, Now choose $g=(b,c)$ where $c\neq f(b)$. Observe that this is possible only because $n\neq 3$.

Let me give you an example suppose in $S_3$, let $f(1)=3$ now $f(3)=1\text{ or }2$ say $1$, >then choose $g=(3,2)$.

Now, observe $$fg(a)=f(g(a))=f(a)=b\\gf(a)=g(f(a))=g(b)=c$$

Hence $fg\neq gf.$. So for every non-identity element $f\in S_n$ there exist another non-identity element $g\in S_n$ s.t. $fg\neq gf$.$\qquad \square$


问:
Chain rule when taking non-dimensionalising an ODE
I have a silly question. So let's say we have:

$$\frac{d^{2}x}{dt^{2}} = kx$$

Now let's say we pick $X = \frac{x}{x_{c}}$ and $T = \frac{t}{t_{c}}$. What I don't understand is, if we plug in $Xx_{c}$ for x, and $Tt_{c}$ for t, how come we get:

$$\frac{x_{c}d^{2}X}{t_{c}^{2}dT^{2}}$$

How come the $t_{c}$ is squared? Can someone do the math and explain it to me?  Our professo r just said. it was. chain rule but I am not sure how

答:
You could write the chain rule down as

$$ \frac{d X}{dT} = \frac{d X}{dx} \frac{d x}{dt} \frac{d t}{dT} $$

and from your relationships $X = \frac{x}{x_{c}}$ and $T = \frac{t}{t_{c}}$ obtain

$$ \frac{d x}{dt} = \frac{d X}{dT} \frac{x_c}{t_c} $$

The calculation for the second derivative is analogous, just take the first derivative of the above, $$ \frac{\bullet}{dt} = \frac{\bullet}{dT} \frac{dT}{dt} = \frac{\bullet}{dT} \frac{1}{t_c}$$ and the $t_c$ term gets squared.


问:
Poisson process with exp waiting times
Let $N(t)$ be a Poisson process with rate $\lambda$. Then $(-1)^{N(t)}$
$N(t)$ wityh rarte $\lambda$ is a process which jumps between two states with between exponential waiting times between jumps.

a) I want to find $E[(-1)^N(t)]$

I get $\sum_{k>=0}(-1)^ke^{\lambda t}\frac{(\lambda t)^k}{k!}$= $e^{-\lambda t}e^{\lambda t}=1$ ??

Also, we can write $E[(-1)^N(t)] =  P{(-1)^{N(t)=1}+(-1)*P{(-1)^{N(t)}=-1}=1$

Can we say that $P{(-1)^{N(t)=1}+P{(-1)^{N(t)}=-1}=1$ How do I now reconcile the two?

b) Next I want to show that $lim_{n \to \infty}P((-1)^{{N(t)}}=1)=1/2$
It makes intuitive sense but how to show it?

Finally I want to find $E(\int_0^{\tau}(-1)^{N(t)}dt)$
Also not so sure how to go about this.

Thanks for the consideration

答:
Part (a) is:  $$\operatorname{E}[(-1)^{N(t)}] = \sum_{k=0}^\infty (-1)^k e^{-\lambda t} \frac{(\lambda t)^k}{k!} = e^{\lambda t} \sum_{k=0}^\infty \frac{(-\lambda t)^k}{k!} = e^{-\lambda t} e^{-\lambda t} = e^{-2\lambda t}.$$

We can see the above is a special case of the probability-generating function of a Poisson distribution: $$\mathcal P_{N(t)}(z) = \operatorname{E}[z^{N(t)}] = e^{\lambda t (z - 1)},$$ where $z = -1$.  Your error is in writing $e^{\lambda t}$ rather than $e^{-\lambda t}$ for the Poisson PMF.

Part (b) is unclear because the expression $$\lim_{n \to \infty} \Pr[(-1)^{N(t)} = 1]$$ has the limiting variable $n$, but you did not define how $n$ is related to $N(t)$.  However, based on part (a), we can see that as $t \to \infty$, $\operatorname{E}[(-1)^{N(t)}] \to 0$, which suggests that $N(t)$ tends to be odd as often as it tends to be even; i.e. $\Pr[(-1)^{N(t)} = 1] = \Pr[(-1)^{N(t)} = -1] = \frac{1}{2}$.  I leave it as an exercise to formally demonstrate this fact.

Part (c) is also unclear.

Title: Poisson process with exp waiting times


问:
$\mathcal{L} \leq \mathcal{X}$, $\dim \mathcal{L} < \infty$, and $\mathcal{M}\leq \mathcal{X}$, then $\mathcal{L} + \mathcal{M}$ is closed
Taken from Conways A Course In Functional Analysis Chapter 5 section 2 problem 1.

Problem Statement: In relation to Claim 2.4, show that if $\mathcal{L} \leq \mathcal{X}$, $\dim \mathcal{L} < \infty$, and $\mathcal{M}\leq \mathcal{X}$, then $\mathcal{L} + \mathcal{M}$ is closed.

I wanted to see if the solution I had below was correct.

For some $x \in \mathcal{X}\backslash \mathcal{M}$: $\mathcal{L} = \{\alpha x: \alpha \in \mathbb{F}\}$. Since $\mathcal{X}$ is a finite dimensional subspace of $\mathcal{X}$, it follows that $\mathcal{L}+ \mathcal{M}$ is also a finite dimensional subspace (with dimension $\dim \mathcal{M}+1$) of $\mathcal{X}$. Thus, it is closed.

答:
No, this doesn't work, primarily for the reason Kavi Rama Murthy points out in the comments: you can't assume $\mathcal{M}$ is finite-dimensional. I also note that you assume or conclude that $\mathcal{L}$ is one-dimensional, which is not OK to conclude, but fine to assume if you set up an induction argument.

Let's suppose $\mathcal{M}$ is a closed subspace of $\mathcal{X}$, and $\mathcal{L} = \operatorname{span}(v) \le \mathcal{X}$. We aim to show $\mathcal{M} + \mathcal{L}$ is closed. If $v \in \mathcal{M}$, then $\mathcal{M} + \mathcal{L} = \mathcal{M}$, which is closed, and hence we are done. Thus, assume without loss of generality that $v \notin \mathcal{M}$.

Suppose $y_n \in \mathcal{M} + \mathcal{L}$ converges to $y \in \mathcal{X}$. Then $y_n = x_n + \alpha_n v$ for some $x_n \in \mathcal{M}$ and some scalar $\alpha_n$. If $\alpha_n$ were unbounded, then by taking subsequences as necessary, we could assume $\alpha_n \to \infty$ or $\alpha_n \to -\infty$, and indeed $\alpha_n \neq 0$. Then,
$$\left\|v + \frac{x_n}{\alpha_n}\right\| = \frac{\|y_n\|}{|\alpha_n|} \to 0,$$
as $y_n$ is convergent and hence bounded. On the other hand, $-x_n / \alpha_n \in \mathcal{M}$, which implies $v \in \overline{\mathcal{M}} = \mathcal{M}$, against assumption.

Therefore, $\alpha_n$ is bounded, and by taking subsequences as necessary, we may assume it is convergent. Let $\alpha$ be its limit. Then,
$$x_n = y_n - \alpha_n v \to y - \alpha v \in \mathcal{M},$$
since $\mathcal{M}$ is closed. As such, we may express:
$$y = \underbrace{y - \alpha v}_{\in \mathcal{M}} + \underbrace{\alpha v}_{\in \mathcal{L}} \in \mathcal{M} + \mathcal{L}.$$

This provides the necessary tool for using induction. Recall that, if $v_1, \ldots, v_{n+1} \in \mathcal{X}$, then
$$\operatorname{span}(v_1, \ldots, v_{n+1}) = \operatorname{span}(v_1, \ldots, v_n) + \operatorname{span}(v_{n+1}).$$
As such, if $\mathcal{L}$ has dimension $n + 1$, then we may express $\mathcal{L} = \mathcal{L}_1 + \mathcal{L}_2$, where $\mathcal{L}_1$ has dimension $n$ and $\mathcal{L}_2$ has dimension $1$. So,
$$\mathcal{M} + \mathcal{L} = (\mathcal{M} + \mathcal{L}_1) + \mathcal{L}_2.$$
If we assume $\mathcal{M} + \mathcal{L}_1$ is closed for any dimension $n$ $\mathcal{L}_1$, then by the result above, $\mathcal{M} + \mathcal{L}$ is closed for any dimension $n + 1$ $\mathcal{L}$. Couple this with the observation that, if $\dim \mathcal{L} = 0$, then $\mathcal{M} + \mathcal{L} = \mathcal{M}$, which is closed, we have proven by induction that $\mathcal{M} + \mathcal{L}$ is closed for any finite-dimensional $\mathcal{L}$.


问:
Normal subgroup of set of all affine transformations in $\mathbb R^2$
Let $\mathcal A(\mathbb R^2)$ be the set of all affine transformations of $\mathbb R^2.$ Let $\text {GL}(\mathbb R^2) = \{T_{M} \mid M\in \text {GL}_2(\mathbb R)\} $ be the set of all bijective linear transformations
where $T_M := T_{M,\boldsymbol{0}}$.

Is $\text {GL}(\mathbb R^2)$ a normal subgroup of $\mathcal A(\mathbb R^2)$ ?

My working:

$$\begin{align}
\text{GL}(\mathbb R^2) \text { is a normal subgroup of } \mathcal {A}(\mathbb R^2)
&\iff \alpha k\alpha^{-1} \in \text{GL}(\mathbb R^2) \text { for all } \alpha \in \mathcal {A}(\mathbb R^2) \text { where } k \in \text {GL}(\mathbb R^2).\\
&\iff \alpha T_{M}\big(T_{A^{-1},-A^{-1}\mathbf b}(\mathbf x)\big) \in \text {GL}(\mathbb R^2)\\
  &\iff \alpha(MA^{-1}\mathbf x-MA^{-1}\mathbf b) \in \text {GL}(\mathbb R^2)\\
  &\iff A(MA^{-1}\mathbf x-MA^{-1}\mathbf b) +\mathbf b  \in \text {GL}(\mathbb R^2)\\
  &\iff AMA^{-1}\mathbf x-(AMA^{-1}-I_{2})\mathbf b \in \text {GL}(\mathbb R^2)\\
  &\iff AMA^{-1} = I_{2}.
\end{align}$$

答:
I'm not quite sure what you meant by the calculation you have there, but the fact is that $\text{GL}(V)$ is not a normal subgroup in $\mathcal{A}(V)$ for any finite-dimensional vector space $V$.

The reason this happens has all to do with the way $\text{GL}(V)$ and $\mathcal{A}(V)$ act on $V$ - in general, a good way to think about any group is to study the actions it has on different sets: in this particular case, there is a natural way to define the action of $\mathcal{A}(V)$ on $V$; namely $T_{M,v}(x)=Mx+v$.

The best way in my opinion to interpret conjugation in groups is by considering what happens to this action once we apply a transformation to the set the group is acting on.

For example, for every $v\in V$, consider the subgroup (called the stabiliser) $\mathcal{A}(V)_v=\{\alpha\in\mathcal{A}(V)\mid \alpha(v)=v\}\leq \mathcal{A}(V)$ of elements that fix $v$ via this action. Well, if we apply some $\beta\in\mathcal{A}(V)$ to $v$ and then calculate the stabiliser you can immediatly see what happens: $$
\mathcal{A}(V)_{\beta(v)}=\{\alpha\in\mathcal{A}(V)\mid\alpha(\beta(v))=\beta(v)\}=\{\alpha\in\mathcal{A}(V)\mid \beta^{-1}\alpha\beta(v)=v\}=\beta\mathcal{A}(V)_v\beta^{-1}
$$ we get the conjugate subgroup of the stabiliser (this is obviously valid in general).

In this particular example, we have that $0$ is (by definition) fixed by every linear transformation, namely $\text{GL}(V)\leq \mathcal{A}(V)_0$. Thus if we apply, for example, a translation $T_{0,v}$ for some non zero vector $v$, we get $$
\mathcal{A}(V)_{T_{0,v}(0)}=\mathcal{A}(V)_v=T_{0,v}\mathcal{A}(V)_0T_{0,-v}
$$ Which means $\text{GL}(V)$ can not be a normal subgroup, because if it were then $T_{0,v}\text{GL}(V)T_{0,-v}=\text{GL}(V)$ and hence $\text{GL}(V)$ would be a subgroup of $\mathcal{A}(V)_v$ where $v$ is non zero, which is clearly not the case: there are obvsiouly linear transformations which don't fix $v$ for any non zero $v$.

I'm aware my answer was a bit overload for this exercise, but I think this route was a bit more instructive that doing the mindless calculations :)


问:
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal?
Let $(B, +, \cdot)$ be a ring (not necessarily unital!) with the property that every $x \in B$ satisfies $x \cdot x = x$.

How does one show that the kernel of any non-zero homomorphism of rings $h:B\rightarrow \mathbb{Z}_2$ is a maximal ideal of $(B, +, \cdot)$?

I'm looking for an elementary proof, not requiring anything more than the definitions of a ring, of an ideal, and of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.

答:
Since $h:B \to \mathbb{Z}_2$ is a non-zero homomorphism, $\ker h$ is a proper ideal of $h.$ If $xy\in \ker h$ then $h(xy)=h(x)h(y)=0$ in $\mathbb{Z}_2$ so either $x$ or $y$ is in $\ker h,$ hence $\ker h$ is a prime ideal of $B.$

Now suppose $I$ is an ideal of $B$ that strictly contains $\ker h.$ Pick $x\in I\setminus \ker h.$ For any $y\in B$ we have $x(y-xy)=0\in \ker h$ and since $\ker h$ is a prime ideal we have $y-xy \in \ker h.$ Thus $y = (y-xy) +xy \in I$ and we conclude that $\ker h$ is maximal.


问:
Difference between a function and a section of a fibre bundle
Suppose $E \rightarrow B$ with projection map $\pi$ and fibre F is a fibre bundle, with a section $\sigma$. How is $\sigma$ different from a function $f:B \rightarrow F$? The standard answer I find everywhere is that $\sigma$ is only a function if E is the trivial bundle $B \times F$, or that in general no such function exists with 'global structure', but I do not see why the former is true, and I do not understand what the latter means. A section seems like it is associating an element of the fibre F to every point in B. How is this not a function, even if E has some sort of non trivial topology? Why should 'twists' or other aberrations from a product in the space E matter, since the fibre over any point in B is the same?

答:
Long story short and summarizing the answers, since I think the OP had my same doubt, a function $f : B \rightarrow F$ takes any base point $m$ and maps it to an element of the same, fixed, topological space $F$. The intuition behind a section $\sigma : B \rightarrow E, \pi(\sigma(m))=m$, is that it maps any point $m$ to a different space, namely the fiber $\pi^{-1}(m)$, which are different copies of F, one for each base point $m$.

Only in particular cases ( e.g. when the bundle is trivial ) the two operations are equivalent in a well defined way.


问:
Statements number system, negating of a sentence.

Why is my answer incorrect or why it must be equality symbol that is coming in?

答:
"If $9>10$ then $3^2=5$" is the same as "$9\le 10$ or $3^2=5$".

If you say "$9$ is not greater than $10$", it means $9$ is either less than $10$ or it is equal to $10$, therefore $\sim(9>10)$ is the same as $9\not >10$ which is the same as $9\le 10$.


问:
Homomorphism of quasi coherent sheaves
I was trying to solve the problems of Liu's book and wanted to show that if both $F$ ,$G$ are coherent then $Hom(F,G)$ is also coherent... but I realised that I really need to understand the meaning and definition of the sheaf of homomorphisms.... so I was tryinng to understand these:

Let $X$ be a scheme, $F$, $G$ two quasi-coherent $O_X$-modules and $U=\operatorname{Spec} A$ be an affine open subset of $X$. Suppose that
$F|_U =\widetilde{M}$ and $G|_U=\widetilde{N}$.
I really need some help to show if $\mathcal{Hom}_U(\widetilde{M},\widetilde{N})$ is isomorphic to $\operatorname{Hom}_A(M,N)^\sim$

Is $\mathcal{Hom}(\widetilde{M},\widetilde{N})$ also a quasi coherent sheaf?

答:
Note that $Hom_U(\tilde{M},\tilde{N})$ identifies naturally with $Hom_A(M,N)$. Indeed, given $f : \tilde{M} \rightarrow \tilde{N}$, its action on the global sections is a $A$-linear map $M \rightarrow N$. Conversely, given $g: M \rightarrow N$, we can consider the maps $g_{D(a)}: M_a=\tilde{M}(D(a)) \rightarrow N_a=\tilde{N}(D(a))$ which are the localizations of $g$ at $a$; they do glue together to form a sheaf morphism $\tilde{g}: \tilde{M} \rightarrow \tilde{N}$.

Now, let $H$ be the sheaf which is the internal Hom of $\tilde{M}$ and $\tilde{N}$ over $U$. By definition, if $V \subset U$ is principal (ie is $D(a)$), $H(V)$ is the set of homomorphisms $\tilde{M}_{|V} \rightarrow \tilde{N}_{|V}$, so by the above identifies with the set of homomorphisms $\tilde{M}(V) \rightarrow \tilde{N}(V)$, ie of homomorphisms $M_a \rightarrow N_a$. However, $\tilde{Hom_A(M,N)}(V)$ is $Hom(M,N) \otimes \mathcal{O}(V)=Hom(M,N)_a$.

The question thus becomes: how to compare $Hom(M,N)_a$ and $Hom(M_a,N_a)$?

It’s easy to see that we have a $A$-linear map $Hom(M,N)_a \rightarrow Hom(M_a,N_a)$.

You can check that this map is injective if $M$ is finitely generated – and that it isn’t in general if $M$ isn’t (hint: consider the case $M,N$ free). It is bijective if $M$ is free.

If we have an exact sequence $A^m \rightarrow A^n \rightarrow M \rightarrow 0$ (ie $M$ finitely presented) then we have a natural morphism between the exact sequences $$0 \rightarrow Hom(M,N)_a \rightarrow Hom(A^n,N)_a \rightarrow Hom(A^m,N)_a$$ and $$0 \rightarrow Hom(M_a,N_a) \rightarrow Hom((A_a)^n,N_a) \rightarrow Hom((A_a)^m,N_a).$$

The two rightmost vertical arrows are isomorphisms, so $Hom(M,N)_a \rightarrow Hom(M_a, N_a)$ is an isomorphism.

As a conclusion, the inner Hom of $\tilde{M},\tilde{N}$ is equal to $\tilde{Hom_A(M,N)}$ when $M$ is finitely presented. When $M$ is only finitely generated, the latter is a subsheaf of the former.

Title: Homomorphism of quasi coherent sheaves


问:
Infinite product of cos and convergence
I am recently self-studying Complex Analysis and came up with a question with regard to infinite products.

I am trying to show that $\prod_{k=1} {\cos(z/2^k)}$ converges. My first thought is to use the complex representation of cosine, $\sum (-1)^n z^2n/(2n)!$ But I just don't know how to get to the form $\prod (1+b_n)$ for which has a theorem for convergence.

Thanks for all the advice.

Title: Infinite product of cos and convergence


问:
Finding x in a sequence
The question is:

To find the limiting sum of the series $5$+$5x$+$5x^2$+$...$$=$$10$ and then solve the equation to find "$x$", I am aware how to find the limiting sum but I don't have an idea on how to solve for $x$

So far I have done:
($\frac{5x}{5}$) $=$($\frac{5x^2}{5x}$) as per: ($\frac{T2}{T1}$)=($\frac{T3}{T2}$)(aka to find the common ratio $r$)

leading me to:

$25x^2$=$25x^2$

but something tells me I'm on the wrong track, am I?

答:
We can see that the sum of this infinite geometric progression has a limit, which implies that its common ratio, $x$, has the following property: $$0\leq\left|x\right|<1$$

We can find the limiting sum: $$\lim_{n\to\infty}\frac{5(1-x^n)}{1-x}=\color{blue}{\frac{5}{1-x}}\quad( x^n\to0 \text{ as }n\to\infty)$$

And since the limiting sum equals $10$, we have $$\frac{5}{1-x}=10\implies \color{blue}{x=\frac{1}2}$$

Note: As John Douma's comment said, you can also divide both sides of the equation by $5$ and get
$$1+x+x^2+\cdots=2$$ and similarly, $$\frac{1}{1-x}=2\implies x=\frac{1}2$$

Title: Finding x in a sequence


问:
Proving the Fourier equation: $\frac{1}{2} L - x = \frac{L}{\pi}\sum^\infty_{n=1}\frac{1}{n} \sin\frac{2n\pi x}{L}$
The following question is about Fourier series, specifically about extension of the Fourier Series to arbitrary intervals:

"Show that:

$$\frac{1}{2} L - x = \frac{L}{\pi}\sum^\infty_{n=1}\frac{1}{n} \sin\frac{2n\pi x}{L},\quad 0<x<L$$"

The question also came with the following hint:

"Note that the series on the right-hand side has the period L. Therefore, it should be viewed as a Fourier series on an interval of length L; in this case, it is $[0, L]$. Think about how
to reduce this scenario into those we have discussed before (e.g. the classic case on $[-\pi, \pi])$ by a suitable change of variable"

I thought to use the change of variable: $t = \frac{2 \pi x}{L} - \pi$, so that as x runs from $0$ to $L$, $t$ runs from $-\pi$ to $\pi$.

After substituting in the variable, the right-hand side reduced to:

$$\frac{L}{\pi}\sum^\infty_{n=1}\frac{1}{n} \sin(nt + n\pi),\quad -\pi<t<\pi$$

Furthermore, since:

\begin{align*}
\sin(nt + n\pi) &= \sin(nt)\cos(n\pi) + \cos(nt)\sin(n\pi)\\
&= \sin(nt)(-1)^n + 0 \quad \text{since n is an integer}
\end{align*}

Hence, the above expression simplifies to:

$$\frac{L}{\pi}\sum^\infty_{n=1}\frac{(-1)^n}{n} \sin(nt),\quad -\pi<t<\pi$$

I got stuck here. I'm not sure if there is some identity that is just evading me so that this reduces to the desired expression $\frac{1}{2}L - x$, and if not, I'm not sure how to get from here to the required expression.

Any help would be appreciated, thank you!

答:
It comes to the expansion of $f(x)=L/2-x$ which is periodic with a period of $L$ in to a fourier series:
$$F(x)A_0/2+\sum_{1}^{n} [A_n\cos(2 n\pi x/L)+B_n \sin(2n\pi x/L)].~~~~(1)$$
Where $A_n=\frac{2}{L} \int_{0}^{L} f(x) \cos(2n\pi x/L) dx,~~ B_n=\frac{2}{L} \int_{0}^{L} f(x) \sin(2n\pi x/L) dx$
The integration by parts gives
$$A_n=\frac{2}{L} \int_{0}^{L} (L/2-x) \cos(2n\pi x/L) dx=\frac{L(1-\cos 2 n \pi)-n\pi \sin 2n \pi}{2 \pi^2 n^2}=0, n\in I ~~~~(2)$$
as $\cos 2n \pi=1, \sin 2n \pi =0.$
Next $$B_n=\frac{2}{L} \int_{0}^{L} (L/2-x) \sin(2n\pi x/L) dx=\frac{L \cos(n\pi)}{n^2\pi^2}[n\pi \cos (n \pi)-\sin(n\pi)]=\frac{L}{n\pi}~~~~(3),$$
as $\cos(n\pi)=(-1)^n$ and $\sin n \pi =0$.
So finally, putting (2,3) in (1), we have
$$\frac{L}{2}-x=\frac{L}{\pi}\sum_{n=1}^{\infty} \frac{\sin (2n \pi x/L)}{n}.$$


问:
Why in Cantor's Intersection Theorem the infinite intersection must contain only one point?
I am going through GF Simmons' Intro to Topology and Modern Analysis.

$F_n$ is a decreasing sequence of non-empty closed subsets of the Metric Space.

It seems that the condition $d(F_n)$->$0$ given in the hyothesis, ensures that the $$F=\bigcap_{n=1}^\infty F_n\neq\varnothing $$ rather F contains exactly one point is not very clear.

Why does "the supremum of the distances between any two points in the closed sets" converging to "$0$"  implies that the set contains exactly one point?

答:
It is because the $$d(A)=sup\{d(x,y)|x,y\in A \}$$
If this converges to $0$ it means that d(x,y) converges to $0$ for the class of closed sets $F_n$ as n->$\infty$.

We know that $d(x,y)=0$ only for x=y.Therefore there must exist only one point in the set $F=\bigcap_{n=1}^\infty F_n$.


问:
$X$ is $\mathscr{G} (\subset \mathscr{F})$ measurable if $E[XH]=0$ for all bounded $H$ such that $E[H|\mathscr{G}]=0$.
Suppose that we have a probability space $(\Omega, \mathscr{F},P)$ and a sub-$\sigma-$algebra $\mathscr{G}$ and we want to show a mapping $X$ is $\mathscr{G}-$measurable. Why does it suffice to show that $E[XH]=0$ for all bounded random variable $H$ such that $E[H|\mathscr{G}]=0$?

This result is implicitly used in the proof below from Protter's Stochastic Integration and Differential Equations but I am having difficulty showing it.

答:
Answer assuming that $EX^{2} <\infty$: Let $H=XI_{|X| \leq N} -E(XI_{|X| \leq N}|\mathcal G)$. Then $H$ is bounded and $E(H|\mathcal G)=0$. Hence $E(HX)=0$. This gives $EX^{2}I_{|X| \leq N}=E[XE(XI_{|X| \leq N}|\mathcal G)] $. Leting $N \to \infty$ we get $EX^{2}=E[XE(X|\mathcal G)] $.  Now expand $E[X-E(X|\mathcal G)]^{2}$ and use Jensen's inequality to see that  this is  $ \leq 0$. Hence $X=E(X|\mathcal G)$ a.s. and $X$ is a.s. and so $X$ is  $\mathcal G$ measurable.

Jensen's inequality is used as follows: $E[E(X|\mathcal G)]^{2}\leq E [E(X^{2}|\mathcal G)]=EX^{2}$


问:
probability of a player drawing a certain suit in a card game
4 players are playing cards. The cards are dealt out equally to them. If player 1 and player 2 have a total of 6 spades among them, what is the probability of player 3 having 2 of the remaining 7 spades?

Assuming this is a normal deck of cards

(7C2 x 39C11)/52C13

I got a probability of 0.055427 that player 3 has 2 of the remaining 7 spades.

答:
Player $1$ and $2$ have total of $6$ spades out of $26$ cards dealt to them. We already know that.

So there are $7$ spades in $26$ cards distributed to player $3$ and $4$.

Probability of player $3$ being dealt $2$ spades $= \displaystyle \frac {{7 \choose 2} \times {19 \choose 11}}{26 \choose 13}$


问:
Understanding big O notation
I'm not a mathematician by any stretch and I'm trying to translate some maths terms into simple maths terms. Please don't laugh, I do consider this complicated!

The equations in question are

```<code>O(n) and O(n ^ 2)
</code>```

Now, I have read up on Wiki about this but it has been written (IMO) for people who already understand it!

I believe n ^ 2 translates to the power of, in this case it is also the equivalent of squaring it (i.e. n * n).

However, I can't get my head around O in terms of what it is describing. Wiki says it's the limiting behaviour. So, does this mean O is more of a description than a function or command? In my understanding, the following 2 equations are the same

```<code>O(n^2)

n^2
</code>```

答: The formally correct $O$-notation has been explained in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=31517&start=20 . Namely, suppose we have been given a positive $g$ defined in a punctured neighborhood of $x_0$. Now $O_{x_0}(g)$ is the class of all functions $f$ such that the ratio $f/g$ is bounded in some punctured neighbourhood of $x_0$. This definition and notation is more rigorous than for example the one given in some university's computer science courses.</p>

OwnerDisplayName: Dave Rook

Title: Understanding big O notation


问:
Multinomial type finite sum
In a problem related to the study of the Weil-Petersson volume of the moduli space of bordered Riemann surfaces of genus $g$ with $m$ geodesic boundaries, all of length $\ell > 0$, I've encountered the following sum:
$$f(g,m,\ell) \equiv \sum_{j=0}^{3g+m-2} \ell^j \sum_{|\boldsymbol{\alpha}| = j} \prod_{i=1}^m \frac{\alpha_i!}{(2\alpha_i)!},$$
where the $\alpha_i$ are positive integers, $|\boldsymbol{\alpha}| = \sum_{i=1}^m \alpha_i$ and $3g+m>2$. Since $g$ will be fixed I will abbreviate $f(g,m,\ell) \rightarrow f(m,\ell)$. I am most interested in the behavior of $f(m,\ell)$ as $m \rightarrow \infty$, and perhaps (but not necessarily) in the regime $\ell \ll 1$ (but fixed). In particular it would be useful to have bounds on $f$. A very simple upper bound (valid for all $g,m,\ell$) is
$$f(m,\ell) \leq \sum_{j=0}^\infty \ell^j \sum_{|\boldsymbol{\alpha}| = j} \prod_{i=1}^m \frac{1}{\alpha_i!} = \sum_{j=0}^\infty \frac{(\ell m)^j}{j!} = e^{\ell m}.$$
Now I am wondering whether better bounds exist, perhaps when $l \ll 1$ is a small positive number. Clearly for any fixed $m$ we have $f(m,\ell) = 1 + \mathcal{O}(\ell)$ as $\ell \rightarrow 0$. For instance, might there be a $\ell_c$ such that for all $\ell < \ell_c$, $f(m,\ell) = 1 + o(1)$ as $m \rightarrow \infty$? If not, can we find an explicit lower bound $f(m,\ell) \geq g(m,\ell)$ for all $\ell$ and $m > m_0(\ell)$ with $g$ a diverging function as $m \rightarrow \infty$?

My attempts to far: What I have tried so far to tackle this problem is to formulate a (crude) lower bound starting from $$f(m,\ell) \geq \sum_{j=0}^{3g+m-2} \ell^j \sum_{|\boldsymbol{\alpha}| = j} \prod_{i=1}^m \frac{1}{(2\alpha_i)!} = \sum_{j=0}^{3g+m-2} \ell^j \frac{1}{2^m (2j)!} \sum_{k=0}^m \binom{m}{k}(m-2k)^{2j},$$
where this last sum was calculated by @metamorphy. With this I've attempted to proceed by approximating the $j$-sum in a perhaps not so rigorous fashion, namely $$\sum_{j=0}^{3g+m-2} \frac{\left[ \sqrt{\ell}(m-2k) \right]^{2j}}{(2j)!} \approx \sum_{j=0}^{\infty} \frac{\left[ \sqrt{\ell}(m-2k) \right]^{2j}}{(2j)!} = \cosh(\sqrt{\ell}(m-2k)).$$ I expect this approximation to hold because in the left-hand sum the terms in the series certainly become small when $j \gtrsim (e/2) \sqrt{\ell} m$ (for any $k$, and assuming $(e/2) \sqrt{\ell} m$ is large), which, when $\ell$ is small enough compared to 1, is much smaller than the highest index $j = 3g+m-2$ we are summing until. So I expect
$$f(m,\ell) \overset{\approx}{\geq} \frac{1}{2^m} \sum_{k=0}^m \binom{m}{k} \cosh\left[ \sqrt{\ell} (m-2k) \right] \geq \frac{e^{\sqrt{\ell} m}}{2^{m+1}} \sum_{k=0}^m \binom{m}{k} \exp\left( -2 \sqrt{\ell} k \right) = \frac{\cosh^m(\sqrt{\ell})}{2},$$
which indeed diverges as $m \rightarrow \infty$ for any $\ell > 0$.

答:
Too long for a comment.

Starting from @metamorphy's elegant solution, considering
$$T_j=\frac{1}{2^m}\sum_{k=0}^m\binom{m}{k}(m-2k)^{2j}$$ what we can see is that
$$T_j=m \, P_{j-1}(m)$$ and the first of these polynomials are
$$\left(
\begin{array}{cc}
 1 & 1 \\
 2 & 3 m-2 \\
 3 & 15 m^2-30 m+16 \\
 4 & 105 m^3-420 m^2+588 m-272 \\
 5 & 945 m^4-6300 m^3+16380 m^2-18960 m+7936 \\
 6 & 10395 m^5-103950 m^4+429660 m^3-893640 m^2+911328 m-353792
\end{array}
\right)$$

The coefficient of the largest power is $(2j-1)!!$ and the constant term seems to be the tangent numbers (look at sequence $A000182$ in $OEIS$)

Title: Multinomial type finite sum


问:
How to evaluate probability distribution given the parameter is greater than a certain value?
Let X ~ Binomial(n, p)

$H_{0}$: p >= .5

$H_{1}$: p < .5

How do we evaluate the p-value?

ie. $P(X \ge x \mid p >= .5)$

I have seen settings where this is just interpreted as $P(X \ge x \mid p=.5)$, but I'm not sure how this is the same thing.

答:
your system of Hypothesis is equivalent to the following

$$
\begin{cases}
\mathcal{H}_0: p=\frac{1}{2}, \\
\mathcal{H}_1: p<\frac{1}{2},
\end{cases}$$

Thus the p-value is calculated as follows:

$$\mathbb{P}\Bigg[X\leq k|p=\frac{1}{2}\Bigg]$$

Example: to test the Hypothesis System above stated, we use a random sample of size $n=15$ observing $k=4$ successes.

The p-value is $5.92\%$ as well represented in the following drawing


问:
Total probability theorem or Bayes theorem?
How do I solve this problem??

The probability of receiving more than 300 mm of rain in each month is given in the following table. If a monthly rainfall record selected at random is found to have more than 300 mm of rain, what is the probability the record is for July? for December?

Jan 0.02
Feb 0.05
Mar 0.10
Apr 0.40
May 0.60
Jun 0.75
Jul 0.80
Aug 0.70
Sep 0.50
Oct 0.20
Nov 0.05
Dec 0.02

答:
Let $R$ be the event that  in the month selected  we have more than $300$ mm of rain, and let $J$ be the event the selected month is July. We want $\Pr(J|R)$. We have
$$\Pr(J|R)=\frac{\Pr(J\cap R)}{\Pr(R)}.$$

Calculate. We have $\Pr(J\cap R)=\frac{1}{12}(0.8)$.

To calculate $\Pr(R)$, multiply $\frac{1}{12}$ by the probability of rain in each month, and add up.

Note that we have the multiplier $\frac{1}{12}$ on top, and in each term at the bottom. So they cancel, and our answer is just $0.8$ divided by the sum of the $12$ given probabilities.


问:
Prove the formula of combination number
$$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{x+k}=\frac{n!}{x\left( x+1 \right) \cdots \left( x+n \right)}}
\\
$$
I recently read an article and found this formula above. I tried to prove it, but only proved the following result. I don't know how to prove this more general combinatorial equation.
$$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{m+k+1}=\int_0^1{\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \left( -1 \right) ^kx^{m+k}\text{d}x}=\int_0^1{x^m\left( 1-x \right) ^n}\text{d}x}=\frac{m!n!}{\left( m+n+1 \right) !}}
$$

答:
HINT:consider by binomial theorem$$x^m{(1-x)}^n=\binom{n}{0}x^m-\binom{n}{1}x^{m+1}+\binom{n}{2}x^{m+2}..$$

Now integrate with limits $0$ to $1$

For integration part see beta function

Title: Prove the formula of combination number


问:
Letters of the word "ZAMBEZIA"
We will take 3 letters from the letters of "ZAMBEZIA". What is the probability of getting 2 vowels and 1 consonant from the taking? I kinda confused whether I should do it immediately like:
Z, A, M, B, E, Z, I, A;
has 4 vowels and 4 consonants, and a total of 8 letters, so the probability is
$$ P(A) = C(4, 2) . C(4, 1)/C(8, 3) $$
or:
letters in ZAMBEZIA is Z, A, M, B, E, I;
has 3 vowels and 3 consonants, and a total of 6 letters, so the probability is
$$P(A) = C(3, 2) . C(3, 1)/C(6, 3)$$
I don't know which one.

答:
One should take
$$S = \{ Z, A, M, B, E, Z, I, A \} $$

This is a multiset, also denoted as

$$S = \{ 2\cdot A, B, E, I, M, 2\cdot Z \} $$

with multiplicity of elements mentioned. It is different from the set

$$S' = \{ A, B, E, I, M, Z \} $$

where each element occurs only once.

Title: Letters of the word "ZAMBEZIA"


问:
Prerequisites/lecture notes for V. Arnold's PDE
I could not understand Lectures on Partial Differential Equations pass the first chapter, even if I have taken a course in pure math PDE (proving that convolving with the heat kernel is actually convergent, mean value property, maximum principle, etc), and another two courses in Lagrangian and Hamiltonian physics.

Arnold's verdict that it's good for undergrad students seems too optimistic. I think one reason for its difficulty is that it has too little explicit calculation and examples. Many statements are like implicit exercises I can't solve.

So what are prerequisites for studying V. Arnold's PDE? Maybe there is a set of lecture notes that expand on this textbook to make it readable, and supply with exercises and solutions?

答:
Vladimir Arnold generally wrote his books at the standard of his own students, who were Russian. Different countries/regions have different standards for what counts as undergraduate-level preparation. Arnold's book would not be amiss in a graduate course on PDEs.

Skimming through the first chapter, Arnold  seems to assume that the reader is fairly comfortable with differential geometric language, such as transversality. That is consistent with what I know of his other books as well. I would treat differential geometry/topology as one of the prerequisites for this book.


问:
Limit of a continued fraction
Given the continued fraction:
$$f(x,N)=\left[2,3,4,...N,x\right]$$
$$f(x,N)=\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{...+\cfrac{1}{x}}}}}$$
is it possible to find an expression for the integral:
$$g(x,N)=\int f(x,N)dx$$
as function of $N$ and $x$?
Thanks.</p>

答:
UPDATE

Let's observe that for :
$$0+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{...+\cfrac{1}{N+\cfrac{1}x}}}}}}$$
the $n$-th convergent is known (from MathWorld, see too OEIS A052119) using Bessel functions :

\begin{align}
A_n&=\frac{I_n(-2)K_1(2)-I_1(-2)K_n(2)}{I_2(-2)K_1(2)-I_1(-2)K_2(2)}\\
&=2\left(I_n(-2)K_1(2)+I_1(2)K_n(2)\right)\\
&=2\left((-1)^n I_n(2)K_1(2)+I_1(2)K_n(2)\right)\\
\\
B_n&=\frac{I_n(-2)K_0(2)-I_0(-2)K_n(2)}{I_1(-2)K_0(2)-I_0(-2)K_1(2)}\\
&=2\left(-I_n(-2)K_0(2)+I_0(2)K_n(2)\right)\\
&=2\left((-1)^{n-1} I_n(2)K_0(2)+I_0(2)K_n(2)\right)\\
\end{align}
(there was a sign error in MathWorld in the above expression, see too OEIS entries A001053 and A001040)

This means that we have (needing the $N+1$-th convergent for the $N$ at the end) :
$$[0;1,2,\dotsc,N]=\frac{A_{N+1}}{B_{N+1}}$$

From this we may get the next convergent using Lord_Farin's helpful link :

$$[0;1,2,\dotsc,N,x]=\frac{xA_{N+1}+A_{N}}{xB_{N+1}+B_{N}}$$

To get your continued fraction you only need to revert this and subtract $1$ obtaining :
$$[0;2,\dotsc,N,x]=\frac{xB_{N+1}+B_{N}}{xA_{N+1}+A_{N}}-1$$

Allowing the computation of :
\begin{align}
g(x,N)&=\int f(x,N)\;dx\\
&=\int \frac{xB_{N+1}+B_{N}}{xA_{N+1}+A_{N}}-1\;dx\\
&=\int \frac{B_{N}A_{N+1}-A_{N}B_{N+1}}{A_{N+1}(xA_{N+1}+A_{N})}+\frac {B_{N+1}}{A_{N+1}}-1\;dx\\
&=\frac{B_{N}A_{N+1}-A_{N}B_{N+1}}{A_{N+1}^2}\log(A_{N+1}\;x+A_{N})+\frac {B_{N+1}-A_{N+1}}{A_{N+1}}x\\
\\
g(x,N)&=\boxed{\displaystyle (-1)^{N+1}\frac{\log(A_{N+1}\;x+A_{N})}{A_{N+1}^2}+\frac {B_{N+1}-A_{N+1}}{A_{N+1}}x}\quad\text{(after simplification)}\\
\end{align}

I'll let you reverify all this,

Title: Limit of a continued fraction


答:
$$S_{n,x}=\sum_{k=0}^{n} {n \choose k}\frac{(-1)^k}{x+k}= \sum_{k=0}^{n} \int_{0}^{1} {n \choose k}(-1)^k t^{x+k-1} dt=\int_{0}^{1} t^{x-1}(1-t)^n dt~~~~(1)$$
$$S_{n,x}= \frac{\Gamma(x) \Gamma(n+1)}{\Gamma(n+x+1)}~~~~~~~(2)$$ If $x$ is an integer, then
$$S_{n.x}=\frac{(x-1)!~ n!}{(x+n)!}=\frac{(x-1)!}{x(x+1))(x+2)...(x+n)}~~~~(3)$$

Here we have use $\beta$ function

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

Note: OP needs to correct the result.

Title: Prove the formula of combination number

Body:
$$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{x+k}=\frac{n!}{x\left( x+1 \right) \cdots \left( x+n \right)}}
\\
$$
I recently read an article and found this formula above. I tried to prove it, but only proved the following result. I don't know how to prove this more general combinatorial equation.
$$
\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \frac{\left( -1 \right) ^k}{m+k+1}=\int_0^1{\sum_{k=0}^n{\left( \begin{array}{c}
 n\\
 k\\
\end{array} \right) \left( -1 \right) ^kx^{m+k}\text{d}x}=\int_0^1{x^m\left( 1-x \right) ^n}\text{d}x}=\frac{m!n!}{\left( m+n+1 \right) !}}
$$


问:
Show that if $\phi \oplus \psi \leq c \implies \inf_{\gamma \geq 0} \int c - \phi \oplus \psi d\gamma = 0 $
In the book Optimal Transport for Applied Mathematicians, the original problem for Optimal Transport is written as:
$$
\min_\gamma \quad \int_{X \times Y} c d\gamma \quad + \quad
\sup_{\phi,\psi} \quad \int_X \phi d\mu + \int_Y \psi d\nu - \int_{X \times Y}\phi(x) - \psi(y) d\gamma
$$

Where $\gamma \in \mathcal P(X\times Y)$ and $\mu,\nu \in \mathcal P(X),\mathcal P(Y)$, respectively.

Then, the author writes the dual as:

$$
\sup_{\phi,\psi} \quad \int_X \phi d\mu +  \int_Y \psi d\nu  \quad +
 \quad \inf_\gamma \quad \int_{X \times Y} c(x,y) - \phi(x) - \psi(y) d\gamma
$$

Finally, the author then claims the following:
$$
\inf_\gamma \quad \int_{X \times Y} c(x,y) - \phi(x) - \psi(y) d\gamma = 0, \quad
 = \begin{cases}
  0, & \text{if } \phi \oplus  \psi \leq c \quad \text{on }X\times Y \\
  -\infty, & \text{otherwise}.
\end{cases}
$$

My question is why if $\phi \oplus \psi \leq c \implies \inf_{\gamma \geq 0} \int c - \phi \oplus \psi d\gamma = 0 $ ?

Since $\gamma $ is a probability measure, is not like I can make the measure as small as I'd like.

For the case where $\gamma$ is just a positive measure, then I can see how we could make $\gamma=0$. But I don’t see how this would be valid assuming that we are dealing with probability measures.

答:
If you look at the book you can see that the infimum is taken over $\mathscr{M}_+(X\times Y)$ - meaning the space of positive finite measures.
Therefore you can just choose $\gamma$ to be the zero measure.


问:
Find all differentiable functions $f : [0, +\infty) \rightarrow \mathbb{R}$ such that $f(1) = 1$ and $f(x)f(y) \leq f(xy)$
The exercise

Find all differentiable functions $f : [0, +\infty) \rightarrow \mathbb{R}$ such that:

$f(1) = 1$

$f(x)f(y) \leq f(xy)$ for all $x, y \geq 0$

My try

I have found that the functions $x \mapsto x^\alpha$ for $\alpha \in [1, +\infty)$ are solutions. The constant function $x \mapsto 1$ is also solution.

As $f(0)^2 \leq f(0)$ one also has that $f(0) \in [0, 1]$.

If I take a solution $f$, my idea would be to try proving it is of the form $f(x) = x^\alpha$ for some $\alpha$. Necessarily this $\alpha$ would be equal to $f'(1)$. So I set $\alpha := f'(1)$ and considered $g(x) = \frac{f(x)}{x^\alpha}$ for $x > 0$. One can show $g$ is derivable and $g'(1) = 0$ and $g$ verifies the functional equation (for $x, y > 0$). So the goal would be to show that $g = 1$, but I've not managed to prove it yet. Any help is welcome folks!

答:
First note that $f(1)=1$, and $f$ differentiable, implies that continuous, and hence
$f(x)>0$, in $(1-\delta,1+\delta)$, for some $\delta>0$.

Next, if $x\in (0,\infty)$, then $x_n=x^{1/n}\in(1-\delta,1+\delta)$, for $n$ sufficiently large, and hence
$$
f(x_n\cdot x_n)\ge \big(f(x_n)\big)^2\quad\Longrightarrow\quad
f(x_n\cdot x_n\cdot x_n)\ge f(x_n\cdot x_n)f(x_n)\ge \big(f(x_n)\big)^3
 \\ \quad\Longrightarrow\quad f(x)\ge f(x_n^n)\ge\big(f(x_n)\big)^n>0.
$$
Thus $f(0,\infty)\subset (0,\infty)$.

Set next $g(x)=\log f(\mathrm{e}^x)$.
Then $g:\mathbb R\to\mathbb R$, $g(0)=0$ and
$$
g(x+y)\ge g(x)+g(y) \quad \text{for all $x,y\in\mathbb R$.}
$$
The $g$ is also differentiable as a composition of such and let
$$
c=\lim_{h\to 0} \frac{g(h)-g(0)}{h}=\lim_{h\to 0} \frac{g(h)}{h}.
$$
Then
$$
g(x+h)\ge g(x)+g(h) \quad\Longrightarrow\quad \frac{g(x+h)-g(x)}{h}-\frac{g(h)}{h}=
\left\{\begin{array}{ccc}\text{non-negative if $h>0$}, \\ \text{non-positive if $h<0$.} \end{array}\right.
$$
and since $\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$ exists, then
$$
\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}=\lim_{h\to 0}\frac{g(h)}{h}=c
$$
which implies that $g(x)=cx$ and
$$
f(x)=\mathrm{e}^{g(\log x)}=\mathrm{e}^{c\log x}=x^c.$$

Finally, as $f$ is defined and it is also differentiable at $x=0$, then $c\ge 1$.


问:
Legendre polynomials n=1 calculation by hand
Considering the Legendre polynomials:
$$P_n(x) = \sum_{m=0}^{n}a_{n,m}{x^m}$$
I know that: $P_0=1$ and $P_1=x$.

However given $P_0$ if I want to find $P_1$ by hand:
$$\langle P_1 | P_0 \rangle = 0 = \int_{-1}^{1}(a_0+a_1x)dx = 2a_0 = 0 \longrightarrow a_0=0$$
Imposing normalization condition:
$$\langle P_1 | P_1 \rangle = 1 = \int_{-1}^{1}a_1^2x^2 dx = \frac{2}{3}a_1^2 = 1 \longrightarrow a_1= \sqrt{ \frac{3}{2}} $$
So from my computations $P_1= \sqrt{ \frac{3}{2}} x$.

Where am I wrong?

答:
$P_1' = x$ is not normalized since

$$\langle P_1' | P_1' \rangle = \int_{-1}^{1} x^2 dx = \dfrac{2}{3} $$

$P_1 = \sqrt{\dfrac{3}{2}}x$ is normalized.

$$\langle P_1 | P_1 \rangle = 1 $$

Both are correct upto scalar multiple. It depends which one you need.


问:
Is $x\mapsto x^2$ (map) from $\mathbb{R}\to \mathbb{R}$ (real numbers), the same as $f(x)=x^2$ and not injective?
For example in: $f(x)=x^2$  from the set of real numbers real numbers to real numbers is not an injective function because of this kind of thing:

$f(2) = 4$ and
$f(-2) = 4$
This is against the definition $f(x) = f(y), x = y$, because $f(2) = f(-2)$ but $2 ≠ -2$

In other words there are two values of A that point to one B.

But I just don't understand if the notation is the same? Can I use the same example to prove this?

答:
Indeed I would say that the two notations are equivalent.

You can either write $f(x)=x^2$ or $x\mapsto x^2$ to express the same idea: $x$ is mapped to its square.

The difference between the two is that the first gives an explicit name to the function, namely $f$, while the second does not.

Specifying the domain is something else. Both can be used on the whole of $\mathbb{R}$ or restricted to some subdomain.


答:
This is just a difference in notation.

What is $f:\mathbb{R}\to\mathbb{R}$, such that $f(x) = x^2$? It is a function that maps elements $x \in \mathbb{R}$ to elements $x^2\in\mathbb{R}$.

In other words, $f(x)=x^2$ from $\mathbb{R}$ to $\mathbb{R}$ is the map:

$f:\mathbb{R}\to\mathbb{R},x\mapsto x^2$.

Title: Is $x\mapsto x^2$ (map) from $\mathbb{R}\to \mathbb{R}$ (real numbers), the same as $f(x)=x^2$ and not injective?

Body:
For example in: $f(x)=x^2$  from the set of real numbers real numbers to real numbers is not an injective function because of this kind of thing:

$f(2) = 4$ and
$f(-2) = 4$
This is against the definition $f(x) = f(y), x = y$, because $f(2) = f(-2)$ but $2 ≠ -2$

In other words there are two values of A that point to one B.

But I just don't understand if the notation is the same? Can I use the same example to prove this?


问:
Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$
Tried to evaluate the integral

$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that

\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\ln^2x}{(x+1+i)(x+1-i)} \, dx
 + \int_0^1 \frac{\ln^2x}{(x+1+ix )(x+1-ix )} \, dx\\
 &= -2\operatorname{Im}\operatorname{Li}_3\left(-\frac{1+i}2\right)
 -2\operatorname{Im} \operatorname{Li}_3(-1-i)
\end{align}

which is equal to $ \frac{5\pi^3}{64}+\frac\pi{16}\ln^22$.

It is perhaps unnecessary, though, to resort to evaluation in complex space. I would like to work out an elementary derivation of this integral result.

Body:
\begin{align}J&=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx\\
&=\int_0^\infty \frac{\ln^2x}{x^2+2x+2} \, dx\\
&\overset{x=\sqrt{2}t}=\frac{1}{\sqrt{2}}\int_0^\infty \frac{\ln^2(\sqrt{2}t)}{t^2+\sqrt{2}t+1} \, dt\\
&=\frac{1}{\sqrt{2}}\underbrace{\int_0^\infty \frac{\ln^2(\sqrt{2})}{t^2+\sqrt{2}t+1} \, dt}_{=\dfrac{\pi\ln^2 2}{8\sqrt{2}}}+\underbrace{\sqrt{2}\int_0^\infty \frac{\ln(\sqrt{2})\ln t}{t^2+\sqrt{2}t+1} \, dt}_{=0}+\frac{1}{\sqrt{2}}\int_0^\infty \frac{\ln^2 t}{t^2+\sqrt{2}t+1} \, dt\\
&=\dfrac{\pi\ln^2 2}{16}+\frac{1}{\sqrt{2}}\int_0^\infty \frac{\ln^2 t}{t^2+\sqrt{2}t+1} \, dt\\
K&=\int_0^\infty \int_0^\infty \frac{\ln^2 (tx)}{(t^2+\sqrt{2}t+1)(x^2+\sqrt{2}x+1)} \, dt\,dx\\
&\overset{u(x)=tx}=\int_0^\infty \int_0^\infty \frac{t\ln^2 u}{(t^2+\sqrt{2}t+1)(u^2+\sqrt{2}tu+t^2)} \, dt\,du\\
&=\int_0^\infty \ln^2 u\left[\frac{(u+1)\ln\left(\frac{t^2+\sqrt{2}tu+u^2}{t^2+\sqrt{2}t+1}\right)}{2(1-u)(1+u^2)}-\frac{\arctan\left(\frac{u+\sqrt{2}t}{u}\right)}{1+u^2}-\frac{\arctan\left(\sqrt{2}t+1\right)}{1+u^2}\right]_{t=0}^{t=\infty}du\\
&=\int_0^\infty \ln^2 u\left(-\frac{\pi}{1+u^2}-\frac{(1+u)\ln u}{(1-u)(1+u^2)}+\frac{\frac{\pi}{2}}{1+u^2}\right) du\\
&=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\int_0^\infty\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du\\
&=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du-\int_1^\infty\frac{(1+v)\ln^3 v}{(1-v)(1+v^2)}dv\\
&\overset{u=\frac{1}{v}}=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du-\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du\\
&=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1\frac{(1+u)\ln^3 u}{(1-u)(1+u^2)}du\\
&=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\int_0^1 \frac{2v\ln^3 v}{1+v^2}dv\\
&\overset{u=v^2}=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\int_0^1 \frac{\ln^3 u}{1+u}du\\
&=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\left(\int_0^1 \frac{\ln^3 u}{1-u}du-\int_0^1 \frac{2v\ln^3 v}{1-v^2}dv\right)\\
&\overset{u=v^2}=-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-2\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\left(\int_0^1 \frac{\ln^3 u}{1-u}du-\frac{1}{8}\int_0^1 \frac{\ln^3 u}{1-u}du\right)\\
&==-\frac{\pi}{2}\int_0^\infty \frac{\ln^2 u}{1+u^2}du-\frac{135}{64}\int_0^1 \frac{\ln^3 u}{1-u}du\\
&=-\frac{\pi}{2}\times \frac{\pi^3}{8}-\frac{135}{64}\times -\frac{\pi^4}{15}\\
&=\frac{5\pi^4}{64}
\end{align}

On the other hand,
\begin{align}K&=2\underbrace{\left(\int_0^\infty\frac{\ln x}{x^2+\sqrt{2}x+1}\right)^2}_{=0}+2\sqrt{2}\left(J-\frac{\pi\ln^2 2}{16}\right)\underbrace{\int_0^1 \frac{1}{x^2+\sqrt{2}x+1}dx}_{=\frac{\pi}{2\sqrt{2}}}\end{align}

Therefore,

$\displaystyle \boxed{J=\frac{5\pi^3}{64}+\frac{\pi\ln^2 2}{16}}$

NB:
I assume that,

\begin{align} \int_0^\infty \frac{\ln^2 x}{1+x^2}dx&=\frac{\pi^3}{8}\\
\int_0^\infty \frac{\ln^3 x}{1-x}dx&=-6\zeta(4)\\
\displaystyle\zeta(4)&=\frac{\pi^4}{90}\end{align}


问:
How to prove the Delta Method?
As exercise 9.14 of "Measure theory and probability theory" by Krishna and Soumendra, I'm trying to demonstrate the delta method.

We have a random variable sequence $X_n$, a divergent sequence of real number $a_n \rightarrow +\infty$ and a function $H$ differentiable in $\theta$, so $H'(\theta)=c$, all such that:

\begin{gather}
a_n \left( X_n - \theta\right) \longrightarrow^d Z  \ \ \ \text{(converge in distribution)}
\end{gather}

I want to show that

\begin{gather}
a_n \left( H(X_n) - H(\theta) \right) \longrightarrow^d cZ  \ \ \
\end{gather}

The book suggest to use the Taylor's expansion
\begin{gather}
H(X)-H(\theta) = c(X - \theta) + R(X)(X - \theta)
\end{gather}

where $R(x)\rightarrow0$. Now considering the claim

\begin{gather}
a_n \left( c(X_n - \theta) + R(X_n)(X_n - \theta) \right) \longrightarrow^d cZ  \ \ \
\end{gather}

it's clear that $a_nc(X_n - \theta)$ tends to $cZ$ by Slutsky theorem but how to show that the remainder part tends to $0$? The book seems to suggest that $R(X_n)$ may be stochastically bounded that would be sufficient, but how to demonstrate it?

答:
Here (in Italian, as you are) you can find the complete proof, but I want suggest another approach:

Using Cramér Rao's inequality we have, in general, the following lower bound for Variance of estimators

$$\mathbb{V}[T]\geq \frac{\Big[\frac{\partial}{\partial\theta}\mathbb{E}_{\theta}[T]\Big]^2}{nI(\theta)}$$

Where $I(\theta)$ is the Fischer Information

If we consider $T$ restricted to the class of unbiased estimator for

$\theta$, we have

$$\mathbb{V}[T]\geq \frac{1}{nI(\theta)}$$

$g(\theta)$, we have

$$\mathbb{V}[T]\geq \frac{\Big[g'(\theta\Big]^2}{nI(\theta)}=\mathbb{V}[\hat{\theta}]\times \Big[g'(\theta)\Big]^2$$

Title: How to prove the Delta Method?


问:
Coordinate offset in linear PDE
I have a linear PDE that is not quite traditional of the form $$ F_{x}(x+y,y) - F_{y}(x+y,y) = 0$$ with $F(x+0,0) = f(x).$

I have tried using the method of characteristics in the standard way but it doesn't seem to apply to the present problem. Is this solvable by classical means such as with the method of characteristics and not merely guessing?

答:
It can be solved by classical means after a change of coordinates. Setting $u=x+y$, $v=y$, the PDE becomes
$$
F_x(u, v) - F_y(u,v) = 0
$$
with $$
\begin{aligned}
F_x &= F_u u_x + F_v v_x = F_u \\
F_y &= F_u u_y + F_v v_y = F_u + F_v
\end{aligned}
$$
according to the chain rule. Thus, we end up with $-F_v(u,v) = 0$, which can be solved straightforwardly as $F(u,v) = g(u)$ for some arbitrary function $g$. Along the boundary $y=0$, we therefore have $F(x,0) = g(x)$, while we must have $F(x,0) = f(x)$ according to the problem statement. This gives us $g = f$. Finally, the solution reads $$F(u,v) = f(u) \, ,$$ or equivalently, $F(x+y, y) = f(x+y)$.

Note: The boundary condition $F(x+0,0) = f(x)$ along $y=0$ is equivalent to the boundary condition $F(u,v) = f(u)$ along $(u,v)=(x,0)$ as written above, which in turn is the same as imposing $F(u,v) = f(u-v)$ along $(u,v)=(x,0)$, with identical final result.

Title: Coordinate offset in linear PDE


问:
Evaluate $\int_{0}^{a} x^3 (ax-x^2)^{\frac 32} dx$
The integral can be written as
\begin{align}
\int_0^a (a-x)^3 (ax-x^2)^{\frac 32} dx &=\int_0^a (a-x)^3 x^{\frac 32} (a-x)^{\frac 32} dx\\
&=\int_0^a (a-x)^{\frac 92} x^{\frac 32} dx
\end{align}

How do I solve further?

答:
Hint:

Put $x=az, dx=a\ dz$  to reach at Beta function

Alternatively,  set $x=a\sin^2t$   to reach at this   $$B(m+1,n+1)=\dfrac{m!n!}{(m+n+1)!}$$


问:
Inequality $\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq 1$
For $x,y,z \in [2,\infty)$, prove that

$\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq  1$

I tried to group the terms and prove that $\frac{xy+z}{x+yz} - \frac{y}{3}\leq \frac{1}{3}$

答:
using the lemma:if $a,b\ge 2$ then $ab\ge a+b$

$$\sum_{cyc} \frac{xy+z}{x+\color{red}{yz}}-\sum \frac{x}{3}$$ $$\le \sum_{cyc} \frac{xy+z}{x+\color{red}{y+z}}-\sum\frac{x}{3}=1-\dfrac{1}{6}\Big( \dfrac{{(x-y)}^2+{(y-z)}^2+{(z-x)}^2}{x+y+z}\Big)\le 1$$


问:
Free online mathematical software
What are the best free user-friendly alternatives to Mathematica and Maple available online?

I used Magma online calculator a few times for computational algebra issues, and was very much satisfied, even though the calculation time there was limited to $60$ seconds.

Very basic computations can be carried out with Wolfram Alpha. What if one is interested in integer relation detection or integration involving special functions, asymptotic analysis etc?

Thank you in advance.

Added: It would be nice to provide links in the answers so that the page becomes easily usable. I would also very much appreciate short summary on what a particular software is suitable/not suitable for. For example, Magma is in my opinion useless for doing the least numerics.

答: Don't forget GAP which is a system for computational discrete algebra, with particular emphasis on Computational Group Theory. Sometimes, I really can't go further without using it while solving and guessing the desired results in Group theory problems. Try it. It makes you feeling better. :)</p>

Title: Free online mathematical software


问:
A function whose value is either one or zero
First I apologize in advance for I don't know math's English at all and I haven't done math in almost a decade.

I'm looking for a function whose "domain/ensemble of definition" would be ℝ (or maybe ℤ) and whose "ensemble/domain of variation" would be ℕ{0, 1}
that would look like something this awesome ascii graph...

```<code>          ^f(x)
          |
          |
1________ | __________
0________\./__________>x
         0|
          |
          |
          |
</code>```

f(x) is always 1, but 0 when x = 0

Actually I need this for programming, I could always find other ways to do what I wanted with booleans so far but it would be much better in most cases to have a simple algebra formula to represent my needs instead.

I just hope it's not too complicated a function to write/understand and that it exists of course. Thanks in advance guys.

答:
I found this question relevant in social science when you are testing for the effect  of a variable but are uncertain whether it is its mere presence which is significant, or whether the degree of the presence should be used. I have used the following approach in Excel to convert degree of presence ($A_2 \ge 1$) to just presence ($A_2=1$).

The following simple nested 'IF' statement will return 1 if $A_2$ is a positive number, $0$ for $0$, and $-1$ for a negative number.

=IF(A2=0,0,IF(A2>0,1,-1))


问:
Spivak Calculus Chapter 6 Exercise 9
The (a) part is clear:

Suppose that $f$ is not continuous at $a$. Prove that for some number
$\varepsilon>0$ there are numbers $x$ arbitrarily close to $a$ with
$|f(x)-f(a)|>\varepsilon$.

It's not the case that $f$ continuous. Therefore it's not the case that $$\forall\varepsilon>0,\exists\delta>0,\forall x(|x-a|<\delta \rightarrow |f(x)-f(a)|<\varepsilon)$$

which is logically equivalent to

$$\exists\varepsilon>0,\forall\delta>0,\exists x(|x-a|<\delta \mbox{ and } |f(x)-f(a)|\geq\varepsilon)$$

There are $\varepsilon>0$ such that $|f(x)-f(a)|>\varepsilon$. Let $\varepsilon'=\frac12\varepsilon$. Then we have $|f(x)-f(a)|\geq\varepsilon>\varepsilon'$

The (b) part is not so much:

Conclude that for some number $\varepsilon>0$ either there are
numbers $x$ arbitrarily close to $a$ with $f(x)<f(a)-\varepsilon$ or
there are numbers $x$ arbitrarily close to $a$ with
$f(x)>f(a)+\varepsilon$

From (a) we have $|f(x)-f(a)|>\varepsilon$, which implies $f(a)-\varepsilon>f(x)>f(a)+\varepsilon$.

But $(f(a)+\varepsilon, f(a)-\varepsilon)=(-\infty,f(a)-\varepsilon) \cap (f(a)+\varepsilon,+\infty)$ is an empty set, $f(x)$ can't be in there.

How does it follow that that $f(x)\in (-\infty,f(a)-\varepsilon) \cup (f(a)+\varepsilon,+\infty)$?

Also it seems like a contradiction, does it mean that $f$ is actually continuous at $a$ (which is nonsense, all functions are continuous for all $a$)?

答: There exists a sequence of distinct points $x_n \to a$ such that $|f(x_n)-f(a)| >\epsilon$ for all $n$. The sets $\{n: f_n(x) >f(a)+\epsilon\}$ and $\{n: f_n(x) <f(a)-\epsilon\}$ cannot both be finite. (If they are both finite then $\mathbb N$ would be  a finite set). If there are infinitely many points in the first set then there are points arbitrarily close to $a$  with $f(x) >f(a)+\epsilon$. If there are infinitely many points in the second set then there are points arbitrarily close to $a$  with $f(x) <f(a)-\epsilon$.

Title: Spivak Calculus Chapter 6 Exercise 9


问:
Is it possible to have more vectors than the dimension of the vector space to be linearly independent
By linear independence I understand that each vector in the vector space is not a scalar multiple of the other i.e. any one vector can be scaled independently without having to scale the other vectors.

But I seem to lack some understanding of this concept as I can think of three vectors in $\mathbb{R}^2$ that I find are linearly independent as in the below image (I could not make a better version of the graph).

Are these three vectors linearly dependant? If so how?

答: For two vectors, your idea of linear independence is OK. But that is not the correct definition for more than two vectors.  This link may provide more information.</p>


问:
Is $\int^a\frac{\sqrt{x^3-a^3}}{\sqrt{x-a}}dx $ equal to $0$?
Obviously this is not a nice integral for us to just analytically compute.

My question is whether this integral is $0$ or whether it is undefined and how we would should what it is? Of course, if $x=a$ then the integrand is undefined and if $x<a$ then the integrand is not real. I'm not sure if this has any relevance to this problem?

答:
As said in comments and answers, you simply need to compute
$$I=\int\sqrt{t^2+at+a^2}\,dt$$
Start completing the square
$$t^2+at+a^2=\left(t+\frac{a}{2}\right)^2+\frac{3 a^2}{4}$$ So, let $$t=\frac{a \left(3 u-\sqrt{3}\right)}{2 \sqrt{3}}\implies I=\frac{3}{4} a^2\int \sqrt{u^2+1}\,du$$  which is well known. One integration by parts gives
$$I=\frac{3}{8} a^2 \left(u\sqrt{u^2+1} +\sinh ^{-1}(u)\right)$$ Back to $t$
$$I=\frac{1}{4} (2 t+a) \sqrt{t^2+at+a^2}+\frac{3}{8} a^2 \sinh ^{-1}\left(\frac{2 t+a}{a\sqrt{3} }\right)+C$$

Integrated between $a$ and $x$ as @J.G. did
$$J=\int_a^x\sqrt{t^2+at+a^2}\,dt=\frac{1}{4} (2 x+a) \sqrt{x^2+ax+a^2}+$$ $$\frac{3}{8} a^2 \sinh ^{-1}\left(\frac{2 x+a}{a\sqrt{3} }\right)-\frac{3}{8} a^2 \left(2 \sqrt{3}+\sinh ^{-1}\left(\sqrt{3}\right)\right)$$


问:
Why $p^i/p^{i+1}$ is a vector space over $O_k/p$
Let $O_K$ be a number ring of number field $K$ and $p$ be a prime ideal of $O_K$. Consider the chain
$$p\subseteq p^2\subseteq ...\subseteq p^v$$
If we choose $a\in p^i\setminus p^{i+1}$, and $b=(a)+p^{i+1}$, so, $p^{i+1}\subset b\subseteq p^i$ so, $b=p^i$.

So, $p^i/p^{i+1}$ is generated by a single element, $a\;mod\;p^{i+1}$. But why it generate $p^i/p^{i+1}$ over $O_K/p$ is what I am not getting.

This is a specific proof form Newurich, page-35, prop-6.1.

答: Obviously $P^n/P^{n+1}$ is an $O/P$-module. But since $O$ is a Dedekind ring, $P$ is maximal, so that 1) $O/P$ is a field ; 2) there is no ideal which is strictly "sandwiched" between $P^n$ and $P^{n+1}$. This means that $P^n/P^{n+1}$ is a vector space of dimension 1 over the residual field $O/P$.


答: Yes, writing $x\mapsto x^2$ signifies the same function as writing $f(x)=x^2$, but without giving it a name. Some times you don't need to refer back to that function later, and need all the letters for other things, so having nameless (anonymous) functions can be useful.


问:
Minimal number of edges in an directed graph that ensures the existence of a cycle
I need to find the minimal number of edges in an directed graph with $n$ vertices, that ensures the existence of a cycle in the graph, and proof that for every directed graph with $n$ vertices and that number of edges (or more) there is a cycle.

I tried to find it for small values of $n$ and find a pattern, but even this didn't work.

答: Hint: show that you can put one edge between every pair of vertices without creating a cycle.</p>


答:
No it is impossible.We know that 2 linearly independent vectors form a basis in $R^2$

Assume you are right and  there exist 3 linearly independent vectors at $R^2$ such that
$v_1,v_2,v_3$ so there must not be scalars different than 0 such that

$c_1v_1+c_2v_2+c_3v_3=0$

we know that $\{v_1,v_2\}$ forms a basis so there exists $a,b$ such that

$c_3av_1+c_3bv_2=v_3$ $\Rightarrow$
$c_1v_1+c_2v_2+c_3av_1+c_3bv_2=(c_1+c_3a)v_1+(c_2+c_3b)v_2=0$

according to fact that $\{v_1,v_2\}$ forms a basis

$(c_1+c_3a)v_1+(c_2+c_3b)v_2=0$ $\Rightarrow$ $c_1+c_3a=0 ,c_2+c_3b=0$

$c_2+c_3b=0$ $\Rightarrow$ $c_3=\frac{-c_2}{b}$

$c_1+c_3a=0$ $\Rightarrow$ $-c_3=\frac{-c_1}{a}$

so there exists a $c_3$ which is not $0$ that is a contradiction so your assumption is wrong

Title: Is it possible to have more vectors than the dimension of the vector space to be linearly independent

Body:
By linear independence I understand that each vector in the vector space is not a scalar multiple of the other i.e. any one vector can be scaled independently without having to scale the other vectors.

But I seem to lack some understanding of this concept as I can think of three vectors in $\mathbb{R}^2$ that I find are linearly independent as in the below image (I could not make a better version of the graph).

Are these three vectors linearly dependant? If so how?


问:
Integral of $\int(4-2x)^\frac{1}{3}dx$
I solved this integral then I did $\frac{d}{dx}$ of $F(x)$ and saw that it's not the same, so I did something wrong in my integration process.
$$\int(4-2x)^\frac{1}{3}dx$$

What I did is $$F(x) =\frac{(4-2x)^\frac{4}{3}}{-2\times\frac{4}{3}} + C$$
Thanks!

答:
Hint:
use this formula :$\displaystyle \int (ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}$

in your question $\displaystyle a=-2 , b=4,n=\frac{1}{3}$

so answer will be $\displaystyle \frac{3\cdot(4-2x)^\frac{4}{3}}{-8}$

Title: Integral of $\int(4-2x)^\frac{1}{3}dx$


问:
Bounding an Error on the inverse.
Suppose $f(x) = x^p + $ lower order terms. Then Asymptotically $f^{-1}(x) \sim x^{\frac{1}{p}} $ for large $|x|$. How can we bound the error in this asymptotic approximation in terms of $|x|$

答:
$f^{-1}(f(x))$ is supposed to be $x$. You can look at different kinds of errors, the two simplest are the absolute difference and the relative difference:
$$E_1(x) = x- f^{-1}(f(x)),\qquad E_2(x) = 1- \frac{f^{-1}(f(x))}{x}$$
If you want to find an asymptotic expansion of the error at infinity you need to look at $E(1/x)$ for $x$ close to $0$. Then you can do a Taylor expansion. This is nicer if we use $E_2$:
$$E_2(1/x)=1-x\sqrt[p]{\frac1{x^p}+\sum_{k<p} a_k\frac1{x^k}}=1-\sqrt[p]{1+\sum_{k<p}a_kx^{p-k}}$$
Here the $a_k$ are the coefficients in $f(x) = x^p + \sum_{k<p} a_k x^k$. This term is $0$ at $x=0$, the first derivative is some function you can easily derive. At $0$ the derivative evaluates to:
$$-\frac1p a_{p-1}$$
so you have
$$E_2(1/x) = 0 -\frac{a_{p-1}}{px} + O(\frac1{x^2})$$
continue in this way.

Title: Bounding an Error on the inverse.


问:
Why does $\frac{1}{z}$ not have an antiderivative but $\frac{1}{z^2}$ does?
On the complex plane $\mathbb C-\{0\}$, it is stated that $\frac{1}{z}$ does not have an antiderivative. But apparently $\frac{1}{z^2}$ does and I find this confusing because both functions have a pole at 0 so that means any closed curve around the origin is not 0 right?

答: The fact that a function has a pole does not mean that its integral along a closed curved around the pole is not zero. This depends on the value of the residue of the function at this pole (the residue at $z=0$ of $f(z)=1/z$ is non-zero, but the residue of $f(z)=1/z^2$ is zero).


问:
Binary vs. Ternary Goldbach Conjecture
Is there an "understandable" explanation of why the ternary Goldbach conjecture is tractable with current methods, while the binary Goldbach conjecture seems to be out of scope with current techniques? As a non-expert it seems to me that these conjectures should be equally hard to prove, which must of course not be true since these statements are not equivalent, but they are of the same "nature" and I don't see why it is such a difference that a number is the sum of two or of three primes...</p>

答: The "current methods" have almost always been refinements of Hardy-Littlewood circle method. Terry Tao has a blog post where he describes how the circle method applies to the problem, and why experts think this method alone will not yield the even Goldbach conjecture. </p>

Title: Binary vs. Ternary Goldbach Conjecture


问:
How does $(abc) = (ac)(ab)$?
I know that permutations in the symmetric group, permutations are the finite products of transpositions. This is given:
$$
(abc) = (ac)(ab) \\
(abcd) = (ad)(ac)(ab) \\
\vdots  \\
(a_1a_2 \cdots a_k) = (a_1a_k)(a_1a_{k-1})\cdots(a_1a_2)
$$

However, I'm lost on how $(abc) = (ac)(ab)$ occurs. I know $(abc)$ takes $a$ to $b$, $b$ to $c$, and $c$ to $a$, but how does the two 2-cycles do the same thing? Is it not $(ac)$ takes $a$ to $c$, then $(ab)$ takes $a$ to $b$, so that gives $cab$? Same with the second line, $(abcd) = (ad)(ac)(ab)$.

Thank you in advance!

答:
$(a b c) =a\mapsto b, b\mapsto c, c\mapsto a$

For $(a c)(a b) $ we need to start the operation from rightmost, so

$(a b)  = a\mapsto b, b\mapsto a, c\mapsto c$, when we apply $(a c)$ on this we get

$a\mapsto b\mapsto b, b\mapsto a\mapsto c, c\mapsto c\mapsto a$, so the composite map is

$a\mapsto b, b\mapsto c, c\mapsto  a = (abc)$

Title: How does $(abc) = (ac)(ab)$?


问:
Prove if$~\vert x \vert < z~$ and $~\vert y \vert < z~$ then $~\Big\vert \dfrac{x+y}{2}\Big\vert + \Big\vert \dfrac{x-y}{2}\Big\vert < z~$
Let $x~$,$y~$,$z~\in\mathbb{R}$

Prove that : if$~\vert x \vert < z~$ and $~\vert y \vert < z~$ then $~\Big\vert \dfrac{x+y}{2}\Big\vert + \Big\vert \dfrac{x-y}{2}\Big\vert < z~$

I know how to prove it by separating cases, but I need a Hint for a direct proof.

答:
Without loss of generality you can assume that $x^2 \ge y^2$. Then
$$
\Bigl(|x+y| + |x-y|\Bigr)^2 = (x+y)^2 + (x-y)^2 + 2|(x+y)(x-y)| = 4x^2 < 4 z^2.
$$

More precisely we have, using that $\max(a, b) = \frac 12 (a+b+|a-b|$:
$$
\Bigl(|x+y| + |x-y|\Bigr)^2 = (x+y)^2 + (x-y)^2 + 2|(x+y)(x-y)| \\
= 2x^2 + 2y^2 + 2|x^2-y^2| = 4\max(x^2, y^2)
$$
and therefore
$$
\left\vert \frac{x+y}{2}\right\vert + \left\vert \frac{x-y}{2}\right\vert = \max(|x|, |y|) \, .
$$


问:
Indian National Olympiad $2005$ Problem $5$ (Not able to understand this problem)
Let $x_1$ be a given positive integer. A sequence $\{x_n\}_ {n\geq 1}$ of positive integers is such that $x_n$, for $n \geq 2$, is obtained from $x_ {n-1}$ by adding some nonzero digit of $x_ {n-1}$.

Prove that

a) the sequence contains an even term;

b) the sequence contains infinitely many even terms

Now i did not understand the question and what it is asking to prove ,
what is meaning of "$x_n$, for $n \geq 2$, is obtained from $x_ {n-1}$ by adding some nonzero digit of $x_ {n-1}$"

So can somebody pls give me some example ?

thankyou

答:
Hint-Notation- $a_1 a_2 a_3 \dots a_n$ is an $n$ digit number, that is $23$ is a $2$ digit number with $a_1=2$ and $a_2=3, 92938$ is a $5$ digit number with $a_1=9,a_2=2,a_3=9,a_4$ $=3$ and $a_5=8$

If you start with $x_1=a_1a_2a_3 \dots a_k$, then if $a_k$ is even then we are done, else show that after few steps you will get $x_n=b_1b_2 \dots b_k$ where $b_1,b_2,b_3 \dots b_{n-1} $ are all odd. Then if $b_n $ is $odd$, then $x_{n+1}$ is even else $x_n$ is even

Repeated application of the above procedure will give you infinitely many even numbers


问:
Minimum step procedure to solve this non-linear system
I would like to find out the easiest way  (i.e. minimal number of steps required) to solve this non-linear system w.r.t variables $x_1, x_2, y_1, y_2$ ($a,b,c,d$ are constants):

$$
\begin{cases} \frac{x_1x_2}{y_1y_2}=a \\ \frac{x_1}{y_1} + \frac{x_2}{y_2}=b \\ x_1+x_2=c \\ y_1+y_2=d \end{cases}
$$

I solved my numerical problem through different approaches, also exploiting non-linear optimization routines but I would like to see if there are different ways to approach this problem that I did not figure out.

Best regards

Body:
Simplify the problem using $x_2=c-x_1$ and $y_2=d-y_1$. So, what remains is
$$\frac{x_1 (c-x_1)}{y_1 (d-y_1)}=a\tag 1$$
$$\frac{c-x_1}{d-y_1}+\frac{x_1}{y_1}=b\tag 2$$  From $(1)$
$$\frac{ (c-x_1)}{ (d-y_1)}=a\,\frac {y_1}{x_1} \tag 3$$ Plug in $(2)$
$$a\,\frac {y_1}{x_1}+\frac{x_1}{y_1}=b\tag 4$$  Let $t=\frac {y_1}{x_1}$ and solve
$$ a t+\frac 1 t=b\implies a t^2-bt+1=0$$


问:
Jensen's Formula and the number of zeros.
I am doing a question that requires me to find the number of zeros in a disk when $f$ is non-constant, bounded and analytic on $D$. I saw that the Wikipedia page of Jensen's formula has the following information:

Jensen's formula can be used to estimate the number of zeros of analytic function in a circle. Namely, if f is a function analytic in a disk of radius R centered at $z_0$ and if |f| is bounded by M on the boundary of that disk, then the number of zeros of f in a circle of radius r < R centered at the same point $z_0$ does not exceed

Does anyone know how this formula is derived?

答:
We can obviously assume $z_0=0$, we also assume that $f$ doesn’t vanish on the circle of radius $r$ (not a big deal, we can take $r’> r$ small enough going to $r$).

Let $r < t < R$. Let $Z_t$ be the “set” of zeroes of $f$ in the disk centered at $z_0$ with radius $t$ (counted with multiplicity). We will make $t$ go to $R$ – we can choose $t$ such that $f$ does not vanish on the circle centered at $z_0$.

The exact Jensen formula, combined with the inequality $|f| \leq M$, gives $$\log{\frac{M}{|f(0)|}} \geq -\sum_{z \in Z_t}{\log{\frac{|z|}{t}}}=\sum_{z \in Z_t}{\log{\frac{t}{|z|}}} \geq \sum_{z \in Z_r}{\log{\frac{t}{|z|}}} \geq \sum_{z \in Z_r}{\log{\frac{t}{r}}}\geq |Z_r|\log{\frac{t}{r}}.$$

Now just take $t \rightarrow R$.


答:
Let $f(z)$ be analytic function such that $f(0)\ne0$, and let $z_1,z_2,\dots,z_m$ denote the zeros of $f(z)$ satisfying $|z_k|<R$. Then Jensen's formula gives

$$
\int_0^1\log|f(Re^{2\pi it})|\mathrm dt=\log\left|f(0)\cdot{R\over z_1}\cdot{R\over z_2}\cdots{R\over z_m}\right|
$$

Now, let $|f(z)|\le M$ for $|z|\le R$, then we have

$$
\log\left|{R\over z_1}\cdot{R\over z_2}\cdots{R\over z_m}\right|\le\log M-\log|f(0)|
$$

Let $0<r<R$, so we can classify $z_1,z_2,\dots,z_m$ into two classes:

$$
\log\left|{R\over z_1}\cdot{R\over z_2}\cdots{R\over z_m}\right|
=\sum_{|z_k|\le r}\log\left|R\over z_k\right|+\sum_{|z_j|>r}\log\left|R\over z_j\right|
$$

Because $\log|R/z_k|$ is at least $\log R/r$, we have

$$
N(r)\log\frac Rr\le\sum_{|z_k|\le r}\log\left|R\over z_k\right|
$$

where $N(r)$ denotes the number of zeros of $f(z)$ satisfying $|z|\le r$. Now, by the transitivity of inequalities, we conclude that the number of zeros of $f(z)$ within $|z|\le r$ satisfies the following inequality:

$$
N(r)\le{1\over\log R/r}\log\left|M\over f(0)\right|
$$

Hope this will help you!

Title: Jensen's Formula and the number of zeros.

Body:
I am doing a question that requires me to find the number of zeros in a disk when $f$ is non-constant, bounded and analytic on $D$. I saw that the Wikipedia page of Jensen's formula has the following information:

Jensen's formula can be used to estimate the number of zeros of analytic function in a circle. Namely, if f is a function analytic in a disk of radius R centered at $z_0$ and if |f| is bounded by M on the boundary of that disk, then the number of zeros of f in a circle of radius r < R centered at the same point $z_0$ does not exceed

Does anyone know how this formula is derived?


问:
Possible error in textbook about weakly dominated strategy of a simple game
This is from Jurgen Weibull's textbook on game theory, page 10, where an example is introduced.

The author claims that the player's third strategy is not weakly dominated by other two pure strategies.

This is obviously false. The third strategy gives a payoff of 1. Whereas the first strategy gives a payoff of 3. $$3 > 1.$$

This means that the third strategy is strictly dominated by the first one and strict dominance implies weak dominance.

Is this an error in the textbook or did I misinterpret something?

答:
$A$ dominates $B$ if I prefer to play $A$ over $B$ regardless of the actions of my opponent. This is not true here, as if the column player plays R, I would rather play the third action over the first, so the first does not dominate the third. The same goes with the second and the third.

As the textbook says, there is mixed dominance here: I prefer to mix the first two actions with equal probabilities over (payoff 1.5) over playing the third one (payoff 1).


问:
On conditional expectation
I need some help with the following problem:

Let X be a continuos random variable that takes values from a to b, where a and b are finite. X follows a given distribution Function 'F'. Suppose that g(X) is a continuous and differentiable function. E.g., g=k*X, where k is a positive constant.

I am trying to compute E(g(X)|X<=x0), where x0<b. I know X<=x0 because a given event has occurred that excludes some interval of values from the support of the original distribution function.

In the example above, is the solution the integral from a to x0 of g(x).f(x)?

Title: On conditional expectation

Body:
$$f_{X|X<x_0}(t)=\frac{1}{F_X(x_0)}f_X(t)\mathbb{1}_{[a;x_0]}(t)$$

thus your expctation becomes

$$\mathbb{E}[g(X)|X<x_0]=\frac{1}{F_X(x_0)}\int_{a}^{x_0}g(x)f_X(x)dx$$


问:
Spherical coordinate transform
If my integral is $$\int_{B(0,R)}x^2+y^2 dxdydz$$

is this $$\int_0^R\Biggr(\int_0^{2\pi}\Biggr(\int_0^{\pi}(r\cos\phi\sin\theta)^2+(r\sin\phi\sin\theta)^2\vert r^2\sin\theta\vert d\theta d\phi dr$$

the proper transformation into spherical coordinates? Since we have $f=x^2 +y^2$, and $g=(r\cos\phi\sin\theta,r\sin\phi\sin\theta,r\cos\theta)$, then $f(g(r,\phi,\theta)=(r\cos\phi\sin\theta)^2+(r\sin\phi\sin\theta)^2$ and the determinant of the Jacobian of $g$ is $r^2\sin\theta$.

答:
Yes, it is. But your integrand can be made even a bit easier, since:

$$(r\cos\phi\sin\theta)^2+(r\sin\phi\sin\theta)^2=r^2\sin^2\theta(\cos^2\phi+\sin^2\phi)=r^2\sin^2\theta$$

so your whole integrand is $\;r^4\sin^3\theta\;$ , since $\;\sin\theta\ge0\;$ for $\;0\le\theta\le\pi\;$ .

Also observe that your $\;\theta,\,\phi\;$ are interchanged with what most people (non-mathematicians, indeed) are used to, since $\;\theta\;$ is usually the azimut and $\;\phi\;$ the vertical angle, and for you and many mathematicians is exactly the other way around. No problem, but it's something that must be taken into account.

Title: Spherical coordinate transform


答: Notice that for $x\rightarrow -x$ we get the same inequality.Symmetrically it follows for $y$.Now WLOG $x\ge y$,:: $x,y\ge 0$ Hence $$|\frac{x+y}{2}|+|\frac{x-y}{2}|=\frac{x+y}{2}+\frac{x-y}{2}=x<z $$

Title: Prove if$~\vert x \vert < z~$ and $~\vert y \vert < z~$ then $~\Big\vert \dfrac{x+y}{2}\Big\vert + \Big\vert \dfrac{x-y}{2}\Big\vert < z~$

Body:
Let $x~$,$y~$,$z~\in\mathbb{R}$

Prove that : if$~\vert x \vert < z~$ and $~\vert y \vert < z~$ then $~\Big\vert \dfrac{x+y}{2}\Big\vert + \Big\vert \dfrac{x-y}{2}\Big\vert < z~$

I know how to prove it by separating cases, but I need a Hint for a direct proof.


问:
Why do the French count so strangely?
Today I've heard a talk about division rules. The lecturer stated that base 12 has a lot of division rules and was therefore commonly used in trade.

English and German name their numbers like they count (with 11 and 12 as exception), but not French:

```<code>  # |   English |           German | French
-----------------------------------------------
  0 |      zero |            null  | zero
  1 |       one |            eins  | un
  2 |       two |            zwei  | deux
  3 |     three |            drei  | trois
  4 |      four |            vier  | quatre
  5 |      five |            fünf  | cinq
  6 |       six |           sechs  | six
  7 |     seven |          sieben  | sept
  8 |     eight |            acht  | huit
  9 |      nine |            neun  | neuf
 10 |       ten |            zehn  | dix
 11 |    eleven |             elf  | onze
 12 |    twelve |           zwölf  | douze
 13 | thir|teen |       drei|zehn  | treize
 14 | four|teen |       vier|zehn  | quatorze
 15 |  fif|teen |       fünf|zehn  | quinze
 16 |  six|teen |       sech|zehn  | seize
 17 |seven|teen |       sieb|zehn  | dix-sept
            18 and 19 are "regular"
 20 |    twenty |          zwanzig | vingt
 21 |twenty-one |  ein|und|zwanzig | vingt et un
 22 |twenty-two | zwei|und|zwanzig | vingt-deux
            23 - 69 are "regular"
 70 |  seven|ty |         sieb|zig | soixante-dix = 60 + 10
              ....
 80 |   eigh|ty |         acht|zig | quatre-vingts = 4*20 ?!?!
 81 |eighty-one |  ein|und|achtzig | quatre-vingt-un = 4*20 + 1
      ...
</code>```

So my question is:

Why do French count so strangely after 79?

(Are there other languages that count similar? What's the historic / mathematical reason for this system?)

Related Questions

Why is a full turn of the circle 360°?

What could be better than base 10?

答:
Many languages have (at least relicts of) non-decimal counting, very often vigesimal (because we have 20 fingers plus toes), but also many other systems. I recommend an old Gutenberg project of mine, The Number Concept

Note for example that the Danish word for 55 is femoghalvtreds "five more than half the third twenty-block"

Title: Why do the French count so strangely?


问:
Differentiability and convex function
If $E\subseteq \mathbb R^d$ is convex and $f:E\to\mathbb R$ is a convex function, is it true that :

For all $x\in E$, if $f$ has partial derivatives at $x$ along all directions $u$ such that $x+tu\in E$ for small enough $t>0$ (i.e., $f$ is Gateaux-differentiable at $x$), then $f$ is differentiable (in the Fréchet sense) at $x$ ?

If $f$ is differentiable on $E$, then its gradient is continuous on $E$, even at boundary points ?

When $E$ is open, the answer is YES to both, but I am mostly interested in what happens at the boudary.

NB : when I say that $f$ is differentiable (in the Fréchet sense) at $x\in E$, I mean that there exists a vector $g\in\mathbb R^d$ such that
$$\frac{f(y)-f(x)-g^\top (y-x)}{\|y-x\|} \xrightarrow[y\to x]{} 0$$
In that case, there is only one such vector that is parallel to $E$, i.e., in the span of $E-E$, and that's what I call the gradient of $f$ at $x$.

答:
This is a counterexample for the first point:

Take $E = [0,\infty)^2$ and set
$$
f(x,y) = \frac{-x\,y}{\sqrt{x^2 + y^2}}.
$$
Then, $f$ is convex and differentiable at all points in $E \setminus \{0\}$. At $0$, all the partial derivatives exist. However, $f$ is not differentiable in $0$.

Edit: And this disproves the second point: Again, $E = [0,\infty)^2$ and
$$f(x,y) = x^2 \, \exp(-y/x^2).$$
Then, $f$ is convex and everywhere differentiable. However,
$$f'(0,0) = (0,0),\quad f'(x,0) = (2\,x, -1)$$
for $x > 0$.

Title: Differentiability and convex function


问:
Constructing a rhombus $ABCD$ with $A$ a given point, $B$ on a given circle, and $C$ on a given line, such that $\angle A=60^\circ$

Given a point $A$, a circle $b$ and a line $c$, construct a rhombus
$ABCD$ such that $B \in b$, $C \in c$ and $ \measuredangle A = 60^\circ $.

Hello! I saw this geometry problem in a textbook and I tried to solve it, but I've made no progress so far. The thing is, I don't even know how to attack such a problem.

Any help appreciated!

答:
Let us observe a rhombus $ABCD$ with $\measuredangle A = 60^\circ.$ Its diagonal $AC$ has the length $|AC|=|AB|\sqrt 3,$ the angle at $A$ in $\triangle ABC$ is of measure $30^{\circ}.$

Therefore, $C$ is an image of $B$ in a transformation $$\sigma=\rho \circ \theta,$$
where the common center of the rotation $\rho$ and the homothety $\theta$ is $A,$ the angle of rotation is $30^{\circ}$ and a coefficient of homothety is $\sqrt 3.$

Then $C$ is an intersection point of $b$ and $\sigma(c).$

For further solutions, we can consider an oppositely oriented angle of rotation.

The enclosed picture shows one solution, a second point $C'$ is marked but the rhombus is not constructed.


问:
Fiber of associated G-bundle
I would like to understand why, given a principal $G$-bundle $(P,M,G)$ and an associated $G$ bundle $P_F=P \times F / \ \sim $, the fiber of $P_F$ is isomorphic to $F$, if this is true.

I was thinking something on these lines. We want to have a map from the fiber $\pi_{P_F}^{-1}(m)$ to $F$. To construct this, let $x \in \pi_{P_F}^{-1}(m)$. We can fix an element $p \in \pi_P^{-1}(m)$ on the principal bundle $P$ and write $x=[p,f(x)]$ for some $f$. Note that $p$ does not depend on $x$, only $f$ does according to the choice made.

Fixing $p$, we have built a map $u : \pi_{P_F}^{-1}(m) \rightarrow F$ such that $x \rightarrow f(x)$.

Does it make sense this approach ? Can it be completed to a proof ?

Further, if this statement is true, can we say rigorously that $P_F$ is a $G-$bundle with fiber $F$ ? I am a bit confused here because if this was the case $F$ would be isomorphic to $G$ (because in any principal G-bundle the fibers should be isomorphic the the structure group G), which actually would make the previous statement trivial... Can somebody help me putting together the pieces of the puzzle and correct me if I am making wrong statements ?

答: This follows from the fact that the action of $G$ is free, let $p:P\rightarrow M$ the projection of the principal bundle. Denote by $p_F:P_F\rightarrow M$ the projection of the associated bundle, that is the quotient of $P\times F$. Let $m\in M$, and $y\in p^{-1}(m)$. Let $y,y'\in F$, we denote by $[x,y], [x,y']$ the image of $[x,y]$ and $[x,y']$ in $P\times F/\simeq$; $p_F([x,y])=p_F([x,y'])$ if and only if there exists $g\in G$ such that $g.(x,y)=(g.x,g.y)=(g.x,g.y)=(x,y')$. This implies that $g=Id_G$ and $y=y'$. Thus the map $F\rightarrow p_F^{-1}(m)$ defined by $y\rightarrow [x,y]$ is injective. It is also surjective since $G$ acts transitively on $p^{-1}(x)$. For every $x'\in p^{-1}(m), x'=g.x, [x',y]=[x,g^{-1}.y]$.

Title: Fiber of associated G-bundle