问:
Probability of (1 + min(X,Y))/(1 + min(X,Z))?
I have been trying to derive the probability $\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right]$, where X, Y, and Z are independent and follow exponential distribution with parameters $λ_x$, $λ_y$, and $λ_z$, respectively. c here is a constant. What I did is briefly as follows:

$
\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right] \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)\\
 = \Pr \left[ {X < Y,X < Z} \right]\\
 + \Pr \left[ {\frac{{1 + X}}{{1 + Z}} < c} \right]\Pr \left[ {X < Y,X > Z} \right]\\
 + \Pr \left[ {\frac{{1 + Y}}{{1 + X}} < c} \right]\Pr \left[ {X > Y,X < Z} \right]\\
 + \Pr \left[ {\frac{{1 + Y}}{{1 + Z}} < c} \right]\Pr \left[ {X > Y,X > Z} \right]            \;\;\;\;\;\;\;\;(2),$

from which I can obtain a closed-form expression. However, when I do simulation to verify my analytic result, I get

where the red curve is from analysis. I don't know why there is a gap between the results.
Do you think that going from (1) to (2) is problematic?

答:
You can calculate the probability by integrating the joint pdf like this

$\Pr \left[ {\frac{{1 + \min \left( {X,Y} \right)}}{{1 + \min \left( {X,Z} \right)}} < c} \right] = \int\limits_{z = 0}^{z = \infty } {\int\limits_{x = 0}^{x =  - 1 + c + c*z} {\int\limits_{y = 0}^{y = \infty } {{f_{X,Y,Z}}\left( {x,y,z} \right)dydxdz} } }  + \int\limits_{z = 0}^{z = \infty } {\int\limits_{x =  - 1 + c + c*z}^{x = \infty } {\int\limits_{y = 0}^{y =  - 1 + c + c*z} {{f_{X,Y,Z}}\left( {x,y,z} \right)dydxdz} } } $

This is for $c>1$


问:
Is the Dirac delta a function?
Is Dirac delta a function? What is its contribution to analysis?

What I know about it:
It is infinite at 0 and 0 everywhere else. Its integration is 1 and I know how does it come.

答:
The delta distribution is not a function from $\mathbb{R} \to \mathbb{R}$. For any such function $f$ that is zero except at finitly many points, as the delta distribution is, $\int f= 0$ (this holds for both the riemann and the lebesgue integral). Yet $\int \delta = 1$...

You can distributions as a function that maps functions (well, rather some functions) to real values though, i.e as a function $\mathbb{R}^\mathbb{R} \to \mathbb{R}$ (since it doesn't may every function to a real value, the domain is not really $\mathbb{R}^\mathbb{R}$ but a subset thereof).

The delta distribution is particularly simply, it just evaluates the function at $0$, i.e. $$
  \delta(f) = f(0) \text{.}
$$ Instead of $\delta(f)$, people often write $\int f \delta$, and that way you get $$
  \int f \delta = f(0) \text{.}
$$
But this is just notation - just as the $df$ and $dx$ in $\frac{df}{dx}$ aren't real numbers, yet sometimes you may treat them as if they were, $\delta$ isn't a function - yet sometimes it behaves like one. For example, you may use partial integration to compute $\delta'$ by doing $$
  d'(f) = \int f \delta' = - \int f' \delta \text{.}
$$
Thus, $\delta'$ is the distribution which, given a function $f$, evaluates the derivative of $f$ at $0$ (and multiplities by $-1$), i.e. returns $-f'(0)$. This is, in fact, the very definition the derivative of a distribution btw.

This syntax also allows you to treat every function (well, actually every interable function) $h$ as a distribution. Following the path laid out by the integral notation for distributions, you get $$
  h(f) = \int h f = \int h(x) f(x) \, dx \text{.}
$$

Note that this allows you to assign a derivative to a lot of functions which aren't otherwise differentiable. If you convert them into a distribution first, you can then set $$
  h'(f) = \int h' f = -\int h f' \text{.}
$$
provided that $f$ is suitable smooth.

This leaves the question of which $f$ are actually allowed here, i.e. which functions are mapped to real values open. You'll generally want them to be differentiable arbitrarily often, but that still leaves multiple choices. You'll have to read up on the theory of distributions to understand all the details.

ContentLicense: CC BY-SA 4.0

Title: Is the Dirac delta a function?


问:
The easiest way possible to evaluate $\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$
I want to solve this integral the easiest way possible, how should I do it?
$$\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$$

I tried with substitution, but it didn't work out. And I think that it isn't possible to use integration-by-parts method - at least I couldn't find a way to do it.

答: Using $1-\cos x= 2\sin^2 \frac x2$, $$I=\int \ln(\cos x) \cdot \frac 12 \csc^2 \frac x2 dx $$ Substitute $t=\cot \frac x2 \implies dt =-\frac 12\csc^2 \frac x2 dx$ : $$I=-\int\ln(\cos x) dt $$ Now, there’s a golden chance to use $$\cos x = \frac{1-\tan^2 (x/2)}{1+\tan^2(x/2)} = \frac{\cot^2 (x/2) -1}{\cot^2(x/2)+1} $$ $$I=\int \ln\left(\frac{t^2+1}{t^2-1} \right) dt =\int \ln(t^2+1)dt-\int \ln(t^2-1) dt$$ And then these can be integrated by parts easily.


问:
Intermediate Value Theorem and polynomial of odd degree
As a consequence of Intermediate Value Theorem (IVT), it is illustrated in almost all calculus books and notes/courses that every polynomial of odd degree over real numbers have a real root.

I tried to see whether this result has been proved without the use of intermediate value property, and I did not find any link or question related to that on this and other sites.

Can one suggest me whether there is a proof of above mentioned consequence of IVT, which is done without using IVT?

I had seen that the fundamental theorem of algebra have been of interest to many people for proving just using algebraic techniques, but among all the known various proofs of FTA, epsilon amount of analysis is used.

So, my above question is of similar nature, but I did not see any comments on its proofs without IVT.

答:
Complex roots of a polynomial with real coefficients occur in pairs.

Hence?

https://en.m.wikipedia.org/wiki/Complex_conjugate_root_theorem


问:
"Closed form" solution of $x(e^x+a) = b$
Is there a way to express the solution of
$$x(e^x+a) = b$$
for $a,b, x \ge 0$ and in terms of  more or less standard functions?

I already know that if $a=0$, then $x = W(b)$, where $W$ is the Lambert W function.
I tried to get this in a form which can be solved by the W function, but unfortunately it does not seem to work. Maybe there is another approach, or possibly its just not possible.

Maybe you could help?

答:
From a formal point of view, there is a solution in terms of the generalized Lambert function  writing

$$e^{-x}=\frac{x}{b-a x}$$
(have a look at equation $(4)$ in the linked paper).

This being said, it is not very practical and think about numerical methods.


问:
Bound on matrix multiplication
I have a result of the form:

$r = A^\top B^\top B A$

Where $A_{[m \times 1]}$ and $B_{[n \times m]}$, so $A^\top A$ is a scalar.

I want to bound $r$ from below using some property of B, and reach the form:

$r \geq b \cdot (A^\top A)$, where $b$ is a scalar derived from the matrix $B$.

A simple example could be the case where $B$ is diagonal.
Then if $b = \min_i B_{i,i}^2$ (the minimal squared value found in $B$),  then the form above hold.

Is there anything I can use for the case $B$ is non diagonal? something that perhaps includes the $trace$ of $B$?

答:
Without additional information, the best you can say is $M = B^TB$ is a positive semi-definite $n\times n$ matrix;
this means $M$ may have an eigenvalue $0$.

In this case, if $A\neq 0$ is its associated eigenvector, we'd have $r = 0$, which would make $b=0$.

Notice that, as a linear transformation, $B: \Bbb R^m\longrightarrow \Bbb R^n$.
This means that when $m>n$ it's guaranteed that $B$ fails to be injective, and hence that $M$ has nontrivial kernel, that is, $M$ has an eigenvalue $0$.

In other words, you could only hope for $b>0$ when $m\leqslant n$.

In general, $b$ would be $M$'s smallest eigenvalue (all of which are real and non-negative).

Title: Bound on matrix multiplication


问:
Does sequence converge if its differences become strictly monotonously greater?
Consider the following sequence:
$(x_n) = x_1, x_2, ...$

Now, let's say this particular sequence has this property: $x_{n + 1} - x_n > 0, \forall n \in \mathbb{N}.$  Then we say that this sequence grows. It' certanily possible that this sequence converges (take, for example, sequence $x_1, x_1 + 0.1, x_1 + 0.01, ...$, we see that $n$-th term is obviously $x_1 + 111...1$ which is obviously some real number).

But add another property of this sequence; let' also say that differences $x_2 - x_1, x_3 - x_2, ...$ make another strictly growing sequence, in other words $x_{n + 1} - x_n > x_{m + 1} - x_m$ for every $n$ such that $n > m$ (it's like positive 2nd derivative in a way).

My intuition tells me that then $(x_n)$ cannot converge. If true, how to prove this? Thanks.

答:
The proof is simple. $x_n$ converges if and only if for every $\epsilon > 0$ there exists an index $N > 0$ such that $|x_p-x_q| < \epsilon$ for every $p, q > N$. In particular setting $p = q+1$ yields the necessary condition $|x_{q+1}-x_q| < \epsilon$ for every $q > N$.

(You really don't need Cauchy sequences here. Just notice that if a sequence $x_n$ has to converge, then obviously $x_{n+1}-x_n \to 0$).

You are requiring $x_{n+1}-x_n > x_{m+1}-x_m > 0$ for some $m$ and all $n > m$. Now if you set $\epsilon = x_{m+1}-x_m > 0$, the necessary condition is not satisfied, no matter how you choose $N$ large.


问:
How to compute next/previous representable rational number?
An (approximate) non-negative rational number representation is a pair of natural numbers each not greater than some fixed limit M (and of course denominator being non-zero).

With this condition there is finite number of representable rational numbers. This means that for each such number we can name previous and next number in the set (of course except for smallest 0 and largest 1). How to compute them?

In other words lets have a set of numbers

$$
R(M)=\{\frac{p}{q} : p,q\in\mathbb{N},q\neq0,p\leq q\leq M\}
$$

where $M\in\mathbb{N}_+$

and given number $\frac{p_1}{q_1}\in R(M)$.

I want to find numbers $\frac{p_S}{q_S}\in R(M)$ and $\frac{p_L}{q_L}\in R(M)$ (if they exist) such that

$$
\frac{p_S}{q_S}<\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_S}{q_S}<\frac{p}{q}<\frac{p_1}{q_1}
$$

and smilarly

$$
\frac{p_L}{q_L}>\frac{p_1}{q_1} \wedge \neg\exists_{\frac{p}{q}\in R(M)}\frac{p_1}{q_1}<\frac{p}{q}<\frac{p_L}{q_L}
$$

答:
So far I have found algorithm iterating over denominators (inspired by mjqxxxx's comment). Thus it is of $O(M)$ time complexity (however each iteration is short and simple).

In Farey sequence for two consecutive fractions $\frac{p_1}{q_1}<\frac{p_L}{q_L}$ it holds that $p_L q_1 - p_1 q_L=1$. Thus $p_L=\frac{p_1 q_L + 1}{q_1}$. We have $\frac{p_1}{q_1}$ and $M$ and we want to know $\frac{p_L}{q_L}$.

Since $1\leq q_L\leq M$ we can now simply iterate over all natural numbers in range $[1;M]$ and for each see whether $\frac{p_1 q_L + 1}{q_1}$ is a natural number. If so than we have the desired fraction $\frac{p_L}{q_L}$.

It may happen that this will be solved for more than one value in that range. This means that depending on $M$ the solution are different fractions. Thus of interest is the last (I think - I haven't prove that) value and it would be better to iterate from $M$ to $1$ rather than the other way.

(Finding previous fraction can be done in similar way using $p_S=\frac{p_1 q_S – 1}{q_1}$.)

It remains an open question whether indeed the largest $q_L$/$q_S$ which solves the equation in natural numbers is the solution. Also this is still $O(M)$ so maybe something faster can be found?

ContentLicense: CC BY-SA 3.0


答:
To have a better understanding of what is the Delta Dirac "function", it is good to know what is a distribution (but this is not necessary). Let $\Omega\subset\mathbb{R}^N$ be a open set and $\mathcal{D}(\Omega)=C_0^\infty(\Omega)$. We define in $\mathcal{D}(\Omega)$ the following notion of convergence: we say that $\phi_n\to 0$ in $\mathcal{D}(\Omega)$ if

I - $\operatorname{spt}(\phi_n)\subset K\subset\Omega$, where $K$ is a fixed compact ($\operatorname{spt}(\phi_n)$ is the supoort of $\phi_n$),

II - For all $j$, $D_j\phi_n\to 0$ uniformly in $\Omega$.

We denote by $\mathcal{D}'(\Omega)$ the set of linear transformations $F:\mathcal{D}(\Omega)\to\mathbb{R}$ that are continuous with respect to the pseudo topology (the notion of convergence) that we defined, i.e. if $\phi_n\to 0$ in $\mathcal{D}(\Omega)$, then $\langle T,\phi_n\rangle=T(\phi_n)\to 0$. We call the elements of $\mathcal{D}'(\Omega)$ distributions.

Now we are able to define the famous delta Dirac "distribution". It is a distribution $\delta_{x}:\mathcal{D}(\Omega)\to\mathbb{R}$ defined by the formula ($x\in\Omega$) $$\langle\delta_{x},\phi\rangle=\phi(x)$$

You can easily check by using the definition that $\delta_x\in\mathcal{D}'(\Omega)$. Also, any funcion $u\in L^1_{loc}(\Omega)$ de fined a distribution by the formula $$\langle T_u,\phi\rangle=\int_\Omega u\phi$$

On the other hand, we can define a notion of convergence in $\mathcal{D}'(\Omega)$, to wit, $T_n\to T$ in $\mathcal{D}'(\Omega)$ if $\langle T_n,\phi\rangle\to\langle T,\phi\rangle $. You can  check by using this definition that if $h_\epsilon$ is a mollifier sequence, then $T_{h_\epsilon}\to \delta _x$ in the sense of distributions.

From the last paragraph you can see, why usullay is said that the Delta Dirac "function" is one that satisfies $\delta(x)=1$, $\delta(y)=0$ if $y\neq x$ and $\int \delta(x)=1$. It is obvius that in the standard sense, this is impossible, but if we were to take the limit of the sequence $h_\epsilon$ pointwise, we would have something like this.

Also there is a lot of places in mathematics where the distribution of Dirac is used, for example, it is know that the Fundamental solution $\Gamma$ of the Laplace equation satisfies $-\Delta\Gamma=\delta$ in the sense of distributions, hence, if you wanna solve the problem $-\Delta u=f$, you can write (at least formally and with some assumptions) $u=\Gamma\ast f$, where $\ast$ stands for convolution.

There are a bunch of other applications of this distribution, but this place here is to small to write them all.

Title: Is the Dirac delta a function?

Body:
Is Dirac delta a function? What is its contribution to analysis?

What I know about it:
It is infinite at 0 and 0 everywhere else. Its integration is 1 and I know how does it come.


问:
How many size 10 sequences can be formed with terms in $\{A,B,C\}$ and the number of terms A is odd?

How many size 10 sequences are there, all formed with terms belonging to $\{A,B,C\}$, in which the number of terms with A is odd.

My proposed solution:
$$\sum_{m\in \{1,3,5,7,9\}} {10 \choose m} 2^{10-m}=28804$$
That is for each $m\in \{1,3,5,7,9\}$ finding the possibilities for A and then filling the remaining spots with B and C (2 possibilities for each non-A spot).

But the given short answer is 2391484 (no solution provided).

Question: what am I missing? or is the given answer wrong?

答:
Assume that each B or C is worth $1$, each A is worth $x$, and the worth of a sequence is the product of the worths of its letters. When we multiply out $(2+x)^{10}$ distributively we obtain $2^{10}$ terms. Their sum is the sum of all worths of the possible ABC-sequences of length $10$. But we want only sequences with an odd number of $x$'s. We can collect these by forming
$$f(x):={(2+x)^{10}-(2-x)^{10}\over2}\ .$$
When we put $x:=1$ here each appearing word obtains the value $1$. The number $N$ of admissible words therefore is
$$N=f(1)={3^{10}-1^{10}\over2}=29\,524\ .$$


问:
Question from logical reasoning
This is a question from syllogism(reasoning). It reads thus:
Statements: $1$. All $A$ are $B$

$2$. All $B$ are $C$.

Conclusion:  Some $C$ may or may not be $A$.
I am confused about the validity of this conclusion.  The definite conclusion that one can draw from these statements is: Some $C$ are $A$. And there is the possibility that some $C$ are not $A$. Does that mean that the given conclusion is true ? Please suggest.

答:
No. You can't say that all C are defintely A. What you can say is that all A are defintely C.

Consider:

All tigers are mammals

All mammals are animals

What follows? All tigers are animals, yes.  All animals are tigers? No.  But of course, if I have animal, it may or may not be a tiger.

Title: Question from logical reasoning


问:
Reparameterizing quaternion in terms of angle
Why is the red part true?

From: http://graphics.stanford.edu/courses/cs348a-17-winter/Papers/quaternion.pdf

答: HINT. Notice $\sin ^2\theta+\cos^2\theta=1$. What does $\theta\in [0,\pi]$ lead to? Consider its graph $(x=\cos \theta, y=\sin \theta )$ on the plane, with $x^2+y^2=1$, which is exactly an upper semicircle.


问:
Proof on showing function $f \in C^1$ on an open & convex set $U \subset \mathbb R^n$ is Lipschitz on compact subsets of $U$
The question is as follows:

Given:

(1) function $f: U \subset \mathbb R^n ==> \mathbb R$

(2) $U$ is open and convex set

(3) $f \in  C^1$ in $U$

Goal: Show that $f$ is Lipschitz on any compact subset of $U$

By now, I have various ideas come to mind, but I can't connect the dots >_<

Here are my thoughts so far:

(1) Recall definitions:

(i) $f \in C^k$ means all partial derivatives up to (and including) order $k$ exist and continuous.  Here $k = 1$

(ii) a set $U$ is convex if for any 2 points $x, y$ in $U$, the segment joining $x$ and $y$ is totally inside $U$

(iii) function $f$ is Lipschitz if there is a bound $M$ such that $|f(x) - f(y)| \leq M |x-y|$

(2)By a theorem in my book:

if function $f \in C^1$ on an open & convex set $U$, then for any 2
  points $x$ and $y$ in $U$, there is a point $s$ lying on the segment
  joining $x$ and $y$ such that $f(x) - f(y) = Df(s) * (x - y)$

I firstly have a feeling that I may need this theorem in the proof, but I can't see the connection between my desired bound $M$ with $Df(s)$.  So I tried to think of other ideas and ...

(3) It turns out that by all the given information, I think if I can show function $f$ is convex, then I'm done because there is a theorem which said: a convex function on a open, convex set $U$ should be Lipschitz on $U$, thus I think it must also be Lipschitz on any subset of $U$, either that subset is compact or not.

However, how can I prove function $f$ is convex, based on given information?  I have a feeling that I have to use the fact that set $U$ is convex, but then I'm stuck on how to proceed further >_>

*Would someone please help me on this question?
Thank you very much ^_^*

答:
You need to use the Mean Value theorem in several variables, as you already stated it:

For $x,y\in U$ there exists $s$ which lies on the line segment between $x$ and $y$ ( this line segment is contained in $U$, since it is convex), such that:
$$f(x)-f(y)=\nabla f(s)\cdot (x-y)\Rightarrow\text{( Cauchy-Schwarz )}$$
$$|f(x)-f(y)|\leq|\nabla f(s)|\cdot |x-y|$$

Now, take a compact subset $S$ of $U$, and $x,y\in S$. Then take a convex compact set $S'$ such that $S\subset S'\subset U$.This is because we want $s\in S'.$

Now, since $f\in C^1$ , the first partial derivatives are continuous, and therefore, bounded in the compact set $S'$. This implies that $|\nabla f(s)|\leq M$

Concluding, for each $x,y\in S$ we have that :
$$|f(x)-f(y)|\leq M\cdot |x-y|$$
which proves that $f$ is Lipschitz in $S$.


问:
If $u = 0$ on $\partial U$ and $\frac{\partial u}{\partial N} = 0$ on $\partial U$, then $\nabla u = 0$ on $\partial U$?
Suppose that $u$ is a smooth function from an open subset $U\subset\mathbb{R}^{n}$ with smooth boundary to $\mathbb{R}$. Let $N$ be the (outward) normal on $\partial U$.

If we know that $$u = 0 \qquad \text{and} \qquad \frac{\partial u}{\partial N}:= \nabla u \cdot N = 0 \qquad \text{on} \qquad \partial U,$$ can we then deduce that $\nabla u = 0$ on $\partial U$?

I came accross this problem when studying a PDE. I tried to play around with integral equations, but it didn't really get me anywhere. Any suggestions would be greatly appreciated.

EDIT: I read somewhere that $u=0$ on $\partial U$ implies that $\nabla u$ and $N$ are parallel on $\partial U$. But I don't know why this is true.

答:
Let's start with the unit disc $U = \{ (r,\theta): r\leq 1 \} \subseteq \mathbb{R}^2$.

Then $\partial_N u = \partial_r u = 0$, so $\nabla u = \begin{pmatrix} \partial_r u \\ \partial_\theta u \end{pmatrix} = \begin{pmatrix} 0 \\ \partial_\theta u \end{pmatrix}$.

But $u \equiv 0$ on $\partial U = \mathbb{S}^1$, so $u(1,\phi) = u(1,\psi) = 0$ for all $\phi,\psi$, and so $\frac{\partial u}{\partial \theta}(1,\phi) = 0$ for all $\phi$. So what you recall seems correct in this case, and I think in general.

Basically if $u \equiv 0$ on $\partial U$, then the projection of the gradient onto the boundary is necessarily zero, leaving only the normal component.


问:
Best algorithm (Time Complexity) to find Minimum spanning tree of an complete, positive weighted, undirected, graph
Suppose that we have a complete undirected positive weighted graph G={V,E} and I want to find one MST of it. My problem is what is the best (in time complexity) algorithm for it?
The best prime complexity is O(V^2)
The best Kruskal complexity is O(E.Log(E))=~O(V^2. Log(V))
any idea for lower time complexity?</p>

答:
I believe there are deterministic linear time algorithms for sufficiently dense (such as complete) graphs.

One example is Prim's algorithm using a d-ary heap

See end of intro in https://en.wikipedia.org/wiki/Prim%27s_algorithm

And another one using Fibonacci heaps: https://dl.acm.org/doi/10.1145/28869.28874

Edit: Linear in the number of edges (i.e. quadratic in vertex number)


问:
For which $p$, does there exist an anti-derivative?
For positive integer $p$, consider the holomorphic function

$f(z)=\frac{\sin (z)}{z^p}$ , for $z\in \mathbb{C}\setminus \{0\}$ .

For which  values of $p$ , does there exist a holomorphic function $g:\mathbb{C}\setminus \{0\}\to \mathbb{C}$   such that $g'(z)=f(z)$ for $z\in
 \mathbb{C}\setminus \{0\}?$

$(1)$ All even integers .

$(2)$ All odd integers.

$(3)$ All multiples of $3$.

$(4)$ All multiples of $4$.

My thinking :- I am aware of the fact "Every analytic function on a simply connected domain has a primitive (anti-derivative)"

Here , the domain is not simply connected and I am not sure how else to proceed.

Please give a hint. Thank you.

答:
Hint: Consider the series expansion of $\sin$. Note that monomial functions of the form $z \mapsto z^n$ have an anti-derivative (on $\Bbb C \setminus \{0\}$) iff $n \neq -1$.

(Convergence issues aside) This tells you that you need to pick a $p$ such that there is no $1/z$ term. After this, show that all such $p$ actually do give a function with an anti-derivative.


问:
Find the area of $A = \{ \langle x,y\rangle \in \mathbb{R}^2 \mathrel| (x+y)^4<a x^2 y,\ x>0 \}$?
I can't really think of how to set the limits</p>

答:
You can parameterize the equation such as
$$x=x(t)$$
$$y=t*x(t)$$
and by replacing to the original equation
$$\big(x(t)+t\,x(t)\big)^4=a\,x(t)^2\,t\,x(t)\Rightarrow x(t)^4(1+t)^4=a\,t\,x(t)^3\Rightarrow x(t)=\frac{a\,t}{(1+t)^4}$$
The limits are $0$ and $\infty$. By Greens therom the area enclosed by the parametric curve is given by
$$A=\frac 12\int_0^{\infty}\bigg(x\,y'-y\,x'\bigg)dt=\frac 12\int_0^{\infty}\bigg(x\,(t\,x)'-t\,x\,x'\bigg)dt=\frac 12\int_0^{\infty}\bigg(x\,(x+t\,x')-t\,x\,x'\bigg)dt$$
$$A=\frac 12\int_0^{\infty}x^2dt=\frac 12\int_0^{\infty}\bigg(\frac{a\,t}{(1+t)^4}\bigg)^2dt$$
This is an improper integral and
$$A=\lim_{b \to \infty}\frac 12\int_0^b\bigg(\frac{a\,t}{(1+t)^4}\bigg)^2dt=\frac 12\lim_{b \to \infty}\bigg(-a^2\frac{1+7b+21b^2}{105(1+b)^7}+\frac{a^2}{105}\bigg)=\frac{a^2}{210}$$</p>


问:
Equilateral triangle $ABC$ with $D$ on $BC$ such that $BD=\frac13BC$. Prove $9\overline{AD}^2=7\overline{AB}^2$
Equilateral triangle $ABC$, point $D$ on side $BC$ such
that $BD=\frac13BC$. Prove $9\overline{AD}^2=7\overline{AB}^2$

In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such
that $BD=\frac{1}{3}BC$. Prove that
$9\overline{AD}^2=7\overline{AB}^2$.

I started by constructing the median/altitude(because it's an equilateral triangle so it is the same thing) $AE$. $$BE=\frac{BC}{2}$$ $$DE=\frac{BC}{2}-\frac{BC}{3}=\frac{BC}{6}$$ $$\overline{AD}^2=\overline{AE}^2+\frac{\overline{BC}^2}{36}$$ $$\overline{AB}^2=\overline{AE}^2+\frac{\overline{BC}^2}{4}$$ But I got $$9\overline{AD}^2=\overline{AB}^2$$ Am I making a mistake somewhere or is there a problem in the question?

答:
It is enough to apply Stewart's theorem. Assuming $AB=AC=BC=3$ and $AD=DE=EC=1$, then letting $AD=x$, we have $\cos\widehat{ADB}+\cos\widehat{CDA}=0$, hence

$$ \frac{1+x^2-9}{2x}+\frac{4+x^2-9}{4x}=0 $$
and $x=\sqrt{7}$, such that $AD^2=\frac{7}{9}AB^2$ as wanted. As a simpler alternative, if we denote as $M$ the midpoint of $DE$ we have $DM=\frac{1}{6}AB,\,AM=\frac{\sqrt{3}}{2}AB$ and the conclusion follows from the Pythagorean theorem.


问:
How to calculate the angle between a rotating reference point and a static remote point at any degree of rotation?
Summarize the problem

I am trying to work out the relative angle/direction between two points from the reference of the first. I have two points: a rotating point A (my reference position) and a static (remote) point B which can be positioned at any distance from A from 0-360 degrees. I need to calculate the relative direction from A to B at any rotation of A.

Most basic description:

B is at a fixed positioned at 0 degrees (direct) north of A.
When A is facing north B is at 0 degrees (ahead) of A.
When A is facing west, B is 90 degrees (right) of A
When A is facing south B is 180 degrees (behind) of A
When A is facing east, B is 270 degrees (left) of A

As A rotates, how can I calculate the relative direction (in degrees) from A to B?

答:
Suppose $B$ is due north of $A$.
If you measure $A$'s facing angle $\theta$ counterclockwise with north as $0$ then the bearing to $B$ is just that angle.

When $B$ is at an initial bearing angle $\phi$ from $A$ (measured counterclockwise) the whole picture is just rotated by $\phi$ so the facing angle is $\theta - \phi$.


问:
Fixed point iteration method $x_{n+1}=g(x_n)$ doubt .
I have a doubt regarding fixed point iteration method  for $f(x)=0$ for a zero of $f$ in $[a,b]$. We write it in equivalent form like $$g(x)=x$$  Now I studied that result is that iteration Method $$x_{n+1}=g(x_n)$$ is convergent iff  $|g’(x)|<c<1,\forall x\in[a,b]$ .  I think I have read wrong result because I have a counter-example as

$$x_{n+1}=g(x_n)$$ where $$g(x)=1+\frac{2}{x}, x\in[1,100]$$ I check it by using general real analysis method that the sequence $<1+\frac{2}{x_n}>$ is convergent to $2$ for any choices of $x\in [1,100]$, but $$|g’(x)|<c<1$$  is not true for all $x$ in $[1,100]$. Where is my mistake in result ? Is it not if and only if form ? Please help . Thank you .

答:
The condition that $|g'(x)|<1$ over an entire interval containing the fixed-point is a sufficient but unnecessary condition for convergence. Likewise the condition that $|g'(x)|>1$ over the same interval is a sufficient but unnecessary condition for divergence (unless one happens to "stumble" upon the fixed-point from outside the interval).

You can't always use these conditions to test for convergence/divergence. For example, $g$ may not be differentiable, or $|g'(x)|$ may swap between being less than or greater than $1$ over the interval of interest. The latter case is what occurs for your example.

Despite this, your example still manages to converge. This may be attributed to the fact that it satisfies $|g'(x)|<1$ for all $x\in(\sqrt2,\infty)$, and $g(x)\in(\sqrt2,\infty)$ for any $x\in(0,\sqrt2]$. Hence, after the first iteration, you are guaranteed to find yourself in a region where it converges.

This is a good example where one should be more interested in eventual convergence i.e. "will it converge if you start close enough to the fixed-point?" Assuming $g$ is continuously convergent, this boils down to whether or not $|g'(x_\star)|<1$ where $x_\star=g(x_\star)$ is the fixed-point. You can further extend this to allow $|g'(x_\star)|=1$ as long as $|g'(x)|<1$ holds almost everywhere near it, which is a sufficient but unnecessary condition for convergence is you start almost anywhere near it.

Using this sufficient but unnecessary condition for your example:

The fixed-point may be observed to be $2$, and $g'(2)=-1/2$. Hence $x_{n+1}=g(x_n)$ converges provided you start close enough to $2$.

Note that this does not tell you how close you will need to start. But it's an easier condition to verify in some cases.

Some other examples you can try:

$x_{n+1}=\arctan(x_n)$ converges to $x_\star=0$ for all real starting points.

$x_{n+1}=2^{\operatorname{sgn}(x_n)}x_n$ converges to $x_\star=0$ only for negative starting points, where $\operatorname{sgn}$ is the sign function.

$x_{n+1}=-2^{\operatorname{sgn}(x_n)}x_n$ converges to $x_\star=0$ only if your starting point is $0$.

For the last two, if we ignore the lack of differentiability at $x=0$, what is their difference? Is it sufficient in either case to look only at the magnitude of the derivatives?


问:
Can't calculate $I=\int_{-\infty }^{\infty } \frac{\cos x}{x^2+1} \, dx$ using residues
I tried to calculate the integral
$$I=\int_{-\infty }^{\infty } \frac{\cos x}{x^2+1} \, dx$$
using residues and got
$$I=2\pi i \text{ Res}\left(\frac{\cos x}{x^2+1},(x,i)\right)=\pi  \cosh 1\approx 4.8$$
which is nonsense because the graph of $y=\frac{\cos x}{x^2+1}$ is between $-\frac{1}{x^2+1}$ and $\frac{1}{x^2+1}$, the area between these two curves is $2\pi$ and $I\approx 4.8$ is too much.

Mathematica gives $I=\frac{\pi}{e}\approx 1.15573$ which makes much more sense.

My questions are.

Given that my use of residue is wrong, what does $\pi  \cosh 1$    represent?

Why residue is wrong in this example?

答:
What you need to do is to calculate the residue of $\dfrac{e^{iz}}{z^2+1}$
then get the real part of it, since $\cos(z)$ is not a bounded function on the upper complex plane but $e^{iz}$ is.


问:
$P[X=Y]=0$ if $X,Y$ are i.i.d. with continuous c.d.f.
I am having lots of trouble proving the following statement:

Let $X,Y$ be two real valued random variables on a probability space $(\Omega,\mathcal{F},P)$. These two variables are independent and identically distributed, iid, in the sense that they share the same distribution $P_X=P_Y$ and the cumulative distribution function $F_X(x)=P_X[X\leq x]$ is continuous. The probability measure $P$ might not necessarily be absolutely continuous wrt a sigma-finite measure. Therefore there might not be a Radon-Nikodym derivative with respect to a sigma-finite measure. If there was a Radon-Nikodym derivative with respect to a sigma-finite measure, it is easier to show what I want.

I need to prove that, under the above conditions $P[X=Y]=0$

Any insights?

Best regards,

Juan Manuel

答:
For any increasing sequence $-\infty=x_0<x_1<\ldots < x_n=\infty$, we have
$$\tag0P[X=Y]\le \sum_{i=1}^n P[x_{i-1}<X\le x_i,x_{i-1}<Y\le x_i]=\sum_{i=1}^n (F_X(x_i)-F_X(x_{i-1}))^2.$$
Since $F_X\colon\mathbb R\to[0,1]$ is continuous and at least covers $(0,1)$, we can pick $x_i$ so that $F_X(x_i)=\frac in$. Then $(0)$ gives us $P[X=Y]\le \frac 1n$.</p>


问:
Given $16$ pens to be distributed between $4$ people ,then find the number of distributions with the given assumptions.
Given $16$ pens to be distributed between $4$ people assuming the fourth  person gets at least $3$ pens and the three other ones should not get more than $5$ pencils,in how many ways it's possible?

We are looking for the integral solutions to the equation $b_1+b_2+b_3+b_3+b_4=16$ such that $b_4 \ge 3$ and $b_1,b_2,b_3 \le 5$,this is given by :

$$[x^{16}]\left[\sum_{3 \le k \le 16}^{}x^k \right]\left[\sum_{0 \le k \le 13}^{}x^k \right]^3$$
$$=[x^{13}]\left[\sum_{0 \le k \le 13}^{}x^k \right]\left[\sum_{0 \le k \le 13}^{}x^k \right]^3$$
$$=[x^{13}](1-x^{14})^4(1-x)^{-4}$$
$$=[x^{13}]\sum_{k=0}^{4}\binom{4}{k}(-1)^kx^{14k}\sum_{k \ge 0}^{}\binom{k+3}{k}x^{k}$$
$$=\binom{16}{13}$$

However the answer is not this,so where was I wrong?

Body:
I'm not sure where you went wrong, but here is a solution using a generating function.
$$\begin{align}
f(x) &= (x^3+x^4+x^5+\dots)(1+x+x^2+x^3+x^4+x^5)^3 \\
&= x^3 \cdot \frac{1}{1-x} \cdot \left( \frac{1-x^6}{1-x} \right)^3 \\
&= x^3 (1-x^6)^3 (1-x)^{-4} \\
&= x^3 (1-3x^6+3x^{12}-x^{18}) \sum_{i=0}^{\infty}\binom{4+i-1}{i} x^i
\end{align}$$
So
$$[x^{16}]f(x) = \binom{4+13-1}{13} -3\cdot \binom{4+7-1}{7} + 3 \cdot \binom{4+1-1}{1} = \boxed{212}$$


答:
What you remember is correct. Indeed, if the the boundary is smooth you can parametrize it with a smooth curve, say $s:\mathbb{R}\rightarrow\mathbb{R}^n$. (I'll be considering $\nabla u$ only on the boundary, even if I won't write it explicitly).

Let's call $v(t) = u(s(t))$. Now $v\equiv0$, because $u$ is $0$ on the boundary. Thus $\dot{v} \equiv0$. Computing the derivative we get $$\dot{v}(t)= \nabla u(s(t))\cdot \dot{s}(t).$$ Recall that the vector $\dot{s}(t)$ lives in the tangent space to the curve in $s(t)$. This means that the projection of $\nabla u(t)$ to the tangent space is $0$ (for every $t$) and it allows us to conclude that if $\nabla u(t)$ has a nonzero component, this must be in the same direction of $N_{s(t)}$ (the normal to the surface in $s(t)$).

Title: If $u = 0$ on $\partial U$ and $\frac{\partial u}{\partial N} = 0$ on $\partial U$, then $\nabla u = 0$ on $\partial U$?

Body:
Suppose that $u$ is a smooth function from an open subset $U\subset\mathbb{R}^{n}$ with smooth boundary to $\mathbb{R}$. Let $N$ be the (outward) normal on $\partial U$.

If we know that $$u = 0 \qquad \text{and} \qquad \frac{\partial u}{\partial N}:= \nabla u \cdot N = 0 \qquad \text{on} \qquad \partial U,$$ can we then deduce that $\nabla u = 0$ on $\partial U$?

I came accross this problem when studying a PDE. I tried to play around with integral equations, but it didn't really get me anywhere. Any suggestions would be greatly appreciated.

EDIT: I read somewhere that $u=0$ on $\partial U$ implies that $\nabla u$ and $N$ are parallel on $\partial U$. But I don't know why this is true.


问:
Determining convergence of a series with an arbitrary index
I am given this infinite series:

$$\sum_{n=a}^{\infty} \frac{6x^n}{\sqrt[5]{n}}$$

and asked to determine its interval of convergence. I applied the Ratio Test and found that the series converges for values of $-1\leq x < 1$. However, I am a little unsettled by the index of summation. So far, all the problems I have done involved indices such as $n=0$ and $n=1$. The index here starts from an arbitrary number $a$, I assume that's what they mean. Does this change anything regarding my problem? Would it be correct to say that since the Ratio Test involves taking a limit an infinity, it does not matter where we start the summation?

答:
Providing that the terms $u_0, u_1, \dots, u_{a-1}$ are defined, the radius of convergence of the power series $S(x) = \sum_{k=0}^\infty u_n x^n$ is the one of the power series $S_a(x)= \sum_{k=a}^\infty u_n x^n$ as we have

$$S_a(x) = S(x) - P_a(x)$$ were $P_a(x) = \sum_{k=0}^{a-1} u_n x^n$ is a polynomial that is defined on all $\mathbb C$. Therefore $S(x)$ converges if and only if $S_a(x)$ converges.


问:
Inequality with local martingale (find sharp estimate for $\mathbb{P}(M_t\geq \alpha)$)
I am considering a local martingale $M_t$ with continuous sample paths (i.e. $M_t\in\mathcal{M}_{C,\,\text{loc}}$) and with quadratic variation $\langle M\rangle_t$ (and also $M_0=\langle M\rangle_0$ a.s.). I am required to find a sharp estimate for $\mathbb{P}(M_t\geq \alpha)$ for $\alpha\in\mathbb{R}$. I am told to consider two points (with $\beta>0$):

i) Prove that if $\langle M\rangle_t\leq t$, then $\mathbb{E}\left(\mathrm{e}^{\beta M_t-\frac12\beta^2t}\right)\leq1$; and

ii) Prove that $\mathbb{P}\left(\mathrm{e}^{\beta M_t}>\mathrm{e}^{\alpha\beta}\right)\leq \mathrm{e}^{\frac12\beta^2t-\alpha\beta}$. (Typo corrected, thanks @UBM !)

Point i) suggests that somewhere in the proof, I must claim that non-negative local martingales are supermartingales, but I am not sure how to go about using this claim. Point ii) suggests to me that I should be doing some form of Markov inequality with i) (although I can't really see it) and minimising the right hand side with respect to $\beta$, is this correct? Any guidance is greatly appreciated.

答:
The Doleans-Dade exponential of the process $\beta M^t, 0 \leq t \leq T$ is

$$\mathscr E (M)_t:=\exp \left(\beta M_t -\frac{1}{2} \beta^2\langle M \rangle_t \right), 0 \leq t \leq T$$ which solves the SDE
$$\mathscr E (M)_t = 1 + \int_0^t \mathscr E (M)_s dM_s.$$
Thus, since $M$ is a continuous local martingale, so is $\mathscr E (M).$ Also note that is an exponential so it is a non-negative process and therefore a supermartingale. Since $E\mathscr E (M)_0 = 1,$ we must have
$$E\mathscr E (M)_t \leq 1 \quad \text{ for all } t \in [0,T]. \tag{1}$$
So part $(i)$ follows from (1) and the hypothesis $\langle M \rangle_t \leq t.$

For part $(ii),$ first note that part $(i)$ is equivalent to
$$E[e^{\beta M_t}] \leq e^{\frac{1}{2}\beta^2 t} \tag{2}.$$

Thus, using the Markov inequality and (2) we have
$$P(e^{\beta M_t} > e^{\alpha \beta}) \leq \frac{E[e^{\beta M_t}]}{e^{\alpha \beta}} \leq e^{\frac{1}{2} \beta^2 t} e^{-\alpha \beta}.$$


问:
Question on Hausdroff Compactification
Let $\tilde{X}$ be the set of Hausdroff Compactifications of a space $X$.

I define a relation as follows

Let $\tilde{X_1}$ $\leq$ $\tilde{X_2}$ iff there exists a continuous onto map from $\tilde{X_2}$ to $\tilde{X_1}$ which leaves the points of $X$ fixed.

I am trying to show it is antisymmetric i.e if $\tilde{X_2}$ is related to $\tilde{X_1}$ and $\tilde{X_1}$ is related to $\tilde{X_2}$ then $\tilde{X_1}$ = $\tilde{X_2}$.

Please help in this.

Also, if $X$ is locally compact and hausdroff then I am trying to show that $X_\infty$ i.e the one point compactification of $X$ is the minimal element ie. if $\tilde{X}$ is any Hausdroff Compactification of $X$ then there exists a continuous onto map from $\tilde{X}$ to $X_\infty$ which leaves the points of $X$ fixed.

Kindly explain how to justify?

答:
For antisymmetry, assume that we have continuous maps $f: \tilde{X}_2 \to \tilde{X}_1$ and $g: \tilde{X}_1 \to \tilde{X}_2$ that fix $X$. We prove that the compositions $gf$ and $fg$ are the respective identities. By symmetry we just prove this for $gf: \tilde{X}_2 \to \tilde{X}_2$. By assumption, $gf$ agrees with $\operatorname{id}_{\tilde{X}_2}$ on the dense subset $X \subset \tilde{X}_2$. Since $X$ is dense in $\tilde{X}_2$ and since $\operatorname{id}_{\tilde{X}_2}$ and $gf$ are continuous, this implies that $gf = \operatorname{id}_{\tilde{X}_2}$. Similarly $fg = \operatorname{id}_{\tilde{X}_1}$, proving that $\tilde{X}_1 \cong \tilde{X}_2$.

I am not quite sure how to prove that $X_\infty$ is minimal without additional assumptions on the Hausdorff compactifications of $X$ that are allowed. However, the following argument works if we assume that we only consider Hausdorff compactifications $\tilde{X}$ of $X$ such that $X$ is open in $\tilde{X}$.

We claim that for any such Hausdorff compactification $\tilde{X}$, the canonical map $p:\tilde{X} \to X_\infty = X \cup \{\infty\}$ given by
\begin{aligned}
x \mapsto \begin{cases} X & x \in X; \\ \infty & x \notin X;\end{cases}
\end{aligned}
is a continuous map. An open subset of $X_\infty$ is either given by $U \subseteq X$ open in $X$, or by $X\setminus C \cup \{\infty\}$ for $C \subseteq X$ closed and compact. We have
$$
p^{-1}(X \setminus C \cup \{\infty\}) = \tilde{X} \setminus C,
$$
which is open since $C \subseteq \tilde{X}$ is compact and thus closed (as $\tilde{X}$ is Hausdorff). We have
$$
p^{-1}(U) = U \subseteq X \subset \tilde{X},
$$
and since $X \subset \tilde{X}$ is open by assumption it follows that $U \subset \tilde{X}$ is also open.

Title: Question on Hausdroff Compactification


问:
Function measurable iff the components are?
If $\boldsymbol{f}:X\to\mathbb{R}^n$ is a $\mu$-measurable function, I think it is quite easy to see that its components $f_i$ also are. In fact, the projection $\pi_i:\mathbb{R}^n\to\mathbb{R}$, $\boldsymbol{x}\mapsto x_i$ is continuous and therefore Borel measurable, and $f_i=\pi_i\circ\boldsymbol{f} $. Therefore the counterimage through $\pi_i$ of a Borel subset of $\mathbb{R}$ is a Borel subset of $\mathbb{R}^n$, whose counterimage through $\boldsymbol{f}$ is a $\mu$-measurable subset of $X$.

I cannot prove the converse, i.e. that, if $f_1,\ldots,f_n$ are $\mu$-measurable functions, then $\boldsymbol{f}=(f_1,\ldots,f_n)$ is, but I would not be amazed if it were true. Is it and, if it is, how can it be proved? I thank any answerer very much.

答:
Although your question seems to have been solved, you mentioned in a comment that you were looking for something more general than this answer. So I'll add something more general here.

Consider a measurable space $(X, \Sigma_X)$ and bunch of $n$ measurable spaces $(Y_i, \Sigma_Y^i)$ for $1 \leq i \leq n$. The $\sigma$-algebras $\Sigma_Y^i$ can be combined into a product $\sigma$-algebra denoted as follows:

$$\Sigma_Y = \otimes_{i=1}^n \Sigma_Y^i$$

Now, consider functions $f_1, f_2, \dots, f_n$ where the type of each function is $f_i : X \to Y_i$. Suppose every $f_i$ is measurable w.r.t. $\Sigma_X$ and $\Sigma_Y^i$. You want to show that the vector valued function $f = (f_1, f_2, \dots, f_n)$ is measurable w.r.t. $\Sigma_X$ and $\Sigma_Y$.

To show this, we will invoke the $\pi$-System Sufficiency lemma from below. With this, it is sufficient to show only that the preimages of all elements of a $\pi$-system generating $\Sigma_Y$ are measurable w.r.t. $\Sigma_X$.

Well, luckily for us, it's quite easy to find such a $\pi$-system. By definition, the product measure $\Sigma_Y$ is defined as the $\sigma$-algebra generated by the following $\pi$-system:

$$\{Y_1 \times Y_2 \times \dots \times Y_n : Y_i \in \Sigma_Y^i\}$$

Thus, applying our lemma, given any $Y_1, \dots, Y_n \in \Sigma_Y$, it suffices to show:

$$f^{-1}(Y_1 \times Y_2 \times \dots \times Y_n) \in \Sigma_X$$

The nice thing about this is that since our set is a "rectangle" i.e. a product of measurable sets, we can decompose the inverse image into individual components. Such a decomposition could not be done for an arbitrary set in $\Sigma_Y$, which shows why the application of the $\pi$-System Sufficiency lemma was an essential step. So let's decompose:

$$f^{-1}(Y_1 \times Y_2 \times \dots \times Y_n) = \\
f_1^{-1}(Y_1) \cap f_2^{-1}(Y_2) \cap \dots \cap f_n^{-1}(Y_n) = \\
\bigcap_{i = 1}^n f_i^{-1}(Y_i)$$

Since each $f_i$ is itself measurable, then each $f_i^{-1}(Y_i) \in \Sigma_X$ for all $i$. And since all $\sigma$-algebras are closed under finite intersections, then we have:

$$\bigcap_{i = 1}^n f_i^{-1}(Y_i) \in \Sigma_X$$

Which completes the proof.

Lemma: $\pi$-System Sufficiency

Suppose we have sets $X, Y$ with respective $\sigma$-algebras $\Sigma_X, \Sigma_Y$. Suppose $A$ is a $\pi$-system over $Y$ such that $\sigma(A) = \Sigma_Y$. Then any function $f : X \to Y$ is measurable w.r.t. $\Sigma_X$ and $\Sigma_Y$ iff $\forall E \in A : f^{-1}(E) \in \Sigma_X$.

For now, I'll leave the proof of this up to an interested reader, but may come back to this and fill in the details. Basically, the strategy is to apply the $\pi$-system lemma from here, and show that the class of $Y$ subsets whose preimage is measurable w.r.t. $\Sigma_X$ is a d-system. Or you can probably find a proof somewhere on Google :)

Notes

The question alludes to a measure $\mu$. This implies the existence of a measure space over which $\mu$ is defined. However, the above result only needs a measurable space, not a measure space. That is, we don't need a measure at all, just a $\sigma$-algebra. Of course, as mentioned here, any measurable space can be turned into a measure space easily enough, so this is just a minor point.


问:
Prove that $m\leq 4$
This question has been cross-posted on Math.Overflow.

There are positive integers $a,b,c,d_i$, s.t. $\sqrt{a+\sqrt{b}+\sqrt{c}}=\sum_{i=1}^m \sqrt{d_i}$, and for any $i\ne j$, $\sqrt{d_i/d_j}$ is not a rational number. Prove that, $m\le 4$.

I tried some algebraic transformations, such as
$$
a+\sqrt{b}+\sqrt{c}=\left( \sum_{i=1}^m{\sqrt{d_i}} \right) ^2
$$
But they didn't work.

Would it be related to Galois Theory?

Can anyone help?

答: This has already been answered by @Fedor Petrov on Math.Overflow. And I think this can be accepted as a nice answer.</p>

Title: Prove that $m\leq 4$


答:
You had an error. It should be $\overline{BE}^2=\frac14\overline{AB}^2$and

$$\overline{AD}^2=\overline{AE}^2+\frac{\overline{BC}^2}{36}
=\overline{AB}^2- \frac14 \overline{AB}^2+\frac{\overline{AB}^2}{36} = \frac79\overline{AB}^2$$

Title: Equilateral triangle $ABC$ with $D$ on $BC$ such that $BD=\frac13BC$. Prove $9\overline{AD}^2=7\overline{AB}^2$

Body:
Equilateral triangle $ABC$, point $D$ on side $BC$ such
that $BD=\frac13BC$. Prove $9\overline{AD}^2=7\overline{AB}^2$

In an equilateral triangle $ABC$, $D$ is a point on side $BC$ such
that $BD=\frac{1}{3}BC$. Prove that
$9\overline{AD}^2=7\overline{AB}^2$.

I started by constructing the median/altitude(because it's an equilateral triangle so it is the same thing) $AE$. $$BE=\frac{BC}{2}$$ $$DE=\frac{BC}{2}-\frac{BC}{3}=\frac{BC}{6}$$ $$\overline{AD}^2=\overline{AE}^2+\frac{\overline{BC}^2}{36}$$ $$\overline{AB}^2=\overline{AE}^2+\frac{\overline{BC}^2}{4}$$ But I got $$9\overline{AD}^2=\overline{AB}^2$$ Am I making a mistake somewhere or is there a problem in the question?


问:
Generating cyclic subgroups of matrices
I'm confused about how to go about generating the cyclic subgroups of given matricies.

In a question I'm doing I have to find what matrices does the intersection of 2 cyclic subgroups of matrices contain both unsure how to do so.

Q)  Which matrices does H ∩ K contain? What is |H ∩ K|? When K denotes the (cyclic) subgroup of $GL(2, \Bbb C$) generated by X and when H denotes the (cyclic) subgroup generated by Y.

X=$\bigl( \begin{smallmatrix} \theta & 0 \\ 0 & \theta^{-1} \end{smallmatrix} \bigr)$ and Y= $\bigl( \begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix} \bigr)$ where $\theta = e^{\frac {πi}{3}}$

Is there a method for finding the cyclic subgroup? I found this Is this group of matrices cyclic? but I'm not sure if it applies here. I think I can find the intersection if I could just know how to get the matrices

答: A cyclic group (in particular, a subgroup of some other group) is a group generated by some element (in our case, matrix) $A$. This means that such group includes elements $e$ (the identity element, in our case the identity matrix), $A$, $A^2$, $A^3$ and so on until we get $e$ again. So write down powers of $X$ and $Y$ until you get the identity matrix, and you'll get all the matrices from $K$ and $H$ respectively.

Title: Generating cyclic subgroups of matrices


问:
Does the limit exist if a function approaches a limit where it is discontinuous??

If the function is discontinuous at $x=0$ and$f(0)=4.5$ . And as $x$ approaches $0$, the value of function nears $3$.

Does the limit in this case exist?

Imho ,it does and equals $3$ but one of my friend disagreed.

I'd be grateful if someone even hints at my error.

Thank you.

答: The limit exists, and is $3$. The fact that the limit is not the value of the function there is what tells you the function isn't continuous.


问:
Find $c$ in equation system with 2 equations and 4 variables
Determine all values of $c$ in the following system $$\left\{\begin{matrix}
a+2b+5c+10d=211\\
a+b+c+d=52
\end{matrix}\right.$$

a,b,c,d are positive integers.

I've done

$R_1-R_2\Rightarrow b+4c+9d=159\Rightarrow 4c+9d=159-b$

Diofantic equation $4c+9d=1\Rightarrow c=-2,\ d=1 \Rightarrow c=2b-318,\ d=159-b$

$4$ and $9$ has GCD=$1$. $c=2b-318+9n,\ d=159-b-4n$

$a+b+c+d=52\Rightarrow a+b+(2b-318+9n)+(159-b-4n)=52\Rightarrow a+2b+5n-211=0$

I have counted as far as I can. Maybe I did something completely wrong. Do you have any tips on how I can continue, or how I should do?

It feels like all values of $c$ can be up to 50.

答:
This simple procedure for finding a integer lattice basis (given some linear homogeneous diophantine equations) does not seem widely taught. One needs to know how to use matrices.  There is a ton of  material written on basis reduction; it is assumed that we know how to find a basis to begin with...

$$
\left(
\begin{array}{rrrr}
1&2&5&10 \\
1&1&1&1 \\
\end{array}
\right)
\left(
\begin{array}{rrrr}
-1&2&-3&8 \\
1&-1&4&-9 \\
0&0&-1&0 \\
0&0&0&1 \\
\end{array}
\right) =
\left(
\begin{array}{rrrr}
1&0&0&0 \\
0&1&0&0 \\
\end{array}
\right)
$$

$$
\left(
\begin{array}{rr}
-3&8 \\
4&-9 \\
-1&0 \\
0&1 \\
\end{array}
\right)
\left(
\begin{array}{cc}
2&3 \\
1&1 \\
\end{array}
\right) =
\left(
\begin{array}{rr}
2&-1 \\
-1&3 \\
-2&-3 \\
1&1 \\
\end{array}
\right)
$$

The Gram matrix of the initial basis is
$$
\left(
\begin{array}{cc}
26&-60 \\
-60&146 \\
\end{array}
\right)
$$

The vectors in the reduced basis are much closer to orthogonal, and the Gram matrix is closer to diagonal. Note that basis reduction for a two dimensional lattice, as here, is easily found (by hand) using Gauss reduction of the quadratic form associated with the Gram matrix, keeping track of the two by two matrix that accomplishes Gauss reduction as $P^TH_1P = H_2$
$$
\left(
\begin{array}{cc}
10&2 \\
2&20 \\
\end{array}
\right)
$$

In the lattice given by making the system homogeneous, a reduced basis (as rows)
is
$$  ( 2,-1,-2,1) , \; \; \; ( -1,3,-3,1) \; \; .  \;  $$
Solving the original system gives one solution $( -107, 159, 0, 0) $ so that all integer solutions are found by
$$  (\; \; \; -107 + 2s-t,\; \; \; 159 -s+3t,\; \; \; -2s - 3t, \; \; \;s+t \; \; \;)$$
Then it is just inequalities about $s,t$ to get all four elements non-negative

Here is is blown up. We can see that there are many lattice points inside or on the black quadrilateral. If $a,b,c,d$ must be positive, we count those strictly inside. If non-negative, we also include any lattice points   on the boundary.

As $c = -2s - 3t,$ we see that  the line at with $2s + 3t = -40 $ (parallel to an edge)  narrowly misses the quadrilateral, so that we know $c \leq 39 .$ A more careful calculation of $c = -2s-3t$ along the boundary shows that positivity of $a,b,c,d$ demands $c \leq 37,$ achieved when $s=40, t = -39$ and $(a,b,c,d) = (12,2,37,1) \; . \; \;$
The nearby (boundary) point $s= 39, t=-39$ in the diagram gives $(a,b,c,d) = (10,3,39,0) \; , \; \;$  which violates strict positivity.

Another nearby (boundary) point $s= 38, t=-38$ in the diagram gives $(a,b,c,d) = (7,7,38,0) \; , \; \;$  which also violates strict positivity.


问:
prove that $n(n+1)$ is even using induction

Prove that $n(n+1)$ is even using induction

The base case of $n=1$ gives us $2$ which is even.

Assuming $n=k$ is true,

$n=(k+1)$ gives us $ k^2 +2k +k +2$

while $k(k+1) + (k+1)$ gives us $k^2+2k+1.$

whats is the next step to prove this by induction? I can't seem to show

$ k^2 +2k +k +2$ = $k^2+2k+1$

答:
$n(n+1)$ is an even number.

Take any $n\in\mathbb N$, then $n$ is either even or odd.

Suppose $n$ is even $\implies n= 2m$ for some $m\implies n(n+ 1) = 2m(n+ 1)\implies n(n+ 1)$ is even.

Suppose $n$ is odd $\implies n+1$ is even $\implies n+1 = 2m$ for some $m \implies n(n+1) = 2nm\implies n(n+1)$ is even.


问:
Path between vertices with an odd degree

Is it true that in an undirected graph with exactly $2$ vertices with an odd degree there is a path that connects them and why?

It must be true but I cannot prove it.

答:
graph

this is true as you can see in the above picture of graph. This is the only case of 2 vertices of odd degree so they must be connected.


问:
Fourier Transform of a compactly supported distribution : how to write the definition in a "better way"?
I am really troubled by the following Theorem :

*Theorem : If $u\in \mathcal{E}'(\mathbb R^d)$ (space of compactly supported distributions), then its Fourier transform is a function, and :
$\hat{u}(\xi)=\langle u,e^{-ix\cdot\xi}\rangle$

For a distribution of $u\in \mathcal{S}'(\mathbb R),$ we define the Fourier transform of $u$ as a distribution, defined for all $\phi \in \mathcal{S}(\mathbb R^d)$ by $\langle\hat u,\phi\rangle=\langle u, \hat\phi \rangle$

Moreover $\mathcal{E}'(\mathbb R^d) \subset \mathcal{S}'(\mathbb R^d)$.

So.. If I follow the definition that is given to me I'm supposed to have, for $u \in \mathcal{E}'(\mathbb R^d),$ $\hat{u}$ as a distribution.

Edit : Evaluating $\hat{u}$ in a real or complex number then doesn't make any sense to me. A distribution of this type has a Schwartz function as argument , and not a complex number to my opinion. So I don't understand how can the symbolism "$\hat{u}(\xi)"$ have a sense in the theorem ? It can also simply be a misunderstanding on my part of the rôle of $\hat{u}$,

A bit stuck with this. Also, if it's just a notation problem, I'd also be interested on how you can easily understand/read this theorem, as this makes a nonsense to me.

答:
Another way of saying it : (To be confirmed)

Meaning of the theorem : There exists a $L_1^\text{Loc}(R^d)$ function , such as $\hat{u}$=$T_f$ and :

$$\forall \varphi \in D(R^d),\langle\hat{u},\varphi\rangle=\langle T_f, \varphi \rangle, \text{ where } f:\xi \rightarrow \langle u,e^{-ix\cdot\xi}\rangle$$

So for example , if $u$ is a compactly supported distribution , Let's say $u=T_g$ then :

\begin{align}
& \forall \varphi \in D(R^d),\,\langle\hat{u},\varphi\rangle=\langle T_f,\varphi\rangle \\[6pt]
= {} & \int \langle u,e^{-ix\cdot\xi}\rangle \varphi(\xi) \, d\xi \\[6pt]
= {} & \int \varphi(\xi)\int g(x) e^{-ix\cdot\xi} \, dx \,d\xi
\end{align}


问:
5 digit numbers from 1 3 3 0 0
How many 5 digit number positive integers can be made from the following numbers: $1,3,3,0,0$?

First I calculate all possible combinations $$\frac{5!}{2!2!}=30$$
Then I need to remove all 5 digit numbers startting with $0$ and $00$. They are not 5 digit numbers positive integers.

Starting with $00$ $$\frac{3!}{2!}=3$$
Starting with $0$ $$\frac{4!}{2!}=12$$
$12-3=9$, else I would remove numbers starting with $00$ double.

$30-3-9=18$

Is $18$ the correct answer, or where did i thought wrong?

答: Working from the other direction, the first digit is either $1$ or $3$, and then there are no restrictions on the remaining digits which can be permuted in $4!/2!/2!=6$ and $4!/2!=12$ ways respectively. Thus I arrive at $18$ ways, and you are correct.

Title: 5 digit numbers from 1 3 3 0 0


问:
How many times would I be expected to repeat an outfit?
Imagine I own 5 bottoms, 3 tops, and 2 hats. Bottoms, tops, hats are denoted as $X_b$, $X_t$, $X_h$.

An "outfit", $X$, is defined as a particular choosing of 1 pair of bottoms, 1 top, and 1 hat.

Question

For a given number of days, $N$, how many times would I be expected to repeat an outfit if I picked 1 pair of bottoms, 1 top, and 1 hat at random each day?

Note: I am referring to the probability any outfit is repeated, not a particular one.

My Work

There are 30 outfits, since $5 *3 * 2 = 30$. Thus, for a particular outfit to be repeated at least once in $N$ days, that probability is the compliment of the probability this particular outfit is NOT repeated once in 30 days. In other words, we want the complement that "a given outfit is chosen, and then on $N-1$ days, some other outfit is worn.":

$$EQ1: \ 1 - [\frac{1}{30} \times (\frac{29}{30})^{N-1}] = $$

But since every outfit is equally likely, we can multiply $EQ1$ by 30:

$$EQ2: 30 \times (1 - [\frac{1}{30} \times (\frac{29}{30})^{N-1}]) = $$
$$EQ2: 30 - (\frac{29}{30})^{N-1}= $$

However, I know this is wrong because it yields a probability above 1!

答:
Your EQ1 is wrong.  You also need to subtract the chance that the outfit is never worn because then that outfit is not repeated.  You also need a factor $N$ for the number of days you could have worn it in the term for wearing it once.  So the chance a particular outfit is not repeated is
$$1-\left(\frac {29}{30}\right)^{N}-N\cdot\frac 1{30}\left(\frac {29}{30}\right)^{N-1}$$
For $N=30$ this gives about $0.2642$ chance a given outfit is not repeated.  You can use linearity of expectation to get an expected number of outfits repeated at least once of almost $8$.  You expect about $11$ outfits never to have been worn and another $11$ to have been worn once each.

To get the chance that no outfit is repeated, you have $30$ choices the first day to avoid a repeat, $29$ the next day, and so on.  The total chance is $0$ if $N \gt 30$, otherwise it is
$$\frac {30!}{30^N(30-N)!}$$
for $N=30$ this comes out about $1.288 \cdot 10^{-12}$


问:
Conditions on $\alpha$ such that the function is integrable.
Let $P \in \mathbb{R}[X]$ be a polynomial. For wich values of $\alpha \in \mathbb{R}$ is the function
$$ f: (0,+\infty) \rightarrow \mathbb{R}: f(x) = \frac{P(\sin x)}{x^\alpha} $$ integrable?

So we have to look into the behavior of this function for $x \rightarrow 0$ and $x \rightarrow + \infty$.

I find for $x \rightarrow 0$ that $\sin x = \Theta(x)$ so $P(\sin x) = \Theta(x^t)$, where $t$ is the lowest degree of $f$. It is evident that $x^\alpha = \Theta(x^\alpha)$. Therefore I find that $f(x) = \Theta(x^{t-\alpha})$ for $x \rightarrow 0$.

Hence $f$ is integrable in the neighborhood of $0$ when $\alpha - t < 1 \iff \alpha < t+1$.

Let us now look for $x \rightarrow \infty$. I think that $P(\sin x )= O(1)$ because $\sin x = O(1)$. Again, it is evident that $x^\alpha = \Theta(x^\alpha)$. Therefore I find that for $x \rightarrow +\infty$: $f(x) = O(x^{-\alpha}) $. So, if $\alpha >1$, we find that $f$ is integrable for $x\rightarrow +\infty$.

My question: am I correct until this point because I find it weird that I had to work with the lowest degree of $f$. Now because our last expression was only an $O$-expression, I have to check if $f$ is integrable if $\alpha \leq 1$. My intuition says that $f$ won't be integrable but I am not able to do that.

Thanks in advance!

答:
For $x\rightarrow 0$, your proof is perfectly right. There is nothing strange about looking at the valuation (= the lowest degree with a non-zero coefficient). In fact as $|\sin|<1$, the $\sin^k$ that weighs the most is the one with the lowest $k$.

However I think you should work with equivalents : it avoids verification. You can say $P(\sin x) \sim a_k (\sin x)^k$ where $k$ is the valuation of $P$. So $\frac{P(\sin x)}{x^\alpha} \sim \frac{a_k}{x^{\alpha-k}}$. We recognize a classic Riemann integral, so we directly get that the function is integrable around $0$ if and only if $\alpha -k < 1$ (pay attention to the fact that I can say that because we work with positive functions : $|\sin x|$,...).

As for $x\rightarrow +\infty$, I think the function is not integrable when $\alpha \leqslant 1$ (I think you wrote $\alpha > 1$ instead).
Proof : I make the proof for $\alpha = 1$, the case $\alpha > 1$ is then immediate.

I start with an instructive example : P = X.

Let's prove that $\frac{\sin x}{x}$ is not integrable.
There are different ways to do it.
For example we can remark that for $k\in \mathbb{N}$:

$\int_{k\pi}^{(k+1)\pi} \frac{|\sin x|}{x} \geqslant \frac{\int_{k\pi}^{(k+1)\pi} |sin|}{(k+1)\pi} = \frac{\int_{0}^{\pi} |sin|}{(k+1)\pi} = \frac{2}{(k+1)\pi}$.

As $\sum \frac{1}{k}$ diverges, we get that $\frac{\sin x}{x}$ is not integrable

This example shows how to procede for the general case.

For $P$ a polynome, we just need to show that $\int_{0}^{\pi} |P(\sin)|$ is not zero or negative, and it can be done easily either by calculating the integral, or by induction on the degree of $P$.
Then we get that the function is not integrable.

Hope it helped.


问:
Integration and dot product
Let $f : \mathbb{R}^n \rightarrow \mathbb{R}$ continuously differentiable.
We assume that there exists $L > 0$ such that
\begin{equation}
   \|∇f(x)-∇f(x')\| \leq  L\|x − x'\| \qquad \forall (x,x') \in \mathbb{R}^n \times \mathbb{R}^n.
\end{equation}

Show that
\begin{equation}
  |f(x + h) − f(x) − \langle ∇f(x), h\rangle | \leq \frac{L}{2} \|h\|^2 \qquad \forall (x, h) \in \mathbb{R}^n \times \mathbb{R}^n.
\end{equation}

I do not understand part of the demonstration of which here is:

As $f$ is continuously differentiable, we have from Taylor's formula at zero order with residual in integral form
\begin{equation}
  f (x + h) = f(x) + \int^1_0 \langle ∇f(x + th), h \rangle dt. \tag{1}
\end{equation}

Me, when I apply Taylor's formula at zero order with residual in integral form, we get
\begin{equation}
  f (x + h) = f(x) + \int_x^{x+h} ∇f(t) dt. \tag{2}
\end{equation}
What is the integration process that allows us to leave (2) to (1) ?

答: $\nabla f(t)$ isn't a number. The output of $f$ is a number, so $f(x+h)$ will surely be a number. That means we must have the thing we integrate be a number! The Taylor's theorem you use is perhaps a naive generalization of the one from single variate calculus, but does not hold in multivariate calculus. The form your text states is the correct generalization.

Title: Integration and dot product


答:
Being as don't care about multiple repeats, then you can use the Inclusion Exclusion principle.  See https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle.

Say $N=3$.  On the second day there is a 1/30 chance you had a repeat.  On the third day, there is a $1/30$ chance you had a repeat of either day 1 or day 2, but then there's a chance that day 1 and day 2 are repeats of each other, thus the chance of a new repeat on day 3 is $1/30 + 1/30 - (1/30)^2$.  I believe the formula you want is then
$$
\sum_{i=1}^{N-1} \sum_{k=1}^i (-1)^{k+1}\binom{i}{k}\left(\frac{1}{30}\right)^k.
$$
Of course, this formula is not valid once the double sum passes 30.

Title: How many times would I be expected to repeat an outfit?

Body:
Imagine I own 5 bottoms, 3 tops, and 2 hats. Bottoms, tops, hats are denoted as $X_b$, $X_t$, $X_h$.

An "outfit", $X$, is defined as a particular choosing of 1 pair of bottoms, 1 top, and 1 hat.

Question

For a given number of days, $N$, how many times would I be expected to repeat an outfit if I picked 1 pair of bottoms, 1 top, and 1 hat at random each day?

Note: I am referring to the probability any outfit is repeated, not a particular one.

My Work

There are 30 outfits, since $5 *3 * 2 = 30$. Thus, for a particular outfit to be repeated at least once in $N$ days, that probability is the compliment of the probability this particular outfit is NOT repeated once in 30 days. In other words, we want the complement that "a given outfit is chosen, and then on $N-1$ days, some other outfit is worn.":

$$EQ1: \ 1 - [\frac{1}{30} \times (\frac{29}{30})^{N-1}] = $$

But since every outfit is equally likely, we can multiply $EQ1$ by 30:

$$EQ2: 30 \times (1 - [\frac{1}{30} \times (\frac{29}{30})^{N-1}]) = $$
$$EQ2: 30 - (\frac{29}{30})^{N-1}= $$

However, I know this is wrong because it yields a probability above 1!


答:
Another way to puzzle through:  There are $3$ ways to arrange the digits $1,3,3$.  Given one of these arrangements, say $313$ there are $3$ places to put two $0$'s:

$$3X1X3X$$

Think of each $X$ as a basket and now you have two zeros to put in the three baskets.  There are $3$ ways to put both zeros in one basket.  Then there are three ways to put the zeros in different baskets.  So for each of the three arrangements of non-zero digits, there are $6$ ways to toss in the zeros.  So $18$ is the answer.

Title: 5 digit numbers from 1 3 3 0 0

Body:
How many 5 digit number positive integers can be made from the following numbers: $1,3,3,0,0$?

First I calculate all possible combinations $$\frac{5!}{2!2!}=30$$
Then I need to remove all 5 digit numbers startting with $0$ and $00$. They are not 5 digit numbers positive integers.

Starting with $00$ $$\frac{3!}{2!}=3$$
Starting with $0$ $$\frac{4!}{2!}=12$$
$12-3=9$, else I would remove numbers starting with $00$ double.

$30-3-9=18$

Is $18$ the correct answer, or where did i thought wrong?


问:
Probability of getting 6 heads in a row from 200 flips and intuition about this high value
A few days ago I had an argument with a friend about this question :

What is the probability of getting 6 heads in a row from 200 flips ?

I argued it is high probability (significantly bigger than a half) while he argued it is low probability.

When I tried to give exact formula I failed, so we checked the web were the answer was about 84%, yet he is still not convinced so from this I have two questions:

What is the exact formula for $k$ Heads in a row (consecutive) out of $n$ coin flips?

(Not a mathematical) How to convince my friend that 6 in a row have high probability? Meaning what is the intuition behind the question ?

答:
A bit late, I ran some trials to see.

I did 5000 trials on a sim I wrote in python.
Other readers have done 33 sets of 6 tosses, which is not what you asked for.
In the case of TTTHHH HHHTTT TTTHHH, that method would say there are 0 streaks of 6.

Here is the code I used:
def toss100():
numStreak = 0
truestreak = 0

falsestreak = 0
for x in range(200):

```<code>     flip = coin()
     #print(flip)

     if (flip == True):
         falsestreak = 0
         truestreak += 1
     elif (flip == False):
         falsestreak += 1
         truestreak = 0

     #print("heads streak: ", falsestreak)
     #print("tails streak: ", truestreak)
     if (falsestreak == 6 or truestreak == 6):
         numStreak += 1
         falsestreak = 0
         truestreak = 0

return numStreak
</code>```

total = 50000
streaks = 0
for i in range(total):
streaks += toss100()
print ("Total Streaks: ", streaks)
print("Streaks per 200 tosses: ", streaks/50000)

And the output: Total Streaks:  155647
Streaks per 200 tosses:  3.11294

Hope this helps.

Body:
A bit late, I ran some trials to see.

I did 5000 trials on a sim I wrote in python.
Other readers have done 33 sets of 6 tosses, which is not what you asked for.
In the case of TTTHHH HHHTTT TTTHHH, that method would say there are 0 streaks of 6.

Here is the code I used:
def toss100():
numStreak = 0
truestreak = 0

falsestreak = 0
for x in range(200):

     flip = coin()
     #print(flip)

     if (flip == True):
         falsestreak = 0
         truestreak += 1
     elif (flip == False):
         falsestreak += 1
         truestreak = 0

     #print("heads streak: ", falsestreak)
     #print("tails streak: ", truestreak)
     if (falsestreak == 6 or truestreak == 6):
         numStreak += 1
         falsestreak = 0
         truestreak = 0

return numStreak

total = 50000
streaks = 0
for i in range(total):
streaks += toss100()
print ("Total Streaks: ", streaks)
print("Streaks per 200 tosses: ", streaks/50000)

And the output: Total Streaks:  155647
Streaks per 200 tosses:  3.11294

Hope this helps.


问:
Proving if linear functions are isometries
I have a difeomorfism between two spaces $f:M\rightarrow S$ in $\mathbb{R}^3$, that is defined by a 3 x 3 matrix. I have to determine that this difeomorfism is not an isometry. Our method for proving that a difeomorfism is not an isometry is very long but is meant for a standard difeomorfism. I seem to remember that is $\mathbb{R}^2$, the determinant of the matrix acted as a scale factor; can the same be said in $\mathbb{R}^3$? In other words, if the matrix has a determinant not equal to 1, can I say, without further calculation that it is not an isometry, and similarly, if the determinant is 1 (or maybe -1), can I say that it must be an isometry?

Many thanks :)

答:
If the determinant of a matrix is not $+1$ or $-1$, it is definitely not an isometry--because isometries would preserve lengths, and hence volumes. (However, they may reverse orientation, and so $-1$ is allowed!)

But there are matrices with determinant 1 that are not isometries! Take the linear map which scales the $x$-axis up by a factor of 4, and scales both $y$-- and $z$-axes down by a factor of 2. This has a total determinant of $4 \cdot (1/2) \cdot (1/2) = 1,$ but is not an isometry.

To check if a matrix is an isometry, you need the columns to form an orthonormal basis--so all have norm 1 and are mutually orthogonal.


问:
Is there any analytical approximation to this integral?
I have an integral:$$ I(x) = x\int_{x}^{\infty}K_{5/3}(\xi)\,\rm{d}\xi~,$$
where $K$ is modified Bessel function.
I need to evaluate this integral for tens of thousand different $x$ ultimately to use it for some numerical calculation. In the extreme limits of $x\gg 1$ or $x\sim 0$, the approximations are:
$$\lim_{x\rightarrow 0}I(x) = \frac{4\pi}{\sqrt{3}\Gamma(1/3)}\left(\frac{x}{2}\right)^{1/3}~,$$
and
$$\lim_{x\rightarrow \infty}I(x) = \sqrt{\frac{\pi x}{2}}e^{-x}~.$$

However, I would like to know if there are any approximate expressions $I_\text{approx}(x)$ valid for all $x$ such that $$\frac{|I_\text{approx}(x)-I(x)|}{I(x)}\lesssim 0.1~,$$
for all $x$.

答:
By http://dlmf.nist.gov/10.29.E1 (or http://functions.wolfram.com/03.04.20.0006.01), we obtain
$$
\int_x^\infty  {K_{5/3} (t)\mathrm{d}t}  =  - \int_x^\infty  {(K_{1/3} (t) + 2K'_{2/3} (t))\mathrm{d}t}  = 2K_{2/3} (x) - \int_x^\infty  {K_{1/3} (t)\mathrm{d}t} .
$$
By the change of integration variables $t = \frac{2}{3}s^{3/2}$ and http://dlmf.nist.gov/9.6.E15, this becomes
$$
2K_{2/3} (x) - \pi \sqrt 3 \int_{(3x/2)^{2/3} }^\infty  {\operatorname{Ai}(s)\mathrm{d}s} ,
$$
where $\operatorname{Ai}$ is the Airy function. By http://dlmf.nist.gov/9.10.E1 and then by http://dlmf.nist.gov/9.6.E15 and http://dlmf.nist.gov/9.6.E16, this further equals to
\begin{align*}
& 2K_{2/3} (x) - \pi ^2 \sqrt 3 \left( {\operatorname{Ai}((3x/2)^{2/3} )\operatorname{Gi}'((3x/2)^{2/3} ) - \operatorname{Ai}'((3x/2)^{2/3} )\operatorname{Gi}((3x/2)^{2/3} )} \right)
\\ & = 2K_{2/3} (x) \!-\! \pi\! \left( {(3x/2)^{1/3} K_{1/3} (x)\!\operatorname{Gi}'((3x/2)^{2/3} ) + (3x/2)^{2/3} K_{2/3} (x)\!\operatorname{Gi}((3x/2)^{2/3} )} \right)\!.
\end{align*}
Here $\operatorname{Gi}$ denotes the Scorer function (see http://dlmf.nist.gov/9.12).


问:
$P[X=Y]=0$ if $X,Y$ are i.i.d. with continuous c.d.f.
I am having lots of trouble proving the following statement:

Let $X,Y$ be two real valued random variables on a probability space $(\Omega,\mathcal{F},P)$. These two variables are independent and identically distributed, iid, in the sense that they share the same distribution $P_X=P_Y$ and the cumulative distribution function $F_X(x)=P_X[X\leq x]$ is continuous. The probability measure $P$ might not necessarily be absolutely continuous wrt a sigma-finite measure. Therefore there might not be a Radon-Nikodym derivative with respect to a sigma-finite measure. If there was a Radon-Nikodym derivative with respect to a sigma-finite measure, it is easier to show what I want.

I need to prove that, under the above conditions $P[X=Y]=0$

Any insights?

Best regards,

Juan Manuel

答:
Here is a big machinery argument. Since the cdf is continuous, the probability of each point is zero. let $D$ be the diagonal on $\mathbb{R}^2$. By Fubini's theorem, you get </p>

<p>$$P(X=Y)=\int_\mathbb{R^2}1_DdP_{(X,Y)}=\int_\mathbb{R}\int_\mathbb{R}1_D(x,y)~dP_X(x)~ dP_Y(y) =\int_\mathbb{R}0~dP_Y(y)=0.$$</p>

Body:
Here is a big machinery argument. Since the cdf is continuous, the probability of each point is zero. let $D$ be the diagonal on $\mathbb{R}^2$. By Fubini's theorem, you get

$$P(X=Y)=\int_\mathbb{R^2}1_DdP_{(X,Y)}=\int_\mathbb{R}\int_\mathbb{R}1_D(x,y)~dP_X(x)~ dP_Y(y) =\int_\mathbb{R}0~dP_Y(y)=0.$$


问:
Finding the equation of a line in a plane, concurrent with two lines that are themselves determined by pairs of planes
I need to find a vectorial line equation that is contained in the plane $\pi : x - y + z = 0$ and the line t is concurrent with the lines
$$ r : \left\{ \begin{array}{lcr} x+y+2z-2 & = & 0 \\ x-y & = & 0 \end{array} \right. \qquad s : \left\{ \begin{array}{lcr} x-z+2 & = & 0 \\ y & = & 0 \end{array} \right.$$

I found the vectorial line equation of r and s, but I don't know what to do after this.

$$\begin{align}
r : (\phantom{-}2,2,-1) + \alpha(\phantom{-}2,2,-2) \\
s : (-1,0,\phantom{-}1) + \beta(-1,0,-1)
\end{align}$$

答:
If the line $r\subseteq\pi\equiv x-y+z=0$, then the vector that defines the direction of $r$ is perpendicular to the orthogonal vector of $\pi$. Let $dir(r)=Span( \underline u )$ be the direction of $r$ generated by the vector $\underline u$ and $\underline v=(1,-1,1)$ the orthogonal vector to $\pi\implies\langle \underline u,\underline v\rangle=0$, so we can take $\underline u=(0,1,1)$.

This means that $r$ has an equation like $\begin{cases}x=\alpha\\y=\beta+t\\z=\gamma+t \end{cases}$ and if it has to be concurrent with $h\equiv\begin{cases}x+y+2z-2=0\\x-y=0 \end{cases}$ and $k\equiv\begin{cases}x-z+2=0\\y=0 \end{cases}$, we'll have that a generic point $r \ni P=(\alpha,\beta+t,\gamma+t)$ satisfies the equation of $h$ and $k$. In conclusionn, we can find the line $r$ solving the following system:
$$\begin{cases}\alpha+\beta+t+2\gamma+2t-2=0\\\alpha-\beta-t=0\\\alpha-\gamma-t+2=0\\\gamma+t=0 \end{cases}.$$


问:
Representation of the multiplier algebra of a crossed product
If $G$ is a locally compact group and it acts on a C*-algebra $A$, then there is a canonical *-homomorphism (injection) from the reduced crossed product $A\rtimes G$ to the adjointable operators on the Hilbert $A$-module $A \otimes L^2(G),$ $B(A \otimes L^2(G)).$ Is there a canonical *-homomorphism from the multiplier algebra $M(A\rtimes G)$ to $B(A \otimes L^2(G))$?

答:
The answer is yes, and it is indeed a special case of a much more general result:

If $B$ and $C$ are  C*-algebras, $E$ is a right Hilbert module over $C$,    and $\pi :B\to L(E)$ is a non-degenerate
*-homomorphism, then $\pi$ extends uniquely to a *-homomorphism $\tilde\pi $ from the multiplier algebra $M(B)$ to $L(E)$.

The argument is based on the fact that the non-degeneracy of $\pi $ guarantees that the subset of $E$ formed by the
elements of the form
$$
  x=\sum_{i=1}^n \pi (b_i)x_i,
  $$
with $b_i\in  B$, and $x_i\in  E$,
is dense in $E$.  Therefore, given $m$ in $M(B)$, one may define
$$
  \tilde\pi (m)x =  \sum_{i=1}^n \pi (mb_i)x_i.
  $$

In case $\pi $ fails to be non-degenerate, the uniqueness of $\pi $  may no longer be guaranteed, but the
existence results persists.   To see why, one may restrict each $\pi (b)$ to the essential space
$$
  E_{\text{ess}}:= \overline{\text{span}}\{\pi (b)x: b\in  B, \ x\in  E\},
  $$
and then the representation becomes non-degenerate and the above applies,


问:
Function that is $o(x)$ but not $O(x^2)$.
Maybe I might be not very precise in the definition but I think

$f(x) \in O(x)$ means $\exists C$ such that $\lvert f(x)\rvert \le
C\lvert x\rvert$

$f(x) \in o(x)$ means $\forall C \exists \delta$ such that $x\in
(-\delta,\delta) \implies \lvert f(x)\rvert \le C\lvert x\rvert$.

It seems that $f\in O(x^2) \implies f\in o(x)$ but I think the converse might not be true. I was wondering if there were an easy example of a function with $f\in o(x)$ but $f\not\in O(x^2)$

答:
Be precise when you write $o(x)$ or $O(x)$ because it supposes you have a point of reference.

The definition of $f(x) \underset{x\sim a}{=} O(x)$ is  $\exists K\in \mathbb{R}, \exists \eta >0, \forall x\in [a-\eta, a+\eta], |f(x)|\leqslant K|x|$.

And the definition of $f(x) \underset{x\sim a}{=} o(x)$ is $\frac{f(x)}{x} \underset{x\rightarrow a}{\longrightarrow} 0$.

So for example :

if we take $a=0$, $f(x)=O(x^2) \Rightarrow f(x) = o(x)$, but the inverse is false, taking for example $f(x)=x^{3/2}$

but if we take $a=+\infty$, $f(x) = O(x^2)$ does not imply $f(x) = o(x)$ : for example you can take $f(x)=x^2$ : $x^2 = O(x^2)$ but we clearly don't have $x^2 = o(x)$. Conversly $f(x) = o(x) \Rightarrow f(x)=O(x^2)$.


问:
Infinite paths that connect two vertices?
This is a follow-up to another question concerning infinite paths which was admittedly ill-posed. I hope this question is posed better.

The graph $N$ with vertex set $V(N) = \mathbb{N}$ and $(x,y) \in E(N)$ iff $x < y$ contains paths of arbitrary length connecting 0 and an appropriate $n$ .

The graph $Q$ with vertex set $V(Q) = \mathbb{Q}$ and $(x,y) \in E(Q)$ iff $x < y$ contains paths of arbitrary length connecting 0 and 1.

Of course, both graphs contain infinite paths, starting from 0, but ending nowhere.

It's more or less obvious, that $N$ doesn't contain a path of infinite length connecting 0 and an $n\in \mathbb{N}$ (because all $n$ are finite).

But it's hard (for me) to "see" and get a feeling why $Q$ doesn't contain a path of infinite length connecting 0 and 1: each finite path between 0 and 1 is a finite subset of $E(Q)$: $\lbrace (0,q_1),(q_1,q_2),...,(q_n,1)\rbrace$. Why cannot there be an infinite subset of that kind?

Is it impossible to define "that kind", i.e. the
  property "being a path
  connecting 0 and 1", or can we define
  it (in second order logic maybe) but
  prove, that no infinite subset of $E(Q)$ with this property exists?

答:
Why cannot there be an infinite subset of that kind?

Of what kind? Hans, you are still refusing to be any more precise about what you want this definition to do, and until you start talking with some precision I don't know what there is to say beyond "the standard definitions do not allow you to speak of an infinite path connecting two points because infinite paths do not have two endpoints" or the other things I already said in response to your last question.


问:
Character table-related notation $\mathbf Z_m$ in Curtis–Reiner
This is a question regarding a seemingly unexplained notation in Curtis–Reiner: Methods of Representation Theory, volume 1. In subsection §09D on the character table, on page 215, between equations (9.28) and (9.29), the notation $\mathbf Z_m(C_i)$ is used, but as far as I can tell, $\mathbf Z_m$ was never introduced. Could someone please explain to me what it means?

答: The following definition has been suggested to me. Consider the action of $G$ on $Z_m$; this yields a representation $G\to\mathrm{GL}_{z_m}(K)$. Then $\mathbf Z_m$ is the $K$-linear extension $KG\to M_{z_m}(K)$. This indeed makes the paragraph cited above make sense.


问:
Determinant of a 3x3 matrix by permutation of rows in sagemaths (with CoCalc)
In order to solve a mathematical problem, I would have liked to know how to calculate the determinant of a 3x3 matrix with permutation of rows in sagemaths on Co Calc. Here is a type of statement, the values change each time the page is reloaded, hence the interest of programming an algorithm to find the result of the exercise by changing the values of the corresponding matrix.

Thanking you in advance

Body:
Use this code to show $\det A=\det B$

R.<a,b,c> = QQ[]
A = matrix(R, [[a,b,c],[-8,-62,-38],[15,2,55]])
B = matrix(R, [[15,2,55],[a,b,c],[-8,-62,-38]])
det(A)
det(B)


问:
$\lim_{n\to\infty} \sqrt[n(n+1)]{ \prod_{r=0}^n {n\choose r}} $
I have been trying to figure out the following limit:

$$\lim_{n\to\infty} \sqrt[n(n+1)]{\prod_{r=0}^n {n\choose r}}=\mathscr L$$

It’s quite natural to take the natural log of both sides: $$\ln \mathscr L = \lim_{n\to\infty} \frac{\sum_{r=0}^n \ln {n\choose r}}{n(n+1)} \\ =\lim_{n\to\infty} \frac{(n+1)\ln(n!) - 2\sum_{r=0}^n \ln(r!)}{n(n+1)}\\  =\lim_{n\to\infty} \frac{(n+1)\ln(n!) - 2\ln(\color{green}{n! \times (n-1)! \times (n-2)! \dots \times 2!\times 1!})}{n(n+1)}$$ I believe the green term is called the superfactorial of $n$, but I barely know anything about it. The last thing I tried was to write $$n! \times (n-1)! \times (n-2)! \dots 2!\times 1! =n^1 \times (n-1)^2 \times (n-2)^3 \dots \times 2^{n-1} \times 1^n = \prod_{r=1}^n r^{n+1-r}$$ but that doesn’t help. Any ideas on how to evaluate this limit?

答:
In fact, by the Stolz thorem, one has
\begin{eqnarray}
\ln \mathscr L &=& \lim_{n\to\infty} \frac{\sum_{r=0}^n \ln {n\choose r}}{n(n+1)}\\
&=&\lim_{n\to\infty} \frac{\sum_{r=0}^n \ln {n\choose r}-\sum_{r=0}^{n-1} \ln {n-1\choose r}}{n(n+1)-(n-1)n}\\
&=&\lim_{n\to\infty} \frac{\sum_{r=0}^{n-1} \ln {n\choose r}-\sum_{r=0}^{n-1} \ln {n-1\choose r}}{2n}\\
&=&\lim_{n\to\infty} \frac{\sum_{r=0}^{n-1} \ln \frac{{n\choose r}}{{n-1\choose r}}}{2n}\\
&=&\lim_{n\to\infty} \frac{\sum_{r=0}^{n-1} \ln \frac{n}{n-r}}{2n}\\
&=&\lim_{n\to\infty} \frac{\ln \frac{n^n}{n!}}{2n}\\
&=&\lim_{n\to\infty} \frac{n\ln n-\ln n!}{2n}\\
&=&\lim_{n\to\infty} \frac{n\ln n-(n\ln n-n+O(\ln n))}{2n}\\
&=&\frac12.
\end{eqnarray}
Here the Stirling approximation
$$ \ln n!=n\ln n-n+O(\ln n) $$
is used. So
$$ \mathscr L=\sqrt{e}. $$


问:
Finding the intersecting points of two strictly monotonic increasing functions
I need to prove this:
Let $f$ be a strictly monotonic continuous real-valued function defined on $[a,b]$ such that $f(a)<a$ and $f(b)>b$, then $\exists$ exactly one c $\in$ $(a,b)$ such that $f(c)=c$

My attempt

I could figure out from the given data that $f$ is strictly increasing. I tried to construct a new function:

$g(x)=f(x)-x$

and I got $g(a)<0$  and  $g(b)>0$. I now know for sure that there exits at least one such c $\in$ $(a,b)$ such that $f(c)=c$. But how do I conclude there is exactly one such $c$?

Please suggest. Thanks in advance.

EDIT
I may have chosen the wrong one but there are other options to choose from:

$\exists$ exactly one c $\in$ $(a,b)$ such that $f(c)=c$

$\exists$ $c_1, c_2$ $\in$ $(a,b)$ such that $f(c_1)=c_1$ and $f(c_2)=c_2$

$\exists$ no such c $\in$ $(a,b)$ such that $f(c)=c$

$\exists$ infinitely many such c $\in$ $(a,b)$ such that $f(c)=c$

答:
NOT TRUE.

Take
$$
f(x)=x+\frac{1}{2}\sin x.
$$
Then $f'(x)>0$ and $f(x)=x$, whenever $x=k\pi$.


答:
The series expanision of the hyperfactorial is
$$\log (H(n))=n^2 \left(\frac{\log (n)}{2}-\frac{1}{4}\right)+\frac{1}{2} n \log (n)+\log \left(A
   \sqrt[12]{n}\right)+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right)$$ where $A$ is Glaisher constant.

Use Stirling also.

Title: $\lim_{n\to\infty} \sqrt[n(n+1)]{ \prod_{r=0}^n {n\choose r}} $

Body:
I have been trying to figure out the following limit:

$$\lim_{n\to\infty} \sqrt[n(n+1)]{\prod_{r=0}^n {n\choose r}}=\mathscr L$$

It’s quite natural to take the natural log of both sides: $$\ln \mathscr L = \lim_{n\to\infty} \frac{\sum_{r=0}^n \ln {n\choose r}}{n(n+1)} \\ =\lim_{n\to\infty} \frac{(n+1)\ln(n!) - 2\sum_{r=0}^n \ln(r!)}{n(n+1)}\\  =\lim_{n\to\infty} \frac{(n+1)\ln(n!) - 2\ln(\color{green}{n! \times (n-1)! \times (n-2)! \dots \times 2!\times 1!})}{n(n+1)}$$ I believe the green term is called the superfactorial of $n$, but I barely know anything about it. The last thing I tried was to write $$n! \times (n-1)! \times (n-2)! \dots 2!\times 1! =n^1 \times (n-1)^2 \times (n-2)^3 \dots \times 2^{n-1} \times 1^n = \prod_{r=1}^n r^{n+1-r}$$ but that doesn’t help. Any ideas on how to evaluate this limit?


问:
Two limits without L’Hopital
$$\lim_{x \to \infty}\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}$$

$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}$$

For the first function I intended using $\lim_{x\to \infty}\frac{\ln x}{x}=0$ but I couldn’t figure it out. For the second limit I have no idea. Please help me solve them.

Title: Two limits without L’Hopital

Body:
Note

$$\lim_{x \to \infty}\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}
=\lim_{x \to \infty}\frac{\ln x^3+ \ln(1-\frac5{x^2}+\frac3{x^3})}{\ln x^5+ \ln(1-\frac6{x^3}+\frac6{x^5})}=\lim_{x \to \infty}\frac{3\ln x }{5\ln x }= \frac35
$$

$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}
=\lim_{x\to \infty}\frac{\ln e^{3x}+\ln(1+e^{-3x})}{x}
= \lim_{x\to \infty}\frac{{3x}}{x}=3
$$


Title: Two limits without L’Hopital

Body:
$$\lim_{x \to \infty}\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}$$

$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}$$

For the first function I intended using $\lim_{x\to \infty}\frac{\ln x}{x}=0$ but I couldn’t figure it out. For the second limit I have no idea. Please help me solve them.

Body:
Hints:

$$\frac{\ln(x^3-5x+3)}{\ln(x^5-6x^2-6)}=\frac{3\ln(x)+\ln(1-5x^{-2}+3x^{-3})}{5\ln(x)+\ln(1-6x^{-3}-6x^{-5})}.$$

$$\lim_{x\to \infty}\frac{\ln(1+e^{3x})}{x}=\lim_{t\to \infty}\frac{\ln(1+t^3)}{\ln(t)}.$$


答:
We have

$$\sum_{k=1}^n k \ln(n+1-k) = \sum_{t = n+1-k, 1\le t \le n} (n+1-t) \ln t = S_1 - S_2$$
where
$$ S_1 = \sum_{t=1}^n  (n+1) \ln t = (n+1) \ln (n!) $$
may be evaluated from Stirling's formula
and $$S_2 = \sum_{t=1}^n t \ln t.$$
Hence it's sufficient to evaluate $S_2$. We have
$$\frac{S_2}{n} = \sum_{t=1}^n \frac{t}n (\ln \frac{t}n + \ln n) = \sum_{t=1}^n \frac{t}n \ln \frac{t}n + \frac{\ln n}{n} \sum_{t=1}^n t =  n S_3 + \frac{\ln n}{n} \frac{n(n+1)}{2},$$
where $$S_3 = \frac1n \sum_{t=1}^n \frac{t}n \ln \frac{t}n \to \int_{0}^1 x \ln x dx = - \frac14.$$


问:
a holomorphic extension of a complex function
Let $f:D(2+2i,4)\to \mathbb{C}$ with $f^{(n)}(1+i)=\frac{1}{n!}, \ \forall n\in \mathbb{N}_0$. Then $f$ extended holomorphicly on every disk $D(2+2i, 2(k+2)),\ k\in\mathbb{N}.$ My work:
By using the Identity theorem I got $f(z)$ which is holomorphic everywhere (I guess) on $\mathbb{C}$. So, $f$ extended holomorphically on every disk $D(2+2i, 2(k+2)),\ k\in\mathbb{N}.$
Is it correct? However, I don't know by hypothesis if $f$ is holomorphic anywhere else on complex plane, except point $1+i$. Does it affect that?

答:
Identity Theorem implies that
$$
f(z)=\sum_{n=0}^\infty \frac{(z-1-i)^n}{(n!)^2},
$$
and $f$ is entire analytic.


答: For the first, divide the numerator and denominator by $\ln x$, giving$$\lim_{x\to\infty}\frac{3+\tfrac{\ln\left(1-\frac{5}{x^2}+\frac{3}{x^3}\right)}{\ln x}}{5+\tfrac{\ln\left(1-\frac{6}{x^3}-\frac{6}{x^5}\right)}{\ln x}}=\frac{3+0/\infty}{5+0/\infty}=\frac35.$$For the second, subtract out $3$, giving$$3+\lim_{x\to\infty}\frac{\ln(1+e^{-3x})}{x}=3+\frac{0}{\infty}=3.$$

Title: Two limits without L’Hopital


答:
i) just factor $x^3$ in the argument of the logarithm of the numerator and then use log properties to break it into a sum.
In the denominator do the same but factoring $x^5$. The result should follow easily.

ii) just factor $e^{3x}$ and then break the logarithm of the product as the sum of the logarithms. The result also here follows.

Title: Two limits without L’Hopital


问:
Distances between two complexes when using Persistence Homology
I am using Persistence Homology to look at two different facebook networks. I can generate a distance matrix between individuals and then create the usual barcodes and persistence diagrams according to topological data analysis methods.

Usually I create an alpha complex or Rips complex from the data and then compute the corresponding topological summaries. But I was wondering what methods exist to compute the distance between two different complexes? So say I compute the complexes for two very similar networks, I would need a metric to define which networks are more similar and which ones are less similar.

Of course trying to match topological complexes might be hard, because that is like an isomorphism problem. But are there other good ways, or consensus ways to describe the distance between two objects or observations based upon persistence homology? I can google for references to this question, but since I don't come from a pure math background it can be hard to tell which papers are better than others. If anyone has any good references on this question, that would be helpful as well.

答:
Often data come in the form of point clouds (i.e., finite metric spaces), which can be used to construct simplicial complexes later.  One theoretically-attractive metric one can put on the collection of compact metric spaces is the Gromov-Hausdorff distance.  Many of the stability results concerning the persistent homology pipeline use this distance on the input metric spaces.  More generally, one can consider Gromov-Wasserstein distances from optimal transport by equipping point clouds with some prior probability distribution (e.g., uniform).  Wasserstein metrics and bottleneck distance are also used to compare persistence diagrams/barcodes, so if you want to measure the distance between two point clouds by their barcode summary, this is the typical use-case.

Also, many of the objects in TDA come as filtered objects, including the filtered complexes that are used to compute persistent homology.  In this case, the interleaving distance may be another option to compare two filtered objects, though this is more traditionally used to compare persistence modules specifically.


问:
Calculating $\sqrt{28\cdot 29 \cdot 30\cdot 31+1}$
Is it possible to calculate $\sqrt{28 \cdot 29 \cdot 30 \cdot 31 +1}$ without any kind of electronic aid?

I tried to factor it using equations like $(x+y)^2=x^2+2xy+y^2$ but it didn't work.

Body:
\begin{align}
&\text{Let }x=30
\\ \\
\therefore&\ \ \ \ \ \sqrt{(x-2)(x-1)x(x+1)+1}
\\ \\
&=\sqrt{[(x-2)(x+1)[(x-1)x]+1}
\\ \\
&=\sqrt{(x^2-x-2)((x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x)^2-2(x^2-x)+1}
\\ \\
&=\sqrt{(x^2-x-1)^2}
\\
&=x^2-x-1
\\
&=30^2-30-1
\\
&=\boxed{869}
\end{align}


问:
$X, Y, Z$ square-integrable rvs. Let that $Z$ is $\sigma(Y)$-measurable and $X$ is not $\sigma(Y)$-measurable. Prove the equivalence of statements
Let $X, Y, Z$ are square-integrable random variables. Let also that $Z$ is $\sigma(Y)$-measurable and $X$ is not $\sigma(Y)$-measurable.

Are the following statements equivalent?

$X-Z$ not correlate with any square-integrable $\sigma(Y)$-measurable random variable $W$

$X-Z$ is independent with any square-integrable $\sigma(Y)$-measurable random variable $W$

I tried to solve the problem through a conditional expectation, and used the fact that if $\mathbb E(X \mid Y) = const$ then $X$ and $Y$ are independent. It is also true in the opposite direction.
But it seems that this does not work in general, and I am at a dead end.

答:
The two conditions are not equivalent.  The idea of the example below is taken from Dilip Sarwate's comment on this question.

Let $(X,Y)$ be uniform on the ball $B(0,1)$ in $\mathbb{R}^{2}$.  That is, if $h$ is any bounded, continuous function in $\mathbb{R}^{2}$, then
\begin{equation*}
\mathbb{E}(h(X,Y)) = \omega_{2}^{-1} \int_{B(0,1)} h(x,y) \, dx \, dy,
\end{equation*}
where $\omega_{2}$ is the area of $B(0,1)$.  Notice that $X$ and $Y$ are not independent.

Let $Z = 0$.  Observe that (2) fails since $X - Z = X$ is not independent of $Y$, even while $Y$ is a square integrable $\sigma(Y)$-measurable random variable.

Recall that $W$ is $\sigma(Y)$-measurable if and only if there is a Borel function $k : \mathbb{R} \to \mathbb{R}$ such that $W = k(Y)$.  It follows that (1) holds.  Indeed, given such a square integrable $W$, we can write
\begin{equation*}
\mathbb{E}((X - Z)W) = \mathbb{E}(XW) = \omega_{2}^{-1} \int_{B(0,1)} x k(y) \, dx \, dy = \omega_{2}^{-1} \int_{-1}^{1} \left[ \int_{-\sqrt{1 - y^{2}}}^{\sqrt{1 - y^{2}}} x \, dx \right] k(y) \, dy = 0.
\end{equation*}

This gives an example where (1) holds, but (2) does not.


问:
The easiest way possible to evaluate $\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$
I want to solve this integral the easiest way possible, how should I do it?
$$\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx$$

I tried with substitution, but it didn't work out. And I think that it isn't possible to use integration-by-parts method - at least I couldn't find a way to do it.

答:
Integrating by parts is maybe the easiest

\begin{align}
\int_{}^{}\frac{\ln{(\cos{x})}}{1-\cos{x}}dx
&=-\int \ln(\cos x)d(\cot \frac x2)\\
&=-\cot \frac x2\ln(\cos x) -\int (1+\sec x)dx\\
&= -\cot \frac x2\ln(\cos x) -x -\ln(\tan x+\sec x) +C
\end{align}


问:
5 digit numbers from 1 3 3 0 0
How many 5 digit number positive integers can be made from the following numbers: $1,3,3,0,0$?

First I calculate all possible combinations $$\frac{5!}{2!2!}=30$$
Then I need to remove all 5 digit numbers startting with $0$ and $00$. They are not 5 digit numbers positive integers.

Starting with $00$ $$\frac{3!}{2!}=3$$
Starting with $0$ $$\frac{4!}{2!}=12$$
$12-3=9$, else I would remove numbers starting with $00$ double.

$30-3-9=18$

Is $18$ the correct answer, or where did i thought wrong?

答:
In the simplest of terms (as a check), without using permutations and combinations:  You draw from the available choices randomly to calculate all combinations.

The first number is one of 3-choices (1, 3, or 3).
The second number is one of 4-choices (all that remain).
The third is one of the 3-that remains.
Then one of 2.
And finally, the last choice.

There are 4 different ways to arrange the number 13300 (first 3 first, second 3 first, and the same with the zeros in combination.

$$3\cdot 4\cdot 3\cdot 2\cdot 1 = 72$$

Then, to account for the double $0$s and double $3$s, divide by 2 twice.

$$\frac{72}{2\cdot 2} = 18$$

Title: 5 digit numbers from 1 3 3 0 0


问:
Function evaluation from area
Can we find a smooth continuous $  f(z)$ if

$$ \int_1^3 f(x) dx = \dfrac{26}{3},\; \int_2^4 f(y) dy=\dfrac{56}{3} ?$$

EDIT1:

What I actually wanted to ask was:

If $( f,g )$ are functions inverse of each other then find $f(x)$

$$ \int_1^3 f(x) dx + \int_2^4 g(y) dy=10 $$

Answer is known as any function as being "obvious" but how to go about showing the same?

Comments appreciated.

答: The $26$ is a dead give-a-way.  It's $3^3-1^3$.  The $3$ in the denominator clinches it: $f(x) = x^2$.  Then just check the other integral.

Title: Function evaluation from area


问:
Difference between vector and pseudovector
There is a definition of a pseudovector on wikipedia.
On the top right there is this image with explanation:

A loop of wire (black), carrying a current I, creates a magnetic field
B (blue). If the position and current of the wire are reflected across
the plane indicated by the dashed line, the magnetic field it
generates would not be reflected: Instead, it would be reflected and
reversed. The position of the wire and its current are "true" vectors,
but the magnetic field B is a pseudovector.

OK I understand the issue with mirror reflection but then I can say the opposite:

We have a loop of wire in a magnetic field B which creates current I in the wire.  If the position and magnetic field are reflected across the plane indicated by the dashed line, the current it generates would not be reflected: Instead, it would be reflected and reversed. The position of the wire and its magnetic field are "true" vectors, but the current I is a pseudovector.

Here it is how it would look:

So I came to an opposite conclusion - B is true vector and I is pseudovector. Where is my logic wrong?

答: $\vec I$ represents true physical motion, so it must be reflected when you reflect the coordinates.  We don't know about $\vec B$ until we chase through the situation.


问:
Fresnel integral: stationary phase approx.
The following integral describes the propagation of light (in certain cases)
$$
U_i(x,y,z) =
\frac{e^{ikz}}{i\lambda z}
 \int_{-\infty}^\infty d\xi \int_{-\infty}^\infty d\eta  \;
U_0(\xi, \eta,0) \,
\exp\left({i \frac{k}{2z}\left[
(x-\xi)^2 + (y-\eta)^2
\right]
}\right)
$$
In the limit $z\to 0$ it is said that the quadratic phase factor act like a Dirac $\delta$ distribution. While this is intuitively clear (in the context of optics) I don't see how it works mathematically.

$k=2\pi/\lambda > 0$ is the wave number,

$\lambda>0$ is the wavelength,

$(x,y,z)$ is a position, and so is $(\xi, \eta, 0)$. Hence, they are all real numbers.

$U_0$ is a real valued function, which represent the electric field. It is a "well-behaved" function (i.e. finite, and smooth). Usually we consider that it differs from zero only on a finite surface $\Sigma$ -- the aperture.

答:
We can write $U_i$ as a convolution between two function, $U_0$ and $h(x,y,z) = \ldots$
$$
U_i(x,y,z) = U_0(x,y,0) * h(x,y,z)
$$

Next, we Fourier transform both sides (in 2D)  and use the Fourier theorem for convolutions. This yields
$$
\tilde U_i(f_x, f_y, z) = \tilde U_0(f_x,f_y, 0) \cdot \tilde h(f_x,f_y,z)
$$
Doing the calculation, we get
$
\tilde h(f_x,f_y,z) = e^{ikz} e^{-i\pi \lambda z (f_x^2 + f_y^2)}
$.

Now, we take the limit  $z\to 0$ and obtain $\tilde h(f_x,f_y,0) = 1$.

Finally, we use the inverse 2D Fourier transform on both sides as well as the Fourier theorem for convolutions. This yields
$$
U_i(x,y,z) = U_0(x,y) * \mathcal{F}^{-1}\{1\}
= U_0(x,y) * \delta(x,y)
$$


问:
Solving Dual Problem using Simplex Method?
Let's say I have a primal problem as follows:

```<code>(P) Max 2x1 + 3x2 + 5x3
s.t
x1 + 2x2 + 3x3 <= 8
x1 - 2x2 + 2x3 <=6
x>=0
</code>```

Then we obtain the dual as follows:

```<code>(D) Min 8w1 + 6w2
s.t
w1 + w2 >= 2
2w1 - 2w2 >= 3
3w1 + 2w2 >= 5
w>=0
</code>```

After obtaining the dual problem, when I try to apply Simplex method on the dual problem, I have realized that the table is already optimal which leads to the optimal objective value 0. But applying simplex to the primal problem, it is not possible to get the same solution. How is that possible? Is it not possible to apply Simplex method to the dual problem?

答: Optimal solutions are $x=(7,0.5,0)$ and $w=(1.75,0.25)$, both with objective value $15.5$.  Your dual formulation is correct.  Yes, you can apply the simplex method to the dual problem, but $0$ is not optimal.


问:
Composition of Partial Isometry
Let $T$ be a linear operator in $H$, a Hilbert space.

An operator $T \in L(H)$ is said to be a partial isometry if the restriction of $T$ to $ker(T)^{\perp}$ is an isometry. I would like to prove that given a partial isometry $T$, then $T^{2}$ is a partial isometry if and only if $(T^{*}TTT^{*})$=$(TT^{*}T^{*}T)$. All I could do so far was to prove the classic result : $T=TT^{*}T$ iff $T$ is a partial isometry iff $T^{*}=T^{*}TT^{*}$, that might be useful.

Thank you for help :)

答:
You can prove this using the following observation: Let $T$ be a bounded linear operator on a Hilbert space. Then $T$ is a partial isometry if and only if the spectrum of $T^{\ast} T$ consists only of $\{0,1\}$. Furthermore, let $A$ be a bounded, self-adjoint operator. Then $\sigma(A) \subset \{0,1\}$ if and only if $A = A^2$; this follows from the spectral theorem. </p>

<p>So our general strategy to prove that $T$ is an isometry should be to prove the equivalent relation $T^{\ast} T = (T^{\ast} T)^2$. If you apply this directly, you get your "classic" result $T = TT^{\ast} T$. So all that remains for you to prove the partial isometry condition for $T^2$ is to check all this instead with $T^* T^* TT = (T^*T^* TT)^2$. </p>

Title: Composition of Partial Isometry

Body:
You can prove this using the following observation: Let $T$ be a bounded linear operator on a Hilbert space. Then $T$ is a partial isometry if and only if the spectrum of $T^{\ast} T$ consists only of $\{0,1\}$. Furthermore, let $A$ be a bounded, self-adjoint operator. Then $\sigma(A) \subset \{0,1\}$ if and only if $A = A^2$; this follows from the spectral theorem.

So our general strategy to prove that $T$ is an isometry should be to prove the equivalent relation $T^{\ast} T = (T^{\ast} T)^2$. If you apply this directly, you get your "classic" result $T = TT^{\ast} T$. So all that remains for you to prove the partial isometry condition for $T^2$ is to check all this instead with $T^* T^* TT = (T^*T^* TT)^2$.