问:
Comparison of the largest eigenvalues of two symmetric matrices
Let $$A=\left(\begin{array}{ccccc|cccc}0&\cdots&0&0&0&0&0&a\\\vdots&\cdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\0&\cdots&0&0&0&0&0&a\\0&\cdots&0&0&b&0&0&0\\0&\cdots&0&b&0&b&0&0\\\hline 0&\cdots&0&0&b&0&\frac{1}{\sqrt{2}}&0\\0&\cdots&0&0&0&\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\a&\cdots&a&0&0&0&\frac{1}{\sqrt{2}}&0
\end{array}\right),$$ be an $n\times n$ matrix, where $a=\sqrt{\frac{6}{7}}$ and $b=\frac{1}{\sqrt{2}}$.

Let $B$ be same as $A$ except the second diagonal block which is $\left(\begin{array}{ccc}0&\sqrt{\frac{2}{3}}&\sqrt{\frac{8}{21}}\\\sqrt{\frac{2}{3}}&0&0\\\sqrt{\frac{8}{21}}&0&0\end{array}\right).$ With help of matlab I have verified that largest eigenvalue of $A$ is greater than that of $B$. How to prove it? I thought of doing it with calculating determinant of partitioned matrices but couldn't succeed.

答:
Not an answer, just too long for a comment.

Well, I decided to try and brute-force the computation of the characteristic polynimial for $A$ and got the following:

let $D_n$ be the determinannt of $tI_n-A$ (asumme $n\geq6$ so $A$ will contain the required $a$'s), so:

$$D_n = tD_{n-1} -a^2t^{n-6}(t^4-6b^2t^2+b^4)$$

and:

$$D_6 = t^6+(b-a^2)t^4 -3b^2t^3+(b^4-2b^3)t^2+(3a^2b^2-ab^4)t-a^2b^4$$


问:
Reference request: The set of all walks of finite length in a finite, strongly connected digraph is countable.
Let $\mathcal{G}$ be a finite, strongly connected directed graph. It seems evident to me that the set of all walks of finite length through $\mathcal{G}$ is countable, since walks can be enumerated first by path length, and for each path length by the set of all initial vertexes, and for every such set by the set of connected vertexes, and so on and so forth, in the same way that a language containing only finite strings can be enumerated.

Is my intuition here correct?

Is this result so obvious as to reasonably not require a reference in a doctoral thesis?

If a citation is needed, is there a reference that provides this result or a stronger result that implies it?

答:
The result is, in fact, obvious. Define a "walk of length $n$" to be a sequence of nodes $v_1, v_2, ..., v_{n + 1}$ such that there is an edge between $v_1$ and $v_2$, $v_2$ and $v_3$, etc.

Clearly, this is an infinite subset of the set of all sequences of nodes. And the set of all sequences of elements of a finite, non-empty set is always countably infinite.

Thus, the set of all walks will be countably infinite.

One would assume that any math or CS Ph.D. would realize this immediately, but you should ask your advisor whether you need a proof/citation, not the internet. Your advisor's opinion is the one that matters, not mine.

ContentLicense: CC BY-SA 4.0


问:
Differentiating under the integral sign.
Let $f=f(x, \lambda, z):\mathbb{R}^n\times \mathbb{R}_{+}\times \mathbb{R}^n$ and let
$$F(\lambda, z) = \int_{\mathbb{R}^n} |f(x, \lambda, z)|^p dx$$
be a functional defined for $1<p<+\infty.$ Then if $\lambda^{*}, z^{*}$ are minimizes of $F,$ it must be the case that
$$\partial_{\lambda} F(\lambda^{*},z^{*}) = 0\implies p\int|f|^{p-2} f \partial_{\lambda }f = 0 $$
$$\partial_{z}F(\lambda^{*},z^{*}) = 0 \implies p\int|f|^{p-2} f \partial_{z}f = 0.$$

Are these first-order conditions correct?

答:
If $F(\lambda,z)$ this is strict enough to garanty that the partial derivative and the integration are interchangeable. That stems from a set of theorems and corollares of real and complex Analysis which is part of Mathematics. It is hard to decide how much in depth the set has to go and depends on the schools of Mathematics one tends to prefer.

As shown in the question neither the domain nor the integration infinitesimals are important. Is this cases applies the stem function concept and the simplifications based thereon.

The $p-2$ originates from the fact that the partial derivative is for the absolute function the derivative poses one -1 and the other -1 from the chain role for the potence of $p$. The $f$ originates from the inner derivative of the absolute function and the partial derivative of f $\partial_{\lambda,z}f$ is from the chain rule.

You may attach the integration domain back to the conditions and they remain still correct.

So the main step is
$$\partial_{\lambda,z} \int \rightarrow \int \partial_{\lambda,z}$$
is the identity of equality under the condition
$$F(\lambda,z)=\int_{\Bbb R^{n}} \vert(f(x,\lambda,z))\vert^{p}dx$$
This define a space of functions and is famous mathematics definition and condition set. An example for text to dive deeper is Locally integrable function where this is generalization with the name absolute p-integrable function space on the $\Bbb R^{n}$.

Would not the complicated integral there the following equation holds by simple rules for partial derivative of functions:
$$\partial_{\lambda,z}\vert(f(x,\lambda,z))\vert^{p}=p\vert(f(x,\lambda,z))\vert^{p-2}f(x,\lambda,z)\partial_{\lambda,z}f(x,\lambda,z)$$
As explain above without the use of a formulare.

The last problem part is $1<p<2$. In this open interval the first factor has negative potence. This might be cause to some extra discussion whether the conditions hold depending on the inner structure of $f$. As You imagine the convergence of $f$ depending on $x$ has to be faster than a limiting border function. The restriction to the $\Bbb R^{n}$ limits what might happen further.

The main step holds in general under the condition and is valid pointwise locally and for $\Bbb R^{n}$ it is valid in $\Bbb R^{n}$ for $(f,p)$. There are som examples discussed in the linked page and the references therein.
$$\partial_{\lambda,z} \int = \int \partial_{\lambda,z}$$
A set of important functions is Lp_space $L^p$ and this is valid on the Banach space. This is a better starting point L-infinity and Sobolev_space with the scalar product as inner product.

The main knowledge is the condition is stronger than the named first-order conditions to take away from this for the restriction $1<p<\infty$.


问:
Question abour linear maps proofs
As example, consider the following statement: Let $V$ and $W$ be finite-dimensional with $\dim V\ge\dim W\ge 2$. Show that $A=\{T\in\mathcal{L}(V,W):T\ \text{not surjective}\}$ is not a subspace of $\mathcal{L}(V,W)$.

To prove the statement, we can pass by considering a basis of $V$ and $W$ and construct linear applications "sending" the basis of $V$ to $W$. My question is the following:

Why do we pass by it? Why we can't construct linear maps considering only elements of each vector space (to prove this type of statements)? My question might be vague, but I don't really understand why considering basis's it proves, let's say, the "general" case? Of course, knowing the basis of vector space we can construct each element of that vector space. But I am quite confused about the fact that we "send" a basis to another one... If someone could give a clear explanation of this, I would really appreciate it. Thank you in advance!

答:
WLOG, take $V = k^n$ and $W = k^m$ where $k$ is the underlying field, with basis vectors $e_i$ as usual.

Consider the linear maps $f_i : V \to W$ defined by $f_i(e_i) = e_i$ and $f_i(e_j) = 0$ for all $j \neq i$. These maps are not surjective because the image of $f_i$ is the 1-dimensional vector space spanned by $e_i$, and $\dim(W) \geq 2 > 1$, so the image of $f_i$ cannot be all of $W$.

Note: the maps $f_i$ are only well-defined because $n \geq m$. For suppose that $n < m$. Then one defining equation for $f_m$ is $f_m(e_m) = e_m$. But $e_m$ is not a basis vector of $k^n$, since $n < m$. I was advised to make my use of the assumption $\dim(V) \geq \dim(W)$ explicit, so I have done so.

But now consider $\sum\limits_{i = 1}^m f_i$. This map is indeed surjective.

Thus, the set of non-surjective maps is not closed under addition and is not a subspace of $\mathcal{L}(V, W)$.

Your question is why we can assume, WLOG, that $V = k^n$ and $W = k^m$. This is because every $j$-dimensional vector space is isomorphic to $k^j$.

Why do we specify a basis? Because it makes the math easier. Once we pick the basis, it's easy to define linear maps in terms of the basis. Once we can specify maps in terms of the basis, it's easy to find non-surjective linear maps whose combination is surjective.

How would you know from looking at the problem that you should use a basis? The fact that the problem specifies "finite-dimensional" is a dead giveaway. "Finite-dimensional" means "has a finite basis". We can't use the fact that the vector spaces are finite-dimensional unless we somehow make use of their finite bases.

Title: Question abour linear maps proofs


问:
Solving $y'' - 2y' + y = \delta(t-2)$ for y(0) = 0, y'(0) = 0 by using Laplace Transforms. Need help finishing the problem.
So I'm working on this question and I've taken the Laplcace transforms for everything and separated it as $Y(s) = F(s)G(s)$, and then I've put it into the convolution equation to get $y(t)$. My problem is, I don't understand how to fully/properly integrate this when I have the $\delta(T-2)$ inside the integral. Can anyone explain what to do? I'm adding a picture of what I have until now that I am sure is correct so that there's somewhat of an explanation of my train of thought? (the u and v' at the bottom were if I was going to solve with integration by parts, which I also am not sure of how to use here).
But yeah, please help and tell me where I'm going wrong...

答:
There is nothing wrong with what you did.
\begin{align*}
y(t) & = \int_0^t \delta(T-2) (t-T)e^{t-T} \ dT \\
& = \begin{cases}
0 & \text{if } t < 2 \\
(t-2)e^{t-2} & \text{if } t \ge 2,
\end{cases}
\end{align*}
where we made use of the sifting property of the delta function
$$ \int_a^b \delta(x-x_0) f(x) \ dx = \begin{cases}
f(x_0) & \text{if } a < x_0 < b \\
0 & \text{otherwise}.
\end{cases}$$


问:
Amicable Pairs (How many have been found so far?)
My question is simply, how many amicable pairs have been found so far using super computers?

I have been trying to find any kind of answer online that is up to date, but after searching in depth I can't find any academic papers (or google results for that matter) that date past 2009.

Have people given up on finding these numbers because they have minimal use in mathematics?

答:
There are 1.2 billion amicable pairs known and listed at sech.me/ap/
The host of the website encourages anyone to join the search for new pairs
as volunteers in the BOINC project (much like the GIMPS project).

Volunteers are finding new pairs daily.

ContentLicense: CC BY-SA 3.0


问:
How to use MVT to prove the following problem
Let $f(x)$ be continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b),|f'(x)|\leqslant1$. prove that for any $x_1,x_2\in[a,b]$ establish$$|f(x_1)-f(x_2)|\leqslant\frac{b-a}{2}.$$
I want to use MVT.But I don't know how did the $\frac{1}{2}$.

答:
Same solution probably written differently and in a more intricate way but it came to me like that before reading MartinR solution:

Suppose wlog $x<y$ and per absurdum that $|f(y)-f(x)|>(b-a)/2$. In such a case we would have:

$|f(y)-f(x)|>(b-a)/2 \rightarrow y-x>(b-a)/2 \rightarrow (x-a)+(b-y)<(b-a)/2$ [1].

Now this is an identity:

$f(y)-f(x)=-[(f(x)-f(a))+(f(b)-f(y))] [2]$

Looking at the right member of [2] and applying MVT +  [1]:

$|[(f(x)-f(a))+(f(b)-f(y))]|<|(f(x)-f(a))|+|(f(b)-f(y))|<(x-a)+(b-y)<(b-a)/2$

But instead from the absurd hypothesis the left member of [2]:

$|f(y)-f(x)|>(b-a)/2$

Which is than a contradiction.

It is the same solution as MartinR, only with the argument presented in slightly a different way ...


问:
Is my this step correct?
Can you please tell me whether i did correct or wrong in the following equation.

$\text{P}\left[(\sum_{k=1}^{M}V_k)<x\right]$

=
$\left[\sum_{k=1}^{M}\text{P}(V_k<x)\right]$-------(1)

where $\text{P}$ is probability, $V_k$ is random variables.

Is my step in eq.(1) correct.

答: This is clearly wrong. Suppose that $M=2$, $x=1$, and each $V_k$ is $0$ or $1$, each with probability $\frac12$. Then the lefthand side is $\frac14$, and the righthand side is $\frac12+\frac12=1$.

Title: Is my this step correct?


问:
Checking monotonicity of a sequence with derivative of its function
I have a sequence $a_n = \frac {3+n^2}{1-n}$

I assume this sequence is contained in the function $f$ such that $a_n=f(n)$

Now I take the derivative of $f(n)$, which gives me $\frac{-(n+1)(n-3)}{(1-n)^2}$

I find that $(1-n)^2$ is positive for all $n>1$

I also find that, for $n>3$, the numerator will always be negative.

So $\frac{negative}{positive} = negative$

Since $f'(n)$ is negative for all $n>3$, it means the function $f(n)$ is eventually decreasing.

Which should also imply that the sequence $a_n=\frac {3+n^2}{1-n}$ is eventually decreasing, meaning it is a monotonic sequence.

However, the textbook answer says this sequence is not monotone.

What am I doing wrong?

答: Looks like, based on your analysis, that the sequence is monotonic for $n\ge4: a_4, a_5, a_6, \cdots$, but the sequence $a_2,a_3,a_4 \cdots$ is not monotonic.  The term $a_1$ is either infinite or undefined.


问:
Calculate limit of $\Gamma$ function for special values.
I would like to calculate the limit without any software but have no idea how to do it.

$$f(n) = \lim_{c \rightarrow 0} \frac{\Gamma(-n + c) + \Gamma(-n - c))}{2}$$
$$n = 0, 1, 2, ...$$

Wolfram in some way claculates it, for example:
$$f(0) = - \gamma$$
$$f(1) = \gamma - 1$$
$$f(2) = \frac{3 - 2 \gamma}{4}$$
$$f(3) = \frac{6 \gamma - 11}{36}$$
$$(...)$$

It seems that the solutions will be somethink like that:
$$f(n) = \frac{(a - \gamma)(-1)^{n}}{b}$$

答:
Using series expansions for the gamma function we have
$$
\begin{aligned}
f(n)
&=\frac{1}{2}\lim_{\epsilon\to 0}(\Gamma(-n+\epsilon)+\Gamma(-n-\epsilon))\\
&=\frac{(-1)^n}{2 n!}\lim_{\epsilon\to 0}(\tfrac{1}{\epsilon}(1+\psi(n+1)\epsilon)-\tfrac{1}{\epsilon}(1-\psi(n+1)\epsilon)+\mathcal O(\epsilon))\\
&=\frac{(-1)^n}{2 n!}\lim_{\epsilon\to 0}(\tfrac{1}{\epsilon}+\psi(n+1)-\tfrac{1}{\epsilon}+\psi(n+1)+\mathcal O(\epsilon))\\
&=\frac{(-1)^n}{n!}\lim_{\epsilon\to 0}(\psi(n+1)+\mathcal O(\epsilon))\\
&=\frac{(-1)^n}{n!}\psi(n+1),
\end{aligned}
$$
where $\psi(z)$ is the Digamma function. Then using specialized values for $\psi(n+1)$ we find
$$
\begin{aligned}
f(n)
&=\frac{(-1)^n}{n!}(H_n-\gamma),
\end{aligned}
$$
where $H_n$ is the harmonic number and $\gamma=0.57721\dots$ is the Euler-Mascheroni constant.

We have for the first several values of $n$:
$$
\left(
\begin{array}{cc}
 n & f(n)\\
 0 & -\gamma  \\
 1 & \gamma-1 \\
 2 & -\frac{1}{2} \left(\gamma-\frac{3}{2} \right) \\
 3 & \frac{1}{6} \left(\gamma -\frac{11}{6}\right) \\
 4 & -\frac{1}{24} \left(\gamma-\frac{25}{12} \right) \\
 5 & \frac{1}{120} \left(\gamma -\frac{137}{60}\right) \\
\end{array}
\right)
$$


答:
$a_2 = -7$.

$a_3 = \frac{12}{-2}=-6$

$a_4=\frac{19}{-3}$

Hence $a_2 < a_3$ but $a_3 > a_4$, hence it is not monotonoe. Monotonic function is monotone over all the indices, it's not about becoming monotonic eventually.

Title: Checking monotonicity of a sequence with derivative of its function

Body:
I have a sequence $a_n = \frac {3+n^2}{1-n}$

I assume this sequence is contained in the function $f$ such that $a_n=f(n)$

Now I take the derivative of $f(n)$, which gives me $\frac{-(n+1)(n-3)}{(1-n)^2}$

I find that $(1-n)^2$ is positive for all $n>1$

I also find that, for $n>3$, the numerator will always be negative.

So $\frac{negative}{positive} = negative$

Since $f'(n)$ is negative for all $n>3$, it means the function $f(n)$ is eventually decreasing.

Which should also imply that the sequence $a_n=\frac {3+n^2}{1-n}$ is eventually decreasing, meaning it is a monotonic sequence.

However, the textbook answer says this sequence is not monotone.

What am I doing wrong?


答: You are fully right, the sequence is decreasing for $n \geq 3$. However $a_2=-7$ and $a_3=-6$, it means it is not decreasing for all $n \geq 1$.


问:
Laurent Series and annular regions
I'm a bit stuck on this question:

Identify the annular regions  centered on $z = 0$ in which the function :

$$f(z) = \frac{1}{z(z+1)}$$

is analytic.  Find the Laurent series about $z=0$ which is valid for each of the regions.

I've found the Taylor series for $\displaystyle \frac{1}{z+1}$ which gives me a principle part of $\displaystyle \frac{1}{z}$, so I assume the region will then be $0 < |z| < 1$.  I'm not sure how to progress from there. Do I then find the Taylor series for $\displaystyle \frac{1}{z}$?

Thanks in advance.

答:
Dear Bow, you're not asked to find "Taylor series" for $1/z$ or for $1/z(z+1)$. You are asked to find the Laurent series - which are allowed to have negative powers of $z$ and which is just $1/z$ added to each term of the Taylor expansion of $1/(z+1)$. You will find out that the Taylor expansion around $z=0$ for $1/(z+1)$ converges only somewhere - you know where, right? - and the similar Laurent expansion will converge in the same region - you know why?

Then, this region will not cover the whole plane. So you ambitiously try to find a similar Laurent expansion for the whole rest of the plane, assuming that there is one Laurent expansion that works everywhere in that region. How do you do that? Well, substitute $z=1/z'$. So any Laurent expansion in $z$ may be converted to a simple Laurent expansion in terms of $1/z'$. Do you see why?

Well, $f(z') = 1/[(1/z')(1+1/z')] = \frac{z^{\prime 2 }}{(z'+1)}$. Now, try to expand this around $z'=0$ which is equivalent to $z=\infty$, isn't it? This will only converge for small enough values of $z'$, right? So it will only converge for large enough absolute values of $z$, right?

I don't want to write the whole solution verbatim - even though I almost did it - because your question seems as a homework and it would be both cheating and it would teach you nothing if you just copied it. But in this particular case, the annuli are either disks or the rest of the plane - so there is no "proper" annulus with two circular boundaries.

Title: Laurent Series and annular regions


答:
Using results from gamma function with negative argument

$$\Gamma (-n + c) = \frac{ a_n}{c} + b_n + O(c)$$

with
$$b_n~=~-\dfrac{(-1)^n}{n!}\cdot\gamma~-~\dfrac{S_1(n+1,~2)}{n!^2}$$

thus your limit is
$$
f(n)=\lim_{c\rightarrow 0}\frac{\Gamma (-n + c) + \Gamma (-n - c)}{2}= b_n
$$

For $n=0, 2, \dots, 5$:

$$
\begin{array}{l}
 f(0) =  -\gamma  \\
f(1) = -1+\gamma  \\
f(2) = \frac{3}{4}-\frac{\gamma }{2} \\
f(3) = -\frac{11}{36}+\frac{\gamma }{6} \\
f(4) = \frac{25}{288}-\frac{\gamma }{24} \\
f(5) = -\frac{137}{7200}+\frac{\gamma }{120} \\
\end{array}
$$

Title: Calculate limit of $\Gamma$ function for special values.

Body:
I would like to calculate the limit without any software but have no idea how to do it.

$$f(n) = \lim_{c \rightarrow 0} \frac{\Gamma(-n + c) + \Gamma(-n - c))}{2}$$
$$n = 0, 1, 2, ...$$

Wolfram in some way claculates it, for example:
$$f(0) = - \gamma$$
$$f(1) = \gamma - 1$$
$$f(2) = \frac{3 - 2 \gamma}{4}$$
$$f(3) = \frac{6 \gamma - 11}{36}$$
$$(...)$$

It seems that the solutions will be somethink like that:
$$f(n) = \frac{(a - \gamma)(-1)^{n}}{b}$$


问:
Simplify the expression $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$
Simplify the expression $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$.

Once I put it into a calculator, it gives me that $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1$, something which afterwards is easily proved. My question is, without a calculator, how can you work out that the expression is equal to $1$?

答:
Let $\phi=\frac{1+\sqrt{5}}2$ the golden ratio, it verifies the equation $\begin{cases}\phi^2=\phi+1\\\phi^{-1}=\phi-1\end{cases}$

In particular $\phi^3=2\phi+1=\sqrt{5}+2$ and similarly $\sqrt{5}-2=\phi^{-3}$

Therefore $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=\sqrt[3]{\phi^3}-\sqrt[3]{\phi^{-3}}=\phi-\phi^{-1}=1$


问:
Are all integrals of the from “polynomial1 divided by polynomial 2” solvable?
I'm asking whether all integrals of the form "polynomial 1 of degree n" divided by "polynomial 2 of degree m" are solvable for all values of n and m and for all coefficients of polynomial 1 and 2. If not, is there a formula that tells you which integral is unsolvable and which is solvable?</p>

答:
$$ \int \frac{p_{1}()}{p_{2}()} $$
This all depends on the characteristics of $p_{1}, p_{2}$ and their relationship, degree of $p_{1}$ is $n$, and degree of $p_{2}$ is $m$

If $n \gt m $, use $ P = D \dot Q + R$ to re-express and integrate to get the new function, which would be a polynomial of degree $n-m+1$, the new polynomial $n-m+1$ is unique, can be solvable or not

But if the Reminder on division is not a number, but another polynomial, the process here is different involves partial fraction and the results is not always polynomial but some logarithmic function

If $n \gt m $, result is not a polynomial, let's not speak of solving


答:
Use the formula $(a-b)^3 = a^3 - b^3 - 3ab(a-b)$. Let $a = \sqrt[3]{\sqrt{5} + 2}$ and $b = \sqrt[3]{\sqrt{5} - 2}$. Plug in the values of $a$, $b$. We can get (some calaulation steps are omitted)

$$(\sqrt[3]{\sqrt{5} + 2} - \sqrt[3]{\sqrt{5} - 2}) = 4 - 3 \cdot(\sqrt[3]{\sqrt{5} + 2} - \sqrt[3]{\sqrt{5} - 2})$$

which means

$$(a-b)^3 = 4 - 3 \cdot (a-b)$$

Then let $x = a-b$. The rest it to find the root of polynomial $x^3 = 4 - 3x$, which has only one real root.

Body:
Simplify the expression $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}$.

Once I put it into a calculator, it gives me that $\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1$, something which afterwards is easily proved. My question is, without a calculator, how can you work out that the expression is equal to $1$?


问:
What is the maximum possible number of solutions of the equation $ f (z) = az + b$, as $a $ and $b$ vary over complex numbers?
Suppose $f : \mathbb{C} \to\mathbb{ C} $ is a holomorphic function such that the real part of $f''(z)$  is strictly positive for all $ z \in \mathbb{C}$.
What is the maximum possible number of solutions of the equation $ f (z) = az + b$, as $a $ and $b$ vary over
complex numbers?

I have tried a lot but still confused.

答:
The part you may be missing is that if $f$ is entire then $f''$ is entire as well. You are told that it has positive real part, but the only entire functions with positive real part are constants.

So $f''(z)=c$ for all $z$, for some fixed $c \in \mathbb{C}$. Therefore $f'(z)$ is... and therefore $f(z)$ itself is...

EDIT: by popular demand, let's proceed. Since $f''$ is constant, $f'$ is linear and $f(z)=Az^2+Bz+C$ is a polynomial of degree $2$ ($A$ must be positive, so the degree is not less than that). The equation $f(z)=az+b$ therefore has exactly two solutions, counted with multiplicity. You can also say that it has at most two solutions.

Indeed, you can take $f(z)=z^2$ (which has $f''(z)=2$, satisfying the requirement of positive real part), and $a=0$, $b=1$. The equation is now $z^2=1$ which has exactly two distinct solutions.


问:
Variance of Hypergeometric
In Introduction to Probability by Blitzstein and Hwang, #48 of Chapter 4 asks the reader to first find $E \binom{X}{2}$ "by thinking" instead of using any "complicated calculations" and then says to use that value to compute $Var(X)$.  Over here $X$ ~ $HG(w, b, n)$.

I found $E \binom{X}{2} = \frac{\binom{w}{2}}{\binom{N}{2}} \cdot \binom{n}{2}$ by "thinking," but that did not yield the variance that they provided as a solution, which is in particular $Var(X) = \frac{N-n}{N-1} npq$ where $N = w + b, p = \frac{w}{N}$ and $q = 1-p$. I found by noting that the probability any two selected items have tag w (I could rigorize this with indicator variables, but since it is very uniform, I'm hoping this is clear enough).

I can't figure out whether my solution is wrong because I computed $E \binom{X}{2}$ wrong or if I made some mistake in my (incredibly painful) computation. As such, I was wondering if someone could confirm that my computation for $E \binom{X}{2}$ is correct.

答:
Your computation of $\mathbb E[\binom {\mathbf X}2]$ is correct. From it, we get $\mathbb E[\mathbf X(\mathbf X-1)] = \frac{w(w-1) \cdot n(n-1)}{N(N-1)}$. To get $\text{Var}[\mathbf X] = \mathbb E[\mathbf X^2] - \mathbb E[\mathbf X]^2$, we should add $\mathbb E[\mathbf X] = \frac{ w n}{N}$ and subtract $\mathbb E[\mathbf X]^2 = \frac{w^2 n^2}{N^2}$.

To simplify the algebra, note that all three of these terms have a factor of $\frac{wn}{N} = np$ in them, which we also want in the final answer. This leaves us with
$$
   \text{Var}[\mathbf X] = \frac{wn}{N} \left(\frac{(w-1)(n-1)}{N-1} + 1 - \frac{w n}{N}\right).
$$
Combining all three terms inside the parentheses is the painful part, but it should give $\frac{N^2 - Nn - Nw + nw}{N(N-1)}$ after some terms cancel, which factors as $\frac{(N-n)(N-w)}{N(N-1)}$.

Now, we get
$$
   \text{Var}[\mathbf X] = \frac{N-n}{N-1} \cdot n \cdot \frac wN \cdot \frac{N-w}{n} = \frac{N-n}{N-1} npq.
$$

Title: Variance of Hypergeometric


问:
Solution to operator equation, surjectivity
Suppose $T:V\to W$, where $V,W$ are banach spaces and $Tf = k$ (for instance $T$ might be an integral operator). They say that the equation has solution when $T$ is injective and so $T^{-1}$ exists. Why the surjectivity is neglected in the litterature? Is it related to the properties of infinite-dimensional spaces?

In the finite dimensional case of a function $f:R\to R$, the bijectivity is required for the existence of $f^{-1}$.

答: If $T$ is injective, the solution may not exist for all $k\in W$. However, if the solution exists, it is unique. As a result, injectivity of $T$ assures the uniqueness (but not existence), whereas the surjectivity of $T$ assures the existence: for all $k\in W$ the solution $Tf = k$ exists (but may not be unique).</p>


问:
Heron's Formula
Heron's Formula gives the area of a triangle when the length of all three sides are known. There is no need to calculate angles or other distances in the triangle first to determine its area. The formula is given by

$Area=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{c}$, $a, b, c$ are sides of the triangle and $p$ is the perimeter of the triangle. The following is my concern:

If one of the sides of the triangle is greater than $p$, then the Area will not be a real number (which shouldn't be true).

Example.  Let the sides of a triangle be 175 metre, 88 metre and 84 metre, then $p=173.5$. Therefore,
$Area=\sqrt{-1991498.0625}$, which not a real value.

Therefore, the following is my question:

Why shouldn't the area be expressed as $Area=\sqrt{|p(p-a)(p-b)(p-c)|}$?

答: In addition to the correct answers' "that's not a triangle!", the imaginary values are still suggestive.  Kendig wrote an American Mathematical Monthly article titled "Is a 2000-Year-Old Formula Still Keeping Some Secrets?" exploring ways to make sense of those values.  (Currently) you can read the entire article here.

Title: Heron's Formula


问:
For a transitive action of a path-connected group, does every path lift?
Let $G$ be a Hausdorff, path-connected group acting transitively on a Hausdorff space $X$. Assume the action is continous (i.e. $(g,x) \mapsto g \cdot x$ is continuous) and transitive. It follows easily that $X$ is also path connected.

Given any path $x_t$ in $X$, must there exist a path $g_t$ in $G$ such that $x_t = g_t \cdot x_0$, all $t \in [0,1]$.

I'm inclined to think the answer is "no". In this case, I'd also be very interested to hear of additional hypotheses which make the answer into "yes". Thanks.

答:
If $G$ is a Lie group and the stabilizer is a closed subgroup then the map $G \rightarrow X$ sending $g \mapsto gx_0$ is a principal $H$-bundle and so satisfies path lifting.

More generally, it is a theorem of Steenrod that $G \rightarrow X$ is a fiber bundle (hence, in particular, satisfies path lifting) if there exists a local section of this map around $x_0$ and the stabilizer is a closed subgroup.

I can't seem to find literature on whether or not $G \rightarrow X$ might be a fibration without actually being a fiber bundle... so I don't know. I also can't find a counterexample for when $G \rightarrow G/H$ fails to be a fibration when $H$ is closed (the trouble seems to be coming up with groups $G$ that are far from being Lie groups.)


问:
Lagrange multipliers to optimize sum of cubes of roots of a quadratic

If $\lambda \geq -2$ such that the sum of cubes of roots of the equation $x^2 + (2-\lambda )x + (10-\lambda) = 0$ is minimum, then the magnitude of difference of root is..?

Let roots be $\alpha $ and $\beta$, we wish to maximize $f(\alpha,\beta) = \alpha^3 + \beta^3$ under the constraint of $ \alpha + \beta = 2- \lambda $ and $\alpha \beta= 10 -\lambda$ , combining the constraints we get:

$$ \alpha + \beta + \alpha \beta = 8$$

Now apply lagrange multiplier equation, we get:

$$ 3 \alpha^2 = \mu ( 1 + \beta)$$

And,

$$ 3 \beta^2 = \mu ( 1 + \alpha)$$

Taking the ratio of the two equations:

$$ \frac{\alpha^2 }{\beta^2} = \frac{1+\beta}{1 + \alpha} \tag{1}$$

Substracting one on both sides:

$$ \frac{\alpha + \beta}{\beta^2} =  \frac{-1}{1+\alpha} \tag{2}$$

This leads to:

$$ \alpha + \beta + \alpha^2 + \alpha \beta  = - \beta^2$$

Hence,

$$ -12 = \alpha^2 + \beta^2$$

Now.. I am not sure what to do. One way is to ditch the lagrange multiplier approach and directly express the objective function in terms of $\lambda$ and minimize but is there a way to proceed further from the information we have?

答:
This is a solution without Lagrange Multiplier method. See my edit in the end for Lagrange Multiplier method.

$\alpha + \beta = \lambda-2, \alpha\beta = 10 - \lambda$

$f(\alpha, \beta) = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3 \alpha \beta(\alpha + \beta)$

So, $f(\alpha, \beta) = (\lambda-2)^3 - 3 (10-\lambda)(\lambda-2) = \lambda^3 - 3 \lambda^2 - 24 \lambda + 52$

Now we take derivative with respect to $\lambda$ and equate to $0$,

$\lambda^2 - 2\lambda-8 = 0 \implies \lambda = -2, 4$

Second derivative test confirms a local minima at $\lambda = 4$. We also observe that the function $f$ is monotonically decreasing for $\lambda \lt - 2$ and is monotonically increasing for $\lambda \gt 4$. So in the given domain $\lambda \geq - 2$, absolute minima occurs at $\lambda = 4$.

And so we have,
$\alpha + \beta = \lambda - 2 = 2, \alpha \beta = 10 - \lambda = 6$

Solving, $\alpha = 1 \pm i \sqrt 5, \beta = 1 \mp i \sqrt5$, which are complex roots of the quadratic for $\lambda = 4$.

For Lagrange Multiplier method,

Say $\alpha + \beta = x, \alpha \beta = y$

You found $\alpha^2 + \beta^2 = - 8$. We also have $y = 8 - x$

$x^2 = - 8 + 2y = - 8 + 16 - 2x \implies x^2 + 2x - 8 = 0 \ $
(where $x = \alpha + \beta$)

Solving $x = 2, y = 6; x = - 4, y = 12$.

We test $\alpha^3 + \beta^3 = x^3 - 3 xy$ for both and obtain $x = 2, y = 6$ to be the minima.

$(\alpha-\beta)^2 = x^2 - 4y = - 20 \implies \alpha-\beta = \pm \sqrt{-20}$


问:
Show that $T(O)$ is not open in $Y$
How can I show that $T(O)$ is not open in $Y$? where we have that $O = [{x \in X : ||x||_X < 1}]$.

$X:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$ and $||x||_X = \sum_{n=1}^{\infty}|x_n|$

$Y:[ {x = (x_1, x_2, x_3, . . .) : x_n ∈ \mathbb{K}, \sum_{n=1}^{\infty}|x_n|< \infty}]$ and $||x||_Y = sup_{n \in \mathbb{N}} |x_n|$

I was personally thinking of using the Open Mapping theorem. If $X$ and $Y$ are Banach spaces and $T : X \rightarrow Y$ is a surjective continuous linear operator, then T is an open map.

I have easily proven that $T$ is surjective and bounded.

I was thinking of proving that $0$  isn't an interior point by assuming there is a $Y$ ball around $0$ contained in $O$ and trying to find a contradiction by finding an element in the ball that's not in $T(O)$,by using a sequence of the form m(0,0,...,1,0,0,...) but this leads me nowhere... Also why does this not contradict the Open Mapping theorem?

Any help would be grateful. Thanks in advance.

答:
It's clear that $O$ is open in $X$. So if the open mapping theorem applied, then $T(O)$ would have to be open in $Y$. Since the exercise asks you to prove that it's not, clearly the open mapping theorem doesn't apply. Since $T$ is trivially surjective, one of the spaces involved is not Banach. (One of them is carrying the wrong norm, but you should show this in detail.)

To show that $T(O)$ is not open in the space $Y$, you need to show that for no $\epsilon > 0$ we have $B(0,\epsilon) \subset T(O)$ (i.e. that 0 is not an interior point). Note that $T(O)$ is just $O$, but the points have a different norm. Can you find an element $y$ for which the $Y$-norm of $y$ is less than $\epsilon$ (so $y \in B(0,\epsilon)$), but for which the $X$-norm is greater than 1 (so $y \notin T(O)$)?

The element $m \cdot e_k$ you're suggesting would have the same norm in both spaces. You need to look for an element where the two norms disagree.

Title: Show that $T(O)$ is not open in $Y$


问:
Minimum number of subsets
For some integer $n$, consider $S=\{1, \dots, n\}$. Let $S_1, \dots, S_k$ be subsets that satisfy the property that each $x\in S$ is in exactly 3 of the $S_i$'s and so that the intersection of these three subsets is $x$. What is the minimum value of $k$ (the number of subsets)?

I thought that if each $x$ was in 2 of the sets, $k$ would be at least $2*\sqrt{n}$, but was not sure how to proceed for the case of 3 sets.

答:
This is the Fano plane:

How might this figure relate to the question?

Title: Minimum number of subsets


问:
$W^1(μ,\frac{δ_{x_0}+δ_{x_1}}{2})=\frac{1}{2}(∫d(x_0,y)μ(dy)+∫d(x_1,y)μ(dy))$?
I was looking at this post to try to understand a bit better the Wasserstein distance. In the thread that $$W^1(μ,δ_{x_0})=∫d(x_0,y)μ(dy).$$ But I was wondering if this remains true for $$W^1(μ,\frac{δ_{x_0}+δ_{x_1}}{2})=\frac{1}{2}(∫d(x_0,y)μ(dy)+∫d(x_1,y)μ(dy)).$$ I know we have the $\le$-inequality because of the triangle inequality but I am not sure if we have the equality.

答: No, usually not. For example, if $\mu=\frac 1 2(\delta_{x_0}+\delta_{x_1})$, then the left side is of course $0$, while the right side is $\frac 1 2 d(x_0,x_1)$.


问:
Online tool for making graphs (vertices and edges)?
Anyone know of an online tool available for making graphs (as in graph theory - consisting of edges and vertices)? I have about 36 vertices and even more edges that I wish to draw.
(why do I have so many? It's for pathing in a game)

Only tool available to me right now is MS Paint and it would be very messy.

Edit:

I'm not actually looking for a 'standard' - just some way to neatly draw a set of vertices and edges, preferably without having to write code. I was planning to screenshot the pic and use it in a question on gamedev but I managed to solve the problem for now. Thanks for all the answers. I still prefer GeoGebra because it accessible online, neat, accurate, aligns elements to a grid, has consistent shapes, and has a high degree of customization available.

答:
If you need a really quick solution to draw graphs quickly for pedagogical examples you can use https://graphonline.ru/en although its not ideal.

But it works and you can just click the nodes where you need them and join them up.

ContentLicense: CC BY-SA 2.5


问:
Finding a point on a bezier surface using De Casteljau's algorithm
Given $16$ control points $(x,y,z)$ of a bicubic bezier patch, how do I use De Casteljau's algorithm to generate a point $(s,t) = (0.5, 0.2)$ on the surface?

As far as I understand, this kind of bezier patch can be thought of $4$ curves along the $s$ parameter. To find the $z$ I mentioned above I have to solve each of the $4$ $s$-curves for the value $s=0.5$. How do I do this? Using $4$ points means I have a $n=3$ bezier curve with a starting point and an ending point as well as two respective points for the tangents. Solving this I should have $4$ new points to make up a new curve which I can evaluate at $t=0.2$.

Where do I start and what do I need to do? How is the Bernstein Polynomial related to this?

You can find the points here: http://i.stack.imgur.com/FYE2u.png

Please excuse me my not-so-good english skills or wrong usage of mathematical terminologyas it's not my first language.

答:
There's a good explanation starting on this web page.

I recommend that you read the three sections on Bezier surfaces in "Unit 8".

The process you described in your question is roughly correct.

Let's call the 16 points:
$$P_{11}, P_{12}, P_{13}, P_{14}$$
$$P_{21}, P_{22}, P_{23}, P_{24}$$
$$P_{31}, P_{32}, P_{33}, P_{34}$$
$$P_{41}, P_{42}, P_{43}, P_{44}$$

And let $\phi_1$, $\phi_2$, $\phi_3$, $\phi_4$ be the Bernstein polynomials of degree 3.

Suppose we want to evaluate a point on the surface with parameter values $(s,t) = (0.5, 0.2)$.

The first thing we do is use the 16 points four at-a-time to construct four curves. We start with the points of the first row: $P_{11}$, $P_{12}$, $P_{13}$, $P_{14}$. We can use them to construct a Bezier curve
$$
C_1(s) = \phi_1(s)P_{11} + \phi_2(s)P_{12} + \phi_3(s)P_{13} + \phi_4(s)P_{14} =
\sum_{j=1}^4 \phi_j(s)P_{1,j}
$$
Then, on this curve, we can calculate the point $Q_1 = C_1(0.5)$ at value $s=0.5$.
$$
Q_1 = C_1(0.5) = \sum_{j=1}^4 \phi_j(0.5)P_{1,j}
$$
You can calculate $Q_1$ by the normal de Casteljau algrorithm, or any other method.

We do the same things with the other three rows of points. For each row, we construct a Bezier curve, and then we calculate the point at parameter value $s=0.5$. We get points
$$\text{second row:}\quad  Q_2 = C_2(0.5) = \sum_{j=1}^4 \phi_j(0.5)P_{2,j}$$
$$\text{third row:}\quad   Q_3 = C_3(0.5) = \sum_{j=1}^4 \phi_j(0.5)P_{3,j}$$
$$\text{fourth row:}\quad  Q_4 = C_4(0.5) = \sum_{j=1}^4 \phi_j(0.5)P_{4,j}$$
We can write all this more tidily as:
$$
Q_i = C_i(0.5) = \sum_{j=1}^4 \phi_j(0.5)P_{ij}  \quad (i=1,2,3,4)
$$

Now we just do the Bezier thing in the other direction -- we use the points $Q_1$, $Q_2$, $Q_3$, $Q_4$ to construct a Bezier curve Q(t) given by
$$
Q(t) = \phi_1(t)Q_1 + \phi_2(t)Q_2 + \phi_3(t)Q_3 + \phi_4(t)Q_4 =
\sum_{i=1}^4 \phi_i(t)Q_i
$$
Then our final surface point is the point at parameter value $t=0.2$ on the curve $Q(t)$:
$$
X(0.5, 0.2) = Q(0.2) = \sum_{i=1}^4 \phi_i(0.2)Q_i
$$
Again, you can calculate this point using the de Casteljau algorithm or any other method.

To understand how all this relates to the surface equation, let's substitute the expressions for the $Q_i$ into our last equation. We get
$$
X(0.5, 0.2)  = \sum_{i=1}^4 \phi_i(0.2)Q_i  =
\sum_{i=1}^4 \phi_i(0.2)\left(\sum_{j=1}^4 \phi_j(0.5)P_{ij}\right)
$$
Rearranging this gives:
$$
X(0.5, 0.2)  = \sum_{i=1}^4 \sum_{j=1}^4 \phi_j(0.5) \phi_i(0.2) P_{ij}
$$
You should recognize the right-hand side as the formula for a Bezier surface.

Here's a picture:

The blue curves are $C_1$, $C_2$, $C_3$, $C_4$, the red dots are $Q_1$, $Q_2$, $Q_3$, $Q_4$, and the red curve is $Q$.


问:
PDF of volume of tetrahedron with random coordinates

Question

What is the probability distribution function (PDF) of the absolute volume of a tetrahedron with random coordinates?

The 4 random tetrahedron vertices in $\mathbb{R}^3$ are
$$ \mathbf{\mathrm{X}_1} =(x_1^1,x_1^2,x_1^3),\;\;   \mathbf{\mathrm{X}_2}=(x_2^1,x_2^2,x_2^3),\;\;    \mathbf{\mathrm{X}_3}=(x_3^1,x_3^2,x_3^3),\;\;    \mathbf{\mathrm{X}_4} =(x_4^1,x_4^2,x_4^3)$$
where $x_i^j$ are independent standard normal distributed variables
$$x_i^j\sim\mathcal{N}(0,1)$$
The non-oriented volume of a random tetrahedron instance is
$$V=\frac{1}{6}\left| ( \mathbf{\mathrm{X}_1}-  \mathbf{\mathrm{X}_4})\cdot \left(( \mathbf{\mathrm{X}_2}-  \mathbf{\mathrm{X}_4} ) \times (  \mathbf{\mathrm{X}_3}-  \mathbf{\mathrm{X}_4} )\right) \right|  \tag{1} $$ $$=\frac{1}{6}\left|x_1^2 x_2^3 x_3^1 - x_1^1 x_2^3 x_3^2
+ x_1^3 x_2^1 x_3^2- x_1^2 x_2^1 x_3^3
+ x_1^1 x_2^2 x_3^3- x_1^3 x_2^2 x_3^1
+ x_1^3 x_2^2 x_4^1- x_1^1 x_2^2 x_4^3
+ x_1^1 x_2^3 x_4^2- x_1^2 x_2^3 x_4^1
+ x_1^2 x_2^1 x_4^3- x_1^3 x_2^1 x_4^2
+ x_2^3 x_3^2 x_4^1 -x_2^1 x_3^2 x_4^3
+ x_2^1 x_3^3 x_4^2- x_2^2 x_3^3 x_4^1
+ x_2^2 x_3^1 x_4^3- x_2^3 x_3^1 x_4^2
+ x_1^2 x_3^3 x_4^1- x_1^1 x_3^3 x_4^2
+ x_1^3 x_3^1 x_4^2- x_1^2 x_3^1 x_4^3
+ x_1^1 x_3^2 x_4^3- x_1^3 x_3^2 x_4^1
\right|$$

Known relations

The expectation value of $V$  is
$$\mathbb{E}[V]=\frac{2}{3}\sqrt{\frac{2}{\pi}}\tag{2}$$
A proof can be found in a Math Stack Exchange post.

The variance of $V$ is
$$\mathbb{Var}[V]=\mathbb{E}[V^2]-(\mathbb{E}[V])^2=\frac{2}{3}-\frac{8}{9\pi}\tag{3}$$
where $\mathbb{E}[V^2]$ can be calculated by multiple integration.

Approximate relations based on empirical data

The remaining part contains only unproven statements that could give indications of the true solution.

The probability distribution of empirical data of $V$ can be fitted quite well with a function of the form
$$f(V)=\text{exp}\left(-\left(\frac{V}{c_2}\right)^{c_1}\right)c_3\tag{4}$$
where $c_1,c_2,c_3$ are fit parameters. As a PDF must fulfill the conditions
$$\int_0^\infty f(V) \mathrm{d}V=1\ \,\, \text{and}\ \int_0^\infty V f(V)\mathrm{d}V=\mathbb{E}[V]$$
the fit parameters $c_2$ and $c_3$ in eq.(4) can be expressed in dependence of $c_1$
$$c_2=\mathbb{E}[V]\frac{\Gamma(1/c_1)}{\Gamma(2/c_1)}\ ,\;\; c_3=\frac{c_1}{c_2\Gamma(1/c_1)}$$
with $\Gamma$ being the Gamma function. Only $c_1$ remains to be fitted. The best fit is for $c_1\approx\pi/4$, i.e. $c_1\approx 0.7854, c_2\approx 0.3491, c_3\approx 2.4944$. However it is not known whether eq.(4) is the true form of the PDF at all. It just models well experimental data.

答:
Since the coordinates of the vertices are i.i.d. as standard normal variables, then
the coordinates of the difference of two vertices are i.i.d normal variables with
$0$ mean and variance $2$.

But the differences taken wrt the same vertex are no longer uncorrelated
and this fact complicates the treatment.

Herewith I am providing, for the moment, the PDF-CDF of the volume of a
random tetrahedron with a vertex at the origin, and the other three vertices
having independent standard normal coordinates.

This  Mathworld article indicates that the PDF of the product of three normal independent variables
can be expressed through a Meijer G-Function.   To find by this way the distribution of the sum
of the triplets appearing in the algebraic definition of the triple product is therefore unviable.

However the distribution of each vertex has the nice property of spherical
symmetry and we are going to exploit that via the trigonometric version of the triple product.

a) Polar distribution

Using Cartesian and Spherical (geographic convention) coordinates in parallel
$$
\left\{ \matrix{
  x = r\cos \phi \cos \theta  \hfill \cr
  y = r\cos \phi \sin \theta  \hfill \cr
  z = r\sin \phi  \hfill \cr}  \right.\quad \left| \matrix{
  \; - \pi /2 \le \phi  \le \pi /2 \hfill \cr
  \; - \pi  < \theta  \le \pi  \hfill \cr}  \right.\quad dx\,dy\,dz
\leftrightarrow r^{\,2} \cos \phi \,dr\,d\theta \,d\phi
$$
we can write the spatial distribution of a vertex as
$$
\eqalign{
  & p_p ({\bf V}) dV = \,{\cal N}_{\sigma ^{\,2} } \,(x){\cal N}_{\sigma ^{\,2} } \,(y){\cal N}_{\sigma ^{\,2} } \,(z)\,dx\,dy\,dz =   \cr
  &  = \left( {{1 \over {\sigma \sqrt {2\pi } }}} \right)^{\,3} e^{\, - \,{1 \over 2}\left( {{r \over \sigma }} \right)^{\,2} }
 r^{\,2} \cos \phi \,dr\,d\theta \,d\phi  =   \cr
  &  = \left( {{1 \over {\sqrt {2\pi } }}} \right)^{\,3} e^{\, - \,{1 \over 2}\left( {{r \over \sigma }} \right)^{\,2} }
 \left( {{r \over \sigma }} \right)^{\,2} \,d\left( {{r \over \sigma }} \right)\,\cos \phi \,d\theta \,d\phi =  \cr
  &  = \left( {{1 \over {\sigma \sqrt {2\pi } }}} \right)^{\,2} {\cal N}_{\sigma ^{\,2} } \,(r)\,dA_{\,r} \,dr
 = {1 \over {2\pi }}{\cal N}_1 \,(r/\sigma )\,dA_{\,r/\sigma } \,d\left( {{r \over \sigma }} \right)\, =   \cr
  &  = \left( {{1 \over {2^{3/2} \Gamma \left( {3/2} \right)}}} \right)e^{\, - \,{1 \over 2}
\left( {{r \over \sigma }} \right)^{\,2} } \left( {{r \over \sigma }} \right)^{\,2}
 \,d\left( {{r \over \sigma }} \right)\,\cos \phi \,d\theta \,d\phi  =   \cr
  &  = {1 \over {4\pi }}\,\chi _3 \,(r/\sigma )\,d\left( {{r \over \sigma }} \right)\,\,\cos \phi \,d\theta \,d\phi  =   \cr
  &  = {1 \over {2\pi }} \,\,((r/\sigma )^{\,2} )\,\left( {{r \over \sigma }} \right)d\left( {{r \over \sigma }} \right)
 \,\,\cos \phi \,d\theta \,d\phi  =   \cr
  &  = \,{1 \over {4\pi }}\chi _3 \,(r/\sigma )\,{{dA_{\,r/\sigma } } \over {(r/\sigma )^{\,2} }}\,d\left( {{r \over \sigma }} \right)
 = {1 \over {4\pi }}\,\chi _3 \,(r/\sigma )\,d\Omega \,d\left( {{r \over \sigma }} \right) \cr}
$$
where:

we generalize to the case of zero mean and generic variance $\sigma ^{\,2} $;

$dA$ is the surface area element;

$d\Omega$  the solid angle in steradians;

$\chi _3$ and $\chi _3^2$ are respectively the chi and chi-square distributions.

The radial distribution is instead
$$
\eqalign{
  & p_r (r)dr = \int_{\phi  =  - \pi /2}^{\pi /2} {\int_{\theta  =  - \pi }^\pi
  {\left( {{1 \over {\sigma \sqrt {2\pi } }}} \right)^{\,3} e^{\, - {1 \over 2}\left( {{r \over \sigma }} \right)^{\,2} }
 r^{\,2} \cos \phi \,dr\,d\theta \,d\phi } }  =   \cr
  &  = 4\pi r^{\,2} \left( {{1 \over {\sigma \sqrt {2\pi } }}} \right)^{\,3} e^{\, - {1 \over 2}
\left( {{r \over \sigma }} \right)^{\,2} } dr
 = 4\pi \left( {{r \over \sigma }} \right)^{\,2}
 \left( {{1 \over {\sqrt {2\pi } }}} \right)^{\,3} e^{\, - {1 \over 2}\left( {{r \over \sigma }} \right)^{\,2} }
 d\left( {{r \over \sigma }} \right) =   \cr
  &  = \left( {{{\left( {r/\sigma } \right)^{\,2} } \over {2^{\,1/2} \Gamma \left( {3/2} \right)}}} \right)
e^{\, - {1 \over 2}\left( {{r \over \sigma }} \right)^{\,2} } d\left( {{r \over \sigma }} \right)
 = 2\left( {{r \over \sigma }} \right)^{\,2} {\cal N}_1 \,(r/\sigma )\,\,d\left( {{r \over \sigma }} \right) =   \cr
  &  = \chi _3 \,(r/\sigma )\,d\left( {r/\sigma } \right)
 = 2\left( {r/\sigma } \right)\chi _3^2 \,\,((r/\sigma )^{\,2} )d\left( {r/\sigma } \right) \cr}
$$

b) Cross product

Passing to unitary variance for simplicity,
we can compute the distribution of the modulus ($c$) of the cross product of two vectors (two vertices) by fixing one vector $\bf v$ of modulus $v$ and
then the set of vectors having a component $c /v$ normal to the first, and thus lying over the cylindrical shell of radius
$c/v, (c+dc)/v$ around $\bf v$, and  integrate over $v$.
That is
$$
\eqalign{
  & p_c (c)dc
 = \int\limits_v {p_r (v)\,dv{\cal N}_1 \,(c/v){{dc} \over v}{\cal N}_1 \,(0)2\pi {c \over v}}  =   \cr
  &  = \int\limits_v {2v^{\,2} {\cal N}_1 \,(v)\,\,dv{\cal N}_1 \,(c/v){{dc} \over v}
{\cal N}_1 \,(0)2\pi {c \over v}}  =   \cr
  &  = {{4\pi } \over {\sqrt {2\pi } }}cdc\int\limits_v {{\cal N}_1 \,(v)\,\,{\cal N}_1 \,(c/v)} \;dv =   \cr
  &  = {2 \over {\sqrt {2\pi } }}cdc\int_{v = 0}^\infty
  {e^{\, - \,{1 \over 2}\left( {v^{\,2}  + c^{\,2} /v^{\,2} } \right)} \,dv}  \cr}
$$
and the corresponding CDF being
$$
\eqalign{
  & P_c (c) = {2 \over {\sqrt {2\pi } }}\int_{t\, = \,0}^c {\int_{v\, = \,0}^\infty
  {t\,e^{\, - \,{1 \over 2}\left( {v^{\,2}  + t^{\,2} /v^{\,2} } \right)} dt\,dv} }  =   \cr
  &  = {2 \over {\sqrt {2\pi } }}\int_{v\, = \,0}^\infty
  {e^{\, - \,{1 \over 2}\left( {v^{\,2} } \right)} dv
 \int_{t\, = \,0}^c {t\,e^{\, - \,{1 \over 2}\left( {t^{\,2} /v^{\,2} } \right)} dt\,} }  =   \cr
  &  = {1 \over {\sqrt {2\pi } }}\int_{v\, = \,0}^\infty
  {v^{\,2} e^{\, - \,{1 \over 2}\left( {v^{\,2} } \right)} dv
 \int_{t\, = \,0}^c {2\,\left( {{t \over v}} \right)\,e^{\, - \,
 {1 \over 2}\left( {t^{\,2} /v^{\,2} } \right)} d\left( {{t \over v}} \right)\,} }  =   \cr
  &  = {2 \over {\sqrt {2\pi } }}\int_{v\, = \,0}^\infty
  {v^{\,2} e^{\, - \,{1 \over 2}\left( {v^{\,2} } \right)} dv
 \int_{u/2\, = \,0}^{\left( {c^{\,2} /v^{\,2} } \right)/2}
 {\,e^{\, - \,{1 \over 2}u} d\left( {{u \over 2}} \right)\,} }  =   \cr
  &  = \sqrt {{2 \over \pi }} \int_{v\, = \,0}^\infty
  {v^{\,2} e^{\, - \,{1 \over 2}\left( {v^{\,2} } \right)} dv
 \left( {1 - e^{\, - {1 \over 2}\,\left( {c^{\,2} /v^{\,2} } \right)} } \right)}  =   \cr
  &  = \sqrt {{2 \over \pi }} \left( {\int_{v\, = \,0}^\infty
  {v^{\,2} e^{\, - \,{1 \over 2}\left( {v^{\,2} } \right)} dv}
  - \int_{v\, = \,0}^\infty  {v^{\,2}
 e^{\, - \,{1 \over 2}\left( {v^{\,2}  + c^{\,2} /v^{\,2} } \right)} dv} } \right)  \cr
  &  = 1 - \sqrt {{2 \over \pi }} \int_{v\, = \,0}^\infty
  {v^{\,2} e^{\, - \,{1 \over 2}\left( {v^{\,2}  + c^{\,2} /v^{\,2} } \right)} dv}  \cr}
$$

c) Dot product

For the modulus (absolute value) $q$ of the dot product of two vectors , we fix again a vector $\bf v$
and integrate over a plane normal to it at distance $q/v$, and double the result
to include the symmetric plane,
which means
$$
\eqalign{
  & p_d (q)dq = 2\int\limits_v {2v^{\,2} {\cal N}_1 \,(v)\,dv{\cal N}_1 \,(q/v){{dq} \over v}}  =   \cr
  &  = 4dq\int\limits_v {v\,{\cal N}_1 \,(v)\,\,{\cal N}_1 \,(q/v)dv}  =   \cr
  &  = {2 \over \pi }dq\int_{v\, = \,0}^\infty
  {v\,e^{\, - \,{1 \over 2}\left( {v^{\,2}  + q^{\,2} /v^{\,2} } \right)} dv}  \cr}
$$
and
$$
\eqalign{
  & P_d \left( q \right)
 = \int_{t\, = \,0}^q {{2 \over \pi }dt\int_{v\, = \,0}^\infty
  {v\,e^{\, - \,{1 \over 2}\left( {v^{\,2}  + t^{\,2} /v^{\,2} } \right)} dv} }  =   \cr
  &  = {2 \over \pi }\int_{v\, = \,0}^\infty
  {v\,e^{\, - \,{1 \over 2}\left( {v^{\,2} } \right)} dv
\int_{t\, = \,0}^q {e^{\, - \,{1 \over 2}\left( {t^{\,2} /v^{\,2} } \right)} dt} }  =   \cr
  &  = \sqrt {{2 \over \pi }} \int_{v\, = \,0}^\infty
  {v^{\,2} \,e^{\, - \,{1 \over 2}\left( {v^{\,2} } \right)} \,
{\rm erf}\left( {{q \over {\sqrt 2 v}}} \right)dv}  \cr}
$$

d) Triple product

Now it is easy to combine  the results above and obtain for
the volume $v$ of a parallelepiped , so leaving apart the factor of $1/6$
$$
\eqalign{
  & p_t (v)dv = 2\int\limits_c {p_c (c)dc{\cal N}_1 \,(v/c){{dv} \over c}}  =   \cr
  &  = 2\int_{c\, = \,0}^\infty  {{2 \over {\sqrt {2\pi } }}c\,dc{\cal N}_1 \,(v/c){{dv} \over c}
\int_{t = 0}^\infty  {e^{\, - \,{1 \over 2}\left( {t^{\,2}  + c^{\,2} /t^{\,2} } \right)} \,dt} }  =   \cr
  &  = {2 \over \pi }dv\int_{c\, = \,0}^\infty  {e^{\, - \,{1 \over 2}\left( {v^{\,2} /c^{\,2} } \right)} dc
\int_{t = 0}^\infty  {e^{\, - \,{1 \over 2}\left( {t^{\,2}  + c^{\,2} /t^{\,2} } \right)} \,dt} }  \cr}
$$
and the CDF
$$
\eqalign{
  & P_{\,t} (v) = {2 \over \pi }\int_{t = 0}^v {dt\int_{c = 0}^\infty
  {e^{\, - \,\,{1 \over 2}\left( {{{t^{\,2} } \over {c^{\,2} }}} \right)} dc\int_{u = 0}^\infty
  {e^{\, - \,{1 \over 2}\left( {u^{\,2}  + c^{\,2} /u^{\,2} } \right)} \,du} } }  =   \cr
  &  = {2 \over \pi }\int_{c = 0}^\infty
  {dc\int_{t = 0}^v {e^{\, - \,\,{1 \over 2}\left( {{{t^{\,2} } \over {c^{\,2} }}} \right)} dt
 \int_{u = 0}^\infty  {e^{\, - \,{1 \over 2}\left( {u^{\,2}  + c^{\,2} /u^{\,2} } \right)} \,du} } }  =   \cr
  &  = \sqrt {{2 \over \pi }} \int_{c = 0}^\infty  {\,\,c\;{\rm erf}\left( {{v \over {\sqrt 2 \,c}}} \right)dc
 \int_{u = 0}^\infty  {e^{\, - \,{1 \over 2}\left( {u^{\,2}  + c^{\,2} /u^{\,2} } \right)} \,du} }  \cr}
$$

which are definitely less complicated than expected, and allow in case to work out some approximations.

Note that all the  CDF formulas above have been checked by numerical simulation, and besides that
all of them correctly evaluate to 1 at $\infty$.
Since they are obtained by exact integration of the PDF's, also the latter should be correct.

-- update --

I just discovered  thanks to A0, that the inner integral above is discussed in many other
posts, e.g. in this, and that it is simply
$$
\int_{x = 0}^\infty  {e^{\, - \,{1 \over 2}\left( {x^{\,2}  + c^{\,2} /x^{\,2} } \right)} \,dx}
  = \sqrt {{\pi  \over 2}} \,e^{\, - \,\,\sqrt {c^{\,2} } }
$$
By that, for the triple product (volume of a parallelepiped) we get

$$
\eqalign{
  & p_t (v)\,dv = dv\sqrt {{2 \over \pi }} \int_{t\, = \,0}^\infty
  {e^{\, - \,{1 \over 2}\left( {v^{\,2} /t^{\,2}  + 2t} \right)} dt\,}   \cr
  & P_{\,t} (v) = \int_{t = 0}^\infty  {\,\,t\;e^{\, - \,\,t} \,
{\rm erf}\left( {{v \over {\sqrt 2 \,t}}} \right)dt\,}  \cr}
$$

e) General distribution

Concerning a Tetrahedron (parallelepiped) in a general position, i.e. the scheme considered above plus a translation
of the origin, the numerical simulation suggests that the $P_{\,t} (v)$ above converts to $P_{\,t} (2 v)$.

I am thriving to find a justification of that.


答:
I have updated this answer with a purely combinatorial description, since the original problem was combinatorial in flavor.

Choose $k$ so as small as possible so that $\binom{k}3\ge n$. Identify the numbers $\{1,\dots,\binom{k}3\}$ with the subsets of $\{1,\dots,k\}$ of size $3$. Then, for each $i\in\{1,\dots,k\}$, let $S_i$ be the set of numbers in $\{1,\dots,n\}$ whose corresponding subset contains $i$. Clearly, any $x\in\{1,\dots,n\}$, corresponding to the set $\{i,j,h\}$, is contained in exactly three subsets $S_i,S_j,S_h$, and the intersection of these subsets is $x$.

This is optimal for the reason lulu said in a comment; in order for such a collection of sets to exist, you you need $\binom{k}3\ge n$. Asymptotically, we get $k\sim \sqrt[3]{6n}$.

This generalizes to any multiplicity of intersections. In particular, $2\sqrt{n}$ for the two-way intersection problem you mentioned is not optimal, and you can obtain $\sim\sqrt{2n}$ by letting $\{1,\dots,\binom{k}2\}$ correspond to pairs of numbers in $\{1,\dots,k\}$.

Title: Minimum number of subsets

Body:
For some integer $n$, consider $S=\{1, \dots, n\}$. Let $S_1, \dots, S_k$ be subsets that satisfy the property that each $x\in S$ is in exactly 3 of the $S_i$'s and so that the intersection of these three subsets is $x$. What is the minimum value of $k$ (the number of subsets)?

I thought that if each $x$ was in 2 of the sets, $k$ would be at least $2*\sqrt{n}$, but was not sure how to proceed for the case of 3 sets.


问:
Expected Value of a person to find own seat as on ticket

Given 50 people who bought airplane tickets. There are seats listed on those tickets, but all passengers except the last one sit in a random seat. The last one decides to take the place on the ticket. How many people will change their seat? Find the expected value if the disturbed person has to find their own seat on the ticket

My approach to the above formulation is taking a random variable X which denotes whether the person finds the listed seat on the ticket which is 1 else 0. The last person chooses from remaining 49 seats, thus probability is $\frac{1}{49}$. Now this will displace the person sitting on the random seat so the probability he can choose his correct seat is $\frac{1}{48}$. Now this process can be repeated as it will displace another person and we get $\frac{1}{47}$ which will displace another person and we get $\frac{1}{46}$ and so on. Thus expected value is the summation of all of them = $\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+...$. This is done so as to find people changing their seat.

Can you please let me know if the approach to the above formulation is correct?

答: I do not know if I am correct but we can solve it intuitively like supposing two cases i.e. the last person sits on the remaining seat which conforms to the mentioned seat in the ticket or the last person sits on seats excluding the remaining seat. Thus probability of this event is 1/2 same can be done for 49 people as the last person sits on the correct seat and rest of the remaining people are disturbed so we get expected value close to 1/2*49=24.5</p>


问:
What is the thought process on defining a new function in Intermediate Value Problems?
I have no clue on where my lecturer is getting the function g from. I'm assuming it somehow comes from the absolute value condition:

Is there a way to solve this problem without defining a new function?

答:
The idea behind the proof is that we want to compare the values of $f$ at points that are a distance of $\frac 1 2$ apart. So let’s consider $f$ at pairs of points $(a,b)$ where $b=a+ \frac 1 2$and $a$ goes from $0$ to $\frac 1 2$.

If $f(0)=f(\frac 1 2)$ then we are done. So let’s suppose $f(\frac 1 2) > f(0)$ so that initially $f(b) = f(\frac 1 2) > f(0) = f(a)$. But when we slide our pair of points to the other end of the interval we have $f(b) = f(1) < f(\frac 1 2) = f(a)$, so $f(a)$ has “overtaken” $f(b)$.  We can argue from continuity that there must be some intermediate point where $f(b)=f(a)$. And if $f(\frac 1 2) < f(0)$ we can use the same argument starting at $a=\frac 1 2$ and sliding backwards to $a=0$.

The function $g$ is just introduced to formalise this argument.

You might like to consider why the property of continuity is important by trying to find a non-continuous function that does not have this property.


问:
If A is a positive definite matrix, how to show that the product of any two diagonal elements, $A_{ii} A_{jj}$, is greater than $|A_{ij}|^2$?
This seems like a basic linear algebra proof but I am stumped.

Let $A \in \mathbb{R}^{n \times n} $ be a positive definite matrix, defined as $u^TAu > 0$ for all $u \in \mathbb{R}^{n}\setminus\{0\}$.

How does one show:

$A_{ii} A_{jj} > |A_{ij}|^2$ when $i \ne j$

?

I have tried using different values of $u$ in the definition of positive definite but have not achieved that inequality. Eg I tried $u = [0, ...,  A_{jj}, ..., 1, ..., 0]^T $, where the $A_{jj}$ and $1$ are at the $i$th and $j$th positions, respectively, which yields

$u^TAu = A_{ii}A_{jj} + |A_{ij}|^2 + A_{ji}A_{ii} + A_{jj}A_{ij} > 0$

which has the terms I'm looking for, but does not appear useful for proving the inequality .

答: Let $u=xe^i+ye^j$ so $u^TAu=x^2[A_{ii}+2zA_{ij}+z^2A_{jj}]>0,\,z:=y/x$ for all $x,\,y\in\Bbb R\setminus\{0\}$. Therefore, the discriminant $(2A_{ij})-2A_{ii}A_{jj}<0$.


答:
Here is how you would go about coming up with this yourself. I will rewrite the problem statement for clarity since the problem was provided in a linked image.

Problem. If $f : [0,1] \to \mathbb{R}$ is a continuous function such that $f(0) = f(1)$. Show that there exist $p,q \in [0,1]$ with $|p-q| = \frac{1}{2}$ and $f(p) = f(q)$.

You know (presumably?) you want to apply the Intermediate Value Theorem, which guarantees the existence of a particular argument of the function with a given value. Here we want two values, $p,q$, but they are related via $|p-q|=\frac{1}{2}$, so that either $p - q = \frac{1}{2}$ or $q - p = \frac{1}{2}$. From the symmetry in the problem we can assume without loss of generality that it is the latter, so $q = p + \frac{1}{2}$. Thus, we may rewrite the conclusion as saying that we want a point $p$ such that both $p,p+\frac{1}{2} \in [0,1]$ and $f(p) = f(p+\frac{1}{2})$. This reduces it to the existence of a single point $p$.

The conditions $p,p+\frac{1}{2} \in [0,1]$ forces $p \in [0,\frac{1}{2}]$, and we can rewrite the equality as $f(p) - f(p+\frac{1}{2}) = 0$. Now the conclusion matches the form of the Intermediate Value Theorem, but the function in question is $g(x) = f(x) - f(x+\frac{1}{2})$, with $g : [0,\frac{1}{2}] \to \mathbb{R}$. Consequently, it is this $g$ we need to check is continuous and that $g(0),g(\frac{1}{2})$ have opposite signs, hence your lecturer's proof.

Title: What is the thought process on defining a new function in Intermediate Value Problems?

Body:
I have no clue on where my lecturer is getting the function g from. I'm assuming it somehow comes from the absolute value condition:

Is there a way to solve this problem without defining a new function?


问:
Simultaneously solenoidal and irrotational vector
Can we have a vector field $\vec{V}(\vec{r})$ such that $\nabla\times\vec{V}=\vec{0}$ and $\nabla\cdot\vec{V}=0$

答:
Cheap answer: sure just take a constant vector field so that all derivatives are zero.

A more interesting answer: a vector field in the plane which is both solenoidal and irrotational is basically the same thing as a holomorphic function in the complex plane. See here for more information on that.

So to get an example of an irrotational and solenoidal vector field in 3d space (which I guess is where you want to work judging from your cross product notation above), you can take any holomorphic function $f$ on $\mathbb{C}$ and form its "Polya vector field", basically $\overline f$ viewed as a vector field on $\mathbb{R}^2$. Then copy that vector field on every horizontal plane in 3d space.

Example: $f(z)=z^2$ is a holomorphic function. In terms of standard $(x,y)$ coordinates, this is $(x,y) \mapsto (x^2 -y^2 ,2xy)$. The associated Polya vector field is $(x,y) \mapsto (x^2-y^2,-2xy)$. Now make it a vector field on $\mathbb{R}^3$ by making it independent of $z$ and zeroing the third coordinate:
$$\vec V(x,y,z) = (x^2-y^2,-2xy,0).$$


问:
Contradiction of the Open Mapping Theorem?
I have been given the following: $T:X\to Y$, where

$X=Y=${$x=(x_1,x_2,…),∑_i|x_i|<∞$} , where $X$ and $Y$ are normed linear spaces (not Banach).

They have the following norms $\|x\|_X = \sum_i|x_i|$ and $\|x\|_Y = \sup_{n\in\mathbb{N}}|x_n|$.

Now I have already shown that T is bounded, surjective and that for the set $O = \{x\in X : \|x\|_X < 1\}$ , we have that $T(O)$ is not open in $Y$. Now for this last statement, I am trying to show that this doesn't contradict the Open Mapping Theorem. I am not sure how to show this? Any help is appreciated.

My definition of the open mapping theorem in this case is:

Let $X$ and $Y$ be Banach spaces and
let $T ∈ B(X, Y )$. If the mapping $T$ is surjective, then $T$ is an open mapping.

答:
It's clear that O is open in X. So if the open mapping theorem applied, then T(O) would have to be open in Y. Since the exercise asks you to prove that it's not, clearly the open mapping theorem doesn't apply. Since T is trivially surjective, one of the spaces involved is not Banach.

HINT: One of them is carrying the wrong norm. The space $Y$ is just $\ell^1$ with the $\ell^\infty$ norm. You should look for a sequence which converges in the $\ell^\infty$ norm, and whose limit is in $\ell^\infty$ but not in $\ell^1$. This gives you a Cauchy sequence without a limit.


答:
Let $\alpha\subseteq\{1,\ldots,n\}$ be an index set, and let $A[\alpha]$ be the submatrix of $A$ whose rows and columns are labelled by $\alpha$. Then $A[\alpha]$ is positive definite. Indeed, let $x\in \mathbb R^n$ with $x[\alpha]\neq 0$ and $x[\alpha^C]=0$. Thus,
$x[\alpha]^TA[\alpha]x[\alpha]=x^TAx>0$. Since $x$ is arbitrary, $A[\alpha]$ is positive definite.

In particular, if $\alpha=\{i,j\}$, where $i\neq j$ (say $i<j$), then $A[\alpha]=\begin{bmatrix} A_{ii}& A_{ij}\\ A_{ij}& A_{jj}\end{bmatrix}$ is positive definite, and so it has positive determinant, that is,
$A_{ii}A_{jj}-|A_{ij}|^2=\det(A)>0$. (This proof still works for complex positive definite matrices, and one also gets a corresponding inequality for positive semidefinite matrices.)

Title: If A is a positive definite matrix, how to show that the product of any two diagonal elements, $A_{ii} A_{jj}$, is greater than $|A_{ij}|^2$?

Body:
This seems like a basic linear algebra proof but I am stumped.

Let $A \in \mathbb{R}^{n \times n} $ be a positive definite matrix, defined as $u^TAu > 0$ for all $u \in \mathbb{R}^{n}\setminus\{0\}$.

How does one show:

$A_{ii} A_{jj} > |A_{ij}|^2$ when $i \ne j$

?

I have tried using different values of $u$ in the definition of positive definite but have not achieved that inequality. Eg I tried $u = [0, ...,  A_{jj}, ..., 1, ..., 0]^T $, where the $A_{jj}$ and $1$ are at the $i$th and $j$th positions, respectively, which yields

$u^TAu = A_{ii}A_{jj} + |A_{ij}|^2 + A_{ji}A_{ii} + A_{jj}A_{ij} > 0$

which has the terms I'm looking for, but does not appear useful for proving the inequality .


问:
Which practicable solutions exist to fair distribution of a finite resource over a finite time horizon?
Imagine we have $X$ amount of some resource available over a fixed, known, and finite time period $T$.

An unknown but finite number $N$ of users can each at arbitrary times request some share $Y = y/X$ of the resource $X$. We do not know in advance when future requests arrive and each user may request several times during $T$.

On one hand we wish to have spent all of $X$ by the end of $T$. On the other hand, we do not wish to run out of $X$ before the end of $T$.

How can we find a fair strategy for granting $Y$ to each user, i.e. determining the fraction $y$? I would like to adapt $y$ over the time period $T$ to adapt to observed demand. For example: if the demand turns out to be lower than expected early in the period, we will want to increase $y$ to increase the probability that we spend all of $X$. Or, if the demand turns out to be higher, we will want to decrease $y$ to decrease the probability that we run out of our resource way before the end of $T$.

I am interested in identifying in some sense optimal or close to optimal approaches for solving this problem in practice. I am convinced this is such a common problem that someone must have treated it many times over the years and I am guessing an answer may be some kind of dynamic programming?

答: The number of requests in a time seems a Poisson($\lambda$) distribution ($\lambda$ expected events in 1 unity of time), the optimal strategy would be $Y=\frac{X}{\lambda T}$. If you don't know the real parameter of the population, you could estimate it using the sample mean (which is the maximum likelihood estimation of the real parameter). This value will be getting closer to the real parameter over time, so you could use it to adapt your $y$.


问:
Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$
$\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

I guess it is convergent, so I apply comparsion test for this.

$\frac{log^2}{n^2} < \frac{n^2}{n^2} = 1$

So it is bounded by 1 and hence it is convergent.

Can I do in this way? It kind of make sense, but I have never deal with constant in this case

Usually we do bounding using another term depend on $n$?

答: No, you can't do it that way. You have shown that the terms are bounded by $1$, but that's also true of the series $1+1+1+\dots$; does that series converge?</p>


问:
How to represent a linear solution as an affine solution
I have gotten the following general solution for some system

$Ax=w$

Therefore our general solution is $x=\begin{pmatrix}3\\
1-\gamma\\
3+\gamma\\
\gamma
\end{pmatrix}=\begin{pmatrix}3\\
1\\
3\\
0
\end{pmatrix}+\gamma\begin{pmatrix}0\\
-1\\
1\\
1
\end{pmatrix}$

I understand that an affine space is a linear space + some sort of offset from the origin. Isn't this exactly what I already have? How would I find the affine map for this solution set?

答:
Your answer is correct. In general you can express an affine subspace $\Lambda$ as
$$\Lambda\equiv P_0+W=P_0+\text{dir}(\Lambda),$$
where $P_0$ is a point and $W$ is the vector space associated to your affine subspace $\Lambda$, so $W$ contains the linear combinations of the vectors that generate $\Lambda$ (and its "direction")  and it determines its dimension.

In your example, the affine subspace is a line in $\mathbb R^4$, since $\text{dim}(W)=\text{dim}(\Big\langle\begin{pmatrix} 0\\-1\\1\\1\end{pmatrix} \Big\rangle)=1$.


问:
irreducible characters of a group extension
Let $G$ be a group and $M$ is a minimal normal subgroup of $G$. Suppose that $M$ is an elementary abelian $2$-group and $M=C_G(M)$, so that $G/M$ acts on $M$. The action is faithful and assume that is also irreducible. Suppose that furthermore $G/M$ is a non abelian simple group (for my purposes, $G/M$ can be $A_5$ or $PSL_2(8)$, by the way).

I want to know some information about the irreducible complex characters of $G$, for example the degree of real characters.

My approach is check all the irreducible representations of $G/M$ over $GF(2)$ (that are known), build the semidirect product of $G/M$ by $M$ on GAP and print the character table. Unfortunately, this may not be the case since the group $M$ may not have a complement in $G$. It would be possible to calculate the second cohomology group $H^2(G/M,M)$ with GAP, check that is trivial (if it is true) and conclude that $M$ has a complement in $G$ (the extension splits, namely). There is a package in GAP, cohomolo, maintained by D. Holt, that does this kind of calculation, but I'm struggling to load it (I'm having some technical problems) and I wandering if there a more theoretical approach.

答:
Well, if you really only interested in $G/M = A_5$ or ${\rm PSL}(2,8)$, then I can tell you what you want to know.

$A_5$ has just two nontrivial  irreducible modules over ${\rm GF}(2)$, both of dimension $4$, and they both have trivial $2$-cohomology.

${\rm PSL}(2,8)$ has four such modules, of dimensions $6,8$ and $12$. Of these, only the one of dimension $6$ (which is the natural module) has a nonsplit extension, and that one is unique.

You can access all of these groups except for the split extension $2^{12}:L_2(8)$ from the perfect groups database in GAP.

For example, the nonsplit extension $2^6\!\cdot\! L_2(8)$ can
be accessed by:

```<code>> gap> G := PerfectGroup( IsPermGroup,32256,2);
  L2(8) N 2^6
</code>```

and then you can just compute its character table directly.

Body:
Well, if you really only interested in $G/M = A_5$ or ${\rm PSL}(2,8)$, then I can tell you what you want to know.

$A_5$ has just two nontrivial  irreducible modules over ${\rm GF}(2)$, both of dimension $4$, and they both have trivial $2$-cohomology.

${\rm PSL}(2,8)$ has four such modules, of dimensions $6,8$ and $12$. Of these, only the one of dimension $6$ (which is the natural module) has a nonsplit extension, and that one is unique.

You can access all of these groups except for the split extension $2^{12}:L_2(8)$ from the perfect groups database in GAP.

For example, the nonsplit extension $2^6\!\cdot\! L_2(8)$ can
be accessed by:

> gap> G := PerfectGroup( IsPermGroup,32256,2);
  L2(8) N 2^6

and then you can just compute its character table directly.


问:
Can $U(n)$ be constructed as a countable union of proper Lie subgroups of $U(n)$?
Can $U(n)$ be constructed as a countable union of proper Lie subgroups of $U(n)$ for any $n$?

As I see it, the obvious answer is "no", since the dimension of the Lie algebra for any proper subgroup of $U(n)$ must be strictly less than $\dim\mathfrak{u}(n)$, meaning that the dimension of the subgroup must be strictly less than the dimension of $U(n)$, and a smooth manifold of dimension $d$ cannot be covered by a countable union of submanifolds of dimension $k<d$.

Is this intuition correct, or am I missing something critical?

(Also, if any of my terminology here is in error or seems awkward, I'd appreciate a heads up. With thanks.)

答: Your intuition is correct. More generally, no differentiable manifold can be written as a countable union of closed submanifolds of smaller dimension. This, together with the fact that $U(n)$ is connected (which imples that no subgroup of $U(n)$ has the same dimension of $\dim U(n)$), is enough to prove what you want to prove.


问:
Prove that $2\cdot r=l$ for radius r of circle inscribed in a square of length l.
The title says it all, however i would like to elaborate the whole problem.

Problem:
Consider an equilateral triangle inscribed in a circle. Suppose this circle is inscribed in a square.
Let $A_{c}$, $A_{s}$, $A_{t}$ be area of circle, square and triangle respectively.
Find $A_{t}/A_{s}$?

Partial Solution:

By Definition,

$A_{c}=\pi\cdot r^2$,
$A_{s}=l^2$,
$A_{t}=(b \cdot h)/2$
where r is the radius of circle, l is the length of the side of square, b and h are base length and height of triangle respectively.

Intuitive solution (more assumptions):

Let

$r = l/2 $

$b = l$

$h=l/2$

then $A_{t}/A_{s}= 1/4$

The assumptions are not justifiable but I think they give a hint for approaching the actual solution.

I am not sure about the assumption no. 2 and 3 but the first looks promising for an inscribed circle in a square.

I would appreciate any ideas for proving it.

答: The first one is certainly true. The segment from the center of the circle to a point of tangency is perpendicular to one side, so parallel to another side, and half its length. (Consider the other radius parallel to that second side.)</p>


问:
A Thousand Points
A 1000 points are graphed in the co-ordinate plane. Explain why it is possible to draw a straight line in the plane so that half the points on one side of the line and half are on the other.

This question was taken from Precalculus by Stewart,Watson.

答:
$1000$ points define a finite number of directions. Choose a line that has a different direction (so that it cannot contain two points).

Then by translating the line, you can leave any number of points on a side. On the picture, the green line has a different direction than the $15$ pairs of points (check the directions rose).

Title: A Thousand Points


答: You can't do that way as Gerry mentioned. However, you know that $\log^2n\leq\sqrt n$ for $n$ big enough and that's why for big $n$ you have $\frac{(\log n)^2}{n^2}\leq \frac{1}{n\sqrt{n}}$. For the latter series do converge.</p>

Title: Is this series convergent? $\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

Body:
$\sum_{i=1}^{\infty} \frac{(\log n)^2}{n^2}$

I guess it is convergent, so I apply comparsion test for this.

$\frac{log^2}{n^2} < \frac{n^2}{n^2} = 1$

So it is bounded by 1 and hence it is convergent.

Can I do in this way? It kind of make sense, but I have never deal with constant in this case

Usually we do bounding using another term depend on $n$?


问:
Sketch the force vectors
A particle, of mass m and with electric charge Q, falls vertically with a constant velocity v0. Under these conditions, the air resistance force can be considered as $R_{ar} = k \cdot v$, where $k$ is a constant and $v$ is the velocity. The particle then penetrates a region where a uniform and constant magnetic field B acts, perpendicular to the plane of the paper and entering it, as shown in the figure. The speed of the particle is then changed, acquiring, after a certain period of time, a new value vl, constant.
(remember that the intensity of the magnetic force is $|F_M| = |q| |v| |B|$, in SI units for v perpendicular to B)

Sketch the force vectors (Weight, $R_ {ar}$ and $F_M$) that act on the particle, in the presence of the B field, in the situation where the speed becomes the speed $v_L$. Represent, by a dashed line, the direction and direction of $v_L$

Attemp: I had only been able to express the value of the particle velocity vl, in the region where the B field operates as a function of m, g, k, B and Q.

The tangential velocity has a constant modulus, so we have an MRU: the tangential acceleration is zero.

Because it is a uniform circular motion, there will be a constant centripetal acceleration: $ac = \frac{V^2}{R}$

With this there is a constant centripetal force ($F_c = m \cdot \frac{V^2}{R}$) with direction pointing towards the center of the path. And this does not influence the value of V

$$v = \frac{mg}{\sqrt{k^2+q^2B^2}}$$

答:
The formula is correct, but the justification is wrong. Let's suppose that you have $\vec v_L$ making an angle $\theta$ with respect to $\vec g$ The air resistance will be opposite of $\vec v_L$, the magnetic force is perpendicular to $\vec v_L$.

If the particle would move on a circular trajectory with a constant velocity, the magnetic force would be always perpendicular to the trajectory and constant in magnitude, the air resistance will be always tangent (and constant in magnitude), and the gravity will be constant, but always pointing down. Therefore the sum of them will never be constant.

The other option you have is that at some point the particle will move in a straight line. With the requirement that the sum of the forces is zero, you get a constant velocity. If we write the sum of the forces along $v_L$ and perpendicular to that, we get:
$$mg\cos\theta=kv_L\\mg\sin\theta=qv_LB$$
Squaring the equations and adding them together:
$$m^2g^2=(k^2+q^2B^2)v_L^2$$
In the initial phase, without the magnetic field, the air resistance completely cancels gravity, so $$kv_0=mg$$
Putting it all together:
$$v_L=\sqrt{\frac{m^2g^2}{\frac{m^2g^2}{v_0^2}+q^2B^2}}=\frac{v_0}{\sqrt{1+\left(\frac{qv_0B}{mg}\right)^2}}$$

Title: Sketch the force vectors


问:
Prove that $\underline{\int_a^b}f(x)dx=\underline{\int_a^c}f(x)dx+\underline{\int_c^b}f(x)dx.$
Let $f:[a,b]\to \mathbb{R}$ a bounded function. Prove that $$\underline{\int_a^b}f(x)dx=\underline{\int_a^c}f(x)dx+\underline{\int_c^b}f(x)dx,$$ for $a<c<b$.

Let $ \epsilon> 0 $ be given. So, there exists a partition $ \mathcal {P} $ such that $$L (P, f) <\underline {H} (f) + \epsilon,$$ where $ \underline {H} (f) $ is the lower integral. Let's take the following two partitions, $\mathcal {P} _1 = \{a = x_0, x_1, \ldots, x_ {n + 1} = c \}$ and  $\mathcal {P} _2 = \{c = x_ {n + 1}, x_ {n + 2}, \ldots, x_ {2n} = b \}$. Let's call $\mathcal {P} '= \mathcal {P} \cup \{c \}$. Then, $$\underline {H} (a,c)+\underline {H} (c,b)\leq L(\mathcal{P}_1,f)+L(\mathcal{P}_2,f)=L(\mathcal{P}',f)\leq L(\mathcal{P},f).$$
Since $L (P, f) <\underline {H} (f) + \epsilon,$ then $$\underline {H} (a,c)+\underline {H} (c,b)\leq \underline {H} (f) + \epsilon$$
Since $ \epsilon $ is arbitrary, we have one inequality, and the other is almost similar. Is my proof correct? Although I have something weird that is the $ \alpha $ and here it is $ dx $.

答:
Take any partition $P$ of $[a,b]$. If the point $c$ is included, let $P'=P$. Otherwise add the point $c$ to form the refined partition $P'\supset P$.  This induces a partition $P'_1$ of $[a,c]$ and a partition $P'_2$ of $[c,b]$ such that $P' = P'_1 \cup P'_2$,

Hence,

$$L(P,f) \leqslant L(P',f) = L(P'_1,f) + L(P'_2,f) \leqslant  \underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx$$

The first inequality holds because $P'$ is  a refinement of $P$. The last inequality holds because the lower integral is the supremum taken over lower sums, e.g.,

$$\underline{\int_a}^cf(x) \, dx = \sup \{L(Q,f): Q \text{ is a partition of  } [a,c] \}$$

Since this holds for every partition $P$ of $[a,b]$, it follows that

$$\underline{\int_a}^bf(x) \, dx = \sup \{L(P,f): P \text{ is a partition of  } [a,b] \} \leqslant \underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx $$

Note also that if  $P_1$ and $P_2$ are partitions of $[a,c]$ and $[c,b]$, respectively, then $P = P_1 \cup P_2$ is a partition of $[a,b]$ , and

$$L(P_1,f) + L(P_2,f) = L(P,f) \leqslant \underline{\int_a}^bf(x) \, dx$$

Taking suprema over $P_1$ and $P_2$ on the left it follows that

$$\underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx \leqslant \underline{\int_a}^bf(x) \, dx $$

Together both inequalities prove the desired result,

$$\underline{\int_a}^bf(x) \, dx =\underline{\int_a}^cf(x) \, dx +\underline{\int_c}^bf(x) \, dx $$


问:
$f'(x)=f(x)$ and $f(0)=0$ implies that $f(x)=0$ formal proof
How can I prove that if a function is such that $f'(x)=f(x)$ and also $f(0)=0$ then $f(x)=0$ for every $x$. I have an idea but it's too long, I want to know if there is a simple way to do it. Thanks!
Obviously in a formal way.</p>

答:
Here is a different approach, more suited for calculus students who do not know how to solve differential equations yet.

It's easy to prove by induction that since $f'=f$, the function $f$ must have derivatives of all orders and $f = f' = f'' = ...$ for all $x$, and hence $f(0) = f'(0) = f''(0) = \ldots$.

Then, by Taylor's Theorem, we expand $f(x)$ in series around 0 to get

$$
f(x) = \sum_{n=0}^\infty \frac{x^n}{n!} f^{(n)}(0) = \sum_{n=0}^\infty \frac{x^n}{n!} \cdot 0
     = 0.
$$


问:
Finding a Maclaurin Series solution to a 2. order linear differential equation using sigma notation
I am stuck at the following question:

Find in the form of sigma noation, the Maclaurin series solution to:

$$\frac {d^2y}{dx^2}-x \frac {dy}{dx}-y=0$$

given that $y(0)=1$ and $y'(0)=0$.

What I have tried:

We may assume that the equation can be written as:
$$\sum_{k=0}(k+1)(k+2) a_{k+2}x^k-x\sum_{k=0}(k+1) a_{k+1}x^k-\sum_{k=0}a_{k}x^k=0$$

This is how my textbook solved a similar example and it is the method my book expects me to use.
I then tried to find a recurrence relation (this is also the way my textbook solved a similar example):

*:

$$(k+1)(k+2) a_{k+2}-(k+1) a_{k+1}-a_{k}=0$$

Since $y(0)=1$ and $\frac y'(0)=0$ and since a maclaurin series as the form $y(0)+xy'(x)+\frac {x^2}{2!}y''(0)+...$ we can say that $a_0$=$y(0)$=$1$ and $a_1$=$\frac{y'(0)}{2!}$=$0$.

from *:
$$(0+1)(0+2) a_{0+2}-(0+1) a_{0+1}-a_{0}=0$$
$$2a_{2}-a_{1}-a_{0}=0$$
$$2a_{2}-a_{0}=0$$
$$a_{2}=\frac {a_{0}}2$$

using * repeatetly:

$$(1+1)(1+2) a_{1+2}-(1+1) a_{1+1}-a_{1}=0$$
$$6a_{3}-2a_{2}-a_{1}=0$$
$$6a_{3}-2a_{2}=0$$
$$6a_{3}=2a_{2}$$
$$6a_{3}=a_{0}$$
$$a_{3}=\frac{a_{0}}6$$

and $a_4=\frac {a_0}{12}$, $a_5=\frac {a_0}{40}$...

Since $a_n$ is the factor of every $n$th term of the maclaurin series:
$$y=a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4+a_5x^5...$$
$$y=a_0 + \frac {a_{0}}2 x^2 + \frac{a_{0}}6 x^3 + \frac {a_0}{12} x^4+ \frac {a_0}{40} x^5...$$

But I do not see a pattern to write it using sigma notation. In fact, I know that it is wrong since my textbook says that the solution is:

$$y = \sum_{n=0}\frac {x^{2n}}{2^nn!}$$

which is something different.

Could anybody explain how to arrive at the correct answer and tell me what I did wrong? I would be really nice!

答:
The key is the leading $x$ in the series for the first derivative:

$$\begin{align}
0&=\sum_{k\ge0}(k+1)(k+2) a_{k+2}x^k-x\sum_{k\ge0}(k+1) a_{k+1}x^k-\sum_{k\ge0}a_{k}x^k\\[1ex]
&=\sum_{k\ge0}(k+1)(k+2)a_{k+2}x^k-\sum_{k\ge0}(k+1)a_{k+1}x^{k+1}-\sum_{k\ge0}a_kx^k\\[1ex]
&=\color{red}{\sum_{k\ge0}(k+1)(k+2)a_{k+2}x^k}-\sum_{k\ge1}ka_kx^k-\color{blue}{\sum_{k\ge0}a_kx^k}\\[1ex]
&=\color{red}{2a_2}-\color{blue}{a_0}+\sum_{k\ge1}\bigg((k+1)(k+2)a_{k+2}-(k+1)a_k\bigg)x^k\\[1ex]
&=\sum_{k\ge0}\bigg((k+1)(k+2)a_{k+2}-(k+1)a_k\bigg)x^k
\end{align}$$

From here, you have the recurrence

$$\begin{cases}a_0=1\\a_1=0\\a_{k+2}=\frac1{k+2}a_k&\text{for }k>1\end{cases}$$

so that

$$\{a_k\}_{k\ge0}=\left\{1,0,\frac12,0,\frac18,0,\frac1{48},0,\frac1{384},\ldots\right\}$$

The odd-indexed terms vanish, and for the even-indexed ones ($k=2n$) you have

$$a_{2n}=\frac{a_{2(n-1)}}{2n}=\frac{a_{2(n-2)}}{2^2n(n-1)}=\cdots=\frac{a_0}{2^nn!}$$


问:
Asymptotic curves of hyperboloid of one sheet
There is an exercise in Manfredo do Carmo's book,page172,that aks about the asymptotic curves of $x^2+y^2-z^2=1$. I used the formula $e(u')^2+2fu'v'+g(v´)^2=0$ where $e,f,g$ are the coefficients of the second fundamental form.I got this

$$ (u^2+v^2-1)((u´)^2+(v')^2)=u(u')^2-2uu'vv'+v(v')^2$$

but I have no idea how to proceed to find $u$ and $v$. Any ideas?

答: First of all, what parametrization are you using? Second of all, there is a fundamental geometric fact you should know about the hyperboloid of one sheet (or the hyperbolic paraboloid, saddle surface) — it is doubly ruled. Through each point of the surface there are two lines contained in the surface.</p>


问:
Double Hodge star property
Let $\star$ be the Hodge star operator. Prove that

$$\star \star \omega \enspace = \enspace (-1)^{p(n-p)} \cdot \omega$$

for some $\omega \in \Lambda^p$.

What I have so far is the following: Without loss of generality, I assume $\omega = dx^{i_1} \wedge \ldots \wedge dx^{i_p}$. Then I have

\begin{align}
 \star \omega \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \cdot \varepsilon^{i_1 \ldots i_p}{}_{j_{p+1} \ldots j_n} \cdot dx^{j_{p+1}} \wedge \ldots \wedge dx^{j_n} \\
\end{align}

Renaming the indices yields

\begin{align}
 \star \omega \enspace &= \enspace \frac{\sqrt{\det g}}{(n-p)!} \cdot \varepsilon^{i_1 \ldots i_p}{}_{k_1 \ldots k_{n-p}} \cdot dx^{k_1} \wedge \ldots \wedge dx^{k_{n-p}} \\
\end{align}

It follows that

\begin{align}
 \star \star \omega \enspace &= \enspace \frac{\det g}{(n-p)! \; p!} \cdot \varepsilon^{i_1 \ldots i_p}{}_{k_1 \ldots k_{n-p}} \cdot \varepsilon^{k_1 \ldots k_{n-p}}{}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{\det g}{(n-p)! \; p!} \cdot \Big( (-1)^{p(n-p)} \cdot \varepsilon_{k_1 \ldots k_{n-p}}{}^{i_1 \ldots i_p} \Big) \cdot \varepsilon^{k_1 \ldots k_{n-p}}{}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{\det g}{(n-p)! \; p!} \cdot (-1)^{p(n-p)} \cdot (n-p)! \cdot \delta^{i_1 \ldots i_p}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{\det g}{p!} \cdot (-1)^{p(n-p)} \cdot p! \cdot dx^{i_1} \wedge \ldots \wedge dx^{i_p}\\
&= \enspace \det g \cdot (-1)^{p(n-p)} \cdot \omega
\end{align}

which is nearly what I want, but how do I get rid of that $\det g$? Any suggestions -- or errors that I missed?

${}$

P.S.: I know that this question has already been asked on this platform, but I cannot follow all of the steps in these cases, and I would really appreciate knowing what went wrong in my approach.

答:
I will assume you are in the Riemannian setting, where the metric is positive definite and so on.

The notation $\varepsilon^{i_1\dots i_p}{}_{j_{p+1}\dots j_n}$ is kind of weird, because the permutation symbol $\varepsilon_{i_1\dots i_n}$ is "not a tensor" in the sense that you cannot manipulate its indices at will. Instead, they satisfy the identities

$$
g_{i_1j_1}\cdots g_{i_nj_n}\varepsilon^{j_1\dots j_n}
= (\det g)\varepsilon_{i_1\dots i_n},
$$
$$
g^{i_1j_1}\cdots g^{i_nj_n}\varepsilon_{j_1\dots j_n}
= \frac{1}{\det g}\varepsilon^{i_1\dots i_n}.
$$
Even when they satisfy this "weird rule", their product is a tensor
$$
\varepsilon_{i_1\dots i_n}\varepsilon^{j_1\dots j_n}
= \delta^{j_1\dots j_n}_{i_1\dots i_n}.
$$

Since you cannot raise and lower the indices of the $\epsilon$s, I suggest you always write them with either all indices up or all indices down. Then your Hodge dual looks like
$$ (\star\omega)_{a_{p+1}\dots a_n} = \frac{\sqrt{\det g}}{(n-p)!} \omega_{c_1\dots c_p}g^{a_1c_1}\cdots g^{a_pc_p} \varepsilon_{a_1\dots a_p a_{p+1}\dots a_n} $$
instead. Then, using the second identity I wrote, you have

\begin{align}
 \star \star \omega \enspace
&= \enspace \frac{\det g}{(n-p)! \; p!} g^{i_1 r_1}\cdots g^{i_pr_p}\cdot \varepsilon_{r_1\dots r_pk_1 \ldots k_{n-p}} \cdot g^{s_1k_1}\cdots g^{s_{n-p}k_{n-p}}\cdot \varepsilon_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{1}{(n-p)! \; p!} \varepsilon^{i_1\dots i_ps_1 \ldots s_{n-p}} \cdot \varepsilon_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{1}{(n-p)! \; p!} \delta^{i_1\dots i_ps_1 \ldots s_{n-p}}_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{1}{(n-p)! \; p!} (-1)^{p(n-p)}\delta^{s_1 \ldots s_{n-p}i_1\dots i_p}_{s_1 \ldots s_{n-p}\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace \frac{1}{p!}(-1)^{p(n-p)}\delta^{i_1\dots i_p}_{\ell_{n-p+1} \ldots \ell_n} \cdot dx^{\ell_{n-p+1}} \wedge \ldots \wedge dx^{\ell_n}\\
&= \enspace (-1)^{p(n-p)} \cdot \omega
\end{align}

Alternatively, you can do the whole derivation using the tensors $\def\sgn{\mathrm{sgn}\,}$
$$E_{i_1\dots i_n} = \sqrt{\det g}\,\varepsilon_{i_1\dots i_n}$$
$$E^{i_1\dots i_n} = \frac{1}{\sqrt{\det g}}\,\varepsilon^{i_1\dots i_n}$$
instead of the permutation symbols. Then the Hodge star looks like
$$ (\star\omega)_{a_{p+1}\dots a_n} = \frac{1}{(n-p)!} \omega_{c_1\dots c_p}g^{a_1c_1}\cdots g^{a_pc_p} E_{a_1\dots a_p a_{p+1}\dots a_n} $$
and you can freely raise and lower the indices of the $E$s using the metric.

Note: in the semi-Riemannian case, all the $\sqrt{\det g}$ get replaced with $\sqrt{|\det g|}$ and one of the Levi-Civita tensors (conventionally the one with upper indices) has to incorporate a factor of $\sgn g$ in order to obtain one from the other raising an lowering indices. In this case the formula you want is
$$\star\star=(-1)^{k(n-k)}(\sgn g)$$
and the Levi-Civita tensors satisfy
$$E_{j_1\dots j_n}E^{i_1\dots i_n}
= (\sgn g)\delta_{j_1\dots j_n}^{i_1\dots i_n}$$

Title: Double Hodge star property


答:
What you did is wrong since you have to compare with the general term of a converging series.

You can try Cauchy's Condensation Test

$$\frac{2^nn^2\log^22}{2^{2n}}=\log^2\frac{n^2}{2^n}$$

and now it's easy to check that $\,\displaystyle{\sum_{n=1}^\infty\frac{n^2}{2^n}}\;$ converges (for example, the n-th root test)


答:
So my first thought is draw any line.  If more points are on one side than the other than just slide the line parallel up or down to cross more points to increase the number of points on one side and less the number of points on the other.

The problem is we don't know that if we shove the line it won't intersect two or more points at a time or that as we nudge it it won't "immediately" intersect more.

But as there are only finitely many points and uncountably many slopes a line can take we can easily construct a line so that it intersects no points and as we slide it.... it can still intersect two points at a time.

Well, there are $1000$ points.  SO there are ${1000\choose 2} = \frac {1000\times 1001}2 = 500500$ pairs of points so we can find a line whose slope isn't equal to any of the lines formed by connecting two points.  Thus if we slide this line parrallelly, it will never intersect two points.  So we can just slide it until it crosses on point at a time until we get $500$ above and $500$ below.

But we still have the issue of how can we know that we can cross one point and not "immediately" hit another.

Well, when the line hits a single point there are $999$ other points not on the line.  $999$ is a finite number so if we compare all the points and their distance to the line.  There must be some minimum positive value. And when we slide off the point at less than that minimum value, we will slide of our current point an not "immediately* hit another.

So that's it.  We make a line with a slope that is no parallel to any pair of points.  We count how many points are above and below it.  We slide it in the direction to make it more balanced. Because the line is not parallel to any pair as we slide it it will intersect and pass individual points one at a time.  And we can slide it until we adjust the numbers perfectly.

Title: A Thousand Points

Body:
A 1000 points are graphed in the co-ordinate plane. Explain why it is possible to draw a straight line in the plane so that half the points on one side of the line and half are on the other.

This question was taken from Precalculus by Stewart,Watson.


问:
What is the derivative of digamma function?
I want to know $(\frac{\Gamma'(\alpha)}{\Gamma(\alpha)})'$. My goal is to show $\alpha $ times this derivative of digamma is greater than 1.

The background of question is to show $\bar{x}$ is not asymptotically efficient for Gamma($\alpha$,1), because the ratio of Var$\bar{x}$ and Cramer-Rao Lower Bound is greater than 1. I can show that this ratio is $\alpha $ times this derivative of digamma. But I don't know how to show it is >1.

答:
First note that by definition of the polygamma function:
$$
\alpha \partial_\alpha\frac{\Gamma^\prime(\alpha)}{\Gamma(\alpha)}=\alpha \partial_\alpha^2\log\Gamma(\alpha)=\alpha\psi^{(1)}(\alpha).
$$
Now calling on this proof detailing inequalities of polygamma functions we find for all $\alpha>0$:
$$
\frac{1}{\alpha}+\frac{1}{2\alpha^2}\leq\psi^{(1)}(\alpha)\leq \frac{1}{\alpha}+\frac{1}{\alpha^2}.
$$
Multiplying the entire inequality by $\alpha$ and only considering the lower bound then gives
$$
\alpha\psi^{(1)}(\alpha)\geq1+\frac{1}{2\alpha}>1,
$$
which is the desired result.


问:
Test if coin is fair with significance/confidence of 95%
Please take a look at the next $10$ mins of this lecture starting here:

https://youtu.be/rYefUsYuEp0?t=1147

We are trying to find a number $\xi$ such that:

$P(|S_n-n \cdot (1/2)| \le \xi) \approx 0.95$

The professor finds $\xi=31$ but he says that he "pretends" $S_n$ is approximately standard normal.
Why $S_n$? I apply the CLT literally and so I "pretend" that

$$Z_n = \frac{(S_n - n/2)}{(1/4)\cdot\sqrt{n}}$$  is standard normal.

By "pretend" I just mean the usual i.e. I get the right to use the standard normal tables

(as justified by the CLT).

In the standard normal table I looked for $0.975$ and I found the Z-score number $1.96$.

When I did all the computations I got $\xi = 15.495$ which is twice less. Why?

I am interpreting it this way: if I make 1000 coin tosses and if the number of heads I observe is no more than 15 away from 500, then I conclude with certainty of about 95% that my coin is fair.

Am I incorrect conceptually? Or did I mess up the calculations?

Or is there something else here which I am not taking into account?

A deviation/difference of 31 given 1000 coin tosses seems too much to me just intuitively. But intuition can lie.

Also, not sure why he says $S_n$ is approximately standard normal.

It should be $Z_n$, right?

Maybe the lecturer is oversimplifying just for presentation purposes,

and that's why he gets 31 and not 15.49. Is it indeed so?

答:
The lecturer does not actually claim $S_n$ is approximately standard normal, only that it is approximately normal.  See time index 24:40 in the linked video.

That said, this statement is itself inaccurate.  With increasing $n$, the distribution of $S_n$ does not tend toward a normal distribution--it remains binomial with parameters $n$ and $p$.  Rather, the CLT says that it is $(S_n - n/2)/\sqrt{n}$ that tends to a normal random variable, and this is because $S_n = B_1 + B_2 + \cdots + B_n$ where $B_i \sim \operatorname{Bernoulli}(p)$ are the underlying iid variables in the theorem.

However, this inaccuracy is not fatal.  We can still appeal to the CLT to obtain an estimate of the boundary of the rejection region; i.e., critical value $\xi$ for the test statistic $|S - 500|$.  Keep in mind, the use of a normal approximation here is motivated by convenience, not necessity.  We could use the exact distribution of $S$ to find the smallest integer $\xi$ such that $$\Pr[|S - 500| \le \xi] = \sum_{x = 500 - \xi}^{500 + \xi} \binom{1000}{x} 2^{-1000} \ge 1 - \alpha = 0.95.$$  In this way, the Type I error of the test will be at most $\alpha$.  Such a calculation is easily tractable with a suitable computer program; for instance,

$$\begin{array}{c|c}
\xi & \Pr[|S - 500| \le \xi] \\
\hline
0 & 0.025225 \\
1 & 0.0755744 \\
\vdots & \vdots \\
30 & 0.946322 \\
31 & 0.953709 \\
32 & 0.960221 \\
\vdots & \vdots \\
500 & 1.
\end{array}$$
But if we do not have such a program, then it is the normal approximation to the binomial that we may use.  The idea is that $S$ is approximately normal with mean $\mu = np = 500$, and standard deviation $\sigma = \sqrt{np(1-p)} = 5 \sqrt{10} \approx 15.8114$.  Then $$\begin{align}
\Pr[|S - 500| \le \xi] &\approx \Pr\left[\frac{- \xi}{15.8114} \le \frac{S - \mu}{\sigma} \le \frac{\xi}{15.8114}\right] \\
&\approx \Phi\left(\frac{\xi}{15.8114}\right) - \Phi\left(\frac{- \xi}{15.8114}\right) \\
&= 1 - 2\Phi\left(-\frac{\xi}{15.8114}\right), \end{align}$$ where $\Phi$ is the CDF of the standard normal distribution.  Setting this to $1 - \alpha$ then gives us $$
\xi = -(15.8114)\Phi^{-1}(\alpha/2)$$ where $\Phi^{-1}$ is the inverse CDF or quantile function.  For $\alpha = 0.05$, $\Phi^{-1}(\alpha/2) \approx -1.95996$, therefore $$\xi \approx 30.9898 \approx 31.$$


问:
What is the correct notation for an intersection of generalized intersections which share only some predicates?
Let $Q$ be an index set that indexes two families of sets $\{S_i\}_{i \in Q}$ and $\{F_i\}_{i \in Q}$.
Let $<_A$ and $<_B$ be partial orders on $Q$.

Let $i \in Q$.
Suppose I want to compute
$$
\left(\bigcap \{S_j : (j <_A i) \land (j <_B i)\}\right) \cap
\left(\bigcap \{F_j : (j <_A i) \land (j >_B i)\}\right).
$$

What is a more compact way to represent this?
Maybe the following?
$$
\left(\bigcap_{j <_B i} S_j\right)
\cap
\left(\bigcap_{j >_B i} F_j\right)
: j<_Ai
$$
Also, is it okay to leave out that $j \in Q$ for brevity?

答:
If things like this occurred often enough, I’d probably introduce notation for $\{j\in Q:j<_Xi\}$ and $\{j\in Q:j>_Xi\}$ for any relation $<_X$ that might appear, perhaps $\downarrow_Xi$ and $\uparrow_Xi$. Then I’d write it

$$\bigcap_{j\in\downarrow_Ai\cap\downarrow_Bi}S_j\cap\bigcap_{j\in\downarrow_Ai\cap\uparrow_Bi}F_j\,.$$

I definitely would not leave that $j<_Ai$ hanging as you have done; without introducing new notation I’d make it

$$\bigcap_{\substack{j<_Ai\\j<_Bi}}S_j\cap\bigcap_{\substack{j<_Ai\\j>_Bi}}F_j\,.$$


答:
Note that, by the integral test

$$ \int_{1}^{\infty} \frac{(\ln(x))^2}{x^2}dx = -{\frac { \left( \ln  \left( x \right)  \right) ^{2}}{x}}-2\,{\frac {
\ln  \left( x \right) }{x}}-\frac{2}{x}
\Big|_{x=1}^{\infty} =2 . $$

the series converges.

Main result: Suppose $f$ is continuous, positive, decreasing function on $[1,\infty)$ and let $a_n=f(n)$. Then

$(a)$ if $\int_{1}^{\infty}f(x) dx$ is convergent, then $\sum_{n=1}^{\infty} a_n $ is convergent.

$(b)$ if $\int_{1}^{\infty}f(x) dx$ is divergent, then $\sum_{n=1}^{\infty} a_n $ is divergent.


问:
Showing that a positive definite homogenous polynomial attains a minimum on the unit n-sphere
Let $$p(X)=\sum_{r}a_{r_1r_2...r_n}{x_1}^{r_1}{x_2}^{r_2}...{x_n}^{r_n}$$
be a homogeneous polynomial of degree $r$ and suppose that $p(X)>0$ for all non-zero $X\in\mathbb{R}^n$.

Then $p$ attains a non-zero minimum on the unit n-sphere.

I have not gotten very far nor do I have a clear strategy in mind but this is my attempt so far:

Let $S=\{X\hspace{1mm}|\hspace{1mm}|X|=1\}$ and $T=\{p(X)\hspace{1mm}|\hspace{1mm}X\in S\}$.

It is quite easy to show that $T$ is bounded and it is also non-empty, thus inf$(T)=\alpha$ exists. Here I though the right way to go was to assume that $\alpha=0$ and show that it leads to a contradiction and then use the fact that $\alpha$ is the infimum of $T$ and non-zero to get to the desired conclusion. However, I can only get as far as showing that if $\varepsilon$ is given there is some $X_0\in S$ such that $\alpha<P(X_0)<\alpha+\varepsilon$, and from here I cant see a way to a contradiction if I assume $\alpha=0$.

If anyone can lead me in the right direction or give me a solution, that would be great.

答:
The unit $n$-sphere $S^n$ is compact, therefore every continuous real-valued function on $S^n$ has a minimum and a maximum. Therefore the value $\alpha=\inf T$ is actually a minimum, meaning there is $x_0\in S^n$ such that $\alpha=p(x_0)>0$.

Your proof is also on the right track. Continuing from $\alpha<p(x_0)<\alpha+\varepsilon$ and assuming $\alpha=0$, then there is, for each $n$, a point $x_n$ such that $p(x_n)<\frac{1}{n}$. Again, using compactness of $S^n$, there is a point $x_\infty$ such that for each $\delta>0,N\in\mathbb N$, there is $m>N$ such that $|x_\infty-x_m|<\delta$. From here use that $p(x)$ is a continuous function and try to show that $p(x_\infty)=\alpha=0$.