问:
Mean value by stochastic matrix
Let $u^{(0)}:=v \in R^d$ and $u^{(k+1)}:=Hu^{(k)}, H \in R^{d,d}$   being a symmetrical matrix with nonnegative coefficients such that the sum of its row elements is 1 for each row. I need to prove the following: $$\, \frac{1}{d}\sum_{j=1}^{d}u_j^{(k)}=\frac{1}{d}\sum_{j=1}^{d}v_j$$

It is clear that $u^{(k+1)}=H^{k+1}u^{(0)}.$ I am wondering if a property of symmetrical matrix should be used. While $H$ is a symmetrical matrix, $H^{k+1}$ must not be one. Since the elements of each row of $H$ sum up to 1,the convex combination of the components of the vector $v$ gives a value situated between the smallest and largest component of $v$. I do not know how to proceed. Can somebody provide some hint or prove the statement ?

答:
You can just make the calculations:
\begin{align*}
\sum_{i=1}^d u_i^{(k)} = & \sum_{i=1}^d \sum_{j=1}^d H_{ij} u_j^{(k-1)}=\sum_{j=1}^d \sum_{i=1}^d H_{ij} u_j^{(k-1)} = \sum_{j=1}^d u_j^{(k-1)}\underbrace{\sum_{i=1}^d H_{ij}}_{=1} = \sum_{j=1}^d u_j^{(k-1)}
\end{align*}

if the sum of the elements is kept from one iteration to the next, it will always be equal to the sum of the elements in $u^{(0)}=v$. I don't see the need for requiring that $H$ has positive elements. It is also not necessary that $H$ is symmetric, it is only necessary that the sum of the elements is each column is 1.

Title: Mean value by stochastic matrix


问:
Evaluate $ 2\pi \int_0^1 \ln(x^2+2)\sqrt{1+\frac{4x^2}{(x^2+2)^2}}\,dx $
I've been trying to solve this this problem for so long but haven't had much luck, can someone please help me out?

The original question was:

Find the area of the part of the surface $z = \ln(x^2+y^2+2)$ that lies above the disk $x^2 + y^2 \le 1$ correct to four decimal places by expressing the area in terms of a single integral.

I set up the integral as:

$$
2\pi \int_0^1 \ln(x^2+2)\sqrt{1+\frac{4x^2}{(x^2+2)^2}}\,dx.
$$

edit 1: Sorry for being unclear, I need a numerical value at the end, I've tried solving it by parts but it hasn't gotten me anywhere. I could just put it into a calculator and get an answer but I'm more curious about how to actually set it up and the steps to solve this problem.

答:
This is not a solid of revolution. The proble asks the area of the surface $z\log \left(x^2+y^2+2\right)$ inside the cylinder $x^2+y^2=1$

$z=\log \left(x^2+y^2+2\right)$ and $D:x^2+y^2\le 1$

$$S=\iint\limits_{D} \sqrt{z_x^2+z_y^2+1}\,dA$$

$$z_x=\frac{2 x}{x^2+y^2+2};\;z_y=\frac{2 y}{x^2+y^2+2}$$

$$S=\iint\limits_{D} \sqrt{z_x^2+z_y^2+1}\,dA$$

$$S=\iint\limits_{D}\sqrt{\frac{4 x^2}{\left(x^2+y^2+2\right)^2}+\frac{4 y^2}{\left(x^2+y^2+2\right)^2}+1}\,dA$$
$$S=\iint\limits_{D}\sqrt{1+\frac{4 \left(x^2+y^2\right)}{\left(x^2+y^2+2\right)^2}}\,dA$$
Changing to polar coordinates
$x=r\cos t;\;y=r\sin t$ and $Jacobian = r$ we get
$$S=\int _0^1\int _0^{2 \pi } r \sqrt{\frac{4 r^2}{\left(r^2+2\right)^2}+1}\,dtdr$$
$$S=\int _0^1\int _0^{2 \pi }\frac{r }{r^2+2}\sqrt{r^4+8 r^2+4}\,dtdr$$
$$S=\frac{1}{2} \pi  \left(2 \sqrt{13}-4+\log \left(\frac{1}{162} \left(95 \sqrt{13}+343\right)\right)-4 \sqrt{2} \cot ^{-1}\left(\sqrt{2}\right)+4 \sqrt{2} \cot ^{-1}\left(\sqrt{26}\right)\right)\approx 3.5618$$

ContentLicense: CC BY-SA 4.0


问:
Why is $\cos(\theta) < \dfrac{\sin(\theta)}{\theta}$ for $\theta \in (0,\dfrac{\pi}{2})$
I read that $ 0 < \cos(\theta) < \dfrac{\sin(\theta)}{\theta} < \dfrac{1}{\cos(\theta)}$ for $\theta \in (0,\dfrac{\pi}{2})$. I am confused about the middle part. I don't find it intutive (even from the graph). Please explain me why this is so.

答: Hint : For $0 < \theta < \frac{\pi}{2}$, $$\cos(\theta) < \frac{\sin(\theta)}{\theta} \Longleftrightarrow \theta < \tan(\theta)$$


问:
Derivation of exact-fit polynomial equation
I was reading this Wikipedia page and I understood the math until the $\vec{y} = \mathbf{X}\vec{\beta}+\vec{\varepsilon}$ equation (I understand that equation). After that, I cannot follow the math. Also, I currently am only concerned about exact fit polynomials, so all matrices that will be inverted will be square.

Can anybody explain the derivation? Thanks!

答:
The error term is $y-Xb$. What follows is the standard procedure for minimizing this error.

$$\|\epsilon\|^2=\|y-Xb\|^2=(y-Xb)^\top(y-Xb)=y^\top y-2b^\top X^\top y+b^\top X^\top Xb$$
Differentiating wrt $b$ gives the minimum, $$-2X^\top y+2X^\top Xb=0$$ hence $b=(X^\top X)^{-1}X^\top y$.


问:
How is this an open cover for any of the projective complex spaces?
A little confused how this set of equivalence classes forms an open cover. We'll start with the definition of an n-dimensional complex projective space:

Let $X=(\mathbb{C}^{n+1}\backslash \{0 \}) / \sim$

Where $x\sim y$ iff $x=\lambda y$

$\rightarrow \mathbb{C}P^n=X/\sim$

Let's take an arbitrary element from our punctured complex metric space:

Let $z=(z_0,z_1,...z_n)$ $\in \mathbb{C}^{n+1}\backslash \{0 \}$

Denote $\bar{z}$ the equivalence class of $z$, then define the collection of points

$U_i=(\bar{z}|z_i\neq 0)$

Here's what I read that completely perplexes me:

$\{U_i|0\leq i\leq n\}$ (allegedly) forms an open cover of the projective space $\mathbb{C}P^n$. How exactly can this be true? My first attempt at understanding was for $n=1$. In this case we essentially have two vectors, correct? In order to "cover" this projective plane would it not need to be a circle of points?

答:
Let $p : C^{n+1} \setminus \{0\} \to \mathbb CP^n$ denote the quotient map.

The sets $V_i = \{ z \in \mathbb C^{n+1} \mid z_i \ne 0 \}$ are open in $\mathbb C^{n+1}$, thus open in $\mathbb C^{n+1} \setminus \{0\}$ since $0 \notin V_i$. Clearly $\bigcup_{i=0}^n V_i = \mathbb C^{n+1} \setminus \{0\}$. We have $U_i = p(V_i)$, thus the $U_i$ cover $\mathbb CP^n$.  We claim that $p^{-1}(U_i) = p^{-1}(p(V_i)) = V_i$ which proves that that $U_i$ is open in $\mathbb CP^n$.

The inclusion  $\supset$ is trivial. So let $z \in p^{-1}(p(V_i))$, i.e. $p(z) \in p(V_i)$. This means that $p(z) = p(w)$ for some $w \in V_i$, thus $z = \lambda w$ for some $\lambda \in \mathbb C \setminus \{0\}$. But $w_i \ne 0$, hence $z_i = \lambda w_i \ne 0$ which shows $z \in V_i$.

Let us prove that each $U_i$ is homeomorphic to $\mathbb C^n$.

The restriction $p_i : V_i \to U_i$ of $p$ is a quotient map. In fact. let $U \subset U_i$ such that $p_i^{-1}(U)$ is open in $V_i$. We have $p_i^{-1}(U) = p^{-1}(U) \cap V_i = p^{-1}(U)$ because $p^{-1}(U) \subset p^{-1}(U_i) = V_i$. Since $V_i$ is open in $\mathbb C^{n+1} \setminus \{0\}$, we see that $p^{-1}(U)$ is open in $\mathbb C^{n+1} \setminus \{0\}$, thus $U$ is open in $\mathbb CP^n$. Hence $U$ is also open in $U_i$.

Let $\mathbb C^n_i$ denote the set of all $z \in \mathbb C^{n+1}$ such that $z_i = 1$. This is a complex hyperplane, in particular it is a homeomorphic copy of $\mathbb C^n$. Clearly $\mathbb C^n_i \subset V_i$. It therefore suffices to show that the restriction $q_i : \mathbb C^n_i \to U_i$ of $p_i$ is a homeomorphism. Define $r_i : V_i \to  \mathbb C^n_i, r_i(z) = \frac{1}{z_i}z$, which is a continuous map.  We have $(q_i \circ r_i)(z) = q_i(r_i(z)) = q_i(\frac{1}{z_i}z) = p(\frac{1}{z_i}z) = p(z) = p_i(z)$, i.e. $q_i \circ r_i = p_i$. Since $p_i$ is surjective, also $q_i$ must be surjective. It is also injective since $q_i(z) = q_i(z')$ (which is the same as $p(z) = p(z')$) means that $z  = \lambda z'$ for some $\lambda \in \mathbb C \setminus \{0\}$. Thus $1 = z_i = \lambda z'_i = \lambda 1 = \lambda$, therefore $z = z'$. This shows that $q_i$ is a continuous bijection. We shall show that $q_i$ is a quotient map which completes the proof since bijective quotient maps are homeomorphisms. So let $U \subset U_i$ be such that $q_i^{-1}(U)$ is open in $\mathbb C^n_i$. Then $r_i^{-1}(q_i^{-1}(U)) = (q_i \circ r_i)^{-1}(U) = p_i^{-1}(U)$ is open in $V_i$ and by 1. $U$ is open in $U_i$.


问:
Inverse function to $\cos(\sqrt{x})$ for $x\in [\pi^2; 4\pi^2]$.
I am having trouble with inverse function to $f(x)=\cos(\sqrt{x})$ for $x\in [\pi^2; 4\pi^2]$.

For $x\in [0;  \pi^2]$ it should simply be $\arccos^2(x)$, but I do not completely understand, what to do, if I want to find inverse for $x\in [\pi^2; 4\pi^2]$.

I would guess it would be some $f^{-1}(x)=-\arccos^2 (x) . a + b$?
I may be wrong.

Thanks for help.

答:
Solve

$$y=\cos(\sqrt x)$$ for $x\in[\pi^2,4\pi^2]$.

We can write

$$\sqrt x=2k\pi\pm\arccos(y)$$ and look for solutions in $[\pi,2\pi]$. As the arc cosine is in $[0,\pi]$, we have

$$\sqrt x=2\pi-\arccos(y)$$ and finally

$$x=(2\pi-\arccos(y))^2.$$


答: $f(x)=x\cos x-\sin x \implies f'(x)=\cos x-x\sin x-\cos x <0, \in (0,\pi/2).$So $f(x)$ is a decreasing function. Then $x>0 \implies f(x) <f(0)=0 \implies x \cos x-\sin x<0 \implies \cos x<\frac{\sin x}{x}, x\in (0,\pi/2).$

Title: Why is $\cos(\theta) < \dfrac{\sin(\theta)}{\theta}$ for $\theta \in (0,\dfrac{\pi}{2})$

Body: I read that $ 0 < \cos(\theta) < \dfrac{\sin(\theta)}{\theta} < \dfrac{1}{\cos(\theta)}$ for $\theta \in (0,\dfrac{\pi}{2})$. I am confused about the middle part. I don't find it intutive (even from the graph). Please explain me why this is so.


问:
Help with proof reading: Chapter 12.1, theorem 4 from Abstract Algebra 3rd Edition Dummit & Foote
I'm having trouble understanding the proof of this theorem, which can be found on page 460 of Dummit & Foote

To prove 1), we may without loss of generality assume $N \neq \{0\}$. And as $R$ is principal, $\varphi(N) = (a_\varphi)$ for all $\phi \colon M \rightarrow R$. So we set $\sum = \{(a_\varphi) \colon \varphi \in Hom(M,R) \}$. This is non-empty and has a maximal element, which we denote $(a_\nu) = (a_1)$. Let's say $a_1 = \nu(y)$.

We can show that $a_1$ is not zero, and that $a_1$ divides $\varphi(y)$ for every $\varphi \in Hom(M,R)$. Letting $\pi$ be the natural projection onto the $i$-th coordinate of the basis, we get $\pi_i(y) = a_1 b_i$. We define $y_1 = \sum_{i=1}^n b_i x_i$ where $x_1, \ldots, x_n$ is the basis of $M$. We can show that $a_1 y_1 = y$. Now, we want to show the below.

I'm having trouble understanding the proof for $b)$. It write for arbitrary $n \in N$, $n = \nu(n) y_1 + (n - \nu(n) y_1)$. What I'm trouble having see is

-Why is $n-\nu(n) y_1 \in N$?

-Why do we need $N \cap \text{ker } \nu$, instead of just $\text{ker } \nu$?

Thanks for your help!

EDIT: For my second point, my proof says to (to show that the sum is direct. We do know that N is the sum of the two modules at this point) suppose $x \in Ra_1 y_1 \cap (N \cap \text{Ker } \nu)$. Then we write $x = ra_1y_1 \in N$, and $\nu(x) = 0$.

Then $0 = \nu(x) = \nu(r a_1 y_1) = ra_1 \nu(y_1) = ra_1$, so we get $r = 0$ as $a_1 \neq 0$. My problem is that in this line, it seems $x \in N$ seems unnecessary.

EDIT EDIT: well $x$ must be in $N$ since we want that whole thing to be $N$. It's all good now I think.

答:
We have found $\nu\colon N\to R$ with image $(a_1)$, and taken $y\in N$ with $\nu(y)=a_1$. Given $n\in N$ we have $\nu(n)\in(a_1)$, say $\nu(n)=ta_1$, so that
$$ \nu(n)y_1 = ta_1y_1 = ty \in N. $$
Thus both $n$ and $\nu(n)y_1$ lie in $N$, so their difference $n-\nu(n)y_1$ does too.

In general a decomposition $M=Rm\oplus K$ will not pass to the submodule $N$, so we will not in general have $N=(Rm\cap N)\oplus(K\cap N)$. The statement (b) says that we do have it in our special situation, with $m=y_1$ and $K=\mathrm{Ker}\,\nu$.

Once we have such a compatible decomposition, we can pass to the case of the free module $M\cap K$, of rank one less that $M$, and its submodule $N\cap K$, and apply induction.


问:
Motivation behind this eccentric Ramanujan Identity
I just visited the MathJaX page due to the Math.SE website showing some problems while loading the page. I saw some demo math equations samples at this page, when this identity actually caught my attention:

$$ \dfrac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\dfrac{e^{-2\pi}} {1+\dfrac{e^{-4\pi}} {1+\dfrac{e^{-6\pi}} {1+\dfrac{e^{-8\pi}} {1+\ldots} } } } $$

My question is how does one come up with such an identity. Could there be any motivation behind this or is this just brute force calculations and then discovering it as an identity.

答:
As with Mariano, I too have no idea what sort of mathematical sorcery Ramanujan used, but what you have there is a special case of the Rogers-Ramanujan continued fraction (sans the $\sqrt[5]{q}$ factor).

Letting

$$R(q)=\cfrac{\sqrt[5]{q}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cdots}}}$$

the question amounts to how one might arrive at

$$R(\exp(-2\pi))=\sqrt{\phi\sqrt{5}}-\phi$$

As noted in this paper, Ramanujan had derived an explicit formula for $R(q)$ for certain special values of $q$ in his "lost notebook":

$$\begin{align*}R\left(\exp(-2\pi\sqrt{n})\right)&=\frac1{4t_n^2}\left((1-\phi t_n)\sqrt{1-t_n}-\sqrt{(1-t_n)(1+\phi t_n)^2-4\phi t_n}\right)\times \\&\left(\sqrt{(1-t_n)\left(1-\frac{t_n}{\phi}\right)^2+\frac{4 t_n}{\phi}}-\left(1+\frac{t_n}{\phi}\right)\sqrt{1-t_n}\right)\end{align*}$$

where $t_n=-\dfrac{G_{n/25}}{G_{25n}}$ and $G_n$ is the Ramanujan $G$-function (a.k.a. Ramanujan(-Weber) class invariant),

$$G_n=2^{-\frac14}q^{-\frac1{24}}\prod_{k=0}^\infty (1+q^{2k+1})=2^{-\frac14}q^{-\frac1{24}}(-q;q^2)_\infty$$

and $q=\exp(-\pi\sqrt n)$. (the details for proving it are a bit involved, but they can be seen in the linked paper, or here.)

Your case corresponds to $n=1$. Since $G_{1/n}=G_{n}$ (proven in this paper), $t_n=-1$, and making this substitution into the special formula for $R\left(\exp(-2\pi\sqrt{n})\right)$ yields the desired identity.

ContentLicense: CC BY-SA 3.0


问:
Showing that $EF = 0$ and $EF^t = I_4$ without using determinants but images and kernels
We have two $4\times 4$ matrices, $E$ and $F$. And $EF = 0$, and $EF^t = I_4$, where $F^t$ denotes the transpose of $t$ and $I_4$ the identity matrix in $M_4(\mathbb{R})$.

By using determinants, you're able to prove this statement false since the determinant of $EF$ and $EF^t$ are the same, yet the determinants of the corresponding $0$ and identity matrix in $\mathbb{R}^4$ are different. However, how would you go about approaching this in terms of images and kernels?

Thanks!

Edited for clarity--

答:
So we have two matrices (operators) $\,E\,,\,F\,$ of order $\,4\times 4\,$ over the reals, and we can think of them as operators (matrices).

Hints: The following are facts you should check you understand and can prove that follows from the basics and/or from the given data:

$$1)\;\;\;x\in\ker(EF)\iff E(Fx)=0\iff Fx\in\ker E,\;\text{and from here we get that}$$
$$\; EF=0\implies \text{Im}\,F\le\ker E$$

$$2)\;\;\;\text{row rank of $\,F\,=$ column rank of $\,F\,=\,$ row rank of $\,F^t\,$ }$$

$$3)\;\;\;\dim\text{Im}\,F=\text{ column rank}\ldots etc.$$


问:
Projection map: change of coordinates
Let $X\subset\mathbb{P}^n$ be a projective variety,and let $P\in\mathbb{P}^n$ be a point not in $X$.

By a change of coordinates we can assume that $P=[0:\cdots:1].$

Question. what change of coordinates are we talking about? How can you write it explicitly?

Let $H\cong\mathbb{P}^{n-1}\subset\mathbb{P}^n$ be a linear subspace of codimension $1$ tha does not contain $P$; again by a change of coordinates we can assume that $H=\{x_n=0\}$.

Question. what change of coordinates are we talking about? How can you write it explicitly?

Thanks!

答:
Let $\pi:\mathbb A^{n+1} \setminus 0 \to \mathbb P^n$ be the projection map, and let $v_0,..., v_n$ be a basis for the vector space structure $V$ induced on affine space by your choice of coordinates downstairs (the spans $\langle v_i \rangle$ are obtained as
$$
\langle v_i \rangle = \pi^{-1}(0:\cdots : 0 : \underbrace{1}_{i\text{-th entry}} : 0 : \cdots : 0),
$$

so just pick any nonzero vector therein). Then in particular $\pi^{-1}(P) = \langle v_n \rangle$. Similarly you can lift a given point $[a_0:\cdots: a_n]$ to the one-dimensional subspace $\langle a_0v_0 + \cdots + a_n v_n \rangle$. I leave it to you to write down a matrix that interchanges these subspaces.

The second change of coordinates is similar: choose a basis for $\pi^{-1}(H)$. Since $P \not \in H$, the union of this basis with $\left\{ v_n \right\}$ is a basis for $V$. The same is true if we start with $\pi^{-1}\left(\left\{ x_n =0 \right\}\right)$, choose a basis, and adjoin $v_n$. Thus there is a change of basis matrix for $V$ that will carry $H$ to $\left\{ x_n =0 \right\}$.

Title: Projection map: change of coordinates


答:
On the picture, it seems obvious that the sine is shorter than the arc $AB$, and if you wind the tangent on the circle, you end up past $A$. A stronger argument is that the area of the triangle $OBC$ exceeds that of the sector $OBA$.

Hence

$$\frac\sin\theta<\theta$$ and $$\theta<\tan\theta.$$


问:
Superadditive games and the worth of the grand coalition
I'm studying a first course in Game Theory, I don't understand why in superadditive games, the grand coalition has the highest payoff among all coalitional structures.

If by contraddiction we assume that in a superadditive game $(G,v)$ there exist $S\subset N$ such that $v(S)>v(N)$, then by superadditivity $v(N)\geq v(S)+v(N - S)>v(N)+v(N-S).$

Is there something that force $v(N-S)$ to be positive or null?

答:
In the first answer, it was claimed that a super-additive game has only a non-negative range and that the grand coalition must have the greatest value. Both claims are not correct. Of course, if the game has non-negative coalition values, then the grand coalition must have greatest value. And yes, almost all relevant economical games are mapping into the non-negative range. However, the definition of super-additivity allows that the value of a coalition can be negative, but then the grand coalition cannot have anymore the greatest value. As a counter-example consider the following three-person game

$v(\emptyset)=0,v(\{1\})=-1,v(\{2\})=0,v(\{3\})=0,v(\{1,2\})=7,v(\{1,3\})=7,v(\{2,3\})=13,v(N)=12.$

This game is super-additive and the greatest value of $13$ is assigned to coalition $\{2,3\}$. Furthermore, the only convention imposed on a TU game is that the empty set will be assigned with the zero value.


问:
What is the $n$th Cyclotomic Polynomial for a given $n$
Are there any formula to compute the $n$th cyclotomic polynomial for a given $n$? I know how to compute it for small $n$, but for large $n$, I need a general formula..

答:
Here is a general formula:
$$
\Phi_{n}(x)=\prod _{d\mid n}(x^{d}-1)^{\mu \left({\frac {n}{d}}\right)}
$$
where $\mu$ is the Möbius function. It follows by using Möbius inversion on
$$
x^{n}-1=\prod _{d\mid n}\Phi_{d}(x)
$$
This also gives a recursive formula for $\Phi _n$:
$$
\Phi_n(x) = \frac{x^{n}-1}{\prod _{d\mid n, d\ne n}\Phi_{d}(x)}
$$
Note that both formulas express $\Phi_n$ as a rational function which simplifies to a polynomial.

It is unlikely that there is a general formula for the coefficients of $\Phi _n$ in monomial form.


问:
Is the difference of two recursively enumerable sets, reducible to $K$?
Is the difference of two recursively enumerable sets, reducible to $K$?

$W_x/W_y=\{z|z \in W_x  \&  z \notin W_y\}$

$K=\{x|\Phi_x(x) \downarrow\}$

$W_x= \text{dom}(\Phi_x)$

答:
No.

Let $\omega$ denote the set of natural numbers. $K$ is c.e. but incomputable. If a set $A$ and its complement $\bar{A} = \omega - A$ is c.e., then $A$ must be computable. Hence $\bar{K} = \omega - K$, the complement of $K$, is not a c.e. set.

It is clear that $\omega$, the set of natural numbers, is c.e. (it is computable). $K$ is c.e. Thus

$\omega - K$ is a different of two c.e. sets. $\omega - K = \bar{K}$ which as mentioned above is not c.e.

Suppose $\omega - K = \bar{K}$ was many to one reducible to $K$; in notations $\bar{K} \leq_m K$. Since $K$ is c.e., $\bar{K}$ would have to c.e. (see the little lemma under the line). This a contradiction.

If if two set $A \leq_m B$ (are many to one reducible) and $B$ is c.e., then $A$ is c.e.

By definition of $A \leq_m B$, there is a computable function $f$ such that

$x \in A$ if and only if $f(x) \in B$.

Since $B$ is c.e.,  $B = W_n$ for some $n$. Define a new partial computable function

$\Psi(x) = \begin{cases}
0 & \quad \text{if }\Phi_n(f(x)) \downarrow \\
\uparrow & \quad \text{otherwise}
\end{cases}$

Since $\Psi$ is partial computable, it has a index $p$. That is, $\Psi = \Phi_p$. Then $A = W_p$ since $x \in A$ if and only if $f(x) \in B$ if and only if $\Phi_n(f(x)) \downarrow$ if and only if $\Phi_p(x) \downarrow$. $A$ is c.e.


问:
Finding a bijection of the set of binary strings of length $n$
Let $S$ be the set of binary strings of length $n$, i.e. each $x\in S$ is a string whose each entry is a $0$ or a $1$. Let $W(x)$ be the Hamming weight of $x\in S$ i.e. $W(x)$ is the number of ones in $x$. I am looking for a formula for a bijection $f: S\to S$ which shuffles a lot of vertices based on Hamming weight. For example, say $x\in S$ with $W(x)=w$ is mapped to say $y$ with $W(y)=w-2$ and $y$ in turn is mapped to $z$ with $W(z)=w-5$ and so on and finally the low weight strings are mapped back to $x$. This is just an example of what I'm looking for.

Bijections that I'm not looking for:

Swaps. For example, say we choose $2^n/3$ vertices using some rule and call this the set $A$, which has $1/3$ of the total strings, but each $x\in A$ is mapped to a vertex where ones and zeros are flipped. So in this case, low weight strings are swapped with high weight strings but this is not what I'm looking for. I need a good mixing of strings of different weights.

Shifts. Say we choose all strings of a particular weight $w$ and call this the set $B$. Now say each $x\in B$ is mapped to $y\in B$ where $y$ is obtained from $x$ by shifting each bit of $x$ by a constant amount. This bijection clearly maps strings of weight $w$ to strings of weight $w$ and hence this is not what I'm looking for because again I 'm looking for a good mixing of strings of different weights.

答:
Here is one approach which may or may not be what you're looking for:

Choose an arbitrary binary string to be the key. Then bit-wise xor with the key is a bijection.

More concretely, take a binary string $x$, and flip all the bits where the corresponding entry of the key is a 1. For instance, if $n=3$ and the key is $101$, then this bijection is "flip the first and the third bit".

If the weight of the key is even, then this bijection sends even-weighted strings to even-weighted strings, and vice versa. It can only change the weight by at most $W(\text{key})$. But if $W(\text{key})\approx \frac n2$, this shouldn't be too bad.


问:
Generalisation of IMO 1990/P3:For which $b $ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?

For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n+1$?

It was from my LTE/ Zsigmondy handout.

By taking examples, it looks like for $b= 2^k-1 , 2$ it's not true .

Here's my progress:

I got $b=4,5,6,8,9$ works ( $2,3,7$ doesn't )

$n$ is odd : If not then $4\mid b^n-1$ , but $b^n \equiv 0,1 \mod 4$

If $b+1$ is a power of $2$ and $n^2\mid b^n+1$, then $n=1$ : Let $p$ smallest prime dividing $n$ ( note that it can't be $2$) . Then $n^2\mid b^{2n}-1 \implies p\mid b^{2n}-1 \implies p\mid b^{\gcd(2n,p-1)}-1 \implies p\mid b^2-1\implies p|b-1 \text{(since $b+1$ is power of $2$)}\implies p\mid b^n+1 \implies p|2.$ Contradiction.

Any hints? Thanks in advance!

答:
Pretty famous problem!

As you have already showed that numbers of the form $2^k-1$ are not a validate candidate, we consider the case when $b+1$ is not a power of $2$.

If $b+1$ is not a power of $2$, then there exists an odd prime divisor $p$ of $b+1$.

 Can you show Also, $p^2\mid b^p+1$ (LTE?)

 By Zsigmondy's theorem, there exists a primitive prime divisor $q$ of $b^p+1$.

 $p\neq q$

 Can you show $q^2\mid (b^p)^q+1$? (LTE again)

 So we get $(pq)^2\mid b^{pq}+1$ and thus, similarly moving ahead, we create an infinite family of $n$'s such that $n^2|b^n+1$.

PS: Woah! Amazing proof for the case when $b+1$ is a power of $2$! Like thinking about the least prime dividing $n$ was amazing :O


问:
Since differential forms of odd rank are anti-commutative, what does that mean for multivariable integration?
For differential forms $\alpha\in\Omega^p$ and $\beta\in\Omega^q$, they satisfy the relation
$$\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha$$
Thus, odd ranked forms are anti-commutative. At first, this made me think that changing the order of definite integration of a multivariable function would change the sign of the result, but this  is not the case because
$$\int_a^b\int_c^df(x,y)\,dx\,dy = \int_c^d\int_a^bf(x,y)\,dy\,dx$$
What are some results of the anti-commutativity property in (definite or indefinite) integration? Additionally, how is the canceled out in the above example?

答:
I asked this question a few days ago, and I think I have stumbled upon the answer. We integrate forms over an oriented manifold (usually a sub-manifold). Let's say we have the following iterated integral:
$$
\int_a^b\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y
$$
The manifold in this case is the region
$$R=\left\{(x,y)\in\mathbb{R}^2\ \middle|\ a\leq y\leq b,\ g(y)\leq x\leq h(y)\right\}$$
which can be equivalently written as
$$R=\left\{(x,y)\in\mathbb{R}^2\ \middle|\ h^{-1}(x)\leq y\leq g^{-1}(x),\ \inf_{[a,b]}{g}\leq x\leq \sup_{[a,b]}{h}\right\}$$
Because of the anti-commutativity of 1-forms,
$$
\int\!\!\!\int_R f\ \mathrm{d}x\wedge\mathrm{d}y = -\int\!\!\!\int_R f\ \mathrm{d}y\wedge\mathrm{d}x
$$
To avoid picking up a minus sign, we must flip the orientation of the region so that
$$
\int\!\!\!\int_R f\ \mathrm{d}x\wedge\mathrm{d}y = \int\!\!\!\int_{R'} f\ \mathrm{d}y\wedge\mathrm{d}x
$$
where $R'$ has the opposite orientation of $R$. How do we express orientation in iterated integration? Let's look at a 1-dimensional example. Recall that
$$
\int_a^b f\ \mathrm{d}x = -\int_b^a f\ \mathrm{d}x
$$
The interval of integration $[a,b]$ is the same on both sides of this equation. The difference is the orientation of that interval. Since we have determined two expressions for the region $R$ earlier, we can say that
$$
\int_a^b\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y = \int_{\inf_{[a,b]}\!g}^{\sup_{[a,b]}\!h}\!\!\!\int_{h^{-1}(x)}^{g^{-1}(x)} f\ \mathrm{d}y\,\mathrm{d}x
$$
Here, the order of differential forms is flipped which picks up a minus sign, but the orientation of the manifold (region) also flipped which cancelled it out. Let $c=\inf_{[a,b]}g$ and $d=\sup_{[a,b]}h$. Then, $a=\inf_{[c,d]} h^{-1}$ and $b=\sup_{[c,d]} g^{-1}$ and we can write
$$
\int_{\inf_{[c,d]} h^{-1}}^{\sup_{[c,d]} g^{-1}}\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y = \int_{\inf_{[a,b]}\!g}^{\sup_{[a,b]}\!h}\!\!\!\int_{h^{-1}(x)}^{g^{-1}(x)} f\ \mathrm{d}y\,\mathrm{d}x
$$
By flipping one of the integrals on the right, we recover the original orientation of $R$ and recover the minus sign picked up from flipping our 1-forms:
$$
\int_{\inf_{[c,d]} h^{-1}}^{\sup_{[c,d]} g^{-1}}\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y = -\int_{\inf_{[a,b]}\!g}^{\sup_{[a,b]}\!h}\!\!\!\int_{g^{-1}(x)}^{h^{-1}(x)} f\ \mathrm{d}y\,\mathrm{d}x
$$
In short, when we change order of integration, we automatically change the orientation of the manifold we are integrating because the canonical place to put the limits in iterated integration are such that this is the case.


问:
Constructing triangle $\triangle ABC$ given median to the side $c$ and angles $\alpha$ and $\beta$

Constructing triangle $\triangle ABC$ given median to the side $c$ and angles $\alpha$ and $\beta$

I started with the median. Then I constructed a circle to each side of the median, such that the circle is a set of all points which are the vertices of angle $\alpha$ and $\beta$ above the given segment (the median) respectively. Now, if I only could create a line segment $AB$, such that it passes the point $C_1$ (the endpoint of the median and the middle point of side $AB$), each endpoint is on a different circle (to get the correct angles), and the line segments $AC_1$ and $C_1B$ are the same length ($C_1$ is the endpoint of a median), I would be done. Any ideas?

答: So, based on the suggestion of @Blue, the point is to find the midpoint of the centers of the circles $S$, construct a line segment $SC_1$, and then a line perpendicular to $SC_1$ passing through $C_1$. The two intersections with the circles are our points $A$ and $B$.


问:
Closed form for $ \int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$
I've been looking at

$$\int\limits_0^\infty  {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$

It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:

$$\displaystyle \int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$

$$\int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$

$$\int\limits_0^\infty  {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$

So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $\dfrac{1}{4}$, $\dfrac{1}{3}$ and $\dfrac{2}{5}$. Is there any way of finding it? I'll work on it and update with any ideas.

UPDATE:

The integral reduces to finding

$$\int\limits_{ - \infty }^\infty  {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt} $$

With $a =\dfrac{n+1}{m}$ which converges only if

$$0 < a < 1$$

Using series I find the solution is

$$\sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} $$

Can this be put it terms of the Digamma Function or something of the sort?

答:
Integrate over a closed contour that is in the shape of a wedge in the 1st quadrant, having an angle of $\pi/5$.  That is, consider

$$\oint_C dz \frac{z^2}{z^{10}+1}$$

where $C$ is that wedge contour.  As mentioned above, the contour splits into 3 pieces:

$$\oint_C dz \frac{z^2}{z^{10}+1} = \int_0^R dx \frac{x^2}{x^{10}+1} + i R \int_0^{\pi/5}d\phi \, e^{i \phi} \frac{R^2 e^{i 2 \phi}}{R^{10} e^{i 10 \phi}+1}+ e^{i 3 \pi/5} \int_R^0 dx \frac{x^2}{x^{10}+1} $$

As $R \to \infty$, the second integral vanishes as $1/R^7$ (Note that the denominator never vanishes).  The rest is then just equal to $i 2 \pi$ times the residue at the pole $z=e^{i \pi/10}$.  Thus, combining the integrals, we have

$$\left (1-e^{i 3 \pi/5}\right) \int_0^{\infty} dx \frac{x^2}{x^{10}+1} = i 2 \pi \frac{e^{i 2 \pi/10}}{10 e^{i 9 \pi/10}} $$

Doing the algebra, I get

$$\int_0^{\infty} dx \frac{x^2}{x^{10}+1} = \frac{\pi}{10 \cos{(\pi/5)}}$$


问:
Given that the equation $ax + 4 = 3x - b$ has more than 1 solution, find the value of $(4a + 3b)^{20202020}$
Find the value of $(4a + 3b)^{20202020}$

given that $ax + 4 = 3x - b$ has more than one solution

答:
Welcome to MSE. $$ax+4=3x-b\\ax-3x=-b-4\\x(a-3)=-(b+4)$$when this equation has more than 1 solution ? .....   when both sides get zero.

$$x\times 0=0$$ has more than one solution

can you take over?


问:
Study suggestions
I am a 16 year old high school student with too much free time. For the past two years I have been teaching myself university level maths in my free time. I have studied the basic first year subjects such as multivariable calculus, linear algebra and ordinary differential equations. My question is what should I study next?

I have tried more pure subjects such as real analysis, abstract algebra and number theory and additionally more applied subjects such as nonlinear dynamics, chaos theory and topology, but they either didn't spark my interest or were too difficult. I have even written a few papers that never really worked out.I posted one here I feel as if I need to regain my curiosity for mathematics and look for something beyond what academics present at seminars for children (referring to topology). Therefore, what is a subject that is seldom presented to an undergraduate but still is an valid field in mathematics?

答: Are you good in programming? Maybe you could learn Matlab and alike if you're interested in numerical mathematics too</p>

Title: Study suggestions


问:
Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates
How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?

I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$
and this function in spherical coordinates is
\begin{align*}
&\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi
\end{align*}
but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.

答:
The set $E$ is a ball of radius ${1\over2}$, centered at $\left({1\over2},0,0\right)$. The integrand
$$f(x,y,z):={yz\over x^2+y^2+z^2}$$
satisfies $f(x,-y,z)\equiv-f(x,y,z)$. This means that $f$ is odd with respect to the symmetry plane $y=0$ of $E$. It is then obvious that the requested integral has value $0$.


问:
functional equation:$f(f(x))=6x-f(x)$

Prove that there exists a unique function  $f:R^{+}\rightarrow R^{+}$ $$f(f(x))=6x-f(x)$$

My Try

Define $a_{k+1}=f(a_k)$ then we have  the recursive relation $$a_{k+2}+a_{k+1}-6a_k=0$$ whose  characteristic equation is $$x^{k+2}+x^{k+1}-6x^k=0$$ $$x^2+x-6=0 \Rightarrow x=-3 ,x=2$$ ie $$a_k=c_1 {(-3)}^k+c_2{(2)}^k$$ .As $x>0 \Rightarrow a_0>0\Rightarrow 2c_2>3c_1$

i am  stuck now as i have not been able to find $c_1,c_2$

答:
After OP's work: For $a_{k+2}+a_{k+1}-6a_k=0$, take $a_k=t^k$ to get $t_{1,2}=2,-3$.
Choose only positive root to write $a_k=C 2^k \implies a_0=C=x$ (by assumption), next $a_1=C. 2=2x$. By assumption $f(x)=a_1.$ So you get $f(x)=2x.$

Note: here $$a_0=x, a_1=f(x),a_2=ff(x), a_3=fff(x),....,a_k=f^{k}(x).$$

Title: functional equation:$f(f(x))=6x-f(x)$


问:
Minimize a function
I am trying to minimize the function $f(x) = \sqrt[x]{c\times x!}$ with respect to $x$($c>0$ is a constant), for $x>0$.
I have tried plotting the function to examine its behavior, and it looks to me that if we consider values of $x>0$, this function should have a global minimum. However, I don't know how to solve this minimization problem analytically, as the derivative is a rather strange looking function.

答:
As @themathandlanguagetutor, consider
$$f(x)=\Big[c \,\Gamma (x+1)\Big]^{\frac{1}{x}}$$ and assume $c>0$. Using logarithmic differentiation, we have
$$\frac{f'(x)}{f(x)}=\frac 1 {x^2}\Big[x\, \psi (x+1)-\log (\Gamma (x+1))-\log (c) \Big]$$ So, we are looking for the zero of function
$$g(x)=x\,\psi (x+1)-\log (\Gamma (x+1))-\log (c)$$
$$g'(x)=x \,\psi ^{(1)}(x+1) \quad >0 \quad \quad\forall x >0$$ So, only one solution.

Now, $\big[x\,\psi (x+1)-\log (\Gamma (x+1))\big]$ is very linear as soon as $x > 1$ and a good approximation of it  is
$$x\,\psi (x+1)-\log (\Gamma (x+1))=x+\frac{1}{2} (1-\log (2 \pi  x))-\frac{1}{6
   x}+O\left(\frac{1}{x^2}\right)$$ So, keeping the first and second terms, as an approximation, we need to solve
$$x+\frac{1}{2} (1-\log (2 \pi  x))=\log(c) \implies x=-\frac{1}{2} W_{-1}\left(-\frac{e}{\pi  c^2}\right)$$ where $W_{-1}(.)$ is the second branch of Lambert function.

This would be the $x_0$ for Newton method.

Below are some results

$$\left(
\begin{array}{ccc}
 c & x_0 & \text{solution} \\
 2 & 1.20556 &  1.39437 \\
 3 & 1.81582 &  1.93161 \\
 4 & 2.19930 &  2.29176 \\
 5 & 2.48314 &  2.56345 \\
 6 & 2.70899 &  2.78165 \\
 7 & 2.89662 &  2.96393 \\
 8 & 3.05712 &  3.12043 \\
 9 & 3.19732 &  3.25750 \\
 10 & 3.32177 &  3.37942 \\
 20 & 4.12296 &  4.16825 \\
 30 & 4.58111 &  4.62144 \\
 40 & 4.90271 &  4.94016 \\
 50 & 5.15051 &  5.18600\\
 60 & 5.35202 &  5.38606 \\
 70 & 5.52178 &  5.55470 \\
 80 & 5.66842 &  5.70041 \\
 90 & 5.79746 &  5.82868 \\
 100 & 5.91266 &  5.94323
\end{array}
\right)$$
To give an idea about the convergence, let us try for $c=10^4$; the iterates would be (with a ridiculous number of figures)
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & 10.819975929577196143 \\
 1 & 10.836092659850328733 \\
 2 & 10.836092115380538490 \\
 3 & 10.836092115380537870
\end{array}
\right)$$

So, we have a simple way to locate the minimum. Now, computing $f(x)$ does not make any problem.

Edit

Knowing the approximation $x_0$ we can improve a lot using one single iteration of Newton, Halley or Householder method. The simplest one will be
$$x_1=x_0-\frac {g(x_0)}{g'(x_0)}$$ The above table will now be
$$\left(
\begin{array}{ccc}
 c & x_1 & \text{solution} \\
 2 & 1.39889 &  1.39437 \\
 3 & 1.93246 &  1.93161 \\
 4 & 2.29214 &  2.29176 \\
 5 & 2.56368 &  2.56345 \\
 6 & 2.78180 &  2.78165 \\
 7 & 2.96405 &  2.96393 \\
 8 & 3.12052 &  3.12043 \\
 9 & 3.25758 &  3.25750 \\
 10 & 3.37948 &  3.37942 \\
 20 & 4.16828 &  4.16825 \\
 30 & 4.62146 &  4.62144 \\
 40 & 4.94018 &  4.94016 \\
 50 & 5.18601 &  5.18600\\
 60 & 5.38607 &  5.38606 \\
 70 & 5.55470 &  5.55470 \\
 80 & 5.70042 &  5.70041 \\
 90 & 5.82869 &  5.82868 \\
 100 & 5.94323 &  5.94323
\end{array}
\right)$$

In other words, for practical applications, one single iteration of Halley or Householder methods will give the solution.

For illustration, for the case of $c=10^4$, here are the values ofhe first iterate as a function of the order of the method
$$x_1^{(2)}=\color{red}{10.836092}6598503$$
$$x_1^{(3)}=\color{red}{10.83609211}48391$$
$$x_1^{(4)}=\color{red}     {10.83609211538}11$$
$$x_1^{(\infty)}=\color{red}{10.83609211538054}$$

It must be noticed that $f(x_0)\times f''(x_0) <0 \quad \forall c$ which, by Darboux theorem, means that we shall face one overshoot of the solution along the path to convergence. At the opposite $f(x_1)\times f''(x_1) >0 \quad \forall c$ which will avoid it.

Title: Minimize a function


问:
Integral identity with square of Jacobi polynomial
This has stumped me for a while: I have a function $\zeta_k^S(x)$ that can be expressed using Jacobi polynomials $P_k^{(\alpha,\beta)}(x)$:

$\zeta_k^S(x):=\displaystyle\sum_{j=0}^k\frac{(-1)^jx^{2j+2}{k \choose j}(4S-k-1)!}{(j+2)!(1+x^2)^{2S}(4S-k-j-1)!}=\frac{x^2k!}{\Gamma(3+k)(1+x^2)^{2S}}P_k^{(2,-4S-2)}(1+2x^2)$

where $S\in\{3,\frac{7}{2},4,\frac{9}{2},\ldots\}$ and $k\in\{0,1,\ldots,2S-1\}$ and  using

$\frac{\Gamma(m-n)}{\Gamma(n)}=\Bigg\{\begin{array}{cc} (-1)^m\frac{n!}{(n-m)!} & : m\leq n\\ 0 & : m>n \end{array}\ \ \ \ \ \ \ \ \ $ for $n\in\{-1,-2,-3,\ldots\}\ \ $.

I want to show the following identity, which I believe to be true:

$\mathcal{I}_k :=8\pi S\displaystyle\int_0^{\infty}\frac{\zeta_k^S(x)\cdot\zeta_k^S(x)\cdot x\ dx}{(1+x^2)^2}=\frac{4\pi S}{(4S-2k-1)(4S-k)(4S-k+1)(k+1)(k+2)}\ \ $.

But I'm at a loss for how. Here's what I have tried:

$\bullet$ Doing the integral explicitly. I get the following sum:

$\mathcal{I}_k=\displaystyle\sum_{j,m=0}^k\frac{4\pi S(-1)^{j+m}{k \choose j}{k \choose m}(4S-k-1)!^2(j+m+2)!(4S-j-m-2)!}{(j+2)!(m+2)!(4S-k-j-1)!(4S-k-m-1)!(4S+1)!}\ \ $.

I've tried to simplify this and also to use proof by induction, but I can't seem to relate $\mathcal{I}_{k+1}$ to $\mathcal{I}_k$ in a useful way.

$\bullet$ Using Rodriguez' formula $P_k^{(\alpha,\beta)}(x)=\frac{(-1)^k}{2^kk!}(1-x)^{-\alpha}(1+x)^{-\beta}\frac{d^k}{dx^k}\Big((1-x)^{\alpha+k}(1+x)^{\beta+k}\Big)$ and the expression for the derivative of the Jacobi polynomial, together with partial integration. I make some headway, but fail in the end.

$\bullet$ Again using proof by induction, but this time through the following recurrence relation

$2n(n+\alpha+\beta)(2n+\alpha+\beta-2)P_n^{(\alpha,\beta)}=(2n+\alpha+\beta-1)\Big((2n+\alpha+\beta)(2n+\alpha+\beta-2)x-\alpha^2-\beta^2\Big)P_{n-1}^{(\alpha,\beta)}-2(n-\alpha-1)(n+\beta-1)(2n+\alpha+\beta)P_{n-2}^{(\alpha,\beta)}$

This almost works, but I haven't found a way to deal with the extra factor of $x$ in front of $P_{n-1}^{(\alpha,\beta)}$, and this messes up my relation between $\mathcal{I}_k$, $\mathcal{I}_{k-1}$ and $\mathcal{I}_{k-2}$.

$\bullet$ Looking at recurrence relations between hypergeometric functions, in which $\zeta_k^S(x)$ can also be expressed, without finding any useful ones.

$\bullet$ Using the expression in this paper relating products of Jacobi functions to linear combinations. However it ends up being too hairy for me.

Any ideas?

答:
Let me first rewrite the statement in a bit more understandable form.

Under the restrictions you have on $S$ and $k$, you want to show that
$$\int_0^{\infty}\frac{\left[P_k^{(2,-4S-2)}(1+2x^2)\right]^2x^5dx}{(1+x^2)^{2+4S}}=\frac{(k+1)(k+2)}{2(4S-2k-1)(4S-k)(4S-k+1)}.\tag{1}$$

To compute the integral on the left, rewrite it as
\begin{align}
\int_0^{\infty}\frac{\left[P_k^{(2,-4S-2)}(1+2x^2)\right]^2x^5dx}{(1+x^2)^{2+4S}}=\frac{1}{2}\int_0^{\infty}\frac{\left[P_k^{(2,-4S-2)}(1+2y)\right]^2y^2dy}{(1+y)^{2+4S}}=\\
=2^{4S-2}\int_1^{\infty}(1-t)^2(1+t)^{-4S-2}\left[P_k^{(2,-4S-2)}(t)\right]^2dt,\tag{2}
\end{align}
where we used the change of variables $y=x^2$ to obtain the 1st equality and $t=1+2y$ for the 2nd one. Note that (2) looks very much like $\|P_k^{(2,-4S-2)}\|^2$ with respect to the usual orthogonality measure $w(t)=(1-t)^2(1+t)^{-4S-2}$ for Jacobi polynomials $P_k^{(2,-4S-2)}$, except that we integrate from $1$ to $\infty$ instead of from $-1$ to $1$. Actually, the latter integral would diverge because of values of $S$ but this hints nevertheless that (2)  should be a "good" quantity.

Indeed, let us replace $w(t)P_k^{(2,-4S-2)}(t)$ in (2) using Rodrigues formula for Jacobi polynomials:
$$w(t)P_k^{(2,-4S-2)}(t)=\frac{1}{(-2)^k k!}\frac{d^k}{dt^k}\left(1-t\right)^{2+k}\left(1+t\right)^{-4S-2+k}$$
Integrate $k$ times by parts (it is rather clear that all the boundary terms will vanish) so that all derivatives act on the second polynomial in the product:
\begin{align}&\int_1^{\infty}(1-t)^2(1+t)^{-4S-2}\left[P_k^{(2,-4S-2)}(t)\right]^2dt=\\
&=\frac{1}{2^k k!}
\int_1^{\infty}\left(1-t\right)^{2+k}\left(1+t\right)^{-4S-2+k}\frac{d^k}{dt^k}\left(P_k^{(2,-4S-2)}(t)\right)dt=\\&=\frac{A_k}{2^k}\int_1^{\infty}\left(1-t\right)^{2+k}\left(1+t\right)^{-4S-2+k}dt,\tag{3}
\end{align}
where $A_k$ denotes the leading coefficient in $P_k^{(2,-4S-2)}(t)=A_kt^k+\ldots$ This coefficient is explicitly given by
$$A_k=\frac{(4S-k-1)!}{(-2)^k k!\,(4S-2k-1)!}.\tag{4}$$
In addition, the integral in (3) can also be explicitly calculated (e.g. setting $u=\frac{t-1}{t+1}$ one obtains a beta function integral):
$$\int_1^{\infty}\left(1-t\right)^{2+k}\left(1+t\right)^{-4S-2+k}dt=(-1)^k2^{-4S+1+2k}
\frac{(4S-2k-2)!\,(k+2)!}{(4S-k+1)!}.\tag{5}$$
Now combining (2), (3), (4) and (5), after careful elementary simplification we obtain (1). $\blacksquare$


问:
Find the Maximum Likelihood Estimator for $\theta^2$
Let $X_{1}, ..., X_{n}$ represent a random sample from $U(0, \theta)$

How to find the MLE of $\theta^2$ and does this estimator unbiased? Does the MLE satisfy the regularity conditions? Really confused on this question.

答:
I'll give you some pointers to solve the question. First you need to calculate the likelihood function L. For the uniform distribution in your case you should obtain:

$L(\theta) = \frac{1}{\theta^{n}}$ for $x_{1},\, x_{2},\, \dots,\, x_{n} \in (0,\, \theta)$

and $0$ elsewhere.

Consider the loglikelihood where the likelihood is not zero:

$\log(L(\theta)) = \frac{-n}{\theta}$ for points where $L(\theta)$ is non-zero.

Now notice that this term is always negative, so we conclude that the loglikelihood is decreasing. This implies that smallest possible value for $\theta$ will be the MLE, since any value higher than that must have a lower likelihood, since the loglikelihood (and thus the likelihood) is decreasing.

Now, the lowest value possible for $\theta$ is actually the highest value you observed in your sample. (It couldn't be lower than the highest observed value in the sample, since than the highest observed value would not lie in the interval $(0, \theta)$.)

Now notice that the function which squares positive numbers is an injective function and thus a bijection on an appropriate image. Then you can use that the general property that if $A$ is an MLE for $a$, then $g(A)$ is an MLE for $g(a)$, where g is a bijection.

For regularity conditions it depends on what they are. You didn't add the conditions in the question, but usually the problem with the uniform distribution and regularity is that the domain on which the PDF of the uniform distribution is defined, depends on $\theta$.


问:
Can there exist a Non real Hermitian matrix Which is Diagonizable over R
We know any Hermitian matrix is Diagonizable over C always .
And If the entries are real then it is a symmetric matrix so it is Diagonizable over R as well as C

So here my Question is.. Does there exist any Hermitian matrix with at least one non real entry such that The matrix is Diagonizable over R

答: If the hermitian matrix $A$ is diagonalizable over the reals, then there are real matrices $D$ (diagonal) and $S$ (invertible) such that $A = SDS^{-1}$. Hence $A$ is a product of real matrices. And $A$ is real.


问:
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

Solve:   $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$

My Try

$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$

I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!

答:
Like Barry Cipra,

replacing $\sin x=s,\cos x=c,\cos2x=2c^2-1,\sin2x=2sc$

$$3(2sc)+4(3c^2-1)-2c+6s-6=0$$

$$\iff4c^2-c(1-3s)+3s-5=0$$

$$\implies c=\cdots=-1,\dfrac{5-3s}4$$

If $\cos x=c=-1, x=(2n+1)\pi$

Else $4\cos x+3\sin x=5$

Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$

See also: Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$


问:
Using Cauchy's Convergence Criterion to prove divergence of series
Problem:

Using Cauchy's Convergence Criterion I have to prove that:

$$\sum_{n=1}^\infty \frac{1}{\sqrt{n(n+1)}}$$
diverges.

Attempt:

Cauchy's Convergence Criterion:

$\forall \epsilon \gt 0  \, \, \exists$ an integer $N$ s.t $|a_n + a_{n+1} + ... + a_{m}| \lt \epsilon$ if $m\ge n \gt N$.

So I have to prove that for any $\epsilon \le |a_n + a_{n+1} + ... + a_{m}| $ . But I am completely stuck here. Any guidance will be much appreaciated.

<s>EDIT: Further Attempt

Consider

$$|\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+ ... + \frac{1}{\sqrt{m(m+1)}}|$$

Which is less or equal to $\frac{1}{n+1}$, so we get:

$$|\frac{1}{n+1}|\le|\frac{1}{\sqrt{n(n+1)}}+\frac{1}{\sqrt{(n+1)(n+2)}}+ ... + \frac{1}{\sqrt{m(m+1)}}|$$
$$\downarrow$$
So if $\epsilon \gt 0$ we can choose $N$ so that
$$|\frac{1}{N+1}| \ge \epsilon$$

Thus proving that the series diverges?</s>

答:
Hint :
$$\frac{1}{\sqrt{n(n+1)}} \geq \frac{1}{n+1}$$

and now, you can use the Cauchy criterion to prove that the harmonic series $\sum \frac{1}{n+1}$ diverges.


答:
$x^2+y^2+z^2-x=0 \implies (x-\frac{1}{2})^2 + y^2 + z^2 = (\frac{1}{2})^2$

So it is a sphere with radius $\frac{1}{2}$ centered at $(\frac{1}{2},0,0)$.

In spherical coordinates, $x = \rho \cos \theta \sin \phi, \, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi \,$ where $\theta$ is the azimuthal angle and $\phi$ is the polar angle (just opposite of your convention).

$x^2 + y^2 + z^2 = \rho^2$ so substituting in the equation of our sphere,

$\rho^2 = \rho \cos \theta \sin \phi \implies \rho = \cos \theta \sin \phi$.

Now the part that you have to be careful about is the bounds of azimuthal angle. Please consider a circle in XY plane with center at $(\frac{1}{2},0)$ and radius of $\frac{1}{2}$ and equation of $r = \cos \theta$. $\theta$ varies betweeen  -$\frac{\pi}{2}$ and $\frac{\pi}{2}$ (range of $\pi$ instead of $2 \pi$ for a sphere centered at the origin). It is same for this sphere.

So the integral becomes

$I = \displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi  \, d\rho \, d\phi \, d\theta$

And based on the given function, its integral above $XY$ plane and below will cancel each other.

If I was integrating over the part of the sphere only in the first octant, the integral will be

$I = \displaystyle \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi  \, d\rho \, d\phi \, d\theta = \frac{1}{72}$

Title: Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates

Body:
How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?

I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$
and this function in spherical coordinates is
\begin{align*}
&\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi
\end{align*}
but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.


问:
How do I find $g(x)$ from $f(g(x))$ and $f '(x)$?
Given $f(g(x)) = x$, and $f '(x) = 1 + [f(x)]^2$, how do I find $g '(x)$?

I have tried, subbing $g(x)$ into $f'(x)$ and got that $f'(g(x)) = 1+x^2$ but I am not sure if I am doing the right thing. It seems like a very simple problem but I'm having trouble with it.

答:
Since
$$
\frac{df}{dx} = 1 + f^2
$$
we have
$$
\int \frac{df}{1+f^2} = \int dx
$$
Then
$$
\arctan{f} = x + C \iff f = \tan \left( x + C \right)
$$

From which we immediately get
$$
g(x)=\arctan x-C
$$
Thus
$$
g'(x) =\dfrac{1}{1+x^2}
$$


问:
Existence of extreme points
Let $X$ be a reflexive, smooth but not strictly convex Banach space. Since $X$ is not strictly convex, it is clear that there exists a maximal face $C \subseteq S_X$ which is not a singleton.

I do not understand the following assertion: "Since $X$ is reflexive, there exists an extreme point $e \in$ ext$(C) \subseteq $ ext$(B_X)$".

Why can we assert that the singleton $\{e\}$ is a face of $C$ using the reflexivity of $X$?

答:
If $X$ is a reflexive Banach space, by Dieudonne's theorem, its closed unit ball is weakly compact. Besides, since $C$ is a maximal face of $S_X$, there exists $x^* \in S_{X^*}$ such that
$$ C = (x^*)^{-1}(\{1\}) \cap B_X, $$

Moreover, $(x^*)^{-1}(\{1\})$ is also a weakly compact subset since $x^*$ is weakly continuous, then, $C$ is weakly compact.

Now, $C$ is under the hypotheses of Krein-Milman theorem, which guaranteed the existence of extreme points for $C$.

Title: Existence of extreme points


问:
Number Theory Square Root of Non-Prime Mod
I'm trying to figure out $\sqrt{15}$ of $Z^*_{35}$ and $Z^*_{17}$? I do know how to compute the sqaure root for $Z^*_{17}$ since its a mod of odd prime and I got the answer 7, but I don't know how to compute it for the case $Z^*_{35}$ and any help or idea would be much appreciated. Thanks!

答:
Note that $15$ and $35$ have the common factor $5$, and so we set out to eliminate it.

Start with

$x^2\equiv15\bmod35$

Because of the common factor we know that $x\equiv5y$ for some $y$.  So then

$25y^2\equiv15\bmod35$

And now divide by the factor of $5$.  Since the modulus shares this factor we have to divide it by $5$, too, thus

$5y^2\equiv3\bmod7$

Multiplying by $5^{-1}\equiv3\bmod 7$ gives

$y^2\equiv9\equiv2\bmod7$

which should be easy to solve.  Then back substitute $x\equiv 5y\bmod 35$ to finish.


问:
Understanding the term "double roots" of a quadratic equation
We know that when a quadratic equation $ ax^2 + bx + c = 0 $ has zero discriminant value, then the quadratic equation has only "one root". But why do some mathematician call it "double root" and also that the equation still has two roots, but their roots are double. When we draw the equation in the coordinate plane, we can see that the graph of the equation will just touch the $ x $-axis at only one point.

The second question is: We know that the sum of the roots of a quadratic equation $ ax^2 + bx + c = 0 $ is $ -\dfrac{b}{a} $. But when the discriminant of that equation is zero, the sum is also $ -\dfrac{b}{a} $. It doesn't make sense. For example, if we have quadratic equation $ x^2 + 4x + 4 = 0 $, we can factor it as $ \left(x + 2\right)^2 = 0 $ and so $ x = -2 $. Thus, the sum is just $ -2 $. But with "sum of the roots" formula, the sum is $ -4 $. Which one is correct? Is this the reason we call it double root?

Thanks in advance :)

答:
Consider the assertion “the polynomial $(x-a)(x-b)$ has two roots”. Is it true? Well, yes if $a\ne b$, but no if $a=b$. If $a=b$, then that polynomial is $(x-a)^2$ and then we say that $a$ is a double root. Then if we count the roots with their multiplicity, (a double root is counted twice, a triple root is counted three times, …), a polynomial with degree $n$ will always have $n$ roots (at least over the complex numbers).

And, going back to quadratics, if we say that the roots of $(x-a)^2$ are $a$ and $a$, then $a+a=2a$ and $(x-a)^2=x^2-2ax+a^2$ and indeed the sum of the roots is minus the coefficient of $x$.


问:
Approximation of $H^1_0$-function from below
Say $v \in H^1_0(\Omega)$ for some bounded Lipschitz domain $\Omega \subset \mathbb{R}^n$ satisfying $|v(x)| \leq 1$ for almost any $x \in \Omega$. I wonder whether there exists a sequence $( v_j )_{j \in \mathbb{N}} \subset \mathcal{C}^\infty_0(\Omega)$ with $|v_j(x)| \leq 1$ for almost any $x \in \Omega$ and all $j \in \mathbb{N}$ such that $v_j \to v$ in $H^1(\Omega)$.

I considered setting $v = 0$ near the boundary and mollifying the resulting function. This should yield convergence in $L^2$ but only local convergence of the gradients in $L^2$. I then tried to use a partition of unity, but it seems that I then loose the compact support.

I am happy about any kind of suggestion. Thanks in advance.

答:
Yes, this is possible.

Sketch of a proof:

This can be done with the help of the accepted answer to this question, but it requires some modifications.

Let $\varepsilon>0$.
In the method in the linked answer one can also choose a function
$\phi:\Bbb R\to\Bbb R$ such that
$\phi(x)=x$ for $x\in [-1,1]$
and $|\phi(x)|<1+\varepsilon$ for all $x\in \Bbb R$.

Then there is a smooth approximation $u_\varepsilon \in C_0^{\infty}(\Omega)$ that satisfies
$\|u_\varepsilon - v\|_{H_0^1(\Omega)}<\varepsilon$.
( Note that the $C_0^{\infty}(\Omega)$-property is not explicitly stated there, but will automatically happen if we use the same construction.)

Finally, we can define
$$v_j := \frac1{1+1/j} u_{1/j},$$
which satisfies your assumptions.


问:
Solving quadratic form $\mathbf{x}^\mathrm{T}\mathbf{A}\mathbf{x} = c$ for $\mathbf{x}$
This is a simple question I hope, is there an easy way to solve:
$$
\mathbf{x}^\mathrm{T}\mathbf{A}\mathbf{x} = c
$$
for $\mathbf{x}$? (Assume $\mathbf{A}$ is positive definite).

Geometrically the solution should be an ellipse with axes given by the eigenvectors of $\mathbf{A}$ I think.

答:
In general, you'll have more than one solution, if $c>0$. Recall that $A$ define an inner product; as particular case, for $A=I_n$, how would you solve $\lVert x\rVert_2^2=c$?

One way to do it would be to compute the square root of $A$ (well defined as $A$ is positive definite) $S$, then solve $\lVert y\rVert_2^2=c$; the solutions are of the form $y=Sx$, and it remains to solve the system ($S$ is invertible) to get $x$.


问:
Prove or disprove that the set  is closed in metric space ( [ −1 , 1 ] , ∞ ) where ∞(  ,  ) = sup −1≤  ≤1 | () − ()
Let  = {  ∈ [−1 , 1] ∶ 2 ≤ (0) ≤ 4 }

My idea is to use the fact that for each closed set  in (  , ) , the preimage −1[  ] is closed in metric space (,).  We have the set A is all the continuous function in the interval [-1,1], how can we use this in the proof?

By looking at similar questions, is the boundary set is A itself? Otherwise , how do we find the boundary set?

答: Let $(f_n)_{n\in \Bbb N}$ be a convergent sequence of members of $A,$ converging to $g.$  For any $n\in \Bbb N$ we have $|f_n(0)-g(0)|\le d(f_n,g).$ And $d(f_n,g)\to 0,$ so $f_n(0)-g(0)\to 0.$ So $g(0)=\lim_{n\to\infty}f_n(0)\in [2,4]$ because every $f_n(0)\in [2,4]$ and $[2,4]$ is closed in $\Bbb R.$ So $g\in A.$


问:
Generalized solution to batteries and torch problem
I know this question has been asked here a couple of times.

We have four charged batteries, four uncharged batteries, and a radio which needs two charged batteries to work. We do not know which batteries are charged and which ones are uncharged. What is the least number of attempts that suffices to make sure the radio will work? (An attempt consists of putting two batteries in the radio and checking if the radio works or not).

It also has a brilliant solution using graph theory, but I'm interested in general problem, as it can be applicable to variety of other things.

My question is: What approach would be optimal to take if there were $n$ number of batteries, out of which $m$ batteries are charged, and most important: $k$ working batteries were needed start the radio?

EDIT: What I'm trying to do is bring some of the optimization to the program that I have. It's not about batteries, but this analogy helps me reason about it.
I'm not only interested in the number of combinations (even though it's really helpful), but also the exact combinations.

I was thinking about example $k = 6$, and $n = 48$, and let's say $m=28$. I figured that dividing the batteries into groups of 8 like:

```<code>1 2 3 4 5 6
1 2 3 4 7 8
1 2 5 6 7 8
3 4 5 6 7 8
</code>```

would require 2 of those not working in order not to fire, and then divided the complete pile into 6 smaller piles, then I thought about combining the piles in order to maximize the number of the broken batteries required to drop the combination set, and this is where I know I'm not doing something optimal and that it must exist mathematical approach to the problem.

I also figured out based on Daniels answer that this is not a trivial thing, and I am willing to accept the answer which points me to the right direction. (with $k,n,m$ from example above or with general case)

答:
The solution in the linked question readily generalizes via $k$-uniform hypergraphs, in which edges are $k$-element subsets of the vertex set. The problem thus can be equivalently formulated in the following way.

What is the minimum number of edges that a $k$-uniform hypergraph on $n$ vertices has to have to ensure that its complement doesn't contain a clique of size $m$?

The answer to this question is called the Turán number $T(n,m,k)$. Unfortunately, its value seems to be unknown currently for basically any $k > 2$; for some bounds, see e.g. Section 9 of this survey.


问:
Can we take the exponential function of a vector?
Main question:

If $x=(x_1,x_2)\in \mathbb R^2$, can we have the exponential function of the vector, i.e.
$$
e^{i \pi x}=e^{i \pi (x_1,x_2)} \quad? \tag 1
$$

Follow-up question:

Can we in some way "split" $e^{i \pi (x_1,x_2)}$?
I'm thinking of the rule the $e^{ia}e^{ib}=e^{i(a+b)}$, where $a,b\in \mathbb R$.

A test, if we use the standard basis and write the vector as
$$
x=(x_1,x_2)=\hat e_1 x_1+\hat e_2x_2 \tag 2
$$
Can we now write $(1)$ as
\begin{align}
e^{i \pi (x_1,x_2)} &=
e^{i \pi (\hat e_1 x_1+\hat e_2x_2)}  \tag 3\\
&=
e^{i \pi (\hat e_1 x_1)}
e^{i \pi (\hat e_2 x_2)}
\quad? \tag 4
\end{align}
Is this legitimate?

答:
That is equivalent to saying that $\exp(e_i)$ is defined for each $i \in \{1,2\}$ where $e_i's$ are basis vectors. When it comes to analytic functions, I often think of its Taylor expansion.

$$
\exp(x): = 1 + x + x^2/2 + \cdots
$$

If $x$ is a real number, the definition can be used to find the corresponding function value. It can also be applied to matrices for which exponentiation is well defined (in this case the $1$ in the right-hand side of the equation should be replaced by the identity I). However, for a vector $x$, $x^2$ is not defined in general. If you define the exponentiation of vectors, it would make sense, but you can't in reality as you know.


问:
Equivalent formulations of non-degenerate $C^*$-subalgebra?
Let $A$ be a $C^*$-algebra, faithfully represented on the Hilbert space $H$. Thus $A \subseteq B(H)$.

In the literature, I have seen the following $3$ definitions being used to say that $A$ is a non-degenerate $C^*$-subalgebra of $B(H)$:

(1) $[AH]= H$. Here $[AH]= \overline{\operatorname{span}}\{ah: a \in A, h \in H\}$.

(2) $\forall x \in H\setminus \{0\}: \exists u \in A: ux \neq 0.$

(3) $AB_0(H)$ is dense in $B_0(H)$. Here $B_0(H)$ are the compact operators on $H$.

It is easy to see that $(1) \iff (2)$. However, do we also have $(2) \iff (3)$ ? If not, do we at least have an implication or are these concepts not related, although having the same name?

答:
Note that the finite rank maps are dense in the space of compact operators on a Hilbert space. Furhter note that the finite rank maps are spanned by the rank $1$ maps. These can all be written as $v\otimes w^*$, which denotes the map $x\mapsto (x,w)\cdot v$.

Now suppose $(1)$ holds, then for any $v\in H$ there is a sequence $a_n(h_n)\to v$ with $a_n\in A$, $h_n\in H$. Then
$$a_n\circ (h_n\otimes w^*) = (a_n(h_n))\otimes w^* \to v\otimes w^*$$
where convergence is in the operator norm. This means that the rank $1$ maps lie in the closure of $AB_0(H)$, hence all of $B_0(H)$ is in the closure.

If we suppose $(3)$ holds we must be able to approximate any compact operator by elements of $AB_0(H)$. In particular for any $x$ lying in the range of a compact operator you must have sequences $a_n,k_n, x_n$ so that $(a_n(k_n(x_n))\to x$. This implies $(1)$, since every element is in the range of some compact operator.


问:
Relative Grading of a Series of Scores Against Highest and Lowest Scores
I apologize in advance for posting a question that may be very basic math. Please bare with me .

I have an online questionnaire with hundreds of multiple choice questions where my students get 1 hour to solve as many as they can (it is technically impossible to solve all).

I collect in a spreadsheet the number of correct answers each student was capable of answering during the 1 hour time session:

```<code>Student   | Correct Answers | Score
----------+-----------------+-------
Student A |       130       |  100
Student B |        70       |
Student C |        30       |
Student D |        50       |
Student E |        60       |
Student F |        20       |
Student G |        30       |
Student H |        40       |
Student I |        30       |
Student J |        30       |
Student K |        20       |    0
</code>```

What I would like to do is to grade each student relative to the highest and lowest scores in their specific group so that the student who correctly answered 130 questions will get 100 and the student who answered only 20 will get zero.

I am trying to come up with the right excel formula to do the job but i'm lacking the mathematical understanding of how to calculate this as a function.

I guess my question is if someone can explain in laymen terms the mathematical logic to achieve this kind of task.

Thank you,

答:
So essentially you have a function which maps the min value $(20)$ to $0$ and  max value $(130)$  to $100$. This function needs to be strictly increasing in your original score

There is no one answer to this question, and it depends on which gives you results that you are happy with. Some easy choices are

Linear - Just use the function

$$\text{Adjusted Score} = \frac{\text{Actual Score} - 20}{130-20} \times 100$$

This is an easy conversion, and is linear - it is equally sensitive at each score to change

Logarithmic - Using the function

$$\text{Adjusted Score} = 100\log_{10}\left(1 + 9\times\frac{\text{Actual Score} - 20}{130-20}\right) $$

This is more involved, but doesn't differentiate scores at the higher end very much - even if they have 5 points different in their actual score, they might have very small difference in their adjusted scores

Exponential - You could fit an exponential curve to not differentiate lower scores but instead increase differentiation at higher scores

It's really up to you, and what you see fit. Play around with different functions and see which scheme best highlights the point you want to get across to your students with their grades


问:
What does it mean for phase lines to be qualitatively the same?
I have the following phase lines for a differential equation
$$\dfrac{dy}{dt}=y^{2}+a$$

And it was mentioned that the phase lines for $a < 0$ are qualitatively the same, and the phase lines for $a > 0$ are qualitatively the same. However, I feel confused. What does it mean for phase lines to be qualitatively the same? And why isn't $a=0$ considered to be qualitatively the same?

答: An answers with actual formula transformations is possible here. If you consider $$y(t)=cu(ct)$$ for some fixed parameter $c=\sqrt{|a|}$ (for $a\ne 0$), then $$y'(t)=c^2u'(ct)$$ on the one hand and $$y(t)^2+a=c^2u(ct)^2+a$$ on the other. So for $a>0$ the equation cancels to a normal form $$u'=u^2+1,$$ for $a<0$ it cancels to a normal form $$u'=u^2-1$$ and $a=0$ is a special case that requires no transformation to a normal form.


问:
Finding $A,B,C,D \subseteq \{1,2,...n\}$ such that $A \cup B \cup C \cup D = \{1,2,...,n\}$
I have this combinatorial question:

Find the number of $(A,B,C,D)$ of sets $A,B,C,D \subseteq \{1,2,...,n\}$ such that $A \cup B \cup C \cup D = \{1,2,...,n\}$

I started by saying:

We choose the $k$ elements that are in $A$ which is $n \choose k$ and then we distribute all the $n-k$ elements  to the rest of the 3 sets.

Which is obviously wrong, because even if we choose $k$ to be in $A$, there still can be a set which has common elements of $A$.

I'll be happy if anyone could give me a direction.

EDIT:

Perhaps the answer is $4^n$? For every element in $N$ we choose if it is in $A, B, C$ or $D$

答:
For each element $k$ we can choose if it is in $A, B, C, D$. This amounts to $16$ possible choices for $k$.

For only one of these choices, we have that $k \notin A \cup B \cup C \cup D$.

Since we have such a consideration for each $1 \le k \le n$, the solution is $15^n$.

In general, when distributing elements over sets, if counting from the point of view of sets is hard, a good first reflex is to try and approach the problem from the point of view of the elements. Often this makes for an easy solution.


问:
Finding the determinant of a matrix with $0$s on the diagonal, $1$s in the first row and column, and $x$ elsewhere.
How do you attempt to solve the deterimnant of
$$D_n=\begin{vmatrix}
0 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & x & x & \cdots & x \\
1 & x & 0 & x & \cdots & x \\
1 & x & x & 0 & \ddots & \vdots \\
\vdots & \vdots & \vdots & x & \ddots & x \\
1 & x & x & \cdots & x & 0
\end{vmatrix}$$

My approach is to try and create a similar matrix with finite dimensions (something like a $4\times 4$) and then try and simplify it, but I always get stuck in the process of simplifying...

In general, how do you approach these types of questions?

答:
The first approach is to compute the first few values.
Here they are (WA):
$$
\begin{array}{c}
n& 1 & 2 & 3 & 4 & 5
\\
D_n&0 & -1 & 2x & -3x^2 & 4x^3
\end{array}
$$
The pattern is clear. Try a proof by induction.


Title: How do I find $g(x)$ from $f(g(x))$ and $f '(x)$?

Body:
Given $f(g(x)) = x$, and $f '(x) = 1 + [f(x)]^2$, how do I find $g '(x)$?

I have tried, subbing $g(x)$ into $f'(x)$ and got that $f'(g(x)) = 1+x^2$ but I am not sure if I am doing the right thing. It seems like a very simple problem but I'm having trouble with it.

Body:
Note that
$$[f(g(x))]'=f'(g(x))g'(x)=[x]'=1$$
By assumption $f'(g(x))=1+[f(g(x))]^2=1+x^2$ and hence
$$f'(g(x))g'(x)=(1+x^2)g'(x)=1\implies g'(x)=\frac1{1+x^2}$$


答:
By Determinant of a specially structured matrix ($a$'s on the diagonal, all other entries equal to $b$)

we know that the determinant is
$$
(-1)^{n-1}(n-1)x^{n}
$$
for the matrix, where the one's are replaced by $x$. But multiplying the first row and column by $x$, we can obtain such a matrix from the given one. The determinant changes by a factor $x^2$ then. So the determinant of our matrix then is
$$
(-1)^{n-1}(n-1)x^{n-2}.
$$

Title: Finding the determinant of a matrix with $0$s on the diagonal, $1$s in the first row and column, and $x$ elsewhere.

Body:
How do you attempt to solve the deterimnant of
$$D_n=\begin{vmatrix}
0 & 1 & 1 & 1 & \cdots & 1 \\
1 & 0 & x & x & \cdots & x \\
1 & x & 0 & x & \cdots & x \\
1 & x & x & 0 & \ddots & \vdots \\
\vdots & \vdots & \vdots & x & \ddots & x \\
1 & x & x & \cdots & x & 0
\end{vmatrix}$$

My approach is to try and create a similar matrix with finite dimensions (something like a $4\times 4$) and then try and simplify it, but I always get stuck in the process of simplifying...

In general, how do you approach these types of questions?


问:
Induction step other than $1$
Let $k \in \mathbb{N}$. I'm trying to prove that the following implication is true only when $k=1$:
$$[\exists n_0 \ \ \ P(n_0)] \land[\forall n\ge n_0 \ \ \ P(n) \implies P(n+k)] \implies [\forall n\ge n_0 \ \ \ P(n) ]$$
Obviously in the case $k=1$ we have the axiom of induction. How can we construct counterexamples for the other cases $k\not = 1$? Is it possible to give only one counterexample which works for all $k\not = 1$?

答: Fix your $k, n_0$ and let $P(n)$ be true iff $n \equiv n_0 \pmod{k}$. Then $P(n) \implies P(n+k)$, but $P(n)$ is only true for all $n \geq n_0$ if $k=1$.

Title: Induction step other than $1$


问:
Prove the series is absolutely convergent.
Let $\{a_k\}_{k=1}^\infty$ be a sequence in $\mathbb C$ with $a_k≠0$ for all $k \in \mathbb N$ such that $\lim_{k\to \infty}a_k=0$.

How do you prove that the series $\sum_{i=0}^∞ a_1 a_2 \cdot\cdot\cdot \,a_i$ is absolutely convergent.

答:
Take $n\in \Bbb N$ such that $k>n\implies |a_k|<1/2.$

For $i\in \Bbb N$ let $b_i=\prod_{j=1}^ia_i.$

Let $B=\sum_{i=1}^n b_i.$

For $j\ge 1$ let $c_j=\prod_{i=1}^j a_{j+n}.\;$ Then $$ j\ge 1 \implies |c_j|<2^{-j}$$ and also $$\sum_{i=1}^{\infty}b_i=B+b_n\sum_{j=1}^{\infty} c_j.$$


问:
Numerical Analysis or Numerical Linear Algebra Book with Available Solutions?
I am a beginning graduate student in Mathematics soon and I am planning to self-study Numerical Analysis and Numerical Linear Algebra. I know there are already reference-request questions about this, but I am looking for some more specific books.

I am looking for books in the following categories:

Books with difficult problems, and solutions in the back (ideal)

Books with difficult problems

Books with solutions in the back

Books with separate solution manuals

Problem books (ideally with solutions in the back, or solution manuals)

Could you please recommend me some textbooks, and tell me in which category they are? Thank you very much, any recommendations are immensely appreciated!

答:
I was an algebra major student and later a postgraduate student in topological algebra. We had two main books with linear algebra problems (but, as far as I remember, non-Numerical). The books contained a lot of problems and, in the back, hints for some of (I guess, harder) problems. Maybe there are separate books containing full solutions for these problems.

[Kos] This is an English translation of the second edition of [Kos2]. This books contains relatively hard problems. One of its three chapters (about a hundred pages)  is devoted to linear algebra and geometry. Unfortunately, the second edition of the book contains
sufficiently many misprints, which are hopefully fixed in [Kos2].

[Pro] This books has about 350 pages and two thousands linear algebra problems.

References

[Kos] Exercises in algebra: a collection of exercises in algebra, linear algebra and geometry, ed. Kostrikin A.I., Gordon and Breach Publ., 1996.

[Kos2] A collection of linear algebra problems, 3-edn, corrected and extended, ed. Kostrikin A.I., Moscow, Phys.-mat. lit, 2001, in Russian (Сборник задач по линейной алгебре, под редакцией А.И. Кострикина, издание третье, исправленное и дополненное, Москва, Физико-математическая литература, 2001). I did not find an English translation.

[Pro] Proskuryakov I.V., A collection of linear algebra problems, in Russian (Проскуряков И.В. Сборник задач по линейной алгебре). I did not find an English translation.


问:
Sum of decomposition coefficients of collinear points is $1$, but here seems to be a violation.
Below is a proposition in a text on analytic geometry:

Given two points $A$ and $B$, $A\ne B$, $O$ an arbitrary point, if point $C$ is collinear with $A$ and $B$, then vector $\overrightarrow{OC}$ can be decomposed by $\overrightarrow{OA}$ and $\overrightarrow{OB}$, and the sum of decomposition coefficients is $1$. That is, $\overrightarrow{OC}=a\overrightarrow{OA}+b\overrightarrow{OB}$ and $a+b=1$.

The text further states that the conclusion holds even if $O$ is collinear with $A$ and $B$ (I checked the proof and could find no problem), only the decomposition coefficients are not unique (but this seems to have nothing to do with my question).

Now, I put $O$ on line $\ell=\overline{AB}$. Take two points $C_1$ and $C_2$ on $\ell$, so we have $$\overrightarrow{OC_1}=a_1\overrightarrow{OA}+b_1\overrightarrow{OB} \qquad \qquad\overrightarrow{OC_2}=a_2\overrightarrow{OA}+b_2\overrightarrow{OB}$$ with $a_1+b_1=a_2+b_2=1$ by the proposition.

Next, let's consider $\overrightarrow{OC_1}+\overrightarrow{OC_2}$. Since $O$, $C_1$, $C_2$ are all on line $\ell$, vectors $\overrightarrow{OC_1}$ and $\overrightarrow{OC_2}$ are collinear with $A$ and $B$, so is their sum. We have $$\begin{align}
\overrightarrow{OC_1}+\overrightarrow{OC_2} &=a_1\overrightarrow{OA}+b_1\overrightarrow{OB}+a_2\overrightarrow{OA}+b_2\overrightarrow{OB} \\
&=(a_1+a_2)\overrightarrow{OA}+(b_1+b_2)\overrightarrow{OB}
\end{align}$$
However, the sum of the decomposition coefficients of $\overrightarrow{OC_1}+\overrightarrow{OC_2}$ is $a_1+a_2+b_1+b_2=1+1=2$, not $1$ any more, a violation of the proposition.

The proposition of the text should be correct, so where was I wrong in applying the proposition? How to reconcile this contradiction?

Thank you for your help.

答:
My bad. In the case where $O,A,B,C_1,C_2$ are collinear, the decomposition rule should be "there exists a decomposition where the sum of decomposition coefficients is $1$" instead of the sum always being $1$. I believe the ambiguity of the subject of "can be decomposed" is at fault here.

To illustrate this, consider the case where all these points are on the real line, so they can be represented as real directed numbers. Let $O = 0, A = 2, B = 3$. Then $\overrightarrow {OA} = 2, \overrightarrow {OB}=3$.

Take the case $C_1 = A =2$. Then a decomposition could be $\overrightarrow {OC_1} = 1\cdot\overrightarrow {OA} + 0 \cdot\overrightarrow {OB}$, with sum of decomposition coefficients $1$. Another decomposition could be $\overrightarrow {OC_1} = 0\cdot\overrightarrow {OA} + \frac23\cdot\overrightarrow {OB}$, with sum of decomposition coefficients $\frac23$. We can even decompose it as $\overrightarrow {OC_1} = -2\cdot\overrightarrow {OA} + 2\cdot\overrightarrow {OB}$, with sum of decomposition coefficients $0$. This shows that the decomposition is not unique. Most decompositions violate the "rule", but there still exists a decomposition with sum of decomposition coefficients $1$.

For example, if we have $C_2 = B = 3$, we can say:

$$\overrightarrow {OC_1} = 1\cdot\overrightarrow {OA} + 0 \cdot\overrightarrow {OB}$$
$$\overrightarrow {OC_2} = 0\cdot\overrightarrow {OA} + 1 \cdot\overrightarrow {OB}$$
$$5=\overrightarrow {OC_1}+\overrightarrow {OC_2}  = 1\cdot\overrightarrow {OA} + 1 \cdot\overrightarrow {OB} = -2\cdot \overrightarrow {OA} + 3 \cdot\overrightarrow {OB}$$

where the latter decomposition satisfies the rule but not the former.


问:
Does $\|Ax\|=\|Bx\| \enspace \forall x\in D(A)=D(B)$ imply $\langle Ax,Ay \rangle = \langle Bx,By \rangle$
I used this with additional assumption of A and B being closed in here. But i would like to see the proof or some counter-examples</p>

答: Yes. If $\|Az\|=\|Bz\|$ for all $z\in D(A)$ (where $X$ is a space with scalar product $(.,.)$ - assumed from the question) then we can put $z=x-y$ and we get $\|Az\|^2=\|Ax\|^2-2(Ax,Ay)+\|Ay\|^2=\|Bx\|^2-2(Bx,By)+\|By\|^2=\|Bz\|^2$. Using our assumption again (i.e. $\|Ax\|=\|Bx\|$ and $\|Ay\|=\|By\|$), we arrive to $(Ax,Ay)=(Bx,By)$ for every $x,y\in D(A)$.


问:
Total Variation Distance Triangle Inequality
I have the following definition for the total variation distance between two probability measures on a metric space $X$.  $d(\mu,\nu) = \inf\{\mathbb{P}(V\neq W) : \text{$V:\Omega\rightarrow X$ has distribution $\mu$, $W:\Omega\rightarrow X$ has distribution $\nu$}\}.$  How do I check that the triangle inequality holds for this function?

答:
The total variation distance is a special case of minimal metric. So, triangle inequality follows from this fact.

For example, in the paper "Metric distances in spaces of random variables and their distributions", V. M. Zolotarev, section 1.4, there is a proof.


问:
Embedding of finite groups in small $\operatorname{GL}(n,F)$
How does we obtain a small (not necessarily smallest) value of $n$ such that a $p$-group of order $p^m$ can be embedded in $\operatorname{GL}(n,\mathbb{F}_p)$?

[For embedding of a finite group $G$ in $S_n$, we can compute a small value of $n$ by looking at the minimum index of a subgroup $H$ containing no proper normal subgroup of $G$. Here, this value of $n$ need be necessarily smallest.]

答:
An easy to embed $p$-group is the Sylow $p$-subgroup of $\operatorname{GL}(n,p)$. It has $m = n(n−1)/2$, the largest possible $m$.  A hard to embed $p$-group is the cyclic group of order $p^m$.  It embeds in $\operatorname{GL}(n,p)$ only if $n ≥ p^{m−1}+1$.

Easy groups have minimal $n = Θ(\sqrt{m})$ and hard groups have minimal $n = Θ(p^m)$, so there is a very wide range of bounds based on the structure of the group.

In my experience, the single worst feature of a $p$-group for a small embedding is its exponent.  Cyclic groups are nearly regular anyways, so there is very little room to be "worse" asymptotically.

The proof of the exponential embedding lower bound is just a calculation with Jordan canonical forms, and applies to all fields of characteristic $p$.


问:
Showing that $EF = 0$ and $EF^t = I_4$ without using determinants but images and kernels
We have two $4\times 4$ matrices, $E$ and $F$. And $EF = 0$, and $EF^t = I_4$, where $F^t$ denotes the transpose of $t$ and $I_4$ the identity matrix in $M_4(\mathbb{R})$.

By using determinants, you're able to prove this statement false since the determinant of $EF$ and $EF^t$ are the same, yet the determinants of the corresponding $0$ and identity matrix in $\mathbb{R}^4$ are different. However, how would you go about approaching this in terms of images and kernels?

Thanks!

Edited for clarity--

答:
This is not really talking about images nor kernels, but :
$0 = 0E^tF^t = EFE^tF^t = (EF^tE^t)^tF^t = (E^t)^tF^t = EF^t = I_4$, contradiction.</p>


问:
Constructing triangle $\triangle ABC$ given median to the side $c$ and angles $\alpha$ and $\beta$

Constructing triangle $\triangle ABC$ given median to the side $c$ and angles $\alpha$ and $\beta$

I started with the median. Then I constructed a circle to each side of the median, such that the circle is a set of all points which are the vertices of angle $\alpha$ and $\beta$ above the given segment (the median) respectively. Now, if I only could create a line segment $AB$, such that it passes the point $C_1$ (the endpoint of the median and the middle point of side $AB$), each endpoint is on a different circle (to get the correct angles), and the line segments $AC_1$ and $C_1B$ are the same length ($C_1$ is the endpoint of a median), I would be done. Any ideas?

答: You can also employ this approach: draw a triangle $ABC$ with $\widehat{A}=\alpha$ and $\widehat{B}=\beta$, consider the length of the median $MA$ and apply a homotethy to $ABC$ in such a way that the length of $MA$ becomes the wanted length.


问:
Definite integral as limit of sum $\int_{a}^{b}\sin(x)dx$
I learn calculus and get stuck. I need help

We have $\int_{a}^{b}\sin(x)dx$, need to calculate using integral sums, so we split $[a,b]$ into $n$ equal parts: $dx \to h=\frac{b-a}{n}$. So

$$\sigma_{n} =\sum_{i=1}^{n}h\times \sin(a+i\cdot h)$$
Multiply both side to $2\sin(\frac h2)$ and use formula $\cos(\beta)-\cos(\alpha)=2\cdot \sin(\frac{\alpha+\beta}{2})\cdot \sin(\frac{\alpha-\beta}{2})$ where $\alpha=a+(i+\frac{1}{2})\cdot h$ ,$\beta=a+(i-\frac{1}{2})\cdot h$

We will get this: $$ \sum_{i=1}^{n}\sin(a+i\cdot h)=\frac{1}{2\cdot \sin(\frac{h}{2})} \cdot \sum_{i=1}^{n}2\sin(a+i\cdot h)\cdot \sin(\frac{h}{2})=\frac{1}{2\cdot \sin(\frac{h}{2})} \cdot \sum_{i=1}^{n}[\cos(a+(i-\frac{1}{2})\cdot h)-\cos(a+(i+\frac{1}{2})\cdot h)]$$

and then the book says that it (the sum $1$ to $n$) is equal to $$\frac{\cos(a+\frac{1}{2}h)-\cos(a+(n+\frac{1}{2})h)}{2\sin(\frac{h}{2})} $$
and I can't understand this.

Attaching Image for the Same

答:
Use the sum of finite geometric series and $e^{it}=\cos t+i\sin t:$
$$S=\sum_{k=1}^n \sin(a+kh)= \Im \sum_{k=1}^{n} e^{i(a+kn)}=\Im e^{ia} \sum_{k=1}^{n} e^{ihk}= \Im e^{i(a+h)}\frac{e^{inh}-1}{e^{ih}-1}.$$
$$\implies S=\Im e^{i(a+h)} e^{inh/2} \frac{\sin(nh/2)}{\sin (h/2)}=\frac{\sin (a+h+nh/2) \sin(nh/2)}{\sin (h/2)}.$$


问:
Combination of a piecewise defined function and a square function
Let's say I have two functions, $f(x)=x^2$, and:

$$g(x)=
\begin{cases}
- 4 & x \leq 0\\
|x- 4| & x > 0
\end{cases}$$

Now I have to write combinations $f\circ g$ and $g\circ f$.

I think $f\circ g$ should be:

$$f\circ g(x)=
\begin{cases}
16 & x \leq 0\\
(x- 4)^2 & x > 0
\end{cases}$$

and $g\circ f$ should be $g\circ f(x)= |x^2- 4|$, considering that $x$ will always be positive, but now I don't get how would I graph these composite functions, and also how to check continuity at $x=0$, especially of $g\circ f$ since I just eliminated half part of it.

答:
$f(x) = x^2$

$$
g(x) = \begin{cases}
         -4 & \text{if \(x \leq 0\)}\\
         |x - 4| & \text{if \(x > 0\)}
         \end{cases}
$$

You correctly found that
\begin{align*}
(f \circ g)(x) & = f(g(x))\\
               & = \begin{cases}
                   f(-4) & \text{if \(x \leq 0\)}\\
                   f(|x - 4| & \text{if \(x > 0\)}
                   \end{cases}
                   \\
              & = \begin{cases}
                  16 & \text{if \(x \leq 0\)}\\
                  |x - 4|^2 & \text{if \(x \geq 0\)}
                  \end{cases}
                  \\
              & = \begin{cases}
                  16 & \text{if \(x \leq 0\)}\\
                  (x - 4)^2 & \text{if \(x \geq 0\)}
                  \end{cases}
\end{align*}

Since
\begin{align*}
\lim_{x \to 0^+} (f \circ g)(x) & = 16\\
\lim_{x \to 0^-} (f \circ g)(x) & = 16\\
\end{align*}
we obtain
$$\lim_{x \to 0} (f \circ g)(x) = 16$$
For a function to be continuous at $x = 0$, the limit as $x$ approaches $0$ must exist and the function must be equal to its limit when $x = 0$.  Since $(f \circ g)(0) = 16$,
$$\lim_{x \to 0} (f \circ g)(x) = (f \circ g)(0)$$
Thus, the function $f \circ g$ is continuous at $x = 0$.

Since $x^2 > 0$ unless $x = 0$, where $f(0) = 0$, we obtain
\begin{align*}
(g \circ f)(x) & = g(f(x))\\
               & = g(x^2)\\
               & = \begin{cases}
                   -4 & \text{if \(x^2 \leq 0\)}\\
                   |x^2 - 4| & \text{if \(x^2 > 0\)}
                   \end{cases}
                   \\
              & = \begin{cases}
                  -4 & \text{if \(x = 0\)}\\
                  |x^2 - 4| & \text{otherwise}
                  \end{cases}
\end{align*}

Observe that
\begin{align*}
\lim_{x \to 0^+} (g \circ f)(x) = 4\\
\lim_{x \to 0^-} (g \circ f)(x) = 4
\end{align*}
Hence,
$$\lim_{x \to 0} (g \circ f)(x) = 4$$
Thus, the limit of the function exists at $x = 0$.  However, $(g \circ f)(0) = -4$.  Since
$$\lim_{x \to 0} (g \circ f)(x) \neq (g \circ f)(0)$$
the function $g \circ f$ is not continuous at $x = 0$ because it is not equal to its limit.  Since the continuity could be removed by defining $(g \circ f)(0) = 4$, which would require defining $g(x) = 4$ when $x = 0$, the function $g \circ f$ is said to have a removable discontinuity at $x = 0$.


问:
How do you prove $ \sum_{n=1}^{\infty}{a_n} = \sum_{n=1}^{\infty}{a_{2n}} + \sum_{n=1}^{\infty}{a_{2n-1}} $
How do you prove that by adding two series, both convergent, one
$$ \sum_{n=1}^{\infty}{a_{2n}} $$

and the other

$$ \sum_{n=1}^{\infty}{a_{2n-1}} $$

You get

$$\sum_{n=1}^{\infty}{a_n}$$

答:
If $E_N=\sum_{n=1}^Na_{2n}$, $O_N=\sum_{n=1}^Na_{2n-1}$ and $S_N=\sum_{n=1}^Na_{n}$, then:

$$S_{2N}=E_N+O_N$$
$$S_{2N-1}=E_{N-1}+O_N$$

(I.e. the sum of the first $2N$ elements of the sequence is the sum of the first $N$ odd-indexed and the first $N$ even-indexed elements; the sum of the first $2N-1$ elements of the sequence is the sum of the first $N$ odd-indexed and the first $N-1$ even-indexed elements.)

As $\sum_{N=1}^{\infty}a_{2n}$ and $\sum_{N=1}^{\infty}a_{2n-1}$ converge, this means that the partial sums $E_N, O_N$ converge, and let:

$$\sum_{N=1}^{\infty}a_{2n}=\lim_{N\to\infty}E_n=E$$
$$\sum_{N=1}^{\infty}a_{2n-1}=\lim_{N\to\infty}O_n=O$$

where $E,O\in\mathbb R$ are the two infinite series' sums.

Now, that means that:

$$\lim_{N\to\infty}S_{2N}=\lim_{N\to\infty}(E_N+O_N)=\lim_{N\to\infty}E_N+\lim_{N\to\infty}O_N=E+O$$
$$\lim_{N\to\infty}S_{2N-1}=\lim_{N\to\infty}(E_{N-1}+O_N)=\lim_{N\to\infty}E_{N-1}+\lim_{N\to\infty}O_N=E+O$$

Now, I suppose you know that, if for a sequence we can find out that the subsequence of odd-indexed elements and the subsequence of even-idexed elements both converge to the same value, then the whole sequence converges to the same value. Thus, $\lim_{N\to\infty}S_N=E+O$ as well, i.e. the series $\sum_{n=1}^{\infty}a_n=\lim_{N\to\infty}S_N$ converges to $E+O=\sum_{n=1}^{\infty}a_{2n}+\sum_{n=1}^{\infty}a_{2n-1}$.


问:
Question regarding minima and maxima related to its relation with values?
Let us say that you took a graph of $y = x^2$.

Values of y for this parabolic graph is 1,4,9 for values of x =1,2,3.

Values for for its slope by differentiation is 2,4,6 for x = 1,2,3.$\frac{dy}{dx}$

I took its differentiated values as velocity and not differentiated (I.e y=x^2) as speed.

Q1 It travels 1m in 1sec and then 4m in 2sec.
Whereas velocity is 2m in 1sec and 4m in 2sec.
How is that possible?
Since it would also mean that it traveled 2m in 1sec which is not right.

Q2 $x^3-3x=y$
When differentiated, slope is $3x^2-3$.
Values of x from this quadratic = 1 and -1.

When I put these values in the 1st equation ,
I get +2 and -2.

My sir said these are values of minima and maxima function.

What is the difference of these values with if I had directly put values in the differentiated equation.
For which I get =0 for both.Why are these not values of function of maxima and minima?

Q3 why are differentiated values equal to 0.
Y= X^2 , its $\frac{dy}{dx}$ =2x =0, Why not equal to tan theta which is the slope.

答:
Answer to Question 1:

If you know, the acceleration will be
$$a=\dfrac{\textrm dv}{\textrm dt}= 2$$
So, the system will be accelerated. The formula $\text{speed = distance/time}$ is not valid when $a\neq 0$. The correct formula is
$$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$$
And this holds correctly for your example.

Speed at $0$th second is $0$ and speed at $1$st second is $2$. So, average speed is $\frac{0+2}{2}=1$ as expected.
$$\left[\text{Average}=\frac{\text{Sum of all values}}{\text{Number of values}}\right]$$

Answer to Question 2 and 3:

You misunderstood things. Maxima and minima are the maximum and minimum values that a function can have, not the differentiated function in a given interval. So, $y=x^3-3x$ can be at least $-2$ or at most $2$ in the interval $[0,1]$, not $y=3x^2-3$.

This is the rule to set the differentiated function to $0$. $\dfrac{\textrm dv}{\textrm dt}$ is already the slope, so why would we equate it to slope?

Note that at the maximum and minimum values, the slope is zero. To find the maxima and minima, we thus equate it to zero. Just remember it for now, it is a rule.

Hope this is clear. Ask anything if required :)


答:
A simple way to verify your answer in mathematics is to substitute many small numbers in the formula. You can then verify your answer by hand calculation.

In your case, what is the answer for $n=1$ if you list all the collection of sets? Does it match your 'guessed' formula for $n=1$? What about $n=2$? and so on.

Now coming back to your problem.

In Polya's words:"If you can't solve a problem, then there is an easier problem you can solve: find it."

Let's try a simpler problem: How many pairs $A,B$ can you find such that $ A \cup B=\{1,2,3 \cdots n\}$?

In this case, clearly each of the $n$ elements have a choice of being in $A$, $B$ or both. In other words, each of the $n$ elements have $3$ choices independently. So the answer would be $3^n$.

Now can you solve your problem?

Title: Finding $A,B,C,D \subseteq \{1,2,...n\}$ such that $A \cup B \cup C \cup D = \{1,2,...,n\}$

Body:
I have this combinatorial question:

Find the number of $(A,B,C,D)$ of sets $A,B,C,D \subseteq \{1,2,...,n\}$ such that $A \cup B \cup C \cup D = \{1,2,...,n\}$

I started by saying:

We choose the $k$ elements that are in $A$ which is $n \choose k$ and then we distribute all the $n-k$ elements  to the rest of the 3 sets.

Which is obviously wrong, because even if we choose $k$ to be in $A$, there still can be a set which has common elements of $A$.

I'll be happy if anyone could give me a direction.

EDIT:

Perhaps the answer is $4^n$? For every element in $N$ we choose if it is in $A, B, C$ or $D$


问:
What does it mean if $\mathbb{E}[X^2] = \mathbb{E}[X]^2$?
I was wondering if there was any significance to assuming $\mathbb{E}[X^2] = \mathbb{E}[X]^2$? If X is normally distributed, this would correpsonding to it having zero variance (which I guess makes it a dirac delta function, or something similar?)

But is there any other intepretation of this? In particular, if we have three random variables $X_1,X_2,X_3$ such that $X_1+X_2+X_3=1$, what assumption are we making if we say $\mathbb{E}[X_iX_j] = \mathbb{E}[X_i]\mathbb{E}[X_j]$. Is there any context in which doing this can make sense? If $i \neq j$, then that just means $X_i$ and $X_j$ are independent (at least for two of the three values of $i,j$, since as the sum is 1, they aren't really all independent). But can it make sense for $i=j$?

For a bit more context, I was thinking in the context of say a system of chemical reaction equations.

You can write down a forward equation that describes the time evolution of the probability distribution. You can then calculate expected values and end up with something that looks like

$$\frac{d\mathbb{E}[X_1]}{dt} = a\mathbb{E}[X_1]+b\mathbb{E}[X_2]+c\mathbb{E}[X_1X_3]+d\mathbb{E}[X_1X_2]$$

$$\frac{d\mathbb{E}[X_2]}{dt} = e\mathbb{E}[X_3]+f\mathbb{E}[X_1X_2]+g\mathbb{E}[X_2X_3]-h\mathbb{E}[X_1^2]$$

If instead you wrote down a deterministic system for the same system of reactions using the law of mass action, you would arrive at essentially the same system, but without the expected values. But that's comparable to $\mathbb{E}[X_1^2]$ being replaced by $\mathbb{E}[X_1]^2$. I was wondering if there was any way of thinking about the relationship between these two in terms variances/ covariances of the random variable?

答:
If $E(X^2)=E^2(X)$ then the variance is $0$ and the random variable is actually a constant.

If $E(XY)=E(X)E(Y)$ then the Covariance between them is $0$ which means that they are uncorrelated (but possibly dependent!)


答:
The condition for equality in Holder's /Cauchy-Schwarz inequality says $E[X^{2}]=(EX)^{2}$ only if $X$ is a constant random variable.

If $E[X_iX_j]=(EX_i)( EX_j)$ for all $i$ and $j$ then each $X_i$ is a constant.

Title: What does it mean if $\mathbb{E}[X^2] = \mathbb{E}[X]^2$?

Body:
I was wondering if there was any significance to assuming $\mathbb{E}[X^2] = \mathbb{E}[X]^2$? If X is normally distributed, this would correpsonding to it having zero variance (which I guess makes it a dirac delta function, or something similar?)

But is there any other intepretation of this? In particular, if we have three random variables $X_1,X_2,X_3$ such that $X_1+X_2+X_3=1$, what assumption are we making if we say $\mathbb{E}[X_iX_j] = \mathbb{E}[X_i]\mathbb{E}[X_j]$. Is there any context in which doing this can make sense? If $i \neq j$, then that just means $X_i$ and $X_j$ are independent (at least for two of the three values of $i,j$, since as the sum is 1, they aren't really all independent). But can it make sense for $i=j$?

For a bit more context, I was thinking in the context of say a system of chemical reaction equations.

You can write down a forward equation that describes the time evolution of the probability distribution. You can then calculate expected values and end up with something that looks like

$$\frac{d\mathbb{E}[X_1]}{dt} = a\mathbb{E}[X_1]+b\mathbb{E}[X_2]+c\mathbb{E}[X_1X_3]+d\mathbb{E}[X_1X_2]$$

$$\frac{d\mathbb{E}[X_2]}{dt} = e\mathbb{E}[X_3]+f\mathbb{E}[X_1X_2]+g\mathbb{E}[X_2X_3]-h\mathbb{E}[X_1^2]$$

If instead you wrote down a deterministic system for the same system of reactions using the law of mass action, you would arrive at essentially the same system, but without the expected values. But that's comparable to $\mathbb{E}[X_1^2]$ being replaced by $\mathbb{E}[X_1]^2$. I was wondering if there was any way of thinking about the relationship between these two in terms variances/ covariances of the random variable?


问:
Where does this particular series converge?
I have the following series:

$$\sum_{n=0}^\infty (-1)^{n}\frac{x^{b+n}}{b+n}.$$

The series seems to converge in the open interval $(0,1)$. But can you tell to which function it converges?

Please note that here $b$ is any positive real number less than $1$ and $x$ lies in the open interval $(0,1)$ and $n$ belongs to the set of non-negative integers.

Any help will be appreciated.

答:
$$\sum_{n=0}^\infty(-1)^nx^{b-1+n}=\frac{x^{b-1}}{x+1}$$ so that

$$\sum_{n=0}^\infty(-1)^n\frac{x^{b+n}}{b+n}=\int_0^x\frac{t^b}{t+1}dx.$$

This is an incomplete Beta integral.


答: Schaum's Outline of Numerical Analysis.</p>

Title: Numerical Analysis or Numerical Linear Algebra Book with Available Solutions?

Body:
I am a beginning graduate student in Mathematics soon and I am planning to self-study Numerical Analysis and Numerical Linear Algebra. I know there are already reference-request questions about this, but I am looking for some more specific books.

I am looking for books in the following categories:

Books with difficult problems, and solutions in the back (ideal)

Books with difficult problems

Books with solutions in the back

Books with separate solution manuals

Problem books (ideally with solutions in the back, or solution manuals)

Could you please recommend me some textbooks, and tell me in which category they are? Thank you very much, any recommendations are immensely appreciated!

Body: Schaum's Outline of Numerical Analysis.


问:
Find the Maximum Likelihood Estimator for $\theta^2$
Let $X_{1}, ..., X_{n}$ represent a random sample from $U(0, \theta)$

How to find the MLE of $\theta^2$ and does this estimator unbiased? Does the MLE satisfy the regularity conditions? Really confused on this question.

答:
First calculate the likelihood

$$L(\theta)=\frac{1}{\theta^n}\cdot \mathbb{1}_{(X_{(n)};\infty)}(\theta)$$

Where $X_{(n)}$ is the maximum of $X_i$.

It is self evident that $L$ is stricly decreasing in $\theta$ and as per the fact that $X_{(n)}$ is not included in the domain $L$ has not a maximum.

But the correct definition of ML Estimator is the argsup, not the argmax...so $\hat{\theta}$ can exist and not belonging to $\theta$ domain. It must belong to its euclidean closure.

Concluding, $\hat{\theta}_{ML}=X_{(n)}$ and using invariance property of ML Estimators:

$$\widehat{\theta^2}=(X_{(n)})^2$$


答:
We consider the series, $\sum b_n$ where, $b_n=\begin{cases} a_n, 2|n\\0,2\not | n\end{cases}$  and

$\sum c_n$ where, $c_n=\begin{cases} a_n, 2\not | n\\0,2|n\end{cases}$

Then, $\sum b_n=\sum a_{2n}$ and $\sum c_n=\sum a_{2n-1}$

We have now, $\sum b_n +\sum c_n=\sum (b_n+c_n)=\sum a_n$ which proves the result.

Title: How do you prove $ \sum_{n=1}^{\infty}{a_n} = \sum_{n=1}^{\infty}{a_{2n}} + \sum_{n=1}^{\infty}{a_{2n-1}} $

Body:
How do you prove that by adding two series, both convergent, one
$$ \sum_{n=1}^{\infty}{a_{2n}} $$

and the other

$$ \sum_{n=1}^{\infty}{a_{2n-1}} $$

You get

$$\sum_{n=1}^{\infty}{a_n}$$


问:
Circles $C_1,C_2$ have radii $3$ and $7$ respectively. Point $Q$ is chosen on $C_2$ such that $PQ$ is tangent at $Q$. Find $PQ$.

Circles $C_1,C_2$ have radii $3$ and $7$ respectively. The circles intersect at distinct $A$ and $B$. A point $P$ outside $C_2$ lies on the line determined by $A$ and $B$ at a distance $5$ from center $C_1$. Point $Q$ is chosen on $C_2$ such that $PQ$ is tangent at $Q$. Find $PQ$.

What I Tried: Here is a picture in Geogebra :-

I have no good idea on how to start this. I cannot, for example, find the lengths of $C_1C$ and $CC_2$ from pure deduction, ($C_1$ and $C_2$ are centres of the circles) and even if Pythagorean Theorem may look like they can work, I cannot understand how, because I don't have many required lengths. I also tried similarity but in fact I could not find any similar triangles here.

Can anyone help me? Thank You.

答:
There has be a better way, but here is a solution by bashing Pythagoreas' Theorem. By extracting the information of lengths of radii from the circles, we end up with the following figure (here $c=7$, but the length of $PQ$ is independent of $c$):

By Pythagoreas' Theorem:

$$C_1C = \sqrt{9-a^2}, \quad CC_2 = \sqrt{c^2-a^2}, \quad CP = \sqrt{25 - (9-a^2)} = a+b$$
$$(a+b)^2 = 16+a^2$$
$$PQ = \sqrt{PC_2^2-c^2}=\sqrt{((a+b)^2+(c^2-a^2))-c^2}=\sqrt{(a+b)^2-a^2} = \sqrt{16}=4$$