问:
non-Borel set with Tonelli like property
I am looking for an example of a non-Borel set $C \subseteq \mathbb{R}^2$ such that :

i) $C_x = \{y: (x,y) \in C \} \in \mathcal{B} (R)$ and $C^y= \{ x : (x,y) \in C \} \in \mathcal{B}(R)$ for all $x,y \in \mathbb{R}$

ii) The functions $x \mapsto \lambda (C_x)  $ and $y \mapsto \lambda (C^y) $ are Borel measurables and
$\int \lambda (C_x) d\lambda(x) = \int \lambda(C^y) d \lambda(y)$

I tried the area under the graph of the Cantor function but does not work.

答:
Consider the function $f: \mathbb{R} \rightarrow \mathbb{R}^2$ defined by $f(x)=(x,x)$.
It is immediate that $f$ is continuous and so Borel-measurable.

Let $E\subseteq \mathbb{R}$ be a non-Borel subset of $\mathbb{R}$. Define $C=\{(x,x)\in \mathbb{R}^2 : x \in E \}$.

Claim: $C$ is a non-Borel subset of $\mathbb{R}^2$.

Proof: Suppose $C$ is a Borel  subset of $\mathbb{R}^2$. Since $f$ is Borel-measurable, and $E=f^{-1}(C)$, we have that $E$ is a Borel subset of $\mathbb{R}$. Contradiction. $\square$

Now, note that, for all $x,y \in \mathbb{R}$, $C_x = \{y: (x,y) \in C \} =\{ x \}\in \mathcal{B} (R)$ and $C^y= \{ x : (x,y) \in C \} = \{ y \} \in \mathcal{B}(R)$.

Moreover, The functions $x \mapsto \lambda (C_x)=0  $ and $y \mapsto \lambda (C^y)=0 $ are are Borel measurables and
$$\int \lambda (C_x) d\lambda(x) = 0 = \int \lambda(C^y) d \lambda(y)$$


问:
Proving a function has bounded derivatives
The function in question is:

$ \mathrm{e}^\frac{1}{\left(4x+3\right)^2-1}  $   where $ x∈ (-1, -1/2) $

I've tried graphing the derivatives and it seems that they explode as $ x \rightarrow -1/2 $

答:
Note that
$$f'(x)=\frac{-8(4x+3)}{\left((4x+3)^2-1\right)^2}e^{\frac{1}{(4x+3)^2-1}} $$
and continuing the differentiation, we can see the $n$-th derivative is
$$f^{(n)}(x)=\frac{P_n(x)}{\left((4x+3)^2-1\right)^m}e^{\frac{1}{(4x+3)^2-1}} $$
where $P_n$ is some polynomial and $m$ is a positive integer. So, the $n$-th derivative is continuous in $(-1,-1/2)$ and to show it's bounded, it suffices to show the limits
$$\lim_{x\to (-1)^{+}}f^{(n)}(x),\lim_{x\to (-1/2)^{-}}f^{(n)}(x) $$
exist. Substitute $t=-\frac{1}{(4x+3)^2-1}\to\infty$ as $x\to (-1)^{+}$:
$$\lim_{x\to (-1)^{+}}
\frac{e^{\frac{1}{(4x+3)^2-1}}}{\left((4x+3)^2-1\right)^m}=\lim_{t\to\infty}(-t)^me^{-t}=(-1)^m\lim_{t\to\infty}\frac{t^m}{e^t}=0$$
So the first limt is $P_n(-1)\cdot 0=0$. The second limit is handled in exactly the same way.

ContentLicense: CC BY-SA 4.0


问:
How do the Bruhat cells of based on upper triangular matrices relate to the cells of lower triangular matrices?
Let $B_+$ be the subgroup of $GL(n)$ of upper triangular matrices and let $B_-$ be the subgroup of lower triangular matrices. Let $\pi \in S_n$ and $P_{\pi}$ be the corresponding permutation matrix.

Is there any relation between the sets $B_+P_{\pi}B_+$ and $B_-P_{\pi}B_-$? Are they equal to eachother? Disjoint? Thanks in advance

答: Since any Borel ($B_+$ or $B_-$) contains the maximal torus, the sets you've written down are not disjoint. But by my comment, they do not coincide either (although all Borels are conjugate to each other - this also holds for more general groups beyond $GL_n$).


问:
Confusion in interpreting why $A\subseteq B$ implies $A\cap X\subseteq B\cap X$
I'm working on the beginnings of Set Theory and I need help understand the proof the question below:

Prove that $A ⊆ B$ and $A \cap B^\complement=\emptyset$ are equivalent.

The proof is:

$$ A⊆ B\implies A∩ B^\complement ⊆ B ∩ B^\complement$$
As $B ∩ B^\complement = \emptyset$,
$$A ∩ B^\complement \subseteq \emptyset$$
$$A ∩ B^\complement = \emptyset$$

Therefore, both statements are equivalent.

My confusion lies in the step $A∩ B^\complement ⊆ B ∩ B^\complement$. What is the logic for this? How can we "add" the intersection of $B^\complement$ to each side of the subset?

答: In general, say you have sets $A,B,C$ such that $A \subseteq B$. Then $A \cap C \subseteq B \cap C$. This is because if for arbitrary $x$, $x \in A \cap C$, $x \in C$ and $x \in A \subseteq B$, so $x \in B$ too. Hence $x \in B \cap C$.


问:
Parametrization of a boundary of a closed disk
The  question that I am referring to is:

Let $D$ be the closed disc bounded by the circle with equation

$$(x-6)^{2} + y^{2} = 36$$

I am asked to find the parameterization of the boundary.

I am a bit confused on how to do this. I know that $x^2 + y^2 = 36$ which means $r = 6$.

Since $x = r\cos(\theta)$ and $y = r\sin(\theta)$, how would the $(x-6)^2$ be parameterized?

答:
HINT

In general, if we are given the circumference
\begin{align*}
(x-a)^{2} + (y-b)^{2} = r^{2}
\end{align*}

we can parametrize it as follows

\begin{align*}
\begin{cases}
x = a + r\cos(\theta)\\\\
y = b + r\sin(\theta)
\end{cases}
\end{align*}

where $\theta\in[0,2\pi]$. Can you apply it to the example proposed?


答:
It suffices to study $\exp(1/(x-1))$ on the set $(-1,0)$, or equivalently $f(x)=\exp(\frac{-1}x)$ on the set $(0,1)$. The result for your function then follows by repeated application of chain rule. This function  $f$ actually extends to a bounded and smooth function on $\mathbb R$, but it will be enough for your exercise to prove that
$$ f^{(n)}(x) = \frac{p_n(x)}{x^{2n}} f(x)$$
where $p_1$ is some polynomial of degree $n-1$,
which is proved inductively in many places, for instance this Wikipedia page. This and the fact that $0\le f(x) \le C_n x^{2n}$ for every $n$ gives the boundedness near 0; the function is smooth away from zero so in particular bounded near 1.

Title: Proving a function has bounded derivatives

Body:
The function in question is:

$ \mathrm{e}^\frac{1}{\left(4x+3\right)^2-1}  $   where $ x∈ (-1, -1/2) $

I've tried graphing the derivatives and it seems that they explode as $ x \rightarrow -1/2 $


问:
Eigenvalues of an almost diagonal matrix
I know that the eigenvalue of a diagonal matrix is simply the values in the diagonal. However, if I have a matrix of the following form:

$$
\begin{bmatrix}
a & b & 0 & 0 \\
b & c & 0  & 0 \\
0 & 0 & d & e \\
0 & 0 & e & f
\end{bmatrix}.
$$
Is there a closed form way to express the eigenvalues of this matrix? I can derive the eigenvalue of the smaller blocks along the diagonal, but how does it relate to the overall matrix?

答: If you have a block matrix $$\begin{bmatrix}A&0\\0&B\end{bmatrix},$$ its characteristic polynomial is $p_A(x)p_B(x)$.


答:
The eigenvalues of a block diagonal matrix are the eigenvalues of each block. The corresponding eigenvectors are the eigenvectors of each block padded with zeros. For example:

The eigenvalues of the matrix $$A = \begin{bmatrix}4 & 3 \\ 3 & 4 \end{bmatrix}$$ are $7$ and $1$, and the corresponding eigenvectors are respectively $$\begin{bmatrix}1/\sqrt{2}\\ 1/\sqrt{2} \end{bmatrix} \quad \text{and} \quad \begin{bmatrix}1/\sqrt{2}\\ -1/\sqrt{2} \end{bmatrix}.$$

The eigenvalues of the matrix $$B = \begin{bmatrix}2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2\end{bmatrix}$$ are $2+\sqrt{2}$, $2$, and $2-\sqrt{2}$ and the corresponding eigenvectors are respectively $$\begin{bmatrix}1/2 \\ -1/\sqrt{2} \\ 1/2\end{bmatrix}, \quad \begin{bmatrix}-1/\sqrt{2} \\ 0 \\ 1/\sqrt{2}\end{bmatrix}, \quad \text{and} \quad \begin{bmatrix}1/2 \\ 1/\sqrt{2} \\ 1/2\end{bmatrix}.$$

The eigenvalues of the matrix $$\begin{bmatrix}A & 0 \\ 0 & B \end{bmatrix} = \begin{bmatrix}4 & 3 & 0 & 0 & 0 \\ 3 & 4 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 1 & 2 \end{bmatrix}$$ are $7$, $1$, $2+\sqrt{2}$, $2$, and $2-\sqrt{2}$, and the corresponding eigenvectors are respectively $$\begin{bmatrix}1/\sqrt{2}\\ 1/\sqrt{2} \\ 0 \\ 0 \\ 0\end{bmatrix}, \quad \begin{bmatrix}1/\sqrt{2}\\ -1/\sqrt{2} \\ 0 \\ 0 \\ 0\end{bmatrix}, \quad \begin{bmatrix}0 \\ 0 \\ 1/2 \\ -1/\sqrt{2} \\ 1/2\end{bmatrix}, \quad \begin{bmatrix}0 \\ 0 \\ -1/\sqrt{2} \\ 0 \\ 1/\sqrt{2}\end{bmatrix}, \quad \text{and} \quad \begin{bmatrix}0 \\ 0 \\ 1/2 \\ 1/\sqrt{2} \\ 1/2\end{bmatrix}.$$

Title: Eigenvalues of an almost diagonal matrix

Body:
I know that the eigenvalue of a diagonal matrix is simply the values in the diagonal. However, if I have a matrix of the following form:

$$
\begin{bmatrix}
a & b & 0 & 0 \\
b & c & 0  & 0 \\
0 & 0 & d & e \\
0 & 0 & e & f
\end{bmatrix}.
$$
Is there a closed form way to express the eigenvalues of this matrix? I can derive the eigenvalue of the smaller blocks along the diagonal, but how does it relate to the overall matrix?


问:
uniformizer on elliptic curve in Silverman's book
I'd like to ask remark 1.1 in chapter 2 of Silverman's "Arithmetic of elliptic curves".

Let $K$ be a field and $C$ be a curve with a smooth point $P$, after proving $\bar{K}[C]_P$ is a discrete valuation ring in proposition 1.1, Silverman mentions that if $P\in C(K)$, then $K(C)$ contains uniformizers for $P$. In another word, there are uniformizers that are defined over K at $P$.

I want to know how to prove this and if there are some references of algebraic curves in Silverman's style in chapter II.

Thanks.

答:
The short answer: Let $P = (x_P,y_P)$. Since $P$ is a smooth point on $C$, $\left.\displaystyle \frac{dy}{dx}\right|_P$ is well defined on $C$; note that the value may well be $\infty$. Then either $\left.\displaystyle \frac{dy}{dx}\right|_P = 0$ or $\left.\displaystyle \frac{dy}{dx}\right|_P \neq 0$ (the latter case includes the case $\left.\displaystyle \frac{dy}{dx}\right|_P = \infty$.)

If $\left.\displaystyle \frac{dy}{dx}\right|_P = 0$, then $x-x_P$ is a uniformizer. Otherwise $y-y_P$ is a uniformizer. I won't prove this for you because proving this on your own is a necessary step towards being able to understand the rest of the book, and so I suggest you do so.

To answer your reference request, I recommend Lorenzini, "An Invitation to Arithmetic Geometry."


问:
Find eigenvectors of $A$
Let $A=\begin{pmatrix} 0&0&0&0\\1&0&0&-2\\0&1&0&1\\0&0&1&2 \end{pmatrix}$

I find eignevalues $-1,0,1,2$ and $0$ has multiplicity of $2$.

But I only find one eigenvector of $0$ and all the other eigenvalues have eigenvectors, so anyway $A$ has four eigenvectors and diagonalizable.  But I think $0$ must have $2$ eigenvectors and one of eigenvalue can't have eigenvector. What's wrong with me?

I corrected matrix!!!!!

Then the C.P is $x(x^3-2x^2-x+2)=x(x+1)(x-1)(x-2)$ and the eigenvalues that I got is right.

答: $0$ is not an eigenvalue of multiplicity $2$. This would give you $5$ eigenvalues for a $4 \times 4$ matrix, which is not possible. So you cannot find more than $1$ eigenvector for the eigenvalue $0$.</p>

ContentLicense: CC BY-SA 3.0

Title: Find eigenvectors of $A$


问:
How to calculate the number of connecting lines amongst corners given the number of edges
Example:

A shape with 4 edges has 2 connecting lines in the middle.

A shape with 5 edges has 5 connecting lines in the middle.

A shape with 6 edges has 9 connecting lines in the middle.

I can see the increment is

4->5 (2 lines to 5 lines; difference of 3)

5->6 (5 lines to 9 lines difference of 4).

It seems to increment that way. What is the mathematical property of this and what is the formula to calculate this given any number of edges?

答:
This is existentially consequential. My interest is that universal peer-to-peer civic discourse becomes intractably complex as population size increases. Ancient Athens could have a democracy in which each citizen's voice could be heard in the agora, whereas in a country with a population of $3. 3 \cdot 10 ^ 8$, citizens can only vote with the voice of each asymptotically approaching zero.

The number of peer communication paths basically is 1/2 the square of the number of communicators. For asymmetrical communication leader and follower communication, the number of communication paths is the same as the number of members of the the group, which is a much smaller number, and, in particular, for each communicatee, it is 1.

Large groups are mathematically inconsistent with peer-to-peer civic discourse. No town hall can realistically accommodate $(3.3 \cdot 10 ^ 8) ^ 2 / 2$ communication paths. Is there even a networking super-computer that can multiplex that many communication lines?


问:
Problem about a recursive sequence $a_n = \sqrt{\frac{a_{n-1}}{a_{n-2}}}$ with initial conditions $a_0=8\text{ and }a_1=\frac{1}{2\sqrt{2}}$
I am trying to solve the recurrence relation
$$a_n = \sqrt{\frac{a_{n-1}}{a_{n-2}}}$$ with initial conditions
$$a_0=8\text{ and }a_1=\frac{1}{2\sqrt{2}}$$

Can anyone please help me with the technique to approach this problem?

答:
Take logs and it becomes
$b_n = \dfrac{b_{n-1}+b_{n-2}}{2}
$.
This has been investigated here.

Note that
$a_0a_1^2 = 1$.
That is not a coincidence.


问:
upper bound for $\frac{x+y}{ax+y}$
I need an upper bound for
$$\frac{x+y}{ax+y}$$

I know that $$1\leq a< 2$$
$x\geq 0 $ and $y\geq 0 $ . This upper bound can include just $a$ and constant numbers not $x$ or $y$.

thanks a lot.

答: Hint: If $x=0$, then $\frac{x+y}{ax+y}=1$. Suppose $x\neq0$. Let $t=\frac{y}{x}$. Then $\frac{x+y}{ax+y}=\frac{t+1}{a+t}=1-\frac{a-1}{a+t}$.</p>

Title: upper bound for $\frac{x+y}{ax+y}$


问:
How do I find the volume of $\cos x$ about the $y$-axis?
Evaluate the volume formed by rotating the region bounded by $y=0$ and $y=\cos x$ for $\frac{\pi}{2}\le x\le\frac{3\pi}{2}$ about the y-axis.

I tried making $x=\cos^{-1}y$ then solving for $$\int_{{\frac{\pi}{2}}}^{\frac{3\pi}{2}}\cos^{-1}y\ dy$$ but clearly that is undefined.
How do I solve this?

答:
One way would be to use cylindrical shells:  $-2\pi\int_{\pi/2}^{3\pi/2}x\cos x\rm dx$, where the minus sign is because $\cos$ is negative on $(π/2,3π/2)$.

The integral can be solved by integration by parts: $ x\sin x=\int \sin x\rm dx+\int x\cos x\rm dx$, so $\int x\cos x \rm dx=x\sin x+\cos x+C$.


问:
I don't understand why the partial derivative of $ F $ with respect to $v'$ is like this, does $y'$ depend on $v'$?
I don't understand why the partial derivative of $ F $ with respect to $v'$ is like this, does $y'$ depend on $v'$?

Image attached:

答:
It is a notation from Physics.

Consider that $F$ is a function of three variables $F(u,v, \dot{v})$. (Do not interpret $\dot{v}$ as related to $v$).

Consider that $f$ is a function of three variables $f(x,y, y')$. (Do not interpret $y'$ as related to $y$).

Now, given the formula:

\begin{align*}
F(u,v, \dot{v})&= f\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)(x_u+x_v\dot{v}) =\\
&=(x_u+x_v\dot{v})f\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)
\end{align*}

We differentiate it, still considering that $\dot{v}$ is an independent variable. We have

\begin{align*}
\frac{\partial F}{\partial \dot{v}}(u,v, \dot{v})&= x_v f\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)+ \\
&+(x_u+x_v\dot{v})\frac{\partial f}{\partial y'}\left(x(u,v),y(u,v), \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)\frac{\partial}{\partial\dot{v}}\left ( \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)
\end{align*}
Or writing in a simpler way
$$\frac{\partial F}{\partial \dot{v}}= x_v f + (x_u+x_v\dot{v})\frac{\partial f}{\partial y'}\frac{\partial}{\partial\dot{v}}\left ( \frac{y_u+y_v\dot{v}}{x_u+x_v\dot{v}} \right)$$


问:
Rate of convergence of roots
Suppose I have a sequence of smooth functions $f_n$ with unique roots $x_n$, such that $f_n$ converges pointwise to a smooth function $f$ with unique root $x$. Also assume that $x_n\to x$. Is there a way to describe the rate of convergence of $x_n$ to $x$, possibly in terms of derivatives of $f_n$ and $f$?

答:
Instead of keeping things general, start with a simple example.

Let the sequence of smooth functions $(f_n)_{n\in \mathbb{N}}$ be defined by $f_n(x) =  (x-\frac{1}{n})(x + \frac{1}{n})(x-\frac{2}{n})(x+\frac{2}{n})$. This sequence has a limit $f(x) = x^4$, which has a unique root $x^{\ast}=0$. This adheres to the parameters of your question ($f_n$ has multiple roots and all of them converge to a single root of $f$ as $f_n$ converges to $f$).

The sequence $f_n$ has $4$ roots $ \frac{1}{n},-\frac{1}{n},\frac{2}{n},-\frac{2}{n}$. Taking derivatives of $f_n$ gives $f_n^{\prime}(x) = 4x^3 - \frac{10}{n^2}x$, $f_n^{\prime \prime}(x) = 12x^2 - \frac{10}{n^2}$, $f_n^{\prime \prime \prime}(x) = 24x$.

In this example, can you explain the rate at which the roots converge depending on the rate at which the derivatives of the sequences converge?

Title: Rate of convergence of roots


问:
A polyomial problem to show that Q (x)/$x^k $ is strictly positive

I am extremely sorry I couldnot type as it requires much time to do MathJax.

This problem Question 8 was asked in an entrance examination at the 10+2 level.
My approqch for part (a) was  to use differentiation and find Q'(x) but it is not so ....helping me to arrive at the result.
Any hints or help is appreciable.

Thank you

答:
(Fill in the gaps as needed. If you're stuck, show what you've considered and explain why you're stuck.)

Hint: Show that $ |x^k | > |\sum a_i x^{k+i}|$.


问:
Difference in definition of limit points and cluster points of a set
I was being taught the Limit points and Cluster points by my teacher. He mentioned that there's a difference in between them as stated in S. Kumaresan's Topology of Metric Spaces:

Taking the metric space as the set of Real numbers,

Limit points: Let $A\subseteq\mathbb{R}$. A point x $\in\mathbb{R}$ is a limit point of $A$ iff $\forall\epsilon>0, \left(x-\epsilon,x+\epsilon\right)\cap A\neq\phi$

whereas

Cluster points: A point $c$ is said to be the cluster point of a set $A\left(\subseteq\mathbb{R}\right)$ if $\forall\epsilon>0$, $\left(c-\epsilon,c+\epsilon\right)\cap A-\{c\}\neq\phi$.

A simple example would explain the difference between these two:

Let $A=\{1,2,3,4,5\}$. Then according to these definitions of limit and cluster points, the set of all the limit points of $A$ is $\{1,2,3,4,5\}$. But, there is no cluster point of this set.

But in Rudin, the limit point is defined as:

Limit points: A point $x$ is said to be a limit point of a set $A\left(\subseteq\mathbb{R}\right)$ if $\forall\epsilon>0$, $\left(x-\epsilon,x+\epsilon\right)\cap A-\{x\}\neq\phi$.

which is the same as the definition of cluster points in S. Kumaresan's book.

So, going by this definition, we find that there is no limit or cluster points of the set $A$ in my example.

I'm a little curious to know if there is any difference in the set of cluster points and limit points? And if so, then why is it not mentioned in Rudin?

(Note: The above definitions are not the exact words of the books though. The original definitions took any metric space instead of only the set of Real numbers)

There's a possible similar question posted here but I couldn't get it completely.

答: Being honest I always get confused by the names given to these kind of points (especially since the terms are not universal), but what there is no doubt is that for $A \subseteq \mathbb R$ we have $$\overline{A} = \{x \in \mathbb R : (x-\varepsilon,x+\varepsilon) \cap A \neq \varnothing \textrm{ for all } \varepsilon>0\}$$ and that $$A' = \{x \in \mathbb R : (x-\varepsilon,x+\varepsilon) \cap (A\setminus \{x\}) \neq \varnothing \textrm{ for all } \varepsilon>0\}.$$ Note that $x \in A'$ if and only if $x \in \overline{A \setminus \{x\}}$, and that $\overline{A} = A \cup A'$. So, if $A$ is a finite set of points, $A' = \varnothing$ and then $\overline A = A$.


答:
Hint:

$\dfrac{x+y}{ax+y}=\dfrac{ax+y-(a-1)x}{ax+y}=1-\dfrac{(a-1)x}{ax+y}$

What's the minimum value of $\dfrac{(a-1)x}{ax+y}$?

Since $x,y \ge 0$ and $a \ge 1$, minimum value of the $\dfrac{(a-1)x}{ax+y}$ is $0$.

Title: upper bound for $\frac{x+y}{ax+y}$

Body:
I need an upper bound for
$$\frac{x+y}{ax+y}$$

I know that $$1\leq a< 2$$
$x\geq 0 $ and $y\geq 0 $ . This upper bound can include just $a$ and constant numbers not $x$ or $y$.

thanks a lot.


问:
How does one evaluate $\int \frac{\sin(x)}{\sin(5x)} \ dx$

The below problem is taken from Joseph Edwards book Integral Calculus for beginners.

How does one show:
$$5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\} $$

Splitting $\sin{(5x)}$ as $\sin{(4x+x)}$ doesn't seem to be of much help since then we have a big term in the denominator after expansion.

答:
Note

\begin{align}
&\frac{\sin x}{\sin5x}=\frac1{4\left(\cos2x- \cos\frac{2\pi}{5}\right) \left(\cos2x- \cos \frac{4\pi}{5}\right)}\\=&\ \frac1{2\sqrt5}\bigg(\frac1{\cos2x- \cos\frac{2\pi}{5} }
-\frac1{\cos2x- \cos\frac{4\pi}{5} }
\bigg)\\
=&\ \frac1{4\sqrt5}\bigg(\frac1{\sin(\frac\pi5+x)\sin(\frac\pi5-x) }-\frac1{\sin(\frac{2\pi}5+x)\sin(\frac{2\pi}5-x) }
\bigg)\\
=&\ \frac15\sin\frac\pi5\left(\cot(\frac\pi5+x)+\cot(\frac\pi5-x) \right)-\frac15\sin\frac{2\pi}5\left(\cot(\frac{2\pi}5+x)+\cot(\frac{2\pi}5-x) \right)
\end{align}
Then, apply $(\ln \sin t)’= \cot t$ to integrate
$$\int \frac{\sin x}{\sin5x}= \frac15\sin\frac{2\pi}5\>\ln \frac{\sin(\frac{2\pi}5-x) }{\sin(\frac{2\pi}5+x) } - \frac15\sin\frac\pi5\>\ln \frac{\sin(\frac\pi5-x) }{\sin(\frac\pi5+x) }+C
$$


问:
Taylor expansion of $\log(z^2-3z+2)$ at z = 0
Find the Taylor series of the function $$f(z) = \log(z^2-3z+2) \quad \forall z \in \mathbb{C} \backslash \left\{1,2 \right\} $$ at $z = 0$ and calculate the radius of convergence for the resultant series.

Body:
$$f(z) = \log(z^2-3z+2) \implies f'(z)=\frac{2 z-3}{z^2-3 z+2}=\frac{1}{z-1}+\frac{1}{z-2}$$
So, making the story short,
$$f'(z)=-\sum_{n=0}^\infty (1+2^{-(n+1)})\,z^n$$ Integrate termwise
$$f(z)=-\sum_{n=1}^\infty \frac{\left(1+2^{-(n+1)}\right) }{n+1} z^{n+1}+K=-\sum_{n=0}^\infty \frac{1+2^{-n}}{n}z^n+K$$ and since $f(0)=\log(2)$ then $K=\log(2)$.


答:
I don't know if it could be the correct method, but I show you what I was thinking of.
So, imagine to plug arbitrary numbers inside the inequality. You'd notice that the sum (with sigma symbol) is always positive (because it starts with 1 and approaches to positive infinity). Additionally, it has an absolute value, so it is surely always positive.
the fraction is always positive too, because it has "1+[always positive stuff]". It approaches to 0 from the right (to prove that, imagine to plug every positive or negative number and see what would happen). Therefore, it's always positive. zero isn't included because of the sign of the inequality "i.e <, which means "is less than and not equal to [something]".

Moreover, x is inside the absolute value too, so you could surely say Q(X) is always positive.

EDIT (because I forgot the main part of this exercise): imagine to plug arbitrary values inside "k" variable (because k is a variable), you'd notice that it doesn't matter what you plug into it, because you'd get a positive value.
secondly, if x is negative, it's not a problem because in the inequality you have the absolute values.

Title: A polyomial problem to show that Q (x)/$x^k $ is strictly positive

Body:
I am extremely sorry I couldnot type as it requires much time to do MathJax.

This problem Question 8 was asked in an entrance examination at the 10+2 level.
My approqch for part (a) was  to use differentiation and find Q'(x) but it is not so ....helping me to arrive at the result.
Any hints or help is appreciable.

Thank you


问:
Acturial Studies: what is the formula for compound discount with simpe discount over the final fractional period?
What is the formula for compound discount with simple discount over the fractional period. I could not find the formula in my book and when I looked online the ones that had the formula where answers to a question in my book which I did not want to see the answer in order to get the formula. The formulas that I know are: Note that $d$ stands for the discount rate.

$a(t)^{-1}=(1-d)^t$ for compound discount

$a(t)^{-1}=(1-dt)$

$d_{m}=\frac{i_{m}}{1+i_{m}}$

I am wondering why no one has answered am I asking a question that does not make sense?

Edit: my question is in response to the following:

Method A assumes simple interest over final fractional periods, while Method B assumes simple discount over final fractional periods. The annual effective rate of interest is 20%. Find the ratio of the present value of a payment made in $1.5$ years computed under Method A to that computed under Method B

答:
Let d be the discount rate, t be the total time of investment, and k be the largest integer such that t > k .  First we calculate the compound interest over time k, then multiply that by the simple interest for time t-k.  Thus your accumulation function is:

$ a(t) = (1 -d)^{-k} (1-d(t-k))  $

Similarly, the accumulation function for compound interest with interest rate i :

$ a(t) = (1 + i)^k (1 + it) $

Bonus points if you don't forget to take $ (a(t))^{-1} $ for present value in your question like I did.


问:
Small question about derivative
how to derive $\int_0^1 G(t,s) e(s)ds$ with respect to $t$

Where $G(t,s)$ is a Green function and $e:(0,1)\rightarrow \mathbb{R}$ continuous and $e\in L(0,1)$

Please help me

Thank you

答: You can commute the derivative and the integral operators in this case. See here.</p>

Title: Small question about derivative


问:
Notation for dot product vs multiplication
Is there a difference in the dot (position, size, etc) used in the vector dot product vs the one use for multiplication?

For typesetting here we use \cdot for both, i.e. $$2\cdot 3$$ or $$\mathbf u\cdot \mathbf v$$ Would that that be correct or is there an more proper alternative?

For instance, in addition to \cdot ($\cdot$) there is also \bullet ($\bullet$).

答:
Note that typical multiplication is really a special case of dot product in one dimension. In other words, if $x = (x_k)_{k=1}^n$ and $y = (y_k)_{k=1}^n$ with $x,y \in \mathbb{R}^n$, then
$x \cdot y = \sum_{k=1}^n x_k y_k$, so if we are doing this in 1D, we have $x\cdot y = x_1y_1$.

This will explain why the notation is the same.


问:
Correlated Implies Non-Zero Conditional Expectation?
Suppose $X$ and $Y$ are correlated, and that they each have mean zero. As such, $E(XY) \neq 0$.

Is it true that $E(X|Y) \neq 0?$

The way I define the conditional expectation $E(X|Y)$ is the function $h(Y)$ such that
$$
E\left((X - h(Y)) \cdot g(Y)\right) = 0 \text{ for all functions } g(\cdot).
$$

I'm unable to get anywhere with this, even though it seems like a pretty simple question. Can anyone lend a hand?

答:
Hint:

In your definition of $\ E(X|Y)\ $ take $\ g(Y)=Y\ $, giving $\ E\big(h(Y)Y\big)=$$E(XY)\ne0\ $.  Is this possible if $\ h(Y)=0\ $?


问:
Solving $\lim\limits_{n\to\infty}n\Big[\Big(1+\frac{1}{n}\Big)^{m}-1\Big]$ without l'Hôpital.
Using l'Hôpital's rule we can easily see the limit is $m$. I'm wondering whether there is any other way of solving the limit withOUT using l'Hôpital's.

EDIT: I'm avoiding l'Hôpital because I encountered the limit in a point of the book where the author has not yet introduced that topic, and I'm trying to figure out what he had in mind for solving this limit. So far the chapters have been an introductory one (properties of real numbers, set theory, etc...), a chapter for sequences, and series. I'm solving the limit as part of the Raabe criterion for series.

答:
Expanding $(1+\frac{1}{n})^m$ using binomial theorem we get the limit.
\begin{align*}
\lim_{n\rightarrow \infty}n((1+1/n)^m-1) = \lim_{n\rightarrow \infty}n\biggl(\sum_{i=1}^m {m \choose i}\frac{1}{n^i}\biggr)=m
\end{align*}


问:
Line of Intersection of Planes (Closest Point)
I found  the line of intersection of  two planes and found the parametric equations for that line of intersection.

But how do I find a point on that line of intersection that is closest to a certain point that is given to me  in the problem?

For example my parametric equations I found for the line of intersection of the planes, 2x + 10y + 2z= -2 and 4x + 2y - 5z = -4 are

x=-2-6t
y=2t
z=-4t

and I need to find a point one the line of intersection that is closest to point (12,14,0).

Would I need to use lagrange multipliers?

答:
No, you wouldn't need to use Lagrange multipliers.

Find $t$ that minimizes the square of the distance from $(-2-6t,2t,-4t)$ to $(12,14,0)$,

which is $(-2-6t-12)^2+(2t-14)^2+(-4t)^2=56(t+1)^2+336$,

which is minimized when $t=-1$.


答:
HINT:

Note that use of Bernoulli's Inequality reveals for $n>1$

$$\frac mn\le \left(1+\frac1n\right)^m-1\le \frac{m/n}{1-\frac mn}$$

And you can finish now.

Title: Solving $\lim\limits_{n\to\infty}n\Big[\Big(1+\frac{1}{n}\Big)^{m}-1\Big]$ without l'Hôpital.

Body:
Using l'Hôpital's rule we can easily see the limit is $m$. I'm wondering whether there is any other way of solving the limit withOUT using l'Hôpital's.

EDIT: I'm avoiding l'Hôpital because I encountered the limit in a point of the book where the author has not yet introduced that topic, and I'm trying to figure out what he had in mind for solving this limit. So far the chapters have been an introductory one (properties of real numbers, set theory, etc...), a chapter for sequences, and series. I'm solving the limit as part of the Raabe criterion for series.


问:
Does the Prufer p-group contain direct sums of cyclic groups?
Let $G$ be the Prufer $p$-group.  Then my question is, does $G$ contain subgroups which are isomorphic to a finite or countable direct sum of finite cyclic groups?

Of course, some direct sums of cyclic groups are themselves cyclic groups, but I'm interested direct sums of cyclic groups which are not cyclic.

答: No. As stated on Wikipedia, every proper subgroup of the Prufer $p$-group is finite cyclic. The Prufer $p$-group itself also is not a nontrivial direct sum (of any abelian groups whatsoever), because it has no nontrivial idempotent endomorphisms (its endomorphism ring is the $p$-adic integers).


答:
Do a variable substitution:

$x=1+\frac{1}{n}$

$n=\frac{1}{x-1}$

So the limit becomes

$$\lim\limits_{x\to1^+}\frac{x^m-1}{x-1}$$

$$\lim\limits_{x\to1^+}(x^{m-1}+x^{m-2}+...1)$$

$$=m$$


问:
Diagonalize a quasi-uniform tridiagonal matrix
How to diagonalize the following matrix?

$$
A=\begin{pmatrix}
0  & a& 0 & 0 & 0 & \cdots \\
a& 0  & b & 0 & 0 & \cdots \\
0 & b & 0  & b & 0 & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\cdots & 0 & 0 & b & 0  & b \\
\cdots & 0 & 0 & 0 & b & 0  \\
\end{pmatrix}
$$

答:
Let me write in answer box. Apply the row operations $[(-b/a)R_1+R_3 \to R_3]$, $[-R_3+R_5 \to R_5]$, ..., $[-R_{2k+1}+R_{2k+3} \to R_{2k+3}]$ Denote this operations by the matrices $C_i$ and their corresponding columns operations are denoted by $C_i^T$. Then (let the matrix in question is $A$)

$C_sC_{s-1}...C_1AC_1^T...C_s^T$ is the following matrix

\begin{bmatrix}
A &  & & & & & \\
 & 0 &  &  &  &  &\\
 &  & B & & & & \\
 &  & & 0 & & & \\
 &  & & & B &  & \\
 &  & & & & 0 & \\
 &  & & & & & B \\
&  & & & & &  & . \\
&  & & & & & & & .
\end{bmatrix}

where $A=\begin{bmatrix}
0 & a \\
a & 0
\end{bmatrix}$, and $B=\begin{bmatrix}
0 & b \\
b & 0
\end{bmatrix}$ and $0$ is $2 \times 2$ zero matrix. Then apply the following row operations

$R_{4k+2}+R_{4k+1}=R_{4k+1}$ and their corresponding column operations $C_{4k+2}+C_{4k+1}=C_{4k+1}$simultaneously. Let these row operations denoted by $E_k$. Then

$E_t...E_1(C_sC_{s-1}...C_1AC_1^T...C_s^T)E_1^T...E_t^T$ is the following matrix

\begin{bmatrix}
A &  & & & & & \\
 & 0 &  &  &  &  &\\
 &  & B & & & & \\
 &  & & 0 & & & \\
 &  & & & B &  & \\
 &  & & & & 0 & \\
 &  & & & & & B \\
&  & & & & &  & . \\
&  & & & & & & & .
\end{bmatrix}

where

where $A=\begin{bmatrix}
2a & a \\
a & 0
\end{bmatrix}$, and $B=\begin{bmatrix}
2b & b \\
b & 0
\end{bmatrix}$ and $0$ is $2 \times 2$ zero matrix. Then apply the following row operations

$(-1/2)R_{4k+1}+R_{4k+2}=R_{4k+2}$ and their corresponding column operations simultaneously. Let these row operations denoted by $F_k$. Then

$F_t^T...F_1^T(E_t...E_1C_sC_{s-1}...C_1AC_1^T...C_s^TE_1^T...E_t^T)F_1^T...F_t^T$ is the diagonal matrix whose diagonal entries are $2a,-a/2,0,0,2b,-b/2,0,0,2b,-b/2,0,0,...$


问:
coarser/finer relation for relative topology and norm topology
Suppose $p\in [1, \infty]$. Show the relative topology on $l^p(\mathbb{R})$ induced from the product topology on $\mathbb{R}^{\omega}= \Pi_{l=1}^{\infty} \mathbb{R}$ is coarser than the $p$-norm topology on $l^p(\mathbb{R})$, but the relative topology on $l^p(\mathbb{R})$induced by the box topology on $\mathbb{R}^{\omega}$ is finer than the $p$-norm topology on $l^p(\mathbb{R})$.

Attempt:
I know that any basic open set $U$ in the relative product space is an union of the form $\Pi_{i=1}^{\infty} O_i$, where $O_i \in \mathbb{R}$. So it's like a box. And it's clear geometrically that we can always find a $p$-ball $B(x,r)$ such that it's contained in $U$ for every $x\in U$. But I'm not sure how to construct such $r$.

Thanks!

答:
A basic open set containing a point $x_0$ in the product topology is of the form $\{x: |p_1(x)-p_1(x_0)| <r_1, ...., |p_N(x)-p_N(x_0)|<r_N\}$ where $p_1,p_2,..$ are the projection maps. If $r=\min \{r_1,r_2,..,r_N\}$ then the ball of radius $r$ in $\ell^{p}$ around $x_0$ is contained in this basic neighborhood. This proves the first part.

Consider any ball $B(x_0,r)$ in $\ell^{p}$. $\{x: |p_i(x)-p_i(x_0)|<\frac r {2^{i/p}} \forall i\}$ is an open set containing $x_0$ in the Box topology which is contained in $B(x_0,r)$ .


问:
1D Wave Equation Problem Separation of Variables
I need to solve the following 1D Wave Equation problem using Separation of Variables, but I cannot figure it out.
\begin{align}
u_{tt} &= u_{xx}\\
u_x(t,0) &= u_x(t,1) = 0\\
u(0,x) &= x(1-x)\\
u_t(0,x) &= 0
\end{align}
I have done the following work to solve it, but do not know how to find the values of $D$, $A$, $B$, $d_k$, or $b_k$.

Assume $u(t,x) = w(t)v(x) \longrightarrow w''(t) = \lambda w(t), v''(x) =\lambda v(x)$

Case 1: $\lambda = 0$
\begin{align}
v(x) &= Cx+D\\
v'(x) &= C\\
v'(0) &= v'(1) = C = 0
\end{align}
$D$ can be anything.

\begin{align}
u(t,x) &= D(At+B) +\sum_{k=1}^{\infty}{d_k\cos(k\pi t)\sin(k\pi x) + b_k \sin(k\pi t)\sin(k\pi x)}\\
u(0,x) &= x(1-x) = DB+\sum_{k=1}^{\infty}{d_k\sin(k\pi x)}\\
u_t(0,x) &= 0 = DA+\sum_{k=1}^{\infty}{b_kk\pi\sin(k\pi x)}
\end{align}
This is as far as I have gotten and I cannot seem to be able to find $D$, $A$, $B$, $d_k$, or $b_k$. Any help is greatly appreciated.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\on{u}_{x}\pars{t,x} =
-\pi\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\,n\sin\pars{n\pi x}}$ already satisfies the boundary conditions.

\begin{align}
&\implies \on{u}\pars{t,x} =
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}\cos\pars{n\pi x} + \on{f}\pars{t}\label{1}\tag{1}
\\[5mm] &
\mbox{Note that}
\\ &\
\left.\begin{array}{rcl}
\ds{\int_{0}^{1}\cos\pars{m\pi x}\cos\pars{n\pi x}
\,\dd x} & \ds{=} &
\ds{{1 \over 2}\,\delta_{mn}}
\\
\ds{\int_{0}^{1}\cos\pars{n\pi x}\,\dd x} & \ds{=} &
\ds{\delta_{n0}}
\end{array}\right\}\,,\qquad
m, n \in \mathbb{N}_{\,\geq\ 0}
\label{2}\tag{2}
\end{align}

$\ds{\on{u}\pars{t,x}}$ must satisfies the above differential equation:
\begin{align}
&\sum_{n = 1}^{\infty}
\ddot{\on{a}}_{n}\pars{t}\cos\pars{n\pi x} + \ddot{\on{f}}\pars{t}
\\[2mm] = &\
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{t}
\pars{-n^{2}\pi^{2}}\cos\pars{n\pi x}
\label{3}\tag{3}
\end{align}

Integrate -see (\ref{2})- both members of (\ref{3}) over
$\ds{x \in \pars{0,1}}$:
$$
\implies\ddot{\on{f}}\pars{t} = 0\implies
\on{f}\pars{t} = bt + c\,,\quad b, c = \mbox{constants}
$$

With (\ref{2}) and (\ref{3}):
\begin{align}
&\ddot{\on{a}}_{n}\pars{t} = -n^{2}\pi^{2}\on{a}\pars{t}
\\[2mm] &
\implies
\on{a}_{n}\pars{t} = \on{a}_{n}\pars{0}\cos\pars{n\pi t} +
\dot{\on{a}}_{n}\pars{0}\,{\sin\pars{n\pi t} \over n\pi}
\end{align}

$\ds{\on{u}\pars{t,x}}$ is reduced to
\begin{align}
\on{u}\pars{t,x} & =
\sum_{n = 1}^{\infty}\bracks{\on{a}_{n}\pars{0}\cos\pars{n\pi t} +
\dot{\on{a}}_{n}\pars{0}\,{\sin\pars{n\pi t} \over n\pi}}
\\[2mm] & \cos\pars{n\pi x} + bt + c
\\[5mm]
\on{u}_{t}\pars{t,x} & =
\sum_{n = 1}^{\infty}\bracks{%
-n\pi\on{a}_{n}\pars{0}\sin\pars{n\pi t} +
\dot{\on{a}}_{n}\pars{0}\cos\pars{n\pi t}}
\\[2mm] & \cos\pars{n\pi x} + b
\end{align}

\begin{align}
x\pars{1 - x} = \on{u}\pars{0,x} & =
\sum_{n = 1}^{\infty}\on{a}_{n}\pars{0}
\cos\pars{n\pi x} + c\label{4}\tag{4}
\\[5mm]
0 = \on{u}_{t}\pars{0,x} & =
\sum_{n = 1}^{\infty}\dot{\on{a}}_{n}\pars{0}\cos\pars{n\pi x} + b
\end{align}

With the last equation and (\ref{2}), I found
$\ds{\dot{\on{a}}_{n}\pars{0} = 0}$ and $\ds{b = 0}$.

Integrate both members of (\ref{4}) over $\ds{x \in \pars{0,1}}$:
$$
\int_{0}^{1}x\pars{1 - x}\dd x = \int_{0}^{1}c\,\dd x
\implies c = {1 \over 6}
$$

With (\ref{2}) and (\ref{4}):
$$
\underbrace{2\int_{0}^{1}x\pars{1 - x}\cos\pars{n\pi x}
\,\dd x}
_{\ds{-\,\bracks{1 + \pars{-1}^{n}}{2 \over n^{2}\pi^{2}}}} = \on{a}_{n}\pars{0}
$$

Finally,
$$
\bbx{\on{u}\pars{t,x} =
{1 \over 6} - {1 \over \pi^{2}}
\sum_{n = 1}^{\infty}
{\cos\pars{2n\pi t}\cos\pars{2n\pi x} \over n^{2}}} \\
$$


问:
Generallizing independent weighings solution for coin weighing puzzle
There is a puzzle that goes something like "you have $n$ coins with $m$ uses of s balance scale allowed; find out coin with the different weight" (normally given with $n=12$ and $m=3$).

Normally I see the solution given in a dependent form (ie. weight coins 1-4 against 5-8, if they balance do this, else do that). I understand how to generalize this dependent form for any $n$ such that I can minimize $m$.

However, I recently came upon an independent solution to this problem for $n=12$ and $m=3$. It relied on using the weightings as bits in a ternary number that would then identify the final coin. I copied the Wikipedia solution at the bottom.

I, however, don't understand how to generalize this independent solution (or really fully understand this one). How would one generalize this?

```<code>\// A light    /\\ A heavy
/\/ B light    \/\ B heavy
//\ C light    \\/ C heavy

\/- D light    /\- D heavy
-\/ E light    -/\ E heavy
/-\ F light    \-/ F heavy

\\- G light    //- G heavy
-\\ H light    -// H heavy
\-\ I light    /-/ I heavy

/-- J light    \-- J heavy
-/- K light    -\- K heavy
--/ L light    --\ L heavy

--- M either lighter or heavier (13-coin case),
    or all coins weigh the same (12-coin case)

1st weighing:  left side: ADGI, right side: BCFJ
2nd weighing:  left side: BEGH, right side: ACDK
3rd weighing:  left side: CFHI, right side: ABEL
</code>```

答:
First, you need to figure out the optimal number of weighings as a function of $n$. For any $k\in \mathbb N$, let
$$
n(k)=\tfrac12(3^k-1)
$$
It turns out that $k$ weighings are necessary and sufficient to find the fake among $n(k)$ coins. Conversely, given $n$ coins, then $k$ weighings are required, where $k$ is the smallest integer for which $n(k)\ge n$. From now on, $k$ will denote this smallest number of necessary weighings for the current $n$ under consideration.

A strategy for finding the fake among $n$ coins with $k$ weighings can be thought of as a $k\times n$ matrix whose entries are either $+1,-1$, or $0$. Each row represents a weighing, and each column represents a coin. During each weighing, the coins corresponding to $+1$ are weighed against those corresponding to $-1$, with the $0$ coins left off the scale. This weighing strategy must satisfy these two conditions, which are also sufficient:

The sum of each row is zero, meaning an equal number of coins are in each pan for each weighing.

For any two columns, $v$ and $w$, $v\neq w$ and $v\neq -w$. Indeed, if $v=w$, the strategy could not distinguish the case where $v$'s coin is fake and heavy from the case where $w$ is fake and heavy. Similarly, if $v=-w$, you cannot distinguish $v$ being heavy from $w$ being light.

For each $n\ge 3$, I will show how to construct a valid $k\times n$, weighing matrix, $A_n$.

First, let us deal with the special case when $n$ is of the form $n(k)-1$ for some $k\in \mathbb N$, i.e. $n\in \{3,12,39,\dots\}$. The base case is
$$
A_{n(2)-1} = A_3=\begin{bmatrix}+&-&0\\0&+&-\end{bmatrix}
$$
Then, $A_{n(k)-1}$ is constructed recursively as follows:
$$
A_{n(k)-1}=\left[\begin{array}{c|c|c|c|c|c}
 +&-&0&+\cdots +   &-\cdots - &0\cdots 0
\\\hline
\begin{array}{c}0\\\vdots\\0\end{array} &
\begin{array}{c}+\\\vdots\\+\end{array} &
\begin{array}{c}-\\\vdots\\-\end{array} &
A_{n(k-1)-1} & A_{n(k-1)-1} & A_{n(k-1)-1}
\end{array}\right]
$$
You can prove by induction on $k$ that $A_{n(k)-1}$ is a valid weighing matrix, with rows summing to zero and columns pairwise distinct from their selves and their negatives. Furthermore, you can prove that $A_{n(k)-1}$ has no columns of all zeroes.

For example, here is the construction of $A_{12}=A_{n(3)-1}$ from the base case $A_3$:
$$
A_{n(3)-1}=\begin{bmatrix}
+&-&0&+&+&+&-&-&-&0&0&0\\
0&+&-&
\color{red}+&\color{red}-&\color{red}0&
\color{green}+&\color{green}-&\color{green}0&
\color{blue}+&\color{blue}-&\color{blue}0
\\
0&+&-&
\color{red}0&\color{red}+&\color{red}-&
\color{green}0&\color{green}+&\color{green}-&
\color{blue}0&\color{blue}+&\color{blue}-
\end{bmatrix}
$$

Next, we construct $A_{n(k)}$, for each $k\ge 2$. Since $A_{n(k)-1}$ has no columns of zeroes, we can construct $A_{n(k)}$ by appending a column of zeroes to $A_{n(k)-1}$.

Finally, for any other $n$ not yet covered before, we construct $A_n$ as follows. Every integer $n$ can be written in the form $n=2n'+n''$, where $n'$ and $n''$ are integers whose absolute difference is at most one (e.g, $14 = 2\cdot 5+4$). Then
$$
A_n = \left[\begin{array}{c|c|c}
++\cdots + & --\cdots -& 00\cdots 0
\\\hline &&\\
A_{n'}&A_{n'}&A_{n''}
\\ &&
\end{array}\right]
$$There is one small caveat; when $n$ is of the form $n(k)+1$, e.g. $n\in \{14,41,122,\dots\}$, then matrices $A_{n'}$ and $A_{n''}$ have different numbers of rows. This is fixed by appending a row of zeroes to $A_{n''}$.

For this induction to work, you need base cases $A_{3},A_{4},\dots,A_{8}$. We already found $A_3$, and $A_4$ is obtained from $A_3$ by adding a column of zeroes. I leave it as a last puzzle to the reader to find valid matrices $A_n$ for $n\in \{5,6,7,8\}$. These will have $3$ rows and $n$ columns.


问:
probability that the number of roulette spins until the sum of the spins exceeds a given number

A roulette wheel is labelled $1$ through $n$, with each number having
the same probability of being selected and each spin of the roulette is
independent of the other spins. We spin the roulette until the
sum of the numbers selected is strictly greater than $n$. Let $S$ be the number of spins it takes for the sum to be strictly greater than $n$. For an integer $m$
such that $0 \leq m \leq n$, find the probability that $\operatorname{Pr}(S \geq m+1)$?

I understand the continuous version (a sequence $x_j$ is sampled from a uniformly distribution on the interval $[0,1]$ and you continue until the sum is more than 1) has expected value $e$ for the number of samples to get to a sum that exceeds 1. That I understand but in this case, I'm lost on what is multiplied to $(1/n)^m$ (representing the first $m$ roulette spins) as my calculations don't give $e$ as a limit.

How can one obtain an expression for the probability that $S$ is at least $m+1$? In particular, what is multiplied to $(1/n)^m$?

EDIT: Previous problem incorrectly stated $S$ to be the sum of the spins, when it should be the number of spins for the sum to exceed $n$

答:
We have : $$
P(S \geq m+1) = P(X_1 + ... + X_m \leq n)
$$

where $X_i$ are iid uniformly distributed in $\{1,2,...,n\}$. Therefore, it is sufficient to find the RHS.

The answer is just $\sum_{1 \leq i_1,...,i_m \leq n} 1_{i_1+...+i_m \leq n} P(X_1 = i_1,X_2 = i_2,...,X_m = i_m)$, which just becomes $\frac {K}{n^m}$, where :
$$
K = |\{(i_1,i_2,...,i_m) \in \{1,2,...,n\}^m : i_1+i_2+...+i_m \leq n\}|
$$

We claim that the set given in the definition of $K$ is in bijection with the set $\{(i_1,i_2,...,i_m) \in \{0,1,...,m\}^m : i_1+i_2+...+i_m \leq n-m\}$. The bijection is given by taking a tuple and subtracting $1$ from each entry. It is easy to see, as $n>m$ that the bijection holds.

Now, we adjoin an index $i_{m+1} = n-m - (i_1+...+i_m)$, so we get that the above set is in bijection with :
$$
\{(i_1,i_2,...,i_m,i_{m+1}) \in \{0,1,...,m\}^{m+1} : i_1+i_2+...+i_m +i_{m+1} = n-m\}
$$

by projection onto the first $m$ coordinates.

This last set is calculated using stars and bars. Indeed, imagine a set of $n$ blanks, each filled with a $|$ or a $\circ$ (ball), so that there are exactly $m$ of $|$ and $n-m$ of $\circ$.
$$
\underbrace{|\circ||\circ\circ\circ|\circ||...|\circ| \circ \circ}_{\text{$n$ blanks}}
$$

The bars separate the circles into $m+1$ parts, each containing $i_1,...,i_{m+1}$ number of circles, which total to $n-m$ circles, and each of which is non-negative. In the example above, $i_1 = 0,i_2 = 1,i_3  =0 , i_4 = 3$ and so on.

It is easy to see that the number of ways of placing these bars and circles is the same as the number of elements in the set mentioned earlier. But we are basically placing $m$ bars into $n$ slots : the answer to this is just $\binom{n}{m}$.

Thus, we basically have $K = \binom{n}{m}$, with $Pr(S \geq m+1) = \frac{\binom{n}{m}}{n^m}$.

The expectation of $S$ is equal to $\sum_{i=0}^n \binom ni n^{-i} = (1 + \frac 1n)^n \to e$ as $n \to \infty$ (i.e. as we approach the uniform distribution)


问:
Functions on quotient topologies
I'm having an issue of (seemingly) conflicting information from textbooks involving mappings between topological spaces. If $ \pi : S \rightarrow S/\sim $ is the projection of a topology S into a quotient over the relation $ \sim $, the topology of $ S $ is transferred to the quotient by requiring that all sets $ V \in S / \sim \, $ are open if $ \pi^{-1} (V) $ are open in $ S $. I'm assuming that this mapping is probably defined such that $ \pi^{-1}(V) := \{ x \in S \, | \, \pi(x) = [x] \; \;  \forall [x] \in V \subset S/\sim \}$.

In Tu's manifolds (page 72), this condition constitutes a topology because:

$$ \pi^{-1} \Big( \bigcup_\alpha V_\alpha \Big) = \bigcup_\alpha \pi^{-1} (V_\alpha) \\
\pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) = \bigcap_\alpha \pi^{-1} (V_\alpha) $$

for an indexed collection of sets $ V_\alpha \in S/\sim $. However, Baker's Introduction to Topology (page 22) states that for a mapping between sets $ f:X \rightarrow Y $ and a collection of indexed sets $ U_\alpha \in X $,

$$ f \Big( \bigcap_\alpha U_\alpha \Big) \subseteq \bigcap_\alpha f(U_\alpha ) $$

and that equality of these two sets is satisfied only when $ f $ is one-to-one. However, when this is applied to the mapping $ \pi^{-1} $ above, it is not only clear that $ \pi^{-1} $ isn't one-to-one, it isn't even well defined as a function, because it's very probable that many elements of $ S $ will map to a single equivalence class in $ S/\sim $, making the inverse multi-valued. Additionally, I can't seem to find any flaw in the following proof that the missing $ \subseteq $ relationship is true all the time:

$$ \text{Let } x \in \bigcap_\alpha \pi^{-1} (V_\alpha) \implies x \in \pi^{-1}(V_\alpha) \quad \forall \alpha \\
\implies \exists [x] \in V_\alpha \ni \pi(x) = [x]  \quad \forall \alpha \implies [x] \in \bigcap_\alpha V_\alpha \\
\implies x \in \pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) \implies \bigcap_\alpha \pi^{-1} (V_\alpha) \subseteq \pi^{-1} \Big( \bigcap_\alpha V_\alpha \Big) $$

So is my proof valid, and Baker is simply wrong? I highly doubt that, it seems I'm missing something critical on why Tu is still correct, and Baker's theorem doesn't apply to this situation? I've seen a similar proof to the above posted here before, I guess my main question is why Baker's one-to-one restriction on the equality isn't killing my proof and as a result Tu's definition.

答:
Preimages of maps behave differently, in some sense "nicer" if you want, than images. One of my professors always used to say

preimages are good, images are evil.

What he meant is that, given a map $f\colon X\to Y$ with a collection of subsets $A_i \subset X, B_i \subset Y$ for an index set $I$, preimages respect unions and intersections $$f^{-1}\left(\bigcap_i B_i\right) = \bigcap_if^{-1}(B_i)\quad \text{and}\quad f^{-1}\left(\bigcup_i B_i\right) = \bigcup_if^{-1}(B_i)$$
the images, in general, do not:

$$f\left(\bigcap_i A_i\right) \subseteq \bigcap_i f(A_i)$$

And, as you already figured, they are equal iff $f$ is injective.

However, you still seem to be confusing a few things. The projection $\pi$ is for obvious reasons not injective but it's surjective and it's certainly well defined. You are confusing preimages $$\pi^{-1}([s])\subset S$$ of elements $[s]\in S/{\sim}$ under $\pi$, which are subsets of $S$, with the inverse function $$\pi^{-1}\colon S/{\sim}\to S.$$ These are entirely different things.

The projection $\pi$ projects all elements $a,a' \in S$ that are equal $a\sim a'$ with respect to the equivalence relation $\sim$ onto their common equivalence class $[a]$ in $S/{\sim}$.

Title: Functions on quotient topologies


问:
Determine local extrema of 2 functions
So I have 2 functions, $f(x)=cos(\frac{2}{x})$ and $g(x)=x^{\frac{1}{3}}$

For f(x), how many local extrema it has? It is very weird because near 0 it is many up and down. So do they belong to local maximum and minimum?

For g(x) in the interval (-3,2). does it have any local extrema? Because in this interval there is global maximum at g(2) and global minimum at g(-3).

答:
$f$ has infinitely many local maxima and infinitely many local minima as you have observed.

$g$ is strictly increasing in the given interval so it has no local maximum of minimum at any point of the open interval $(-3,2)$.

Title: Determine local extrema of 2 functions


问:
Geometric proof
Let the three sides of a triangle be $a,b$ and $c$. If the equation

$$a^2+b^2+c^2=ab +bc+ac$$

holds true, then the triangle is an equilateral triangle.

How do we prove this? An answer or even the slightest hint will be appreciated.

答:
Note that
$$a^2 + b^2 + c^2 = ab + bc + ca \implies (a-b)^2 + (b-c)^2 + (c-a)^2 = 0$$
I trust you can finish it off from here.</p>

Title: Geometric proof


问:
How do I convert this to spherical coordinates?
The task is to set up the integral in spherical coordinates. The given integral is: $$\int_0^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\int_{\sqrt{2x^2+2y^2}}^{\sqrt{8}} z^2-x^2-y^2 dzdydx $$
I'm new to spherical coordinates so I get confused as to what my bounds for $\phi$ would be? I'm getting $\phi=[0,\arctan(\frac{1}{\sqrt{2}})]$ Is that correct?

答:
This figure showing the integration region should help:

$- \pi/2 \leq \theta \leq + \pi /2$

$0 \leq z \leq \sqrt{8}$

$0 \leq \phi \leq \tan^{-1} (1/\sqrt{2})$


问:
Confusion in interpreting why $A\subseteq B$ implies $A\cap X\subseteq B\cap X$
I'm working on the beginnings of Set Theory and I need help understand the proof the question below:

Prove that $A ⊆ B$ and $A \cap B^\complement=\emptyset$ are equivalent.

The proof is:

$$ A⊆ B\implies A∩ B^\complement ⊆ B ∩ B^\complement$$
As $B ∩ B^\complement = \emptyset$,
$$A ∩ B^\complement \subseteq \emptyset$$
$$A ∩ B^\complement = \emptyset$$

Therefore, both statements are equivalent.

My confusion lies in the step $A∩ B^\complement ⊆ B ∩ B^\complement$. What is the logic for this? How can we "add" the intersection of $B^\complement$ to each side of the subset?

答:
If every element of $A$ is in $B$.  Then every element that is in both $A$ and $X$ is in both $B$ and in $X$.

==========

It's direct  from definitions.

$A \subseteq B$ means for all $x$, if $x \in A$ then $x \in B$.

And for all $x$, if $x \in M\cap N$ then $x \in M$ and $x \in N$.

So if $A\subseteq B$ then.....

If $x \in A \cap X$ then $x \in A$ and $x\in X$.  And since $A\subseteq B$ and $x \in A$ we know $x \in B$.  And as $x\in X$ then $x \in B$ and $x \in X$.  So $x \in B\cap X$.

So if $x \in A\cap X$ then $x \in B\cap X$.  And that, by definition means $A\cap X \subset B\cap X$.

So $A\subseteq B \implies A\cap X \subseteq B\cap X$.


问:
What is the Levi-Civita connection of product of semi-Riemann manifolds
Let $(M_i,g_i)$ be semi-Riemann manifold for $i=1,2$, and $ M=(M_1 \times M_2, g_1 +g_2)$ is the product semi-Riemann manifold, with  $ \pi_i: M \rightarrow M_i$ be the canonical projection.

If $X_i$ is a vector field on $M_i$, then define $\overline{X_i}$ as a vector field on $M$ such that $\pi_i.\overline{X_i}=X_i$ and $\pi_j.\overline{X_i}=0$ for $ j \ne i$; in other words, $\overline{X_1}=(X_1,0)$ and $\overline{X_2}=(0,X_i)$. Then, $[\overline{X_1},\overline{X_2}]=0$.

Now, let $\nabla^i$ be the Levi-Civita connection on $M_i$ and $\nabla$ the Levi-Civita conection on $M$. Then we need to show that $\nabla_{\overline{X_i}}\overline{Y_i}=\overline{\nabla^i_{X_i}Y_i}$ and $\nabla_{\overline{X_1}} \overline{Y_2}=0$.

I am not sure how can I use work on the Levi-Civita connection operator on ordered pairs..

Any hint is appreciated.

答:
With each projection $\pi_i\colon M_1\times M_2 \to M_i$, pull back the Levi-Civita connection $\nabla^i$ of $(M_i,g_i)$ to a connection $\pi_i^*(\nabla^i)$ in the pull-back bundle $\pi_i^*(TM_i) \to M_1\times M_2$. Then form the direct sum bundle $$\pi_1^*(TM_1) \oplus \pi_2^*(TM_2) \to M_1\times M_2,$$and equip it with the direct sum connection $\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2)$. This is the Levi-Civita connection of $(M_1\times M_2 , \pi_1^*g_1 + \pi_2^*g_2)$, since $$\pi_1^*(TM_1) \oplus \pi_2^*(TM_2) \cong T(M_1\times M_2).$$Let $(x^i)$ and $(y^\mu)$ be local coordinates for $M_1$ and $M_2$, respectively, giving local coordinates $(x^i \circ \pi_1, y^\mu \circ \pi_2)$ for $M_1\times M_2$. Any vector field $X$ on $M_1 \times M_2$ may be locally written as $$X_{(x,y)} = \underbrace{\sum_i X^i(x,y) \frac{\partial}{\partial (x^i\circ \pi_1)}\bigg|_{(x,y)}}_{\mbox{$\doteq X_1(x,y)\in T_xM_1$}} + \underbrace{\sum_\mu X^\mu(x,y) \frac{\partial}{\partial (y^\mu \circ \pi_2)}\bigg|_{(x,y)}}_{\mbox{$\doteq X_2(x,y)\in T_yM_2$}},$$and similarly for a second vector field $Y$.

The problem here is that $X_1$ is not a vector field on $M_1$ because it depends also on the point $y \in M_2$ --- likewise for $X_2$ not being a vector field on $M_2$. They are, in fact, sections of $\pi_1^*(TM_1)$ and $\pi_2^*(TM_2)$ (this implies that they are particular vector fields on the full manifold $M_1\times M_2$).

For simplicity, let's omit the $\pi$'s and denote the coordinate vector fields by $\partial_i$ and $\partial_\mu$ only (either on each $M_i$ or on the full product $M_1\times M_2$ --- there will be a slight abuse of notation in what follows, but note the use of letters in different alphabets to indicate these objects live, a priori, in different spaces).

The business with pull-back bundles is the way to work around this "bug".

Using the characteristic properties of direct sums and pull-backs, i.e.,

$(\nabla' \oplus \nabla'')_X(\psi'\oplus \psi'') = \nabla'_X\psi' + \nabla_X''\psi''$

$(F^*\nabla)_X(Y \circ F) = \nabla_{F_\ast X}Y$

one has $$(\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2))_XY  = (\pi_1^*(\nabla^1))_{X_1}Y_1 + (\pi_2^*(\nabla^2))_{X_2}Y_2,$$and each term in the right side above is computed in the same way --- let's do the first: $$\begin{align*} (\pi_1^*(\nabla^1))_{X_1}Y_1 &= (\pi_1^*(\nabla^1))_{X_1}\left( \sum_i Y^i \partial_i\right) \\ &= \sum_i X_1(Y^i) \partial_i + \sum_i Y^i (\pi_1^*(\nabla_1))_{X_1}\partial_i \\ &= \sum_i X_1(Y^i)\partial_i + \sum_{i,j} Y^iX^j (\pi_1^*(\nabla^1))_{\partial_j}\partial_i \\ &= \sum_k X_1(Y^k)\partial_k + {\color{red}{\sum_{i,j} X^iY^j \nabla^1_{\partial_i}\partial_j}} \\ &= \sum_k X_1(Y^k)\partial_k + \sum_{i,j,k} X^iY^j (\Gamma^1)_{ij}^k\partial_k \end{align*}$$In the term in red, the abuse of notation $x^i \equiv x^i \circ \pi_1$ kicked in. Similarly, one gets $$(\pi_2^*(\nabla^2))_{X_2}Y_2 = \sum_{\lambda} X_2(Y^\lambda)\partial_\lambda + \sum_{\mu,\nu,\lambda} X^\mu Y^\nu (\Gamma^2)_{\mu\nu}^\lambda \partial_\lambda.$$So, how does one find the Christoffel symbols $\Gamma_{ab}^c$ for this connection? If all $a,b,c$ are not all in the "same alphabet" (i.e., all mid-alphabet latin letters, or all greek letters), we get zero. Else, we get $$(\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2))_{\partial_i}\partial_j = (\Gamma^1)_{ij}^k \partial_k \qquad \mbox{and} \quad (\pi_1^*(\nabla^1)\oplus \pi_2^*(\nabla^2))_{\partial_\mu}\partial_\nu = (\Gamma^2)_{\mu\nu}^\lambda \partial_\lambda.$$

Those results coincide with what you get using the coordinate-version of the Koszul formula $$\Gamma_{ab}^c = \frac{1}{2} \sum_d g^{cd}(\partial_ag_{bd} + \partial_b g_{ad} - \partial_dg_{ab}),$$since we have (in block form) that $$\begin{pmatrix} ((g_1)_{ij})_{i,j} & 0 \\ 0 & ((g_2)_{\mu\nu})_{\mu,\nu} \end{pmatrix}^{-1} = \begin{pmatrix} ((g_1)^{ij})_{i,j} & 0 \\ 0 & ((g_2)^{\mu\nu})_{\mu,\nu} \end{pmatrix}. $$


问:
The sixth number system
In high school, I learned there are 5 number systems, namely:

Natural numbers ($\Bbb N$)

Integers ($\Bbb Z$)

Rational numbers ($\Bbb Q$)

Real numbers ($\Bbb R$)

Complex numbers ($\Bbb C$)

I remember one time our teacher told us that there is a sixth number system denoted by $\Bbb H$ and that is used by video games developers. Is it right that there is such number system?

If so, are there other number systems? And why haven't we studied them at university?

答: There are also quaternions and octonions.  See the book by Conway and Smith on this subject.  The former corresponds to four-dimensional numbers, the latter to eight-dimensional numbers.  </p>

Title: The sixth number system


问:
How large should n be to guarantee the error in approximating ln(.5) with ln(1+x) centered at a = 0 is less than .0001?
I am asked to find the number of terms such that the error of approximating ln(.5) < 0.001 using Taylor's inequality. I can do these using the remainder theorem from the alternating series test, but we are required to use Taylor's inequality.

I am struggling right now to find M such that
$ |f^{n+1}(x)| \le M$.

I have found the derivative
$|f^{n+1}(x)|= \frac{n!}{(1+x)^{n+1}}$. My problem is with finding M. I know that $x \in (-.5, 0)$, and the derivative (absolute value) is decreasing. If I plug -.5 into the derivative to get M, M seems to just blow up and isn't bounded.

Is there any other way to find M in this situation?

答:
Making it general, we have (for $|x|<1$)
$$\log(1+x)=\sum_{n=1}^\infty (-1)^{n+1}\,\frac {x^n} n=\sum_{n=1}^p (-1)^{n+1}\,\frac {x^n} n+\sum_{n=p+1}^\infty (-1)^{n+1}\,\frac {x^n} n$$ and you want to know $p$ such that
$$R_p=\frac {x^{p+1}}{p+1} \leq 10^{-k}$$ that is to say
$$(p+1) \geq x^{p+1} 10^k$$
In the real domain, the only analytical solution of the equation is
$$p=-\frac{W\left[-10^k \log (x)\right]}{\log (x)}-1$$ where $W(.)$ is Lambert function whicch is widely used (on the search bar here, just type Lambert and you will find $3392$ entries). For sure, after computing $p$, you will use $\lceil p\rceil$.

For $k=5$ and $x=\frac 12$, this will give $p=11.9183$, then $\lceil p\rceil=12$.

Let us check
$$R_{11}=\frac{1}{49152}\sim 2.03\times 10^{-5} > 10^{-5}$$
$$R_{12}=\frac{1}{106496}\sim 9.39\times 10^{-6} < 10^{-5}$$

If you do not have access to this function, you can use the expansion for large values of $t$ as given in the linked page
$$W(t)= L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(2L_2^2-9L_2+6)}{6L_1^3}+\cdots$$ where $L_1=\log(t)$ and $L_2=\log(L_1)$.

From a coding point of view, there is big advantage to know in advance the number of terms to be added since, in the calculation loop, you will not have any more IF test (this is a very expensive operation).

Suppose that, for the same $x=\frac 12$, we want $k=100$. We should find $p=322.854$  then $\lceil p\rceil=323$. Checking again
$$R_{322}=4.81\times 10^{-100} >10^{-100}$$ $$ R_{323}=9.03\times 10^{-101} <10^{-100} $$

Welcome to the world of the magnificent Lambert function !


问:
On a curious equation regarding odd perfect numbers
Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

Since $\gcd(p^k, \sigma(p^k))=1$, then we essentially get the equation
$$\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)(p-1)}{p^k - 1},$$
where $\sigma(x)$ is the classical sum of divisors of the positive integer $x$, and $D(x)=2x-\sigma(x)$ is the deficiency of $x$.

We derive
$$\frac{\sigma(m^2)}{p^k}=\sigma(m^2) - D(m^2)(p-1) = \sigma(m^2) - 2(p-1)m^2 + (p-1)\sigma(m^2)$$
$$=2(1-p)m^2 + p\sigma(m^2)=\Bigg(I(m^2)-\frac{2(p-1)}{p}\Bigg)\cdot{pm^2},$$
where $I(x)=\sigma(x)/x$ is the abundancy index of $x$.

Here is my question:

Is it then possible to get a positive lower bound on the quantity
$$I(m^2)-\frac{2(p-1)}{p}?$$

Note that I know the following facts:

(1) $\sigma(m^2)/p^k \geq 315$ by a result of Broughan, Delbourgo, and Zhou.

(2) $2(p-1)/p < I(m^2) \leq 2p/(p+1)$ follows from $(p+1)/p \leq I(p^k) < p/(p-1)$

Body:
On OP's request, I am converting my comment into an answer.

FYI :

$I(m^2)-\dfrac{2(p-1)}{p}$ can be written as

$$I(m^2)-\frac{2(p-1)}{p}=\frac{\sigma(m^2)}{m^2}-\frac{2(p-1)}{p}=\frac{2p^k}{\sigma(p^k)}-\frac{2(p-1)}{p}$$
$$=\frac{2p^k(p-1)}{p^{k+1}-1}-\frac{2(p-1)}{p}=\frac{2p^{k+1}(p-1)-2(p-1)(p^{k+1}-1)}{p(p^{k+1}-1)}=\frac{2(p-1)}{p(p^{k+1}-1)}$$

Let $f(k,p):=\dfrac{2(p-1)}{p(p^{k+1}-1)}$.

Then, we have $$\dfrac{\partial f(k,p)}{\partial k}\lt 0,\qquad \dfrac{\partial f(k,p)}{\partial p}\lt 0$$
and
$$\lim_{k\to\infty}f(k,p)=\lim_{p\to \infty}f(k,p)=0$$


问:
A faster way to find the term $x^0$ in the expansion of $(1-\frac{x}{3})^{5}(1+\frac{2}{x})^{3}$
Now I am actually looking for a method in the general case when you have a product of two binomials. The method I use is equating coefficients; I have provided my example solution to this question as an answer. Now this is a pretty good method I reckon but I wonder if someone out there has a faster, more efficient solution to these types of questions that they would like to share (please?). The thing that annoys me is plugging huge sums of numbers into the calculator that takes more time than the answer itself.</p>

答:
My way:

Find the general term for both binomials separately

To find the independent term, consider the possible expansions by inspection then plug in the appropriate values into the general terms

Use calculator to sum up values (very lengthy process if there are lots of cases)


问:
What the probability of rolling a number?
A die has been "loaded" so that the probability of rolling any even number is $\frac{3}{15}$ and the probability of rolling any odd number is $\frac{2}{15}$. (Assume the die is six sided with each side numbered one through six. Enter your probabilities as fractions.)

What is the probability of rolling a $2$, given that an even number is rolled?

I tried to solve it by assuming that:

$A$ is the event of rolling an even number

$B$ Is the event of rolling a $2$
Then I used the conditional probability form:
$$P(A|B) = \frac{P(A \cap B )}{P(B)},$$
where
$P(A) = \frac{3}{15}$ is given, $P(B) = \frac13 \cdot\frac{3}{15}$.

I am not pretty sure of $P(B)$ and I do not know how to find $P(A \cap B )$.
Any help please?

答:
(a) Intuitively (per @lulu's Comment): Even numbers are equally likely and 2 is one of three even numbers, so $1/3.$

(b) Formally, by conditional probability:
$$P(2 | E) = \frac{P(2 \cap E)}{P(E)} = \frac{P(2)}{P(E)}
= \frac{3/15}{9/15} = \frac 13.$$

(c) By simulation of ten million rolls of the unfair die in R: Also, $1/3.$ With ten million rolls one can expect about three place accuracy. (Brackets indicate conditioning; | means or.)

```<code>set.seed(1202)
prob = c(2,3,2,3,2,3)/15
roll = sample(1:6, 10^7, rep=T, p=prob)
table(roll)/10^7
roll
        1         2         3         4         5         6
0.1334890 0.1999361 0.1335797 0.1997803 0.1332513 0.1999636

mean(roll[roll==2 | roll==4 | roll==6] == 2)
[1] 0.3334046  # aprx 1/3
</code>```

Body:
(a) Intuitively (per @lulu's Comment): Even numbers are equally likely and 2 is one of three even numbers, so $1/3.$

(b) Formally, by conditional probability:
$$P(2 | E) = \frac{P(2 \cap E)}{P(E)} = \frac{P(2)}{P(E)}
= \frac{3/15}{9/15} = \frac 13.$$

(c) By simulation of ten million rolls of the unfair die in R: Also, $1/3.$ With ten million rolls one can expect about three place accuracy. (Brackets indicate conditioning; | means or.)

set.seed(1202)
prob = c(2,3,2,3,2,3)/15
roll = sample(1:6, 10^7, rep=T, p=prob)
table(roll)/10^7
roll
        1         2         3         4         5         6
0.1334890 0.1999361 0.1335797 0.1997803 0.1332513 0.1999636

mean(roll[roll==2 | roll==4 | roll==6] == 2)
[1] 0.3334046  # aprx 1/3


问:
Define a sequence $b_{0},b_{1}, b_{2},.... \in R$ recursively by $b_{0} = 0$ and $b_{n+1}=(b_{n})^2+ \frac{1}{4}$....
Define a sequence $b_{0},b_{1}, b_{2},.... \in R$ recursively by $b_{0} = 0$ and $b_{n+1}=(b_{n})^2+ \frac{1}{4}$.Prove that, for all $n \leq 1$, $0 < b_{n} <1.$

I'm not sure how to approach this question, I've tried to use proof by induction, but I'm lost.

答:
Hint: It might be easier to prove (by induction) that $0 < b_n < \tfrac{1}{2}$ for all integers $n \ge 1$.

You already know that $0 < b_1 = \tfrac{1}{4} < \tfrac{1}{2}$. Can you show that $0 < b_n < \tfrac{1}{2}$ implies $0 < b_{n+1} = b_n^2+\tfrac{1}{4} < \tfrac{1}{2}$?


问:
Surjectivity of function in the functional equation $f \big(xf (y)-f(x)\big)=2f (x)+xy $
I came across the following question while studying functional equations.

Let $f: \mathbb{R} \to \mathbb{R} $ be a function satisfying
$$f \big(xf (y)-f(x)\big)=2f (x)+xy $$
$\forall x, y \in \mathbb {R} $, prove that $f $ is a surjective function.

I really have no idea that how to prove the aforementioned as I am very new to functional equations.
The author has provided a very short proof without mentioning much details, making it very confusing and difficult to understand.

A pedantic proof would be highly helpful and will really be appreciated.

Thank you

Best regards !

答:
If we fix $x = 1$, then the equation becomes
$$f(f(y) - f(1)) = 2f(1) + y,$$
for any $y$. Intuitively, the right hand side can be made to be any value we want. The left hand side is $f$ of some expression of $y$ (it doesn't really matter what). So, we can choose a value of $y$ to make the right hand side whatever we want, then we have $f$ of an expression of $y$ that produces this value. This is what makes $f$ surjective.

If we want to prove it formally, fix $b \in \Bbb{R}$. Let $x = 1$ and $y = b - 2f(1)$. Then our equation turns into
$$f(f(b - f(1)) - f(1)) = 2f(1) + b - 2f(1) = b.$$
If we let $a = f(b - f(1)) - f(1)$, then we have constructed an $a \in \Bbb{R}$ such that $f(a) = b$, proving surjectivity.


Title: A faster way to find the term $x^0$ in the expansion of $(1-\frac{x}{3})^{5}(1+\frac{2}{x})^{3}$

Body: Now I am actually looking for a method in the general case when you have a product of two binomials. The method I use is equating coefficients; I have provided my example solution to this question as an answer. Now this is a pretty good method I reckon but I wonder if someone out there has a faster, more efficient solution to these types of questions that they would like to share (please?). The thing that annoys me is plugging huge sums of numbers into the calculator that takes more time than the answer itself.</p>

Body:
I will not claim  this is faster ....

as $${(1+\frac{2}{x})}^3=\frac{{(x+2)}^3}{x^3}$$

we have to find the coefficient of $x^3$ in $$f(x)={(x+2)}^3{(1-x/3)}^5=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4..+a_8x^8$$

Notice that  $$f'''(x)=6a_3+24a_4x+..$$
$$f'''(0)=6a_3$$ From here can you finish it?


问:
What is the cardinal $\aleph_2^{\aleph_1}$
I'm looking for the cardinality of the set of functions : $$\mathbb{R}\rightarrow F(\mathbb{R},\mathbb{R})$$

Where the latter is the set of all functions from the reals to the reals.

Is it : $$\aleph_2^{\aleph_1}=2^{\aleph_1\aleph_1}=2^{\aleph_1}=\aleph_2$$ ?

答:
Following the OP, below I'll write "$F(A,B)$" for the set of functions from $A$ to $B$.

You seem to be assuming that $\vert\mathbb{R}\vert=\aleph_1$ and $\vert\mathbb{R}^\mathbb{R}\vert=\aleph_2$ - or perhaps more snappily, that $2^{\aleph_0}=\aleph_1$ and $2^{(2^{\aleph_0})}=\aleph_2$. This is not provable from the usual axioms of set theory alone - see here.

However, modulo this you are correct: the cardinaity of the set in question is indeed $2^{(2^{\aleph_0})}$, or more snappily $\beth_2$. This can be informally motivated by $$\vert F(\mathbb{R}, F(\mathbb{R},\mathbb{R}))\vert= ((2^{\aleph_0})^{2^{\aleph_0}})^{2^{\aleph_0}}=2^{\aleph_0\cdot 2^{\aleph_0}\cdot 2^{\aleph_0}}=2^{2^{\aleph_0}}.$$

Of course this sort of manipulation requires justification: we have to argue that cardinal exponentiation matches up with function set formation, and that cardinal exponentiation follows the appropriate basic rules. If you haven't seen these, rigorously interpreting and justifying the above calculation is a good exercise.


问:
Maximum of a function outcome of implicit function theorem
I am trying to prove the next:

If the point $(a, b)$ satisfies the equation $F(x, y) = xy − \log\frac{x}{y} = 0, (x, y > 0),$
then there exists a $C^1$ function $y = f(x)$ which passes through this point and is unique in its neighborhood. Furthermore, there is one point $(a, b)$ such that the corresponding function
$y = f(x)$ has a maximum at $a.$

$F$ is a function of class $C^1$ in $(0,\infty)\times(0,\infty)$ because is sum of functions on such class. By hypothesis, $F(a,b)=0$ and $Det(\frac{\partial F}{\partial y}(a,b))\neq 0.$ Then Implicit Function Theorem ensures that there is a function f of class $C^1$ defined on a neigborhood of $a$ such that $F(x,f(x)) = 0$ in such neigborhood and $b=f(a).$

To prove uniqueness is enough to consider $F(x, f(x))=0=F(x, g(x))$ in such neigborhood and check the cases when $f\neq g.$

I have problems proving that $f$ has a maximum; I have tried to use implicit differentiation and equals to zero to get a critic point but I do not find something useful.

Any kind of help is thanked in advanced.

答:
Implicit differentiation will give you
$$f'(x) = \frac{\frac1x-y}{\frac1y+x},$$
and so we'll have a critical point at $(a,b)$ precisely when $\dfrac1a-b = 0$, i.e., when $ab=1$. But then substituting into $F(a,b)=0$ gives $\log a = 1/2$, or $a=e^{1/2}$ and $b=e^{-1/2}$.
(I did not stop to check that $a$ is a local maximum of $f$.)


答:
Using the binomial theorem,
\begin{align*}
\left( 1 - \frac{x}{3} \right)^5 &= \sum_{i=0}^{5} \binom{5}{i} 1^{5-i} \left(\frac{-x}{3} \right)^i  \\
    &= \sum_{i=0}^5 \binom{5}{i} (-3)^{-i} x^i  \text{ and }  \\
\left( 1 + \frac{2}{x} \right)^3 &= \sum_{j=0}^{3} \binom{3}{j}1^{3-j} \left(\frac{2}{x} \right)^j  \\
    &= \sum_{j=0}^{3} \binom{3}{j} 2^j x^{-j}
\end{align*}
So \begin{align*}
\left( 1 - \frac{x}{3} \right)^5 \left( 1 + \frac{2}{x} \right)^3
    &= \left( \sum_{i=0}^5 \binom{5}{i} (-3)^{-i} x^i \right) \left( \sum_{j=0}^{3} \binom{3}{j} 2^j x^{-j} \right)  \\
    &= \sum_{i=0}^5 \sum_{j=0}^{3} \binom{5}{i} \binom{3}{j} (-3)^{-i} 2^j x^{i-j}
\end{align*}
The only contributions to $x^0$ have $i - j = 0$, so $i = j$.  The range of $i$ is $[0,5]$ and the range of $j$ is $[0,3]$.  The intersection of these two intervals is $[0,3]$, so we have four terms contributing to the coefficient of $x^0$:
$$  \sum_{j=0}^{3} \binom{5}{j} \binom{3}{j} (-3)^{-j} 2^j  $$
(It seems that we have done a lot of work to get here.  With practice, you can jump straight to this expression.)
Call the four terms $a_j = \binom{5}{j} \binom{3}{j} (-3)^{-j} 2^j$ for $j \in [0,3]$.  Notice
$$ a_0 = \binom{5}{0} \binom{3}{0} (-3)^{-0} 2^0 = 1  \text{.}  $$
Then
\begin{align*}
a_1 &= a_0 \cdot \frac{5}{1} \cdot \frac{3}{1} \cdot \frac{-1}{3} \cdot 2 = -10  \text{,}  \\
a_2 &= a_1 \cdot \frac{4}{2} \cdot \frac{2}{2} \cdot \frac{-1}{3} \cdot 2 = \frac{40}{3}  \text{, and}  \\
a_3 &= a_2 \cdot \frac{3}{3} \cdot \frac{1}{3} \cdot \frac{-1}{3} \cdot 2 = \frac{-80}{27}
\end{align*}
This says we can compute the next term from the previous one.  The second and third factors are the old mental arithmetic method for calculating rows of Pascal's triangle by multiplying by successive fractions having decreasing numerators and increasing denominators.

Then the coefficient of $x^0$ is
$$  1 - 10 + \frac{40}{3} - \frac{80}{27} = \frac{37}{27}  \text{.}  $$


问:
Can a function be differentiable everywhere on its domain, but not Lipschitz on its domain?
If we suppose that a function is differentiable at every point on its domain (but place no more restrictions than that), does it follow that the function is Lipschitz on its domain? I think that it does not, and I suspect that it is because we do not require that the function is continuously differentiable. But I'm not sure how to prove my thought - specifically, I can't come up with a counterexample.

If I am right that it is not Lipschitz, can we say if it will be locally Lipschitz at every point?

答:
If $f\colon\Bbb R\longrightarrow\Bbb R$ is the function defined by $f(x)=x^2$, then $f$ is differentiable and even a $C^\infty$ function, but it is not Lipschitz continuous.

On the other hand, if you define$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x^2\sin\left(\frac1{x^2}\right)&\text{ if }x\ne0\\0&\text{ if }x=0,\end{cases}\end{array}$$then $f$ is not Lipschitz continuous on any neighborhood of $0$. This has to do with the fact that $f'$ is unbounded (which is stronger than being discontinuous).


问:
Finding the limit when denominator = 0
I'm having trouble solving this limit:

$$\lim_{x \to -2^-} \frac{1}{(x + 2)^2}$$

I can't find a way to rationalize the denominator. Also, is there a way to do it without plugging in -2.001 and stuff or graphing it?

EDIT:

I realized after asking this question that it doesn't matter if you take the above limit from the right, left, or both. It's always $+\infty$. Here's an equation that gives $+\infty$ from the right and $-\infty$ from the left: $$\lim_{x \to 3^+} \frac{x - 4}{x - 3}$$

How do I (algebraically) determine if it is positive or negative?

答:
Maybe this way of thinking about it will seem a little more intuitive to you:

Let $\varepsilon > 0$, and consider the limit

$$\lim_{x \rightarrow -2^{-}} \frac{1}{(x+2)^2}
= \lim_{\varepsilon \rightarrow 0} \frac{1}{((-2-\varepsilon)+2)^2}
= \lim_{\varepsilon \rightarrow 0} \frac{1}{\varepsilon^2}.$$

Now the expression $\frac{1}{\varepsilon^2}$ can be made arbitrarily large by choosing $\varepsilon$ small enough, and so the limit does not exist.

Title: Finding the limit when denominator = 0


问:
What's a better way to integrate this?
$$ \int \frac{1}{x^2 + z^2} dx $$

I tried substitution and also by parts. By parts is getting messy and I am not sure if I am getting the right answer. I am trying to figure out an easier way or the proper way to integrate this. Could someone please show me?

答:
The right substitution will do the job for you. Set $x = z \tan(t)$. We then have $dx = z \sec^2(t) dt$. Hence,
$$\int \dfrac{dx}{x^2+z^2} = \int \dfrac{z \sec^2(t) dt}{z^2(\tan^2(t)+1)} = \dfrac1z \int \dfrac{\sec^2(t) dt}{\sec^2(t)} = \dfrac{t}z + \text{const} = \dfrac{\arctan(x/z)}z + \text{const}$$
We made use of the following two facts in the above derivation.

$\dfrac{d(\tan(t))}{dt} = \sec^2(t)$

$\tan^2(t)+1 = \sec^2(t)$

Title: What's a better way to integrate this?


问:
Applying Cauchy's Integral Formula for $\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} d z$
So, here is a problem:

$\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} d z$

the answer is $2i$

I cant understand what I am doing wrong:

Cauchy's Integral Formula:

$f(a)=\frac{1}{2 \pi i} \oint_{\gamma} \frac{f(z)}{z-a} d z$

In this case:

$f(a) = \{a=2\} = \exp(\frac{1}{2-1}) = \exp(1)$

Thus

$\oint_{|z|=4} \frac{e^{1 /(z-1)}}{z-2} d z = 2 \pi i exp(1)$

???

答: You cannot apply Cauchy's integral formula here because $\exp\left(\frac1{z-1}\right)$ is not analytic on $\overline{D(0,4)}$. So, it is natural that it does no provide the correct answer. I suggest that you try the residue theorem instead.


问:
Difficult coin weighing puzzle: 14 coins, 1 fake (heavier or lighter), 3 pre-determined weighings
This recent question reminds me of a coin weighing puzzle I learned many years ago.  It is one of the hardest puzzles of this kind that I know.  I will post my solution in a few days, and meanwhile hopefully someone may enjoy it.  (My apologies if this is a repeat, but I searched and could not find this exact version.)

There are $14$ suspect coins, $13$ of which are good and have the same weight, and the last one is bad and have a different weight (heavier or lighter).  In addition, you have a $15$th coin that is known to be good.

You want to find which suspect coin is bad, and as much as possible (see below), whether it is heavier or lighter.  There are therefore $28$ possible answers: $14$ suspects $\times \{heavier, lighter\}$.

You are allowed $3$ weighings on a balance.  Now of course, $3$ weighings only give you $3^3 = 27$ possible outcomes, so you cannot fully distinguish all $28$ answers.  The requirement is that:

$26$ of the $27$ outcomes must lead to a unique answer (which coin is bad and whether it is heavier or lighter)

while the last outcome must lead to knowing which coin is bad, but without knowing if it's heavier or lighter (i.e. it lumps together the $2$ answers for that coin).

The above puzzle would be difficult enough, but here's the final twist: What coins to use in a weighing cannot depend on the results of previous weighings.

To be more precise, label the suspect coins ABCDEFGHIJKLMN and the known-to-be-good coin X.  Before you begin, you must write down what two subsets of coins are involved in each of the $3$ weighings, e.g. ABCDX-EFGHN, IJKL-MNAB, CDEFGH-IJKLMN.  This way, your second weighing IJKL-MNAB is pre-determined and cannot depend on the result of the first weighing ABCDX >/=/< EFGHN, etc.  (Indeed, you can now do the $3$ weighings in any order.)

Can you find such a set of $3$ pre-determined weighings that meets the requirement?

HINT #1: The outcome $(=, =, =)$, i.e. all $3$ weighings being equal, can only happen if the bad coin is not used in any weighing at all.  This corresponds to the 2nd bullet of the requirement.  I.e. in any correct solution, there is exactly one coin that is unused in any weighing, and the outcome $(=,=,=)$ maps to this coin being bad, but without knowing if the coin is heavier or lighter.

HINT #2: Let the $28$ answers be $S = \{A+, A-, B+, B-, ..., N+, N-\}$ where $+$ and $-$ mean heavier and lighter respectively.  Meanwhile, the $27$ outcomes form a $3 \times 3 \times 3$ cube, which we can denote $T = \{-1, 0, +1\}^3$, where $-1, 0, +1$ denote the left side of the balance being lighter, equal, or heavier.  We need to find a mapping $f: S \to T$ with these properties:

Hint #1 already shows that $f(N+) = f(N-) = (0,0,0)$.

The remaining $26$ answers and $26$ outcomes must map bijectively.

Pre-determined weighings $\implies f(A+)$ and $f(A-)$ are related in a certain way.  How?

What other constraints do we need on $f$?

答:
There is a very simple description of an optimal predetermined weighing strategy for any number of coins $n\ge 1$. This uses the balanced ternary system, which I describe now.   Every positive integer $n$ can be written uniquely in the form
$$
n=\sum_{i=0}^\infty b_i3^i,\qquad b_i\in\{-1,0,+1\}\text{ for }i\in\mathbb N, \text{only finitely many $b_i\neq 0$.}
$$
For example, $25=1\cdot 3^3+0\cdot 3^2+(-1)\cdot 3^1+1.$ Using $+$ as the symbol for the digit $1$ and $-$ for the digit zero, we would write $25$ in balanced ternary, with infinitely many leading zeroes, as
$$
25=\cdots000+0-+
$$
Now, consider the following transformation on this infinite sequence of $\pm$s and $0$s; negate each symbol which has an odd number of zeroes to its right. The result to the above example is
$$
25\bowtie\cdots 000\color{red}-0-+
$$
I call this the twisted ternary representation of $25$. Next, arrange all of these infinite sequences into an infinite matrix, where the digits which are were negated when converting to twisted ternary are highlighted in red.

$$
\def\r{\color{red}}
\begin{matrix}
0  & \bowtie & \cdots & 0 & 0 & 0 & 0\\
 1 & \bowtie & \cdots & 0 & 0 & 0 & +\\
 2 & \bowtie & \cdots & 0 & 0 & + & -\\
 3 & \bowtie & \cdots & 0 & 0 & \r - & 0\\
 4 & \bowtie & \cdots & 0 & 0 & + & +\\
 5 & \bowtie & \cdots & 0 & + & - & -\\
 6 & \bowtie & \cdots & 0 & \r - & \r + & 0\\
7  & \bowtie & \cdots & 0 & + & - & +\\
 8 & \bowtie & \cdots & 0 & \r - & 0 & -\\
 9 & \bowtie & \cdots & 0 & + & 0 & 0\\
10 & \bowtie & \cdots & 0 & \r - & 0 & +\\
11 & \bowtie & \cdots & 0 & + & + & -\\
12 & \bowtie & \cdots & 0 & \r - & \r - & 0\\
13 & \bowtie & \cdots & 0 & + & + & +\\
14 & \bowtie & \cdots & + & - & - & -\\
\vdots &&\vdots &&&\vdots
\end{matrix}
$$
To find the weighing strategy for $n$ coins, number the coins from $0$ to $n-1$. For each column of that matrix, weigh the coins corresponding to row labels of the $+$'s in that columns, against the coins corresponding to $-$'s (ignoring the infinitely many leading columns whose entries $0$ to $n-1$ are all zero). You may also need to add the reference coin to one side to equalize these groups.

For your problem of $n=14$, the weighings are (where $R$ denotes reference coin):

$1,4,7,10,13\quad $ vs $\quad 2,5,8,11,R$

$2,4,6,11,13\quad $ vs $\quad 3,5,7,12,R$

$5,7,9,11,13\quad $ vs $\quad 6,8,10,12,R$.


问:
What is the probability (is my solution correct? If not solve it please)
Probability that a shooter hits the mark with a shot is 0,78. Find the probability if from five shots the shooter hits the mark at least two times.
My solution is: Let D: the event that it hits the mark at least rwi times and E: the shooter hits the mark only one time and B: the shooter never hits the mark (1-0,78) then: P(D)=1-P(B)-P(E) =1-(0.22)^5 - (5)(0,78)(0,22)^4 its multiple by 5 cuz there are 5 possibilities to choose one shot .</p>

答:
Assuming shots are independent, let the number $X$ of hits in $n = 5$ shots be
distributed $X \sim \mathsf{Binom}(n=5,\, p = 0.78).$ You seek $P(X\ge 2) = 1 - P(X\le 1).$ In R statistical software, where pbinom is a binomial CDF, this can be computed as follows:

```<code>1 - pbinom(1, 5, 0.78)
[1] 0.9903487
</code>```

Or you can use the formula for the binomial PDF:

$$1 - P(X=0) - P(X=1) = 1 - (1-.78)^5 + 5(.78)(1-.78)^4 = 0.9903487.$$

```<code>p = 0.78; 1 - (1-p)^5 - 5*p*(1-p)^4
[1] 0.9903487
</code>```

Body:
Assuming shots are independent, let the number $X$ of hits in $n = 5$ shots be
distributed $X \sim \mathsf{Binom}(n=5,\, p = 0.78).$ You seek $P(X\ge 2) = 1 - P(X\le 1).$ In R statistical software, where pbinom is a binomial CDF, this can be computed as follows:

1 - pbinom(1, 5, 0.78)
[1] 0.9903487

Or you can use the formula for the binomial PDF:

$$1 - P(X=0) - P(X=1) = 1 - (1-.78)^5 + 5(.78)(1-.78)^4 = 0.9903487.$$

p = 0.78; 1 - (1-p)^5 - 5*p*(1-p)^4
[1] 0.9903487


问:
Elementary function $f:\mathbb{R}\to \mathbb{R}$ so that $f(0)=0, f(n)=2^n$ for $n\in \mathbb{N}$
Is there any elementary function $f:\mathbb{R}\to \mathbb{R}$ such that $f(0)=0$ and $f(n)=2^n$ for every positive interger $n$?

By elementary functions, I mean functions that are sum, product, inverse, or composition of exponential, polynomial, rational, trigonometric functions.

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

答:
The function $f(x) = \exp(\log(2)x)$ satisfies $f(n) = 2^n$ for all integers $n \ge 1$ and $f(0)=1$.

To get what you want, we could try to find a function $g$ such that $g(0)=1$ and $g(n)=0$ for all integers $n \ge 1$ (and then take $f-g$). Such a function $g$ is easy to find: just take $g(x) = \sin\big(\frac{\pi}{2} 2^x\big)$.

Then $(f-g)(0) = 1-\sin(\frac{\pi}{2})=0$ and for $n \ge 1$, $(f-g)(n) = 2^n - \sin(\pi 2^{n-1}) = 2^n$.

So $x \mapsto 2^x - \sin(\pi 2^{x-1})$ is a solution.


问:
How are continued fractions useful?
On Wolfram Alpha, I see continued fractions being listed in the results. Although I understand continued fractions, and how they can be used for approximations, what is a better approximation than a decimal representation of the number?

For example,

31.999999999664

tells me so much more than

So, I do not get why continued fractions are important. Are they supposed to tell me something which I am missing all these days?

答: Continued fractions are useful for converting decimals into fractions, or reducing fractions to their common divisor.  Often finding fractions between X and Y, like 8/11 for the logrithms of the chords of the heptagon, means that (in this case), you only need to consider 11 cases.</p>

Title: How are continued fractions useful?


问:
$A$ is nilpotent matrix and satisfies the equation $A^5=3A^3-2A.$ Prove $A=0.$
I'm stuck at this question , I had an idea how to solve it but I think it's not good tho.

The Question :

$A$ is nilpotent and $A^5=3A^3-2A,$ prove $A=0.$

( $A^k = 0 :$ for certain  $k\in \mathbb{N}$)

My try : I tried changing the equation to the form $\Rightarrow A=\frac{A^3(3I-A^2)}{2}$ and then multiplying both sides by $A^{k-3}$ and proceed with $\Rightarrow  A^{k-2} = 0$.

But I'm not sure how to proceed with it now.

答:
Hint If $A^n=0$ and $n \geq 2$ then, multiply your relation by $A^{n-2}$ to get
$$A^{n+3}=3A^{n+1}-2A^{n-1}$$

Show that $A^{n-1}=0$. Repeat.

P.S. If you learned about the minimal polynomial, then the following is a shorter alternate solution: The minimal polynomial of $A$ must divide some $X^n$ and $X^5-3X^3+2X$, and hence it must divide their GCD which is $X$.


问:
Understanding roots of complicated equations
This is a very broad question apparently, but is there any systematic way to obtain (sort of) sharp estimates for roots of complicated equations? Or maybe bounds for the roots?

For example, consider the equation $$x = A - B\dfrac{e^x}{1+e^x}$$where $A,B$ are large positive integers with $B>A$. It is easy to see that this equation has a unique root, by observing that the function $f(x)=x - A + B\dfrac{e^x}{1+e^x}$ is increasing in $x$. But I want to get some pretty sharp estimates of where the root $x^*$ lies.

I have looked at several numerical examples and it seems that $x^*$ is very close to $\log(A/(B-A))$. But I cannot really prove it.

So I am asking, given a complicated equation, what is a good way to get a good estimate on where the root must lie?

答:
Unfortunately, it doesn't exist a general method, because every equation can easily be slightly different from each other.

I show you what I'd do in this case,

First things first, I'd see the entire equation as a rational one. How do I do that? I simply found a common denominator. Then, solve a system of two equations (numerator and denominator). In order to obtain the right result, you must ask yourself: "which solution satisfies both equations at the same time?" Seeing this on a number line can help a lot.

Tell me if you have any doubt.


答: Hint If $A$ is nilpotent but $A \neq 0$, then there is some $v \in \Bbb F^n$ such that $A v \neq 0$ but $A^2 v = 0$. Now, multiply both sides of the given equation, $$A^5 = 3 A^3 - 2 A ,$$ by $v$.

Title: $A$ is nilpotent matrix and satisfies the equation $A^5=3A^3-2A.$ Prove $A=0.$

Body:
I'm stuck at this question , I had an idea how to solve it but I think it's not good tho.

The Question :

$A$ is nilpotent and $A^5=3A^3-2A,$ prove $A=0.$

( $A^k = 0 :$ for certain  $k\in \mathbb{N}$)

My try : I tried changing the equation to the form $\Rightarrow A=\frac{A^3(3I-A^2)}{2}$ and then multiplying both sides by $A^{k-3}$ and proceed with $\Rightarrow  A^{k-2} = 0$.

But I'm not sure how to proceed with it now.


问:
Estimate of Bernoulli numbers from the contour integral
One knows (see https://mathworld.wolfram.com/BernoulliNumber.html)
that the Bernoulli number $B_n$ is $B_n=\frac{n!}{2i\pi}\int_{\mathcal C}\frac z{e^z-1}\frac{\mathrm dz}{z^{n+1}}$, where $\mathcal C$ is a closed contour included in $\{z\in\mathbb C\mid |z|<2\pi\}$. With this formula, is it possible to obtain the estimate: $|B_{2n}|\sim4\sqrt{\pi n}\left(\frac n{\pi e}\right)^n$? The proof I know depends on the zeta function, that I do not want to use.

答:
Note that the estimate is supposed to be
$\lvert B_{2m} \rvert \sim 4\sqrt{\pi m}\left(\frac{m}{\pi e}\right)^{2m}$, with exponent $2m$, not $m$. And by Stirling's approximation $n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n$, we can rewrite it as $\frac{\lvert B_{2m} \rvert}{(2m)!} \sim 2 \left(\frac{1}{2 \pi}\right)^{2m}$.

The contour integral you gave is equivalent to $\frac{z}{e^z-1} = \sum_{n=0}^\infty \frac{B_n}{n!} z^n $. We'll be applying a typical sort of method in combinatorics and number theory, relating asymptotics of a sequence to the singularities of a generating function. The radius of convergence is $2 \pi$ since the singularities nearest to the origin are at $z = \pm 2 \pi i$. From the standard theorem about the radius of convergence of power series, we can already conclude immediately that
$$
  \frac{1}{2 \pi} = \limsup_n \sqrt[n]{\lvert B_n \rvert / n!}.
$$
This is not as strong as the desired result, unfortunately - we need something better.

To improve the previous estimate, consider the function left over after taking $\frac{z}{e^z-1}$ and subtracting off the poles at $z = \pm 2 \pi i$, which are simple poles with residues $\pm 2 \pi i$. That is, consider $\frac{z}{e^z-1} - \frac{2 \pi i}{z - 2 \pi i} - \frac{-2 \pi i}{z + 2 \pi i}$. This has the power series $\sum_{n=0}^\infty \left( \frac{B_n}{n!} + \frac{1}{(2 \pi i)^n} + \frac{1}{(-2 \pi i)^n} \right) z^n$. And now the singularities nearest to the origin are $z = \pm 4 \pi i$. So
$$
  \frac{1}{4 \pi} = \limsup_n \sqrt[n]{\left\lvert \frac{B_n}{n!} + \frac{1}{(2 \pi i)^n} + \frac{1}{(-2 \pi i)^n} \right\rvert}.
$$
In particular, $\frac{B_{2m}}{(2m)!} + (-1)^m \frac{2}{(2 \pi)^{2m}} = O \left( \left( \frac{1}{4 \pi} + \varepsilon \right)^{2m} \right)$ for every $\varepsilon > 0$. This is a stronger form of the desired result $\frac{\lvert B_{2m} \rvert}{(2m)!} \sim 2 \left(\frac{1}{2 \pi}\right)^{2m}$.


问:
An upper bound of the form $A\cdot n\log n$ for the $n$th prime number
It is well known that if $p_n$ is the $n$th prime number, then
$$p_n>n\log n$$
and in fact, $p_n\sim n\log n$. I wanted to flip this inequality. That is, I wanted to find a real number $A$ such that
$$p_n<A\cdot n\log n$$
From my calculations, $A$ should be greater than $1$, but less than $2$ (remember, as small as possible). I calculated the value of $A$ to be less than $1.33$. But, my computer is not much fast, so I couldn't find the smallest possible $A$. And note that $p_n<1.33n\log n$ doesn't work for all $n$, but it (probably) works for all $n>40000$. So my question is, is it true that
$$p_n<1.33\,n\log n\quad\forall n>40000$$
and if yes, how can it be proved?

This inequality is indeed very good; since 1 is very near to 1.33. Any help would be appreciated.

Update: Weakening the bounds from this article gives
$$p_n<1.37n\log n$$
I don't know if this would help, but a somewhat related procedure can probably prove this.

答:
From (1.7) in the paper cited by @JimmyK4542, it follows that
$$
p_n  < 1.1756n\log n
$$
for all $n\geq 40000$. In fact, it also follows that
$$
p_n  < 1.2232n\log n
$$
for all $n\geq 20$.


问:
Is $\pi$ periodic in any numeral system?
Is $\pi$ periodic in any base-k numeral system, where k is integer ? And what is the status of this problem?</p>

答: If $\pi$ were periodic in any base, then it would be rational, and therefore periodic in every base.  This does not happen.</p>


问:
Line of Intersection of Planes (Closest Point)
I found  the line of intersection of  two planes and found the parametric equations for that line of intersection.

But how do I find a point on that line of intersection that is closest to a certain point that is given to me  in the problem?

For example my parametric equations I found for the line of intersection of the planes, 2x + 10y + 2z= -2 and 4x + 2y - 5z = -4 are

x=-2-6t
y=2t
z=-4t

and I need to find a point one the line of intersection that is closest to point (12,14,0).

Would I need to use lagrange multipliers?

答:
If this exercise is within the context of analytical geometry, I would rather suggest the following approach:

Let $P = \begin{pmatrix}-2 \\ 0 \\ 0 \end{pmatrix}$, $d = \begin{pmatrix}-6 \\ 2 \\ -4 \end{pmatrix}$ and $Q = \begin{pmatrix}12 \\ 14 \\ 0 \end{pmatrix}$. Then the line is
$P + td$ and $PQ = Q-P = \begin{pmatrix}14 \\ 14 \\ 0 \end{pmatrix}$.

You find the closest point to Q on the line by orthogonally projecting $PQ$ onto the line:

$$\left(PQ \cdot \frac{d}{|d|}\right)\frac{d}{|d|} = \frac{PQ \cdot d}{|d|^2}d =-d$$

Hence, $P-d = \begin{pmatrix}4 \\ -2 \\ 4 \end{pmatrix}$ is the closest point.


问:
Explicit construction of a nested local base in the weak*-topology
Consider the $C^*$-algebra $C([a,b])$ of complex continuous functions in a closed interval of $\mathbb{R}$. I want to construct a countable nested local base in the state space of the algebra with respect to the $weak*$-topology. Such a base exists since the algebra is separable. How can we construct it explicitly?

答:
Theorem.  Let $X$ be a normed space.

If the algebraic dimension of $X$ is  uncountable then the origin of the dual space $X'$ does not admit a countable
neighborhood base relative to the weak$^*$ topology.

If $X$ is separable then every bounded subset of $X'$ is metrizable relative to the weak$^*$ topology.

Proof.  (1) Suppose by contradiction that $\{V_n\}_n$ is a countable neighborhood base of the origin.  By definition of
the weak$^*$ topology,  for each $n$  there is a finite set $F_n\subseteq X$, and a positive number $\varepsilon _n$,  such that the set
$$
  U(F_n, \varepsilon _n):= \{f\in  X': |f(x)|<\varepsilon _n,  \forall  x\in  F_n\}
  $$
is contained in $V_n$.

By hypothesis we have that
$$
  \text{span}\left(\bigcup_nF_n\right) \subsetneq X,
  $$
so we may pick some vector $y$ in $X$
not lying in the linear span of  $\bigcup_nF_n$.  Observing that
$U(\{y\},1)$ is a neighborhood of the origin in $X'$, we will reach a contradiction by proving that
$$
  V_n\not\subseteq U(G,1),
  $$
for any given $n$.  In fact, since $y\notin \text{span}(F_n)$, we may use Hahn-Banach to produce a continuous linear
functional $f$ on $X$ vanishing on $F_n$, and such that $f(y)=1$.    It then follows that any multiple of $f$ lies in
$U(F_n, \varepsilon _n)$, and hence also in
$V_n$, but $\lambda f$ is not in $U(G,1)$ if $|\lambda |\geq 1$.

(2)  Assuming that $X$ is separable,
choose a dense subset  $\{x_n\}_n$ of the unit ball of $X$.  Given any $f$ and $g$ in $X'$, define
$$
  d(f,g) = \sum_n 2^{-n}\min\{1,  |f(x_n)-g(x_n)|\}.
  $$
It is easy to see that $d$ is a metric on $X'$ and moreover that a net $\{f_i\}_i$ converges to some $f$ according to
this metric iff
$$
  f_i(x_n)\ {\buildrel i\to\infty \over \longrightarrow}\ f(x_n),\quad\forall n.
  \tag 1
  $$

On the other hand,  it is well known that a bounded net $\{f_i\}_i$ converges to  $f$ in the weak$^*$ topology iff (1)
holds.

So if $S$ is a bounded subset of $X'$, the nets within  $S$ converge weak$^*$ iff they converge according to $d$,
thus proving that the weak$^*$ topology on $S$ is metrizable by $d$.
QED

It is well known that the algebraic dimension of any infinite dimension Banach space  is uncountable,  so the Theorem
above applies to $C([0, 1])$ and hence the origin in $C([0, 1])'$ does not admit a countable neighborhood base relative
to the weak$^*$ topology.

On the other hand, the state space of  $C([0, 1])$ is a bounded set, so part (2) of the above Theorem applies.


问:
How to simplify $x^a(u+v)^b$ and $(x+y)^a(u+v)^b$ when $a, b$ are binary digits?
Say $a, b \in \{0,1\}$ and everything else is, say, a real number (though I don't think it should matter). Consider the following two expressions

$$
(\mathrm{i})\,\, x^a(u+v)^b \qquad \qquad  (\mathrm{ii})\,\, (x+y)^a(u+v)^b
$$

Is there a way to simplify any of these expressions?

For example, we can see
\begin{align*}
x^a(u+v)^b = \begin{cases}
1 & \text{if $ab = 00$}\\
u+v & \text{if $ab = 01$}\\
x & \text{if $ab = 10$}\\
xu + xv & \text{if $ab = 11$}.
\end{cases}
\end{align*}

I'd like to distribute the $x^a$ somehow. Same goes for (ii).

答: Hint: Find a linear function such that $f(0) = 1$ and $ f(1) = x$. Then, $f(a) = x^a$.


问:
A question on extension of valuations / absolute values
I'm reading "Fourier analysis on number fields" by Ramakrishnan and Valenza. Now I come to page 160: extension of absolute values. In the text the authors write: Let $K/L$ be a finite extension of global fields of degree $n$ (say number fields). Let $\mathscr{P}(K),\mathscr{P}(L)$ be the set of their places, respectively. An absolute value on $K$ gives rise to an absolute value on $L$ via restriction, so we obtain a map:

$r:=r_{K/L}:\mathscr{P}(K)\rightarrow \mathscr{P}(L), v\mapsto u.$

If $r(v)=u$, then $v$ is said to lie over $u$. And then there's a series propositions talking about the analysis of the fibers of this $r$.

What made me puzzled is: I've already known the basic fact on the extension of absolute values: i.e. the following theorem---If $K/L$ is an algebraic extension and $L$ is equipped with an absolute value $|\cdot|_K$, then there's a unique way to extend this extension to $K$, namely the formula $|\alpha|:=|N_{L/K}(\alpha)|_K,\forall \alpha \in L$. So I once thought that for an extension of number fields $L/K$ and a fixed place $u$ of $K$, there's a unique way to extend the absolute value to $(L,v)$ and obtain a unique extension of absolute values over local fields $L_v/K_u$. But according to the book I believe there's something wrong here: there are different places lying over $u$ in general.

Can anyone tell me what's wrong in my understanding of these two facts? Thanks in advance!

答: Your issue is that the fact you are recalling, i.e. that $|\cdot|_K$ has a unique extension to $L$, applies when $K$ a complete valued field (or more generally a "henselian" valued field, but I suspect the former is where you saw it come up). So the comment will not apply to $K$ a global field.