问:
How $ \mathbb {Z}_2 \oplus \mathbb {Z}_2 \oplus \mathbb {Z}_3 \oplus \mathbb {Z}_5 \approx \mathbb {Z}_2 \oplus \mathbb {Z}_6 \oplus \mathbb {Z}_5$
In the book Contemporary abstract algebra by Joseph Gallian in page 166 the following things are said :

By using results above in an iterative fashion, one can express the same group (up to isomorphism) in many different forms. For example, we have
$$ \mathbb {Z}_2 \oplus \mathbb {Z}_2 \oplus \mathbb {Z}_3 \oplus \mathbb {Z}_5 \approx \mathbb {Z}_2 \oplus \mathbb {Z}_6 \oplus \mathbb {Z}_5 \approx \mathbb {Z}_2 \oplus \mathbb {Z}_{30} \\(*) $$

Similarly,
$$ \mathbb {Z}_2 \oplus \mathbb {Z}_2 \oplus \mathbb {Z}_3 \oplus \mathbb {Z}_5
\approx
\mathbb {Z}_2 \oplus \mathbb {Z}_6 \oplus \mathbb {Z}_5 \approx
\mathbb {Z}_2 \oplus \mathbb {Z}_3 \oplus \mathbb {Z}_2 \oplus \mathbb {Z}_5
\approx
 \mathbb {Z}_6 \oplus \mathbb {Z}_{10}\\ (**)$$
Thus, $ \mathbb {Z}_2 \oplus \mathbb{Z}_{30} \approx \mathbb{Z}_6 \oplus \mathbb{Z}_{10} $

I am not understanding how they are talking about the isomorphisms between these product groups with the help of the results above. I have tried a lot but can't get through it. Please help. It's really important in my self study.

The results above were :

Theorem 8.2 Let G and H be finite cyclic groups. Then $G \oplus H$ is cyclic if and only if $|G|$ and $|H|$ are relatively prime.

Corollary 1 An external direct product $G_1 \oplus G_2 \oplus ... \oplus G_n$ of a finite number of finite cyclic groups is cyclic iff $|G_i|$ and $|G_j|$ are relatively prime when $i \ne j$.

Corollary 2 Let $ m = n_1n_2 ...n_k $. Then $\mathbb Z_m$ is isomorphic to $\mathbb Z_{n_1} \oplus \mathbb Z_{n_2} \oplus ... \oplus \mathbb Z_{n_k}$ iff $n_i$ and $n_j$ are relatively prime when $i \ne j$.

答: I think I see it now, as $Z_2 \oplus Z_3 \approx Z_6$ so it is replaced by $Z_6$ and again as $Z_5 \oplus Z_6 \approx Z_{30}$ that's why it is replaced by $Z_{30}$ and doing so on in the next cases too.


问:
Evaluate $\int x \cos x^2 dx$
Hay I have hit this in my book

Evaluate $\int x \cos x^2 dx$.

I got $x^2 \sin(x^2) / 2 $

But I used a online calculator to check it and it is giving me $\sin(x^2)/2 $

Where dose my X go?

ContentLicense: CC BY-SA 3.0

Title: Evaluate $\int x \cos x^2 dx$

Body:
Let $u=x^2$. Then $du = 2xdx$. Hence
$$
\int x \cos(x^2)\ dx = \frac{1}{2}\int \cos(u)\ du = \frac{\sin(u)}{2} + C = \frac{\sin(x^2)}{2} + C.
$$


问:
Solution to 2nd order ode
So my problem starts with applying the Laplace transform to a PDE.
After apply the Laplace Transform, my PDE becomes a 2nd order ODE:
$$\ddot{x}+\frac{2}{t}\dot{x}-\frac{p}{k}x = -\frac{A}{k}$$
where p,A and k are positive constants.
This does not look like a 2nd order ODE with constant coefficients so wondering what type of ODE this is and how do I solve this?

答:
Your ODE is of the type in $y(x)$
$$xy''+2y'-axy=-cx \implies xy''+2y'-x(ay+c)=0~~~~(1)$$
Let $ay-c=z$, then we get a homogenous 2nd order ODE:
$$\frac{x}{a}z''+\frac{2}{a}z'-xz=0 \implies z''+\frac{2}{x}z'-2az=0~~~~(2)$$
Next Let $z(x)= u(x)/x$, we get
$$u''(x)-2au(x)=0 \implies u(x)=C_1 e^{x\sqrt{2a}}+C_2 e^{-x\sqrt{2a}}$$
$$\implies z(x)=\frac{1}{x}[C_1 e^{x\sqrt{2a}}+C_2 e^{-x\sqrt{2a}}]$$
Finally
$$y(x)=\frac{1}{ax}[C_1 e^{x\sqrt{2a}}+C_2 e^{-x\sqrt{2a}}]+\frac{c}{a}$$

ContentLicense: CC BY-SA 4.0

Title: Solution to 2nd order ode


问:
Issue with distance point-to-line
I am having a small issue with the distance calculation of a point-to-line and I cannot find the mistake.

I have the line:

$$x + 2y + 3 = 0$$

And I am trying to compute its distance to point $(x_0, y_0) = (0,0)$. According to the well-known formula:

$$ d = \frac{\left|ax_0+by_0+c\right|}{\sqrt{a^2 + b^2}} = \frac{\left|3\right|}{\sqrt{5}} = 1.3416$$

Now, I try to compute the same distance on a different manner. The line intersects the $y$ axes at $p=(0, -1.5)$. I am trying to solve the following triangle (sorry for the cheap plotting):

The distance computed before should correspond to the length of the red segment (perpendicular to the line that goes through the desired point), which could also be computed as:
$$ d = 1.5 \cdot cos(30^{\circ}) = 1.299$$

Both results for the length of the red segment do not match. What am I doing wrong?

Thank you in advance!

答:
THe equation of  line:

$y=-\frac 12 x-\frac32$

Equation of red line:

$y=2x$

So the coordinates of intersection is : $(x=-\frac 35, y=\frac 65)$

The distance is:

$d=\sqrt {(-\frac 35)^2+(\frac{6}5)^2}=1.3416$

Title: Issue with distance point-to-line


问:
Some tablets of a company are refilled by a machine

Some tablets of a company are refilled by a machine that each contain 25 mg of medication. The amount that goes into the tablets is normally distributed, with mean 25 mg and a standard deviation of 5 mg.

What is the probability that the total weight of medication in 20
randomly selected tablets will be at least 500 mg?

I was wondering if such a question would require the Central Limit Theorem, or can it be done some other way?

Thank you.

答:
The central limit theorem does not need to be invoked, because you are told that the medication amount in each capsule is normally distributed.  Therefore, the total medication contained in ten capsules is also normally distributed, without the need for the CLT, because the sum of independent normal random variables is also normal.

Using your notation, $$X_i \sim \operatorname{Normal}(\mu = 10, \sigma = 0.5), \quad i = 1, 2, \ldots 10,$$ hence $$S_{10} = \sum_{i=1}^{10} X_i \sim \operatorname{Normal}(\mu_S = 100, \sigma_S = 0.5 \sqrt{10}).$$  It follows that $$\Pr[S_{10} \ge 90] = \Pr\left[\frac{S_{10} - \mu_S}{\sigma_S} \ge \frac{90 - 100}{0.5 \sqrt{10}} \right].$$ What we did in the step above was standardize $S_{10}$.  The LHS of the inequality is now a standard normal random variable $Z$ with zero mean and unit variance; thus $$\Pr[S_{10} \ge 90] = \Pr[Z \ge -6.32456].$$  The probability that a standard normal random variable is greater than $-6.32456$ standard deviations below its mean of $0$ is almost equal to $1$; using a computer, it is $$0.999999999873018570526456\ldots.$$

Think about why this answer makes sense.  If the standard deviation of a single capsule is $0.5$ mg, and you have $10$ capsules, in order for the total medication amount to be $10$ mg below the total mean, ($90$ is $10$ below the mean of $100$), each capsule would on average need to contain at least $2$ standard deviations below the average amount per capsule.  Since being $2$ or more standard deviations below the mean happens only about $2.27\%$ of the time, it should not be surprising that the answer is almost $1$.


问:
Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$, where $f_n(x)=\frac{1-|x|^n}{1+|x|^n}$ .

For each $n\in \Bbb N$ and $x\in \Bbb R$, define
$$\begin{align} f_n(x)=\frac{1-|x|^n}{1+|x|^n}_. \end{align} $$
Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$.

For $x\in (-1,1)$, $f_n(x)\to 1$, and for $x>1$ or $x< -1$, $ f_n(x)=\frac{1-|x|^n}{1+|x|^n} \to -1$. Define the limit function $\begin{align} f(x)=\begin{cases}
1,  & \text{if $x\in (-1,1)$ } \\
-1, & \text{if $x>1$ or $x< -1$}
\end{cases}_. \end{align}$ Then $f_n \to f$ pointwise on $\Bbb R$.

Let $\varepsilon_n=\sup\{\left|f(x)-f_n(x) \right|:x\in\Bbb R \}$, how can I get a contradiction by showing that $\varepsilon_n$ do not converge to $0$?

答:
Your function $f$ is not complete: for $|x|=1$ we have $f_n(x) = 0 \to 0,$ hence $f(x)=0.$

The limit function $f$ is not continuous, but all $f_n$ are continuous ! Conclusion ?


问:
Existence of solution for a linear system mod 2
Let $A$ be a (skew-) symmetric matrix over $\mathbb{Z}/2$.
(In fact, I would take $A$ as the linking matrix of an oriented framed link in $S^3$ or the matrix representing the intersection form on a closed smooth 4-manifold. The following statement however does seem to hold in general.)
I am interested in the following linear system over $\mathbb{Z}/2$,
$$a_{i1}x_1+a_{i2}x_2\cdots+a_{in}x_n=a_{ii},\quad i=1,\cdots,n.$$

This system is known to always have a solution. (c.f. Saveliev's Lectures on the Topology of 3-Manifolds.)
But I cannot see why is this true unless $A$ is nonsingular over $\mathbb{Z}/2$.
Is there a general method to deal with these kinds of linear systems?

答: The $\mathbb{Z}_2$ intersection form on a closed smooth $4$-manifold is always nondegenerate by Poincare duality.


问:
How to use stars and bars to count how many terms there are in a polynomial expansion?
So here's what I understand from this.

The number of terms in the expansion (before combining like terms) of any polynomial is given by ${k^{n}}$ where ${k}$ is the number of variables and ${n}$ is the degree of the polynomial. For example, ${(x + y + z)^4}$ is given by ${3^{4}=81}$.

To find the number of terms after combining, I can use the binomial theorem. This gives me ${n + 1}$ terms for an ${n}$th-degree binomial. For example, ${(x+y)^{12}}$ has 13 terms.

But the thing is, this doesn't work with trinomials and higher. My teacher instructed me to use the stars and bars method to find terms for such problems. I understand how to plug values in but don't quite understand how it works.

For example, the number of terms in the trinomial ${(x+y+z)^{7}}$ is given by ${n + k - 1 \choose k - 1 }={9 \choose 2} = {36}$. How exactly does this work? It seems relatively simple if you use the whole "distribute ${n}$ pieces of candies among ${k}$ number of children" type of problems, but I don't quite get it for this specific case.

答: When you expand $(a+b+c+....)^n$, the sum of the exponents in every term must be $n$, and every possible term whose exponents have a sum of $n$ will be present. For example, in$(x+y+z)^7$, some possible terms are $x^2y^3z^2, xy^4z^2, x^7, y^2z^5$, plus every other term whose exponents add to $7$. Thus you are "distributing" the exponent $n$ among the $k$ terms in the bracket, which is why the calculation is the same as for distributing $n$ candies among $k$ children.


答:
A simple argument

As all $f_n$ are continuous, if the converge to $f$ was uniform, $f$ would have to be continuous. As it is not the case, the convergence is not uniform.

Title: Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$, where $f_n(x)=\frac{1-|x|^n}{1+|x|^n}$ .

Body:
For each $n\in \Bbb N$ and $x\in \Bbb R$, define
$$\begin{align} f_n(x)=\frac{1-|x|^n}{1+|x|^n}_. \end{align} $$
Prove that $\{f_n\}$ converges pointwise but not uniformly on $\Bbb R$.

For $x\in (-1,1)$, $f_n(x)\to 1$, and for $x>1$ or $x< -1$, $ f_n(x)=\frac{1-|x|^n}{1+|x|^n} \to -1$. Define the limit function $\begin{align} f(x)=\begin{cases}
1,  & \text{if $x\in (-1,1)$ } \\
-1, & \text{if $x>1$ or $x< -1$}
\end{cases}_. \end{align}$ Then $f_n \to f$ pointwise on $\Bbb R$.

Let $\varepsilon_n=\sup\{\left|f(x)-f_n(x) \right|:x\in\Bbb R \}$, how can I get a contradiction by showing that $\varepsilon_n$ do not converge to $0$?


答:
U-substitution allows us to "reverse"/invert the process one uses with the chain rule in differentiating.

Challenge: Try differentiating your answer, and then differentiating the given solution, and see which one gives you the expression in the integrand, below, in $(1)$?

Now, we have the following integral.

$$\int x \cos(x^2 )dx \tag{1}$$

Using u - substitution:

Let $\color{blue}{\bf u = x^2};\;$ then $\,du = 2x dx \implies \color{red}{x\,dx = \frac 12 du}$

Substitute: $$\int \color{red}{\bf x } \cos(\color{blue}{\bf x^2 })\color{red}{\bf \,dx} \quad = \quad \color{red}{\bf \frac 12} \int \cos \color{blue}{\bf u} \color{red}{\bf\,du} \quad = \quad \frac 12 \sin u + C$$

Back substituting: $$\frac 12 \sin u + C\; =\;\frac 12 \sin(x^2) + C \;= \;\frac{\sin(x^2)}{2} + C$$

Title: Evaluate $\int x \cos x^2 dx$

Body:
Hay I have hit this in my book

Evaluate $\int x \cos x^2 dx$.

I got $x^2 \sin(x^2) / 2 $

But I used a online calculator to check it and it is giving me $\sin(x^2)/2 $

Where dose my X go?


答:
When you expand $(x_1+x_2+...+x_k)^n$ and combine the terms, the sum will look like this: $$\sum_{0\leq\lambda_1,\lambda_2,...,\lambda_k\leq n, \lambda_1+\lambda_2+...+\lambda_k=n}x_1^{\lambda_1}x_2^{\lambda_2}...x_k^{\lambda_k}\cdot c_i$$ where $c_i$s are random constants.

So the number of terms after you expand and combine is the number of ordered triplets $(\lambda_1,\lambda_2,...,\lambda_k)$ with the property that $0\leq\lambda_1,\lambda_2,...,\lambda_k\leq n$ and $\lambda_1+\lambda_2+...+\lambda_k=n$ which gives you stars and bars. (you distribute $n$ candies to $k$ children)

Title: How to use stars and bars to count how many terms there are in a polynomial expansion?

Body:
So here's what I understand from this.

The number of terms in the expansion (before combining like terms) of any polynomial is given by ${k^{n}}$ where ${k}$ is the number of variables and ${n}$ is the degree of the polynomial. For example, ${(x + y + z)^4}$ is given by ${3^{4}=81}$.

To find the number of terms after combining, I can use the binomial theorem. This gives me ${n + 1}$ terms for an ${n}$th-degree binomial. For example, ${(x+y)^{12}}$ has 13 terms.

But the thing is, this doesn't work with trinomials and higher. My teacher instructed me to use the stars and bars method to find terms for such problems. I understand how to plug values in but don't quite understand how it works.

For example, the number of terms in the trinomial ${(x+y+z)^{7}}$ is given by ${n + k - 1 \choose k - 1 }={9 \choose 2} = {36}$. How exactly does this work? It seems relatively simple if you use the whole "distribute ${n}$ pieces of candies among ${k}$ number of children" type of problems, but I don't quite get it for this specific case.


问:
Maximizing a harmonic mean via SOCP
$$ \begin{array}{ll} \underset {x} {\text{maximize}} & \left(\sum\limits_{i=1}^m \dfrac{1}{a_i^T x + b_i}\right)^{-1} \\ \text{subject to} & Ax > b \end{array} $$ where $a_i^T$ is the $i$-th row of the matrix $A$. Transform this optimization problem into a second-order cone program (SOCP).

The solution is maximize $1^Tt$ subject to $t_i(a_ix+b_i) \ge 1, i = 1,2...,m$ and $t \ge 0$.

Can someone explain to me why are these problems equivalent? I don't see how transformation has been carried out. I think, there has been made some change of variables. Like $t_i = 1/(a_ix+b_i)$. However, if so how should one get $t_i(a_ix+b_i) \ge 1$. I learnt that by the change of variable one can obtain an equivalent problem if the transformation that one uses is one-to-one and its image covers the domain of the original function. I don't see it here. Also how was the $^{-1}$ (inverse) sign be eliminated in the "transformed" problem (SOCP).

答:
The objective of maximizing
$$\mathrm{maximize}\ \frac{1}{\sum_i \frac{1}{a_i^Tx+b}}$$
is equivalent to minimizing the denominator
$$\mathrm{minimize}\ {\sum_i \frac{1}{a_i^Tx+b}}$$
which is equivalent to minimizing
$$\mathrm{minimize}\ {\sum_i t_i}\quad \mathrm{subject\ to}\quad t_i\geq\frac{1}{a_i^Tx+b}, i=1,\ldots,n$$
where $t_i$ are new auxiliary variables. The last step holds because in the optimal solution the constraints in "subject to" will actually be satisfied with equality (if not, one could decrease some $t_i$ and have a better objective). This is just a special variant of the more general rule that
$$\mathrm{minimize}_{x}\ f(x)$$
is the same as
$$\mathrm{minimize}_{t,x}\ t\quad \mathrm{subject\ to}\ t\geq f(x).$$

Have a look at the MOSEK Moseling Cookbook for examples of this kind in conic quadratic optimization.

Title: Maximizing a harmonic mean via SOCP


问:
Lyapunov Function showing a unstable critical point of system of ODE to be stable
I am considering the following system of ODE:

$$x'=y, y'=x-x^3$$

where (0,0) is a critical point of it.

By considering its locally linear system, I found that this critical point is an unstable saddle point. And I believe that this conclusion is true since I had verified it by plotting the phase portrait.

However, when I consider the following Lyapunov Function,
$$V(x,y)=\frac{1}{4}(u^2-1)^2+\frac{1}{2}v^2$$
upon computation, the derivative of V by the system is constant zero, so it is negative semidefinite.
Would this fact show that (0,0) is a stable critical point?

Thank you very much for reading the post.

答:
While it is true that solutions of the ODE follow the level curves of $V$, the point $(0,0)$ is also a saddle point of
$$
V=\frac12y^2+\frac14-\frac12x^2+\frac14x^4,
$$
thus your claim is wrong/not applicable.


问:
How to evaluate $\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2$ in a particular way.
How to evaluate:
$$\sum _{n=1}^{\infty }\left(\frac{H_n^2+H_n^{\left(2\right)}}{n}\right)^2,$$
without splitting the expression into more sums.

Here $H_n^{\left(m\right)}=\sum _{k=1}^n\frac{1}{k^m}$ is the harmonic number of order $m$.

If one just wants to evaluate it if we split we have,
$$2\sum _{n=1}^{\infty }\frac{H_n^2H_n^{\left(2\right)}}{n^2}+\sum _{n=1}^{\infty }\frac{H_n^4}{n^2}+\sum _{n=1}^{\infty }\frac{\left(H_n^{\left(2\right)}\right)^2}{n^2},$$
Then making use of this results one only has to compute
$$\sum _{n=1}^{\infty }\frac{\left(H_n^{\left(2\right)}\right)^2}{n^2}$$

But I'd like to know if its possible to evaluate the series without splitting or expanding the terms.

答:
To find the desired series we must first consider the following integral.
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx$$
To evaluate it one can make use of the following trilogarithm identity.
$$\operatorname{Li}_3\left(\frac{x}{x-1}\right)=-\operatorname{Li}_3\left(x\right)-\operatorname{Li}_3\left(1-x\right)+\zeta \left(3\right)+\frac{1}{6}\ln ^3\left(1-x\right)$$
$$+\zeta \left(2\right)\ln \left(1-x\right)-\frac{1}{2}\ln \left(x\right)\ln ^2\left(1-x\right)$$
Using it on the previous integral yields:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(x\right)}{x}\:dx-\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(1-x\right)}{x}\:dx$$
$$+\zeta \left(3\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:dx+\frac{1}{6}\int _0^1\frac{\ln ^5\left(1-x\right)}{x}\:dx+\zeta \left(2\right)\int _0^1\frac{\ln ^3\left(1-x\right)}{x}\:dx$$
$$-\frac{1}{2}\int _0^1\frac{\ln \left(x\right)\ln ^4\left(1-x\right)}{x}\:dx$$
$$=-\frac{81}{2}\zeta \left(6\right)+2\zeta ^2\left(3\right)+12\sum _{k=1}^{\infty }\frac{H_k}{k^5}-\sum _{k=1}^{\infty }\frac{H_k^2}{k^4}-\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{k^4}-2\sum _{k=1}^{\infty }\frac{H_k^{\left(3\right)}}{k^3}$$
The series remaining can be calculated quite easily and a nice thing to know is that to evaluate  them one does not have to cross paths with the series in the body of the OP.

Thus:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{581}{24}\zeta \left(6\right)-4\zeta ^2\left(3\right)$$
Now what is left to do is to consider the following generating function.
$$\sum _{k=1}^{\infty }\frac{x^{k-1}}{k}\left(H_k^2+H_k^{\left(2\right)}\right)=-2\frac{\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}$$
Which can be found along other generating functions in the book (Almost) Impossible Integrals, Sums, and Series, page $\#285$.

Using it on the previously found integral means that,
$$\int _0^1\frac{\ln ^2\left(1-x\right)\operatorname{Li}_3\left(\frac{x}{x-1}\right)}{x}\:dx=-\frac{1}{2}\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2$$
Thus:
$$\sum _{k=1}^{\infty }\left(\frac{H_k^2+H_k^{\left(2\right)}}{k}\right)^2=\frac{581}{12}\zeta \left(6\right)+8\zeta ^2\left(3\right)$$


问:
Finding two skew symmetric matrices with same exponential value
I've have the following problem.

I need to come with an example for two skew symmetric matrix $A,B$ such that $\exp(2\pi A)$ = $\exp(2\pi B)$

I tried matrix $2x2$ but I guess this is useless to just try, I probably need to construct the matrix by some thm.

Now, Only I know is that the function $F$ defined by $F(A) = \exp(A)$ has the next properties:

derivative near 0 is the identity matrix

$F$ itself is diffeomorphism of neighborhood contains 0 to neighborhood of $I_n$

How can I proceed from here?

答:
Hint: If we let $J = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$, then $\exp(\theta J) = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos\theta\end{bmatrix}$.

Can you find two numbers $\theta_1 \neq \theta_2$ such that $\exp(\theta_1 J) = \exp(\theta_2 J)$?


问:
Trace of a matrix
What is the trace of $e^{A}$ where A is a $4 \times4$ matrix

$$\begin{bmatrix}0 & 0 & 0 & t\\
 0 & 0 &-t & 0\\
 0 & t & 0 & 0\\
-t & 0 & 0 & 0 \end{bmatrix}$$

where $t=\dfrac{\pi}{4}$.

答:
Looking at the basis $(e_1,e_4,-e_2,e_3)$, $A$ is similar to a block-diagonal matrix made of two identical blocks
$$
B=\pmatrix{0&t\\-t&0}.
$$
Note that $A^n$ is similar to the block-diagonal matrix made of two copies of $B^n$, so $\mbox{tr} \,A^n=2\,\mbox{tr}\,B^n$ for every $n\in \mathbb{N}$. Then an easy induction shows that
$$
B^{2n}=\pmatrix{(-1)^nt^{2n}&0\\ 0&(-1)^nt^{2n}}\qquad B^{2n+1}=\pmatrix{0&(-1)^nt^{2n+1}\\ (-1)^{n+1}t^{2n+1}&0}.
$$
Therefore
$$
\mbox{tr}\,e^A=\mbox{tr}\sum_{n=0}^{+\infty}
\frac{1}{n!}A^n=\sum_{n=0}^{+\infty}
\frac{1}{n!}\mbox{tr}A^n=\sum_{n=0}^{+\infty}
\frac{2}{n!}\mbox{tr}B^n=4\sum_{n=0}^{+\infty}
\frac{(-1)^nt^{2n}}{(2n)!}=4\cos t=2\sqrt{2}.$$</p>

Title: Trace of a matrix


问:
Philosophy of Lebesgue integration
This is a question about Lebesgue's theory of integration. What is the motive behind giving a meaning to the integral to as larger a class of functions as possible? What is the motive behind looking for a "good" integration theory? So far I am accustomed to mathematical theories that involve constructing new subsets and finding patterns among those subsets in solving some concrete questions. But, analysis seems to be a little different in that it is primarily based on giving meaning to operations on infinitely many quantities. In this context, I find it hard to understand how Lebesgue might have thought about coming up with a new theory when Riemman's theory was already in place. What exactly is the fault of the analysis that it can't integrate Dirichlet's function using Riemann's theory?

In some instances, Riemman's theory of integration gives a finite value to some improper integrals, but Lebesgue theory does not. How do we know beforehand that Lebesgue theory does not give a different value than the Riemann's to some, say, a number-theoretic question? In which case, what does "the value of integral" mean?

Is it that once the theory is developed, it might have applications in the future? I am sorry my questions are incoherent and may be meaningless too, but I request to understand my difficulty in accepting two theories of integration in mathematics like one accepts Newton's theory of gravity and Einstein's theory of relativity in Physics. I urge you to resolve my difficulty in understanding.

答: If $f\colon[a,b]\longrightarrow\Bbb R$ is Riemann-integrable, then it is also Lebesgue-integrable, and the integrals are equal. So, in the context of the Lebesgue integration theory, there are more functions that you can integrate. Also, more natural statements about integrals are true in the context of Lebesgue integration. For instance, if $f$ is integrable and if $g\colon[a,b]\longrightarrow\Bbb R$ is such that $|g|\leqslant|f|$, then $g$ is integrable too (assuming that $g$ is measurable): this holds for the Lebesgue integral, but not for the Riemann one. Also, a natural distance from a function $f$ to a function $g$ in the context of a theory of integration is $\int_a^b|f(x)-g(x)|\,\mathrm dx$. With respect to this distance, the space of Riemann-integrable functions is not complete: there are Cauchy sequences that don't converge. But the space of Lebesgue-integrable functions is complete, and this is important for Analysis; it allows use to build new functions as limits of existing ones.

Title: Philosophy of Lebesgue integration


问:
Set of constants such that trigonometric equation always sums to zero

Consider the equation:

$$ \sum_{i=1}^n \cos( x + a_i) = 0$$
Where the above identity holds for all $x$ $\in$ R, how much solutions set of constants are possible to make the equation zero i.e: sequence of $a_i$ for a given $n$? Find all such sequences.

A question I came up with myself, I'll post my work about the question and hopefully,  someone else can help me finish.

Consider,

$$ \sum_{i=1}^n \cos(x + a_i) = 0 \tag{1}$$

Differentiate with $x$ (remember defined for all $x \in R$

$$ \sum_{i=1}^n \sin(x + a_i) = 0 \tag{2}$$

Combining (1) and (2),

$$ \sum_{i=1}^n e^{j(x+a_i)} = 0$$

For the identity to hold for all $x$,

$$ \sum_{i=1}^n e^{ja_i} = 0$$

If we consider complex numbers as vectors, then a trivial solution to the above equation is by application of polygon rule of vector addition. Eg: for n=3, find three-unit complex numbers that form a triangle, n=4 find four-unit complex numbers forming a square. If so, then we can think of the sequence of $ a_i$ as an arithmetic progression characterized by a common difference and a starting term $a_1$. I.e:

$$ \sum_{i=1}^n e^{j(a_1 + (n-1)d)} = 0$$

I'm having a bit of difficulty figuring out what the common difference should be...

Other than that, are there 'other kind' of solutions to the problem statement?

答:
Partial answer

For $n=2$ there is a unique solution up to the rotation moving one of the element to $1$. Same thing for $n=3$.

For $n =2p$ even, the are an infinite number of solutions up to a rotation: it is sufficient to be able to pair elements diametrically opposed.

For $n=2p+1 \gt 3$, just take $3$ elements positioned on an equilateral triangle plus an even number of elements positioned as just above.

Need to see what kind of other solutions exist...


问:
Fourier analysis convergence argument help
Suppose $g \in L^{1}([0,1])$ and define:

$\hat{g}(n)=\int g(x)e^{-2\pi inx}dx$ for $n\in\mathbb{Z}$.

I want to show that if $g\in L^{1}([0,1])$

and $\{\hat{g}(n)\}\in\ell^{1}(\mathbb{Z})$

then $S_{N}g(x)=\underset{|n|\leq N}{\sum}\hat{g}(n)e^{2\pi inx}$ converges on $[0,1]$.

Attempt so far:

We know that $\sum|\hat{g_n}|<\infty$.

Being absolutely convergent, can define $g_0:\mathbb{R}\rightarrow C$ as:
$$g_0(x)=\sum^{\infty}_{n=-\infty}\hat{g}e^{2\pi inx}=lim_{N \rightarrow \infty}S_Ng(x)$$
We can observe that the limit is uniform in $x$ because:
$|g_0(x)-S_Ng| \leq \sum_{|n|>N}\hat{g}(n)$ for $x \in [0,1]$

Which leads to: $$\lVert g_0 - S_N g \rVert_{\infty}\leq \sum_{|n|>N}\hat{g}(n)\rightarrow 0$$ as $N \rightarrow \infty$

So,

$$lim_{N \rightarrow \infty} \lVert \frac{S_0g + S_1g+...+S_{N-1}g}{N} \rVert_{\infty}=0$$

Is the above attempt legitimate? Any clarifications and all criticisms are welcome.

答: You have proved  the Fourier series converges uniformly to some function $g_0$. To show that $g=g_0$ a.e. it is enough to show that $\hat g(n)=\hat{g_0}(n)$ for all $n$. For this write down  the definition of $\hat{g_0}(n)$ and interchange the integral and the limit (which is permissible by uniform convergence).


问:
Find all polynomials in a vector space orthogonal to another polynomial with Gram Schmidt possibly
Here I am given the following.

Define an inner product in $\mathcal{P}_2$ by: $$\left
=p(-1)q(-1)+p(0)q(0)+p(1)q(1).$$ a) Starting with the basis $\left\{1,x,x^2\right\}$ of $\mathcal{P}_2$, use the Gram-Schmidt process to find an orthonormal basis for $\mathcal{P}_2$ with respect to this inner product.

I already did this. The orthonormal basis is $\left\{\frac{1}{\sqrt{3}},\frac{1}{\sqrt{2}}(x),\sqrt{\frac{3}{2}}\left(x^2-\frac{2}{3}\right)\right\}$.

b) Find all the polynomials in $\mathcal{P}_2$ that are orthogonal to $x^2-1$.

I'm stuck here. I took the inner product of $ax^2+bx+c$ and $x^2-1$ which gives be that $c=0$ but how can I get what $a$ and $b$ are?

答:
You are done: simply verify that both $x$ and $x^2$ give zero inner product with $x^2-1$, and therefore so do any linear combination of them, that is, $a$ and $b$ can be arbitrary.

Also, the perpendicular of a nonzero vector in a 3d inner product space should be a 2d subspace, i.e. the degree of freedom of variables $a,b,c$ is $2$, which is clearly the case now.


问:
Solution for a second order differential equation
I am looking for a hint for the solution of $y''=\frac{6}{x}y$ without using the power series. Thanks for any help.

答:
Let $y=z \sqrt{6x}$ to make
$$4x^2\, z''+4x\,z'-(1+24x)z=0$$ which is very close to Bessel differential equation.

It would even be more obvious using $x=t^2$.


问:
Prove $|(a+b) \log (a+b) - a \log a| \leq b \log (\frac{1}{b})$
Can you prove or point me where I can reference the following assertion?

$|(a+b) \log (a+b) - a \log a| \leq b \log (\frac{1}{b}); a \in [0,1), b \in (0,1], a+b \in [0,1]$

答:
We can rewrite the given inequality as,

$$\left\lvert\log{\frac{(a+b)^{a+b}}{a^{a}}}\right\rvert \leq log{(\frac{1}{b})^{b}}$$

Now since $$a+b \geq a $$so $$(a+b)^{(a+b)}\geq a^{a}$$
hence, $$\left\lvert\log{\frac{(a+b)^{a+b}}{a^{a}}}\right\rvert =log{\frac{(a+b)^{a+b}}{a^{a}}}$$
so thee inequality is now,
$$log{\frac{(a+b)^{a+b}}{a^{a}}}\leq log{(\frac{1}{b})^{b}}$$
since logarithm is an increasing function,
$$\frac{(a+b)^{a+b}}{a^{a}}\leq (\frac{1}{b})^{b}$$
$$\frac{(a+b)^{a+b}}{a^{a}}\leq \frac{1}{b^{b}}$$
$$\frac{(a+b)^{a+b}}{a^{a}}\geq b^{b}$$
$$(a+b)^{a}(a+b)^{b} \geq a^{a}b^{b}$$
Now for $$(a+b) \in [0,1],a \in [0,1),b \in (0,1]$$
$$(a+b)^{a} \geq a^{a}\ and\ (a+b)^{b} \geq b^{b}$$
so$$(a+b)^{a}(a+b)^{b} \geq a^{a}b^{b}$$
is true.
Therefore the given inequality is true.


问:
A problem pertaining to functional equations with conditions $P(2020) = 2019$ and (b) $(P(x) + 1)^2 = P(x^2 + 1)$ for all $x\in\mathbb{R}$

Determine all polynomials $P(x)$ with real coefficients satisfying the
following conditions:

$P(2020) = 2019$

$(P(x) + 1)^2 = P(x^2 + 1)$ for all
$x\in\mathbb{R}$

I wrote down $p(x$) as $a_nx^n +\ldots+a_0$ and then started applying substituting $2020$ and tried to get awkward results. I know this is absolutely wrong. Kindly provide a rigorous answer

答:
I define $(x_n)$ by:  $\ x_0=2020 \ $ and $\ \forall n \in \mathbb N \ , \ x_{n+1}=x_n^2 +1$.

We can show by induction that:  $\ \forall n \in \mathbb N \ , P(x_n)=x_n-1$

$P(x)-(x-1)$ is a polynomial and has an infinity of roots.

And we can conclude that $P(x)=x-1$.


问:
Understanding Hom-bundle on vector bundles (Atiyah)
I am trying to make sense of the following statment (from Atiyah's K-theory book)

Suppose $V$ and $W$ are vector spaces, and that $E=X \times V, F=X \times W$ are the corresponding product bundles. Then any homomorphism $\phi:E \to F$ determines a map $\Phi:X \to \text{Hom}(V,W)$ by the formula $\phi(x,v) = (x,\Phi(x)v).$

I guess I don't really understand what this is trying to say. Is it that a homomorphism between product bundles over a common base space, induces a map between the associated vector spaces? I guess I am used to seeing a map given explicitly rather than implicitly.

(Tangentially Related)

答:
In general, if $E$ and $F$ are arbitrary vector bundles over $X$, then by restricting to the fibers over $x \in X$, any vector bundle map $\phi: E \to F$ gives a linear map $\phi_x:E_x \to F_x$.  In the case of a product bundle, all fibers of $E$ are $V$ and all fibers of $F$ are $W$.  So every $x \in X$ gives us a map $\phi_x: V \to W$.  Conversely, given a map $\Phi: X \to Hom(V,W)$ we get a map of vector bundles by $(x,v) \mapsto (x, \Phi(x)(v))$.

Is it that a homomorphism between product bundles over a common base space, induces a map between the associated vector spaces?

A map of product bundles induces many maps between the associated vector spaces; you get one for each $x \in X$ and these maps vary continuously (or smoothly) depending on what type of maps of vector bundles you want to consider.  Indeed, maps of product bundles are nothing but maps between the vector spaces parametrized by the base space.


问:
Show that for each $m\in \mathbb{N}$, there exists $n\in \mathbb{N}$ such that $x^2\equiv 1 \pmod n$ has at least $m$ solutions?
I am trying to solve the following problem:

Show that for each $m\in \mathbb{N}$, there exists $n\in \mathbb{N}$ such that $x^2\equiv 1 \pmod n$ has at least $m$ solutions?

I figured that this is equivalent to showing that there is a sequence $n_1,n_2,n_3,\dots$ such that

$$x^2\equiv 1 \pmod {n_1}\qquad x^2\equiv 1 \pmod {n_2}\qquad x^2\equiv 1 \pmod {n_3} \qquad\dots$$

Has an increasing number of solutions. By computation, I figured that this sequence could be

$$n_1=2\qquad n_2 =2\cdot 3\qquad n_3 = 2\cdot 3 \cdot 5\qquad n_4 = 2\cdot 3 \cdot 5 \cdot 7 \qquad \dots $$

Now comes my trouble: I can't figure out why "adding" another prime in each step of the sequence makes more solutions appear. Can you help me?

答:
let $p$ be a prime number

$x^2\equiv 1 \pmod p$ $\Rightarrow$ $x^2-1\equiv (x-1)(x+1)\equiv 0 \pmod p$

$(x-1)(x+1)\equiv 0 \pmod p$ $\Rightarrow$ $x \equiv 1$ or $x \equiv -1$

Let $n=p_1p_2...p_n$ and $x^2\equiv 1 \pmod n$

$x^2 \equiv 1 \pmod {p_1} $ $\Rightarrow$ $x \equiv \pm1 \pmod{p_1} $

$x^2 \equiv 1 \pmod{p_2}$ $\Rightarrow$ $x \equiv \pm1 \pmod{p_2} $

$\vdots$

$x^2 \equiv 1 \pmod{p_n}$ $\Rightarrow$ $x \equiv \pm1 \pmod{p_n} $

we know that for $a_i \in \{-1,1\}$ $(a_1,a_2,...,a_n)=(x \pmod{p_1},x \pmod{p_2},...,x \pmod{p_n})$ shows a solution for $x^2 \equiv 1 \pmod{n}$

there are $n$ primes and $2$ options for per prime so there are $2^n $ solutions and for all $m \in N^+$ there exists a $n \in N$ such that $2^{n}<m<2^{n+1}$ which proves your statement


问:
Convergence of Bernoulli distribution using Central Limit Theorem
while I was studying the Central Limit theorem, this problem showed up:

If $0<p = 1-q < 1$ and $x > 0$, then
$$
\sum_{k \in K} \binom{n}{k}p^kq^{n-k} \to 2 \int_{0}^{x} \frac{1}{\sqrt{2\pi}}e^{\frac{-1}{2}u^2} du,
$$
where $K := \left\{k: np - x\sqrt{npq} \leq k \leq np+x\sqrt{npq}\right\}$.

The book sugests to apply the theorem to a sequence of random variables with the Bernoulli distribuition. Can somebody help me with this? Doesn't look hard, but I cant understand even the first steps.

Thanks in advance!!

答: Let $(X_i)$ be an i.i.d sequence where $X_n$ takes the value $1$ with probability $p$ and  the value $0$ with probability $q$. Let $S_n=X_1+X_2+\cdots+X_n$. Note that $EX_n=p$ and $EX_n^{2}=p$ so the variance of $X_n $ is $p-p^{2}=pq$. By CLT $P(-x \leq \frac {S_n-np} {\sqrt {npq}} \leq x) \to \int_{-x}^{x} \frac  1{\sqrt {2\pi}} e^{-x^{2}/2} dx=2\int_{0}^{x} \frac  1{\sqrt {2\pi}} e^{-x^{2}/2} dx$. Now just re-write the event on the left.


问:
Proving that $GL(n,\mathbb R)$ is an open subset of $M_n(\mathbb R)$ [the direct way]
Definition:

$GL(n,\mathbb R)$ is the set of all $n \times n$ invertible matrices like $A$ such that $det(A)\neq 0$.

$M_n(\mathbb R)$ is simply the set of all $n\times n$ matrices.

Question:

Prove that $GL(n,\mathbb R)$ is an open subset of $M_n(\mathbb R)$ using the definition of an open set

Note: I know the proof that gets help from the continuity of $det$ function and the fact that $\mathbb R - \{0\}$ is open. But, I want a proof which works just with the direct definition of open subsets (which can be found here).

Any idea?

答:
Expanding on 3 computational parts of @MichaelBurr 's proof above:

1.

We want bounds for:
$$\det(N)=\sum_{\sigma\in S_n}(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}(M_{i,\sigma(i)}+\delta\epsilon_{i,\sigma(i)}).$$

Let $S$ be a non-empty subset of $\{1,2,...,n\}$.$$\prod_{1\leq i\leq n}(a_i+b_i)=\sum_{S\subseteq\{1,2,...,n\}}\left(\prod_{i\in S}a_i\right)\left(\prod_{j\in S^c}b_j\right)$$

For $S\ne\emptyset$,
\begin{align}
\prod_{1\leq i\leq n}(M_{i,\sigma(i)}+\delta\epsilon_{i,\sigma(i)})
&=\sum_{S\subseteq\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)\\
&=\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)+\prod_{1\leq i\leq n}M_{i,\sigma(i)}\end{align}
Also since $K,\delta>0$
\begin{align}
\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)&\le\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}K\right)\left(\prod_{j\in S^c}\delta\right)\\
&=\sum_{S\subset\{1,2,...,n\}}K^{|S|}\cdot\delta^{(n-|S|)}\\
&=\sum_{i=|S|\\{1\le i\le n-1}}K^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}\\
&\le\sum_{i=1}^nK^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}\\
\end{align}
where, $|S|$ is the cardinality of $S$. Let $A=\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)$.
\begin{align}\therefore
\det(N)
&=\sum_{\sigma\in S_n}(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}(M_{i,\sigma(i)}+\delta\epsilon_{i,\sigma(i)})\\
&=\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)\right\}+\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}\prod_{1\leq i\leq n}M_{i,\sigma(i)}\right\}\\
&=\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}\sum_{S\subset\{1,2,...,n\}}\left(\prod_{i\in S}M_{i,\sigma(i)}\right)\left(\prod_{j\in S^c}\delta\epsilon_{j,\sigma(j)}\right)\right\}+\det(M)\\
&=\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}A\right\}+\det(M).
\end{align}
Using
$$-\sum_{\sigma\in S_n}A\le\sum_{\sigma\in S_n}\left\{(-1)^{\operatorname{sgn}(\sigma)}A\right\}\le\sum_{\sigma\in S_n}A$$
and
$$\sum_{\sigma\in S_n}A\le\sum_{\sigma\in S_n}\sum_{i=1}^n K^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}\le n!\sum_{i=1}^n K^i\cdot\delta^{(n-i)}\cdot{\binom{n}{i}}=n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}$$
we finally get,
$$\det(M)-n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}\leq\det(N)\leq \det(M)+n!\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}.$$

2.

We want a simpler form of the bound obtained above, preferably in terms of $\delta$.

$$K^{n-i}\le
\begin{cases}
K^{n-1}(\ge1),  & \text{if $K\ge1$} \\
1(>K^{n-1}) & \text{if $0<K<1$}
\end{cases}\mathrm{~~and~~}\delta^i\le\delta~~(\mathrm{since~}0<\delta<1).$$
Therefore,$$\sum_{i=1}^n\binom{n}{i}\delta^iK^{n-i}\le\sum_{i=1}^n\binom{n}{i}\delta\max\{1,K^{n-1}\}=\delta\max\{1,K^{n-1}\}\sum_{i=1}^n\binom{n}{i}\le\delta\max\{1,K^{n-1}\}2^n$$
Finally,
$$\det(M)-n!\delta\max\{1,K^{n-1}\}2^n\leq\det(N)\leq \det(M)+n!\delta\max\{1,K^{n-1}\}2^n.$$

3.

Now, to show that $N\in GL_n(\mathbb{R})$ it is enough to show that $\det(N)>0$ or $\det(N)<0$. We get this by choosing $\delta$ such that
$$\det(M)-n!\delta\max\{1,K^{n-1}\}2^n>0\mathrm{~~or~~}\det(M)+n!\delta\max\{1,K^{n-1}\}2^n<0$$
$$\Rightarrow\frac{\det(M)}{n!\max\{1,K^{n-1}\}2^n}>\delta\mathrm{~~or~~}\frac{\det(M)}{n!\max\{1,K^{n-1}\}2^n}<-\delta$$
$$\Rightarrow\frac{|\det(M)|}{n!\max\{1,K^{n-1}\}2^n}>\delta.$$


问:
$f$ convex, $M$ subspace of $\mathbb{R}^n$. Then $x = argmin_{x \in M} f(x)$ $\iff$ $\nabla f(x) \perp M$
I am trying to solve the following exercise, which I found in some online lecture notes on convex functions (https://sites.math.washington.edu/~ddrusv/crs/Math_516/bookwithindex.pdf, page 18):

Let $f: \mathbb{R}^n \to \mathbb{R}$ be a smooth convex function and $M \subset \mathbb{R}^n$ be a linear subspace.
Then $x = argmin_{x \in M} f(x)$ $\iff$ $\nabla f(x) \perp M$.

I am able to prove one implication but not the other. I use the fact that, since $f$ is convex, we have
$$
f(y) \geq f(x) + \nabla f(x)^\top (y-x) \quad \forall y \in M.
$$
Then, if $\nabla f (x) \perp M$, we have $\nabla f(x)^\top (y-x) = 0$ and hence $f(y) \geq f(x)$ for all $y \in M$.

I have problems with the other implication.

I would like to use the above convexity inequality together with the fact that
$$
f(y) \geq f(x) \quad \forall y \in M
$$
but I do not come up with a smart way to put them together. Can someone help please?

答:
The convexity of $f$ is not needed for the implication in the other direction. Assume that $x \in M$ and
$$
 f(x) = \min \{ f(y) \mid y \in M \} \,.
$$
If $h \in M$ then the function
$$
 \Bbb R \ni t \mapsto \phi(t) = f(x + th)
$$
is differentiable and has a minimum at $t = 0$. It follows that
$$
 0 = \phi'(0) = \nabla f(x)^T \cdot h \, .
$$
This holds for all $h \in M$, so that $\nabla f(x) \perp M$.


问:
If $p$ is a polynomial approximating a continuous function $h$, why can we assume $\left\|p\right\|_\infty\le\left\|h\right\|_\infty$?
Let $K\subseteq\mathbb R$ be compact and $h:K\to\mathbb R$ be continuous and $\varepsilon>0$. By the Stone-Weierstrass theorem, there is a polynomial $p:K\to\mathbb R$ with $\left\|h-p\right\|_\infty<\varepsilon$.

Why can we choose $p$ such that $\left\|p\right\|_\infty\le\left\|h\right\|_\infty$?

I'm not sure whether it's really crucial that $p$ is a polynomial. In general, if $K$ is a compact Hausdorff space, $\mathcal C$ is a subalgebra of $C(K)$ with $1\in\mathcal C$ and $f:K\to\mathbb R$ is continuous with $c:=\left\|f\right\|_\infty\ne0$, then there is a $\tilde g\in\mathcal C$ with $$\left\|\frac fc-\tilde g\right\|_\infty<\frac\varepsilon2.$$ Now $g:=c\tilde g\in\mathcal C$ and $\left\|f-g\right\|_\infty<\varepsilon$. But only if $\left\|\tilde g\right\|\le1$, we obtain $\left\|g\right\|_\infty\le c=\left\|f\right\|_\infty$ ...

答: $\|p\|\leq \epsilon +\|h\|$. Let $q=\frac {\|h\|} {\epsilon+\|h\|} p$. The $q$ is a polynomial and $\|q\| \leq \|h\|$. Now $\|q-h\| \leq \|q-p\|+\|p-h\|$. Can you finish?


问:
Exterior derivative of $(dx^{i_1} \land ... \land dx^{i_p})$
I am currently working on differentialforms. We introduced the exterior derivative, as follows:

The exterior derivative is the operator $d: \Lambda^p \rightarrow \Lambda^{p+1}$ with the properties:

$$1. \quad d( \alpha + \beta ) = d\alpha + d\beta$$
$$2. \quad d^2 = 0$$
$$3. \quad \text{on 0-forms}: df = \tfrac{\partial f}{\partial x^i} dx^i$$
$$4. \quad \text{for a function $f$ and a form $\omega$}: d(f\omega) = (df) \land \omega + f d\omega$$

Now, one of the problems I have to work on is the following:

Show that $d( dx^{i_1} \land ... \land dx^{i_p} ) = 0$.

I am afraid, that I have no clue on how to show this. Can anyone provide a hint or a solution?

答: Hint: Try to proceed by induction on $p$. Start with $d(dx^i)=0$ (property 2), use this to show that for $i<j$ you get $d(x^idx^j)=dx^i\wedge dx^j$, which implies the statement for $p=2$ and so on.


问:
What real number is exactly one less than its cube?
And does it have any of the special properties that the golden ratio (one less than its square) has?</p>

答:
If you multiply everything out and solve the resulting cubic function, you should get

$$\frac{\sqrt[3]{12\sqrt{69} + 108} - \sqrt[3]{12\sqrt{69} - 108}}{6}$$

But I don't know if this number has any  special properties yet.


答:
For this for any m we need to have:

$x=\underbrace{p_1p_2p_3p_4 \cdot\cdot\cdot}_{m\space times} ±1$

Therefore:

$x^2=P(t= \underbrace{p_1p_2p_3p_4 \cdot\cdot\cdot}_{m\space times}) +1$

Where P(t)represent a polynmial having factors $p_1p_2p_3...$.

In this case any prime factor $(p_i)$ can be considered as n such that :

$x^2\equiv 1 \mod (n)$

Body:
I am trying to solve the following problem:

Show that for each $m\in \mathbb{N}$, there exists $n\in \mathbb{N}$ such that $x^2\equiv 1 \pmod n$ has at least $m$ solutions?

I figured that this is equivalent to showing that there is a sequence $n_1,n_2,n_3,\dots$ such that

$$x^2\equiv 1 \pmod {n_1}\qquad x^2\equiv 1 \pmod {n_2}\qquad x^2\equiv 1 \pmod {n_3} \qquad\dots$$

Has an increasing number of solutions. By computation, I figured that this sequence could be

$$n_1=2\qquad n_2 =2\cdot 3\qquad n_3 = 2\cdot 3 \cdot 5\qquad n_4 = 2\cdot 3 \cdot 5 \cdot 7 \qquad \dots $$

Now comes my trouble: I can't figure out why "adding" another prime in each step of the sequence makes more solutions appear. Can you help me?


问:
2D asteroid problem
I don't know how to solve the next problem, and if anyone could explain to me step by step how it is solved I would thank you a lot. I know it's not hard, but I'm not seeing how to do it. Thank you in advance!

An asteroid passes next to a planet with mass m, and position (0, 0). The asteroid has an initial velocity (Vx, Vy) and a starting position (Px, Py). Considering that gravity is 1, and ignoring the size of both objects, get the formula to calculate the asteroid position in the future.

答:
Here is the coordinate system and the force diagram. At first, it's a good idea to forget about any initial velocities and just focus on the forces. There's only one force acting, which is the gravity on the asteroid. According to Newton's law of universal gravitation, it can be calculated as
$$
\vec{F_g} = G \frac{m_{\text{asteroid}} m_{\text{planet}}}{r^2} \vec{r}
$$
where $G=1$ is a constant (set to $1$ by the problem statement) and $\vec{r}$ is a  unit vector pointing from the asteroid to the planet (overlapping with the blue vector in the picture). $r$ is just the distance between the two objects, so $r^2 = x ^2 + y^2$.

Next, we want to split this into $x$ and $y$ components. The angle that the blue vector makes (let's call it $\theta$) with the horizontal satisfies
$$
\tan \theta = \frac{x}{y}
$$
Now, the components of the force are
$$
\begin{cases}
x\text{-component}: F_x = -\sin \theta |\vec{F_g}| = -\sin \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2} \\
y\text{-component}: F_x = -\cos \theta |\vec{F_g}| = -\cos \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2}  \\
\end{cases}
$$
To wrap up the analysis with the forces, we apply Newton's second law of motion to the asteroid: $\sum \vec{F} = m_{\text{asteroid}} \vec{a}$. Since gravity is the only force, the previous expressions give the $x$ and $y$-components of the acceleration:
$$
\begin{cases}
x\text{-component}:-\sin \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2} = m_{\text{asteroid}} a_x \\
y\text{-component}:-\cos \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2} =  m_{\text{asteroid}} a_y  \\
\end{cases}
$$
Here, $a_x$ and $a_y$ are the components of acceleration. The mass of the asteroid gets cancelled,and we set $m_{\text{planet}}=m$, and now we have the acceleration components
$$
\begin{cases}
a_x = \frac{d^2 x}{dt^2} = -\frac{m \sin \theta  }{x ^2 + y^2} \\
a_y = \frac{d^2 y}{dt^2} = - \frac{m \cos \theta}{x ^2 + y^2}  \\
\end{cases}
$$
But we're not done yet. We have to form the equations of motion. So far, we only have commented on the second derivative of position. We have to incorporate the initial conditions as well (we know the initial values of $P_x$ and $P_y$, and also the initial values of the first derivative of these). So now we have the coupled second-order differential equations
$$
\begin{cases}
\frac{d^2 x}{dt^2} = -\frac{m \sin \theta  }{x ^2 + y^2} \\
\frac{d^2 y}{dt^2} = - \frac{m \cos \theta}{x ^2 + y^2}  \\
\end{cases}
\qquad
\text{with}
\quad
\begin{cases}
\frac{d x}{dt} (0) = V_x \\
\frac{d y}{dt} (0) = V_y \\
\end{cases}
\quad
\text{and}
\quad
\begin{cases}
x (0) = P_x \\
y(0) = P_y \\
\end{cases}
$$
This can be solved numerically with a computer. When implementing the solution algorithm to a computer, one must be a bit careful with that, but there are several methods. The simplest is Euler's method, but there you have to choose a really small time step. You would have to make hundreds if not thousands of steps to have a decent accuracy. Another way to solve would be some version of Runge-Kutta.

Here's a quickly put together Python3 code that could serve as an example on how to numerically solve this problem:

```<code>import math
import sys

m = 10.0
delta_t = 0.001 # Don't put this to zero!
V_x = 10.0
V_y = -3.0
P_x = 2.0
P_y = 3.0
T_max = 500
total_iterations = int(T_max / delta_t) + 1

def calculateAcceleration(x,y):
    global m
    theta = math.atan2(x,y)
    acceleration_x = -m*math.sin(theta)/(x**2 + y**2)
    acceleration_y = -m*math.cos(theta)/(x**2 + y**2)
    return [acceleration_x,acceleration_y]

def iterate(x,y,velocity_x, velocity_y):
    global delta_t
    acceleration_vector = calculateAcceleration(x,y)
    acceleration_x = acceleration_vector[0]
    acceleration_y = acceleration_vector[1]
    # Using first-order Euler method
    velocity_x_new = acceleration_x + delta_t*acceleration_x
    velocity_y_new = acceleration_y + delta_t*acceleration_y
    x_new = x + velocity_x_new*delta_t
    y_new = y + velocity_y_new*delta_t

    return [x_new,y_new,velocity_x_new,velocity_y_new]

def fly():
    global V_x, V_y
    global P_x, P_y
    global delta_t
    global total_iterations

    iteration_number = 0
    time = 0

    # Initial values
    x = P_x
    y = P_y
    velocity_x = V_x
    velocity_y = V_y

    while iteration_number < total_iterations:
        iteration_number += 1
        new_parameters = iterate(x,y,velocity_x, velocity_y)
        x       = new_parameters[0]
        y       = new_parameters[1]
        velocity_x  = new_parameters[2]
        velocity_y  = new_parameters[3]
        print(x,y)
        input()
fly()
</code>```

Title: 2D asteroid problem

Body:
Here is the coordinate system and the force diagram. At first, it's a good idea to forget about any initial velocities and just focus on the forces. There's only one force acting, which is the gravity on the asteroid. According to Newton's law of universal gravitation, it can be calculated as
$$
\vec{F_g} = G \frac{m_{\text{asteroid}} m_{\text{planet}}}{r^2} \vec{r}
$$
where $G=1$ is a constant (set to $1$ by the problem statement) and $\vec{r}$ is a  unit vector pointing from the asteroid to the planet (overlapping with the blue vector in the picture). $r$ is just the distance between the two objects, so $r^2 = x ^2 + y^2$.

Next, we want to split this into $x$ and $y$ components. The angle that the blue vector makes (let's call it $\theta$) with the horizontal satisfies
$$
\tan \theta = \frac{x}{y}
$$
Now, the components of the force are
$$
\begin{cases}
x\text{-component}: F_x = -\sin \theta |\vec{F_g}| = -\sin \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2} \\
y\text{-component}: F_x = -\cos \theta |\vec{F_g}| = -\cos \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2}  \\
\end{cases}
$$
To wrap up the analysis with the forces, we apply Newton's second law of motion to the asteroid: $\sum \vec{F} = m_{\text{asteroid}} \vec{a}$. Since gravity is the only force, the previous expressions give the $x$ and $y$-components of the acceleration:
$$
\begin{cases}
x\text{-component}:-\sin \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2} = m_{\text{asteroid}} a_x \\
y\text{-component}:-\cos \theta \frac{m_{\text{asteroid}} m_{\text{planet}}}{x ^2 + y^2} =  m_{\text{asteroid}} a_y  \\
\end{cases}
$$
Here, $a_x$ and $a_y$ are the components of acceleration. The mass of the asteroid gets cancelled,and we set $m_{\text{planet}}=m$, and now we have the acceleration components
$$
\begin{cases}
a_x = \frac{d^2 x}{dt^2} = -\frac{m \sin \theta  }{x ^2 + y^2} \\
a_y = \frac{d^2 y}{dt^2} = - \frac{m \cos \theta}{x ^2 + y^2}  \\
\end{cases}
$$
But we're not done yet. We have to form the equations of motion. So far, we only have commented on the second derivative of position. We have to incorporate the initial conditions as well (we know the initial values of $P_x$ and $P_y$, and also the initial values of the first derivative of these). So now we have the coupled second-order differential equations
$$
\begin{cases}
\frac{d^2 x}{dt^2} = -\frac{m \sin \theta  }{x ^2 + y^2} \\
\frac{d^2 y}{dt^2} = - \frac{m \cos \theta}{x ^2 + y^2}  \\
\end{cases}
\qquad
\text{with}
\quad
\begin{cases}
\frac{d x}{dt} (0) = V_x \\
\frac{d y}{dt} (0) = V_y \\
\end{cases}
\quad
\text{and}
\quad
\begin{cases}
x (0) = P_x \\
y(0) = P_y \\
\end{cases}
$$
This can be solved numerically with a computer. When implementing the solution algorithm to a computer, one must be a bit careful with that, but there are several methods. The simplest is Euler's method, but there you have to choose a really small time step. You would have to make hundreds if not thousands of steps to have a decent accuracy. Another way to solve would be some version of Runge-Kutta.

Here's a quickly put together Python3 code that could serve as an example on how to numerically solve this problem:

import math
import sys

m = 10.0
delta_t = 0.001 # Don't put this to zero!
V_x = 10.0
V_y = -3.0
P_x = 2.0
P_y = 3.0
T_max = 500
total_iterations = int(T_max / delta_t) + 1

def calculateAcceleration(x,y):
    global m
    theta = math.atan2(x,y)
    acceleration_x = -m*math.sin(theta)/(x**2 + y**2)
    acceleration_y = -m*math.cos(theta)/(x**2 + y**2)
    return [acceleration_x,acceleration_y]

def iterate(x,y,velocity_x, velocity_y):
    global delta_t
    acceleration_vector = calculateAcceleration(x,y)
    acceleration_x = acceleration_vector[0]
    acceleration_y = acceleration_vector[1]
    # Using first-order Euler method
    velocity_x_new = acceleration_x + delta_t*acceleration_x
    velocity_y_new = acceleration_y + delta_t*acceleration_y
    x_new = x + velocity_x_new*delta_t
    y_new = y + velocity_y_new*delta_t

    return [x_new,y_new,velocity_x_new,velocity_y_new]

def fly():
    global V_x, V_y
    global P_x, P_y
    global delta_t
    global total_iterations

    iteration_number = 0
    time = 0

    # Initial values
    x = P_x
    y = P_y
    velocity_x = V_x
    velocity_y = V_y

    while iteration_number < total_iterations:
        iteration_number += 1
        new_parameters = iterate(x,y,velocity_x, velocity_y)
        x       = new_parameters[0]
        y       = new_parameters[1]
        velocity_x  = new_parameters[2]
        velocity_y  = new_parameters[3]
        print(x,y)
        input()
fly()


问:
Continuity of the approximations
Let $(X,\|\cdot\|)$ be a Banach space, $r>0$ and $B\subset X$ non-empty. Assume that $f:B\longrightarrow X$ is continuous, and for each $f(x)\in f(B)$ there exists a unique $y_{x}\in C$ such that $\|f(x)-y_{x}\|\leq r$, for certain (we can assume even convex and compact) $C\subset f(B)$, $C\neq f(B)$.

So, we can define the function $g:B \longrightarrow X$ by $g(x):=y_{x}$, with $y_{x}$ as above. As $y_{x}$ is unique, $g$ is well defined. My question is: Can we prove that $g$ si continuous?

Many thanks in advance for your comments and/or suggestions.

答:
Since $C\subset f(B)$ then the uniqueness of $y_x$ implies that $\|x-y\|>r$ for any distinct points $x,y\in C$. Since $C$ is compact, it is finite. Thus the set $f(B)$ is contained in the union of a finite family $\mathcal D$ of pairwise disjoint closed balls of radius $r$ centered at points of $C$. For each subset $C’$ of $C$, $g^{-1}(C’)=\bigcup\{ f^{-1}(D): D\in\mathcal D$ and $D$ is centered at some point of $C’\}$. Since the map $f$ is continuous and the family $\mathcal D$ is finite, the set $g^{-1}(C’)$ is closed and thus the map $g$ is continuous.

But the map $g$ is continuous even when we do not require $C\subset f(B)$. Indeed, let $x\in B$ be any point and $\{x_n\}$ be any sequence of points of $B$ converging to $x$. Since $C$ is compact, a sequence $\{g(x_n)\}$ has a cluster point $y\in C$. Then $f(x_n)$ is contained in a closed ball $B_0$ of radius $r$ centered at $y$ for infinitely many $n$. By the continuity of $f$, $f(x)\in B_0$, so $g(x)=y$. Suppose to the contrary that $\{g(x_n)\}$ does not converge to $y$. Then there exists a neighborhood $U$ of $y$ and a subsequence $\{x_{n_k}:k\in\Bbb N\}$ of $\{x_n\}$ such that $g(x_{n_k})\in C\setminus U$ for each $k\in\Bbb N$. Since $C\setminus U$ is compact, a sequence $\{g(x_{n_k}):k\in\Bbb N \}$ has a cluster point $y’\in C\setminus U$. Then similarly to the above we can show that $g(x)=y’\ne y$, a contradiction.

On the other hand, if we relax the compactness of $C$ then the function $g$ can be discontinuous. For instance, let $X=\ell_2$ with the standard basis $\{e_n\}$ (for each natural $n$, $n$-th coordinate of $e_n$ is $1$, and the other coordinates are zeroes). Let $C=\{0\}\cup \{e_1+(1+1/n)e_n: n\in\Bbb N\}$, $B=\{e_1\}\cup \{e_1+e_n/n:n\in\Bbb N\}$, $r=1$, and $f$ be the identity map. Then $g(e_1)=0$, $\{e_1+e_n/n\}$ converges to $e_1$, but a sequence $\{g(e_1+e_n/n)\}= \{e_1+(1+1/n)e_n\}$ is not even Cauchy.

Title: Continuity of the approximations


问:
Find an orthogonal basis for the subspace described by $x − y − z = 0$.
Find an orthogonal basis for the subspace described by $x − y − z = 0$.

First, find two linearly independent vectors in the span of the subspace above.

$v_1=\begin{pmatrix} 1\\ 1\\0 \end{pmatrix}$ $v_2=\begin{pmatrix} 1\\ 0\\1 \end{pmatrix}$

Then find the projection of $v_2$ on to $v_1$. this equals $\frac{v_2⋅v_1}{v_1⋅v_1}v_1=\begin{pmatrix} 1/2\\ 1/2\\0 \end{pmatrix}$

Now I need to find the orthogonal component. $\begin{pmatrix} 1\\ 0\\1 \end{pmatrix}-\begin{pmatrix} 1/2\\ 1/2\\0 \end{pmatrix}=\begin{pmatrix} 1/2\\ -1/2\\1 \end{pmatrix}$

Therefore, the orthogonal basis for the subspace is $\begin{pmatrix} 1\\ 1\\0 \end{pmatrix},\begin{pmatrix} 1/2\\ -1/2\\1 \end{pmatrix}$

Questions: Did I do this correctly? Do my steps make sense? Alternative solutions?

答:
Looks good to me. You could explicitly check that

They are orthogonal

They span the space you want them to span

just to be in the safe side. Arithmetic errors can happen to anyone. Checking things is a good habit to have.

A common next step is to make basis vectors of unit length, but that would be an orthonormal basis, so it doesn't seem like your exercise requires that.


问:
Convergence radius of power series on specific set
Let $p,q$ be primes such that $3 \leq p < q$ and $\langle p, q \rangle = \lbrace mp+nq ~|~ m,n\in\Bbb{N} \rbrace$.

Compute the radius of convergence of :
$$ S(z) = \sum_{s \in \langle p,q \rangle}{z^s}. $$

My first thought was to write $s$ as $mp+nq$ and then use the convergence test :
$$ \frac{\lvert z^{(m+1)p+(n+1)q} \rvert}{\lvert z^{mp+nq} \rvert} \longrightarrow \lvert z^{pq} \rvert. $$
So the radius is $1$.
But I'm pretty sure that it's not correct to use the test on two variables like above.

答:
$\langle p,q \rangle$ is a subset of the positive integers. For $|z| < 1$
$$
\sum_{s \in \langle p,q \rangle}{|z|^s} \le \sum_{n=1}^\infty |z|^n = \frac{1}{1-|z|} < \infty
$$
so that the radius of convergence is at least one. On the other hand, $\langle p,q \rangle$ is an infinite set, so that the series diverges for $z=1$.

Therefore the radius of convergence  is exactly one.

Your proof is not correct because $mp+nq$ and $(m+1)p+(n+1)q$ are not necessarily two consecutive exponents of the series.


问:
$ x\cos y + y \cos x = \pi$ , how do I find $y''(0)$ in easiest way possible?
I considered brute force differentiating but that is very hard. I also tried expanding the cosines and '$y$' as a Taylor series but I  don't think that helps much either

答:
Well well, it's been over six months and I had recently found a better way to do this.

$$ x \cos y + y \cos x = \pi$$

Now, rewrite as:

$$ Q(x,y) = x \cos y + y \cos x - \pi$$
r
D.w.r.t.x and use implicit function theorem

$$ \frac{dQ}{dx} = \left[  \cos y  - y \sin x \right] +y' \left [ -x \sin y + \cos x \right] $$

Now notice that $\frac{dQ}{dx} = G(x,y,y')$ , use the implicit function theorem yet again:

$$ \frac{dG}{dx} =( \left[ - y \cos x \right] + y' \left[-\sin y - \sin x \right] )+ (\left[ - \sin y - \sin x \right] + y' \left[ -x \cos y \right])y' +\left[ (-x \sin y + \cos x) \right] y'' $$

Hence, our three implicit equations relating the function and it's derivatives are:
$$ \left[  x \cos y + y \cos x - \pi \right]=0 \tag{0}$$
$$ \left[  \cos y  - y \sin x \right] +y' \left [ -x \sin y + \cos x \right] =0\tag{1} $$
$$( \left[ - y \cos x \right] + y' \left[-\sin y - \sin x \right] )+ (\left[ - \sin y - \sin x \right] + y' \left[ -x \cos y \right])y' +\left[ (-x \sin y + \cos x) \right] y'' =0 \tag{2}$$

Plugging $ x = 0$,

$$ y- \pi = 0 \tag{0}$$

$$ \left[ \cos(y) \right] + y' =0$$

$$ ( -y -y' \sin y) + (- \sin y  )y' + \left[ 1 \right] y''=0$$

Simplfying,

$$ y= \pi$$
$$ y'=1$$
$$y''=\pi$$

Refer here


问:
Does convergence to normal distribution (central limit theorem) imply convergence in probability and expectation?
Let $X_n$ be a sequence of random variables, and let $a_n$ and $b_n$ be sequences of real numbers. Assume that we have
$$
\frac{X_n-a_n}{b_n} \to \mathcal{N}(0,1)
$$
in distribution, where $\mathcal{N}(0,1)$ denotes the standard normal distribution.

My question is does our assumptions imply that $X_n/a_n \to 1$ in probability and $\mathbb{E}[X_n]/a_n\to 1$? What about $\mathbb{E}{X_n^2}$?

Further questions:
If the answer to the questions are all negative, is there any general sufficient condition on $a_n$ and $b_n$ that CLT implies convergence in probability or expectation?

I ask this question because convergence in probability or expectation is often easier to prove than convergence in distribution, so I wander when the latter actually implies the former.

答:
Counter-example for first part: $X_n=-X$ for all $n$ where $X\sim N(0,1)$ and $a_n=0, b_n=1$.  [$X_n$ does not tend to $0$ in probability and $EX_n^{2}$ does not tend to $0$].

CLT theorem gives easy examples where the hypothesis is satisfied but $X_n$ does not even have finite mean.


问:
Find $\int\cosh^\frac{1}{2}u~du$ and/or $\int\frac{1}{\cosh^\frac{1}{2}\theta}~d\theta$
I am trying to find a function of the form $y=f(x)$ such that the volume of the solid generated by the function between any two points around the $x$ axis is numerically equal to its length bewteen the same $2$ points. Using integrals, we can express this as an equation:
$$\pi\int y^2dx=\int\sqrt{\left(\frac{dy}{dx}\right)^2+1}\quad dx$$
ie
$$\pi y^2=\sqrt{\left(\frac{dy}{dx}\right)^2+1}$$
Rearranging, we get the differential equation
$$\frac{dy}{dx}=\sqrt{\pi^2y^4-1}$$
I tried to solve this first by separating the variables:
$$\int\frac{1}{\sqrt{\pi^2y^4-1}}~dy=x+c$$
So the next step is obviously to find
$$I=\int\frac{1}{\sqrt{\pi^2y^4-1}}~dy$$
Now here lies the problem. I tried to go about this using $2$ methods, each using different substitutions:

1.

Let $\pi y^2=\cosh u$:
$$\sinh u\frac{du}{dy}=2\pi y\implies dy=\frac{1}{2\pi y}\sinh{u}~ du$$
$$\implies I=\frac{1}{2\pi}\int\frac{1}{y}du=\frac{1}{2\sqrt{\pi^3}}\int\cosh^\frac{1}{2}u~du$$
How can I find the following?
$$\int\cosh^\frac{1}{2}u~du$$
2.

Using the substitutions $\pi y^2=p$ followed by $p=\cosh \theta$ gives us the result
$$I=\frac{1}{2\sqrt{\pi}}\int\frac{1}{\cosh^\frac{1}{2}\theta}~d\theta$$
Again, how I can I find the following?
$$\int\frac{1}{\cosh^\frac{1}{2}\theta}~d\theta$$

Thanks for your help.

答:
As mentioned in comments, there is no solution with elementary functions. However, if you have Byrd and Friedman – the great big book of elliptic integrals and functions – this problem falls very easily.

The second integral $\int\cosh^{-1/2}\theta\,d\theta$ is the slightly easier one to tackle, for B&F 296.00 gives
$$\int\frac1{\sqrt{\cosh\theta}}\,d\theta=\sqrt2F\left(\cos^{-1}\frac1{\sqrt{\cosh\theta}},m=\frac12\right)+K$$
Substituting $\cosh\theta=\pi y^2$ gives
$$I=\frac1{\sqrt{2\pi}}F\left(\cos^{-1}\frac1{y\sqrt\pi},\frac12\right)+K$$


问:
Can we use symbols from the language model our universe is made of?
In my question here, within my response to part 1, I used the symbol "$+$", which was not a part of the provided $\mathcal L$. Is this something that is allowed?

Thank you!

答:
See the very first paragraphs of Leary's textbook [page 4]regarding structures and languages:

we have to be able to look at some symbols in a particular formal language and then take those symbols and relate them in some way to a mathematical structure. Different interpretations of the symbols will lead to different conclusions as regards the truth of the formal statement.

We have to take care of the difference between the structure, i.e. the domain of mathematical objects which is "talked about" and which is the subject matter
of the discussion, and the language with which we "talk about" the objects and their properties.

We have a language $\mathcal L$ with the function symbol $S$ (for the successor) and the relation symbol $<$.

When we interpret the language into e.g. the structure $\mathbb N$ of natural numbers, we can say that $S(n)$ is interpreted with $n+1$, because "we know" what does it mean for a natural number $n$ to add one to it.

But you cannot use the symbol $+$ in the formulas of the language $\mathcal L$, because it is not parte of $\mathcal L$.


问:
What is the difference between regulator and stabilization
What is the difference between regulator and stabilization in control theory

don't they both minimize the disturbance to the system?

could answer be elaborated from the view of state and output?

答:
If I understand your question properly, we have two things to consider.

Stability

This is the ability for a system to operate under a variety of conditions without self destructing. There are two categories of interest; 1) the ability for the system to return to equilibrium after an initial displacement away from equilibrium and, 2) the ability for the system to produce a bounded output for a bounded input. There are differences between time-varying and nonlinear systems.

Regulator

In this system the reference input is zero (since it is missing). It is desired to keep the output as near to zero as possible in the system.

So, in both cases, we certainly want them to remain stable, but one allows two types of inputs, while the other wants to have a constant zero input. One can ask why they defined two different concepts for this and that is a valid question, but they needed a system that didn't have any inputs as opposed to the general case I suppose.


问:
Finding minimum value of $\tan A+\tan B$, given $A+B=30^\circ$

If $A>0$, $B>0$ and $A+B=30^\circ$, find the minimum value of $\tan A+\tan B$.

A similar question has been asked here. Many good answers are posted there. I understood that.

One particular answer by @Bill Kleinhans intrigued me as it was short and straightforward. But I didn't fully understand that answer. It uses Jensen's inequality, which I know, but don't know how to use that to get the answer here.

Not posting this as a comment there because Bill has been away close to three years now.

答:
Bill's idea may be this..

consider  $f(x)=\tan x$  $f''(x)=2\sec^2 x \tan x>0$ for $x\in (0,\pi/3)$

.By jensen $$f(x)+f(y)\le \frac{f(x)+f(y}{2}\Rightarrow \frac{\tan A+\tan B}{2}\ge {\tan(\frac{A+B}{2})}=\tan 15^o$$


问:
proving equation including the zeta-function
how do I prove that
$$\frac{ζ（2n）}{Π^{2n}}-\frac{ζ(2n-2)}{3!Π^{2n-2}}+\frac{ζ（2n-4）}{5！Π^{2n-4}}-……+\frac{(-1)^{n-1}ζ(2)}{(2n-1)!Π^2}+\frac{(-1)^nn}{(2n+1)!} = 0$$
maybe i can use $\sum_{n=1}^∞ \frac{z^{2n-1}}{(2n-1)!}{(-1)^{n-1}} = sinz$

but i have no idea and really need your help

Thank you!

答: You need to use  $$\frac{\pi^2}{\sin^2 (\pi z)} = \sum_n \frac1{(z+n)^2}= z^{-2}+\sum_{k\ge 1} 2\zeta(2k)z^{2k-2} $$ Then multiply both sides by $\sin^2(\pi z)$'s Taylor series


问:
How to determine whether a problem belongs to the poisson distribution?
I was solving my university past paper and came about this question:

```<code>A microbiologist wants to estimate the concentration of a certain type of bacteriumin a wastewater sample.
She puts a 0.5 mL sample of the waste-water on a microscope slide and counts 39 bacteria.
Estimate the concentration of bacteria per mL, in this waste-water.
</code>```

Ater searching my textbook I figured out that this problem belongs to Poisson distribution.But my question is how are we supposed to identify to which distribution the problem belongs to?
I know the general rule to identify questions that belong to Poisson distribution:They should have small success probability and large sample size.Also the mean would be equal to the variance.Any help would be helpful for my End sem exam.

答:
First, your reasoning that "They should have small success probability and large sample size. Also the mean would be equal to the variance." is wrong, this isn't related to the Poisson distribution, you can have other distributions sharing these properties and you can have a Poisson distribution without these properties so I don't advise you to look for these as they are unrelated. Similarly, having small success probabilities and large sample size is unrelated.

Second, a Poisson distribution is when you are given on average how much an event occurs in a certain interval, and you want to find the probability that it occurs another number of times (different that than the mean) in the interval. The interval could be time, distance, weight, etc... anything.
Moreover, the events should be independent.

Note that you are always given the mean explicitly or given enough information to find the mean yourself.

To illustrate this by example, assume you are given the following:

A call center receives 4 calls per hour, what is the probability it
receives 6 calls per hour?

Here the events are independent since a customer calling doesn't affect the probability of another calling. You are given the mean (4 calls per hour) and you want to find the probability that a value other than the mean (6 calls) happens, so you use the Poisson distribution.

Note that they might not explicitly state the same intervals, so for example one variation of this problem could be the following:

A call center receives 5 calls per hour, what is the probability it
receives 15 calls in the next 90 minutes?

Here you would next to convert what is given to match the mean, so 15 calls per 90 minutes is the same as 10 calls per an hour, then you work.

Going back to your example:

A microbiologist wants to estimate the concentration of a certain type of bacteriumin a wastewater sample.
She puts a 0.5 mL sample of the waste-water on a microscope slide and counts 39 bacteria.
Estimate the concentration of bacteria per mL, in this waste-water.

This problem can be solved with simple reasoning by just doing $\frac{39}{0.5}$ as if you have x amount of y/2 total you will have 2x amount in y (just multiplying both sides by 2, no distributions involved)


问:
How to continue the induction demonstration?
Be $\{A_n\}_{n \in \mathbb{N}}$  a sequence of connecteds ones such that $A_n \cap A_{n+1} \neq \emptyset \ \forall n \in {N}.$ Prove that $\cup_{n\in {N}}A_n $ is connected.

I tried to define $B_n$ as being the union of $A_1$ to $A_n$. By induction, these $B_n$ are related (as we know?). We can assume that $B_1$ is not empty and herefore contains a point $"a"$. So, all $B_n$ are related and have a point in common, and therefore the union is related. But I cannot mathematically express this demonstration.

答:
Take $X \subset \bigcup_{n \in \mathbb{N}}A_n$ open and closed. Then, for $n \in \mathbb{N}$, $X \cap A_n$ and $X^\complement \cap A_n$ are open subset of $A_n$ so $X \cap A_n$ is equal to $A_n$ or is empty.

Assume that $X \cap A_n = A_n$ for one $n \in \mathbb{N}$ : then $X \cap A_{n-1}$ and $X \cap A_{n+1}$ are nonempty so $X \cap A_{n-1} = A_{n-1}$ and $X \cap A_{n+1} = A_{n+1}$. By induction, $X = \bigcup_{n \in \mathbb{N}}A_n$.

If otherwise $X \cap A_n$ is empty for all $n$, then $X$ is empty.


问:
Problem in counting principles from Precalculus by Openstax
Here is the question:
The set $S$ consists of 900,000,000 whole numbers, each being the same number of digits long. How many digits long is a number from S?(Hint: use the fact that a whole number cannot start with the digit 0).

 What I don't understand is how to take into account the fact that there are $r$ digits in the number, when the first digit can only be between 1-9, and the rest $(r-1)$ digits are between $(0-9)$

 this is how I tried to solve the problem:

1) $S_{size} = 900000000$

2) $9\cdot_{10}N_{r-1} = 900000000$

3) $9\cdot_{10}N_{r-1} = 900000000$

4) $_nC_r = \frac{n!}{r!(n-r)!}$

5) $_10C_r-1 = \frac{10!}{(r-1)!(10-(r-1))!}$

6) $9\cdot \frac{10!}{(r-1)!(10-(r-1))!} = 900000000$

7) $\frac{3,628,800}{(r-1)!(11-r))!} = 100,000,000$

答:
The cardnality of $S$ is $9 \times 10^8$.

In the first digit, there are $9$ options since it can't begin with $0$, after which we have $10$ options for ther remaining $8$ positions.

The length of the digit is $8+1=9$.


问:
Quotient rings of Gaussian integers $\mathbb{Z}[i]/(2)$, $\mathbb{Z}[i]/(3)$, $\mathbb{Z}[i]/(5)$,
I am studying for an algebra qualifying exam and came across the following problem.

Let $R$ be the ring of Gaussian Integers. Of the three quotient rings
  $$R/(2),\quad R/(3),\quad R/(5),$$
  one is a field, one is isomorphic to a product of fields, and one is neither a field nor a product of fields. Which is which and why?

I know that $2=(1+i)(1-i)$ and $5=(1+2i)(1-2i)$, so neither $(2)$ nor $(5)$ is a prime ideal of $R$. Then (I think) these same equations in $R/(2)$ and $R/(5)$, respectively, show that neither is an integral domain. Regardless, I can show that 3 is a Gaussian prime, hence $(3)$ is maximal in $R$ and $R/(3)$ is the field. But if I am correct about the others not being integral domains, I fail to see how either could be a product of fields.

I hope that this can be answered easily and quickly. Thanks.

答:
Now is a very good time for a quick foray into the ideal-theoretic version of Sun-Ze (better known as the Chinese Remainder Theorem). Let $R$ be a commutative ring with $1$ and $I,J\triangleleft R$ coprime ideals, i.e. ideals such that $I+J=R$. Then

$$\frac{R}{I\cap J}\cong\frac{R}{I}\times\frac{R}{J}.$$

First let's recover the usual understanding of SZ from this statement, then we'll prove it. Thanks to Bezout's identity, $(n)+(m)={\bf Z}$ iff $\gcd(n,m)=1$, so the hypothesis is clearly analogous. Plus we have $(n)\cap(m)=({\rm lcm}(m,n))$. As $nm=\gcd(n,m){\rm lcm}(n,m)$, if $n,m$ are coprime then compute the intersection $(n)\cap(m)=(nm)$. Thus we have ${\bf Z}/(nm)\cong{\bf Z}/(n)\times{\bf Z}/(m)$. Clearly induction and the fundamental theorem of arithmetic (unique factorization) give the general algebraic version of SZ, the decomposition ${\bf Z}/\prod p_i^{e_i}{\bf Z}\cong\prod{\bf Z}/p_i^{e_i}{\bf Z}$.

(How this algebraic version of SZ relates to the elementary-number-theoretic version involving existence and uniqueness of solutions to systems of congruences I will not cover.)

Without coprimality, there are counterexamples though. For instance, if $p\in\bf Z$ is prime, then the finite rings ${\bf Z}/p^2{\bf Z}$ and ${\bf F}_p\times{\bf F}_p$ (where ${\bf F}_p:={\bf Z}/p{\bf Z}$) are not isomorphic, in particular not even as additive groups (the product is not a cyclic group under addition).

Now here's the proof. Define the map $R\to R/I\times R/J$ by $r\mapsto (r+I,r+J)$. The kernel of this map is clearly $I\cap J$. It suffices to prove this map is surjective in order to establish the claim. We know that $1=i+j$ for some $i\in I$, $j\in J$ since $I+J=R$, and so we further know that $1=i$ mod $J$ and $1=j$ mod $I$, so $i\mapsto(I,1+J)$ and $j\mapsto(1+I,J)$, but these latter two elements generate all of $R/I\times R/J$ as an $R$-module so the image must be the whole codomain.

Now let's work with ${\cal O}={\bf Z}[i]$, the ring of integers of ${\bf Q}(i)$, aka the Gaussian integers. Here you have found that $(2)=(1+i)(1-i)=(1+i)^2$ (since $1-i=-i(1+i)$ and $-i$ is a unit), that the ideal $(3)$ is prime, and that $(5)=(1+2i)(1-2i)$. Furthermore $(1+i)$ is obviously not coprime to itself, while $(1+2i),(1-2i)$ are coprime since $1=i(1+2i)+(1+i)(1-2i)$ is contained in $(1+2i)+(1-2i)$. Alternatively, $(1+2i)$ is prime and so is $(1-2i)$ but they are not equal so they are coprime. Anyway, you have

${\bf Z}[i]/(3)$ is a field and

${\bf Z}[i]/(5)\cong{\bf Z}[i]/(1+2i)\times{\bf Z}[i]/(1-2i)$ is a product of fields.

Go ahead and count the number of elements to see which fields they are. However, ${\bf Z}[i]/(2)={\bf Z}[i]/(1+i)^2$ is not a field or product of fields, although the fact that its characteristic is prime (two) may throw one off the chase. In ${\bf Z}[i]/(1+i)^2$, the element $1+i$ is nilpotent. Since this ring has order four, it is not difficult to check that it is isomorphic to ${\bf F}_2[\varepsilon]/(\varepsilon^2)$, which is not a product of fields since $\epsilon\leftrightarrow 1+i$ is nilpotent and products of fields contain no nonzero nilpotents.


问:
Images and preimages of groups using group homomorphism
I'm trying to solve this problem from my group theory course:

Let $\phi:G_1\rightarrow G_2$ be a group homomorphism. Also let
$N\unlhd G_1$ (normal subgroup) and $K\leq G_2$.

a) Is $\phi(N)$ a normal subgroup of $G_2$?

b) Is $\phi^{-1}(K)=\{x\in G_1:\phi(x)\in K\}$ a subgroup of $G_1$?

c) If $K$ is normal, is $\phi^{-1}(K)$ a normal subgroup of $G_1$?

The work I've done so far:

a) Bein $y\in G_2$, if $\phi(N)$ is normal in $G_2$ then it must verify $y\phi(N)y^{-1}= \phi(N)$. I've seen that, if $\phi$ is surjective, then $\exists x\in G_1$ so that $\phi(x)=y$, so we can see that
$$y\phi(N)y^{-1}=\phi(x)\phi(N)\phi(x)^{-1}=\phi(xNx^{-1})=\phi(N).$$
So I've proven $\phi(N)\unlhd G_2$ just in the case $\phi$ is surjective (I'm not sure if I need to prove that $\phi$ being non-surjective implies $\phi(N)$ being not normal, or if it is trivial).

b) For this second question, I've seen that, given $x\in G_1$ so that $\phi(x)\in K$, then $\phi(x)^{-1}=\phi(x^-1)\in K$ (since $K$ is group). Also that $\phi(e_1)=e_2\in K$. It obviously verifies associativity, and, given two $x,y\in\phi^{-1}(k)$, then also $\phi(xy)=\phi(x)\phi(y)\in K$, so I conclude that $\phi^{-1}(K)$ is subgroup of $G_1$.

c) We have that, for any $g\in G_1$, (assuming of course $K$ normal subgroup of $G_2$), then
$$\phi(g)K\phi(g^{-1})=K$$
We apply $\phi^{-1}$ to the expression above and we get that
$$\phi^{-1}\left(\phi(g)K\phi(g^{-1})\right)=g\phi^{-1}(K)g^{-1}=\phi^{-1}(K),$$
so we conclude that $K\unlhd G_2$ implies $\phi^{-1}(K)\unlhd G_1$ (Is this statement true even when $\phi$ is not surjective? I think it is but I may be wrong).

Are these solutions correct? If not, why? Any help will be appreciated, thanks in advance.

答:
Statement (a) is false, but you can't prove that $\phi(N)$ is not normal: it may or may not be. How do you show it's false? Take your favorite example of a group $A$ with a nonnormal subgroup $B$; then $B$ is normal in $B$, but its image under the inclusion map $B\to A$ is not normal in $A$. On the contrary, if $G_2$ is abelian, then $\phi(N)$ would be normal in $G_2$ no matter whether $N$ is normal in $G$ or not.

Statement (a) is indeed true under the additional assumption that $\phi$ is surjective.

Your proof of (b) is correct.

For (c) your argument is wrong: there is no $\phi^{-1}$ map in general. The notation $\phi^{-1}(K)$ just denotes $\{x\in G_1:\phi(x)\in K\}$, but does not imply the inverse map exists. And you should prove that $\phi^{-1}(AB)=\phi^{-1}(A)\phi^{-1}(B)$, in case you just consider the inverse image. But you're using $\phi^{-1}(\phi(g))=g$ and this invalidates the proof.

Set $H=\phi^{-1}(K)$ for simplicity and consider $g\in G$; for $x\in H$, you have
$$
\phi(gxg^{-1})=\phi(g)\phi(x)\phi(g)^{-1}
$$
and this belongs to $K$, because $\phi(x)\in K$ by assumption and $K$ is normal. Therefore $gxg^{-1}\in H$.


问:
How to find $\lim_{t \rightarrow 0} (ax^t + by^t)^{\frac{1}{t}}$ with different methods?
Suppose $a,b,x,y$ are all positive and $a+b=1$. Compute $\lim_{t \rightarrow 0} (ax^t + by^t)^{\frac{1}{t}}$.

I tried to put this into a form where L'Hôpital could be useful, but I was unable to do so. How would one solve this limit?

Thank you for any answers.

答:
Take the log of $(ax^t+by^t)^\frac{1}{2}.$
We have
\begin{align*}
\frac{d}{dt}(\log(ax^t+by^t))&=\frac{a(\log x\cdot x^t)+b(\log y\cdot y^t)}{ax^t+by^t}.
\end{align*}
Plugging in $t=0$ gives
\begin{align*}
\lim_{t\to0}\log (ax^t+by^t)^\frac{1}{t}&=f'(0)\\
&=a\log x+b\log y,
\end{align*}because $a+b=1.$
So the original limit is
\begin{align*}
e^ {a\log x+b\log y}&=e^{\log x^a}\cdot e^{\log y^b}\\
&=x^ay^b.
\end{align*}


问:
solutions to quadratic matrix equation $XFX - F^{-1}X^{-1}F = 0$
What is the solution for the matrix $X$ in the following quadratic matrix equation?

$$ XFX - F^{-1}X^{-1}F = 0 $$

where $F$ is a $N \times N$ discrete Fourier transform (DFT) matrix, so it's unitary ($F^{-1} = F^\dagger$). Based on similar quadratic matrix equations it seems that $X$ would be expressed in terms of fractional powers of $F$. I have access to the eigenvectors of $F$ and hence all fractional powers of $F$ thanks to the discrete fractional Fourier transform (DFRT) in Candan et al. (2000).

Found an almost trivial solution $X = F^{-1/3}$ by guessing and checking.

Substituting $X = F^{-1/3}$ into the equation results in $F^{1/3} - F^{1/3} = 0$ which is true hence $X = F^{-1/3}$ is a valid solution.

Are there other nontrivial solutions? $F^{-1/3 + 4n}$ for integer $n$ is another obvious one due to the periodicity of the fractional Fourier transform.

答:
The equation is equivalent to
$$
(XF)^3=XF(XFX)F=XF(F^{-1}X^{-1}F)F=F^2.
$$
Therefore the solutions are given by $X=CF^{-1}$ where $C^3=F^2$.

Remark. Since the eigenvalues of $F$ are $\pm1$ and $\pm i$, the matrix $C$ (which is a cubic root of $F^2$) in general doesn't commute with $F$ (which is a square root of $F^2$). Therefore $CF^{-1}$ is not necessarily an inverse cubic root of $F$.


问:
Limit Involving a Tetration of the Variable
We wish to evaluate this challenging limit
$$
\lim_{x\to 1}\left(\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}
$$

My Observation : As an initial guess, I shall consider $f_{n}(x)=x^{x^{x{^{.^{.^{.}}}}}}$ with $n$ being the number of $x$. To simplify this problem I considered the case of $n=1$, it is easy to see that the limit is $1$, to be sure of my assertion I further considered the case for $n=2$, we receive :
$$
\left(\frac{x^{x}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}
$$
We thus simplify this expression :
\begin{align*}
\left(\frac{x^{x}-1}{x-1}\right)^{\displaystyle\frac{1}{\sqrt{x-1}}}&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{\exp(x\ln(x))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{\exp(x(x-1+o(x-1)))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(\frac{1+x(x-1)+o(x(x-1))-1}{x-1}\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}\ln\left(x+o(x)\right)\right) \\
&=\exp\left(\frac{1}{\sqrt{x-1}}(x-1+o(x-1))\right) \\
&=\exp(\sqrt{x-1}+o(\sqrt{x-1}))
\end{align*}
Where :$$\displaystyle\lim_{x\to 1}\exp(\sqrt{x-1}+o(\sqrt{x-1}))=1$$

Now the big question is how can we extend this for the general case that is for $n\geq3$? Any help would be much appreciated.

Body:
$$a_n=\left(\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}\right)^{ \frac{1}{\sqrt{x-1}}}\implies \log(a_n)=\frac{1}{\sqrt{x-1}}\log\left(\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}\right)$$

Using @Ninad Munshi's answer, we then have that, around $x=1$, for an infinite tetration
$$x^{x^{x{^{.^{.^{.}}}}}}=1+(x-1)+(x-1)^2+\frac{3}{2} (x-1)^3+\frac{7}{3} (x-1)^4+O\left((x-1)^5\right)$$ that is to say
$$\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}=1+(x-1)+\frac{3}{2} (x-1)^2+\frac{7}{3} (x-1)^3+O\left((x-1)^4\right)$$
$$\log\left(\frac{x^{x^{x{^{.^{.^{.}}}}}}-1}{x-1}\right)=(x-1)+(x-1)^2+\frac{7}{6} (x-1)^3+O\left((x-1)^4\right)$$
$$\log(a_n)=(x-1)^{1/2}+(x-1)^{3/2}+\frac{7}{6} (x-1)^{5/2}+O\left((x-1)^{7/2}\right)$$

Now, let us do the same for the discrete cases where $n$ is small
$$\log(a_2)=(x-1)^{1/2}+\frac{1}{6} (x-1)^{5/2}+O\left((x-1)^{7/2}\right)$$
$$\log(a_3)=(x-1)^{1/2}+(x-1)^{3/2}+\frac{1}{6} (x-1)^{5/2}+O\left((x-1)^{7/2}\right)$$
$$\log(a_4)=(x-1)^{1/2}+(x-1)^{3/2}+\frac{7}{6} (x-1)^{5/2}+O\left((x-1)^{7/2}\right)$$

This is a quite fast convergence to order $O\left((x-1)^{7/2}\right)$.

Pushing to $O\left((x-1)^{9/2}\right)$, the coefficient of  $(x-1)^{7/2}$  will succesively be $\left\{-\frac{1}{8},\frac{7}{24},\frac{19}{24},\frac{43}{24}\right\}$ (no $\cdots$).


问:
GCD identity for univariate polynomials
I would like to know if my solution to the following question concerning the gcd of two polynomial correct>

Question:

Suppose $f(x)$, $g(x)$ and $h(x)$ are polynomials in $F[x]$.  Let $gcd(f(x),g(x)h(x))=1$.  Show that $gcd(f(x),g(x))=gcd(f(x),h(x))=1$

Proof:  Since $gcd(f(x),g(x)h(x))=1$, then there exists some polynomials $u(x)$ and $v(x)$ in $F[x]$ where
 $f(x)u(x)+(g(x)h(x))v(x)=1$. $(*)$

Applying associativity of multiplication in $(*)$ gives $(g(x)h(x))v(x)=g(x)(h(x)v(x))$ then $f(x)u(x)+(g(x)h(x))v(x)=f(x)u(x)+g(x)(h(x)v(x))=1$ which implies $gcd(f(x), g(x)=1$.

Applying commutativity to $(*)$, then again associativity of multiplication to $(*)$ gives $(g(x)h(x))v(x)=(h(x)g(x))v(x))$ and $(h(x)g(x))v(x))=h(x)(g(x)v(x))$ in $(*)$
 implies that $f(x)u(x)+(g(x)h(x))v(x)=f(x)u(x)+(h(x)g(x))v(x)=f(x)u(x)+h(x)(g(x)v(x))=1$, which gives $gcd(f(x), h(x))=1$ also.  Hence we have $gcd(f(x), g(x))=gcd(f(x), h(x))=1$

Thank you in advance

答:
The gcd $(a,b)\,$ is characterized by the following fundamental gcd Universal Property

$$ c\mid a,b\,\color{#0a0}\Leftarrow\!\color{#c00}\Rightarrow\, c\mid (a,b)\quad \text{[gcd Universal Property]}\qquad$$

For $\,c = (a,b)\,$ arrow $\color{#0a0}{(\Leftarrow)}$ shows $\,(a,b)\mid a,b,\,$ i.e. $\,(a,b)\,$ is a common divisor of $\,a,b,\,$ and the opposite arrow $\color{#c00}{(\Rightarrow)}$ shows that $\,(a,b)\,$ is divisible by every common divisor of $\,a,b,\,$ so $\,(a,b)\,$ is a greatest (degree) common divisor  (unique if we normalize it to be monic, i.e. lead coef $=1).\,$

Therefore generally $\,(f,g)\mid f,g\,\Rightarrow\, (f,g)\mid f,gh\,\color{#c00}\Rightarrow\, \color{#90f}{(f,g)}\mid \color{#0a0}{(f,gh)}$.

Further, in your particular case we are given that $\,\color{#0a0}{(f,gh)\!=1}\,$ thus  $\color{#90f}{(f,g)}\! =\! 1$

Your use of the Bezout identity to deduce this is essentially repeating the Bezout-based proof of the gcd Universal Property in this particular case. Instead you should abstract out and prove this basic property (whose proof is easier), then invoke it as above (by name). This will work more generally in rings where there is no Bezout identity (e.g. $\,\Bbb Z[x]\,$ or $\,\Bbb Q[x,y]\:\!)\,$ but the gcd universal property still holds true (it is the definition of the gcd in general domains).

Title: GCD identity for univariate polynomials


答:
$$\tan A+\tan B=\dfrac{\sin(A+B)}{\cos A\cos B}=\dfrac{2\sin(A+B)}{\cos(A-B)+\cos(A+B)} =\dfrac1{\cos(2A-30^\circ)+\dfrac{\sqrt3}2}$$

So, we need to maximize $\cos(2A-30^\circ)+\dfrac{\sqrt3}2$

Now  $-30^\circ<2A-30^\circ<30^\circ\implies\cos30^\circ<\cos(2A-30^\circ)\le1$

Title: Finding minimum value of $\tan A+\tan B$, given $A+B=30^\circ$

Body:
If $A>0$, $B>0$ and $A+B=30^\circ$, find the minimum value of $\tan A+\tan B$.

A similar question has been asked here. Many good answers are posted there. I understood that.

One particular answer by @Bill Kleinhans intrigued me as it was short and straightforward. But I didn't fully understand that answer. It uses Jensen's inequality, which I know, but don't know how to use that to get the answer here.

Not posting this as a comment there because Bill has been away close to three years now.


问:
Solving finite and infinite nested square roots of 2 - yet another interesting approach
Consider the following consecutive equalities:

$\sqrt2=2\cos(\frac{1}{4})\pi$

$\sqrt{2-\sqrt2}=2\cos(\frac{3}{8})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2}}}=2\cos(\frac{5}{16})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}=2\cos(\frac{11}{32})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}=2\cos(\frac{21}{64})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}}=2\cos(\frac{43}{128})\pi$

Here is the beautiful part

$\frac{3}{8} = \frac{1}{4}+\frac{1}{8}$

$\frac{5}{16} = \frac{3}{8}-\frac{1}{16}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}$

$\frac{11}{32} = \frac{5}{16}+\frac{1}{32}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}$

$\frac{21}{64} = \frac{11}{32}-\frac{1}{64}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}$

$\frac{43}{128} = \frac{21}{64}+\frac{1}{128}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{128}$

If extended to infinity $\sqrt{2-\sqrt{2-\sqrt{2...}}} = 2cos(\frac{1}{3})\pi$ (we can calculate the sum of infinite Geometric progression and we get there result as $1\over3$)

We can easily guess that if a nested square root has $n$ $2$’s then a denominator of the corresponding cosine argument is $2^{n+1}$.

Most beautiful part is getting the numerator which is $2^{n+1} \over 3$ approximated to nearest integer.[or approximation to floor and making it as very next odd number(if the floor is even number)--> and this is better ]

Subsequent numbers having numerator are as follows

$\frac{2^3}{3}=3;\frac{2^4}{3}=5;\frac{2^5}{3}=11;\frac{2^6}{3}=21;\frac{2^7}{3}=43;\frac{2^8}{3}=85;\frac{2^9}{3}=171;\frac{2^{10}}{3}=341...$

Therefore for the finite number of nested square roots of 2 like this, it is the easy way to evaluate the angle inside the cosine function!

One more exciting part is association with integer sequence of type Jacobsthal sequence $a(n-1) + 2a(n-2)$, with $a(0) = 0, a(1) = 1$.

Still more interesting part is Fibonacci like pattern from above integer sequence

My question is can we link geometric progression to simplify different patterns of the finite or infinite nested square roots of 2, like this?

To expand my question to understand

Let us consider finite nested square roots of 2 in simplified way

$\sqrt{2-\sqrt2}$ as $n\sqrt2(1-)$ and $\sqrt{2-\sqrt{2+\sqrt2}}$ as $n\sqrt2(1-1+)$

If we have finite nested square roots of 2 such as $n\sqrt2(1-2+)$ repeated signs inside nested radical 'n' times and infinitely, how can we get the angle?. What I have shown is simplest example. Let us consider $n\sqrt2(3-1+7-)$ or $n\sqrt2(2-3+4-1+)$ and so on. Is there any easier way to find the cosine angle?

答:
I don't quite understand the phrasing of your question, and in particular I don't quite understand if you're asking for a proof of these identities or not. Here is a proof. We can reverse-engineer the result we need by asking what relation between two angles $\alpha_n, \alpha_{n+1}$ is implied by

$$\sqrt{2 - 2 \cos \alpha_n} = 2 \cos \alpha_{n+1};$$

here $\alpha_n$ and $\alpha_{n+1}$ are two consecutive angles among the sequence of angles you've listed. Squaring and using the cosine double-angle formula $\cos 2 \theta = 2 \cos^2 \theta - 1$ gives

$$2 - 2 \cos \alpha_n = 4 \cos^2 \alpha_{n+1} = 2 + 2 \cos 2 \alpha_{n+1}$$

which gives $\cos \alpha_n = - \cos 2 \alpha_{n+1}$. If $\alpha_i \in \left[ 0, \frac{\pi}{2} \right]$, which holds for all of your angles, then this gives

$$\alpha_{n+1} = \frac{\pi - \alpha_n}{2}.$$

It's not hard to see that this generates all of the angles you've listed, starting from $\alpha_0 = \frac{\pi}{4}$, and that this sequence converges to the unique fixed point $\alpha_{\infty} = \frac{\pi}{3}$ of the map $\alpha \mapsto \frac{\pi - \alpha}{2}$. We have the explicit closed form

$$\alpha_n = \frac{\pi}{3} - \frac{\pi}{12} \left( - \frac{1}{2} \right)^n$$

which can be proven by rewriting the recurrence relation as $\alpha_{n+1} - \frac{\pi}{3} = \left( \alpha_n - \frac{\pi}{3} \right) \left( - \frac{1}{2} \right)$. Adding these two fractions gives

$$\alpha_n = \left( \frac{2^{n+2} - (-1)^n}{3 \cdot 2^{n+2}} \right) \pi$$

and then we note that $2^{n+2} \equiv (-1)^{n+2} \equiv (-1)^n \bmod 3$ so the numerator is always divisible by $3$, and dividing it by $3$ gives $\frac{2^{n+2} - (-1)^n}{3}$ which is the closest integer to $\frac{2^{n+2}}{3}$ as you observed.


问:
Taylor expanding implicit curves

$$ x \cos y + y \cos x = 0$$

Suppose I wished to write the above as polynomial around $x=0$, we can see by substitution of $x=0$ that $y=\pi$ around that point from the original equation. Hence, could we write the equation as:

$$ x \left[ \pi  +\frac{d y}{dx} \frac{ (y- \pi)^2}{2!} .. + stuff \right] + y \left[ 1 - \frac{x^2}{2!} .. stuff \right]=0 $$

Is the above equation correct? What would be the rationale/ intuition behind expanding $y$ individually in the equation?

答:
$y(0)=\pi$,then D.w.r.t. $x$ would lead to
$$x\cos y+y \cos x=0 \implies \cos y -xy'\sin y+y'\cos x-y\sin x
\implies y'(0)=0$$
Similarly, go on D. w.t. $x$ to get $y''(0)$, $y'''(0)$ etc.
Then the series will be $$y(x)=\pi-x+y''(0)\frac{x^2}{2!}+y'''(0)\frac{x^3}{3!}+...$$

Title: Taylor expanding implicit curves


答:
Let $\{a_n\}$ be a sequence of consecutive nested square roots of $2$, that is $a_n=\sqrt{2\pm a_{n-1}}$ for each $n\ge n_0$.
Assuming $a_{n-1}=2\cos x_{n-1}$ for $x_{n-1}\in [0,\pi]$, we obtain $a_n=2\cos x_n$, where $x_n=x_{n-1}/2$, if  $a_n=\sqrt{2+a_{n-1}}$ and $x_n=(\pi-x_{n-1})/2$, if $a_n=\sqrt{2-a_{n-1}}$. This way the sequence of signs transforms to a sequence of angles. The latter looks close to a geometric series only in special cases. When all signs are “$+$” then $\{x_n\}$ is a geometric series. When all signs are “$-$ then $(\pi/3-x_n)=-(\pi/3-x_{n-1})/2$. That is, $\{\pi/3-x_n\}$ is a geometric series.

Title: Solving finite and infinite nested square roots of 2 - yet another interesting approach

Body:
Consider the following consecutive equalities:

$\sqrt2=2\cos(\frac{1}{4})\pi$

$\sqrt{2-\sqrt2}=2\cos(\frac{3}{8})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2}}}=2\cos(\frac{5}{16})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}=2\cos(\frac{11}{32})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}=2\cos(\frac{21}{64})\pi$

$\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{2}}}}}}=2\cos(\frac{43}{128})\pi$

Here is the beautiful part

$\frac{3}{8} = \frac{1}{4}+\frac{1}{8}$

$\frac{5}{16} = \frac{3}{8}-\frac{1}{16}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}$

$\frac{11}{32} = \frac{5}{16}+\frac{1}{32}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}$

$\frac{21}{64} = \frac{11}{32}-\frac{1}{64}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}$

$\frac{43}{128} = \frac{21}{64}+\frac{1}{128}$ = $\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}+\frac{1}{128}$

If extended to infinity $\sqrt{2-\sqrt{2-\sqrt{2...}}} = 2cos(\frac{1}{3})\pi$ (we can calculate the sum of infinite Geometric progression and we get there result as $1\over3$)

We can easily guess that if a nested square root has $n$ $2$’s then a denominator of the corresponding cosine argument is $2^{n+1}$.

Most beautiful part is getting the numerator which is $2^{n+1} \over 3$ approximated to nearest integer.[or approximation to floor and making it as very next odd number(if the floor is even number)--> and this is better ]

Subsequent numbers having numerator are as follows

$\frac{2^3}{3}=3;\frac{2^4}{3}=5;\frac{2^5}{3}=11;\frac{2^6}{3}=21;\frac{2^7}{3}=43;\frac{2^8}{3}=85;\frac{2^9}{3}=171;\frac{2^{10}}{3}=341...$

Therefore for the finite number of nested square roots of 2 like this, it is the easy way to evaluate the angle inside the cosine function!

One more exciting part is association with integer sequence of type Jacobsthal sequence $a(n-1) + 2a(n-2)$, with $a(0) = 0, a(1) = 1$.

Still more interesting part is Fibonacci like pattern from above integer sequence

My question is can we link geometric progression to simplify different patterns of the finite or infinite nested square roots of 2, like this?

To expand my question to understand

Let us consider finite nested square roots of 2 in simplified way

$\sqrt{2-\sqrt2}$ as $n\sqrt2(1-)$ and $\sqrt{2-\sqrt{2+\sqrt2}}$ as $n\sqrt2(1-1+)$

If we have finite nested square roots of 2 such as $n\sqrt2(1-2+)$ repeated signs inside nested radical 'n' times and infinitely, how can we get the angle?. What I have shown is simplest example. Let us consider $n\sqrt2(3-1+7-)$ or $n\sqrt2(2-3+4-1+)$ and so on. Is there any easier way to find the cosine angle?


问:
transform weibull distributed data to change the scale
For my study I need to provide a data set that includes three weibull distributed samples with the same shape, but different scale.

In the best case, I want to do a transformation onto one sample, to get as a result a sample with the same shape, but another scale. Is this possible?

(Additional:

I assume, I need to find an equation, that I can solve for the scale, and then use it as transformation, but I am a little lost, to find the equation to start with. I have tried the PDF, but I am not able to solve it for the scale (nor is the software I used).

The lecturer does not allow to generate a random sample and I do not want to lie. Besides I assume I am expected to learn something by this, but I am not able to see it until now.)

答:
If $X\sim Weib(\lambda=1,k)$ we have
$$\mathbb P(X<t)=1-e^{-t^k}.$$

Your goal, if I have well understood is to arrive to a generic Weibull distribution, where,
$$\mathbb P(f(X)<t)=1-e^{-(t/\lambda)^k}$$
applying a function $f$ on $X$. Now, consider that for any random variable $Y$ with distribution $F_Y$ and any $a>0$ we have
$$F_{aY}(t)=\mathbb P(aY<t)=\mathbb P(Y<t/a)=F_Y(t/a)$$


问:
Dynamic system show that it has only one equilibrium point
I have the following dynamic system $A\cdot\ddot{x}+B\cdot\dot{x}+ C\cdot x=0 $ and A,B,C are positive definite matrices.
I have to show that it has only one equilibrium point and that this point is asymptotically stable. Any help will be appreciated!

UPDATE
Following the comment about treating A,B,C as scalars
$\dot{x_1}=x_2 = f_1(x_2)\\
\dot{x_2}=(-B\cdot x_2-C\cdot x_1)/A = f_2(x_1,x_2)$

$\longrightarrow \dot{x}=A(x)|_{x=0} \cdot x \\ $

$ A(x)=\begin{bmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \\
\end{bmatrix}=\begin{bmatrix}
0 & 1\\
-C/A&-B/A\\
\end{bmatrix}$
$det(A-\lambda\cdot I)= 0 \longrightarrow det(\begin{bmatrix}
-\lambda & 1\\
-C/A& -\lambda -B/A\\
\end{bmatrix}) =0 \longrightarrow \lambda^2+\lambda B/A+C/A=0$\

But I don't seem to get any idea on how to generalize it.

答:
If $x_1$ and $x_2$ are equilibrium points then

$$
\cases{
C\cdot x_1 = 0\\
C\cdot x_2 = 0
}\Rightarrow C\cdot(x_1-x_2)=0
$$

but due to the nature of $C$ follows $x_1-x_2=0\Rightarrow x_1 = x_2$


问:
Prove $xy\notin H\cup K$ if $x\in H\backslash K$ and $y\in K\backslash H$ ($H,K$ subgroups)
I need to prove this statement for some exercise from my group theory course:

Being $G$ group, and $H,K\leq G$. If $H\nsubseteq K$ and $K\nsubseteq
> H$, then  $$x\in H\backslash K, y\in K \backslash H \implies xy\notin H\cup K.$$

I don't know if this is trivial. It seem trivial to me, but I'm not sure. How can I prove this?

答:
Suppose that $xy\in H\cup K$.

Case 1: $xy\in H$. Since $x\in H$, we have $y=x^{-1}(xy)\in H$; which is a contradiction since $y\notin H$.

Can you complete the proof for Case 2 where $xy\in K$?


答: Hint: Note that $y=x^{-1}\cdot xy$ and $x=xy\cdot y^{-1}$.

Title: Prove $xy\notin H\cup K$ if $x\in H\backslash K$ and $y\in K\backslash H$ ($H,K$ subgroups)

Body:
I need to prove this statement for some exercise from my group theory course:

Being $G$ group, and $H,K\leq G$. If $H\nsubseteq K$ and $K\nsubseteq
> H$, then  $$x\in H\backslash K, y\in K \backslash H \implies xy\notin H\cup K.$$

I don't know if this is trivial. It seem trivial to me, but I'm not sure. How can I prove this?