问:
Kernel of ring homomorphism from a polynomial ring over field to a field is maximal ideal or zero ideal
Question If $E,F$ are fields and $\beta:F[x]\rightarrow E$ a homomorphism of rings. Show that the kernel of $\beta$ is a maximal ideal or a zero ideal.

I just wonder where does the zero ideal case arise? I haven't seen any necessity to discuss the situation of zero ideal. Any help would be appreciated!

答: Let $E$ be the field of fractions of $F[x]$ and $\beta$ the obvious embedding.


问:
Linear transformation from the set of continuous functions $C([0,1])$ to $\mathbb{R^2}$.

Let $\varepsilon : C([0,1]) \to \mathbb{R^2}$ be a mapping $$f\mapsto (f(0), f(1))$$ show that $\varepsilon$ is a linear transformation from the set of continuous functions $C([0,1])$ to $\mathbb{R^2}$.

For $\varepsilon$ to be linear transformation it would need to satisfy $\varepsilon(x+y) = \varepsilon(x) + \varepsilon(y)$ and $\varepsilon(cx) = c\varepsilon(x)$. However, I'm not sure how to approach this since we're not given any excplicit function to show this? Any hints would be appreciated...

答:
You need just the definition of function addition and multiplacation of function by a scalar. In $C([0,1])$ functions have both same domain and codomain, so you have permission for addition and scalar multiplication.
$\epsilon(f+g)=((f+g)(0),(f+g)(1))=?$
$\epsilon(cf)=((cf)(0),(cf)(1))=?$, $c\in\mathbb{R}$

ContentLicense: CC BY-SA 4.0


问:
Calculating an improper integral.
There is an improper integral which has to be calculated and I have no idea about it. The problem is in the argument of cosine being just $x$ and not $ax$. If it were $ax$, the problem would have been trivial. Here it is: $$\int_{0}^{+\infty} \frac{a x \cos(x)-\sin(ax)}{x^2} \, dx, \qquad a>0$$
No complex analysis is allowed to use, only calculus knowledge, for example, differentiating integrals depending on a parameter.

答:
Too long for comments but this is an interesting integral.
$$I=\int\frac{a x \cos(x)-\sin(ax)}{x^2} \, dx=a\int\frac{ \cos(x)}x \, dx-\int \frac{\sin(ax)} {x^2}\,dx$$
$$\int\frac{ \cos(x)}x \, dx=\text{Ci}(x)$$
$$\int \frac{\sin(ax)} {x^2}\,dx=a\int \frac{\sin(y)} {y^2}\,dy$$
One integration by parts gives
$$\int \frac{\sin(y)} {y^2}\,dy=\text{Ci}(y)-\frac{\sin (y)}{y}$$ Back to $x$
$$I=a \,\text{Ci}(x)-a \,\text{Ci}(a x)+\frac{\sin (a x)}{x}$$
Now, considering
$$J(t)=\int_t^\infty\frac{a x \cos(x)-\sin(ax)}{x^2} \, dx$$
$$\lim_{t\to \infty } \, J(t)=0$$

$$J(t)=a(1-\log (a))+\frac a2 \sum_{n=1}^\infty (-1)^{n+1}\,\frac{a^{2n}-(2n+1) } {n\,(2n+1)!} t^{2n}$$

Title: Calculating an improper integral.


问:
Determining a Determinant given some scalar products

We are given $x,y,z$ vectors in $\mathbb R^3$. We know that $x^Tx=3, y^Ty=3, z^Tz=3, x^Ty=2, y^Tz=2, z^Tx=1$. Furhermore we are given that
for some matrix $A\in \mathbb R^{3\times 3}$, we have $$Ax=y+z, Ay=x+z, Az=x+y$$ Determine whether the vectors $x,y,z$ are linearly
dependant and compute the determinant of $A$.

I cannot manage to solve this, neither of the questions... Any thoughts?

答:
For the first part, multiply
$\alpha x^T+\beta y^T+\gamma z^T=0$
by $x$, $y$, and $z$ to obtain
$$\begin{align}
3\alpha+2\beta+\gamma &=0\\
2\alpha+3\beta+2\gamma &=0\\
\alpha+2\beta+3\gamma &=0\\
\end{align}.$$
Show that the only solution is $(0,0,0)$, hence $\{x,y,z\}$ is a basis.  (Notice that the coefficient matrix is the Gram-matrix of $x$, $y$, and $z$.)

The matrix of $A$ in respect to that basis is
$$\begin{pmatrix}
0 & 1 & 1\\
1 & 0 & 1\\
1 & 1 & 0
\end{pmatrix}.$$
Now compute the determinant.


问:
Ergodicity and constant Lyapunov exponents on attractors
The Wikipedia article  on Lyapunov exponents says the following:

For a [real] dynamical system with evolution equation $\dot{x_i} = f_i(x)$ in an n–dimensional phase space, the spectrum of Lyapunov exponents $\{\lambda_1, \lambda_2, \dots, \lambda_n \}$ in general, depends on the starting point $x_0$. However, we will usually be interested in the attractor (or attractors) of a dynamical system, and there will normally be one set of exponents associated with each attractor. The choice of starting point may determine which attractor the system ends up on, if there is more than one.

In the same paragraph it says:

The set of Lyapunov exponents will be the same for almost all starting points of an ergodic component of the dynamical system.

Am I to conclude from this that once a system has settled on an attractor, it is always ergodic? In other words, after the transients of the system, the system will simply visit all states on the attractor with an equal frequency?

答:
Am I to conclude from this that once a system has settled on an attractor, it is always ergodic? In other words, after the transients of the system, the system will simply visit all states on the attractor with an equal frequency?

For every definition of attractor that I am aware of, an attractor is minimal and inseparable, i.e., it does not have any true subset that is an attractor itself. If two trajectories starting on an attractor would visit disjoint sets of points, these sets would be separate attractors. Now, once two trajectories have visited the same point, they are identical except for a time shift on account of the system being deterministic.

However there is some fine print to be had here, namely what exactly do we mean by attractor, inseparable, after transients, visit, equal frequency, etc.
For any non-pathologic system, it doesn’t matter which exact definition of ergodic or attractor you use, however, there are systems exhibiting weak ergodicity breaking (see, e.g., this paper). The phase space of these systems can be segmented into regions that are connected, but the expected waiting time for going from one region to another is infinite. Depending on your definitions, all the regions form one attractor, but the system is not ergodic (even if on the attractor).

Finally, even for not-pathologic system, you need to pay attention what exactly you mean with equal frequency, as the attractor is not equally dense everywhere in phase space – but then the attractor has measure zero with respect to a measure of the phase space.


问:
Proving $\tan(90^\circ- x) = \cot x$
How do you prove $\tan(90^\circ-x)=\cot x$ using the compound angle formulae?

答:
An alternative approach that avoids infinities and limits:

$$\tan(90^\circ-x)=\tan(45^\circ+(45^\circ-x))=\frac{1+\tan(45^\circ-x)}{1-\tan(45^\circ-x)}=\frac{1+\left(\frac{1-\tan x}{1+\tan x}\right)}{1-\left(\frac{1-\tan x}{1+\tan x}\right)}$$
$$=\frac{2}{2\tan x}=\frac{1}{\tan x}=\cot x$$
as required.

Title: Proving $\tan(90^\circ- x) = \cot x$


问:
Critical points of the function $f(x,y) = xe^y$ subject to the constraint $x^2+y^2=2$
To find critical points of the function $f(x,y) = xe^y$ subject to the constraint $x^2+y^2=2$.

Let $g(x,y) = x^2+y^2-2$. We use the method of Lagrange Multipliers:

$$f_x = \lambda g_x \implies e^y = 2\lambda x , f_y = \lambda g_y \implies xe^y = 2\lambda y.$$

From the above, we get $x^2 = y$ and using $x^2+y^2=2$ we have $y = 1,-2$. Thus the critical point is $(1,1)$.

Is the answer and method correct?

答: In your solution, when you say that $x^2 = y$, you are assuming that $\lambda \ne 0$. You should also cover this case. Even if you can immediately dismiss it, it should be mentioned. Also, you seem to be missing critical points... Why is $(-1,1)$ not a critical point?


问:
Probability distribution estimation
The joint frequency distribution of two discrete random variables $X$ and $Y$ is given in the table below:

$y$

1
2
3
4

1
0.10
0.05
0.02
0.02

$x$
2
0.05
0.20
0.05
0.02

3
0.02
0.05
0.20
0.04

4
0.02
0.02
0.04
0.10

Using $\text{Var}(X)=\sum [x_i−\mathbb E(X)]^2 \cdot \mathbb P(X=x_i)$, we can define the conditional variance of $X$ for a given value of $Y$: $\text{Var}(X|Y=y)$. For example: $\text{Var}(X|Y=1)=\sum[x_i−\mathbb E(X|Y=1)]^2⋅\mathbb P(X=x_i|Y=1)$. Find $\text{Var}(X|Y=1)$.

In order to solve this I estimated this tables below
|       | y     |   |   |   |   |   |
|-  |:-:    |-  |-  |-  |-  |-  |
| x     |       | 1     | 2     | 3     | 4     |   |
|   | 1     | 0.10  | 0.05  | 0.02  | 0.02  | 0.19  |
|   | 2     | 0.05  | 0.20  | 0.05  | 0.02  | 0.32  |
|   | 3     | 0.02  | 0.05  | 0.20  | 0.04  | 0.31  |
|   | 4     | 0.02  | 0.02  | 0.04  | 0.10  | 0.18  |
|   |   | 0.19  | 0.32  | 0.31  | 0.18  |   |

1
2
3
4

X | Y = 1
0.53
0.26
0.11
0.11

X | Y = 2
0.16
0.63
0.16
0.06

X | Y = 3
0.06
0.16
0.65
0.13

X | Y = 4
0.11
0.11
0.22
0.56

W   = E(X | Y = y)

1.79
2.13
2.84
3.22

So next should be estimation of of E(W). In last table I estimated half of this but I can't find how to estimate other half with values 0.19, 0.32, 0.31, 0.18 So can anybody help me how to find this ?

答: Are you trying to calculate $\mathrm{Var}(X|Y=1)$? To calculate that, you can use the table (with rows $X | Y = y$ for $y = 1, 2, 3, 4$) you have computed. Since we are interested in the distribution of $X$ conditional on $Y = 1$, you would be interested in the first row. Indeed, you can then simply imagine those four entries as defining your desired distribution, and you can then use the usual formulas to calculate the expected value (e.g. $\mathbb{E}[X|Y=1] = 0.53(1) + 0.26(2) + 0.11(3) + 0.11(4) = 1.82$), and the variance (i.e. using the first formula you wrote).

Title: Probability distribution estimation


问:
How to solve explicitly the Dirichlet problem (in dimension 2) with boundary data $f(e^{i\theta}) := ( \sin(\theta) - \cos(2\theta))^2$

Solve explicitly the Dirichlet problem on the unit ball $B_1(0) \subseteq \mathbb{R}^2$
\begin{cases}
\Delta u = 0 \\
u|_{\partial B_1(0)} = f
\end{cases}
with $f(e^{i\theta}) := ( \sin(\theta) - \cos(2\theta))^2$

Using the theorem I studied, the solution inside the unit ball is of the form
$$u(re^{i\phi})=\frac{1}{2\pi} \int_0^{2\pi} \frac{1-r^2}{1-2r\cos({\theta - \phi}) + r^2} \ f(e^{i\theta})d\theta$$

so I need to evaluate the following integral:
$$\int_0^{2\pi} \frac{( \sin(\theta) - \cos(2\theta))^2}{1-2r\cos({\theta - \phi}) + r^2} \ d\theta$$

I would appreciate any hint or other ways to solve this problem.

答:
In this situation, it is easier to deal with the following expression for the Poisson Kernel:
$$P_r(\vartheta)=\sum_\mathbb Z e^{in\vartheta}r^{|n|}$$
Using this, and
$$u(re^{i\phi})=\frac{1}{2\pi}\int_0^{2\pi} P_r(\phi-\vartheta)f(e^{i\vartheta})$$
The result follows easily:
$$u(re^{i\phi})=\sum_\mathbb{Z} \left(\frac{1}{2\pi}\int_0^{2\pi}f(e^{i\vartheta})e^{-in\vartheta}d\vartheta\right)r^{|n|}e^{in\phi}$$
It is not hard to see that
$$f(e^{i\phi})=1+\sin(\phi)-\frac12\cos(2\phi)-\sin(3\phi)+\frac12\cos(4\phi)$$
Splitting real and immaginary part we get (since $u$ is real):
$$u(re^\phi)=\frac{1}{2\pi}\int_0^{2\pi}f(e^{i \vartheta})d\vartheta+ \sum_0^\infty
 \left(\frac{1}{\pi}\int_0^{2\pi}f(e^{i\vartheta})\cos(n\vartheta)d\vartheta\right)r^n\cos(n\phi)+
\sum_{n=1}^\infty \left(\frac{1}{\pi}\int_{n=1}^{2\pi}f(e^{i\vartheta})\sin(n\vartheta)d\vartheta\right)r^n\sin(n\phi)$$

Using the usual orthogonality relations between $\sin(nx),\cos(mx)$, one gets
$$u(re^{i\phi})=1+r\sin(\phi)-\frac12r^2\cos(2\phi)-r^2\sin(3\phi)+r^4\frac12\cos(4\phi)$$


答: You missed the other critical point. Once you get $y=1$, you get $x^2=1$, $x=\pm 1$, so the critical points are $(1,1)$ and $(-1,1)$.

Title: Critical points of the function $f(x,y) = xe^y$ subject to the constraint $x^2+y^2=2$

Body:
To find critical points of the function $f(x,y) = xe^y$ subject to the constraint $x^2+y^2=2$.

Let $g(x,y) = x^2+y^2-2$. We use the method of Lagrange Multipliers:

$$f_x = \lambda g_x \implies e^y = 2\lambda x , f_y = \lambda g_y \implies xe^y = 2\lambda y.$$

From the above, we get $x^2 = y$ and using $x^2+y^2=2$ we have $y = 1,-2$. Thus the critical point is $(1,1)$.

Is the answer and method correct?


问:
Math blogs, pros and cons for writers?
I regularly read blogs by three mathematicians, and occasionally run into others. Definitely they help me a lot studying mathematics.

But now I am more interested in the writers' perspective, and I wonder whether writing a math blog help study mathematics. What are the pros and cons for the writer himself, especially for a first-year graduate student?

A lot of users have tons of experience writing blogs. I really want to hear your opinion on this.

Thanks!

Maybe I just start writing some of my own thoughts: cons: 1. writing a blog consumes lots of time. pros: 1. writing a blog forces you to think deeply and clearly; 2. it helps you learn how to communicate mathematics; 3. it might help you connect to others who share your passion.

答:
What do you want to do with your math blog? What can a math blog do for you? These are the questions you want to be asking before starting up a math blog.

Some people (such as myself) write a math blog much like people write other personal blogs. They may discuss their day-to-day lives (or the mathematically related pats of them) and the mathematics may be a recurring theme, but mathematical exposition might not always be the overall focus.

Others use math blogs to organize and disseminate mathematical exposition. One use is writing and posting notes on some mathematics you are doing at the moment. Terence Tao's blog is an example where you can catch glimpses of his research in some of his posts.

Some students use blogs to store notes on their studies, but I think this practice is meaningful if and only if you intend for others to view and benefit from or comment on the notes. Abstract Nonsense in one such blog. If you don't care for others reading and interacting with your notes, then it's probably best to keep them offline.

Some blogs are topic-oriented, like Sketches of Topology. They are useful for getting people interested in your research and finding people who share your interests to interact with.

Some blogs have an explicitly educational/organizational purpose. Secret Blogging Seminar is organized by a group of mathematicians who discuss topics that they or the community think are important. There is a blog at http://392c.wordpress.com/ that is devoted to a specific course.

Finally, many blogs are there because mathematicians want to reach out to the broader mathematical and scientific community, for reasons varying from research collaboration, idea-bouncing, community feedback, commenting on current events in the field, and so on. Terence Tao and Tim Gowers and others have blogs that exemplify this sort of use, as well as combining all of the above.

What you should do is see for yourself what needs you want to fill via a math blog. Many blogs are an eclectic mix of all the purposes I have described above, and some do other things as well, like Polymath. Always keep in mind, though: much like mathematics is no spectator sport, mathematical blogging is really a team sport. To get the most out of a mathematics blog, I think you should try to reach out to the community and make new friends and connections.

ContentLicense: CC BY-SA 3.0

Title: Math blogs, pros and cons for writers?


问:
solving $\int_{-\infty}^\infty \frac{\alpha y\cdot e^{-\alpha\sqrt{x^2 + y^2}}}{\sqrt{x^2 + y^2}}dx $
Trying to solve the following integral,
$$\int_{-\infty}^\infty  \frac{\alpha y\cdot e^{-\alpha\sqrt{x^2 + y^2}}}{\sqrt{x^2 + y^2}}dx $$
where $\alpha$ is a constant.

this is an integral over $dx$ of the partial derivative of the original function
$ 1-e^{-\alpha\sqrt{x^2+y^2}}$ by $dy$:

$$ \frac{d(1-e^{-\alpha\sqrt{x^2+y^2}})}{dy} $$

Thanks!

答: $x=y\sinh t$ leads to the modified Bessel's (a.k.a. Macdonald's) $K_0$: assuming $\alpha,y>0$, $$\int_{-\infty}^\infty\frac{e^{-\alpha\sqrt{x^2+y^2}}}{\sqrt{x^2+y^2}}\,dx=\int_{-\infty}^\infty e^{-\alpha y\cosh t}\,dt=2K_0(\alpha y).$$


问:
Determine next in sequence when many patterns visible
In a sequence given to find the next number that follows what happens if we can see 2 different ways to produce the next number? Given that the approach is based on inductive reasoning, unless the sequence is some specific well known math sequence how do we choose which is the most sensible choice?

Example:

If we have:

$3, 6, 18, 36, 108, 216, X?$

We could claim the following:

Since each number of the given sequence above is exactly divided by the 2 previous ones i.e. 18 is divided exactly by 6 and 3, 36 by 18 and 6, 108 by 36 and 18 and 216 divided exactly by 108 and 36 then $X=432$ since it is exactly divided by $108$ and $216$

Since we have $6 = 3*2, 18 = 6*3, 36=18*2, 108=36*3, 216=108*2$ then $X=648=216*3$

Is there some reason (2) is more valid than (1)?

答:
The most sensible choice

Unless you define exactly what you mean by sensible, the question you're asking isn't well-defined. The Fibonacci sequence is as good as any sequence starting like $\{1,1,2,3,5,8,13,-9383948,...\}$. In your case, $X$ can be absolutely any number and there would be no objective rule for ranking the options.

You can create a particular definition of sensible. For example, you could say "If it's an arithmetic or geometric sequence, those are sensible choices". Then you could use that rule to distinguish different options. But then your question answers itself: to rank the options you need to create a subjective rule to do that. The validity of your rule is again, subjective.


问:
Column and Row Space of a Matrix
From Bernard Kolman Linear Algebra I have encountered this example.
Find a basis for column space:
$$A=\begin{bmatrix}
1 &-2 &0 &3 &4\\
3 & 2 & 8 &1&4\\
2 & 3 & 7 & 2 & 3\\
-1 & 2 & 0 & 4 & -3
\end{bmatrix}.$$In the question he transpose it and reduced to row echelon form.
$$A^{\mathbf{T}}=\begin{bmatrix}
1&3&2&-1\\
-2&2&3&2\\
0&8&7&0\\
3&1&2&4\\
-4&4&3&-3\end{bmatrix}\equiv \begin{bmatrix}
1&0&0&\frac{11}{24}\\
0&1&0&\frac{-49}{24}\\
0&0&1&\frac{7}{3}\\
0&0&0&0\\
0&0&0&0\end{bmatrix}$$and he took firt three rows and again transpose it so he said that it is baiss for column space. But why? Why we transpose it? And why we don't just reduced it to row echelon form? Because in the previous sections, we reduced the matrix and take the pivot's in order to find a basis. I just don't understand. Thanks.

答:
Consider, say,$$A=\begin{bmatrix}1&1\\1&1\end{bmatrix}.$$Its column space is $\left\{\left[\begin{smallmatrix}x\\x\end{smallmatrix}\right]\,\middle|\,x\in\Bbb R\right\}$, right? But the reduced row echelon form of $A$ is $\left[\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right]$, and the column space of this matrix is $\left\{\left[\begin{smallmatrix}x\\0\end{smallmatrix}\right]\,\middle|\,x\in\Bbb R\right\}$, which is not the previous one.

But we do not have the same problem with rows. The row space of a matrix and the row space of its reduced row echelon form are the same.

Title: Column and Row Space of a Matrix


问:
Proof involving consecutive coefficients

If $a_1,a_2,a_3,a_4$ are four consecutive coefficients in the expansion of $(1+x)^n$, prove that $$\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}=\frac{2a_{2}}{a_{2}+a_{3}}$$

My solution:

Let $a_1 x^{r-1},a_2 x^{r},a_3 x^{r+1},a_4 x^{r+2} = {n\choose{r-1}} x^{r-1},{n\choose{r}} x^{r},{n\choose{r+1}} x^{r+1},{n\choose{r+2}} x^{r+2}$.

I sub this into the required proof and I have to do some laboured, long algebra process.

From my experience, someone on MSE will have a much more concise proof than this, so I am asking to enlighten myself.

答:
Hints:

Write the terms in the equality to be proved as follows:

$$\frac{a_k}{a_k+a_{k+1}}=\frac1{1+\cfrac{a_{k+1}}{a_k}}$$

Evaluate now each term separatedly. For example:

$$\frac{a_4}{a_3}=\frac{\binom n{r+2}}{\binom n{r+1}}=\frac{n!(r+1)! (n-r-1)!}{n!(r+2)!(n-r-2)!}=\frac{n-r-1}{r+2}\implies1+\frac{a_4}{a_3}=\frac{n+1}{r+2}\implies$$

$$ \frac1{1+\frac{a_4}{a_3}}=\frac{r+2}{n+1}$$

After you do this with the left side (and they have they same denominator), do the same with the right side...and it follows at once that they are the same


问:
Runge-Kutta method for a 2nd order nonlinear ODE/ CMC surface meridian
I'm trying to solve numerically the following IVP nonlinear second-order differential equation:
$$f''(x)=A\left(1+f'(x)^2\right)^{3/2}+\frac{\left(1+f'(x)^2\right)}{f(x)}$$

where $A$ is a constant and the problem has the initial conditions (IVP) $f(0)=R$ and $f'(0)=0$, where $R$ is also a constant.

I'm used to applying the RK4 method for first-order differential equations, but is it possible to solve it by the RK4 method as well? If so could someone provide me a good reference to understand this specific application for method? I also open to any other useful numerical method that can solve this IVP. Thanks in advance.

答:
You can just write your equation as a system of two first order equations and apply RK to that system.

$$
\begin{cases}
f' = &g \\
g' = &A(1+g^2)^{3/2} + \dfrac{1+g^2}{f}\\
\end{cases}
$$

with initial conditions $f(0) = R, g(0)=0$. If the method is appropriate to this particular problem is another matter. There seems to be a singularity near $x=1$.


问:
Central Limit Theorem for an Exponential Distribution
I came across the following question online (stats.libretexts.org).

De Anza statistics estimated that the amount of change daytime students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.

Find the probability that the average of the 25 students was between \$0.80 and $1.00. Graph the situation, and shade in the area to be determined.

My approach:

Since it is about the mean, we have to use the central limit theorem. Since E(x) = 0.88, and we have an exponential distribution, the standard deviation is 0.88 as well.

Hence, we get:

$$P\left(\frac{(0.8-0.88)}{0.88 / \sqrt(25)} ≤ Z ≤ \frac{(1-0.88)}{0.88 / \sqrt(25)}\right) = P(-0.4545 ≤ Z ≤ 0.68181) = 0.4276$$

Yet, the solution says the corresponding probability is 0.1882. Where is my mistake?

答:
It is not a good idea to edit the post inserting another question. I was just to reply to your first question when you edited and posted the new one.

First problem

The result in your textbook is wrong. Anyway, I do not know why you calculated (correcty) an appoximation of the result, given that for an exponential distribution it is very easy to get the exact result:

$X_i\sim Exp(25/22)$

$Y=\sum_{i=1}^{25}X_i\sim Gamma(25;25/22)$

$\frac{25}{11}Y\sim \chi_{(50)}^2$

Thus $P(0.8<\overline{X}_{25}<1)=P(20<Y<25)=P\Big(\frac{25}{11}20<\chi_{(50)}^2<\frac{25}{11}25\Big)=0.656-0.236=0.420$

Second problem

they assumed normality.

In effect,

$P(X>12)=P\Bigg(Z>\frac{12-10.53}{2}\Bigg)=1-\Phi(0.735)\approx 0.231170$


答:
As Forester hinted, as you are interested in the coefficient, you should drop the $x$ monomials.

All that is needed here is basic operations on binomial coefficients:
$$
\binom{n-1}{k-1} + \binom{n-1}{k} = \binom{n}{k}
$$

In your case, you chose:
$$
a_i = \binom{n}{r+i-2}
$$

Then: $a_i + a_{i+1} = \binom{n+1}{r+i-1}$ and $\frac{a_i}{a_i + a_{i+1}} = \frac{r+i-1}{n+1}$.

Checking that the equation you're looking for holds is fairly straightforward:
$$
\frac{a_1}{a_1 + a_2} + \frac{a_3}{a_3 + a_4} = \frac{2r+2}{n+1} = 2\frac{a_2}{a_2 + a_3}
$$

Title: Proof involving consecutive coefficients

Body:
If $a_1,a_2,a_3,a_4$ are four consecutive coefficients in the expansion of $(1+x)^n$, prove that $$\frac{a_{1}}{a_{1}+a_{2}}+\frac{a_{3}}{a_{3}+a_{4}}=\frac{2a_{2}}{a_{2}+a_{3}}$$

My solution:

Let $a_1 x^{r-1},a_2 x^{r},a_3 x^{r+1},a_4 x^{r+2} = {n\choose{r-1}} x^{r-1},{n\choose{r}} x^{r},{n\choose{r+1}} x^{r+1},{n\choose{r+2}} x^{r+2}$.

I sub this into the required proof and I have to do some laboured, long algebra process.

From my experience, someone on MSE will have a much more concise proof than this, so I am asking to enlighten myself.


问:
What's wrong with this Kuhn-Tucker optimization?
The function $u(x,y,z) = xyz$ is to be maximized, under constraints: $ 0 \le x \le 1, y \ge 2, z \ge 0 $ and $ 4 - x - y - z \ge 0 $

Now I'm not quite sure how to translate the x-constraint into proper inequalaties, but my first shot would be $ h_1 = x \ge 0$ and $h_2 = 1-x \ge 0 $.

So I set up the Lagrangian like:
$$ L = xyz + \lambda_1 x + \lambda_2(1-x) + \lambda_3(y-2) + \lambda_4 z + \lambda_5(4-x-y-z) $$

... resulting in FOC like:

[... irrelevant FOC omitted ...]

$$ \frac{\delta L}{\delta z} = xy + \lambda_4 - \lambda_5 = 0 $$
$$ \frac{\delta L}{\delta \lambda_4} = z \ge 0, \lambda_4 \ge 0, z \lambda_4 = 0$$

Now, as $u=xyz$, neither $x,y,z$ can be $0$ in maximum; thus, $\lambda_4 = 0$. Also, as $u$ is isotone in all $x,y,z$, the fifth constraint has to be effective in maximum, so $\lambda_5 = 0 $ either. So plugging into the first FOC from above, $xy = 0$ - so either $x$ or $y$ has to be $0$, which cannot be a maximum though!

So am I making a mistake in the way I set up the Lagrangian here? In the derivatives? Or can't I apply this method here at all? If so, why not?

Thanks for any hint!

答:
Since the fourth constraint: $z\geq0$ will not bind at the maximum, you are correct that $\lambda_4=0$ (by the complimentary slackness condition). However, since the 5th constraint will bind, $\lambda_5\geq0$.

You are left with $xy=\lambda_5$ rather than $xy=0$.


问:
Way to tackle probability odds problems
I have checked out other questions, but still haven't got satisfactory answer as conditions in those questions were slightly different.
Disclaimer: Only for educational purposes.
We have a fair dice with 10 faces (1-10) and $100. Each roll is completely random. If the number you chose doesn't show up you make 10% profit on your bet, but if the number shows up you lose your entire bet.
How to tackle this problem? What are the ways to optimize profits and limit losses?</p>

答:
Expanding on my comment, the expected remaining cash $C_{n+1}$ after round $n$ is:

$$C_{n+1}=0.9\times B_n\times(1.1)+0.1\times B_n\times 0+(C_n-B_n)=0.99B_n+C_n-B_n$$
so
$$C_{n+1}=C_n-0.001B_n,$$
so as user2661923 says, you're better off betting small amounts $B_n$ because on average $0.001B_n$ will be subtracted from your cash pot every round. Of course you could make big bucks if you're lucky, but then again you can also lose everything that way too.

Title: Way to tackle probability odds problems


问:
Playing around with the Nested Interval Theorem
The Nested Interval Theorem says that:

If $(I_n)$ is a sequence of closed and bounded non-empty intervals in $\mathbb{R}$ with $I_1 \supseteq I_2 \supseteq ...$ then $$\bigcap\limits_{i= 1}^{\infty} I_i \neq \varnothing$$
Moreover, if $\text{len}(I_n) \to 0$ as $n\to\infty$, then $\displaystyle \bigcap\limits_{i= 1}^{\infty} I_i$ is a singleton.

I'm curious about a few questions:

If $(I_n)$ is a sequence of closed non-empty intervals in $\mathbb{R}$ with $I_1 \supseteq I_2 \supseteq ...$, and $\text{len}(I_n) \to 0$ as $n\to\infty$. Also, $$\bigcap\limits_{i= 1}^{\infty} I_i \neq \varnothing$$

Does this imply that $I_n$ is bounded $\forall n\in\mathbb{N}$? What if we remove the condition that $\text{len}(I_n) \to 0$ as $n\to\infty$?

If, in addition to the above, we say that $$\bigcap\limits_{i= 1}^{\infty} I_i = \{x\}$$ what more can we deduce? I suppose we can say that any sequence $x_n\in I_n, \forall n$ is such that $\lim x_n = x$.

If $(I_n)$ is a sequence of bounded non-empty intervals in $\mathbb{R}$ with $I_1 \supseteq I_2 \supseteq ...$, and $\text{len}(I_n) \to 0$ as $n\to\infty$. Also, $$\bigcap\limits_{i= 1}^{\infty} I_i \neq \varnothing$$

Does this imply that $I_n$ is closed $\forall n\in\mathbb{N}$? What if we remove the condition that $\text{len}(I_n) \to 0$ as $n\to\infty$?

If, in addition to the above, we say that $$\bigcap\limits_{i= 1}^{\infty} I_i = \{x\}$$ what more can we deduce?

答:
Regarding your first qestion: "Does this imply that $_$ is bounded?", no.
Suppose that it is the case that we have some sequence of closed intervals such that $I_1\supset I_2...,$ where len$(I_n)\to 0$ as $n\to \infty$ and,
$$\bigcap_{n=1}^{\infty}{I_n}\neq\emptyset$$
If this would imply that $I_n$ is bounded $\forall n\in\mathbb{N}$, the we could just construct some closed unbounded interval $I_0$ such that $I_0\supset I_1$. Then all the previous assumtions holds for the sequence of intervals $(J_n)$, where $J_n=I_{n-1}$ but $(J_n)$ is not bounded for all natural numbers.

The same logic can be applied to your question if it implies that the intervals are closed.

To your other question about a sequence $x_n\in I_n$. If len$(I_n)\to 0$ as $n\to \infty$ Then yes, such a sequence would converge to $x$.

To show this, let $\varepsilon$ be given, then there is some $N\in \mathbb{N}$ such that if $n\geq N$ the len$(I_n)<\varepsilon$. Since $x,x_n\in I_n \implies |x-x_n|<\varepsilon$.


问:
Commuting idempotents repel each other
From a Banach algebra $A$ there are elements $e,f$ satisfying $e^2=e,f^2=f,ef=fe,e\ne f$, prove that $\|e-f\|\ge1$;

I can prove$\|e+f|\ge 1$by$\|e-f\|=\|e^2-f^2\|=\|(e-f)(e+f\|\le\|e-f\|\cdot\|e+f\|$, how to substitute$-
$for$+$?

答:
Letting
$$
  e'=e-ef, \quad\text{and}\quad f'=f-ef,
  $$
we have that $e'$ and $f'$ are idempotent elements satisfying
$$
  e'f'=f'e'=0, \quad\text{and}\quad  \|e'-f'\|=\|e-f\|.
  $$
We may therefore suppose, without loss of generality, that the originally given $e$ and $f$ satisfy $ef=fe=0$.

In this case observe that
$$
  B=\text{span}\{e, f\}
  $$
is a closed unital sub-algebra of $A$,  with unit $1_B=e+f$.  Letting
$u=e-f$,
we then have that $u^2=1_B$, so the spectrum of $u$ (relative to $B$) is contained in $\{1, -1\}$.
Therefore
$$
  \|e-f\| = \|u\| \geq  \sup\big\{|\lambda |: \lambda \in \sigma_B(u)\big\} = 1.
  $$

Title: Commuting idempotents repel each other


问:
Lebesgue integrable implies measurable.
Suppose we define Lebesgue integral like this, but without assuming that $f$ is measurable. Suppose there exists a finite integral $\int_Xfd\mu$. Does it follow that $f$ is measurable?

UPD. $f$ is integrable, if $\Xi(f) = \xi(f)$.

答:
An attemp to develop Behnam Esmayli's idea.

$f$ is Lebesgue integrable $\implies (\forall\epsilon>0)(\exists P_{\epsilon}: E = \sqcup_{i=1}^{\infty}E_i)\quad U(P_{\epsilon}, f) - L(P_{\epsilon}, f) < \epsilon$, where
$U(P_{\epsilon},f) = \sum_{i=1}^{\infty}\sup_{E_i}f(x)\mu(E_i), \quad L(P_{\epsilon},f) = \sum_{i=1}^{\infty}\inf_{E_i}f(x)\mu(E_i)$.

Define for $\epsilon = 1, \frac{1}{2}, ..., \frac{1}{n}, ... \quad h_n(x) = \sum_{i=1}^{\infty}\inf_{E_i}f(x)*\chi_{E_i}(x)$, $h_n(x) \leq h_{n+1}(x) \leq f(x)$, $h_n(x)$ is measurable.

Similarly $g_n(x) = \sum_{i=1}^{\infty}\sup_{E_i}f(x)*\chi_{E_i}(x).$ Then $(\exists N \in \mathbb {N})(\forall x \in E) \quad f(x) -h(x) < g(x) - h(x) < \frac{1}{n}$, thus $h_n(x) \rightrightarrows f(x)$, and since $h_n(x)$ is measurable, $f(x)$ is also measurable.

Any critics, please.

Title: Lebesgue integrable implies measurable.


Body:
$$\frac{a_1}{a_1+a_2}=\frac{\binom{n}{r-1}}{\binom{n+1}{r}}=\frac{n!(n-r+1)!r!}{(r-1)!(n-r+1)!(n+1)!}=\frac{r}{n+1}$$

Similarly, $$\frac{a_3}{a_3+a_4}=\frac{r+2}{n+1}$$

So, summing gives:

$$\frac{2(r+1)}{n+1}$$

which equals $$\frac{2a_2}{a_2+a_3}$$


问:
Mapping of a Lens-shaped region by a Möbius Transformation
Consider the 'lens' described by $\{z:|z-i|<\sqrt{2}\ \text{and}\ |z+i|<\sqrt{2} \}$ . We want to map this to the upper right quadrant using a Möbius transformation.

The two circles meet at $z=1,-1$ and so if we use the map $f:z \mapsto  \frac{z-1}{z+1}$ we should send the lens to a sector centered on the origin.

I have two questions:

How do we find the angle sweeped out by the sector? As $f$ is conformal, it will be equal to the angle at which the two circles meet; but how is this calculated? And is there some general method for working this out?

After finding this angle, how do we work out how the sector is rotated relative to say the real axis after applying $f$? Again, is there some general way of doing this also?

After establishing this, the question becomes simple as we can use a power map to adjust to the correct angle and then a rotation to position the sector.

After responses to 1. and 2. I think my approach is sound, though I'd be interested in hearing if it isn't also.

答:
The centers $\pm i$ of the circles and the circle intersection points $\pm1$ form a square, so the circles intersect at right angles (the sides of the square are tangents). Therefore, a power map is not needed here. (The general method of course involves some acrtan manipulations with these four points)

For the second question, the lens is symmetric with respect to the real axis and your transform leaves the real axis invariant, therefore the image is also symmetric with respect to the real axis. So you should rotate by $45^\circ$ in the end.


问:
Gibbs' inequality; continuous case
I'm reading alternative proof of Gibbs' inequality written in wikipedia, which states that

Suppose that $P=\{p_1,...,p_n\}$ be probability distribution.

Then for any other probability distribution $Q=\{q_1,...,q_n\}$, $-\sum p_i\log{p_i} \leq -\sum p_i\log{q_i}$

I have two question.

First, the proof uses Jensen's inequality to claim that $\sum p_i \log \frac{q_i}{p_i} \leq \log \sum p_i \frac{q_i}{p_i}$.

But why does it hold? I think Jensen's inequality just says that $\sum \log{(p_i \frac{q_i}{p_i})} \leq \log \sum p_i \frac{q_i}{p_i}$

Second, can we apply Gibbs' inequality in the continuous case?
i.e., does $-\int f(x)\log{f(x)} dx \leq -\int f(x)\log{g(x)} dx$ still holds for probability density function $f, g$?

I think since we can apply Jensen's inequality in the continuous case, we can still argue that the continuous case holds. Nevertheless, I cannot find any mention of Gibbs' inequality in the continuous case; they only deal with the discrete case. Is there any problem to deal with the continuous case? Or can I use Gibbs' inequality in the continuous case as I write?

Thanks.

答:
Partial answer for the first part using "elementary stuff" that OP has asked for in comments:

symbol
explantion

$p_i$
weights

$\log$
concave function

$q_i/p_i$
points in the domain where the concave function is defined

In the quoted inequality $$\sum p_i \log \frac{q_i}{p_i} \leq \log \sum p_i \frac{q_i}{p_i},$$

LHS: weighted average of $\log(q_i/p_i)$'s.

RHS: $\log$ of weighted average of $q_i/p_i$'s.

Note that $\log$ is concave, so Jensen's inequality would give the desired result.

The second part is addressed in the linked blog article, and I'm not going retype his/her arguments.

Title: Gibbs' inequality; continuous case


问:
Counting deltahedra with $2n$ faces
A deltahedron is, according to Wikipedia,

a polyhedron whose faces are all equilateral triangles.

There is only one deltahedron with four faces: the tetrahedron.

Likewise, there is only one deltahedron with six faces: the triangular bipyramid

There are two with eight faces:

the octahedron

the biaugmented tetrahedron (not convex)

And at least five with 10 faces:

the pentagonal bipyramid

the augmented octahedron (contains coplanar faces)

the three triaugmented tetrahedra

I'm looking to see if anyone has computed the number of deltahedra with $2n$ faces. I did a cursory OEIS search, but didn't find anything.

Is there a known formula or recurrence relation for the number of deltahedra that can be constructed from $2n$ equilateral triangles? If so, is there a corresponding OEIS entry?

Title: Counting deltahedra with $2n$ faces

Body:
I applied OEIS A000577 to proof the count of 2n faces deltahedra.

I used 3 4-iamonds (A000577(4)=3) to proof only one deltahedron with four face. I used 12 6-iamonds (A000577(6)=12) proof only one deltahedron with six faces. Also, I used 66 8-iamonds (A000577(8)=66) proof only two deltahedron with eight faces. In the same manners, I used 448 10-iamonds (A000577(10)=448) proof only five (5) deltahedron with ten faces. Recently, I have spent two years (year 2018 -2020) and used 3334 12-iamonds (A000577(12)=3334) to construct and proof thirteen (13) deltahedra with twelve faces.

They are D12(0,4,4,0), D12(1,3,3,1), D12(1,4,1,2), D12(2,0,6,0), D12(2,1,4,1), D12(2,2,2,2)1, D12(2,2,2,2)2, D12(2,2,3,0,1), D12(2,3,1,1,1)1, D12(2,4,0,0,2), D12(3,1,1,3), D12(3,1,2,1,1), D12(4,0,0,4). Where D12 stands for a deltahedron with 12 faces. For example, D12(0,4,4,0) denote that 12-deltahedron with 8 vertices: 4 of degree 4 and 4 of degrees 5 and none of degree 3 and 6.
Since having known the count of thirteen (13) deltahedra with twelve faces, by single-face augmentation and multiple-face augmentation methods (Naoya Tsuruta, Enumeration of Deltahedral Graphs With Up to 10 Vertices, @2014 ISGG) we discovered forty-seven (47) graphs of deltahedra with fourteen faces.
They are D14(0,3,6,0), D14(0,4,4,1), D14(0,5,2,2), D14(0,6,0,3), D14(1,2,5,1), D14(1,3,3,2), D14(1,4,1,3), D14(2,1,4,2)1, D14(2,1,4,2)2, D14(2,2,2,3)1, D14(2,2,2,3)2, D14(2,2,2,3)3, D14(2,3,0,4), D14(3,0,3,3)1, D14(3,0,3,3)2. D14(1,4,2,1,1)1, D14(1,4,2,1,1)2, D14(1,5,1,0,2), D14(2,1,5,0,1), D14(2,2,3,1,1)1, D14(2,2,3,1,1)2, D14(2,2,3,1,1)3, D14(2,3,1,2,1)1, D14(2,3,1,2,1)2, D14(2,3,1,2,1)3, D14(2,3,2,0,2)1, D14(2,3,2,0,2)2, D14(3,1,2,2,1)1, D14(3,1,2,2,1)2, D14(3,1,3,0,2), D14(3,2,0,3,1)1, D14(3,2,0,3,1)2, D14(3,2,1,1,2)1, D14(3,2,1,1,2)2, D14(4,0,2,1,2), D14(4,1,0,2,2), D14(2,2,4,0,0,1), D14(2,3,2,1,0,1)1, D14(2,3,2,1,0,1)2, D14(2,4,0,2,0,1), D14(2,4,1,0,1,1), D14(3,1,3,1,0,1), D14(3,2,1,2,0,1)1, D14(3,2,1,2,0,1)2, D14(3,2,2,0,1,1)1, D14(3,2,2,0,1,1)2, D14(4,0,2,2,0,1). Please note that D14(2,1,4,2)1 and D14(2,1,4,2)2 share the same numbers of vertices and edges, that is isomorphism graph.

Currently, I am working on counting deltahedra with sixteen (16) faces.  Since having known the count of forty-seven (47) graphs of deltahedra with fourteen (14) faces, by single-face augmentation and multiple-face augmentation methods we discovered 226 graphs of deltahedra with sixteen faces including twelve isomorphism graphs of  D16(3,2,1,1,1). As per Prof. Naoya Tsuruta’s paper (Random Realization of Polyhedral Graphs as Deltahedra, J. for Geometry Graphics 227-236, 2015) there exists 233 graphs including 154 Deltahedral Graphs and 79 Non-Deltahedral Graphs. I will continue construct and confirm with Prof. Tsuruta for all graphs of deltahedra with sixteen (16) faces.


Title: Commuting idempotents repel each other

Body:
From a Banach algebra $A$ there are elements $e,f$ satisfying $e^2=e,f^2=f,ef=fe,e\ne f$, prove that $\|e-f\|\ge1$;

I can prove$\|e+f|\ge 1$by$\|e-f\|=\|e^2-f^2\|=\|(e-f)(e+f\|\le\|e-f\|\cdot\|e+f\|$, how to substitute$-
$for$+$?

Body:
Let $E=e(1-f)$ and $F=(1-e)f$; then $E^2=E$, $F^2=F$, $EF=FE$ and $E\ne F$. So, by the same proof as in the question, $1\le\|E+F\|$. Then
$$1\le\|e(1-f)+f(1-e)\|=\|e+f-2ef\|=\|(e-f)^2\|\le\|e-f\|^2$$


问:
Tough Number Theory mod problem
If the product of the factors of $30^{12}$ that are congruent to 1 mod 7 can be expressed as $2^{a} \cdot 3^{b} \cdot 5^{c},$ find $a+b+c$.

I tried using mod rules to simplify but I got a very strange answer which was divisible by 7 and did not fit the answer criteria. I noted that 15 is 1 mod 7 and even 8 is 1 mod 7, but I do not know how to go ahead from there.

EDIT: I also noted that 6 is -1 mod 7 and 30^12 is 1 mod 7. But that didn't help either.

答:
Since $30=2\cdot 3\cdot 5$, the factors of $30^{12}$ are of the form $2^x\cdot3^y\cdot5^z$ where $x,y,z$ are between $0$ and $12$.

Since $7$ is prime, we know from Fermat's little theorem that $a^6 \equiv 1 \bmod 7$ for $a$ coprime to $7$. So, if we have $2^x\cdot3^y\cdot5^z\equiv 1 \bmod 7 $ for some particular $x,y,z$ we can also say that $2^{x+6}\cdot3^y\cdot5^z\equiv1 \bmod 7 $ and similarly $2^x\cdot3^{y+6}\cdot5^z$, $2^x\cdot3^y\cdot5^{z+6}$ etc.

Directly calculating powers of $2\bmod 7$ we find the cycle is actually only length $3$: $2^0\equiv 1,2^1\equiv 2,2^2\equiv 4,2^3\equiv 8\equiv 1$. So we can focus on $x\in\{0,1,2\}$ and infer other solutions by adding multiples of $3$ to the base $x$ value.

We can also check that powers of both $3$ and $5$ cycle through all $6$ coprime residue values in sequence; $3^y\equiv \{1,3,2,6,4,5,1,\ldots\}$ and $5^z\equiv \{1,5,4,6,2,3,1,\ldots\}$ (starting from zero in both cases).

For $x=0$, which also holds for $x = 3,6,9,12$ giving $2^x\equiv 1 \bmod 7$, we can have $(y,z) = (0,0), (0,6), (0,12), (6,0), (6,6), (6,12), (12,0),(12,6), (12,12)$ and also $(y,z) = (1,1), (2,2), (3,3), (4,4), (5,5)$ and the values formed by adding $6$ to either $y$ or $z$ or both. This gives  $9+4\cdot 5 = \fbox{29}$ options for each of the $\fbox{5}$ eligible values of $x$ with the sum of $(y+z)$ over all these options at $108+30+2\cdot 60+90 = \fbox{348}$ and total value of all eligible $x$ at $0+3+6+9+12=\fbox{30}$.

For $x\in \{1,4,7,10\}$, $\fbox{4}$ values totalling $\fbox{22}$ with $2^x\equiv 2\bmod 7$, we can have $(y,z)$ formed from $(\{0,6,12\},\{2,8\})$, $(\{4,10\},\{0,6,12\})$, and also $(1,3), (2,4), (3,5), (5,1)$ with the $+6$ options for $y$ and $z$. Overall $6\cdot 2 + 4\cdot 4 = \fbox{28}$ options for a total $(y+z)$ of $66+78+24+2\cdot 48+72 = \fbox{336}$ per $x$ value.

For $x\in \{2,5,8,11\}$, $\fbox{4}$ values  totalling $\fbox{26}$ with $2^x\equiv 4\bmod 7$, we can have $(y,z)$ formed from $(\{0,6,12\},\{4,10\})$, $(\{2,8\},\{0,6,12\})$, and also $(1,5), (3,1), (4,2), (5,3)$ with the $+6$ options for $y$ and $z$. This also gives $\fbox{28}$ options per $x$ value for a total $(y+z)$ of $\fbox{336}$.

Across all options then we have $5\cdot 348 + 4\cdot 336  + 4\cdot 336 = 4428$ total for $(y+z)$ and $29\cdot 30 + 28\cdot 22+ 28\cdot 26 = 2214$ total for $x$.

This gives use the desired $(a+b+c)$ total  of $2214+4428 = 6642$.

Title: Tough Number Theory mod problem


问:
$\mathcal{A}\perp_\mathcal{G}\mathcal{B}\wedge\mathcal{H}\subseteq\mathcal{G}\implies\mathcal{A}\perp_\mathcal{H}\mathcal{B}$?
If $\mathcal{A}\perp_\mathcal{G}\mathcal{B}$ and $\mathcal{H}\subseteq\mathcal{G}$, is it the case that $\mathcal{A}\perp_\mathcal{H}\mathcal{B}$? Here $\mathcal{A}$, $\mathcal{B}$, $\mathcal{G}$ and $\mathcal{H}$ are sub-$\sigma$-algebras of some common probability space $\left(\Omega,\mathcal{F},P\right)$ and $\cdot\perp_\square\cdot$ is the conditional independence relation.</p>

答:
I'd like to describe a slight variation on Davide's answer that i find crisper.

Suppose $\mathcal{A}$ and $\mathcal{B}$ are dependent fields that are conditionally independent given $\mathcal{G}$ (see, e.g., the Wikipedia article on Conditional Independence) and let $\mathcal{H}:=\left\{\emptyset,\Omega\right\}$. Then independence conditional on $\mathcal{H}$ is the same as unconditional independence, hence the desired result.


问:
Show that a polynomial is inside $\Bbb Q^*$
Observe the following polynomial over the polynomial ring $\Bbb Q[x]$:
$$f(x)=x^5+1$$
We can rewrite $f$ as $f(x)=(x-1)(x^4-x^3+x^2-x+1)$. Now for showing that $f$ is not reducible, $(x-1) \in \Bbb Q[x]^*$. Now I know that $\Bbb Q[x]^*=\Bbb Q^*$ but I can't use it...

答:
Your factorisation is slightly off; it should be
$$f(x)=(x+1)(x^4-x^3+x^2-x+1)$$
The existence of a linear factor already shows by definition that $f$ is reducible.


问:
A reflection question about a complex function
Let's say we have an entire complex function $f(z)$ such that:

$f(z)$ is real when $z$ is real

$f(z)$ is purely imaginary when $z$ is purely imaginary.

SO basically this entire function maps real/im axis to itself.
The question asks me to show this function is an odd function and I have no clue.

答: Let $f_1$ be the restriction of $f$ to $\{\Im(z)\ge 0\}$. By Schwarz's reflection, this admits an analytic extension to $\mathbb{C}$ defined in $\Im(z)\le 0$ by $g(z)=\overline{f_1(\bar z)}$. By the identity principle, $g=f$. Since $f$ is purely imaginary on $i\mathbb{R}$, this equation tells us that $f(z)+f(-z)$ is zero on $i\mathbb{R}$. By the identity principle again, $f(z)+f(-z)\equiv 0$, which implies that $f$ is odd.


问:
Finding trace of matrix from given conditions
I have a question that says,

Let A be a matrix such that $A = [a_{ij}]$ such that $a_{ij}=u_iv_j$ $\forall$ $1 \leq i\leq n$ and $1 \leq j\leq n$, and $u_i, v_j \in \mathbb{R}$ satisfies $A^5=16A$, then find $Tr(A)$, $($Here, $Tr(A)$ denotes the trace of $A$$)$

I have tried it as follows:

$$A^5=16A$$

$$\implies A^{-1}A^5=16A^{-1}A$$
$$\implies IA^4=16I$$
$$\implies A^4=16I$$

But, in the question, it is given that $a_{ij}=u_iv_j$, what are $u, v$ here? The answer is stated to be $2$, but I cannot proceed further. Any help on what $u, v$ are and how to proceed would be helpful. I think that the question might have been incomplete, but if anyone can help anyone, it would be nice. I am aware that another query exists regarding the same question, but since that is a year and a half old, and doesn't help me much, I decided to post another query.

答:
The fact that $a_{ij} = u_iv_j$ means that $A$ has rank $\leq 1$ (as $A = U^TV$). Note that this implies that $A$ cannot be inverted unlike what you wrote. Then, we can proceed by looking at the eigenvalues of $A$ (as the trace of $A$ is the sum of these eigenvalues).

The fact that $A$ has rank $0$ or $1$ means that $0$ is an eigenvalue. $A^5 = 16 A$ could be used to find the remaining eigenvalues but we're still missing some information: taking $A=0$ satisfies all your hypotheses but does not yield the answer you expect. This might be solved by having more information on $U$ and $V$ (in particular that they are not $0$).


问:
What is rank of $f(A)$, where $f$ is the minimal polynomial of $A$?
If $f(x)$ is minimal polynomial of the $4\times 4$ matrix
$$A=\begin{pmatrix}
0 &0 &0& 1\\
1 &0 &0 &0\\
0 &1 &0 &0\\
0 &0 &1 &0
\end{pmatrix}$$

Then what is rank of $f(A)$? I think $f(A)$ will be a zero matrix so its rank is 0. Am I right?

答: Yes, that's exactly right...  $\phantom{ok I'll add a little random text here to avoid automatic conversion}{}{}$</p>


问:
Good Notation that clearly distinguish between a vector and its components, a linear transform and its matrix
When we are doing calculations with vectors/matrices, we often need the following things:

A finite dimensional inner product space $V$ with inner product $\langle\cdot,\cdot\rangle.$

A basis $\{\mathbf e_j\}$

A matrix $M$

A linear transform $T$ that matrix $M$ represents in this given basis.

A vector $v\in V.$

Its components $v_j$ in the basis.

The quadratic form value $\langle v,Tv\rangle.$

If the basis is orthonormal, we can also write $\langle v,Tv\rangle=\sum M_{ij} v_i v_j.$

So far, everything is clear. There are no ambiguities. However, sometimes, we just say $V=\mathbb R^n,$ and this is where confusion in notation starts:

First of all, $T$ now has a matrix $A$ in the standard basis, so $\langle v,Tv\rangle= v^T A v.$ Note that now, $v^T$ makes sense as a row vector.

If we represent the components of $v$ in basis $\{\mathbf e_j\}$ by another column vector $v',$ then we write $v'^T M v'=\langle v,Tv\rangle.$

Now, there are several caveat of such notations:

$v,v'$ represents the same vector in $\mathbb R^n$, so one feels reluctant to use two different symbols for them.

The essential issue is, by letting $V=\mathbb R^n,$ we have lost the notation to distinguish between components of a vector in some basis and the vector itself which is independent of the choice of basis - both the vector itself and the list of components in some basis are written as "column vectors". If I just write $$ \begin{bmatrix} 1\\3 \end{bmatrix} $$ then you don't know if this is meant to be a vector in the space $V$ or just the component of some vector in some basis. Every time I write down such a vector I have to say what I mean again and again.

Of course, the above issue can be resolved by defining explicitly an isomorphism $\varphi: \mathbb R^n \to \mathbb R^n,$ given by $v \mapsto v'$. But this does not seem to be necessary every time, and it further complicates the notation.

Index notation like $\sum M_{ij} v_i v_j$ is not always possible to use, for example when $v_j$ are given as specific numbers like $\begin{bmatrix} 1\\3 \end{bmatrix},$ it is clumsy to define a new variable $v_j$ that equals these constant numbers.

In summary, in an abstract vector space, we can easily use notation to distinguish the vector $v\in V$ itself and its components written as something like $ \begin{bmatrix} 1\\3 \end{bmatrix} $ in some basis. However, if we work with $V=\mathbb R^n,$ the distinction in notation between vector itself and components is lost.

Question: Are there any strategies to make such a distinction clear? What notation will prevent this ambiguity?

答:
If $\beta=(\beta_1,\cdots,\beta_n)$ is an ordered basis of the vector space $V$ (over $\mathbb{R}$), and $v$ is an element of $V$ such that
$$
v=v_1\beta_1+\cdots+v_n\beta_n
$$
we write $[v]_\beta=(v_1,\cdots,v_n)$. (For convenience let me use row vectors.)

This notation works for both abstract $V$ and the concrete $\mathbb{R}^n$.

See for instance Linear Algebra by Friedberg-Insel-Spence.

Notes.

Consider $\mathbb{R}^2$ and the element $v=(1,1)$.

Under the standard basis $\beta=\{(1,0), (0,1)\}$, you have
$$
[v]_\beta=[(1,1)]_{\beta}=(1,1)
$$

Under the basis $\alpha=\{(2,0),(0,3)\}$, you have
$$
[v]_\alpha=[(1,1)]_{\alpha}=(\frac 12,\frac13)
$$


问:
How to simplify a partial sum obtained by Legendre polynomials
Context

I am working on an electrostatics problem. I have undertaken Fourier analysis. By $k$, I denote a natural number $k=0,1,2,\ldots$. I have obtained the following partial sum in terms of the even number $2k$.

$$S{(2k)}=\frac1{2^{2k}}\sum_{m=0}^{k}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\,\frac{1}{2k-2m+1}.$$

The boundary problem that this question originates from has odd symmetry. Therefore, I expect all even Fourier coefficients--except potentially $2k=0$--to be null. This expectation is verified by the correct answer below.

Question

Can the above expression be simplified further?

答:
Since (a known formula which easily follows from e.g. Rodrigues') $$P_n(x)=\frac{1}{2^n}\sum_{m=0}^{\lfloor n/2\rfloor}(-1)^m\binom{n}{m}\binom{2n-2m}{n}x^{n-2m},$$ the given expression is equal to $\int_0^1 P_{2k}(x)\,dx=\frac12\int_{-1}^1 P_{2k}(x)\,dx$.

This is $1$ for $k=0$ and $0$ for $k>0$.


问:
How to calculate the percentage of increase/decrease with negative numbers?
I feel like an idiot for asking this but i can't get my formula to work with negative numbers

assume you want to know the percentage of an increase/decrease between numbers

```<code>2.39      1.79       =100-(1.79/2.39*100)=>  which is 25.1% decrease
</code>```

but how would i change this formula when there are some negative numbers?

```<code>6.11      -3.73      =100-(-3.73/6.11*100) which is 161% but should be -161%
</code>```

the negative sign is lost.. what  I am missing here?

also

```<code>-2.1       0.6       =100-(-3.73/6.11*100) which is 128.6% ??? is it?
</code>```

答:
Conventional Formula

The conventional formula for computing the relative growth between two values $a$ and $b$ is $\displaystyle \frac{b-a}{a}$.

For example, if $a = 50$ and $b = 60$, the relative growth is $\displaystyle \frac{60-50}{50} = 0.2 = 20\%$.

So far, so good. But what if $a$ and $b$ have different signs? For example, $a = -10$ and $b = 20$?

The conventional formula would return a negative growth of $-300\%$, which does not make much sense.

Adjusted Formula

Believe it or not, there is no universally-accepted formula for the relative growth of signed values.

The one used by most statisticians is $\displaystyle \frac{b-a}{|a|}$, where $|a|$ is the absolute value of $a$ ($-a$ when $a$ is negative).

This formula is perfectly valid, but usually leads to counter-intuitive results. For example, the relative growth between $-10$ and $20$ is $300\%$, and the relative growth between $-20$ and $20$ is $200\%$. Both pairs of values end at the same exact value ($20$), yet the absolute growth for the first pair ($30$) is lower than the absolute growth for the second ($40$), while the relative growth is greater for the first than the second. How can that be?

Interpretation

To better understand what is going on, it usually helps to think of the relative growth between two values of different signs as being composed of two separate parts: the relative growth from the first value to zero, plus the relative growth from zero to the second value. For example, the relative growth between $-10$ and $20$ is equal to the relative growth between $-10$ and $0$, plus the relative growth between $0$ and $20$.

The relative growth between $-10$ and $0$ is $\displaystyle \frac{0-(-10)}{|-10|} = \frac{10}{10} = 1 = 100\%$ according to our previous formula.

In fact, the relative growth between any negative value and $0$ is always equal to $100\%$, which actually makes sense when one comes to think about it. But things become a bit more tricky when we need to compute the relative growth between $0$ and a positive value, like $20$ in our previous example. There, we cannot simply use our formula, because it would lead to a division by zero. Instead, we have to compute this relative growth in relation to the previous one, by computing the ratio of two absolute growths.

For example, when going from $-10$ to $20$, we gain $10$ in absolute terms between $-10$ and $0$, then $20$ in absolute terms again between $0$ and $20$. Therefore, we gained twice as much going from $0$ to $20$ than we did when going from $-10$ to $0$. And as we have seen earlier, the relative growth between $-10$ and $0$ is $100\%$, therefore the relative growth between $0$ and $20$ should be twice that amount, or $200\%$, and the relative growth between $-10$ and $20$ should be the sum of our two relative growths, or $100\% + 200\% = 300\%%$, which matches what our adjusted formula gave us in the first place. In other words, we found two ways of getting to the same result, but this should not come as a surprise if you look at the original equation.

Indeed, if $a$ is negative, we can rewrite $\displaystyle \frac{b-a}{|a|}$ as $\displaystyle \frac{b-a}{-a}$. Then, we can split the fraction as $\displaystyle \frac{b}{-a} + \frac{-a}{-a}$ and further simplify it into $\displaystyle \frac{b}{-a} + 1$, or even $\displaystyle 1 + \frac{b}{-a}$. This addition has two addends, $1$ and $\displaystyle \frac{b}{-a}$. The former is the relative growth from the first value to zero (always equal to $1$), and the latter is the ratio of the absolute growth between the first value and $0$ and the absolute growth between $0$ and the second value, which we can write as $\displaystyle \frac{b - 0}{0 -a}$.

When we look at the formula through that angle, we realize that a relative growth between two values of different signs is only affected by the ratio between the two values $\displaystyle \frac{b}{-a}$, unlike the relative growth between two values of the same sign, which is the ratio between the difference of the two values and the first value $\displaystyle \frac{b-a}{a}$. But this is where the adjusted formula becomes misleading, because we should not look at the former as a ratio between two values. Instead, we should look at it as a ratio between two absolute growths $\displaystyle \frac{b - 0}{0 -a}$.

Make no mistake though: this is not a trivial arithmetic rewriting. Instead, by writing $\displaystyle \frac{b}{-a}$ as $\displaystyle \frac{b - 0}{0 -a}$, we clearly communicate the fact that our fraction is a ratio between two absolute growths, instead of a ratio between two values. And this is what explains that the relative growth between $-10$ and $20$ ($300\%$) is greater than the relative growth between $-20$ and $20$ ($200\%$).

As explained earlier, to grow from $-10$ to $20$, you first need to grow to from $-10$ to $0$, then grow from $0$ to $20$, and the first growth component to $0$ is always equal to $100\%$. But when you grow from $0$ to $20$, while this absolute growth ($20$) is the same whether you started from $-10$ or $-20$, the relative growth is twice as much if you started from $-10$ than if you started from $-20$.

In other words, what the relative growth of values of different signs really measures is the relative growth after you have crossed the zero line, because the relative growth before you cross the zero line is always the same ($100\%$).


问:
Radius of convergence of a power series given by the composition of 2 analityc functions.
I know that the composition of 2 analytic functions is analytic, and I conjectured that the radius of convergence of the resulting power series should be the smallest of the 2, but I'm not sure and I wouldn't know how to prove that.

Any help will be much appreciated. Thank you

答:
Consider
$$
\cosh z = \sum\limits_{n = 0}^\infty  {\frac{{z^{2n} }}{{(2n)!}}}
$$
and
$$
\sqrt {1 + z}  = \sum\limits_{n = 0}^\infty  {\binom{1/2}{n}z^n } .
$$
Tha radii of convergence is $\infty$ and $1$, respectively. The composition
$$
\cosh \sqrt {1 + z}  = \sum\limits_{n = 0}^\infty  {\frac{{(1 + z)^n }}{{(2n)!}}}
$$
is an entire function, since the series on the right-hand side converges for all complex $z$. Thus the radius of convergence of the power series of $\cosh \sqrt {1 + z}$ at the origin is $+\infty$ and not $1$ as it supposed to be based on your conjecture.


答:
$a_{ij}=u_iv_j$ is equivalent to saying $A=uv^T$. Hence $$A^2=uv^Tuv^T=(v^T u)A,\quad\ldots\quad A^n=(v^T u)^{n-1} A$$

Given that $A^5=16A$ then shows $v^T u=\pm2$ or $\pm2i$, or $A=0$.

Note that the trace of $A$ is $\sum_iv_iu_i=v^Tu$.

[Note that $A$ is not invertible, so one cannot multiply by $A^{-1}$.]

Title: Finding trace of matrix from given conditions

Body:
I have a question that says,

Let A be a matrix such that $A = [a_{ij}]$ such that $a_{ij}=u_iv_j$ $\forall$ $1 \leq i\leq n$ and $1 \leq j\leq n$, and $u_i, v_j \in \mathbb{R}$ satisfies $A^5=16A$, then find $Tr(A)$, $($Here, $Tr(A)$ denotes the trace of $A$$)$

I have tried it as follows:

$$A^5=16A$$

$$\implies A^{-1}A^5=16A^{-1}A$$
$$\implies IA^4=16I$$
$$\implies A^4=16I$$

But, in the question, it is given that $a_{ij}=u_iv_j$, what are $u, v$ here? The answer is stated to be $2$, but I cannot proceed further. Any help on what $u, v$ are and how to proceed would be helpful. I think that the question might have been incomplete, but if anyone can help anyone, it would be nice. I am aware that another query exists regarding the same question, but since that is a year and a half old, and doesn't help me much, I decided to post another query.


问:
Obtaining a Lagrangian from a NLPDE
The variational approximation is often used when working with optical solitons. Similarly, the Noether theorem is often used to find conserved quantities. Both of these methods require a Lagrangian. However, not every soliton-equation has a Lagrangian (hence methods like Hasegawa's).

Is there any way to check if a given non-linear partial differential equation corresponds to the Euler Lagrange equation of a Lagrangian?

Take, for example, the equation:

$(u_{t}+6uu_{x})_{x}+3 \sigma^{2}u_{yy}=0$

which is known as the dispersionless KP equation. What should I do to know if it corresponds to the Euler-Lagrange equation of a Lagrangian?

答:
Indeed, this is not always an easy task. There are some mathematical tools aimed at retrieving Lagrangian densities from (partial) differential equations such as the semi-inverse method, see Ref. (1). Note that there might be several possibilities to derive consistent variational principles, see e.g. Ref. (2).

In variational calculus, the choice of variables and coordinates plays a crucial role. One important aspect relates to the meaning of $(x,y)$ as spatial or material coordinates. Another aspect relates to the fact that the dispersionless KP equation is equivalent to the Khokhlov–Zabolotskaya equation up to a linear change of coordinates. To illustrate the first feature, let's tackle the particular case $\sigma \to 0$ of the inviscid Burgers' equation $u_t + 6uu_x = 0$. This PDE is the Euler-Lagrange equation corresponding to the variational principle (3)
$$
\delta \iint_{\Bbb R^2} v(x,t)^2 \text{d} x\, \text{d} t = 0 \, ,
$$
where $v = \dot x$ is the Eulerian velocity field ($x$ denotes Eulerian spatial coordinates).
In facts, we have $$
\delta (v^2) = 2v\, \delta v = 2v\, \delta \dot x = 2v\left((\delta x)_t + v (\delta x)_x\right) ,
$$
and the stationary-value condition becomes
$$
\iint_{\Bbb R^2} \delta (v^2)\, \text{d} x\, \text{d} t = -6\iint_{\Bbb R^2} \left(u_t + 6uu_x\right)\delta x\,  \text{d} x\, \text{d} t = 0
$$
with $u=\frac13 v$.

J.-H. He, "Variational Principle for the Generalized KdV-Burgers Equation with Fractal Derivatives for Shallow Water Waves", J. Appl. Comput. Mech. 6(4) (2020), 735-740. DOI:10.22055/JACM.2019.14813

J.K. Hunter, Y. Zheng, "On a completely integrable nonlinear hyperbolic variational
equation", Physica D 79 (1994), 361-386. DOI:10.1016/S0167-2789(05)80015-6

V.L. Berdichevsky, Variational Principles of Continuum Mechanics I: Fundamentals, Springer, 2009. DOI:10.1007/978-3-540-88467-5

Title: Obtaining a Lagrangian from a NLPDE


问:
Calculating individual probabilities for an unknown distribution
I encountered the following problem:

According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.

Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.

How can we solve this problem, as it is explicitly stated that the distribution is unknown? In that case we would always have to use the central limit theorem to determine the probability that the mean is over 12 hours, but not that an individual takes over 12 hours, no?

The solution given there is:

"No. I would not be totally surprised because the probability is 0.2312".

Unfortunately, I have no idea how to approach this problem, as I always thought one need to know the distribution when you try to calculate the individual probabilities and not sums or means.

The problem is Question 7.2.3 from:

https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Book%3A_Introductory_Statistics_(OpenStax)/07%3A_The_Central_Limit_Theorem/7.E%3A_The_Central_Limit_Theorem_(Exercises)

答:
Perhaps better to say something like

"No. I would not be totally surprised because if I assumed a normal distribution and a standard deviation of $2$ then the probability an individual taking more than $12$ hours or higher would be $0.2312$. Those assumptions are likely to be wrong, but are enough to judge whether I would be surprised by at least one individual taking over $12$ hours."

In fact, with those assumptions, I would be surprised that only one of the $36$ took more than $12$ hours.


问:
A prime number divides only the numerator
I need to prove that there is a prime number $q$ which divides the numerator of $\frac{x^{\ell e}-1}{(x^{\ell}-1)(x^{e}-1)}$ and doesn't divide the denominator. Here, $\ell,e,x$ are integers, and $x$ is of the form $p_1^{p_2^{\ell-1}}\cdot p_2^{p_1^{e-1}}$, where $p_1,p_2$ are primes distinct from $q$, and $\ell =p_1$, $e=p_2$.

Forgot something important: $\ell e>\ell+e$.

Attempt:
I have tried using the next sentence to make it easier for me:
$$(x^{\ell}-1,x^{e}-1)=x^{(\ell,e)}-1$$ Since $(\ell,e)=1,$ we get: $(x^{\ell}-1,x^{e}-1)=x^{(\ell,e)}-1=x-1$.
To prove that there is a prime $q$ divides only the numerator, we shall check whether the number $\frac{x^{\ell e}-1}{(x^{\ell}-1)(x^{e}-1)}$ is integer, and after division of the numerator by the denominator, there is $q$ in the factorization.

Without any success from here.

答:
We can write:

$$\frac{N=(x^l)^e-1}{D=(x^l-1)(x^e-1)}=\frac{(x^l)^{e-1}+(x^l)^{e-2}+\cdot \cdot \cdot +1}{x^e-1}$$

If $e=2k$ is even then:

$x^l+1|(x^l)^{e-1}+(x^l)^{e-2}+\cdot \cdot \cdot +1$

Therefore:

$x^l+1|N$

Also we have:

$D=x^e-1=(x^k-1)(x^k+1)$

If $l\neq k=\frac e2$ then $x^l+1$ divides N but does not divide D.So $x^l+1$ must be primes .For example for $l=2$ and  $x=2$ gives $q=5$ so if e=6 , for instance, we get:

$D=(2^2-1)(2^6-1)=189$

$N=2^{12}-1=4095$.

The fraction is symmetric for and l and e so we may change e with l in above argument; the condition is:

$L\neq \frac e2$

Or:

$e\neq \frac l2$

That is one of l or e must be even.

Update:

Suppose e and l are odd, then it is possible to have a prime satisfying the condition for example for x=2, l=3, e=5 we get:

$\frac ND=\frac{2^{15}-1}{(2^3-1)(2^5-1)}=\frac{151}{7\cdot 31}$

So prime we are looking for is $151$.

Also for certain values of x, e and l, the numerator my be divisible by denominator. The condition must be discussed in separate question. For example for x=2,l=2, e=5 we get:

$\frac ND=31$

So 31 divides numerator.

Or with x=3, l=2 and e=5 we get:

$\frac ND=4 \times 61$

That means prime 61 only divides the numerator.


问:
Composite vs Composition: which one is the correct word?
So, I am a bit confused about the word I should use to refer to

$$f\circ g.$$

I have always used ''composition'' but I have seen some people (perhaps everyone who uses the term correctly) say composite.

I have added the reference-request topic in case you want to suggest a book explaining this sort of questions (I have no idea if such a book exists).

答:
By definition, $f \circ g$ is the function such that for all $x$, $(f \circ g)(x)=f(g(x))$. The RHS is the composition of $f$ and $g$. The LHS is the composite function: the result of meshing $f$ and $g$ into a single function. Performing the composite function $f \circ g$ on any given input  is the same as performing the functions $g$ and $f$ one after the other.

An analogy can be drawn with matrices. If $A$ and $B$ are two square matrices, then $A \times B$ is the single matrix that represents the linear transformation that results after applying $B$ and $A$ in succession.


答:
We can rewrite the sum as
$$
\eqalign{
  & S(k) = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
 \left( \matrix{  2k \cr   m \cr}  \right)\left( \matrix{  4k - 2m \cr   2k \cr}  \right)
{1 \over {2k - 2m + 1}}}  =   \cr
  &  = {1 \over {2^{\,2k} }}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
 \left( \matrix{  2k \cr   m \cr}  \right){{\left( {4k - 2m} \right)^{\,\underline {\,2k\,} } }
 \over {\left( {2k} \right)!}}{1 \over {\left( {2k - 2m + 1} \right)}}}  =   \cr
  &  = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k {\left( { - 1} \right)^{\,m}
 \left( \matrix{  2k \cr   m \cr}  \right)
\left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} } }  \cr}
$$

Now the Falling factorial
$$
p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} }
$$
is a polynomial in $m$ with the following characteristics.

$$
\left\{ {\matrix{
   {k = 0\quad  \Rightarrow \quad } & \matrix{
  p(m,0) = \left( { - 2m} \right)^{\,\underline {\, - 1\,} }
  = {1 \over {\left( { - 2m + 1} \right)^{\underline {\,1\,} } }}
 = {1 \over { - 2m + 1}}\quad  \Rightarrow  \hfill \cr
   \Rightarrow \quad p(0,0) = 1 \hfill \cr}   \cr
   {1 \le k\quad  \Rightarrow \quad } & \matrix{
  p(m,k) = \left( {2\left( {2k - m} \right)} \right)^{\,\underline {\,2k - 1\,} }  \hfill \cr
  {\rm polynomial}\,{\rm in}\,{\rm m}\,{\rm ofdegree}\;2k - 1 \hfill \cr
  {\rm with}\,{\rm zeros} \in \left[ {k + 1,\;2k} \right] \hfill \cr}   \cr
 } } \right.
$$

Therefore we have
$$
S(0) = 1
$$
and for $1 \le k$
$$
S(k)\quad \left| {\;1 \le k} \right. = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^k
 {\left( { - 1} \right)^{\,m} \left( \matrix{  2k \cr  m \cr}  \right)p(m,k)}
$$
Since $p(m,k) = 0$ for $k+1 \le m \le 2m$ we can extend the sum to $2k$ and multiply the summand by $1 = (-1)^{2k}$
$$
\eqalign{
  & S(k)\quad \left| {\;1 \le k} \right.\quad
  = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k}
 {\left( { - 1} \right)^{\,m} \left( \matrix{  2k \cr   m \cr}  \right)p(m,k)}  =   \cr
  &  = {1 \over {2^{\,2k} \left( {2k} \right)!}}\sum\limits_{m = 0}^{2k}
 {\left( { - 1} \right)^{2k - \,m} \left( \matrix{  2k \cr   m \cr}  \right)p(m,k)}  \cr}
$$
Here we recognize that the sum represents the finite difference (unitary step)
of order $2k$ of $p(m,k)$, and it is known (re. to the Newton series)
that the difference of a polynomial , of order greater than the degree of the same is identically null.

In conclusion $$S(k)= \delta _{\,k, \, 0} =\binom {0}{k}$$

Title: How to simplify a partial sum obtained by Legendre polynomials

Body:
Context

I am working on an electrostatics problem. I have undertaken Fourier analysis. By $k$, I denote a natural number $k=0,1,2,\ldots$. I have obtained the following partial sum in terms of the even number $2k$.

$$S{(2k)}=\frac1{2^{2k}}\sum_{m=0}^{k}(-1)^m\binom{2k}{m}\binom{4k-2m}{2k}\,\frac{1}{2k-2m+1}.$$

The boundary problem that this question originates from has odd symmetry. Therefore, I expect all even Fourier coefficients--except potentially $2k=0$--to be null. This expectation is verified by the correct answer below.

Question

Can the above expression be simplified further?


问:
What is the fifth digit from the left of $12345678987654321\times625$

We have been calculated $$12345678987654321\times625$$

Then what is the fifth digit from the left of this number?

I can rewrite it as  $$12345678987654321\times625={\underbrace{111111111}_{9\text{times}}}^2\times25^2=(111111111\times20+111111111\times5)^2=(2222222220+555555555)^2=2\underbrace{77777777}_{8 \text{times}}5^2$$
But I can't find fifth digit from the left of this square number.

答:
For similar problems with more digits (to prevent multiplying out as easily as in this case) you could adopt the following method.

$123456 \times 625 = 77160000$

$123457 \times 625 = 77160625$

So the answer is $0$.


问:
Variational derivation of wave equation (Euler-Lagrange equation)
In elasticity, the kinetic energy $T$ and the potential energy $V$ are
$$
T(u_t) = \frac{1}{2}\rho{u_t}^2 \qquad\text{and}\qquad V(u_x) = \frac{1}{2} E {u_x}^2 ,
$$
where $u$ is the displacement, $u_t$ the velocity, and $u_x$ the strain. The symbols $\rho>0$ and $E>0$ denote respectively the mass density and Young's modulus. The corresponding Lagrangian is $\mathcal{L} = T-V$. Using the principle of least action, how do we show that the 1D wave equation $$\rho u_{tt} = E u_{xx}$$ is the corresponding Euler-Lagrange equation?

If this wave equation comes from $\frac{\text d}{\text d t} \frac{\partial \mathcal{L}}{\partial \dot q} = \frac{\partial \mathcal{L}}{\partial q}$, what is $q$? On the left-hand side, one would like $\dot{q} = u_t$, while on the right-hand side, ${q} = u$ gives zero. How to resolve this inconsistency? I am aware of this exercise from Evans, but would like to tackle the particular case of the wave equation in elasticity.

答:
After I read some books dedicated to variational calculus, I came up with the following answer, which is perfectly consistent with the accepted answer by @user10354138. From the Lagrangian density $$\mathcal L(u,u_t,u_x) = T(u_t)-V(u_x)$$ defined in OP, we formulate the principle of stationary action $\delta \mathcal A = 0$,
where $\mathcal A = \iint \mathcal L \,\text{d}x\text d t$ is the action, and $\delta \mathcal A$ denotes its first variation w.r.t. $u$, $u_t$, $u_x$.
Using integration by parts and the commutativity of variation and differentiation, we find
\begin{aligned}
\delta A &= \iint_{\Bbb R^2} \left(\frac{\partial \mathcal L}{\partial u}\delta u + \frac{\partial \mathcal L}{\partial u_t}\delta u_t + \frac{\partial \mathcal L}{\partial u_x}\delta u_x\right) \text{d}x\text d t \\
&= \iint_{\Bbb R^2} \left(\frac{\partial \mathcal L}{\partial u} - \frac{\partial}{\partial t}\frac{\partial \mathcal L}{\partial u_t} - \frac{\partial}{\partial x}\frac{\partial \mathcal L}{\partial u_x}\right) \delta u\, \text{d}x\text d t \, ,
\end{aligned}
where natural boundary conditions have been assumed.
This gives us the Euler-Lagrange equation $$
\frac{\partial \mathcal L}{\partial u} - \frac{\partial}{\partial t}\frac{\partial \mathcal L}{\partial u_t} - \frac{\partial}{\partial x}\frac{\partial \mathcal L}{\partial u_x} = 0
$$
and therefore the wave equation $\rho u_{tt} - E u_{xx} = 0$.


问:
Finding Pythagorean triples, with a conjecture.
A right triangle with integer side lengths ($a , b , c$ where $a < b < c$)
can be found (to a certain degree) with the inner radius ($r$), the formulas for the area of a triangle and Pythagoras theorem.

We use these to find a relationship between $r$ and the sides of $a$ and $b$.

$a^2 + b^2 = c^2$

$a^2 =  c^2 - b^2$

$a^2 = (c + b)(c -b)$

We now can sub 2 expressions that are equivalent to $c + b$ and $c- b$ into the theorem:

Inner radius formula:

$r = ( a +  b - c)/2$

$2r = a + b - c$

$2r - a =  b - c$

$a - 2r = c - b$

Inner radius multiplied by semi perimeter formula (A formula for area in right triangles):

$ ab/2 = r (a + b + c)/2$

$ ab = r(a + b + c)$

$ab/r = a + b + c$

$ab/r - a = b + c $

Now we can substitute for $c - b$ and $b + c$.

$a^2  = (ab/r - a)(a - 2r)$

which when rearranged is:

$b=\frac{2r\left(a-r\right)}{a-2r}$

If we substitute in any value of $a$ we can create a function in terms of $b$ and $r$ where $b = b(r)$

For example, lets write this for the Pythagorean triple, where $a = 3.$

$b(r) = 2r(3 - r)/ 3 - 2r$,

We can conjecture that there is only one positive integer value of $b$ where $r$ is also an integer within the domain of $0 < r < a/2$, which means we can find a direct solution for both $b$ and $r$ by graphing.

https://www.desmos.com/calculator/zzxbryrahc

Note: If you look at when $r$ is $1$, you'll see that $b$ is $4$, which is true in the case of a Pythagorean triple with side lengths of ($3,4,5$).

My question is if there's a way to decrease the number of possibilities more than just $ 0 < r < a/2$  this method would become nearly impossible without the use of a graphing calculator for larger Pythagorean triples, Any other way to improve on this would be much appreciated.

答:
You have
$$b=\frac{2r\left(a-r\right)}{a-2r}$$

You can rewrite this as:

$$b = r - \frac{a}2 + \frac{a^2}{2(a-2r)}$$

For this to be an integer, you want $a^2$ to be divisible by $a-2r$. So if $a$ is given, look for the divisors $d|a^2$ for which $d<a$, and then let $r=\frac{a-d}2$. Of course you want this to be an integer too, so you also want $d$ to have the same parity are $a$.

Note that not all of these divisors will necessarily work, not only because these don't always give integer solution due to those halves in the expression, but also because there is no guarantee that even with integer $a$, $b$ and $r$ that they actually give a Pythagorean triangle.

This is all easy to do even for very large $a$ provided you have its prime factorization.

This also suggests that there will be several values of $r$ which could work with larger $a$, so that your conjecture is probably false even when restricted to only Pythagorean triangles.

Edit:

Note that Pythagorean triangles always have integer inradius, so your conjecture essentially says that no two Pythagorean triangles have the same shortest side. Here are two (primitive) Pythagorean triangles with the same shortest side: $(20, 21, 29)$ and $(20, 99, 101)$. Here is another pair sharing a shortest side with an odd length: $(105, 208, 233)$ and $(105,608,617)$.


问:
Mean Value Theorem with second derivative without using Lagrange Remainder

For $h > 0$ let $I = [x - h, x + h]$ and  $$g(x) = \frac{f(x + h) - 2f(x) + f(x - h)}{h^2}.$$
If $f \in C^2(I)$, show that $\exists \zeta \in (x - h, x + h) \colon g(x) = f''(\zeta)$.

My solution thus far:

Using the Mean Value Theorem, $\exists b \in (x, x + h)$ and $\exists a \in (x - h, x) \colon$
$$\frac{f(x + h) - f(x)}{h} = f'(b) \text{ and } \frac{f(x) - f(x - h)}{h} = f'(a)$$
$\implies \exists \zeta \in (x - h, x + h) \colon$
$$\frac{f'(b) - f'(a)}{b - a} = \frac{f(x+h) - 2f(x) + f(x-h)}{(b - a) \cdot h} = f''(\zeta).$$
How can I go from $(b - a) \cdot h$ to $h^2$ without using Lagrange Remainder?

答:
Denote with $M(x):= \frac{f(x+h)-f(x)}h$, then you have $g(x) = \frac{M(x)-M(x-h)}h$. By the mean value theorem you get a $\xi\in (x-h,x)$ with
$$g(x) = M'(\xi)=\frac{f'(\xi+h)-f'(\xi)}{h}$$
apply the mean value theorem again to get a $\zeta\in ( \xi,\xi+h)\subseteq (x-h,x+h)$ with
$$g(x) = f''(\zeta)$$


问:
Two basic logic questions
I'm pretty sure that these two logical moves are okay, but I just wanted to be sure.  If there are any subtleties (e.g. they are okay in first-order logic but not in more complex systems), please let me know.

1) Can you universally instantiate a negated proposition?  I.e. can you move from:

$ \left ( \forall x \right )\left ( x \rightarrow Y \right ) $

to

$ \neg b \rightarrow Y $?

2) Can you use rules of replacement within the scope of an operator?  I.e. can you move from:

$ \left ( \forall x \right )\left ( x \vee Y \right ) $

to

$ \left ( \forall x \right ) \neg \left (\neg x \& \neg Y \right ) $?

I think these two moves are okay but just want to be sure.  Thanks!

答:
I suppose You should only be allowed to instantiate with variables and constants.

Yes, you can. This is a shortcut to
$(\forall x)\phi(x)\vdash \phi(x)$ entails $\phi(x)$ by specialization, entails $\psi(x)$ by whatever rules you have in mind, entails $(\forall x)\psi(x)$ by generalization

Title: Two basic logic questions


答: One may say that $f\circ g$ is a composite function, the word "composite" being an adjective, or that $f\circ g$ is a composition of functions.

Title: Composite vs Composition: which one is the correct word?

Body:
So, I am a bit confused about the word I should use to refer to

$$f\circ g.$$

I have always used ''composition'' but I have seen some people (perhaps everyone who uses the term correctly) say composite.

I have added the reference-request topic in case you want to suggest a book explaining this sort of questions (I have no idea if such a book exists).


问:
Let $x_0$ be a transcendental number, $x_{n+1}=\frac{3-x_n}{x_n^2+3x_n-2}$. What is the limit of $x_n$?
Let $x_0$ be a transcendental number, $$x_{n+1}=\frac{3-x_n}{x_{n}^{2}+3x_{n}-2}$$
What is the limit of $x_{n}$?

Choose $x_0=\pi$, and is seems that the limit of $x_n$ is $-1$. But what is the proof for this $\pi$ and other numbers? Let
$$f(x)=\frac{3-x}{x^{2}+3x-2}$$
The following may be helpful.
$$f'(x)=\frac{(x-7)(x+1)}{(x^{2}+3x-2)^2}$$
$$f(x)-x=\frac{-(x-1)(x+1)(x+3)}{x^{2}+3x-2}$$
$$f(x)+1=\frac{(x+1)^{2}}{x^{2}+3x-2}$$.

答:
Let $f(x) = \frac{3-x}{x^2+3x-2}$. If $\lim x_n$ exists, then $L = \lim x_{n+1}=\lim x_n$, so set $$L=f(L)$$

There's three solutions to this: $L = -3, -1, 1$. In order to find the correct one, note that for a small neighborhood around $-3$, you have $|f(x)+3|>|x+3|$, and around $1$, you have $|f(x)-1|>|x-1|$. For both $-3$ and $1$, the difference will be made even bigger. Around $-1$ on the other hand, you have $|f(x)+1|<|x+1|$, so the difference is becoming smaller (this is not a rigorous proof but more of an intuitive one).

Thus, for "most" $x_0$, it will converge to $-1$. The only way it will converge to $-3$ or $1$ is if it converges exactly in a finite number of iterations. But for that to be true, it has to be a solution to $$f^n(x_0) = -3$$
(or $1$)
for some $n$, meaning that it must be algebraic. Therefore, for all transcendental, the limit will be $-1$.


问:
Runge-Kutta method for a 2nd order nonlinear ODE/ CMC surface meridian
I'm trying to solve numerically the following IVP nonlinear second-order differential equation:
$$f''(x)=A\left(1+f'(x)^2\right)^{3/2}+\frac{\left(1+f'(x)^2\right)}{f(x)}$$

where $A$ is a constant and the problem has the initial conditions (IVP) $f(0)=R$ and $f'(0)=0$, where $R$ is also a constant.

I'm used to applying the RK4 method for first-order differential equations, but is it possible to solve it by the RK4 method as well? If so could someone provide me a good reference to understand this specific application for method? I also open to any other useful numerical method that can solve this IVP. Thanks in advance.

答:
I had worked on this problem before and am glad to share it with you.

Recognize the slope $\phi=y'$ and main curvature $\kappa_1:$

$$ \dfrac{y''}{(1+y'^2)^{\frac32}}=A +\dfrac{1}{y\sqrt{1+y'^2}} \tag1 $$

Let $y'= \tan \phi,\text{so that }\cos \phi =\dfrac{ 1}{\sqrt{1+y'^2}} =u \;$

$$\kappa_1= A+ \dfrac{u}{y} \tag2  $$

With proper sign

$$ \kappa_1+\kappa_2 = A = 2 H \tag3 $$

where $\kappa_2$ is second curvature perpendicular to the principal meridional direction.

It is an important surface of constant mean curvature CMC when internal pressure acts on a soap bubble as a minimal surface area of revolution for a given enclosed volume.

Btw It can be formulated using calculus of variations nicely.

Having a constant mean curvature=H, they are also known as  CMC Delaunay  surfaces.

You can directly integrate to first order ODE. Its alternate form first degree:

$$\dfrac { \sin \phi\; d \phi}{dy}-\dfrac{u}{y} \to
-\dfrac { du}{dy}-\dfrac{u}{y} = 2H  \tag4 $$

Integrating,

$$ u= - H y+\dfrac{c}{y}=\dfrac{1}{\sqrt{1+y'^2}} \tag 5 $$

$$ \boxed{\cos \phi = \dfrac{c}{y}- H y} \tag 6 $$
is the differential equation of DeLaunay surfaces. Vertical tangent at $ y=\sqrt{c/H}.$

Characteristics of shape:

When $ H= 0, $ it is a Catenary; when $ H=1 , $ it is a Circle.

The meridians $(x,y)$ are calculated numerically from above ODEs and plotted below using RK4 available with  Mathematica. Initial values $ R=1, y_0'= 0$

An interesting fact about Unduloids: If a conic rolls on a straight line then the Unduloids form as loci of their foci.

The second order RK4 is handled similar to first order, one more initial condition should be supplied.


答: "Composition" is a noun: The composition of two surjective functions is surjective. "Composite" can act as an adjective, so it requires a noun to modify: The composite function is surjective as it results from composing two surjective functions. But it can also serve as a noun: Given two surjective function, their composite is surjective.</p>


问:
Show that $\sqrt[n]{b}-\sqrt[n]{a}<\sqrt[n]{b-a}$
Let $a, b \in \mathbb{R}$ with $0 \leq a<b$ and let $n \in \mathbb{N}$ with $n \geq 2$

a) Show that $\sqrt[n]{b}-\sqrt[n]{a}<\sqrt[n]{b-a}$.

b) Show that $\frac{a-1}{n} \geq \sqrt[n]{a}-1$

I tried Induction and Mean value theorem, but Failed.

答:
Note as written it is false, since letting $a=0$ yields $\sqrt[n]{b} < \sqrt[n]{b}$, so you need $\le$, not $<$.

The second one you can write as $a - na^{1/n} \ge 1 - n$ so defined $f(x) = x-nx^{1/n}$ and find its minimum.

In a similar way, you can define $f(a,b) = (b-a)^{1/n} - b^{1/n} + a^{1/n}$ and minimize that over $0 \le a < b \in \mathbb{R}$ to prove that $f(a,b) \ge 0$.


问:
Sums of roots of unity
If the integral linear combination of some $n$th roots of unity has magnitude 1, does this necessarily imply that this linear combination is some root of unity as well? More precisely,

Let $\zeta_1, \ldots \zeta_k$ be $n$th roots of unity. If
  $$|\sum_{i=1}^k n_i \zeta_i| = 1,$$
  where $n_i \in \mathbb{Z}$, does this imply that $\sum_{i=1}^k n_i \zeta_i$ is an $n$th root of unity? What about if the $n_i$ are Gaussian integers?

答:
Yes. To be more specific, I will prove the following: Let $\zeta$ be an $n$-th root of unity and let $\alpha = \sum a_k \zeta^k$ for some integers $a_k$. If $|\alpha|=1$, then $\alpha$ is a root of unity.

The key will be the following theorem of Kronecker: Let $\beta$ be an algebraic integer, and suppose that all the Galois conjugates of $\beta$ have absolute value $1$. Then $\beta$ is a root of unity. We discussed this over on MO. (I will restate the condition on Galois conjugates more explicitly below.)

It is clear that $\alpha$ is an algebraic integer, so the key is to see that all of its Galois conjugates also have norm $1$. Very explicitly, we want to see that, for any $r$ relatively prime to $n$, we have $|\sum a_k \zeta^{kr}|=1$.
Let's write $\alpha^{(r)}$ for $\sum a_k \zeta^{kr}$.

Here is the point. We have
$$|\alpha|^2 = \alpha \overline{\alpha} = \left( \sum a_k \zeta^k \right) \left( \sum a_{\ell} \zeta^{- \ell} \right).$$
Multiplying this out gives a big polynomial in $\zeta$ with integer coefficients; call this polynomial $N$.
The same computation shows that $|\alpha^{(r)}|^2 = N(\zeta^r)$.

So our hypothesis is that $N(\zeta) = 1$ and we want to show that $N(\zeta^r)=1$ as well. Since $N(\zeta)=1$, we see that $\zeta$ is a root of $N-1$. Since the $n$-th cyclotomic polynomial is irreducible, this implies that $\zeta^r$ is also
a root of $N-1$, and we are done.

Conceptually, this argument worked because complex conjugation was in the center of $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$. An analogous argument shows that, if $K$ is any CM field, and $\alpha$ in $K$ has norm $1$, then all the conjugates of $\alpha$ have norm $1$.

Title: Sums of roots of unity


问:
This special random subset of uniformly distributed numbers is still uniformly distributed?
I asked similar question in A special random subset of uniformly distributed numbers is still uniformly distributed?

Here, I slightly change my random number generation method, and want to see whether the sampled numbers are still uniformly distributed.

Assume that I have a value range [1,1000].

Goal: I want to have 10 numbers randomly sampled from [1,1000].

case1:

```<code>I sample 20 numbers, a1, ..., a20 from [1,1000].

Then I sample b1, ..., b10 from [a1, a2, ..., a20].

b1, ..., b10 are what I want.
</code>```

case2:

```<code>I sample 20 numbers, a1, ..., a20 from [1,1000].

Then I sort a1, ..., a20.

For notation simplicity, I assume that after sorting, a1< ...< a20.

Then, for each consecutive pair of numbers,

I uniformly at random select one number.

Eventually, I can get 10 numbers as well.
</code>```

I know the numbers in case1 are uniformly distributed.

Does anyone have idea whether the numbers in case2 are uniformly distributed?

答:
Let us do a much simplified calculation. We want to choose $2$ numbers out of the integers $1$ to $10$. In Scheme 1, we choose $4$ numbers, then choose $2$ of them. In Scheme 2, we choose $4$, order them, pick one from the first group of two, and one from the second group.

Suppose that we are choosing the $4$ without replacement. With Scheme 1, the probability of ending up with the set $\{9,10\}$ is $\frac{1}{45}$. With Scheme 2, it is $0$.

With replacement gets more complicated, because even Scheme 1 bifurcates. But in various explicit calculations, I get different probabilities.

Remark: There have been several related posts. In all of them, the version of Scheme 2 that is used produces a tamer, more "evenly" distributed result than true random sampling.

Reminds me of an anecdote, perhaps not true, about when bridge tournaments switched from manually shuffled decks (then replicated for the other table), to hands generated by a random number generator. Veteran bridge players complained that the computer was dealing "wilder" than normal hands, for example hands with more voids. They were sort of right: manually shuffled decks produce hands not uniformly distributed over all possible hands.


答:
For the first one : you have $\sqrt[n]{b}-\sqrt[n]{a}=\sqrt[n]{a}(\sqrt[n]{b/a}-1)$ and $\sqrt[n]{a-b}=\sqrt[n]{a}\sqrt[n]{b/a-1}$ so you can divide by $\sqrt[n]{a}$ on both side (if $a \ne 0$, otherwise it is trivial).

You should now compare $f(x)= \sqrt[n]{x}-1$ and $g(x)=\sqrt[n]{x-1}$ for $x \geq 1$. Since $f(1)=g(1)=0$ and $f'(x) \leq g'(x)$ for $x \geq 1$ you have your answer.

For the second part see the second answer.

Body:
Let $a, b \in \mathbb{R}$ with $0 \leq a<b$ and let $n \in \mathbb{N}$ with $n \geq 2$

a) Show that $\sqrt[n]{b}-\sqrt[n]{a}<\sqrt[n]{b-a}$.

b) Show that $\frac{a-1}{n} \geq \sqrt[n]{a}-1$

I tried Induction and Mean value theorem, but Failed.


问:
Given R.V $X_i\geq0$ where $X_i$ is a sequence if $\mathbb{E}[\sqrt{X_n}]\to\infty$ then for sure the following doesn't hold $X_n\underset{a.s}{\to}0$

Given R.V $X_i\geq0$ where $X_i$ is a sequence if $\mathbb{E}[\sqrt{X_n}]\to\infty$
then for sure the following doesn't hold $X_n\underset{a.s}{\to}0$

Prove or disprove

I think this is not true but I can't find the right sequence does'nt converge

答:
Let $X$ be any positive random variable with $\mathbb{E}[\sqrt{X}]=\infty$ and define $X_n := X \cdot 1_{\{X \geq n\}}$, then
$$\mathbb{E}[\sqrt{X_n}] = \infty$$
for all $n \in \mathbb{N}$, although $X_n \rightarrow 0$ almost surely.


答:
The following is a counterexample:

Consider the probability space $([0,1],\mathfrak{B}([0,1]),\lambda|_{[0,1]}$) and set $X_n=n^41_{(0,\frac{1}{n})}$.

Then $E(\sqrt{X_n})=n$ but the sequence $(X_n)$ converges to the constant $0$ pointwise (so in aprticular almost surely).

Body:
Given R.V $X_i\geq0$ where $X_i$ is a sequence if $\mathbb{E}[\sqrt{X_n}]\to\infty$
then for sure the following doesn't hold $X_n\underset{a.s}{\to}0$

Prove or disprove

I think this is not true but I can't find the right sequence does'nt converge


问:
Creating an algebraic formula
I need to create a formula for a piece of software and can't quite work out how to form it as my Maths is a little rusty.  I would appreciate any help.

I have two values X & Y where each can be between 0 and 1.0

Their relationship is shown below in the picture but has the equation

$y=√(1−x^2)$

I have added an image showing their relationship.

I need a formula that returns ~1 as X & Y comes equal to each other and tends to 0 when they aren't (i.e X=1,Y=0).  Basically to know (in terms of $[0,1]$) when the point on the line is in the most upper right of the line.

Apologies if I haven't explained well enough.

Thanks

Title: Creating an algebraic formula

Body:
Let $f=1-|x-y|=1-\left|x-\sqrt{1-x^2}\right|$

When $x=0$ and $y=1$ then $f=1-|0-1|=1-|-1|=1-1=0$

When $x=1$ and $y=0$ then $f=1-|1-0|=1-|1|=1-1=0$

When $x=y=\frac{\sqrt2}2$ then $f=1-\left|\frac{\sqrt2}2-\frac{\sqrt2}2\right|=1-|0|=1-0=1$


问:
Is it possible to find a monogenic field, which has a non monogenic subfield?
Let $L$ be a monogenic field, that is $\mathcal{O}_L=\mathbb{Z}[\alpha]$. Let $K \subset L$ be a subfield, does it necessarily have this monogenic property? Do we have necessarily $\mathcal{O}_K=\mathbb{Z}[\beta]$?

My attempt: We can assume that $L=\mathbb{Q}[\alpha]$ and $\mathcal{O}_L=\mathbb{Z}[\alpha]$ and $K=\mathbb{Q}[\beta]$, where $\beta=g(\alpha)$ and $g$ is a monic polynomial and both of them are algebraic integers. I want to try to show $\mathbb{Q}[\beta] \cap \mathbb{Z}[\alpha] \subseteq  \mathbb{Z}[\beta]$. But I can not, and I have some doubt if this is true or not.

答: It would be easy to find $K$ first and $L \supset K$ later. According to this answer, $K = \mathbb{Q}(\sqrt{17}, \sqrt{41})$ is not monogenic. However, since $K/\mathbb{Q}$ is abelian, by Kronecker-Weber theorem, it is contained in some cyclotimic field $L = \mathbb{Q}(\zeta_{n})$ (I think $n = 17 \times 41$ or $n = 4 \times 17 \times 41$ would work, but I'm not sure), which is monogenic.


答:
Whilst not directly related to your method, some points I'd like to give are that this is the case of Fermat's last Theorem with $n=2$ and so you could use part of the proof from that to find values, which stems from the fact that we can say:
$$a^2+b^2=c^2\Rightarrow (a/c)^2+(b/c)^2=1$$
so it is equivalent to trying to find integer coordinates $(a/c,b/c)$ on a unit circle.

There is also Euclid's formula for which two positive integers $m,n$ are coprime and give:
$$\begin{pmatrix}a\\b\\c\end{pmatrix}=\begin{pmatrix}m^2-n^2\\2mn\\m^2+n^2\end{pmatrix}$$
Although this will not find all of the pythagorean triples that exist.

Title: Finding Pythagorean triples, with a conjecture.

Body:
A right triangle with integer side lengths ($a , b , c$ where $a < b < c$)
can be found (to a certain degree) with the inner radius ($r$), the formulas for the area of a triangle and Pythagoras theorem.

We use these to find a relationship between $r$ and the sides of $a$ and $b$.

$a^2 + b^2 = c^2$

$a^2 =  c^2 - b^2$

$a^2 = (c + b)(c -b)$

We now can sub 2 expressions that are equivalent to $c + b$ and $c- b$ into the theorem:

Inner radius formula:

$r = ( a +  b - c)/2$

$2r = a + b - c$

$2r - a =  b - c$

$a - 2r = c - b$

Inner radius multiplied by semi perimeter formula (A formula for area in right triangles):

$ ab/2 = r (a + b + c)/2$

$ ab = r(a + b + c)$

$ab/r = a + b + c$

$ab/r - a = b + c $

Now we can substitute for $c - b$ and $b + c$.

$a^2  = (ab/r - a)(a - 2r)$

which when rearranged is:

$b=\frac{2r\left(a-r\right)}{a-2r}$

If we substitute in any value of $a$ we can create a function in terms of $b$ and $r$ where $b = b(r)$

For example, lets write this for the Pythagorean triple, where $a = 3.$

$b(r) = 2r(3 - r)/ 3 - 2r$,

We can conjecture that there is only one positive integer value of $b$ where $r$ is also an integer within the domain of $0 < r < a/2$, which means we can find a direct solution for both $b$ and $r$ by graphing.

https://www.desmos.com/calculator/zzxbryrahc

Note: If you look at when $r$ is $1$, you'll see that $b$ is $4$, which is true in the case of a Pythagorean triple with side lengths of ($3,4,5$).

My question is if there's a way to decrease the number of possibilities more than just $ 0 < r < a/2$  this method would become nearly impossible without the use of a graphing calculator for larger Pythagorean triples, Any other way to improve on this would be much appreciated.