问:
Combination with Restriction and Repetition
I have a number $x$, let's say $5$, and I want to sort the number out into $4$ digits so that the sum of the digits is equal to $5$, but the value of each digit cannot exceed $3$. $0$ would be an acceptable digit. In addition, numbers like $3200$ and $3020$ both counts as the same number, as I'm looking for the series of digits, not the specific order. For $x=5$, the answer should be $4$, as the values would be $3200, 2210, 3110, and\ 2111$. How could I make an equation for this, for all x values $0 \le x \le 12$?

答:
Let $p(n,k,s)$ be the number of partitions of $n$ into at most $k$ parts of size at most $s$.  By conditioning on whether there is a part of size $s$ or not, you can obtain a recurrence relation
$$p(n,k,s)=p(n-s,k-1,s)+p(n,k,s-1)$$
with boundary conditions
\begin{align}
p(0,k,s)&=1\\
p(n,k,s)&=0 &&\text{if $n>ks$}
\end{align}
You want to compute $p(n,4,3)$ for $n\in \{0,1,\dots,12\}$.
For example,
$$p(1,4,3)=p(1,4,1)=p(0,3,1)+p(1,4,0)=1+0=1$$
and
$$p(2,4,3)=p(2,4,2)=p(0,3,2)+p(2,4,1)=1+p(1,3,1)+p(2,4,0)=1+1+0=2.$$


问:
Invertibility of matrix.
If a matrix $A$ is invertible then can we prove that $(I+DA^{-1})$ is also invertible. Or, what are the necessary and sufficient condition for it to be invertible? Given that $D$ is a small perturbation of $A$ and $I$ is the identity matrix.

答: If $D = A + B$, $I+D A^{-1} = 2 I + B A^{-1}$.  So if $\| B A^{-1}\| < 2$ (where $\| \cdot \|$ is any submultiplicative matrix norm), $I + D A^{-1}$ is invertible.

ContentLicense: CC BY-SA 4.0

Title: Invertibility of matrix.


问:
In acute $\triangle ABC$, $D$ is midpoint of $BC$, $DE \perp AB$, $DF \perp AC$, and $CE = BF$. Prove "geometrically" that $AB = AC$.
Really seemingly simple geometry problem:

In acute $\triangle ABC$, $D$ is midpoint of $BC$. $DE \perp AB$, $DF \perp AC$. $CE = BF$. Prove that $AB = AC$

I am looking for a more "geometrical" solution. We can apply cosine theorem on $\triangle BCF$ and $\triangle BCE$ to prove that $\cos B = \cos C$. But the computation is annoying. Is there a more pretty solution?

答:
To show $AB=AC$, it's enough to show $EB = FC$ and $AE=AF$. To show the former, I suggest thinking about showing congruence of $\triangle EGB$ and $\triangle FGC$, where $G$ is the intersection of $EC$ and $FB$. Argue that $EG$ = $FG$ and $BG = CG$.

Maybe you can do something similar to show $AE=AF$ - the triangles $\triangle AEC$ and $\triangle AFB$ in particular.


问:
Show that $\lim_{n \to \infty} \left(1+\frac{a_n}{n}\right)^n= e^a$ when $a_n \to a$
I have seen a proof that shows
$$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x$$
by looking at the Taylor series expansion of $\ln(1+x)$ at $x=0$.

To prove a theorem, my textbook uses the fact
$$\lim_{n \to \infty} \left(1+\frac{a_n}{n}\right)^n = e^a$$
when $a_n \to a$.

How can I prove this?

答:
Write$\left(1+\cfrac{a_n}{n}\right)^n$ as
$\left(\left(1+\cfrac{a_n}{n}\right)^{\cfrac{n}{a_n}}\right)^{\large{a_n}}$.

We know $\lim_{u\to \infty}\left(1+\dfrac1u\right)^u=e$. therefor $\lim_{n\to \infty}\left(1+\cfrac{a_n}{n}\right)^{\dfrac{n}{a_n}}=e$. and the original limit equal to $e^a$

Edit: As @DannyPak-KeungChan mentioned it is possible that value of $a$ be equal to $0$. in this case we should plug in this value in the original limit to get $\lim_{n \to \infty} \left(1+\frac{0}{n}\right)^n = 1=e^0$.


问:
Given joint pdf find $f(z)$, where $Z=X+Y$
Let $f_{X,Y}(x,y)=\frac{1}{8}$ for $-2<x<2$, $0<y<2$. Find $f(z)$ where $Z=X+Y$.

I am having difficulty solving the above problem. My attempt at a solution relies on a convolution formula, which says that the $pdf$ of $Z=X+Y$, given the joint $pdf$ $f(x,y)$ is given by $$f_Z(z)= \int_{s+t=z} f(s,t)ds=\int_{-\infty}^{\infty} f(s,z-s)ds$$. So first note that $f(s,z-s)=\frac{1}{8}$ if $-2<s<2$ and $0<z-s<2$. Then note that $0<z-s<2$ is the same as $z-2<s<z$. But these regions depend on the value of $z$. Hence,
  $$\begin{equation}
    f_Z(z)=
    \begin{cases}
      \int_{-2}^{z} \frac{1}{8}ds, & \text{if}\ 0\leq z \leq2\\
      \int_{z-2}^{2} \frac{1}{8}ds, & \text{if}\ 2<z\leq4
    \end{cases}
  \end{equation}$$. After evaluating the integrals, I get

$$\begin{equation}
    f_Z(z)=
    \begin{cases}
      \frac{z-2}{8}, & \text{if}\ 0\leq z \leq2\\
      \frac{4-z}{8}, & \text{if}\ 2<z\leq4
    \end{cases}
  \end{equation}$$.

I have solved similar problems like this one but all of them have been defined over the intervals $0<x<1$ and $0<y<1$. In this problem, however, we have the intervals $(-2,2)$ for $x$ and $(0,2)$ for $y$.I have graphed these regions but I have no intuition on how to divide it. Any help would be appreciated. Thanks.

答:
First note that when $X\in[-2..2]$ and $Y\in[0..2]$, then $X+Y\in[-2..4]$, thus this is the support for the pdf of $Z$.

You have obtained that: $$f_{\small Z}(z)=\int_\Bbb R \tfrac 18\mathbf 1_{-2\leqslant s\leqslant 2, 0\leqslant z-s\leqslant 2}\,\mathrm d s$$

With a little rearangment, and the afformentioned note, we shall obtain:$$f_{\small Z}(z)=\tfrac 18\mathbf 1_{-2\leqslant z\leqslant 4}\int_\Bbb R \mathbf 1_{\max(-2,z-2)\leqslant s\leqslant \min(2,z)}\,\mathrm d s$$

So, $-2=z-2$ and $2=z$ are the points we partition the support for $Z$'s pdf. ($z=0$ and $z=2$, obvs.)

$$f_{\small Z}(z)=\tfrac 18\left(\mathbf 1_{-2\leqslant z\lt 0}\int_{-2}^{z}\mathrm d s+\mathbf 1_{0\leqslant z\lt 2}\int_{z-2}^z\mathrm d s+\mathbf 1_{2\leqslant z\leqslant 4}\int_{z-2}^2\mathrm d s \right)$$

Of course those integrals are trivial.

$$f_{\small Z}(z)=\tfrac{z+2}{8}\mathbf 1_{-2\leqslant z\lt 0}+\tfrac{2}{8}\mathbf 1_{0\leqslant z\lt 2}+\tfrac {4-z}{8}\mathbf 1_{2\leqslant z\leqslant 4}$$

Or if you prefer:$$f_{\small Z}(z)=\begin{cases}(z+2)/8&:& -2\leqslant 2<0\\1/4&:&0\leqslant z\lt 2\\(4-z)/8&:&2\leqslant z\leqslant 4\\0&:&\textsf{elsewhere}\end{cases}$$

NB: This pdf is in the shape of a trapezoid.

ContentLicense: CC BY-SA 3.0


问:
Using Spherical coordinates find the volume:
Inside the surfaces $z=x^2+y^2$ and $z=\sqrt{2-x^2-y^2}$

I integrated over the ranges:

$0 \leq \theta \leq 2\pi$

$ 0 \leq \phi \leq \frac{\pi}{2}$

$0 \leq r \leq \sqrt{2}$

I get $\frac{\pi}{2}(4\sqrt{2} -4).$

There answer is the same except a $-\frac{7}{2}$ instead of the 4 at the end. Obviously I'm missing a 1\2 but I seems, I can not find it.

答:
A related problem. The title of your problem asks for the use of spherical coordinates. You got the wrong integral limits. Note that, $ \rho $ (I am using $\rho$ instead of $r$) is bounded from above by two different surfaces. The sphere when $ 0 \leq \phi \leq \frac{\pi}{4} $ and the parabolic when $\frac{\pi}{4}  \leq \phi \leq \frac{\pi}{2} $. So, we have

$$ V = \left(\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{\sqrt{2}}+\int_{0}^{2\pi}\int_{\pi/4}^{\pi/2}\int_{0}^{\frac{\cos(\phi)}{\sin^2(\phi)}}\right)\rho^2\sin(\phi)d\rho d\phi d\theta. $$

I leave it here for you to do the calculations.

that, for $0\leq \rho \leq \frac{\cos(\phi)}{\sin^2(\phi)}$, we used the equation $z=x^2+y^2$ to get . I leave it here for you to do the calculations.

Notes:

1) To get $\rho = \frac{\cos(\phi)}{\sin^2(\phi)}$ use the equation $z=x^2+y^2$ and the spherical coordinates of $x,y,$ and $z$.

2) The integral

$$ \int \frac{\cos^3(\phi)}{\sin^5(\phi)}d\phi = \int \cot^3(\phi) \csc^2(\phi) d\phi=-\frac{1}{4}\cot^4(x). $$


问:
How to build a Cross-correlated Covariance matrix by solving an equation with Covariance and Variance expression of an unknown random variable?
Firstly, I want to give you a maximum of informations and precisions about my issue. If I can't manage to get the expected results, I will launch a bounty, maybe some experts or symply people who have been already faced to a similar problem will be able to help me

I have 2 covariance matrices known $\text{Cov}_X$ and $\text{Cov}_Y$ that I want to cross-correlate.

1) For this, I have performed a diagonalisation of each Fisher matrix $F_X$ (FISH_eigen_sp_flat in code) and $F_Y$ (FISH_eigen_xc_flat in code) associated of Covariance matrices $\text{Cov}_X$ and $\text{Cov}_Y$.

So, I have 2 different linear combinations that are uncorraleted, i.e just related by eigen values ($\dfrac{1}{\sigma_i^2}$) as respect of their combination.

Then, I get the diagonal matrices $D_X$ and $D_Y$. I can't build a "global" Fisher matrix directly by summing the 2 diagonal matrices since the linear combination of random variables is different between the 2 Fisher matrices.I have eigen vectors represented by $X$ (with $D_X$ diagonal) and $Y$ matrices (with $D_Y$ diagonal matrix)

That's why I think that I could perform a "global" combination of eigen vectors where I can respect the MLE estimator as eigen value :

$$\dfrac{1}{\sigma_{\hat{\tau}}^{2}}=\dfrac{1}{\sigma_1^2}+\dfrac{1}{\sigma_2^2}\quad(1)$$

because $\sigma_{\hat{\tau}}$ corresponds to the best estimator from MLE method (The only inconvenient in this approach for me is that I respect it only on a linear combination of random variables, and not just one, if someone could tell me if am wrong or not about this)

So, I thought a convenient linear combination that could allow to achieve it would be under the form :

$$M = X + Y + Z$$

with $$Z =aX + bY$$

where $a$ and $b$ have to be determined and such that :

$$\text{Var}(X+Y+Z)=\text{Var}(X)+\text{Var}(Y)$$

Therefore, I am looking for a way to find $Z$ (not null) that could verify :

$$\text{Var}(X+Y+Z)=\text{Var}(X+Y)+\text{Var}(Z)+2\,\text{Cov}(X+Y,Z)$$

$$= \text{Var}(X)+\text{Var}(Y)+2\,\text{Cov}(X,Y)+\text{Var}(Z)+2\,\text{Cov}(X,Z)+2\,\text{Cov}(Y,Z)=\text{Var}(X)+\text{Var}(Y)$$

So, we can remove on both sides the terms $\text{Var}(X)+\text{Var}(Y)$ to get the equation :

$$2\,\text{Cov}(X,Y)+\text{Var}(Z)+2\,\text{Cov}(X,Z)+2\,\text{Cov}(Y,Z)=0$$

It follows with solution under the form $Z=a\,X+b\,Y$ :

$$2\,\text{Cov}(X,Y)+a^2\,\text{Var}(X)+b^2\,\text{Var}(Y)+2ab\,\text{Cov}(X,Y)+2\,\text{Cov}(X,aX+bY)+2\,\text{Cov}(Y,aX+bY)$$

$$=2\,\text{Cov}(X,Y)+a^2\,\text{Var}(X)+b^2\,\text{Var}(Y)+2ab\,\text{Cov}(X,Y)+2\,\text{Cov}(X,aX)+2\,\text{Cov}(X,bY)$$

$$=2\,\text{Cov}(X,Y)+a^2\,\text{Var}(X)+b^2\,\text{Var}(Y)+2ab\,\text{Cov}(X,Y)+2a\,\text{Var}(X)+2b\,\text{Cov}(X,Y)+2a\,\text{Cov}(X,Y)+2b\,\text{Var}(Y)=0\quad(2)$$

Finally, If I fix a value for $a$, I have to solve this equation like a second order equation with $b$ as unknown parameter (below b=x=unknown) :

$$\alpha\,x^2+\beta x + \delta$$

with :

$\alpha=\text{Var}(Y)$

$\beta=2\,\text{Cov}(X,Y)+2a\,\text{Cov}(X,Y)+2\,\text{Var}(Y)$

$\delta=2\,\text{Cov}(X,Y)+a^2\,\text{Var}(X)+2a\,\text{Var}(X)+2a\,\text{Cov}(X,Y)$

To avoid complex solutions of this second order equation, I have to respect :

$$\beta^2-4\,\alpha\delta > 0$$

2) We could for example take $a=-1$ and try to find the b unknown like this (more precision, I think that b unknow parameter is surely a matrix or a vector : if someone could confirm it since I take the scalar value $a=1$, this would fine to tell it). This choice of $a=-1$ may seem to be arbitral, it is not only the unique solution for this issue but I have not any error concerning my issue.

Therefore, by taking $a=-1$, I have implemented a function to compute varX, varY and CovXY in Python :

```<code># Solution Z = aX + bY ; Taking a = -1 => solve unknown b represented by b_coef :
b_coef = np.zeros((7,7))
VarX = np.zeros((7,7))
VarY = np.zeros((7,7))
CovXY = np.zeros((7,7))

# Compute covariance between vectors of matrix
def compute_Cov(A,B):
  C = np.zeros((7,7))
  for i in range(7):
    C[0:7,i]= np.mean(A[0:7,i]*B[0:7,i]) - np.mean(A[0:7,i])*np.mean(B[0:7,i])
  return C
</code>```

I can this way compute the variance of X (compute_Cov(X,X)), Y (compute_Cov(Y,Y) and covariance(X,Y) (compute_Cov(X,Y)).

Diagonalisation) By diagonalising the 2 Covariance matrices (inverse of Fisher matrices FISH_sp_flat and FISH_xc_flat) representing the variances on each parameters that I want to compute the standard deviations, I think I build a linear combination of these parameters that is uncorrelated :

This way, I can sum respectively for each combination the eigen values like this :

```<code># Get eigen values (marginalized error since we handle Covariance error)
# and eigen vectors giving matrix "P" with convention : F = P D P^-1
eigen_sp_flat, eigenv_sp_flat = np.linalg.eig(np.linalg.inv(F_X))
eigen_xc_flat, eigenv_xc_flat = np.linalg.eig(np.linalg.inv(F_Y))

# Sum of FISH_eigen sp and xc
FISH_eigen_sp_flat = np.linalg.inv(np.diag(eigen_sp_flat))
FISH_eigen_xc_flat = np.linalg.inv(np.diag(eigen_xc_flat))

# MLE method (Maximum Likelihood Estimator) : Sum of FISH_eigen sp and xc : 1/sig^2 = 1/sig1^2 + 1/sig^2
FISH_eigen_sum = FISH_eigen_sp_flat + FISH_eigen_xc_flat
</code>```

3) Once linear combination of parameters and Var(X), Var(Y), Cov(X,Y) computed, I build my final covariance matrix like this :

```<code># Sum of passing matrix X+Y+Z with X = eigenv_sp_flat, Y = eigenv_xc_flat and Z to infer
eigenv_final = eigenv_sp_flat + eigenv_xc_flat + a*eigenv_sp_flat + b_coef*eigenv_xc_flat

# Covariance and Fisher matrix final
COV_final = np.dot(eigenv_final, np.linalg.inv(FISH_eigen_sum))
COV_final = np.dot(COV_final, np.linalg.inv(eigenv_final))
FISH_final = np.linalg.inv(COV_final)
</code>```

The issue is that I get a better FoM (1389) (Figure of Merit = 1/area(w0,wa) which is the inverse area of 1 C.L contour for the joint parameters (w0,wa)) compared to a simple sum betwen the 2 starting Fisher matrix (1235): I would expect a more significant improvement and some anomalies appears also in the fnal constraints (for example, for a random variable, it doesn't respect the equation(1) ( $\dfrac{1}{\sigma_{\hat{\tau}}^{2}}=\dfrac{1}{\sigma_1^2}+\dfrac{1}{\sigma_2^2}\quad(1)$), i.e I have no gain for $\sigma_{\hat{\tau}}^{2}$ compared to the smallest individual $\sigma$ ($\sigma_1$ or $\sigma_2$).

Anyone could confirm me if my approach is correct, especially the calculation of a new basis that cheks V(X+Y+Z) = Var(X)+Var(Y) which seems for me essential ?

I hope that I have been enough clear in my explanations : the issue is simple and a little tricky in the same time.

EDIT 1: I realized that the condition of orthogonality ($F.T^T\neq I_d$) on the final building of eigen vectors is not respected in my code.

So, this characteristic of orthogonality is essential if I want to check the condition $\text{Var}(X+Y+Z)=\text{Var}(X)+\text{Var}(Y)$ where $Z =ax + bY$.

Therefore, I think this condition of orthogonality adds a complementary condition on the choice (and so the computation) of coefficient $a$ and $b$ into $Z =ax + bY$.

That's why I have asked another question about this criterion of orthogonality on : Building an orthogonal basis from the sum of different orthogonal basis.

Hope this will help you

Anyone could see how to translate this condition to determine a single value for $a$ and for $b$ ?

Don't hesitate to ask further information.

Any help/fix/suggestion is welcome. Sorry if this is a little bit long to read.

EDIT 2 : I have slightly modified the function compute_Cov : Does it make sense to compute variance $\text{Var}(X)$, $\text{Var}(Y)$, and $\text{CoVar}(X,Y)$ with $X$, $Y$ covariance matrices like this ? :

```<code># Compute covariance between vectors of matrix
def compute_Cov(A, B):
  C = np.zeros(7)
  for i in range(7):
    C[i]= np.cov(A[0:7,i]*B[0:7,i])
  return C

VarX = compute_Cov(X,X)
VarX = compute_Cov(Y,Y)
CoVarX = compute_Cov(X,Y)
</code>```

I have serious doubts about what I do since the beginning, anyone could see clearer ?

答:
You are trying to find $u, v, \text{and }w$ such that:

$$
u^2 + \sigma_X u v + \sigma_Y u w + \rho_{XY} \sigma_X \sigma_Y = 0 \\
u \geq 0 \\
-1 \leq v, w \leq +1,
$$

where $u = \sigma_Z$, $v = \rho_{X,Z}$ and $w = \rho_{Y,Z}$. Once you know $u, v, \text{and }w$, it is straightforward to find $a \text{ and } b$ in $Z = aX + bY$.

This needs to be solved numerically. You might have to preclude edge cases ($\rho_{XY} = \pm 1$).

An example: $\sigma_1 = \sigma_X = 0.1, \sigma_1 = \sigma_Y = 0.2$ and $\rho_{XY}$ varies from -0.9 to 0.9. $a$ and $b$ from minimising $(u^2 + \sigma_X u v + \sigma_Y u w + \rho_{XY} \sigma_X \sigma_Y)^2$ subject to the constraints look like this. The optimal objective values are of the order of $10^{-15}$, so practically $0$.


问:
Big Omega and Big Omega Infinity Differences
In a book it is written that by writing $f(n)=\overset{\infty}{\Omega}(g(n))$ we mean that there exists a positive constant $c$ such that $f(n) \ge c\times g(n) \geq 0$ for infinitely many integers $n$. Whereas a Big Omega definition says that there exist positive constants $c$ and $n_0$ such that $0 \leq c\times g(n) \leq f(n)$ for all $n \geq n_0$.

My question is why is it not right to say that these definitions are equal when for the Big Omega definition there is a $n \geq n_0$ condition for $n$, which in my opinion tells that the definition must be true for infinitely many $n$'s that are just greater or equal to $n_0$.

答:
Suppose $f(n)=1$ and $g(n)=0$ when $n$ is even. Suppose $f(n)=0$ and $g(n)=1$ when $n$ is odd. With $c=1,$ the sets $\{n\in \Bbb N: f(n)\ge cg(n)\ge 0\}$ and $\{n\in \Bbb N: g(n)\ge cf(n)\}$ are both infinite. But we do NOT have $f=O(g)$ nor $g=O(f).$

The Landau "big-$O$" notation is only concerned with the absolute values of the functions, and may be used when the variable $n$ can take non-integer values, and may be used when the variable approaches $r\in \Bbb R.\,$

If it is understood that $n$ must be an integer then "$f(n)=O(g(n))$ as $n\to \infty$"  means $$\exists n'\in \Bbb N\, \exists c>0\,\forall n\in \Bbb N \,(n>n'\implies |f(n)|\le c|g(n)|\,).$$

If the variable can take any real value, it is customary to use $x,$ not $n$. So "$f(x)=O(g(x))$ as $x \to\infty$" means $$\exists x'\in \Bbb R^+\,\exists c>0\,\forall x\in \Bbb R\,(x>x'\implies |f(x)|\le c|g(x)|\,).$$

If $r\in \Bbb R$ then "$f(x)=O(g(x))$ as $x\to r$" means $$\exists \delta >0 \, \exists c>0 \,\forall x\in \Bbb \,(0<|x-r|<\delta \implies |f(x)|\le c|g(x)|\,).$$ For example $(\sin x)^2/x= O(x)$ as $x\to 0.$ In some contexts $x$ may be restricted to values only $>r.$


问:
Proving that $|\sin(z)|^2 = \sin(x)^2+\sinh(y)^2$
Goal: Prove $|\sin(z)|^2 = \sin(x)^2+\sinh(y)^2$

I have
\begin{align*}
\sin(x+iy)
&=\sin(x)\cos(iy)+\cos(x)\sin(iy)\\
&=\sin(x)\cosh(y)+i\cos(x)\sinh(y)\\
|\sin(z)|^2
&=\sin(x)^2 \cosh(x)^2+\cos(x)^2\sinh(y)^2
\end{align*}

How do I proceed?

答:
How to proceed:

Replace $\cos^2$ with $1-\sin^2$ and $\cosh^2$ with $1+\sinh^2$

(Also note that $\cos(iy)=\cosh(\color{red}y)$.)


问:
Why is $x^4-3x^2+18$ irreducible over the $3$-adics?
Let $f = x^4 - 3x^2 + 18 \in \mathbb{Q}_3[x]$. Since there is an LMFDB page of an extension defined by this polynomial, I assume that $f$ is irreducible. Could you verify if this is true and also why that is (not) the case?

It is not an Eisenstein polynomial (as $3^2$ divides $18$, the constant coefficient of $f$). I also tried the substitution $y = x^2$ and asked myself if $g = y^2 - 3y + 18$ is irreducible. Its reduction mod $3$ is $y^2$ which is not irreducible, so I cannot say whether $g$ is irreducible or not.

答:
Well, here’s an argument that uses a somewhat non-elementary fact about $p$-adic field extensions, namely that if the residue-field extension degree is $f$ and the ramification degree is $e$, then the total degree $n$ is $n=ef$. This $f$ is also the degree over the base field of the maximal unramified intermediate extension.

I’m about to show that if $\alpha$ is a root of $F(x)=x^4-3x^2+18$, then the extension $\Bbb Q_3(\alpha)\supset\Bbb Q_3$ has $f$ and $e$ both equal to $2$, so that $[\Bbb Q_3(\alpha):\Bbb Q_3]=4$. This shows that your polynomial $F$ is irreducible.

You’ve already seen that $G(x)=x^2-3x+18$ has $\alpha^2=\gamma$ for a root, and as I’ve pointed out in a comment, $\gamma/3$ is a root of $x^2-x+2=g(x)$. Use the Quadratic Formula to see that to get roots of $g$, you need $\sqrt{-7}$, and since $x^2+7\equiv x^2+1\pmod3$, adjoining $\sqrt{-7}$ is same as adjoining $i$ to $\Bbb Q_3$, giving the quadratic unramified extension of $\Bbb Q_3$. So $f=2$ for the extension $\Bbb Q_3(\gamma)\supset\Bbb Q_3$.

Since the roots of $g(x)$ are $3$-adic units (the constant term of the minimal polynomial is a unit), we have $v_3(\gamma)=1$, as @RobertIsrael has shown by different methods. This means that $v_3(\alpha)=v_3(\sqrt\gamma\,)=1/2$, which in turn means that the ramification degree $e$ of $\Bbb Q_3$ over $\Bbb Q_3$ is at least two. Thus we have $f\ge2$, $g\ge2$, and their product $n$, the degree of the minimal polynomial for $\alpha$ over $\Bbb Q_3$, is $\le4$. It follows that all three inequalities are equalities, and $F$ is irreducible.


问:
Find all positive integer solutions to the equation $(2x-1)^3 + 16 = y^4$
How to approach this kind of questions ?

I just knew that y is odd and wrote the equation as $(2x-1)^3 = (y^2 + 4)$(y+2)(y-2)
I don't know what to do after this . Any hint would be appreciated.

答:
There are no such solutions.

HINT: First note that $y$ must be odd. For odd $y$ though, note that $y^2+4$ and $y^2-4$ are relatively prime so each must be a cube. They differ by only 8 though. So $y^2+4$ cannot be any larger than 27. That narrows down possibilities you need to check considerably.


问:
Proof Verification: $\frac{x}{x+1} \leq \ln(x+1)$ for $x > 0$ with equality iff $x=0$
Proof Verification: $\frac{x}{x+1} \leq \ln(x+1)$ for $x > -1$ with equality iff $x=0$

My Attempt:

We know that $\frac{1}{(x+1)^2} \leq \frac{1}{x+1}$, for $x > 0. $At $x=0 $ they are the same.

$$\int\frac{1}{(x+1)^2}dx \leq \int\frac{1}{x+1}dx$$
$$\frac{-1}{x+1} +C_1 \leq \ln(x+1) + C_2$$

We conclude that $C_1 =1, C_2 =0$ f0r the $x=0$ case from the original equality to hold.

$$\frac{-1}{x+1}+\frac{x+1}{x+1}-1 +1 \leq \ln(x+1)$$
$$\frac{x}{x+1} \leq \ln(x+1)$$

Is this approach right?

答:
Your answer is ok. But it is better to integrate
$$ \frac{1}{(t+1)^2} \leq \frac{1}{t+1}$$
from $0$ to $x$; namely
$$ \int_0^x\frac{1}{(t+1)^2}dt \leq \int_0^x\frac{1}{t+1}dt. $$
Then you will get the desired inequality soon instead of concluding $C_1=C_2=0$.


问:
How prove this $\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$
let $a,b\in R$,and such that
$$\left(\sqrt{a^2+b^4}-a\right)\left(\sqrt{b^2+a^4}-b\right)\le a^2b^2$$

prove that $$a+b\ge 0$$

I think this is very beatifull problem, have you nice methods? Thank you,

I have see this problem
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$
then we  have $$x+y=0$$

This problem have some nice methods,
let $f(x)=\ln\left({x+\sqrt{x^2+1}}\right)$, then $f(x)$ is increasing and is odd funciotion, and
$$\ln\left({y+\sqrt{y^2+1}}\right)=-\ln{-y+\sqrt{y^2+1}}=-f(-y)$$
so
$$f(x)+f(-y)=0\Longrightarrow f(x)=f(-y)$$
then $x+y=0$

and I have see this problem
$$\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1$$
then we have $$x+y=0$$
solution: let $$x=\dfrac{1}{2}\left(u-\dfrac{1}{u}\right),y=\dfrac{1}{2}\left(v-\dfrac{1}{v}\right)$$

then we have
$$\dfrac{(uv-1)((u+v)^2uv+(u-v)^2)}{u^2v^2}=0$$

since $$(u+v)^2uv+(u-v)^2\ge 0$$
so $$uv=1$$
then $$x+y=\dfrac{1}{2}\left(u+v-\dfrac{1}{u}-\dfrac{1}{v}\right)=0$$

答:
Step 1: Multiply by the conjugate, we get that

$$ b^4 a^4 \leq a^2 b^2 ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b), $$
$$ a^2 b^2 \leq ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b)$$.

This gives us the chain of inequalities

$$ ( \sqrt{a^2 + b^4} -a ) ( \sqrt{ b^2 + a^4} - b) \leq a^2 b^2 \leq ( \sqrt{a^2 + b^4} +a ) ( \sqrt{ b^2 + a^4} + b)$$

Take the extreme ends, we get $0 \leq a \sqrt{b^2 + a^4} + b \sqrt{a^2 + b^4} $.

Step 2: Consider cases

If $a, b <0$, then the terms in the LHS are clearly greater than $b^2 \times a^2$, hence the inequality is not true.

If $a,b \geq 0$, then clearly $a+b \geq 0.$

Hence, we may assume that $ a \leq 0 \leq b$, and we want to show that $ -a \leq b$.  Now, because we are used to dealing with positive (non-negative reals), let me replace $a$ with $-a$ (not necessary, but simplifies considerations later)

Step 3: With this substitution, the inequality in step 1 gives us

$$ a\sqrt{b^2 + a^4} \leq b \sqrt{a^2 + b^4}$$

Since the LHS is non-negative, we may square it to obtain

$$a^2 (b^2 + a^4) \leq b^2 (a^2 + b^4 \Rightarrow a^6 \leq b^6 \Rightarrow a \leq b.$$

But this is what we want to show in Step 2, hence we are done. (remember we substituted $a$ for $-a$.)

Step 1 gives you another way to prove your equality case. Namely, we get that

$$ 0 = y \sqrt{x^2+1} + x \sqrt{y^2+1} $$

Hence, we have $ - y \sqrt{x^2 + 1} = x \sqrt{y^2+1} \Rightarrow y^2(x^2+1)=x^2(y^2+1) \Rightarrow y^2=x^2$. Then check that $y=x$ is not a valid solution (unless $y=x=0$), hence we must have $y=-x$.

This seems much more direct than your approach, and is motivated by considering conjugates.

The inequality in step 1 can also be obtained directly, by expanding and showing that

$$ 0 \leq \sqrt{a^2+b^4} \sqrt{b^2+a^4} + ab - a^2b^2 \leq a \sqrt{b^2 + a^4} + b \sqrt{a^2 + b^4}$$

However, this is not immediately obvious from the question. It's more of 'on hindsight'.

I'd be interested in seeing how the equality case can be obtained through direct expansion. (I don't see how to do this as yet.)


问:
Evaluating limits of the type $\left.[e^{ikx}\Phi(x)]\right|_{x=-\infty}^{x=\infty}$
How to evaluate limits of the type:
$$
\left.[e^{ikx}\Phi(x)]\right|_{x=-\infty}^{x=\infty}
$$

Such terms occur while trying to integrate by parts when taking the Fourier Transform of the Poisson Equation. For potentials such as $\Phi(x)=\frac{1}{x}$ this is quite evidently $0$, but I'm not sure how to deal with cases such as the solution to the 2D Poisson Equation: $\Phi(\mathbf{x})=\ln|\mathbf{x}|$. A similar limit occurs from the integral representation of the delta function:
$$
\delta(k)=\int_{-\infty}^{\infty}dx e^{ikx}  \qquad \implies \qquad ik\delta(k) = \left.[e^{ikx}]\right|_{x=-\infty}^{x=\infty}
$$

答:
Let $$f_n(y)=\int_{-n}^n e^{2i\pi yx}dx$$
Then $f_n \to \delta$ in the sense of distributions, which means (by definition) that for all $\phi \in C^\infty_c(\Bbb{R})$  $$\lim_{n\to \infty} \int_{-\infty}^\infty f_n(y) \phi(y)dy = \phi(0)$$
In fact it stays true for any $\phi\in C^1(\Bbb{R})$ which is $L^1$ and whose derivative is $L^1$.

This is how we define and compute the Fourier transforms of things like $1$ and $\log |x|$.


问:
Prove Inequalities $\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$ Using Calculus
I am often asked to prove inequalities with natural logarithms. An example question I am stuck on is this:

Let $f(x) = \ln(\frac{1+x}{1-x})$ for $-1 < x < 1$. Prove that

$$\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$$ with equality iff $x=0$.

My general approach has been to try to find a function which upper and lower bounds the function $1/x$ such that the integral of the function will result in the upper and lower bound. I know that the definition of $\ln(x) = \int_{1}^{x} \frac{1}{t}dt$.

Example I want to prove (1)$f_1(x) < \ln(x) <f_2(x)$ . If I can find (2)$F_1(x) < 1/x < F_2(x)$, then by taking the integral I can get the original inequality. Problem is I find it difficult to come up with the (1) relationship.  Espically when it gets complicated.

Are there better ways / tricks that can help me solve these proofs more easily? What would be your approach to this problem?

Body:
Let

$$g(x)=\frac{x(2-x)}{1-x}- \ln\frac{1+x}{1-x}
$$
$$g’(x)= \frac{x(x^2-x+2)}{(1-x)^2(1+x)},\>\>\>
g’’(x) =\frac{2(3x^2+1)}{(1-x)^3(1+x)^2}
$$
Note $g(0)=g’(0)=0$ and $g’’(x)>0$, which means that $g(x)$ is convex with the minimum at $g(0)=0$. Thus, $g(x)\ge 0$, or
$$\ln\frac{1+x}{1-x}  \le \frac{x(2-x)}{1-x}$$
Similarly, let

$$h(x)=\ln\frac{1+x}{1-x} - \frac{x(2+x)}{1+x}
$$
$$h’(x)= \frac{x(x^2+x+2)}{(1-x)(1+x)^2},\>\>\>h’’(x) =\frac{2(3x^2+1)}{(1-x)^2(1+x)^3}$$
Note $h(0)=h’(0)=0$ and $h’’(x)>0$, which means that $h(x)$ is convex with the minimum at $h(0)=0$. Thus, $h(x)\ge 0$, or
$$ \frac{x(2+x)}{1+x}\le \ln\frac{1+x}{1-x}  $$


问:
how to integrate $ \int_\frac74^2 \lvert 1-x\rvert dx $ manully?
I calculated it on https://www.integral-calculator.com/
$$
\int_{7/4}^{2}\left\lvert 1 - x\right\rvert\mathrm{d}x
$$
and got the answer but I could not get the answer right when substituting.
Also, it would be very helpful if there were another way to do the integration instead of the solution given by the site

答: $\lvert 1-x\rvert = -1+x$ when $x\geq1$

Body: $\lvert 1-x\rvert = -1+x$ when $x\geq1$


问:
understanding inverse complex log in $e^z=-2$
I can see $z = ln(2)+i\pi(2k+1)$, but I can't grasp how to get there using the formula for $e^z=w$, $z =  log(w) + 2ik\pi$ as $log(-2)$ doesnt make sense.

答:
Write -2 in its polar form which is $2e^{i \pi}$.
You get $e^z = 2e^{i \pi}$. Then apply the formula you mentioned.


问:
relation between convergence and norms of sub-vectors in the sequence
Assume we have a sequence $\mathbf{x}_{n} \in l_{2}$, where $l_{2}$ is a space of square-summable sequences. Then, consider the following statements

Statement A: $\mathbf{x}_{n}\to \mathbf{c}  \in l_{2}$ ,i.e. the sequence $\mathbf{x}_{n}$ converges in $l_{2}$ and

Statement B: For any $\varepsilon > 0$ there exists $n_{0}$ and $K$ such that for all $n > n_{0}$ we have  $\sup_{n>n_{0}} \sum_{k>K} x_{n,k}^{2} < \varepsilon$

What is the relation between statements?

If Statement A then Statement B or

Statement A iff Statement B

答:
Let me clarify the notation a little bit. Let $\mathbf x = (\mathbf x_n)_{n \in \Bbb N}$. So it is $\mathbf x$ that is a sequence in $\ell_2$. For each $n, \mathbf x_n$ is an element of $\ell_2$, that is, it is $\mathbf x_n = (x_{n,k})_{k \in \Bbb N}$, a sequence in $\Bbb R$.

Being in $\ell_2$ each $\mathbf x_n$ is square-summable. That is, for each $n$, there is some real $s < \infty$ such that $s = \sum_{k=0}^\infty x_{n,k}^2$. Or:

For every $n \in \Bbb N$, and every $\varepsilon > 0$, there is a $K > 0$ such that for all $K' \ge K$, $$\left|s_n - \sum_{k=0}^{K'}x_{n,k}^2\right| < \varepsilon$$

But $s_n - \sum_{k=0}^{K'}x_{n,k}^2 = \sum_{k>K'} x_{n,k}^2 \ge 0$, so the inequality can be rewritten as $$\sum_{k>K'} x_{n,k}^2 < \varepsilon$$
But also, since everything is positive, if it is true for $K' = K$, it will also be true for any higher $K'$, hence being square-summable is equivalent to

For every $n \in \Bbb N$ and for every $\varepsilon > 0$, there is a $K$ such that $\sum_{k>K} x_{n,k}^2 < \varepsilon$

Now, as stated every $n$ can have a different $K$. If you demand that the same $K$ holds for every $n$:

For every $\varepsilon > 0$, there is a $K$ such that for every $n \in \Bbb N,
\sum_{k>K} x_{n,k}^2 < \varepsilon$

This says that the $\mathbf x_n$ are uniformly square-summable with respect to $n$.

At this point we are very close to your statement B. The only difference (other than a trivial index variance) being your only requiring $K$ to work for all $n$ above some $n_0$. But that condition serves no purpose. because if a particular $K'$ works for all $n > n_0$, then so do all $K > K'$. And for each $n \le n_0, \mathbf x_n$ is also square summable, so there is some $K_n$ that works for it. Picking the max of $K'$ and all the $K_n$ for $n \le n_0$ gives a $K$ which works for all $n$.

So B is the condition that the $\mathbf x_n$ are uniformly square-summable. It should be clear at this point that it isn't directly related to convergence of $\mathbf x$ in $\ell_2$. Both A and B can hold without the other holding.

Letting $\mathbf x_n =((-1)^n)_{k\in \Bbb N}$ give an example of B holding when A does not. Counter-examples going the other way are more problematic to construct, and I'm out of time. Maybe later.


问:
Finding a point on a Fermat related curve
I am trying to find solutions to the equation $ w^5+x^5=y^5+z^5 $. To do this I am performing a substitution $w=r+\sqrt{s}, x=r-\sqrt{s}, y=t+\sqrt{u}, z=t-\sqrt{u}$. This results in the equation: $r^5+10r^3s+5rs^2=t^5+10t^3u+5tu^2$.

After trying various values of r and t I stumbled upon a promising candidate: $r=1, t=-4$. This results in the equation $(s+1)^2+(2u+32)^2=820$, which has solutions in s and u.

I want s to be a square, which will yield a near solution to the original equation (albeit still involving $\sqrt{u}$). Setting $x=s^2+1, y=2u+32$ results in the curve $y^2 +(x^2+1)^2=820$. This would be an elliptic curve, if there could be found a rational point on it.

So my question is: Does the curve $y^2 +(x^2+1)^2=820$ have a rational point?

Edit: The first test case had an obvious solution that I missed. This resulted in the nice equation: $6^5-4^5=(-4-\sqrt{-10})^5+(-4+\sqrt{-10})^5$. As player3236 pointed out, no value of u in this equation will be positive, so the solutions will always involve an imaginary component. This will happen any time t is chosen to be negative. However, if t is positive there is potential for u to also be positive.

If r=1, t=16, the new equation is $(x^2+1)^2+838860=y^2$. Does this new equation have rational solutions? Wolfram Alpha couldn't find any this time.

Edit 2: I did find a point on the curve with $x=\frac{931}{34}$

答:
Let us consider the first curve, $y^2+(x^2+1)^2= 820$. Finding points on this curve is equivalent to finding points on the curve
$$
\begin{aligned}
E'\ &:\ &V^2 &= -((U+5)^2+1)^2+820\ ,\text{ i.e.}\\
E'\ &:\ &V^2 &= -U^4 - 20U^3 - 152U^2 - 520U + 144\ .
\end{aligned}
$$
Let us write the above in the form $V^2 =aU^4+bU^3+cU^2+dU+q^2$, with obvious coefficients $a,b,c,d; q$. Let $Q$ be $Q=2q$.
Now Ian Connell, Elliptic Curve Handbook, Proposition 1.2.1, page 105 gives a birational passage to the elliptic curve with equation
$$
E''\ :\ Y^2+a_1XY+a_3Y=X^3+a_2X^2+a_4X+a_6\ ,$$
where:
$$
\begin{aligned}
a_1 &= d/q\ ,\\
a_2 &= c-d^2/Q^2\ ,\\
a_3 &= Qb\ ,\\
a_4 &= -Q^2a\ ,\\
a_6 &= a_2a_4 = a(d^2-Q^2c)\ .
\end{aligned}
$$
A point $(X,Y)$ on the curve $E''$ induces a point $(U,V)$ on $E'$ by the formulas:
$$
U = \frac 1Y\left(Q(X+c)-\frac 1Q\cdot d^2\right)\ ,\qquad
V = -q+\frac 1Q\cdot U(UX-d)\ .
$$
In our case, the curve $E''$ is the following curve, sage code:

```<code>sage: var('x,y');
sage: def f(x,y):    return -y^2 - (x^2+1)^2 + 820
sage: f(x+5, y).expand()
-x^4 - 20*x^3 - 152*x^2 - y^2 - 520*x + 144
sage: a, b, c, d, q = -1, -20, -152, -520, 12
sage: Q = 2*q
sage: a1, a2, a3, a4 = d/q, c-d^2/Q^2, Q*b, -Q^2*a
sage: a6 = a2*a4
sage: E = EllipticCurve(QQ, [a1, a2, a3, a4, a6])
sage: E.rank()
2
sage: E.torsion_points()
[(0 : 1 : 0), (52 : 4100/3 : 1)]
</code>```

This gives a lot of points on the curve. The following code constructs some of them:

```<code>solutions = []
P1, P2 = E.gens()
T = E.torsion_points()[1]
for j1 in [-3..3]:
    for j2 in [-3..3]:
        for n in [0..1]:
            R = j1*P1 + j2*P2 + n*T
            if R == E(0):    continue
            X, Y = R.xy()
            if Y == 0:    continue
            U = (Q*(X+c) - d^2/Q) / Y
            V = -q + U*(U*X - d)/Q
            U5, V = abs(U+5), abs(V)
            if (U5, V) in solutions:    continue
            solutions.append( (U5, V) )
            u = V/2 - 16
            # if u < 0:    continue
            print(r"x &= \frac{%s}{%s}\\"
                  % (U5.numer(), latex(factor(U5.denom()))))
            print(r"y &= \frac{%s}{%s}\\"
                  % (V.numer(), latex(factor(V.denom()))))
            print(r"u &= \frac{%s}{%s}\\"
                  % (latex(factor(u.numer())), latex(factor(u.denom()))))
            if u.is_square():    print(r"&SQUARE\\")
            print("  &\\qquad y^2 + (x^2+1)^2 - 820 = %s\n\\\\[4mm]\n%%"
                  % f(U5, V))
</code>```

We have then the following solutions:
$$
\begin{aligned}
x &= \frac{13036658315510032988204253801235}{13 \cdot 29 \cdot 37 \cdot 21701 \cdot 62233 \cdot 87697 \cdot 5563471867729}\\
y &= \frac{2405690869878559044594514807781718914294941723532442030205314708}{13^{2} \cdot 29^{2} \cdot 37^{2} \cdot 21701^{2} \cdot 62233^{2} \cdot 87697^{2} \cdot 5563471867729^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 23 \cdot 39733 \cdot 1952241098341 \cdot 3791563425833816372596018279215226374188039}{13^{2} \cdot 29^{2} \cdot 37^{2} \cdot 21701^{2} \cdot 62233^{2} \cdot 87697^{2} \cdot 5563471867729^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{162458182555112323161}{5 \cdot 97 \cdot 73623006391638001}\\
y &= \frac{23822580678926521285740950989704151999428}{5^{2} \cdot 97^{2} \cdot 73623006391638001^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 71 \cdot 333589 \cdot 2236160247547007 \cdot 6164477497937567}{5^{2} \cdot 97^{2} \cdot 73623006391638001^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{9520442060365}{3642799992361}\\
y &= \frac{365511913152330277363122252}{3642799992361^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 103 \cdot 28702827158601638691367}{3642799992361^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{230878603791}{5 \cdot 6829 \cdot 1874317}\\
y &= \frac{102280146333948447788892}{5^{2} \cdot 6829^{2} \cdot 1874317^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 29 \cdot 2998141 \cdot 6366883182061}{5^{2} \cdot 6829^{2} \cdot 1874317^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{93542279748835}{27917 \cdot 898599293}\\
y &= \frac{15387605729678347093142160468}{27917^{2} \cdot 898599293^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 23 \cdot 683 \cdot 1559 \cdot 553099907 \cdot 6744379151}{27917^{2} \cdot 898599293^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{4992964600942606254201}{5 \cdot 8233 \cdot 186581 \cdot 262351875749}\\
y &= \frac{112597556643703829003523268375206881581624068}{5^{2} \cdot 8233^{2} \cdot 186581^{2} \cdot 262351875749^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 71 \cdot 687289 \cdot 35949192861661 \cdot 2470089037887599849437}{5^{2} \cdot 8233^{2} \cdot 186581^{2} \cdot 262351875749^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{14100446200928414271395695884163165}{10133 \cdot 9131338913 \cdot 32850086210006457589}\\
y &= \frac{163409560140746617280818877377336839053725151518595253277111416731532}{10133^{2} \cdot 9131338913^{2} \cdot 32850086210006457589^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 13^{2} \cdot 131 \cdot 40253120493723100804553179 \cdot 7419158906364631246083380484383646761}{10133^{2} \cdot 9131338913^{2} \cdot 32850086210006457589^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{61101860425670949880707892899}{5^{2} \cdot 265205909 \cdot 1913532975314529661}\\
y &= \frac{2465433439082350072102436786861216960772068168407137748332}{5^{4} \cdot 265205909^{2} \cdot 1913532975314529661^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 19 \cdot 9348263 \cdot 115636877949077 \cdot 18964570146285293 \cdot 1723484690850703801}{5^{4} \cdot 265205909^{2} \cdot 1913532975314529661^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{3182700450859025}{22129 \cdot 147875300509}\\
y &= \frac{305926196130319251970804940340252}{22129^{2} \cdot 147875300509^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 43 \cdot 131 \cdot 5527 \cdot 58995650061058355531771}{22129^{2} \cdot 147875300509^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{640372149}{5^{2} \cdot 409 \cdot 11969}\\
y &= \frac{57258245706223668}{5^{4} \cdot 409^{2} \cdot 11969^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 105506277038649083}{5^{4} \cdot 409^{2} \cdot 11969^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{33025}{37 \cdot 3673}\\
y &= \frac{528512197188}{37^{2} \cdot 3673^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 23 \cdot 983 \cdot 691079}{37^{2} \cdot 3673^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{108983619}{5^{2} \cdot 850529}\\
y &= \frac{3950171837892948}{5^{4} \cdot 850529^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 13^{2} \cdot 134581 \cdot 10509997}{5^{4} \cdot 850529^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{141166512970975}{13 \cdot 29 \cdot 90641 \cdot 2835253}\\
y &= \frac{267192592960622599823692451292}{13^{2} \cdot 29^{2} \cdot 90641^{2} \cdot 2835253^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 103 \cdot 871062163457 \cdot 18493060659611}{13^{2} \cdot 29^{2} \cdot 90641^{2} \cdot 2835253^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{300465457943068995736805109}{5^{2} \cdot 548139169 \cdot 4846452536454329}\\
y &= \frac{83584824775100492091781590892844195239295177451886092}{5^{4} \cdot 548139169^{2} \cdot 4846452536454329^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 349 \cdot 214147 \cdot 5094064139 \cdot 20327786573 \cdot 1859326778033444504699597}{5^{4} \cdot 548139169^{2} \cdot 4846452536454329^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{2665287830263248801195859415}{113 \cdot 5472031792894910653554197}\\
y &= \frac{7989458932698213513688381614156914391236133060029901852}{113^{2} \cdot 5472031792894910653554197^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 127 \cdot 1023257 \cdot 3429708645773522945507 \cdot 476274917191165521198937}{113^{2} \cdot 5472031792894910653554197^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{201902890715811}{5 \cdot 4729 \cdot 2901654253}\\
y &= \frac{126894617470983770240467695828}{5^{2} \cdot 4729^{2} \cdot 2901654253^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 93166237 \cdot 4886427649 \cdot 13035867311}{5^{2} \cdot 4729^{2} \cdot 2901654253^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{504185}{152821}\\
y &= \frac{608447007108}{152821^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 131 \cdot 229 \cdot 1157449}{152821^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{261}{5 \cdot 13}\\
y &= \frac{96972}{5^{2} \cdot 13^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 19 \cdot 503}{5^{2} \cdot 13^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{1415}{661}\\
y &= \frac{12271452}{661^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 13 \cdot 6577}{661^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{5260848291}{5 \cdot 217539397}\\
y &= \frac{17744529175451385108}{5^{2} \cdot 217539397^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 20939 \cdot 2804953 \cdot 7783379}{5^{2} \cdot 217539397^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{46203128231239359785}{37 \cdot 1337256503359283713}\\
y &= \frac{69953560999420955768703715813623276985668}{37^{2} \cdot 1337256503359283713^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 1187 \cdot 135867929869169674775218266100668241}{37^{2} \cdot 1337256503359283713^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{5659195370905602649178527671}{5 \cdot 249389877977 \cdot 7234008048184553}\\
y &= \frac{2327274688594258349538778705315778236547654610669652515588}{5^{2} \cdot 249389877977^{2} \cdot 7234008048184553^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 13^{2} \cdot 23 \cdot 4446191 \cdot 4885823 \cdot 11581429 \cdot 300089788549 \cdot 21414576369482434643}{5^{2} \cdot 249389877977^{2} \cdot 7234008048184553^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{218157575240795}{13 \cdot 29 \cdot 110077841033}\\
y &= \frac{352116840259517020019969772}{13^{2} \cdot 29^{2} \cdot 110077841033^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 1559 \cdot 58909 \cdot 29811990671960835851}{13^{2} \cdot 29^{2} \cdot 110077841033^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{30159}{5 \cdot 37 \cdot 277}\\
y &= \frac{75115465692}{5^{2} \cdot 37^{2} \cdot 277^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 19 \cdot 103 \cdot 419 \cdot 2719}{5^{2} \cdot 37^{2} \cdot 277^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{5}{1}\\
y &= \frac{12}{1}\\
u &= \frac{-1 \cdot 2 \cdot 5}{1}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{9}{5}\\
y &= \frac{708}{5^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 23}{5^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{1850795}{432241}\\
y &= \frac{3946471955028}{432241^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 911 \cdot 42897937}{432241^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{25093501670280879}{5 \cdot 97 \cdot 857 \cdot 15377 \cdot 1317793}\\
y &= \frac{1906726072369386038223938515453212}{5^{2} \cdot 97^{2} \cdot 857^{2} \cdot 15377^{2} \cdot 1317793^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11446381 \cdot 7935036462375784894868837}{5^{2} \cdot 97^{2} \cdot 857^{2} \cdot 15377^{2} \cdot 1317793^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{10812125034210473313993622054805}{23072673329 \cdot 143581662104612874209}\\
y &= \frac{287075176351657878630882446552860447116555520044022707438613452}{23072673329^{2} \cdot 143581662104612874209^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 13 \cdot 503 \cdot 10753 \cdot 559223123 \cdot 13447826686304359 \cdot 6062603250107171605870871959}{23072673329^{2} \cdot 143581662104612874209^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{1705233582536073760995190741133872101}{5 \cdot 244327342969 \cdot 266515540930558827574357}\\
y &= \frac{362431472808647895360320065190973444447130394211867699577642447822619532}{5^{2} \cdot 244327342969^{2} \cdot 266515540930558827574357^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 29 \cdot 45932966323 \cdot 794963514281 \cdot 91794463398902240064343 \cdot 599400942698438604785969}{5^{2} \cdot 244327342969^{2} \cdot 266515540930558827574357^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{3154741859903558524046951461789436053287025}{613 \cdot 359837 \cdot 709913 \cdot 7599836289694012095782190757}\\
y &= \frac{38930269299271168229289261505326698150375319537776856686256422554566753985971200683908}{613^{2} \cdot 359837^{2} \cdot 709913^{2} \cdot 7599836289694012095782190757^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 23 \cdot 71 \cdot 33073 \cdot 77591 \cdot 159469 \cdot 96643537 \cdot 37692226417 \cdot 175669559711 \cdot 3736240723150976619930114531647866139}{613^{2} \cdot 359837^{2} \cdot 709913^{2} \cdot 7599836289694012095782190757^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
\end{aligned}
$$

On the second curve,
$$
y^2+(x^2+1) = 838860\ ,
$$
we can more or less do the same. First, the curve is birational equivalent to...

```<code>var('x,y');
def f(x,y):    return -y^2 + (x^2+1)^2 + 838860
f(x + 931/34, y).expand()
a, b, c, d, q = 1, 1862/17, 2601439/578, 808030727/9826, 1369043/1156
Q = 2*q
a1, a2, a3, a4 = d/q, c-d^2/Q^2, Q*b, -Q^2*a
a6 = a2*a4
E = EllipticCurve(QQ, [a1, a2, a3, a4, a6])
rk = E.rank()
print("E is an elliptic curve of rank %s" % rk)
print("Generators:")
for P in E.gens():
    print("\t", P.xy())
for T in E.torsion_points():
    if T == E(0):    continue
    print("\t", T.xy())
</code>```

And we get again a curve of rank two!

```<code>E is an elliptic curve of rank 2
Generators:
     (-1369043/578, 0)
     (-43316325/28322, -41113249906388/614068441)
     (-867917/578, -106213097760/1369043)
</code>```

A better form is...

```<code>sage: E.minimal_model()
Elliptic Curve defined by y^2 = x^3 - x^2 - 209715*x over Rational Field
</code>```

Then some points on $y^2 = (x^2+1)^2 + 838860 are:

```<code> solutions = []
 P1, P2 = E.gens()
 T = E.torsion_points()[1]
 for j1 in [-2..2]:
     for j2 in [-2..2]:
         for n in [0..1]:
             R = j1*P1 + j2*P2 + n*T
             if R == E(0):    continue
             X, Y = R.xy()
             if Y == 0:    continue
             U = (Q*(X+c) - d^2/Q) / Y
             V = -q + U*(U*X - d)/Q
             UU, V = abs(U + 931/34), abs(V)
             if (UU, V) in solutions:    continue
             solutions.append( (UU, V) )
             print(r"x &= \frac{%s}{%s}\\"
                   % (UU.numer(), UU.denom()))
             print(r"y &= \frac{%s}{%s}\\"
                   % (V.numer(), V.denom()))
             print("  &\\qquad (x^2+1)^2 + 838860 - y^2 = %s\n\\\\[4mm]\n%%"
                   % ((UU^2+1)^2 + 838860 - V^2))
</code>```

$$
\begin{aligned}
x &= \frac{4640129248052012171295041}{56328368752930789692104}\\
y &= \frac{21729172452104012184427503136290818755933810040513}{3172885126366149767519655678919343059119946816}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{312505635338264918831323}{5635665178453622365027}\\
y &= \frac{101930516183122284005580603951145069243469807068}{31760722003634699217295111109664688832710729}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{7697107706380753981821512221}{186853640262015669462073894}\\
y &= \frac{67355326049228509693898695429792231527200858728844325683}{34914282879166763220702909129407082141202455516323236}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{275273488143971163955319670170828849327}{8630446951340393789955958138130027033}\\
y &= \frac{102015582307789500427230911928915609414149466194469828884544309431564059167628}{74484614579900697494445334607256438244445639280727623918729596137310783089}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3533796460790528465936090156525696152253306912169010481}{142147419283298414256273589168805617982980986867368204}\\
y &= \frac{22336879382547877312447713567495156125129879126864659964611288771826011654660047981168198612554244346982363633}{20205888808901837866902182036543346606858038621737120342109726603597659188209906227548837538265889310185616}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{66981948434235}{4493946651194}\\
y &= \frac{19038090760503750938241601891}{20195556503777767101625636}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{9827459237}{1028878515}\\
y &= \frac{974459905490087541356}{1058590998628605225}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{369725522175}{86671374244}\\
y &= \frac{6881632670450531112199441}{7511927113343506571536}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{10885438126521930467}{10771530642536438565}\\
y &= \frac{106267543905924773766136574076877880671436}{116025872383101461045183783794019259225}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3106512218127349796388060274811685}{494034186048468200762637842991614}\\
y &= \frac{223760671475152068058777277411257298563715773166193274657640742778051}{244069776984572492223657898435483022217396770912501687097274324996}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3382515311}{218686124}\\
y &= \frac{45283088549214408113}{47823620830143376}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3017}{143}\\
y &= \frac{20832748}{20449}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{408358573}{11636677}\\
y &= \frac{207929610753260668}{135412251602329}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{64609139182230553631}{1409763762048977264}\\
y &= \frac{4555779364380348768652797926075775093953}{1987433864786485387900613108388925696}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{544306536341}{6280755390}\\
y &= \frac{298503660402510832364131}{39447888269014052100}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{1215}{7}\\
y &= \frac{1476956}{49}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{931}{34}\\
y &= \frac{1369043}{1156}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
\end{aligned}
$$

Note: I'll try to look closer, and see why we get gurves of rank two...

Body:
Let us consider the first curve, $y^2+(x^2+1)^2= 820$. Finding points on this curve is equivalent to finding points on the curve
$$
\begin{aligned}
E'\ &:\ &V^2 &= -((U+5)^2+1)^2+820\ ,\text{ i.e.}\\
E'\ &:\ &V^2 &= -U^4 - 20U^3 - 152U^2 - 520U + 144\ .
\end{aligned}
$$
Let us write the above in the form $V^2 =aU^4+bU^3+cU^2+dU+q^2$, with obvious coefficients $a,b,c,d; q$. Let $Q$ be $Q=2q$.
Now Ian Connell, Elliptic Curve Handbook, Proposition 1.2.1, page 105 gives a birational passage to the elliptic curve with equation
$$
E''\ :\ Y^2+a_1XY+a_3Y=X^3+a_2X^2+a_4X+a_6\ ,$$
where:
$$
\begin{aligned}
a_1 &= d/q\ ,\\
a_2 &= c-d^2/Q^2\ ,\\
a_3 &= Qb\ ,\\
a_4 &= -Q^2a\ ,\\
a_6 &= a_2a_4 = a(d^2-Q^2c)\ .
\end{aligned}
$$
A point $(X,Y)$ on the curve $E''$ induces a point $(U,V)$ on $E'$ by the formulas:
$$
U = \frac 1Y\left(Q(X+c)-\frac 1Q\cdot d^2\right)\ ,\qquad
V = -q+\frac 1Q\cdot U(UX-d)\ .
$$
In our case, the curve $E''$ is the following curve, sage code:

sage: var('x,y');
sage: def f(x,y):    return -y^2 - (x^2+1)^2 + 820
sage: f(x+5, y).expand()
-x^4 - 20*x^3 - 152*x^2 - y^2 - 520*x + 144
sage: a, b, c, d, q = -1, -20, -152, -520, 12
sage: Q = 2*q
sage: a1, a2, a3, a4 = d/q, c-d^2/Q^2, Q*b, -Q^2*a
sage: a6 = a2*a4
sage: E = EllipticCurve(QQ, [a1, a2, a3, a4, a6])
sage: E.rank()
2
sage: E.torsion_points()
[(0 : 1 : 0), (52 : 4100/3 : 1)]

This gives a lot of points on the curve. The following code constructs some of them:

solutions = []
P1, P2 = E.gens()
T = E.torsion_points()[1]
for j1 in [-3..3]:
    for j2 in [-3..3]:
        for n in [0..1]:
            R = j1*P1 + j2*P2 + n*T
            if R == E(0):    continue
            X, Y = R.xy()
            if Y == 0:    continue
            U = (Q*(X+c) - d^2/Q) / Y
            V = -q + U*(U*X - d)/Q
            U5, V = abs(U+5), abs(V)
            if (U5, V) in solutions:    continue
            solutions.append( (U5, V) )
            u = V/2 - 16
            # if u < 0:    continue
            print(r"x &= \frac{%s}{%s}\\"
                  % (U5.numer(), latex(factor(U5.denom()))))
            print(r"y &= \frac{%s}{%s}\\"
                  % (V.numer(), latex(factor(V.denom()))))
            print(r"u &= \frac{%s}{%s}\\"
                  % (latex(factor(u.numer())), latex(factor(u.denom()))))
            if u.is_square():    print(r"&SQUARE\\")
            print("  &\\qquad y^2 + (x^2+1)^2 - 820 = %s\n\\\\[4mm]\n%%"
                  % f(U5, V))

We have then the following solutions:
$$
\begin{aligned}
x &= \frac{13036658315510032988204253801235}{13 \cdot 29 \cdot 37 \cdot 21701 \cdot 62233 \cdot 87697 \cdot 5563471867729}\\
y &= \frac{2405690869878559044594514807781718914294941723532442030205314708}{13^{2} \cdot 29^{2} \cdot 37^{2} \cdot 21701^{2} \cdot 62233^{2} \cdot 87697^{2} \cdot 5563471867729^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 23 \cdot 39733 \cdot 1952241098341 \cdot 3791563425833816372596018279215226374188039}{13^{2} \cdot 29^{2} \cdot 37^{2} \cdot 21701^{2} \cdot 62233^{2} \cdot 87697^{2} \cdot 5563471867729^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{162458182555112323161}{5 \cdot 97 \cdot 73623006391638001}\\
y &= \frac{23822580678926521285740950989704151999428}{5^{2} \cdot 97^{2} \cdot 73623006391638001^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 71 \cdot 333589 \cdot 2236160247547007 \cdot 6164477497937567}{5^{2} \cdot 97^{2} \cdot 73623006391638001^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{9520442060365}{3642799992361}\\
y &= \frac{365511913152330277363122252}{3642799992361^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 103 \cdot 28702827158601638691367}{3642799992361^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{230878603791}{5 \cdot 6829 \cdot 1874317}\\
y &= \frac{102280146333948447788892}{5^{2} \cdot 6829^{2} \cdot 1874317^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 29 \cdot 2998141 \cdot 6366883182061}{5^{2} \cdot 6829^{2} \cdot 1874317^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{93542279748835}{27917 \cdot 898599293}\\
y &= \frac{15387605729678347093142160468}{27917^{2} \cdot 898599293^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 23 \cdot 683 \cdot 1559 \cdot 553099907 \cdot 6744379151}{27917^{2} \cdot 898599293^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{4992964600942606254201}{5 \cdot 8233 \cdot 186581 \cdot 262351875749}\\
y &= \frac{112597556643703829003523268375206881581624068}{5^{2} \cdot 8233^{2} \cdot 186581^{2} \cdot 262351875749^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 71 \cdot 687289 \cdot 35949192861661 \cdot 2470089037887599849437}{5^{2} \cdot 8233^{2} \cdot 186581^{2} \cdot 262351875749^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{14100446200928414271395695884163165}{10133 \cdot 9131338913 \cdot 32850086210006457589}\\
y &= \frac{163409560140746617280818877377336839053725151518595253277111416731532}{10133^{2} \cdot 9131338913^{2} \cdot 32850086210006457589^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 13^{2} \cdot 131 \cdot 40253120493723100804553179 \cdot 7419158906364631246083380484383646761}{10133^{2} \cdot 9131338913^{2} \cdot 32850086210006457589^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{61101860425670949880707892899}{5^{2} \cdot 265205909 \cdot 1913532975314529661}\\
y &= \frac{2465433439082350072102436786861216960772068168407137748332}{5^{4} \cdot 265205909^{2} \cdot 1913532975314529661^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 19 \cdot 9348263 \cdot 115636877949077 \cdot 18964570146285293 \cdot 1723484690850703801}{5^{4} \cdot 265205909^{2} \cdot 1913532975314529661^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{3182700450859025}{22129 \cdot 147875300509}\\
y &= \frac{305926196130319251970804940340252}{22129^{2} \cdot 147875300509^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 43 \cdot 131 \cdot 5527 \cdot 58995650061058355531771}{22129^{2} \cdot 147875300509^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{640372149}{5^{2} \cdot 409 \cdot 11969}\\
y &= \frac{57258245706223668}{5^{4} \cdot 409^{2} \cdot 11969^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 105506277038649083}{5^{4} \cdot 409^{2} \cdot 11969^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{33025}{37 \cdot 3673}\\
y &= \frac{528512197188}{37^{2} \cdot 3673^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 23 \cdot 983 \cdot 691079}{37^{2} \cdot 3673^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{108983619}{5^{2} \cdot 850529}\\
y &= \frac{3950171837892948}{5^{4} \cdot 850529^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 13^{2} \cdot 134581 \cdot 10509997}{5^{4} \cdot 850529^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{141166512970975}{13 \cdot 29 \cdot 90641 \cdot 2835253}\\
y &= \frac{267192592960622599823692451292}{13^{2} \cdot 29^{2} \cdot 90641^{2} \cdot 2835253^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 103 \cdot 871062163457 \cdot 18493060659611}{13^{2} \cdot 29^{2} \cdot 90641^{2} \cdot 2835253^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{300465457943068995736805109}{5^{2} \cdot 548139169 \cdot 4846452536454329}\\
y &= \frac{83584824775100492091781590892844195239295177451886092}{5^{4} \cdot 548139169^{2} \cdot 4846452536454329^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 349 \cdot 214147 \cdot 5094064139 \cdot 20327786573 \cdot 1859326778033444504699597}{5^{4} \cdot 548139169^{2} \cdot 4846452536454329^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{2665287830263248801195859415}{113 \cdot 5472031792894910653554197}\\
y &= \frac{7989458932698213513688381614156914391236133060029901852}{113^{2} \cdot 5472031792894910653554197^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 127 \cdot 1023257 \cdot 3429708645773522945507 \cdot 476274917191165521198937}{113^{2} \cdot 5472031792894910653554197^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{201902890715811}{5 \cdot 4729 \cdot 2901654253}\\
y &= \frac{126894617470983770240467695828}{5^{2} \cdot 4729^{2} \cdot 2901654253^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 93166237 \cdot 4886427649 \cdot 13035867311}{5^{2} \cdot 4729^{2} \cdot 2901654253^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{504185}{152821}\\
y &= \frac{608447007108}{152821^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 131 \cdot 229 \cdot 1157449}{152821^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{261}{5 \cdot 13}\\
y &= \frac{96972}{5^{2} \cdot 13^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 19 \cdot 503}{5^{2} \cdot 13^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{1415}{661}\\
y &= \frac{12271452}{661^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 13 \cdot 6577}{661^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{5260848291}{5 \cdot 217539397}\\
y &= \frac{17744529175451385108}{5^{2} \cdot 217539397^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 20939 \cdot 2804953 \cdot 7783379}{5^{2} \cdot 217539397^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{46203128231239359785}{37 \cdot 1337256503359283713}\\
y &= \frac{69953560999420955768703715813623276985668}{37^{2} \cdot 1337256503359283713^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 1187 \cdot 135867929869169674775218266100668241}{37^{2} \cdot 1337256503359283713^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{5659195370905602649178527671}{5 \cdot 249389877977 \cdot 7234008048184553}\\
y &= \frac{2327274688594258349538778705315778236547654610669652515588}{5^{2} \cdot 249389877977^{2} \cdot 7234008048184553^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11 \cdot 13^{2} \cdot 23 \cdot 4446191 \cdot 4885823 \cdot 11581429 \cdot 300089788549 \cdot 21414576369482434643}{5^{2} \cdot 249389877977^{2} \cdot 7234008048184553^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{218157575240795}{13 \cdot 29 \cdot 110077841033}\\
y &= \frac{352116840259517020019969772}{13^{2} \cdot 29^{2} \cdot 110077841033^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 1559 \cdot 58909 \cdot 29811990671960835851}{13^{2} \cdot 29^{2} \cdot 110077841033^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{30159}{5 \cdot 37 \cdot 277}\\
y &= \frac{75115465692}{5^{2} \cdot 37^{2} \cdot 277^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 19 \cdot 103 \cdot 419 \cdot 2719}{5^{2} \cdot 37^{2} \cdot 277^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{5}{1}\\
y &= \frac{12}{1}\\
u &= \frac{-1 \cdot 2 \cdot 5}{1}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{9}{5}\\
y &= \frac{708}{5^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 23}{5^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{1850795}{432241}\\
y &= \frac{3946471955028}{432241^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 911 \cdot 42897937}{432241^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{25093501670280879}{5 \cdot 97 \cdot 857 \cdot 15377 \cdot 1317793}\\
y &= \frac{1906726072369386038223938515453212}{5^{2} \cdot 97^{2} \cdot 857^{2} \cdot 15377^{2} \cdot 1317793^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 11446381 \cdot 7935036462375784894868837}{5^{2} \cdot 97^{2} \cdot 857^{2} \cdot 15377^{2} \cdot 1317793^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{10812125034210473313993622054805}{23072673329 \cdot 143581662104612874209}\\
y &= \frac{287075176351657878630882446552860447116555520044022707438613452}{23072673329^{2} \cdot 143581662104612874209^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 5 \cdot 13 \cdot 503 \cdot 10753 \cdot 559223123 \cdot 13447826686304359 \cdot 6062603250107171605870871959}{23072673329^{2} \cdot 143581662104612874209^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{1705233582536073760995190741133872101}{5 \cdot 244327342969 \cdot 266515540930558827574357}\\
y &= \frac{362431472808647895360320065190973444447130394211867699577642447822619532}{5^{2} \cdot 244327342969^{2} \cdot 266515540930558827574357^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 13 \cdot 29 \cdot 45932966323 \cdot 794963514281 \cdot 91794463398902240064343 \cdot 599400942698438604785969}{5^{2} \cdot 244327342969^{2} \cdot 266515540930558827574357^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
x &= \frac{3154741859903558524046951461789436053287025}{613 \cdot 359837 \cdot 709913 \cdot 7599836289694012095782190757}\\
y &= \frac{38930269299271168229289261505326698150375319537776856686256422554566753985971200683908}{613^{2} \cdot 359837^{2} \cdot 709913^{2} \cdot 7599836289694012095782190757^{2}}\\
u &= \frac{-1 \cdot 2 \cdot 23 \cdot 71 \cdot 33073 \cdot 77591 \cdot 159469 \cdot 96643537 \cdot 37692226417 \cdot 175669559711 \cdot 3736240723150976619930114531647866139}{613^{2} \cdot 359837^{2} \cdot 709913^{2} \cdot 7599836289694012095782190757^{2}}\\
  &\qquad y^2 + (x^2+1)^2 - 820 = 0
\\[4mm]
%
\end{aligned}
$$

On the second curve,
$$
y^2+(x^2+1) = 838860\ ,
$$
we can more or less do the same. First, the curve is birational equivalent to...

var('x,y');
def f(x,y):    return -y^2 + (x^2+1)^2 + 838860
f(x + 931/34, y).expand()
a, b, c, d, q = 1, 1862/17, 2601439/578, 808030727/9826, 1369043/1156
Q = 2*q
a1, a2, a3, a4 = d/q, c-d^2/Q^2, Q*b, -Q^2*a
a6 = a2*a4
E = EllipticCurve(QQ, [a1, a2, a3, a4, a6])
rk = E.rank()
print("E is an elliptic curve of rank %s" % rk)
print("Generators:")
for P in E.gens():
    print("\t", P.xy())
for T in E.torsion_points():
    if T == E(0):    continue
    print("\t", T.xy())

And we get again a curve of rank two!

E is an elliptic curve of rank 2
Generators:
     (-1369043/578, 0)
     (-43316325/28322, -41113249906388/614068441)
     (-867917/578, -106213097760/1369043)

A better form is...

sage: E.minimal_model()
Elliptic Curve defined by y^2 = x^3 - x^2 - 209715*x over Rational Field

Then some points on $y^2 = (x^2+1)^2 + 838860 are:

 solutions = []
 P1, P2 = E.gens()
 T = E.torsion_points()[1]
 for j1 in [-2..2]:
     for j2 in [-2..2]:
         for n in [0..1]:
             R = j1*P1 + j2*P2 + n*T
             if R == E(0):    continue
             X, Y = R.xy()
             if Y == 0:    continue
             U = (Q*(X+c) - d^2/Q) / Y
             V = -q + U*(U*X - d)/Q
             UU, V = abs(U + 931/34), abs(V)
             if (UU, V) in solutions:    continue
             solutions.append( (UU, V) )
             print(r"x &= \frac{%s}{%s}\\"
                   % (UU.numer(), UU.denom()))
             print(r"y &= \frac{%s}{%s}\\"
                   % (V.numer(), V.denom()))
             print("  &\\qquad (x^2+1)^2 + 838860 - y^2 = %s\n\\\\[4mm]\n%%"
                   % ((UU^2+1)^2 + 838860 - V^2))

$$
\begin{aligned}
x &= \frac{4640129248052012171295041}{56328368752930789692104}\\
y &= \frac{21729172452104012184427503136290818755933810040513}{3172885126366149767519655678919343059119946816}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{312505635338264918831323}{5635665178453622365027}\\
y &= \frac{101930516183122284005580603951145069243469807068}{31760722003634699217295111109664688832710729}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{7697107706380753981821512221}{186853640262015669462073894}\\
y &= \frac{67355326049228509693898695429792231527200858728844325683}{34914282879166763220702909129407082141202455516323236}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{275273488143971163955319670170828849327}{8630446951340393789955958138130027033}\\
y &= \frac{102015582307789500427230911928915609414149466194469828884544309431564059167628}{74484614579900697494445334607256438244445639280727623918729596137310783089}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3533796460790528465936090156525696152253306912169010481}{142147419283298414256273589168805617982980986867368204}\\
y &= \frac{22336879382547877312447713567495156125129879126864659964611288771826011654660047981168198612554244346982363633}{20205888808901837866902182036543346606858038621737120342109726603597659188209906227548837538265889310185616}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{66981948434235}{4493946651194}\\
y &= \frac{19038090760503750938241601891}{20195556503777767101625636}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{9827459237}{1028878515}\\
y &= \frac{974459905490087541356}{1058590998628605225}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{369725522175}{86671374244}\\
y &= \frac{6881632670450531112199441}{7511927113343506571536}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{10885438126521930467}{10771530642536438565}\\
y &= \frac{106267543905924773766136574076877880671436}{116025872383101461045183783794019259225}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3106512218127349796388060274811685}{494034186048468200762637842991614}\\
y &= \frac{223760671475152068058777277411257298563715773166193274657640742778051}{244069776984572492223657898435483022217396770912501687097274324996}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3382515311}{218686124}\\
y &= \frac{45283088549214408113}{47823620830143376}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{3017}{143}\\
y &= \frac{20832748}{20449}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{408358573}{11636677}\\
y &= \frac{207929610753260668}{135412251602329}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{64609139182230553631}{1409763762048977264}\\
y &= \frac{4555779364380348768652797926075775093953}{1987433864786485387900613108388925696}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{544306536341}{6280755390}\\
y &= \frac{298503660402510832364131}{39447888269014052100}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{1215}{7}\\
y &= \frac{1476956}{49}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
x &= \frac{931}{34}\\
y &= \frac{1369043}{1156}\\
  &\qquad (x^2+1)^2 + 838860 - y^2 = 0
\\[4mm]
%
\end{aligned}
$$

Note: I'll try to look closer, and see why we get gurves of rank two...


答: When $z=x^2+y^2=r^2$ and $z=\sqrt{2-x^2-y^2}=\sqrt{2-r^2}$ intersect each other, you will have $$r=1$$ This means that $z=1$ and $\tan(\phi)=1$. So the range of changing for $\phi$ will be $\pi/4\leq\phi\leq\pi$. While typing, I saw @Mhenni did it completely, so I am ending. :-)

Title: Using Spherical coordinates find the volume:

Body:
Inside the surfaces $z=x^2+y^2$ and $z=\sqrt{2-x^2-y^2}$

I integrated over the ranges:

$0 \leq \theta \leq 2\pi$

$ 0 \leq \phi \leq \frac{\pi}{2}$

$0 \leq r \leq \sqrt{2}$

I get $\frac{\pi}{2}(4\sqrt{2} -4).$

There answer is the same except a $-\frac{7}{2}$ instead of the 4 at the end. Obviously I'm missing a 1\2 but I seems, I can not find it.


问:
Simultaneous diagonalizability of infinitely many endomorphisms in infinite dimension
Let $V$ be a vector space and let $(f_i)_{i \in I}$ be a family of endomorphisms of $V$.

Question.
Suppose that each $f_i$ is diagonalizable and that $f_i$ commutes with $f_j$ for any two indices $i$ and $j$.
Is the family $(f_i)_{i \in I}$ simultaneously diagonalizable?

I’m using here the following notion of simultaneous diagonalizability.

The family $(f_i)_{i \in I}$ is simultaneously diagonalizable if we have the decomposition
$$
    V = \bigoplus_{(\lambda_i)_{i \in I}} \operatorname{Eig}( (f_i)_{i \in I}, (\lambda_i)_{i \in I} )
  $$
where $\operatorname{Eig}( (f_i)_{i \in I}, (\lambda_i)_{i \in I} ) = \{ v \in V \mid \text{$f_i(v) = \lambda_i v$ for every $i \in I$}\}$ is the common eigenspace of the family $(f_i)_{i \in I}$ for the eigenvalues $(\lambda_i)_{i \in I}$.
Equivalently, $V$ admits a basis consisting of common eigenvectors of the $f_i$.

It is (well-)known that the the answer to the question is yes if

$I$ is finite;

$V$ is finite-dimensional; or more generally

if the linear subspace of $\operatorname{End}(V)$ spanned by the $f_i$ is finite-dimensional.

But I don’t know what happens in general if both $I$ is infinite and $V$ is infinite-dimensional.

答:
Surprisingly (to me anyway), this is false! Let $V$ be the vector space $k^X$ of functions on the Cantor set $X = \{ 0, 1 \}^{\mathbb{N}}$. $V$ is in fact a commutative $k$-algebra with respect to pointwise multiplication, so we can define commuting families of operators on $V$ via pointwise multiplication. Consider the countable family of functions given by each coordinate projection $p_i : \{ 0, 1 \}^{\mathbb{N}} \to \{ 0, 1 \} \subset k$; via pointwise multiplication these give a commuting family of endomorphisms $f_i$. Each $f_i$ is idempotent and hence diagonalizable; the only eigenvalues are $0$ and $1$, and $V$ splits up into a direct sum of the $0$-eigenspace and $1$-eigenspace, which consist of the kernel and image of $f_i$ (the functions supported on the sequences whose $i^{th}$ coordinate is $0$, and the functions supported on the sequences whose $i^{th}$ coordinate is $1$, respectively).

A similar argument shows that any finite subset of the $f_i$ are simultaneously diagonalizable. However, the $f_i$ are not simultaneously diagonalizable. The problem is that a simultaneous eigenvector for all of the $f_i$ must be a function $X \to k$ which is supported at a single sequence $x \in X$ in the Cantor set, whose coordinates are determined by the eigenvalues of the $f_i$ acting on this eigenvector. So the subspace of $V$ spanned by simultaneous eigenvectors for all of the $f_i$ consists of the functions with finite support, which is not all of $V$.

This counterexample comes out of thinking about what goes wrong if you try to push the usual argument for $I$ finite to the case that $I$ is infinite. The usual argument is to decompose into eigenspaces for $f_1$, say, then consider the action of $f_2$ on each eigenspace and decompose into eigenspaces for that action, and so forth. But if you try to repeat this "and so forth" argument countably many times you may find that you are "cutting up" the eigenspaces "too finely." This is very vague but trying to make it rigorous leads naturally to the above example, if we think of the minimal nontrivial amount of "cutting up" each new endomorphism can do as cutting the eigenspaces in "half." The Cantor set is, somewhat loosely speaking, the universal object which can be cut in half infinitely many times.

Edit, 6/13/22: Actually there's a much easier way of thinking about this counterexample. It just comes from wondering: what if instead of a direct sum decomposition we had a direct product decomposition

$$V \cong \prod_{(\lambda_i)_{i \in I}} \text{Eig}((f_i)_{i \in I}, (\lambda_i)_{i \in I})$$

and then trying to construct the minimal infinite such thing. One way to say this is, the desired statement can't possibly always hold because it's not stable under passing to dual spaces.


答:
The best way to get a grasp of the situation (in my opinion) would be to graph the integrand over the given interval:

The vertical lines represent the bounds of your integration, and the red graph is $f(x) = |1-x|$. Recalling that the integral just gives the signed area under a curve, if you can find the area of the green trapezoid, you can find the value of the integral.

(Of course, you can also just note as Adam did that, on $(7/4,2)$, $|1-x| = x-1$, and so you can perform a more standard integral that way. I prefer this geometric explanation, but the algebraic idea is also nice.)

Title: how to integrate $ \int_\frac74^2 \lvert 1-x\rvert dx $ manully?

Body:
I calculated it on https://www.integral-calculator.com/
$$
\int_{7/4}^{2}\left\lvert 1 - x\right\rvert\mathrm{d}x
$$
and got the answer but I could not get the answer right when substituting.
Also, it would be very helpful if there were another way to do the integration instead of the solution given by the site


问:
Problem to understand a solution about polyhedron.
in a tutorial on of our "teachers" written down this exercise and I do not understand the solution:

Given is an polyhedron in standard form $K=\{ x \in R^n | Ax =b,x \geq 0 \}$ with matrix $A\in R^{n-1xn}% and $rank(A) =n-1$.
=> O is not empty and has at most two vertices.

My question would be, why is P not empty and why has P at most two vertices, I actually don't understand it?

Consider $n=2$ $A =(y_1,y_2)$ so we have $K=\{ x \in R^n| (y_1,y_2)x =b,x \geq 0 \}$ and this is a line. How can a line have two vertices?
There has to be a $c\in R^2$ with $c^Tx < c^Ty$ for all $y  \in K$ while $y\neq x$ and because its a line that should no be possible?

答:
A line segment has two vertices, namely the endpoints.  You could also have a ray, which has one endpoint (which is a vertex), or a line, which has none.

However, the teacher is wrong in one respect: the polyhedron could be empty.  For example, this is the case if all entries of $A$ are nonnegative and $b$ has some negative entries.


问:
Moment generating function applied in $2t$
I am having some trouble with this problem, adapted from Grimmet&Welsh:

If $X + Y$ and $X - Y$ are independent, show that
\begin{align}
M\left(2t\right) = M\left(t\right)^{3}M\left(-t\right),
\end{align}
where $X,Y$ are independet r.v. with mean $0$, variance $1$ and $M(t)$ finite.

How to prove it? Does $X$ and $Y$ needs to have normal distribution? Thank you!

答:
Hints:

$M(2t) = E[e^{2tX}]$

$M(t) = E[e^{tX}] = E[e^{tY}]$

$M(-t) = E[e^{-tY}]$

$2X = (X+Y) + (X-Y)$

If $U$ and $V$ are independent random variables, then $E[f(U)g(V)] = E[f(U)] E[g(V)]$.


问:
How to test the series for convergence involving repeating sequences in the denominator and numerator?
I have to test the following factorial for convergence:

$$\sum_{n=0}^\infty \frac{7\cdot15\cdot23\cdot\cdot\cdot\cdot\cdot(8n-1)}{8\cdot17\cdot26\cdot\cdot\cdot\cdot\cdot(9n-1)}$$

What I have attempted to do is use the ratio test as such:

$$\lim_{n\to 0}\lvert (\frac{8n+7}{9n+8})(\frac{9n-1}{8n-1})\rvert = 1$$

This means that the use of this test is inconclusive. I also feel as though the steps I made in this is incorrect. Something about my work does not seem right, I cannot piece it together.

答: Your ratio is wrong.  Note that $9n-1$ and $8n-1$ are factors in $a_n$ as well as $a_{n+1}$, so they should cancel out when you take $a_{n+1}/a_n$.


问:
Prove that any group of order $27$ is not simple
I'm stuck on this problem from my abstract algebra course:

Prove that if $G$ is a group with $|G|=27$, then $G$ is not simple.

First I noticed $|G|=27=3^3$. I thought I can use a statement I saw on the text book:

Given $H\leq G$ with $G$ finite and $|G:H|=p$ being $p$ the minimum prime number that divides $|G|$, then $H\unlhd G.$

This would prove that $G$ has a non-trivial normal subgroup, and that would mean $G$ is not simple. But in order to use this I need to prove first that my group $G$ has some subgroup of order $3^2$ (If I'm not wrong, this isn't trivial). So if my reasoning is right, I need to prove that any group of order $27$ has some subgroup of order $3^2$, and my problem will be solved. Am I right? How can I prove this last statement? Any help will be appreciated, thanks in advance.

答:
A well-known fact is that any $p$-group of order $p^n$ contains subgroups of order $p^i$ for each $i\le n$.

This is more than enough, since we can then take a subgroup of index $p$, which, by another well-known fact must be normal.

Thus no $p$-group with $n\gt1$ is simple.

There's an induction proof of the first fact that relies on the additional well-known fact that $p$-groups have nontrivial center (which can be seen by looking at the class equation).  Couple this with the fourth well-known fact that the center is always a normal subgroup, and we have a different proof.


问:
Why are the graph of $f(x) = (x^2)^\frac{1}{6}$ and the graph of $f(x) = x^\frac{1}{3}$ not the same?
Why is the graph of $f(x) = (x^2)^\frac{1}{6}$ not the same graph as $f(x) = x^\frac{1}{3}$?

Shouldn't they be the same because when you apply the exponent rules to the first equation, you get the same result as the second equation?

By the way, I am an Algebra 2 student in High School.

答:
When you square the quantity $x$, the negative sign disappears, and it doesn't come back if you raise it to the $\frac{1}{6}$ power. If you just take $\frac{1}{3}$ power, then the negative stays negative.

$((-1)^2)^\frac{1}{6}=1^\frac{1}{6}=1$, while $(-1)^\frac{1}{3}=-1$

Therefore, what you see on the first graph is the absolute value of the second graph.

This hyperlink contains a graph of the two equations. Note that when $x<0$ the red solid line is the blue dotted line reflected about the x-axis.


问:
Higher homotopy of knot theory space
Knot theory is the study of the path components of $\operatorname{Emb}(S^1 , \mathbb{R}^3)$. I know sometimes in geometric topology, these function spaces have very complicated path components, but relatively simple higher homotopy. Are the homotopy types of the path components of any particular knots known? And more generally, do we know the higher homotopy groups of $\operatorname{Emb}(S^1 , \mathbb{R}^n)$?

I imagine this is probably approached through embedding calculus, so I offer a meagre first step: if $n>3$ the codimension of the spaces is enough that the embedding tower converges. Moreover, it is well known that the first stage of the embedding tower is immersions, which in our case is equivalent to formal immersions. If I am not mistaken, this space $\operatorname{fimm}(S^1, \mathbb{R}^n)$ is equivalent to $\Omega_{free} S^{n-1}$ because both $S^1$ and $\mathbb{R}^n$ are parallelizable and the latter contractible.

Could anyone outline the known work or provide references?

答:
Tyrone: Plenty is known about the case you suggest. The space of long knots just removes some ambiguity.

Remember that there is a fiber sequence $\text{Emb}_*(S^1, S^3) \to \text{Emb}(S^1, S^3) \to (TS^3 \setminus 0)$, the last map sending $\gamma$ to $\gamma'(0)$ and the first is the space of embeddings with prescribed values for $\gamma(0)$ and the derivative there.

The first space is seen to be equivalent to the space of long knots in $\Bbb R^3$ by a standard fiber sequence relating the space of embeddings with specified first derivative to the space of embeddings with specified germ. So a full understanding of the space of long knots gives rise to a full understanding of the space of all knots. The former is more natural and easier to compute in terms of.

To my recollection Budney has completely finished these calculations (perhaps in the paper you cite).

Title: Higher homotopy of knot theory space


问:
Number of ways to write $2n$ as sum of two primes is unbounded
For all $k>0$, we are to show that

there exists an even number $2n > 0$ that can be written as sum of two primes in at least $k$ ways.
($8=3+5=5+3$ counts as one way)

I tried to use pigeonhole principle to "fill" odd numbers in the lower half of $2n$ with primes, i.e. to show $\pi(2n)-[\frac{n}{2}]$ can be arbitrarily large, but it turns out to diverge to negative infinity instead.

Of course, requiring the entire half to be primes is an overkill, but I don't know how to rephrase the problem into a manageable form.

It probably has something to do with Prime Number Theorem, but I am not sure.

Thank you for any help.

答: You're on the right track with the pigeonhole principle, but your counting is wrong; you want to get a 'quadratic-like' term to show unboundedness. Try this approach instead: consider how many unordered pairs of primes $\leq n$ there are; call this $f(n)$. Now, note that there are only $n$ even numbers less than $2n$. If each of these could be written as the sum of primes in $\leq C$ different ways for some constant $C$, what would that say about $f(n)$? Can you find a contradiction from here?


Title: How to test the series for convergence involving repeating sequences in the denominator and numerator?

Body:
I have to test the following factorial for convergence:

$$\sum_{n=0}^\infty \frac{7\cdot15\cdot23\cdot\cdot\cdot\cdot\cdot(8n-1)}{8\cdot17\cdot26\cdot\cdot\cdot\cdot\cdot(9n-1)}$$

What I have attempted to do is use the ratio test as such:

$$\lim_{n\to 0}\lvert (\frac{8n+7}{9n+8})(\frac{9n-1}{8n-1})\rvert = 1$$

This means that the use of this test is inconclusive. I also feel as though the steps I made in this is incorrect. Something about my work does not seem right, I cannot piece it together.

Body:
You have
$$a_n =\frac {\prod_{i=1}^n(8i-1) } {\prod_{i=1}^n(9i-1) }=\left(\frac{8}{9}\right)^n\,\frac{\Gamma \left(\frac{8}{9}\right)}{\Gamma \left(\frac{7}{8}\right)}\,\frac{\Gamma \left(n+\frac{7}{8}\right)}{\Gamma \left(n+\frac{8}{9}\right)}$$
$$\frac {a_{n+1}}{a_n}=\frac{8 n+7}{9 n+8} \quad \to \frac{8}{9}$$


问:
prove inequality $\frac{\sin \pi x}{\pi x} \geq \frac{1-x^2}{1+x^2}$ for all R
Let $F(x) = \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} $, is even, so just prove $x > 0$.

When $ x \geq 1 $, $\frac{\sin\pi x}{\pi x} \leq 1$, so
$$ \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} \\= -\frac{\sin\pi (x-1)}{\pi(x-1)} \frac{x-1}{x} - \frac{1-x^2}{1+x^2} \\ \geq \frac{1-x}{x} - \frac{1-x^2}{1+x^2} \\ = \frac{(x-1)^2}{x(1+x^2)} \\ \geq 0 $$

So how to prove it when $0 < x <1 $?

Body:
Equivalent problem: Let $0 < y < \pi$. Prove that $\frac{\sin y}{y} \ge \frac{\pi^2 - y^2}{\pi^2 + y^2}$.

It suffices to prove that, for all $y$ in $[0, \pi]$,
$$\sin y - y \frac{\pi^2 - y^2}{\pi^2 + y^2} \ge 0.$$
Denote LHS by $g(y)$. We have the following result. The proof is given at the end.

Fact 1: If $y\in (0, \pi)$ satisfying $g'(y) = 0$, then $g(y) \ge 0$.

Now, recall that a continuous function on a bounded and closed interval achieves its minimum
at either endpoints or the interior points with zero derivative.
Thus, by Fact 1, noting also that $g(0) = g(\pi) = 0$, we have $g(y) \ge 0$ on $[0, \pi]$.
We are done.

$\phantom{2}$

Proof of Fact 1:
From $g'(y) = 0$, we have $\cos y = \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}$.
We have,
\begin{align}
&(\sin y)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & 1 - (\cos y)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & 1 - \left(\frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & \left(1 + \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)\left(1 - \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)
- \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & \frac{2\pi^2(\pi^2 - y^2)}{(\pi^2 + y^2)^2}\cdot\frac{2y^4 + 6\pi^2y^2}{(\pi^2 + y^2)^2}
- \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}
\Big[2\pi^2(2y^2 + 6\pi^2) - (\pi^2 - y^2)(\pi^2 + y^2)^2\Big]\\
=\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[y^6+ \pi^2y^4+(-\pi^4+ 4\pi^2)y^2 + ( - \pi^6 + 12\pi^4)\Big]\\
\ge\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[\pi^2y^4+(-\pi^4+ 4\pi^2)y^2 + ( - \pi^6 + 12\pi^4)\Big]\\
\ge\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[ 2 \sqrt{\pi^2y^4 \cdot (- \pi^6 + 12\pi^4)} + (-\pi^4+ 4\pi^2)y^2\Big]\tag{1}\\
=\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[ 2\pi \sqrt{- \pi^2 + 12} + (-\pi^2+ 4)\Big]\pi^2 y^2\\
\ge\ & 0
\end{align}
where we have used AM-GM inequality in (1).

We are done.


问:
Convergence of $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$

Does $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$ converge?

I've tried splitting it into $\int_{x=0}^1 x^8 e^{-\sqrt x} dx$ + $\int_{x=1}^\infty x^8 e^{-\sqrt x} dx$. The first one is a continuous function over a bounded interval so that converges however I'm struggling with the second one. I have tried using limit comparison with $e^{- \frac{\sqrt x}{2}}$ as a test function but ended up going in circles. I have also tried direct comparison and integration by parts.

Thanks

Morgzy

答:
Let $g(x)=\sqrt{x}-10\ln(x)$. Then $g$ is increasing for $x>400$.

$g(e^9)=e^{4.5}-90>0$. So $g$ is positive and increasing for $x>e^9>400$.

So, for $x>e^9$ we have that
\begin{eqnarray}
\sqrt{x}-10\ln(x)>0\\
\ln(x^{10})<\sqrt{x}\\
x^{10}<e^{\sqrt{x}}\\
x^8e^{-\sqrt{x}}<\frac{1}{x^2}
\end{eqnarray}

You should be able to finish from there.


问:
Relationships of Eigenvalues in Algebraic Closure
Suppose that $k$ is a field, and $A \in M_n(k)$  is a matrix that becomes diagonalizable over $\overline{k}$, the algebraic closure of $k$. Let $\lambda_1, \ldots, \lambda_n$ denote the (not necessarily distinct) eigenvalues of $A$ in $\overline{k}$.

Must each $\lambda_i$ be in $k$?

Must $\sum_i \lambda_i$ be in $k$?

Must $\sum_i \lambda_i^2$ be in $k$?

Must $\sum_{i<j} \lambda_i\lambda_j$ be in $k$?

Must $\sum_{i<j} (\lambda_i - \lambda_j)$ be in $k$?

I think #1 is almost certainly false, otherwise the algebraic closure wouldn't be necessary. For #2, I believe that it is true, since the sum of the diagonal entries (all elements of $k$) is equal to the sum of the eigenvalues. I am not even sure how to start for the rest.

答:
Hint: the characteristic polynomial of a matrix, $\chi_A(x) := \det(xI-A)$, has as roots the eigenvalues of $A$.  Can you express your sums in terms of this polynomial's coefficients?

(Think in terms of symmetric polynomials.  If your sums are not symmetric, you should look for counter-examples, perhaps with $k=\mathbb{R}$.)

Further notes: If $\lambda_1,\ldots,\lambda_k$ are the eigenvalues of $A$, then
$$\chi_A(x)= (x-\lambda_1)\cdots (x-\lambda_k).$$
The constant coefficient is then $\pm \lambda_1 \cdots \lambda_k$, and the $x$-coefficient is $$\pm\bigg((\lambda_2 \cdots \lambda_k)+(\lambda_1\lambda_3 \cdots \lambda_k)+(\lambda_1\lambda_2 \lambda_4 \cdots \lambda_k) + \cdots + (\lambda_1 \lambda_2 \cdots \lambda_{k-1}\bigg)$$
(It's easiest to ignore the exact signs.)  There are general formulas for these coefficients in terms of the roots of our polynomial, called the elementary symmetric polynomials.  It is a fundamental result that any symmetric polynomial (that is, a polynomial that remains unchanged by any permutation of it's inputs) can be written as a sum of symmetric polynomials.

For your problem, you'll want to use that the value of any symmetric polynomial lies in the field generated by the values of your elementary symmetric polynomials.  But the elementary symmetric polynomials show up as the coefficients of $\chi_A$, so they lie in the base field.


问:
The relationship between $\alpha^i\otimes \alpha^j$ and $v^iw^j$
In the book An introduction to manifolds by Tu, Loring W, there's an exercise:

Let $e_{1}, \ldots, e_{n}$ be a basis for a vector space $V$ and let $\alpha^{1}, \ldots, \alpha^{n}$ be its dual basis in $V^{\vee}$. Suppose $\left[g_{i j}\right] \in \mathbb{R}^{n \times n}$ is an $n \times n$ matrix. Define a bilinear function $f: V \times V \rightarrow \mathbb{R}$ by
$$
f(v, w)=\sum_{1 \leq i, j \leq n} g_{i j} v^{i} w^{j}
$$
for $v=\sum v^{i} e_{i}$ and $w=\sum w^{j} e_{j}$ in $V$. Describe $f$ in terms of the tensor products of $\alpha^{i}$ and $\alpha^{j}$, $1 \leq i, j \leq n$

And the answer says
$$
f=\sum g_{ij}\alpha^i\otimes \alpha^j
$$

I am getting confused about these notations. What is the relationship between $\alpha ^i \otimes \alpha ^j$ and $v^i w^j$? Is  $\alpha ^i \otimes \alpha ^j$ equal to $v^i w^j$? Hope someone could help me with clarifying these notations.

答:
For two linear functions $\gamma_1,\gamma_2 : V \to \mathbb R$ we define the bilinear map $\gamma_1 \otimes \gamma_2 : V \times V \to \mathbb R$ such that $(\gamma_1 \otimes \gamma_2)(v,w) = \gamma_1(v)\gamma_2(w)$. So, for any $(v,w) \in V \times V$ we have
$$f(v,w) = \sum_{1 \leq i, j \leq n} g_{ij} v^{i} w^{j} = \sum_{1 \leq i, j \leq n} g_{ij} (\alpha^i \otimes \alpha^j)(v,w) = \bigg( \sum_{1 \leq i, j \leq n} g_{ij} \alpha^i \otimes \alpha^j \bigg)(v,w)$$ and then $$f = \sum_{1 \leq i, j \leq n} g_{ij} \alpha^i \otimes \alpha^j.$$


问:
How do you solve this limit? $ \lim_{x\to \infty} x(e^{2/x} - 1 ) $
I was going through studying with L'Hospital's rule, and went through this question. I've tried hard to solve it, but always couldn't go through all the way to the answer...

$$\lim_{x\to \infty} x(e^{2/x} - 1 ) $$

Body:
$$\lim_{x\to \infty} x(e^{2/x} - 1 ) $$
$$\lim_{x\to \infty} \frac{(e^{2/x} - 1 )}{\frac{1}{x}} $$
$$\lim_{x\to \infty} \frac{-\frac{2e^{\frac{2}{x}}}{x^2}}{-\frac{1}{x^2}}$$ ($\frac{\infty}{\infty}\Rightarrow$L'hopital)
$$\lim_{x\to \infty} 2e^{\frac{2}{x}}$$
$$2\cdot 1$$
$$2$$


答:
HINT

\begin{align*}
\exp(x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \ldots & \Rightarrow \exp\left(\frac{2}{x}\right) = 1 + \frac{2}{x} + \frac{2^{2}}{2x^{2}} + \frac{2^{3}}{6x^{3}} + \ldots\\\\
& \Rightarrow \exp\left(\frac{2}{x}\right) - 1 = \frac{2}{x} + \frac{2^{2}}{2x^{2}} + \frac{2^{3}}{6x^{3}} + \ldots\\\\
& \Rightarrow x\left[\exp\left(\frac{2}{x}\right) - 1\right] = 2 + \frac{2^{2}}{2x} + \frac{2^{3}}{6x^{2}} + \ldots
\end{align*}

BONUS

If we make the change of variable $u = 1/x$, we get that
\begin{align*}
\lim_{x\to\infty}x\left[\exp\left(\frac{2}{x}\right) - 1\right] = \lim_{u\to 0;u\in(0,\infty)}\frac{\exp(2u) - 1}{u - 0} = (\exp(2u))'\Biggr|_{u = 0} = 2
\end{align*}

Title: How do you solve this limit? $ \lim_{x\to \infty} x(e^{2/x} - 1 ) $

Body:
I was going through studying with L'Hospital's rule, and went through this question. I've tried hard to solve it, but always couldn't go through all the way to the answer...

$$\lim_{x\to \infty} x(e^{2/x} - 1 ) $$


答:
The simplest way is by considering $x= \frac{1}{\frac{1}{x}}$. Therefore we get

$$
\begin{aligned}
\lim_{x \to \infty}x\left(e^{\frac{2}{x}}-1\right) &= \lim_{ x \to \infty}\frac{e^{\frac{2}{x}}-1}{\frac{1}{x}} \\
&=\lim_{ x \to \infty}\frac{\frac{-2}{x^2}e^{\frac{2}{x}}}{\frac{-1}{x^2}} \\
&=\lim_{x \to \infty} 2e^{\frac{2}{x}} \\
&= 2
\end{aligned}
$$
the second of equation of which is due to L'Hospital's rule.

Alternative solution without using L'Hospital's rule:

Simply by the linear approximation: $\lim_{ x \to \infty}\frac{e^x-1}{x}=1$, which is denoted by $e^x-1 \sim x$. Subtituting with $x = \frac{2}{t}$, we have $e^{\frac{2}{t}}-1 \sim \frac{2}{t}$, hence

$$
\lim_{x \to \infty} x\left(e^{\frac{2}{x}}-1\right)=\lim_{x \to \infty}x \cdot \frac{2}{x} = 2.
$$


答: I solve this limit as $\lim\limits_{x\to\infty}x\left(1+\frac2x+O\left(\frac1{x^2}\right)-1\right)=2.$


问:
Help solving a nonlinear first order ODE
Is it possible to find an exact solution for the following ODE?
$$ \frac{1}{\sqrt{1+y'^2}} + \frac{Ay}{2}-\frac{B}{y}=0$$
Given that $A$ and $B$ are positive constants. Any help is appreciated.

答:
I do not think that you could obtain $y(x)$ but swithching variables, you easuation is
$$\frac{1}{\sqrt{1+\frac{1}{x'^2}}} + \frac{Ay}{2}-\frac{B}{y}=0$$ leading to
$$x'= \pm \sqrt{\frac{\left(A y^2-2 B\right)^2}{\left(A y^2-2 B\right)^2-4 y^2}}=\pm \frac{A y^2-2 B } {\sqrt{(A y^2-2y-2 B)(A y^2+2y-2 B) } }$$ which is separable.

The  result would involve elliptic integrals of the first an second kinds.


问:
Is a negative Hessian (of a 1-D function of 2 variables) at some point sufficient to indicate a saddle point?
Is a negative Hessian (of a 1-D function of 2 variables) at some point sufficient to indicate a saddle point, or do you separately need to know that the point is a critical point?</p>

答: Take $f(x,y) = x^2y^2$. Then $\det(Hf) = -12x^2y^2$. This is negative for all $\{(x, y) \in \mathbb{R}^2: xy \neq 0\}$ and clearly these points are not all saddle points.


问:
Inductive proof that sum of reciprocals of odd natural numbers diverges?
I am trying to prove via induction that for any natural number $N$, there exists an $n$ such that:

$$
\sum_{i=1}^{n} \dfrac{1}{2i-1} > N
$$

Or

$$
1 + \dfrac{1}{3} + \dfrac{1}{5} + ... + \dfrac{1}{2n-1} > N
$$

I've started by attempting to use induction on $N$.  The basis step is easy.  Let $N=1$, and it is clear that $1 + \dfrac{1}{3} > N$, so $n = 2$ suffices to show the basis is true.

For the inductive step, I've started with this:

For some $k > 0$, show that

$$
1 + \dfrac{1}{3} + \dfrac{1}{5} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2(n+1)-1} + ... + \dfrac{1}{2(n+k)-1} > N + 1
$$

Now I've tweaked this inequality in a bunch of different ways, but I haven't been able to get closer to proving that it's true for a certain $k$.

The closest I've got is trying to show that for a given $k$,

$$
\dfrac{1}{2(n+1)-1} + ... + \dfrac{1}{2(n+k)-1} > 1
$$

Can anyone point me in the right direction?

答:
$$
\frac{1}{2(n+1)-1} > \frac{1}{2(n+2)-1} > \cdots > \frac{1}{2(n+k)-1}.
$$

Therefore
$$
\frac{1}{2(n+1)-1} + \frac{1}{2(n+2)-1} + \cdots + \frac{1}{2(n+k)-1}
> \frac{k}{2(n+k)-1}.
$$

The right-hand side will never be greater than $1.$ But if you can make it greater than, say, $\frac14,$ all you need to do is repeat the process four times and you will have a sum of terms that is greater than $1.$


答:
2., 3., 4. are symmetric polynomial in the eigenvalues, hence can be expressed using the elementary symmetric polynomials, which occur as coefficients in ...?

and 5: Consider the $2\times 2$ real matrix $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$

Title: Relationships of Eigenvalues in Algebraic Closure

Body:
Suppose that $k$ is a field, and $A \in M_n(k)$  is a matrix that becomes diagonalizable over $\overline{k}$, the algebraic closure of $k$. Let $\lambda_1, \ldots, \lambda_n$ denote the (not necessarily distinct) eigenvalues of $A$ in $\overline{k}$.

Must each $\lambda_i$ be in $k$?

Must $\sum_i \lambda_i$ be in $k$?

Must $\sum_i \lambda_i^2$ be in $k$?

Must $\sum_{i<j} \lambda_i\lambda_j$ be in $k$?

Must $\sum_{i<j} (\lambda_i - \lambda_j)$ be in $k$?

I think #1 is almost certainly false, otherwise the algebraic closure wouldn't be necessary. For #2, I believe that it is true, since the sum of the diagonal entries (all elements of $k$) is equal to the sum of the eigenvalues. I am not even sure how to start for the rest.


答:
the first two terms are both bigger then $1/4.$   Next two terms, denominators 5,7, each reciprocal is bigger then $1/8$ sum is bigger then $1/4.$  Next denominators 9,11,13,15, each bigger than $1/16,$ sum is bigger than $1/4.$  Then denominators 17, 19, 21, 23, 25, 27, 29, 31, each reciprocal bigger than $1/32,$  the sum of eight terms is bigger than $1/4.$

And so on

Title: Inductive proof that sum of reciprocals of odd natural numbers diverges?

Body:
I am trying to prove via induction that for any natural number $N$, there exists an $n$ such that:

$$
\sum_{i=1}^{n} \dfrac{1}{2i-1} > N
$$

Or

$$
1 + \dfrac{1}{3} + \dfrac{1}{5} + ... + \dfrac{1}{2n-1} > N
$$

I've started by attempting to use induction on $N$.  The basis step is easy.  Let $N=1$, and it is clear that $1 + \dfrac{1}{3} > N$, so $n = 2$ suffices to show the basis is true.

For the inductive step, I've started with this:

For some $k > 0$, show that

$$
1 + \dfrac{1}{3} + \dfrac{1}{5} + ... + \dfrac{1}{2n-1} + \dfrac{1}{2(n+1)-1} + ... + \dfrac{1}{2(n+k)-1} > N + 1
$$

Now I've tweaked this inequality in a bunch of different ways, but I haven't been able to get closer to proving that it's true for a certain $k$.

The closest I've got is trying to show that for a given $k$,

$$
\dfrac{1}{2(n+1)-1} + ... + \dfrac{1}{2(n+k)-1} > 1
$$

Can anyone point me in the right direction?


问:
Continuous homomorphisms between Lie groups are smooth
I am reading the proof of the fact that continuous homomorphisms between Lie groups are smooth from these lecture notes:  https://warwick.ac.uk/fac/sci/maths/people/staff/weiyi_zhang/ma4e0.pdf

The result is Theorem 2.6.11 in the notes and is on page 30.

After showing that $p$ is smooth, bijective and a group homomorphism, how does it follows that $dp$ has constant rank and hence must be bijective?

答: If $p:G\to H$ is any smooth homomorphism between Lie groups, then $dp$ has constant rank.  Indeed, for each $g\in G$, $$p\circ L_g=L_{p(g)}\circ p$$ since $p$ is a homomorphism (where $L_g$ denotes left translation by $g$).  Taking the differential at $e$ on both sides, we get $$dp_g\circ d(L_g)_e=d(L_{p(g)})_e\circ dp_e.$$ Since $L_g$ and $L_{p(g)}$ are diffeomorphisms, this means $dp_g$ and $dp_e$ have the same rank.