问:
Is it possible to prove that $\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^a}\geq 0$ for $a\geq 1$ and $x\in [0,\pi]$?
I was trying to prove that
$$\sum_{k=1}^{\infty}\frac{\sin(kx)}{k^a}\geq 0,$$

for $0\leq x\leq \pi$, and $a\geq1$.

For $a$ being an odd integer, this is not really a problem, as the sums may then be expressed using Bernoulli polynomials. However, it seems that the assumption is true for all (real) values of $a$, not just integers, so I tried to split the sums and estimate the remainders, but unfortunately to no avail.

Does anyone have an idea on how to tackle this problem?

Thank you very much in advance.

答:
We can assume that $x\neq 0$. Then your sum is
$$
\Im \sum\limits_{k = 1}^\infty  {\frac{{e^{ixk} }}{{k^a }}}  = \Im \operatorname{Li}_a (e^{ix} ) = \frac{1}{{\Gamma (a)}}\Im \int_0^{ + \infty } \!\!\frac{{t^{a - 1} }}{{e^{t - ix}  - 1}}dt = \frac{\sin x}{{\Gamma (a)}}\int_0^{ + \infty } \!\!\frac{{t^{a - 1} e^t }}{{e^{2t}  - 2e^t \cos x + 1}}dt ,
$$
which has the desired property since $
e^{2t}  - 2e^t \cos x + 1 > e^{2t}  - 2e^t  + 1 = (e^t  - 1)^2  > 0$ for $t>0$. Here $\operatorname{Li}_a$ denotes the polylogarithm, see http://dlmf.nist.gov/25.12.ii.


问:
Prove that a subsequence converges.

Suppose that $b_n^{2} \rightarrow4$. Prove that $b_n$ has a
subsequence that converges to $2$ or $-2$.

I'm not really sure on the approach, I see that simply taking the root yields something similar, but doesn't say much about a subsequence and doesn't really seem like a proof.

答:
If $b_n \ge 0$ then
$$
 |b_n^2 -4| =  |b_n+2| \cdot |b_n-2| \ge 2 |b_n-2| \implies |b_n-2| \le \frac 12 |b_n^2 -4|
$$
and if $b_n < 0$ then
$$
 |b_n^2 -4| =  |b_n-2| \cdot |b_n+2| \ge 2 |b_n+2|\implies |b_n+2| \le \frac 12 |b_n^2 -4| \, .
$$
At least one of these cases must occur infinitely often, i.e. there is a subsequence $(b_{n_k})$ of non-negative elements (which converges to $+2$) or there is a subsequence $(b_{n_k})$ of negative elements (which converges to $-2$).

ContentLicense: CC BY-SA 4.0

Title: Prove that a subsequence converges.


答:
Since $(b_n^2)$ is convergent, $(b_n^2)$ is bounded, hence $(b_n)$ is bounded. Bolzano-Weierstraß gives: there is a subsequence $(b_{n_k})$ which is convergent. Let us denote the limit of this subsequence by $b$.

Then $ b_{n_k}^2 \to b^2$ as $k \to \infty.$

But $ (b_{n_k}^2)$ is a subsequence of $(b_n^2)$, hence  $ b_{n_k}^2 \to 4$ as $k \to \infty.$

This gives $b^2=4.$

Title: Prove that a subsequence converges.

Body:
Suppose that $b_n^{2} \rightarrow4$. Prove that $b_n$ has a
subsequence that converges to $2$ or $-2$.

I'm not really sure on the approach, I see that simply taking the root yields something similar, but doesn't say much about a subsequence and doesn't really seem like a proof.


问:
continuity with polar coordinates same as sequential continuity?
I got a function:

$f(x,y)= \begin{cases}\frac{x^3-3\,x\,y^2}{x^2+y^2} &(x,y) \neq 0\\
0&(x,y) = 0
\end{cases}$

substitution polar coordinates ($x = r\,\cos(\varphi)$, $y = r\,\sin(\varphi)$) I get:

$f(x,y) = r\,\cos(3\,\varphi)$. Supposly this has to be continuous, because:

for $r_n \xrightarrow{} 0$ the function approaches zero: $\underset{n \to \infty}{\lim} r_n \,\cos(3\varphi) = 0$.

However: If I chose $(x_n,y_n) = (\frac{1}{n},\frac{1}{n})\xrightarrow{} (0,0)$, I also get:

$\underset{n\to\infty}{\lim} f(x_n,y_n) = \underset{n\to\infty}{\lim} -\frac{1}{n} = 0 $. Does this also proof continuity at $0$?

答:
$f(x,y)=\frac{x^3-3xy^2}{x^2+y^2}$

Use the change of variables given by $x=r\cos \theta,y=r \sin \theta$
notice that When $(x,y)\to(0,0)$ then $r\to 0$.Remplazing our change of variables and symplifying the expression you should get.

$$\lim_{r \to 0} r\left(cos^3 \theta -3\sin^2\theta \cos \theta\right)=0$$

Which is  unique for different $\theta$, and then $f$ is continuous at $(0,0)$


问:
Probably a really easy question...does the following ring exist?
This may be an easy couplet of questions but I have never seen a ring of this form ever brought up:

Does the following exist as a ring:

$R=\mathbb{Z}[x]/\{x^2=0; 7x = 0\}$

2.Does $R$ have the form $R=\{ax+b; b \in \mathbb{Z}; a \in \mathbb{F}_7\}$,
where $a'x+b'$ and $ax+b$ are distinct elements as long as either $a' \not = a$ or $b'\not = b$. Put another way, for example, $6x$ and $13x$ are the same element in $R$, but $6$ and $13$ are distinct elements in $R$.

If the answer to either is NO, then why not.

Thanks for looking at this!

答:
Starting with an abelian group $G$, there is a construction called the trivial extension $\mathbb Z\ltimes G$ which creates a ring whose underlying abelian group is $\mathbb Z\times G$ and whose multiplication is defined by

$(n, a)(m, b)=(nm, n\cdot b+m\cdot a)$

This is isomorphic to your ring via the map $R\to \mathbb Z\ltimes C_7$ given by

$\sum_{i=0}^k n_ix^i + I\mapsto (n_0, n_1\cdot c)$

Here I have used $I$ to denote your ideal $(x^2, 7x)$, and I have chosen $c$ to be a generator of the additively written cyclic abelian group $C_7$.


问:
What does XY represent in an equation that contains the square of X and Y?
For instance, an equation of an ellipse has equation of the form $\frac { x^2} {2^2}+\frac {y^2} {1^2}=1 $ is an ellipse centered at the origin with a horizontal major axis.
However, I noticed that equations of ellipses that contain xy term don’t have horizontal major axises and they look like a rotated ellipse.

my question

How does xy term affect the graph ? And does it have this affect to all other kind of graphs ? like hyperbola,circles,...
And if there is a Geometrical interpretation for this Term.

答:
$$f(x,y)=ax^2+by^2+cxy$$

Let apply the rotation $\begin{cases}x=\cos(t)u+\sin(t)v\\y=-\sin(t)u+\cos(t)v\end{cases}$

Then $f(u,v)=\underbrace{(\cdots)}_Au^2+\underbrace{(\cdots)}_Bv^2+\Big(\underbrace{c\,\cos(2t)-(a-b)\sin(2t)}_C\Big)uv$

If we take $t=\frac 12\tan^{-1}(\frac c{a-b})$ then $C=0$ and we get $$f(u,v)=Au^2+Bv^2$$

Which is an ellipse with axis parallel to coordinate axis in $(u,v)$ base.

So as long as the sign of $A,B$ is unchanged then the effect of the $xy$ term in the equation is some rotation combined with some distortion. Since it changes also the values of $A,B$ the ellipse get elongated either in one coordinate or the other.

But beware, it may happen that if $c$ is large enough compared to $a,b$ then one or both coefficients $A,B$ will change sign.

So instead of an ellipse $|A|u^2+|B|v^2=cst$ you might get an hyperbola $|A|u^2-|B|v^2=cst$ (try it on Desmos, https://www.desmos.com/calculator/orpaclx455 make $c$ vary).


问:
Number of elements of prime order $p$ in $\mathrm{SL}_2(\mathbb{F}_q)$
Let $p,q$ be prime numbers.
It is known that there is one element of order $2$ in $\mathrm{SL}_2(\mathbb{F}_q)$.

How many elements of order $p>2$ are there in $\mathrm{SL}_2(\mathbb{F}_q)$?

ContentLicense: CC BY-SA 3.0

Body:
$(p-1)q(q+1)/2$ if $p|q-1$.

$(p-1)q(q-1)/2$ if $p|q+1$.

$p^2-1$ if $p=q$.


问:
Third-order Runge Kutta method
Prove that the third-order Runge-Kutta method reproduces the Taylor series of the solution up to and including terms inh3for any differential equations.

$$x(t+h) =x(t) + \frac19 (2 K_1 + 3 K_2 + 4 K_3)$$

where

$$\begin{aligned} K_1 &= h f(t,x)\\ K_2 &= h f \left(t + \frac12 h, x + \frac12 K_1 \right)\\ K_3 &= h f \left(t + \frac34 h, x + \frac34 K_2 \right)\end{aligned}$$

I am really struggling with proving this. I need any help for that. Thank you.

答:
The Butcher tableau for the given method is
$$
\begin{array}{c|c}c&A\\\hline&b\end{array}
\qquad=\qquad
\begin{array}{c|ccc}
0&\\
\frac12&\frac12\\
\frac34&0&\frac34\\
\hline
&\frac29&\frac39&\frac49
\end{array}
$$
For 3-stage methods you get order conditions (see for instance Butcher 2008)
\begin{align}
\text{order $1$}&:& b_1+b_2+b_3&=1\\
\text{order $2$}&:&  b_2c_2+b_3c_3&=\tfrac12\\
\text{order $3$}&:&  b_2c_2^2+b_3c_3^2&=\tfrac13\\
 & &  b_3a_{32}c_2&=\tfrac16\\
\end{align}

It can be easily checked that these conditions are satisfied.

Title: Third-order Runge Kutta method


问:
Solving integral of a function of derivative
I'm having problems in solving the following integral:

$$\int_0^z s(1-s)\left[\frac{A'(s)}{A(s)}\right]^2 ds - \int_0^z s(1-2s)\left[\frac{A'(s)}{A(s)}\right]ds$$

Where $A'(s)$ denote the first derivative of $A(s)$ with respect to s.Here $A(s)$ is an extreme value generator, for extreme value copulas, but that shouldn't be relevant for the purpose.

This integral should be simplified as follows:

$$\int_0^z \frac{s(1-s)}{A(s)}dA'(s) -z(1-z)\frac{A'(z)}{A(z)}$$

I actually tried several times, changing measure for logs and so on, but I don't see such solution!
Does anybody have any idea? Probably I'm forgetting some rule from calculus, but I did not find anything on integral of function of derivatives.
As always, any help would be very much appreciated. Many thanks!

答:
The solution is as simple as that:
Let's integrate by part the first component w.r.t. $g(s) = (s-s^2) A'(s) $ and $f'(s) = \frac{A'(s)}{A^2(s)}$, from which we get,
$$=-(s-s^2)\frac{A'(s)}{A(s)} |_0^z  + \int_0^z [(1-2s)\frac{A'(s)}{A(s)} + (s-s^2)\frac{A''(s)}{A(s)} ]ds - \int_0^z (1-2s)\frac{A'(s)}{A(s)}$$
From which, after simplification, letting $A''(s)ds = dA'(s) $, and considering that $A(0)=1$ (by the bundary condition in comment above) we get
$$\int_0^z \frac{s(1-s)}{A(s)}dA'(s) - z(1-z) \frac{A'(s)}{A(s)}$$

Which is the desired expression.

$\square$


问:
Integral $\int \frac{\cos x }{2+\sin 2x} dx$
I have tried to find antiderivative of
$$ \frac{\cos x\  }{2+\sin 2x} $$
using the variable change $t= \cos x -\sin x$ with sin $\sin2x=2\sin x\cos x $. But i don't come up to its closed-form result as shown below.

How can I find its antiderivative? Thanks in advance

答:
Substitute $t=x-\frac\pi4$ to integrate
\begin{align}
\int \frac{\cos x\ }{2+\sin 2x}dx=&\int \frac{\cos (t+\frac\pi4)\ }{1 +2\cos^2t }dt\\
=&\frac1{\sqrt2}\int \frac{\cos t}{3 -2\sin^2t }dt
-\frac1{\sqrt2}\int \frac{\sin t}{1 +2\cos^2t }dt\\
=& \frac1{2\sqrt3}\tanh^{-1}(\sqrt{\frac23}\sin t)+\frac12\tan^{-1}(\sqrt2\cos t)+C
\end{align}


问:
Numerically evaluating $\iiint_{C} \frac{ {\rm d}x \, {\rm d}y \, {\rm d}z}{x^2 +y^2 + z^2}$

Numerically evaluate $$\iiint_{C} \frac{ {\rm d}x \, {\rm d}y \, {\rm d}z}{x^2 +y^2 + z^2}$$ where $$C = [-1,1] \times [-1,1] \times [-1,1] = [-1,1]^3$$

I do not expect an analytic solution. I would be satisfied with a numeric value. But even this is not so easy.

How to deal with the singularity at the origin? One idea is to first integrate over the unit ball, which can be done analytically, then use monte carlo to integrate over the rest irregular region. But I am not sure whether this will converge fast enough.

答:
Here is a triple integral attempt in spherical coordinates. It does not lead to an analytic answer and I had to evaluate the integral with the help of WolframAlpha.

We consider the cube is made of $8$ unit cubes in $8$ octants. The unit cube in the first octant is bound by,

$x = 0, x = 1, y = 0, y = 1, z = 0,  z= 1$. We again divide this into two parts by plane $x = y$ and integral in the sub-region bound by $y = 0, x = 1, x = y, z = 0, z = 1$ can be evaluated as below using spherical coordinates -

$x = \rho \cos \theta \sin \phi, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi$. $x^2+y^2+z^2 = \rho^2$.

$0 \leq \theta \leq \frac{\pi}{4}$.

i) when we are bound by plane $z = 1$

$z = 1 \implies \rho = \sec \phi$

To find the upper bound of $\phi$, let's look at the intersection of plane $z = 1, x = 1$ which is given by,

$\rho = \sec\phi = \sec\theta \csc\phi \implies \phi = \arctan(\sec\theta)$

So, $0 \leq \phi \leq \arctan(\sec\theta)$

$I_1 = \displaystyle \int_{0}^{\pi/4} \int_0^{\arctan(\sec\theta)} \int_0^{\sec\phi} \sin\phi \ d\rho \ d\phi \ d\theta \ \approx 0.319755$

ii) when we are bound by plane $x = 1$

$x = 1 \implies \rho = \sec \theta \csc \phi$

$\arctan(\sec\theta) \leq \phi \leq \frac{\pi}{2}$

$I_2 = \displaystyle \int_0^{\pi/4}
 \int_{\arctan(\sec\theta)}^{\pi/2} \int_0^{\sec\theta \csc\phi} \sin\phi \ d\rho \ d\phi \ d\theta \approx 0.639510$

So considering symmetry, the integral over the entire cubic region evaluates to,

$I = 16(I_1 + I_2) \approx 15.348$.


问:
Spiralling Sin Wave Equation
I am looking to plot a function which has the following shape in python shown in the image below: (apologies for the poor quality). Is there a function that has this form? The only way I can describe it is a sinusoidal spiralling function.

答: An equation on parametric form :

Title: Spiralling Sin Wave Equation

Body: An equation on parametric form :


问:
Summation simpilification process
Here is my summation:

$$\sum_{n=0}^{N/2-1}\frac1{25+nB}=A$$
where $0.01\le A\le2$, $10\le N\le2000$ and I need to find the $B$ for different values of $N$ I calculate this summation online (check here)
But the Digamma function makes the output function complex and it is not easy to solve with a calculator or even by hand. I need to calculate the result of this summation with hand easily. Is there any way to convert my summation to simple formula? Something like this:

\begin{align*}
\sum^n_{k=1}k&=\frac{n(n+1)}{2}\\[5pt]
\sum^n_{k=1}k^2&=\frac{n(n+1)(2n+1)}{6}\\[5pt]
\end{align*}

答:
Exact Solution:

Consider the function:
$$
I'(x) = \sum_{k=0}^n\frac{1}{x+kb}
$$

If we integrate both sides:
$$
I(x) + C = \sum_{k=0}^n\log\left(x+kb\right) = \log\prod_{k=0}^n\left(x+kb\right) = \log\frac{(x+nb)!}{(x-1)!}
$$

In other words:
$$
I(x) + C = \log\Gamma(x+nb+1) - \log\Gamma(x)
$$

If we take the derivative on both sides once more:
$$
I'(x) = \psi(x+nb+1) - \psi(x),\quad\text{where}\quad
\psi(x) = \frac{d}{dx}\log\Gamma(x)
$$

Where $\psi(x)$ is known as digamma function. And that's why your summation depends on the digamma function, because its the derivative of $I(x)$ at $x=25$. Thus, there's no way to convert the sum to a simple formula, as that would be equivalent of saying there exists a simple formula for the digamma function. However, there's the possibility of using an approximation.

Approximate Solution:

Asymptotic expansion of the gamma function can be written like this:

$$\ln\Gamma(x) = x\ln x - x + \frac{1}{2}\log\frac{2\pi}{x} + O\left(\frac{1}{x}\right)$$
$$\psi(x) = \ln x - \frac{1}{2x} + O\left(\frac{1}{x^2}\right) $$

Thus, an approximate solution is:
$$
I'(x) = \ln (x+nb+1) - \ln (x) + \frac{1}{2x} - \frac{1}{2(x+nb+1)} + O\left(\frac{1}{x^2}\right)
$$
$$
I'(x) = \ln\left(1 + \frac{nb+1}{x}\right) + \frac{nb+1}{2x(x+1+nb)} +  O\left(\frac{1}{x^2}\right)
$$
$$
I'(x) = \ln\left(1 + \frac{nb+1}{x}\right) + O\left(\frac{1}{x^2}\right)
$$

=]

To translate this to your situation, let $A = I'(25)$, $b=B$, $x=25$ and $n = N/2-1$, and thus you get:
$$
A \approx \ln\left(1 + \frac{1-B}{25} + \frac{NB}{50}\right)
$$

=]

Title: Summation simpilification process


问:
Proof of Weak Law of Large Numbers with non-zero covariance

Let $ X_{1}, \cdots, X_{n} $ be a sequence of dependent random variables having the same finite mean $ \mu=\mathrm{E}\left(X_{1}\right), $ the same finite variance $ \sigma^{2}=\operatorname{Var}\left(X_{1}\right), $ and $ \left|\operatorname{Cov}\left(X_{i}, X_{j}\right)\right| \leq \frac{1}{(i-j)^{4}} $ for $ i \neq j . $ Show that $ \bar{X} $ converges in probability to $ \mu . $

I am stuck on the upper bound of the sum of the covariances: $\sum_{i=1}^{n} \sum_{j \neq i,j=1}^{n}|{\rm Cov}(X_i,X_j)|\le2 \cdot \sum_{k=1}^{n-1}(\frac{n-k}{k^4}) $

How to show that $n^2$ is a proper upper bound of the sum?

答: Since $\left\lvert\operatorname{Cov}(X_i,X_j)\right\rvert\leq (i-j)^{-4}\leq 1$, summing over all $n(n-1)$ ordered pairs $(i,j)$ with $i\neq j$ gives the easy upper bound $n(n-1)<n^2$.


问:
Change of bases (arithmetic)
How can we check if a certain fraction in a certain base is infinite dizime.

If it is a infinite dizime how can we quickly find the representation of that fraction in that base.
The method I have to find its representation is quite slow in some cases and I would like to know if there is a faster method.

The method I have used consistes: Let's say I want to change to base 5. I check if the number ($z$) is between $a*5^n<z<(a-1)*5^n,(0 \le a<5)$ then I subtract $(a-1)*5^n$ and move to a lower base of $5$.

答:
A fraction has an infinite repeat when the denominator (in lowest terms) has a prime factor that does not divide into the base.  In base $10$ the fractions that terminate are ones with denominators of the form $2^a5^b$.  If there is any other prime in the factorization of the denominator, the fraction will terminate.  If it terminates, it will have $\max(a,b)$ places because that is the smallest power of $10$ that $2^a5^b$ divides into.

Your approach to converting a whole number to another base works fine.  More commonly taught is to divide the number by $5$ and keep the remainder as the ones digit of the converted number.  Divide the quotient by $5$ and keep the remainder as the fives digit.  Keep going until you are done.  It is described here.

Title: Change of bases (arithmetic)


问:
Can a uniform convergent power series be invariant under rearrangement
Suppose $S$ is a power series, uniformly convergent. Now under rearrangement, does the sum stays the same? I know, if the series is absolute convergence then this is true, but under uniform convergence does it holds? I am assuming, the sum always converges, as we taking the domain to be the radius of convergence.

答:
Assume wlog the power series centered at $0$ (otherwise translate); then convergence at only one point $|z_0|=r>0$ implies absolute convergence on the disc $|z|<r$, so the question is kind of moot except for the boundary disc $|z|=r$ where the result is not true.

One can have a power series with radius $1$, uniformly convergent on $|z|=1$ but not absolutely convergent there (eg $\sum_{n \ge 1}\frac{e^{in \log n}}{n}z^n, |z| \le 1$ or for real coefficients just take the real part so
$\sum_{n \ge 1}\frac{\cos (n \log n)}{n}z^n, |z| \le 1$).

A convergent but non absolutely convergent series of complex numbers cannot be arbitrarily rearranged by the usual Riemann rearrangement theorem applied to the real or imaginary parts as one those at least cannot be absolutely convergent

(this last part is independent of any uniform convergence conditions or the fact the series is a power series as it applies to any point where the series is convergent but not absolutely convergent - the non trivial part is the existence of uniformly convergent power series that are not absolutely convergent on the boundary circle of convergence).


问:
Convex set also has convex closure (alternative proof)
While studying convex analysis, i tried to write a proof for corollary 2 which i do not know if it is indeed correct.

First statement(theorem):

Let $S$ be a convex set in $R^{n}$ with a nonempty interior. Let $x_{1} \in \operatorname{cl} S$ and $x_{2} \in$ int
S. Then $\lambda x_{1}+(1-\lambda) x_{2} \in$ int $S$ for each $\lambda \in(0,1)$. I understood the proof regarding this one.

Corollary 1:

Let $S$ be a convex set. Then int $S$ is convex.

Answered here: Convex set also has convex interior (Corollary)

Corollary 2:

Let $S$ be a convex set with a nonempty interior. Then cl $S$ is convex.

My attempt for corollary 2(alternative proof(?)):

For cl $S$ to be convex, one has to let $x_{1}$, $x_{2}$ $\in$ cl $S$ and show that $\lambda x_{1}+(1-\lambda) x_{2} \in cl(S)$.

Let $x_{1}$, $x_{2}$ $\in$ $int(S)$ $\implies$ $\lambda x_{1}+(1-\lambda) x_{2} \in$ int $S$ by corollary 1.

Furthermore, since $\operatorname{int}(S) \subset S \subset \operatorname{cl}(S)$ we conclude that $x_{1}, x_{2} \in cl(S)$ and that $\lambda x_{1}+(1-\lambda) x_{2} \in$ cl $S$. Therefore, $cl(S)$ is convex.

Original proof:

Let $x_{1}, x_{2} \in \operatorname{cl} S .$ Pick $z \in$ int $S$ (by assumption, int $\left.S \neq \varnothing\right) .$ By the
theorem, $\lambda x_{2}+(1-\lambda) z \in$ int $S$ for each $\lambda \in(0,1) .$ Now fix $\mu \in(0,1) .$ By the
theorem, $\mu \mathbf{x}_{1}+(1-\mu)\left[\lambda \mathbf{x}_{2}+(1-\lambda) \mathbf{z}\right] \in$ int $S \subset S$ for each $\lambda \in(0,1) .$ If we take
the limit as $\lambda$ approaches $1,$ it follows that $\mu \mathrm{x}_{1}+(1-\mu) \mathrm{x}_{2} \in cl(S),$ and the proof is complete.

From the book: Nonlinear Programming (Theory and Algorithms)

Can someone help me? I dont know if my proof is correct.

Thanks in advance, Lucas

答:
Your proof is not correct. You take $x_1,x_2 \in cl(S)$ and later you asume that $x_1,x_2 \in int(S).$

You can not asume that $x_1,x_2 \in int(S).$


问:
Correct way to write a symbol for the Jacobian matrix
Context: I'm writing an essay on the subject of hydrodynamics which deals with numerical methods to solve a system of partial differential equations (shallow water equations or Saint-Venant equations).
These can be written as a system of two equations or, alternatively, in a single line using vector notation:

\begin{equation}
    \frac{\partial U}{\partial t}+\frac{\partial F}{\partial U}\frac{\partial U}{\partial x}  = S(U)
\end{equation}

Where $F$ and $U$ are two vector functions with two components each. It's not important for the scope of this question to know what they represent physically, the important part is that both $F$ and $U$ are functions of $x$ and $t$ ($F=F(x,t)$, $U=U(x,t)$), but $F$ can be written as well as a function of the components of $U$ ($F=F(U)$).
I want to write the system in an even more compact way using the Jacobian:

\begin{equation}
    \frac{\partial U}{\partial t}+J\frac{\partial U}{\partial x}  = S(U)
\end{equation}

Where $J$ is the Jacobian matrix of the vector function $F$ not with respect to the independent variables $(x,t)$, but with respect to the components of $U$: i.e., $J$ is the matrix of all the partial derivatives of the components of $F$ with respect to $U$:

\begin{equation}
J =\frac{\partial F}{\partial U} = \begin{vmatrix}\frac{\partial F_1}{\partial U_1} && \frac{\partial F_1}{\partial U_2} \\ \frac{\partial F_2}{\partial U_1} && \frac{\partial F_2 }{\partial U_2}\end{vmatrix}
\end{equation}

Which is the correct notation I should use to write it? $J(F)$? $J(F(U))$? $J_U(F)$? I'm really confused because the Jacobian matrix is usually calculated using $(x,y,z)$ as independent variables, but this time I want to signal that I'm deriving with respect to another set of variables!

答:
I usually write $J_F (U)$

Inside the brakets I always put the variables we are deriving with respect to.

I put the F down below because it is the function I am deriving. Hope this helps!


答:
By completing the square, you can group $xy$ with another term, for instance

$$x^2 + xy + y^2 = 5 \iff (x^2 + xy + \tfrac14 y^2) + \tfrac34 y^2 = 5 \iff (x + \tfrac12y)^2 + \tfrac34 y^2 = 5.$$

This means that:

The transformation $(x,y) \mapsto (x + \frac12y, y)$ takes points satisfying $x^2+xy+y^2=5$ to points satisfying $x^2 + \frac34 y^2 = 5$: an axis-aligned ellipse we understand.

The inverse transformation $(x,y) \mapsto (x - \frac12y, y)$ takes the axis-aligned ellipse $x^2 + \frac34 y^2 = 5$ to the ellipse $x^2 + xy + y^2 = 5$.

The transformation $(x,y) \mapsto (x - \frac12y, y)$ is a skew transformation that preserves points on the $x$-axis but shifts points above it to the left (and points below it to the right). It's a convenient way to draw the ellipse $x^2 + xy + y^2 = 5$ in Paint or similar software, which has an axis-aligned-ellipse tool and a skew tool (but no arbitrary ellipse tool).

Here's a diagram of the axis-aligned ellipse $x^2 + \frac34 y^2 = 5$ and the transformed ellipse $x^2 + xy + y^2 = 5$:

Title: What does XY represent in an equation that contains the square of X and Y?

Body:
For instance, an equation of an ellipse has equation of the form $\frac { x^2} {2^2}+\frac {y^2} {1^2}=1 $ is an ellipse centered at the origin with a horizontal major axis.
However, I noticed that equations of ellipses that contain xy term don’t have horizontal major axises and they look like a rotated ellipse.

my question

How does xy term affect the graph ? And does it have this affect to all other kind of graphs ? like hyperbola,circles,...
And if there is a Geometrical interpretation for this Term.


问:
Area of intersection between two circles
Suppose you have 2 circles that intersect each other in such a way that each circle passes through the other's center. What is the area between the circle(or common area) i.e. area between the centres of the circles?

答:
One approach is to set up a Boolean function $f$  as

$$f(x,y) = \left\{
  \begin{array}{ll}
    1 & \text{if } x^2+y^2<r^2 \text{ and } (x-r)^2+y^2<r^2 \\
    0 & \text{else}.
  \end{array}\right.
$$

The area can then be expressed as

$$\int_0^r \int_{-r}^r f(x,y) \, dx \, dy.$$

Here's how to do this computation in Mathematica.

```<code>Integrate[
  Boole[x^2 + y^2 < r^2 && (x - r)^2 + y^2 < r^2],
 {x, 0, r}, {y, -r, r},
 Assumptions -> r > 0
]

(* Out: -((3*Sqrt[3] - 4*Pi)*r^2)/6 *)
</code>```

Body:
One approach is to set up a Boolean function $f$  as

$$f(x,y) = \left\{
  \begin{array}{ll}
    1 & \text{if } x^2+y^2<r^2 \text{ and } (x-r)^2+y^2<r^2 \\
    0 & \text{else}.
  \end{array}\right.
$$

The area can then be expressed as

$$\int_0^r \int_{-r}^r f(x,y) \, dx \, dy.$$

Here's how to do this computation in Mathematica.

Integrate[
  Boole[x^2 + y^2 < r^2 && (x - r)^2 + y^2 < r^2],
 {x, 0, r}, {y, -r, r},
 Assumptions -> r > 0
]

(* Out: -((3*Sqrt[3] - 4*Pi)*r^2)/6 *)


问:
Natural isomorphism $f^{-1}k_{YZ}\rightarrow k_{Xf^{-1}Z}$
For $f:X\rightarrow Y$ continuous map, and $Z$ a closed subset of $Y$. I want to constract the natural isomorphism $f^{-1}k_{YZ}\rightarrow k_{Xf^{-1}Z}$. Where $f^{-1}k_{YZ}$ is the pullback of the sheaf $k_{YZ}:=(\iota_{Z*}\circ\iota_Z^{-1})k_Y$ and $k_Y$ is the constant sheaf with stalk $k$. Similarly $k_{Xf^{-1}Z}:=(\iota_{f^{-1}Z*}\circ\iota_{f^{-1}Z}^{-1})k_X$ ($\iota$, is the inclusion).

The fact that both sheaves agree on the stalks is straightforward, but I am having difficulties trying to find the natural map.

Any hints?

答:
For convenience let $W$ denote $f^{-1}(Z)$.

First we show that $f^{-1}(k_Y)\cong k_X$. Let $T$ be the topological space with one point, and let $g:Y\rightarrow T$ be the unique continuous map. Then $k_Y$ is the pull-back $g^{-1}(k_T)$. So $f^{-1}k_Y\cong f^{-1}g^{-1}(k_T)\cong (g\circ f)^{-1}(k_T)$ is the constant sheaf on $X$.

Now by the adjunction between the pull-back and the push-forward we know that to have a morphism $f^{-1}(k_{YZ})\rightarrow k_{X,f^{-1}Z}$ is the same as to have a morphism $\iota_{Z*}\circ\iota_Z^{-1}(k_Y)\rightarrow f_*\circ\iota_{W*}\iota^{-1}_W\circ f^{-1}(k_Y)$. But $f\circ\iota_W=\iota_Z\circ f$, so the right-hand side is $\iota_{Z*}\circ f_*\circ f^{-1}\circ\iota_Z^{-1}(k_Y)$. Hence from the canonical adjunction morphism $\operatorname{id}\rightarrow f_*\circ f^{-1}$ we obtain the canonical morphism $\iota_{Z*}\circ\iota_Z^{-1}\rightarrow \iota_{Z*}\circ f_*\circ f^{-1}\circ\iota_Z^{-1}$.

Hope this helps.


问:
First deriative, multiplied by first derivative
I have one question. If I multiply 2 first deriatives, will I receive second derivative? Example:
$$\frac{\partial a}{\partial t} = \dot{a}$$
Is it true?:
$$\frac{\partial\dot{a}}{\partial t}=\ddot{a}$$

答:
Imagine $f(x)=x^2$, then we have that $f'(x)=2x$ and

$$ 2=f''(x)\neq\left(f'(x)\right)^2=4x^2$$


问:
How to find the first term common to 2 AP
Suppose there are 2 APs, AP1 and AP2 where the first terms are a1 and a2 respectively and the common difference is d1 and d2. How do you find a mathematical formula to define the AP of the common terms, i.e. their first term and the common difference, without doing random guessing

let's suppose the first term is A.P’s 3,7,11,... and the second 2,9,16,...

I know the common difference of the AP of common terms is d=LCM(d1,d2) = LCM(4,7)=28 but what about the first term.

Since the common term must have the value and are defined by their own respective APs equation

a1+(n1-1)d1= a2+(n2-1)d2

and so

3+(n1-1)4= 2+(n2-1)7

4n1-4=7n2

4(n1-1)=7n2

Now what? One thing I did was find the first n1 value divisible by 7(cause it's on the other side, LHS) and the first n2 value divisible by 4 (cause it's on the other side, the RHS). And it seems to work for me, but what if I couldn't take 4 out of the bracket like this time.

Is there are better way?

I think I need to go to Fundamental theorem of Arithmetics because I vaguely remember doing something like this

答:
Suppose you have one arithmetic progression $a+mc$ and another $b+nd$ where $a,b$ are the sequences' initial terms, $m,n$ are the indices of the sequences, and $c,d$ are the differences of the sequences respectively.  To simplify notation, I will treat sequences as starting from index zero, but this is easily adjusted if you insist on sequences starting from index 1.

I assume that by "first term common to both" you do not require that their index be the same in each sequence, that you are instead simply looking for the smallest term which occurs in both sequences at some point but not necessarily at the same index.

Now... the terms which occur in the first progression are all $a$ more than some integer multiple of $c$.  Written in terms of modular equivalences, that would be that if $x$ is some term in the first sequence we have that $x\equiv a\pmod{c}$.  Similarly, if $x$ were also in the second sequence we would have $x\equiv b\pmod{d}$

So, our task is to solve the simultaneous set of equivalencies $\begin{cases}x\equiv a\pmod{c}\\x\equiv b\pmod{d}\end{cases}$ for the smallest $x$ which is greater than equal to both $a$ and $b$.

This is precisely the type of problem that the Chinese Remainder Theorem is meant to solve.

Note that it might not be possible to find a solution for certain $a,b,c,d$ in the case that $\gcd(c,d)>1$.  For instance the arithmetic progressions $2m$ and $1+2n$ generate the even numbers and odd numbers respectively... there is no overlap.

In the case of $\gcd(c,d)=1$ we can via the Extended Euclidean Algorithm find the coefficients $j,k$ for Bézout's Identity satisfying $jc+kd=1$ which are guaranteed to exist.

Armed with this knowledge, letting $x\equiv bjc+akd\pmod{cd}$, we can see then that $x\equiv a\pmod{c}$ since $bjc+akd\equiv 0+akd\equiv ajc+akd\equiv a(jc+kd)\equiv a(1)\equiv a\pmod{c}$.  Similarly, we see that this choice of $x$ satisfies $x\equiv b\pmod{d}$.

Now, all that remains is to find the smallest such $x$ which is simultaneously greater than or equal to $a$ and $b$.

Adjusting this to allow for $c,d$ which are not coprime follows similar principles but the value of $\gcd(c,d)$ appears in more calculations.  Follow the link to see how it changes.

As for your specific example, we have the sequences $3+4m$ and $2+7n$.

We see that $\gcd(4,7)=1=2\cdot 4 + (-1)\cdot 7$

We see then that the numbers common to both are of the form $x\equiv 2\cdot 2\cdot 4 + 3\cdot (-1)\cdot 7\pmod{4\cdot 7}$, or simplifying $x\equiv 16 -21\equiv -5\equiv 23\pmod{28}$

That is to say, the numbers which are common to both sequences are $23,23+28,23+2\cdot 28, 23+3\cdot 28,23+4\cdot 28,\dots$ or otherwise written $23,51,79,107,\dots$

The smallest of which which is larger than both $a$ and $b$ (larger than $3$ and $2$) is $23$ itself.


问:
Proving that there are 3 solutions at most $f(x)=e^x-(ax^2+bx+c)$

Prove that the function $f(x)=e^x-(ax^2+bx+c)$ has 3 solutions at most .

$a$,$b$ and $c$ are constants.

This is the information given about the function, I tried a couple of things and I am not sure if what I did is right.

First the function is continuous and differentiable since $e^x$ is continuous and differentiable and $ax^2+bx+c$ is a polynomial so it is also continuous and differentiable therefore we can use rolles theorem.

I tried doing a couple of derivatives such as $f'(x)=e^x-2ax-b$ and $f''(x)=e^x-2a$ and lastly $f'''(x)=e^x$

so the third derivative has no solution , the second one is $x=ln(2a)$ since the second derivative has only 1 solution , then the first derivative has 2 at most and the original has 3 at most.

Is it the right way to solve it? am I missing something? thank you for the help!
By solution I mean f(x)=0

答:
The arguments you make need to be a little more precise, but the idea is correct!

Let the function $f(x)=e^x-(ax^2+bx+c)$ that is a continuous and differentiable function in $\mathbb{R}$ and let its derivative function $f'(x)=e^x-2ax-b$ that is continuous and differentiable in $\mathbb{R}$. The function $f(x)$ is in the conditions described by Rolle's Theorem on any closed and bounded interval we want.

Suppose $f(x)$ has $n$ distinct real roots with $n>3$. Let be the distinct real roots, which we call ordered from least to greatest, $x_1,x_2,x_3,\ldots ,x_n\in \mathbb{R}$. In such a case,

$$f(x_1)=f(x_2)=f(x_3)=\cdots =f(x_n)=0$$

By Rolle's Theorem, there exists at least one value in the interval $(x_1,x_2)$ where $f'(x)=0$; at least one value in the interval $(x_2,x_3)$ where $f'(x)=0$; ... at least one value in the interval $(x_{n-1},x_n)$ where $f'(x)=0$.

We have just proved the existence of at least $n-1$ roots of $f'(x)$ with $n>3$ in a correct way. That is, there exist at least $3$ roots of the equation $e^x-2ax-b=0$. Let be the distinct real roots $x_1',x_2',x_3',\ldots, x_{n-1}'\in \mathbb{R}$. In such a case,

$$f'(x_1')=f'(x_2')=\cdots =f'(x_{n-1}')=0$$

By Rolle's Theorem (applied to $f'$), there exists at least one value in the interval $(x_1,x_2)$ where $f''(x)=0$; at least one value in the interval $(x_2,x_3)$ where $f''(x)=0$,... at least one value in the interval $(x_{n-2},x_{n-1})$ where $f''(x)=0$.

We have just proved the existence of at least $n-2$ roots of $f''(x)$ with $n>3$. That is, there exist at least $2$ roots of the equation $f''(x)=e^x-2a=0$.

On the other hand, we know how to solve algebraically this equation which, as it turns out, has only one real solution other than if $a>0$:

$$e^x-2a=0\Leftrightarrow e^x=2a \Leftrightarrow x=\ln (2a)$$

And if $a\leq 0$, it doesn't exist any real solution to the equation $e^x-2a=0$ because $e^x>0, \,\, \forall x\in \mathbb{R}$.

This is an absurd thing according to what we have found above.

Therefore, the absurdity comes from considering that $f(x)$ has more than three distinct real roots. We conclude that the number of roots of $f(x)$ cannot exceed the number of three.


问:
Can I rewrite this solution to a log growth question with less steps and text?
I am working on a textbook exercise:

Given the log growth formula of $\frac{1000}{1+9e^{-0.6t}}$ find how long it takes for the population to reach 900.

The solution provided is 7.3 and I was able to arrive at this. But, reading over my solution I 'feel' like it's got too many steps and text and I just wondered if there was a simpler route to get my answer.

Here's my working:

$$900=\frac{1000}{1+9e^{-0.6t}}$$
$$1000=(900)(1+9e^{-0.6t})$$
$$\frac{1000}{900}-1=9e^{-0.6t}$$

(here is where I think I'm doing thing's 'wrong'. I have a fraction of a fraction on the left side, is there a better way to denote this? Can I somehow use a negative exponent instead of dividing by 9 for example? Or anything else? Or is what I have sound?)

$$\frac{\frac{1000}{900}-1}{9}=e^{-0.6t}$$
$$ln(\frac{1000}{900}-1)-ln9=-0.6t$$

(Again, here I feel like this expression is too cluttered, is there a simpler, less text way to express it?)
$$\frac{ln(\frac{1000}{900}-1)-ln9}{-0.6}=t$$
$$t=7.3$$

More generally, I'm wondering if I'm missing any opportunities to simplify my fraction expressions, either using rules of logs that I've missed or any other means?

Body:
Your solution looks OK,
and there is almost always sevaral other ways, for example,

\begin{align}
\frac{1000}{1+9e^{-0.6t}}&=900
\tag{1}\label{1}
,\\
\frac{10}{1+9e^{-0.6t}}&=\frac91
\tag{2}\label{2}
,\\
\frac{10-9}{1+9e^{-0.6t}-1}
&=\frac91
\tag{3}\label{3}
,\\
\frac{1}{9e^{-0.6t}}
&=\frac91
\tag{4}\label{4}
,\\
e^{0.6t}&=81
\tag{5}\label{5}
,\\
0.6t&=4\ln 3
\tag{6}\label{6}
,\\
t&=\tfrac{20}3\,\ln3
\approx 7.324
\tag{7}\label{7}
.
\end{align}


答:
Your sum is
$$
A = \psi \left( {\left( {\tfrac{N}{2} - 1} \right)B + 26} \right) - \psi (25).
$$
It is not possible to express your sum in terms of elementary functions. Nevertheless, one can obtain good upper and lower bounds for $B$. It is known that
$$
\log \left( {x + \tfrac{1}{2}} \right) < \psi (x + 1) < \log (x+1)
$$
for all $x>0$. Thus
$$
\log \left( {\left( {\tfrac{N}{2} - 1} \right)B + \tfrac{{51}}{2}} \right) < A + \psi (25) < \log \left( {\left( {\tfrac{N}{2} - 1} \right)B + 26} \right),
$$
i.e.,
$$
\frac{{2e^{A + \psi (25)}  - 52}}{{N - 2}} < B < \frac{{2e^{A + \psi (25)}  - 51}}{{N - 2}},
$$
with $\psi(25)=3.198742512\ldots$. For example, with $A=1$ and $N=500$, we get
$$
0.163 \ldots  < B < 0.165 \ldots .
$$
You can go beyond this, by using asymptotic approximations for the inverse of the digamma function (cf. https://doi.org/10.1007/s11139-014-9659-3). In particular,
$$
B \approx \frac{{2e^{A + \psi (25)}  - 51}}{{(N - 2)}} - \frac{1}{{12(N - 2)e^{A + \psi (25)} }}.
$$
With $A=1$ and $N=500$, this gives $B=0.16506787525\ldots$, whereas the true value is $B = 0.16506787531\ldots$.

Title: Summation simpilification process

Body:
Here is my summation:

$$\sum_{n=0}^{N/2-1}\frac1{25+nB}=A$$
where $0.01\le A\le2$, $10\le N\le2000$ and I need to find the $B$ for different values of $N$ I calculate this summation online (check here)
But the Digamma function makes the output function complex and it is not easy to solve with a calculator or even by hand. I need to calculate the result of this summation with hand easily. Is there any way to convert my summation to simple formula? Something like this:

\begin{align*}
\sum^n_{k=1}k&=\frac{n(n+1)}{2}\\[5pt]
\sum^n_{k=1}k^2&=\frac{n(n+1)(2n+1)}{6}\\[5pt]
\end{align*}


问:
Equation of the tangent line??
I was wondering if any of you knew how to find the equation of the line tangent to the curve $y = x^3-5x+2$ at point $(2,0)$ in standard form.

I personally got $7x - y - 14 = 0$. However, that is apparently incorrect.

Does anyone know what I may be doing wrong? Thank you in advance!

答:
This is correct. To check it, you need first to find the slope of the tangent line: from $y=x^3-5x+2$, we have $y'=3x^2-5$. For $x=2$, this gives $y'=12-5=7$.

Therefore, the slope-point equation of the tangent line is $y=0+7(x-2)$ which is equivalent to $y=7x-14 \iff 7x-y-14=0$.

Title: Equation of the tangent line??


问:
Hermitian Matrix multiplied by complex numbers
I have a question regarding self-adjoint(hermitian) matrices, and their properties when multiplied by an imaginary number.
The matrix in question is:
$$\mathbf{A}  = c
\begin{bmatrix}
0&0&0&0\\
0&0&2i&0\\
0&-2i&0&0\\
0&0&0&0
\end{bmatrix}\\
\textit{where} \; \mathbf{A}^\dagger=\mathbf{A} $$
This should be pretty simple to answer to, but I am having a brain fart at the moment: if we assume $c=\frac{i}{2}$, then:
$$\mathbf{A}  =
\begin{bmatrix}
0&0&0&0\\
0&0&-1&0\\
0&1&0&0\\
0&0&0&0
\end{bmatrix}\\
\textit{where} \; \mathbf{A}^\dagger\neq\mathbf{A} $$
Do hermitian operators stop to be so if multiplied by a complex number?

答: When $A$ is Hermitian, we have $(cA)^\dagger-cA=c^\dagger A^\dagger-cA=(c^\dagger-c)A$. Therefore $cA$ is Hermitian if and only if $A=0$ or $c$ is real.


答:
Your answer is correct, but can be written more explicitly as $y=7x-14$. If $f(x)=x^3-5x+2$, then $f'(x) = 3x^2-5$. In general, the equation of the tangent to the point $(a,f(a))$ is given by
$$
y-f(a) = f'(a)(x-a) \, .
$$
Here, $a=2$, $f(a)=0$, and $f'(a)=f'(2)=7$. Hence, the equation of the tangent line is
$$
y=7(x-2) \implies y=7x-14 \, .
$$

Title: Equation of the tangent line??

Body:
I was wondering if any of you knew how to find the equation of the line tangent to the curve $y = x^3-5x+2$ at point $(2,0)$ in standard form.

I personally got $7x - y - 14 = 0$. However, that is apparently incorrect.

Does anyone know what I may be doing wrong? Thank you in advance!


问:
Sturm-Liouville Problem

Possible Duplicate:

  Sturm-Liouville Problem

How could one prove that there are at most countably many eigenvalues of the Sturm-Liouville problem $−Lu=ju$, $j$ = eigenvalue, and $u$ is in $C^2[a,b]$?

I have put some more thought and I am still confused about this problem.
Suppose I have $-Lu=\lambda u$ where $-Lu=(pu')'+qu.$  How can you justify that if $\left\{ c_{\alpha}\right\} _{\alpha\in\Gamma}$
is a set of non-zero real numbers such that $\mathbb{\sum}_{j=1}^{N}c_{\alpha_{j}}^{2}\leq1$
for every choice of $N\geq1$ and every choice of distinct $\alpha.$
Then $\Gamma$ is a countable set. How to proceed from there? Is there an easy solution? Thank you very much for help.

答:
$\sum_{j=1}^N c_{\alpha_j}^2\le 1$ implies $0< c_\alpha^2\le 1$ for each $\alpha\in\Gamma$.

Let $A_n$, $n\ge 1$, be the subset of $S=\{c_\alpha\}_{\alpha\in\Gamma}$ defined by
$$A_n = \left\{c_\alpha\in S : \frac {1}{n+1} < c_\alpha^2 \le \frac 1 n\right\}$$
$A_n$ must contain $n$ elements at most, otherwise
$$\sum_{c_\alpha\in A_n} c_\alpha^2 > \sum_{c_\alpha\in A_n} \frac{1}{n+1} = \frac{\text{Card}(A_n)}{n+1} \ge 1$$
Evidently $\cup_n A_n = S$; $S$ is a countable union of finite sets, so it is countable.

Title: Sturm-Liouville Problem


问:
Determine the number of digits in $(2^{120})(5^{125})$.
Determine the number of digits in $(2^{120})(5^{125})$.

This is a bonus question I found in a Grade 12 math textbook, and I'm curious on how to solve it. What I find strange is that this question is in the  "Power Functions" section of "Polynomial Functions"; yet, this seems to have little to do with functions.

Any ideas as to how to solve this? I have tried to, but can't find a solution without a computer or calculator.

Body:
Observe that
$$2^{120}5^{125}=2^{120}5^{120}5^5=10^{120}\cdot3125$$
So the number is $3125$ followed by $120$ zeroes.


答:
You can indeed save some work.

Write
$$
\frac{1000}{900} -1 = \frac{10}{9} -1 = \frac{1}{9}
$$
as soon as it appears.

Then you reach
$$
\frac{1}{81} = e^{-0.6t}
$$
with just one natural logarithm to calculate.

Title: Can I rewrite this solution to a log growth question with less steps and text?

Body:
I am working on a textbook exercise:

Given the log growth formula of $\frac{1000}{1+9e^{-0.6t}}$ find how long it takes for the population to reach 900.

The solution provided is 7.3 and I was able to arrive at this. But, reading over my solution I 'feel' like it's got too many steps and text and I just wondered if there was a simpler route to get my answer.

Here's my working:

$$900=\frac{1000}{1+9e^{-0.6t}}$$
$$1000=(900)(1+9e^{-0.6t})$$
$$\frac{1000}{900}-1=9e^{-0.6t}$$

(here is where I think I'm doing thing's 'wrong'. I have a fraction of a fraction on the left side, is there a better way to denote this? Can I somehow use a negative exponent instead of dividing by 9 for example? Or anything else? Or is what I have sound?)

$$\frac{\frac{1000}{900}-1}{9}=e^{-0.6t}$$
$$ln(\frac{1000}{900}-1)-ln9=-0.6t$$

(Again, here I feel like this expression is too cluttered, is there a simpler, less text way to express it?)
$$\frac{ln(\frac{1000}{900}-1)-ln9}{-0.6}=t$$
$$t=7.3$$

More generally, I'm wondering if I'm missing any opportunities to simplify my fraction expressions, either using rules of logs that I've missed or any other means?


问:
Reducing modulo 5 does not make the equations equivalent?
Edit 2: I was trying to solve $2^{n} + 5^{n} - 65 = p^{2}$ for positive integer solutions

$$2^{n} + 5^{n} - 65 \equiv p^{2} \mod 5$$

has different solutions from

$$2^n \equiv p^2 \mod 5$$

Why is this so? Does this mean that both equations are not equivalent? Is reducing modulo $5$ a bad idea for finding integer solutions of $n$ and $p$?

Note: There are more solutions (including the solutions for the previous equation) not shown for $2^n \equiv p^2 \mod 5$, but it still stands that there are suddenly more solutions.

Edit: Equation 2 has too many solutions, including solutions where $n$ is not $0$. Here are some of them:

答:
Here, $5^n \equiv 0 \mod 5$, except for $n = 0$.

Therefore, we have to make a distinction between $n = 0$ and $n \ne 0$.

Moreover, $\phi(5) = 4$.

Therefore,

$$ 2^n \mod 5 \equiv 2^{n \mod 4} \mod 5 $$.

As $n = 0$ is a special case, we have to consider $n = 0, 1, 2, 3, 4$.

Concerning $p$: if $p$ is a solution, then $p+5$ is a solution too. Then, for $p$, we have only to consider its value modulo 5.

At this point, it is easy to consider the different cases.

If $n = 0$:

$$p^2 = 2 \mod 5$$

It is easy to check that $p^2$ is equal to $0, 1, 4$ modulo 5. So, no solution

If $n \ne 0$:

$$p^2 \equiv 2^n \mod 5$$

$$n = 1 \implies p^2=2 \mod 5 \implies no\,solution\,in\,p$$
$$n = 2 \implies p^2=4 \mod 5 \implies p=2\,or\,p=3$$
$$n = 3 \implies p^2=3 \mod 5 \implies no\,solution\,in\,p$$
$$n = 4 \implies p^2=1 \mod 5 \implies p=1\,or\,p=4$$


问:
Is $f$ is onto ? True/false
Let $G$ be an abelian group  and  $x \in G$.Define map  $f: G \to G$ by $f(x)=x^{-1}$

Now my question is that

Is  $f$ is onto ? True/false

My attempt : I think $f$ will  not onto because $\frac{1}{ x }\notin G$

Take $G= \mathbb{Z}$  and  $x = 2$   but $\frac{1}{2} \notin \mathbb{Z}$

答:
This map $f$ is always onto/surjective. Your example does not make any sense. If $\mathbb{Z}$ is your group under addition, then inverse of $2$ is -2, NOT $1/2.$

The reason why the map is always surjective is the following: for every element $x \in G$, there is another element namely $x^{-1} \in G.$

Another way: You just argue using the fact $ (x^{-1})^{-1}= x.$

Title: Is $f$ is onto ? True/false


答:
Your answer is correct. However, in the hope that this will be helpful to you later on, I am going to provide you with a standard algorithm that helps you determine the tangent plane at a certain point on a level set of $F : \mathbb{R}^n \rightarrow \mathbb{R}.$ Assume that $F$ is "nice enough" (for example, take it to be $C^1$). Consider the level set $F(x_1, \dots, x_n) = c,$ where $c \in \mathbb{R}.$ Then it is well known fact that the normal at a point $y \in F^{-1}(c)$ is given by the gradient of $F$ at that point, $\nabla F (y)$ (for a proof, consider a curve on the level set passing through your point, $\gamma$; then you have that $F(\gamma(t)) = c$ and just use the chain rule to obtain this lemmata). It follows that your tangent plane at $y$ is given by $\nabla F(y) \cdot x = 0$ (or a plane parallel to it, rather).

In this particular case, we have $F(x, y) = y - x^3 + 5x - 2,$ so that $\nabla F(x, y) = (-3 x^2 + 5, 1).$ In particular, $\nabla F(2, 0) = (-7, 1).$ Thus the equation of the tangent at $(2, 0)$ will be $-7x + y + c = 0.$ Since we want $(2, 0)$ to be in this plane, we get $c = 14.$ Hence the answer: $-7 x + y + 14 = 0.$ I hope this helps. :)

Title: Equation of the tangent line??


问:
How to remember which function is concave and which one is convex?
I always struggle to remember when a function is convex and concave:

Do you have a particular trick to help you remember this?

My trick is based on the Spanish phrase "No cabe", pronounced nô ˈka.βe, which sound just like "concave". "No cabe" means it does not fit. Thus, whilst you can put something into a convex function (e.g. think of a bowl), you cannot put something into a concave function. Hence the relation.

I am curious on what other, perhaps more efficient methods people use.

答: Well, I think of it as they both are in a smiling contest, concave smiles with the v and convex with the v and x. However, the fact that the function convex smiles more makes the concave sad.</p>


答: Whether $G$ is abelian or not, $f(x^{-1})=(x^{-1})^{-1}=x$, so $f$ is surjective.  That is, in a group every element has an inverse.

Title: Is $f$ is onto ? True/false

Body:
Let $G$ be an abelian group  and  $x \in G$.Define map  $f: G \to G$ by $f(x)=x^{-1}$

Now my question is that

Is  $f$ is onto ? True/false

My attempt : I think $f$ will  not onto because $\frac{1}{ x }\notin G$

Take $G= \mathbb{Z}$  and  $x = 2$   but $\frac{1}{2} \notin \mathbb{Z}$


问:
Proving that either $2^n-1 $ or $ 2^n+1$ is not prime
Not sure what approach to take with this:

Prove that at least $2^n-1 $ or $ 2^n+1$ is composite $\forall$ $n>2$

答:
Hint: $\ 3\nmid a,\ \,3\mid (a\!-\!1)a(a\!+\!1)\!\stackrel{\color{#0a0}{\rm Remark}}{\Rightarrow} 3\mid a\!-\!1\ \ {\rm or}\ \  a\!+\!1,\,$ so $\,a>4\Rightarrow a\!-\!1\ \ {\rm or}\ \ a\!+\!1\,$ not prime.

$\color{#0a0}{\rm Remark\!:}\, $ More generally, any sequence of $\,n\,$ consecutive naturals has an element divisible by $\,n.\,$ This has a simple inductive proof: $ $ shifting such a sequence  by one does not change its set of remainders mod $\,n,\,$ since it replaces the old least element $\:\color{#C00}a\:$ by the new greatest element $\:\color{#C00}{a+n}$
$$\begin{array}{}& \color{#C00}a, &\!\!\!\! a+1, &\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1  & \\
                  \to & &\!\!\!\! a+1,&\!\!\!\! a+2, &\!\!\!\! \cdots, &\!\!\!\! a+n-1, &\!\!\!\! \color{#C00}{a+n} \end{array}\qquad$$

Since $\: \color{#C00}{a\,\equiv\, a\!+\!n}\pmod n,\:$ the shift does not change the remainders in the sequence. Thus the remainders are the same as the base case $\ 0,1,2,\ldots,n-1\ =\: $ all $ $ possible remainders $ $ mod $\,n.\,$ Therefore the sequence has an element with remainder $\,0,\,$ i.e. an element divisible by $\,n.$


问:
What does the term "unbiased estimator" mean?
In my textbook for my statistics class, it says that $s^2$, sample variance is a "unbiased estimator" for population variance, $\sigma^2$. Does this mean that when we use $s^2$ as a point estimator for $\sigma^2$, it precisely equals $\sigma^2$? So $s^2$ is not even an estimation/approximation for $\sigma^2$? What does unbiased estimator mean?

Thank You!

答:
$S^2$ is unbiased estimator for the population variance $\sigma^2$ because, as per definition

$$\mathbb{E}[S^2]=\sigma^2$$

there are other and most important properties of an estimator, i.e. consistency, sufficiency, efficiency, etc etc.


答:
No, equality does not hold in general:
$$
f''(x):=\lim_{h \to 0}\frac{f'(x+h)-f'(x)}{h} \neq \left(\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\right)^2=f'(x)^2 \, .
$$

Title: First deriative, multiplied by first derivative

Body:
I have one question. If I multiply 2 first deriatives, will I receive second derivative? Example:
$$\frac{\partial a}{\partial t} = \dot{a}$$
Is it true?:
$$\frac{\partial\dot{a}}{\partial t}=\ddot{a}$$


问:
Irreducible Polynomials and Ring Homomorphism

Help and hints for parts (c) and (d) ! Thanks :)

答:
For $c$: Which elements of $\mathbb{Z}_7$ are squares?

For $d$: Remember that $\mathbb{Z}_7[x]/\langle x^2-3\rangle\cong\mathbb{Z}_7[\sqrt{3}]$. Can you find homomorphisms $\mathbb{Z}[x] \to \mathbb{Z}_7[x] \to\mathbb{Z}_7[x]/\langle x^2-3\rangle$?


问:
Sum of all possible surjective functions.
For my combinatorics homework I have the following problem:

Let $A = \{a_1, a_2, ..., a_m\}$ and $B = \{b_1, b_2, ... , b_n\}$ with $m \geq n$. Show that the amount of surjective functions $f$ from $A$ to $B$ is equal to:
\begin{equation}
\sum_{i=0}^n (-1)^i {n \choose i} (n-i)^m
\end{equation}

My first thought was that, because these functions must be surjective, for every element $b \in B$ there must be a unique $a \in A$ such that $f(a) = b$. Therefore we must choose $n$ elements in $A$ that have a unique image on $B$ and the rest of the elements in $A$ can be sent to a random element in $B$. So i was thinking about:
\begin{equation}
{m \choose n} \cdot n! \cdot n^{m-n}
\end{equation}
as the amount of all possible surjective functions $f$ from $A$ to $B$.

My second thought was that the combination is unnecessary and therefore the formula must be:

\begin{equation}
n! \cdot n^{m-n}
\end{equation}
Is this correct? If so, how can i prove that this is the same as the summation given at the beginning?

I hope anybody can help me with this problem.

Thanks in advance!

答:
The first formula follows from the inclusion-exclusion principle in the following way: denote by $F_i,$ $1 \leq i \leq n$ the set of all functions from $A$ to $B$ such that the cardinal of their images is exactly $n - i.$ Let us count these. Let $1 \leq i \leq n.$ We must first choose $i$ of the $n$ elements of $B$ that are not in the range of $f \in F_i.$ This ammounts to $\binom{n}{i}$ choices. Moreover, once we have chosen the range of $f$, we have exactly $(n - i)^m$ choices for $f,$ so that the total number is equal to $\binom{n}{i} (n - i)^m.$ Note that the surjective functions are exactly the complement of the union of the $F_i'$s. This cardinal is easily computed with the inclusion-exclusion principle and it is equal to $\sum_{1 \leq i \leq n} \binom{n}{i} (-1)^{i - 1} (n - i)^m.$ Thus the number of surjective functions is $$n^m - \sum_{1 \leq i \leq n} \binom{n}{i} (-1)^{i - 1} (n - i)^m = \sum_{0 \leq i \leq n} (-1)^i \binom{n}{i} (n - i)^m.$$

The number of surjective functions from $A$ to $B$ can also be approximated directly in a similar manner to the one you were trying to use. The method you described is more or less precisely correct. Let $f: A \rightarrow B$ be surjective. We go through the elements of $B$ one by one, starting with $b_1.$ Since $f$ is surjective, there must be $a_j \in A$ such that $f(a_j) = b_1,$ so we have $m$ choices of $a$ for $b_1.$ Similarly, for $b_2$ we will have $m - 1$ choices and so on. We get $\binom{m}{n} n!$ so far. The other elements in $A$ can be mapped to anything in $B,$ so we have another $n^{m - n}$ choices, for a total of $\binom{m}{n} n! n^{n - m}$ so far. The thing is we may allow for repetitions if we use this method, so this will not be our final answer. I hope this helps. :)


问:
Suppose $f$ is differentiable at $p_0$, show that total derivative of $f$ at $p_0$ is unique
Let $R\subseteq \Bbb{R}^m$ be an open subset, and let $f\colon R\to\Bbb{R}^n$ be a function. Let $p_0\in R$ be a given point. Suppose $f$ is differentiable at $p_0$. Show that the total derivative of $f$ at $p_0$ is uniquely determined.

Hint given was to work with 2 linear transformations and show that the limit of this map is the zero map, not sure what to do with this hint.

答:
Let $\lambda, \mu: \mathbb{R}^m \to \mathbb{R}^n$ linear transformations such that
$$
\label{eq1}
\lim_{h \to \mathbf{0}} \frac{\Vert f(p_0+h)- f(p_0) -\lambda (p_0)\Vert}{\Vert h \Vert}=0 \quad \quad (1)
$$
and
$$
\lim_{h \to \mathbf{0}} \frac{\Vert f(p_0+h)- f(p_0) -\mu (p_0)\Vert}{\Vert h \Vert}=0. \quad \quad (2)
$$

Set $d(h)=f(p_0+h)-f(p_0).$ Hence we have
\begin{eqnarray*}
\lim_{h \to \mathbf{0}} \frac{ \Vert \lambda (h)- \mu(h) \Vert}{\Vert h \Vert} &=& \lim_{h \to \mathbf{0}} \frac{ \Vert \lambda (h)-d(h)+d(h)- \mu(h) \Vert}{\Vert h \Vert} \\
 &\leq & \lim_{h \to \mathbf{0}} \frac{ \Vert \lambda (h)- d(h) \Vert}{\Vert h \Vert}+\lim_{h \to \mathbf{0}} \frac{ \Vert d (h)- \mu(h) \Vert}{\Vert h \Vert} = 0.
\end{eqnarray*}
Where the last equality is a consequence of (1) and (2).

Now, it should be clear that $\lim_{h \to \mathbf{0}} \frac{ \Vert \lambda (h)- \mu(h) \Vert}{\Vert h \Vert}=0. \quad \quad (3)$

Now lets prove that $\lambda (x) = \mu (x)$ for all $x \in \mathbb{R}^m.$

The equality holds for $x = \mathbf{0}$. So, suppose $x \neq \mathbf{0}.$

We shall prove that $\Vert \mu (x)- \lambda (x) \Vert < \varepsilon$ for all $\varepsilon >0$. So pick an arbitrary positive $\varepsilon.$ By (3), there's a $\delta>0$ such that
$$0<\Vert h \Vert < \delta \Longrightarrow \frac{\Vert \lambda (h) - \mu (h) \Vert}{\Vert h \Vert}<\frac{\varepsilon}{\Vert x \Vert}.$$

Now pick a $t \in \mathbb{R}$ such that $0 < \Vert tx \Vert <\delta.$ Hence for $h=tx$ we have by linearity of $\lambda$ and $\mu,$ $$\frac{\varepsilon}{\Vert x \Vert}>\frac{\Vert \lambda (tx)- \mu (tx) \Vert}{\Vert tx \Vert}=\frac{|t|\Vert \lambda (x)- \mu (x) \Vert}{|t|\Vert x \Vert}=\frac{\Vert \lambda (x)- \mu (x) \Vert}{\Vert x \Vert}=\frac{\Vert \lambda (x)- \mu (x)\Vert }{\Vert x \Vert}.$$

Now, multiply the last inequality by $\Vert x \Vert.$


问:
Hodge star operator independent of orthonormal basis
I'm reading the book on differential geometry by Gockeler and Schucker and I came across a proof which I don't understand regarding the Hodge star operator. In the book, they define the Hodge star operator on an $n$-dimensional oriented vector space by picking some positively oriented orthonormal basis $\{e_i\}_{i=1}^n$ and defining the map on the forms induced by the duals:
$$*:\Lambda^p V\rightarrow \Lambda^{n-p}V$$
$$ *(e^{i_1}\wedge\cdots\wedge e^{i_p})=\epsilon_{i_1\ldots i_n}\eta^{i_1i_1}\cdots\eta^{i_pi_p}e^{i_{p+1}}\wedge\cdots\wedge e^{i_n} $$
Where $\epsilon$ is the Levi-Civita symbol and $\eta$ is a pseudo-Riemannian metric, which is diagonal with $r$ diagonal entries equal to $1$ and $s$ equal to $-1$.

They then claim that this definition does not depend on the choice of oriented orthonormal basis, and they explain it as follows:

Where equation $3.16$ is the equation written above and equation $3.9$ is the equation which defines the elements of $O(r,s)$:
$$(\Lambda ^{-1})^t\eta \Lambda=\eta$$
I don't really understand their argument, and I'd appreciate any help in understanding it, even for the simple case of $\eta$ being the identity matrix (so just the usual Euclidean metric). I came across this related post, but it didn't contain a complete solution I could understand.

I tried proving it myself, by taking another positive orthonormal basis $\{x_i\}_{i=1}^n$, and writing $e_i=Ax_i$ for some matrix $A\in SO(r,s)$ and plugging this into the formula above, but I'm not sure on how to proceed from here.

Thanks in advance.

答:
If you're familiar with index notation, it possible, though tedious, to verify this using a direct computation. Here's an alternate route which uses the basis-independent definition.

First, we note that there exists a canonical top form $\omega\in\Lambda^nV$ defined by $\omega=e^1\wedge\cdots\wedge e^n$ where $e^i$ is any oriented orthonormal basis. This is independent of the choice of basis, since
$$
Ae^1\wedge\cdots\wedge Ae^n=\det(A)e^1\wedge\cdots\wedge e^n
$$
and every element of $SO(V)$ has unit determinant.

Additionally, since $\eta$ defines an inner product $\langle\ ,\ \rangle$ on $V$, we can define an inner product of $\Lambda^pV$ by
$$
\langle e^{i_1}\wedge\cdots\wedge e^{i_p},e^{j_1}\wedge\cdots\wedge e^{j_p}\rangle=\eta^{i_1j_1}\cdots\eta^{i_pj_p} \\
i_1<\cdots<i_p,\ \ \ j_1<\cdots<j_p
$$
Where $e^i$ is any orthonormal basis i.e. we define the basis $\{e^{i_1}\wedge\cdots\wedge e^{i_p}:i_1<\cdots<i_p\}$ to be orthonormal. This definition is independent of the choice of basis, since $\langle Au,Av\rangle=\langle u,v\rangle$ for $A\in SO(V)$.

The Hodge star can then be defined by
$$
\mu\wedge(\star\nu)=\langle\mu,\nu\rangle\omega\ \ \ \ \ \forall\mu,\nu\in\Lambda^pV
$$
Since everything in this expression is basis-independent, it suffices to show that $\star$ is well-defined. One way of doing this is by showing that this expression is uniquely satisfied by your expression for $\star(e^{i_1}\wedge\cdots\wedge e^{i_p})$ and the rest follows from linearity.


问:
The class number of $\mathbb{Q}(\sqrt{82})$ is even
Let $K = \mathbb{Q}(\sqrt{82})$.
I want to show that the class number $h_K$ of $K$ is even using the fact that the equation
$$ X^2 - 82Y^2 = \pm2 $$
has a solution in $\mathbb{Z}_p$ for all primes $p$ but no solution in $\mathbb{Z}$.

This is Exercise 2.2 in Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory, which has already been asked here; my question is about the answer of Matt Emerton.
I thought about leaving a comment there but the question is from 2012 and Emerton has not been online since 2014 so I thought it best to ask a new question.

I don't see what links the local solutions and the class number.
The fact that there is no global solution tells us that the prime $\mathfrak{p}_2$ given by $\mathfrak{p}_2^2 = (2)$ is not principal.
Can we deduce that its order in the class group is even from the local solutions?

It turns out that the class group is cyclic of order $4$ and generated by $\mathfrak{p}_3 = (3, \sqrt{82}+1)$.
However, it seems to me (from the formulation of the exercise and the answer linked above) that the parity of $h_K$ can be deduced from the existence/nonexistence of solutions of the equation only,  without computing the full class group.

答:
The fact that $x^2 - 82y^2 = \pm 2$ does not have integral solutions implies
that the class number is not $1$. If you know in addition that $2$ ramifies in the quadratic subfield you can even conclude that the class number is even.

The fact that $x^2 - 82y^2 = \pm 2$ has local solutions everywhere (the sign doesn't matter since $9 + \sqrt{82}$ has norm $-1$) implies that the ideal class generated by the prime ideal above $2$ is in the principal genus and thus must be a square. This actually implies that the class number is divisible by $4$. Local solvability, by the way, follows almost immediately from $10^2 - 82 = 2 \cdot 3^2$ and $18^2 - 82 = 2 \cdot 11^2$.

Consider, on the other hand, the equation $x^2 - 10y^2 = 2$. As above, the fact that it does not have an integral solution implies that the class number of ${\mathbb Q}(\sqrt{10})$ is even. But this time, the equation is not solvable everywhere locally, e.g. in ${\mathbb Z}_5$; this is essentially because reduction modulo $5$ gives a contradiction. In particular, the ideal class of the prime ideal above $2$ is not a square.

The approach to genus theory via local solvability is, if I recall it correctly, presented in Hecke's excellent book on algebraic numbers.


问:
When does the function $f(y) = 1 + y + y^2$ show mainly the units in a quadratic number ring? Is a corresponding equation solvable in integers?
Motivated by a strange coincidence in polynomial geometry I decided to learn some elementary algebraic number theory recently and armed with "Quadratische Zahlkörper", a recent german introductory text by Franz Lemmermeyer, I tried to understand the following phenomenon, which I had found empirically:

Let $f(y) = 1 + y + y^2$, $x = a + \sqrt{m} \, b$ be an element of the maximal order in $\mathbb{Q(\sqrt{m})}$, so $m \in \mathbb{Z\backslash\{0,1}\}$ is squarefree, $N(x) = a^2 - m b^2$ is the algebraic norm of $x$ and $D(x) = 4mb^2$ the discriminant of $x$, while there are the usual restrictions on $a$ and $b$ according to the residue class of $m \mod 4$.Then:

Assertion:

$N(f(x^2))$ is a square integer, if and only if $x$ is a unit of the maximal order or if $ab = 0$.

This result is certainly useless, if one wants to find units, but I (with no clue in number theory or algebraic geometry) found it so peculiar, that I started to call it "disuseful" for fun. Before giving the things I managed to find/prove, I will ask the things, which would be helpful to me:

Perspective: This/similar phenomena will have been discovered before and I would appreciate appropriate references, preferably textbooks to point me to other investigations of this type.

Help to discuss the solutions to
$$ K^2 = (4mb^2 - 1)^2 + l(a^2 - mb^2 + 1)$$
in integers only, see the end of the question; doing so will give a necessary condition to find valid solutions of $N(f(x^2)) = K^2$, if there are any more.

Results:

An elementary computation shows $$N(f(x^2)) = N(f(x))N(f(-x)) = (f(N(x)) + D(x))^2 - D(x)(N(x) - 1)^2 \, ,$$
from which one infers or directly computes:
$$a = 0 \ \Rightarrow \ N(f(x^2)) = (f(-N(x)))^2 \, ,$$
$$ b = 0 \ \Rightarrow \ N(f(x^2)) = (f(N(x)))^2 \, ,$$
$$N(x) = 1 \ \Rightarrow \ N(f(x^2)) = (D(x) + 3)^2 \, ,$$
$$N(x) = -1 \ \Rightarrow \ N(f(x^2)) = (D(x) - 1)^2 \, .$$
Looking at a number of examples corresponding to the above and not corresponding to them for positive as well as negative values of $m$ and various residue classes, the observations were always confirmed, but I did not try a systematic brute force search up to now. Indeed, everything up to this point is elementary algebraic manipulation only.

Proof idea:

Let $K$ be a hypothetical strictly positive integer fulfilling
$$K^2 \ = \ (f(N(x)) + D(x))^2 - D(x)(N(x) - 1)^2 \, ,$$
then the case $f(N(x)) + D(x) = 0$ leads to squares if and only if $m \, = \, -1$. For $a \, \neq \, 0; \, m \, = \, -1$ we obtain the equation
$$(2b^2 - 1)^2 \ = \ n^2 + (n+1)^2 \quad (\star) $$ from this. Applying the general parametrization for Pythagorean Triples
$$ (c(u^2 - v^2)^2 \, + \, (2cuv)^2 \ = \ (c(u^2 + v^2))^2 \, ,$$
where $c,u,v$ are nonvanishing integers, and $u, v$ can be chosen with opposite parity, we obtain the Pell equation $s^2 - 2t^2 \ = \ \pm 1$ in integers from the right hand side of $(\star)$, showing $c \, = \, 1$, which implies, that $b$ cannot be even, as $2b^2 - 1$ must then equal $u^2 + v^2$. Therefore $(\star)$ leads to
$$ 2*b^2(b^2 - 1) \ = \ n(n + 1) \, ,$$
which has the integer solution $n \, = \, b^2 \, = \, 3$, but $b$ is then no integer, and further developments show there is no integer solution for any odd integer $b$.

To approach the general case, we use the geometric sum formular on $f(N(x))$ and some hopefully correct manipulations to arrive at
$$(N(x) - 1)^2 \left( K^2 - (D(x) - 1)^2 - D(x)(N(x) + 1)(N(x) + 3) \right) \ = \ N(x) \ \left( (N(x)^5 - 2N(x)^2) - (N(x) - 2) \right) \, ( \star \star).$$
As $x$ is not a unit, we can divide by $(N(x) - 1)^2$ to get
$$ K^2 - (D(x) - 1)^2 - D(x)(N(x) + 1)(N(x) + 3) \ = \ N(x)(N(x) + 1)(N(x)^2 + N(x) + 2)\, ,$$
which shows, that $N(x) + 1$ must divide $K^2 - (D(x) - 1)^2$. This leads to the equation in the beginning, where $l$ is an integer, which constitutes a necessary condition for solutions different from those already found to exist. If the discriminant were allowed to be one, $l$ could be chosen to be a square multiple of $(N(x) + 1)$. This not being the case, I tried to investigate, whether the parametrization of the Pythagorean triples allows a solution; this seems impossible, because $m$ is squarefree and $u^2 + 1$ is not a square, but at the moment I am stuck.
I have already checked, whether the sides of the equation $(\star \star)$ agree as polynomials, given an appropriate choice of $K^2$, they do not, but as far as I am aware, this would constitute only a sufficient, not a necessary condition for an integer solution, and some might still be hiding there.

答:
There are at least some counterexamples, see equation $(\circ \circ)$ at the bottom and the following discussion.

First I will adress the diophantine equation of the question, though it turned out not to be necessary for the solution of the general problem proposed below, because it is an interesting variant of the solution of the equation for Pythagorean tripels.

I managed to find one particular solution relevant to the original problem and an abundance of solutions for the diophantine equation as such in the case $m \ = \ -1$. In this particular case only, $(4mb^2 - 1)^2$ may be regarded as the square of a sum of an even and an odd square, so it can be expressed as the side of the well-known parametrization equation of Pythagorean triples, which contains only one square. By this Pythagorean parametrization, we can then regard it as the sum of two new squares and write down a solution for the diophantine equation as such:
$$ (u + v)^2 = u^2 + v^2 + 2uv \, ,$$
where $u$, $v$ are integers, where one, say $u$, is even and the other, therefore $v$, is odd and $u,v$ may be multiplied each by another arbitrary integer to obtain yet more solutions. To solve our original problem, we have to account for our special circumstances, which forces the following choices, where we integrate $m = -1$ from the start:
$$ u^2 + v^2 = (1 + 4b^2) \, , \ v = (a^2 + b^2 + 1)^2 \, , \ l = 2u \, ,$$
which in turn lead to a consistency condition for $u$ expressible as the following equation:
$$u^2 \ = \ -a^4 - 2a^2 - (2a^2 - 6)b^2 + 15b^4 \, .$$
The particular solution I found to this one is $a = b = 1$, leading to $u = 4$, which gives $49 = 25 + 24$ as solution to our diophantine equation according to the given parametrization. This solution does not solve our original equation though, so it does not lead to $N(f(x^2))$ being a square. After all, the condition was indeed only necessary, not sufficient. I left the matter there to look for a different approach, which follows.

From now on $N(x) \, = \, N$ and $D(x) \, = \, D$ will be used, because there is no ambiguity in their use below and it improves readability in the long equations.

If we add $(N + 1)(N + 3)$ to the last equation in the question and rearrange, converting some of the terms along the way and putting everything except for $K^2$ on the right hand side, we get:
$$K^2 \ = \ (D-1)^2 \, + \, (N + 3)(D - 1)(N + 1) \, + \, (N^2 + 3)(N + 1)^2 \, . $$
The right hand side is now a not quite quadratic form in $D - 1$ and $N + 1$, which shows how to proceed. We use $N + 3 \, = \, N + 1 + 2$ and $N^2 + 3 = N^2 + 2 + 1$ to rearrange again and obtain
$$K^2 \ = \ (D+N)^2 \, + \, (N^2 + D + 1)(N + 1)^2 \, . $$
We can now turn the right hand side into a square by adding $(D + 4N)(N + 1)^2$, giving
$$K^2 \, + \, (D + 4N)(N + 1)^2 \ = \ (D + N + (N + 1)^2)^2 \, . \ (\circ)$$
At this point, it is expedient to introduce the trace $T(x) \, = \, T \, = \, 2a$ of the algebraic number, as this shows, that we have two squares on the left and shows the special position of the conditions leading to squares from the question. We have $T^2 \, = \, D + 4N$ by the general definitions of norm, trace and discriminant of numbers from a quadratic field, so we get
$$K^2 \, + \, T^2(N + 1)^2 \ = \ (D + (T^2 - D)/4 + (N + 1)^2)^2 \, . \ (\circ \circ) $$
If $T \, = \, 0 \, = \, a$ or $N \, = \, -1$, we obviously always get a true equation in integers, and the case $D \, = \, 0 \, = \, b$ is dealt with affirmatively by putting $c \, = \, 1$, $u \, = \, T/2$ and $v \, = \, N + 1$ in the parametrization of the Pythagorean Triples.

When fitting the case $N \, = \, 1$ into the Pythagorean framework, I found a very small amount of counterexamples, but there may be more under other circumstances:
For $N \, = \, 1$ the choice of $c \, = \, 1, \ u \, = \, a(N + 1), \ v \, = \, 1$ is appropriate to fulfill equation $(\circ \circ)$, in this case the following equation has to be true:
$$ 1 \, + \, a^2 (N+1)^2 \ = \ 3mb^2 \, + \, a^2 \, + \, (N+1)^2 \, , $$
where the right hand side is just one of many ways to write the right hand side of $(\circ \circ)$. Using $mb^2 \, = \, a^2 - N$, we can show, that this is equivalent to
$$ (a^2 - 1)(N+3)(N-1) \ = \ -3(N-1) \, .$$
Here it is obvious, that the equation is always fulfilled for $N \, = \, 1$, but there are some additional solutions:  Dividing by $N-1$, we find $N \, = \, -4$ and $a^2 \, = \, 4$ for possible integer solutions, which leads to $8 \, = \, mb^2$. As $m$ is squarefree, this equation has the solutions $m \, = \, 2 \, = \, |b|$.

We therefore find, that $m \, = \, |a| \, = \, |b| \, = \, 2$ leads to the solution $K \, = \, 35$ of our original problem, corresponding to the Pythagorean Triple $35^2 + 12^2 \, = \, 37^2$.

As there is potentially more flexibility in the choice of $c, u, v$, there may still be more counterexamples. Anyway, I have found a sufficiently satisfactory answer for my original question.


问:
What are the practical implications of the eigenvalues of a skew-symmetric matrix being purely imaginary or zero?
I analyse real-world data by decomposing asymmetric square matrices with zero diagonals into a symmetric and a skew-symmetric part, and treating the eigenstructure of the skew-symmetric part as providing canonical variates, in a method proposed by Gower.  I find this provides often strikingly interpretable results and am working towards encapsulating it with additional analyses in a piece of software.  For instance, simple rotation of the axes often produces instantly interpretable patterns.  The first two eigenvalues typically, but not always, account for 98% of the variance in the matrix.

Does the fact that the eigenvalues are purely imaginary have any practical consequences that may not be apparent to a non-mathematician long used to working with real positive definite matrices?

答:
The differential equation
$$
\dot x(t) + Ax(t)=0
$$
is norm-conserving (energy-conserving) for skew-symmetric $A$ in the sense that
$$
\|x(t)\|_2= \|x(0)\|_2
$$
for all $t$. This might be compared to $|\exp(it)|=1$ for all $t$.

If $A$ would be positive definite the $\|x(t)\|_2$ would decay exponentially fast.


问:
$X=\mathbb{E}[X\mid\mathcal{F}] $ implying $\mathcal{F}$-measurability of $X$
For a random variable $X$ we know, that if $X$ is measurable w.r.t some filtration $\mathcal{F}$ it satisfies $X=\mathbb{E}[X\mid\mathcal{F}]$.

I wonder whether one can reverse this property and state, if $X=\mathbb{E}[X\mid\mathcal{F}]$ then $X$ is $\mathcal{F}$-measurable. This would accumulate to $X$ is $\mathcal{F}$ measurable if and only if $X=\mathbb{E}[X\mid\mathcal{F}]$.

I guess it makes sense by definition of the conditional expectation:

Definition of the conditional expectation

The conditional expectation of $X$ given $\mathcal{F}$, denoted by $\mathbb{E}[X\mid\mathcal{F}]$, is defined as any random variable Y which satisfies

$Y$ is $\mathcal{F}$-measurable

$\int_A X d\mathbb{P} = \int_A Y d\mathbb{P}$ for all $A \in \mathcal{F}$

Does the logic work? Thanks in advance!

答: Since $\mathbb E[X\mid \mathcal F]$ is $\mathcal F-$measurable, the answer is obviouly yes.


问:
Edge-coloring with constraints (electronics)
My work is in electronics modeling, the sought procedure could be helpful in efficiently formulating circuit equations by current and voltage sources only.

For a given circuit topology, assign color/type I (current source) or V (voltage source) to components, such that

no node/vertex is only connected to type I edges

no possible mesh/loop/cycle with type V only

ground is considered type V

some components could have a fixed or preferred type.

My first approach would be constructive and trying to swap types when conflicts arise.

Are there any heuristics known? I only could find problems like minimum number of types/colors, but not the constraints as described above.

Example

Minimum spanning tree (Prim's algorithm) with netlist

(node 1 node 2 weight, node numbers see example)

```<code>0 1 1
1 2 1
2 3 2
3 0 3
</code>```

chooses the spanning (V-type) tree as

```<code>0 1
1 2
2 3
</code>```

The trick is to apply the correct weights, and these depend on the nature of component (V, I, R, L, C). Then MST gives the correct V branches.

For example, if we have a preference for C-> V (weight = 1) and L->I (weight = 3) and an assumed R||C, the branch preference becomes V, so gets the lowest cost.

答:
Solved with quarague's hint using a spanning tree and correct weighting, see the example above.

Thanks to all who contributed!


问:
What kind of differential equation is $y'=2(\frac{y+2}{x+y-1})^2$
I have an equation in the form $y'=2(\frac{y+2}{x+y-1})^2$. Does this type of equation have a name? I'm struggling to find any information about it or the solution method. My book says this can be solved similarly to the equation $y'=\frac{y+2}{x+y-1}$. That is, with substitutions we can turn this into a homogenous equation and then separate the variables. However, since the expression is inside a function this makes things difficult.

Body:
Hint: Rewrite it as
$$y'=2\left(\frac{y+2}{x+y-1}\right)^2 = 2\left(\frac{y+2}{(x-3)+(y+2)}\right)^2$$
then, let $y_1 = y+2$ and $x_1 = x-3$.


问:
Find the flux of $F(x,y,z)=(xy^2,yz^2+xze^{\sin(z^2)},zx^2+e^{x^2})$ through $S=\{(x,y,z):x^2+y^2+z^2=9,x\geq 0\}$
Find the flux of $F(x,y,z)=(xy^2,yz^2+xze^{\sin(z^2)},zx^2+e^{x^2})$ through $S=\{(x,y,z):x^2+y^2+z^2=9,x\geq 0\}$ with a non negative $x$ normal.

My attempt:

We want to use gauss. First we need to close $S$, so we define $S_2=\{(x,y,z):y^2+z^2\leq3^2,x=0\}$.

Now $S_1=S \cup S_2$ is a closed surface. $F\in C^1$ so using gauss:

$$\iint\limits_{S_1}F\cdot dr=\iiint\limits_{V}div(F)dxdydz=\iiint\limits_{V}x^2+y^2+z^2dxdydz=_{\text {spherical coordinates}}\int_0^\pi \int_{-\frac{\pi}{2}}^\frac{\pi}{2}\int_0^3r^4\sin\phi drd\phi d\theta=$$

$$48.6\pi\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\sin\phi d\phi=0$$

Now using additive of integrals:

$$\iint\limits_{S}F\cdot dr=-\iint\limits_{S_2}F\cdot dr$$

We know the normal of $S_2\rightarrow \vec n_2=(1,0,0)$

$$-\iint\limits_{S_2}F\cdot \vec n_2 ds=-\iint\limits_{S_2}xy^2$$

Im stuck here. Maybee i've misplayed all of this. Anyway, any help will be appreciated.

答:
You have a mistake in the order of your integral because we have

$x \geq 0$ i.e. $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

$0 \leq \phi \leq \pi$.

So the integral should be,

$\displaystyle \int_0^{\pi} \int_{-\pi/2}^{\pi/2}\int_0^3 r^4 \sin\phi \ dr \ d\theta \ d\phi = \frac{486 \pi}{5}$

Now to apply divergence theorem, we closed the surface with disk $y^2+z^2 = 9$ at $x = 0$. So to find flux through the spherical surface, we must subtract flux through the disk at $x=0$. Please note the outward unit normal vector to the disk $y^2+z^2 = 9$ at $x = 0$ is $(-1, 0, 0)$. But the $x$ component of the vector field is $0$ at $x = 0$. So there is no flux through the disk. Hence your final answer is as calculated above.


问:
Let $C[a,b]$ be the set of continuous functions with the sup norm. Show that $\varphi:C[a,b] \to \Bbb{R}, \varphi(f) = f(a)$ is continuous.

Let $C[a,b]$ be the set of continuous functions from $a$ to $b$ with the sup norm $d(f,g) = \sup\{|f(x)-g(x)| : a\leqslant x\leqslant b\}, f,g \in C[a,b]$. Show that $\varphi:C[a,b] \to \Bbb{R}, \varphi(f) = f(a)$ is continuous.

I'm lost on how to show this. This seemed easy at first, but I cannot find any definition to use here. $\varphi$ eats continuous functions and outputs the continous function evaluated at $a$, but how can I show this rigorously?

Body: Given $\varepsilon>0$, take $\delta=\varepsilon$ and then\begin{align}\|f-g\|<\delta&\iff\sup_{x\in[a,b]}\bigl|f(x)-g(x)\bigr|<\varepsilon\\&\implies\bigl|f(a)-g(a)\bigr|<\varepsilon\\&\iff\bigl|\varphi(f)-\varphi(g)\bigr|<\varepsilon.\end{align}


问:
Solution method for $y'=f(\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2 })$
What kind of a solution method will work for this equation? My book hints, that this may be solved with the method used for $y'=\frac{a_1 x + b_1 y + c_1}{a_2 x + b_2 y + c_2 }$. That is, first I substitute $x=\xi+h$ and $y=\eta +k$. This lead to $$y'=f\Biggr(\frac{a_1 \xi + b_1 \eta + (a_1h+b_1k+c_1)}{a_2 \xi + b_2 \eta +(a_2h+b_2k+ c_2) }\Biggr)$$

Then we can solve the linear system for the variables $h,k$ so that we are left with $$y'=f\Biggr(\frac{a_1 \xi + b_1 \eta}{a_2 \xi + b_2 \eta }\Biggr)$$

then making the substitution $\eta = u\xi$ leads to $$\eta'=f\Biggr(\frac{a_1 \xi + b_1 u\xi}{a_2 \xi + b_2 u\xi }\Biggr)$$

and $$u'\xi+u= f\Biggr(\frac{a_1 +u}{a_2 +u }\Biggr)$$

which can then be transformed into an equation with variables separated, by reducing $u$ and taking the reciprocal of the right side. This is the equation $$\frac{1}{\xi}d\xi= \biggr(f\Biggr(\frac{a_1 +u}{a_2 +u }\Biggr)-u\Biggr)^{-1}du$$

Is this right?

答:
You missed $b_1, b_2$ but otherwise looks alright.

$$\frac{1}{\xi}d\xi= \biggr(f\Biggr(\frac{a_1 +b_1u}{a_2 +b_2u }\Biggr)-u\Biggr)^{-1}du$$
And so,
$$\log(x-h) = \int \biggr(f\Biggr(\frac{a_1 +b_1u}{a_2 +b_2u }\Biggr)-u\Biggr)^{-1}du, u = \frac{y-k} {x-h}$$


答:
$$y'=2\left(\frac{y+2}{x+y-1}\right)^2$$
After the change of variables $\quad\begin{cases}
y=Y-2\\
x=X+3
\end{cases}\quad$ the ODE becomes homogeneous :
$$\frac{dY}{dX}=2\left(\frac{Y}{X+Y} \right)^2$$
The usual change to solve homogeneous ODES is $\quad Y=X\:u(X)$

Title: What kind of differential equation is $y'=2(\frac{y+2}{x+y-1})^2$

Body: I have an equation in the form $y'=2(\frac{y+2}{x+y-1})^2$. Does this type of equation have a name? I'm struggling to find any information about it or the solution method. My book says this can be solved similarly to the equation $y'=\frac{y+2}{x+y-1}$. That is, with substitutions we can turn this into a homogenous equation and then separate the variables. However, since the expression is inside a function this makes things difficult.


问:
Do you need reduced row echelon form when finding the null space?
I have heard several people say that you need to put a matrix in reduced row echelon form to find a basis for the null space of a matrix. But why does it not suffice to simply go down to row echelon form (not necessarily reduced row echelon form) and then get a basis from there?

答:
It is not needed to convert a matrix to its row reduced echelon form (rref) or even its row echelon form. If you can find the null space i.e., the linear combination of the columns that maps them to the zero vector just by looking at the original unreduced form, then you do not necessarily have to convert the matrix to rref. Gaussian elimination and Jordan back-substitution (methods used to convert a matrix to its rref) preserve the null space of the matrix, so it does not matter if you identify the null space using the original matrix or its rref.

We convert a matrix to its rref because its easier (most of the times) to identify the null space using the rref than using the original form.


问:
Solving 3-SAT with Linear Programming?
Suppose we have a set of indices $I = \{1, \dots, n\}$ and a corresponding set of boolean variables $\{X_1, \dots, X_n\}$.  Suppose further that we have a 3-CNF expression with $m$ clauses, with indices $J = \{1, \dots, m\}$, where the $j$th clause is
\begin{equation}
\left (
\bigvee_{i \in P(j)}X_i
\right)
\vee
\left (
\bigvee_{i \in N(j)} \neg X_i
\right).
\end{equation}
The sets $P(j)$ and $N(j)$ describe the variables that appear in the $j$th clause, and they satisfy the property $|P(j)| + |N(j)| \leq 3$ for each $j \in J$.

Now, suppose we solve the following linear program:
\begin{equation}
\begin{aligned}
\rho=\max_{\substack{x_i, \ \forall i \in I \\ z_j, \ \forall j \in J}} & && \sum_{j\in J} z_j \\
\text{s.t.}\ \  & && z_j \geq x_i, \ \ \ \forall i \in P(j), \forall j \in J, \\
& && z_j \geq 1- x_i, \ \ \ \forall i \in N(j), \forall j \in J, \\
& && z_j \leq \sum_{i \in P(j)} x_i + \sum_{i \in N(j)} (1 - x_i), \ \ \ \forall j \in J, \\
& && 0 \leq x_i \leq 1, \ \ \ \forall i \in I, \\
& && 0 \leq z_j \leq 1, \ \ \ \forall j \in J. \\
\end{aligned}
\end{equation}

We know that the optimal value $\rho$ satisfies $\rho \leq m$.

Will it ever be true that the 3-CNF is not satisfiable but the optimal value of the corresponding linear program is $m$?

If yes, can you provide an example of such a 3-CNF?

答:
I believe I've found my example.  Consider the following 3-CNF with $n = 3$ and $m = 8$:
\begin{equation}
(X_1 \vee X_2 \vee X_3) \wedge
(X_1 \vee X_2 \vee \neg X_3) \wedge
(X_1 \vee \neg X_2 \vee X_3) \wedge
(X_1 \vee \neg X_2 \vee \neg X_3) \wedge
(\neg X_1 \vee X_2 \vee X_3) \wedge
(\neg X_1 \vee X_2 \vee \neg X_3) \wedge
(\neg X_1 \vee \neg X_2 \vee X_3) \wedge
(\neg X_1 \vee \neg X_2 \vee \neg X_3)
\end{equation}
This 3-CNF is not satisfiable.  However, for the corresponding linear program, the solution $z_1 = \dots = z_8 = 1$ and $x_1 = x_2 = x_3 = 0.5$ is feasible, with an objective value of $\rho = 8 = m$.

I would still be interested in a simpler example, if someone can provide one.

Title: Solving 3-SAT with Linear Programming?


问:
If a two variable smooth function has two global minima, will it necessarily have a third critical point?
Assume that $f:\mathbb{R}^2\to\mathbb{R}$ a  $C^{\infty}$ function that has exactly two minimum global points. Is it true that $f$ has always another critical point?

A standard visualization trick is to imagine a terrain of height $f(x,y)$ at the point $(x,y)$, and then imagine an endless rain pouring with water level rising steadily on the entire plane.

Because there are only two global minima, they must both be isolated local minima also. Therefore, initially the water will collect into two small lakes around the minima.

Those two points are connected by a compact line segment $K$. As a continuous function, $f$ attains a maximum value $M$ on the set $K$. This means that when the water level has reached $M$, the two lakes will have been merged.

The set $S$ of water levels $z$ such that two lakes are connected is thus non-empty and bounded from below. Therefore it has an infimum $m$.

It is natural to think that at water height $m$ there should be a critical point. A saddle point is easy to visualize. For example the function (originally suggested in a deleted answer) $f(x,y)=x^2+y^2(1-y)^2$ has a saddle point at the midway point between the two local minima at $(0,0)$ and $(0,1)$. But, can we prove that one always exists?

Follow-ups:

Does the answer change, if we replace $\Bbb{R}^2$ with a compact domain? What if $f$ is a $C^\infty$ function on a torus ($S^1\times S^1$) or the surface of a sphere ($S^2$). Ok, on a compact domain the function will have a maximum, but if we assume only isolated critical points, what else is implied by the presence of two global minima?

Similarly, what if we have local minima instead of global?

If it makes a difference you are also welcome to introduce an extra condition (like when the domain is not compact you could still assume the derivatives to be bounded - not sure that would be at all relevant, but who knows).

答:
$
\def\norm#1{\lVert#1\rVert}
$The answer to the question as stated is no as Martin showed, but is yes if we add the condition that $f(x)→∞$ as $\norm{x}→∞$. Martin's example pushes the saddle point 'to infinity', which would be blocked by this condition. And we do not need global minima, nor even continuous derivatives!

Theorem. Take any differentiable $f : ℝ^2→ℝ$ such that $f$ has at least two local minima and $f(x)→∞$ as $\norm{x}→∞$. Then $f$ has a third stationary point.

Proof. Let $a,b$ be two (distinct) local minima of $f$. Let $L$ be the straight line segment from $a$ to $b$, and let $m$ be the maximum value of $f$ on $L$ by EVT (extreme value theorem). For each $k∈ℕ$ let $T(k)$ be a regular tiling of $ℝ^2$ by (closed) hexagons each with diameter $2^{-k}$ such that $a,b$ are respectively in the interior of some hexagonal tile $A,B$. Define the height of each tile $H$ in $T(k)$ to be the minimum value of $f$ on $H$, which exists by EVT. Note that if any tile $H$ has height no greater than that of all its neighbouring tiles, then $f$ has a local minimum on $H$, so we can assume that every tile besides $A$ or $B$ has height greater than that of some neighbour. Impose an enumeration on the tiles in $T(k)$ (say in hexagonal rings outward from $A$). For any tiles $G,H$, we say that $G$ is higher than $H$ (and that $H$ is lower than $G$) iff either ( $G$ has height higher than $H$ ) or ( $G,H$ have the same height but $G$ is after $H$ in the enumeration ). Note that for each tile $H$ there are only finitely many tiles lower than $H$ (since $f(x)→∞$ as $\norm{x}→∞$).

Then from any tile $H$ we can reach $A$ or $B$ via a downhill path, defined as a connected sequence of tiles each of which is higher than the next, because iteratively moving to a lower tile must terminate eventually. Thus there is a good tile, defined to be a tile of height at most $m$ from which we can reach both $A$ and $B$ each via a downhill path, because $L$ passes through a finite connected sequence of tiles from $A$ to $B$, and that sequence has consecutive tiles $I,J$ such that there is a downhill path from $I$ to $A$ and a downhill path from $J$ to $B$, so either $I$ or $J$ is a good tile. Let $M(k)$ be the lowest good tile, and let $O(k)$ be the centre of $M(k)$. Note that the second tiles of any downhill paths $P,Q$ from $M(k)$ cannot be adjacent, otherwise the higher one of those tiles would be a good tile lower than $M(k)$.

Observe that $O$ is a bounded sequence because each term is within distance $1$ from some point in $\{ x : x∈ℝ^2 ∧ f(x) ≤ m \}$, and the latter is bounded. Thus by BZ (Bolzano-Weierstrass) there is some strictly increasing sequence $i : ℕ→ℕ$ and point $c∈ℝ^2$ such that $\lim_{k→∞} O(i(k)) = c$.

From now let us assume that $f$ has only two local minima. By the local minimum of $f$ at $a$, there is some closed annulus $D$ around $a$ with inner radius $r$ and outer radius $s$ with $0<r<s<|L|$ such that $f{↾}D ≥ f(a)$. Let $u = \min_{x∈D} f(x)$. Then $u > f(a)$, otherwise $f$ has a local minimum in $D$ different from $a$ and $b$. And for all sufficiently large $k$ every downhill path from a good tile in $T(k)$ must pass through some tile contained within $D$, and so $M(k)$ has height at least $u$. Symmetrically, there is some $v > f(b)$ such that $M(k)$ has height at least $v$ for all sufficiently large $k$. Since $i$ is strictly increasing, we thus have $f(c) = \lim_{k→∞} f(O(i(k))) ≥ \max(u,v)$ and hence $c∉\{a,b\}$.

If $f$ is stationary at $c$, then we are done. Otherwise, there is some nonzero linear $g : ℝ^2→ℝ$ such that $f(c+t) ∈ f(c)+g(t)+o(\norm{t})$ as $t→⟨0,0⟩$, and hence for some sufficiently large $k$ we have that $M(k)$ and its neighbours are sufficiently close to $c$ that those neighbours lower than $M(k)$ are consecutive around $M(k)$ and number at most four. [1]

Let $P$ be a downhill path from $M(k)$ to $A$ and $Q$ be a downhill path from $M(k)$ to $B$, and let $R,S$ be the second tiles of $P,Q$ respectively. We now have two cases (up to symmetry):

(Case 1)

There is only one neighbour $X$ of $M(k)$ between $R$ and $S$:

$X$ must be lower than $M(k)$. If $X$ is higher than $R$ or $S$, then the combined path $P{+}Q$ can be altered to pass through $X$ instead of $M(k)$, so one of $X,R,S$ would be a good tile. If $X$ is lower than both $R$ and $S$, then since $X$ has a downhill path to $A$ or $B$, respectively $S$ or $R$ would be a good tile.

(Case 2)

There are two lower neighbours $X,Y$ of $M(k)$ between $R$ and $S$:

If $X$ is higher than $R$ or $Y$ is higher than $S$, then we can insert $X$ or $Y$ respectively into $P{+}Q$, which yields an instance of Case 1. If $X$ is lower than $R$ and $Y$ is lower than $S$, then by symmetry we can assume that $X$ is lower than $Y$, and so since $X$ has a downhill path to $A$ or $B$, respectively $S$ or $R$ would be a good tile.

In both cases, this contradicts minimality of $M(k)$. Therefore $c$ is indeed the point we are looking for.

−−−−−−−

[1] This part is kind of painful to prove rigorously, but it should be clear from a diagram.


问:
Explain why it is not necessary to use numerical method to find the two solutions of the equation.
This question I havent been able to understand. I am self studying and have no teacher to ask so I have been stuck for the past hour and don't get this is the full question the part I don't get is 14 c

Any help is much appreciated. Thank you!

答: Hint: If a cubic polynomial $f(x)$ has exactly two roots, then one of them must be a double root, so that it is also a root of the derivative $f'(x)$, which is a quadratic polynomial.


问:
Riemann-Stieltjes integral of a general function f with respect to the greatest integer function.
I want to prove $\sum_{n\leq x} f(n)= \int_{y}^{x}f(t)d[t]$ where the interval of integration is [y,x]. Here n is an integer.
My attempt:- By Riemann Stieltjes integration's definition, we have
$\int_{y}^{x}f(t)dt = \lim_{n \to +\infty}\sum_{ k=1}^{n-1} f(c_{k})([x_{k+1}]-[x_{k}])  $ where is $c_{k}$ is in $[x_{k}, x_{k+1}]$.
I do not know how to proceed further. Kindly guide me further.

答: I will assume that $f$ is a continuous function on the interval $[a, b].$ Consider a partition $P = (x_0 = a < x_1 < \dots < x_{n - 1} < x_n)$ and a system of intermediate points $\xi = (\xi_1, \dots, \xi_n).$ The the associated Riemann-Stieltjes sum is simply $S(f; g, P, \xi) := \sum_{1 \leq i \leq n} f(\xi_i)(g(x_i) - g(x_{i - 1})) = \sum_{1 \leq i \leq n} f(\xi_i) (\lfloor x_i \rfloor - \lfloor x_{i - 1} \rfloor).$ We consider a sequence of partitions $P_n = (x_0^n < \dots < x_N^n)$ with $\Vert P_n \Vert \to 0$ and any associated system of intermediate points $\xi_n.$ It follows that for every integer $k$ between $a$ and $b$ there is a unique interval $[x^n_{i_k - 1}, x^n_{i_k}]$ such that $k \in [x^n_{i_k - 1}, x^n_{i_k}]$ and $\lfloor x^n_{i_k} \rfloor - \lfloor x^n_{i_kk - 1} \rfloor = 1$ for $n$ sufficiently large (if $n$ happens to be one of the ends of this interval, then it would be found in two intervals of this type; however, there is only one that also verifies the second condition since $\Vert P_n \Vert$). Then $$S(f;g, P_n, \xi_n) = \sum_{1 \leq i \leq N} f(\xi^n_i) (\lfloor x^n_i \rfloor - \lfloor x^n_{i - 1} \rfloor) = \sum_{1 \leq i \leq N} f(\xi^n_{i_k}).$$ Since $\xi^n_{i_k}$ is always in the interval that contains $k,$ $\Vert P_n \Vert \to 0$ and $f$ is continuous, we get that $f(\xi_{i_k}^n) \to f(k)$ as $n \to \infty.$ It follows that $$\int_a^b f d\lfloor t \rfloor = \sum_{n \in \mathbb{Z} \cap [a, b]} f(n),$$ where this sum is taken to be $0$ if the intersection is empty. I hope this helps. :)


问:
Levy-Ito decomposition intuition
So a Levy process $(X_t)_{t\geq0}$ can be decomposed into three parts

$$X_t = \mu t + \sigma^2B_t + L_\nu(t)$$ where $L_\nu(t)$ is "a compound Poisson process with Levy measure $\nu$".

I know the Levy measure of a set $A$ is the expected number of steps of $X_t$ having step size in $A$ in a time interval t. I.e. $\nu(A) = \mathbb{E}[N(1,A)]$ where $N(t,A)$ is the Poisson random measure of a set of counts $A$ in a time interval $t$. In the paper I'm looking at, $$L_\nu(t) = \int_{|h|>1} h N_\nu (t,dh) + \int_{|h|\leq 1} h \big(N_\nu(t,dh) - t\nu(dh)\big).$$

I understand the first two terms of the decomposition, linear drift and Brownian motion, but not the third. Roughly, the third component seems to contribute information about how fast the process is moving. But I don't understand the details -- like, why are we adding $L_\nu(t)$ rather than scaling somehow? And why are we integrating over $h$ in this way?

答:
$L_\nu(t)$ tracks the (partially compensated) sum of the jumps of $X$ that have occurred up to time $t$. Because $\nu(h:|h|>1)<\infty$, there are finitely many jumps in $[0,t]$ of magnitude $>1$. Thus the integral $\int_{|h|>1}hN_\nu(t,dh)$ is just a (random) finite sum.
As for the other integral, the Lévy measure $\nu$ satisfies $\int_{|h|\le 1}h^2\nu(dh)<\infty$, but it may happen that $\nu(h:|h|\le 1)=\infty$ or even $\int_{|h|\le 1}|h|\nu(dh)=\infty$. The first infinity signals that $N_\nu(t,\{|h|\le 1\})=\infty$ a.s.; i.e., that there are infinitely many small jumps up to time $t$. The second integral in your formula for $L_\nu(t)$ is, more accurately,
$$
\lim_{n\to\infty}\int_{1/n<|h|\le 1}h\left(N_\nu(t,dh)-t\nu(dh)\right).
$$
The compensating term $t\nu(dh)$ is just the mean of $N_\nu(t,dh)$. The  integral above converges absolutely, and the $n\to\infty$ limit can be understood to hold in the $L^2$ sense. The $L^2$ convergence is made possible by $\int_{|h|\le 1}h^2\nu(dh)<\infty$.
The point of the compensation is that although $\int_{|h|\le 1}|h|N_\nu(t,dh)$ may diverge, it does so like $t\int_{|h|\le 1}|h|\nu(dh)$, so if the singularity is subtracted off then convergence can occur.

The Poisson process $N_\nu$ is determined by the jumps of $X$:
$$
N((u,v]\times A]=\#\{s>0: u<s\le v, X_s-X_{s-}\in A\}
$$

(What is your source for information about Lévy processes?)

Title: Levy-Ito decomposition intuition


问:
Norm with maximum of coefficients on polynomial ring

Given a non-archimedean normed field $(F,|\cdot|)$, show we can extend the norm to $F[x]$ by setting $|a_0+a_1x+\ldots+a_nx^n|\mapsto \max(|a_0|,|a_1|,\ldots,|a_n|).$

I am having trouble showing the multiplicativity of this extended norm. We have to show that
\begin{equation*}
\max_k \left\vert \sum_{i+j=k} a_ib_j \right\vert=\max_i |a_i|\cdot \max_j |b_j|.
\end{equation*}
Since $|-|$ is non-archimedean, we have
$$
\left\vert \sum_{i+j=k} a_ib_j \right\vert\leq \max_{\substack{i,j \\ i+j=k}} |a_i | |b_j| \leq \max_{i,j}|a_i| |b_j|=\max_i |a_i|\cdot \max_j |b_j|.
$$
Taking $\max_k$ on both sides gives one equality of what we need to show. However, I need to inspect when equalities occur.

Any hints?

答:
Let
$$
f=a_0+a_1x+a_2x^2+\dots+a_nx^n
$$
and
$$
g=b_0+b_1x+b_2x^2+\dots+b_mx^m.
$$
Let $r$ be the smallest index such that $|a_r|=|f|$ and $s$ be the smallest index such that $|b_s|=|g|$.

Now, consider $|fg|$.  In particular, let's look at $|(fg)_{r+s}|$.  As you note,
$$
(fg)_{r+s}=\sum_{i=0}^{r+s}a_ib_{r+s-i}.
$$
We observe that $|a_rb_s|=|f||g|$.  Moreover, for $i\not=r$, $|a_ib_{r+s-i}|<|f||g|$ by the assumptions on $r$ and $s$ (either $i<r$ or $r+s-i<s$).  One of the properties of non-Archimedean norms states when the norm of a sum is equal to the maximum norm of the summands, this is exactly the current case.


问:
Fourier series of a piecewise function, is my coefficient $a_n$ correct?
Here is a piecewise function extended periodically with period $2\pi$.
$$f(x)=
    \begin{cases}
        \pi + x & -\pi<x<-\frac{\pi}{2} \\
        x & -\frac{\pi}{2}< x<\frac{\pi}{2} \\
        \pi - x & \frac{\pi}{2}< x<\pi \\
    \end{cases}
$$

I know to find the $a_n$ coefficients of a Fourier Series I can rewrite this in the form
$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cdot\cos(nx)dx\\
=\frac{1}{\pi}\left(\int_{-\pi}^{-\frac{\pi}{2}}(\pi+x)\cdot\cos(nx)dx+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x\cdot\cos(nx)dx+\int_{\frac{\pi}{2}}^{\pi}(\pi-x)\cdot\cos(nx)dx\right)$$

I then note that the middle integrand is an odd function overall, and is being evaluated over such an interval that the result would be zero, so it can be discarded. Then, after evaluating the outer two integrals and using the facts that $\sin(n\pi)=0\:\forall\, n$, $\cos(-x)=\cos(x)$, and $\sin(-x)=-\sin(x)$, I discover a number of cancellations and end up with the following.
$$a_n=\frac{2}{\pi n^2}\left(\cos(n\pi)-\cos(\frac{n\pi}{2})\right)$$

When I iterate through the $n$ inside the cosine functions only, I find that $a_n$ is always one of $\{\frac{-2}{\pi n^2},\frac{4}{\pi n^2},\frac{-2}{\pi n^2},\frac{0}{\pi n^2}\}$ where these values correspond to the multiples $\{1\cdot n,2\cdot n, 3\cdot n,4\cdot n\}$ respectively.

I can't find a way to express this so that I can rid myself of the cosine functions within the expression, I have only begun studying this and we can usually turn this into something like $\frac{2}{\pi n^2}(-1)^{n}$ or other similar expressions.

I've double and triple checked my integrations and I'm 99.99% sure I've made no mistakes on that front, any help would be appreciated.

答:
If you read my comment, we have

$$\cos n\pi-\cos\frac{n\pi}2=(-1)^n-(-1)^{n/2}=\begin{cases}-1,&n\;\text{ is odd}\\{}\\0,&n=2\pmod 4\\{}\\2,&n=0\pmod 4\end{cases}$$