答:
I'll give it a naive try.

Let's attempt to prove it by induction, and lets set $n=0$ to prove our base case. Since all natural numbers are bigger than $1$ (or are $1$), we know that there are no natural numbers in $(0,1)$. Base case proved.

Now let's set $n=k+1$, instead of $n=k$. Then we need to show that $(k+1,(k+1)+1)$ reduces to $(k,k+1)$ somehow. Now, for every element in $x\in(k+1,(k+1)+1)$, really $x\in(k,k+1)+1$, making the union of all sets of the form $(n,n+1)$ inductive if $1\in{(0,1),(1,2),(2,3),...}$, which it's not. So because $1$ is not in this set, $1+k$ for $k=0,1,2,...$ is also not in this set by the definition of inductive, so we cannot get the natural numbers from these sets by the definition of  a natural number.

Title: How to prove rigorously that, for any natural number $n$, the open interval $(n, n+1)$ contains no natural number?

Body:
A subset $E$ of $\mathbb{R}$, the set of real numbers, is said to be inductive if $1 \in E$, and for every $x \in E$ the number $x+1 \in E$ also.

The set $\mathbb{N}$ of natural numbers is defined to be the intersection of all the inductive sets of real numbers.

Since the sets $\{ x \in \mathbb{R} | x > 0 \}$ and $\{ x \in \mathbb{R} | x \geq 1 \}$ are inductive, therefore we can conclude that all the natural numbers are $\geq 1$.

Now using this machinery how can we (use induction to) prove the following statement?

For any natural number $n$, there exists no natural number in the open interval $(n, n+1)$.


问:
Dominated Convergence & Differential Equation Limits
can someone kindly help me with these few questions? :)

$\displaystyle \frac{dy}{dx} + \left( x+\frac1{x} \right) y = 1$ with $y(1)=0$.

I do not get what limits to take in the integration factor and/or when integrating both sides. I managed the integration/method well - just the limits are causing some problems!

Let $K(x,t)$ be a continuous function defined for $x\in [a, +\infty)$ and $t\in [c,d]$ where $a$, $c$, $d$ are fixed real numbers. What does it mean to say that $K(x,t)$ has dominated convergence on $[a, +\infty) \times  [c,d]$?

Find $\displaystyle \frac{d}{dt} \left( \int_1^{+\infty} \frac 1x e^{-xt} dx \right)$ for $t>0$, indicating clearly why your manipulations are justified.

I managed to do the integration (by taking the $\displaystyle \frac{d}{dt}$ inside the integral). I’m not sure how to justify it, because you can’t find $\displaystyle \max \left(\frac1{x} e^{-xt}\right)$ via differentiation.

p.s. this is exam prep not homework! :)

答:
Here are some hints, but you can find answers to all three questions in any calculus book :-)

I do not get what limits to take in the integration factor and/or when integrating both sides.

Since we are talking about an ordinary differential equation with prescribed initial value, I'm not quite sure what you mean by "limits to take in the integration factor". If you are talking about "separation of variables and integration", then the integration part is about finding the antiderivative of the integrant, not about doing a definite integral.

What does it mean to say that K(x,t) has dominated convergence...

I know the term "dominated convergence" from one context only, the Lesbegue dominated convergence theorem, see Wikipedia.

I managed to do the integration (by taking the ddt inside the integral). I’m not sure how to justify it...

This is called the Leibniz integral rule (Wikipedia), or more generally differentiation under the integral sign. Every analysis course will cover some theorem that states sufficient conditions such that the switch of differentiation and integration is valid, depending on the notion of integral (Cauchy, Riemann, Lesbegue integral...). You simply need to look up which version was used in your class and therefore which conditions you need to prove (or at least tell us in your question, so that we can help you with that).

ContentLicense: CC BY-SA 3.0


问:
Bijection $\mathbb N \to \mathbb Z$ using floor function. Define bijection $\mathbb Z \to \mathbb Q$
After defining the bijection
$$
F: \mathbb{N}\to \mathbb{Z}$$ using floor function,
and by defining bijections:

$$\mathbb{N} \to \mathbb{N} \setminus \{ 0 \}\\
\mathbb{Q} \to \mathbb{Q} \setminus \{ 0 \}
$$

define bijection:

$$\mathbb{Z} \to \mathbb{Q} \setminus \{ 0 \}$$.

After a few attempts I got first one $F: \Bbb N \to \Bbb Z$ , $F(x) = (-1)^x \lfloor(x/2)\rfloor$ but I'm stuck with other ones.
I'm new into that course at university - if I made some mistakes - I'm sorry.

答:
$F : \mathbb N \to \mathbb Z$ defined as $F(n)=(-1)^n \lfloor {n+1 \over 2} \rfloor$.

$G : \mathbb N \to \mathbb N \backslash \{0\}$ defined as $G(n)=n+1$.

$H : \mathbb Q \to \mathbb Q \backslash \{0\}$ defined as $H(n)=n+1$ for $n \in \mathbb N$ and identity function otherwise.

You can find very nice detailed answers here.

ContentLicense: CC BY-SA 4.0


问:
Considering all the permutations of the following word, what is the total number of words in which the the following condition is true? ${}$
Among all the permutations of the letters of the word " JANLOKPAL", the number of words in which "O" never lies between "J" and "N" is?

I first thought of subtracting the cases in which O does occur between J and N, and then subtracting it from total possible permutations, but I can't seem to arrive at the correct answer.

Thank you.

答:
Let's count the words which contains blocks JON or NOJ. _J_O_N_ has four spaces around it and we have to put six more letters. One letter can be filled in four ways. Now our block looks like e.g., _J_O_A_N_. We have five spots available for second letter. We go on filling, as the available spots keeps increasing by $1$.

Number of words containing JON or NOJ (keeping in mind to not overcount the two repeated letters) is
$$2 \times \frac{4\cdot5\cdot6\cdot7\cdot8\cdot9}{2!2!}$$

Subtracting this from total, desired result is
$$\frac{9!}{2!2!}-\frac{9!}{2!2!\times 3}=\frac{2}{3}\times\frac{9!}{2!2!}=60480$$

Looking at the last expression, we realize that we could have simply ignored other letters. Since from $3!$ permutations of $J,O,N$, only four were desired, the straightforward answer is
$$\frac{4}{3!}\times\frac{9!}{2!2!}=\frac{2}{3}\times\frac{9!}{2!2!}=60480$$


问:
Why does not noise accumulate in a stochastic differential equation as it would in a random walk?
I had a question about stochastic differential equations. I come from statistics, so I am more familiar with say random walks, which are of course, discrete. So if I had a process like where $x(t)$ traces the random walk evolution of a variable over time, then that might look like this recursively:

$$
x(t+1) = x(t) + \epsilon \sim Binomial(p,n)
$$

where $p,n$ are parameters for the chance of success and number of trials, respectively. If I ran this simulation multiple and plotted time $t$ versus $x(t)$, I would get a fan or cone shaped plot with all trajectories starting at the same point, but then progressively spreading out with each time step.

My question is, when I look at plots of stochastic differential equations, I don't see this same kind of fanning behavior. When talking about an SDE of the form:

$$
du = f(u, p, t)dt + g(u, p, t)dW
$$

In this case, $f$ is the deterministic part of the process and $g$ is the noise process, usually Brownian motion. This is of course a continuous time equation. Here is a plot of a stochastic version of the lorenz equation that conforms to the SDE form above--taken from the Julia Differential Equations documentation.

Again, my question is, why don't we see the trajectory fan out over time. It seems like the trajectory does wiggle, but it does not diffuse as in a random walk. I was just trying to understand what are the forces at play that prevent the noise process to overwhelm the deterministic part of the process over time?

答:
The chaotic attractor of the Lorenz system is, as the name says, attracting. Thus any deviation introduced by the stochastic term is corrected towards the double spiral.

You would still get the trajectory fan if you were to compute and plot several paths with different random walks. Again, any deviation from the surface of the spirals will be quickly corrected, but inside that surface you these random movements stay or are even slightly increased by the dynamic.

As one can see, in the outer rim of the spirals where the velocity of the Lorenz system is high the paths are almost smooth, while in the slower inner parts the noise terms have more influence.


答:
My interpretation is same as what N.F. Taussig said that all are permissible arrangements except where $O$ is somewhere between $J$ and $N$.

Here is how I counted -

Total permissible arrangements $ = \displaystyle \frac{9!}{2! \ 2!}$

We first make all bad arrangements. So we start with placing $ \_ J \_ O \_ N \_ $ . That gives us $4$ spaces and we either choose one of the places to place two $L$ together or we choose two of $4$ places for one L each. Now we have a string of $5$ letters and that gives us $6$ spaces. We now place two $A$ the same way. Then we have a string of $7$ letters and we place $K$ in one of the $8$ places and finally $P$ in one of the $9$ places.

We have same number of arrangements for $N \ O \ J$.

Lastly, we subtract from total arrangements to find number of good arrangements.

So the answer is

$\displaystyle \small \frac{9!}{2! \ 2!} - 2 \cdot \bigg[{4 \choose 1} + {4 \choose 2}\bigg] \cdot \bigg[{6 \choose 1} + {6 \choose 2}\bigg] \cdot 8 \cdot 9 = 60480$

Title: Considering all the permutations of the following word, what is the total number of words in which the the following condition is true? ${}$

Body:
Among all the permutations of the letters of the word " JANLOKPAL", the number of words in which "O" never lies between "J" and "N" is?

I first thought of subtracting the cases in which O does occur between J and N, and then subtracting it from total possible permutations, but I can't seem to arrive at the correct answer.

Thank you.


问:
Showing a distribution is not complete for a parameter $\theta$
I am trying to show that a normal distribution with parameters $\mu = 0$ and variance $\theta$ is not complete. I am looking for a function $u(x)$ that is not equal to 0 such that $\mathbb E(u(x)) = 0$.

I have done some research on this problem and I have found that $\bar{X}$ and $S$ (sample standard deviation) are independent and can help yield me a function that will give me $\mathbb E(u(x)) = 0$. I know that $\bar{X}$ is normally distributed with mean 0 and variance $\theta/n$. I know that $S^2$ has a chi-squared distribution. I was trying to take an expectation $\mathbb E(\bar{X} S^2)$ to yield a value of 0, but I am not sure how exactly to do that.

Also, would this approach be correct in showing that it is not complete for $\theta$?

EDIT:
An iid sample is taken and theta > 0.

答:
I think your question is not accurate. The question is, show that the family of distributions $\{f(x,\theta); 0<\theta<\infty\}$ where $X$ is distributed as $\operatorname{Normal}(0, \theta)$ is NOT complete.

A test for completeness can be done by finding a function $u(x)$ not equal to the zero-function such that $E[u(x)]=0$ $ \forall \theta$.

For this normal family this is rather easy, since you know the expected value of X is $0$ and it does not matter who $\theta$ is. Just use $u(x)=x$.

And you do not need a sample at all. But if you want to, use it and use the joint distribution. The same will hold.


问:
There exists a square matrix $A$ with rational entries and dimension $2019 \times 2019$ such that $A^3 + 6A^2 - 2I = 0$?
a) There exists a square matrix $A$ with rational entries and dimension $2019 \times 2019$ such that $A^3 + 6A^2 - 2I = 0$?

b) And a square matrix $A$ with rational entries and dimension $2019 \times 2019$ such that $A^4 + 6A^3 - 2I = 0$?

-

Im trying to answer this questions. For the first one, i thought the following:

$A^2\cdot\dfrac{(A + 6I)}{2} = I \Longrightarrow det(A)^2 = \dfrac{2^{2019}}{det(A+6I)}$

Then, because the matrix has rational entries, the determinant of $A$ must be rational, and then:

$det(A + 6I) = k^2\cdot(2^{2n + 1})$, where $k \in \mathbb{Q^*}$, $n \in \mathbb{Z}$.

And we obtain $det(A) = \dfrac{2^{1009-n}}{k}$.

I was hopping i could get a contradiction out of this but i'm not able to find one. Any hints? Thanks! (On the question b) i tried the same thing but didn't work).

答:
THIS IS WRONG, PLEASE CHECK EDIT

(a) Let,
$$
A = \begin{pmatrix}
0 & 0 & 2 \\
1 & 0 & 0 \\
0 & 1 & -6
\end{pmatrix}
$$
Then a matrix that'll work is,
$$
\begin{pmatrix}
A & 0 \\
0 & I_{2016}
\end{pmatrix}
$$
(b) Let,
$$
B = \begin{pmatrix}
0 & 0 & 0 & 2 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & -6
\end{pmatrix}
$$
Then a matrix that'll work is,
$$
\begin{pmatrix}
B & 0 \\
0 & I_{2015}
\end{pmatrix}
$$
The $A$ and $B$ are called companion matrices of the corresponding polynomials.

EDIT : As pointed out in the comments, the above solution is wrong. I sort of typed it in a hurry, and didn't check if it actually worked or not. I apologise for that. I'm keeping this for context of the comments.

Having said that, as pointed out by @greg in the comments, $I \otimes A$ does work for (a) and (b) is impossible, as pointed out by @JyrkiLahtonen.


问:
Relation between end point of focal chord of parabola
Consider a parabola $y^2 = 4ax$ , parameterize it as $x=at^2$ and $y=2at$, then it is found that if we have a line segment passing through focus, with each points having value of $t$ as $t_1$ and $t_2$ for the parameterization, then it must be that:

$$ t_1 \cdot t_2 = -1$$

Hope for hints.

答:
Hint:

Let the parameters be $p,q$. So the eq of the line is $(y-2ap)(ap^2-aq^2)=(x-ap^2)(2ap-2aq)$. What point does this pass through so that it is a focal chord?


答:
We can count directly.

The word JANLOKPAL contains nine letters, of which one is a J, two are As, one is an N, two are Ls, one is an O, and one is a K.  We wish to count those permutations of the letters of the word JANLOKPAL in which the O does not appear somewhere between the J and the N.

We have nine positions to fill.  Choose two of the nine positions for the As, two of the remaining seven positions for the Ls, one of the remaining five positions for the K, one of of the remaining four positions for the P.  That leaves three positions to fill with J, N, and O.  Since the O cannot be placed between the J and the N, it must fill either the first or last of these positions, giving us two choices for placing the O.  Once we have placed the O, there are no further restrictions.  Hence, we can place the J and N in the remaining two positions in $2!$ ways.  Thus, there are
$$\binom{9}{2}\binom{7}{2} \cdot 5 \cdot 4 \cdot 2 \cdot 2! = 60,480$$
permutations of the letters of the word JANLOKPAL in which the O does not appear somewhere between the J and the N.


问:
How many sets can we get by taking interiors and closures?
Consider the following problem. I am looking for the maximum number of different sets that we can generate by one set $B \subseteq \mathbb{R}$ by taking a finite amount of closures and interiors. For example $\{0\}$ generates the sets $\{0\}$ and $\emptyset$. At first I thought the answer was 3 (we can only generate B, the closure of B and the interior of B) because I thought $\overline{B^\circ}=\overline{B}$ and $\overline{B}^\circ=B^\circ$. But then I looked at the example $B=\mathbb{Q}$ and found that $\overline{\mathbb{Q}}^\circ=\mathbb{R}^\circ=\mathbb{R}\neq\emptyset=\mathbb{Q}^\circ$.

This is not what I would expect and I clearly need another method to find that maximum.

答:
Trying with the set $B=(0,1)\cup(1,2)\cup\{3\}\cup([4,5]\cap\mathbb{Q}$ (as suggested by Hagen von Eitzen) gives me that the answer has to be at least 8.

Taking first the closure followed by the interior followed by the closure and so on, gives me 4 extra possibilities.

Starting with the interior instead gives me 3 more possibilities.

Including the set I have started with, I have 8 possibilities. The only question that is left: is it the maximum? I guess so, but to prove it I think a need to find certain relations between the closures and interior so that I can relate them to the two sequences I found with this example. More precisely (if we denote taking an interior by I and taking a closure by C) we have to proof that:

CICIC = CIC and ICIC = IC

Any suggestions for proofing this?


问:
Find this limit without using L'Hopital's rule or Taylor Series
I want to solve $$\lim_{x\to 1^{-}}{\frac{\sqrt[4]{\cos{\frac{\pi x}{2}}}}{\arccos(\frac{2}{\pi}\arcsin(x))}}$$
without using differentiation in any context. Basically without using L'Hopital's rule and Taylor's theorem.

I have managed to simplify it somehow by substituting $x = -\frac{2}{\pi}y +1$:
$$\lim_{x\to 1^{-}}{\frac{\sqrt[4]{\cos{\frac{\pi x}{2}}}}{\arccos(\frac{2}{\pi}\arcsin(x))}} = $$
$$\lim_{y\to 0^{+}}{\frac{\sqrt[4]{\sin{y}}}{\arccos(\frac{2}{\pi}\arcsin(-\frac{2}{\pi}y +1))}}=$$
$$\lim_{y\to 0^{+}}{\frac{\sqrt[4]{\sin{y}}}{\sqrt[4]{y}}}\lim_{y\to 0^{+}}{\frac{\sqrt[4]{y}}{\arccos(\frac{2}{\pi}\arcsin(-\frac{2}{\pi}y +1))}} =$$
$$1 \lim_{y\to 0^{+}}{\frac{\sqrt[4]{y}}{\arccos(\frac{2}{\pi}\arcsin(-\frac{2}{\pi}y +1))}}$$

However I am having trouble simplifying further.

I am allowed to use the following facts:
$$\sin{x} = x + o(x), \quad x \to 0$$
$$\tan{x} = x + o(x), \quad x \to 0$$
$$\arcsin{x} = x + o(x), \quad x \to 0$$
$$\arctan{x} = x + o(x), \quad x \to 0$$
$$\ln{(1+x)} = x + o(x), \quad x \to 0$$
$$\cos{x} = 1 - \frac{x^2}{2} + o(x^2), \quad x \to 0$$
$$a^x = 1+ x\ln{a} + o(x), \quad x \to 0$$
$$(1+x)^a = 1 + ax + o(x), \quad x \to 0$$
$$\cos{x}\sim 1, \quad x \to 0$$
$$if \quad \lim_{x \to a}{g(x) \ln{f(x)}} = A \quad then, \ \lim_{x \to a}{(f(x))^{g(x)}} =
\begin{cases}
e^A \ &if \ A \in R \\
\infty  \ &if \ A = \infty \\
0  \ &if \ A = -\infty
\end{cases}
$$

Any hints or help will be very much appreciated.

答:
From $$ \cos x = 1 - \frac{x^2}{2} + O(x^4) \qquad \text{for }x\to 0 $$
we can conclude
$$ x = \arccos\Big(1-\frac{x^2}{2} + O(x^4)\Big) \qquad \text{for }x\to 0  $$
$$ \sqrt{2t} = \arccos\Big(1-t + O(t^2)\Big) \qquad \text{for }t\to 0^+  $$
$$ \arccos(1-t) = \sqrt{2t\big(1 + O(t)\big)} \qquad \text{for }t\to 0^+  $$
$$ \arccos(y) = \sqrt{2(1-y)\big(1 + O(1-y)\big)} \qquad \text{for }y\to 1^-  $$
$$ \arccos(y) = \sqrt{2(1-y)} +  O\big((1-y)^\frac32\big)  \qquad \text{for }y\to 1^-  $$
Since $$\arcsin x = \frac{\pi}{2}-\arccos x $$ we have
$$ \arcsin(x) = \frac{\pi}{2} - \sqrt{2(1-x)} +  O\big((1-x)^\frac32\big)  \qquad \text{for }x\to 1^-  $$
$$ \frac{2}{\pi}\arcsin(x) = 1 - \frac{2}{\pi}\sqrt{2(1-x)} +  O\big((1-x)^\frac32\big)  \qquad \text{for }x\to 1^-  $$
\begin{align} \arccos \big(\frac{2}{\pi}\arcsin(x)\big) &= \arccos\Big(1 - \frac{2}{\pi}\sqrt{2(1-x)} +  O\big((1-x)^\frac32\big) \Big) = \\
&= \sqrt{\frac{4}{\pi}\sqrt{2(1-x)} + O\big((1-x)^\frac32\big)} + O\big((1-x)^\frac34\big) = \\
&= \sqrt{\frac{4\sqrt{2}}{\pi}} (1-x)^\frac14 +  O\big((1-x)^\frac34\big)  \qquad \text{for }x\to 1^- \end{align}
Similarily, because $$ \cos(x) = \sin(\frac{\pi}{2}-x)$$
we have
$$ \cos(x) = (\frac{\pi}{2}-x) + O\big((\frac{\pi}{2}-x)^3\big) \qquad \text{for } x\to \frac{\pi}{2}$$
$$ \cos(\frac{\pi x}{2}) = \frac{\pi}{2}(1-x) + O\big((1-x)^3\big) \qquad \text{for } x\to 1$$
\begin{align} \sqrt[4]{\cos(\frac{\pi x}{2})} &= \sqrt[4]{\frac{\pi}{2}(1-x) + O\big((1-x)^3\big)}\\
&= \sqrt[4]{\frac{\pi}{2}(1-x)} \sqrt[4]{1+O\big((1-x)^2\big)} =\\
&= \sqrt[4]{\frac{\pi}{2}} (1-x)^\frac14 + O\big((1-x)^\frac94\big) \qquad \text{for } x\to 1^- \end{align}
In total, we have
\begin{align} \frac{\sqrt[4]{\cos(\frac{\pi x}{2})}}{\arccos \big(\frac{2}{\pi}\arcsin(x)\big)} &= \frac{\sqrt[4]{\frac{\pi}{2}} (1-x)^\frac14 + O\big((1-x)^\frac94\big)}{\sqrt{\frac{4\sqrt{2}}{\pi}} (1-x)^\frac14 +  O\big((1-x)^\frac34\big)} = \\
&= \frac{\pi^\frac34}{2\sqrt{2}} + O\big((1-x)^\frac12\big)\end{align}


问:
integral operator norm
I have to find the norm of the Volterra integral operator, but I don't know if I did this correct or not. Can anyone help?
$$A:(C[0,1],\mathbb{R})\rightarrow (C[0,1],\mathbb{R})$$
where,
$$(A\phi)(x)=\int_0^x f(y)dy$$
My solution:
\begin{align*}
\|A\|&=\sup{\frac{|(A\phi)(x)|}{\|\phi\|_{\infty}} \text{ s.t. } \phi \in (C[0,1],\mathbb{R}), \phi \neq 0 }\\
&=\sup{|(A\phi)(x)|}=\sup{|\int_0^x \varphi(y)dy|, \|\varphi\|_{\infty}=1}\\
&=\sup(x)
\end{align*}

答:
There's some good work here, but there are mistakes. The first is that you'll have to use inequality signs, since you brought the absolute values in and then supped out the $\varphi$. Also, you are taking the sup over $x$ and $x\in [0,1]$, so that last sup is actually $1$. You should also be clarifying that you're taking the sup in both $x$ and $\varphi$.

Here's your work with these fixes implemented: for any $x\in [0,1]$ and $\varphi\in C([0,1],\mathbb{R})$ with $\|\varphi\|_\infty=1,$ we have the chain of inequalities
$$|(A\varphi)(x)|=\left|\int\limits_0^x \varphi(y)\, dy\right|\leq \int\limits_0^x |\varphi(y)|\, dy\leq \int\limits_0^1 |\varphi(y)|\, dy\leq \sup_{x\in [0,1]}|\varphi(x)|\int\limits_0^1\, dy=1.$$  Since the left-hand side is independent of both $x$ and $\varphi$, we can take sups over both of them and get that $$A=\sup_{\|\varphi\|_\infty=1}\|A\varphi\|_{\infty}\leq 1.$$ On the other hand, if we take $\varphi(x)=1,$ then $\|A\varphi\|_{\infty}=1$, and thus $$\|A\|=1.$$

Title: integral operator norm


问:
Circle and Rectangle: Solving Systems of Nonlinear Equation
Hello Good day everyone,

Can everyone help me with this problem.

From a circular piece of metal sheet with diameter $20 \text{cm}$ a rectangular piece with perimeter $28\text{cm}$ is to be cut. Find the dimensions of the rectangular piece.

I tried answering this one but I have doubts if I am going to use the system with equation $x^2+y^2=20^2$ and $x+y=14$. And as I evaluate the system, it leads to a imaginary numbers.

Thank you for the response!

答:
lets say that both the circle and rectangle are at the same centre, my answer is best understood if visualized, in the rectangle the length " $b$ " is horizontal and the breath " $a$ " is vertical

we know that the perimeter $2(a+b)$ of the rectangle is $28$, then $a+b = 14$ and the circle has a diameter of $28$, then it's area is $\pi \cdot (20/2)^2$
obviously the area of the rectangle has to be less than the circle, since it was cut from it, then $a\cdot b < \pi \cdot 10^2$ and $b = 14-a$

interpreting the maths from my diagram, the circle $!(1,2,3,4,5,6,7,8,9)$ is cut into $9$ parts, according to my theory we would use the information to extract the dimension of the rectangle

so that area of the circle $!(1,2,3,4,5,6,7,8,9)$ equals area of the rectangle $!(9)$ plus area of the eclipse $!(1,2,3,5,6,7)$ plus area of the second eclipse $!(1,3,4,5,7,8)$ minus area of the sector $!(1)$ minus area of the sector $!(3)$ minus area of the sector $!(5)$ minus area of the sector $!(7)$

hehe,  because a cicrcle has a perfect regular shape, the area of the sector $!(1)$ equals sector $!(3)$ equals sector $!(5)$ equals sector $!(7)$, this sector is the sector of some eclipse

we have to find the two radius of the eclipse $!(1,2,3,5,6,7)$ which is
$$ R_{a} = \sqrt{10^2-(\frac{b}{2})^2 } $$
$$ R_{b} = \frac{20-b}{2}$$
the radius of the second eclipse $!(1,3,4,5,7,8)$ is
$$ r_{a} = \frac{20-a}{2}$$
$$ r_{b} = \sqrt{10^2-(\frac{a}{2})^2}$$
the radius of the sector at the edge is also calculated, sector $!(1)$=$!(3)$=$!(5)$=$!(7)$
$$ ( R_{a} - \frac{  a}{2} )$$ and $$ ( r_{b} - \frac{ b}{2} )$$
area of circle $!(1,2,3,4,5,6,7,8,9)$ = area of rectangle $!(9)$ + area of eclipse $!(1,2,3,5,6,7)$ + area of eclipse $!(1,3,4,5,7,8)$  - 4 * area of sector $!(1)$

$$ \pi \cdot 10^2 = a b + \pi \cdot R_{a} \cdot R_{b} + \pi \cdot r_{a} \cdot r_{b} - 4 \cdot \frac{1}{4} \cdot \pi \cdot ( R_{a} - \frac{ a}{2}) \cdot ( r_{b} - \frac{ b}{2}) $$

$$ \pi \cdot 10^2 = a b + \frac{ \pi  (20-b) \cdot \sqrt{10^2-\frac{b}{2}^2} }{2} + \frac{ \pi  (20-a) \cdot \sqrt{10^2- \frac{a}{2}^2 }}{2}  - 4 \cdot \frac{1}{4} \cdot \pi \cdot (  \sqrt{10^2-\frac{b}{2}^2 } - \frac{a}{2} )\cdot (  \sqrt{10^2-\frac{a}{2}^2 } - \frac{b}{2} )$$

$$ \pi \cdot 10^2 = a b + \frac{ \pi (20-b) \cdot \sqrt{10^2-\frac{b}{2}^2} }{2} + \frac{ \pi  (20-a) \cdot \sqrt{10^2-\frac{a}{2}^2 }}{2}  -  \pi \cdot   \sqrt{10^2-\frac{b}{2}^2}  \cdot   \sqrt{10^2-\frac{a}{2}^2 } + \pi \cdot \frac{b}{2} \cdot \sqrt{10^2-\frac{b}{2}^2}  + \pi \cdot \frac{a}{2} \cdot \sqrt{10^2-\frac{a}{2}^2}  - \pi \cdot \frac{a \cdot b}{4} $$

$$ \pi \cdot 10^2 - a b +  \pi \cdot \frac{a  b}{4}  + \pi \cdot   \sqrt{10^2-\frac{b}{2}^2}  \cdot   \sqrt{10^2-\frac{a}{2}^2 } =  \frac{ \pi (20-b) \cdot \sqrt{10^2-\frac{b}{2}^2} }{2} + \frac{ \pi  (20-a) \cdot \sqrt{10^2-\frac{a}{2}^2} }{2}  + \pi \cdot \frac{b}{2} \cdot \sqrt{10^2-\frac{b}{2}^2}  + \pi \cdot \frac{a}{2} \cdot \sqrt{10^2-\frac{a}{2}^2}  $$

$$  (\pi\cdot 10^2 - a b +  \pi \cdot \frac{a \cdot b}{4}  + \pi \cdot   \sqrt{10^2-\frac{b}{2}^2}  \cdot   \sqrt{10^2-\frac{a}{2}^2 } )^2 = (  \frac{ \pi  (20-b) \sqrt{10^2-\frac{b}{2}^2} }{2} + \frac{ \pi  (20-a) \cdot \sqrt{10^2-\frac{a}{2}^2} }{2}  + \pi \cdot \frac{b}{2} \cdot \sqrt{10^2-\frac{b}{2}^2}  + \pi \cdot \frac{a}{2} \cdot \sqrt{10^2-\frac{a}{2}^2} )^2  $$

$$  (\pi \cdot 10^2 - a b +  \pi \cdot \frac{a  b}{4} + \pi \cdot  \sqrt{10^2-\frac{b}{2}^2} \cdot \sqrt{10^2- \frac{a}{2}^2 }{2} )^2 - (  \frac{ \pi  (20-b)  \sqrt{10^2-\frac{b}{2}^2 }}{2} + \frac{ \pi  (20-a) \cdot \sqrt{10^2-\frac{a}{2}^2 }}  + \pi  \frac{b}{2} \cdot \sqrt{10^2-\frac{b}{2}^2}  + \pi \cdot \frac{a}{2} \cdot \sqrt{10^2-\frac{a}{2}^2} )^2  = 0 $$

$$ -\frac{a \cdot \sqrt{100-\frac{a^2}{4}} \cdot b \cdot \sqrt{100-\frac{b^2}{4} } \cdot \pi^2 }{2} -\frac{a^2  b^2 \pi^2}{8} -50 a b  \pi^2 +2 a\cdot \sqrt{100-\frac{a^2}{4}}\cdot b\cdot \sqrt{100-\frac{b^2}{4}}\cdot \pi +\frac{a^2 b^2 \pi}{2} + 200 a b\pi-a^2 b^2 = 0 $$

$$  \frac{ -a  b \cdot  ( \sqrt{400-a^2}\cdot \sqrt{400-b^2}\cdot \pi^2+a\cdot b\cdot \pi^2+400\cdot \pi^2-4\cdot \sqrt{400-a^2}\cdot \sqrt{400-b^2}\cdot \pi-4 a b\pi-1600 \pi+8a b ) }{8} = 0 $$

$$ \sqrt{400-a^2}\cdot \sqrt{400-b^2}\cdot \pi^2+ab \pi^2+400 \pi^2-4\sqrt{400-a^2}\cdot \sqrt{400-b^2}\cdot \pi-4 a b \pi-1600\pi+8 a b = 0 $$

$$ \sqrt{400-a^2}\cdot \sqrt{400-b^2} \cdot (\pi^2-4\cdot \pi)+(a b+400) \pi^2+(-4 ab-1600) \pi+8 a b = 0 $$

$$ \sqrt{400-a^2}\cdot \sqrt{400-b^2} \cdot (\pi^2-4\cdot \pi) = -(a b+400) \pi^2+(4 a b+1600)\pi  -8 a b  $$

$$ (  \sqrt{400-a^2}\cdot \sqrt{400-b^2} \cdot (\pi^2-4 \pi) )^2 = ( -(a  b+400)\pi^2+(4 a b+1600) \pi  -8 a b )^2 $$

$$ (  \sqrt{400-a^2}\cdot \sqrt{400-b^2} \cdot (\pi^2-4 \pi) )^2 - ( -(a b+400) \pi^2+(4 a b+1600) \pi  -8a b )^2 = 0 $$

$$  -400b^2\pi^4-800ab\pi^4-400a^2\pi^4+3200b^2\pi^3+6400ab\pi^3+3200a^2\pi^3-16a^2b^2\pi^2-6400b^2\pi^2-19200ab\pi^2-6400a^2\pi^2+64a^2b^2\pi+25600ab\pi-64a^2b^2 = 0 $$
$$ -16( 25b^2\pi^4+50ab\pi^4+25a^2\pi^4-200b^2\pi^3-400ab\pi^3-200a^2\pi^3+a^2b^2\pi^2+400b^2\pi^2+1200ab\pi^2+400a^2\pi^2-4a^2b^2\pi-1600ab\pi+4a^2b^2 ) = 0 $$
$$ 25b^2\pi^4+50ab\pi^4+25a^2\pi^4-200b^2\pi^3-400ab\pi^3-200a^2\pi^3+a^2b^2\pi^2+400b^2\pi^2+1200ab\pi^2+400a^2\pi^2-4a^2b^2\pi-1600ab\pi+4a^2b^2  = 0 $$
$$ 25a^2\pi^4+50(14-a)a\pi^4+25(14-a)^2\pi^4-200a^2\pi^3-400(14-a)a\pi^3-200(14-a)^2\pi^3+(14-a)^2a^2\pi^2+400a^2\pi^2+1200(14-a)a\pi^2+400\pi(14-a)^2\pi^2-4(14-a)^2a^2\pi-1600(14-a)a\pi+4(14-a)^2a^2 = 0$$
$$ 4900\pi^4-39200\pi^3+a^4\pi^2-28a^3\pi^2-204a^2\pi^2+5600a\pi^2+78400\pi^2-4a^4\pi+112a^3\pi+816a^2\pi-22400\pi a\pi+4a^4-112a^3+784a^2 = 0 $$
$$1.303233786730186a^4-36.49054602844518a^3+1334.140307507042a^2-15101.89079431095a+35635.48524426483 = 0$$
$$[a=3.188302332738908,a=10.81169766726106,a=27.28084114342327i+7.000000000000011,a=7.000000000000011-27.28084114342327i]$$
therefore the dimension of the rectangle is $ 3.18 $ by $10.81$


问:
how to create an environment for the discrete metric?
Given the discrete metric. I have to show the different environments (amount of shape $\{x ∈ X: d(x,x_0) < r\}$) for different $x_0$ and $r$.
My  thoughts: $r$ must equal to $1$ because we have the discrete metric, so the environment must look like $E(x, x_0) < 1$. Now whats happening with the "different" $x_0$?

答:
I hope I'm understanding you right, but pay attention to the $\textbf{name}$ of this metric, you are talking abut the discrete metric, which is corresponding with the discrete topology (the open sets are all the subsets of $X$).

This metric is not the euclidean metric, it is pretty hard to visualise this metric, but it is possible to visualise the open balls:

Let $B(x,r)$ be the open ball with radius $r$ at $x\in X$.  $$B(x,r):=\{y\in X\mid d(x,y)<r\}$$

Hence:
$\forall x\in X:$

If $0\leq r<1$, $B(x,r)=\{x\}$. Why? Because $B(x,r):=\{y\in X\mid d(x,y)<r\}$ and $0\leq r<1$, but the metric is discrete, hence $\forall x\neq y\in X: d(y,x)=1$, but $0\leq r<1$!, which means $B(x,r)=\{y\in X\mid d(x,y)<r<1\}=\{x\}$

If $1\leq r$, $B(x,r)=X$. Why? Because $B(x,r):=\{y\in X\mid d(x,y)<r\}$ and $r\geq1$, but the metric is discrete, hence $\forall x\neq y\in X: d(y,x)=1$, and in particular $d(y,x)\geq 1$


问:
The Language of the Set Theory (with ZF) and their ability to express all mathematics
Accordingly, the Language of Set Theory (in this case using $ZF$ axioms) is built up with the aim to express all mathematics. Now, I know that, for example, the construction of the numbers ($\mathbf{ \mathbb{N},\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}}$) is got right from the axioms and definitions from the Language of Set Theory, so everything is inside the same theory. But, when we are dealing with especial structures like for example Euclidian Geometry, we need to state new axioms, and, apparently, it doesn't make any sense to add this axioms to our original set theory because it might be possible to find a different model, maybe completly artificial, in which some axiom of Euclidian geometry migth not hold or result in a contradiction. But Euclidian Geometry and all the different theories that one might think of are part of mathematics. So my point is that it is impossible to develop all of mathematics just from the axioms of set theory. Does it make sense? Please tell me if I'm wrong and if I'm missing something. Also I have these questions:

In building these new theories (like euclidean geometrie, etc), each theory uses the language of set theory. But also we add new axioms. So, are we talking of a different language? Are we talking of a different theory? Is the language of set theory a metalanguage in this new language? I'm totally confused.

答:
First of all, let me point that the axioms of $\sf ZF$ were given to give a reasonable axiomatization for the concept of "set". That's all. And don't feel bad that you're confused, this is a very confusing point.

To address you actual question, in mathematics we have a theory, which is what we work with - e.g. Euclidean geometry - and we have a meta-theory which is the mathematical theory in which we work. This meta-theory allows us to perform certain things (for example, formalize the concept of a proof or a model).

In set theory we can describe what does it mean for something to be a language, that is a collection of symbols with certain rules. It's really just a set which the correct cardinality. And we can write the rules for creating formulas and sentences in that language, so we can write down the axioms -- for example the axioms of Euclidean geometry, or even the axioms of $\sf ZF$ itself.

Next we can describe what does it mean to be an interpretation for the language. It's a pair, a non-empty set (the universe of the structure) and the function which interprets every symbol in the language into a relevant element/subset/relation on that universe. And with these we can ask whether certain sentences are true in this structure. And we can ask whether or not there is a structure to the language of geometry in which the axioms of geometry.

But all this happens inside a universe of set theory. We don't add to the meta-theory, we work inside the meta-theory. We use the meta-theory like an ambulance. It takes us to the hospital, but we don't perform brain surgery inside it.

So really what we do is define some sets which represent our language, the axioms, the model, and we prove things on these sets which in turn end up to be proofs in and about Euclidean geometry. Or group theory. Or lattice-ordered monoids. It doesn't matter.

But the huge point, which I can't stress enough, is that $\sf ZF$ is the foundational theory. It's like a planet. When want to build a structure we don't add planets to the solar system, we just build a structure inside the planet Earth, using material that Earth has provided us with.

The same happens with $\sf ZF$ or $\sf ZFC$. The theory provides us a planet, and materials to do mathematics and to construct mathematical models and universes for different theories. All using sets and $\in$.

Other links of interest:

When we say, "ZFC can found most of mathematics," what do we really mean?

How to write $\pi$ as a set in ZF? (In particular the comments to the answers are relevant to your question)


问:
Expand $X(t) = e^{W(t)}$
Let's take $X(t) = e^{W(t)}$ where $W$ is a Wienner process

I want to prove that $dX(t) = \frac12X(t)dt + X(t)dW(t), \;\;\;X(0) = 1$

My work so far

Idea of proof is very simple: I want to show that $X$ is an Ito process i.e. is in format of:

$$X(t) = X(0) + \int_0^t\mu(s)ds + \int_0^t\sigma(s)dW(s)$$

because then we know that $$dX(t)=\mu(t)dt + \sigma(t)dW(t)$$

It's easy to see that the wanted equation will be satisfied when I choose $\mu(s) = \frac12 X(s)$, $\sigma(s) = X(s)$.

So there is only one thing left - prove that with such $\mu$ and $\sigma$ specified $X$ is an Ito process, so

$$X(t) = X(0) + \int_0^t\mu(s)ds + \int_0^t\sigma(s)dW(s)$$

$$X(t) = 1 + \int_0^t \frac12 X(s) ds + \int_0^t X(s) dW(s)$$

$$X(t) = 1 + \frac12 \int_0^t e^{W(s)} ds + \int_0^t e^{W(s)} dW(s)$$

It's very easy to calculate that:

$$\frac12 \int_0^t e^{W(s)} ds = \frac12 e^{W(t)} - \frac12$$

But I have no idea how can we calculate

$$\int_0^t e^{W(s)}dW(s)$$

I tried substitution $u = W(s)$ but then what exactly is $\frac{dW(s)}{ds}$ ?

Could you please give me a hand with calculating so ?

答:
$$\int_0^t e^{W(s)}dW(s)=\int_0^t e^{W(s)}\frac{dW(s)}{ds}ds$$
making the substitution $u=W(s)$ gives:
$$\int_{W(0)}^{W(t)} e^udu=e^{W(t)}-e^{W(0)}$$
I am not sure if this is what you were looking for so let me know

If we work backwards from the result you are trying to prove:
$$dX(t)=\frac12X(t)dt+X(t)dW(t)$$
we are able to rewrite this as:
$$\frac{dX(t)}{X(t)}=\frac12dt+dW(t)$$
now integrate both sides and we get:
$$\ln X(t)=\frac t2+W(t)+C\tag{1}$$
which seems like a contradiction as in your definition we have:
$$X(t)=e^{W(t)}\Rightarrow \ln X(t)=W(t)$$

Title: Expand $X(t) = e^{W(t)}$


问:
Question on the order of an element in a group
We know that in a finite group of order say $g$, an element of the group will have order of element $m\leq g$. However, is it necessarily true that at least one element in the group $\textbf{must}$ have order of element $g$?

答:
No.

Consider the Klein four group. It has four elements and yet no element of order four.


答: No. If $G$ has order $m$ and $H$ has order $n$, then $G\times H$ has order $mn$. But if $m$ and $n$ are not relatively prime, then no element of $G\times H$ has order $mn$.

Title: Question on the order of an element in a group

Body: We know that in a finite group of order say $g$, an element of the group will have order of element $m\leq g$. However, is it necessarily true that at least one element in the group $\textbf{must}$ have order of element $g$?


问:
Integrating $ x \phi(x, \mu, \sigma) $
I am trying to compute the following integral:
$$\int x \cdot \phi (x) dx $$
... where $\phi(x)$ is the normal density function with mean $\mu$ and stdev $\sigma$.
Owen (1980), see https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions, states the solution as:

$$\int x \phi(a+bx) dx = - \frac{1}{b^2}(\phi(a+bx)+a \Phi(a+bx)) + C $$

In order to test out the solution, I have tried to compute the integral for a couple of example values using the built in integrate function in R. One example is:

```<code>mu <- 10
sigma <- 5
x <- 15

tst <- function(x, mu, sigma){
  return(x * dnorm(x, mu, sigma))
}

integrate(tst, mu = mu, sigma = sigma, lower = -Inf, upper = x, abs.tol = 0)$value
> 7.203594

</code>```

... where the result of roughly 7.2 makes intuitively good sense. However, when I use Owen's function, I get a completely different result. Again, in R (where I have assumed that $a = \mu$ and $b = \sigma$):

```<code>-(1/sigma^2) * (dnorm(mu+sigma*x)+mu*pnorm(mu+sigma*x))
> -0.4
</code>```

I am probably missing something or using Owen's integral improperly since I would expect to get the same results using the integrate function in R and Owen's formula. Can you help me identify why the two are not equal?

Title: Integrating $ x \phi(x, \mu, \sigma) $

Body:
Using
$$\phi(x)=\frac{1}{\sqrt{2 \pi }}e^{-\frac{(x-\mu )^2}{2 \sigma ^2}}$$
$$I=\int x  \,\phi(a+bx)\,dx=\frac{1}{\sqrt{2 \pi }}\int x e^{-\frac{(a +bx-\mu )^2}{2 \sigma ^2}}\,dx$$ Let $a+bx=t$ to make
$$I=\frac{1}{b^2\sqrt{2 \pi }}\int (t-a)\,e^{-\frac{(t-\mu )^2}{2 \sigma ^2}}\,dt$$
$$\int (t-a)\,e^{-\frac{(t-\mu )^2}{2 \sigma ^2}}\,dt=\int t\,e^{-\frac{(t-\mu )^2}{2 \sigma ^2}}\,dt-a\int e^{-\frac{(t-\mu )^2}{2 \sigma ^2}}\,dt$$
$$\int t\,e^{-\frac{(t-\mu )^2}{2 \sigma ^2}}\,dt=\sqrt{\frac{\pi }{2}} \mu  \sigma \, \text{erf}\left(\frac{t-\mu }{\sqrt{2} \sigma
   }\right)-\sigma ^2 e^{-\frac{(t-\mu )^2}{2 \sigma ^2}}$$ Back to $x$
$$I=\frac{\sigma  (\mu -a) }{2
   b^2}\text{erf}\left(\frac{a+b x-\mu }{\sqrt{2} \sigma }\right)-\frac{\sigma ^2 }{\sqrt{2 \pi } b^2}e^{-\frac{(a+b x-\mu )^2}{2 \sigma ^2}}$$


问:
Let $m$ and $n$ be integers such that $\gcd (m, n) = 1$. Show that $\gcd (m + n, m - n) = 1$ or $2$
Let m and n be integers such that $\gcd (m, n) = 1$. Show that
$\gcd(m + n, m - n) = 1$ or $2$

答:
Hint

If $d$ divides $a$ and $d$ divides $b$

then $d$ divides both $(a+b)$ and $(a - b).$


问:
Quotient space of Polynomial algebras C[X]
How can I prove that the quotient $C[X]/I$ is finite-dimensional, where $C[X]$ is the algebra of polynomials and $I=P_n(x)C[x]$ is ideal generated by a polynomial of degree n? And how can I find the dimension of this quotient?
I know that $C[X]/P_2(x)C[x]$ is isomorphic to complex plane

答:
The division algorithm: for any $a,b\in\mathbb{C}[x]$, there exist $q,r\in\mathbb{C}[x]$ such that $a=bq+r$ with $\deg(r)<\deg(b)$ or $r=0$.

As an example, let's consider $\mathbb{Q}[x]/(x^2+1)$.  The set of all possible remainders when dividing by $x^2+1$ is $\{ax+b : a,b\in\mathbb{Q}\}$.  (These are all attained and inequivalent modulo $x^2+1$.)  This is a two-dimensional vector space over $\mathbb{Q}$.  The multiplication is multiplication modulo $x^2+1$ (equality modulo $x^2+1$, so $x^2=-1$):
$$
(ax+b)(cx+d)=acx^2+(ad+bc)x+bd=(ad+bc)x+(bd-ac).
$$


问:
efficient way to find one extreme point of a Zonotope in high dimension
Given one set of generators $G \subset \mathbb{R}^d$, one can easily generate a Zonotope using ConvexHull $Z = CH(\{\sum_{i \in S} i| S \subseteq G\})$.

Now my interest is to find one vertex of $Z$, is there any efficient way to do so?

答:
Choose a generic $c\in\Bbb R^d$ (that is, $\langle x,c\rangle\not=0$ for all $x\in G$) and consider the subset

$$G_c:=\{x\in G\mid \langle x,c\rangle > 0\}\subseteq G.$$

The point $v_c:=\sum_{x\in G_c} x$ is then a vertex of $Z$ (and every vertex is of this form).


问:
Prove equality of absolute values (integrals)
Can someone help me with this proof?

Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function defined on a non-empty closed interval.

Prove: $|\int_{a}^{b}f(x)dx|=\int_{a}^{b}|f(x)|dx$ $\iff$ $f$ is either non-negative or non-positive.

(Hint: See $f$ as the difference between its positive part and its negative part)

So far I've got:

$\impliedby:|\int_{a}^{b}f(x)dx| = |\int_{a}^{b}f_+(x)dx -\int_a^bf_-(x)dx|\leq\int_{a}^{b}f_+(x)dx+\int_a^bf_-(x)dx=\int_{a}^{b}|f(x)|dx$

We have equality if either $\int_{a}^{b}f_+(x)dx=0$ or $\int_{a}^{b}f_-(x)dx=0$.

答: If $\left|\int_a^bf(x)\,\mathrm dx\right|=\int_a^b|f(x)|\,\mathrm dx$, then $\int_a^bf(x)\,\mathrm dx=\pm\int_a^b|f(x)|\,\mathrm dx$. Suppose that$$\int_a^bf(x)\,\mathrm dx=\int_a^b|f(x)|\,\mathrm dx.$$Then $\int_a^b|f(x)|-f(x)\,\mathrm dx=0$. But $|f|-f$ is a continuous non-negative function, and therefore$$\int_a^b|f(x)|-f(x)\,\mathrm dx=0\iff|f|-f=0\iff f=|f|.$$The case in which $\int_a^bf(x)\,\mathrm dx=-\int_a^b|f(x)|\,\mathrm dx$ is similar: you can deduce that $f=-|f|$.


Title: Relation between end point of focal chord of parabola

Body:
Consider a parabola $y^2 = 4ax$ , parameterize it as $x=at^2$ and $y=2at$, then it is found that if we have a line segment passing through focus, with each points having value of $t$ as $t_1$ and $t_2$ for the parameterization, then it must be that:

$$ t_1 \cdot t_2 = -1$$

Hope for hints.

Body:
Chord passing through $(at_1^2,2at_1)$ and $(at_2^2,2at_2)$ is
$$y-2at_2=\frac{2at_1-2at_2}{at_1^2-at_2^2}(x-at_2^2)$$
$$y-2at_2=\frac{2}{t_1+t_2}(x-at_2^2)$$

It should pass through focus $(a,0)$
$$\Rightarrow -at_2=\frac{1}{t_1+t_2}\,a(1-t_2^2)$$
$$\Rightarrow t_1t_2=-1$$


答:
Here is the complete answer to your question.
A group of order $n$ has an element of order $n$ if and only if the group is cyclic. This statement is an easy exercise.


问:
$\sum_{i=0}^{k} a^i$
$\sum_{i=0}^{k} a^i= \frac{a^{k+1}-1}{a-1} = \Theta(a^k)$

I suppose this sum is true for $a>1$ but if I have $0<a<1$ what the result would be?

答: To prove $\sum_{i=0}^ka^i=\frac{a^{\color{blue}{k}+1}-1}{\color{blue}{a}-1}$, use induction. You needn't assume constraints on $a$, except $a\ne1$. But if $|a|<1$, it may be convenient to change the signs of the numerator & denominator, writing the result as $\frac{1-a^{k+1}}{1-a}$. In terms of $\Theta$ notation, note the small-$a$ case achieves a $O(1)$ result with a $k\to\infty$ limit.

Title: $\sum_{i=0}^{k} a^i$


答:
For any $\;1\neq a\in\Bbb C\;$ , it in fact must be

$$\sum_{i=0}^k a^i=\frac{a^{k+1}-1}{a-1} $$

You can easily prove this with say induction on $\;k\;$ ...

Title: $\sum_{i=0}^{k} a^i$

Body:
$\sum_{i=0}^{k} a^i= \frac{a^{k+1}-1}{a-1} = \Theta(a^k)$

I suppose this sum is true for $a>1$ but if I have $0<a<1$ what the result would be?


问:
Prove that $G(n) = \sum_{d|n} \frac{n}{d}\tau(d)$ is multiplicative.
I don't know how to prove this because it seems to me that $f(d)=\frac{n}{d}\tau(d)$ is not multiplicative.

答:
The (Dirichlet) convolution of two multiplicative functions is multiplicative, i.e.
$$
f(n)=\sum_{d|n}g(n/d)h(d)
$$
is multiplicative if $g$, $h$ are.  Both $g(n)=n$ and $\tau(n)=\sum_{d|n}1$ (if that's what you mean by $\tau$) are multiplicative, so their convolution is as well.


问:
Equivalent definitions of abelian variety
While studying abelian varieties I've seen the following definitions:

An abelian variety over a field $k$ is a complete group $k$-variety.

An abelian variety over a field $k$ is a smooth, connected, proper $k$-group scheme X.

Since a complete $k$-variety is nothing but a proper, geometrically integral $k$-scheme, I wonder if "smooth + connected = geometrically integral". I would like to know if this is true and also why these definitions are equivalent.

答:
for any group variety over a field, smoothness is equivalent to being geometrically reduced.it is true because you can check smoothness after a base change to algebraic closure, over an algebraically closed field smooth locus of a reduced variety is non-empty open $U$(there are several ways to see this) and then because you have a group scheme you can translate $U$ by elements of $A(k)$ and again this would be an open smooth subvariety and obviously union of this translations is the whole space.

also always we have Smooth+connected implies geometrically integral: smooth obviously implies reduced and if you take a point on the intersection of two irreducible components the variety can't be smooth at that point(because for example, the local ring at that point could not be regular).

I'm not sure why do you think separated condition has anything to do with other hypothesis because it is needed in both definitions.


问:
derivative of the supremum of a set of n functions
Let $\mathcal{S} = \{f_1(x),f_2(x),\dots,f_n(x)\}$ be a set composed by $n$ differentiable functions, where $x \in \mathbb{R}^m$.

Let $h(x) = \sup \mathcal{S}$

Is it possible to compute $\frac{\partial h}{\partial x}$ ?

I know that if $n=2$ then $h(x) = \frac{f_1(x) + f_2(x) + |f_1(x) - f_2(x)|}{2}$ that is differentiable assuming $f_1(x)\ne f_2(x)$
but is it possible to generalize this?

答:
Your approach, by expressing the maximum of two functions in terms of an absolute value of their difference, does demonstrate that $h(x) = \sup S(x)$ is differentiable at points $x$ where all the underlying function values $f_1(x),\ldots,f_n(x)$ are distinct.

We can see this by repeatedly applying your expression for two values:

$$ \max \{f_1(x),f_2(x)\} = \frac{f_1(x)+f_2(x)+|f_1(x)-f_2(x)|}{2} $$

to obtain the maximum of several values:

$$ \max \{f_1(x),f_2(x),\ldots,f_n(x)\} = \max \{ f_1(x), \max\{f_2(x),\ldots,f_n(x)\} \} $$

so that if all the $f_i(x)$ are distinct, then at each stage we obtain subexpressions where the absolute value is taken of the difference of unequal values.  Hence the supremum $h(x) = \sup S$ will be differentiable at such $x$.

To illustrate the idea (as well as how cumbersome these formulas become), for $n=3$ we would get this expression:

$$ h(x) = \frac{f_1(x) + \frac{f_2(x) + f_3(x) + |f_2(x) - f_3(x)|}{2} + |f_1(x) - \frac{f_2(x) + f_3(x) + |f_2(x) - f_3(x)|}{2}|}{2} $$

As I tried to sketch out in the Comments, requiring all the $f_i(x)$ to be distinct will exclude some points where $h(x)$ is actually differentiable.  The condition can be weakened to requiring only that some $f_i(x)$ is strictly greater than all the other $f_j(x)$, $j\neq i$.  In that case by continuity there will be an open neighborhood of $x$ in which $h$ agrees identically with $f_i(x)$. Hence:

$$ h'(x) = f_i'(x) $$

Finally we note that $h(x)$ might sometimes be differentiable also at points $x$ where the two greatest function values are equal, and it will be impossible to rule this out without further information.

For example, $h(x) = \sup \{x^3,-x^3\} = |x^3|$ has derivative zero at the "crossover point" $x=0$.


问:
Why is the space of continuous functions on $[0,1]$ a manifold?
I am reading some machine learning(neural network) related math, and it uses the manifold $M$ of continuous functions on $[0,1]$.(Every function is a point of $M$)

Why is $M$ a manifold?

BTW, it also says its dimension is infinite because a system of parameters for a continuous function defined on $[0,1]$ is the set of rational numbers $\mathbb Q \cap [0,1]$. I think if I know how the atlas is constructed I can deduce this myself.

答:
Since the function is continuous, to determine the function, all you have to do is determine it on some dense subset of $[0,1]$. If this is not clear, let me know and I can add an explanation why. Now $\mathbb{Q} \cap [0,1]$ is such a dense (countable) subset. Take any ordering of these, I’ll denote it as $q_1,q_2,...$.

Then the map that sends $f \mapsto (f(q_1),f(q_2),...) \in \mathbb{R}^{\mathbb{N}}$ is injective. I’m not sure what definition of infinite dimensional manifold you use but this will help build the chart. Also you should check that this is map is continuous but it follows from the topology of uniform convergence quite easily. (If you prefer the compact open topology, this coincides with uniform convergence here).


问:
Finding the minimizer of the variational problem $J[y]= \int_{0}^{2} \left( 1-y'^2 \right)^2dx $ with $y(0)=0$ and $y(2)=0$.
This is a question from a math contest.

Find the minimizer of $J[y]= \int_{0}^{2} \left( 1-y'^2  \right)^2dx$ with the conditions: $y(0)=0$ and $y(2)=0$.

Now as $F$ is $\geq 0$, so to minimize we let $y'^2=1$, $i.e. y'= \pm 1 $ $\implies y= \pm x +c$

but then the initial conditions are not satisfied. So is the only minimizer is $y \equiv  0$?

Also for finding the extremal, using the Euler's condition we have:

$$ \frac{\partial F}{\partial y}- \frac{d}{dx} \left[ \frac{\partial F}{\partial y'} \right] =0 $$
$$ i.e. y''(3y'^2 -1)=0 $$
But I am unable to solve the differential equation.

Any help is highly appreciated.

答:
Perhaps they allow for continuous and piece-wise differentiable maps?
Then $y(x)=x$, $x\in[0,1]$ and $y(x)=2-x$, $x\in [1,2]$, seems a good candidate. Since you may approximate this arbitrarily well by a $C^1$ function the inf of the functional is zero but is not achieved on $C^1$ functions.


答:
The common interpretation of the claim that you can do maths in $ZF$ is that given a model of $ZF$ you can produce in it models of all relevant objects of study in mathematics. So, as you mention, given a model of $ZF$, you can construct in it a set which models the natural numbers, a set which models the real numbers, and so on. That does not mean that you add any new symbols for addition or multiplication, but rather that you show that there is set in the universe of sets of the $ZF$ model together with identifiable extra structure that turns it (while it's still just a set) and all that extra structure (still all just sets, since that is all you have in $ZF$) into a model of, say, the natural numbers satisfying Peano + Induction. Similarly for other number systems.

In the same way you can construct a model of Euclidean geometry as well as models of non-Euclidean geometry all within the confines of a $ZF$ model. It is in this sense that "mathematics can be done within $ZF$". Of course, it is possible that some aspects of mathematics can't be captured by $ZF$. For instance, category theory requires a tower of $ZF$ models with suitable properties (see Grothendieck universes).

You are in a sense correct that when modeling, say, the real numbers inside $ZF$, the language of set theory can be thought of as becoming a meta-language and the language of the reals becomes the actual language. But this is not how it's thought of in practice. In practice, when you deal with the reals for the purposes of real analysis you couldn't care less that they can be modelled inside $ZF$. Instead, you just work with them axiomatically, assuming a background language and model of (naive) set theory. The difficulties arise when one realizes that there are many different models of $ZF$ (unless of courser there aren't any models at all) and these may affect what happens in the reals. The study of the effects of certain axioms of set theory on the resulting models of the reals inside is part of axiomatic set theory and can get quite complicated. You can also have a look at topos theory where similar things happen. In topos theory the idea is to axiomatize some basic properties of the category of sets in such a way that such a construct admits an internal logic. Many toposes allow for internal constructions of natural numbers, real numbers etc. And funny things may happen. For instance, there is a topos in which the internal notion of real numbers is such that all functions from the reals to the reals are continuous.

Title: The Language of the Set Theory (with ZF) and their ability to express all mathematics

Body:
Accordingly, the Language of Set Theory (in this case using $ZF$ axioms) is built up with the aim to express all mathematics. Now, I know that, for example, the construction of the numbers ($\mathbf{ \mathbb{N},\mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}}$) is got right from the axioms and definitions from the Language of Set Theory, so everything is inside the same theory. But, when we are dealing with especial structures like for example Euclidian Geometry, we need to state new axioms, and, apparently, it doesn't make any sense to add this axioms to our original set theory because it might be possible to find a different model, maybe completly artificial, in which some axiom of Euclidian geometry migth not hold or result in a contradiction. But Euclidian Geometry and all the different theories that one might think of are part of mathematics. So my point is that it is impossible to develop all of mathematics just from the axioms of set theory. Does it make sense? Please tell me if I'm wrong and if I'm missing something. Also I have these questions:

In building these new theories (like euclidean geometrie, etc), each theory uses the language of set theory. But also we add new axioms. So, are we talking of a different language? Are we talking of a different theory? Is the language of set theory a metalanguage in this new language? I'm totally confused.


问:
Find which two bottles are contaminated

King Bubba III has 1024 potions, 1022 of which are just water while two of them are magical gives him superpowers (the special potions are colorless, odorless, etc.). Bubba, who is an intellectual, has a machine that tells him whether a particular sample of liquid is magical or not. However, he has calculated that he can mix the potions and still use the machine - if he takes a set of potions and inputs it into his machine, the machine will return positive if there is magic in any of the bottles and negative if none of the bottles are magic. There is, however, two limitations on the machine - firstly, because mixing too many potions dilutes magic, he can mix at most 32 potions an receive an accurate result from the machine, and secondly, the machine will return all the results at once, in 7 days (which means Bubba cannot base future tests off of the result of past tests). What is the minimal amount, up to a factor of 2, of times he needs to use the machine and identify both magical potions?

I think the answer actually should just be $1024$, because I've tested many other strategies you can use but still we must consider the least optimal case, which makes reducing this number significantly rather difficult. (there is a slight optimization you can use to reduce the number of uses to $1023$ but I really dont care

I believe the a similar, well-known variant of this problem has been posted here and here, but I'm not sure how the ideas can be cross applied - there are extra restrictions in this problem.

答:
We can prove that at least $57$ tests are necessary, using an information-theoretic argument. Assuming the two bottles were chosen uniformly at random, the probability $p$ of a any particular test returning negative would be
$$
p={1024-32 \choose 2}\Big/{1024 \choose 2}
$$
It follows that the entropy of a single test is $H(p)=-p\log p-(1-p)\log (1-p)\approx 0.333579$ (Wolfram|Alpha calculation). This means that every test gives roughly one third of a bit of information on average. Since there are $\log_2 \binom{1024}2\approx 19$ bits of uncertainty, it should take at least $19/ 0.333579\approx 57$ tests to eliminate all uncertainty.

We can make this reasoning precise. Let $E_1,E_2,\dots,E_n$ denote the results of the $n$ tests, let $X$ denote the unknown pair of bottles (chosen uniformly at random, and suppose there exists a function $f$ for which
$$
X=f(E_1,\dots,E_n)
$$
Then these two entropy inequalities are what you need to make my reasoning precise:

For a random variable (or vector) $Y$, and a deterministic function $f$, $H(f(Y))\le H(Y)$.

For a joint distribution $(E_1,\dots,E_n)$, $H((E_1,\dots,E_n))\le H(E_1)+\dots +H(E_n)$.

This implies that $H(X)\le n H(E_1)$, so you need at least $H(X)/H(E_1)$ tests.

Title: Find which two bottles are contaminated


问:
How to translate a "formal" definition of the distributive property?
I have seen sometimes a formal representation of the distributive property over an operator defined as follows:

A function $f$ is said to distribute over the operator $\bigoplus$ if
there is an operator $\bigotimes$ such that  $f(x\bigoplus y)=f(x)\bigotimes f(y)$

When we see a definition like this, how do we "translate" it to a more concrete example?

E.g. multiplication distributes through addition and hence:

$x\cdot (y + z) = x \cdot y + x \cdot z$

So in this case what is the $f$ what is the $\bigotimes$ and what is the $\bigoplus$?

答:
Let $f(t):=x\cdot t$, $\oplus:=+$, $\otimes:=+$.

Then $f(y\oplus z)=f(y)\otimes f(z)$.

Edit

Although I have answered the question as set, I think that the definition given isn't quite the one I'd expect to see. It makes sense, but forces us to treat the $x$'s individually. To get something more uniform I would have preferred a definition like this:

The function $\mu: (x,y)\mapsto \mu(x,y)$ is said to right-distribute over the binary $\alpha:(x,y)\mapsto\alpha(x,y) $ if there exists a  binary $\alpha':(x,y)\mapsto\alpha'(x,y) $ such that
$$
\mu(x,\alpha(y,z))=\alpha'(\mu(x,y),\mu(x,z)).
$$

Then in the example we'd take multiplication for $\mu$, and addition for each of  $\alpha, \alpha'$.


问:
Definition of Dirac mass : $\delta (A)=1$ if $0\in A$ and $0$ otherwise or $\int_{\mathbb R}f(x)\delta (dx)=f(0)$
The way I define $\delta $ mass is $$\delta (A)=\begin{cases}1&0\in A\\0&\text{otherwise}.\end{cases}$$

However, in my lecture of measure, they define it as $$\int_{\mathbb R}f(x)\delta (dx)=f(0).$$

Are both definition the same ? If using one definition rather than an other ?

答:
Suppose your first definition.

Step 1: Suppose $f$ is simple, i.e. $f(x)=\sum_{i=1}^n a_i\boldsymbol 1_{A_i}.$ Since, at most one $A_i$ contain $0$, $$\int_{\mathbb R}f(x)\delta (\mathrm d x)=\sum_{i=1}^n a_i\delta (A_i)=f(0).$$

Step 2: If $f\geq 0$ there is a sequence $(f_n)$ of simple function s.t. $f_n\nearrow f$. The claim follow using Monotone convergence theorem.

Step 3: If $f$ is measurable, then write it as $f^+-f^-$ where $f^+(x)=f(x)\vee 0$ and $f^-(x)=-(f(x)\wedge 0)$, and apply the Step 2. The second definition follows.

Suppose your second definition

In particular, if $A$ is measurable, then so is $f(x)=\boldsymbol 1_A(x)$. Therefore $$\delta (A)=\int_{A}\delta (\mathrm d x)=\int_{\mathbb R}\boldsymbol 1_A(x)\delta (\mathrm d x)=f(0)=\begin{cases}1&0\in A\\ 0&0\notin A\end{cases}.$$
The first definition follows.

Remark Measurability of the function and of the set can be understood as Borel-measurable or Lebesgue-measurable.


问:
Losses/Profits Calculations
I have a case study (see the picture) and i am struggling to understand the calculations. The case study is on Credit Risk.

The picture shows the data from a bank in the last 12 months including write offs, loses and profits.

I need to understand how the below are calculated ?

Total Losses

Profits

My first thought for profit is the below relationship :

Profit = Annual Income - Total Losses

However, when i use data from January, for example, the outcome is not correct.

$£288,000 - £229.677 = £58,323 $

Any help it will be appreciated :)

Below is a further description of the variables:

Accounts Opened in a month

Average Risk score, for the accounts opened in the month

Annualised Income (Income over a 12 month period)

Average Overdraft Amount Granted

Average Write-Off Balance &

Annualised Losses (Losses over a 12 month period)

Antonis

答: Profit = Annual income + Average Granted Overdraft Limit + Average Write-Off Balance - Total Losses</p>

Title: Losses/Profits Calculations


问:
Finding the Closed Form of a Recursion
I wanted to find the closed form of the recursion: $$a_n = \frac{1}{n+1} + \frac{n-1}{n+1} a_{n-1}$$ where $a_0 = 0$.

So far, my progress has been to multiply both sides by $\frac{n(n+1)}{2}$ which gives $$\frac{n(n+1)}{2} a_n = \frac{n}{2} + \frac{n(n-1)}{2} a_{n-1}.$$

Now, if we let $b_n = \frac{n(n+1)}{2} a_n$, the recursion becomes: $$b_n = \frac{n}{2} + b_{n-1}.$$

How do I continue on from here? I feel like I have to figure out a closed form for $b_n$ and use that to compute $a_n$, but I can't seem to figure out a way to do so.

Any help would be greatly appreciated.

答:
One can easily see that $a_0=0, a_n=\frac{1}{2}$ for $n\ge 1$ is the solution. However, the OP seems to be interested in general methods which could help solve this recurrence.

(In the following let's assume we are solving it for $n>0$ and with $a_1$ given, otherwise there won't be much more to add than what we've seen so far.)

There is one general observation that might help solve it: it is a linear (though not a homogenous) recurrence. This means that any solution $a_n$ of it will be possible to express as one particular solution (e.g. $b_n=\frac{1}{2}$) plus a solution of the corresponding homogenous recurrence ($c_n=\frac{n-1}{n+1}c_{n-1}$).

The last one is a telescoping recurrence: $c_n=\frac{n-1}{n+1}c_{n-1}=\frac{(n-1)(n-2)}{(n+1)n}c_{n-2}=\ldots=\frac{(n-1)(n-2)\cdots 1}{(n+1)n(n-1)\cdots 3}c_1=\frac{2}{(n+1)n}c_1$ - as all the other factors in the numerator and the denominator will cancel out. Thus, the solution for $a_n$ is: $$a_n=b_n+c_n=\frac{1}{2}+\frac{2}{(n+1)n}(a_1-\frac{1}{2})$$

Title: Finding the Closed Form of a Recursion


问:
Modelling random points on a circle
Consider this problem.

Suppose 3 (distinct) points are uniformly and independently distributed on a circle of unit length (smaller than a unit circle!). This is really circle and not disc. Call one of these points $B_{circ}$.

Let $Z$ be the random variable denoting the distance of the point $B_{circ}$ to its neighbour in the anti-clockwise direction.

Let $W$ be the minimum distance of the point B to its nearest neighbour, either clockwise or anti-clockwise, whichever is smaller.

Let $M$ be the minimum distance between any 2 of the points.

Is the following model correct?

Let the other points be $A_{circ}$ and $C_{circ}$. They are all necessarily distinct (almost surely, but there's no measure theory here) because the probability that any 2 or all 3 of them are the same is zero. Map them all into unit half-open interval $[0,1)$, which is bijective with the circle, and call them respectively $B,A,C$. These are iid uniform(0,1) random variables, much like how the original $B_{circ}$, $A_{circ}$ and $C_{circ}$ are iid uniform(over the circle) random variables.

In terms of the order of $A,B,C$, there are exactly $2=2!=(3-1)!$ cases here because of circular permutation. The cases are $A<B<C$ and $C<B<A$.

These cases may also be respectively described as $B-A > 0$ and $B-C > 0$.

$Z=(B-A) 1_{B-A > 0} + (B-C) 1_{B-C>0}$

4.1. Alternatively, $Z=\max\{B-A,B-C\}=B+\max\{-A,-C\}=B-\min\{A,C\}$. Basically, the idea behind '$\max\{B-A,B-C\}$' is that exactly 1 of these quantities is going to be positive while the other is negative.

$W=\min\{Z,U\}$, where $U$ is similar to $Z$: $U=\max\{C,A\}-B=\max\{C-B,A-B\}$.

5.1. We have that $Z+U=|C-A|$ and then $W=\min\{B-\min\{A,C\},|C-A|-(B-\min\{A,C\})\} = \min\{Z,|C-A|-Z\}$.

$M=\min\{W,|C-A|\} = \min\{Z,|C-A|-Z,|C-A|\} = \min\{B-\min\{A,C\},|C-A|-(B-\min\{A,C\}),|C-A|\}$

These questions are all related, but I hope I made each self-contained

Random points on a circle

More random points on a circle

Most random points on a circle

Modelling random points on a circle

Remodelling random points on a circle: Arc length between points distributed on circle is uniform?

答:
I see a fundamental problem with this approach.

It is true that there are really only two sequences of three points on the circle: $A,B,C$ are encountered in that sequence either clockwise or counterclockwise.
But when you assign numerical coordinates to the three points via the bijection between the circle and $[0,1)$, and then start dealing with arithmetic operations and relations among the three numbers, you change the topology of the problem.

Here is an example. In one case you could have $A = 0.4,$ $B = 0.6.$
In another case you could have $A = 0.9,$ $B = 0.1.$
The two cases are isometric on the circle: $B$ is $0.2$ units counterclockwise from $A$. But numerically they are very different: $B - A = 0.2$ and $A < B$ in one case,
$B - A = -0.8$ and $B < A$ in the other case.

Perhaps the most obvious contradiction is, if $A,B,C$ truly are iid uniform on $[0,1)$, then there is a non-zero probability that $A < C < B,$
and likewise for three other cases that you have ruled out in the question.

It is possible to exclude the other four cases from consideration, but to do so you have to give up the iid assumption and restrict the support of $(A,B,C)$ to a subset of the cube $[0,1)^3.$ The distribution of $(A,B,C)$ is then uniform on that subset.

Then to compute probabilities correctly you have to understand the shape of that subset of the cube.
I think this makes the problem much harder than it needs to be.

Title: Modelling random points on a circle


问:
How to estimate the magnitude of least positive integer solution for Legendre-Diophantine equation?
We know that there is a famous theorem to determine whether there is an integer solution to the Legendre-Diophantine equation with three variables.

Legendre theorem of ternary quadratic form

The indeterminate (Diophantine) equation

$$ax^2+by^2+cz^2=0$$

whose coefficients a, b and c are pairwise coprime integers,
square-free and not all of the same sign, has a non-zero rational
solution if and only if all following three congruences are solvable:
\begin{align*}
  u^2\,&\equiv-bc\pmod{|a|},\\
  v^2\,&\equiv-ca\pmod{|b|},\\
  w^2&\equiv-ab\pmod{|c|},\\
  \end{align*}

https://encyclopediaofmath.org/wiki/Legendre_theorem

However, I was baffled by a Legendre equation $x^2 + 5 y^2 = 4389 z^2$.

(ps: $4389=3\cdot7\cdot11\cdot19$)

With the help of the above theorem and software, we found by error that the Diophantine equation $\text{exists}$ a non-zero integer solution, but I applied the software again to search brutally, and found no solution less than $10000$.

[Edit]In the process of judging whether the congruence equation has a solution, I mistakenly think that Jacobian sign is equivalent to Legendre sign.

Now I want to know how to estimate the magnitude of least positive integer solution for this Diophantine equation?

Wolfram Mathematica code:

```<code>In[1]={JacobiSymbol[-a*b, Abs[c]],
       JacobiSymbol[-a*c, Abs[b]],
       JacobiSymbol[-b*c, Abs[a]]} /. {
           a -> 1, b -> 5, c -> -4389}

Out[1]={1, 1, 1}

In[2]=Solve[x^2 + 5 y^2 == 4389 z^2 && 0 < {x, y, z} < 10000, {
      x, y, z}, Integers]

Out[2]={}
</code>```

Edit

$$x^2 + 5 y^2 = 20769 z^2$$

Thanks for Jagy's solution. I also want to know the magnitude between the least positive integer solution and the coefficient of the above Diophantine equation.

```<code>Solve[x^2 + 5 y^2 == 20769 z^2 && 0 < {x, y, z} < 100, {
   x, y, z}, Integers]

{{x -> 17, y -> 64, z -> 1}, {x -> 58, y -> 59, z -> 1},
   {x -> 82, y -> 53, z -> 1}}
</code>```

答:
well, no, Legendre symbol
$$  (-20|11) = (-1|11)(5|11) = (-1|11)(11|5) = (-1|11) = -1 $$

If $x^2 + 5 y^2 $ is divisible by $11,$ it must be divisible by $121 = 11^2.$  Indeed,  writing
$$11^t \parallel a^2 + 5 b^2  $$
to mean $11^t | a^2 + 5 b^2  $ but not $11^{t+1} |a^2 + 5 b^2,  $
the exponent $t$ must be even. Writing $t=2 \tau,$ we must have
$11^\tau | a  $ and $11^\tau | b  $

You need to check each prime that divides one of the coefficients

from page 82, Corollary 1, in Cassels, Rational Quadratic Forms: if $abc$ is squarefree, while $ax^2 + b y^2 + c z^2$ is indefinite, such that there is a nonzero integer solution to $ax^2 + b y^2 + c z^2=0,$  then there is such a solution with $$|a|x^2 + |b| y^2 + |c| z^2 < 4 |abc|.$$ Note that allowing square factors in $abc$ would raise the size of the smallest solution.

IF
$$ x=101 u^2 -492uv -429v^2 \; ,$$
$$ y=188 u^2 +318uv +24v^2 \; ,$$
$$ z=3 u^2 +4uv +3v^2 \; , $$
then
$$ x^2 + 5y^2 = 20769 z^2 \; . $$
which gives infinitely many solutions, including some small ones.

More matrices of coefficients of $u^2,uv,v^2$ The first one gives
$$ x=17 u^2 -640uv -429v^2 \; ,$$
$$ y=64 u^2 +34uv +24v^2 \; ,$$
$$ z= u^2 + 5v^2 \; . $$

```<code> 17   -640    -85 transposed       17     64      1
 64     34   -320 transposed     -640     34      0
  1      0      5 transposed      -85   -320      5

101   -492   -429 transposed      101    188      3
188    318     24 transposed     -492    318      4
  3      4      3 transposed     -429     24      3

 26   -574   -826 transposed       26    193      3
193    532    259 transposed     -574    532      8
  3      8      7 transposed     -826    259      7

 58   -590   -290 transposed       58     59      1
 59    116   -295 transposed     -590    116      0
  1      0      5 transposed     -290   -295      5

 82    366   -858 transposed       82     53      1
 53   -270    -48 transposed      366   -270     -2
  1     -2      6 transposed     -858    -48      6
</code>```

Umm. There are several solutions with $z=1,$  since

j

```<code>agy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./bingo_target
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
1 0 5
Discriminant  20
Target to be represented?
20769
20769 = 3 * 7 * 23 * 43
 xbound  145
 ybound  65
 x = 143, y = 8 target 20769
 x = 143, y = -8 target 20769
 x = 142, y = 11 target 20769
 x = 142, y = -11 target 20769
 x = 137, y = 20 target 20769
 x = 137, y = -20 target 20769
 x = 118, y = 37 target 20769
 x = 118, y = -37 target 20769
 x = 113, y = 40 target 20769
 x = 113, y = -40 target 20769
 x = 82, y = 53 target 20769
 x = 82, y = -53 target 20769
 x = 58, y = 59 target 20769
 x = 58, y = -59 target 20769
 x = 17, y = 64 target 20769
 x = 17, y = -64 target 20769
20769 = 3 * 7 * 23 * 43
</code>```

Body:
well, no, Legendre symbol
$$  (-20|11) = (-1|11)(5|11) = (-1|11)(11|5) = (-1|11) = -1 $$

If $x^2 + 5 y^2 $ is divisible by $11,$ it must be divisible by $121 = 11^2.$  Indeed,  writing
$$11^t \parallel a^2 + 5 b^2  $$
to mean $11^t | a^2 + 5 b^2  $ but not $11^{t+1} |a^2 + 5 b^2,  $
the exponent $t$ must be even. Writing $t=2 \tau,$ we must have
$11^\tau | a  $ and $11^\tau | b  $

You need to check each prime that divides one of the coefficients

from page 82, Corollary 1, in Cassels, Rational Quadratic Forms: if $abc$ is squarefree, while $ax^2 + b y^2 + c z^2$ is indefinite, such that there is a nonzero integer solution to $ax^2 + b y^2 + c z^2=0,$  then there is such a solution with $$|a|x^2 + |b| y^2 + |c| z^2 < 4 |abc|.$$ Note that allowing square factors in $abc$ would raise the size of the smallest solution.

IF
$$ x=101 u^2 -492uv -429v^2 \; ,$$
$$ y=188 u^2 +318uv +24v^2 \; ,$$
$$ z=3 u^2 +4uv +3v^2 \; , $$
then
$$ x^2 + 5y^2 = 20769 z^2 \; . $$
which gives infinitely many solutions, including some small ones.

More matrices of coefficients of $u^2,uv,v^2$ The first one gives
$$ x=17 u^2 -640uv -429v^2 \; ,$$
$$ y=64 u^2 +34uv +24v^2 \; ,$$
$$ z= u^2 + 5v^2 \; . $$

 17   -640    -85 transposed       17     64      1
 64     34   -320 transposed     -640     34      0
  1      0      5 transposed      -85   -320      5

101   -492   -429 transposed      101    188      3
188    318     24 transposed     -492    318      4
  3      4      3 transposed     -429     24      3

 26   -574   -826 transposed       26    193      3
193    532    259 transposed     -574    532      8
  3      8      7 transposed     -826    259      7

 58   -590   -290 transposed       58     59      1
 59    116   -295 transposed     -590    116      0
  1      0      5 transposed     -290   -295      5

 82    366   -858 transposed       82     53      1
 53   -270    -48 transposed      366   -270     -2
  1     -2      6 transposed     -858    -48      6

Umm. There are several solutions with $z=1,$  since

j

agy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./bingo_target
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2
1 0 5
Discriminant  20
Target to be represented?
20769
20769 = 3 * 7 * 23 * 43
 xbound  145
 ybound  65
 x = 143, y = 8 target 20769
 x = 143, y = -8 target 20769
 x = 142, y = 11 target 20769
 x = 142, y = -11 target 20769
 x = 137, y = 20 target 20769
 x = 137, y = -20 target 20769
 x = 118, y = 37 target 20769
 x = 118, y = -37 target 20769
 x = 113, y = 40 target 20769
 x = 113, y = -40 target 20769
 x = 82, y = 53 target 20769
 x = 82, y = -53 target 20769
 x = 58, y = 59 target 20769
 x = 58, y = -59 target 20769
 x = 17, y = 64 target 20769
 x = 17, y = -64 target 20769
20769 = 3 * 7 * 23 * 43


问:
Homotopy of a Continuous Proper Map to a Smooth Proper Map
Working on a problem in Lee's Intro to Smooth Manifolds (2nd edition, problem 8, chapter 6).  The problem states:

Prove that a proper continuous map between smooth manifolds is
homotopic to a proper smooth map between said manifolds.

Let $M$ and $N$ be the manifolds with $F:M \to N$ a proper continuous map.  Since $F$ is continuous we automatically have the Whitney Approximation theorem: we can embed $N \subseteq \mathbb{R}^n$ for some $n$ and that $F$ is $\delta$-close to a smooth map $\widetilde{F}: M \to \mathbb{R}^n$.  Because $\widetilde{F}(M)$ can fit into some tubular neighborhood $U \subseteq \mathbb{R}^n$ of $N$, we have there is a mapping $r: U \to N$ which is a retraction and a submersion into $N$.  We can then take our homotopy to be $H: M\times [0,1] \to N$ given by

$$
H(x,t) \;\; =\;\; r \left ( (1-t) F(x) + t\widetilde{F}(x) \right ).
$$

The theorem shows that $F$ is homotopic to a smooth map $r\circ \widetilde{F}$, but we want to show that $r\circ\widetilde{F}$ is proper.

I'm somewhat lost as to proof strategies and need some help.  Lee hints that we need to show that $\tilde{F}$ is proper, and hopefully $r$ can easily show to be proper.  Where I'm getting stuck is in picking an arbitrary $K \subseteq N$ which is compact; how do we show $\widetilde{F}^{-1}(K)$ is compact as well?  I think we can pick a sequence in $\widetilde{F}^{-1}(K)$ and find a convergent subsequence.  Somewhere in there I'm assuming we need to use the fact that $||F(x) - \widetilde{F}(x)|| < \delta(x)$ for all $x \in M$, given $\delta:M \to \mathbb{R}_+$.

Any help is appreciated on this.  Thanks in advance!

答:
Recall Theorem 6.24 and Proposition 6.25: Let $M\hookrightarrow \Bbb R^n$ be an embedded smooth submanifold of dimension $m$. For each $x\in M$ define $N_xM:=\big\{v\in T_x\Bbb R^n\big| v\perp T_xM\big\}$ and $NM:=\big\{(x,v)\in \Bbb R^n\times \Bbb R^n:x\in M,v\in N_xM\big\}$.

Also, define $\pi\colon NM\to M$ as $(x,v)\longmapsto x$. Then, $NM$ is a $k$ dimensional smooth submanifold of $\Bbb R^n\times \Bbb R^n$ and $\pi$ is a submersion.

Define a map $E\colon NM\to \Bbb R^n$ as $(x,v)\longmapsto x+v$. Then at any point $(x,0)\in NM$ the map $dE$ is non-singular, i.e., for each $x\in M$ we have $\delta>0$ such that $E$ maps diffeomorphically $V_\delta(x):=\big\{(x',v')\in NM: |x-x'|<\delta,|v'|<\delta\big\}$ onto an open neighborhood of $x$ in $\Bbb R^n$.
Now, let $$\rho(x):=\sup\big\{\delta\leq 1: E\text{ maps }V_\delta(x)\text{ diffeomorphically }$$$$\text{onto an open neighborhood of }x\text{ in }\Bbb R^n\big\}.$$ Then, $\rho\colon M\to (0,\infty)$ is continuous and $0<\rho \leq 1$.

Next, define $V:=\left\{(x,v)\in NM:|v|<\frac{1}{2}\rho(x)\right\}$. Then, $E$ maps diffeomorphically $V$ onto an open subset $U$ of $\Bbb R^n$ with $M\subseteq U$.

Now, define $r\colon U\to M$ by $r:=\pi\circ E^{-1}$. So, $r$ is a retraction and submersion.

Observation: Let $y,y'\in U$ with $|y-y'|<\epsilon$ for some $\epsilon>0$. Write, $E^{-1}(y)=(x,v)\in V$ and $E^{-1}(y')=(x',v')\in V$, i.e. $r(y)+v=x+v=y$ and $r(y')+v'=x'+v'=y'$. So, $$|x-x'|=\big|(y-y')-(v-v')\big|\leq |y-y'|+|v-v'|$$$$\leq \epsilon+|v|+|v'|\leq \epsilon+\frac{1}{2}\rho(x)+\frac{1}{2}\rho(x')\leq \epsilon+1$$ as $\rho\leq 1$. Therefore, $$|y-y'|<\epsilon\implies \big|r(y)-r(y')\big|\leq\epsilon+1.$$

$\textbf{Theorem:}$ Now, let $F\colon N\to M$ be a continuous proper map. Then, there is a proper map $H\colon N\times [0,1]\to M$ such that $H(-,0)=F$ and $H(-,1)\colon N\to M$ is smooth proper map.

$\textbf{Proof:}$ Let $r\colon U\to M$ be as given above. Let $\delta(x):=\sup\big\{\varepsilon \leq 1:B_\varepsilon(x)\subseteq U\big\}$ for each $x\in M$. Then, $\delta:M\to (0,\infty)$ is a continuous function using similar argument just like to show $\rho$ is continuous. Also, $\delta\leq 1$. Let $\widetilde \delta:=\delta\circ F$ and choose a smooth function $\widetilde F:N\to \Bbb R^n$ such that $\left|\widetilde F(y)-F(y)\right|<\widetilde\delta (y)$ for each $y\in M$. In particular, $\left|\widetilde F-F\right|\leq 1$. Define, $$H(p,t):=r\left((1-t)F(p)+t\widetilde F(p)\right)\text{ for }(p,t)\in N\times [0,1].$$
This is well-defined because $\left|\widetilde F(p)-F(p)\right|<\delta\big(F(p)\big)$ i.e. $\widetilde F(p)\in B_{\delta\big(F(p)\big)}\big(F(p)\big)\subseteq U$.

Now, for $(p,t)\in N\times [0,1]$ we have $$\left|(1-t)F(p)+t\widetilde F(p)-F(p)\right|\leq t\cdot\left|\widetilde F(p)-F(p)\right|\leq 1$$ $$\implies\big|H(p,t)-r\circ F(p)\big|=\big|H(p,t)-F(p)\big|\leq 2.$$

Now, consider a $\textbf{proper}$ embedding $N\hookrightarrow \mathbb R^k$ for some $k$. Note that any proper map from a topological space to a locally compact, Hausdorff space is a closed map. Hence, we can think $N$ as a closed subset of $\mathbb R^k$. With induced metric on $N$ from $\mathbb R^k$ and with usual distance metric on $[0,1]$, consider the product metric on $N\times [0,1]$. So, a closed and bounded subset of $N\times [0,1]\subseteq_\text{closed}\Bbb R^{k+1}$ is compact by Heine–Borel theorem. Similarly, considering the proper embedding $M\hookrightarrow \Bbb R^n$ and giving induced metric on $M$ from $\Bbb R^n$, we can say any closed-bounded subset of $M$ is compact.

So, to prove $H$ is proper, it is enough to show $H^{-1}(C)$ is closed and bounded in $N\times [0,1]$ for any compact set $C$ is $M$. Let $C$ be a compact subset of $M$. Now, $H^{-1}(C)$ is closed in $N\times [0,1]$ as $H$ is continuous and $C$ is closed in $M$. Hence, we only need to show $H^{-1}(C)$ is bounded subset of $N\times [0,1]$. Now, notice that $\big\{(p_n,t_n)\big\}\subseteq N\times [0,1]$ is bounded if and only if $\{p_n\}\subseteq N$ is bounded. Also, $F$ is proper implies $\big\{F(p_n)\big\}$ is bounded if and only if $\{p_n\}$ is bounded, for any sequence $\{p_n\}$  in $N$. Finally, note that $\big| H(p,t)-F(p)\big|\leq 2$ for all $(p,t)\in N\times [0,1]$. So, if there were an unbounded sequence $\big\{(p_n,t_n)\big\}\subseteq  H^{-1}(C)$, then $\{p_n\}$, and hence $\big\{F(p_n)\big\}$ would be unbounded, a contradiction, as $C$ is a bounded subset of $\Bbb R^n$.  So, $ H$ is a proper map. $\square$


答:
I think all of $a_n = \frac{1}{2}$

You can easily prove it using induction also.

$a_1 = \frac{1}{2}$

Assume $a_m = \frac{1}{2}$

To prove $a_{m+1} = \frac{1}{2}$

$a_{m+1} = \frac{1}{m+2} + \frac{m+1-1}{m+2} a_m$

$a_{m+1} = \frac{1}{m+2} + \frac{m}{m+2} \frac{1}{2}$

$a_{m+1} = \frac{1}{2}$

Since the relation is true for $n=1$ and assuming the result for $a_m$, we can prove it for $a_{m+1}$, hence the result is true by induction.

Also we can get the same from your equation,

$b_n = \frac{n}{2} +b_{n-1}$

$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$

$b_n = \frac{n}{2} +\frac{n-1}{2}+ \dots + \frac{n-k+1}{2} + b_{n-k}$

Substitute k = n

$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$

$b_n = \frac{n}{2} +\frac{n-1}{2}+ \dots + \frac{1}{2} + b_{0}$

$b_n = \frac{n}{2} + \frac{n-1}{2}+b_{n-2}$

$b_n = \frac{n + n-1 +n-2 + \dots +1}{2}+0$

$b_n = \frac{n(n+1)}{4}$

$\frac{n(n+1)}{2} a_n = \frac{n(n+1)}{4}$

$a_n =\frac{1}{2}$

Title: Finding the Closed Form of a Recursion

Body:
I wanted to find the closed form of the recursion: $$a_n = \frac{1}{n+1} + \frac{n-1}{n+1} a_{n-1}$$ where $a_0 = 0$.

So far, my progress has been to multiply both sides by $\frac{n(n+1)}{2}$ which gives $$\frac{n(n+1)}{2} a_n = \frac{n}{2} + \frac{n(n-1)}{2} a_{n-1}.$$

Now, if we let $b_n = \frac{n(n+1)}{2} a_n$, the recursion becomes: $$b_n = \frac{n}{2} + b_{n-1}.$$

How do I continue on from here? I feel like I have to figure out a closed form for $b_n$ and use that to compute $a_n$, but I can't seem to figure out a way to do so.

Any help would be greatly appreciated.


问:
Some questions about SAGE
I search a programmable calculator like PARI/GP. In the net, I came across
 SAGE. I have some questions about the download of SAGE

Is it really absolutely free ?

Is it faster than PARI/PG ?

Has it more features (for example galois groups of polynomials of high
degrees) ?

Which space is needed on disk ?

Does it have tar.gz-files or zip.files. In the first case, I do not have
the supporting programs. The best case would be : .exe (executable)

Perhaps, someone has SAGE and can answer my questions .

答:
Yes, it is totally free. No ads, no paying for extra bits, nothing like that. It's open source, even, so you can download all the source code you would ever want. (They even encourage you too, and improve it along the way).

Is it faster than PARI - in short, no. PARI is actually included inside SAGE (along with many other free things like maxima, etc.). SAGE interfaces to PARI, and this comes at a cost. But SAGE gets to use the same optimized algorithms as PARI, and I've never noticed  that any difference in speed is noticeable.

It has many features, and these features are expanding. It can do many things with Galois groups, but I have not played much with these.

SAGE is huge because it's designed to be completely self-sufficient. Thus it includes many packages that a competent computer-using mathematician already has, and takes up roughly 4 gigs on file (their site says 7 gigs are necessary to run it). There are really two options here: someone can run a sage server and you can access it, so that it takes up zero space on disk but someone has to have a dedicated server; or you can have all 4 gigs on your own disk. If this is the case, you'll actually need to run it on linux. If you have Windows/Mac, they provide a virtual image already installed on a red hat emulator that can be used with other (also free) software like virtualbox.

Since you run Windows, you won't be getting either a tar.gz or a zip, but rather a .ova file that can be used in virtualbox (which runs on Windows via an .exe). On their site, they link you to virtualbox too.

If you just want to try it out, you can try their free notebook, but it's pretty slow because anyone can use it (and many do). Of course, if you want to try it out, just go ahead and download it and install it. It does not take very long, and their documentation for installation is pretty good.

As a disclaimer, I am a SAGE developer as well (in the contributing to their code through their trac server because it's open source sense, not in the I-helped-design-and-come-up-with-the-idea sense).

Title: Some questions about SAGE


问:
Plane given 3 homogeneous points
I am studying a slide from this lecture that says given three homogeneous points $v_1, v_2, v_3$ in 3D space, one can define the plane containing them as $\mathbf{\pi}=(v_1 \times v_2) \times (v_2 \times v_3)$. Does anyone know how this is derived or can forward me to the resources to understand this? I know of other methods to solving for $\mathbf{\pi}$ such as solving:
$$\begin{bmatrix} v_{1}^T\\ v_{2}^T\\ v_3^T\end{bmatrix}\mathbf{\pi}=0$$

However, I have never seen this formula before and am not sure how to derive/prove it.

For some context this is the slide, although most of the details on it are unimportant for understanding this problem.

答:
What you are seeing is the intersection (meet) of two lines. Each line is defined from the connection (join) of two points. The $\times$ operator here isn't an actual cross product, because the cross product isn't defined for 4-vectors. I did not find where in the video they define the details of $\times$, but being an overview I doubt they went into this.

It is important to note that with planar homogeneous coordinates the line connecting two points is found using the vector cross product $\times$.

You can try it out, to find the $(a,b,c)$ coordinates of the line connecting the points $(x_1,y_1,1)$ and $(x_2,y_2,1)$ is

$$ \pmatrix{x_1 \\ x_1 \\ 1} \times \pmatrix{ x_2 \\ y_2 \\ 1} = \pmatrix{ y_1-y_2 \\ x_2 - x_1 \\ x_1 y_2 - x_2 y_1 }$$

making the equation of the line

$$ (y_1-y_2) x + (x_2-x_1) y + (x_1 y_2 - x_2 y_1) = 0 $$

Conversely, the point where two lines intersect is also found by the vector cross product $\times$. For example, two lines $u x + v y + w = 0$ and $p x + q y + r = 0$ meet at a point

$$ \pmatrix{ u \\ v \\ w} \times \pmatrix{p \\ q \\ r} = \pmatrix{ r v-q w \\ p w - r u \\ q u - p v } $$ with location $$ \pmatrix{x \\ y} = \pmatrix{ \frac{r v-q w}{q u - p v} \\ \frac{p w - r u}{q u - p v}}$$

So somehow in computer vision, the idea of some magical operator with symbol $\times$ exists that can be used to define the line where two points join ($v_1 \times v_2$) and the plane where two lines meet ($ (v_1 \times v_2) \times (v_2 \times v_3)$).

Note that the jump from 2D to 3D isn't trivial because of the way lines are defined.

In 2D we often use the $(a,b,c)$ coordinates for $ax + b y + c = 0$, where the vector $(a,b)$ is perpendicular to the line.

In 3D we often use Plücker coordinates $(\vec{e}, \vec{r}\times\vec{r})$ where the vector $\vec{e}$ is parallel to the line.

The realization I had is that a 2D line is a projection of a 3D plane that is perpendicular to the xy plane. A 3D plane is defined by the normal vector and the negative distance $(\vec{n},-d)$ and you get the exact 2D line coordinates if $\vec{n}=(a,b,0)$ and $d=-c$.

This is also evident that in 2D lines are dual to points, but in 3D lines are dual to lines, and points and dual to planes. So the $\times$ operator can be defined between dual objects (as well as the $\cdot$ operator for incidence) like points and lines in a plane, but not for points and lines in 3D.

So there is some trickery and magic going on here and they are glossing over some major details to get to a point.

Title: Plane given 3 homogeneous points


问:
Topological group homeomorphic
Exists a topological group homeomorphic to $(0, 1)$? and to $[0, 1]$?
I believe that with includes the extremes is it, to contrary than when they aren't included.
I tried to use the definition of topological group, but I'm not sure how to distinguish the two cases because of the extremes.

答:
$[0,1]$ is not homeomorphic to a topological group. To see why, note that the left multiplication action of a topological group on itself is transitive, and it follows that the complements of any two points are homeomorphic. But the complement of each endpoint of $[0,1]$ is connected whereas the complement of each interior point is disconnected.

As said in the comment of @PatrickNicodemus, $(0,1)$ is homeomorphic to the topological group $\mathbb R$.

Title: Topological group homeomorphic


答:
As mentioned by @Patrick, $(0,1)$ is homeomorphic to ${\mathbb R}$.

However $[0, 1]$ is not homeomorphic to any topological group
because every topological group $G$ is homogeneous,  that is,  for any two points $g$ and $h$ in $G$,  there is a
homeomorphism of $G$ sending $g$ to $h$,  namely left-multiplication by $hg^{-1}$.

However there is no homeomorphism
from $[0,1]$ to itself sending $0$ to $1/2$.

Title: Topological group homeomorphic

Body:
Exists a topological group homeomorphic to $(0, 1)$? and to $[0, 1]$?
I believe that with includes the extremes is it, to contrary than when they aren't included.
I tried to use the definition of topological group, but I'm not sure how to distinguish the two cases because of the extremes.


问:
$U$ is a subspace of $V$ and $\psi:V\to W$ is a linear map. Show that $\dim(\psi(U))\leqslant\dim(U)$.
This is where I have gotten to so far, by I am under the impression that I have made an assumption with the restriction. Is this correct?

By the Rank Nullity Theorem, $\operatorname{rank}(\psi)+\operatorname{null}(\psi)=\dim(V)$.

Now $\operatorname{null}(\psi)\geqslant0\implies\dim(V)-\dim(\operatorname{Im}(\psi))\geqslant0\iff\dim(V)\geqslant\dim(\operatorname{Im}(\psi))$.

Consider now $\psi|_U:U\longrightarrow\operatorname{Im}(\psi|_U)$.

And so $\dim(U)\geqslant\dim(\operatorname{Im}|_U(\psi))=\dim(\psi(U))\blacksquare$.

答: Apply the rank-nullity theorem to $\psi|_U$:\begin{align}\operatorname{rank}(\psi|_U)+\operatorname{null}(\psi|_U)=\dim U&\implies\operatorname{rank}(\psi|_U)\leqslant\dim U\\&\iff\dim\psi(U)\leqslant\dim U.\end{align}


问:
Riemann-Stieltjes Integral
Let $$\alpha(x) = \Biggl\{ \begin{array}{cc} 0 & x=0, \\\ 2 & 0<x\leq 1 \\\ 1 & 1<x \leq 2 \end{array}$$

Let $f(x)$ be continuous on $[0,2]$. Let $P = \{0,0.5,1,1.5,2\}$. For this partition, find the sum $S(P,t)$ approximating the integral

$$\int\limits_{0}^{2} f(x) d \alpha(x)$$

for every $t$ for which this sum is defined. What is the value of the integral? Justify your answer by referring to the definition of the integral only.

I know the definition but cannot seem to apply to it to cases. This is bad, I know, but can you go through this example with me. Once I understand how to apply it I'm sure I'll be able to figure out the rest of the questions! :) Oh... I did draw $a(x)$. It's the whole finding $L(P)$ and $U(P)$ with sums that I find difficult.

答:
A little hint: You need to apply the definition :-) If you say you cannot to that, I have to agree with Theo and Qiaochu that this is a strong indication that you either don't know, or don't understand, the definition.

In that case, you should try to understand the most simple example that was mentioned in your class. For every definition in every math class, you should come up with the most simple example possible, and some nontrivial example.

The first step in applying the definition is a choice of a partition of the integration interval. Choose some numbers $x_1, x_2, x_3 \in [0, 2]$ and write the sum that approximates the integral. Then choose a refinement of this partition and write down the sum again. Maybe you'll notice that it is possible to choose a sequence of partitions such that you can evaluate the sums and determine the limit that is the integral...

Addendum: Ok, here is another hint: For $\alpha$ differentiable, we know that
$$
\int f(x) d \alpha(x) = \int f(x) \alpha'(x) dx
$$
So, if $\alpha$ is a jump function, that is constant with the notable exeption of a jump, it is possible to simplify the integral. For example, define
$$
\alpha(x) := 0 \; \text{for} \; x \lt 0
$$
and
$$
\alpha(x) := 1 \; \text{for} \; x \ge 0
$$

Then we can rewrite the integral
$$
\int_{-a}^{a} f(x) d\alpha(x) =  \int_{-a}^{-\epsilon} f(x) \alpha'(x) dx + \int_{-\epsilon}^{\epsilon} f(x) d\alpha(x) + \int_{\epsilon}^{a} f(x) \alpha'(x) dx
$$
for an arbitrarily small $\epsilon \gt 0, \epsilon \lt a$. Since the differential of $\alpha$ vanishes for the first and the last term, we get
$$
\int_{-a}^{a} f(x) d\alpha(x) =  \int_{-\epsilon}^{\epsilon} f(x) d\alpha(x)
$$
So, it is actually enough to consider what happens in an $\epsilon$ neighborhood of 0, in order to calculate the integral.

Now, any summand in U(P) and in L(P) will be zero except the one containing the difference of $\alpha(\epsilon) - \alpha(-\epsilon)$, so we have
$$
U(P) = (\alpha(\epsilon) - \alpha(-\epsilon)) \sup_{ x \in [-\epsilon, \epsilon]} f(x) = \sup_{ x \in [-\epsilon, \epsilon]} f(x)
$$
and
$$
L(P) = (\alpha(\epsilon) - \alpha(-\epsilon) ) \inf_{ x \in [-\epsilon, \epsilon]} f(x) = \inf_{ x \in [-\epsilon, \epsilon]} f(x)
$$
Since $f$ is continuous, we get in the limit $\epsilon \to 0$:
$$
U(P) = L(P) = f(0)
$$

Title: Riemann-Stieltjes Integral


问:
Where do $p$-adic numbers and $p$-Sylow theory both appear?
Both $p$-adic numbers and $p$-Sylow theory are by design "arithmetic" ways of "localizing," so it stands to reason they might be in cahoots in certain contexts. Are they? </p>

答:
Here's one reasonably simple link:  In terms of Sylow theory, a nice home for generalization is the land of profinite groups, where you have directly analogous Sylow theorems (with directly analogous proofs).  Among profinite groups there is the canonical $\widehat{\mathbb{Z}}$, the profinite completion of $\mathbb{Z}$ and free profinite group of rank 1.  The decomposition
$$
\widehat{\mathbb{Z}}\cong \prod_p \mathbb{Z}_p
$$
shows that the $p$-adic integers are simply the $p$-Sylow subgroup of this very canonical group.</p>


问:
How to estimate the growth of the following sequence.
$a_1=1,\forall n\ge 2(a_n=\sqrt{n+a_{n-1}})$.I want to find the smallest natural number n such that $a_n>10$.I find that the sequence is increasing,and I guess:For natural number k,there are only 2k items $a_{j+1},a_{j+2},\cdots,a_{j+2k}$ such that $k\le a_i<k+1,i=j+1,j+2,\cdots,j+2k$，but I have no idea how to prove it.I tried Mathematical induction but I failed ,Please help!I found the answer is 91  using python,but I can't prove it yet.I think the problem differs from Asymptotic behavior of the sequence $a_n=\sqrt{n+a_{n-1}}$ i focus on the specific properties while that problem cares about asymptotic behavior.

答:
I find another way to solve the problem.
Suppose $n\ge 91$,$$a_n = \sqrt{91+a_{n-1}}=\sqrt{91+\sqrt{90+a_{n-2}}}>\sqrt{91+\sqrt{90}}>10.$$Suppose $n \le 90$, $a_{90}=\sqrt{90+a_{98}}>a_{98}\Rightarrow a_{98}<10$.hence,$a_{90}=\sqrt{90+a_{98}}<\sqrt{90+10}=10$.
Though I know how to solve the problem, i wonder how to prove my guess that there are only 2k items in the sequence such that  integer part of them are k.


问:
application to Lusin's Theorem
If $f:\mathbb{R}\rightarrow \mathbb{R}$ is Lebesgue measurable, then there exists a sequence of continuous functions $\{f_n \}$ converges pointwise to $f$  almost everywhere on $\mathbb{R}$.

My work:

I take $n\in\mathbb{N}$ be fixed, then by Lusin's theorem there exists a closed subset $F_n $ such that $m(\mathbb{R} - { F_n}) < \frac{1}{n}$ and a continuous function $f_{n}=f$ on $F_n$. Then, we can define $F=\cup_{n=1}^{\infty}F_n$ so that for any $x\in \mathbb{R} - F$, there exists $N \in \mathbb{N}$ such that
$|f-f_n| =0 < \epsilon$ for any $\epsilon >0$ on $F_n$ for all $n> \mathbb{N}$.
I think this shows the pointwise convergence on $\mathbb{R}$ a.e as $m(\mathbb{R} - { F_n}) < \frac{1}{n}$.

I think I need to select $F_n$ to be increasing sequence of closed sets, right?

答:
I think that we are talking about the Lebesgue-meausre $m$. This argumentation of yours does not work then (because of many reasons), but most importantly, because $\mathbb{R}$ has infinite measure. Check the requirements of Lusin's theorem.

Try something else: It is well known that there exists a sequence of continuous functions $(f_n)_{n \in \mathbb{N}}$ such that
$$
\int_\mathbb{R} \lvert f_n - f \rvert~\mathrm{d}x \overset{n \rightarrow \infty}{\longrightarrow} 0.
$$
You can prove this using approximation of $f$ by a series of characteristic functions that themselves you approximate by continuous functions.

Now choose a subsequence $(f_{k_n})_{n \in \mathbb{N}}$ such that
$$
\int_\mathbb{R} \lvert f_{k_n} - f \rvert~\mathrm{d}x \leq 2^n
$$
for all $n \in \mathbb{N}$.
Therefore
$$
\lim_{m \rightarrow \infty} \sum_{n = 1}^m \int_\mathbb{R} \lvert f_{k_n}  - f \rvert~\mathrm{d}x < \infty.
$$
By the monotone convergence theorem:
$$
\lim_{m \rightarrow \infty} \sum_{n = 1}^m \int_\mathbb{R} \lvert f_{k_n}  - f \rvert~\mathrm{d}x = \int_{\mathbb{R}} \sum_{k = 1}^\infty \lvert f_{k_n} - f \rvert~\mathrm{d}x < \infty
$$
Clearly, we then must have
$$
\sum_{k = 1}^\infty \lvert f_{k_n}(x) - f(x) \rvert < \infty
$$
a.e., so $f_{k_n} \overset{n \rightarrow \infty}{\longrightarrow} f$ pointwise a.e.. Thus $(f_{k_n})_{n \in \mathbb{N}}$ is your desired sequence.

Title: application to Lusin's Theorem


问:
Z transform causality
$$\dfrac{z(z-1\cos a)}{ z^2-2z\cos a+1}$$

For this transform, I have

Zeros: $0, cos a $

Poles: $\cos a \pm j\sin a$

The ROC: $z>1$

The system is stable
My question is, is it causal??

答:
The system is causal.

Let, $X(z) = \dfrac{z(1-z\cos a)}{z^2-2z\cos a +1} \Rightarrow \dfrac{X(z)}z =  \dfrac{1-z\cos a}{z^2-2z\cos a +1}$

You've found the poles as, $\cos a \pm j \sin a = e^{\pm ja}$.

So let,

$\dfrac{X(z)}{z} =  \dfrac{1-z\cos a}{(z-e^{j a})(z-e^{-ja})} =  \dfrac{k}{z-e^{ja}}+ \dfrac{k^*}{z-e^{-ja}} $

Now by inspection, $k =\dfrac{1-e^{ja}\cos a}{e^{ja} - e^{-ja}} = \dfrac{1- e^{ja}(\frac{e^{ja} + e^{-ja}}{2})}{e^{ja} - e^{-ja}} = -\dfrac{e^{ja}}{2}$

and $k^* = -\dfrac{e^{-ja}}{2}$.

So,

$X(z) = -\dfrac12\left[\dfrac{ze^{ja}}{z-e^{ja}} + \dfrac{ze^{-ja}}{z -e^{-ja}}\right]$

ROC: $|z|>1$ and here $|e^{\pm ja} | = 1$. So clearly, $|z| > |e^{\pm ja}|$.

Thus

$X(z) = -\dfrac12 \mathcal Z\left\{[e^{ja}(e^{jan}) + e^{-ja}e^{-jan}]u(n) \right\} = \mathcal Z \left\{-\cos (an+a)u(n)\right\}$

and the system is causal.

For the question now being edited, follow the same partial fraction method. You'll end up getting, $k = k^* = \frac 12$. And it'll reduce to

$$X(z) = \frac{z}{z-e^{ja}}+\frac{z}{z-e^{-ja}} = \mathcal Z \{ \cos (an) u(n) \}$$

This is again causal.

Title: Z transform causality


答:
Consider the $\exp$ function, and the ordinary addition and multiplication/. You have
$$\exp(x+y)=\exp(x)\cdot\exp(y).$$

Title: How to translate a "formal" definition of the distributive property?

Body:
I have seen sometimes a formal representation of the distributive property over an operator defined as follows:

A function $f$ is said to distribute over the operator $\bigoplus$ if
there is an operator $\bigotimes$ such that  $f(x\bigoplus y)=f(x)\bigotimes f(y)$

When we see a definition like this, how do we "translate" it to a more concrete example?

E.g. multiplication distributes through addition and hence:

$x\cdot (y + z) = x \cdot y + x \cdot z$

So in this case what is the $f$ what is the $\bigotimes$ and what is the $\bigoplus$?