问:
Increasing and bijective function

$f$ is a function: $E \to F$ such that for any $a \in E \, and \,  b \in E$, we have the property:

$a < b$ if and only if $f(a) < f(b)$

What is characteristics of such function f ? Prove it.

I think that $f$ should be a bijective increasing function but I'm not sure whether such condition is too strong or not. Moreover, if it is such a case, how to prove the part: if $f(a) < f(b)$ then $a < b$ ?

答:
$f$ must be injective, since if we have $x \neq y$, this means one is strictly larger than the other, and so one of their images under $f$ is also strictly larger than the other, so $f(x) \neq f(y)$.

However it does not need to be surjective, for example $f = arctan$.

Title: Increasing and bijective function


问:
A set of $n$ orthonormal vectors provides a basis for $\mathbb R^n$
Let $\{v_1,...,v_n\} \subseteq \mathbb R^n$ be a set of mutually orthogonal, non-zero vectors of length $1$.

How does one prove that this set is a basis of $\mathbb R^n$?

I have already proved that they are linearly independent, but I still need to prove that they span $\mathbb R^n$. I have also read that a set of $n$ linearly independent vectors in $\mathbb R^n$   is a basis, but I couldn't find a proof of this theorem.

答: Any linearly independent set with cardinality $n$ forms a basis for  $\mathbb{R}^n$. Even it is true for any $n$ dimensional vector space.

ContentLicense: CC BY-SA 3.0

ContentLicense: CC BY-SA 4.0


答:
What is the characteristics of such function $f$?

This kind of function is called a monotonically increasing function. These functions are automatically injective. Not necessarily they are bijective. Think about a simple example: $\mathbb{A}=[1, 2]$ and function $f$ from $\mathbb{A}$ into $\mathbb{R}$ that maps $x$ into $x^2.$

Title: Increasing and bijective function

Body:
$f$ is a function: $E \to F$ such that for any $a \in E \, and \,  b \in E$, we have the property:

$a < b$ if and only if $f(a) < f(b)$

What is characteristics of such function f ? Prove it.

I think that $f$ should be a bijective increasing function but I'm not sure whether such condition is too strong or not. Moreover, if it is such a case, how to prove the part: if $f(a) < f(b)$ then $a < b$ ?


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

答:
We can build an Equilateral triangle between the points, whose side length is $r$:

picture source

So we know that the points of intersection are $\sqrt{3}r$ apart, and the angle at them is $60^\circ$, by building a rhombus between the dots and centers we know that the angle that "opens" the area is $120^\circ$:

We can now calculate half the area in question as a circular segment:

$$S=2\left[\frac{r^2}{2}\left(\frac{2\pi}{3}-\sin\left(\frac{2\pi}{3}\right)\right)\right]=r^2\left(\frac{2\pi}{3}-\frac{\sqrt{3}}{2}\right)$$

$S$ is the total area in question


问:
Fitting data that follows specific equation
Given the following equation:

$y = \frac{80-x}{100-x}$

If I had real world data that followed this exact equation, how would I back out an equation to always predict the correct $y$ value given an $x$ value?

A second degree polynomial comes close but is not exact. I would hope for the resulting equation to be simple, given the simplicity of the equation to generate the data.

Sample Data:

x, y

0.000,  0.800

10.000, 0.778

20.000, 0.750

30.000, 0.714

40.000, 0.667

50.000, 0.600

60.000, 0.500

70.000, 0.333

80.000, 0.000

答:
$$y=\frac{ax+b}{x+c}\quad\implies\quad xy=-cy+ax+b$$
Approximate $a,b,c$ are computed thanks to linear regression :

The above method is valid because no criteria of fitting is specified in the question.

If a particular criteria of fitting was requiered non-linear regression should be necessary. But the result would be very close to the above result because the data is  almost not scattered.


问:
A Numerical Sequences proof.
Let $ 0<a<1 $ be a real number, and let $ a_n \in \left ( -1,0 \right )$ defined by the relation $ \sqrt[n]{a} = 1+a_n$ ,   $ n \in \mathbb{N}$.

Show the following inequality:

$ |a_n| \le \frac{1}{n} \bigl( \frac{1-a}{a}\bigr)$,    $ n \ge 1$.

This problem was proposed by my professor of Mathematical Analysis $2$. The following is my personal solution, different from the one expected.

答:
By hypothesis it's $ \sqrt[n]{a} = 1 + a_n$ , in particular $ a_n=\sqrt[n]{a}-1<0$ $\to$ $|a_n|=1-\sqrt[n]{a}$.

Second member can be viewed as a division of polynomials:

$$ 1 - \sqrt[n]{a} = \frac{1 - a}{1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}} }$$

Note that terms in denominator,

$$1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}}$$

are all $\ge$ $a$: being $ 0<a<1$, indeed, elevating $a$ to an exponent $ \in \left [ 0,1 \right]$ (note that exponents of $ a$ in denominator are $\frac{1}{n}$, $\frac{2}{n}$, ... , $\frac{n-1}{n} \le 1$) will get a value  $\ge a$, i.e. closer to $1$.
There are $n$ terms in denominator (note that all powers of $ a $ from $0$ to $ n-1 $ are present, and $ n-1+1=n$).                                              Then it's

$$ |a_n| = 1 - \sqrt[n]{a} = \frac{1 - a}{1 + \sqrt[n]{a} + \sqrt[n]{a^2} + \sqrt[n]{a^3} + ... +\sqrt[n]{a^{n-1}} } \le \frac{1-a}{na}$$

from which follows: $ |a_n| \le \frac{1}{n} \bigl( \frac{1-a}{a} \bigr)$

$\square$

Title: A Numerical Sequences proof.


问:
Prove that for two real numbers $a,b$, if for any $e>0$ they can be bounded by $s,s'$: $s'\geq a\geq s,s'\geq b\geq s$ and $s'-s<e$, then $a=b$.
I don't understand the proof of lemma presented in the title. I found it in the book "Differential and integral calculus" by G. M. Fichtenholz, it's lemma 2, section 2.8, chapter 1. I don't think this book was ever published in English, I'm using Polish version, so I will translate the original statement and proof. The lemma states: "Let $\alpha, \beta$ be real numbers. If for any number $e > 0$, $\alpha$ and $\beta$ can be bounded by the same rational numbers $s, s'$,
$$s' \geq \alpha \geq s,\quad s' \geq \beta \geq s,$$
with a difference less than $e$,
$$s' - s < e,$$ then $\alpha$ and $\beta$ must be equal.

The lemma will be proved by reduction to absurdity. Let $\alpha > \beta$. We can put two rational numbers  $r, r'$ between $\alpha$ and $\beta$(this was stated and proved in another lemma),
$$\alpha > r' > r > \beta.$$
Then for any $s', s$, between which are $\alpha$ and $\beta$:
$$s' > r' > r > s, \quad s' - s > r' - r > 0,$$
so $s' - s$, contrary to lemma assumptions, can't be smaller than $e = r' - r$. This contradiction proves correctness of the lemma."

I don't understand the idea of this proof. I think that in proofs by contradiction, one has to assume the proposition to be false which leads to contradiction. I don't see such assumption there. Personally I belive that the only thing that was proved, was that necessary conditions of the lemma weren't met. If it indeed is a valid proof, please explain it to me, how from the given statements it follows that $\alpha = \beta$.

答:
Give two propositions $P,Q$, the statement $P\rightarrow Q$ is equivalent to $\neg Q \rightarrow \neg P$. This means that it's equivalent to prove:

if $P$ is true, then $Q$ is true

if $Q$ is false, then $P$ is false

On your lemma, you can identify $P$ as "for any two real numbers [...] and $s-s'<e$" and $Q$ as $a=b$. Then, proving that lemma is equivalent to proving that that whenever $\alpha\neq \beta$ ($Q$ is false), we can find two real numbers and $e>0$ such that for every $s,s'$ verifying $s'\geq \alpha\geq s$ and $s'\geq \beta \geq s$, it must be $s-s'>e$ ($P$ is false).

Then, let's assume that $\alpha\neq\beta$. Because $\mathbb{R}$ is a totally ordered field, either $\alpha>\beta$ or $\alpha<\beta$. We can assume without loss of generality that $\alpha>\beta$; because if $\alpha<\beta$, the proof would be the same by just "exchanging" the variables $\alpha$ and $\beta$. The proof you posted starts from here on.

PS: I don't really know if you needed this much information, if this wasn't really your question or if it's ok. Let me know if there is something else.


问:
Number of possible inequalities for $n$ variables.
Let's say there are two variables $x$ and $y$. Here are the possible inequalities:

(i) $x<y$

(ii) $x>y$

(iii) $x=y$

So, there are $3$ possible inequalities. But let's say there are $n$ variables $x_1,x_2,x_3,x_4,\cdots,x_n$. How many possible distinct inequalities are there?

I assume this problem requires combinatorics , but I am not sure how to apply them in this problem.

Edit: Maybe this problem can be related with coin arranging (point me if I am wrong)

First of all , we can arrange coins (where order matters).But ,we will also allow for the coins to stack on top of each other.In the stacks (There can be multiple stacks) , order doesn't matter.Stacks will be represented with parenthesis.

For $1$ coin , there is $1$ way to arrange them.

$1$ $[x_1]$

For $2$ coins , there are $3$ ways to arrange them.

$12$ $[x_1 < x_2]$

$21$ $[x_1 > x_2]$

$(12)$ $[x_1 = x_2]$

For $3$ coins , there are $13$ ways to arrange them.

$123$ $[x_1 < x_2,x_2 < x_3,x_3 > x_1]$

$132$ $[x_1 < x_2,x_2 > x_3,x_3 > x_1]$

$213$ $[x_1 > x_2,x_2 < x_3,x_3 > x_1]$

$231$ $[x_1 > x_2,x_2 < x_3,x_3 < x_1]$

$312$ $[x_1 < x_2,x_2 > x_3,x_3 < x_1]$

$321$ $[x_1 > x_2,x_2 > x_3,x_3 < x_1]$

$(12)3$ $[x_1 = x_2,x_2 < x_3,x_3 > x_1]$

$3(12)$ $[x_1 = x_2,x_2 > x_3,x_3 < x_1]$

$1(23)$ $[x_1 < x_2,x_2 = x_3,x_3 > x_1]$

$(23)1$ $[x_1 > x_2,x_2 = x_3,x_3 < x_1]$

$2(13)$ $[x_1 > x_2,x_2 < x_3,x_3 = x_1]$

$(13)2$ $[x_1 < x_2,x_2 > x_3,x_3 = x_1]$

$(123)$ $[x_1 = x_2,x_2 = x_3,x_3 = x_1]$

I thought of thinking of stacks as variables that are equal.And a variable being left/right to another variable corresponding to a varialbe less than/ greater than another variable.So my thought is..

number of ways to arrange $n$ coins = number of distinct inequalities for $n$ variables.(Hope this helps)

答:
As Mike Earnest commented, these are ordered Bell numbers or Fubini numbers, and OEIS A000670 shows many ways to calculate them, though not a closed form.

My preference would be with a recursion, letting $f(n,k)$ be the number of orderings of $n$ variables involving $k$ distinct values. When you introduce the $n$th variable, you can

either introduce it to an ordering of $n-1$ variables with $k$ distinct values by matching it to one of the existing distinct values

or introduce it to an ordering of $n-1$ variables with $k-1$ distinct values by matching it to one of the $k$ gaps between, before or after the existing distinct values

So $$f(n,k)=k\big(f(n-1,k) + f(n-1,k-1)\big)$$

and you can start with $f(1,1)=1$ and $f(1,k)=0$ for $k\not=1$, or from $f(0,0)=1$ and $f(0,k)=0$ for $k\not=0$.

You then get a table like this

$$\begin{matrix}
 & k& 0& 1& 2& 3& 4& 5& 6& 7& 8& & \text{Sum}\\
n& & & & & & & & & & & & \\
0& & 1& 0& 0& 0& 0& 0& 0& 0& 0& & 0\\
1& & 0& 1& 0& 0& 0& 0& 0& 0& 0& & 1\\
2& & 0& 1& 2& 0& 0& 0& 0& 0& 0& & 3\\
3& & 0& 1& 6& 6& 0& 0& 0& 0& 0& & 13\\
4& & 0& 1& 14& 36& 24& 0& 0& 0& 0& & 75\\
5& & 0& 1& 30& 150& 240& 120& 0& 0& 0& & 541\\
6& & 0& 1& 62& 540& 1560& 1800& 720& 0& 0& & 4683\\
7& & 0& 1& 126& 1806& 8400& 16800& 15120& 5040& 0& & 47293\\
8& & 0& 1& 254& 5796& 40824& 126000& 191520& 141120& 40320& & 545835\\
\end{matrix}$$

with the final column giving the result you want


问:
If a line and its points are removed from a projective geometry, is the resulting is affine geometry?
I saw that it's written in a couple of places, but couldn't find proper proof for this.

I saw it for example in:

here and here

I would like to find a reference to a proof of this (as I'm sure some book or a thing like that addresses the problem) or the proof itself.

I hope it's legit to ask for this.

答:
Let $E$ be a vector space over the field $\Bbb K$, and let

$$ \mathcal P(E) = \big\{F\subseteq E\;\; : \;\; F \text{ subspace of  } E, \;\dim(F)=1\big\} \;=\;
\big\{\text{vector (straight) lines of } E\big\} $$

be the projective space induced by $E$.

If $H$ is a hyperplane of $E$, i.e. $\mathcal P(H)\:$ is, by definition, a hyperplane of $\mathcal P(E)$, we want to prove that

$$ \mathcal A := \mathcal P(E)\setminus\mathcal P(H) \tag{1}$$

is an affine space contained in $\mathcal P(E)$, with associated translation space $H$, also denoted $\;\mathcal A(H)$.

Since $H$ is a hyperplane of $E$, we have $\;\text{codim}(H)=1$, hence there exists $\;a\in E, \;a\neq0,$ such that

$$ E=H\oplus \langle a\rangle. \tag{2}$$

For each $v\in E$, we have $\;v=h+\lambda a\;$ for suitable and uniquely determined $\;h\in H\;$ and $\;\lambda\in\Bbb K$; as the vectors $\;h\in H\;$ are obtained for $\;\lambda= 0$, we can write

$$ \mathcal P(E) = \mathcal P(H)\,\dot\cup\,\big\{\big\langle h+\lambda a\big\rangle\;\; :\;\;h\in H, \;\lambda\in\Bbb R, \;\lambda\neq 0\big\}, \tag{3}$$

where $\;\dot\cup\;$ denote the disjoint union.

Now, we observe that the set $\,\mathcal A\,$ after the disjoint union sign in $(3)$ is equal to

$$ \mathcal A = \big\{\big\langle h+a\big\rangle\;\;:\;\;h\in H\big\}. $$

Indeed, being $\;\lambda\neq 0,\;$ we have

$$ \big\langle h+\lambda a\big\rangle = \big\langle \lambda\lambda^{-1}h+\lambda a\big\rangle = \big\langle \lambda\big(\lambda^{-1}h+a\big)\big\rangle = \big\langle\lambda^{-1}h+a\big\rangle, $$

and obviously $\;\lambda^{-1}h\;$ describes $\;H\;$ as $\;h\;$ describes $\;H$.

The points of $\;\mathcal A(H)\;$ are therefore by definition the projective points lying in $\; \mathcal P(E)\setminus \mathcal P(H)\;$, i.e. $(1)$ holds.

Finally, $\;\mathcal A(H)\;$ is an affine space through the structure defined by the function

$$ \phi\; : \; \mathcal A(H) \times \mathcal A(H) \;\;\longrightarrow\;\; H $$

defined by

$$ \phi\big(\big\langle h_1+a\big\rangle, \big\langle h_2+a\big\rangle\big) := h_2-h_1, $$

as it is easy to verify.

As a last observation, it can also be easily verified that the preceding construction of $\;\mathcal A(H)\;$ does not depend on the choice of $\;a\;$ in $(2)$. In fact, if $\,a'\neq0\,$ is another vector satisfying $(2)$, and denoted $\,\mathcal A'(H)\,$ the affine space so obtained, the function

$$ f\;:\; \langle h+a\rangle \longmapsto \langle h+a'\rangle\; $$

from $\;\mathcal A(H)\;$ to $\;\mathcal A'(H)\;$ is an affine isomorphism.


问:
How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?
This is a multiple choice question, where using calculators is not allowed. Candidates have, on average, two minutes to solve the problem.

PROBLEM:

How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?

CHOICES:

(A) $4$ (B) $5$ (C) $6$ (D) $7$ (E) $8$

What I have tried was not feasible:

Using differentiation, I found that the curve is bounded between $y=\frac{9}{8}$ and $y=-2$.

When $x<-8$, $\frac{1}{4}x<-2$ , and when $x>\frac{9}{2}$, $\frac{1}{4}x>\frac{9}{8}$. So, any solution to the equation $y=\sin(x)+\cos(2x)=\frac{1}{4}x$ must lie in the interval $(-8,\frac{9}{2})$.

I do not know how to proceed from these two points. Even I know, that takes a long time.

Also, I do not know if using the fact that the curve is periodic will really help.

By Desmos, I took a look to the graph, the line cuts the curve $7$ times. So the correct choice is $D$.

Any help on how to solve this would be appreciated. THANKS!

答:
This answer does not pretend to be rigorous but just suggests an approach to solve the problem in two minutes (you need a peace of paper and a pencil).

You have already almost done the job. The function is periodic with period $2\pi$ and has on this period two minima $(\frac\pi2,0),(\frac{3\pi}2,-2)$
and two maxima: $(\arcsin(\frac14),\frac98),(\pi-\arcsin(\frac14),\frac98)$.

Now just draw a picture, putting on the plot approximate positons of maxima and minima and connecting them with straight lines. We know that  $\arcsin(\frac14)\approx\frac\pi{12}$, but in fact the value does not matter as the figure tells you!

The only doubt is about the leftmost intersection and it is resolved by observing
$$
-8<-\frac{5\pi}2\iff \pi<\frac{16}5.
$$


问:
Prove that $\displaystyle \lim_{k→\infty} \|f−f∗g_k\|_1=0$
Let $(g_i)$ be  a sequence of functions such that $g_k: \mathbb{R}^n \to [0,\infty)$, $\text{supp}\, (g_k) \subset \overline B(0, 1/k)$ and $\int g_k (x) \, dx =1$ for all $k=1, 2, 3, \ldots$. Let  $f \in L^1(\mathbb{R}^n)$. Show that
$$
\lim_{k \to \infty} \|f - f*g_k\|_1 =0.
$$

What I have tried:

We have the following lemma: Let $1≤p≤∞$, $f∈L_1$, $g\in L_p$. Then $f∗g\in L_p$ and $\|f∗g\|_{L_p}≤\|f\|_{L_1}\|g\|_{L_p}$. Because $f\in L_1$ and $g_k\in L_1\; \forall \,k$ we can calculate:
$$\lim_{k \to \infty} \|f - f*g_k\|_1 =\lim_{k \to \infty} \int_{\mathbb{R}^n}f(x) -\int_{\mathbb{R}^n}f(y)g_k(x-y)\,dy\,dx $$
$$\leq \lim_{k \to \infty} \int_{\mathbb{R}^n}f(x)-\int_{\mathbb{R}^n}f(y)\,dy\int_{\overline B(0, 1/k)}g_k(x-y)\,dy\,dx$$
$$=\int_{\mathbb{R}^n}f(x)\,dx-\int_{\mathbb{R}^n}f(y)\,dy=0$$
Does this make any sense or is it in the right direction? Any help is appreciated.

答:
The norm of $f$ in $L^{1}$ is not $\int f(x)dx $. It is $\int |f(x)|dx$ and absolute value is absolutely essential!.

First consider the case when $f$ is continuous with compact support. $\|f-f*g_k\|_1=\int \int |f(x)-\int f(x-y)g_k(y) dy|dx \leq \int \int |f(x)-f(x-y)|g_k(y) dydx $. The inside integral can be taken over $\|y\| <\frac  1k$. Also, the integral w.r.t. $x$ can be taken from $-T-1$ to $T+1$ if $f$ has upport in $[-T,T]$.  Using uniform continuity of $f$ we see easily that $\|f-f*g_k\|_1 <\epsilon$ provided $k$ is sufficiently large.

For the general case you have to use the fact that $f$ can be approximated in $L^{1}$ norm by continuous functions with compact support. You will also need the inequality $\|f_1*f_2\|_1 \leq \|f_1\|_1 \|f_2\|_1$. This part of the proof is fairly straightforward and I will leave the details to you.


问:
Row of $A = BC$ as a linear combination of the rows of $C$
How to prove that any row of a matrix $A = BC$ can be written as a linear combination of the rows of $C$. I am able to visualize it with an example, but finding it difficult to prove with notations of vector space. </p>

答:
By definition of matrix multiplication, $A_{i,j}=\sum_k B_{i,k}C_{k,j}$.

So a row of $A$ looks like $[\sum_k B_{i,k}C_{k,1},\sum_k B_{i,k}C_{k,2},\dots,\sum_k B_{i,k}C_{k,n}]$.

But that is the same as $$\sum_k[ B_{i,k}C_{k,1}, B_{i,k}C_{k,2},\dots, B_{i,k}C_{k,n}]\\=\sum_k B_{i,k}[ C_{k,1}, C_{k,2},\dots, C_{k,n}]$$

Clearly that row vector is the $k$'th row of $C$, so this last expression is a linear combination of the rows of $C$.


问:
A Bit More Than Positive Definite Matrix.
A positive definite (take as implying symmetric) square matrix $A$ satisfies for all $x$

$$
0< \langle A x ,x \rangle.~~~~~~~~~~~~~~~~~(1)
$$

it turns out that this $\implies$ its eigenvalues are positive. $\textbf{Q1:}$ Does the reverse implication hold?

Consider now a different condition for $A$ which I will call elliptic, $A$ is elliptic if for all $x$ and some $c>0$.

$$
c|x|^2\leq \langle A x ,x \rangle.~~~~~~~~~~~(2)
$$

$\textbf{Q2:}$ how different is $(1)$ to $(2)$ what extra properties do we need of $A$ to get $(2)$ from $(1)$?

答:
Write $A'=\frac{A+A^T}{2}$. Then $A'$ is symmetric and $A-A'$ is skew-symmetric; thus $\langle x,\,Ax\rangle = \langle x,\,A'x\rangle$ for any real vector $x$.

Now, $A'$ is diagonal is some orthonormal basis, which implies that $\{\langle x,\,Ax\rangle,\,\|x\|=1\}$ is exactly the interval between the lowest and the highest eigenvalue of $A'$, hence the conclusion.


问:
Epsilon delta proof sketch of $\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}$
I need to prove that $\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}$.
My first step was re-writing this as: $\frac{1}{1+\sqrt{x}}$. So now for the sketch of the proof i got:

Let $\epsilon>0$. Note that for all $x\in\mathbb{R}^+$ with $|x-1|<\delta$:
$\left|\frac{1}{1+\sqrt{x}}-\frac{1}{2}\right|=\frac{|1-\sqrt{x}|}{2|1+\sqrt{x}|}=\frac{|1-\sqrt{x}|}{2(1+\sqrt{x})}$. (since $|1+\sqrt{x}|\geq 0$ we can omit the absolute value symbols).

Now, observe that: $|x-1|<\delta \Leftrightarrow -\delta<x-1<\delta\Leftrightarrow -\delta+1<x<\delta+1\Leftrightarrow\sqrt{-\delta+1}-1<\sqrt{x}-1<\sqrt{\delta+1}-1$. And thus $|\sqrt{x}-1|<\sqrt{\delta+1}-1$. (Note that has to be $\delta<1$ or the root would be negative)

Now observe that:
$\frac{|1-\sqrt{x}|}{2(1+\sqrt{x})}\leq \frac{1}{2}|1-\sqrt{x}|<\frac{1}{2}(\sqrt{\delta+1}-1)<\sqrt{\delta+1}-1<\epsilon.$

So, we get that $\delta+1<(\epsilon+1)^2\Leftrightarrow\delta<\epsilon^2+2\epsilon$.

For me,this is a bit confusing. Normally i would get a more elegant delta definition, so i don't know if what i did was true. Another problem for me was that normally i could use the $|x-1|<\delta$ directly, and maybe got another restriction. But this time i only used that $|\sqrt{x}-1|<\sqrt{\delta+1}-1$. Can somebody clarify, if what i did was right / a way to do this more elegantly?

答:
As $x=1$ is not considered, you are allowed to simplify the expression as

$$\frac1{1+\sqrt x}$$ when clearly tends to $\dfrac12$.

Now solving

$$\frac1{1+\sqrt x}-\frac12=\pm \epsilon$$ (the function is monotonic decreasing) we have

$$x=\left(1-\frac2{1\pm2\epsilon}\right)^2$$

and the proof works with

$$\delta\le\min\left(1-\left(1-\frac2{1+2\epsilon}\right)^2, \left(1-\frac2{1-2\epsilon}\right)^2-1\right)$$ which is the tightest bound.


答:
Caution: $A$ is positive definite , iff

$$(1) \quad 0< \langle A x ,x \rangle$$

for all $x \ne 0.$

$Q2:$

$(1)$ and $(2)$ are equivalent:

It is clear that $(2)$ implies $(1)$.

Now supoose that $(1)$ holds. The function $f(x):=\langle A x ,x \rangle$ is continuous and $K:=\{x \in \mathbb R^n: ||x||=1\}$ is compact. Hence there is $c>0$ such that $f(x) \ge c$ for all $x \in K.$ This gives $(2)$

Title: A Bit More Than Positive Definite Matrix.

Body:
A positive definite (take as implying symmetric) square matrix $A$ satisfies for all $x$

$$
0< \langle A x ,x \rangle.~~~~~~~~~~~~~~~~~(1)
$$

it turns out that this $\implies$ its eigenvalues are positive. $\textbf{Q1:}$ Does the reverse implication hold?

Consider now a different condition for $A$ which I will call elliptic, $A$ is elliptic if for all $x$ and some $c>0$.

$$
c|x|^2\leq \langle A x ,x \rangle.~~~~~~~~~~~(2)
$$

$\textbf{Q2:}$ how different is $(1)$ to $(2)$ what extra properties do we need of $A$ to get $(2)$ from $(1)$?


问:
Spherical surface in cartesian coordinates
In the context of optics the so called biconic surface
is used to describe optical elements. The biconic surface has several parameters, however, I am only interesed in the case where it reduces to the form
$$
z = \frac{c(x^2 + y^2)}{1 + \sqrt{1 - c^2 (x^2 + y^2)}}
$$
It is said that this equation defines a spherical surface with radius $R=1/c$. Unfortunately, I do not see how this is derived. How do I get from
$$
x^2 + y^2 + z^2 = R^2
$$
to the above relation?

答:
Yes, it is a spherical surface with radius $R=\frac{1}{c}$.

Write the first equation as
$$z=\frac{c \left(x^2+y^2\right)}{\sqrt{1-c^2 \left(x^2+y^2\right)}+1}$$
$$\sqrt{1-c^2 \left(x^2+y^2\right)}+1=\frac{c \left(x^2+y^2\right)}{z}$$
$$\sqrt{1-c^2 \left(x^2+y^2\right)}=\frac{c \left(x^2+y^2\right)}{z}-1$$
square both sides
$$1-c^2 x^2-c^2 y^2=1+\frac{c^2 x^4+2 c^2 x^2 y^2+c^2 y^4-2 c x^2 z-2 c y^2 z}{z^2}$$
simplify, multiply by $z^2$ and move everything in one side
$$c^2 x^4+2 c^2 x^2 y^2+c^2 x^2 z^2+c^2 y^4+c^2 y^2 z^2-2 c x^2 z-2 c y^2 z=0$$
Simplify by $c$
$$c x^4+2 c x^2 y^2+c x^2 z^2+c y^4+c y^2 z^2-2 x^2 z-2 y^2 z=0$$
collect $x^2$ and $y^2$
$$x^2 \left(c x^2+c y^2+c z^2-2 z\right)+y^2 \left(c x^2+c y^2+c z^2-2 z\right)=0$$
and finally
$$\left(x^2+y^2\right) \left(c x^2+c y^2+c z^2-2 z\right)=0$$
as $x^2+y^2=0$ has no non trivial solutions we get
$$c x^2+c y^2+c z^2-2 z=0$$
divide by $c$
$$x^2+y^2+z^2-\frac{2 }{c}z=0$$
Which can be written as
$$x^2+y^2+\left(z-\frac{1}{c}\right)^2=\left(\frac{1}{c}\right)^2$$
which is the equation of a sphere with canter $\left(0,0,\frac{1}{c}\right)$ and radius $R=\frac{1}{c}$.

Hope this is useful


问:
Primitive of $z$ conjugated
So I tried to understand complex analysis better, but I stumbled on this problem. I'm certain that a primitive (anti-derivative) of the function $z\mapsto \bar{z}$ doesn't exist, since the integral on the unit circle doesn't vanish (I hope that this is correct, anyway). However, someone gave me this "proof":

$$\int \bar{z} ~dz=\int \overline{Re(z)+i\cdot Im(z)}~dz=\int Re(z)-i\cdot Im(z)~dz=\int Re(z)~dz-i\int Im(z)~dz$$ $$=Re\int z~dz-i\cdot Im\int z~dz=\overline{Re\int z~dz+i\cdot Im\int z~dz}=\overline{\int z~dz}.$$

Since $z$ has primitive $z^2/2+C$ (for some constant $C$), we would get a primitive for $\overline{z}$, which doesn't make sense. Where does this go wrong? Thanks in advance!

答:
This is wrong:
$$
 \int_\gamma \operatorname{Re}(z) \, dz = \operatorname{Re} \left( \int_\gamma z \, dz\right)
$$
and also the same with the imaginary parts. Generally, for a curve $\gamma:[a, b] \to D \subset \Bbb C$ and a continuous function $f: D \to \Bbb C$,
$$
 \operatorname{Re} \left( \int_\gamma f(z) \, dz\right) = \operatorname{Re} \left( \int_a^b f(\gamma(z))\cdot \gamma'(t) \, dt  \right)
= \int_a^b \operatorname{Re}\left(f(\gamma(z)) \cdot \gamma'(t)\right) \, dt
$$
and
$$
\int_\gamma \operatorname{Re} (f(z)) \, dz = \int_a^b \operatorname{Re}\bigl(f(\gamma(z))\bigr) \cdot \gamma'(t) \, dt
$$
and these expressions are not equal in general.

In our case, since $\bar z = 1/z$ on the unit circle,
$$
\int_\gamma \bar{z} \, dz = \int_\gamma \frac 1z \, dz = 2 \pi i \ne 0.
$$

Title: Primitive of $z$ conjugated


问:
Open subspaces of locally compact Hausdorff spaces are locally compact
Let $X$ be locally compact and Hausdorff. I want to prove the following:

If $Y\subset X$ is open then $Y$ is locally compact.

I have proved that closed subsets of $X$ are locally compact, but how can one prove this?

I also want to use these two lemmas to conclude the following:

If $Y\subset X$ is locally closed then $Y$ is locally compact.

答:
The previous proofs were too complicated and rather confusticate the problem.

Instead you can simply follow your nose:

We follow the definition of a locally compact space as it should be:*

Every point has a neighborhood basis of compact subsets.

Having settled our grounds, let's get straight to the proof:

Consider a locally compact space and an open subspace $X\subseteq Y$.

Let $x\in U\subseteq X$ be a neighborhood of $x\in X$.

Assume wlog that $U$ is open in $X$ (e.g. $U=X\cap V$).

So also $U$ open in $Y$ as by assumption $X\subseteq Y$ open!

By assumption $Y$ is locally compact.

So we may sandwich this in $x\in L\subseteq U$ with $L$ a compact nbhd of $x\in Y$.

Since this took place in $L\subseteq X$, the restricted neighborhood remains $L\cap X=L$.

This defines the desired compact nbhd.

*Two small remarks for common agreement:

Generally these are not open and possibly not closed.

Hausdorffness does not play any role in our proof. It only serves to characterise (for the original space $Y$) that the existence of single compact neighborhood may be sufficient for local compactness.


问:
Submersion map from $S^2$ to $S^2$ is a diffeomorphism
If $f:S^2\mapsto S^2$ be a submersion. Show that $f$ is a diffeomorphism.

I got that $f$ is surjective using $S^2$ is connected and compact.(Theorem :  If $X$ is compact and $Y$ is connected, then every submersion $f:X→Y$ is surjective)

答:
To say that $f:S^2\to S^2$ is a submersion means that at any $s\in S$ the tangent map $T_sf:T_sS^2\to T_sS^2$ is surjective, hence bijective since source and target of that linear map have same dimension (namely 2).

We deduce that $f$ is a local diffeomorphism.

But a local diffeomorphism (or even local homeomorphism) with compact source is a covering space, so that $f$ is actually a differentiable covering map.

And now for the coup de grâce: since $S^2$ is simply connected a connected covering map must be one sheeted, so that $f$ is a bijective local diffeomorphism.

Its inverse $f^{-1}$ is also differentiable (by some avatar of the implicit function theorem), and finally we can proudly assert:

Yes, $f$ is a diffeomorphism.


问:
fiber factors of affine schemes
Let $X,Y \neq \emptyset$ be sufficiently nice schemes over, say, a field $k$. Assume that $X \times_k Y$ is affine. Does it follow that $X$ and $Y$ are affine?

Perhaps this works with Serre's criterion.

答: I think running your argument in reverse works. Suppose, say, $X$ was not affine. Then there is a coherent sheaf $\mathcal{F}$ on $X$ with $H^1(X, \mathcal{F}) \neq 0$. Choose any coherent sheaf $\mathcal{G}$ on $Y$ with a nonzero global section (e.g. the structure sheaf). Then the Kunneth formula shows that $H^1(X \times_k Y, p_1^*(\mathcal{F}) \otimes p_2^*(\mathcal{G})) \neq 0$, so $X\times_k Y$  cannot be affine.</p>

Title: fiber factors of affine schemes


问:
calculating the determinant of an $n \times n$ integer matrix
I want to write a polynomial algorithm for calculating the determinant of an $n \times n$ integer matrix. There are various codes in different programming languages on the web but unfortunately I am not good at understanding codes which are written by other people. Normally, what steps will it take to calculate the determinant of an $n \times n$ integer matrix, then I can write the algorithm and compute its time complexity on my own. </p>

答:
One method is to use Gaussian elimination to reduce the matrix to upper triangular form.  The determinant is then the product of the diagonal matrix elements.  The problem with this method when matrix elements are integers is that Gaussian elimination will produce fractions in general.  For large matrices, the denominators of these fractions can blow up rather fast.

One way around this is to work modulo a prime, $p$.  This is equivalent to working over the finite field $\mathbf{F}_p.$  Since division can be performed in a finite field, you don't have to worry about the fraction problem.  Of course the answer you obtain will only be correct modulo $p.$  If you happen to know that your determinant is between $0$ and $p,$ then you're done at this point.  Hadamard's determinant bound can be used to place an upper bound on the absolute value of the answer.  If this upper bound is $h,$ then as long as $p>2h,$ you will be able to recover the answer.  (You need $2h$ because the answer might be negative.)

Hadamard's bound also gets big very quickly, and most of the time is far bigger than the actual determinant, which means that you may have to use a $p$ that is impractically large.  An alternative is to perform the calculation modulo several primes $p_1,\ldots,p_k$ whose product is larger than $2h.$  The Chinese remainder theorem can then be used to reconstruct the answer.

A different method is to use Dodgson condensation.  All of these suggestions are taken form Henri Cohen's book, A Course in Computational Algebraic Number Theory, which I highly recommend.  If this isn't enough detail to get you started, let me know, and I'll try to expand on this answer.


问:
Showing $\cos^3x-3\cos x \sin^2 x = \cos 3 x$
Why are these equal?
$$\cos^3x-3\cos x \sin^2 x = \cos 3 x$$
Which additional theorems apply?

答:
Hint: use the compound angle formula shown below to derive the 'triple angle formula' for $\cos(3x)$:
$$
\cos(\alpha + \beta)=\cos\alpha \cos \beta - \sin\alpha\sin\beta \, .
$$
Then, rewrite the RHS using the triple angle formula, and simplify your answer.


问:
If $X$ and $Z$ are independent: $\mathbb{E}(Y|X) = \mathbb{E}(Y) \implies \mathbb{E}(Y|X,Z) = \mathbb{E}(Y|Z)$?
My question is, given that $X$ and $Z$ are independent is it true that

$$
\mathbb{E}(Y|X) = \mathbb{E}(Y) \implies \mathbb{E}(Y|X,Z) = \mathbb{E}(Y|Z)
$$

I do not seem to be able to prove it but I cannot come up with a counterexample or disprove it.

Does the answer change if on top of $X$ and $Z$ being independent we also impose that $\mathbb{E}(Y|Z) \neq \mathbb{E}(Y)$?

答:
No it is not true

A classic example is where the following triples are equally probable

```<code>X  Y  Z
0  0  0
0  1  1
1  0  1
1  1  0
</code>```

so $\mathbb{E}(Y|X) = \mathbb{E}(Y) = \mathbb{E}(Y|Z)=\frac12$ but $\mathbb{E}(Y|X,Z) \equiv X+Z \mod 2$ so is $0$ or $1$

Body:
No it is not true

A classic example is where the following triples are equally probable

X  Y  Z
0  0  0
0  1  1
1  0  1
1  1  0

so $\mathbb{E}(Y|X) = \mathbb{E}(Y) = \mathbb{E}(Y|Z)=\frac12$ but $\mathbb{E}(Y|X,Z) \equiv X+Z \mod 2$ so is $0$ or $1$


Title: Showing $\cos^3x-3\cos x \sin^2 x = \cos 3 x$

Body:
Why are these equal?
$$\cos^3x-3\cos x \sin^2 x = \cos 3 x$$
Which additional theorems apply?

Body:
$$\cos(3x)=\mathrm{Re}(e^{3ix})=\mathrm{Re}\left[(\cos(x)+i\sin(x))^3\right]$$

so $$\cos(3x)=\mathrm{Re}\left[\cos^3(x)+3i\cos^2(x)\sin(x)+3i^2\cos(x)\sin^2(x) + i^3\sin^3(x)\right]$$

so

$$\cos(3x)=\cos^3(x)-3\cos(x)\sin^2(x)$$


答:
Your method is fine; the bounds you on $\delta$ are "loose" in the sense that they do not correspond to the largest symmetric interval about $x = 1$ that guarantees $|1/(1+\sqrt{x}) - 1/2| < \epsilon$, but they are still valid for showing that the limit exists and equals $1/2$.  In other words, your bound $\delta < \epsilon^2 + 2\epsilon$ approaches $0$ faster than the more complicated bound $$\delta < \begin{cases} \frac{8\epsilon}{(1+2\epsilon)^2}, & 0 < \epsilon < 1/2, \\ 1, & \epsilon \ge 1/2, \end{cases} $$  (note that the quantity $8\epsilon/(1-2\epsilon)^2$ is always strictly greater for $\epsilon > 0$).

That said, some of the reasoning in your proof can be simplified as suggested in the comments. But there's nothing mathematically wrong with your approach.

Title: Epsilon delta proof sketch of $\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}$

Body:
I need to prove that $\lim_{x\to1}\frac{1-\sqrt{x}}{1-x}=\frac{1}{2}$.
My first step was re-writing this as: $\frac{1}{1+\sqrt{x}}$. So now for the sketch of the proof i got:

Let $\epsilon>0$. Note that for all $x\in\mathbb{R}^+$ with $|x-1|<\delta$:
$\left|\frac{1}{1+\sqrt{x}}-\frac{1}{2}\right|=\frac{|1-\sqrt{x}|}{2|1+\sqrt{x}|}=\frac{|1-\sqrt{x}|}{2(1+\sqrt{x})}$. (since $|1+\sqrt{x}|\geq 0$ we can omit the absolute value symbols).

Now, observe that: $|x-1|<\delta \Leftrightarrow -\delta<x-1<\delta\Leftrightarrow -\delta+1<x<\delta+1\Leftrightarrow\sqrt{-\delta+1}-1<\sqrt{x}-1<\sqrt{\delta+1}-1$. And thus $|\sqrt{x}-1|<\sqrt{\delta+1}-1$. (Note that has to be $\delta<1$ or the root would be negative)

Now observe that:
$\frac{|1-\sqrt{x}|}{2(1+\sqrt{x})}\leq \frac{1}{2}|1-\sqrt{x}|<\frac{1}{2}(\sqrt{\delta+1}-1)<\sqrt{\delta+1}-1<\epsilon.$

So, we get that $\delta+1<(\epsilon+1)^2\Leftrightarrow\delta<\epsilon^2+2\epsilon$.

For me,this is a bit confusing. Normally i would get a more elegant delta definition, so i don't know if what i did was true. Another problem for me was that normally i could use the $|x-1|<\delta$ directly, and maybe got another restriction. But this time i only used that $|\sqrt{x}-1|<\sqrt{\delta+1}-1$. Can somebody clarify, if what i did was right / a way to do this more elegantly?


问:
Easy to read partial differential equations book?
I'm looking for an easy to read undergraduate book on partial differential equations, ideally something that is not much harder than a multivariable calculus/ordinary differential equations book.

I am preparing for a course which is using the text by Walter Strauss, but I found this text a bit difficult to read. Specifically, I found that many of the derivations were missing steps that were not obvious to me or provided little justification for the manipulations. When I searched for introductory books however the Strauss book seems to be recommended.

Google seems to recommend, among others:

Partial Differential Equations for Scientists and Engineers by Farlow

Introduction to Partial Differential Equations by Peter Olver

I have ruled out:

Partial Differential Equations: An Introduction by Walter Strauss

An Introduction to Partial Differential Equations by Michael Renardy

Partial Differential Equations by Fritz John

Partial Differential Equations by Lawrence C Evans

My background is having read A First Course in Differential Equations with Modelling Applications by Dennis Zill. Would I be better off reading the extended version of this book (Differential Equations with Boundary Value Problems)? I was a little bit hesitant because I was not sure how relevant the book is to PDEs. The chapters I have not read are Fourier Series, Boundary Value Problems in Rectangular Coordinates, Boundary Value Problems in Other Coordinate Systems, Integral Transforms and Numerical Solutions of Partial Differential Equations.

EDIT: I ended up finishing about half the book, which is everything covered in my course, Chapters 1 through 7.

答:
I agree with the advice to continue with the Zill book, and to fill in steps in math books such as Strauss, but if you want something really easy, I have some lecture notes on differential equations at http://bterrell.net
that introduce various topics in ODE and PDE by focusing on some applications.</p>


问:
Spectral radius of a block circulant matrix
Let $A$ the block matrix given by the blocks:
$$\tilde{A}=\begin{pmatrix}
0&-\mu&0&...&0&-\mu\\
-\mu&0&-\mu&...&0&0\\
0&-\mu&0&...&0&0\\
...&...&...&...&...\\
0&0&0&...&0&-\mu\\
-\mu&0&0&...&-\mu&
\end{pmatrix}\quad\quad A^*=\mathbf{diag}(-\mu,-\mu,...,-\mu) $$
So $$
A=\begin{pmatrix}
\tilde{A}&A^*&0&...&0&A^*\\
A^*&\tilde{A}&A^*&0&...&0\\
0&A^*&\tilde{A}&...&...&0\\
...&...&...&...&...&...\\
A^*&0&...&0&A^*&\tilde{A}
\end{pmatrix} $$
Does anyone know some trick to calculate spectral radius of $A$? Maybe I can use the matrix norm defined by $||A||_1 = \max_{1\le j \le n} \sum_{i=1}^n |a_{ij}|$

答:
By Gerschgorin disc theorem, $\rho(A)\le4|\mu|$. Since $Ae=-4\mu e$ (where $e$ is the vector of ones), we conclude that $\rho(A)=4|\mu|$.

Actually we can obtain all eigenvalues of $A$ explicitly. Let $S_r$ be the real symmetric $r\times r$ matrix
$$
\pmatrix{0&1&0&\cdots&0&1\\
1&0&1&\ddots&\vdots&0\\
0&1&\ddots&\ddots&0&\vdots\\
\vdots&\ddots&\ddots&\ddots&\ddots&0\\
0&\cdots&0&\ddots&0&1\\
1&0&\cdots&0&1&0}.
$$
This matrix is circulant and its eigenvalues admit an explicit formula:
$$
\lambda_k^{(r)}=e^{2\pi ik/r}+e^{2\pi ik(r-1)/r}=2\cos\left(\frac{2k\pi}{r}\right),\ k=0,1,\ldots,r-1.
$$
Now suppose that $\tilde{A}$ (or $A^\ast$) has $m$ rows and $A$ has $n$ block rows.
Then $A=-\mu(I_n\otimes S_m+S_n\otimes I_m)$. Therefore the eigenvalues of $A$ are  given by
$$
-\mu(\lambda_k^{(m)}+\lambda_l^{(n)})
=-2\mu\left(\cos\left(\frac{2k\pi}{m}\right)+\cos\left(\frac{2l\pi}{n}\right)\right)
$$
where $k\in\{0,1,\ldots,m-1\}$ and $l\in\{0,1,\ldots,n-1\}$.


问:
Function $f: \Bbb{R} \to \Bbb{R}$ that's not continuous using the definition that the preimage is not open.

Let $f: \Bbb{R} \to \Bbb{R}$, $$f(x)=\begin{cases} 0, &x \leqslant 0, \\ 1, &x>0.\end{cases}$$ $f$ is not continuous, but by definition there exists an open set $V \subset \Bbb{R}$ whose preimage $f^{-1}(V)$ is not open in $\Bbb{R}$. Give an example of such an open set $V$.

I'm not sure  I understand the definition correctly I've tried plotting two real lines and finding a closed set in the domain such that the image would be open, but cannot seem to find it. I feel like $f^{-1}(V)=[-\frac12, \frac12]$ would have worked, but I'm not sure what will the image of this be since $f$ was defined piecewise. Any hints would be appreciated.

答:
$f$ takes only two values. For $V=(-1/2,1/2)$ which is open in the codomain, $f^{-1}(V)=(-\infty,0]$ which is not open in the domain.

Note that for judging continuity of $f$, we need to look at preimages of open sets in the codomain. You may build a non-open set $S$ in the domain whose image $f(S)$ in the codomain is open, but the inverse image of $f(S)$ is not necessarily equal to $S:~S\subseteq f^{-1}(f(S))$. This is what precisely happens when you take $S=[-1/2,1/2],f(S)=\{0,1\}$ but $f^{-1}(f(S))=\Bbb R$ which is open in the domain.


问:
How can I show whether the series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $ converges or diverges?
While revising for exams, I came across a question where at the end of it, we had to determine whether the below series was convergent or divergent:

$$\sum_{n=1}^\infty \frac{(-1)^n}{n(2+(-1)^n)} $$

Unfortunately, other than knowing that as the series $ \sum_{n=1}^\infty \frac{(-1)^n}{n} $ converges, this series will most likely converge as the $(2 + (-1)^n)$ terms are bounded, I can't see a way of using this to determine if it converges or not.

Here's a list of some of the other ideas I've tried, which don't seem to get me anywhere:

Generalising the series - By this, I mean replacing the $-1$ for $z$ and then try and use the ratio test to see whether $-1$ is inside or outside or on the boundary of the circle of convergence - however, when I did so, I believe that the ratio test is inconclusive for all values of $|z|$, which while disappointing, is an interesting feature of the series.

Summation by parts - While this is usually a nice tool to use for when dealing with awkward sums, I can't see any good choice of sequences $ (a_n) $ and $(b_n)$ to choose to use.

Summing consecutive terms - By this, I mean evaluating the series by looking at the sum of

$$\sum_{k=1}^\infty \frac{(-1)^{2k}}{2k(2+(-1)^{2k})} + \frac{(-1)^{2k-1}}{(2k-1)(2+(-1)^{2k-1})}$$

$$ \Rightarrow \sum_{k=1}^\infty \frac{1}{2k} - \frac{1}{3(2k-1)}$$

$\quad$ but then as we are left with parts of the harmonic series this doesn't seem like a nice way to  $\quad$evaluate it.

Other than that, I'm completely out of ideas, so any new ones (or ways to make my old ones work) would be much appreciated!

Body:
Hint:
$$\sum_{n=2k-1}^{2k} \frac{(-1)^n}{n(2+(-1)^n)}=-\frac{1}{2k-1}+\frac{1}{6k}=-\frac{4k+1}{6k(2k-1)}.$$


答: Hint: Try and calculate $F^{-1}(V)$ where $V$ is any open set that includes $0$, but does not include $1$.

Title: Function $f: \Bbb{R} \to \Bbb{R}$ that's not continuous using the definition that the preimage is not open.

Body:
Let $f: \Bbb{R} \to \Bbb{R}$, $$f(x)=\begin{cases} 0, &x \leqslant 0, \\ 1, &x>0.\end{cases}$$ $f$ is not continuous, but by definition there exists an open set $V \subset \Bbb{R}$ whose preimage $f^{-1}(V)$ is not open in $\Bbb{R}$. Give an example of such an open set $V$.

I'm not sure  I understand the definition correctly I've tried plotting two real lines and finding a closed set in the domain such that the image would be open, but cannot seem to find it. I feel like $f^{-1}(V)=[-\frac12, \frac12]$ would have worked, but I'm not sure what will the image of this be since $f$ was defined piecewise. Any hints would be appreciated.


问:
Can the tangent line be defined independently of the derivative?
The graph of the function $f:x \mapsto x^{1/3}$ has a 'vertical tangent' at $x=0$:

Although this idea is certainly geometrically sound, from what I understand the tangent line is defined by the derivative, not vice versa. In other words, the tangent line to a function at the point $(a,f(a))$ is simply the line given by the equation
$$
y - f(a) = f'(a)(x-a) \, ,
$$
where $f'(a)$ is of course defined as a limit. Since $f'(0)$ does not exist in this case, I'm unsure if we can truly say that the graph has a vertical tangent. The intuitive idea of a tangent 'just touching' the curve breaks down when we consider, for instance, the graph of a linear function, where the tangent touches the graph of the function itself at infinitely many points. Nevertheless, I have heard people say that the tangent line is fundamentally a geometric concept. Although the slope of the tangent line 'agrees' with the derivative if the derivative exists, there are instances where the tangent line is a meaningful concept even when the derivative doesn't exist. If this be the case, then what is the formal definition of a tangent?

答:
Your geometric intuition is sound. The problem about taking the derivative is an artifact of the coordinate system used to define it. This is solved in differential geometry where a space is covered by an atlas of compatible coordinate charts. This way, we can choose the right coordinate chart for the problem at hand. For example, to find the 'vertical tangent', we swap the coordinate axes, and find that this tangent is now 'horizontal'. Of course, the terms 'horizontal' and 'vertical' relate to our choice of chart. In more invariant geometric language, what we want to understand is how the curve actually curves in the plane. And what we see is that the location of your 'vertical tangent' is better described as the location of an inflection point

In differential geometry, you will see that both the geometric and derivational notion of a tangent dovetail exactly. In fact, this occurs so naturally, that in some treatments of differential geometry, it's the derivations - aka the derivative - that takes the starring role - as the development is simpler.

(In more sophisticated treatments of differential geometry for singular spaces or infinite-dimensional spaces it's seen that they diverge again, though of course there remains an intimate relationship. For example, in Michors & Kriegl conveniant calculus for infinite dimensional spaces it is the geometric tangent along curves that takes the starring role and the derivational is dispensed with. The same goes for Neeb's development of calculus in locally convex spaces (again it's the infinite-dimensional context that is important). In contrast, in Souriau's diffeology and developed by Isglesias-Zemmour (important for both infinite-dimensional and singular spaces and vastly more general than the foregoing), a notion of smoothness is introduced that is inspired from topology. Here, Christensen after correcting a definition by Hector and Laubinger, shows that there are distinct geometric and derivational tangent vectors. They are not the same here, but of course they are in many natural cases).


问:
Range of: $\sqrt{\sin(\cos x)}+\sqrt{\cos(\sin x)}$
Range of:
$$\sqrt{\sin(\cos x)}+\sqrt{\cos(\sin x)}$$

Any help will be appreciated.

答:
A related problem. Note that, we can restrict our work on the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$, since the function is periodic. Now, the function achieves its max at the point $x=0$ and its min at $x=-\frac{\pi}{2},\frac{\pi}{2}$. To see this, let's
write the function as
$$ y(x) = \sqrt{\sin(\cos x)}+\sqrt{\cos(\sin x)}=g(x)+h(x).$$

Note that, $g(x),h(x)> 0$. So, we can just maximize each function on its own. That's implies
$ g(x) $ achieves its max at $x=0$ and $h(x)$ attains its max at the same point $x=0$. You can use the derivative test to see this. So the max of the function $y(x)$ is
:
$$ y(0) = \sqrt{\sin(\cos 0)}+\sqrt{\cos(\sin 0)}= 1+\sqrt{\sin 1}. $$

The min can be calculated as:

$$y\left(\frac{\pi}{2}\right)= y\left(-\frac{\pi}{2}\right) = \sqrt{\cos(1)}$$

So the range is:

$$ \sqrt{\cos(1)} \leq y\leq  1+\sqrt{\sin 1}.$$


问:
Derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ without qoutient rule
I'm trying to find derivative of  $\frac{\cos t-\sin t}{\cos t+\sin t}$ in a different way: there is a trick to find derivative of the form $\frac{ax+b}{cx+d}$:
$$\left(\frac{ax+b}{cx+d}\right)'=\frac{ad-bc}{(cx+d)^2}$$
I'm trying to use it for taking derivative of $\frac{\cos t-\sin t}{\cos t+\sin t}$ too. the result of  is $\frac{-2}{(\cos t+\sin t)^2}$. but I don't know how to use this method for the function

答:
Inspired by J.G.'s answer we can rewrite $\frac{\cos t-\sin t}{\cos t+ \sin t}$ as $\frac{1-\tan t}{1+\tan t}$. Let's call it  $f(t)$:

$$f(t)=\frac{-\tan t+1}{\tan t+1}$$
Then by chain rule and by using the trick I mentioned in my question we have :
$$\frac{df}{dt}=\frac{df}{d(\tan t)}\times\frac{d(\tan t)}{dt}=\frac{-2}{(1+\tan t)^2}\times\sec^2 t=\frac{-2}{(\cos t+\sin t)^2}$$


问:
A criterion of total boundness of a uniform space
Is the following true:

Conjecture A uniform space $(U;F)$ is totally bounded iff for every entourage $E$ of this space there exists a finite set $B\subseteq U$ and a natural $n$ such that $E^n[B] = U$.

If not, could you provide a counter-example?

答:
We are dealing with topological groups. I heard the following. Banaszczyk called a topological group $G$ to be weakly precompact (or weakly bounded), if for each neighborhood $U$ of the unit of the group $G$ there exist a finite subset $F$ of $G$ and a number $n$ such that $U^nF=G$. And the group of all monotonically increasing homeomorphisms of the unit segment (probably, endowed with the pointwise topology) is weakly bounded, but not totally bounded.

PS. We are writing a paper on the relations of different types of boundness in topological groups.  If you are interested in it, then I can post here a link to the paper after we shall write it.


问:
$x^4-mx^3-2mx^2+2m^2x=0$ roots in arithmetic progression
For which real values of $m$ roots of equation
$$x^4-mx^3-2mx^2+2m^2x=0$$
are in arithmetic progression?

I managed to find a solution using a lot of casework, i.e. factoring the equation into
$$x(x-m)(x^2-2m)=0$$
and then finding $m$ for which it has one root, three distinct roots and four distinct roots but I think there may be an easier solution and I'm just overcomplicating it. In addition, I'm not quite sure if I should include the one root case.

Could someone guide me to a better solution? Thanks.

答:
WLOG the roots are $a-3d, a-d,a+d, a+3d$

As $(a-3d)(a-d)(a+d)(a+3d)=0,$

Case$\#1:$

If $a-3d=0,$ the rest of the roots become $2d,4d,6d$

If $2d=m,4d\cdot6d=2m\implies\dfrac m{12}=\left(\dfrac m2\right)^2\iff m(3m-1)=0$

If $4d=m,2d\cdot6d=2m\implies\left(\dfrac m4\right)^2=\dfrac m6\implies m=?$

If $6d=m,2d\cdot8d=2m\implies\left(\dfrac m6\right)^2=\dfrac m8\implies m=?$

Case$\#2:$   $a+3d=0$ Replace $d$ with $-d$ in Case$\#1$

Case$\#3:$   If $a-d=0, a=d$   the rest of the roots become $-2d,2d,4d$

$$x^4-mx^3-2mx^2+2mx=x(x-4d)(x-2d)(x+2d)$$

$$\iff x^3-mx^2-2mx+2m=(x-4d)(x^2-4d^2)=x^3-4dx^2-4d^2x+16d^3$$

$$\implies m=4d\ \ \ \ (1),2m=16d^3\ \ \ \ (2),-2m=-4d^2\ \ \ \ (3)$$

If $d\ne0,$

$(2)/(3)\implies-1=\dfrac{16d^3}{-4d^2}=-4d, d=?\ \ \ \ (4)$

$(2)/(1)\implies2=\dfrac{16d^3}{4d}=4d^2\implies d^2=?$ but this does not satisfy $(4)$

Case$\#4:$   $a+d=0$ Replace $d$ with $-d$ in Case$\#3$


问:
Prove that the difference of an integer and its reverse order is divisible by 9
Define $u$ to be an integer with a decimal representation $\overline{a_{n}a_{n - 1}a_{n - 2}\cdots a_{2}a_{1}a_{0}}$. Then, define $u'$ to be the integer whose decimal representation is the reverse order of the decimal representation $\overline{a_{0}a_{1}a_{2}\cdots a_{n-2}a_{n-1}a_{0}}$.

Writing $u$ and $u'$ in expanded form,
\begin{align*}
u &\;=\; (10^{n})(a_{n}) + (10^{n - 1})(a_{n - 1}) + (10^{n - 2})(a_{n - 2}) + \cdots + (10^{2})(a_{2}) + (10^{1})(a_{1}) + (10^{0})(a_{0}) \\ u' &\;=\; (10^{0})(a_{n}) + (10^{1})(a_{n - 1}) + (10^{2})(a_{n - 2}) + \cdots + (10^{n - 2})(a_{2}) + (10^{n - 1})(a_{1}) + (10^{n})(a_{0})
\end{align*}

By evaluating $u - u'$, we get $\displaystyle\sum_{k = 0}^{n}(10^{n - k} - 10^{k})(a_{k})$. For $u - u'$ to be divisible by 9, $(10^{n - k} - 10^{k})$ must be divisible by 9. Solving for the remainder of $(10^{n - k} - 10^{k})$ when divided by 9,

\begin{align*}
r &\;\equiv\; (10^{n - k} - 10^{k}) \bmod 9 \\ r &\;\equiv\; (10^{n - k} \bmod 9) - (10^{k} \bmod 9) \\ r &\;\equiv\; \left((10 \bmod 9)^{n - k} \bmod 9\right) - \left((10 \bmod 9)^{k} \bmod 9\right) \\ r &\;\equiv\; \left(1^{n - k} \bmod 9\right) - \left(1^{k} \bmod 9\right) \\ r &\;\equiv\;1 - 1 \\ r &\;\equiv\; 0
\end{align*}

Since the remainder is zero, this implies that the difference of an integer and its reverse order is divisible by 9.

Is my proof correct, by any chance? I got curious about this when I used the calculator out of boredom.

I think the part where I solved for the remainder seems weird, but is there a chance that it is correct, though notationally wrong? If it is wrong, what should I do?

答:
Nice observation. You proof is correct but is perhaps longer than necessary.

Here is a more general fact:

Take an integer $u$ and write its decimal representation. Reorder the digits at will, to get another integer $u'$. Then $u-u'$ is divisible by $9$.

Indeed, if $u=(a_{n}a_{n - 1}a_{n - 2}\cdots a_{2}a_{1}a_{0})_{10}$, then $u \equiv a_{n}+a_{n - 1}+a_{n - 2}+\cdots a_{2}+a_{1}+a_{0} \bmod 9$, because $10^k \equiv 1 \bmod 9$. The same holds for $u'$, hence the result.


问:
Proof limit value with $\epsilon/ \delta$
I have to find the limit value for $f(x)=\sqrt{1+x}$ for $x \rightarrow0$. And then show with $\epsilon /\delta$ that I have found the right limit value.
I have found the limit value to $\sqrt{1+x} \rightarrow1$ for $x \rightarrow0$, by:
$$\lim_{x \rightarrow 0} \sqrt{1+x}
=\sqrt{\lim_{x \rightarrow 0} (1+x)}=\sqrt{\lim_{x \rightarrow 0} (1)+\lim_{x \rightarrow 0}(x)}=\sqrt{1+\lim_{x \rightarrow 0}(x)}=1$$

But I got a problem when I have to show it by $\epsilon /\delta$. When I write:
$$|f(x)-1|\Rightarrow|\sqrt{1+x}-1|$$
I can't see what I can use as a bound easier to find a $\delta$. Can anyone help me?

答:
Hint.

What you need is essentially the continuity of the function $y\mapsto\sqrt{y}$ at $y=1$.

So you need an estimate
$$
|\sqrt{y}-\sqrt{1}|<\epsilon
$$

But observe that
$$
|\sqrt{y}-\sqrt{1}|=\frac{|y-1|}{\sqrt{y}+1}\le |y-1|
$$


问:
Finding the stationary point with implicit differentiation
I want to find the stationary points of the curve $y^3+3xy^2-x^3=3$

I differentiate to get
$\frac{x^2-y^2}{y^2+2xy}$

and so $x^2-y^2=0$ but I have two unknowns and I'm not sure how to solve.

答:
Yeah you're on the right track... you just gotta keep going.
\begin{align} \frac{x^2-y^2}{y^2+2xy} = 0\ \\
\\
\implies y(y+2x) \neq 0\quad \text{and}\quad (x+y)(x-y) = 0\\
\\
\implies (y \neq 0\quad \text{and}\quad y \neq -2x)\quad \text{and}\quad (x=y\quad \text{or}\quad x = -y).\\
\end{align}
$$$$
Now substitute $x = y$ back into the original equation $\ y^3+3xy^2-x^3=3\ $ to find the coordinate of the stationary point corresponding to when $x = y$.

Then do a similar back-substitution for when $\ x = -y\ $ to find the coordinate of the other stationary point.


问:
How to get from one inequality to another
I am trying to add an extra step in between the inequalities (1) and (2) in a proof for the below lemma to try to understand how (2) was achieved. I have managed to understand how to get to the first sum in (2), but I can't seem to get a middle step in between the second sum in (1) and the second sum in (2). Could someone please explain how we get
$$\sum^{\infty}_{n=N+1}r^n|c_n| \leq \frac{M}{N}\sum^{\infty}_{n=N+1}r^n .$$

Lemma:
Presume that the Abel means $A_r = \sum^{\infty}_{n=1}r^nc_n$ of the series $\sum^{\infty}_{n=1}c_n$ are bounded as $r$ tends to $1$ (where $r<1)$. Then, if $c_n= O(1/n)$, we have that the partial sums $S_N=\sum^N_{n=1}c_n$ are bounded.

Proof:
Let $r =1-\frac{1}{N}$ and pick $M$ s.t. $n|c_n|\leq M$. Now, we find an approximation for the difference $$S_N-A_r = \sum^N_{n=1}(c_n -r^nc_n) - \sum^{\infty}_{n=N+1}r^nc_n$$
using $$1-r^n = (1-r)(1+r+...+r^{n-1}) \leq n(1-r).$$
Our approximation is:

\begin{equation*}
    \begin{split}
        |S_N-A_r| &\leq \sum^N_{n=1}|c_n|(1-r^n)+ \sum^{\infty}_{n=N+1}r^n|c_n| \;\; (1)\\
        &\leq \sum^N_{n=1}|c_n|n(1-r)+ \\
        &\leq M\sum^N_{n=1}(1-r) + \frac{M}{N}\sum^{\infty}_{n=N+1}r^n \;\; (2)\\
          \end{split}
\end{equation*}

答:
For all $n \geq N + 1$, we have :

\begin{equation*}
    \begin{split}
        |c_n| &= n \frac{|c_n|}{n} \\
        &\leq n \frac{|c_n|}{N}\\
        |c_n| &\leq \frac{M}{N}\\
    \end{split}
\end{equation*}

The first inequality holds because $n \geq N$; the second one holds by choice of $M$. From this it follows that:
$$\sum^{\infty}_{n=N+1}r^n|c_n| \leq \frac{M}{N}\sum^{\infty}_{n=N+1}r^n$$


问:
probability of drawing 2 spades from a pack of cards given certain conditions
I am studying maths as a hobby and have come across this problem:

2 cards are drawn simultaneously from a pack of 52. What is the probability that both are spades if one card in known to be a) black, b) red, c) a spade, d) a king?

I've tried a few approaches.

I worked out all the things I thought might be relevant.

e.g. Total ways of drawing 2 spades is $\binom {13}{2} = 78$, total ways of drawing black cards = $\binom {26}{2} = 325$, total ways of drawing any 2 cards = $\binom {52}{2} = 1326$ but I cannot find how to use these to get the answer.

Then I tried to reason as follows: in c), if we know the first card is a spade the probability of drawing another spade is $\frac {12}{52}$

but none of these approaches lead me to the correct answers.

答:
If we take $(c)$ for example,

The question says we know that one of the cards is a spade, what is the probability that both are spades? We need to apply conditional probability.

If Event $A$ is getting both as spades and Event $B$ is getting at least one spade in draw of $2$ cards then

i) $P(B) = \displaystyle \frac{{52 \choose 2} - {39 \choose 2}}{52 \choose 2} = \frac{15}{34}$

Numerator is number of ways to get at least one spade in a draw of $2$ cards, which is all possible draws of $2$ cards minus draws where none of the two cards is a spade.

ii) $ P(A\cap B) = \displaystyle \frac{{13 \choose 2}}{52 \choose 2} = \frac{1}{17}$

[please note that $P(A \cap B) = P(A)$]

So the desired probability $ \displaystyle P (A|B) = \frac{P(A\cap B)}{P(B)} = \frac{2}{15}$.

Now can you use this to evaluate others? If you get stuck, let me know.

While the above can be simplified a bit more, I have gone ahead and used standard working which is important to get used to for conditional probability.


问:
Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$
How can I solve this question with out find a,b,c,d

$$a+b+c+d=2$$

$$a^2+b^2+c^2+d^2=30$$

$$a^3+b^3+c^3+d^3=44$$

$$a^4+b^4+c^4+d^4=354$$

so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$

If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d

Is there any help?

答:
Here's an approach which involves AM-GM inequality(To bound the numbers).

I have considered the positive values of $(a,b,c,d)$. I can consider $a^2,b^2,c^2$ and $d^2$, since they are all strictly positive.

$\dfrac{a^2+b^2+c^2+d^2}{4} \ge \sqrt{abcd}$

$\dfrac{30}{4} \ge \sqrt{abcd} \implies 56.25 \ge abcd$

Since $a^2+b^2+c^2+d^2=30$ and $56 >|abcd|$, One of the $|a|,|b|,|c|,|d|$ is at most $5$.

Considering $a^4+b^4+c^4+d^4$, one of the $|a|,|b|,|c|,|d|$ is atmost $4$, since $5^4=625$ .

Now we have $a^2+b^2+c^2+d^2=30$, the value of $(|a|,|b|,|c|,|d|)$ is from set  $(1,2,3,4)$.

And also:

Considering $4th$ degree polynomial such that $a,b,c,d$ are the roots of the equation.

$x^4+px^3+qx^2+rx+s=0$

$a+b+c+d=2=-p$

$(a+b+c+d)^2-a^2+b^2+c^2+d^2=2(ab+bc+cd+ad+ac+bd)=q=-13$

Finding $\sum abc$ and $abcd$ with the help of other inequalities and finding the roots of equation is another way to go. (Not quite sure whether constructing polynomial is a great way to go. )

Edit: I have considered the absolute values of $(a,b,c,d)$ . I didn't assume them to be $+ve$.


问:
Are nullhomotopic maps trivial in $Set$?
Let $F$ a contravariant functor $F: Ho(Top) \to Set$. Let $f$ a morphism in $Ho(Top)$ such that $f$ is nullhomotopic, which means that $F(f) = F(c)$, where $c$ is the constant map. Knowing that $F(\left\{x_0 \right\})$ is a one element set, does this imply that $F(f) = 0$? Why?

答:
In the assumptions of Hatcher 4E.2 it is said that the codomain of $h$ is the category of pointed sets; this category has a zero object (=initial and terminal), the (or rather, any) singleton.

Evidently, all cohomology theories are functors to (families of) pointed sets: $h^n(A)$ are abelian groups! (-;

RE the comment: yes, in principle, yes. The notation may vary (a matter of taste I guess?), but the idea is that there is a unique arrow $A \to 0 \to B$ obtained composing the terminal arrow of $A$ and the initial arrow of $B$. This map exists in every category with a zero object, and it is "constant at the basepoint of $B$".

This said, the category of pointed sets behaves in many ways like the category of pointed spaces, for example it supports a smash product structure obtained from the pushout
$$\require{AMScd}
\begin{CD}
A\vee B @>>> A \times B \\
@VVV @VVV \\
0 @>>> A\land B
\end{CD}
$$
rendering ${\sf Set}_*$ a monoidal closed category (beware, yet, that the smash product is not the product: instead, the above pushout is telling you that it measures how far are coproducts and products from being equal: the bigger $A\land B$, the further), and it sits in an adjunction
$$
(-)_+ : {\sf Set} \leftrightarrows {\sf Set}_* : U
$$
rendering the category of pointed sets the category of algebras for a monad on $\sf Set$ (this is a monad of interest for functional programmers, who call it "Maybe": it takes a set $X$ and gives you back a new set $X_+$ with a disjoint point $\bullet$). But these many similarities shouldn't trick you into believing that everything that holds true for pointed spaces transports to pointed sets.


问:
Bijection Basic Lemma
For a finite set $A$, $f : A \to A$ is a bijection if there is an inverse function $g : A \to A$ such that
$$\forall x \in A\quad g(f(x)) = x$$

Proof.
If $f(x) = f(x')$, then $x = g(f(x)) = g(f(x')) = x'$. Therefore, $f$ is one-to-one.

Since $f$ is one-to-one, there must be $\vert A\vert$ elements in the range of $f$. This implies that $f$ is also one-to-one

I am confused on what $f : A \to A$ is supposed to mean. I understand that $f : A \to B$ means that $\forall x$ in the domain $A$ maps to the co-domain $B$, but how does $A$ map to itself in $f : A \to A$? Also, how do we arrive at $f(x) = f(x')$ from our antecedent? To me, the statement, $f(x) = f(x')$, looks like it is not one-to-one.

答:
$f:A\to A$ means that $f$ maps every element of $A$ to $A$. If $f$ is onto, then $f$ is actually a rearrangement of elements of $A$ (because the dimension of $A$ is finite). But $f$ doesn't have to be onto, for example, $f\equiv 0$ is also a map from $A$ to $A$. And of course, $f$ now is not a rearrangement.

The proof actually shows that the map $f$ is injective. And for maps on finite dimensional spaces, injectivity is equivalent to surjectivity. See here.

Title: Bijection Basic Lemma


问:
A triangle; given side and two heights
A triangle $ABC$ is given $BC=8$ and $AE=4$ and $BK=6,4$ are heights. Find $AC$ and $AB$.

So we can find the area of $ABC$: $$S=\dfrac{BC\cdot AE}{2}=16.$$
So $AC=\dfrac{2S}{BK}=\dfrac{32}{6,4}=5.$
How can I find $AB$ from here?

答:
Let $|BC|=a=8$,
$|AE|=h_a=4$,
$|BK|=h_b=6.4$,
$|AC|=b$,
$|AB|=c$.

As you already found, $|AC|=b=5$,
we can use  Heron’s formula
for the area in terms of squared lengths of the sides $a,b,c$:

\begin{align}
16\,S^2&=
4a^2 b^2-(a^2+b^2-c^2)^2
,
\end{align}

which provides two valid solutions,

\begin{align}
c^2&=a^2+b^2-\sqrt{4a^2 b^2-16S^2}
\\
\text{and }\quad
c^2&=a^2+b^2+\sqrt{4a^2 b^2-16S^2}
.
\end{align}

Using known value for $S=16$, we arrive at two solutions,
\begin{align}
c&=\sqrt{41}
\\
\text{and }\quad
c&=\sqrt{137}
,
\end{align}

both fit the given conditions, see the images:

Title: A triangle; given side and two heights


问:
How to show the hermitian adjoint of two operator's kronecker product is equal to their hermitian adjoints' kronecker product?
I mean, the question is simple, asking me to show $(A\otimes B)^* = A^*\otimes B^*$, where * is just the hermitian adjoints. However, it's hard for me to use the definition, aka put them into (x,y) just like the $+$ case, in fact, I am not even sure about the dimension since the kronecker product will be $n^2$ if the original matrices are square with dim $n$.

Thanks for the help.

答:
$(A \otimes B)^*$ is the unique linear operator satisfying
$$
\langle (A \otimes B)^* y, x\rangle = \langle y , (A\otimes B) x\rangle,
$$
for all $x, y \in H\otimes K$.

Now take any two product vectors in $v_1 \otimes w_1, v_2 \otimes w_2 \in H \otimes K$. Then for a product vector we find that
$$
\begin{aligned}
\langle v_1 \otimes w_1, (A\otimes B)(v_2\otimes w_2)\rangle &= \langle v_1 \otimes w_1, (Av_2) \otimes (Bw_2)\rangle \\
&= \langle v_1, A v_2\rangle \langle w_1, B w_2\rangle \\
&= \langle A^* v_1,v_2\rangle \langle B^* w_1,w_2\rangle \\
&= \langle (A^* \otimes B^*)(v_1 \otimes w_1), v_2 \otimes w_2\rangle.
\end{aligned}
$$

So we see that for any product vector the correct identity holds. Then by linearity of the inner product and the fact that we can write any vector as a linear combination of product vectors, we can extend this to all vectors in $H \otimes K$.


问:
Primitive of $z$ conjugated
So I tried to understand complex analysis better, but I stumbled on this problem. I'm certain that a primitive (anti-derivative) of the function $z\mapsto \bar{z}$ doesn't exist, since the integral on the unit circle doesn't vanish (I hope that this is correct, anyway). However, someone gave me this "proof":

$$\int \bar{z} ~dz=\int \overline{Re(z)+i\cdot Im(z)}~dz=\int Re(z)-i\cdot Im(z)~dz=\int Re(z)~dz-i\int Im(z)~dz$$ $$=Re\int z~dz-i\cdot Im\int z~dz=\overline{Re\int z~dz+i\cdot Im\int z~dz}=\overline{\int z~dz}.$$

Since $z$ has primitive $z^2/2+C$ (for some constant $C$), we would get a primitive for $\overline{z}$, which doesn't make sense. Where does this go wrong? Thanks in advance!

答:
First I need to know what $\int\overline{z}\,dz$ means.  I do not know.  Does it mean: find $u$ so that $du = \overline{z}\,dz$.

Assuming that is the meaning, I claim there is no solution.  The problem: find $u$, a complex-valued function of $x,y$, so that $du = (x-iy)(dx+idy)$, that is $du = (x-iy)dx+(ix+y)dy$.  That is,
$$
\frac{\partial u}{\partial x} = x-iy,\qquad
\frac{\partial u}{\partial y} = ix+y
$$
If that is true, then (the partial derivatives are continuous)
$$
i = \frac{\partial^2 u}{\partial x\partial y}
=\frac{\partial^2 u}{\partial y\partial x} = -i
$$
Contradiction.
That is, $(−)+(+)$ is not an exact differential.

Title: Primitive of $z$ conjugated


问:
Set theory, injective and surjective functions task
Prove that if
$$|Z \times Z| = |X \cup Y| $$
then: there exists injective function (one-to-one) $f: Z \rightarrow Y$ OR there exists a surjective function $k : X \rightarrow Z$ .

If Z is finite, I've come with this: If $|X| \geq|Z|$, then there is a surjective  for X to Z.
If $|X| < Z$ then $|Y| >=|Z|$ so there is an injective function from Z to Y.

But I can't express it in the language of set theory.
And I can't figure out when Z is infinite. Any tups?

答:
Usually, when you have to prove something of the form
$$proposition_1 \ \ \mbox{ OR } \ \ proposition_2$$
You can assume that $proposition_1$ is false, and prove that $proposition_2$ is true. We use this strategy to prove this theorem.

First of all, fix a bijection $f: X \cup Y \to Z \times Z$.

Now, assume that "there exists a surjective function $X \to Z$" is false. Let's prove that there exists an injective function $Z \to Y$.

Consider the projection on the first coordinate $\pi: Z \times Z \to Z$
$$\pi (z_1,z_2)=z_1$$
Clearly $\pi$ is surjective. Consider the composition map $\pi \circ f : X \to Z$ (restricted to $X$) defined by
$$( \pi \circ f )(x) = \pi (f(x))$$
By our assumption, $\pi \circ f$ is not surjective. Hence there exists $z \in Z$ such that
$$\forall x \in X, \ \ \pi (f(x)) \neq z$$
Fix such an element $z_0 \in Z$. Then we can say that
$$\forall x \in X, \ \ \forall z \in Z, \ \ f(x) \neq (z_0,z)$$
this is equivalent to
$$\forall x \in X, \ \ \forall z \in Z, \ \ f^{-1}(z_0,z) \neq x$$
this is equivalent to
$$\forall z \in Z, \ \ f^{-1}(z_0,z) \notin X$$
However, for all $z \in Z$ you have $f^{-1}(z_0,z) \in X \cup Y$. This means that necessarily
$$\forall z \in Z, \ \ f^{-1}(z_0,z) \in Y$$
This defines a function $g: Z \to Y$
$$g(z)= f^{-1}(z_0,z)$$
I leave to you to check that $g$ is indeed injective (use the fact that $f^{-1}$ is injective). This completes the proof.


问:
N (even) sided Polygon within Circle, with 1 given chord length, all other sides are equal
enter image description here

Hello,

Right figure (ignore left)

I am having an issue solving this equation for even sided polygons (odd is easy as working out radius is easy)

I have a horizontal line (say 2100mm) (a1 from right image),

and a perpendicular measurement from the large horizontal line to the parallel/opposite side of the polygon (say 500mm) (ai from right image).

E.g.

6 sided polygon (need equation to be n dependent)

length n1 = 2100.
length n2 = n3 = n4 = n5 = n6

Arc of n1n2 = n6n1

arc of n2n3 = n3n4 = n4n5 = n5n6

all intersecting point are on circumference of a circle.

How do I work out length n2,n3,n4,n5,n6

and arcs n2n3,n3n4,n4n5,n5n6 OR n1n2,n6n1

with constraints n1 = 2100, n1 & n4 are parallel and perpendicular distance = 500.

Any help would be greatly appreciated.

答:
The length of a chord corresponding to the aperture angle $2\alpha$ is $r\cos\alpha$. So with your data, and denoting $m:=n-1$,

$$\begin{cases}a_1=r\cos(m\alpha),\\a_i=r\cos(\alpha).\end{cases}$$

So you have to solve the trigonometric equation

$$a_i\cos(m\alpha)=a_1\cos(\alpha)$$ also written

$$a_iT_m(\cos(\alpha))=a_1\cos(\alpha)$$ where $T_m$ is a Chebyshev polynomial.

Unfortunately, the equation is polynomial and for $n>4$ you can't spare a numerical resolution.

If you don't know $a_i$ but the distance between $a_1$ and $a_i$, the resolution will be similar, with a more complicated equation.


问:
A question on localization of polynomials and geometry.
I showed the ideal in (a) is maximal but I couldn't show the localizations are isomorphic, I know that the polynomial y is zero but couldn't get any further.
Section (b) didn't succeed at all.

Exercise 6.13.

(a) Let $R=\mathbb{R}[x,y]/(xy)$ and $P=(x-1)\trianglelefteq R$. Show that $P$ is a maximal ideal of $R$, and that $R_P\cong\mathbb{R}[x]_{(x-1)}$. What does this mean geometrically?

(b) Let $R$ be a ring and $a\in R$. Show that $R_a\cong R[x]/(ax-1)$, where $R_a$ denotes the localization of $R$ at $a$ as in Example 6.5 (c). Can you give a geometric interpretation of this statement?

答:
Here is an "elementary" and "intuitive" explanation for (a): Let $k$ be the field of real numbers, let $A:=k[x,y]/(xy)$ and let $p: B:=k[x,y]\rightarrow A$ be the projection map. Let $C:=Spec(A)$ and let $I:=(x-1) \subseteq B$. In the ring $A$ you have the following equality:

$$ xy=(x-1+1))y=(x-1)y+y$$

hence

$$ y=-(x-1)y+xy \equiv -(x-1)y\text{ in the ring $A$}.$$

Hence the element $y$ is in the ideal $J:=p(I)$. Hence it follows $P:=(x-1,y) \subseteq J$ and hence $J$ is a maximal ideal in $A$. In the localized ring $A_P$ it follows $x$ is a unit hence there is an equality of ideals $(xy)=(y)$ in $A_P$. Hence you get the following calculation:

$$I1.\text{   }A_P \cong (k[x,y]/(xy))_{(x-1,y)} \cong (k[x,y]/(y))_{(x-1,y)} \cong k[x]_{(x-1)} .$$

Geometrically the point $(x-1,y)$ is the point with coordinates $(1,0)$ on the variety $Z(xy)$ defined by the $x$-axis and $y$-axis. Hence when you restrict yourself to the $x$-axis $\mathbb{A}^1_k:=Spec(k[x])$ (you let $y=0$) you get the point $x=1$ on $\mathbb{A}^1_k$.

Question: "I showed the ideal in (a) is maximal but I couldn't show the localizations are isomorphic, I know that the polynomial y is zero but couldn't get any further."

Answer: In the formula I1, when you localize $A$ at $P$ you localize the polynomial ring $k[x]$ in the $x$-variable at the corresponding maximal ideal $(x-1)$. Hence the formula I1 expresses the fact that the localized ring $A_P$ is the local ring $\mathcal{O}_{C,P}$ of the irreducible component $C_x:=Z(x)\cong Spec(k[x])$  of $C$ at the maximal ideal $(x-1)$. The variety $C$ has two irreducible components $C_x,C_y$ and the point $P:=(x-1,y)$ is on the component $C_x$. When you "localize" at $P$ you "throw away" the component $C_y$.

This property holds in general: If $X := \cup X_i$ where $X_i$ are the irreducible components of $X$ and $p\in X_i$ with $p\notin X_j$ for $j\neq i$ then there is an isomorphism of local rings

$$ \mathcal{O}_{X,p}\cong \mathcal{O}_{X_i,p}.$$

In fact, you can always choose an affine open subscheme $U_i:=Spec(A_i) \subseteq X_i$, with $p\in U_i$ with the property there is an isomorphism

$$ \mathcal{O}_{X,p}\cong \mathcal{O}_{U_i,p}$$

of local rings.

Question (b).

Lemma. There is an isomorphism of rings $R_a \cong B:=R[t]/(ta-1)$.

Proof. Define the map $\phi^*: R \rightarrow B$ as the canonical map. It follows $\phi^*(a):=a$ is a unit hence there is a canonical induced map (by the universal property of localization)

$$ \phi: R_a \rightarrow B$$

defined by $\phi(x/a^n):=xt^n.$

Define the map $\psi: B \rightarrow R_a$ by $\psi(f(t)):=f(1/a)$. It follows $\phi, \psi$ are well defined maps that are inverses to each other and induce isomorphism $R_a \cong B$. QED.

A "geometric" interpretation could be the following: Let $k$ be the ring of integers and let $R$ be a $k$-algebra. By definition $R[t]:=R \otimes_k k[t]$ is the tensor product of $R$ and $k[t]$ over $k$. Geometrically the tensor product corresponds to the fiber product:

$$Spec(R[t]) \cong Spec(R) \times_k Spec(k[t])$$

and the ideal $I:=(ta-1)$ defines a hypersurface $Z(ta-1):=Spec(R[t]/I) \subseteq Spec(R) \times Spec(k[t])$. There is a projection map

$$p: Spec(R) \times_k Spec(k[t]) \rightarrow Spec(R)$$

inducing an isomorphism of varieties/schemes

$$ p^*: Spec(R[t]/(ta-1)) \cong Spec(R_a).$$

Hence you have realized the variety/scheme $Spec(R_a)$ as a hypersurface in the product $Spec(R) \times_k Spec(k[t])$. The variety/scheme $Spec(R_a):=D(a) \subseteq Spec(R)$ is an open subvariety/scheme and you have realized $D(a)$ as a hypersurface in the product $Spec(R) \times_k Spec(k[t])$.


问:
How to slice an area in rectangles optimally?
Given a contiguous subset of a chessboard (or, more general, a 2d rectangular grid), how can I algorithmically determine a minimal set of rectangles covering the area?

In this example, the "contiguous subset" is the black area, and I am interested in the three red rectangles. There should be no overlap between the rectangles. The relative sizes of the rectangles is not too important; it would be a bonus if the sizes weren't too different.

答:
If the problem is that small, you can use brute-force search:
list all rectangles contained in the area,
list all sets of rectangles,
keep those that partition the area,
and select the best one.

To reduce the number of candidates,
you can arrange non-overlapping sets of rectangles into a tree
(the root is the empty set, each level adds one rectangle,
the partitions are among the leaves).
In your implementation, you do not need to store the tree,
you just have to explore it.

If the problem is larger, instead of all rectangles,
you can use the maximal rectangles,
their intersections and differences.

If an approximate solution is acceptable,
you can start with a partition into rectangles
(say, all $1\times1$ rectangles),
grow those rectangles as much as possible (in random directions),
remove or cut them (at random) to have a partition.
Iterate many times and keep the best solution.
I think a similar algorithm is used to
simplify boolean expressions
(but overlaps are allowed).


答:
I believe this is an instance of the 'multiplying by 1' trick:

Let $\epsilon > 0$, $\delta := \epsilon$. Then, for $x$ such that $|x|<\delta$ we have:

$$|f(x) - 1| = |\sqrt{x+1} - 1| = |\frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{\sqrt{x+1}+1}| = |\frac{x}{\sqrt{x+1}+1}| \leq |x| < \delta = \epsilon $$

Title: Proof limit value with $\epsilon/ \delta$

Body:
I have to find the limit value for $f(x)=\sqrt{1+x}$ for $x \rightarrow0$. And then show with $\epsilon /\delta$ that I have found the right limit value.
I have found the limit value to $\sqrt{1+x} \rightarrow1$ for $x \rightarrow0$, by:
$$\lim_{x \rightarrow 0} \sqrt{1+x}
=\sqrt{\lim_{x \rightarrow 0} (1+x)}=\sqrt{\lim_{x \rightarrow 0} (1)+\lim_{x \rightarrow 0}(x)}=\sqrt{1+\lim_{x \rightarrow 0}(x)}=1$$

But I got a problem when I have to show it by $\epsilon /\delta$. When I write:
$$|f(x)-1|\Rightarrow|\sqrt{1+x}-1|$$
I can't see what I can use as a bound easier to find a $\delta$. Can anyone help me?


问:
show that there exists a unique solution to the boundary value problem $(3-2e^{-x})u''+(\frac{1}{2000}+e^{-x})u=0$
show that there exists a unique solution to the boundary value problem $(3-2e^{-x})u''+(\frac{1}{2000}+e^{-x})u=0$ for $0\le x \le 4$ with $u(0)=u(4)=0$.

I tried to look at the homogeneous ODE $u''+\frac{\frac{1}{2000}+e^{-x}}{3-2e^{-x}}u=0$ and bound it from above with an ODE of the form $u''+r(x)u=0$ with a constant $r(x)=1+\frac{1}{2000}$, so that I get the solution $u_*=sin(\sqrt{1+\frac{1}{2000}}x)$. if this solution doesn't vanish on the interval [0,4] then the only solution of the original ODE is the trivial solution, but $u_*$ has a zero at $\frac{\pi}{\sqrt{1+\frac{1}{2000}}}\le4$ so I'm not sure what to do next, do I need to bound the ODE with a non-constant $r(x)$?

答:
Hint

Set $y=u'$, then you problem becomes $$\begin{pmatrix}u\\y\end{pmatrix}'= \begin{pmatrix}1&0\\0&\frac{1/2000+e^{-x}}{3-2e^{-x}}\end{pmatrix}\begin{pmatrix}u\\ y\end{pmatrix}.$$
and use Picard-Lindelof theorem.


问:
Totient minimal value for semiprimes
I have two question concerning Euler Totient of semiprimes.

First question : given $N=p_1 * p_2$ and $M=p_3*p_4$ where $p_1,p_2,p_3,p_4$ are prime numbers greater than 5; and $M>N$ this means that also $\phi(M)>\phi(N)$ ?

Second question : what is the minimum value that $\phi(M)$ can have ? I know that Mr. Sierpinski estimated the minimum value involving the Eulero-Mascheroni constant, but do we have a better "low bound" estimationf for the Euler Totient of a semiprime like $M$ ?

Thanks a lot guys

答:
For the first question, the answer is no. You can have $M>N$ and $\varphi(M)<\varphi(N)$.

For example take $M=7\cdot 131=917$. Then $\varphi(M)=6\cdot 130=780$.

And take $N=29\cdot 31=899$. Then $\varphi(N)=28\cdot 30 =840$.

Title: Totient minimal value for semiprimes


问:
Verifying some facts about finite-volume hyperbolic manifolds
Let $M$ be a finite-volume hyperbolic manifold. If we chop off the cusps (are there only finitely many?) to get a compact surface with (finitely many?) boundary components, and then lift to the universal cover, we get hyperbolic space after deleting a family of horoballs with disjoint closures (known as neutered or truncated hyperbolic space). Apparently, it is true that:

The set of basepoints of those horoballs is dense in the boundary sphere.

There is an R > 0 such that no two horoballs in this family are at distance < R from each other.

Any idea why these statements are true? My rough thoughts are as follows (but even if I'm on the right track I don't know how to make the arguments precise). The first statement I think should have to do with $M$ being finite-volume. Maybe the idea is that a fundamental domain for the lattice action must have finite volume, but it won't if there's an open set of points on the boundary sphere that don't correspond to deleted horoballs. The second statement maybe has to do with how there are finitely(?) many boundary components that lift to the horospheres, and since the covering map is a local isometry, you can't find pairs of horospheres that are arbitrarily close.

UPDATE:

I have verified that indeed there are only finitely many cusps, and I have also verified statement 2.

As for statement 1, I tried looking through this book by Kapovich.

He defines a discrete subgroup $\Gamma$ of Isom($\mathbb{H}^n$) to be geometrically finite if every point in its limit set $\Lambda(\Gamma)$ is either a conical limit point or a cusped parabolic point. He defines a point in the limit set to be a cusped parabolic point if its stabilizer in $\Gamma$ is an almost abelian group $A$ containing parabolic elements such that $\Lambda(\Gamma)/A$ is compact. I believe that these cusped parabolic points are my horoball basepoints. (He even comments later that "parabolic fixed points are never conical limit points.")

Later he proves a theorem by Ahlfors which says that if $G \subset \text{Isom}(\mathbb{H}^3)$ is a geometrically finite discrete group, then its limit set either is all of $\hat{\mathbb{C}}$ or has Lebesgue measure zero. This seemed useful so I took a closer look, but in the proof, he says that a.e. limit point of $G$ is a conical limit point. This is troubling to me because this appears to be contrary to statement 1. Surely I am misunderstanding something, but I can't figure out what it is.

答:
It sounds like you are getting off track by focussing on geometrically finite groups.

What you want to look for is a theorem which says that if $M = \mathbb H^n / \Gamma$ has finite volume, equivalently if $\Gamma$ has a finite volume fundamental domain, then the limit set of $\Gamma$ is the entire sphere $S^{n-1}_\infty$ (equivalently, if the limit set is not the entire sphere, then $M$ has infinite volume).

It follows, from the very definition of the limit set, that every orbit of the $\Gamma$ action on $S^{n-1}_\infty$ is dense.

In particular, for each horoball base point $x \in S^{n-1}_\infty$, its entire orbit $\Gamma \cdot x$ consists of horoball base points, and that orbit is dense.


答:
Note that the area can also be expressed as
$$S= 16= \frac12 AC \cdot BC \sin C= \frac12 \cdot 5 \cdot 8 \sin C
$$
which leads to
$$ \sin C = \frac45\implies \cos C= \pm \sqrt{1-\sin^2 C}=\pm \frac35$$
Per the cosine rule
$$AB=\sqrt{ AC^2 +BC^2 -2AC\cdot BC \cos C }
= \sqrt{ 25+64\pm 48}=\sqrt{41},\>\sqrt{137}
$$

Title: A triangle; given side and two heights

Body:
A triangle $ABC$ is given $BC=8$ and $AE=4$ and $BK=6,4$ are heights. Find $AC$ and $AB$.

So we can find the area of $ABC$: $$S=\dfrac{BC\cdot AE}{2}=16.$$
So $AC=\dfrac{2S}{BK}=\dfrac{32}{6,4}=5.$
How can I find $AB$ from here?


问:
Showing a morphism $f : X \to \mathbb{P}^n_A$ is uniquely determined by an invertible sheaf $\mathcal{L}$ on a scheme $X$ generated by global sections
So it is a well-known theorem that given an invertible sheaf $\mathcal{L}$ on a scheme $X$ over a ring $A$, such that $\mathcal{L}$ is generated by global sections $s_0, \dots, s_n$, then there exists a unique morphism $f : X \to \mathbb{P}^n_A$ such that $f^*x_i = s_i$ for all $i$.

I want to prove the uniqueness part of this theorem. The proof I've seen of this starts out more or less in the following way:

We first suppose a morphism $f : X \to \mathbb{P}^n_A$ such that $f^*x_i = s_i$ for all $i$ exists. Then we see that $X$ is covered by the open sets $X_{s_i} = \{x \in X \mid s_i(x) \neq 0\}$ and we can regard $f$ as being glued by the morphisms $$f_i = f|_{X_{s_i}} : X_{s_i} \to D_+(x_i) = \operatorname{Spec} A[x_0, \dots, x_n]_{(x_i)}$$ and this in turn corresponds to a morphism of $A$-algebras
$$f^\sharp : A[x_0, \dots, x_n]_{(x_i)} \to \Gamma(X_{s_i}, \mathcal{O}_X).$$

Then we notice that $x_i$ generates $\mathcal{O}(1)$ on $D_+(x_i)$ and that $x_j = \frac{x_j}{x_i}$ in $A[x_0, \dots, x_n]_{(x_i)}$ for $0 \leq j \leq n$.

And now this is where the argument gets a bit sketchy for me. The proof I've seen then goes on to say that similarly, pulling back via $f^\sharp$ gives $$s_j = f^*_i(x_j) = f_i^\sharp\left(\frac{x_j}{x_i}x_i\right) = f_i^\sharp\left(\frac{x_j}{x_i}\right)s_i$$ where we interpret $\frac{x_j}{x_i}$ as a fraction of $\Gamma(D_+(x_i), \mathcal{O}_{\mathbb{P}^n_A})$.

Then the proof claims that from each morphism $f : X \to \mathbb{P}^n_A$ we get $n+1$ distinguished sections $s_0, \dots, s_n$ from which we can determine the morphism $f_i$. (Call this statement (1))

And then finally the  proof concludes that $f$ is uniquely determined by the data $L, s_0, \dots, s_n$. (Call this statement (2))

Now I don't understand how statements 1 and 2 follow from what is done in the proof up until that point. In particular it is not clear to me how the sections $s_0, \dots, s_n$ can determine the morphism $f_i$, or what it even means for these sections to "determine the morphism" in this context.

答:
First, let us be careful with the notations: if $A$ is a ring, the $i$-th affine patch $D_+(x_i)$ in $\mathbb P^n_A=\operatorname{Proj}A[x_0,...,x_n]$ is given by $\operatorname{Spec}(A[(x_j/x_i)_{j\neq i}])$, not $\operatorname{Spec}(A[(x_j)_{0\leq j \leq n}]_{(x_i)})$.

Let $\mathcal L$ be an invertible sheaf on a scheme $X$, and let $s_0,...,s_n$ be global sections that generate $\mathcal L$ at every point. We want to use the $s_i$ in a natural way to define a morphism $X \to \mathbb P^n_X=\operatorname{Proj}(\mathcal O_X[x_0,...,x_n])$. (Then, by composition, any map $X \to \operatorname{Spec} A$ gives you a morphism $X \to \mathbb P^n_A$.) Intuitively speaking, we want this morphism to "send $x_i$ to $s_i$". Since the $s_i$ are not local sections of $\mathcal O_X$ per se, the formal way to do it is the following.

For each $i$, call $U_i$ the maximal open of $X$ on which $s_i$ generates $\mathcal L$. The $U_i$ cover $X$ by hypothesis. For each $i$ we have a natural morphism of $\mathcal O_X$-modules $f_i\colon \mathcal O_X\to \mathcal L$ mapping $1$ to $s_i$, and the restriction of $f_i$ to $U_i$ is an isomorphism. Let's call $s_j^{(i)}$ the preimage of $s_j$ in $O_X(U_i)$ under the isomorphism $f_i|_{U_i}$.

By the sheaf property, all we need to do is exhibit morphisms $U_i \to \mathbb P^n_X$ which coincide on the pairwise intersections $U_i \cap U_j$. Define a map from $U_i$ to the patch $D_+(x_i)$ of $\mathbb P^n_X$ by mapping $x_j/x_i$ to $s_j^{(i)}/s_i^{(i)}$. These maps glue accordingly since on $U_i \cap U_j$ we have $(s_i^{(j)}/s_j^{(j)})^{-1}=s_j^{(i)}/s_i^{(i)}$, so you have defined a morphism $X \to \mathbb P^n_X$.

This morphism is "determined by the $s_i$" in the sense that follows. Pick another set of global sections $(t_0,...,t_n)$ generating $\mathcal L$, and assume that the resulting morphisms $X \to \mathbb P^n$ are the same. Then it follows immediately from our definition of the morphism that for each $i$, the vanishing sets of $t_i$ and $s_i$ are equal, i.e. there is a unit $\lambda_i\in\mathcal O_X^\times(X)$ such that $s_i=\lambda_i t_i$. By the gluing conditions, we find that for each $i,j$ we have $\lambda_i=\lambda_j$ on $U_i \cap U_j$. Therefore, there exists a global section $\lambda$ of $\mathcal O_X^\times(X)$ such that for each $i$, we have $s_i=\lambda t_i$. This corresponds to the intuitive idea that $(s_0:...:s_n)=(t_0:...:t_n)$ in the projective space.


问:
Runge-Kutta method accuracy
I got a Runge-Kutta method here and I solve this system using it.

So here's  Runge-Kutta stuff
\begin{align}
 k_1 &= f(t_n, y_n)
\\
 k_2 &= f(t_n + h/2, y_n + hk_1/2)
\\
 k_3 &= f(t_n+h, y_n - hk_1 + 2hk_2)
\\\hline
 y_{n+1} &= y_n + h(k_1 + 4k_2 + k_3)/6
\end{align}
where $h$ is step

Here's my test system
\begin{align}
 y'_1 &= -5y_1 - 10y_2 + 14e^{-x}
\\
 y'_2 &= -10y_1 - 5y_2 + 14e^{-x}
\end{align}
with exact solution  $y_1(x) = y_2(x) = e^{-x}$

UPD: The initial condition here is $y_1(0) = y_2(0) = 1$

I need to solve it on $[0;4]$.

Well, I thought I solved it right, because I checked how the exact solution and these approximate solution plots looks like (on the left, on the right I zoomed plot until saw difference)

Also I checked how the plot of the difference between exact solution and approximate solution depending on step (let's call it e/h) looks like.

So $e/h$ it looks like this

But when I checked e/h^4 dependence it looked like these

I showed it to the teacher and she said that my solution is wrong, it's not suppose to be like these! I show my code to her asked for help but she said that she doesn't understand matlab :c

Have I really done something wrong? And if yes what I've done wrong? And if not how to prove that I'm right?

Here's my code btw

Runge-Kutta method

```<code>  function [ res_y ] = RungeKutta(dim, size, grid, step, f1,f2,y1, y2)

    k1=zeros(dim);
    k2=zeros(dim);
    k3=zeros(dim);

    h = step;

    res_y(1,1) = y1;
    res_y(2,1) = y2;

    for i=1: size

       k1(1)= f1(grid(i),y1,y2);
       k1(2)= f2(grid(i),y1,y2);

       k2(1)= f1(grid(i)+h/2, y1+h*k1(1)/2, y2+h*k1(2)/2);
       k2(2)= f2(grid(i)+h/2, y1+h*k1(1)/2, y2+h*k1(2)/2);

       k3(1)= f1(grid(i)+h, y1-h*k1(1)+2*h*k2(1), y2-h*k1(2)+2*h*k2(2));
       k3(2)= f2(grid(i)+h, y1-h*k1(1)+2*h*k2(1), y2-h*k1(2)+2*h*k2(2));

       res_y(1,i+1) = y1 + h*(k1(1) + 4*k2(1) + k3(1))/6;
       res_y(2,i+1) = y1 + h*(k1(2) + 4*k2(2) + k3(2))/6;

       y1 = res_y(1,i+1);
       y2 = res_y(2,i+1);
    end

 end
</code>```

Main method

```<code>    a = 0; b = 4;
    h = 0.1; % step
    t = a:h:b; %grid
    n = 2;
    m = size(t,2);

     hold on;
         plot(t, exp(-t),'b-')
         plot(t, exp(-t),'r--')
     hold off;

    y1=1; y2 = 1;

    f1_ptr = @f1;% out = -5 * y1 - 10 * y2 + (14)*exp(-x);
    f2_ptr = @f2;% out = -10 * y1 - 5 * y2 + (14)*exp(-x);

    res_y = RungeKutta(n,m-1,t,h,f1_ptr, f2_ptr,1,1);

    hold on;
    plot(t,res_y);

    hold off;

%e/h and e/h^4 plots

 fig_a = figure;
set(fig_a,'name','e/h','numbertitle','off')
hold on;

counter = 0;

for h=0.001:0.01:0.1
    y1=1; y2 = 1;
    t = a:h:b;
    m = size(t,2);
    counter = counter + 1;

    result_appr = RungeKutta(n,m-1,t,h,f1_ptr, f2_ptr,y1,y2);
    result_exact = exp(-t);

   result_difference = abs(result_appr(1, :) - result_exact);

    e1(counter) = max(result_difference);
    e2(counter) = max(result_difference);

    hh = h*h*h*h;

    ehh1(counter)=e1(counter)/hh;
    ehh2(counter)=e2(counter)/hh;

end;

h=0.001:0.01:0.1;
    plot(h,e1,'c');
    plot(h,e2,'c');
hold off;

fig_b = figure;
set(fig_b ,'name','e/h^4','numbertitle','off')

hold on;
    plot(h,ehh1,'r')
    plot(h,ehh2,'b')
hold off;
</code>```

f1 function

```<code>function [ out ] = f1( x, y1, y2, alpha, beta )
if nargin == 3
    alpha = 5;
    beta = 10;
end

out = -alpha * y1 - beta * y2 + (alpha + beta - 1)*exp(-x);

end
</code>```

f2 function

```<code>function [ out ] = f2( x, y1, y2, alpha, beta )
if nargin == 3
    alpha = 5;
    beta = 10;
end
out = -beta * y1 - alpha * y2 + (alpha + beta - 1)*exp(-x);

end
</code>```

答:
The method is a classical third-order method as computed and presented by M. Wilhelm Runge in 1901, based on the Simpson quadrature rule.

This means that the global error will scale like $h^3$ and the local truncation error like $h^4$.

Plotting the error as computed against $h$, but in a loglog plot and for a geometric sequence of step sizes, for h = 10.^[-4:0.5:-1], gives a line that has a very nice slope $3$.

Another way to visualize this is to plot the difference of the computed to the exact solution divided by $h^3$. This gives an error profile plot

that nicely and visually converges to the leading coefficient $c(x)$ in $e(x)=c(x)h^3+O(h^4)$. Note that this is the absolute error, and that the solutions fall exponentially to zero, which explains that the error falls in the end. In a plot of the relative error one could expect the curve to continue to grow away from zero.

Title: Runge-Kutta method accuracy


问:
Which one result in mathematics has surprised you the most?
A large part of my fascination in mathematics is because of some very surprising results that I have seen there.

I remember one I found very hard to swallow when I first encountered it, was what is known as the Banach Tarski Paradox. It states that you can separate a ball $x^2+y^2+z^2 \le 1$ into finitely many disjoint parts, rotate and translate them and rejoin (by taking disjoint union), and you end up with exactly two complete balls of the same radius!

So I ask you which are your most surprising moments in maths?

Chances are you will have more than one. May I request post multiple answers in that case, so the voting system will bring the ones most people think as surprising up. Thanks!

答:
That one can count on and on without end. </p>

<p>(Of course, this surprise was a while ago.)</p>

LastEditorDisplayName: anon

Body:
That one can count on and on without end.

(Of course, this surprise was a while ago.)


问:
$X=max\{B_t: 0 \le t \le 1\}$ and $Y=B_2-B_1$ with $(B_t)_{t \ge 0}$ Brownian motion
Let $X=\max\{B_t: 0 \le t \le 1\}$ and set $Y=B_2-B_1$

(i) Prove that $P(X>Y)\ge1/2$.

(ii) Calculate the distribution functions of $X$ and of $Z=|Y|$.

(iii) Prove or disprove that $P(X>|Y|)\ge 1/2$.

While I don't know how to solve (i), I know that $f_Z(t)=2*f_{N(0,1)}(t)$,  $\forall t \ge0$ and zero else. For this reason, I'm inclined to think that (iii) is incorrect.

答:
By the reflection principle you get $P(X>a)=2P(B_1>a)$ for $a>0$. Thus the distribution of $X$ is given by
$$P(X\leq a)=1-2P(B_1>a)=1-2\int_a^\infty\frac{\exp(-\frac{x^2}{2})}{\sqrt{2\pi}}\mathrm dx
$$
and the pdf is $f(x)=\sqrt{\frac{2}{\pi}}\exp(-\frac{x^2}{2})\cdot1_{]0,\infty]}(x)$. Now you know the pdf of two independet random variables $X$ and $Y$, so you can use the convolution of probability distributions to calculate the pdf of $X-Y=X+(-Y)$ (note: $Y$ and $-Y$ have the same distribution). Then you just have to calculate the integral
$$P(X-Y>0)=\int_0^\infty\int_0^\infty\sqrt{\frac{2}{\pi}}\exp\left(-\frac{x^2}{2}\right)\frac{\exp(-\frac{(z-x)^2}{2})}{\sqrt{2\pi}}\mathrm dx\mathrm dz.
$$
This is the solution for (i) and the first part of (ii). The second part of (ii) is already solved by you and (iii) can solved in the same way as above by considering $P(X-|Y|>0)$.


问:
About the limit of a function in two (or more) variables
Let $f:D\subseteq \mathbb R^2\longrightarrow\mathbb R$ and suppose that $x_0$ is a limit point for $D$ such that $\lim_{x\to x_0} f(x)=\ell$.

If we restrict $f$ to any subset  $S\subseteq D$ ($S$ is for example a curve) whereby
  $x_0$ is a limit point, and take the limit for $x\to x_0$
  (along the new restricted domain) the value of the limit  will be $\ell$.

The converse of the above statement is clearly true (take $S=D$).

Now if $\lim_{x\to x_0} f(x)=\ell$, for $x$ varying in the image of any continuous curve $\gamma:[a,b]\longrightarrow D$ (whereby $x_0$ is a limit point), is it true that $\lim_{x\to x_0} f(x)=\ell$ with $x\in D$? Any counterexamples?

答:
The answer is no.

A very famous counter-example is $$
  f \,:\, \mathbb{C}\setminus \{0\} \to \mathbb{C} \,:\, z \to e^{-\frac{1}{z^2}}
$$
$f$ has an essential singularity at $0$, i.e. not only can't you extend $f$ continously to $\mathbb{C}$, $f$ even takes nearly every value of $\mathbb{C}$ (everone except 2 or so) on an arbitrary small neighbourhood of $0$.

Yet, the restriction of $f$ to $\mathbb{R}$ can be extended continously to the whole real line (you get $f(0) = 0$), and this extension is even $C^\infty$ (you get, in fact, $f^{(n)}(0) = 0$ for all $n \in \mathbb{N}$).

You can express that function without using complex numbers by e.g. doing $$\begin{aligned}
  \Rightarrow&&
  z &= e^{\log |z| + i\arg z } \\
  \Rightarrow&&
  z^2 &= e^{2\log |z| + i2\arg z }  \\
  \Rightarrow&&
  \frac{1}{z^2} &= e^{-2\log |z| - i2\arg z } \\
  \Rightarrow&&
  -\frac{1}{z^2} &= e^{-2\log |z| +i\pi - i2\arg z} = \frac{1}{|z|^2}\left(\cos(\pi - 2\arg z) + i\sin(\pi - 2\arg z)\right) \\
  \Rightarrow&&
  e^{-\frac{1}{z^2}} &= e^{\frac{1}{|z|^2}\cos(\pi - 2\arg z)}\left(\cos\sin(\pi - 2\arg z) + i\sin\sin(\pi - 2\arg z)\right)
\end{aligned}$$

This yields (if I haven't made a mistake somehwere...) $$
  f \,:\, \mathbb{R}^2 \to \mathbb{R^2},
  \begin{pmatrix}x\\y\end{pmatrix} \to
  e^{\frac{\cos(\pi - 2\arg (x,y))}{x^2 + y^2}}
  \begin{pmatrix}
     \cos\sin(\pi - 2\arg(x,y)) \\
     \sin\sin(\pi - 2\arg(x,y))
  \end{pmatrix}
$$
where $\arg(x,y)$ is the argument function, i.e. the function that maps a vector $(x,y)$ to the angle between $(x,y)$ and $(1,0)$ (http://en.wikipedia.org/wiki/Argument_%28complex_analysis%29).

Since you're not interested in differentiability, you can pick a simpler counter-example, though. Observe for example that the reason this function has a limit when you approach $x$ via the real line is that the exponent is negative in this case. You can use a similar idea and define $$
  f \,:\, \mathbb{R}^2 \to \mathbb{R^2},
  \left(x,y\right)^T \to
  e^{\frac{y}{x^2 + y^2}}
$$

This function is constant $1$ on $\mathbb{R} \times \{0\}$, but on $\{0\}\times\mathbb{R}^+$ you have $$
  \lim_{y\to 0^+} f(0,y) = \lim_{y\to 0^+} e^{\frac{1}{y}} = \infty \text{.}
$$


问:
Dual space of $L^\infty$ is $L^1$ with the weak-* topology?
A friend of mine found a book in which the author said that the dual space of $L^\infty$ is $L^1$, of course not with the norm topology but with the weak-* topology. Does anyone know where I can find this result? Thanks.</p>

答:
This is a consequence of the Mackey-Arens Theorem, see for instance Reed, Simon: Methods of modern mathematical physics, Theorem V.22. It says:

For a dual pair $(X,Y)$, a loc. conv. topology $\mathcal{T}$ is a dual-$(X,Y)$ topology if and only if it is stronger than the $\sigma(X,Y)$-topology (i.e. the weak topology) and weaker than the Mackey-topology (called the $\tau(X,Y)$-topology).

Now choose $X = L^\infty$ and $Y = L^1$, hence the $\sigma(L^\infty,L^1)$-topology is the dual-$(L^\infty,L^1)$ topology, i.e. the topological (or continuous) dual of $(L^\infty, \sigma(L^\infty,L^1)$) is $L^1$.

To notice that $(L^\infty,L^1)$ is a dual pair, see for instance this wikipedia article or the above textbook.

Since you asked for references, I am certain this result is present in most textbooks on topological vector spaces.

OwnerDisplayName: Camilo Chaparro


问:
$x^4-mx^3-2mx^2+2m^2x=0$ roots in arithmetic progression
For which real values of $m$ roots of equation
$$x^4-mx^3-2mx^2+2m^2x=0$$
are in arithmetic progression?

I managed to find a solution using a lot of casework, i.e. factoring the equation into
$$x(x-m)(x^2-2m)=0$$
and then finding $m$ for which it has one root, three distinct roots and four distinct roots but I think there may be an easier solution and I'm just overcomplicating it. In addition, I'm not quite sure if I should include the one root case.

Could someone guide me to a better solution? Thanks.

答:
From your factorized equation, the roots are $0$, $m$, $-\surd(2m)$, and $\surd(2m)$. If $m<0$, then $m$ lies off the imaginary axis, on which the three roots $0$ and $\pm\,\mathrm i\surd(-2m)$ lie, violating the condition that the roots are in arithmetic progression. Hence $m\geqslant0$. It follows that the arithmetic progression is $-\surd(2m)$, $0$, $\surd(2m)$, $m$. Therefore $m=2\surd(2m)$, giving
$$m=0\quad\text{or}\quad m=8.$$