问:
Diameter of $A = \{f_n \mid n\in \mathbb{N} \}$ where $f_n(x)=x^n$.

Let $C[0,1]$ be the set of continuous functions on $[0,1]$ with the sup-norm. Determine the diameter of the set $A = \{f_n \mid n\in \mathbb{N} \}$ where $f_n(x)=x^n$.

The diameter is clearly $1$. To show this I would need to show that $d(A) \leqslant 1$ and $d(A) \geqslant 1$. For the first case letting $f_n,g_m \in A$ I have that on $[0,1]$ the sup norm gives $$d(f,g) = ||f-g|| = \sup|f(x) -g(x)| = |x^n-x^m| \leqslant 1$$ thus $d(A) \leqslant 1$. However I'm not sure how to show the other way $d(A) \geqslant 1$, how should I approach this?

答: Let $0\leq a<1$ and $a<x<1$. Since $x^n\to0$ as $n\to\infty$ there is some $n\in\Bbb N$ such that $x^n<x-a$, i.e. $x-x^n>a$. Hence $\|f_1-f_n\|>a$ and it follows that $d(A)\geq\sup_{n\in\Bbb N}\|f_1-f_n\|\geq 1$.


答:
Let $F(x)=x^m-x^n,$ where $x\in[0,1]$ ans suppose $n\gt m.$ Then $F'(x)=(m-nx^{n-m})x^{m-1}$ and therefore $F$ has its maximum (this need a little more than the first derivative) at $\left(\frac{m}{n}\right)^{1/(n-m)}.$ This gives an explicit expression the distance between $f_m$ and $f_n$ in your set as $$||f_m-f_n|| =\left(\dfrac{m}{n}\right)^{m/(n-m)}-\left(\dfrac{m}{n}\right)^{n/(n-m)}.$$ Now compute the limit of the distance $n\to\infty$ at $m=1$ to get what you want.

Added: I think here $m=1$ needs a bit more explanation. Note that distance between $f_n$ and $f_m$ gits larger as they gets apart. You can see this by plotting $n=1, 2, 3$ on the same graph (and $n=0,$ if you allowed it). So to make the maximum distance between to we can fix $m=1$ (or any other fixed natural also would work) and let $n\to\infty.$

ContentLicense: CC BY-SA 4.0

Title: Diameter of $A = \{f_n \mid n\in \mathbb{N} \}$ where $f_n(x)=x^n$.

Body:
Let $C[0,1]$ be the set of continuous functions on $[0,1]$ with the sup-norm. Determine the diameter of the set $A = \{f_n \mid n\in \mathbb{N} \}$ where $f_n(x)=x^n$.

The diameter is clearly $1$. To show this I would need to show that $d(A) \leqslant 1$ and $d(A) \geqslant 1$. For the first case letting $f_n,g_m \in A$ I have that on $[0,1]$ the sup norm gives $$d(f,g) = ||f-g|| = \sup|f(x) -g(x)| = |x^n-x^m| \leqslant 1$$ thus $d(A) \leqslant 1$. However I'm not sure how to show the other way $d(A) \geqslant 1$, how should I approach this?


问:
Approximation Distributions
Can someone explain why in distributions:

$ \operatorname{Binomial}(n,p)\approx  \operatorname{Normal}(np, np(1-p))$

$ \operatorname{Binomial}(n,p) \approx  \operatorname{Poisson}(np)$

答:
It's the result of the Central limit Theorem in the version of De Moivre - Laplace

It's the limit distribution of a binomial when $p=\lambda/n$

Title: Approximation Distributions


问:
When can we use the implies arrow?
So I'm a complete beginner to logic and set theory and I've been trying to learn a bit about conditional statements and logic however I'm quite confused on if then statements.
For example if $p\implies q$
I saw that this means that if p is true then q must be true also
My confusion is when can we use this arrow can we only use it if the statement is true?

As an example if I had the statement if $x^2>0
$ then $x<0$ could I write $p\implies q$ where p=$x^2>0
$ and $q=
 x<0$

Or would it be incorrect to do so as this statement is false if this is not true when we see the implies arrow in theorems and proofs how am I to know that the statement is true or false?.

Thanks in advance.

答:
p -> q
if p then q means that if p is true then q has to be true.
in your example x < 0 doesn't necessarily be true if x^2 > 0</p>

Title: When can we use the implies arrow?


问:
Geometric distribution, rolling a dice
We roll a fair dice until we get a number $3$. We win as many dollars as is a number of times we didn't get a $3$. If we get $3$ in a first throw, we loose $y$ dollars. What is the maximum value for $y$ so that this game is profitable for us?

I don't really know where to start. Of course I'd like to compute some kind of expectancy, but how to include this information about a possible outcome of a first roll?

答:
The expections numbers of throws till getting $3$ is obviously $6$, since its a bernoulli distribution with $p(x=3)=\frac{1}{6}$ and $p(x\neq3)=\frac{5}{6}$.

In every throw we did'nt throw $3$ the expection is:

$E(X)=0.2\cdot(1+2+4+5+6)=3.6$

So we expect for $5 \cdot 3.6=18$ so the maximnum $y$ would be $18$

I used 5 and not 6, since its the time of throwing without getting 3.

Title: Geometric distribution, rolling a dice


问:
Grand Prix Race- Differential Equations
Driver A has boon leading archrival B for a while by a steady 3 miles. Only 2 miles from the finish, driver A ran out of gas and decelerated thereafter at ta rate proportional to the square of his remaining speed. One mile later,driver A's speed was exactly halved.If driver B's speed remained constant,who won the race?</p>

答:
Let $A(t)$ be the distances from the finish of $A$ at time $t$ from when A runs out of gas, and the initialy velocity of $A$ and $B$ be $v$. Then $B$ takes a time of $\frac{5}{v}$ to reach the finish.

$A(0)=0 \ , \ A'(0)=-v$

$\frac{\mathrm{d}^2A}{\mathrm{d}t^2}=k(\frac{\mathrm{d}A}{\mathrm{d}t})^2\\$

Seperation of variables to find $\frac{\mathrm{d}A}{\mathrm{d}t}$ followed by integration gives $A(t)=c_1t+\frac{kt^4}{12}+c_2$.

From the boundary conditions, $c_2=2 \ \ c_1=-v$.

Where $A(t)=1, \ \frac{\mathrm{d}A}{\mathrm{d}t}=\frac{v}{2}$. Hence $v=\frac{2kt^3}{3}, \ t^4=\frac{12}{7k}$ and $k^4=0.534v$.

I cannot answer the question without knowing $v$, but the time taken by $a$ is just the solution to $A(t)=0$.

ContentLicense: CC BY-SA 3.0

Title: Grand Prix Race- Differential Equations


问:
Showing that the indicator function forms a basis

Let $E$ a non-empty finite set and $F(E)$ the set of all functions $f: E \to \mathbb{R}$. Define for every $f \in E$ the function $X_f : E \to \mathbb{R}$ with $$x \mapsto \begin{cases} 1 \text{, if $x=f$} \\ 0, \text{otherwise}\end{cases}$$ show that $(X_{f_1}, \dots ,X_{f_{n}})$ is a basis for $F(E)$ if the set $E$ has $n$ elements and $\{f_1,\dots, f_n\} = E.$

The notation here is very confusing it seems that the writer wants me to show that the sequence of the functions forms a basis for $F(E).$ Also isn't this function the indicator function or am I missing something here? Any hints on how to start showing something like this would be appreciated. I know that in order to show that some set forms a basis for a vector space I would need to show that it spans the space and that it's elements are linearly independent.

答:
I suggest to rewrite the problem as:

Let $E$ be a non-empty set, and $F(E)$ be the set of all functions from $E$ to $\mathbb R$. For every $e \in E$ define the function $\chi_e : E \to \mathbb R$ (so $\chi_e \in F(E)$) by sending $e$ to $1$ and every other element distinct from $e$ to $0$. Show that if $E = \{e_1,\dots,e_n\}$ for some $n \in \mathbb Z^+$, then the set $\{\chi_{e_1},\dots,\chi_{e_n}\}$ is a basis for $F(E)$.

Now prove that, for every $f \in F(E)$, $f$ is equal to the function $f(e_1)\chi_{e_1} + \cdots + f(e_n)\chi_{e_n}$  (how do we check that two functions are equal?). Also, if $a_1,\dots,a_n$ are $n$ real numbers such that $a_1\chi_{e_1} + \cdots + a_n\chi_{e_n}$ is the zero function, evaluating $a_1\chi_{e_1} + \cdots + a_n\chi_{e_n}$ in $e_j$ (for $j=1,\dots,n$) shows that $a_j=0$ (why?).

Hope this helps you to clarify the things. And let me know if something that I wrote was not clear to you.


问:
3D Expected Value GEO
Consider two nonparallel planes $X,Y$ that go through to the center of a sphere of radius one. This cuts the sphere into 4 parts $A,B,C,D$. The two parts $A,B$ are on the same side of $X$, such that the volume of $A$ is greater than the volume of $B$. If the expected value of the volume of $A$ can be expressed as $\frac{a}b\cdot\pi$, where $\gcd(a,b)=1$ find $a+b$.

I found this problem in a math competition archive and I was curious about how to solve this. I attempted to find the range of the possible values of $A$ but was unable to.

答:
By symmetry, we can suppose that the line along which $X$ and $Y$ intersect is vertical, and fix $X$ to be in a certain direction, so that we only consider the possible angles of $Y$ as it rotates around our polar axis.

Looking at the sphere from above, it should be clear that the volumes of the pieces only depend on the angles between them, and scale linearly with the central angle between the two pieces. In the following picture, we are looking at the sphere from above, with $X$ the red horizontal line and $Y$ the blue line (which could be rotated at any angle),

If the larger of the two angles between $X$ and $Y$ is $\theta$, then the volume of $A$ will just be the volume of the sphere times the fraction of $360^\circ$ that $\theta$ take up, i.e. $\frac{4\pi}{3}\cdot \frac\theta{2\pi}=\frac{2\theta}{3}$.

Since  $\theta$ is uniform on $[\pi/2,\pi]$ (because the "true" angle between $X$ and $Y$ uis uniform mon $[0,\pi]$ but we always choose the larger half), this means that the expected value of $\theta$ is $\frac{3\pi}4$, and so the expected volume of $A$ is $\frac{\pi}{2}$. Hence, $a=1, b=2, a+b=3$.

Title: 3D Expected Value GEO


问:
If $n$ is even, prove $P(x)$ cannot have an inverse function
If $P(x)$ is a polynomial of degree $n$, how can I prove that if $n$ is even, $P(x)$ cannot have an inverse function.

答:
This answer assumes that you are talking about polynomials over the real numbers.

Hint.

Consider the degree of $P'(x)$.

Then note that every polynomial of odd degree must have one real zero.

Suppose $P$ has an inverse $Q$ so that $P(Q(x))=x$ on $\mathbb{R}$. Take the
derivatives on both sides to get a contradiction.


问:
How to find the amount to added every month or year to get the required amount after certain years?
I want to do a Java application for which after giving the current savings, and the rate of interest and  and required amount after specified  no of years, it has to show how much a person has to earn a month or year to achieve his target.
I'm aware of the formula $$\text{present value}={\text{future value} \over (1+\text{interest rate})^{\text{no. of years}}}$$
Now after
getting present value, should I divide present value by no. of months in those years, to get the amount he has to invest per month ?</p>

答:
Take a look at the geometric mean. It can be used to find average growth rates. From the average rate and the given amount you can easily derive the average increase per time period.

EDIT
Maybe you're looking for this:

$$(1+\text{interest rate})^{\text{no. of years}}=\frac{\text{future value}} {\text{present value}}$$

$$\text{interest rate}=\sqrt[\leftroot{-2}\uproot{2}\text{no. of years}]{\frac{\text{future value}} {\text{present value}}} - 1$$


答:
This may be an overly complicated solution, but:

Without loss of generality $P$ is monic, so $P(x)\to\infty$ as $x\to\pm\infty$. Fix $M$ greater than $|P|$'s value at any of its turning points, as there are finitely many of these. Applying the intermediate value theorem to $1/P$ either side of the smallest inteval containing $P$'s real roots, $P(x)=M$ has roots both side of this integral.

Title: If $n$ is even, prove $P(x)$ cannot have an inverse function

Body: If $P(x)$ is a polynomial of degree $n$, how can I prove that if $n$ is even, $P(x)$ cannot have an inverse function.


问:
What is the percentage of these numbers?
Consider the numbers $N$ of the form:

$2580\cdot x + 2236\cdot y$, with $x$ and $y$ nonnegative integers.

I have to calculate the density of these numbers below $10^5$.

I mean suppose $\rho (10^5)$ is the number of $N$'s below $10^5$.

Which is the value of  $\frac{\rho(10^5)}{10^5}$?

Is the brute force approach useful with a calculator?

$\gcd(2580,2236)=172$

Does it mean that roughly $1/172$ of nonnegative integers below $10^5$ has that form?

Now a question a bit harder:

what percentage of the $N$'s are also congruent to $10^m\pmod {41}$? Is this equivalent to ask what percentage of the multiples of $172$ are congruent to $10^m\pmod {41}$, for some positive $m$?

答:
Bezout's Lemma.  Every number of the form $2580x + 2236y$ is a multiple of $\gcd(2580,2236) =172$.  And every multiple of $172$ can be written of the forms $2580x + 2236y$.

So $N\in \mathbb Z$ is of the form $2580x + 2236y;x,y \in \mathbb Z\iff N$ is a multiple of $172$.

So you need to figure out what percentage of integers are multiples of $172$.

And $\frac 1{172} \approx 0.5814\%$ of all integers are multiples of $172$.

There are $\lfloor \frac {10^5}{172} \rfloor = 581$ multiples of $172$ less than $10^5$.

....

Okay Okay the second qestion:

One out of $41$ numbers are congruent to $K\pmod {41}$ for any $K$.

And so you want all numbers that are $x \equiv 10^5 \pmod {41}$ and $x \equiv 0 \pmod {172}$.

$\gcd(41,10^5)=1$ so by chinese remainder theorem there is one unique solutions $\pmod{41\cdot 172}$.

So the percentage is $\frac 1{41\cdot 172}=0.03387533875338753387533875338753\%$.

There will be around $\frac {10^5}{41\cdot 172} = 14.18$ such numbers less than $10^5$, so there will be either $14$ or $15$ such numbers.

If $k$ is the smallest such number then all the numbers will be of the form $41*172m + k$ for $m =0...... 13$.  That is $41*172 + k, 41*172*2 + k, 41*172*3 + k,   ..... 41*172*13 + k$.   The next number of the form will be $41*172*14 + k$ and that will be a $15$ potential value if $14*172*14 + k \le 10^5$.

To actually solve these numbers:

$10^2 \equiv 100 \equiv 18\pmod {41}$.  $10^3 \equiv 180\equiv 164+16\equiv 16\pmod {41}$.  $10^4 \equiv 160\equiv -4 \pmod {41}$.   So $10^5\equiv -40\equiv 1 \pmod{41}$.

So we need to solve $x\equiv 1\pmod {41}$ and $x \equiv 0\pmod{172}$.  And $21*41-5*172 =1$ so $-21*41 +1 \equiv 0 \pmod {172}$ and $-21*41 + 1 \equiv 1 \pmod {42}$.

So $k = -21*41 + 41*172+1 = 41*151 + 1$.

And $14*41*172 + 41*1451 + 1 > 10^5$ so there are $14$ such numbers.


问:
Max number of nodes in non-binary tree in summation form
Can some body help me to find total number of nodes in this tree network see the picture tree network, here Level-1 one has 4 nodes, and the next level has 3 and so 2, 2 is the minimum number of children a parent can have and there are maximum 3 levels allowed. What is the mathematical way of writing this in a summation form. for example the number of nodes in this example is 41 but how such a number is calculated through formula. If the nodes in the levels are not known then what would be its general formula?

答:
This post explains how to calculate an estimation of the total number of nodes in the tree, not the exact number.

If n is the number of levels in a similar tree, the number of nodes in the final level is n! (n factorial). The number of nodes in all the levels working backwards is as follows:

Level - Total Nodes

n - n! (same as saying n!/1!)

n-1 - n!/2! (total nodes in level n divided by 2)

n-2 - n!/3! (total nodes in level n divided by 2, then by 3)

n-3 - n!/4! (total nodes in level n divided by 2, then by 3, then by
4)

...

root node - n!/n! (equals 1)

If you add up the formulas for total nodes at each level and substitute n!/1! for n! since they are equal, you would get the following formula:

n!/1! + n!/2! + n!/3! + n!/4! + ... + n!/n!

take out the common factor n!

n!(1/1! + 1/2! + 1/3! + 1/4! + ... + 1/n!)

the stuff in the parentheses is similar to the following formula for the constant e:

e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ...

except it's missing the initial number 1, therefore the formula to approximate the total number of nodes in the entire tree is

n!(e-1) where n is the number of levels (depth) in the tree.

If you apply this formula to the tree in the original question, there are 4 levels so 4!(e-1) is about 41.2387, and the actual number of nodes is 41.


问:
When is it true that, if $AB^2-BH^2=AC^2-CH^2$, $AH$ is perpendicular to $BC$?
Consider triangle $\Delta ABC$, and point $H\in BC$.

For what triangles is the following true: if $AB^2-BH^2=AC^2-CH^2$, then $AH$ is perpendicular to $BC$.

(I do not know what the answer is – all triangles, no triangles, all but obtuse triangles...)

答:
Let $\alpha=\angle CHA$. Then we have
\begin{align*}
|AC|^2-|CH|^2&=|AH|^2-2|AH||HC|\cos\alpha\\
|AB|^2-|BH|^2&=|AH|^2-2|AH||BH|\cos(\pi-\alpha)\\
\end{align*}
It follows that $$|HC|\cos\alpha=|BH|\cos(\pi-\alpha)=-|BH|\cos\alpha$$
As $|HC|,|BH|>0$ this implies that $\cos\alpha=0$, i.e. $\alpha=\frac{\pi}{2}$.

Picture:


问:
Is there an expression for the ith term of this sequence $1,1,2,1,2,3,1,2,3,4,1,2,3,4,5...?$
I'm trying to do some work in Excel and if I found a formula for this sequence it would help a lot. I don't particularly need to know why the formula works.

I have found the sequence here .

But there in no nice form for it.

Thanks!

答:
I've pulled this out of a comment, though it doesn't quite answer the question. There are some flexible ways of generating sequences in Excel which don't depend on a closed formula. If you need to reference a sequence within a formula, then you can generate the sequence (eg on another worksheet) and use a "lookup" formula to pick the term you need. There are ways of doing this (using a name for the cells containing the sequence) which can make the final formula easier to read and check.

For example, to generate the sequence you have asked about in this question, put 1 in cells A1 and B1, and use the formula =IF(B1>MAX($A1:A1),1,B1+1) in cell C1. Then copy this formula across the first row. I've left the formula in plain text, so you ought to be able to copy it. Obvious adjustments would apply for this to work in a different row, or in a column.

There are limitations on this technique, of course, but it generates shorter formulae, which can be easier to understand when you come back to them later.


答:
I'm assuming that the domain is $\mathbb{R}. $If $p$ is injective, then it should be increasing or decreasing, but its derivative is a polynomial of odd degree, hence its image contains both positive and negative values.

Remark: it doesn' work for arbitraty rings. In $\mathbb{F}_{2}[x]$, for example, the polynomial $x^2$ is a bijective function.


答:
Recall that a vector space V over a field F is a set where $av_1+bv_2\in V, a,b\in F, v_1,v_2\in V$. A vector space of functions, there is an additional "thing" that is a set X over which the function acts. That is, suppose X is any set, then a vector space V over F is the set of all functions $V=\{f:(f:X\rightarrow F)\}$.

Here, E is the set over which the functions act. Let F be the vector space and R be the field. Then the vector space $F=\{f:(f:E\rightarrow \mathbb R)\}$. Let E have n elements, i.e. $E=\{e_1,...,e_n\}$. The claim is that the set of all $$g_{e_i}(x)=\begin{cases}1&x=e_i\\0&x\ne e_i\end{cases}$$

is a basis of F.

First, show that any vector in F can be written as a linear combination of $g_{e_i}$. Notice that $f(x)=f(e_1)g_{e_1}(x)+...+f(e_n)g_{e_n}(x)$ for any function f. This is because f can take on the finitely many elements in E as inputs, and for each value of e, it has a real valued output. The vector space F consists of infinitely many functions that take the inputs e and outputs a real number for each. Notice the coefficients are real numbers, so they are in $\mathbb R$, the field. You may be thinking that it's weird for the function to depend on its values, but it should be fine by taking a leap of faith.

Next you also have to show that the vectors in the basis are linearly independent, i.e. that $a_1g_{e_1}(x)+...+a_ng_{e_n}(x)=0$ implies that $a=0$. Suppose for the sake of contradiction that $a_j\ne 0$. But then plugging in $e_j$, you get $a_j$, which is not 0. Thus the coefficients must all be 0.

Title: Showing that the indicator function forms a basis

Body:
Let $E$ a non-empty finite set and $F(E)$ the set of all functions $f: E \to \mathbb{R}$. Define for every $f \in E$ the function $X_f : E \to \mathbb{R}$ with $$x \mapsto \begin{cases} 1 \text{, if $x=f$} \\ 0, \text{otherwise}\end{cases}$$ show that $(X_{f_1}, \dots ,X_{f_{n}})$ is a basis for $F(E)$ if the set $E$ has $n$ elements and $\{f_1,\dots, f_n\} = E.$

The notation here is very confusing it seems that the writer wants me to show that the sequence of the functions forms a basis for $F(E).$ Also isn't this function the indicator function or am I missing something here? Any hints on how to start showing something like this would be appreciated. I know that in order to show that some set forms a basis for a vector space I would need to show that it spans the space and that it's elements are linearly independent.


答:
$p\implies q$ is considered to have a value of $true$ in one of these two cases

$p$ is false

or

$p$ is true and $q$ is true

So e.g. if one says this statement:

if the Earth is larger than the Sun, then 5 is even

this statement is true even though both $p$ and $q$ are false here.

But if one says:

if the Sun is larger than the Earth, then 5 is even

this statement is false.

Now to your question. You should write

$p\implies q$

when the implication as a whole is $true$

i.e. in one of the two cases mentioned above.

By the way before studying first-order logic it might be better to learn propositional logic.

Title: When can we use the implies arrow?

Body:
So I'm a complete beginner to logic and set theory and I've been trying to learn a bit about conditional statements and logic however I'm quite confused on if then statements.
For example if $p\implies q$
I saw that this means that if p is true then q must be true also
My confusion is when can we use this arrow can we only use it if the statement is true?

As an example if I had the statement if $x^2>0
$ then $x<0$ could I write $p\implies q$ where p=$x^2>0
$ and $q=
 x<0$

Or would it be incorrect to do so as this statement is false if this is not true when we see the implies arrow in theorems and proofs how am I to know that the statement is true or false?.

Thanks in advance.


问:
$\mathbb{Z}[\sqrt{11}]$ is norm-euclidean
I'm trying to show that $\mathbb{Z}[\sqrt{11}]$ is Euclidean with respect to the function $a+b\sqrt{11} \mapsto|N(a+b\sqrt{11})| = | a^2 -11b^2|$

By multiplicativity, it suffices to show that $\forall x \in \mathbb{Q}(\sqrt{11}) \exists n \in \mathbb{Z}(\sqrt{11}):|N(n-x)| < 1$

For the analogous statement for $\mathbb Z [\sqrt6]$, it worked by considering different cases, so I tried to do the same thing here. Here is what I did so far:

Let $x+y\sqrt{11} \in \mathbb Q (\sqrt{11})$

Case 1: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \frac{1}{\sqrt{11}}$, then we can choose such a $b$ and a $a \in \mathbb Z$ s.t. $|x-a| \leq \frac{1}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \frac{1}{\sqrt{11}}$

Case 2: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{5}{44}}$ Then we have $1 < 11 (y-b)^2 < \frac{5}{4}$, so we can choose $a \in \mathbb Z$ such that $\frac{1}{2} \leq |x-a| \leq 1$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on suppose $\forall b \in \mathbb Z: |y-b| > \sqrt{\frac{5}{44}}$

Case 3: Suppose there exists a $b \in \mathbb Z$ s.t. $|y-b| < \sqrt{\frac{2}{11}}$ Then we can choose $a \in \mathbb Z $ s.t. $1 \leq |x-a| \leq \frac{3}{2}$, then we have $|N(x+y\sqrt{11}-(a+b\sqrt{11}))| < 1$

From now on, we may suppose that $|y-b| > \sqrt{\frac{2}{11}}$.

This is where I'm stuck. I tried choosing $b \in \mathbb Z$ s.t. $\frac{1}{2} \geq |y-b| > \sqrt{\frac{2}{11}}$, but then I run into problems, whether I choose $a \in \mathbb Z$ s.t. $1 \leq |x-a| \leq \frac{3}{2}$ or s.t. $ \frac{3}{2} \leq |x-a| \leq 2$

答:
Division algorithm for Euclidean domain $\mathbb{Z}[\sqrt{11}]$

While Oppenheim (1934) proved that there is a division algorithm for $\mathbb{Z}[\sqrt{11}]$. No examples was given explicitly. Inspired by his proof, one suitable division algorithm was identified. Hence, one could compute in $\mathbb{Z}[\sqrt{11}]$.

Let $R = \mathbb{Z}[\sqrt{11}] = \{ s+t\sqrt{11} : s,t \in \mathbb{Z} \} $ be a quadratic integer ring. Let $\alpha = s+t\sqrt{11} \in R$. Define norm $N: R \setminus\{0\} \rightarrow \mathbb{Z}_{\geq 0}$ by setting $\alpha \mapsto |s^2-11t^2|$, which is well defined in $R$. Proof omitted. Norm $N$ is also well defined in $\mathbb{Q}[\sqrt{11}]$. Proof omitted. There exists $\alpha/\beta \in \mathbb{Q}[\sqrt{11}]$ $=$ $\{ s+t\sqrt{11} : s,t \in \mathbb{Q} \} \supset R $ with a numerator and a denominator $\alpha, \beta \in R$, $\beta \ne 0$. Proof omitted.

Suppose the nearest lattice point to point $\alpha / \beta$ is $m+n\sqrt{11} \in R$ (i.e. $m,n \in \mathbb{Z}$) with $\alpha / \beta = (m+n\sqrt{11}) + (a+b\sqrt{11})$, $a,b \in \mathbb{Q}$, $-\frac{1}{2} \leq a \leq \frac{1}{2}$ and $-\frac{1}{2} \leq b \leq \frac{1}{2}$. We will consider lattice point $(m,n) = m+n\sqrt{11}$ and five other nearby lattice points. They, $(m+x,n+y)$ with $(x,y) \in \{ (0,0), (1,0), (-1,0), (2,0), (2,1), (-5,2) \} $, are candidate quotients $q$ to a division algorithm. By considering only positive values of $a,b$ such that $0 \leq a \leq \frac{1}{2}$ and $0 \leq b \leq \frac{1}{2}$, similar arguments could be made for the three other quadrants by symmetry. With $q=(m+x,n+y)$, rewrite $\alpha = q \beta +r$ as:

\begin{align}
    \alpha / \beta &= q + r/\beta    \\
    &= \big( (m+n\sqrt{11}) + (x+y\sqrt{11}) \big) + \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)
\end{align}

Inequality $N(r) < N(\beta)$ is true $\iff$ $N(r)/N(\beta) < 1$ $\iff$ $N(r/\beta) < 1$ $\iff$ $N \big( (a+b\sqrt{11}) - (x+y\sqrt{11}) \big)$ $= |(x-a)^2 - 11(y-b)^2| < 1$
. Proof omitted.

Case 1: Consider lattice point $(m,n)$, $0 \leq a^2 \leq \frac{1}{4}$ and $0 \leq 11b^2 < 1$. The maximum possible difference occurs when $a^2 = 0$ and $11b^2 = 1^-$. Therefore $|a^2 - 11b^2| < 1$ and we have:

\begin{align}
    0 &\leq a \leq \frac{1}{2}    \\
    0 &\leq b < \sqrt{ \frac{1}{11} } \approx 0.302
\end{align}

Case 2: Consider lattice point $(m+1,\ n)$, $\frac{1}{4} \leq (1-a)^2 \leq 1$ and $1 < 11b^2 < \frac{5}{4}$. The maximum possible difference occurs when $(1-a)^2 = \frac{1}{4}$ and $11b^2 = \frac{5}{4}^-$. Therefore $|(1-a)^2 - 11b^2| < 1$ and we have:

\begin{align}
    0 &\leq a \leq \frac{1}{2}    \\
    \sqrt{ \frac{1}{11} } &< b < \sqrt{ \frac{5}{44} } \approx 0.337
\end{align}

Case 3: Consider lattice point $(m-1,\ n)$, $1 \leq (1+a)^2 \leq \frac{9}{4}$ and $\frac{5}{4} < 11b^2 < 2$. The maximum possible differences are $2^- - 1 < 1$ and $\frac{9}{4} - \frac{5}{4}^+ < 1$. Therefore $|(1+a)^2 - 11b^2| < 1$ and we have:

\begin{align}
    0 &\leq a \leq \frac{1}{2}    \\
    \sqrt{ \frac{5}{44} } &< b < \sqrt{ \frac{2}{11} } \approx 0.426
\end{align}

Case 4: Consider lattice point $(m+2,\ n)$, $\frac{9}{4} \leq (2-a)^2 < 3$ and $2 < 11b^2 \leq \frac{11}{4}$. The maximum possible difference occurs when $(2-a)^2 = 3^-$ and $11b^2 = 2^+$. Therefore $|(2-a)^2 - 11b^2| < 1$ and we have:

\begin{align}
    0.270 \approx 2-\sqrt{3} &< a \leq \frac{1}{2}    \\
    \sqrt{ \frac{2}{11} } &< b \leq \frac{1}{2}
\end{align}

Case 5: Consider lattice point $(m+2,\ n+1)$, $3 < (2-a)^2 \leq 4$ and $3 < 11(1-b)^2 < 4$. The maximum possible difference are $4^- - 3^+ < 1$ and $4 - 3^+ < 1$. Therefore $|(2-a)^2 - 11(1-b)^2| < 1$ and we have:

\begin{align}
    0 &\leq a < 2-\sqrt{3}    \\
    0.397 \approx 1-\sqrt{ \frac{4}{11} } &< b < 1-\sqrt{ \frac{3}{11} } \approx 0.478
\end{align}

Case 6: Consider lattice point $(m+2,\ n+1)$ again with different ranges of $a,b$, $\frac{9}{4} \leq (2-a)^2 < \frac{15}{4}$ and $\frac{11}{4} \leq 11(1-b)^2 < \frac{13}{4}$. The maximum possible differences are $\frac{13}{4}^- - \frac{9}{4} < 1$ and $\frac{15}{4}^- - \frac{11}{4} < 1$. Therefore $|(2-a)^2 - 11(1-b)^2| < 1$ and we have:

\begin{align}
    0.064 \approx 2 - \sqrt{ \frac{15}{4} } &< a \leq \frac{1}{2}    \\
    0.456 \approx 1-\sqrt{\frac{13}{44}} &< b \leq \frac{1}{2}
\end{align}

Case 7: Consider lattice point $(m-5,\ n+2)$, $25 \leq (5+a)^2 < \frac{103}{4}$ and $\frac{99}{4} \leq 11(2-b)^2 < \frac{103}{4}$. The maximum possible difference occurs when $(5+a)^2 = \frac{103}{4}^-$ and $11(2-b)^2 = \frac{99}{4}$. Therefore $|(5+a)^2 - 11(2-b)^2| < 1$ and we have:

\begin{align}
    0 &\leq a < \sqrt{ \frac{103}{4} } - 5 \approx 0.074    \\
    0.470 \approx 2 - \sqrt{\frac{103}{44}} &< b \leq \frac{1}{2}
\end{align}

All boundary cases are ignored, since $\sqrt{ \frac{1}{11} }$, $\sqrt{ \frac{5}{44} }$, $1-\sqrt{ \frac{4}{11} }$, $\sqrt{ \frac{2}{11} }$, $1-\sqrt{\frac{13}{44}}$, $2 - \sqrt{\frac{103}{44}}$, $1-\sqrt{ \frac{3}{11} }$, $2 - \sqrt{ \frac{15}{4} }$, $\sqrt{ \frac{103}{4} } - 5$, $2-\sqrt{3}$ $\notin \mathbb{Q}$.

Since the quadratic integer ring $\mathbb{Z}[\sqrt{11}]$ has (a) a well defined norm and (b) a division algorithm $\alpha = q \beta + r$ such that either $r=0$ or $N(r)<(b)$, for all numerators and denominators $\alpha, \beta \in \mathbb{Z}[\sqrt{11}]$, $\beta \ne 0$ with some quotient and remainder $q, r \in \mathbb{Z}[\sqrt{11}]$, it is a Euclidean domain.    $\blacksquare$


问:
How to go from this summation to this approximation?
https://learn.fmi.uni-sofia.bg/pluginfile.php/194197/mod_resource/content/2/Telephone_numbers.pdf

I'm a high school student investigating on telephone numbers (involution numbers) depicted as T(n). If you read this link, you'll have a good understanding on what I'm talking about. Under the Mathematical Properties subheading in this pdf, there's a section on the summation formula and its approximation. I have no idea how this step works. It mentions that it uses Stirling's approximation. Anyone willing to help? For a certain n value (has to be a natural number) it says it goes from:

$\sum_{k=0}^{[n/2]} \frac{n!}{2^k(n-2k)!k!}$

to

$(\frac{n}{e})^{(\frac{n}{2})} \frac {e^\sqrt n}{(4e)^\frac{1}{4}}$

Note that the summation formula has [n/2], but that's because for odd numbers it's [(n-1)/2]. So if n=5 then k=2 is the largest k.
Maybe it's easier to think of it through two separate formula's according to parity, one with [n/2] and another with [(n-1)/2]

I have no idea how the k disappears. Please help

答:
The following holds:
\begin{align*}
T(n)=\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{n!}{2^kk!(n-2k)!}
\sim\frac{1}{\sqrt{2}}n^{\frac{n}{2}}\exp\left(-\frac{1}{2}n+n^{\frac{1}{2}}-\frac{1}{4}\right)\tag{1}
\end{align*}

The validity of the asymptotic expansion (1) can be obtained in two steps:

The first step is to derive an exponential generating function $\mathcal{T}(z)=\sum_{n=0}^\infty T(n)\frac{z^n}{n!}$.

The generating function $\mathcal{T}(z)$ encodes the wanted information of the coefficients $T(n)$ which can be extracted using complex analytic methods. Here we can apply a technique from Hayman to obtain the analytic expansion.

We start with the first step which is in fact the easy part.

Exponential generating function $\mathcal{T}(z)$:

We obtain from the left-hand side of (1):
\begin{align*}
\color{blue}{\mathcal{T}(z)}&=\sum_{n=0}^\infty T(n)\frac{z^n}{n!}\\
&=\sum_{n=0}^\infty \sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{2^kk!(n-2k)!}z^n\tag{2.1}\\
&=\sum_{n=0}^\infty\sum_{{2k+l=n}\atop{k,l\geq 0}}\frac{1}{2^kk!l!}z^n\tag{2.2}\\
&=\sum_{k=0}^\infty\frac{z^{2k}}{2^kk!}\sum_{l=0}^\infty\frac{z^l}{l!}\tag{2.3}\\
&=\exp\left(\frac{z^2}{2}\right)\exp\left(z\right)\tag{2.4}\\
&\,\,\color{blue}{=\exp\left(\frac{z^2}{2}+z\right)}\tag{3}
\end{align*}

Comment:

In (2.1) we take $T(n)$ from (1) and cancel $n!$.

In (2.2) we write the coefficient of the Cauchy product in the form $$\sum_{k=0}^n a_kb_{n-k}=\sum_{{k+l=n}\atop{k,l\geq 0}}a_kb_l$$

In (2.3) we write the series as product of two series.

In (2.4) we use the series expansion $e^z=\sum_{n=0}^\infty \frac{z^n}{n!}$.

Hayman's method:

We can now apply to the exponential generating function in (3) the technique published by Hayman (1956) to derive the asymptotic expansion in (1). This is shown in detail in this answer.

Notes:

Some more detailed information regarding Hayman's method is given in chapter VIII.5 Admissibility in Analytic combinatorics by P. Flajolet and R. Sedgewick.

A more general case was proven by H. S. Wilf in The Asymptotics of $e^{P(z)}$ and the number of elements of each order in $S_n$
from 1986. In this proof we can also find a nice application of the Lagrange inversion theorem.


问:
Prove that $\inf_{n\in\mathbb{N}}\sup_{k\geq n}(a_k+b_k)\leq \inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$
Let $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ be two sequences of $\color{red}{\overline{\mathbb{R}}:=\mathbb{R}\cup\{-\infty,\infty\}}$. Suppose that the sum $a_n+b_n$ exists for all $n\in\mathbb{N}$. Prove that if the sum $\inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$ exists, then

$$\inf_{n\in\mathbb{N}}\sup_{k\geq n}(a_k+b_k)\leq \inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$$

I want to prove the above inequality without using the notion of limits! Remember that the following operations ARE NOT defined: $(\infty)-(\infty)$, $(-\infty)+(\infty )$, $(\infty )+(-\infty )$ and $(-\infty)-(-\infty )$.

The order relation $\leq $ is defined in the following way: given any $x,y\in\overline{\mathbb{R}}$ we say that $x\leq y$ if, and only if, at least one of the following propositions is true:

If $x,y\in\mathbb{R}$, then $x\leq y$ (here $\leq$ is the usual order of the real numbers)

$y=\infty$

$x=-\infty$

I tried to use the inequalities below, but I couldn't solve that problem.

If the sum $\sup_{k\geq n}a_k+\sup_{k\geq n}b_k$ exists, then $\sup_{k\geq n}(a_k+b_k)\leq \sup_{k\geq n}a_k+\sup_{k\geq n}b_k$ for all $n\in\mathbb{N}$

If the sum $\inf_{n\in\mathbb{N}}a_n+\inf_{n\in\mathbb{N}}b_n$ exists, then $\inf_{n\in\mathbb{N}}a_n+\inf_{n\in\mathbb{N}}b_n\leq \inf_{n\in\mathbb{N}}(a_n+b_n)$

Thank you for your attention!

答:
Let us denote
$$
\newcommand{\newsup}{\mathop{\smash{\mathrm{sup}}}}
\begin{align}
 A &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}a_k \, ,\\
 B &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}b_k \, , \\
 C &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}(a_k+b_k) \, .
\end{align}
$$
If $A = +\infty$ or $B = +\infty$ then $A+B=+\infty$ (since the sum is assumed to be defined), so that $C \le A+B$ holds.

Now consider the case that $A, B \in \Bbb R \cup \{ - \infty \}$. The idea is to show that
$$ \tag{*}
 \forall x, y \in \Bbb R: \left(x > A \text{ and } y > B \implies C \le x + y \right)\, .
$$
It is not difficult to see that $(*)$ implies $C \le A+B$:

If $A = -\infty$  then $C < x + y$ for fixed $y \in \Bbb R$ and all $x \in \Bbb R$, so that necessarily $C = -\infty$.

Similarly if $B= -\infty$.

Otherwise both $A$ and $B$ are real numbers and $(*)$ implies $C \le (A+\epsilon)+( B+\epsilon)$ with arbitrarily small $\epsilon > 0$.

It remains to prove $(*)$: If $x > A$ and $y > B$ then, using the definition of the infimum,
$$
\exists n_1 \in \Bbb N: \newsup_{k\geq n_1}a_k < x \, , \\
\exists n_2 \in \Bbb N: \newsup_{k\geq n_2}b_k < y \, .
$$
It follows that
$$
a_k + b_k \le x+y
$$
for all $k \ge \max(n_1, n_2)$ and therefore
$$
 C \le \sup_{k\geq \max(n_1, n_2)}(a_k+b_k) \le x +y \, .
$$


问:
Natural Deduction proof for $\forall x \neg A \implies \neg \exists xA$
$\forall x \neg A \implies \neg \exists xA$

I won't ask you to solve this for me, but can you please give some guiding lines on how to approach a proof in NDFOL?

There are many tricks that the TA shows in class, that I could not dream of...

P.S. I managed to proof $\neg \exists xA \implies \forall x \neg A$ but could not get on from there.

Thanks!

After the proposed answer, let me see if I got this correct:

$\exists x A \implies \exists x A$ (axiom assumption)

$\exists x A \implies A$ (from 1)

$\exists x A, \forall x \neg A \implies \forall x \neg A$ (axiom assumption)

$\exists x A, \forall x \neg A \implies \neg A$ ($\forall$ extraction, from 3)

$\forall x \neg A \implies \neg \exists x A$ (from 2,4)

Am I correct?

I could not understand the justification going from (1) to (2)

答:
To prove an implication, the general guideline for constructing formal proofs is: Assume the premise and the negation of the consequence, and derive a contradiction.

In your present case:

Assume $\exists x A(x)$. By Existential Instantiation, we have $A(t)$ for some (unspecified but fixed) $t$.

Assume $\forall x \neg A(x)$. By Universal Instantiation, we have $\neg A(t)$.

Now $A(t)$ and $\neg A(t)$ combine into a contradiction $\bot$.

We use Negation Introduction on the open assumption $\exists x A(x)$ to conclude $\neg \exists x A(x)$.

Finally, by Implication Introduction, we conclude the desired $\forall x \neg A(x) \implies \neg \exists x A(x)$ holds without any assumption.

Q.E.D.

Addressing OP's efforts:

As you see, there is a difference in our notations. I have parametrised $A$ as $A(x)$, while you haven't. However, this is crucial for the inference of $(2)$ from $(1)$. The expression $A(t)$ contains a $t$, which we can think of as an arbitrary "witness" of $\exists x A(x)$. It is this witness $t$ we apply Universal Instantiation / $\forall$ Extraction to: Since $\forall x \neg A(x)$, in particular $\neg A(t)$, where $t$ is the witness to $\exists x A(x)$.

This working with witnesses requires some practice, and even then one can sometimes mix things up. They are however crucial for the validity of the reasoning, so be sure to try and derive some more "trivialities" containing existential quantifiers!

A final remark (inspired by the comment by Peter Smith) is that in place of where you wrote "axiom", it's better to write one of "assumption" or "hypothesis", because these words have different meanings in mathematical lingo.


问:
How do I go about calculating the perimeter of a triangle that consists of three vertices of a cube?
I'm given a unit cube. The task is to find the perimeter of triangle ACE. I have no other information, but considering it's a unit cube that should be more than enough.

Am I looking for an answer like this by any chance?

Anyways, cheers!

答:
The question would have been more specific if the annotation of the vertices were given.
So, the edges of the triangle can either be 1. a surface diagonal i.e., the diagonal connecting the running along the surface of the cube, 2. a body diagonal i.e.,  the diagonal connecting the vertices at opposite ends of the cube 3. an edge.
Egde = a =  1

surface diagonal = s = $\sqrt{a^2 + a^2}$ = $\sqrt{1^1 + 1^1}$ = $\sqrt{2}$

body diagonal =$ \sqrt{a^2 + s^2} = \sqrt{1^2 + 2} = \sqrt{3}$

see which of these three are the edges of your triangle and add them.


问:
Linear Maps and component functions
Prove or disprove the following.

Let $f$: $\mathbb{R}^2$ $\to$ $\mathbb{R}^2$.

be a map with component functions $g$ and $h$ such that $f(x,y)$ = $(g(x,y),h(x,y))$. If $f$ is linear, then $g$ and $h$ are linear as well.

What's the best method to prove/disprove the following? Would showing f be linear in terms of g and h work?

答: Let $\{\pi_1,\pi_2\}$ the dual basis of  $\{(1,0), (0,1)\}$.  Then  $g=\pi_1\circ f$ and $h=\pi_2\circ f$.Then use that the composition of linear maps is linear as well.

Title: Linear Maps and component functions


问:
Regarding this integral function
I'm currently stuck with this problem:

Let $g:(a,b]\to \mathbb{R}$ be a continuous function such that the improper integral $\int_a^b|g(x)|dx$ exists. Prove that:

$$G:(a,b]\to \mathbb{R} $$
$$r\mapsto\int_r^b|g(x)|dx$$

Extends to a continuous function $G:[a,b]\to \mathbb{R}$ such that $G(a)=\int_a^b|g(x)|dx$

I don't quite follow where to begin; my first attempt is to try and utilize the definition of continuity and see where to go from there. If that's correct can you give a hint how to begin writing the proof? Should I use the continuity on the limits of the integral? If is that so, how can does one do it? Those are the main questions that I'm struggling with, and that's the main reason I don't know where to begin.

I'll write down the definition I have for improper integral according to this problem:

Let $g:(a,b]\to \mathbb{R}$ such that is a continuous function, the improper integral is:

$$\int_a^bg(x)dx=\lim_{r\to a}\int_r^bg(x)dx$$

Any hints will be appreciated, thank you for your time.

答:
Hint.

If you write
$$
\int_r^bg(x)dx=\int_a^b1_{[r,b]}(x)g(x)dx
$$
where $1_A$ denotes the indicator function of a set $A$,
your limit follows from the dominated convergence theorem.

Title: Regarding this integral function


问:
Filters in Topology (Converge)
Let $X$ be an infinite set, $\mathscr{F}$ the filter on $X$ generated by the filter base consisting of all complements of finite sets. To which points does $\mathscr{F}$ converge if $X$ is given the cofinite topology?

I would suspect the answer to be every point in $X$. My reasoning, though may be wrong, is given below.

Let $\mathscr{C}$ denote the filter base, i.e. $\mathscr{C}:=\{A^c: A\subset X\text{ is finite}\}$ and $\mathscr{U}_x$ denote the set of all nhbds of $x$. The filter $\mathscr{F}$ converges to $x\in X$ iff each $U\in \mathscr{U}_x$ contains some $C\in\mathscr{C}$.

Notice,

$$C\subseteq U\iff U^c\subseteq C^c.$$

By definition of the topology, the set $U^c$ is finite, whereas by definition of the filter base, the set $C^c$ is infinite. Therefore, each $U\in \mathscr{U}_x$ contains some $C\in\mathscr{C}$.

答: Note that $\mathscr{C}$ is already a filter, so $\mathscr{F}=\mathscr{C}$. It’s true that $\mathscr{C}$ converges to every point of $X$, but there is an error in your argument: for any $C\in\mathscr{C}$ the set $C^c$ is by definition finite, not infinite. But since you haven’t actually picked any set $C\in\mathscr{C}$, this doesn’t say anything about a particular $U\in\mathscr{U}_x$. What you really should say here is that if $U\in\mathscr{U}_x$, then $X\setminus U$ is finite, so $U\in\mathscr{C}$. In fact, $\mathscr{U}_x=\{C\in\mathscr{C}:x\in C\}$ for each $x\in X$, and therefore $\mathscr{U}_x\subseteq\mathscr{C}$.

Title: Filters in Topology (Converge)


问:
A theorem about the Poisson Point process.
In the proof of the Levy-Khintchine theorem, I saw a theorem about the Poisson
point process.

The theorem states that if $\Pi$ is a poission point process on $S$ with
intensity measure $\mu.$ Let $f:S\rightarrow\lbrack0,\infty)$ be a measurable
function. And define
$$
Z=\int_{S}f\left(  x\right)  \Pi\left(  dx\right)
$$

We have the following,

$$
E\left[  Z\right]  =\int_{S}f\left(  x\right)  \mu\left(  dx\right)
$$

and if $E\left[  Z\right]  <\infty,$ then,

$$
Var\left(  Z\right)  =\int_{S}f\left(  x\right)  ^{2}\mu\left(  dx\right)
$$

How should I proof this? I'm thinking about started the proof by assuming $f$ is a
simple function, and apply DCT somehow? Are there any proofs for this theorem?

答:
The proof is given in Proposition 19.5 in Lévy Processes and Infinitely Divisible Distributions by Ken Iti Sato. It goes along the lines:

Show that $Z$ follows a Compound Poisson Process with characteristic function
$$
\varphi_Y(z):={\rm E}[e^{izY}]=\exp\left(\int_S(e^{izf(x)}-1)\,\mu(\mathrm dx)\right),\quad z\in\mathbb{R}.
$$

Use the derivatives of the characeristic function to obtain expressions of the first and second moment, i.e.
$$
i{\rm E}[Y]=\frac{\mathrm d}{\mathrm dz}\varphi_Y(z)\bigg|_{z=0}=i\int_Sf(x)\,\mu(\mathrm dx)
$$
and
$$
i^2{\rm E}[Y^2]=\frac{\mathrm d^2}{\mathrm dz^2}\varphi_Y(z)\bigg|_{z=0}=i^2\int_Sf(x)^2\mu(\mathrm dx)+\left(i\int_Sf(x)\,\mu(\mathrm dx)\right)^2.
$$

Use these expressions to find ${\rm E}[Y]$ and ${\rm Var}(Y)$.


问:
I'm trying out this problem of logic but I'm stuck "Simplify ¬(w∨(n∧¬w)) to ¬w∧¬n"
Here is my work so far, I would appreciate the help</p>
```<code> ¬(w∨(n∧¬w))
 ¬w∧¬(n∧¬w) (De Morgan's)
 ¬w∧(¬n∨¬¬w) (De Morgan's)
 ¬w∧(¬n∨w) (Double Negation)
</code>```

答:
```<code>¬(p∨(q∧¬p))
¬p∧¬(q∧¬p)
¬p∧(¬q∨¬¬p)
¬p∧(¬q∨p)
(¬p∧¬q)∨(¬p∧p)
(¬p∧¬q)∨(p∧¬p)
(¬p∧¬q)∨F
¬p∧¬q
</code>```


问:
General solution of $\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}?$

How to find the general solution of $$\cos (2\arctan x) = \frac{{1 - x^2 }}{{1 + x^2 }}$$

答:
You can start by substituting both sides into $\arccos()$ function:
$$\arccos(\cos(2 \arctan x)) = \arccos\Biggl(\frac{1-x^2}{1+x^2}\Biggl)$$ $$2 \arctan x =\arccos\Biggl(\frac{1-x^2}{1+x^2}\Biggl)$$
Then you can take the derivative of both sides: $$\begin{align} \ \frac{2}{1+x^2} &=\ \frac{1}{\sqrt{1-\Biggl(\frac{1-x^2}{1+x^2}\Biggl)^2}} \frac{4x}{(1+x^2)^2} \\
&=\ \sqrt{\frac{(1+x^2)^2}{4x^2}} \frac{4x}{(1+x^2)^2} \\ \end{align}$$
If you organize the right-hand side of the equation, you eventually get:
$$\bbox[yellow] {\frac{2}{1+x^2}=\frac{2}{1+x^2}}$$
This means the equation holds for all real $x$ values.


问:
Examples for existence proofs using the parity of fixed points of involutions
On Wikipedia, I have seen a nice proof of Zagier for the existence of representations of primes as sum of two squares:

http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares#Zagier.27s_.22one-sentence_proof.22

The proof uses the (trivial) fact that the number of fixed points of an involution of a finite set has the same parity as the cardinality of the set itself.

Do you know any other examples for existence proofs using the parity of fixed points of involutions?

答:
A more general fact is that the number of fixed points of the action of a finite $p$-group on a finite set has the same cardinality $\bmod p$ as the set itself. This is used, for example, to prove that finite $p$-groups have nontrivial center and it is also part of the proof of the Sylow theorems.

While this isn't an existence proof, it can also be used to give a short proof of Fermat's little theorem: given a positive integer $a$, consider the action of $\mathbb{Z}/p\mathbb{Z}$ by cyclic permutation on the set of words of length $p$ on an alphabet of size $a$. There are $a$ fixed points and $a^p$ words, hence $a^p \equiv a \bmod p$.


问:
General solution of differential equation of order 3
Please ,how to find that the general solution of  $u'''(t)=e(t) , t\in [0,1]$ is given by
$u(t)=c_0+c_1t+c_2 t^2 +\frac12 \int_0^t (t-s)^2 e(s) ds$

$e:(0,1)\rightarrow \mathbb{R}$, and $e\in L(0,1)$.

i think that it is general homogeneous solution + particular solution ,

the general homogeneous is $c_0+c_1 t+ c_2 t^2$ , but i dont know how to finde that the paricular solution is $\frac12 \int_0^t (t-s)^2 e(s)ds $

Please;

Thank you

答:
It suffices to see that the strange-looking integral is a particular solution. You probably came up with the following type of expression, or at least you should agree it is an obvious way to go:

$$u_p(t)=\int_0^t\int_0^v\int_0^u e(s)dsdudv.\tag{1}$$

Now this is equal to

$$\int_0^te(s)\cdot{\rm Area}(\{(u,v):s\le u\le v\le t\})ds=\int_0^t\frac{(t-s)^2}{2}e(s)ds. \tag{2}$$

How did we get this? First off, the region of points $(u,v)$ such that $s\le u\le v\le t$ (where $s$ and $t$ are fixed) is a right triangle (try plotting some examples to see this) with sides each of length $t-s$, so its area is $\frac{1}{2}(t-s)^2$. But the more substantial formula is the following:

$$\iint\cdots\int_D f(x_0)dx_0dx_1\cdots dx_n=\int f(x_0)\cdot{\rm Vol}(\{(x_0,x_1,\cdots,x_n)\in D\})dx_0 \tag{3}$$

(under suitable hypotheses most likely). This is a "continuous" generalization of a discrete version

$$\sum_{(x,y)\in A}f(x)=\sum_{x\in X}f(x)\,\#\{(x,y)\in A\} \tag{4}$$

(Where $A\subseteq X\times Y$.)

Anyway I presume most of the above is not relevant to you. If you want a way to derive $(2)$ from scratch without knowing ahead of time what to look for, you will need to familiarize yourself with a technique to change double integrals like $\int_0^v\int_0^u e(s)dsdu$ into single integrals (using by-parts integration), and apply it twice to go from $(1)$ to $(2)$.

If you just want to check that $(2)$ is a particular solution, then you can straight-up differentiate the given function three times and check that the result is $e(t)$, using the general formula

$$\frac{d}{dt}\int_0^t f(t,s)ds=f(t,t)+\int_0^tf_t(t,s)ds\tag{5}$$

(and this formula can be derived using the chain rule + fundamental theorem of calculus).

Going from $(1)$ to $(2)$ with by-parts integration: Alright, first let's look at

$$\int_0^v\int_0^u e(s)dsdu. \tag{6}$$

Use by-parts ($X=\int_0^u e(s)ds$ and $Y=u$) to get

$$\int_0^v XdY=[XY]_0^v-\int_0^v YdX=v\int_0^v e(s)ds-\int_0^v ue(u)du=\int_0^v(v-s)e(s)ds. \tag{7}$$

Thus (using by-parts again)

$$\int_0^t\int_0^v\int_0^ue(s)dsdudv=\int_0^t\int_0^v(v-s)e(s)dsdv \tag{8}$$

$$=\int_0^tv\int_0^ve(s)dsdv-\int_0^t\int_0^vse(s)dsdv \tag{9}$$

($dY=vdv$ and $X=\int_0^ve(s)ds$ in the first integral, same by-parts as $(6)$-$(7)$ in second integral)

$$=\left[\frac{t^2}{2}\int_0^te(s)ds-\int_0^t\frac{v^2}{2}e(v)dv\right]-\left[\int_0^t (t-s)se(s)ds\right] \tag{10}$$

$$=\int_0^t\frac{t^2-s^2-2(t-s)s}{2}e(s)ds=\int_0^t\frac{(t-s)^2}{2}e(s)ds. \tag{11}$$

Going from $(1)$ to $(2)$ with reparametrization: The region of integration in ${\bf R}^3$ is

$$D=\{(s,u,v):0\le s\le u\le v\le t\}. \tag{12}$$

Therefore

$$\int_0^t\int_0^v\int_0^ue(s)dsdudv=\iiint_D e(s)dV=\int_0^t\int_s^t\int_u^te(s)dvduds \tag{13}$$

$$=\int_0^t e(s) \left(\int_s^t\int_u^t 1dvdu\right)ds=\int_0^t\frac{(t-s)^2}{2}e(s)ds. \tag{14}$$

Simply checking that the integral expression is a particular solution: differentiating once,

$$\frac{d}{dt}\int_0^t\frac{(t-s)^2}{2}e(s)ds=\frac{(t-t)^2}{2}e(t)+\int_0^t(t-s)e(s)ds. \tag{15}$$

Differentiating a second time,

$$\frac{d}{dt}\int_0^t(t-s)e(s)ds=(t-t)e(t)+\int_0^te(s)ds. \tag{16}$$

Differentiating a third time,

$$\frac{d}{dt}\int_0^te(s)ds=e(t). \tag{17}$$

Hence $u(t)=\int_0^t\frac{(t-s)^2}{2}e(s)ds$ satisfies $u'''(t)=e(t)$.

Proof of $(5)$: let $G(u,t)$ be such that $\frac{dG}{dt}(u,t)=f(u,t)$. Then

$$\frac{d}{dt}\int_0^tf(t,s)ds=\frac{d}{dt}G(t,t)=\frac{dG}{d\,{\small\rm 1st\,coord}}(t,t)+\frac{dG}{d\,{\small\rm 2nd\,coord}}(t,t) \tag{18}$$

$$=\int_0^tf_u(u,s)|_{u=t}ds+f(u,t)|_{u=t}=\int_0^tf_t(t,s)ds+f(t,t). \tag{19}$$


答:
This is a nice start, continue with:

$\neg w \wedge (\neg n \vee w)$

$(\neg w \wedge \neg n) \vee (\neg w \wedge w)$ (distributivity of $\wedge$ over $\vee$)

$\neg w \wedge \neg n$ (because $\neg w \wedge w$ is always false)

Title: I'm trying out this problem of logic but I'm stuck "Simplify ¬(w∨(n∧¬w)) to ¬w∧¬n"

Body:
Here is my work so far, I would appreciate the help</p>
```<code> ¬(w∨(n∧¬w))
 ¬w∧¬(n∧¬w) (De Morgan's)
 ¬w∧(¬n∨¬¬w) (De Morgan's)
 ¬w∧(¬n∨w) (Double Negation)
</code>```


问:
Please how to find indefinite integral $\int x^{x^x}\mathrm dx$?
Please how to find indefinite integral

$$\int x^{x^x}\mathrm dx$$
Thank for any one help me to find it

答:
For your reference:${}{}{}{}{}$

http://hk.knowledge.yahoo.com/question/question?qid=7008010201949


问:
Weakly* continuous function
I was struggling to understand a proof in the Brezis's book on functional analysis.The statement is as follows :
let $E$ be a normed space over $\mathbb{R}$ and $\phi : E^* \to \mathbb{R}$ be weakly*-continuous. There exist $x_0 \in E$ such that $\phi(f) = <f,x_0>$, $\forall f \in E^*$.
The idea is as follows.

The idea is to use the following : if $X$ is a normed space, and $g,f_1,\dots,f_k$ satisfy, $\forall x \in X$
$$[f_j(x) = 0, \forall j =1,\dots, k] \implies g(x) = 0$$
Then $g \in \mathrm{span}\{f_1,\dots,f_k\}$.

Then, to get back to the original problem, we take a neighborhood $V$ of $0_{X*}$, such that $|\phi(f)|< 1$ if $f \in V$, and V is taken as
$$ V = \{f \in E^* | \; |<f,x_i>| < \epsilon, \, \forall i = 1,\dots,k\}$$ for arbitrarly choosen $x_1,\dots,x_k$ in $E$ and $\epsilon >0$. The book claims that
$$[<f,x_i> = 0, \forall i =1,\dots, k] \implies \phi(f) = 0$$
but I don't see why. No other assumptions is made on $E$.

Thank you fellows for your help!

Giovanni D

答: Your last line comes straight from the definition of $\varphi(f)$ and the definition of continuity. Since $\varphi(f)$ is continuous, for every small number (here it is 1), we can find a neighborhood $V$ of $0$ with radius $\varepsilon$ such that for every $f \in V$ we have $\varphi (f) <$ than our small number. In particular, you see that if $f(x_i) = 0$ for every $i$ then the functional $f$ belongs to every neighborhood of the origin, for any arbitrary radius. Hence $\varphi (f) < \delta$ for any $\delta > 0$. Since $\delta$ is arbitrary, this implies that $\varphi (f)=0$.

Title: Weakly* continuous function


问:
Solve $\sigma^2=(1345)(2687)$ in $S_8$
I would like to solve the equation $\sigma^2=(1345)(2687)$ in $S_8$.

If we reason the same way that I did here we get that $\sigma$ is a cycle of length $8$. Here come the problems. How are we sure that we write all the solutions in this case?

We have the following formula:

$(i_1 i_2 i_3 i_4 i_5 i_6 i_7 i_8)^2=(i_1 i_3 i_5 i_7)(i_2 i_4 i_6 i_8)$.

Ok, so this tells us that one sigma is $(1 2 3 4 6 4 8 5 7)$.

But how do we get all of them? I think that we need to fix one of the 4-cycles and then write all the cyclic permutations of the other one. But I can't show that this actually produces all the solutions (I mean a formal argument, not just enumerating everything and seeing that this works). Furthermore, how would I do this if my $\sigma$ were the product of more cycles?

EDIT: I think that one possible argument for why fixing one cycle works is that if we do the procedure with the cyclic permutations we make sure, for instance, that $1$ goes to everything it can go (and the same applies to all the elements, but I picked $1$ for making myself clear). If we started permuting the other cycle too, we would obviously get the same solutions, because we have already sent each of its elements to all the possible images. Is this argument all right? How would this work for more cycles?

答:
You know that $\sigma$ has no odd cycles, since then $\sigma^2$ would have one, too. You can also observe, that $\sigma$ has no fixed points nor 2-cycles, since then $\sigma^2$ will have a fixed point. It's left to rule out case of two 4-cycles, but then $\sigma^2$ would be the product of four transpositions.

So $\sigma$ is indeed an 8-cycle.

Now, let $c = \sigma(1)$. Then our permutation is

$$ 1 \to c \to 3 \to \sigma^2(c) \to 4 \to \sigma^4(c) \to 5 \to \sigma^6(c) \to 1 $$

and it's both clear that $c \in \{2, 6, 7, 8\}$ and that all such $c$ will give rise to four different solutions for $\sigma$.

Title: Solve $\sigma^2=(1345)(2687)$ in $S_8$


答:
As we can cyclically rotate any cycle, we may assume wlog that $i_1=1$. Then it follows that $i_3=3$, $i_5=4$, $i_7=5$.
Consequently, $i_2$ must be one of $2,6,7,8$. For each of the four possible cases, we find exactly one solution, namely $i_4$ must be the next term in the second 4-cycle and so on.

Title: Solve $\sigma^2=(1345)(2687)$ in $S_8$

Body:
I would like to solve the equation $\sigma^2=(1345)(2687)$ in $S_8$.

If we reason the same way that I did here we get that $\sigma$ is a cycle of length $8$. Here come the problems. How are we sure that we write all the solutions in this case?

We have the following formula:

$(i_1 i_2 i_3 i_4 i_5 i_6 i_7 i_8)^2=(i_1 i_3 i_5 i_7)(i_2 i_4 i_6 i_8)$.

Ok, so this tells us that one sigma is $(1 2 3 4 6 4 8 5 7)$.

But how do we get all of them? I think that we need to fix one of the 4-cycles and then write all the cyclic permutations of the other one. But I can't show that this actually produces all the solutions (I mean a formal argument, not just enumerating everything and seeing that this works). Furthermore, how would I do this if my $\sigma$ were the product of more cycles?

EDIT: I think that one possible argument for why fixing one cycle works is that if we do the procedure with the cyclic permutations we make sure, for instance, that $1$ goes to everything it can go (and the same applies to all the elements, but I picked $1$ for making myself clear). If we started permuting the other cycle too, we would obviously get the same solutions, because we have already sent each of its elements to all the possible images. Is this argument all right? How would this work for more cycles?


问:
How to generate a random vector, guaranteed to be within the hemisphere with respect to another vector?
Given a normalized vector N, how can one generate a random direction vector that is guaranteed to be in the hemisphere with respect to N (i.e. the hemisphere where N is exactly in the middle)? The way I am currently doing this is to sample a random direction vector d and dot it with N and keep that vector if the dot product is greater than 0. This method is not guaranteed to generate a vector in the hemisphere that I am interested in as ~50% of the random vectors would have a 0 or negative dot product result.

I saw somewhere that there is a way to transform the randomly generated vector d to place it in the right hemisphere using a transformation matrix but I don't know how to do it. Can someone write a [pseudo]code for how one generate a direction vector using the transformation method?

答: In this case, you can just randomly generate $d$, and if it does not lie in the right hemisphere, because $d \cdot N < 0$, replace $d$ by $-d$.


问:
Conditional density and Uniform distribution
$Y=X+\xi X$, $X \sim \operatorname{Unif}[-1, 1]$, $\xi \sim \operatorname{Unif}[0, 1]$. $X$ and $\xi$ are independent.

I need to find $f_{Y\mid X}(y\mid x)$. How to do it? I couldn't find joint probability. Any hints?
I think the questions of finding the joint density and the conditional density are equivalent, because it is expressed through each other

答:
The main issue is to derive the joint density

Setting

$Y=X+\xi X$ and $Z=X$ the jacobian is $|J|=1/|z|$ thus the joint density results to me

$$f_{XY}(x, y)=\frac{1}{2|x|}\Big[\mathbb{1}_{[-1;0)}(x)\mathbb{1}_{(2x;x)}(y) + \mathbb{1}_{(0;1]}(x)\mathbb{1}_{(x;2x)}(y) \Big]$$


问:
Brownian Motion and hitting times
I am having trouble with a step in the proof of Theorem 2.9.6 of Karatzas and Shreve's Brownian Motion and Stochastic Calculus. I need to prove that, given a Brownian motion $W$,
$$\beta_t:=\inf\{s>t; W_s=0\}$$

is an optional time for every $t\ge 0$.

Here is my attempt: fix $t\ge0$, and let $T \in \mathbb{R}$.  If $T \le t$, $\{\beta_t <T\}=\emptyset\in\mathcal{F}_T$. Otherwise,
$$\{\beta_t<T\}=\bigcup_{s<t'<t}\{W_{t'}=0\}\in \sigma\left( \bigcup_{s<t'<t}\mathcal{F}_{t'}\right)\subseteq \mathcal{F}_t.$$

Is this correct? If not, there is some hint I could follow? Thank you!

答:
I believe I solved the problem. Given $T>t$, we have that

$$\{\beta_t \ge T\}=\bigcap_n A_n,$$

where $A_n=\{W_s\neq 0; s\in[t+1/n,T-1/n]\}$. But since $W$ is continuous and $[t+1/n,T-1/n]$ is closed, we have
$$A_n = \bigcup_{k\ge 1} \bigcap_{q\in[t+1/n,T-1/n]\cap\mathbb{Q}}\{|W_q|\ge1/k\} \in \mathcal{F}_T.$$

Therefore $\{\beta_t<T\}=\{\beta_t \ge T\}^c\in \mathcal{F}_T$.

Title: Brownian Motion and hitting times


问:
A, B, and C Roll Dice
Hello I have the following question:

Three players A, B, & C take turns rolling a pair of dice. The winner is the first player who obtains the sum of 7 (P[7] = 1/6) on a given roll of the dice. If A rolls 1st, then B, then C, then back to A (etc. etc. etc.). What is the P[A wins], P[B wins], P[C wins]?

I know the P[A]= 36/91 with it being a geometric series of (1/6)/[1-(125/216)]
What confuses me is why does the beginning of the series start as: 1/6 + (5/6)^3(1/6)
Why is it cubed and not squared since there are only 2 other possible winners other than A?

答:
A wins if they succeed immediately, or if there are first three(!) failures by A, B, C and then a success by A, etc.

Alternative approach: If A happens to fail on first roll (with probability $\frac 56$) , we essentially start a game with players$(B,C,A)$ instead of $(A,B,C)$.
Therefore $P[B]=\frac 56P[A]$, $P[C]=\frac 56 P[B]$, and $P[A]=\frac 16+\frac56P[C]$. You can solve these three equations in three unknowns

Title: A, B, and C Roll Dice


答:
Round 1 : A wins with probability $1/6$

Round 2 : B wins with probability $1/6 \times (5/6)$

Round 3 : C wins with probability $1/6 \times (5/6)^2$

Round 4 : A wins with probability $1/6 \times (5/6)^3$

etc.

There is a cube because A has to wait three times for his turn to come again.

Title: A, B, and C Roll Dice

Body:
Hello I have the following question:

Three players A, B, & C take turns rolling a pair of dice. The winner is the first player who obtains the sum of 7 (P[7] = 1/6) on a given roll of the dice. If A rolls 1st, then B, then C, then back to A (etc. etc. etc.). What is the P[A wins], P[B wins], P[C wins]?

I know the P[A]= 36/91 with it being a geometric series of (1/6)/[1-(125/216)]
What confuses me is why does the beginning of the series start as: 1/6 + (5/6)^3(1/6)
Why is it cubed and not squared since there are only 2 other possible winners other than A?


问:
Normal subgroup of finite group
Let $G$ be a finite group of order $385$ and $N$ a normal subgroup of order $7$. Prove that:

(a) If $x\in G$ then $x^{7}\in N \Rightarrow x\in N$

(b) if $g$ an element of $G$ then $g\in N \Leftrightarrow g^{62} \in N$.

Any help? I've only managed to prove that $x^{55}\in N $ $ \forall x\in G$.

答: Assume that $x^7\in N$, i.e. $[x]^7=e$ in $G/N$. So the order of $[x]$ divides $7$. On the other hand it divides $|G/N|=5\cdot11$, hence $[x]=e$, i.e. $x\in N$. The second part about $g\in N$ iff $g^{62}\in N$ can be proven in the exact same way.

Title: Normal subgroup of finite group


问:
Fourier transform of Gaussian noise
So I was doing a Fourier transform on a Gaussian distribution, the histogram of which is as follows. And by the way I sampled a million data, as shown in the image.

What is such distribution?
I find Weibull Distribution and Beta Distribution similar to the histogram, but I'm not sure.

答:
The magnitude of the signal follows a Rice distribution:
https://en.wikipedia.org/wiki/Rice_distribution</p>

Title: Fourier transform of Gaussian noise


问:
Can we define a function on $\mathbb{N}$ recursively from all its previous values?
In set theory, the recursion principle states that a function $f$ can be defined in a way where $f\left(\boldsymbol{n}^{+}\right)$ depends on $\boldsymbol{n}$ and $f\left(\boldsymbol{n}\right)$. The complete statement goes as follows:

Let $y_{0}$ be an element of a set $Y$ and $h: \mathbb{N} \times Y \to Y$ be a function. Then there exists a unique function $f: \mathbb{N} \to Y$ such that $f\left(\boldsymbol{0}\right) = y_{0}$ and $f\left(\boldsymbol{n}^{+}\right) = h\left(\boldsymbol{n},f\left(\boldsymbol{n}\right)\right)$ for all $\boldsymbol{n} \in \mathbb{N}$.

I was thinking if there exists a strong version of the recursion principle, that is, a function $f$ can be defined in a way where $f\left(\boldsymbol{n}^{+}\right)$ is defined by all its previous values $f\left(\boldsymbol{0}\right), f\left(\boldsymbol{1}\right), \dots, f\left(\boldsymbol{n}\right)$. My thought was verified in a recent discussion. In fact, this type of definition is used in defining a Fibonacci sequence:
\begin{equation}
\begin{aligned}
&F_0 = 0,\ F_1 = 1\\
&F_{n+2} = F_{n+1} + F_{n},\ n \geq 0.
\end{aligned}
\end{equation}

I would like to seek a proof for a "strong" statement of this idea. My first difficulty was how to express this type of recursion principle formally within the language of set theory. I have come up with a statement to avoid involving $\left\{f\left(\boldsymbol{0}\right),f\left(\boldsymbol{1}\right),f\left(\boldsymbol{2}\right),\dots,f\left(\boldsymbol{n}\right)\right\}$:

Let $y_{0}$ be any element of a set $Y$ and $h: \mathscr{P}\left(\mathbb{N}\times Y\right) \to \mathscr{P}\left(\mathbb{N}\times Y\right)$ be a function. Then there exists a unique function $f: \mathbb{N} \to Y$ such that
\begin{equation*}
f\left(\boldsymbol{0}\right) = y_{0},
\end{equation*}
and
\begin{equation*}
\forall \boldsymbol{n} \geq \boldsymbol{1}, f\vert_{\boldsymbol{n}^{+}} = h\left(f\vert_{\boldsymbol{n}}\right).
\end{equation*}

I am not sure if this is a legitimate statement for what I am seeking. If so, can anyone provide hints of how to prove it? If there is anything wrong with this statement, can anyone propose a correct one?

Theorem:

Let $y_{0}$ be any element of a set $Y$ and $h: \mathscr{P}\left(\mathbb{N}\times Y\right) \to Y$ be a function. Then there exists a unique function $f: \mathbb{N} \to Y$ such that
\begin{equation*}
f\left(\boldsymbol{0}\right) = y_{0},
\end{equation*}
and
\begin{equation*}
f\left(\boldsymbol{n}^{+}\right) = h\left(f_{\vert \boldsymbol{n}^{+}}\right).
\end{equation*}

Proof:

Assume $Y$ is an arbitrary set and $y_{0} \in Y$. Also assume that we have an arbitrary function $h: \mathscr{P}\left(\mathbb{N}\times Y\right) \to Y$. We may define a function $\hat{h}: \mathbb{N} \times \mathscr{P}\left(\mathbb{N}\times Y\right) \to \mathscr{P}\left(\mathbb{N}\times Y\right)$ as
\begin{equation*}
\forall \boldsymbol{n} \in \mathbb{N}, \forall r \in \mathscr{P}\left(\mathbb{N}\times Y\right),\hat{h}\left(\boldsymbol{n},r\right) = r \cup \left\{ \langle \boldsymbol{n}^{+}, h\left(r\right) \rangle \right\}.
\end{equation*}
Note that $\left\{\langle\boldsymbol{0},y_{0}\rangle\right\} \in \mathscr{P}\left(\mathbb{N}\times Y\right)$. According to the weak recursion principle, there exists a unique function $\hat{f}: \mathbb{N} \to \mathscr{P}\left(\mathbb{N}\times Y\right)$ such that
\begin{equation*}
\hat{f}\left(\boldsymbol{0}\right) = \left\{\langle \boldsymbol{0},y_{0} \rangle\right\}
\end{equation*}
\begin{equation*}
\hat{f}\left(\boldsymbol{n}^{+}\right) = \hat{h}\left(\boldsymbol{n},\hat{f}\left(\boldsymbol{n}\right)\right).
\end{equation*}

We first prove that for any $\boldsymbol{m},\boldsymbol{n} \in \mathbb{N}$, $\boldsymbol{m} \leq \boldsymbol{n}$ implies $\hat{f}\left(\boldsymbol{m}\right) \subseteq \hat{f}\left(\boldsymbol{n}\right)$. Fix $\boldsymbol{m}$ and define the following set:
\begin{equation*}
A = \left\{\boldsymbol{n} \in \mathbb{N}: \boldsymbol{m} \leq \boldsymbol{n} \implies \hat{f}\left(\boldsymbol{m}\right) \subseteq \hat{f}\left(\boldsymbol{n}\right) \right\}.
\end{equation*}
First of all, consider the case $\boldsymbol{0}$. In this case, the only situation worth discussing is $\boldsymbol{m} = \boldsymbol{0}$. It is clear that $\hat{f}\left(\boldsymbol{0}\right) \subseteq \hat{f}\left(\boldsymbol{0}\right)$. Consequently, $\boldsymbol{0} \in A$. Next, assume that $\boldsymbol{n} \in A$. There are three cases to discuss for $\boldsymbol{n}^{+}$. If $\boldsymbol{m} \leq \boldsymbol{n}$, then $\hat{f}\left(\boldsymbol{m}\right) \subseteq \hat{f}\left(\boldsymbol{n}\right)$. Also,
\begin{equation*}
\hat{f}\left(\boldsymbol{n}^{+}\right) = \hat{h}\left(\boldsymbol{n},\hat{f}\left(\boldsymbol{n}\right)\right) = \hat{f}\left(\boldsymbol{n}\right) \cup \left\{ \langle \boldsymbol{n}^{+}, h\left(\hat{f}\left(\boldsymbol{n}\right)\right) \rangle \right\}.
\end{equation*}
It is immediately clear that $\hat{f}\left(\boldsymbol{m}\right) \subseteq \hat{f}\left(\boldsymbol{n}^{+}\right)$, and $\boldsymbol{n}^{+} \in A$. The next case to discuss is $\boldsymbol{m} = \boldsymbol{n}^{+}$. It is also immediately clear that $f\left(\boldsymbol{m}\right) \subseteq f\left(\boldsymbol{m}^{+}\right)$, and $\boldsymbol{n}^{+} \in A$. The third case is $\boldsymbol{m} > \boldsymbol{n}^{+}$. It is a trivial case, and $\boldsymbol{n}^{+} \in A$. Then we have proven that $\boldsymbol{n} \in A$ implies $\boldsymbol{n}^{+} \in A$. Thus, $A = \mathbb{N}$.

Next, we shall prove that each $\hat{f}\left({\boldsymbol{n}}\right)$ is a function on $\boldsymbol{n}^{+}$. Similarly, we define the following set:
\begin{equation*}
B = \left\{\boldsymbol{n} \in \mathbb{N}: \hat{f}\left(\boldsymbol{n}\right) \textrm{ is a function on }\boldsymbol{n}^{+}\right\}.
\end{equation*}
From the definition of $\hat{f}$, it is obvious that $\hat{f}\left(\boldsymbol{0}\right)$ is a function on $\boldsymbol{1}$. Then, $\boldsymbol{0} \in B$. Next assume that $\boldsymbol{n} \in B$, that is, for an arbitrary $\boldsymbol{n} \in \mathbb{N}$, $\hat{f}\left(\boldsymbol{n}\right)$ is a function on $\boldsymbol{n}^{+}$. According the the definition of $\hat{f}$, we have
\begin{equation*}
\hat{f}\left(\boldsymbol{n}^{+}\right) = \hat{h}\left(\boldsymbol{n},\hat{f}\left(\boldsymbol{n}\right)\right) = \hat{f}\left({\boldsymbol{n}}\right) \cup \left\{\langle\boldsymbol{n}^{+}, h\left(\hat{f}\left(\boldsymbol{n}\right)\right) \rangle\right\}.
\end{equation*}
It is immediately clear that $\hat{f}\left(\boldsymbol{n}^{+}\right)$ is a function on $\left(\boldsymbol{n}^{+}\right)^{+}$. Then $\boldsymbol{n}^{+} \in B$. We may conclude that $B = \mathbb{N}$.

Then consider the set
\begin{equation*}
f = \bigcup\operatorname{Range}\left(\hat{f}\right).
\end{equation*}
It should be noted that $f_{\vert\boldsymbol{n}^{+}}$ is of the following property:
\begin{equation*}
f_{\vert\boldsymbol{n}^{+}} = \hat{f}\left(\boldsymbol{n}\right),
\end{equation*}
given that for any $\boldsymbol{m},\boldsymbol{n} \in \mathbb{N}$, $\boldsymbol{m} \leq \boldsymbol{n}$ implies $\hat{f}\left(\boldsymbol{m}\right) \subseteq \hat{f}\left(\boldsymbol{n}\right)$, which has been proved earlier.

Assume that $\boldsymbol{n}_{1},\boldsymbol{n}_{2} \in \mathbb{N}$, $\boldsymbol{n}_{1} \neq \boldsymbol{n}_{2}$ and $\boldsymbol{n}_{1},\boldsymbol{n}_{2} \in \operatorname{domain}\left(f\right)$. It is immediately realized that $\boldsymbol{n}_{1},\boldsymbol{n}_{2} \in \operatorname{domain}\left(\hat{f}\left(\boldsymbol{n}_{3}\right)\right)$ where $\boldsymbol{n}_{3} \geq \max{\left(\boldsymbol{n}_{1},\boldsymbol{n}_{2}\right)}$. As $\hat{f}\left({\boldsymbol{n}_{3}}\right)$ is a function on ${\boldsymbol{n}_{3}}^{+}$, it is clear that $\left[\hat{f}\left(\boldsymbol{n}_{3}\right)\right]\left(\boldsymbol{n}_{1}\right) \neq \left[\hat{f}\left(\boldsymbol{n}_{3}\right)\right]\left(\boldsymbol{n}_{2}\right)$. As $\left[\hat{f}\left(\boldsymbol{n}_{3}\right)\right]\left(\boldsymbol{n}_{1}\right) = f\left(\boldsymbol{n}_{1}\right)$ and $\left[\hat{f}\left(\boldsymbol{n}_{3}\right)\right]\left(\boldsymbol{n}_{2}\right) = f\left(\boldsymbol{n}_{2}\right)$, we have $f\left(\boldsymbol{n}_{1}\right) \neq f\left(\boldsymbol{n}_{2}\right)$. So that $f$ is indeed a function.

We would like to argue that $f$ is a function with $f\left(\boldsymbol{0}\right) = y_{0}$ and $f\left(\boldsymbol{n}^{+}\right) = h\left(f_{\vert\boldsymbol{n}^{+}}\right)$. Using the fact $f_{\vert\boldsymbol{n}^{+}} = \hat{f}\left(\boldsymbol{n}\right)$, we have
\begin{equation*}
f\left(\boldsymbol{0}\right) = f_{\vert\boldsymbol{1}}\left(\boldsymbol{0}\right) = \left[\hat{f}\left(0\right)\right] = y_{0}.
\end{equation*}
Also,
\begin{equation*}
f\left(\boldsymbol{n}^{+}\right) = \left[\hat{f}\left(\boldsymbol{n}^{+}\right)\right]\left(\boldsymbol{n}^{+}\right) = h\left(\hat{f}\left(\boldsymbol{n}\right)\right) = h\left(f_{\vert\boldsymbol{n}^{+}}\right)
\end{equation*}
The uniqueness of $f$ can be easily argued from the uniqueness of $\hat{f}$.

答:
That statement is not true. For example, if we take $h$ to be the identity function we'd need an $f$ such that (for example) $f_{\vert\bf 2}=f_{\vert \bf 1}$, which can't happen. We can do nastier things too: suppose $Y=\mathbb{N}$ and consider e.g. $$h(X)=\{(n, y+1): (n,y)\in X\}.$$

Instead, what we want our "update" function to do is tell us just the next value of $f$. So $h$ should be a map $\mathcal{P}(\mathbb{N}\times Y)\rightarrow Y$, and $f$ should satisfy $$f({\bf 0})=y_0,\quad f({\bf n^+})=h(f_{\vert{\bf n^+}}).$$

And in fact this is how I've always seen the recursion theorem presented.

(Here I'm treating $g_{\vert {\bf a}}$ as $\{(x,y): x<{\bf a}$ and $g(x)=y\}$; if you're using the "$\le$"-version, replace "$h(f_{\vert {\bf n^+}})$" with "$h(f_{\vert{\bf n}})$" above.)

Just like with strong/weak induction, we can prove "strong recursion" from "weak recursion, and this is a good exercise. (HINT: given a function $f:\mathbb{N}\rightarrow Y$, consider the function $\hat{f}:\mathbb{N}\rightarrow\mathcal{P}(\mathbb{N}\times Y): {\bf n}\mapsto f_{\vert{\bf n}}$. Instead of building $f$ by strong recursion, you can build $\hat{f}$ by weak recursion and then define $f$ from $\hat{f}$.)


问:
Can all positive real polynomials be expressed as a sum of polynomial squares?
For $P(x)=y$,   where $P(x)$ is some real polynomial strictly above the $x$-axis, can it be expressed as the sum of polynomial squares?

答:
If "sum of linear squares" means
$$
P(x) = \sum_{j=1}^n (a_jx+b_j)^2,
\tag1$$
then the answer is no.  For example $x^4+1$ is not of that form.  Indeed, for anything of the form $(1)$ we have
$$
\lim_{x \to \infty}\frac{P(x)}{x^4+1} = 0
$$


答: If $P(x)\in\mathbb{R}[x]$ is such that $P(x)\geqslant 0$ for $x\in\mathbb{R}$, then $P(x)$ cannot have odd-degree roots in $\mathbb{R}$ (otherwise it would change its sign in a neighborhood of such a root). Hence each of its roots is either even-degree real one, or comes in a pair with its complex-conjugate. In other words, $$P(x)=\prod_{k=1}^n\big((x-a_k)^2+b_k^2\big)$$ for some real $a_k$ and $b_k$ for $1\leqslant k\leqslant n$. But $$(A^2+B^2)(C^2+D^2)=(AC+BD)^2+(AD-BC)^2,$$ hence (by induction on $n$) $P(x)=A^2(x)+B^2(x)$ for some polynomials $A(x)$ and $B(x)$.

Title: Can all positive real polynomials be expressed as a sum of polynomial squares?

Body: For $P(x)=y$,   where $P(x)$ is some real polynomial strictly above the $x$-axis, can it be expressed as the sum of polynomial squares?


问:
Compact subsets of function spaces, geometry
The subset is called compact when every open cover contains a finite subcover. In Euclidean spaces, it is easy to visualize this by imagining some open ball that contains this set, thinking about the subset of this ball that contains the set and divising this subset into the finite collection of sets. Is there a way of some kind of visualization in the case of infinite-dimensional function spaces?</p>

答: There is a generalization of metric space, called continuity space, where the codomain of the metric is replaced by a quantale (a certain partially ordered set with some extra structure, see Flagg's "quantales and continuity spaces"). It it then shown that every topological space is metrizable as long as one interprets metrizable to mean with respect to a continuity space structure. That means that all of the geometric intuition valid for metric spaces can be carried over to arbitrary topological spaces. </p>


答:
Some background: Suppose X is a vector space and $\Phi $ is a family of linear functionals on $X$. This family defines a  topology on $X$, denoted by $\mathcal T_\Phi$ such that $(X,\mathcal T_\Phi)$ is a topological vector space.  The set

$$\mathcal S = \{ f^{-1} [(-ε,ε)] : ε>0, f \in \Phi \}.$$
is a sub-basis for $0 \in X$ and thus, a basic open set containing $0$ is of the form $ \bigcap_{i=1}^n f_i^{-1} [(-ε,ε)]$.

If $g: (X,\mathcal T_\Phi)  \to \mathbb R$ is linear and continuous  then $g \in <\Phi>= \operatorname{span} (\Phi)$.

To see this, let $V= g^{-1} [(-1,1)].$ By the continuity of $g$, $V \in \mathcal T_\Phi$  and so we can find $ε>0,n \in \mathbb N$ and $f_1,f_2,\dots, f_n \in \Phi$ such that
$$ \bigcap_{i=1}^n f_i^{-1} [(-ε,ε)] \subset V.$$
In particular, $Y:= \bigcap_{i=1}^n \ker f_i \subset V$. Notice that $Y$ is a (non trivial) linear subspace of $X$. Furthermore, $Y \subset \ker G$ and thus we infer that $g \in <f_1,f_2, \dots , f_n> \subset <\Phi>$. Indeed, let $x \in Y$. For all $k \in \mathbb N$ one has that $kx \in Y \subset V$ and so $|g(kx) | <1$. In other words, $g(x) =0$.

The weak-star topology is just a special case, if we consider the space $X^*$ and $\Phi = j(X)$, where $j : X \to X^{**}$ is the canonical isometric embedding.

Title: Weakly* continuous function

Body:
I was struggling to understand a proof in the Brezis's book on functional analysis.The statement is as follows :
let $E$ be a normed space over $\mathbb{R}$ and $\phi : E^* \to \mathbb{R}$ be weakly*-continuous. There exist $x_0 \in E$ such that $\phi(f) = <f,x_0>$, $\forall f \in E^*$.
The idea is as follows.

The idea is to use the following : if $X$ is a normed space, and $g,f_1,\dots,f_k$ satisfy, $\forall x \in X$
$$[f_j(x) = 0, \forall j =1,\dots, k] \implies g(x) = 0$$
Then $g \in \mathrm{span}\{f_1,\dots,f_k\}$.

Then, to get back to the original problem, we take a neighborhood $V$ of $0_{X*}$, such that $|\phi(f)|< 1$ if $f \in V$, and V is taken as
$$ V = \{f \in E^* | \; |<f,x_i>| < \epsilon, \, \forall i = 1,\dots,k\}$$ for arbitrarly choosen $x_1,\dots,x_k$ in $E$ and $\epsilon >0$. The book claims that
$$[<f,x_i> = 0, \forall i =1,\dots, k] \implies \phi(f) = 0$$
but I don't see why. No other assumptions is made on $E$.

Thank you fellows for your help!

Giovanni D


问:
Definition of open sets
Open ball, centered at $$ x \in U,$$ with radius r is defined as
$$ B_r \left( x\right)=\left\{x' \in U \mid \mathrm{d} \left(x, x' \right)<r \right\}.$$

Definition of an open set:
$$ U \ \mathrm{is \ open.} \Leftrightarrow \forall x\in U: \exists r>0: B_r(x)\subset U$$

My question is: don't all points of U anyway satisfy the latter condition? Since $$ x \in U, $$ the open ball is a collection of elements in U and thus a subset of U.

答:
Lets take $U=\{x: ||x|_2\leq 1\}$

In this case if you take some point on the edge, for example $x=[1,0]$ if we are on the plane, than you can't find such $r>0$ cause $x+r\cdot[1,0]=[1+r,0]$ which is not in $U$.

In this case $U$ is a close set.

If we had $U=\{x: ||x|_2< 1\}$ it was an open set.

Title: Definition of open sets


问:
Explicitation of an equation
We consider the function $f(x,y)=\arctan(x^2-y^2-1)$. Why into the neighborhood of $(1, 0)$, there exists a $x=g(y)$ but not a $y=h(x)$ that make explicit the equation $f(x,y)=0$? I think that the existance of $x$ follows from Dini's theorem, is it right?

Thank you

答:
Consider $f(x,y)=\arctan\left(x^2-y(x)^2-1\right)=0$

compute the implicit derivative $y'(x)$. We have:

$$\frac{2x-2y(x)y'(x)}{1+\left(x^2-y(x)^2-1\right)^2}=0$$
and solving for $y'(x)$
$$2x-2y(x)y'(x)=0\to y'(x)=\frac{x}{y(x)}$$
so you can see that $y'(x)$ does not exist at $(1,0)$ and this is the reason why the implicit theorem can't be applied.

Title: Explicitation of an equation


问:
Show that If k is odd, then $\Bbb{Q}_{4k}$ is isomorphic to $\Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$ for some $\alpha$

Show that if k is odd, then $\Bbb{Q}_{4k}$ is isomorphic to $\Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$ for some $\alpha: \Bbb{Z}_4 \rightarrow Aut(\Bbb{Z}_k)$. Calculate $\alpha$ explicitly.

We know that $\Bbb{Q}_{4k} = \{ b^k_{2n}, b^k_{2n}a | 0 \leq k < 2n \}$, and that everything outside of the cyclic group $\langle b \rangle$ is of order 4. What confuses me is that we only have one cyclic group in $\Bbb{Q}$, right? However, in order to write it of the form $\Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$, we need to find a normal cyclic group of order k in $\Bbb{Q}_{4k}$, and I can't really see that.

For example, in the group $\Bbb{Q}_{12} = \{e, b, b^2, b^3, b^4, b^5, a, ab, ab^2, ab^3, ab^4, ab^5\}$, we only have $\langle b \rangle$ as the cyclic group. I cannot see any normal cyclic group of order 3 here.

For the second part of the problem, suppose we accept the fact that $\Bbb{Q}_{4k} = \Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_4$ for some $\alpha: \Bbb{Z}_4 \rightarrow Aut(\Bbb{Z}_k)$.

There is a theorem in the textbook that states:

Corollary Let $\bar{m}$ have exponent k in $\Bbb{Z}^×_n$, and let $α : \Bbb{Z}_k → \Bbb{Z}^×_n$ be the homomorphism that takes the generator to $\bar{m}$. Then writing $\Bbb{Z}_n = \langle b \rangle$, $\Bbb{Z}_k = \langle a \rangle$, and identifying $\Bbb{Z}^×_n$ with $Aut(\Bbb{Z}_n)$, we obtain $$\Bbb{Z}_n \rtimes_{α} \Bbb{Z}_k = \{b^ia^j | 0 ≤ i < n, 0 ≤ j < k\},$$ where b has order n, a has order k, and the multiplication is given by $$b^ia^jb^{i'}a^{j'} = b^{i+m^ji'}a^{j+j'}.$$

Moreover, the nk elements $\{b^ia^j | 0 \leq i < n, 0 \leq j < k\}.$

Since we know that $aba^{-1}=b^{-1}$ in quaternionics, we need to find an $\bar{m}$ such that $b^0aba^{-1}= b^{0+m}a^{-1+1} = b^{-1}$. So $\bar{m} = \bar{-1}$, and $\alpha$ must be send the generator to $\bar{-1}$. I guess this is what the question is asking for when it tells us the calculate $\alpha$ "explicitly". Is that correct?

However, I cannot say that until I prove that $\Bbb{Q}_{4k} \cong \Bbb{Z}_k \rtimes_{\alpha} \Bbb{Z}_{4}$, so I was wondering if anybody could help me with that.

Thank you in advance

答:
Let's be clear on one significant point first:

What confuses me is that we only have one cyclic group in Q, right?

No. Remember that any chosen element $g$ in a group $G$ generates a cyclic group $\langle g\rangle$, so in general a group will have many cyclic subgroups. In fact, since any cyclic group has a generator, any cyclic subgroup of some $G$ will be obtained by generating it from some single element $g\in G$. Moreover, a group $G$ has only one nontrivial cyclic subgroup iff $G$ is itself a cyclic group of prime order.

To prove this, observe that if $G$ is not cyclic of prime order, then either

$G$ is cyclic of composite order $n$ generated by say $g$, in which case $\langle g^d\rangle$ is a proper nontrivial subgroup of $G$ when $d\mid n$ is a proper divisor and $\ne1$ hence $\langle g^d\rangle$ and $G$ itself are two distinct nontrivial subgroups of $G$ that are cyclic, or

$G$ is not cyclic, so pick $x\in G\setminus\{e\}$, then pick $y\in G\setminus\langle x\rangle$, in which case $\langle x\rangle$ and $\langle y\rangle$ are two distinct nontrivial cyclic subgroups of $G$.

Now, back to generalized quaternion groups. The group presentation is

$$Q_{4k}=\langle a,b~|~b^{2k}=a^4=1,~b^n=a^2,~a^{-1}ba=b^{-1} \rangle$$

What are some cyclic subgroups? By experimentation, $\langle b\rangle$, $\langle b^2\rangle$, $\langle b^k\rangle$, $\langle a\rangle$, $\langle a^2\rangle$, etc. (At this point, assume $k$ is odd.) What are some subgroups of order $4$? The obvious one is $\langle a\rangle$. What about normal cyclic subgroups of order $k$? Well, $\langle b^2\rangle$ is cyclic of order $k$. We see that $b=a^2(b^2)^{(1-n)/2}\in\langle a,b^2\rangle$, which we can use to prove that $\langle a\rangle\langle b^2\rangle=Q_{4k}$. Furthermore $\langle b^2\rangle\cap\langle a\rangle=\{1\}$. We can check that $a\langle b^2\rangle a^{-1}=\langle b^{-2}\rangle=\langle b^2\rangle$ hence $\langle b^2\rangle$ is normalized by $a$ and $b$ hence normalized by the subgroup generated by $a$ and $b$, which is the full group $Q_{4k}$. So $\langle b^2\rangle$ is normal, cyclic and order $k$.

Thus, $Q_{4k}=\langle b^2\rangle\rtimes\langle a\rangle$. Apply $x\mapsto a^{-1}xa$ to $b^2$ three times to get $ab^2a^{-1}=b^{-2}$, hence we conclude that the homomorphism $\alpha:\langle a\rangle\to{\rm Aut}(\langle b^2\rangle)$ sends $a$ to the inverse map $x\mapsto x^{-1}$ (this is an automorphism of any abelian group), hence sends $a^2$ to the identity and also sends $a^3=a^{-1}$ to the inverse map. In terms of cyclic groups and unit groups,

$$Z_4\to U(k)={\rm Aut}(Z_k):0,2\mapsto 1;~1,3\mapsto -1.$$

However you'll have to judge for yourself from the context of the text and the material what constitutes an explicit bijection for the purposes of this exercise.


问:
Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution?
The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:

$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)

答: Use an Euler substitution $t=\sqrt{x^2+4}+x$ so $\frac{dt}{dx}=\frac{x}{\sqrt{x^2+4}}+1=\frac{t}{\sqrt{x^2+4}}$ and $\frac{dx}{\sqrt{x^2+4}}=\frac{dt}{t}$.


问:
Most efficient method to solve quintic real roots?
I have searched all over to get an idea of which method is preferrable for computers to solve the real roots of a low degree polynomial, specifically a quintic, but have not come up with a simple answer.

Is there a general algorithm which is considered to be the fastest/most efficient for such a low degree?

答:
The easiest approach, if it works, is to use the Rational Root Theorem to find factors of the original quintic, reducing it to a quartic, a cubic, or quadratic.

User Yuri Negometyanov of MSE offered me a unique solution here to a problem I had with
a quintic equation that resulted in multiplying $\space ABC \space $ of a Pythagorean triple. The method defined clear search limits to be used in finding a quadratic divisor $\space C\space $ of the the quintic equation, leaving a cubic that could be solved by other means. The original quintic equation was

$$ABC=2mk^5- 2 m^5k + P=0 \quad\text{ where}\quad  C=m^2+n^2$$

The objective was to find $\space C:\space C|P\space $ and then to find integer solutions to $\space n=\sqrt{C-m^2},\space m\lt\sqrt{C}.\quad$ From there, we have a quadratic "factor" of the quintic. Your particular quadratic may be different.

To be brief, the limits were
\begin{equation}
 \sqrt[3]{2P} < C < \frac{\sqrt{(4P)^{\frac{4}{5}}+1 } + 1}{2}\quad \land \quad C\bigg |\frac{P}{12}
 \end{equation}

and the search for this involved for $P=192720$

$$P=192720\implies \lfloor\sqrt[3]{2(192720)}\rfloor = 72 \le C \le \bigg\lfloor\frac{\sqrt{\big(4(192720)\big)^{\frac{4}{5}}+1 } + 1}{2}\bigg\rfloor=113$$
$$\land\quad
\frac{P}{12}  =\frac{192720}{12}=16060$$
Only two of $16060$ are between $72$ and $113$ and only one of them works
properly with to yield a "solvable" cubic.


问:
How to calculate this limit, in terms of $a, b, k, ..$?
$$f(0) = a \\ f(n+1) = f(n) + f(n-1) \\ g(0) = b \\ g(n+1) = g(n) + g(n-1)$$
$$ \lim_{n \to \infty} {{f(n+k)} \over  {g(n)}}$$
where $a, b, n, k \in \mathbb Z$

答:
Denote $\phi = \frac{1+\sqrt{5}}{2}$, $\varphi = \frac{1-\sqrt{5}}{2} = 1-\phi = -\phi^{-1}$. We will prove that any sequence defined by

$$ x_n=R\phi^n + T\varphi^n $$

satisfies recurrence $x_{n+1}=x_n+x_{n-1}$. Then we will prove the reverse: our sequences $f(n)$, $g(n)$ have form $R\phi^n + T\varphi^n$ for some $R$, $T$. This is analogous to Binet formula for Fibonacci numbers.

Firstly, we'll check the recurrence. You can check that both $\phi$ and $\varphi$ satisfy equation $x^2+x+1=0$. Therefore

$$ x_{n+2} = R\phi^{n+2} + T\varphi^{n+2} = R\phi^n\phi^2 + T\varphi^n\varphi^2 = R\phi^n (\phi + 1) = T\varphi^n (\varphi+1) = R\phi^{n+1} + R\phi^n + T\varphi^{n+1} + T\varphi^n = x_{n+1}+x_n,$$

q.e.d. Any such sequence satisfies

$$ \lim_{n \to \infty} \frac{x_n}{\phi^n} = R, $$

as $|\varphi|<1$, so $\varphi^n \to 0$.

Now, if we will find such $R_f$, $T_f$, that $R_f\phi^0+T_f\varphi^0 = a$ and $R\phi^1 + T\varphi^1 = c$, then, inductively, $f(n) = R_f\phi^n + T_f\varphi^n$. Finding such $R_f$, $T_f$ is just a matter of solving a system of two linear equations. Analogically, we find $R_g$ and $T_g$.

Then, finally

$$ \lim_{n \to \infty} \frac{f(n+k)}{g(n)} = \lim_{n \to \infty} \frac{f(n+k)}{\phi^{n+k}} \cdot \frac{\phi^n}{g(n)} \cdot \phi^k = \frac{R_f}{R_g} \cdot \phi^k $$

I calculated $R_f$ to be $\frac{a\varphi-c}{\varphi-\phi}, so we can plug this values to get

$$ \lim_{n \to \infty} \frac{f(n+k)}{g(n)} = \frac{a\varphi-c}{b\varphi-d} \cdot \phi^k = \frac{c\phi+a}{d\phi+b} \cdot \phi^k $$


问:
Are topological spaces $X,Y$ homeomorphic if each is homeomorphic to a subspace of the other?
The Schroeder-Bernstein theorem states that

If $A$ and $B$ are two sets and there exists injective maps $f:A \to B$ and $g:B \to A$, then there exists a bijection $h:A \to B$.

The statement above is equivalent to the following

Suppose $A_1 \subset A$ and $B_1 \subset B$ are sets such that there exists a bijective map $f:A \to B_1$ and $g:B \to A_1$, then there exists a bijection $h:A \to B$.

So I was thinking about the problem below. I couldn't prove it, but not getting a counterexample either. Any hints will be appreciated.

Now suppose $A \subset X$, $B \subset Y$ are topological spaces, and suppose that there exists a homeomorphism $\alpha:A \to Y$ and $\beta: B \to X$, then can we conclude that $X$ and $Y$ are homeomorphic?

答: No, Suppose $A=[-1,1], X=\mathbb{R}, B=(-2,2)$ and $Y=[-3,3]$. Then this gives a counter example


答:
Put $$x=2t$$

it becomes

$$\int  \frac{dt}{\sqrt{t^2+1}}=$$

$$\sinh^{-1}(t)+C$$

You can use the fact that if

$$F(t)=\ln(t+\sqrt{t^2+1})$$
then
$$F'(t)=\frac{1}{\sqrt{t^2+1}}$$

Title: Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution?

Body:
The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:

$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)


答:
The topological version of Schroeder-Bernstein fails badly. For example, consider $(0,1)$ and $[0,1]$ with their usual topologies, which are of course non-homeomorphic:

$(0,1)$ is homeomorphic to itself, which is a subspace of $[0,1]$.

$[0,1]$ is homeomorphic to $[{1\over 3}, {1\over 2}]$, which is a subspace of $(0,1)$.

In general, the question of when Schroeder-Bernstein holds for a given category of mathematical objects is a quite interesting one; see e.g. this MathOverflow question. For example, while the counterexample above shows that Schroeder-Bernstein fails for metric spaces, it in fact holds for compact metric spaces (and while $[0,1]$ is compact, $(0,1)$ isn't).

Title: Are topological spaces $X,Y$ homeomorphic if each is homeomorphic to a subspace of the other?

Body:
The Schroeder-Bernstein theorem states that

If $A$ and $B$ are two sets and there exists injective maps $f:A \to B$ and $g:B \to A$, then there exists a bijection $h:A \to B$.

The statement above is equivalent to the following

Suppose $A_1 \subset A$ and $B_1 \subset B$ are sets such that there exists a bijective map $f:A \to B_1$ and $g:B \to A_1$, then there exists a bijection $h:A \to B$.

So I was thinking about the problem below. I couldn't prove it, but not getting a counterexample either. Any hints will be appreciated.

Now suppose $A \subset X$, $B \subset Y$ are topological spaces, and suppose that there exists a homeomorphism $\alpha:A \to Y$ and $\beta: B \to X$, then can we conclude that $X$ and $Y$ are homeomorphic?


答: No, consider $X = [0,1]$ and $Y = [0,1] \cup [2,3]$, with their usual subspace topologies. You may let $A = [0,1/3] \cup [1/3,2/3]$ and $B = [0,1]$. $X$ is connected while $Y$ is not.


问:
How to define $f$ so that $\mathbb{R}P^n \cong \mathbb{R}P^{n-1} ~ \cup_f D^n$
I would like to show that $S^n / \{ v \sim -v \}$ is homeomorphic to $\mathbb{R}P^{n-1} ~\cup_f D^n$, where $D^n = \{ x \in \mathbb{R}^n : |x| \leq 1\}$. I have already shown this for $n=2$.

I should define $f$. However, this is where my difficulty lies: when $n=2$, $\mathbb{R}P^{n-1} \cong S^1$, and it is easy to define $f: \partial D^2 \to \mathbb{R}P^1$ as the map taking $z$ to $z^2$ (where we consider $\partial D^2$ and $\mathbb{R}P^1 \cong S^1$ as a subspace of $\mathbb{C}$.) This attaching map identifies three points in $\partial D^2, \mathbb{R}P^1$ with a single point in their adjunction space; namely, $x \in \partial D^2$ with $-x$, along with a pair of antipodal points in $\mathbb{R}P^1$ (how would I word that rigorously?)

What I would like to know is, how can I define $f:\partial D^n \to \mathbb{R}P^{n-1}$ when $n>2$? When $n=2$, I describe $f$ as a map which 'doubles the angle' of a point of $\partial D^2$, but this does not generalize well to higher $n$.

答: It may help to view $\mathbb{RP}^{n-1}$ as the quotient of $S^{n-1}$ by the antipodal map. Along with the identification $\partial D^n = S^{n-1}$, the attaching map is just the 'obvious' map $x \mapsto [x]$. This is expanded upon in Example 2.42 of Hatcher's Algebraic Topology, where he uses this to compute the (cellular) homology of $\mathbb{RP}^n$.


问:
Proof that closed forms are exact using Stokes' theorem.
Terence Tao says that Stokes' theorem could be taken as a definition of the exterior derivative, and in this spirit I am looking for a proof that closed forms are exact using Stokes' theorem. The special case of 1-forms is fairly straightforward, and I'm wondering if there's a similar proof for higher differential forms.

Theorem: A closed differential form $\phi$ on a simply connected domain is exact.

Proof sketch when $\phi$ is a 1-form:
Choose some base point $x_0$ and define $$f(x) = \int_\alpha\phi$$
where $\alpha$ is any path from $x_0$ to $x$. The choice of $\alpha$ does not matter because say we had another path $\alpha'$ from $x_0$ to $x$. Then since our domain is simply connected we can take a surface $S$ who's boundary is $\alpha - \alpha'$. Then using Stokes' theorem:
$$\int_\alpha\phi - \int_{\alpha'}\phi = \int_{\alpha-\alpha'}\phi = \int_Sd\phi = 0$$
since $\phi$ is closed, so the integral is the same over both paths.

Now I claim $\phi = df$. Say $\beta$ is a path from $x$ to $y$, $\alpha$ a path from $x_0$ to $x$, and $\alpha'$ a path from $x_0$ to $y$. Then:
$$\int_\beta\phi = \int_{\alpha + \beta - \alpha'}\phi + \int_{\alpha'}\phi - \int_\alpha\phi = 0 + f(y) - f(x) = \int_{\partial\beta}f$$
since $\alpha + \beta - \alpha'$ is a closed loop and $\phi$ is closed, so by the same logic as in the previous paragraph that integral is 0. Thus, since we are using Stokes' theorem as our definition of the exterior derivative, $\phi = df$ as desired.

Note that I am totally brushing over the issue of whether the exterior derivative is well defined under this definition.

So, is there a way to adapt this proof to higher order forms? One idea I've been playing with is given an $n$-form $\phi$, define the $(n-1)$-form $\varphi$ by:

$$\varphi_x(v_1, \ldots, v_{n-1}) = \int_0^1 \phi_{\alpha(t)}(v_1, \ldots, v_{n-1}, \frac{d\alpha}{dt})dt$$

where $\alpha$ is again a path from $x_0$ to $x$, in this case parameterized by the interval $[0,1]$. The subscript on the forms is the basepoint, and the arguments are tangent vectors at that point. The $v_i$ are tangent vectors at the basepoint $x$, which means for that integral to make sense we need an isomorphism from the tangent space at $x$ to the tangent space at each point along $\alpha$. We can do this by choosing a basis for each tangent space in some continuous manner (we may have to upgrade our "simply connected" qualifier for this to work).

So, can this idea of integrating an $n$-form over a path to get an $(n-1)$-form work as a proof of this theorem? Is there a different way to prove this theorem using Stokes' theorem? I find the above proof for 1-forms very intuitive and I'd like to see a similarly natural proof of this theorem in general.

答:
Okay, I have some ideas towards a proof for 2-forms. Say $\phi$ is a closed 2-form, we will try to construct a 1-form $\varphi$ with $d\varphi = \phi$. We know that integral of $\varphi$ over any closed loop using Stokes' theorem, since the integral of $\phi$ over any surface with that loop for it's boundary is the same. So, say for any pair of points $x$ and $y$ we could choose a main path between them $\alpha_{xy}$, and say we chose the value of $\int_{\alpha_{xy}}\varphi$ for every $x$ and $y$. Then we could determine the value of $\int\varphi$ over any other path $\beta$ with endpoints $x$ and $y$ because we know the integral over $\alpha_{xy}$ and we know the difference between the integral over $\alpha_{xy}$ and the integral over $\beta$. Then by taking limits of paths as their endpoints approach each other we could find the value of $\varphi$ at individual points, so this would be enough information to construct $\varphi$.

But, we have to be careful how we choose our $\alpha$'s and how we choose the values of the integrals over the $\alpha$'s, because not every way of doing this is consistent. I don't know the theory of continuous spaces of paths, and maybe there's a nice way to do this that I don't know about. But here's my idea of how to achieve this: it would be nice if for any points $x$, $y$, $z$, we had $\alpha_{xy} + \alpha_{yz} = \alpha_{xz}$, because then we could choose some scalar field $f$ and let $\int_{\alpha_{xy}}\varphi = f(y) - f(x)$ for all $x$ and $y$. (and $df$ is like the $+C$ because we're integrating.) We can choose such $\alpha$'s in a not so nice way: take some fractal tree that gets arbitrarily close to every point in the space, and define $\alpha_{xy}$ to be the unique path between those two points traveling along the tree. Insert some analysis here to formalize this idea. I think integrating over these paths shouldn't be a problem because they only get weird at their ends, and those ends have arbitrarily small measure so they end up not mattering in the integral. This solution isn't very nice though, I think we'd have to choose $f$ in some discontinuous way if we want $\varphi$ to be continuous.

Now the question is how to extend this to 3-forms. Some kind of tree made of surfaces? I'm guessing there's a nicer solution using the theory of continuous spaces of curves/surfaces.


问:
Find n where $GCD(a_{n}, 14) = 7$
The question was:

Find $n$ where $GCD(a_{n}, 14) = 7$ where $n$ is natural if you knew that $a_{n} = n + 3$

The book solved it by saying that means $a_{n}$ is a multiple of $7$ but not $14$ so $a_{n} = 7a$ and by putting it in the equation

$GCD(7a, 14) = 7$ means $GCD(a,2) = 1$ which means $a = 2k + 1$ and by putting it in $a_{n}$ we find that n = 14a + 4$

The first, I understand that because if it was a multiple of 14 then we would take $14$ as the $GCD$ not $7$

What I didn't understand is that what putting $a_{n} = 7a$ has anything to do with not being a multiple of 14? 14 can be written as $7(2)$ which means $a_{n}$ can be clearly a multiple of 14 too.

答:
The proof is incorrect / incomplete since it gives only a necessary condition for $n$ to be a solution. To fix it you either need to check that the candidate $n$ works or else verify that all inferences are actually bidirectional (see here for more on the insufficiency of unidirectional inferences). Let's do the proof with bidirectional inferences, which will answer your question along the way.

First, $7$ is a common divisor of $\,n+3\,$ and $\,14,\,$ thus a divisor of $\,n+3,\,$ i.e. $\,\color{#0a0}{7\mid n+3},\,$ Using this and $\, 7x = (7y,7z)\color{#c00}\iff x = (y,z)\,$ by  $\rm\color{#c00}{GDL}$ = gcd distributive law we have

$$\begin{align} 7 = (n+3,14) &\iff \ \ \color{#0a0}{7\mid n+3}\ \ \ \ \ \&\ \ \ 7 = (n+3,\ \ 14)\\[.1em]
&\iff n=4+7k\ \ \ \&\ \ \ 7 = (7+7k,\:\!14)\\[.1em]
&\color{#c00}\iff n=4+7k\ \ \ \&\ \ \ 1 = (1+k,\ 2)\ \ \ \rm by\ \color{#c00}{GDL}\\[.1em]
&\iff n=4+7k\ \ \ \&\ \ \ k = 2j\\[.1em]
&\iff n = 4+7(2j)
\end{align}\qquad$$

Title: Find n where $GCD(a_{n}, 14) = 7$