答:
I shall reformulate this into a coloring problem as follows:

Consider a spherical shell $S$ in $\mathbb{R}^3$, the points on whose surface are colored with either one of black $(B)$ or white $(W)$ colors. If more than half of the shell is colored black, show that it is possible to find a line through the center of the shell that joins two black points.

Some notation:

For a point $p$, I shall denote its diametrically opposite point by $p'$. Note that $(p')' = p$. We shall also call $p'$ as the tunnel-complement of $p$.

For every point $p\in S$, I will denote its color by $c(p)$. In general, $c(p)\in\{B,W\}$

We shall complete the proof by way of contradiction.

Suppose it is not possible to dig such a tunnel, i.e. there does not exist $p$ such that $c(p) = c(p') = B$.

So, for every $p\in S$ such that $c(p) = B$, we have $c(p') = W$.

Consider the sets:

$\mathcal{B} = \{p\in S: c(p) = B\}$

$\mathcal{W} = \{p\in S: c(p) = W\}$

$\mathcal{B}':= \{p': p\in \mathcal{B}\} $

$\mathcal{W}':= \{p': p\in \mathcal{W}\} $

We know that $c(p) = W$ for all $p\in \mathcal{B}'$, i.e. all elements diametrically opposite to black elements must be white. It is also clear that $\mathcal{B}\cap\mathcal{W} = \emptyset$ and $\mathcal{B}\cup\mathcal{W} = S$. Also, $|\mathcal{B}| = |\mathcal{B}'|$ and $|\mathcal{W}| = |\mathcal{W}'|$ i.e. the cardinality of a set and its tunnel-complement are the same. This is because there is an obvious bijection between the two, namely $$f:\mathcal{B}\to \mathcal{B}', p\mapsto p'$$ and $$g:\mathcal{W}\to \mathcal{W}', p\mapsto p'$$

We are given that more than half of the shell is colored black, so that $|\mathcal{B}| > |\mathcal{W}|$. Also, $\mathcal{B}' \subset \mathcal{W}$ since all elements opposite to black elements are white. So, $|\mathcal{B}'| \le |\mathcal{W}|$ which gives $|\mathcal{B}| \le |\mathcal{W}|$ from the bijection seen above.

So, we have $|\mathcal{B}| \le |\mathcal{W}|$ and $|\mathcal{B}| > |\mathcal{W}|$ - which is a contradiction. Hence, it is possible to dig the desired tunnel, i.e. we can find a point $p\in S$ such that $p,p'\in\mathcal{B}$.

Title: How to do this counting problem?


问:
Hoeffdings Bound
I have seen the proof of Hoeffding's Bound for bounded RV $X_i\in(m_i,M_i)$

$$\mathbb{P}\bigg\{\sum_{i=1}^{N}(X_i-\mathbb{E}[X_i])\ge t\bigg\}\le \text{exp}\bigg(\frac{-2t^2}{\sum_{i=1}^{N}(M_i-m_i)^2}\bigg)$$

and am able to follow it. As an  exercise in  the textbook High  Dimensional Probability: An Introduction with Application to Data Science you are asked to prove the statement but with a different absolute constant different  from 2. I am not able to see how one can improve the bound. How does one go about showing the inequality with a different absolute constant?

答:
I am assuming you are talking about Exercise 2.2.7 of Vershynin's book. This is not asking you to improve on the constant $2$: it is asking to prove the result for some constant $C>0$ (i.e., larger than 2 is OK if you can't get the optimal one).

Indeed, one can see that the constant $2$ is tight, taking $m_i = -1, M_i = 1$ for all $i$ and considering $X_1,\dots, X_n$ i.i.d. Rademacher random variables (symmetric  Bernoulli).

ContentLicense: CC BY-SA 4.0

Title: Hoeffdings Bound


问:
Generalized Recursion vs. Turing Completeness
$\newcommand{\NN}{\mathbb N}\newcommand{\UU}{\mathcal U}$I'm currently reading through Homotopy Type Theory: Univalent Foundations of Mathematics. In Exercise 1.10, we construct the Ackermann function using (their version of) primitive recursion, which surprised me because I'm used to having the Ackermann function being a function between primitive recursion and computability.

Of course, what makes the exercise possible is that we can recurse over not just $\NN$ but over functions $\NN\to\NN.$ Roughly speaking, the fact we can define $\textsf{ack}(m+1)$ in terms of $\textsf{ack}(m)$ is enough to recursively define the Ackermann function $\textsf{ack}:\NN\to\NN\to\NN.$

Anyways, being able to recurse over arbitrary types seems like a lot of power, and I was wondering exactly how much power.

Is there a total computable function $\NN\to\NN$ which cannot be formed from the above "generalized" primitive recursion?

I suspect one to exist. If there's a finer line between what generalized primitive recursion gives and computable enumerability, I'd be interested in that too.

答:
Your example, at least, doesn't depend specifically on HoTT. The Ackermann function can be written on numerals in System F for similar reasons. In that case, the expressible functions are those provably total in second-order arithemtic.

The correspondence probably depends on the specifics of the type theory, though. So I am also not sure what the answer would be for something like Martin-löf type theory, and it might depend on exactly which type formers you include, 'how many' universes you have, etc. I would guess that univalence is a non-factor in this sort of thing, but I'm not certain.


问:
$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}$
I have to find this limit but I'm not sure how can I do it

Find this limit:

$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}$

I analyzed it by this way:

\begin{align}
\underbrace{\lim_{n \rightarrow \infty}\left ( \frac{2^n}{2^{n+1}+(-1)^{n+1}} \right )}_{\text{This must go to zero}} + \underbrace{\lim_{n \rightarrow \infty} \left ( \frac{(-1)^n}{2^{n+1}+(-1)^{n+1}} \right )}_{\text{This, also must converge to zero}}
\end{align}

Am I correct? How can I proove it?

答:
$$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}
=\frac 1 2\lim_{n \rightarrow \infty} \frac{1+\left(\frac{(-1)}2\right)^{n}}{1+\left(\frac{(-1)}2\right)^{n+1}}\\=\frac1 2$$

We used $$\lim_{n\to\infty}\left(\frac{(-1)}2\right)^{n}=0$$ which is true as $\big|\left(\frac{(-1)}2\right)\big|<1$.(which implies geometric series converges which then implies limit of nth term goes to zero.)


Body:
I have to find this limit but I'm not sure how can I do it

Find this limit:

$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}$

I analyzed it by this way:

\begin{align}
\underbrace{\lim_{n \rightarrow \infty}\left ( \frac{2^n}{2^{n+1}+(-1)^{n+1}} \right )}_{\text{This must go to zero}} + \underbrace{\lim_{n \rightarrow \infty} \left ( \frac{(-1)^n}{2^{n+1}+(-1)^{n+1}} \right )}_{\text{This, also must converge to zero}}
\end{align}

Am I correct? How can I proove it?

Body:
$$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}$$

$$\lim_{n \rightarrow \infty} \frac{\frac{2^n}{2^n}+\frac{(-1)^n}{2^n}}{\frac{2^{n+1}}{2^n}+\frac{(-1)^{n+1}}{2^n}}$$

As $n \to \infty,$ ($\frac{1}{2^n}$ or $\frac{-1}{2^n}) \to0$

Therefore,

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

The mistake in your method is that the first limit is $1/2$ and not $0$


问:
How to do this counting problem?
I know that the pigeonhole principle is to be applied here but I can’t see yet how.

On a certain planet in the solar system Tau Cetus, more than half the surface of the planet is dry land. Show that the Tau Cetans can dig a tunnel straight through the centre of the planet, beginning and ending on dry land.

答:
We can apply the pigeon hole principle by doing something like this:

First, discretize half the surface of the planet into n pigeon holes. Each pigeon hole can accept two items. First, we fill the pigeon holes with dry land. For any pigeon hole with only one item in it, we can now fill it with wet land.

If we have k regions of dry land and k>n, then we'll have to fill an additional (k-n) pigeon holes with a 2nd region of dry land. Now, we have less than n pigeon holes to fill with wet land which corresponds to less than half of the surface being covered with water.

Each pigeon hole containing two areas of dry land corresponds to positions where you can
start digging in dry land and reach another area of dry land. This only occurs when k>n or in other words, when more than half of the surface is dry.

Title: How to do this counting problem?


问:
What's the best way for an engineer to learn "real" math?
I'm an electrical/computer engineering student and have taken fair number of engineering math courses. In addition to Calc 1/2/3 (differential, integral and multivariable respectfully), I've also taken a course on linear algebra, basic differential equations, basic complex analysis, probability and signal processing (which was essentially a course on different integral transforms).

I'm really interested in learning rigorous math, however the math courses I've taken so far have been very applied - they've been taught with a focus on solving problems instead of proving theorems. I would have to relearn most of what I've been taught, this time with a focus on proofs.

However, I'm afraid that if I spend a while relearning content I already know, I'll soon become bored and lose motivation. However, I don't think not revisiting topics I already know is a good idea, because it would be next to impossible to learn higher level math without knowing lower level math from a proof based point of view.

答:
There is no "engineering math" separate from "math". To learn good math as an engineer, you have no other way but to understand math from its nature.

The first course you learn is probably real analysis. In my opinion, no one masters real analysis when the topic is met the first time. Don't worry if you feel you have done the homework and passed the exam but don't have a clear clue of what it is for. Real analysis is the topic that opens the door to modern math, as compared to what you learned at high school. The core technique you need to master is formal reasoning, or proof construction. Learning first-order logic systematically would definitely provide you with a clearer understanding of this technique. Once you have mastered the tool for formal reasoning, you are able to appreciate proofs written in English in a more fundamental way. Without the knowledge of first order logic, long proofs cannot be truly understood, especially when there are many levels of existential and universal propositions. After understanding first-order logic and probably a little set theory, it is the moment you review real analysis and tell why learning the material the first time is not really helpful.

You will definitely learn linear algebra. This is the course you learn linear spaces and transformations between linear spaces. It is the first time you understand that modern math is concerned with the structures of specific types of sets with operations defined on them. Some universities teach this topic with an emphasis on matrices and vectors. I don't quite agree with this type of education for linear algebra. I would recommend learning this topic from the bottom up. That is, understanding clearly the concepts of linear transformations, and then, associate them with matrices and vectors. With the proof capability you learned in real analysis, you should be able to understand and construct proofs more easily in this course.

Then you will need to know how to model the world of uncertainty. Probability theory and statistics are what you need to learn. Again, pay more attention to theory and do not focus on data at the beginning. You need to rigorously construct your understanding of events, probability, random variables, expectation, convergence, etc. Do note that the concept of convergence from real analysis plays a significant role in the analysis of estimators.

With all these basic course learned and understood, you should learn optimization. Optimization tells you how to make the best choice with a definition of cost in mind. It is widely used in signal processing, control, machine learning, computer graphics, etc. In fact, many rigorous engineering papers published on good conferences or journals are optimization in nature.

With these topics learned and practiced in everyday applications, you should be able to deal with most engineering problems in electrical engineering or computer science in a very rigorous way. With these describing languages in mind, you should solve real problems with coding. Now, coding is a technique that connects knowledge and capability. With a good understanding of math, you should be able to write high-quality codes.

ContentLicense: CC BY-SA 3.0


Title: $\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}}$

Body:
$$\lim_{n \rightarrow \infty} \frac{2^n+(-1)^n}{2^{n+1}+(-1)^{n+1}} = \lim_{n \rightarrow \infty} \frac{2^n+(-1)^{n+1}/2}{2^{n+1}+(-1)^{n+1}} +  \lim_{n \rightarrow \infty}\frac{-(-1)^{n+1}/2 + (-1)^n}{2^{n+1}+(-1)^{n+1}}$$

and:

$$\frac{-1-1}{2^n} <\frac{-(-1)^{n+1}/2 + (-1)^n}{2^{n+1}+(-1)^{n+1}} < \frac{1+1}{2^n}$$

thus by the squeeze theorem, the limit is $\frac{1}{2} + 0 = \frac{1}{2}$.


问:
If we know the congruence of the apex and the base of two isosceles, are both congruent triangles? (help with the proof)
Well, there are general criteria for congruence of triangles:

SAS (Side-Angle-Side)

ASA (Angle-Side-Angle)

SSS (Side-Side-Side)

SAA (Side-Angle-Angle)

And there are some criteria for specific triangles:

SSA (Side-Side-Angle) The side in front of the angle doesn't have to be minor of the adjacent one

HC (Hypotenuse-Cathetus) In a right triangle

HA (Hypotenuse-Angle) In a right triangle

And my question is about one case that I know is true in Euclidean geometry (which I would call AB—Apex-Base—):

If two isosceles triangles have in common the apex and the base, both are congruent

It's "obvious" but the proof is not that obvious. Here's my attempt.

Proof

Let △ABC and △A'B'C' be isosceles triangles like the next figure

Now, if AB = A'B' both triangles are congruent by SAS, but suppose that's not the case. Without loss of generality assume that AB<A'B', thus it's possible to mark a point D' in A'B' such that A'D' = AB, likewise it's possible to mark a point E' in A'C' such that A'E' = AC. Join the points D'E' to form the △A'D'E' ≅ △ABC by SAS.

Then we must conclude that D'E' = B'C', but this is a contradiction since D'E'<B'C', so assuming  AB ≠ A'B' is absurd, hence AB = A'B' which derives △ABC ≅ △A'B'C' Q.E.D.

That's what I did, but I feel that the contradiction is not fully handled, especially because I cannot find another argument but visual that D'E' < B'C'.
As I mentioned before, I know this is true in Euclid's (plane) geometry. I'm more interested in a proof (if exist) avoiding parallels (or equivalently the constant sum of the measures of the angles of any triangle), so that the result would be a neutral geometry like the SAA criterion (https://www.jstor.org/stable/27960835?seq=1).

Or I'm trying to prove a result that's not neutral geometry that in some cases D'E' =B'C'?

Edit

Thanks for your answers, they are very useful. But up this moment all of them depends on the parallel postulate which I try to avoid in order to find out if this is a neutral geometry result.

答:
If you know that the sum of angles in a triangle are $180^\circ$, then, if the apex has value $A$, the other two angles are $90^\circ-A/2$. Then the problem is reduced to the angle-side-angle case. The side is the base.

In your case you use the same angles to show $D'E'||B'C'$. Then the triangles are similar, and the ratio of the sides is equal to the ratio of the bases. You know that the bases are the same, so then the sides must be the same. The problem is then reduced to side-side-side case.


答:
Your proof is correct - good job there.

An easier way to see this is the following:

You have an isoceles triangle, so $\angle ABC=\angle ACB$ (for both).

You know $\angle BAC=\angle B'A'C'$ (by hypothesis).

Angles add up to $180^{\circ}$, so $\angle ABC=\angle A'B'C'=90^{\circ}-\angle BAC$.

Therefore, $\triangle ABC$ is similar to $\triangle A'B'C'$.

You additionally know that length $BC=B'C'$.

Well, they're now congruent (by SAA or equivalent).

Title: If we know the congruence of the apex and the base of two isosceles, are both congruent triangles? (help with the proof)

Body:
Well, there are general criteria for congruence of triangles:

SAS (Side-Angle-Side)

ASA (Angle-Side-Angle)

SSS (Side-Side-Side)

SAA (Side-Angle-Angle)

And there are some criteria for specific triangles:

SSA (Side-Side-Angle) The side in front of the angle doesn't have to be minor of the adjacent one

HC (Hypotenuse-Cathetus) In a right triangle

HA (Hypotenuse-Angle) In a right triangle

And my question is about one case that I know is true in Euclidean geometry (which I would call AB—Apex-Base—):

If two isosceles triangles have in common the apex and the base, both are congruent

It's "obvious" but the proof is not that obvious. Here's my attempt.

Proof

Let △ABC and △A'B'C' be isosceles triangles like the next figure

Now, if AB = A'B' both triangles are congruent by SAS, but suppose that's not the case. Without loss of generality assume that AB<A'B', thus it's possible to mark a point D' in A'B' such that A'D' = AB, likewise it's possible to mark a point E' in A'C' such that A'E' = AC. Join the points D'E' to form the △A'D'E' ≅ △ABC by SAS.

Then we must conclude that D'E' = B'C', but this is a contradiction since D'E'<B'C', so assuming  AB ≠ A'B' is absurd, hence AB = A'B' which derives △ABC ≅ △A'B'C' Q.E.D.

That's what I did, but I feel that the contradiction is not fully handled, especially because I cannot find another argument but visual that D'E' < B'C'.
As I mentioned before, I know this is true in Euclid's (plane) geometry. I'm more interested in a proof (if exist) avoiding parallels (or equivalently the constant sum of the measures of the angles of any triangle), so that the result would be a neutral geometry like the SAA criterion (https://www.jstor.org/stable/27960835?seq=1).

Or I'm trying to prove a result that's not neutral geometry that in some cases D'E' =B'C'?

Edit

Thanks for your answers, they are very useful. But up this moment all of them depends on the parallel postulate which I try to avoid in order to find out if this is a neutral geometry result.


问:
What is the professional term for the combination of the selection in n out of the total m elements?
I know the number of combinations is called ${}_nC_r$, but what about all the exact outcomes?

For example: I have $3$ elements $a,b,c$ and for the parameter $2$, I will have outcomes
$$ab,\quad  ac,\quad ba,\quad bc,\quad ca,\quad cb$$

I want to search different implementations of this, but I don't know what term should I input in google.

答:
If I have a set $X$ with $n$ elements, and I wanted to refer one of the possible "outcomes" of choosing $k$ elements (without the order mattering), I would just call it "a $k$-element subset of $X$". Thus, the number of $k$-element subsets of $X$ is $\binom{n}{k}$ or ${}_nC_k$, depending on your notational preferences. I'm not aware of any more specialized term.

I would use the term "$k$-tuple of elements of $X$" to mean an ordered $k$-element subset of $X$ (possibly with repetitions). Thus, the number of $k$-tuples of distinct elements of $X$ is ${}_nP_k$, and the number of $k$-tuples of elements of $X$ is $n^k$.


问:
Is this an alternative to the Pythagoras theorem or a Proof?
$\triangle$ABC, BC = 3 AC = 4

$\triangle$DEF, EF = 3

($\triangle$DEF can be created by using one of the given sides in $\triangle$ABC
and angles of $\triangle$DEF will be the same as $\triangle$ABC where both C and F are right angles.)

(We are Finding AB),

Step 1: Find DF using the ratio of $\angle$E where $\angle$E = $\angle$A.

Ratio of $\angle$E = DF/3

Ratio of $\angle$A = 3/4

DF/3 = 3/4

DF = 9/4

Step 2: Prove that $\triangle$ABC $\sim$ $\triangle$DEF by AA-

$\angle$A =  $\angle$E

$\angle$C = $\angle$F

Therefore: $\triangle$ABC $\sim$ $\triangle$DEF by AA-

Step 3: Find DE using the scaled factor.

DE = AB(3/4)

step 4: Reflect $\triangle$DEF in the x-axis and align BC and DF to get $\triangle$ABD

Diagram of $\triangle$ ABD

(BC)(AD)/2 = A, Equation 1

(BD)(AB)/2 = A, Equation 2

(A represents the area of $\triangle$ABD)

Substitute 1 into 2

(BC)(AD) = (BD)(AB), Equation 3

(AD = AC + DF)

Substitute AB, BC, AC ,DE and DF into 3

3(4 + 9/4) = (AB$^2$)3/4

3(25/4) = (AB$^2$)3/4

(25)3/4 = (AB$^2$)3/4

AB$^2$ = 25

AB = 5

答:
You have a triangle $ABD$ with right angle at $B$ and a line through $B$ perpendicular to $\overline{AD}$ at $C$.

Your equation
$$
3\cdot\left(4+\frac{9}{4}\right)=(AB)^2\cdot\frac{3}{4}
$$
could be generalized to
$$
(BC)\cdot\left(AC+\frac{(BC)^2}{(AC)}\right)=(AB)^2\cdot\frac{BC}{AC}.
$$
If you multiply both sides by $(AC)$ and divide both sides by $(BC)$ you get
$$
(AC)^2+(BC)^2=(AB)^2,
$$
which is the Pythagorean theorem.


问:
How many ways are there to place $10$ different flags on $7$ different flagpoles if order matters and no flagpole is left empty?
How many ways are there to place $10$ different flags on $7$ different flagpoles if the order of the flags on a flagpole is relevant and every flagpole flies at least one flag?

The way I look at it is that if need at least $1$ flag for each one you can remove that from there so now you $3$ flags to put into $7$ flagpoles which is $7 \cdot 6 \cdot 5 =210$.

答:
Let's start by looking at how many ways we can place $7$ of our $10$ flags on our $7$ poles, so that each pole gets exactly one flag. The first pole can have any of the $10$ flags, then the next pole can have any of the remaining $9$, then the next can have $8$, and so on, giving us a total number of $\dfrac{10!}{(10-7)!} = 604$ $800$ possible ways to give each pole a flag.

Now, given each of these possible ways to put $7$ flags on $7$ poles, let's find how many orders of the $10$ flags on $7$ poles we get. When we place the $3$ remaining flags, we have a few choices we can make: we can either place them such that each flag is on their own, such that two flags are placed together, or such that all three flags are placed in one group. Let's run through each of these options:

Each on their own: If each flag is placed on their own, then we get the $210$ possibilities you calculated originally: the first flag gets $7$ options, the second gets $6$, and the last gets $5$.

Two at a time: If we place them so that two of them are together, then the group of two gets $7$ choices and the single flag gets $6$. We also have to consider the $6$ possible ways to divide the flags this way: $3$ possible ways to choose the group of $2$, and $2$ ways to order each group of $2$. All together, this gives us $252$ possible orders.

Three at a time: If we place the three flags all together, then we have $7$ choices for where to put them and $6$ choices for the order, yielding $42$ possible orders.

Adding these three cases together gives us $504$ possible ways to place the three remaining flags. This means that the total number of ways that we can put $10$ flags onto $7$ flagpoles such that each flagpole has at least one flag is $604$ $800 \cdot 504 =  304$ $819$ $200$.

Edit: got some feedback from the comments.


问:
"Natural" choice of representative of left cosets in a symmetric group
Let $n$ be a positive integer and let $n=n_1+\cdots+ n_k$ be a partition of $n$, where each $n_i$ is a positive integer. Let $\Sigma_n$ be the symmetric group on $n$ letters. Then
$$\Sigma_{n_1}\times\cdots\times\Sigma_{n_k}$$ is a subgroup of $\Sigma_n$. My question is:

Is there a "natural" choice of representatives of left cosets of
$\Sigma_{n_1}\times\cdots\times\Sigma_{n_k}$ in  $\Sigma_n$?

答:
tl;dr: No

I tried to begin by making the word "natural" precise. Let $\mathsf{C}$ be the following category:

An object is a pair $(X,P)$ where $X$ is a finite set and $P \subseteq 2^X$ is a partition of $X$.

A morphism $(X,P) \to (Y,P')$ is an injective function $f : X \to Y$ such
that, for all $S \in P$, there exists $T \in P'$ such that $f(S)
   \subseteq T$.

Let $\Gamma : \mathsf{C} \to \mathsf{Set}$ be the functor defined as follows:

Given an object $(X, \{X_1, \dots, X_n\})$, $\Sigma_{X_1} \times \dots \times \Sigma_{X_n}$ can be canonically identified with a subgroup of $\Sigma_X$ via the inclusions $X_i \subset X$, and this subgroup does not depend on the choice of ordering of the $X_i$'s. Thus, it makes sense to define
$$\Gamma(X, \{X_1, \dots, X_n\}) = \Sigma_X/(\Sigma_{X_1} \times \dots \times \Sigma_{X_n} )$$
(this is just the set of left cosets, not necessarily a group).

Given a morphism $f : (X,\{X_1, \dots, X_n\}) \to (Y,\{Y_1, \dots, Y_m\})$, the injective function $f : X \to Y$ defines a map $f_* : \Sigma_X \to \Sigma_Y$ by
$$f_*(\sigma)(y) = \begin{cases} y &: y \notin f(X) \\ f(\sigma(f^{-1}(y))) &: y \in f(X) \end{cases}$$
This map has the property that $f_*(\Sigma_{X_1} \times \dots \times \Sigma_{X_n}) \subseteq \Sigma_{Y_1} \times \dots \times \Sigma_{Y_m}$. This means that if $\sigma_1^{-1} \sigma_2 \in \Sigma_{X_1} \times \dots \times \Sigma_{X_n}$ then $f_*(\sigma_1)^{-1} f_*(\sigma_2) \in \Sigma_{Y_1} \times \dots \times \Sigma_{Y_m}$, so it makes sense to define
$$\Gamma(f)(\sigma (\Sigma_{X_1} \times \dots \times \Sigma_{X_n})) = f_*(\sigma) (\Sigma_{Y_1} \times \dots \times \Sigma_{Y_m}).$$

For any $(X,P) \in \mathsf{C}$, the identity function on $X$ gives a morphism $\varepsilon_{(X,P)} : (X,\{\{x\} : x \in X\}) \to (X,P)$. This defines a natural transformation $\varepsilon : D \to \operatorname{id}_{\mathsf{C}}$, where $D : \mathsf{C} \to \mathsf{C}$ is the functor $(X,P) \mapsto (X,\{\{x\} : x \in X\})$ (as the notation may suggest, $\varepsilon$ is the counit of an adjunction).

Now consider the natural transformation $\Gamma \varepsilon : \Gamma \circ D \to \Gamma$. For any $(X,P) \in \mathsf{C}$, we have that $$\Gamma(D(X,P)) = \Sigma(X,\{\{x\} : x \in X\}) = \Sigma_X \left/ \prod_{x \in X} \Sigma_{\{x\}}\right. = \Sigma_X / \{e\}$$
is naturally isomorphic to $\Sigma_X$. By canonically identifying $\Sigma_X / \{e\}$ and $\Sigma_X$, we get that $(\Gamma \varepsilon)_{(X,P)}$ is precisely the quotient map
$$\Sigma_X \to \Gamma(X,P).$$
A splitting of this quotient map is precisely a choice of representatives for the cosets, so to look for a natural choice of representatives, we could ask: does there exist a natural splitting of this map? In other words:

Question Does there exist a natural transformation $\xi : \Gamma \to \Gamma \circ D$ such that $(\Gamma \varepsilon) \circ \xi = \operatorname{id}_\Gamma$?

Answer No.

Proof. Suppose such a natural transformation $\xi$ exists. Let $X$ be a set with more than one element. Let $A = (X,\{\{x\} : x \in X\})$ and $B = (X,\{X\})$. By naturality, we get the following commutative diagram:

$\require{AMScd}$
\begin{CD}
\Gamma(A) @>{\Gamma(\varepsilon_B)}>> \Gamma(B)\\
@V{\xi_A}VV @V{\xi_B}VV\\
\Sigma_X @>{\Gamma(D(\varepsilon_B)) = \operatorname{id}_{\Sigma_X}}>> \Sigma_X
\end{CD}

Since $\Gamma(B)$ is a singleton set, $\xi_A = \operatorname{id}_{\Sigma_X} \circ \xi_A = \xi_B \circ \Gamma(\varepsilon_B)$ is constant. Then $\operatorname{id}_{\Gamma(A)} = \Gamma(\varepsilon_A) \circ \xi_A$ is constant, but $\Gamma(A) \cong \Sigma_X$! This contradicts the fact that $X$ has more than one element.

Maybe this is too restrictive – why do we demand that $\xi$ be natural with respect to all morphisms in $\mathsf{C}$? A seemingly reasonable but weaker condition would be to ask that $\xi$ is natural with respect to isomorphisms in $\mathsf{C}$ (formally, we can form the subcategory $\mathsf{C}' \leq \mathsf{C}$ which contains only isomorphisms, then restrict everything down to $\mathsf{C}'$). However, even this condition (which is about as little as one could hope for) cannot be satisfied.

Proof. Suppose there is a splitting $\xi$ of $\Gamma \varepsilon$ which is natural with respect to isomorphisms in $\mathsf{C}$. Let $Q = (\{1, \dots, 6\}, \{\{1,2,3\},\{4,5,6\}\})$, let $G = \Sigma_{\{1, \dots, 6\}}$ and let $H = \Sigma_{\{1,2,3\}} \times \Sigma_{\{4,5,6\}} \leq G$. Let $N$ be the normalizer of $H$. By a simple computation, $N = \operatorname{Aut}_{\mathsf{C}}(Q)$, and the induced action of $N$ on $\Gamma(Q) = G/H$ is given by conjugation. The naturality of $\xi$ says that that $\xi_Q : G/H \to G$ is $N$-equivariant (where $N$ is also acting by conjugation on $G$). We now restrict our attention to $\xi_Q|_{N/H} : N/H \to N$. First of all, this is still a morphism of $N$-sets, because $N/H$ and $N$ are closed under conjugation by elements of $N$. By direct computation, $[N : H] = 2$, so $N/H$ is an abelian group. But then the conjugation action of $N$ on $N/H$ is just given by
$$n \cdot (n' H) = (nn'n^{-1})H = (nH)(n'H)(nH)^{-1} = n'H,$$
so this action is trivial! The $N$-equivariance of $\xi_Q|_{N/H}$ then tells us that $\xi_Q(N/H) \subseteq Z(N)$. By direct computation, $N$ has trivial center, which means $\xi_Q|_{N/H}$ is constant. But $\xi_Q$ is injective, and $\lvert N/H \rvert > 1$, so this is impossible.

I make no claim that this is the simplest counterexample! Anyway, I think this argument shows that there can't be anything like a natural choice of representatives for the cosets of $\Sigma_{X_1} \times \dots \times \Sigma_{X_n}$ in $\Sigma_X$.


问:
Does a minimizer of $f(x)= \sum_{i=1}^n 1_{ \{ \langle v_i,x \rangle > 0 \} } \cdot \langle v_i,x \rangle $ makes most of the terms zero?
Consider a collection of  2-d vectors $v_1,v_2,...,v_n$. Assume all of these are unit norm.   Let
\begin{align}
f(x)= \sum_{i=1}^n 1_{ \{ \langle v_i,x \rangle > 0 \} } \cdot \langle v_i,x \rangle
\end{align}
where $1_{ \{ \cdot \} }$ is the indicator function.
We want to minimize  $f(x)$ over all unit norm vectors $x$.   The solution for $n=2$ is given in here.   Note there can be many minimizers.

Questions:   Q1: Is it true that that one of the minimizers of  $f$  makes the most inner product terms non-positive?

In other words,  define a set
\begin{align}
A= \arg \max_{x: \|x\|=1}  | \{v_i: \langle v_i,x \rangle  \le 0 \}| ,
\end{align}
where $|\cdot |$ is the cardinality operator. The set $A$ contains $x$ that make the most inner producs negative.  Then, one of the  minimizers of $f$ is member of a set $A$.

In essence, the question is asking if there exists a minimizer of $f$ that first makes as many as possible of the terms to be equal to zero.

For $n=2$, where we know the solution, this is true.  It might happen that there exists a choice of $v_i$'s such that this is not true (see in the answers).

Q2: Under what conditions on $v_i$'s is this true?

答: Let $n=5$, $\varepsilon$ be a sufficiently small positive number, $v_1=(0,1)$, $v_2=(0,-1)$, $v_3$ equals $v_1$ rotated by an angle $\varepsilon$ clockwise, $v_4$ equals $v_2$ rotated by an angle $\varepsilon$ counter-clockwise, and $v_5=(-1,0)$. Then $A=\{(-1,0)\}$, but $f(-1,0)=1$, whereas $f(1,0)=2\sin\varepsilon<1$.


问:
Are the reals genuinely a subset of the complex numbers?
In Michael Spivak's Calculus, he defines a complex number as an ordered pair of real numbers: $z=(a,b)$ with $a,b \in \mathbb{R}$. The imaginary unit $i$ is then just a shorthand for the ordered pair $(0,1)$. Spivak goes on to say

When complex numbers were first introduced, it was understood that real numbers were, in particular, complex numbers; if our definition is to be taken seriously then this is not true—a real number is not a pair of real numbers, after all.

Although the complex number $(a,0)$ behaves in pretty much the same way as the real number $a$, they are still not identical. There seem to be some non-trivial differences, as well: we can't write $(5,0)>(3,0)$ in the way we can write $5>3$. With this in mind, I ask the following questions:

Can the real numbers be said to be a subset of the complex numbers if complex numbers are defined as ordered pairs?

If the complex numbers are constructed in some other way, then is it meaningful to write $\mathbb R \subset \mathbb C$?

答:
This is actually one of the problems with the idea of building maths purely on sets alone as foundations. If we take it strictly and formally, we get various statements which may or may not be true, subject to just how we have or have not constructed a particular object. For example, it gets even worse than what you are talking about: in a purely set-theoretic construction, natural numbers are sets, too, and thus we can ask whether, say,

$$1 \subseteq 3$$

and the answer to this is "it depends on your set-theoretic construction"!

For me, what I suggest is that this problem is very reminiscent of one often seen in computer programming: in computers, we have something similar going on in that everything we work with - pictures, sound, text, whatever - ultimately gets represented by the same "stuff": bits. And thus, if one does not have safeguards in place, one can try to interpret, say, the bits corresponding to text as a picture, or a picture as text, or conversely. Of course, what you get will be mostly scramble and nonsense, but you can do it, and the computer won't care.

So to deal with this, we need some way to encode that semantic information - that these two pieces of bits are semantically different - into the language in question.

And the way that is handled in computer programming is to use programming languages that require a data type, to discourage the programmer from arbitrarily mixing of different sets of bits that are meant to represent different things. Data typing attaches a semantic tag to each bit of data to say that it should represent a picture or text or a number, say, and then you cannot, in the same program, freely mix the two.

Likewise, this concept is not unheard of in maths - "type theory" explores a whole array of foundational systems and languages that use something very similar, and indeed both of these fields of application are closely related - but it's not the "standard consensus" foundation for maths.

But were we to use a typed foundation, I'd suggest the answer would best be thought of as a "no, but": the real numbers are not a subset of complex numbers, but we have the "type coercion" rule

$$x \mapsto (x, 0)$$

which allows us to "upgrade" a real number, should it be combined with a complex number in an expression, to a complex number. Such rules often feature in programming languages as I just mentioned, too. In general, they must be defined along with the types in question, but are typically based on whether or not "natural" correspondences of the kind you are perceiving here, exist.


问:
Formally assigning probabilities to sample spaces that change over time
As mentioned in the title, how are probabilities assigned to changing sample spaces?

To make my question more clear, take the classic Polya's Urn Scheme, where we have an urn containing $b$ black balls and $r$ red balls. After each draw of a ball from the urn, the ball is replaced with $c$ more balls of the same color back into the urn. Regarding the first draw, I'm clear..since each ball is equally likely to be picked, a uniform probability space $(\Omega, \mathcal{F}, \mathbb{P})$ does the job once $\Omega$ is taken to be the set of all balls in the urn.

After the first draw, or any draw for that matter, the set of balls in the urn is no longer the same as $\Omega$. But any solution to this problem that I've seen does not mention this..and the answer is computed directly. Of course, its intuitively clear what's going on..but is there a more formal way of depicting this? Do we move to a new sample space $(\Omega', \mathcal{F'}, \mathbb{P}')$ after each draw?

What is the general way of working with sample spaces that change over time / after the occurrence of some events?

答:
What I would do is the following:

Fix a probability space $(\Omega,\mathcal{F},P)$ that captures all the states of the world along the urn process. On that space define two series of random variables $B_n$ and $R_n$ to model the number of black and red balls after $n$ steps in the process (e.g. $B_0=b$ $P$-a.s.).

In this setting one would be able to deduce recursive relations for $B_n$ and $R_n$ without changing the underlying probability space. Instead one gets two series of new sample spaces induced from the random variables. Assuming e.g. all of the $B_n$ are real valued, one would now sample in $(\mathbb{R},\operatorname{Borel}(\mathbb{R}),P^{B_n})$. And these spaces might change in $n$.

Each $\omega\in\Omega$ defines a random series of balls drawn from the urn. The specific definition of $\Omega$ and $P$ do not matter, as long as the induced measures in the sample spaces are as expected (e.g. $P^{B_n}$ could be some discrete distribution conditionally depending on $B_{n-1}$ and $R_{n-1}$).

Edit:

To address the question more directly:
With the above approach we do not have to change anything about $\Omega$ in the process. Instead, we define random variables on $\Omega$ and have their induced sample spaces change with $n$ (though it’s really only the induced measure, e.g. $P^{B_n}$, that changes in $n$).


问:
What is the computational complexity to solve a system of linear equations?
Let $A$ be a known $m \times n$ matrix, $x$ an unknown $1 \times n$ column vector, and $b$ a known $1 \times n$ column vector.
For the following cases, what is the time complexity to solve the system of linear equations $Ax = b$? What algorithm achieves that time complexity? Assume that at least one solution exists.

Case 1: m < n.

Case 2: m = n.

Case 3: m > n.

答:
Either Gaussian elimination with partial pivoting or (householder) QR followed by backward/forward substitution run in $O(mn \min(m,n))$ arithmetic operations and produce a solution to a consistent system of linear equations. When properly implemented on computers using floating point arithmetic, these algorithms almost always produce answers about as accurate as you could expect them to.

If you’re a theoretical computer scientist, you may care not just about the number of arithmetic operations but the complexity of performing these arithmetic operations on the possibly increasingly complicated-to-describe rational numbers produced in Gaussian elimination. I’m not aware of the best exact complexity in general, but it is known that linear systems are solved in polynomial time in the bit complexity required describe the input matrix and right-hand side (in any of the “natural” encodings) by using Gaussian elimination. (This question might have some insights you find interesting.)


问:
Wedge product of a 2-form with a 1-form.
(Wedge product of a 2-form with a 1-form).*

Let $\omega$ be a $2-$form and $\tau$ a $1-$ form on $\mathbb R^3$.
  If $X, Y, Z$ are vector fields on $M$, find an explicit formula for
  $(\omega ∧\tau )(X,Y,Z)$ in terms of the values of $\omega$ and $\tau$
  on the vector fields $X,Y,Z.$

Let $x^1,x^2,x^3$ are the coordinate function.$$\omega=f_1dx^1\wedge dx^2+ f_2dx^2\wedge dx^3+ f_3dx^1\wedge dx^3$$
$$\tau=g_1x^1+g_2 x^2+g_3x^3$$
$\omega \wedge \tau=f_1g_3dx^1\wedge dx^2 \wedge x^3+f_2g_1dx^2\wedge dx^3 \wedge dx^1+f_3g_2 dx^1\wedge dx^3\wedge x^2$
$$X=a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3}$$
$$Y=b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}$$
$$Z=c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3}$$

$$\omega \wedge \tau=f_1g_3dx^1\wedge dx^2 \wedge x^3+f_2g_1dx^2\wedge dx^3 \wedge dx^1+f_3g_2 dx^1\wedge dx^3\wedge x^2(X,Y,Z)$$ $$=f_1g_3dx^1\wedge dx^2 \wedge x^3+f_2g_1dx^2\wedge dx^3 \wedge dx^1+f_3g_2 dx^1\wedge dx^3\wedge x^2(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})=f_1g_3dx^1\wedge dx^2 \wedge x^3(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})+f_2g_1dx^2\wedge dx^3 \wedge dx^1(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})+f_3g_2 dx^1\wedge dx^3\wedge x^2(a^1\frac{\partial}{\partial x^1}+ a^2 \frac{\partial}{\partial x^2}+a^3 \frac{\partial}{\partial x^3},b^1\frac{\partial}{\partial x^1}+ b^2 \frac{\partial}{\partial x^2}+b^3 \frac{\partial}{\partial x^3}, c^1\frac{\partial}{\partial x^1}+ c^2 \frac{\partial}{\partial x^2}+c^3 \frac{\partial}{\partial x^3})=f_1g_3a^1b^2c^3+f_2g_1a^2b^3c^1+f_3g_2a^1b^3c^2$$

Am I correct?

答:
Since we have by definition of the wedge product:
$$(f\wedge g)(v_1,v_2,...,v_{k+l})={1\over k!l!}\sum_{\sigma\in S_{k+l}} (sgn \sigma)f(v_{\sigma(1)},v_{\sigma(2)}...,v_{\sigma(k)})g(v_{\sigma(k+1)},...,v_{\sigma(k+l)})$$

There should be a factor of $$1\over2$$ in the answer by Angina Sang.
$$(\omega\wedge\tau)(X,Y,Z)={1\over2} \{\omega(X,Y)\tau(Z)-\omega(X,Z)\tau(Y)
+\omega(Y,Z)\tau(X)\}$$
There will not be a factor of $1\over2$ because we have used (k,l) shuffles.So the correct answer is $$(\omega\wedge\tau)(X,Y,Z)= \{\omega(X,Y)\tau(Z)-\omega(X,Z)\tau(Y)
+\omega(Y,Z)\tau(X)\}$$


问:
Composite Function Type problem
Let $f\left( x \right) = {x^2}$ and $g\left( x \right) = \sin x$ for all $x \in R$. Then the set of all x satisfying $\left( {fogogof} \right)\left( x \right) = \left( {gogof} \right)\left( x \right)$, where $\left( {fog} \right)\left( x \right) = f\left( {g\left( x \right)} \right)$ is

(A) $ \pm \sqrt {n\pi } ,n \in \left\{ {0,1,2,...} \right\}$

(B) $ \pm \sqrt {n\pi } ,n \in \left\{ {1,2,...} \right\}$

(C) $\frac{\pi }{2} + 2n\pi ,n \in \left\{ {..., - 2, - 1,0,1,2,...} \right\}$

(D) $ 2n\pi ,n \in \left\{ {..., - 2, - 1,0,1,2,...} \right\}$

I am not able to solve this problem because it is composite function

答:
The equation is $f(M)=M$ where $M=g\circ g\circ f(x)$. The fixed points of $f$ are $0,1$. Hence we want $g\circ g\circ f(x)=\sin(\sin x^2)=0,1$.

For $0$, you have $\sin x^2=n\pi$ which lies in $[-1,1]$ only for $n=0$. Thus $\sin x^2=0\implies x^2=k\pi,k\in\Bbb Z_{\ge0}$.

For $1$, you have $\sin x^2=2m\pi+\pi/2\not\in[-1,1]\forall m\in\Bbb Z$.

Title: Composite Function Type problem


答:
First, let us assume that the flags are identical and then we can multiply the solution by $10!$ for each arrangement.

The number of ways of distributing $10$ identical flags to $7$ different flagpoles if each flagpole has at least one flag is the number of solutions of the below equation in the positive integers.  A particular solution corresponds to the placement of $6$ addition signs in the $9$ spaces between successive ones in a row of $10$ ones.  For instance,
$$1 1 + 1 1 1 + 1 +1 +1 +1 1 +1$$
corresponds to the solution $x_1 = 2$, $x_2 = 3$, $x_3 = 1$, $x_4 = 1$, $x_5 = 1$, $x_6 = 1$, $x_7 = 1$.  Thus, the number of solutions of the equation in the positive integers is
$$\binom{9}{6}$$
Since the order of the flags(and flags are different) on a flagpole is relevant, we multiply this result by $10!$ to obtain
$$\binom{9}{6} \cdot 10! = 84 \cdot 10! =  304,819,200$$
ways of placing $10$ different flags on $7$ different flagpoles if the order of the flags is relevant and each flagpole flies at least one flag.

Do correct me if I am wrong.

Title: How many ways are there to place $10$ different flags on $7$ different flagpoles if order matters and no flagpole is left empty?

Body:
How many ways are there to place $10$ different flags on $7$ different flagpoles if the order of the flags on a flagpole is relevant and every flagpole flies at least one flag?

The way I look at it is that if need at least $1$ flag for each one you can remove that from there so now you $3$ flags to put into $7$ flagpoles which is $7 \cdot 6 \cdot 5 =210$.


答:
Every pole gets a flag, and you need to distribute 3 extra flags onto poles. That means there's only three cases:

One pole gets 4 flags, everything else has just 1.

One pole gets 3 flags, one other gets 2 flags, everything else has 1.

Three poles get 2 flags each, everything else has 1.

Now, since every single pole is distinguishable, and even the ordering of flags matters on each pole - there are simply no "equivalent cases". Consider ordering the flags by 1. which pole it is flown on; 2. if the same pole, from top to bottom. Every single way of ordering them is distinct.

So, for case 1, there is really just "pick one pole to have the 4 flags", and then all $10!$ ways of ordering the flags on the poles counts. That's just $7\cdot10!$.

Extending through the cases, the answer is simply just $\left(\binom{7}{1}+\binom{7}{1}\binom{6}{1}+\binom{7}{3}\right)\cdot10!$ and done.


问:
Non-linear function with properties related to Gaussian distribution
Let $\mathcal{A}_n$ be the set of functions $f : \mathbb{R} \to \mathbb{R}^n$ such that:

$f$ is continuous and $f$ is differentiable almost everywhere

if $X \sim \mathcal{N}(0, 1)$, then:

$E[f(X)] = \mathbf{0}$

$E[{\lVert f(X) \rVert}^2] = 1$

$E[{\lVert {f'}(X) \rVert}^2] = 1$

(where $\lVert \cdot \rVert$ is the euclidian distance on $\mathbb{R}^n$)

Could you find a non-linear function that belongs to $\mathcal{A}_n$, or prove the following statement:

$$\forall n \in \mathbb{N^*}, \forall f \in \mathcal{A}_n, f \, \text{is linear}$$

After doing some experiments, I'm almost convinced that the last statement is true, but I have no idea how to prove it. If, on the contrary, $A_n$ contains non-linear functions, then these could be used as (multi-dimensional) activation functions in neural networks.

答:
Edit: Thanks for pointing out the mistake, I have corrected it.

No such function exists.  I will consider first the case $n=1$.
Because $E f(X)^2=1$ we can expand $f$ as a (possibly infinite) sum of Hermite polynomials $f(x)=\sum_{i=1} c_i He_i(x)$ (we don't include the constant term $He_0(x)=1$ because $Ef(X)=0$). The crucial property we will need is that these polynomials are orthogonal wrt to the Gaussian density, more precisely
$$EHe_i(X)He_j(X)=\delta_{ij} i!$$

In particular,

$$Ef^2=\sum_{i=1} c_i^2 i!$$

Furthermore, we will use the property
$$He_i'(x)=iHe_{i-1}(x)$$
From this, we have $f'(x)=\sum_i c_iiHe_{i-1}(x)$ so

$$Ef'^2=\sum_{i=1} c_i^2 i^2 (i-1)!=\sum_{i=1} c_i^2ii!$$

Now, the coefficient of $c_i^2$ in this sum is $i i!$, which is strictly larger than the coefficient of $i!$ in $Ef^2$, unless $i=1$. Since the coefficients are all non-negative, the only way these sums could be equal is if $c_i=0$ for $i>1$. Thus $f(x)$ must be of the form $f(x)=c_1 He_1(x)=c_1x$.

For higher dimensions, we can write $f(x)=(f_1(x),\dots, f_n(x))$. By applying the above argument to each component, we conclude that $Ef_i^2(x)\leq Ef_i'^2(x)$ with equality if and only if $f_i$ is linear. Therefore, if $E|f|^2=E|f'|^2$, then each component $f_i$ must be linear.


问:
How find this integral limit

Find this limit $$\lim_{n\to\infty}\dfrac{\displaystyle\int_{0}^{\frac{\pi}{2}}x\left(\dfrac{\sin{(nx)}}{\sin{x}}\right)^4dx}{n^2}.$$

This problem is post my teacher(tian275461, he is very popular in China), and he told me this problem answer is $\ln{2}$, but I can't prove it. Thank you.

答:
With the substitution $y = nx$, we have
\begin{align}
I_n &= \int_0^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y \\
&= \underbrace{\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,1}} +
\underbrace{\int_{\sqrt{n}}^{n\pi/2} \frac{\sin^4 y}{y^3} \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \mathrm{d} y}_{I_{n,2}}.
\end{align}

Since $u \mapsto \frac{u}{\sin u}$ is increasing on $(0, \frac{\pi}{2}]$, we have
$$1 \le \left(\frac{\frac{y}{n}}{\sin \frac{y}{n}}\right)^4 \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4,
\ \forall 0 < y \le \sqrt{n}.$$
Thus, we have
$$\int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3}\mathrm{d} y \le I_{n,1} \le \left(\frac{\frac{1}{\sqrt{n}}}{\sin \frac{1}{\sqrt{n}}}\right)^4 \int_0^{\sqrt{n}} \frac{\sin^4 y}{y^3} \mathrm{d} y.$$
By the squeeze theorem, we have $\lim_{n\to \infty} I_{n,1} = \int_0^{\infty} \frac{\sin^4 y}{y^3}\mathrm{d} y = \ln 2$
(see e.g. Laplace transform:$\int_0^\infty \frac{\sin^4 x}{x^3} \, dx $).

Also, we have
$$I_{n,2} \le \int_{\sqrt{n}}^{n\pi/2} \frac{1}{\sqrt{n}^3} \left(\frac{\pi}{2}\right)^4 \mathrm{d} y$$
and thus $\lim_{n\to \infty} I_{n,2} = 0$.

Thus, $\lim_{n\to \infty} I_n = \ln 2$.

Title: How find this integral limit


问:
Does every frame contain the same number of time-like and space-like vectors?
For a pseudo-Riemannian manifold M of signature $(p,q)$, would it be valid to say that every frame (ordered basis) $e\in LM$ contains $p$ time-like vectors and $q$ space-like vectors?

If this does not hold, is this true for the weaker case of a Lorenzian manifold?

答: No. Consider the usual Minkowski metric with $\eta = \text{diag}(1,-1,-1,-1)$. Then $(1,0,0,0), (1,1,0,0), (1,1,1,0), (1,1,1,1)$ is an ordered basis with 1 time-like vector, 1 null vector, and 2 space-like vectors.


问:
Does distributivity implies commutativity of one operation
Suppose there is a set $S$, equipped with two binary operations, $*$ and $@$, such that S is closed and associative under both the operations. There exist inverses and identity with respect to both the operations. $*$ is distributive under $@$. (Both the operations are not commutative).

Let $a,b,c,d\in S$, and $(a@b)=\alpha$ and $(c@d)=\beta$.

$(a@b)*(c@d)=\alpha *(c@d)=(\alpha *c)@(\alpha *d) = ((a@b)*c)@((a@b)*d)=(a*c)@(b*c)@(a*d)@(b*d)\;\;----(1)$

Also $(a@b)*(c@d)=(a@b)*\beta=(a*\beta)@(b*\beta) = (a*(c@d))@(b*(c@d))=(a*c)@(a*d)@(b*c)@(b*d)\;\;----(2)$

Comparing (1) and (2).
We conclude that $(b*c)@(a*d)=(a*d)@(b*c)$ using identity, inverses and associativity (cancellation property).

This implies that $@$ is commutative operation in S.

This a contradiction. That surprised me.

So the question is if the operations are not commutative then they can't be distributive always because then one of the operation comes out to be commutative? If one of them is commutative then only there is possibility of showing distributive property?

So like distributivity and commutativity are connected?

答:
You're likely running into triviality problems. Note that your conditions are vastly stronger than even a ring, notably that "multiplication" has an identity and inverse.

In any ring, $0$ (the additive identity) must necessarily absorb under multiplication, and therefore cannot have a multiplicative inverse (look at $(a+(-a))\times b$). The proper definition of a ring, of course, looks at the multiplicative group of units, of which $0$ is never one.

You necessarily use cancellation of multiplication, but that is precisely the operation that need not exist in a non-trivial structure.


答:
It takes more than distributivity: the existence of a $*$ identity and inverses is also crucial to your argument.

To see this, define operations $\otimes$ and $\oplus$ on $\Bbb R$ by $x\oplus y=x$ and $x\otimes y=y$ for all $x,y\in\Bbb R$. You can easily verify that both operations are associative and that

$$x\otimes(y\oplus z)=y=(x\otimes y)\oplus(x\otimes z)$$

and

$$(y\oplus z)\otimes x=x=(y\otimes x)\oplus(z\otimes x)\,,$$

i.e., that $\otimes$ distributes over $\oplus$. Clearly neither operation is commutative, however.

Title: Does distributivity implies commutativity of one operation

Body:
Suppose there is a set $S$, equipped with two binary operations, $*$ and $@$, such that S is closed and associative under both the operations. There exist inverses and identity with respect to both the operations. $*$ is distributive under $@$. (Both the operations are not commutative).

Let $a,b,c,d\in S$, and $(a@b)=\alpha$ and $(c@d)=\beta$.

$(a@b)*(c@d)=\alpha *(c@d)=(\alpha *c)@(\alpha *d) = ((a@b)*c)@((a@b)*d)=(a*c)@(b*c)@(a*d)@(b*d)\;\;----(1)$

Also $(a@b)*(c@d)=(a@b)*\beta=(a*\beta)@(b*\beta) = (a*(c@d))@(b*(c@d))=(a*c)@(a*d)@(b*c)@(b*d)\;\;----(2)$

Comparing (1) and (2).
We conclude that $(b*c)@(a*d)=(a*d)@(b*c)$ using identity, inverses and associativity (cancellation property).

This implies that $@$ is commutative operation in S.

This a contradiction. That surprised me.

So the question is if the operations are not commutative then they can't be distributive always because then one of the operation comes out to be commutative? If one of them is commutative then only there is possibility of showing distributive property?

So like distributivity and commutativity are connected?


问:
help calculating $\sum_{k=0}^n k!{n \choose k} ^2= \text{?}$
I am trying to find the number of relations on $[n] = \{1,\ldots,n\}$ s.t. for all $x,y,z\in R.xRz \wedge yRz\to x=y$. One idea I had was choosing $k$ elements from $n$ then choosing another $k$ elements from $n,$ there are $k!$ bijections between these two sets so we just have to sum over $k$ and here I got stuck, tried finding another approach but came up with nothing.

答:
The following function is used in statistical thermodynamics.

$$f(x)=\sum_{k=0}^n {n \choose k} ^2\,k!\,x^k= (-x)^n U\left(-n,1,-\frac{1}{x}\right)$$ where appears the Tricomi confluent hypergeometric function.

So, your case corresponds to $f(1)$ and the values are reported as sequence $A002720$ in $OEIS$. Accoring to $OEIS$, $a_n= n!\, L_n(-1)$ where appear   Laguerre polynomials.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答: In Field theory The Impossibility of Trisecting the Angle and Doubling the Cube

Title: Landmarks of subjects of mathematics


问:
the domain for $\dfrac{1}{x}\leq\dfrac{1}{2}$
What is the domain for $$\dfrac{1}{x}\leq\dfrac{1}{2}$$

according to the rules of taking the reciprocals, $A\leq B \Leftrightarrow \dfrac{1}{A}\geq \dfrac{1}{B}$, then the domain should be simply $$x\geq2$$

however negative numbers less than $-2$ also satisfy the original inequality. When am I missing in my understanding?

答:
Inequalities with fractions require some care. If you write yours in the form
$$
\frac{1}{x}-\frac{1}{2}\le0
$$
you see that it's equivalent to
$$
\frac{2-x}{2x}\le0
$$
or to
$$
\frac{x(2-x)}{2x^2}\le0
$$
Since $2x^2>0$, we can remove the denominator (but keeping the condition that $x\ne0$). So we have the standard
$$
x(2-x)\le0
$$
or
$$
x(x-2)\ge0
$$
whose solution set is $(-\infty,0]\cup[2,\infty)$. Remembering we can't accept $0$ as solution, we end up with
$$
x<0\text{ or }x\ge2
$$

Of course such a long discussion is too much for this simple case, but it should show how things can go wrong with "hasty simplifications". Here, one can simply observe that when $x<0$ the inequality is clearly satisfied and only for $x>0$ one has to do something more: but for $x>0$ one can indeed reverse the fractions, so this becomes $x\ge2$.

This is analogous to irrational inequalities such as $\sqrt{1-x}\ge x-2$. All numbers satisfying
$$
\begin{cases}
1-x\ge0 & \text{(existence of the square root)}\\
x-2<0 & \text{(the RHS is negative)}
\end{cases}
$$
are solutions of the inequality. When $x-2\ge0$, instead, you can square both sides and proceed as usual:
$$
\begin{cases}
1-x\ge(x-2)^2\\
x-2\ge0
\end{cases}
$$

Don't be hasty and inequalities will bite you no more.


问:
Confusion on the definition of accumulation points
I've been trying to learn a bit about limits of sequences and accumulation points to get a better intuition behind the workings for calculus and I got confused on the definitions of limits,limit points and accumulation points of sequences and sets.

My first question is a limit of a sequence the same as the accumulation point and is that the same as the limit point i looked online and its all very vague.
My second confusion is that is the limit of a sequence the same as the limit of a set if not is there some proof or intuitive explanation as to why not?.

I know this is probably a very simple and probably trivial concept for all of you on here but its confused me a lot.
Thanks in advance

答:
There is a difference between limit and limit point.
The concept is defined for sequences and functions but limit point is defined for sets, as mentioned in above answer. A sequence may have limit point but no limit. For example let $\{a_n\}$ is defined as $$1+\frac{1}{n} , (-1)+ \frac{1}{n},... $$
That $a_n=1+\frac{1}{n} $ for odd n's and $a_n=-1+\frac{1}{n} $ for evens.
In this sequence both $1$ and $-1$ are limit point but the sequence is not convergent and there is no limit.


问:
An element is either in a set or not in a set
Is there a name for the following obvious fact?

Given any set $S$ and any object $s$, either $s \in S$ or $s \notin S$.

答:
This is the law of the excluded middle, which is a tautology is classical logic.

$$ \varphi \lor \lnot \varphi \;\;\text{is true for all well-formed formulas $\varphi$} $$


问:
Finding all accumulation points of a sequence
How would I go about finding all accumulation points (or limit points?) of a sequence? I know that I need to find subsequences that converge to that point. But for that, I first need to find that point. How do I do this? And how can I be sure that I found all accumulation points?

The specific exercise is:

Find all accumulation points of the following sequence:

$(1+ (-1)^n) \frac {n + 1} n + (-1)^n$

Any help, especially for the example above, would be appreciated.

答:
The sequence is not convergent but $3$ is the only limit point. You can see for large $n's$
there are infinitely many points of sequence is near $3$.


问:
If $R \subseteq A \times B$ is a surjective relation, as is $S \subseteq B \times A$, are they bijective relations?
Suppose two sets $A$ and $B$, two relations $R \subseteq A \times B$ and $S  \subseteq B \times A$. If both of them are surjective, then both of them are bijective relations.

Is this correct?

Can we disprove this with the following example:

$A =  \{ 1, 2 \}$

$B = \{3\}$

$R = A \times B =  \{  (1, 3) , (2,3) \} $

$S = B \times A =  \{  (3,1), (3,2)    \}$

Both of $R,S$ are certainly surjective but $R$ isn't injective, isn't it?

答:
Definitions:

Taking some notes from the answer from Musa Al-hassy here, I want to establish some common ground since there appears to be confusion among other users and, indeed, these terms while having definitions don't appear to be in common use. At least in my experience anyhow.

We say $R$ is a relation on $A$ to $B$ if $R \subseteq A \times B$.

We say $R$ is a total relation if $\forall a \in A$ $\exists b \in B$ such that $(a,b) \in R$. (Think of a total function, your typical definition of function.)

We say $R$ is a deterministic relation if
$$(\forall a \in A)(\forall b,b' \in B)\Big((a,b) \in R \land (a,b') \in R \implies b = b'\Big)$$
What this says is that $R$ is single-valued, in a sense, if you think of it in terms of functions: whenever $a$ is related to something, if at all, it is only related to one thing. Alternatively, it can be thought of in terms of partial functions (where any given input has at most one output).

We say $R$ is a function if it is total and deterministic.

We define the transpose $R^T$ of a relation $R$ by reversing everything. That is, $(a,b) \in R \iff (b,a) \in R^T$, or
$$R^T = \Big\{ (b,a) \; \Big| \; (a,b) \in R \Big\}$$

We say $R$ is a surjective relation if $R^T$ is a total relation.

We say $R$ is an injective relation if $R^T$ is a deterministic relation.

Naturally, $R$ is a bijective relation if $R$ is both surjective and injective.

Visualization:

Visualizing relations as directed graphs is helpful. In such a graph, $x$ points to $y$ iff $(x,y) \in R$. In such a case, then, the above properties may be visualized as so:

$R$ is a total relation if every node has at least one arrow from it. Examples and non-examples on the sets $A=B=\{1,2,3\}$ will follow this and most properties. (Note that these visual intuitions become murkier in the event $A \ne B$.)

$R$ is a deterministic relation if every node has no more than one arrow from it.

The transpose relation $R^T$ is the same graph for $R$, but with its arrows reversed:

$R$ is a surjective relation if $R^T$ is total. That is, every node in $R^T$ has an arrow going from it. In turn, $R$ is surjective if every node has an arrow going to it. (Again, think about an analogy to surjective functions.)

$R$ is an injective relation if $R^T$ is deterministic; that is, every node in $R^T$ has at most one arrow leaving it. In turn, $R$ is an injective relation if every node in $R$ has at most one arrow going from it. (Think about function injectivity or partial functions.)

Finally, $R$ is a bijective relation if and only if it is both injective and surjective. Thus, every node has precisely one arrow entering and leaving it. Or, put differently, each node has precisely one arrow leaving it, and they all go to different nodes. (Again, think about bijective functions.)

Now, To Your Question:

Your relation $R$ is surjective. Graphically, it is presented as below:

Notice how there are no arrows pointing to $1$ or $2$. While my above diatribes were focused on subsets of $A \times A$ rather than $A  \times B$, there is no issue here, since every element of $B$ in particular has an arrow pointing to it. Indeed, note that

$$R^T = \Big\{ (3,1),(3,2) \Big\}$$

(Note that $R \subseteq A \times B \iff R^T \subseteq B \times A$.) For totality, we would require that, for all $b \in B$ there is $a \in A$ where $(b,a) \in R^T$. Since the only $b \in B$ is $B = \{3\}$, we're good in this respect.

Your other relation is $S = R^T$, coincidentally, graphically presented below:

As you can see, this is surjective, since every element of $A$ has an arrow pointing to it. Indeed,

$$S^T = R = \Big\{(1,3),(2,3)\Big\}$$

so each element of $A = \{1,2\}$ has an arrow pointing from it - and in $S$, every element of $B = \{3\}$ has an arrow pointing to it.

Are these bijective relations, though? Indeed not; the first is not injective. This follows as $R^T$ has the pairs $(3,1),(3,2)$ in it, but $1 \ne 2$.

Thus, you are correct -- you have found a pair of surjective relations on $A \times B$ and $B \times A$ where one is not a bijection.

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$
This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.

The question is: How would you do it?

Body:
\begin{gather*}
I=\int ^{1}_{0}\left(\frac{3x^{3}}{3x( x-1) +1} +1\right) dx=\int ^{1}_{0}\frac{3x^{3} dx}{3x( x-1) +1} +1\\
Let\ I_{1} =\int ^{1}_{0}\frac{x^{3} dx}{3x( x-1) +1}\\
I_{1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3( 1-x)( 1-x-1) +1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3x( x-1) +1}\\
because\int ^{a}_{0} f( x) dx=\int ^{a}_{0} f( a-x) dx\\
Adding\ both,\\
2I_{1} =\int ^{1}_{0}\frac{x^{3} +( 1-x)^{3} dx}{3x( x-1) +1}\\
Apply\ a^{3} +b^{3} =( a+b)\left( a^{2} +b^{2} -ab\right)\\
2I_{1} =\int ^{1}_{0}\frac{3x( x-1) +1dx}{3x( x-1) +1} =1\\
I_{1} =\frac{1}{2}\\
I=3I_{1} +1=\frac{3}{2} +1=\frac{5}{2}\\
\\
\end{gather*}which I hope is the correct answer:)


问:
Using Well Ordering Principle to Prove Backward Induction of the form $2^{n}$
Assume $\Omega\subseteq\mathbb{N}$ such that :

$(1)$ : $2^{n}\in \Omega$, $\forall n\in\mathbb{N}$.

$(2)$ : If $n\in\Omega$ and $\forall n\geq2$, then $n-1\in\Omega$,

My main objective is to :

Show that $\Omega=\mathbb{N}$

I managed to prove this easily by taking two cases for $\phi(m):=2^{m}$ for some $m\in\mathbb{N}$ one for which $m$ is a power of two and other for which it is not and in both cases it can be inferred that $\phi\in\Omega$.

I now wish to prove this $2^{n}$ variation of induction again but this time I was curious to see how the well ordering principle would work. I shall present my strategy :

[Strategy] : We Assume by contradiction that $\Omega$ is a proper subset of $\mathbb{N}$. This means that $\mathbb{N}\setminus \Omega\neq\emptyset$ a non empty subset of $\mathbb{N}$, then by the well ordering principle, we have that there exists a least element $p\in\mathbb{N}$. I now thought of achieving  contradiction by proving that $p+n\notin\Omega$ which contradicts the fact that $2^{p}>p\in\Omega$.

[Question] : The method of proving $p+n\notin\Omega$ seems non-trivial to me even though it might turn out to be trivial. I thought of using induction to prove this but I have no clear argument for base case and inductive step.

Any help is much appreciated as I am new and still learning elements of set theory by myself $:)$

答:
It’s a little tricky, and it isn’t $\Bbb N\setminus\Omega$ whose least element you want to use.

If $\Bbb N\setminus\Omega\ne\varnothing$, let $p\in\Bbb N\setminus\Omega$. We know that $n<2^n$ for each $n\in\Bbb N$, so $p<2^p$, and $2^p\in\Omega$ by $(1)$. Let $S=\{n\in\Bbb N:2^p-n\notin\Omega\}$; $2^p-(2^p-p)=p\notin\Omega$, so $2^p-p\in S$, and therefore $S\ne\varnothing$. Let $m=\min S$.

$2^p\in\Omega$, so $2^p-1\in\Omega$ by $(2)$, and therefore $m>1$. But then $m-1\in\Bbb N\setminus S$, so $2^p-m+1=2^p-(m-1)\in\Omega$, and $(2)$ then ensures that $2^p-m=(2^p-m+1)-1\in\Omega$, contradicting the choice of $m$. This contradiction shows that $\Bbb N\setminus\Omega$ cannot be non-empty after all and hence that $\Omega=\Bbb N$.


Title: Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$

Body:
This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.

The question is: How would you do it?

Body: $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx=\int_{0}^{1} \frac{(x+2)(3(x^2-x)+1)+2x-1}{3(x^2-x)+1}dx =\int_{0}^{1} (x+2+\frac{2x-1}{3(x^2-x)+1})dx=\int_{0}^{1} (x+2+\frac{(x^2-x+\frac{1}{3})'}{3(x^2-x+\frac{1}{3})})dx=[\frac{1}{2}x^2+2x+\ln(x^2-x+\frac{1}{3})]_0^{1}=\frac{5}{2}$


问:
If $B/IB$ is Noetherian and $B/N$ is faithful, is $B$ Noetherian if we don't assume $B$ is finitely generated?
Let $A$ be a commutative ring, $B$ be a $A$ module.Could you give a example of rings which satisfies the following condition, $1,2,3$.

$B$ is not Noetherian as an $A$-module.

If $I$ is an nonzero ideal of $A$, then $B/IB$ is Noetherian as an $A$-module.

If $N$ is nonzero subbed of $B$, $B/N$ is not faithful as an $A$-module.

If we add condition that $B$ is finitely generated over $A$, we can say there is no example which satisfies 1 and 2 and 3. But if we do not assume that, we may be able to find examples which satisfies 1 and 2 and 3.

Thank you for your help.

答:
No such $A$-module exists. Indeed, we can show that if $B$ satisfies conditions 1,2,3 in the question, then $B$ is finitely generated. Let $0\ne x\in B$, and let $N=Ax$, the submodule generated by $x$. Since $N\ne 0$, the quotient $B/N$ is not a faithful $A$-module, by condition 3. Thus, there exists $0\ne a\in A$ such that $aB\subset N$. By condition 2, $B/aB$ is Noetherian, whence $B/aB$ is finitely generated. So, $B/N$ is finitely generated. Thus, in the following exact sequence$$0\to N\to B\to B/N\to 0$$both the left and right terms are finitely generated. Hence, $B$ is finitely generated.

It is known that no finitely generated $A$-modules satisfy conditions 1,2,3. See step 2 in my answer to this question. Also, note that the paragraph "A different approach" can be gently modified to answer this question without using the finitely-generated hypothesis.


问:
$ x= \pi / 2 - 4/ \pi \sum_{n=1}^{\infty} \frac{ \cos(2n-1) x } { (2n-1)^2}$
This is question number 11.10 (b) APostol's Chapter: Fourier series and Fourier integrals.

Prove that $ x= \pi / 2 - 4/ \pi \sum_{n=1}^{\infty} \frac{ \cos(2n-1) x } { (2n-1)^2}$ if $0 \leq x\leq \pi$.

I have proved $a_0/2 = \pi /2$ using $a_0 = 2/ \pi \int_{0}^{\pi} x\, dx$. Also, I  proved $b_n=0$. But while proving $4/ \pi \sum_{n=1}^{\infty} \frac{ \cos(2n-1) x } { (2n-1)^2}$ if $0 \leq x\leq \pi$, I am getting $a_n=2/\pi [ (\cos (n\pi) / n^2) -1/n^2]$.

Can you please tell how to obtain what is given?

答: You have obtained the correct value for $a_n$. Just note that $\cos (n \pi)=(-1)^{n}$ so your answer coincides  with the given answer.


答:
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Direct calculation by partial fractions
\begin{eqnarray*}
\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}\,\dd x &=&\int_{0}^{1}\frac{3x^{3}+3x^{2}-3x+1}{3x^{2}-3x+1}\, \dd x\\
&=&\int_{0}^{1}\left( \frac{2x-1}{3x^{2}-3x+1}+x+2\right)\,\dd x\\
&=&\int_{0}^{1} \frac{2x-1}{3x^{2}-3x+1} \, \dd x +\int_{0}^{1} x \, \dd x+ 2\int_{0}^{1} \, \dd x\\
&\overset{u=x-\frac{1}{2}}{=}&\color{red}{\int_{-1/2}^{1/2} \frac{2(u+1/2)-1}{3(u+1/2)^{2}-3(u+1/2)+1} \, \dd u}+\int_{0}^{1} x \, \dd x+ 2\int_{0}^{1} \, \dd x\\
&=& \color{red}{0}+\frac{1}{2}+2\\
&=&\boxed{\frac{5}{2}}
\end{eqnarray*}

Note, that $\displaystyle f(u):= \frac{2(u+1/2)-1}{3(u+1/2)^{2}-3(u+1/2)+1}$ is odd function and the interval $[-1/2,1/2]$ is symmetric about $0$, so we have $\displaystyle \color{red}{\boxed{\int_{-1/2}^{1/2}f(u) \, \dd u=0}}$.


答:
The integral can be written as $$\int_{0}^{1} \left(x+2+\dfrac{2x-1}{3x(1-x)+1}\right)\text dx=\dfrac 52+\int_0^1\dfrac{2x-1}{3x(1-x)+1}\text dx $$

Now, in this integral use $x \to 1-x$ and note that it will be the negative of it hence this integral will vanish leaving the answer as $\dfrac{5}2$.


Body:
Sligthly differently:  $$I=\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
$$I=3\int_{0}^{1} \frac{x^3}{3x^2-3x+1}+\int_{0}^{1} 1 dx$$
to get $$ J=\int_{0}^{1} \frac{x^3}{3x^2-3x+1}dx~~~~(1)$$
Now use $\int_{0}^{a} f(x)= \int_{0}^{1} f(a-x) dx= 3J+1$
$$J=\int_{0}^{1} \frac{(1-z)^3}{3x^2-3x+1}dx~~~(2)$$
we get $$2J=\int_{0}^{1} 1 dx \implies J=\frac{1}{2} \implies I=\frac{5}{2}$$


答:
At first, we must check whether integrand function is bounded or not. [More specifically, if it is not bounded, we must approach this problem as improper integral..] Since both numerator and denominator are continuous at [0,1], we enough to check whether denominator has 0-value or not. Since $3\left(x-\dfrac{1}{2}\right)^{2}<3x^{2}-3x+1$, we can conclude that denominator does NOT have 0 value. So if we can get indefinite integral of integrand, then we can also calculate the definite integral.

Let's manipulate integrand function so that we can get indefinite integral. At first, I want to divide numerator into denominator.

$3x^{3}+3x^{2}-3x+1$

$=3(x^{3}+1)-3+(3x^{2}-3x+1)$

$=3(x+1)(x^{2}-x+1)-3+(3x^{2}-3x+1)$

$=(x+1)(3x^{2}-3x+3)-3+(3x^{2}-3x+1)$

$=(x+1)(3x^{2}-3x+1)+2(x+1)-3+(3x^{2}-3x+1)$

$=(x+2)(3x^{2}-3x+1)+(2x-1)$

From this calculation, we can manipulate integrand as $x+2+\dfrac{2x-1}{3(x^{2}-x)+1}$.

Indefinite integral of $x+2$ is $\dfrac{x^{2}}{2}+2x$, so we only need to compute indefinite integral of the last term. To integrate it, I will first substitute $x^{2}-x$ as u, so that $u'(x)=2x-1$. So it can be changed as $\dfrac{1}{3} \dfrac{1}{u+\dfrac{1}{3}}$ and its indefinite integral is $\dfrac{1}{3} \log{\left(u+\dfrac{1}{3}\right)}$. Since u(1)=u(0) in this situation, we only need to calculate $\left[\dfrac{x^{2}}{2}+2x \right]^{1}_{0}=\dfrac{5}{2}$.


问:
How to prove this property for linear operator $A,B$ : $AB-BA=I\implies A^kB-BA^k=kA^{k-1}\quad k\in \mathbb N$.
$$
AB-BA=I\implies \forall k\in\mathbb Z^{+} \quad A^kB-BA^k=kA^{k-1}
$$

note: $A,B$ are some linear operators over $V$ i.e. a linear mapping from $V$ to $V$ or $A:V\to V$

thanks @Mvaldi for pointing out that $AB-BA=I$ can't be satisfied for linear operators on finite-dimensional vector spaces.

I want to prove it by induction. But got stuck on the first step.

if we have proved that $A^kB-BA^k=kA^{k-1}$, then

$$
A^{k+1}B=A(BA^k+kA^{k-1})=ABA^k+kA^k
$$

and
$$
B^{k+1}A=(A^kB-kA^{k-1})A=A^kBA-kA^k
$$

thanks @Semiclassical for fixing the equation.

what can I do then.

perhaps we need to find some matrix equation that is equivalent to $AB-BA=I$..

答: Hint: Multiply $A$ from right  $+$ Multiply $A$ from left.


问:
Asymptotic behavior of a Fourier/Laplace transform
I see many results concerning the asymptotics of Fourier transforms. These link in particular the regularity/continuation properties of the function to the polynomial/exponential decay of its Fourier transform. However, these results often hold only in the real variable. I am interested in the Fourier transform "along the imaginary axis" instead.

Let us be more precise. I am interested by the digamma function $\psi = \frac{\Gamma'}{\Gamma}$, and in the function
$$h(\nu) = \exp\left(-\alpha \psi \left( \frac14 \pm \frac{i\nu}{2} \right)\right),$$

where $\alpha$ is a fixed parameter, say $\alpha > 1$. I am interested in the asymptotic behavior of the Fourier transform of $h$ at $+\infty$. More precisely,
$$\widehat{h}(x) = \int_{\mathbb{R}} h(\nu) e^{ix\nu} d\nu.$$

How to get asymptoptics when $x \to +\infty$ in this situation? I have no feeling about what determines it: size? variations? only asymptotics of $h$?

I had many trials, not convincing. Typically, just changing variables, I can get an expression of the shape
$$e^{-\frac{x}{2}} \int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{2xu} du$$

which looks more like a Laplace (?) transform than a Fourier transform. I was motivated by the fact that I am expecting for other reasons an exponential decay as above, so that I am hoping for a polynomial behavior in $x$ for the remaining integral. However, is the growth/decay estimate of this last integral easier to understand than the original one?

So my question could be synthzised into

Do we have $\int_{i\mathbb{R}} e^{-\alpha \psi(u)} e^{xu} du \ll x^A$ for a certain $A$?

答:
Happy new year! Here's a late answer.

It will be shown that, when $h(v)=\exp\left(-\alpha\psi\left(\frac14+\frac{iv}{2}\right)\right)$,

$$\widehat{h}(x)=(2\alpha)^{1/4}\sqrt{\pi}e^{\alpha\gamma}\cdot\frac{e^{-x/2+2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$
where $\gamma$ is the Euler-Mascheroni constant.

On the other hand, it is trivial that when $h(v)=\exp\left(-\alpha\psi\left(\frac14-\frac{iv}{2}\right)\right)$, $\widehat{h}(x)=0$ for $x>0$.

By the substitution $u=\frac14+\frac{iv}{2}$,

$$\begin{align}
\widehat{h}(x)
&:=\int_{\mathbb  R}\exp\left(-\alpha\psi\left(\frac14+\frac{iv}{2}\right)\right)e^{ixv}dv \\
&=-2ie^{-x/2}\int_{\frac14+i\mathbb  R}\underbrace{e^{-\alpha\psi(u)}e^{2xu}}_{:=f(u)}du \\
H(x)&:=\frac i2e^{x/2}\cdot\widehat{h}(x)=\int_{\frac14+i\mathbb  R}e^{-\alpha\psi(u)}e^{2xu}du \\
\end{align}
$$

By residue theorem and considering the exponential decay of $f(z)$, it can be shown that
$$H(x)-\int_{-\frac12+i\mathbb R}f(u)du=2\pi i\operatorname*{Res}_{z=0}f(z)$$
$$H(x)=2\pi i\operatorname*{Res}_{z=0}f(z)+\underbrace{\int_{-\frac12+i\mathbb R}f(u)du}_{:=J_0}$$

Lemma 1: $J_0=O(e^{-x})$.

Proof:

$$\begin{align}
\left|\int_{-\frac12+i\mathbb R}f(u)du\right|
&=\left|\int_{\mathbb R}e^{-\alpha\psi(-1/2+iu)}e^{-x+2xiu}du\right| \\
&\le\int_{\mathbb R}\left|e^{-\alpha\psi(-1/2+iu)}e^{-x+2xiu}\right|du \\
&=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(-1/2+iu)}\right|du \\
&=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(3/2-iu)-\alpha\pi i\tanh(\pi u)}\right|du \quad (1)\\
&=e^{-x}\int_{\mathbb R}\left|e^{-\alpha\psi(3/2-iu)}\right|du \\
&=Ce^{-x} \qquad (2)
\end{align}
$$

$(1)$: By the reflection formula $\psi(1-x)-\psi(x)=\pi\cot(\pi x)$.

$(2)$: The last integral can be considered as a constant $C$ because it converges (as $e^{-\alpha\psi(3/2-iu)}\approx u^{-\alpha}$ for large $|u|$ and $\alpha>1$) and is independent of $x$.

Now, let's focus on the residue at $0$. Trivially,
$$2\pi i\operatorname*{Res}_{z=0}f(z)=\oint_{|z|=R}f(z)dz \qquad R<1$$

The trick here is to take $R=\sqrt{\frac{\alpha}{2x}}$ (I will explain how to come up with this choice of contour on request).

Define $\phi(z)=\psi(z)+\frac1z+\gamma$. We have $\phi(z)=O(|z|)$ as $z\to 0$.

$$\begin{align}
2\pi i\operatorname*{Res}_{z=0}f(z)
&=\oint_{|z|=R}f(z)dz \\
&=\oint_{|z|=R}\exp\left(-\alpha\left(-\frac1z-\gamma+\phi(z)\right)+2xz\right)dz \\
&=e^{\alpha\gamma}\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1+1\right)dz \\
&=e^{\alpha\gamma}\underbrace{\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)dz}_{:=J_1} \\
&+e^{\alpha\gamma}\underbrace{\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1\right)dz}_{:=J_2} \\
\end{align}
$$

Lemma 2: $$J_1=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$

$$\begin{align}
J_1
&=\int^\pi_{-\pi}\exp\left(\frac{\alpha}{R}e^{-i\theta}+2xRe^{i\theta}\right)iRe^{i\theta}d\theta \\
&=i\sqrt{\frac{\alpha}{2x}}\int^\pi_{-\pi}\exp\left(\alpha\sqrt{\frac{2x}{\alpha}}e^{-i\theta}+2x\sqrt{\frac{\alpha}{2x}}e^{i\theta}\right)e^{i\theta}d\theta \\
&=i\sqrt{\frac{\alpha}{2x}}\int^\pi_{-\pi}\exp\left(2\sqrt{2\alpha x}\cos\theta\right)e^{i\theta}d\theta \\
&=2i\sqrt{\frac{\alpha}{2x}}\int^\pi_{0}\cos\theta \, e^{2\sqrt{2\alpha x}\cos\theta} d\theta \qquad (1)\\
&=2i\sqrt{\frac{\alpha}{2x}}\cdot\pi I_1\left(2\sqrt{2\alpha x}\right) \qquad (2)\\
&=2\pi i\sqrt{\frac{\alpha}{2x}}\cdot \frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{\sqrt{2\pi}\sqrt{2\sqrt{2\alpha x}}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \qquad (3) \\
&=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \\
\end{align}
$$

$(1)$: The imaginary part cancels out due to oddness, and the extra factor of $2$ is due to the evenness of the real part.

$(2)$: $I_1$ is the first order modified Bessel function of the first kind.

$(3)$: Due to the well-known asymptotic expansion $I_1(z)=\frac{e^z}{\sqrt{2\pi z}}\left(1+O\left(\frac1z\right)\right)$ for $z\to\infty$.

Lemma 3: $$J_2=O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right)$$

Proof:

$$\begin{align}
|J_2|
&=\left|\oint_{|z|=R}\exp\left(\frac{\alpha}{z}+2xz\right)\left(e^{-\alpha\phi(z)}-1\right)dz\right| \\
&=\left|\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}\left(\exp\left(-\alpha\phi(Re^{i\theta})\right)-1\right)iRe^{i\theta}d\theta\right| \\
&\le R\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}\left|\exp\left(-\alpha\phi(Re^{i\theta})\right)-1\right|d\theta \\
&\le R\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}C|Re^{i\theta}|d\theta \qquad (1)\\
&=CR^2\int^\pi_{-\pi}e^{2\sqrt{2\alpha}\sqrt{x}\cos\theta}d\theta \\
&=CR^2\cdot 2\pi I_0(2\sqrt{2\alpha}\sqrt{x}) \\
&=C\cdot\frac{\alpha}{2x}\cdot 2\pi \cdot \frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{\sqrt{2\pi}\sqrt{2\sqrt{2\alpha x}}}\left(1+O\left(\frac1{\sqrt x}\right)\right) \qquad (2)\\
&=O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right)
\end{align}
$$

$(1)$: As $\phi(z)=O(|z|)$, $\exp\left(-\alpha\phi(z)\right)-1=\exp(O(|z|))-1=1+O(|z|)-1=O(|z|)$.

$(2)$: Due to the well-known asymptotic expansion $I_0(z)=\frac{e^z}{\sqrt{2\pi z}}\left(1+O\left(\frac1z\right)\right)$ for $z\to\infty$.

Therefore, $$2\pi i\operatorname*{Res}_{z=0}f(z)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)+O\left(\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{5/4}}\right)$$
$$\implies 2\pi i\operatorname*{Res}_{z=0}f(z)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$

In conclusion,
$$H(x)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)+O(e^{-x})$$

$$\implies H(x)=i\cdot\frac{\alpha^{1/4}\sqrt{\pi}e^{\alpha\gamma}}{2^{3/4}}\cdot\frac{e^{2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$

As we defined $H(x)=\frac i2e^{x/2}\cdot\widehat{h}(x)$, it can be concluded, eventually,
$$\widehat{h}(x)=(2\alpha)^{1/4}\sqrt{\pi}e^{\alpha\gamma}\cdot\frac{e^{-x/2+2\sqrt{2\alpha}\sqrt{x}}}{x^{3/4}}\left(1+O\left(\frac1{\sqrt x}\right)\right)$$


问:
Differential equations math question?
We have the equation $xy' -4y - x^2 \sqrt{y} =0$

I modify this a little and I have $dx ( 4y-x^2 \sqrt{y}) -xdy=0$

 I am thinking about solving this by integrating factors,but that is quite a long way..is there any short method?

Title: Differential equations math question?

Body:
Substitute $u=\frac{y}{x^4}$.

$u'=\frac{x^4y'-4yx^3}{x^8}=\frac{xy'-4y}{x^5}$

$y'=x^4u'+4\frac{y}{x}$

$x^5u'+4y-4y-x^4\sqrt{u}=0$

$u'=\frac{\sqrt{u}}{x}$

$2\sqrt{u}=\ln{x}+C$

$\sqrt{y}=\frac{1}{2}(x^2\ln{x}+Cx^2)$


问:
How many blocks can be made of a subset of S in a Steiner system
Assume a Steiner system with parameters $t, k, n$, i.e., $S(t,k,n)$. By definition, the system is an $n$-element set $S$ together with a set of $k$-element subsets of $S$ (called blocks) such that each $t$-element subset of $S$ is contained in exactly one block.

Now consider a Steiner system $S(t,k,n)$ and a subset $X \subset S$ with cardinality $|X| = c > k$. The question I am interested in is:

How many blocks of the system consist exclusively of elements of $X$?

An expression given at Wikipedia quantifies how many blocks include a $c$-element subset of S and holds for values of $c \leq t$.

I am wondering whether there is an expression for all values of $c \in \{t+1,\dots,k\}$ for the question stated above. I would think that the answer depends on the construction method. Are there constructions for which one can give such an expression?

答:
If we let $\mathcal{D}$ be an $S(t,k,n)$, in other words, a $t$-$(n,k,1)$ design, then the complementary design $\mathcal{D}^{c}$, obtained by replacing each block with its complement, is also a $t$-design (though not a Steiner system), in other words it is a $t$-$(n,n-k,\lambda^{\prime})$ for some $\lambda^{\prime}$.

The Wikipedia result you give says that if $\mathcal{D}$ is an $S(t,k,n)$ then for each $0 \leq i \leq t$, we have that there are
$$\lambda_{i} = {n-i \choose t-i}/{k-i \choose t-i}$$
blocks of $\mathcal{D}$ containing any fixed $i$-subset of $\{1,\ldots, n\}$.
This is however a special case of a result that holds for any $t$-$(n,k,\lambda)$ design that gives this value as
$$\lambda_{i} = \lambda{n-i \choose t-i}/{k-i \choose t-i}$$

If we want to know how many blocks of $\mathcal{D}$ are contained in some fixed $j$-set $X$, this number will be constant over all possible $j$-sets as long as $n-t \leq j \leq n$.

To see this, we should realize that this is equivalent to asking how many blocks of $\mathcal{D}^{c}$ contain $X^{c}$ (or rather consider $X^{c}$ to be an arbitrary $(n-j)$-subset).
In this case, modifying the formula above with the parameters of $\mathcal{D}^{c}$, we have
$$\lambda^{\prime}_{j} =
\lambda^{\prime}\frac{{n-(n-j) \choose t-(n-j)}}{{n-k-(n-j) \choose t-(n-j)}} =
\lambda^{\prime}\frac{{j \choose j-(n-t)}}{{j-k \choose j-(n-t)}} =
\lambda^{\prime}\frac{{j \choose n-t}}{{j-k \choose j-(n-t)}}$$
where $\lambda^{\prime}$ is given by
$$\lambda^{\prime} = \sum_{i=0}^{t} (-1)^{i}{t \choose i}\lambda_{i}.$$
(See Godsil, Combinatorial Design Theory Lemma 7.3.1)

Godsil actually gives an easier formula:
$$\lambda_{j}^{\prime} = \sum_{i=0}^{j} (-1)^{i}{j \choose i}\lambda_{i}$$
(in both of these formulas the $\lambda_{i}$ are given by the earlier formula $\lambda_{i} = {n-i \choose t-i}/{k-i \choose t-i}$).

Note: if $j < k$ then $X$ cannot contain any blocks of $\mathcal{D}$; and if $k \leq j < n-t$ then the number of blocks contained in $X$ will depend on the choice of the set $X$ and not strictly on $|X|$. (For example if $j=k$ then the number of blocks contained in $X$ will be $1$ if $X$ is a block of $\mathcal{D}$, and $0$ otherwise.)


问:
Factor group of a free group
Let $F[A]$ be the free group on the generating set $A$. Let $C$ be the commutator subgroup of $F[A]$, then show that $F[A]/C$ is a free abelian group with basis $\{aC \mid a \in A\}$. It is trivial that $F[A]/C$ is abelian and generated by $\{aC \mid a \in A\}$, but how can I prove that $\{aC \mid a \in A\}$ is linearly independent?

答:
You can avoid having to prove that by simply showing that $F[A]/C$ and $\{aC\mid a\in A\}$ have the relevant universal property... That is done in egreg’s answer in the question you link to.

To show they are linearly independent, suppose that $a_1,\ldots,a_n\in A$ be pairwise distinct and $\beta_1,\ldots,\beta_n\in\mathbb{Z}$ are such that
$$a_1^{\beta_1}\cdots a_n^{\beta^n}C=eC.$$
Then $a_1^{\beta_1}\cdots a_n^{\beta^n}\in C$.

But in elements of $C$, the sum of the exponents of each $a\in A$ is $0$. This follows by noting that the map from $F[A]$ to $\mathbb{Z}$ obtained by sending $a$ to $1$ and all other elements of $A$ to $0$ must have $C$ in the kernel (since $\mathbb{Z}$ is abelian), and so if $g\in C$ then $g\mapsto 0$. In particular, the sum of the exponents of $a$ in $g$ add up to $0$. This holds for each $a\in A$.

Since $a_1^{\beta_1}\cdots a_n^{\beta_n}\in C$, then $\beta_i=0$ for each $i$. Thus, the set $\{aC\mid a\in A\}$ is linearly independent in $F[A]/C$, as claimed.

Title: Factor group of a free group


问:
Solving differential equations by definite integration
I'm used to solving differential equations through indefinite integration and then making use of the boundary constraints to find the constants. Now I've learned they can also be solved by using definite integrals, but I don't quite understand the approach yet. Take e. g. the ODE $$ C\frac{d^2u(x)}{dx^2}+p=0 $$ where the domain for $x$ is $[0;l]$ and the BCs are $u(l)=u_1$ and $C\frac{du(0)}{dx}:=Cu'(0)=-N$. ($C$, $p$, $l$, $u_1$ and $N$ are all constants). Could anyone explain how to solve the problem using definite integration? (I have the solution here, but I have no idea how it's done ;).

答:
I’m on my phone so I won’t go too in depth. The TLDR is to just integrate from the IC of your independent variable (e.g. $x_0$), to your dependent variable (e.g. $x$).

$C\frac{d^2u}{dx^2}+p=0$

$\int_{x_0}^x(C\frac{d^2u}{dx^2}+p)dx= \int_{x_0}^x0dx$

$(C\frac{du}{dx}+px)\Big|_{x_0}^x=c\Big|_{x_0}^x$

$C(\frac{du}{dx}-\frac{du}{dx}\big|_{x_0}) +p(x-x_0)=0$

$C(\frac{du}{dx}-u’_0) +p(x-x_0)=0$

$\int_{x_0}^x(C(\frac{du}{dx}-u’_0) +p(x-x_0))dx= \int_{x_0}^x 0dx$

$(C(u(x)-u’_0x)+p(\frac{1}{2}x^2-x_0x)) \Big|_{x_0}^x=0 \Big|_{x_0}^x$

$C(u(x)-u_0-u’_0(x-x_0))+p(\frac{1}{2}(x^2-x_0^2)-x_0(x-x_0))=0$.

You can solve for $u$ as it takes a while to LaTeX on the phone. And you’ll have to find someone else on using definite integrals to solve boundary condition problems (e.g. $x(0)=x_0, x(1)=x_1$) as I’m not sure on that front.


答: Unfortunately, there is NO pair of 2 $n \times n$ matrix A,B such that $AB-BA=I_{n}$. To verify it, we can use property of trace. In other words, tr(AB)=tr(BA) for all square matrix A,B.

Title: How to prove this property for linear operator $A,B$ : $AB-BA=I\implies A^kB-BA^k=kA^{k-1}\quad k\in \mathbb N$.

Body:
$$
AB-BA=I\implies \forall k\in\mathbb Z^{+} \quad A^kB-BA^k=kA^{k-1}
$$

note: $A,B$ are some linear operators over $V$ i.e. a linear mapping from $V$ to $V$ or $A:V\to V$

thanks @Mvaldi for pointing out that $AB-BA=I$ can't be satisfied for linear operators on finite-dimensional vector spaces.

I want to prove it by induction. But got stuck on the first step.

if we have proved that $A^kB-BA^k=kA^{k-1}$, then

$$
A^{k+1}B=A(BA^k+kA^{k-1})=ABA^k+kA^k
$$

and
$$
B^{k+1}A=(A^kB-kA^{k-1})A=A^kBA-kA^k
$$

thanks @Semiclassical for fixing the equation.

what can I do then.

perhaps we need to find some matrix equation that is equivalent to $AB-BA=I$..


答:
For a matrix $M$ denote
$$D(M) = M B - B M$$
Let's show that $D$ behaves like a derivation, that is
$$ D(M\cdot  N) = D(M)\cdot N + M \cdot D(N)$$
LHS is $M N B - B M N$, RHS is $(M B- B M)N + M( N B - B N)$, checks OK.

Then let's show that
$$D(A^k) = k A^{k-1}$$ by induction on $k$.  Just check for $k=1$, and then use
$$D(A^{k+1}) = D(A^k \cdot A) = D(A^k) \cdot A + A^k \cdot D(A)$$


问:
Post probability after some event occurs
In a shop total of 40 customers come every month with a standard deviation 1.
In the first 15 days of month only 5 customers came.
What is the expected customer for this 15 days period?

edit:  Customer number per month is gaussian random variable.
Thank you  @obscurans

答:
What is the expected customer for this 15 days period?

In my opinion it's 35.

Here is my explanation:

$X\sim N[40;1^2]$

$$f_{X|X>5}(x)=\frac{1}{1-F_X(5)}f_X(x)=1.12\cdot 10^{-268}f_X(x)=f_X(x)$$

Thus the conditional expected number of customer in the month is always

$$\mathbb{E}[X|X>5]=40$$

that is 35 remaining in the rest of the month


问:
Prove that $f(x) = \frac{1}{x}$ is not uniformly continuous on $(0,1)$
I want to know if this proof is rigorous enough.

```<code>What we need to show here is that ∃ε>0 ∀δ>0 such that |f(x) - f(y)|>ε ,∃x,y∈(0,1) where |x-y|<δ
Let ε=1
    Let δ be given.
        Case 1: δ ≥ 1,
            let x = 1/δ , y = 1/3δ
            then we have |x-y|=|2/3δ|<δ
            further |f(x)-f(y)| = |-2δ|> δ ≥ 1 = ε
        Case 2: δ < 1,
            let x = δ , y = δ/2
            then we have |x - y|=|δ/2| < δ
            further |f(x) - f(y)| = |1/δ - 2/δ| = 1/δ > 1 =  ε
</code>```

Hence in both cases we have shown that there is some  ε for which, no matter how close we pick two points in the domain (0,1) the distance between the images under the function is not within ε of each other.

P.S - There were a few other threads stating this proof, but none were similar to my method plus some requires other theorems I'm yet to learn. Hope this is not a duplicate. Appreciate your time. Thanks in advance.

答:
Your proof is invalid in the specific case of $\delta=1$, as $x=1$ is illegal. You need to handle that separately if you're going to keep the rest of your proof untouched. Alternately, you could instead use $x=\frac1{2\delta}, y=\frac1{4\delta}$ in the first case. Alternately, you could use $x=\frac12, y=\frac14$, not caring about $\delta$ at all.

If you do decide to handle $\delta=1$ separately, you could without issues use $y=\frac1{2\delta}$ in the first case, to make your two cases more symmetric. This is an entirely unnecessary change, but more of an aesthetic concern.


问:
How to Find the Center of a Parallelogram
I want to find the center of a parallelogram in order to use it in my java program. I have four coordinates of the parallelogram and I want to find the center coordinate of the parallelogram. It seems I need to find the intersection point of the diagonal lines that I couldn't figure out of my poor math skills. Given four coordniates of the parallelogram what is the formula for finding the center of the parallelogram. I hope the intersection point of diagonals is the center of the parallelogram, correct me If it's not. If incase I'm wrong what I need is that intersection point not the center. It seems from another view point center of the bounding rectangle is the center of the parallelogram.

Any help would be appreciated

答: Yes, it's the intersection point of the diagonals. In fact, the diagonals bisect one another; so, if $(A,B)$ is one vertex and $(C,D)$ is the opposite, the center of the parallelogram is just $(\frac{A+C}{2}, \frac{B+D}{2}).$</p>


答:
thanks @C.F.G for providing hint.

(Induction). Suppose that we have $AB-BA=I$ and $A^kB-BA^k=kA^{k-1}$. Then:

$$
\begin{aligned}
A^k(AB-BA)&=A^{k+1}B-A^kBA=A^k\\
(A^kB-BA^k)A&=A^kBA-BA^{k+1}=kA^k\\
\implies &A^{k+1}B+(-A^kBA+A^kBA)-BA^k=(k+1)A^k
\end{aligned}
$$


问:
Would a T test be appropriate for determining the significance of an algorithm's results?
Say you have an algorithm which should yield a higher score when given two similar images and a lower value when given two images that do not resemble each other. Two sets of tests are run, one where pairs of similar images are given and the other where pairs of different images are given. Each test set contains 10 tests on complex images. The algorithm yields a higher score, on average, for the images which resemble each other. Although this seems to confirm a hypothesis that the algorithm works, statistical significance needs to be calculated. Would a T test be appropriate for this? If so, should the normal distribution of results from both test sets first be proven?</p>

答:
Comment continued: Here are examples--perhaps helpful. You have said very
little about your data--maybe not.

Two sample Welch t test in R:  P-value near $0;$ highly significant difference between sample mean.
(Difficult to judge normality with such small samples. I generated these 'fake' samples from normal distributions, so I'm sure they're normal.)

```<code>set.seed(102)
x = rnorm(10, 100, 5)
y = rnorm(10, 120, 6)
stripchart(list(x,y), ylim=c(.5,2.5), pch="|")
</code>```

```<code>t.test(x, y)

        Welch Two Sample t-test

data:  x and y
t = -5.5749, df = 15.849, p-value = 4.334e-05
alternative hypothesis:
 true difference in means is not equal to 0
95 percent confidence interval:
   -20.854715  -9.357465
sample estimates:
mean of x mean of y
 103.8186  118.9247
</code>```

Wilcoxon Signed Rank test in R: P-value near $0;$ highly significant difference between sample medians.

```<code>set.seed(121)
x = rgamma(10, 3, .1)
y = rgamma(10, 3, .1) + 20

median(x); median(y)
[1] 21.53694
[1] 54.64572

stripchart(list(x,y), ylim=c(.5,2.5), pch="|")
</code>```

```<code>wilcox.test(x, y)

        Wilcoxon rank sum test

data:  x and y
W = 5, p-value = 0.0002057
alternative hypothesis:
  true location shift is not equal to 0
</code>```

Body:
Comment continued: Here are examples--perhaps helpful. You have said very
little about your data--maybe not.

Two sample Welch t test in R:  P-value near $0;$ highly significant difference between sample mean.
(Difficult to judge normality with such small samples. I generated these 'fake' samples from normal distributions, so I'm sure they're normal.)

set.seed(102)
x = rnorm(10, 100, 5)
y = rnorm(10, 120, 6)
stripchart(list(x,y), ylim=c(.5,2.5), pch="|")

t.test(x, y)

        Welch Two Sample t-test

data:  x and y
t = -5.5749, df = 15.849, p-value = 4.334e-05
alternative hypothesis:
 true difference in means is not equal to 0
95 percent confidence interval:
   -20.854715  -9.357465
sample estimates:
mean of x mean of y
 103.8186  118.9247

Wilcoxon Signed Rank test in R: P-value near $0;$ highly significant difference between sample medians.

set.seed(121)
x = rgamma(10, 3, .1)
y = rgamma(10, 3, .1) + 20

median(x); median(y)
[1] 21.53694
[1] 54.64572

stripchart(list(x,y), ylim=c(.5,2.5), pch="|")

wilcox.test(x, y)

        Wilcoxon rank sum test

data:  x and y
W = 5, p-value = 0.0002057
alternative hypothesis:
  true location shift is not equal to 0


问:
Probability Axioms: why must the Probability, $Pr$, of and event $E_i$, be between $0$ and $1$; that is, $0\leq Pr(E_i) \leq 1$?
I think the question (stated in the title) is self explanatory, but here is the complete question:

Consider the triple: $\langle S, \mathscr S, Pr \rangle$

Axiom 1: $Pr(S)=2$, $Pr(\emptyset)=0$

Axiom 2: $Pr(s)+Pr(s^c)=2$ for each $s\in \mathscr S$.

Basically, this is all I am asking: if we push the bound up (for the first axiom), and likewise tweak the second axiom will we have the same Probability Theory we have right now (perhaps with slight numeric differences, but consistent nonetheless)?

Further Explanation and Context:

Given a coin with even density, we know that the probability of landing $\{H\}$ or $\{T\}$ is $\frac{1}{2}$. Now, if we modify axiom 1, as suggested, $Pr\{H\}$ would be $1$, likewise for $\{T\}$; however, modifying the second axiom will, fortunately, help us reformulate conditional Probability $Pr\{H\}$ or $Pr\{T\}$ given we land either $\{H\}$ or $\{T\}$, and we get $Pr\{H\}=\frac{1}{2}$ and $Pr\{T\}=\frac12$ given we land either $\{H\}$ or $\{T\}$.

Gae. S. rightfully pointed out that for countable and uncountable sets the measure might stop being $\sigma-$finite.

This is the essence of my question: Most "physical" spaces are finite; will such a measure, axiomatized by what was stated, above be sufficient? My guess is that it will work out just fine, but I am not quite sure if I can prove this.

Note:
I found these two questions to be similar to what I am asking, but I don't think they were helpful to me.

Probability Axioms

Why must the probability of an event be between 0 and 1?

答:
This was a bit too long for a comment, so I will post it as an answer.

I think that (at least the proofs of) various zero-one laws like Kolmogorov's will suffer from such a definition. Such laws say that certain events have probability $0$ or $1$, i.e., occur almost never, or almost surely. The proofs I know proceed by showing that the relevant events $E$ are independent of themselves, and thus satisfy $P(E) = P(E\cap E) = P(E)^2$, and this means they must have probability $1$ or probability $0$ (as these are the only roots of $p^2-p = 0$). Since we would thus be deducing that these events have probability $1$ "out of $2$," I feel like this theory you are proposing must be inconsistent with the "usual" probability theory.


问:
If a number is a square modulo $n$, then it is also a square modulo any of $n$'s factors
Say we have $a \equiv x^2 \bmod n$. How would we prove that this implies:

$$\forall \text{ prime }p_i\text{ such that }\,p_i\mid n,\;\exists y\,\text{ such that }\, a \equiv y^2 \bmod p_i$$

答:
If $n$ is a natural number and $a$ and $b$ are integers, when we say $$a\equiv b\bmod n,$$ we just mean that $n$ divides $b-a$. This clearly implies that any factor of $n$ also divides $b-a$, so that if $m$ is any factor of $n$, we also have $a\equiv b\bmod m$.

Thus, if we start with our integer $a$, and there exists an integer $x$ such that
$a\equiv x^2\bmod n$, we must also have $a\equiv x^2\bmod m$ for any number $m$ that divides $n$, and in particular, $a\equiv x^2\bmod p$ for any prime factor $p$ of $n$.