问:
How to find the minimum value of the expression? $\sqrt{(x-0,5)^2+0,75}+\sqrt{(x-\frac{\sqrt{3}}{2})^2+0,25}$?
Help me with this. What is the minimum value of this expression I do not know how to do that. $\sqrt{(x-0,5)^2+0,75}+\sqrt{(x-\frac{\sqrt{3}}{2})^2+0,25}$       ?</p>

答:
[1] Differentiate the function $f(x)$ w.r.t  $x$ which will give you $\frac{d(fx)}{d x}$, and then equate it to zero, and then solve for $x$

[2] now differentiate the same differntiated function again w.r.t to $x$ i,e $\frac{d^2 f(x)}{d x^2}$and plug in the $x$ value from step 1 to see if the value is greater than 0, if it is , it is the minimum value.


问:
Binary representation of a number $n$ with exactly one leading 0 can be turned into a binary representation of $n+1$

Prove that for $\forall n \in \mathbb{N}$, the binary representation of $n$ with exactly one leading 0 can be turned into a binary representation of $n + 1$  by flipping exactly one bit from 0 to 1, and some number of bits from 1 to 0. For example, the binary representation ofn=7 with one leading 0 is 0111, andn=8 has a binary representation 1000. Only one bit,d3, flips from 0 to 1.

My thoughts are to use induction, and then for the induction part, I want to split it into even and odd, but then I don't know what to do.

答:
HINT: Let $b_kb_{k-1}\ldots b_0$ be the binary representation of an $n$ with exactly one leading zero (so $b_k=0$ and $b_{k-1}=1$). Let $m$ be the smallest non-negative integer such that $b_m=0$; clearly $0\le m\le k$. Show that $n+1$ has the binary representation $b_k'b_{k-1}'\ldots b_0'$, where

$$b_i'=\begin{cases}
b_i,&\text{if }m<i\le k\\
1,&\text{if }i=m\\
0,&\text{if }0\le i<m\,.
\end{cases}$$

ContentLicense: CC BY-SA 4.0


答:
Hint

If the binary representation of $\ n\ $ terminates with exactly $\ k\ $ $1$s preceded by a $0$, then the binary representation of $\ n+1\ $ must terminate with exactly  $\ k\ $ $0$s preceded by a $1$.

Title: Binary representation of a number $n$ with exactly one leading 0 can be turned into a binary representation of $n+1$

Body:
Prove that for $\forall n \in \mathbb{N}$, the binary representation of $n$ with exactly one leading 0 can be turned into a binary representation of $n + 1$  by flipping exactly one bit from 0 to 1, and some number of bits from 1 to 0. For example, the binary representation ofn=7 with one leading 0 is 0111, andn=8 has a binary representation 1000. Only one bit,d3, flips from 0 to 1.

My thoughts are to use induction, and then for the induction part, I want to split it into even and odd, but then I don't know what to do.


问:
Show that an equivalence of categories between two $1$-object categories is an isomorphism.
Suppose that $\mathcal{C},\mathcal{D}$ are two categories with only $1$ object, and that there are functors $F:\mathcal{C}\to\mathcal{D}, G:\mathcal{D}\to\mathcal{C}$ form an equivalence of categories.

Question Is it necessarily true that $FG=1_{\mathcal{D}}$ and $GF=1_{\mathcal{C}}$? If no, give a concrete (counter-)example.

Attempt I think the answer to the question is no. What I can show is that $F\widetilde{G}=1_\mathcal{D}, \widetilde{G}F=1_\mathcal{C}$, where $\widetilde{G}:\mathcal{D}\to \mathcal{C}$ is a "conjugate" of $G$. Could anyone provide a
counterexample/proof?

答: You can use the fact that an equivalence of categories is essentially surjective - see https://stacks.math.columbia.edu/tag/0013 Lemma 4.2.19</p>


答: I'm not sure if I understand correctly. If the binary representation of a number $n$ is $011...11$ (with $k$ digits) then $n = 2^ {k-1} - 1$. When you change the leading $0$ for $1$ and all $1$'s for $0$'s you obtain $100.. 00$ which is $2^{k-1}$.


问:
$ \min _{\operatorname{rank}\{\boldsymbol{B}\}=\boldsymbol{k}<r=rank(A)}\|\boldsymbol{A}-B\|_{2} ,A,B \in \mathbb{C}^{m \times n}$ and A is given
I would like to solve the following minimization problem.$$
\min _{\operatorname{rank}\{\boldsymbol{B}\}=\boldsymbol{k}<r=rank(A)}\|\boldsymbol{A}-\boldsymbol{B}\|_{2} ,\boldsymbol{A,B} \in \mathbb{C}^{m \times n}
$$ where
$
A \in \mathbb{C}^{m \times n}, \operatorname{rank}\{A\}=r, 2 \leq r \leq \min \{m, n\}
$ is a given matrix.

My idea at first was to go with SVD and try to rewrite the norm. But I didn't reach an interesting result.

Any suggestions??

答:
The minimum does not exist, but the infimum is zero.

For example: for any $\epsilon > 0$, the matrices
$$
A = \pmatrix{\epsilon \cdot I_k & 0_{k\times (n-k)}\\ 0_{(m-k)\times k}& 0_{(m-k) \times (m-k)}}, \quad
B = \pmatrix{\epsilon \cdot I_r & 0_{r\times (n-r)}\\ 0_{(m-r)\times r}& 0_{(m-r) \times (m-r)}}
$$
satisfy the constraints with $\|A - B\|_2 = \epsilon$ (assuming $\|\cdot\|_2$ refers to the spectral norm).


问:
Dividing over $C$ with parameters in $acl(B)$
Let $M\models T$ and $C\subset B\subset M$ be finite sets. Let $tp(a/d)$ divides over $C$ and $d\in acl(B)$. Can we conclude that $tp(a/B)$ divides over $C$?

Here is my attempt: Let $\phi(x,d)\in tp(a/d)$ be a formula which divides over $C$. Let $\theta(y,\bar{b})$ be a formula witnesses that $d\in acl(B)$ where $\bar{b}\in B$.  Then $\phi(x,y)\wedge \theta(y,\bar{b})$ is a formula witnesses that $tp(a/B)$ divides over $C$.
Is this idea true?

答:
The answer is no: in a general theory, it is possible to have the situation that $\text{tp}(a/B)$ does not divide over $C\subseteq B$, but $\text{tp}(a/\text{acl}(B))$ divides over $C$. Even worse, it is possible that $\text{tp}(a/B)$ does not divide over $C\subseteq B$, but $\text{tp}(a/\text{acl}(C)B)$ divides over $\text{acl}(C)$.

This is a bit shocking to me - I had believed the answer was yes, and I previously wrote a detailed answer with a proof (now deleted), but the proof was incorrect. I subsequently found a counterexample, Gabe Conant provided a simplification, and I wrote it up here.

In the meantime, let me point out some positive facts:

If $\text{tp}(a/B)$ does not divide over $C\subseteq B$, then $\text{tp}(\text{acl}(Ca)/B)$ does not divide over $C$. So acl can be safely added to the left hand side in a general theory.

If $\text{tp}(a/B)$ does not fork over $C\subseteq B$, then we can add acl on both sides: $\text{tp}(\text{acl}(Ca)/\text{acl}(B))$ does not fork over $C$.

So the answer to your question is yes if you replace dividing by forking. As a consequence, the answer is yes in any theory in which forking equals dividing for complete types (e.g. any stable, simple, or o-minimal theory - or if we additionally assume that $C$ is a model, then in any $\mathrm{NTP}_2$ theory).


问:
Mathematical Induction prime help
I recently obtained "What is Mathematics?" by Richard Courant and I am having trouble understanding what is happening with the Prime Number Unique Factor Composition Proof (found on Page 23).

The first part:

I have looked over it many times but I just don't understand what he is doing and why, as  an example, if you remove the first factor from either side of the equation you end up with two essentially different compositions that make up a new smaller integer.

I'm sure it is a simple error on my behalf but I have been stuck on this for a long time so I would appreciate a walkthrough explained clearly or some pointers in the right direction. Thank you.

答: You want to show that every positive integer can be expressed as a product of prime numbers, secondly you want that such a decomposition is unique (except for the order of factors). Call $\mathscr{F}$ the set of positive integers not satisfying your claim, i.e. the set of positive integers that can be written in more than a single way as a prime product. In terms of $\mathscr{F}$, your claim becomes: show that $\mathscr{F}$ is empty. By contradiction, suppose $\mathscr{F}$ non-empty and look for an absurd.Every non-empty set of positive integers has a smallest element, call $m$ the smallest element of $\mathscr{F}$. Using the book notation, suppose $p_1=q_1$ and  cancel the two factors from both sides. Since the two factorizations were supposed to be different, they remain different after that cancellation, producing an element of $\mathscr{F}$, but smallest than $m$, contradicting the definition of $m$ as the smallest element of $\mathscr{F}$. Then $p_1$ is strictly less than $q_1$ or viceversa...</p>

ContentLicense: CC BY-SA 3.0

Title: Mathematical Induction prime help


问:
Find all $x,y \in \mathbb N^*$ such that $\frac{1}{x} + \frac{1}{y}=5$
My proof:

By the way of contradiction, suppose that we found $(x,y) \in \mathbb N^*$ satisfies:$$\frac{x+y}{xy}=5$$
So : $$x+y=5xy \\ \iff x=5xy-y \iff x=y(5x-1)$$
Since $x\in \mathbb N^* $ so the minimum value of $5x$ is $5>1 \implies y\mid x$

By the same method we have:
$$y=x(5y-1) \implies x\mid y$$
$x\mid y, y\mid x \implies x=y$, Now we can write the original equation as :
$$\frac{2x}{x^2}=5 \iff 2=5x \implies x\notin \mathbb N^* $$
Therefore there is no solution to this equation: $$\frac{x+y}{xy} =5$$

Does this considered a proof?

Body:
$$\frac{1}{x} \le 1 \text{ and }\frac{1}{y} \le 1.$$

Thus
$$\frac{1}{x}+\frac{1}{y} \le 2.$$


答:
Your proof is correct. But why make it complicated like that?

Just notice that $x \ge 1$ then $\frac{1}{x} \le 1$. Same for $y$ that $\frac{1}{y} \le 1$.

Hence,
$$\frac{1}{x}+\frac{1}{y}\le 2 <5$$

So, there is no solution.

Title: Find all $x,y \in \mathbb N^*$ such that $\frac{1}{x} + \frac{1}{y}=5$

Body:
My proof:

By the way of contradiction, suppose that we found $(x,y) \in \mathbb N^*$ satisfies:$$\frac{x+y}{xy}=5$$
So : $$x+y=5xy \\ \iff x=5xy-y \iff x=y(5x-1)$$
Since $x\in \mathbb N^* $ so the minimum value of $5x$ is $5>1 \implies y\mid x$

By the same method we have:
$$y=x(5y-1) \implies x\mid y$$
$x\mid y, y\mid x \implies x=y$, Now we can write the original equation as :
$$\frac{2x}{x^2}=5 \iff 2=5x \implies x\notin \mathbb N^* $$
Therefore there is no solution to this equation: $$\frac{x+y}{xy} =5$$

Does this considered a proof?


答:
Take a look this is a proof by contradiction you're suposing that the two factorizations are different and getting an absurd (in this case that one prime is divisible by another prime), thus the asumption the factorizations were different is false.
m is the smallest choice because you're using the principle of well ordering defined on page 18.. "Every non empty set of positive integers has a smallest member" https://en.wikipedia.org/wiki/Well-order this smallest member is the smallest number you're suposing you can factorize into two different prime factors

Title: Mathematical Induction prime help

Body:
I recently obtained "What is Mathematics?" by Richard Courant and I am having trouble understanding what is happening with the Prime Number Unique Factor Composition Proof (found on Page 23).

The first part:

I have looked over it many times but I just don't understand what he is doing and why, as  an example, if you remove the first factor from either side of the equation you end up with two essentially different compositions that make up a new smaller integer.

I'm sure it is a simple error on my behalf but I have been stuck on this for a long time so I would appreciate a walkthrough explained clearly or some pointers in the right direction. Thank you.


问:
Leading term asymptotic integral containing logarithm
I want to get the leading term in the asymptotic of the integral $$\int_{a}^{+\infty}\frac{x^3}{(1+x^2)^{\alpha}}\ln\frac{x+a}{x-a} \,dx$$ for $a\to +\infty$. Here $a>0$ and $\alpha>\frac{3}{2}$ to ensure integrability at infinity.
My main concern is the integrable singularity in $a$. Usually I do an appropriate change of variables to remove it, but I can't seem to find an appropriate one.

答:
First, we make a change $x=ta$. The integral becomes $I(a, \alpha)=a^4\int_{1}^{\infty}\frac{t^3}{(1+a^2t^2)^{\alpha}}\ln\frac{t+1}{t-1} \,dt=a^{4-2\alpha}\int_{1}^{\infty}\frac{t^{3-2\alpha}}{(1+\frac{1}{a^2t^2})^{\alpha}}\ln\frac{t+1}{t-1} \,dt$.

Integral converges; the only (integrable) singularity is at $t=1$.

We can expand the integrand into the series
$$I(a,\alpha)=a^{4-2\alpha}\int_{1}^{\infty}t^{3-2\alpha}\ln\frac{t+1}{t-1}{(1-\frac{\alpha}{a^21!}\frac{1}{t^2}+\frac{\alpha(\alpha+1)}{a^42!}\frac{1}{t^4}+...)} \,dt$$ where all integrals converge ($3-2\alpha<1$ and $\ln\frac{t+1}{t-1}\sim\frac{1}{t}$ at $t\to\infty$)

The mian asymptotics term is $a^{4-2\alpha}\int_{1}^{\infty}t^{3-2\alpha}\ln\frac{t+1}{t-1}dt=a^{4-2\alpha}\int_{1}^{\infty}t^{b}\ln\frac{t+1}{t-1}dt$

where $b=3-2\alpha$.

It seems that the easiest way to evaluate the integral $J(b)=\int_{1}^{\infty}t^{b}\ln\frac{t+1}{t-1}dt=\int_{0}^{1}x^{-b-2}\ln\frac{1+x}{1-x}dx$  ($t=\frac{1}{x}$) is to expand the integrand into the series:

$J(b)=2\int_{0}^{1}x^{-b-2}\left(x+\frac{x^3}{3}+\frac{x^5}{5}+...\right)dx=2\left(\frac{1}{-b}+\frac{1}{3(-b+2)}+\frac{1}{5(-b+4)}+...\right)=$$=\frac{1}{2}\sum_{k=0}^{\infty}\frac{1}{k+\frac{1}{2}}\frac{1}{k-\frac{b}{2}}=\frac{1}{-b-1}\left(\Psi(-\frac{b}{2})-\Psi(\frac{1}{2})\right)$,

where $\Psi$ is digamma function: $\Psi(1+s)=-\gamma-\sum_{k=1}^{\infty}\left(\frac{1}{k+s}-\frac{1}{k}\right)$

Finally, we get the main asymptotics term$$I(a, \alpha)\sim{a}^{4-2\alpha}\frac{1}{2(\alpha-2)}\left(\Psi(\alpha-\frac{3}{2})-\Psi(\frac{1}{2})\right)$$

At $\alpha\to2$ we have to disclose the uncertainty

$I(a,\alpha\to2)\sim\frac{{a}}{2}^{0}\lim_{\epsilon\to0}\frac{1}{\epsilon}\left(\Psi(\frac{1}{2}+\epsilon)-\Psi(\frac{1}{2})\right)=\frac{1}{2}\Psi'(\frac{1}{2})=\sum_{k=1}^{\infty}\frac{2}{(2k-1)^2}$

$$I(a,\alpha=2)\sim\int_{1}^{\infty}t^{-1}\ln\frac{t+1}{t-1}dt=\frac{3}{2}\zeta(2)=\frac{\pi^2}{4}$$

Other asymptotics terms can be evaluated in the same way.


问:
Can a formal sum be 'infinite'?
I'm reading Abstract Algebra (by Doomit and Foote), where a polynomial is defined as the formal sum

$$a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$

with $n \geq 0$.

Is it valid to define a polynomial as $\sum_{j=0}^{\infty} a_jx^j$, for which exists an $n$ such that $m \geq n$ implies $a_m=0$, and said that it is a formal sum?

答:
Yes, it can.

The abstract definition of a polynomial with coefficients in a commutative ring $R$ is  ‘an element of the set $R^{(\mathbf N)}$, i.e. an infinite sequence of elements of $R$ with finite support. I has a structure of an $R$-algebra if addition is defined componentwise and multiplication is defined as a Cauchy product. With this approach, $X$ denotes the sequence $(0,1,0,0,\dots,0,\dots)$, and one easily checks that $X^2=(0,0,1,0,\dots,0,\dots)$, $X^3=(0,0,0,1,0,\dots)$, and so on.

Similarly, one defines the ring of formal power series, denoted $R[[X]]$, as the set $R^{\mathbf N}$, i.e. the set of  infinite sequences without any condition, endowed with addition componentwise and the Cauchy product. The ring of polynomials $R[X]$ is a subalgebra of the $R$-algebra $R[[X]]$.

Title: Can a formal sum be 'infinite'?


答:
It's important to note that your title question

Show that an equivalence of categories between two 1-object categories is an isomorphism.

is different from the question in the body

Question Is it necessarily true that $FG=1_D$ and $GF=1_C$? If no, give a concrete (counter-)example.

depending on exactly how you define "an equivalence of categories".

It is true that every functor between monoids $F: \mathcal{C} \rightarrow \mathcal{D}$ that forms part of an equivalence of categories is an isosmorphism. This is why the statement in the title of your question is often given as a fact.

It is false to claim that, whenever $F: \mathcal{C} \rightarrow \mathcal{D}$ and $G: \mathcal{D} \rightarrow \mathcal{C}$ are a pair of functors forming an equivalence, $G$ is the (strict) inverse of $F$. As you've shown, all we know is that $G$ is some conjugate of the inverse.

To get a concrete counterexample, take some nonabelian group (which we'll treat as the morphism set of our monoid). Let $F$ be the identity and $G$ be conjugation by some element.

Title: Show that an equivalence of categories between two $1$-object categories is an isomorphism.

Body:
Suppose that $\mathcal{C},\mathcal{D}$ are two categories with only $1$ object, and that there are functors $F:\mathcal{C}\to\mathcal{D}, G:\mathcal{D}\to\mathcal{C}$ form an equivalence of categories.

Question Is it necessarily true that $FG=1_{\mathcal{D}}$ and $GF=1_{\mathcal{C}}$? If no, give a concrete (counter-)example.

Attempt I think the answer to the question is no. What I can show is that $F\widetilde{G}=1_\mathcal{D}, \widetilde{G}F=1_\mathcal{C}$, where $\widetilde{G}:\mathcal{D}\to \mathcal{C}$ is a "conjugate" of $G$. Could anyone provide a
counterexample/proof?


问:
Studying set theory using polynomials?
I came up with a way to do set theory using polynomial arthimetic.

To explain, consider two polynomials $P(x)$ and $Q(x)$, then we can observe the following analogies:

1. In Polynomial addition

$$ |P(x)|+|Q(x)|$$

Is analogous to set intersection of sets because the above equation is only true for $x$ values which make both $P$ and $Q$ zero at the same time.

2. In Polynomial multiplication

$$|P(x)| \cdot |Q(x)|$$

Can be thought of as set union because the above equation is true for the numbers which satisfy either $P$ or $Q$

3. In polynomial division

$$ \frac{|P|}{|Q|} $$

The above expression can be thought of a set difference, we remove the zeros of Q which exist also in P.

Analogies to set complements:

Polynomial reciprocal is similar to set inverses, for example

$$ |P(x)^{-1}| = \frac{1}{|P(x)|}$$

Is defined for all values of $x$ except where $P(x)=0$ where it is undefined. To get the whole universal set back, we multiply inverse with regular one:

$$|P(x)| \cdot \frac{1}{|P(x)| }=1$$

Proving set identities using polynomials

$ A \cup (B \cup C)= (A \cup B) \cup C$

This is directly analogous to associativity of polynomial multiplication , considering three polynomials $P,Q,R$:

$$ ( |P| \cdot |Q|) |R|  = |P|( |Q| \cdot |R|)$$

$A \cup B  = B \cup A, A \cap B , B \cap A$

This is directly analogous to communality of polynomial multiplication.

Demorgan's laws

$$ A^c \cup B^c = (A \cup B)^c$$

Easy relation to understand with polynomial way, consider two polynomials $P(x),Q(x)$, then it is trivial that:

$$ \frac{1}{|P|} \cdot \frac{1}{|Q|} = (|PQ|)^{-1}$$

$$ (A \cap B)^c= A^c \cup B^c$$

This one just turns out weird, but it works well for most identities where there isn't addition and multiplication on both sides

My question: Has this idea been studied before? Where there any new insights from thinking of sets like this?

[Note: I  know of generating functions already]

答: A rather elaborate theory related to this was worked out over a period of several decades by the Russian mathematician Vladimir Logvinovich Rvachev (1926-2006) (biography), called called $R$-functions. See Semi-analytic geometry with R-functions by Vadim Shapiro (2007; footnote on p. 3 discusses the origin of the name "R-function").

Title: Studying set theory using polynomials?


问:
Congruence question and prime numbers
Let $p$ be an odd prime number when $p-1=2^{s}\cdot t$, $s\in \mathbb{N}, t\in \mathbb{Z}$ and $t $ is odd. Need to prove that if $(a,p)=1$, then $a^{t}\equiv 1 \pmod{p}$  or $a^{2^{i}\cdot t}\equiv -1 \pmod{p}$, for $0\le i\le s-1$. In addition $a\in \mathbb{Z^{+}}$.

My attempt: suppose that $(a,p)=1$ and $p-1=2^{s}\cdot t$. Since $0\le i\le s-1$ then let $s=i+x$, for $x\in \mathbb{N}$. According to little's Fermat theorem, since $(a,p)=1$ we have that: $a^{p-1}\equiv 1 \pmod{p}$. Recall that $p-1=2^{s}\cdot t$ so by substitution - $a^{p-1}\equiv a^{2^{s}\cdot t}\equiv 1 \pmod{p}$. Now by substitution of $s=i+x$ we get: $a^{2^{i}\cdot 2^{x}\cdot t}\equiv (a^{2^{i}\cdot 2^{x}})^{t}\equiv (a^{2^{i}\cdot t})^{2^{x}}\equiv 1 \pmod{p}$. From here I don't know how to proceed, but I still didn't use the definition of t and p being odd prime numbers. Therefore I will be glad to get some help. Thanks!

答:
Hint

If $\ a^{2^st}\equiv1\pmod{p}\ $, then $\ \left(a^{2^{s-1}t}-1\right)\left(a^{2^{s-1}t}+1\right)\equiv$$0\pmod{p}\ $, so either $\ a^{2^{s-1}t}\equiv1\pmod{p}\ $ or $\ a^{2^{s-1}t}\equiv-1\pmod{p}\ $. If the second alternative holds, you're home.  Otherwise, now play the same game with $\ s-1\ $ in place of $\ s\ $.

Hint for another approach

Here's another approach which you might find easier.  The multiplicative order $\ \sigma\ $ of $\ a\ $ mod $\ p\ $ must be a divisor of $\ p-1=2^st\ $—that is $\ \sigma=2^jv\ $, where $\ 0\le j\le s\ $ and $\ t=vw\ $ with $\ v\ $ and $\ w\ $ both odd. If $\ j=0\ $, what is $\ a^v\pmod{p}\ $? What then is $\ a^{vw}=a^t\pmod{p}\ $? If $\ j\ge1\ $ and $\ i=j-1\ $, what is $\ a^{2^iv}\pmod{p}\ $? What then is $\ a^{2^ivw}=a^{2^it}\pmod{p}\ $?

Title: Congruence question and prime numbers


问:
A more logical approach to a word problem
I was taught in high school algebra to translate word problems into algebraic expressions. So when I encountered this problem I tried to reason out an algebra formula for it

For every dollar Ben spent on bagels, David spent 25 cents less. Ben
paid $12.50 more than David. How much did they spend in the bagel store
together?

To solve this I imagined a series of comparisons when Ben spends $x$, David spends $.75x$. Loop this relationship until $x - .75x \approx 12.50$. Good. Done. $x = 50$, then add David's for the answer. Coming from computers, I would have set this up in code where a loop (recursion) would increase $x$ until the condition $x - .75x = 12.50$ was met, then the "loop counter/accumulator" would be how much Ben spent, i.e., $50$, etc.

I'm a beginner with math, but it seems like there should be a better approach, something with series and sequences or even calculus derivatives, something better than my brute-force computer algorithm. Can someone enlighten? The "answer" given at the site (see link) is its own brute-force and hardly satisfying. I'm thinking there should be something more formal -- at least for the first part that derives $50$.

Update

I think everyone so far has missed my point. Many of you simply re-did the problem again. I'm wondering if there is a more formal way to do this other than just "figuring it out" (FIO). The whole FIO routine is murky. It looks like a limit problem; it looks like a system of equations, but I'm not experienced enough to know exactly. If there isn't, then let's call it a day....

答:
If the difference is $\$12.50$ then it must be the product of dollars times $\$0.25$ so we divide:
$\quad\dfrac{12.50}{0.25}=50\quad$ for Ben and another $\quad 37.50 \quad$ for David for a total of $\quad\$87.50$


答: Hint: Interpret the given expression as the sum of two distances between points in the plane.

Title: How to find the minimum value of the expression? $\sqrt{(x-0,5)^2+0,75}+\sqrt{(x-\frac{\sqrt{3}}{2})^2+0,25}$?

Body: Help me with this. What is the minimum value of this expression I do not know how to do that. $\sqrt{(x-0,5)^2+0,75}+\sqrt{(x-\frac{\sqrt{3}}{2})^2+0,25}$       ?</p>


问:
Eigenvalues of a particular block circulant matrix
I need to compute all the eigenvalues of the following block-circulant matrix for a research. Can anyone help me compute the eigenvalues of the following matrix?
$$\left[\begin{array}{l}2I&-I&0&0&0&...&0&0&-I\\-I&2I&-I&0&0&...&0&0&0\\0&-I&2I&-I&0&..&0&0&0\\.\\.\\.\\0&0&0&0&0&...&-I&2I&-I\\-I&0&0&0&0&...&0&-I&2I\end{array}\right]$$
In above, $I$ denotes the identity matrix of dimension $a$ and there are $r$ block-rows and block-columns, making the entire matrix have dimension $ar \times ar$. Any help will be greatly appreciated!

答:
This matrix can be written as the Kronecker product $M \otimes I_a$, where $M$ is the $r \times r$ circulant matrix
$$
M = \pmatrix{2&-1&0&\cdots&0&-1\\
-1&2&\ddots & \ddots & \vdots & 0\\
0 & \ddots & \ddots & \ddots & 0 & \vdots\\
\vdots &0&\ddots&\ddots&-1&0\\
0&\vdots&\ddots &-1 & 2 & -1\\
-1 & 0 &\cdots & 0& -1 & 2}.
$$
It follows that the eigenvalues of your matrix are simply the eigenvalues of $M$, each repeated with $a$-fold multiplicity.

The eigenvalues of the circulant matrix $M$ can be computed using the formula here to be
$$
\lambda_k = 2 - e^{2 \pi i k/r} - e^{-2 \pi i k/r} = 2(1 - \cos(2 \pi k/r)), \quad k = 0,1,\dots,r-1.
$$


问:
Confusion on the order of $GL_2(\mathbb{F}_p)$
In this question, they show that the order of $GL_2(\mathbb{F}_p)$ is $(p^2-1)(p^2-p)$.

For the first column, there are $p^2$ options, and we need to exclude the $0$ column, so there are $p^2-1$. That's clear to me.

For the second column, there are again $p^2$ possible combinations, and that answer says we need to exclude the $p$ different scalar multiples of the first column. That's not clear to me; for example if the first column is $[1 \quad 1]^t$, then there are $p$ different scalar multiples, namely, $[0\quad0]^t,[1\quad1]^t,[2\quad 2]^t,...,[p\quad p]^t$. However, if the first column is $[1\quad p]^t$, what are the $p$ combinations that need to be excluded and why do they need to be excluded?

答:
As suggested in your comment, we'll take the first column as $[1\quad 4]^t$ and $p>4$.

We should note that the matrix $\begin{bmatrix}1&1\\4&4\end{bmatrix}$ has determinant zero, so is not in the group $GL_2(\mathbb{F}_p)$. Similarly, if the second column is any multiple of the first column, say $n$ times the first column, then we see that the matrix $\begin{bmatrix}1&n\\4&4n\end{bmatrix}$ also has determinant zero, so is not in the group $GL_2(\mathbb{F}_p)$.

Therefore we are excluding one matrix for each $n\in\mathbb{F}_p$, of which there are $p$ options.


问:
Reflection matrix about a plane
How do you construct a matrix in $\mathbb{R}^3$ that reflects about the plane $y=z$? And is there a way to construct a reflection matrix about any plane in general?

答: To answer the first question. Look what the reflection does to the unit vectors in $\mathbb{R}^{3}$. $(1, 0, 0)$ will be mapped to $(1, 0, 0)$, $(0, 1, 0)$ will be mapped to $(0, 0, 1)$ and $(0, 0, 1)$ will be mapped to $(0, 1, 0)$. This defines the linear mapping and therefore will tell you how the matrix looks like.

Title: Reflection matrix about a plane


问:
Prove M is open on the subspace ($A \cup B$).
$(X,T)$ is a topological space, $A,B $ not empty subsets. If $M \subset A \cap B$, and $M$ is simultaneously open in relative topologies $ $ $\left.T\right|_{A}$ and $ $ $\left.T\right|_{B}$. Prove M is open on the subspace $(A \cup B,\left.T\right|_{(A \cup B)})$

Attempt : Trying to reach $M=G \cap (A \cup B)$ from the expression $M= G_1 \cap A$, $M=G_2 \cap  B$, but i couldn't reach it. Also trying to prove $G_1=G_2=G$ so we have reached that. And $ \mathring{A}=A$ in $A \cup B$ topology but nothing..

答:
Let $M=A\cap t_A=B\cap t_B$ where $t_A$ and $t_B$ belong to $T.$

We have $t_A\cap t_B\in T.$

Now $M\subset t_A$  and $M\subset t_B.$ So $M\subset (t_A\cap t_B).$ And $M\subset A\cup B.$ Therefore $$M\subset
(t_A\cap t_B)\cap (A\cup B)=$$ $$=[(t_A\cap t_B)\cap A]\,\cup \, [(t_A\cap t_B)\cap B]=$$ $$=
[t_B\cap (t_A\cap A)]\,\cup \,[t_A\cap (t_B\cap B)]=$$ $$=[t_B\cap M]\,\cup \,[t_A\cap M]\subset$$ $$\subset [M]\cup [M]=M.$$

Therefore $M=(t_A\cap t_B)\cap (A\cup B)\in T_{A\cup B}.$


问:
decresing sequence implies $\lim{n u_n} $ exists?
if $(u_n)$ is a nonincreasing sequence of positive real
numbers; how to prove or disprove  that the sequence $(nu_n)$ admits a limit?
We know  two results

1- if $(u_n)$ is a  sequence of positive real
numbers such that $\sum_n a_n$ converges, and if $\lim\limits_{n \to \infty} n u_n$ exists, then $\lim\limits_{n \to \infty} n u_n = 0$

2- if $(u_n)$ is a nonincreasing sequence of positive real
numbers such that $\sum_n u_n$ converges, then $\lim_{n \rightarrow \infty} n u_n = 0$.

PS :  I corrected the usual notation for a sequence

答:
Let $y_n = n u_n$. Then $L = \lim_{n \rightarrow \infty} y_n$ exists iff $\forall \varepsilon \exists m_0 \forall m \geq m_0 \mbox{ s.t. }|y_m - L| < \varepsilon$, i.e. for any small value, the $y_n$ beyond some point are all closer to $L$ than that value.

This means that $|n u_n - L| < \varepsilon$, so $|u_n - \frac{L}{n}| < \frac{\varepsilon}{n}$, or to rearrange it (and taking advantage of the fact that $u_n > 0$), $\frac{L - \varepsilon}{n} < u_n < \frac{L + \varepsilon}{n}$.

This means that the limit can only exist if $u_n = o(n^{-1})$ , i.e. $u_n$ shrinks faster than $\frac{1}{n}$.


答:
The given matrix is the Kronecker product $C \otimes I$ of the circulant matrix

$$C=\left[\begin{array}{cccccccc}2&-1&0&0&0&...&0&0&-1\\-1&2&-1&0&0&...&0&0&0\\0&-1&2&-1&0&..&0&0&0\\.\\.\\.\\0&0&0&0&0&...&-1&2&-1\\-1&0&0&0&0&...&0&-1&2\end{array}\right]$$

with matrix $I$.

As such, its eigenvalues are all the possible products of the eigenvalues of $C$ which are known to be the Discrete Fourier Transform of the first line of $C$ (see here) and the eigenvalues of $I$, i.e., $1$ (multiple value).

In fact, in this case, the eigenvalues of $C$ can be given simple expressions.

Title: Eigenvalues of a particular block circulant matrix

Body:
I need to compute all the eigenvalues of the following block-circulant matrix for a research. Can anyone help me compute the eigenvalues of the following matrix?
$$\left[\begin{array}{l}2I&-I&0&0&0&...&0&0&-I\\-I&2I&-I&0&0&...&0&0&0\\0&-I&2I&-I&0&..&0&0&0\\.\\.\\.\\0&0&0&0&0&...&-I&2I&-I\\-I&0&0&0&0&...&0&-I&2I\end{array}\right]$$
In above, $I$ denotes the identity matrix of dimension $a$ and there are $r$ block-rows and block-columns, making the entire matrix have dimension $ar \times ar$. Any help will be greatly appreciated!


问:
Derivative of Multivariate Function
Assume $M_\theta = \sum \theta_iM_i, \phi(M_\theta) \text{ is a differentiable function }\mathbb{R}^{n\times m} \rightarrow \mathbb{R}, \theta = [\theta_1 \dots \theta_N]$. I want to find $\frac{\partial\phi}{\partial \theta_i}$

Using the chain rule: $\frac{\partial\phi}{\partial \theta_i}=\frac{\partial\phi}{\partial M_\theta}\frac{\partial M_\theta}{\partial \theta_i} = \frac{\partial\phi}{\partial M_\theta} M_i$. However, the answer should be the dot product $<M_i,\frac{\partial\phi}{\partial M_\theta}>$. I am not sure how to obtain the correct answer?

答:
First of all, $\phi$ is a function of $n\times m$ matrices. There's no such thing as $\partial \phi/\partial M_\theta$; this is a "total" derivative $D\phi$. OK, so let's let $g(\theta_1,\dots,\theta_N) = \sum \theta_i M_i = M_\theta$. Then we're interested in $f(\theta_1,\dots,\theta_N) = \phi(g(\theta_1,\dots,\theta_N))$. The chain rule tells us that
$$\frac{\partial f}{\partial\theta_i} = D\phi(g(\theta))\left(\frac{\partial g}{\partial\theta_i}\right) = D\phi(g(\theta))(M_i).$$
Note that $\phi$, as a map $M^{n\times m}\to\Bbb R$ has a derivative, which is a linear map also mapping $M^{n\times m}\to\Bbb R$. So you apply that linear map to the $m\times n$ matrix $M_i$ and you get the answer. I would never write the notation that you're using. We could write $\phi'(M_\theta)$ instead of $D\phi(M_\theta)$.

You can think of evaluating this linear map on $M_i$ as a directional derivative in the direction of $M_i$, which is given, as you learned in multivariable calculus, the inner product of the gradient of $\phi$ with the direction vector $M_i$.

If you want to give me a specific example of a function $\phi$ you're working with, I can try to illustrate further.

Title: Derivative of Multivariate Function


答:
Kenta S gave two valid examples in their comment: both $u_n = \frac1{\sqrt n}$ and $u_n=1$ have the property that $\lim_{n\to\infty} nu_n = \infty$. It's important to note that, despite the use of the equals sign in that notation, it indicates a specific way that $\lim_{n\to\infty} nu_n$ fails to exist. (Infinity is not a number.)

But other divergent behaviour is also possible. Consider for example
$$
u_n = 2^{-\lfloor \log_2 n\rfloor} = \min\bigg\{ \frac1{2^k}\colon \frac1{2^k} > \frac1n \bigg\},
$$
so that $(u_n) = (1,\frac12,\frac12,\frac14,\frac14,\frac14,\frac14,\frac18,\dots)$. It's not hard to show that $\limsup_{n\to\infty} nu_n = 2$ but $\liminf_{n\to\infty} nu_n = 1$, and so $\lim_{n\to\infty} nu_n$ does not exist.

Title: decresing sequence implies $\lim{n u_n} $ exists?

Body:
if $(u_n)$ is a nonincreasing sequence of positive real
numbers; how to prove or disprove  that the sequence $(nu_n)$ admits a limit?
We know  two results

1- if $(u_n)$ is a  sequence of positive real
numbers such that $\sum_n a_n$ converges, and if $\lim\limits_{n \to \infty} n u_n$ exists, then $\lim\limits_{n \to \infty} n u_n = 0$

2- if $(u_n)$ is a nonincreasing sequence of positive real
numbers such that $\sum_n u_n$ converges, then $\lim_{n \rightarrow \infty} n u_n = 0$.

PS :  I corrected the usual notation for a sequence


问:
Using the trapezoid rule show that $\sum_{k=1}^nk = \frac{n(n+1)}{2}, \forall n \in \mathbb{N}$
$\sum_{k=1}^nk = \frac{n(n+1)}{2}, \forall n \in \mathbb{N}$

I got as far as

$$k = w_kf(x_k) \\
\sum_{k = 1}^n k = \int_1^n w_kf(x_k)dx = \\
\frac{(n-1)}{2(n-1)}[f_o+2(f_1+f_2+f_3+...+f_{n-1}) +f_n] \\
f(x_k) = k \Rightarrow \\
\frac{1}{2}[(n+1)+2(\sum_{k=2}^{n-1}k)]$$

I got stumped on this part so I improvised

$$\frac{1}{2}[(n+1)+2(\frac{(n-1)n}{2}-1)] = \\
\frac{1}{2}[(n+1)+(n-1)n-2] = \\
\frac{(n-1)(n+1)}{2}$$

I have a gut feeling that on $\frac{(n-1)}{2(n-1)}$ it should be $\frac{(n)}{2(n-1)}$ but the formula states it's $\frac{(b-a)}{2N}$ where N is the number of spaces between the points so I don't know.

Help?

答:
The composite trapezoidal rule for the function $f : [a,b] \rightarrow \mathbb{R}$ with uniform stepsize $h = (b-a)/n$ and points $x_i = a+ih$ for $i=0,1,2,\dots,n$ is given by
$$ T_h(f) = \sum_{i=0}^{n-1} \frac{h}{2} \Big[f(x_i) + f(x_{i+1})\Big] = \frac{h}{2} f(x_0) + h \left(\sum_{i=1}^{n-1} f(x_i)\right) + \frac{h}{2} f(x_i) $$
In the case of $f : [0,n] \rightarrow \mathbb{R}$ given by $f(x) = x$ and $h = 1$ we have $$T_1(f) = 0 + \sum_{k=1}^{n-1} k + \frac{1}{2} n.$$
The trapezoidal rule is exact for all linear functions, so
$$ T_1(f) = \int_0^n x^2 dx = \frac{1}{2}n^2.$$
It follows that
$$\sum_{k=1}^n k = T_1(f) + \frac{1}{2} n = \frac{1}{2}n^2 + \frac{1}{2}n = \frac{n(n+1)}{2}.$$


问:
Hessian and metric transformation
I am trying to obtain the relation between the Hessians when I consider the transformation $\tilde{g}=\sigma^{-1}g$, for $(M^n,g)$ semi-Riemannian manifold and $f\in C^\infty(M)$, then if
$$\text{Hess}f(X,Y)=g(\nabla_X\nabla f,Y),\hspace{.5cm}\widetilde{\text{Hess}}f(X,Y)=\text{Hess}f(X,Y)+\text{other things} $$
for this, I initially thought of applying Koszul's formula to get
$$\tilde{g}(\widetilde\nabla_XY,Z)=\sigma^{-2}\big\{g(\nabla_X Y,Z)-\sigma^{-1}\left[X(\sigma)g(Y,Z)+Y(\sigma)g(Z,X)-Z(\sigma)g(X,Y)\right]\big\}$$

and replace in the expression of the Hessian
$$\widetilde{\text{Hess}}f(X,Y)=\tilde{g}\Big(\nabla_X\widetilde\nabla f-\sigma^{-1}\big[g(\nabla\sigma,X)\widetilde\nabla f+g(\nabla\sigma,\widetilde\nabla f)X-g(X,\widetilde\nabla f)\nabla \sigma\big],Y\Big) $$
but I don't know how to continue. If you could guide me I would be very grateful, also with recommended bibliography.

答:
Setting $\sigma^{-1} = {\rm e}^{2\varphi}$, a standard computation gives $$\widetilde{\nabla}_XY = \nabla_XY + X(\varphi)Y+Y(\varphi)X - g(X,Y){\rm grad}_g\varphi.$$Also note that $${\rm d}f(Z) = g({\rm grad}_gf,Z) = \widetilde{g}({\rm grad}_{\widetilde{g}}f,Z) = {\rm e}^{2\varphi}g({\rm grad}_{\widetilde{g}}f,Z)$$leads to ${\rm grad}_{\widetilde{g}}f = {\rm e}^{-2\varphi} {\rm grad}_gf$. Hence: $$\begin{align} {\rm Hess}_{\widetilde{g}}f(X,Y) &= \widetilde{g}(\widetilde{\nabla}_X{\rm grad}_{\widetilde{g}}f, Y) \\ &= {\rm e}^{2\varphi} g(\widetilde{\nabla}_X({\rm e}^{-2\varphi}{\rm grad}_gf), Y) \\&= {\rm e}^{2\varphi} g(X({\rm e}^{-2\varphi}){\rm grad}_gf + {\rm e}^{-2\varphi} \widetilde{\nabla}_X{\rm grad}_gf, Y) \\ &= {\rm e}^{2\varphi}X({\rm e}^{-2\varphi}) {\rm d}f(Y) + g(\widetilde{\nabla}_X{\rm grad}_gf,Y) \\ &= {\rm e}^{2\varphi}(-2{\rm e}^{-2\varphi}X(\varphi)) {\rm d}f(Y) + g(\nabla_X{\rm grad}_gf+ X(\varphi){\rm grad}_gf+{\rm d}\varphi({\rm grad}_gf)X -
{\rm d}f(X){\rm grad}_g\varphi,Y) \\ &= -2{\rm d}\varphi(X){\rm d}f(Y)+{\rm Hess}_gf(X,Y)+{\rm d}\varphi(X){\rm d}f(Y) + g({\rm d}\varphi,{\rm d}f)g(X,Y) - {\rm d}f(X){\rm d}\varphi(Y) \\ &= -2({\rm d}\varphi\odot {\rm d}f)(X,Y) + {\rm Hess}_gf(X,Y) + g({\rm d}\varphi,{\rm d}f)g(X,Y).\end{align}$$Thus $${\rm Hess}_{\widetilde{g}}f = {\rm Hess}_gf + g({\rm d}\varphi,{\rm d}f)g - 2{\rm d}\varphi\odot{\rm d}f.$$Now, to translate this back to $\sigma$, note that $\sigma = {\rm e}^{-2\varphi}$ implies that ${\rm d}\sigma = -2{\rm e}^{-2\varphi}{\rm d}\varphi = -2\sigma{\rm d}\varphi$, so $${\rm Hess}_{\widetilde{g}}f = {\rm Hess}_gf -\frac{1}{2} g({\rm d}\sigma, {\rm d}f)\widetilde{g}+\frac{1}{\sigma}{\rm d}\sigma \odot {\rm d}f.$$If the above is not convenient for you, ${\rm d}\varphi = -(1/2) {\rm d}(\log \sigma)$ also makes the cut, so $${\rm Hess}_{\widetilde{g}}f = {\rm Hess}_gf -\frac{1}{2} g({\rm d}(\log\sigma),{\rm d}f) +  {\rm d}(\log \sigma)\odot {\rm d}f.$$

Title: Hessian and metric transformation


问:
Is the set made up of all points that satisfy $g_{1}(x, y, z)=x^2 + 2y^2=1 $, $\:g_{2}(x, y, z)= xy + xz=2$ compact?
I want to find the least value of a function whose domain is made up of all points that satisfy the following:

$g_1(x, y, z)= x^2 + 2y^2=1$

$g_2(x, y, z)= xy + xz=2$

In my book it is written that "a minimum exists since the domain is a closed set". However, previously there is a theorem that one can only be certain that function obtains a max and min value if the domain is compact, i.e. both closed and bounded. From what I see, the domain is not bounded, how can we then be certain that "a minimum exists"?

Any help would be appreciated.

$\textbf{Edit:}$: I want to find the least distance form the origin, so my function is $f(x, y,z)=^2+^2+^2$

答: Take $g=(g_1,g_2):\mathbb{R}^3 \longrightarrow \mathbb{R}^2$. Your set is clearly closed since $g$ is continuous and $$\{x^2+2y^2=1\}\cap\{xy+xz=2\}\cap \mathbb{R}^3=g^{-1}(1,2)$$ However, $g^{-1}(1,2)$ is not bounded. To see this, define $\vec{r}:A \longrightarrow \mathbb{R}^3 $ by $$\vec{r}(t)=\bigg(\cos(t),\frac{\sin(t)}{\sqrt{2}},2\sec(t)-\frac{\sin(t)}{\sqrt{2}}\bigg)$$ where $A=\big[0,\frac{\pi}{2}\big)$. You can check that $\vec{r}(A)\subseteq g^{-1}(1,2)$. Moreover, $$||\vec{r}(t)||\geq \Bigg|2\sec(t)-\frac{\sin(t)}{\sqrt{2}}\Bigg|\rightarrow \infty$$ as $t\rightarrow \frac{\pi}{2}^{-}$ so $g^{-1}(1,2)$ isn't bounded.


问:
There is a holomorphic branch of the square root of the function $f(z)=(z − 1)(z − 2)$
a) Prove that for adequately large $R$ in $\Omega=\{|z|>R\}$ there is a holomorphic branch of the square root of the function $f(z)=(z − 1)(z − 2)$

b) Choose one of these branches, find the coefficient $c_{-1}$ of the expansion of the function $g$ into the Laurent series in this area, and find the integral along the positively oriented circle:

We are looking for a function $g:\Omega \to \mathbb C$ such that $g^2(z)=f(z)$. I can't think of any statements from the lecture that I could use here...

答:
$\left(z^2(1-\frac {1}{z})(1-\frac {2}{z})\right)^\frac 12\\
z(1-z^{-1})^\frac 12(1- 2z^{-1})^\frac 12$

We have series expansions for each factor that will converge when $|z|>2$

$z(1-\frac 12 z^{-1}-\frac {1}{8}z^{-2} + \cdots)(1-z^{-1}-\frac {1}{2}z^{-2} + \cdots)$

Multiplying these together we can find the first few coefficients.
$z(1-\frac 32 z^{-1} -\frac 18 z^{-2} + \cdots)$

$c_{-1} = -\frac 18$


问:
Calculate the limit using definite integrals: $\lim_{n\to\infty}2n\sum_{k=1}^n\frac1{(n+2k)^2}$
Calculate the limit using definite integrals: $\lim_{n\to\infty}2n\sum_{k=1}^n\frac1{(n+2k)^2}$

Well, I started like this:

$\lim_{n\to\infty}2n\sum_{k=1}^n\frac1{(n+2k)^2}=\lim_{n\to\infty}2n[\frac1{(n+2)^2}+...+\frac1{(n+2n)^2}]$
From this point and on i'm always getting stuck. I know i should use Riemann integral definition but i don't know how can i recognize which function am i looking at? or how can i recognize the correct $\Delta x$? or at least the "borders" of the definite integral?

Say i presume $\Delta x=\frac1n$ (It's incorrect, Should be $\frac 2n$ - But why?), Then: $\lim_{n\to\infty}\frac1n[\frac{2n^2}{(n+2)^2}+...+\frac{2n^2}{(n+2n)^2}] $, Using limit on both of the 'edges' of the sum should give me the 'borders?' of the definite integral?

As you can see i'm pretty much lost in this kind of questions, BTW: An explanation will be of much more use than only an answer to this one.

答:
Factor out an $n^2$ from the denominator of the summand.  You then get

$$\frac{2}{n} \sum_{k=1}^n \frac{1}{(1+2 k/n)^2}$$

This is the form of a Riemann sum; take the limit as $n \to \infty$ and identify the result as a definite integral over a variable $x$.

A Riemann sum takes the form

$$\int_a^b f(x) \, dx = \lim_{n \to \infty} \frac{b-a}{n} \sum_{k=1}^n f\left(a + (b-a)\frac{k}{n}\right)$$

What is $a$? $b$?


问:
How to understand $\Phi: \operatorname{Hom}(V, W)\rightarrow W$?
I am new to linear algebra and I have the difficulty to understand how a linear transformation can be viewed as an element from a space that maps to another element in another space. For example, I encountered a question that requires me to show $\Phi: \operatorname{Hom}(V, W)\rightarrow W$ is a linear transformation, where $V$ and $W$ are finite dimensional vector spaces. However, I don't know how this can be true and specifically how to define such a function so that it could be linear. Can anyone help me with this?

Thanks in advance.

答:
You can essentially think of Linear maps as matrices. Say we have two finite dimensional vector spaces $\mathbb{R}^n\text{ and }\mathbb{R}^m$ and a matrix $A$ of dimension $m\times n$. Elements in $\mathbb{R}^n$ are vectors of dimension $n\times 1$, let such an element be $u$. So now if we multiply the matrix $A$ with $u$ we get an element $v$. Which is:
$$A.u=v$$

Now, the dimension of $v$ is $m\times 1$, which is an element of $\mathbb{R}^m$. So our matrix A acted on an element $u\in \mathbb{R}^n$ to produce an element $v$ in $\mathbb{R}^m$. Check that if a matrix satisfies all the properties of linearity.

Then consider all such matrices between $\mathbb{R}^n \text{ and } \mathbb{R}^m$ as elements of a vector space. Check if they satisfy all the properties of a vector space.

This is not an answer, but I think this example will help you in getting started.


问:
Binominal problem (I think) to calculate strategy for a mobile game - cost of money in a game of ensuring a certain number of events
There is a mobile game where I am trying to calculate the correct strategy that depends on a series of randomized events.

The game is a battle game where you place troops on a field.  When you want to place a troop, you hit a button and one of five random troops will be placed.  For the sake of this question, the troops are a knight, archer, goblin, wizard, and gnome.  Thus, you have a 20% chance of getting one of those particular cards when you push the button.

To push the button and put a troop on the field costs 100 dollars for the first troop, but increases by 10 dollars for each subsequent troop.  Thus the first troop the player puts on the field costs 100 dollars, the second costs 110 dollars, etc.

The question is what is the probability I get x knights on the field if I push the button y times (and how much will it cost).  Put another way, if I want to be, say, 75% sure that I get 3 knights, how many times should I push the button and pay for a troop and how much will it cost?  This is my basic question (and I'd like to know how to calculate it if I want 5 knights, or 7 knights).

If you want to go down the rabbit hole further with me (and I'd be very happy just for an answer for the above) - another rule of the game is if you have the same troop on the field, you can combine it with another troop of the same kind, in which case those two troops will disappear and it will produce one other random card.  Thus, for example, if your first five troops on the field are knight, archer, archer, goblin, goblin; you can combine the two archers and you will get one new random card (say a knight), and you can combine the two goblins and get one new random card (say a goblin).  So you would then have knight, knight, goblin.  It does not cost money to combine cards, but combining them can increase the chance you get a knight because it gives you another free card.

With this complication, what is the probability that I will get 3 knights on the field after pushing the button 15 times (creating 15 troops), and then combining any non-knight troops that match?  Or put another way, if I want to be 75% confident that I will get 3 knights, how many times should I push the button (taking into account I can combine non-knight troops to get new cards)?  Again, is there a formula that could calculate this so I can see what my probability is if I push the button 20 times or another amount?

Hopefully that's clear.  I appreciate any help!

答:
I will only focus on the first question.

As you only care about knights, the question is simply:

each time you push a button, you get a knight with probability $p = 20\%$; what is the probability of getting at least $k$ knights after pressing $n$ times the button?

This is then a very standard exercise in probability. The answer is $P(n, k) = \sum_{i = k}^n \binom n i p^i (1 - p)^{n - i}$.

This formula answers your questions. For example, if you want to get $3$ knights with probability at least $75\%$, then you need to choose $n$ such that $P(n, 3) \geq 75\%$. We make the following table:

```<code>n  P(n, k)
3  0.008000000000000
4  0.027200000000000
5  0.057920000000000
6  0.098880000000000
7  0.148032000000000
8  0.203082240000000
9  0.261802496000000
10 0.322200473600000
11 0.382598451200000
12 0.441654251520000
13 0.498347819827200
14 0.551949011681281
15 0.601976790745089
16 0.648156279111681
17 0.690377525618279
18 0.728658122450928
19 0.763110659600313
20 0.793915281051527
</code>```

which shows that you should choose $n = 19$.

As to the total cost, it is simply a function of $n$, namely $100 \times n + 10 \times n(n - 1)/2 = 5n^2 + 95n$. Thus for $n = 19$ the cost is $3610$.

Note that another more useful information is the expected cost until $3$ knights occur. This is another topic, though.

Body:
I will only focus on the first question.

As you only care about knights, the question is simply:

each time you push a button, you get a knight with probability $p = 20\%$; what is the probability of getting at least $k$ knights after pressing $n$ times the button?

This is then a very standard exercise in probability. The answer is $P(n, k) = \sum_{i = k}^n \binom n i p^i (1 - p)^{n - i}$.

This formula answers your questions. For example, if you want to get $3$ knights with probability at least $75\%$, then you need to choose $n$ such that $P(n, 3) \geq 75\%$. We make the following table:

n  P(n, k)
3  0.008000000000000
4  0.027200000000000
5  0.057920000000000
6  0.098880000000000
7  0.148032000000000
8  0.203082240000000
9  0.261802496000000
10 0.322200473600000
11 0.382598451200000
12 0.441654251520000
13 0.498347819827200
14 0.551949011681281
15 0.601976790745089
16 0.648156279111681
17 0.690377525618279
18 0.728658122450928
19 0.763110659600313
20 0.793915281051527

which shows that you should choose $n = 19$.

As to the total cost, it is simply a function of $n$, namely $100 \times n + 10 \times n(n - 1)/2 = 5n^2 + 95n$. Thus for $n = 19$ the cost is $3610$.

Note that another more useful information is the expected cost until $3$ knights occur. This is another topic, though.


问:
Ehrenfeucht–Fraïssé game with winning strategy restriction
I'm recently working on a proof about EF Game with some restrictions.

Given a structure C. A C-restricted winning strategy for the spoiler is a winning strategy, in which the spoiler only pick the elements in C.

Now I have two structures, A and B, which must be graphs. I need to show that for every m > 2 there exist the structures A and B, so that neither A- nor B-resricted winning strategy for the spoiler in a m round EF game. But there is a winning strategy for the spoiler when m = 2.

Here's my approach:
First: I show that the spoiler always win in 2 moves(with non restricted winning strategy).
Then: I show that when m > 2, the spoiler can only win non-restrictedly. That means if the spoiler play either A- or B-restricted game, he will lose and the duplicator will win.

Although I already have an approach, I can't find the suitable structures for A and B, I have tried almost every graph and none of them works. Though I have found the suitable graphs for m=2

However I have no idea how to continue and have been struggling for days. So any help and tips would be very appreciated! :D

答:
Here's an outline of how to get a pair of graphs which work for every $m$ at once! Of course these graphs have to be infinite, but I think ultimately this will clarify things substantially.

The key point is to note that restricted EF-games can be thought of as generalized embeddings: if $\mathfrak{A}$ embeds into $\mathfrak{B}$, then for every $n$ Duplicator wins the restricted-to-$\mathfrak{A}$ EF-game.

(Here I use "embed" in the strong sense: an embedding is an injection which preserves and reflects edge relationships, so $aEb\color{red}{\iff} f(a)Ef(b)$.)

Now bi-embeddability is actually not as strong a condition as one might think! There are very different structures which are nonetheless bi-embeddable. For example - and this actually takes us very close to the answer to your question - we can show:

There are graphs $\mathfrak{A},\mathfrak{B}$ which are bi-embeddable but such that $\mathfrak{A}$ is connected and $\mathfrak{B}$ is not connected.

(Of course they have to be infinite - for finite graphs, bi-embeddability is the same as isomorphism.)

HINT: think about taking an existing disconnected graph, and adding a new point $*$ and an edge connecting $*$ to each existing point. This shows how you can embed a disconnected graph in a connected graph, and playing around with this you should be able to come up with an example of the type above.

Now as to your actual problem, you want to combine this idea with preventing $2$-equivalence. Think about distance between nodes in a graph - when we adjoined $*$ in the paragraph above, we didn't just get a connected graph, we got a connected graph of diameter $2$ ...


问:
Kraft Inequality Proof error
In every proof of this theorem I see, I get stuck by the same thing.

Once we choose the final node $v_n$ at depth $l_n$ to be the code word of length $l_n$, the subtree rooted at that $v_n$ has no children!

Therefore the number of leaves it has is not equal to the formula $$r^{l_n - l_i}$$ when $i=n$ but it is meant to hold for $1\leq i \leq n$.

Am I missing something or are the arguments in the proofs flawed?

See proof: https://en.wikipedia.org/wiki/Kraft%E2%80%93McMillan_inequality#Proof

答: In this case, "leaves" refers to nodes at depth $\ell_n$: leaves of the original tree, not leaves of the subtree. The subtree rooted at $v_n$, a node at depth $\ell_n$, contains one node at depth $\ell_n$: the node $v_n$ itself. This agrees with the formula $|A_i| = r^{\ell_n -\ell_i}$, since $r^{\ell_n - \ell_n} = r^0 = 1$.

Title: Kraft Inequality Proof error


问:
Find all $x,y \in \mathbb N^*$ such that $\frac{1}{x} + \frac{1}{y}=5$
My proof:

By the way of contradiction, suppose that we found $(x,y) \in \mathbb N^*$ satisfies:$$\frac{x+y}{xy}=5$$
So : $$x+y=5xy \\ \iff x=5xy-y \iff x=y(5x-1)$$
Since $x\in \mathbb N^* $ so the minimum value of $5x$ is $5>1 \implies y\mid x$

By the same method we have:
$$y=x(5y-1) \implies x\mid y$$
$x\mid y, y\mid x \implies x=y$, Now we can write the original equation as :
$$\frac{2x}{x^2}=5 \iff 2=5x \implies x\notin \mathbb N^* $$
Therefore there is no solution to this equation: $$\frac{x+y}{xy} =5$$

Does this considered a proof?

答:
Though here there are obvious direct ad-hoc methods, it's worth emphasis that they are essentially special cases of one case of Lagrange's solution of the general quadratic binary Diophantine equation, namely that completing a square $ $ generalizes to $ $ completing a product as follows

$$\bbox[1px,border:3px solid #c60]{\bbox[8px,border:1px solid #c00]{\begin{align} axy + bx + cy  &\,=\, d\\[.2em]
\!\!\!\iff (ax+c)(ay+b) &\,=\, ad+bc\end{align}}}\qquad$$

Applied here we get $\, 5xy-x-y = 0\iff (5x-1)(5y-1) = 1\,$ and the rest is easy.

Your proof comes very close to rediscovering this general method.

The same method works in general, where there usually is no such easy direct ad-hoc method.


问:
Let $\mathcal{f}: \mathbb{R} \to \mathbb{R}$ be a continuous function and consider $\mathcal{F}=[(x,y)\in \mathbb{R}^2 \mid y>f(x)] $.

Let $\mathcal{f}: \mathbb{R} \to \mathbb{R}$ be a continuous function and consider $\mathcal{F}=\lbrace (x,y)\in \mathbb{R}^2 \mid y>f(x) \rbrace $.Prove that $\mathcal{F}$ is path-connected.

Proof

Let $P=(p_1,q_1),Q=(p_2,q_2)$ be points of $\mathcal{F}$ we should be find a path between $P,$ and $Q$.For $\varepsilon > 0$ Consider $K=(x,f(x)+\varepsilon)\in \mathcal{F}$.

Let $\alpha_{1}:[0,1] \longrightarrow \mathcal{F}$, with
$\alpha_{1}(t)=(p_1+t(x-p_1),q_1+t(f(x)+\varepsilon -q_1))$ which connect $P$ with $(x,f(x)+\varepsilon)$, notice that The path is always well defined since $\mathcal{f}$ is continuous function and $d\left((x,f(x))(x,f(x)+\varepsilon))\right)<2 \varepsilon $ and in fact $\alpha_1$ is continuous since is a linear function of $t$.

Of the same way define $\alpha_{2}:[0,1] \longrightarrow \mathcal{F}$, given by $\alpha_{2}(t)=(x+t(p_2-x),f(x)+\varepsilon+t(q_2-f(x)+\varepsilon))$ wich connect $(x,f(x)+\varepsilon)$ and $Q$.

Now consider $\alpha[0,1]\longrightarrow \mathcal{F}$ given by
$\alpha(t)=\begin{cases}
\alpha_{1}(2t) \, \, \text{ If $t\in[0,\frac{1}{2}$]}&
\\
\alpha_2(2t-1) \, \, \text{If $t\in [\frac{1}{2},1]$}
 \end{cases}$
Which is continuous by pasting lemma and connect $P$ to $Q$

Hence $\mathcal{F}$ is Path-connected

答:
Let $M$ be the maximum value of $f$ on $[p_1,p_2]$. Then the combination of the vertical line segments from $P$ to $(p_1, M+1)$, from $(p_1, M+1)$ to $(p_2,M+1)$ and then down from $(p_2, M+1)$ to $Q$ is the required path.

No weird $\varepsilon$'s or anything. Just go up, it's a safe way to stay inside $\mathcal{F}$.


问:
Finding the Jacobian, change of variables

Let $D = \left \{ (x,y) : x^2 \leq y^2 \leq 2x^2, 1 \leq xy \leq 2\right \}$

It is given that for any $f \in C^0(\mathbb{R}^2, \mathbb{R}):$
$$ \int_Df(\frac{y}{x^2}, xy)|dxdy| = \int_{\Omega} f(u,v) J(u,v)|dudv|$$

For some $\Omega \subset \mathbb{R}^2$. Find $J(u,v)$.

So obviously $J$ is the Jacobian determinant from the change of variables theorem.

So what I tried is first expressing $x,y$ as functions of $u,v$ which gives us:
$$\left\{\begin{matrix}
u= \frac{y}{x^2}\\
v=xy
\end{matrix}\right.
\implies \left\{\begin{matrix}
x=u^{-\frac{1}{3}}v^{\frac{1}{3}}
\\
y=u^{\frac{1}{3}}v^{\frac{2}{3}}
\end{matrix}\right.$$

And then by the version I read of the variable change theorem we should have:
$$ J(u,v) = \begin{vmatrix}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}
\end{vmatrix} =
\begin{vmatrix}
-\frac{1}{3}u^{-\frac{4}{3}}v^{\frac{1}{3}}
 & \frac{1}{3}u^{-\frac{1}{3}}v^{-\frac{2}{3}}\\
\frac{1}{3}u^{-\frac{2}{3}}v^{\frac{2}{3}} & \frac{2}{3}u^{\frac{1}{3}}v^{-\frac{1}{3}}
\end{vmatrix} = -\frac{1}{3u} $$

But apparently this answer is wrong, and the correct answer is $\frac{1}{3u}$. I noticed that if I'd change the order of the columns, i.e thinking of $v$ as the first variable, the answer would be correct. But I don't see why should I do that?

Can anyone point my mistake?

答:
The change-of-variables formula has the absolute value $|J(u,v)|$ (or perhaps your text uses $J$ to denote the absolute value — I can't guess).

Here's a shortcut to compute the Jacobian. Do the Jacobian $\mathscr J$ of $u,v$ in terms of $x,y$:
$$\mathscr J = \left|\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}\end{matrix}\right| = \left|\begin{matrix} -\frac{2y}{x^3} & \frac1{x^2} \\ y & x \end{matrix}\right| = -3\frac y{x^2} = -3u.$$
Then $J = \dfrac1{\mathscr J} = -\dfrac1{3u}$. Regardless, you need the absolute value to proceed to the double integration.


问:
Are there any integers a and b such that $a^2+b^2 = 10^{100}+3$
I try to find  integers $a$ and $b$ such that $a^2+b^2 = 10^{100}+3$, I try some number without any result

答:
hint

$a^2+b^2 $ is odd, so $ a$ and $ b $ have not the same parity.

Assume
$$a=2A \; \text{ and } \; b=2B+1$$

thus

$$4A^2+4B^2+4B=10^{100}+2$$
or

$$2A^2+2B^2+2B=5.10^{99}+1$$
the LHS is even, while the RHS is not.


问:
What is $\;\int xe^{-x^2} \,dx\;?$
What is $$\int xe^{-x^2} dx\quad?$$

I used substitution to rewrite it as $$\int -\dfrac{1}{2}e^u\, du$$ but this is too hard for me to evaluate. When I used wolfram alpha for $\int e^{-x^2} dx$ I got a weird answer involving a so called error function and pi and such (I'm guessing it has something to do with Euler's identity, but I'm fairly certain this is above my textbook's level).

答:
The problem is to find
$$\int xe^{-x^2}\,dx.$$
So we are trying to find the functions $F(x)$ such that
$$F'(x)=xe^{-x^2}.$$
We can easily verify, by differentiating, that $F(x)=-\frac{1}{2}{e^{-x^2}}$ "works," and therefore the general $F(x)$ such that $F'(x)=xe^{-x^2}$ is given by $F(x)=-\frac{1}{2}e^{-x^2}+C$.

The problem of finding a function $G(x)$ such that $G'(x)=e^{-x^2}$ is entirely different, even though the two functions $xe^{-x^2}$ and $e^{-x^2}$ are closely related. It turns out that there is no elementary function whose derivative is $e^{-x^2}$. Roughly speaking, an elementary function is a function built up from the familiar functions by using addition, subtraction, multiplication, division, and composition (substitution).

So $xe^{-x^2}$ and $e^{-x^2}$ are elementary functions. The first has an elementary indefinite integral. The second doesn't. The second function is very important for many applications. There is a function whose derivative is $e^{-x^2}$, and that function is useful. It just happens not to be an elementary function.

Because the function $e^{-x^2}$ is so important, an antiderivative of a closely related function has been given a name, the error function, often written as $\operatorname{erf}(x)$. That is what Alpha was talking about. If you ever study probability or statistics, you will become deeply familiar with the error function.

Back to our problem! We are trying to integrate $xe^{-x^2}$. We already did that earlier, by a "guess and check" method. But that is not entirely satisfactory, so we now use a standard method, substitution.

As in your post, let $u=-x^2$. Then $du=-2x\,dx$, so $x\,dx=-\frac{1}{2}\,du$. Substitute. We get
$$\int xe^{-x^2}\,dx=\int -\frac{1}{2} e^u\,du=-\frac{1}{2}e^u+C=-\frac{1}{2}e^{-x^2}+C.$$
Finding $\int e^u\,du$ was easy. We know that the derivative of $e^t$ with respect to $t$ is $e^t$, so $\int e^t\,dt=e^t+C$.

Try the following closely related problems.

$1.$ Find $\int  e^{5x+17}\,dx$.

$2.$ Find $\int (\cos x)e^{\sin x}\,dx$.

Title: What is $\;\int xe^{-x^2} \,dx\;?$


问:
checking assumptions with a residual plot
As a part of a task I took a regression equation $y(x)$ for a certain sample (1) and used it with the $x$ values of a different sample (2) to produce a residual plot as shown, for residual-regression value

Next I need to determine whether the regression model is still applicable for sample (2).

I use 2 different approaches to determine it,

I calculate the regression model for sample 2 and compare it to the regression model of sample 1, if the coefficients are different for the two, I determine it to be non-applicable.

2.The residuals need to be distributed with a constant variance, so if I get a slope in my trendline, as shown here, I determine it to be non-applicable as a result of a non-constant variance.

Are these two approaches in fact correct?

答:
The model has several parts.

First, what you are saying about the residuals having constant variance is part of the model, but the argument you are making is that the residuals do not have mean 0. That is a different assumption in the model. Another assumption is that the residuals are normally distributed. These are 3 different assumptions built into the statement that the residuals are iid normal with mean 0 and constant variance.
The test for whether the estimated slope in sample 1 is the same as that in sample 2 is fine by finding the difference in the two estimates and dividing by the square root of the sum of the two standard errors squared. Since the two estimates are independent and approximately normal, this is standard normal if the two have the same slope.
If you want to test whether the slope in sample 2 is equal to some fixed value, that is something else. Subtract the fixed value from the estimate and divide by the s.e.


问:
Question relating topology to measure theory
I have a general question.

In point set topology we say given a set $X$, we take $\tau$ to be the set of its subsets that we will call $\textit{open}$. So Then $(X,\tau)$ becomes a topological space where we say $U \in \tau$ if and only if $$(U \in \mathcal{P}(X)) \wedge (U \text{is open in} X)$$

In which case we can define $\tau$ to be standard on $\mathbb{R}$, cofinite, discreet, indiscreet, $p$-adic,ect.

In Real Analysis or more specifically in measure theory, I noticed that we are given a set $X$, a measure $\mu$ and then a sigma algebra $\sigma_X$ whose elements are $\mu-\textit{measurable}$. Is this correct? Then for any $E \in \mathcal{P}(X)$ can we say $E$ is $\mu - \textit{measurable}$ if and only if

$$E \in \sigma_X$$

Where $\sigma_X$ is closed under complements, countable union and intersection whereas $\tau$ is closed under arb union of open arb intersection of closed and finite union of closed and finite intersection of open? Then just as $\tau$ can be defined how one likes, can one define $\mu$ to be different than standard measure and thus elements of $\sigma_X$ would be different based on the measure introduced?

On this same note, do results such as MCT, DCT, BCT, Egorov all hold if we take different topologies on $\mathbb{R}$? On the same note do point set topology results such as compact subset of closed is closed hold if we swap the measure from standard measure? I know for example Heine Borel holds for standard topology only but that's from Topology to Topology I wanna see if changing assumption of one field rules out results in another.

Sorry if this is a silly question. Thanks for any input! all (knowledgable) input welcome!

答:
Often we start even with just a set $X$ and a $\sigma$-algebra $\mathcal{S}$ on $X$, and the combo is called a measurable space. The members of $\mathcal{S}$ are called (plain) measurable. If we actually then define a function $\mu$ on this $\mathcal{S}$ (with codomain $\Bbb R$ or the extended reals or complex numbers or vectors, depending on what type of measure theory you want, obeying additivity etc.), the combination $(X,\mu, \mathcal{S})$ is called a measure space. (One could consider different measures on the same measurable space, or even the space of all measures on the measurable space, and study its structure etc.)

If we have a topology given, like on $\Bbb R$, the default measurable space will often be the Borel $\sigma$-algebra generated by the topology (or open sets, or the intervals etc.) and people are mostly interested in Borel measures defined on it. For the Reals the notion of a Lebesgue-measurable subset is much more general than just being a Borel set: it's initially only defined for Borel sets but the Carathéodory construction allows us to extend it to a larger $\sigma$-algebra: we can add null sets too and mix those in. So $\mu$-measurable set sometimes means more than just a member of the original measurable space. Many, but not all, classic theorems are indeed typical for Euclidean spaces with the standard topology, Borel sets and Lebesgue-measurable sets, or at least most often used there.


问:
sequence limits - proof by definition
I would like to ask how should I show from the definition of sequence limits that if $\lim a_n = 0$ then $\lim \sqrt{a_n} = 0$. I know how to show it for a specific example, but I don't know how to show it "resulting".

答:
The A by @VIVID is fine. I want to mention that it is useful to know more than one equivalent definition of convergence and of non-convergence, e.g.

(i). $\lim_{n\to\infty} a_n=A$ iff for every $r>0$ the set $\{n\in \Bbb N:|a_n-A|\ge r\}$ is finite.

(ii). $[\neg (\lim_{n\to\infty} b_n=B)\,]$ iff there exists $s>0$ such that the set $\{n\in \Bbb N:|b_n-B|\ge s\}$ is infinite.

Suppose $a_n=(b_n)^2$ for all $n\in \Bbb N$ . We can use (ii) with $B=0$ to prove that $$[\,\neg ( \lim_{n\to\infty} b_n=0)\,]\implies \neg (\lim_{n\to\infty} a_n=0).$$

If $s>0$ and  the set $\{n\in \Bbb N:|b_n|\ge s\}$ is infinite then, with $r=s^2>0,$ the set $\{n\in \Bbb N:|a_n|\ge r\}$ is infinite, which by another instance of (ii) implies  $\neg (\lim_{n\to\infty} a_n=0).$

Title: sequence limits - proof by definition


问:
How do I use chain rule for higher derivatives in several variables?
Let $f$ be a function from unit ball $B\subset \mathbb{R}^3$ to $\mathbb{R}$ which is differentiable $n$ times for large $n$.

Let $\mathbf{a}$ be a point(vector) in the unit ball, fix a non-zero vector $\mathbf{u}$ in $\mathbb{R}^3$.

Let $\gamma:(a,b)\rightarrow B$  be the line $\gamma(t)=\mathbf{a}+t\mathbf{u}$.

Then consider $g(t)=f\circ \gamma(t)$.

I want to know expression for $g^{(k)}(t)$. For $k=1$, we can use Chain rule and say
$$
g'(t)=Df_{\gamma(t))}D\gamma_{t}=\nabla f(\mathbf{a}+t\mathbf{u})\cdot \mathbf{u}.
$$
But, I was unable to see the application of Chain rule or anything else to evaluate $g''(t)$ from this one.

How can I proceed for $g''(t)$?

Note: I am using only the definition of $\nabla$ with no properties; one may state it if needed in the evaluation of higher order derivatives.

答: $$g''(t) = D_t (g'(t)) = D_t (\nabla f (\mathbf a + t \mathbf u) \cdot \mathbf u) = \nabla^2 f (\mathbf a + t \mathbf u) \cdot \mathbf u$$


问:
Proof that $\hat{\beta^{t}}X^{t}y = \hat{y}^{t}\hat{y}$
I need to prove that:

$\hat{\beta^{t}}X^{t}y = \hat{y}^{t}\hat{y}$

I know that:

$\hat{\beta}= \left ( X^{t}X \right )^{-1}X^{t}y$

$\hat{y}= X\left ( X^{t}X \right )^{-1}X^{t}y= X\hat{\beta}=Hy$

Here's what I have so far:

$\hat{y}^{t}\hat{y} = \left ( X\hat{\beta} \right )^{t}X\left ( X^{t}X \right )^{-1}X^{t}y= \hat{\beta}^{t}X^{t}Hy$

I'm clearly doing something wrong, any idea what I can change to prove it?

答:
You almost have it. From your last step:

$$ \hat{y}^{t}\hat{y} = \left ( X\hat{\beta} \right )^{t}X\left ( X^{t}X \right )^{-1}X^{t}y $$

$$ = \hat{\beta}^tX^tX\left ( X^{t}X \right )^{-1}X^{t}y $$

$$ = \hat{\beta}^tX^{t}y . $$


问:
Why is this distribution function neither discrete nor continuous?
I'm refreshing some Probability Theory (it's been a while) and came accross the following example:

Let $X$ be a random variable with distributuion function $$F_X(x) = \left\{\begin{array}{l r}0 & \text{if }x<0\\\frac{1}{2}+\frac{x}{2}&\text{if }0\le x<1\\1&\text{if }x\ge 1\end{array}\right.$$
Then this random variable is neither discrete nor continuous, since $$\mathbb{P}(X=0)=\frac{1}{2},\quad\text{and}\quad\mathbb{P}(X=\frac{1}{2})=0.$$

Is this first statement true since $X$ takes these values on all $x<0$? And the second statement true since $X$ takes that value for only 1 value of $x$, out of infinitely many?

It's probably easy, but I'm not sure if I understand it...

答:
A random variable is continuous if the cdf is a continuous function. This is not continuous at x=0.
A random variable is discrete if it can take only a countable number of values. This can take any value in (0.5,1) which is uncountably infinite.

So it is neither discrete or continuous.

To find $P[X=x]$, find $P[X \le x]=F(x)$ and subtract $P[X<x]$, which is the supremum of $F(y)$ for all $y<x$.  Since $F$ is a non-decreasing function, the supremum is the limit as $y$ approaches $x$ from the left.

Since $P[X\le x]=0$ for all $x<0$,
$$P[X=0]=P[X\le 0]-\lim_{x\rightarrow 0^-}P[X \le x]=\frac{1}2$$

On the other hand,

$$P\left[X=\frac{1}2 \right]=
P\left[X \le \frac{1}2 \right]-\lim_{x\rightarrow {\frac{1}2}^-}P\left[X \le x \right]=
 \frac{1}2 +\frac{\frac{1}2}2-\lim_{x\rightarrow {\frac{1}2}^-}\left(\frac{1}2 +\frac{x}2\right)=0$$


问:
Help me with this this system of equations
Help me with this  system of equations
$$a+b = 3 -c$$
$$\frac{1}{a}+\frac{1}{b}= \frac{5}{12}-\frac{1}{c}$$
$$ a^3+b^3 = 45 -c^3$$</p>

答:
First of all notice that $a$ and $b$ appear "symmetrically": if you exchange $a$ with $b$, the system doesn't change. In these cases, setting

$$x=a+b,\quad y=ab$$

usually helps.

Indeed, you can note now that the left hand side of the second equation is just $x/y$, while

$$
a^3+b^3 = x^3 -3xy
$$

The first equation already tells you what $x$ is; so the third equation tells you what $y$ is. Hence the second equation will give you the value (or values) of $c$. Hence you reduce your problem to finding the numbers $a$ and $b$ knowing their sum and product.

A complete solution

We can write
$$
\begin{cases}
x=3-c\\[2ex]
\frac{x}{y}=\frac{5c-12}{12c}\\[1ex]
x^3-3xy=45-c^3
\end{cases}
$$
The third equation becomes
$$(3-c)^3-3(3-c)y=45-c^3$$
or
$$27-27c+9c^2-c^3-3(3-c)y=45-c^3$$
from which, rearranging terms and simplifying,
$$(3-c)y=3c^2-9c-6$$
or
$$y=\frac{3c^2-9c-6}{3-c}$$
Thus
$$\frac{x}{y}=\frac{(3-c)^2}{3(c^2-3c-2)}$$
and the second equation becomes
$$4c(3-c)^2=(c^2-3c-2)(5c-12).$$
It's a matter of computing, now:
$$36c-24c^2+4c^3=5c^3-15c^2-10c-12c^2+36c+24$$
$$c^3-3c^2-10c+24=0$$
It's easy to see that this has the roots $c=-3$, $c=2$ and $c=4$. Let's see what you get from $c=-3$:
$$
\begin{cases}
a+b=6\\[1ex]
\frac{a+b}{ab}=\frac{3}{4}
\end{cases}
$$
or
$$\begin{cases}
a+b=6\\[1ex]
ab=8
\end{cases}
$$
that can be solved through the auxiliary equation $t^2-6t+8=0$. This one has solutions $t=2$ and $t=4$. Thus you get the solutions
$$
\begin{cases}a=2\\ b=4\\ c=-3 \end{cases}
\qquad
\begin{cases} a=2\\ b=4\\ c=-3 \end{cases}
$$
Observing that the system is actually symmetric in $a$, $b$ and $c$, the other solutions are obtained by permuting the first one.


问:
Quantifier Elimination Pure Identity Language in Chang
In Chang & Keisler's Model Theory, quantifier elimination on the theory of the pure identity language is shown. However, I'm confused about the notion of "basic formula" for which any quantified sentence needs to equivalent to a boolean combination of. Atomic formulas are used as these basic formulas for showing QE for the theory of dense linear orders, however the authors note that doing so in the theory of the pure identity language does not permit $\forall xy(x \equiv y)$ to be expressed as an open formula so the authors add the sentences saying there are at least $n$ distinct elements in the model to their basic formulas. (I believe $n$ is the number of variables being quantified over in the original sentence). The question is, how can we use these sentences for QE when they themselves use quantifiers?

I.e., if $\sigma_3$ is the sentence saying there are $3$ distinct elements in our model, how can we perform QE elimination on this sentence?

答:
This is a fairly common phenomenon. Many interesting theories don't admit quantifier elimination, but do if you extend the language. In your example, if you add predicates $\phi_n$ to the pure theory of equality that assert that the universe has at least $n$ elements, you get quantifier elimination. To define the $\phi_n$ in the original language involves quantifiers, but for many purposes the quantifier elimination process produces a formula in the extended language that is just what you need.

E.g., the quantifier elimination procedure for Presburger arithmetic has to add predicates $\delta_n(x)$ asserting that $n \mid x$ to the language for $n = 2, 3, \ldots$. Presburger's quantifier elimination procedure still gives us a decision procedure, because given a term $t$ with no free variables, we can calculate the truth value of $\delta_n(t)$ for any $n$.


Body:
Divergence of  $\lim_{N\to +\infty}[ \int_{1}^{N+1} f(x) dx -\sum_{n=1}^{N}f(n)] $ where $f(x)= sin(log_e x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})$,$ 0<a<1/2$

My try -
$\lim_{N\to +\infty}[ \int_{1}^{N+1} sin(log_e x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})dx - \sum_{n=1}^{N} sin ( log _en) (\frac{1}{n^{a}}-\frac{1}{n^{1-a}})]  $ where $0<a<1/2$

Define S=$\lim_{N\to +\infty}[ \int_{1}^{N+1} sin(log_e x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})dx - \sum_{n=1}^{N} sin ( log _en) (\frac{1}{n^{a}}-\frac{1}{n^{1-a}})]  $

If $a=1/2$ then S is convergent to 0.

If $a<1/2$ then, $(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})>0$ $ \forall x\geq 1$
and $(\frac{1}{n^{a}}-\frac{1}{n^{1-a}})>0$ $ \forall n\geq 1$.

Please prove that S is divergent when $0<a<1/2$

Body:
Integral $I(a)=\int_{1}^{N+1} \sin(\ln x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})dx$  can be evaluated - just make a change $x=e^t$:
$I(a)=\int_{0}^{\ln(N+1)} \sin(t)(e^{-ta}-e^{(a-1)t})e^tdt=\int_{0}^{\ln(N+1)} \sin(t)(e^{t(1-a)}-e^{at})dt$  and $\sin(t)=\frac{e^{it}-e^{-it}}{2i}= \Im({e}^{it})$

Integral with exponents can be easily taken.

Consider the integral as a sum:  $\sum_{k=1}^{N}\int_{\ln(k)}^{\ln(k+1)} \sin(t)(e^{t(1-a)}-e^{at})dt$

So, you get a series with N term as $a_N=\left(\int_{\ln(N)}^{\ln(N+1)} \sin(t)(e^{t(1-a)}-e^{at})dt-\sin(\ln{N})(\frac{1}{N^{a}}-\frac{1}{N^{1-a}})\right)$ and you can investigate its behaviour at $N\to\infty$.

Let's consider $a_N$:

After evaluating the integral we get:

$a_N=\frac{1-a}{1+(1-a)^2}[(N+1)^{1-a}\sin(\ln(N+1))-N^{1-a}\sin(\ln(N))]-\frac{1}{1+(1-a)^2}[(N+1)^{1-a}\cos(\ln(N+1))-N^{1-a}\cos(\ln(N))]-\frac{a}{1+a^2}[(N+1)^{a}\sin(\ln(N+1))-N^{a}\sin(\ln(N))]+\frac{1}{1+a^2}[(N+1)^{a}\cos(\ln(N+1))-N^{a}\cos(\ln(N))]-[\sin(\ln{N})\frac{1}{N^{a}}]+[\sin(\ln{N})\frac{1}{N^{1-a}}]$

Now, let's consider, for example

$\frac{a}{1+a^2}[(N+1)^{a}\sin(\ln(N+1))-N^{a}\sin(\ln(N))]=\frac{a}{1+a^2}[N^a(1+\frac{1}{N})^{a}\sin(\ln{N}+\ln(1+\frac{1}{N}))-N^{a}\sin(\ln(N))]$

Expanding the sine of the sum, as well as using Taylor series expansion for large N we get

$\frac{aN^a}{1+a^2}[(1+\frac{a}{N}+O(\frac{1}{N^2}))\left(\sin(\ln{N})\cos((\frac{1}{N})+O(\frac{1}{N^2}))+\cos(\ln{N})\sin((\frac{1}{N})+O(\frac{1}{N^2}))\right)-\sin(\ln(N))]=\frac{aN^a}{1+a^2}[(1+\frac{a}{N}+O(\frac{1}{N^2}))\left(\sin(\ln{N})\cos(\frac{1}{N})+O(\frac{1}{N^2})+\cos(\ln{N})\sin(\frac{1}{N})+O(\frac{1}{N^2})\right)-\sin(\ln(N))]=\frac{a}{1+a^2}N^a\left(\frac{a\sin(\ln{N})}{N}+\frac{\cos(\ln{N})}{N}+O(\frac{1}{N^2})\right)$

Finally, evaluating all terms we get

$a_N=N^{-a}\left(\frac{(1-a)^2\sin(\ln{N})}{1+(1-a)^2}+\frac{(1-a)\cos(\ln{N})}{1+(1-a)^2}-\frac{(1-a)\cos(\ln{N})}{1+(1-a)^2}+\frac{\sin(\ln{N})}{1+(1-a)^2}+O(\frac{1}{N^2})\right)+N^{a-1}\left(-\frac{a^2\sin(\ln{N})}{1+a^2}-\frac{a\cos(\ln{N})}{1+a^2}+\frac{a\cos(\ln{N})}{1+a^2}-\frac{\sin(\ln{N})}{1+a^2}+O(\frac{1}{N^2})\right)-$$-\sin(\ln{N})\frac{1}{N^{a}}+\sin(\ln{N})\frac{1}{N^{1-a}}$

After grouping all terms we see that they cancel each other, so we get

$a_N=N^{a}O(\frac{1}{N^2})+N^{1-a}O(\frac{1}{N^2})=O(\frac{1}{N^{2-a}})+O(\frac{1}{N^{1+a}})$

But these series converges absolutely $\Rightarrow$ $a_N$ converges.

Please check my calculations.


问:
find $f \in L^2(0,1)$ such that $f = g + K(f)$
I'm preparing for my qualifying exam and the question bank had the following question under functional analysis section and I'm stuck with this problem.

Let $L^2(0,1)$ be the Hilbert space of square integrable functions on the interval $[0,1]$ with Lebesgue measure. Define $K : L^ 2 (0,1) \rightarrow L^ 2 (0,1)$ by $K(f)(t) =\int_0^t (t − s)f(s)\,ds$. Show that
$||K|| < 1$. Given $g \in L^ 2 (0,1)$ find $f \in L^ 2 (0,1)$ such that $f = g + K(f)$.

My idea for the second part of the problem: I feel like it is the kernel of some map and that the question is asking to find an element in the coset of $g$ (the coset that you get after you quotient out by the kernel).

I really have no idea how to proceed. Any help would be appreciated.

答:
Answer for the second part: $\|K\|<1$ implies that $I-K$ is invertible and its inverse is $\sum\limits_{n=0}^{\infty} K^{n}$. [See my answer here: https://math.stackexchange.com/questions/4021302/prove-existence-of-inverse-of-a-bounded-linear-operator/4021335#4021335 ]

Hence, the unique solution of $f=g+Kf$is $f=(I-K)^{-1} g$.


问:
Convergence of a series and its derivative
I have to study punctual, uniform and total  convergence of the following sums
$$u(x)=\sum_{n=1}^{+\infty}u_n(x)=\sum_{n=1}^{+\infty}\dfrac{1}{n^2}\int_0^1\exp(-nxt^4)dt$$
$$v(x)=\sum_{n=1}^{+\infty}u_n'(x)$$
I'm looking for exercises of sequences of functions which involve the study of sequences of functions in integral form but I haven't study Lebesgue integrability yet. I'm trying to solve the exercises with a quite standard theory about convergence of function sequences and function series.

I initially found a little bit confusing the fact that the integrand has both the variables $x$ and $t$.

What I've done is considering $t\in[0,1]$ and $x\in \mathbb R$. For the punctual convergence I discarded the case $x<0$ because in this case the integrand would be, for $n\to \infty$, $\dfrac{\exp(-nxt^4)}{n^2}=\dfrac{1}{e^{nxt^4}n^2}\sim\dfrac{1}{e^{nxt^4}}\to\infty,\forall x\in\mathbb R^-$. At this point I think I can say that since the integral from zero to one can be divergent or can assume a real value, the sum would have a non zero general term so it can't be punctual convergent in $\mathbb R^-$.

For $x\ge 0$ I have $u_n(0)=\dfrac{1}{n^2}\forall n\in\mathbb N$ and I can apply
$$\dfrac{1}{e^{nxt^4}}\le \dfrac{1}{n}\implies u_n\le\dfrac{1}{n^3}\implies \sum u_n\le\sum\dfrac{1}{n^3}$$
so we have punctual, uniform and total convergence in $[0,+\infty)$.

I really have problems with $u'_n(x)$... Shouldn't it be zero?

答:
Your conclusions and general approach for $\sum u_n(x)$ are correct but the arguments have a few errors.

For $x < 0$:

You claim that

$$\displaystyle \frac{\exp(-nxt^4)}{n^2}=\frac{1}{e^{nxt^4}n^2}\sim_{n \to \infty}\frac{1}{e^{nxt^4}},$$

but the symbolism $f(n)\sim g(n)$ means $\lim_{n \to \infty}f(n)/g(n) = 1$, which is not true in this case.

The correct argument for $x < 0$ is

$$\frac{\exp(-nxt^4)}{n^2}=\frac{\exp(n|x|t^4)}{n^2} > \frac{1}{n^2}\frac{(n|x|t^4)^3}{3!} = \frac{n|x|^3t^{12}}{6},$$

and the series diverges, since,

$$u_n(x) = \frac{1}{n^2} \int_0^1\exp(-nxt^4)\, dt > \frac{n |x|^3}{6}\int_0^1 t^{12} \, dt \underset{n \to \infty} \longrightarrow +\infty $$

For $x \geqslant 0$:

The inequality $\frac{1}{e^{nxt^4}}\le \frac{1}{n}$ holds eventually for fixed $x$ and $t$, but this can't be used for uniform convergence.

Uniform convergence holds by the Weierstrass M-test simply because $\exp(-nxt^4) \leqslant 1$ and

$$u_n(x) = \frac{1}{n^2} \int_0^1\exp(-nxt^4)\, dt \leqslant  \frac{1}{n^2}$$

Regarding $v(x)$:

We can switch the order of the derivative and integral  to obtain

$$\frac{d}{dx}\int_0^1\exp(-nxt^4) \, dt = \int_0^1(-nt^4)\exp(-nxt^4) \, dt,$$

and, thus,

$$v(x) = \sum_{n=1}^\infty u'_n(x) = -\sum_{n=1}^\infty \frac{1}{n}\int_0^1t^4\exp(-nxt^4) \, dt $$

Using a similar argument as before, the series diverges for $x < 0$. We also have divergence for $x = 0$, since

$$u'_n(0) = - \frac{1}{n} \int_0^1 t^4 \, dt = - \frac{1}{5n}$$

For $x > 0$, we have

$$u'_n(x) = \frac{1}{n}\int_0^1t^4\exp(-nxt^4) \, dt < \frac{1}{n}\int_0^1 \frac{t^4}{nxt^4} \, dt = \frac{1}{n^2x},$$

and it follows that the series converges pointwise and absolutely, as well as uniformly for $x \in [a,\infty)$ for any $a > 0$.

There is one  remaining question that I will leave to you. Does   $\sum u'_n(x)$ converge uniformly on $(0,\infty)$?


问:
$d$-dimensional set with Lebesgue measure $\frac{1}{d!}$
Let $X = \{(x_1, ..., x_d)\,|\,0 \leqslant x_1\leqslant ... \leqslant x_d\leqslant 1\}$. Show that $X$ has Lebesgue measure $\lambda_d(X)=\frac{1}{d!}$. I assumed that because of Fubini's theorem $$\lambda_d(X) = \int_X 1 dA = \int_0^1\int_{0}^{x_d}\int_{0}^{x_{d-1}}...\int_0^{x_2} 1 dx_1dx_2...dx_d$$ wich would work out to $\frac{1}{d!}$ due to the power rule. I noticed that the actual integral should be $$\int_0^1\int_{x_{d-1}}^{x_d}\int_{x_{d-2}}^{x_{d-1}}...\int_0^{x_2} 1 dx_1dx_2...dx_d$$
wich doesn't telescope or result in the actual measure. How do I show that $\lambda_d(X) =\frac{1}{d!}$?

答: All the $d!$ permutations  of the coordinates  give the same value and their sum is $\int_0^{1}\int_0^{1}...\int_0^{1} dx_1d_2...dx_d=1$.


问:
Baker–Campbell–Hausdorff formula for generators of $SO(3)$
By noting ${e_1 = (1, 0, 0), e_2 = (0, 1, 0), e_3 = (0, 0, 1)}$ and $E_i = [e_i]_\times$ the generators of $SO(3)$, then we have the following commutator properties :

$$
\begin{align}
  E_1 &= [E_2, E_3] \\
  E_2 &= [E_3, E_1] \\
  E_3 &= [E_1, E_2]
\end{align}
$$

Is it possible to find an explicit expansion of the Baker–Campbell–Hausdorff formula for these generators, such as $\log(e^{E_1} e^{E_2}) $ ?

答:
Of course it's possible. Rodrigues solved the problem in 1840.

In your somewhat idiosyncratic notation, $E_1=L_x$ and $E_2=L_y$ physicists learn in Sophomore year. I will, predictably, choose easy angles $\theta= \pi/2$, instead of your freaky $\theta=1$.

You then trivially have a roll and a pitch,
$$
\exp \left ({\pi\over 2} E_1 \right ) = \begin{bmatrix}
 1 &  0 & 0 \\
 0 &  0 & -1 \\
 0          &   1         & 0
\end{bmatrix}    , \qquad \exp  \left({\pi\over 2}  E_2  \right)
= \begin{bmatrix}
 0 &  0 & 1 \\
 0 &  1 & 0 \\
 -1          &   0          & 0
\end{bmatrix}  ~~~\leadsto \\
\exp  \left({\pi\over 2}  E_1  \right)  \exp  \left({\pi\over 2}  E_2  \right) =
\begin{bmatrix}
 0 &  0 & 1 \\
 1 &  0 & 0 \\
 0          &   1         & 0
\end{bmatrix}  \\  = \exp\left (  {2\pi\over 3}    (E_1+E_2+E_3)/\sqrt 3\right ),
$$
Where we have used this formulation of the  Rodrigues formula, for $\theta=2\pi /3$,
$$M\equiv E_1+E_2+E_3\equiv \sqrt{3} K=\begin{bmatrix}
 0 &  -1 & 1 \\
 1 &  0 & -1 \\
-1  &   1 & 0
\end{bmatrix} , \leadsto\\
\exp (\theta K)= I+\sin\theta ~K+ (1-\cos\theta) K^2 , ~~~\leadsto \\
\exp \left ( {2\pi\over 3} M/\sqrt{3}\right ) =I +{1\over 2} M +{1+1/2\over 3} M^2.
 $$

For general angle formulas, you should use the first link, the Rodrigues-Gibbs formula. (After you've appreciated its music, the Original Rodrigues paper is also instructive.)


问:
Differential form equation conversion to derivative equation
Suppose $x=x(t)$. Let $dx$ and $dt$ be 1-forms.

Does there exist a rigorous theorem for converting $dx=f(t)dt$ into the derivative ("fraction") expression $\frac{dx}{dt}=f(t)$?

Is this possible in general for other possibly multivariate functions (or does it require that $x$ is only a function of $t$). Or is this just false?

My guess is that by the linear independence of the basis differential form dt, it is possible to expand in the following manner and simply equate coefficients of $dt$.

$dx=\frac{dx}{dt}dt$

Therefore, $\frac{dx}{dt}=f(t)$.

答:
As I commented, you want to think of a mapping $x\colon\Bbb R\to\Bbb R$. Using $t$ as a coordinate on the domain and $u$ on the range, you pull back $du$ by the mapping $u=x(t)$. Then you get $dx = x^*du = x'(t)\,dt$. Setting this equal to $f(t)\,dt$, you have $x'(t)\,dt = f(t)\,dt$, whence $x'(t)=f(t)$.

You asked about the multivariate situation. Consider, for example, $$dx = P(u,v)\,du + Q(u,v)\,dv.$$ Then this reduces likewise to the equations
$$\frac{\partial x}{\partial u} = P(u,v) \quad\text{and}\quad \frac{\partial x}{\partial v} = Q(u,v).$$
Can you make the analogous argument?


问:
How to show Let $(X, d)$ be a metric space. Prove that if $X$ has a dense finite subset, then $X$ itself is finite?

Let $(X, d)$ be a metric space. Prove that if $X$ has a dense finite
subset, then $X$ itself is finite.

My attempt: if possible, $X$ is infinite and $A$ be a finite dense subset of $X$. Now for any point $x$ in $X$,any neighborhood of $x$ contains at least one point from $A$. Since $X$ contains infinite number if points and $A$ is finite there will be some contradiction which is not coming in my mind. Please I need help.

答: You aren't far off. If $A$ is a subset of the metric space $X$ that is dense and finite, then $A = X$. To see this assume (for a contradiction) that $A$ a proper finite dense subset of the metric space $X$ and let $x \in X \setminus A$.  As $x \not\in A$, $d(x, a) > 0$ for every $a \in A$. But then if we take $\varepsilon = \frac{\min_{a\in A}d(x, a)}{2}$ (which is well-defined and positive because $A$ is finite), there is no element $a \in A$ with $d(x, a) < \varepsilon$, so $A$ is not dense.


问:
How to solve the irrational system of equations?
Solve the system of equations $$\begin{cases} \sqrt{x+2y+3}+\sqrt{9 x+10y+11}=10,&\\[10pt] \sqrt{12 x+13y+14}+\sqrt{28 x+29y+30}=20. \end{cases} $$</p>

答: In this sort of question, repeated squaring is involved. So any big numbers that you start with (like $28$) are going to produce much bigger numbers, and multiple-term sums will blow up unpleasantly. On the other hand, linear substitutions (such as Maazul has suggested) are easily reversed at the end. Substituting $u=t+y+2$ as a further step in Maazul's method reduces the number of terms significantly, although leaving the $28$ diminished only to $27$. So I suggest $4u=4t+y+2$ instead. While this worsens the first equation somewhat, it keeps the lid on the numbers overall since, after dividing the second equation through by $2$, the biggest number that appears in the resulting equations in $t$ and $u$ is $10$. </p>


问:
Is it possible to split the natural numbers into a finite number of sets so that no pair of numbers within a set adds up to a square?
My attempt:

$\lbrace6,19,30\rbrace$ is sufficient to show that two sets are impossible.

Using a computer program with a brute force method I found that separating the numbers $1$ through $85$ into three sets is possible as shown below:

$\lbrace1,4,6,9,13,14,17,18,20,26,28,33,34,37,41,42,49,54,56,57,62,69,70,73,76,78,81,85\rbrace$

$\lbrace2,5,8,10,12,21,22,25,29,30,32,38,40,45,46,48,50,53,58,61,64,65,66,72,74,77,82,84\rbrace$

$\lbrace3,7,11,15,16,19,23,24,27,31,35,36,39,43,44,47,51,52,55,59,60,63,67,68,71,75,79,80,83\rbrace$

but $1$ through $86$ is impossible.

Edit:

WhatsUp in the answer below provides
the set of four numbers: $\lbrace1058, 6338, 10823, 13826\rbrace$ with an explanation of how he got them.  This is a alternative non-brute force way of showing that separating the natural numbers into three sets is impossible.
In a comment of this question a set of five numbers $\lbrace 7442, 28658,148583,177458,763442\rbrace$
is provided by the user Bob Kadylo. This shows that four sets are impossible.

Edit 2:

In a previous version of my post I made a proof showing that the number of sets needed for Natural numbers $1$ to $N$ so that no pair of numbers in the same set sums to a square is no more than $\lfloor\sqrt{2N-1}\rfloor$. I realized that I can do significantly better than this. In order to explain the method that has a smaller upper bound I have to transform the problem into graph theory. An equivalent formulation is to have $N$ vertices labeled from $1$ to $N$. A pair of points are connected iff the two points add up to a square. Then our goal is to color the vertices using the least number of colors so that no two vertices with the same color are connected. The first step in the greedy algorithm for coloring vertices is to make a list of colors with numbers. (ex. RED-1, BLUE-2, GREEN-3, YELLOW-4, etc.) If during the process more colors are required than are on the coloring list, add more colors to the list. The next step is to pick an uncolored vertex and use the lowest color number that isn't connected that the chosen vertex. Repeat the last step until all vertices are colored. The worst case scenario is to use one more color than the degree value of the vertex with the greatest degree (or tied with the greatest degree). If each vertex that is connected to the greatest degree vertex is a different color then the greatest degree vertex has to be a different color from all of those. The vertex with the greatest degree (or tied with the greatest) is $3$. It has degree $\lfloor\sqrt{N+3}\rfloor-1$. Therefore the Number of sets (or colors) required is no more than $\lfloor\sqrt{N+3}\rfloor$. We can do slightly better by using brooke's theorem which states that if a graph is simple, connected, not complete, and not an odd cycle, then the upper bound of the number of colors is equal to the degree of the greatest degree vertex. This means that the new upper bound is $\lfloor\sqrt{N+3}\rfloor-1$ sets. This is the significant improvment from $\lfloor\sqrt{2N-1}\rfloor$ I mentioned at the beginning.

End edits

For each natural number $X$ there are $\lfloor\sqrt{2X-1}\rfloor-\lfloor\sqrt{X}\rfloor$ numbers that are less than $X$ that when summed to $X$ results in a square. The expression: $\lfloor\sqrt{2X-1}\rfloor-\lfloor\sqrt{X}\rfloor$ increases as $X$ gets larger, because of this fact my guess is that separating the natural numbers into a finite number of sets so that no pair of numbers in a set doesn't sum to a square is impossible.

答:
Without bruteforcing, I find the list $\{1058, 6338, 10823, 13826\}$ which shows that three sets is not enough.

My approach:

I start from the equations
\begin{eqnarray}
a + b &=& u^2\\
c + d &=& v^2\\
a + c &=& x^2\\
b + d &=& y^2\\
a + d &=& m^2\\
b + c &=& n^2.
\end{eqnarray}
The matrix has rank $4$, which means that there are just $6 - 4 = 2$ linearly independent relations among $u^2, \dots, n^2$. They are:
$$u^2 + v^2 = x^2 + y^2 = m^2 + n^2.$$

Also, if we assume $a < b < c < d$, then we have $u^2 < x^2 < m^2, n^2 < y^2 < v^2$.

In order to get positive solutions $a, b, c, d$, it suffices to have $u^2 + x^2 > n^2$.

Now I simply take the number $N = 5 \times 13 \times 17 \times 29$, which can be written as the sum of two squares in many ways. Somewhere in the middle, I take out these three:
$$N = 86^2 + 157^2 = 109^2 + 142^2 = 122^2 + 131^2.$$ These are my candidates of $u^2, \dots, n^2$.

It only remains to solve $a, b, c, d$ back from the original equations, which gives the list in the very beginning.

Luckily, the solutions are integers. But even if we got non-integral solutions, we could always multiply all of them by some square to clear the denominators.

I also tried to extend this method to five numbers. It became a bit messy though, so I gave up midway.


问:
Law of large number integral asymptotic results
Let $(X_n)$ be a i.i.d. random variables, following a uniform distribution. Using the weak law of large numbers,

$$\lim_{n +\infty} \int_0^1 ... \int_0^1 \frac{\phi(x_1) + ... + \phi(x_n)}{\psi(x_1) + ... + \psi(x_n)} dx_1 ... dx_n = \frac{\int_0^1 \phi(x) dx}{\int_0^1 \psi(x) dx}$$

where $\phi$ and $\psi$ are continuous and strictly positive functions.

My thoughts so far:

I looked at Fubini's theorem, which has the same functional form as what I'm trying to prove. However I am not supposed to know any of measure theory, so it is not the right way to approach the problem.

The 0-1 bounds makes me think that the uniform distribution has parameter 0 and 1.

The use of the law of large numbers can be intuited since it is a convergence theorem and we want to prove an asymptotic result.

But I don't know where to start.

答:
The l.h.s. is
$$
\lim_{n +\infty} \int_0^1 ... \int_0^1 \frac{\phi(x_1) + ... + \phi(x_n)}{\psi(x_1) + ... + \psi(x_n)} dx_1 ... dx_n = \lim_{n +\infty} \mathbb E\left[ \frac{\phi(X_1) + ... + \phi(X_n)}{\psi(X_1) + ... + \psi(X_n)} \right]
$$
$$
=\lim_{n +\infty} \mathbb E\left[ \frac{\phi(X_1) + ... + \phi(X_n)}{n}\cdot\frac{n}{\psi(X_1) + ... + \psi(X_n)} \right]
$$
Random variables $\phi(X_1),\phi(X_2),\ldots$ are independent and identically distributed with finite expectation since continuous function $\phi$ on $[0, 1]$ is bounded. And the same holds for  r.v.'s $\psi(X_1),\psi(X_2),\ldots$.

By weak LLN,
$$
 \frac{\phi(X_1) + ... + \phi(X_n)}{n}\xrightarrow{p} \mathbb E[\phi(X_1)],
$$
$$
\frac{n}{\psi(X_1) + ... + \psi(X_n)} \xrightarrow{p} \frac{1}{\mathbb E[\psi(X_1)]},
$$
so
$$\frac{\phi(X_1) + ... + \phi(X_n)}{n}\cdot\frac{n}{\psi(X_1) + ... + \psi(X_n)} \xrightarrow{p} \frac{\mathbb E[\phi(X_1)]}{\mathbb E[\psi(X_1)]}=\frac{\int_0^1 \phi(x)\,dx}{\int_0^1 \psi(x)\,dx}
$$
The only question is why the expectation converges to the same limit. And the ratio
$$
 \frac{\phi(X_1) + ... + \phi(X_n)}{n}\cdot\frac{n}{\psi(X_1) + ... + \psi(X_n)}
$$
is strictly positive and bounded above. So, by dominance convergence theorem, expectations converge to the ratio of integrals at the r.h.s. If you knows some theorems in probability about conditions of convergence of expectations, you can invoke uniform integrability which follows from boundedness.


问:
Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$
How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$

What's  the approach to it?

Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education ).I just know simple elementary results of definite and indefinite integration. Substitutions and all those works good. :)

答:
You want to integrate
$$J = \int_0^{\pi/2} \dfrac{dx}{\left(a^2 +(b^2-a^2) \sin^2(x) \right)^2} = \dfrac1{a^4}\int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$$
where $c = \dfrac{b^2-a^2}{a^2}$.
We now want to integrate $I = \displaystyle \int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$. From Taylor series, we have
$$\dfrac1{(1+cy^2)^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k y^{2k}$$
Hence (swapping integral and infinite summation), we get that
$$\int_0^{\pi/2}\dfrac{dx}{(1+c\sin^2(x))^2} = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \int_0^{\pi/2}\sin^{2k}(x)dx$$
From here, we have
$$\int_0^{\pi/2} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{\pi}{2^{2k+1}}$$
Hence,
$$I = \sum_{k=0}^{\infty}(-1)^k (k+1)c^k \dbinom{2k}k \dfrac{\pi}{2^{2k+1}} = \dfrac{\pi}2 \sum_{k=0}^{\infty} (k+1) \dbinom{2k}k \left(-\dfrac{c}4\right)^k$$
Now from Taylor series, we have
$$\sum_{k=0}^{\infty} (k+1) \dbinom{2k}k x^k = \dfrac{1+4x\sqrt{1-4x}-2x-\sqrt{1-4x}}{(1-4x)^{3/2}}$$
Hence,
$$J = \dfrac{\pi}2 \left(\dfrac{1-c\sqrt{1+c}+c/2-\sqrt{1+c}}{a^4(1+c)^{3/2}} \right)$$
Now plug in the value of $c$ and get the value of the original integral $J$.


问:
Calculate $\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$
Calculate $$\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$$

I think that firstly I should use Taylor's theorem, so I have:$$\int_0^\infty \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots}{(x^2+1)^2}dx$$
However I don't know what I can do the next.

答:
Let $f(x)=\frac{\cos(3x)}{x^4+x^2+1}$.  Inasmuch as $f$ is an even function, we can write

$$\begin{align}
\int_0^\infty \frac{\cos(3x)}{x^4+x^2+1}\,dx&=\frac12 \int_{-\infty}^\infty \frac{e^{i3x}}{x^4+x^2+1}\,dx\\\\
&=\pi i \left(\text{Res}\left(\frac{e^{i3z}}{z^4+z^2+1}, z=e^{i\pi/3}\right)+\text{Res}\left(\frac{e^{i3z}}{z^4+z^2+1}, z=e^{i2\pi/3}\right)\right)
\end{align}$$

where we have used the Residue Theorem.

Can you finish now?


答:
We seek to evaluate

$$J = \int_0^\infty \frac{\cos(3x)}{x^4+x^2+1} \; dx
= \frac{1}{2} \int_{-\infty}^\infty
\frac{\cos(3x)}{x^4+x^2+1} \; dx.$$

Following the hint by @MarkViola we observe that
$\sin(3x)/(x^4+x^2+1)$ is odd hence we have

$$J = \frac{1}{2} \int_{-\infty}^\infty
\frac{\cos(3x)+i\sin(3x)}{x^4+x^2+1} \; dx
= \frac{1}{2} \int_{-\infty}^\infty
\frac{\exp(3ix)}{x^4+x^2+1} \; dx
= \frac{1}{2} K.$$

Introduce

$$f(z) = \frac{\exp(3iz)}{z^4+z^2+1}.$$

We  integrate $f(z)$  along a  contour  consisting of  a line  segment
$\Gamma_0$  from $-R$  to  $R$ on  the real  axis  and a  semicircular
contour $\Gamma_1$ in the upper half plane connecting $R$ to $-R$ with
a counterclockwise  arc of radius  $R$. In the limit  $\Gamma_0$ gives
the integral $K.$ For $\Gamma_1$ we apply the ML bound to get

$$\pi R \times \frac{|\exp(3iR\cos\theta-3R\sin\theta)|}{R^4-R^2-1}
\le \frac{\pi R}{R^4-R^2-1} \rightarrow 0
\quad \text{as} \quad R\rightarrow \infty$$

since  $|\exp(3iR\cos\theta)|=1$  and  $\exp(-3R\sin\theta)  \le  1$
where $0\le\theta\le\pi$ and $z=R\exp(i\theta).$

Now note that

$$w^2+w+1 = \frac{w^3-1}{w-1}$$

so the polynomial in $w$ has roots $\exp(2\pi i k/3)$ where $k=1,2.$
Hence the poles of $f(z)$ are at

$$\exp(2\pi i k/6 + 2\pi i j/2)
= \exp(2\pi i k/6 +  \pi i j) = \pm \exp(\pi i k/3).$$

Restricting to  the upper half plane we get with
$$\rho_0 = \exp(\pi i/3)
= \frac{1}{2} + \frac{1}{2}\sqrt{3} i
\quad\text{and}\quad
\rho_1 = \exp(2\pi i/3)
= - \frac{1}{2} + \frac{1}{2}\sqrt{3} i$$

that

$$K = 2\pi i \times \mathrm{Res}_{z=\rho_0} f(z)
+ 2\pi i \times \mathrm{Res}_{z=\rho_1} f(z).$$

Since $K$ is a real number this becomes

$$K = - 2\pi \times \Im(\mathrm{Res}_{z=\rho_0} f(z))
- 2\pi \times \Im(\mathrm{Res}_{z=\rho_1} f(z)).$$

As both poles are simple we find that

$$\mathrm{Res}_{z=\rho_{0,1}} f(z)
= \frac{\exp(3i\rho_{0,1})}{4\rho_{0,1}^3+2\rho_{0,1}}.$$

so that

$$\mathrm{Res}_{z=\rho_0} f(z)
= \frac{\exp(3i\rho_0)}{2\rho_0-4}
= \exp\left(\frac{3}{2} i - \frac{3}{2} \sqrt{3}\right)
\frac{1}{-3+\sqrt{3} i}
\\ = \exp\left(\frac{3}{2} i - \frac{3}{2} \sqrt{3}\right)
\frac{-3-\sqrt{3}i}{12}
\\ = \frac{1}{12} \exp\left(-\frac{3}{2}\sqrt{3}\right)
\exp\left(\frac{3}{2} i\right)(-3-\sqrt{3}i)$$

and

$$\mathrm{Res}_{z=\rho_1} f(z)
= \frac{\exp(3i\rho_1)}{2\rho_1+4}
= \exp\left(-\frac{3}{2} i - \frac{3}{2} \sqrt{3}\right)
\frac{1}{3+\sqrt{3} i}
\\ = \exp\left(-\frac{3}{2} i - \frac{3}{2} \sqrt{3}\right)
\frac{3-\sqrt{3}i}{12}
\\ = \frac{1}{12} \exp\left(-\frac{3}{2}\sqrt{3}\right)
\exp\left(-\frac{3}{2} i\right)(3-\sqrt{3}i).$$

This gives

$$K = - 2\pi
\frac{1}{12} \exp\left(-\frac{3}{2}\sqrt{3}\right)
\\ \times
(\cos(3/2) \times (-\sqrt{3}) + \sin(3/2) \times (-3)
+ \cos(-3/2) \times (-\sqrt{3}) + \sin(-3/2) \times 3)
\\ = - \pi
\frac{1}{6} \exp\left(-\frac{3}{2}\sqrt{3}\right)
\\ \times
(- \cos(3/2) \sqrt{3} - \sin(3/2) \times 3
- \cos(3/2) \sqrt{3} - \sin(3/2) \times 3).$$

With $J = \frac{1}{2} K$ we find at last

$$\frac{\pi}{6}
\exp\left(-\frac{3}{2}\sqrt{3}\right)
\left(\cos\left(\frac{3}{2}\right)\times\sqrt{3}
+\sin\left(\frac{3}{2}\right)\times 3\right).$$

Following the work by @Quanto this becomes

$$\frac{\pi}{\sqrt{3}}
\exp\left(-\frac{3}{2}\sqrt{3}\right)
\left(\cos\left(\frac{3}{2}\right)\times\frac{1}{2}
+\sin\left(\frac{3}{2}\right)\times\frac{\sqrt{3}}{2}\right).$$

or

$$\bbox[5px,border:2px solid #00A000]{
\frac{\pi}{\sqrt{3}}
\exp\left(-\frac{3}{2}\sqrt{3}\right)
\sin\left(\frac{\pi}{6} + \frac{3}{2}\right).}$$

Title: Calculate $\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$

Body:
Calculate $$\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$$

I think that firstly I should use Taylor's theorem, so I have:$$\int_0^\infty \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots}{(x^2+1)^2}dx$$
However I don't know what I can do the next.