问:
Differential equation for rate of decay: $\frac{dx}{dt}=-\frac{8}{3}x$
I need support and clarification for my working out of this differential equation.

I must solve the following, giving $x$ in terms of $t$.

$$\frac{dx}{dt}=-\frac{8}{3}x$$

My approach: Divide both sides by $x$ and multiply both sides by $dt$, then integrate:

$\int \frac{dx}{x}=-\frac{8}{3}\int dt$

$\ln(x)  = -\frac{8}{3}t$

Then multiply both sides by natural log

$$x = e^{-\frac{8}{3}t}$$

Given that $x = 30$ when $t = 0$, find the rate of decay of any given substance ($x$ = substance) from $30$ to $10$, to the nearest minute. How do I calculate this?

答:
$$\frac{x'(t)}{x(t)}=-\frac{8}{3}\to \log x(t)=-\frac{8}{3}t+C$$
$$x(t)=\exp\left({-\frac{8}{3}t+C}\right)$$
as for $t=0$ we have $x=30$ substitute and get
$30=e^C\to C = \log 30$.

So the solution is
$$x(t)=30 e^{-\frac{8 t}{3}}$$
To find when the substance $x$ goes to $10$ we must solve
$$30 e^{-\frac{8 t}{3}}=10$$
$$e^{-\frac{8 t}{3}}=\frac{1}{3}$$
$$-\frac{8 t}{3}=\log\frac{1}{3}\to t=\frac{3 \log 3}{8}\approx 0.41$$
It $t$ is in hours then the time is $25$ minutes.


问:
Given sets $X,Y$, is it possible to find sets $X',Y'$ such that $X',X$ and $Y',Y$ are equinumerous, $X'\cap Y'=X'\cap X=Y'\cap Y=\emptyset$?
Let $X$ and $Y$ be arbitrary sets. Assuming the axioms of ZFC, is it true that there are sets $X'$ and $Y'$ such that $X'$ and $X$ have the same cardinal number, $Y'$ and $Y$ have the same cardinal number, $X\cap X'=\emptyset$, $Y\cap Y'=\emptyset$ and $X'\cap Y'=\emptyset$?

答:
Yes.

First let $X_0=\{X\}\times X$ and $Y_0=\{Y\}\times Y$, and use the axiom of regularity to show that $X_0\cap X=\varnothing=Y_0\cap Y$. (Clearly $|X_0|=|X|$ and $|Y_0|=|Y|$.) Then let $X'=\{0\}\times X_0$ and $Y'=\{1\}\times Y_0$; clearly $X'\cap Y'=\varnothing$, $|X'|=|X|$, and $|Y'|=|Y|$. Finally, use regularity again to show that $X'\cap X=\varnothing=Y'\cap Y$.

ContentLicense: CC BY-SA 4.0


问:
Calculation of $\int_0^\infty \frac{x^{1-a}}{x+c} dx $ using Gamma function
I am supposed to calculate some integrals using Gamma functions.
The simplest of the integrals is

$$ I(c)=\int_0^\infty \frac{x^{1-a}}{x+c} dx  $$

and the two relations I am supposed to use are
$$ \frac{\Gamma(b+1)}{a^{b+1}}=\int_0^\infty t^be^{-at}dt $$
and Euler's reflection formula,
$$ \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)} $$

Try as I might, I have not managed to correlate the formulas to the integral in any significant way (in large part due to that pesky exponential I can't seem to shake). The supposed result of the above integral is $$I(c)=-\frac{\pi}{\sin(\pi a)}c^{1-a}$$
I definitely don't want to outsource my responsibilities (and once I understand how to go about this I'm confident I'll work out the more complicated integrals myself), but I've hit a wall here. I'd appreciate either a solution or a nudge in the right direction.

答:
This is an outline of a solution, and I hope it's detailed enough for you to fill in the gaps.

I assume you have attempted to use the two relations by substituting the first relation into the second one, which would get you something like
$$
\frac{\pi}{\sin(\pi b)} = \Gamma(b+1)\Gamma(-b) = a \int_0^\infty \int_0^\infty s^{-b-1} t^{b} e^{-a(s+t)} ds dt
$$
The hope is to somehow transform the right hand side into some form that is proportional to $I(c)$. This is in fact the correct direction to work towards.

You said that you can't get rid of the pesky exponential factor, but you can probably also guess that the key is to come up with some substitution where one of the new variables is $u = s+t$, so that the integral with respect to $u$ will get rid of the exponential factor. The substitution that will work is in fact $s = u v$, $t = u(1-v)$. Note that the $u$-dependent part of $s^{-b-1} t^{b}$ will simplify to $u^{-1}$. Combined with the Jacobian factor $u$, the only remaining $u$ dependence in the integrand is $e^{-au}$. Thus integrating with respect to $u$ will get rid of the exponential factor.

The remaining integral should be of the form
$$
\int_0^1 v^{-b-1} (1-v)^{b} dv\ .
$$
A substitution along the lines of $x = \frac{v^2}{1-v^2}$ will bring the integral closer to the form in which $I(c)$ is originally written down.

Note that the relation you are trying to prove can be thought of as a special case of the relation
$$
\mathrm{B}(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}\ ,
$$
that is, the case when $x+y = 1$, and the integral $I(c)$ can be seen as a multiple of one of the integral representations of the beta function (see equation (22)),
$$
\mathrm{B}(2-a, a-1) = \int_0^\infty \frac{x^{1-a}}{1+x}dx\ .
$$
The proof outline I sketched above is a specialization of a standard proof the relation between the two functions. In fact the proof of the general case is more natural and suggests the first substitution used above.


问:
Existence of holomorphic function $f$ such that $f^{(n)}(0) = n^{2n}$
I need help with the following problem:

Is there a holomorphic function $f$ in an open disk around $0$ such that
\begin{align}
f^{(n)}(0) = n^{2n}
\end{align}
for all $n \in \mathbb{N}$?

I thought quite long about this and came up with the following idea:

Suppose there exists such a function. Then we can write it as
\begin{align}
f(z) = \sum_{n=0}^{\infty} \frac{n^{2n}}{n!}z^n.
\end{align}
We can compute the radius of convergence $R$ of this power series:
\begin{align}
R = \lim_{n \to \infty} \left| \frac{n^{2n}}{n!} \cdot \frac{(n+1)!}{(n+1)^{2n+2}} \right| = \lim_{n \to \infty}\left( \frac{n}{n+1}\right)^{2n} \cdot \frac{1}{n+1} \leq \lim_{n \to \infty} \frac{1}{n+1} = 0.
\end{align}

This contradicts the assumption that $f$ is holomorphic in an open disk around $0$ or with other words: $R > 0$.

I am not very confident looking at my solution. (Is it true?) Apart from that I was wondering whether or not there is an easier way to find an answer to this question. The problem reminds me of the Identity Theorem or even the generalized Cauchy Integral Formula. Still I failed to see if (and how) those two could have been used to solve this. In order to gain a better understanding of problems of this kind I would be glad if someone has another good idea for this question!

答:
Yes, if a function is holomorphic in an open region $\Omega$ containing $0$, then it equals its Taylor power series expansion in some open disk around $0$.

More precisely, the radius of convergence of the series is the supremum of all $r$ such that the open disk with center $0$ and radius $r$ is contained in $\Omega$.

Since the power series formed with the required coefficients has radius of convergence $0$, you can rightly conclude that no such analytic function exists.


答:
Your solution is fine. Alternatively you can use the Cauchy inequalities for the derivatives (a consequence of Cauchy's integral theorem): If $f$ is holomorphic in $B_R(0)$, $0 < r < R$, and $M = \sup_{|z|=r} |f(z)|$, then
$$
 \frac{f^{(n)}(0)}{n!} \le \frac{M}{r^n}
$$
In our case that gives
$$
 M \ge \frac{r^n n^{2n}}{n!} \ge r^n n^n
$$
for all $n$, which is not possible.

Title: Existence of holomorphic function $f$ such that $f^{(n)}(0) = n^{2n}$

Body:
I need help with the following problem:

Is there a holomorphic function $f$ in an open disk around $0$ such that
\begin{align}
f^{(n)}(0) = n^{2n}
\end{align}
for all $n \in \mathbb{N}$?

I thought quite long about this and came up with the following idea:

Suppose there exists such a function. Then we can write it as
\begin{align}
f(z) = \sum_{n=0}^{\infty} \frac{n^{2n}}{n!}z^n.
\end{align}
We can compute the radius of convergence $R$ of this power series:
\begin{align}
R = \lim_{n \to \infty} \left| \frac{n^{2n}}{n!} \cdot \frac{(n+1)!}{(n+1)^{2n+2}} \right| = \lim_{n \to \infty}\left( \frac{n}{n+1}\right)^{2n} \cdot \frac{1}{n+1} \leq \lim_{n \to \infty} \frac{1}{n+1} = 0.
\end{align}

This contradicts the assumption that $f$ is holomorphic in an open disk around $0$ or with other words: $R > 0$.

I am not very confident looking at my solution. (Is it true?) Apart from that I was wondering whether or not there is an easier way to find an answer to this question. The problem reminds me of the Identity Theorem or even the generalized Cauchy Integral Formula. Still I failed to see if (and how) those two could have been used to solve this. In order to gain a better understanding of problems of this kind I would be glad if someone has another good idea for this question!


问:
Why does the argument that $\mathbb{Q}[x] / \langle x^{2}\rangle \simeq \mathbb{Q}[x] / \langle x^{2} - 1\rangle $ not apply in $\mathbb{Z_2}$?
So I have an exercise asking if $\mathbb{Q}{[x]} / \langle x^{2}\rangle  \simeq \mathbb{Q}{[x]} / \langle x^{2} - 1\rangle $ and if $\mathbb{Z_2}{[x]} / \langle x^{2}\rangle  \simeq \mathbb{Z_2}{[x]} / \langle x^{2} - 1\rangle $. My opinion is yes for $\mathbb{Q}$ and no for $\mathbb{Z_2}$, using contradiction:

Suppose we have an isomorphism $\phi : \mathbb{Q}{[x]} / \langle x^{2}\rangle  \rightarrow \mathbb{Q}{[x]} / \langle x^{2} - 1\rangle $, that implies $\phi(x^{2}) = \phi(0) = 0 \implies \phi^{-1}(0) = \phi(x^{2}) = 0$, which is not possible since $\phi^{-1}(x^{2}) = 1$, it will work in $\mathbb{Q}$ but we have $0 = 1$ in $\mathbb{Z_2}$.

Could someone please explain to me how to do this?

答:
$\text{Nil}(\frac{\Bbb Q[x]}{x^2-1})$,$\text{Nil}(\frac{\Bbb Z_2[x]}{x^2-1})$ are trivial nilradicals because where $\Bbb F$ is a field we must have  $$f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x^2-1\rangle}\big)\iff f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x-1\rangle}\big)\;\land\;f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x+1\rangle}\big)$$ $$\therefore\;f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x^2-1\rangle}\big)\iff f(x)=0+\langle x^2-1\rangle$$ $x\pm 1$ are irreducibles, $\langle x\pm 1\rangle$ are maximal and thus $\frac{\Bbb F[x]}{\langle x\pm 1\rangle}$ are fields but $$X_1:=x+\langle x^2\rangle_{\Bbb Q[x]}\in\frac{\Bbb Q[x]}{\langle x^2\rangle}\;\;\text{and}\;\;X_2:=x+\langle x^2\rangle_{\Bbb Z_2[x]}\in\frac{\Bbb Z_2[x]}{\langle x^2\rangle}$$ are non-trivial nilpotents as $$X_1^2=x^2+\langle x^2\rangle_{\Bbb Q[x]}=\langle x^2\rangle_{\Bbb Q[x]}\;\;\;\;\;X_2^2=x^2+\langle x^2\rangle_{\Bbb Z_2[x]}=\langle x^2\rangle_{\Bbb Z_2[x]}$$

Therefore $\text{Nil}(\frac{\Bbb Q[x]}{x^2-1})$,$\text{Nil}(\frac{\Bbb Z_2[x]}{x^2-1})$ are trivial but $\text{Nil}(\frac{\Bbb Q[x]}{\langle x^2\rangle}),\text{Nil}(\frac{\Bbb Z_2}{\langle x^2\rangle})$ are $\it{non}$-$\text{trivial}$. $$\therefore\;\frac{\Bbb Q[x]}{\langle x^2\rangle}\not\approx\frac{\Bbb Q[x]}{\langle x^2-1\rangle}\;\land\;\frac{\Bbb Z_2[x]}{\langle x^2\rangle}\not\approx\frac{\Bbb Z_2[x]}{\langle x^2-1\rangle}$$


问:
Is my proof of $C\subseteq A$ and $D\subseteq B$ implies $C\cap D \subseteq A\cap B$ correct?
Suppose we want want to prove if $C\subseteq A$ and $D\subseteq B$ then $C\cap D\subseteq A\cup B$. Is my proof correct?

If $x\in C$ and $x\in D$, since $C\subseteq A$ and $D\subseteq B$, it follows that $C\cap D\subseteq A\cap B$.

If $C\cap D\subseteq A\cap B$ and $x\in C$, then ($x\in C\cap D$ and $x\in A\cap B$) or ($x\not\in C\cap D$ and $x\in A\cap B$). Hence $x\in C\cap D$ and $x\in A\cap B$. Hence $x\in C$ and $x\in A$. Therefore $C\subseteq A$.

If $C\cap D\subseteq A\cap B$ and $x\in D$, then ($x\in C\cap D$ and $x\in A\cap B$) or ($x\not\in C\cap D$ and $x\in A\cap B$). Hence $x\in C\cap D$ and $x\in A\cap B$. Hence $x\in B$. Therefore $D\subseteq B$.

答:
Here it is another way to approach it for the sake of curiosity.

Since $C\subseteq A$ and $D\subseteq B$, we have that $C\cap A = C$ and $D\cap B = D$. Hence we conclude that
\begin{align*}
(C\cap D)\cap(A\cap B) & = (C\cap A)\cap(D\cap B) = C\cap D
\end{align*}
which implies that $C\cap D\subseteq A\cap B$.

Hopefully this helps!


问:
How many solutions are there for $3x_1+3x_2+x_3+x_4=30$?

Find how many solutions there are for $$ 3x_1+3x_2+x_3+x_4=30$$

I know how to solve this: $x_1+x_2+x_3+x_4=30$
and I read this link, but I am still not sure about my answer.
I wrote:
$x_4+x_3=30-3(x_1+x_2) $
then I get this: $0 \le x_1+x_2 \le 10$ (that I know to solve: $66$ combination)
and now I'm not sure what to do with $x_3+x_4$.

答:
Warning: DISGUSTING CASEWORK AHEAD

Rewrite the equation as $3(x_1+x_2)+x_3+x_4=30$. Now, let $a=x_1+x_2$ and $b= x_3+x_4$. Thus, we have $3a+b = 30$. So, we can easily figure that each solution pair $(a,b)$ are $(2,24),(3,21),(4,18),(5,15),(6,12),(7,9),(8,6),(9,3)$. Note that $a,b \ge 2$ because each one is the sum of $2$ naturals, which are both $1$ or greater. Now, we examine each case.

For $(2,24)$, we can make $a=2$ with the pair $(1,1)$ and make $b=24$ with pairs $(1,23)$ through $(23,1)$, which is $23$ pairs. Thus for this case we get $23$ solutions by multiplying $1$ way to make $a=2$ with $23$ ways to make $a=24$.

For $(3,21)$, we can make $a=3$ with the pairs $(1,2),(2,1)$ and make $b=21$ with pairs $(1,20)$ through $(20,1)$, which is $20$ pairs. Thus for this case we get $40$ solutions by multiplying $2$ way to make $a=3$ with $20$ ways to make $a=21$.

Now you can see a pattern right? So far our sum is $1*23+2*20$. Let's continue this pattern on, because we can see a pattern in our cases $(2,24),(3,21),(4,18),(5,15),(6,12),(7,9),(8,6),(9,3)$.

Thus, we get a sum of $1*23+2*20+3*17+4*14+5*11+6*8+7*5+8*2 = 324$ solutions.

EDIT: I notice that some other people are including $0$ as a natural number, so here's a revised solution. Since we can have both $x$ be $0$, or one $x$ be $0$ ,that means we may have $a,b = 0 or 1$. Thus, our new cases are $(0,30),(1,27)(2,24),(3,21),(4,18),(5,15),(6,12),(7,9),(8,6),(9,3),(10,0)$. Also, before for case $(2,24)$, we had $1$ way to make $a=2$ using $(1,1)$. However, now that we can use $0$, we add $2$ pairs to get $(1,1),(2,0),(0,2)$. Similar logic can be used to add $2$ pairs to all ways to make $a,b$. Now, let's find our new sum.

For case $(0,30)$ we can have $a=0$ with $(0,0)$ and $b=30$ with $(0,30)$ through $(30,0)$ or $31$ pairs. Thus, the number of ways for this case is $1*31$.

For case $(1,27)$ we can have $a=1$ with $(0,1),(1,0)$ and $b=27$ with $(0,27)$ through $(27,0)$ or $28$ pairs. So we have $2*28$ total ways for this case.

Now that we have the base cases $1*31$ & $2*28$,we may use the logic we used previously to get our sum of $1*31+2*28+3*25+4*22+5*19+6*16+7*13+8*10+9*7+10*4+11*1 = 726$


答:
The first line, properly edited, suffices.

If $x\in C\cap D$, then $x\in C$ and therefore $x\in A$; moreover $x\in D$ and therefore $x\in B$. Thus $x\in A\cap B$.

This concludes the proof. Nothing more is needed. But you shouldn't jump to conclusions too early: you start with some $x\in C\cap D$ and then it disappears.

Oh, and the implication “if $C\cap D\subseteq A\cap B$ then $C\subseteq A$ and $D\subseteq B$” is false, so you're better not trying to prove it.

Example: $C=\{1\}$, $D=\{2\}$, $A=\{3\}$, $B=\{4\}$.

Title: Is my proof of $C\subseteq A$ and $D\subseteq B$ implies $C\cap D \subseteq A\cap B$ correct?

Body:
Suppose we want want to prove if $C\subseteq A$ and $D\subseteq B$ then $C\cap D\subseteq A\cup B$. Is my proof correct?

If $x\in C$ and $x\in D$, since $C\subseteq A$ and $D\subseteq B$, it follows that $C\cap D\subseteq A\cap B$.

If $C\cap D\subseteq A\cap B$ and $x\in C$, then ($x\in C\cap D$ and $x\in A\cap B$) or ($x\not\in C\cap D$ and $x\in A\cap B$). Hence $x\in C\cap D$ and $x\in A\cap B$. Hence $x\in C$ and $x\in A$. Therefore $C\subseteq A$.

If $C\cap D\subseteq A\cap B$ and $x\in D$, then ($x\in C\cap D$ and $x\in A\cap B$) or ($x\not\in C\cap D$ and $x\in A\cap B$). Hence $x\in C\cap D$ and $x\in A\cap B$. Hence $x\in B$. Therefore $D\subseteq B$.


问:
How to evaluate $\lim_{(x,y)\to (0,0)} \frac{x^4+y^4}{2x-y}$?
Find the limit :

$$\lim_{(x,y)\to (0,0)} \frac{x^4+y^4}{2x-y}$$

I have tried using polar substitution but the problem is the denominator. Denominator can go to $0$ for some value of theta and is not dependent on only $r$.

答:
Let $y=2x-t$. Then

$$\frac{x^4+(2x-t)^4}{t}=17x^4/t+P(t,x),$$

where $P(t,x)$ is a polynomial in $t$ and $x$. Now let $t=x^5$. This will cause the limit to blowup. Tracing this back, let $y:=2x-x^5$ and confirm by usual means that it blows up.


答: Hint: let $y=2x-x^4$. Then the smallest order term in both the numerator and denominator will be quartic, so the limit will be non-zero.

Title: How to evaluate $\lim_{(x,y)\to (0,0)} \frac{x^4+y^4}{2x-y}$?

Body:
Find the limit :

$$\lim_{(x,y)\to (0,0)} \frac{x^4+y^4}{2x-y}$$

I have tried using polar substitution but the problem is the denominator. Denominator can go to $0$ for some value of theta and is not dependent on only $r$.


问:
Combinatorics question related to round robin tournament, to prove that at least two teams have the same score

Q. In a football tournament of one round (each team plays each other once, 2 points for a win, 1 point to each team for draw, 0 for
loss), 28 teams compete. During the tournament more than 75% of the
matches end in a draw. Prove that there were two teams who ended up
with the same final score

I have tried this question for a long time, but I haven't been able to come up with an approach that will prove the given result. Could anyone please provide a solution for this question. I have provided some useful information below

Total number of matches is 378. At least 284 matches have ended in a draw and at most 94 matches have ended with a victor.

答:
Each team has to have a different number of "wins-losses". For $k$ "wins-losses" a team would have to have played in at least $k$ decisive games.

Either $14$ or more teams have a positive "wins-losses" or $14$ or more teams have a negative "wins-losses". Without loss of generality suppose the former. Then the number of wins for these teams is at least $1+2+3+...+14=105$ which is greater than $94$.


问:
Existence of an inner point in a nonempty polyhedron
I was reading some notes on polyhedral analysis and encountered a proof that confused me. The proof is in the book named "Integer and Combinatorial Optimization" by Wolsey and Nemhauser.

Let $M = \{l, 2, ... , m\}, M^{=} = \{i \in M: a_ix = b_i, \forall x \in  P\}$ and let $M^{\leq} = M \setminus M^{=}$ where $P$ is a rational polyhedron.

$Definition:$ A point $x \in P$ called an inner point  of $P$ if $a^ix < b_i , \forall i \in M^{\leq}$.

$Proposition:$ Every nonempty polyhedron has an inner point.

$Proof:$ If $M^{\leq} = \emptyset$, every point of $P$ is inner. Otherwise, for $i \in M^{\leq}$ there exists a point $x^i \in P$ with $a^i x^i < b_i$ . Now $ \hat{x}= (1 / |M^{\leq}|)\sum_{i \in M^{\leq}}x^i \in P$ since $P$ is convex. Since $a^i \hat{x} < b_i, \forall i \in M^{\leq}$, $\hat{x}$ is an inner point.

Doesn't the second statement (Otherwise, for $i \in M^{\leq}$ there exists a point $x^i \in P$ with $a^i x^i < b_i$) imply that we already assume that there exists an inner point? If that is the case, why do we show that there exist $\hat{x}$ satifying the condition of an inner point?

答: No. There is a different $x^i$ for every $i.$ But we want the same $x$ for every $i.$ The proof constructs such an $x.$ On the other hand, I don't think the construction works, since while $\hat{x} \in P,$ I don't really see why $\hat{x}$ satisfies all the strict inequalities.


问:
How do I find all $x$ values without using determinants
I want to find all values of $x$ for which the following matrix is invertible.

$$\begin{pmatrix} x - \frac12 & 0 & 0\\ x & x - \frac13 & 0\\ x^2 & x^3 & x + \frac14\end{pmatrix}$$

I know that using lower triangular determinant (at $x$ values where determinant not equal to zero), I can find easily, but I want to find using reduced echelon form. How can I do that?

答:
If $x=\frac12$, than all entries of the first row are zeros, and therefore the matrix is not invertible.

Assume now that $x\ne\frac12$. Then divide the first row by $x-\frac12$, and you will get$$\begin{bmatrix}1&0&0\\x&x-\frac13&0\\x^2&x^3&x+\frac14\end{bmatrix}.$$Now you add to the second row the first row times $-x$ and you add to the third row the first row times $-x^2$; you will get$$\begin{bmatrix}1&0&0\\0&x-\frac13&0\\0&x^3&x+\frac14\end{bmatrix}.$$If $x=\frac13$, the the second row only has zeros, and therefore the matrix is not invertible.

Assume now that $x\ne\frac13$. Then divide the second row by $x-\frac13$. You will get$$\begin{bmatrix}1&0&0\\0&1&0\\0&x^3&x+\frac14\end{bmatrix}.$$Now, add to the third row the second row times $-x^3$. You will get$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&x+\frac14\end{bmatrix}.$$

If $x=-\frac14$, the third row only has zeros, and therefore the matrix is not invertible. Otherwise, divide the third row by $x+\frac14$. Then you will get the identity matrix, which is invertible. So, the original matrix will also be invertible.


问:
Show that $n \ge \sqrt{n+1}+\sqrt{n}$
(how) Can I show that:

$n \ge \sqrt{n+1}+\sqrt{n}$ ?

It should be true for all $n \ge 5$.

Tried it via induction:

$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.

$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$

Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)

Is this the right way?

答: Hint:  $\sqrt{n} + \sqrt{n+1} \leq 2\sqrt{n+1}$.  Can you take it from there?</p>

ContentLicense: CC BY-SA 3.0

Title: Show that $n \ge \sqrt{n+1}+\sqrt{n}$


问:
References for quasi finite and proper implies finite
Does anybody know a reference for a proof of:

Let $f: X \rightarrow Y$ be a quasi-finite proper morphism of varieties. Then $f$ is finite.

Is there one in Hartshorne? I could not find it.

Thanks!

答:
I think this is just a generalization of the fact that finite integral extensions of domains are finitely generated module over the lower integral domain.

Anyway if you need a reference/proof to the statement in your post, you can cite Lemma 02OG of the stacks project.


问:
Calculating edge cover for this Hypergraph?
Let $D= \{a_1,a_2,...,a_n\}$ be a set of constants. For any subset of $D$ of cardinality $3$, we define another set (we call it hyper-edge) containing all $3 \choose 2$ pair-wise combinations (we call them nodes) of the constants. How can I compute the minimum number of hyper edges required to cover all such pair-wise combinations of constants in $D$ in general ?

I did the first non-trivial example:

Example: $D  = \{a_1,a_2,a_3,a_4\}$, In this case you have $4$ possible subsets, each giving us 3 nodes in a hyper edge, and picking any three such hyper-edges or any three such subsets gives us a complete cover. I list all possible hyper edges below:

${a_1a_2 , a_1a_3, a_2a_3}$

${a_1a_2 , a_1a_4, a_2a_4}$

${a_2a_3,a_3a_4,a_2a_4}$

${a_1a_4,a_1a_3,a_3a_4}$

You can take any three of these subsets and it has all the pairwise nodes.

答:
What you are looking for is known in the literature as a covering design. You want $C(n,3,2)$, where $C(v,k,t)$ is the minimum number of $k$-element subsets which cover all $t$-element subsets of a given $v$-element set.

It is proven in the CRC Handbook of Combinatorial Designs (2nd ed.) that
$$
C(n,3,2)=\left\lceil\frac{n}3\left\lceil\frac{n-1}2\right\rceil\right\rceil.
$$
It is easy to prove that at least this many triangles are necessary. There are $n-1$ edges at each vertex, and each triangle covers at most two edges of a given vertex.  Therefore, each vertex has a quota of $\left\lceil\frac{n-1}2\right\rceil$ triangles, for a total quota of $n\cdot \left\lceil\frac{n-1}2\right\rceil$, and each triangle contributes $3$ points to the total quota.

Finding these optimal covering designs is the hard part. When $n\equiv 1$ or $3\pmod 6$, you can use a Steiner triple system of order $n$, whose construction is described in [Skolem]. For the other congruence classes $\pmod 6$, the constructions are described in detail in the CRC Handbook on Combinatorial Designs (2nd ed), chapter $11$, section $5$, subsection $11.41$. They are too complicated for me to describe here.

If you just need some explicit designs for small $n$, they can be at Dan Gordon's web page, or the La Jolla covering repository (these give the same data).

Charles J. Colbourn and Jeffrey H. Dinitz. 2006. Handbook of Combinatorial Designs, Second Edition (Discrete Mathematics and Its Applications). Chapman & Hall/CRC.

Skolem, T. (1958). Some Remarks on the Triple Systems of Steiner. MATHEMATICA SCANDINAVICA, 6, 273-280. https://doi.org/10.7146/math.scand.a-10551


问:
Let T be a tree such that every leaf is adjacent to a vertex of degree at least 3. Show that there are two leaves with a common neighbor.
I came up with a proof for this but I think it might be a little hand-wavy and I can't figure out why. Some advice on how to make this proof clearer will be really appreciated!

Problem: Let $T$ be a tree such that every leaf is adjacent to a vertex of degree at least $3$. Show that
there are two leaves with a common neighbor.

Proof: Suppose by contradiction that there are no two leaves with a common neighbor in T. Then let there be $k$ leaves in T and because each leaf is adjacent to a vertex of degree at least $3$, then there are k vertices in T that has degree at least $3$. For vertices other than the leaves and the leaves' neighbor, there are $V(T)-2k$ of those vertices and because they are not leaves they have degree of at least 2. So the minimum sum of degree of vertices in $T$ is $$\sum_{v\in T}d(v)=k+3k+2\cdot (V(T)-2k)=2V(T)$$ however by degree sum formula there is exactly $\sum_{v\in T}d(v)=2V(T)-2$, a contradiction.

答:
Suppose tree is finite.

If no two leafs have common neighbour, then when deleting all leafs we get a graph $H$ with all vertices of degree $2$ or more in $H$. So $H$ has a cycle. But then $T$ has also a cycle. A contradiction.


问:
Why is $Var(V) = a^2\sigma^2_x+b^2\sigma^2_y+c^2\sigma^2_w +2ab\sigma_{xy}+2ac\sigma_{xw}+2bc\sigma_{yw}$?
Im reviewing some practice questions to study for an exam and I'm stuck on this particular question.

The random variable X has expected value $\mu_x$ and variance $\sigma^2_x<\infty$. The random variable Y has expected value $\mu_y$ and variance $\sigma^2_y<\infty$. The random variable W has expected value $\mu_w$ and variance $\sigma^2_w<\infty$.These random variables are not necessarily
independent. That is, the covariances of the pairs of random variables are given
by $cov(X,Y) = \sigma_{xy}$, $cov(X,W) = \sigma_{xw}$, and $cov(Y,W)=\sigma_{yw}$. Let $V = aX + bY + cW$.

where I believe that $a,b,c$ are constants (the question doesn't specify, but I'm 99% certain they are based off the answers given to me)

Find $E(V)$ and $Var(V)$

$E(V) = E(aX + bY + cW)$ I can apply the property of linearity with expected values to get
$E(aX + bY + cW) = aE(X) + bE(Y) + cE(W) = a\mu_x + b\mu_y + c\mu_w$

which matches the answer given to me.

I struggle with Var(V).
The answer is $Var(V) = a^2\sigma^2_x+b^2\sigma^2_y+c^2\sigma^2_w +2ab\sigma_{xy}+2ac\sigma_{xw}+2bc\sigma_{yw}$
But I struggle to get this answer because I'm used to the random variables being independent from one another, but in this case they're not (at least not necessarily).
so I tried
$Var(V) = E(V^2) - E(V)^2 = E((aX+bY+cW)^2) - E(V)^2 = E(a^2X^2+b^2Y^2+c^2W^2+2aXbY+2aXcW+2bYcW) - (a\mu_x + b\mu_y + c\mu_w)^2$

But this is clearly wrong because this requires the knowledge about the second moment of $X$ and $Y$ and $W$. So what am I doing wrong? Is there an alternative formula I should be using to calculate the answer? I feel like I should be using the covariances given, but I'm not sure how to. Any hints or help would be greatly appreciated.

Thank You!

答: More generally constants $a_i$ and variables $X_i$ of finite variance satisfy$$\begin{align}\operatorname{Var}\sum_ia_iX_i&=E\left(\sum_ia_iX_i\right)^2-\left(E\sum_ia_iX_i\right)^2\\&=\sum_{ij}a_ia_j(EX_iX_j-EX_iEX_j)\\&=\sum_{ij}a_ia_j\operatorname{Cov}(X_i,\,X_j)\\&=\sum_ia_i^2\operatorname{Var}X_i+2\sum_{i>j}a_ia_j\operatorname{Cov}(X_i,\,X_j).\end{align}$$Even more generally but by the same technique,$$\operatorname{Cov}\left(\sum_ia_iX_i,\,\sum_jb_jY_j\right)=\sum_{ij}a_ib_j\operatorname{Cov}(X_i,\,Y_j).$$


问:
For what definition of the continuity is the continuity at infinity possible?
For what definition of the continuity is the continuity at infinity possible?

For example, let $f: \mathbb{R} \rightarrow \mathbb{R}$ with $f(x)=x$. By intuition, since $f$ is always continuous, at infinity it's also continuous.

Formally, we must have $lim_{x \to \infty+}\ f(x)=lim_{x \to \infty-}\ f(x)$ but this is nonsense because you can't approach plus infinity from the left and from the right. It is not a number.

OR

$lim_{a \to \infty}lim_{x \to a+}\ f(x)=lim_{a \to \infty}lim_{x \to a-}\ f(x)$

in which case two-sided limit at infinity is the same as one-sided and therefore it is defined?

答:
For a real function $f$ defined on a norm vector space $X$, the following definition of limit at $\infty$ makes sense (and is used in several math domains):

$$\lim\limits_{x \to \infty} f(x)=l$$ if and only if for all $\epsilon \gt 0$ it exists $M \ge 0$ such that for all $\Vert x \Vert \gt M$ we have $\vert f(x) -l\vert \lt \epsilon$.

Based on this definition, the limit at $\infty$ of $f(x)=1/x$ is equal to zero. In a similar way the map $g(x)=1/\Vert x \Vert $ defined on a Banach space $(X, \Vert \cdot \Vert)$ has for limit $0$ at $\infty$.

See point at infinity.


问:
On the approximation of distributional derivatives by second order differential coefficients
Please, how to prove the following?

If $u \in C(\Omega)$, $\phi \in C_c^\infty(\Omega)$ with support $K$ and $\Delta^2$ denotes the second order differential quotient, then
$$
\lim_{\delta \to 0} \int_K u (\Delta_{\delta e_i}^2\varphi) \ dx = \lim_{\delta \to 0} \int_K (\Delta_{\delta e_i}^2 u)\varphi \ dx.
$$

To recall, the second order differential coefficient of $u$ is
$$
\Delta_h^2u(x_0) = \frac{u(x_0 + h) + u(x_0 - h) - 2u(x_0)}{|h|^2}.
$$

For context, this is the conclusion of the proof of Proposition 1.1 in Fully Nonlinear Elliptic Equations, by Caffarelli and Cabré.

I'm really clueless on how to proceed.

Thanks in advance.

答:
I was able to find the answers due to hints by David Stolnicki

For $h$ small enough so that $(x \pm h) \in \Omega$ we have

\begin{align*}
\int_K u(x)(\varphi(x + h) + \varphi(x - h)) - \varphi(x)(u(x + h) + u(x - h)) \ dx = 0.
\end{align*}
Indeed, making $y = x + h$ we get
$$
\underbrace{\int_K u(x) \varphi(x + h) \ dx}_{\displaystyle = \int_K u(y - h) \varphi(y) \ dy} - \int_K \varphi(x) u(x - h) \ dx = 0,
$$
and analogously for the other terms. Adding and subtracting $2 u(x) \varphi(x)$, taking $h = \delta e_i$ and dividing by $\delta^2$, we get
$$
\int_K u (\Delta_{\delta e_i}^2 \varphi) \ dx = \int_K (\Delta_{\delta e_i}^2 u) \varphi \ dx.
$$
Since the limit as $\delta \to 0$ of the left-hand side exists, so does the right-hand side's, and the equality is proved.


答: You could also root you tree anywhere and take a leaf of maximal depth $\delta$. Since its parent (of depth $\delta - 1$) is of degree 3, there is another node of depth $\delta$, and it is a leaf (because there is no node of depth $\delta + 1$).

Title: Let T be a tree such that every leaf is adjacent to a vertex of degree at least 3. Show that there are two leaves with a common neighbor.

Body:
I came up with a proof for this but I think it might be a little hand-wavy and I can't figure out why. Some advice on how to make this proof clearer will be really appreciated!

Problem: Let $T$ be a tree such that every leaf is adjacent to a vertex of degree at least $3$. Show that
there are two leaves with a common neighbor.

Proof: Suppose by contradiction that there are no two leaves with a common neighbor in T. Then let there be $k$ leaves in T and because each leaf is adjacent to a vertex of degree at least $3$, then there are k vertices in T that has degree at least $3$. For vertices other than the leaves and the leaves' neighbor, there are $V(T)-2k$ of those vertices and because they are not leaves they have degree of at least 2. So the minimum sum of degree of vertices in $T$ is $$\sum_{v\in T}d(v)=k+3k+2\cdot (V(T)-2k)=2V(T)$$ however by degree sum formula there is exactly $\sum_{v\in T}d(v)=2V(T)-2$, a contradiction.


问:
Cardinality != Density?
I was in a discussion where I argued that the density of two sets of the same cardinality could be different in respect to the infinite range of non-negative integers. Does cardinality imply that any set of ${\aleph_0}$ has equal density to any other set of the same ${\aleph_0}$?

Also, does cardinality imply equivalence in set size? Or is a formal ranking of different tiers which cannot be compared?

Ok, got that so far. Expanding question now:

Densities can be different but the count is the same. Infinite == Infinite iff cardinality is equivalent. Why doesn't a different density imply infinite set != infinite set even if cardinality is equivalent?

答: Subsets of the naturals can have different natural densties.  If the natural density is greater than zero, the subset is countably infinite.  Think of all the naturals (density 1) and the even numbers (density 1/2).  But there are infinite subsets with natural density zero and infinite subsets for which the natural density cannot be defined.</p>

Title: Cardinality != Density?


问:
If $f \in C^\infty$, then $f''(x) = \lim_{t \to 0} \frac{f(x + t) + f(x - t) - 2f(x)}{t^2}$
I would like help to show the following:

If $f \in C^\infty$, then $f''(x) = \lim_{t \to 0} \frac{f(x + t) + f(x - t) - 2f(x)}{t^2}$

This is the definition of second order derivative by differential coefficients. The first order case is just the definition of derivative, but I'm having trouble with the second order case.

I'm really confused, don't even know where to start.

Thanks in advance

答:
Apply Taylor's theorem to get

\begin{align}
f(x+t) &=f(x)+f'(x)t+\frac{f''(x)}{2}t^2+O(t^3)\\
f(x-t) &=f(x)-f'(x)t+\frac{f''(x)}{2}t^2+O(t^3)
\end{align}

Thus, you get that

$$
f(x+t)+f(x-t)-2f(x)=f''(x)t^2+O(t^3)
$$
which should allow you to readily finish the proof.


答:
Hint.

The assumption that $f\in C^\infty$ is very strong.

Simply applying the L'Hopital rule (and the chain rule) once to the RHS, you have
$$
\lim_{t\to 0}\frac{f'(x+t)-f'(x-t)}{2t}
$$

You can keep using L'Hotital (because of the luxury that $f\in C^\infty$) or observe that
$$
\frac{f'(x+t)-f'(x-t)}{2t}=\frac12\left(\frac{f'(x+t)-f'(x)+f'(x)-f'(x-t)}{t}\right)
$$
where you can use the definition of $f''(x)$.

Body:
I would like help to show the following:

If $f \in C^\infty$, then $f''(x) = \lim_{t \to 0} \frac{f(x + t) + f(x - t) - 2f(x)}{t^2}$

This is the definition of second order derivative by differential coefficients. The first order case is just the definition of derivative, but I'm having trouble with the second order case.

I'm really confused, don't even know where to start.

Thanks in advance


问:
State on *-algebra: $\phi(|f+g|^2)$ when $\phi(|f|^2),\phi(|g|^2)>0$ for monomials
Let $\mathcal{A}$ be the unital *-algebra generated by $N^2$ projections $u_{i,j}=u_{i,j}^2=u_{i,j}^*$, such that the rows/columns are partitions of unity $\sum_k u_{ik}=1_\mathcal{A}=\sum_k u_{kj}$, and $u_{ik}u_{i\ell}=u_{kj}u_{\ell j}=\delta_{k,\ell}$. That is $\mathcal{A}$ is the algebra of regular functions on the quantum permutation group $S_N^+$. It has, for $N>4$, various completions to a $\mathrm{C}^*$-algebra.

If you want, please just assume that $\mathcal{A}$ is generated by $N^2$ projections. I suspect that the relations beyond that don't make things harder or easier.

We will use the notation $|x|^2=x^*x$. Suppose that there exists a state $\phi:\mathcal{A}\rightarrow \mathbb{C}$, a positive functional such that $\phi(1_{\mathcal{A}})=1$, such that for all monomials $f\in\mathcal{A}$, that $\phi(|f|^2)>0$.

If $g\in\mathcal{A}$ is another monomial, can we say anything useful
about:

$$\phi(|f+g|^2)?$$

Can we say that $\phi(|f+g|^2)\geq 0$ with equality only when $f=-g$?

Is $f=-g$ possible for monomial $f,\,g\in\mathcal{A}$?

Can we conclude that $\phi$ is faithful, that for all $a\in\mathcal{A}$, $\phi(|a|^2)=0\Rightarrow a=0$

答:
@JP, you are going to laugh at this, but the counter examples are always the same!

As you mentioned that the relations are not so important, I will ignore them.  In this  case I presume that the number
$N^2$ is not so crucial either, so I will look instead at just two projections.

In the algebra  $C([0,1],M_2)$, consider the two projections $P$ and $Q$, defined by
$$
  P(t)= \pmatrix{1 & 0 \cr 0 & 0}, \quad \text{and} \quad
  Q(t):= \pmatrix{t & \sqrt{t-t^2}\cr \sqrt{t-t^2} & 1-t},
  $$
for all $t\in  [0,1]$.

By the universal property of $\mathcal A$, there is a *-homomorphism
$$
  \pi :  \mathcal A \to C([0,1],M_2),
  $$
sending  the two projections $p$ and $q$ generating $\mathcal A$, respectively to $P$ and  $Q$.

Fixing any $t_0$ in $(0, 1)$,  consider the state $\phi$ on $\mathcal A$ defined by
$$
  \phi(a) = \langle \pi (a)|_{t_0}e_1, e_1\rangle .
  $$
In other words, $\phi(a)$ is the top-left entry of the matrix $\pi (a)|_{t_0}$.

We claim that $\phi$ is strictly positive on any monomial,  that is,  on any product whose terms lie in the set $\{p, q\}$.  The reason is that the matrices
$$
  \pmatrix{1 & 0 \cr 0 & 0}, \quad \text{and} \quad
  \pmatrix{{t_0} & \sqrt{{t_0}-{t_0}^2}\cr \sqrt{{t_0}-{t_0}^2} & 1-{t_0}},
  $$
have the following two properties

all entries are nonnegative,

the top-left entry is strictly positive,

and   it is easy to see that the set of matrices satisfying these properties is closed under multiplication.

Thus,  if $f$ is a monomial, then $|f|^2$ is also a monomial, whence
$\phi( |f|^2)>0$.

However, $\phi$ is not faithful.  To see this,  consider the element
$$
  a = pqp-t_0p.
  $$
We then have that $\pi (a)|_{t_0}=0$, so also $\phi(a^*a) = 0$.

On the other hand,  one may verify by direct computation that
$$
  \pi (a)|_{t}\neq 0,\quad\forall  t\in  (0, 1)\setminus \{t_0\},
  $$
so $a\neq 0$.


问:
If $p$ is a prime number such that $2^n <p\leq 2^{n+1}$, then $p$ can not divide $a^{2^n}+1$ for all $a\in\mathbb{N}$.
I want to show the following:

If $3\leq m\leq 2^{n+1}$, then $m$ can not divide $a^{2^n}+1$ for all $a\in\mathbb{N}$. The context is induction.

For $n=1$, it's enough calculate $(0^2, 1^2, 2^2)\equiv \pmod 3$ and $(0^2, 1^2, 2^2, 3^2)\equiv \pmod 4$.

Assume that it's true for $n$.

For $n+1:$ Suppose that $3\leq m\leq 2^{n+2}$ but $m\mid a^{2^{n+1}}+1=(a^2)^{2^n}+1.$ By induction hipotesis, we have that $2^{n+1}<m\leq 2^{n+2}.$

Claim: $m$ is prime.

If $m$ is not a prime number, then all it's divisors is at most $\frac{m}{2}\leq 2^{n+1}$. This is a contradiction by the induction hipotesis.

And now I need to prove the question in the title.

答:
If $p\,\big|\,\left(a^{2^n}+1\right)$ then $a$ has order $2^{n+1}\pmod p$.  But that would imply that $$2^{n+1}\,\big|\,(p-1)$$

which contradicts the given inequality.


问:
Solve transcendental equation: At $α\ll 1$, $\left|\cos x + \alpha \frac{\sin x}{x}\right| > 1$, determine the width of of the $k$-th zone at $k\gg1$.
To solve this transcendental equations approximately:

At $\alpha \ll 1$ find the positive solution of inequality:

$\left|\cos x + \alpha \frac{\sin x}{x}\right| > 1$, they are divided into series of zones numbered with the integres: $k = 0, 1, \ldots$. Determine the width of of the $k$-th zone at $k \gg 1$.

My approach: if we multiphy both sides by $x$ will we get any advantage?

I need some hints for this problem: How to procceed next?

答:
By the usual trigonometric identities
$$
\cos(x)+\sin(x)\fracαx=\sqrt{1+\frac{α^2}{x^2}}\cos(x+\phi(x)),~~\phi(x)\approx -\fracαx
$$
If $x\approx k\pi$, $|k|>1$, then the width of the region around that point is approximately the distance of the roots of
$$
\cos(u)=\sqrt{1+\frac{α^2}{(k\pi)^2}}^{-1}\approx 1-\frac12\frac{α^2}{(k\pi)^2}+\frac38\frac{α^4}{(k\pi)^4}
$$
for the roots closest to $u=0$. In first order, this will be $u=\pm\fracα{k\pi}$, as $\cos(u)=1-\frac12u^2+\frac16u^4+...$, so that the width is $\frac{2α}{k\pi}$.

For $k=0,\pm1$, perhaps only for $k=0$, one would have to use another approach.

condensed approach One could condense the several approximations by using $\frac{α}{x}=\tan\phi(x)\approx\tan(\frac{α}{x})$ from the start. Then the inequality reduces to
$$
\left|\cos\left(x-\phi\right)\right|< \cos\left(\phi\right)
\iff
|x-k\pi|<\frac\pi2~\land~\cos\left(x-k\pi-\phi\right)< \cos\left(\phi\right)
$$
This now can be solved easily
$$
0<x-k\pi<2\phi(x)=2\arctan\left(\frac{α}{x}\right)
$$
where the upper bound can be approximated as above.


问:
$A \in M_n(\mathbb{C})$ invertible and $A^2$ is diagonalizable. Prove $A$ is diagonalizable
I'm interested to know if it's true because I saw that if $A^2$ is diagonalizable then $A$ is not necessary Diagonalizable.

I have a feeling it's true but I'm not sure how to prove it.

This is my proof but I'm not sure if it's good to assume that :

My try

By Diagonalizable we know that exist invertible matrix $P$ such that :

$P^{-1}AAP=diag(\lambda_1,...,\lambda_n)$

$AA=Pdiag(\lambda_1,...,\lambda_n)P^{-1}$

Set $D=Pdiag(\sqrt{\lambda_1},...,\sqrt{\lambda_n})P^{-1}$

So : $D^2=(Pdiag(\sqrt{\lambda_1},...,\sqrt{\lambda_n})P^{-1})(Pdiag(\sqrt{\lambda_1},...,\sqrt{\lambda_n})P^{-1})=Pdiag(\lambda_1,...,\lambda_n)P^{-1}=AA$

Here I wonder - $A=D=Pdiag(\lambda_1,...,\lambda_n)P^{-1} \rightarrow A$ is diagonalizable.

答:
a simple approach is to argue by contradiction.  Suppose $A$ is invertible and defective but $A^r$ is diagonalizable for some natural number $r$.

This means there is some non-zero eigenvalue $\lambda_1$ that has algebraic multiplicity $d$ strictly greater than geometric multiplicity.  Via re-scaling $A$ we can assume WLOG that $\lambda_1=1$. (i.e. re-run the argument on $A':=\lambda_1^{-1}\cdot A$ if you prefer.)

via any triangularization (Jordan, Schur, or others) we have

$S^{-1} A S = \displaystyle \left[\begin{matrix}I_d + N & * & \\ \mathbf 0 & *\end{matrix}\right]$

$S^{-1} A^r S = \displaystyle \left[\begin{matrix}(I_d + N)^r & * & \\ \mathbf 0 & *\end{matrix}\right]=\displaystyle \left[\begin{matrix} I_d & * & \\ \mathbf 0 & *\end{matrix}\right]$

where $N\neq \mathbf 0$ is strictly upper triangular.

Thus
$I_d = (I_d + N)^r = I_d + \binom{r}{1}N^1+ \binom{r}{2}N^2+....+ \binom{r}{r-1}N^{r-1}+ \binom{r}{r}N^{r}$

$\implies \mathbf 0 = \binom{r}{1}N^1+ \binom{r}{2}N^2+....+ \binom{r}{r}N^{r}$

but nilpotent $N\neq \mathbf 0$ and its powers are linearly dependent, a contradiction.


问:
Making sense of Wikipedia derivation of the invariance of the spacetime interval
In this page there is a reproduction of an argument (I believe originally due to Landau?) that the interval between two events in spacetime is independent of the observer. I'm trying to make more precise mathematical sense of this argument. Here's how I interpret it: the first observer uses his coordinates $(t, x, y, z)$ to write down a signature (1, 3) metric on spacetime $\eta_1 = dt^2 - dx^2 - dy^2 - dz^2$. Similarly, the second observer uses his coordinates $(t', x', y', z')$ to write down the metric $\eta_2 = dt'^2 - dx'^2 - dy'^2 - dz'^2$.

At each point $p$ of spacetime $M^4$, the following condition on a tangent vector $v_p \in T_p M$ should hold:
$$ \eta_1(v_p, v_p) = 0 \iff \eta_2(v_p, v_p) = 0 \ \ \ \ \ \ \ \ (*)$$
This just says that observers should agree on which vectors are light-like, i.e. what are the possible velocities of light.

It is then claimed that there must be a proportionality $\eta_1 = a \eta_2$ (a priori possibly depending on p, but in reality not, due to homogeneity assumptions). So this leads me to the question: if $\eta_1, \eta_2$ are two linear signature (1, 3) metrics on a 4-dimensional real vector space which satisfy condition $(*)$, must there be a real constant $a$ such that $\eta_1 = a \eta_2$?

答:
In The Structure of Space-Time Transformations, Borchers & Hegerfeldt show that any self-bijection of Minkowski space (of dimension $\geq 3$) which maps light-cones to light-cones must, up to a scaling factor, be an affine Lorentz transformation.

In my question, letting without loss of generality $\eta_1$ be the standard Minkowski metric in Minkowski 4-space $M^4$, there is an isometry $T : (M^4, \eta_1) \to (M^4, \eta_2)$ since they are metrics with the same signature. The hypothesis $(*)$ of the question guarantees that $T$ maps light-cones to light-cones, and so T must be a scaling of a Lorentz transformation, which means $\eta_2$ must be a scaling of $\eta_1$.


问:
Translating "All humans have a father" into propositional logic
Translating:

All humans have a father

Let $Hx$ be x is human and $Fx$ be x has a father. Then should it be $$\forall x(HxFx)$$? or $$\forall x(Hx\rightarrow Fx)$$?

答:
The second formula is ok.

The first one isn’t. It isn’t even a well formed first order logic formula.


问:
Product of test function and function in first order Sobolev space is also in first order Sobolev space

Suppose $1 \leq p \leq \infty$ , $\phi \in C^{\infty}_c (\Omega) $
and $u \in  W^{1,p}(\Omega)$ ,where $Ω$  is an open subset of
$\mathbb{R}^n$. Prove that $\phi u \in W^{1,p} (Ω)$  and $D_{x_i}(\phi u) = \phi D_{x_i} u + u D_{x_i} \phi $ , where $D_{x_i}$ denotes
the weak i-th partial derivative.

I am trying to understand why I need to show the product rule is true for $D_{x_i}(\phi u)$ when the question already states that $\phi$ is infinitely differentiable and $u$ has weak derivatives. Am I asked to show the product rule is valid for the function $\phi u$ or am I missing something? Thanks for any help.

Edit : Follow up question on how does the weak product rule imply $\phi u \in W^{1,p} (\Omega)$ ?

答:
Still, perhaps one should not shy away from seeing how easy the product rule is to prove: with test functions $\varphi,f$ and distribution $u$,
$$
(\varphi u)'(f) \;=\; - (\varphi u)(f') \;=\; -u(\varphi f')
\;=\; -u((\varphi f)'-\varphi'f)
\;=\; u'(\varphi f)+u(\varphi' f)
$$
$$
\;=\; (\varphi u')(f) + (\varphi' u)(f)
\;=\; (\varphi u'+\varphi'u)(f)
$$
Quite innocent! :)

And, of course, the same pattern holds for partial derivatives in several variables.


问:
Why is { : i ∈ I} a set according to Tao's replacement axiom? Terrence Tao Analysis 1 page 60
Replacement axiom. Let  be a set. For any object ∈, and any object , suppose we have a statement (,) pertaining to  and , such that for each ∈ there is at most one  for which (,) is true. Then there exists a set {: (,) is true for some ∈}, such that for any object ,  ∈{: (,) is true for some ∈} ⇔ (,) is true for some ∈.

Using this axiom why is the set { : i ∈ I} a set? I am not sure how to use Tao's axiom of replacement to argue that this is indeed a set, would it be stated as something like:

let I be a set and let P(i, Y) be the statement that for each i ∈ I let Y =  where  is a set?

But in order to use this axiom we need there to be at most one  for each i ∈ I.

couldnt each i ∈ I also get mapped to another set say Z, so the axiom would fail to imply the above is a set?

i am confused!

答:
The purpose of the statement $P(x, y)$ is to associate members of $A$, labeled $x$, with members of $Y$, labeled $y$.

So for the set $\{X_i : i \in I\}$, our "source" set is $I$, whose elements are labeled as $i$, while our "destination" set is $Y$, whose elements are $X_i$ (which don't have to be sets). That means that our statement would be $P(i, X_i)$, since that pairs the elements of $I$ with the corresponding elements of $Y$.

The replacement axiom asks only that our statement has at most one $y$ for each $x$ - or in our case, at most one $X_i$ for each $i$. For example, if my set $I$ were the numbers $1, 2, 3, 4$, then my statement $P(i, X_i)$:

Can only be true for each value of $i$ once - I can't have $P(1, A)$ and $P(1, B)$ where $A \neq B$. (This is one side of the "at most one" part.)

Can skip values of $i$ - I don't have to have any statements of the form $P(3, C)$. (This is the other side of the "at most one" part.)

Can repeat values of $X_i$ for different values of $i$ - I can have $P(2, D)$ and $P(4, D)$. (This comes from the fact that there is actually no restriction on how often $y$ appears in the axiom.)

Also importantly, the Replacement axiom just asks that such a statement exists. It doesn't require that the statement be unique. So when you ask "couldn't each $i$ get mapped to another set" then the answer is "not using the same statement". The point of the axiom is that you meet the conditions however you like, and the conclusion is then considered a foregone result. If you have a statement $P$ that doesn't satisfy the whole "at most one $y$ for each $x$" part, you can't apply the axiom. If you have two statements, $P$ and $P'$ that both satisfy the conditions, then you can pick either one for the axiom, and this is essentially re-indexing the set.


问:
Prove that $\sum_{r=0}^{n} r \binom{n}{r}^2 = \frac{n}{2} \binom {2n}{n}$
My attempt
$$\sum_{r=0}^{n} r \times \frac nr \left[\binom {n-1}{r-1} \right]^2$$
$$=n\sum _{r=0}^{n} \left [\binom {n-1}{r-1}\right]^2$$

The square on the binomial coefficient is throwing me off, otherwise it would have been easier to proceed. I don’t know what to do with that square.

Edit: As pointed out in the comment, i managed to find out $$\sum_{r=0}^n (\binom nr )^2 = \binom {2n}{n}$$

But using this in the above expression isn’t giving me the right answer. Am I supposed to make a modification to the limits in the sigma function?

答:
Symmetry is the key. We obtain
\begin{align*}
\sum_{r=0}^nr\binom{n}{r}^2&=\sum_{r=0}^n(n-r)\binom{n}{r}^2\tag{1}\\
&=n\sum_{r=0}^n\binom{n}{r}^2-\sum_{r=0}^nr\binom{n}{r}^2\\
\end{align*}
where we changed in the right-hand sum of (1) the order of summation $r\to n-r$.
We already know according to OPs calculation:
\begin{align*}
\sum_{r=0}^n\binom{n}{r}^2=\binom{2n}{n}\tag{2}
\end{align*}

From (1) and (2) we obtain
\begin{align*}
\color{blue}{\sum_{r=0}^nr\binom{n}{r}^2}&=\frac{n}{2}\sum_{r=0}^n\binom{n}{r}^2
\color{blue}{=\frac{n}{2}\binom{2n}{n}}
\end{align*}
and the claim follows.


问:
Show that $n \ge \sqrt{n+1}+\sqrt{n}$
(how) Can I show that:

$n \ge \sqrt{n+1}+\sqrt{n}$ ?

It should be true for all $n \ge 5$.

Tried it via induction:

$n=5$: $5 \ge \sqrt{5} + \sqrt{6} $ is true.

$n\implies n+1$:
I need to show that $n+1 \ge \sqrt{n+1} + \sqrt{n+2}$

Starting with $n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 $ .. (now??)

Is this the right way?

答:
To add to your step, observe the following:

$$\sqrt{n}+1 = \sqrt{(\sqrt{n}+1)^2} = \sqrt{n+1+2\sqrt{n}} > \sqrt{n+1+1} = \sqrt{n+2}.$$

The "$>$" part comes from your assumption $n \ge 5$, so $2\sqrt{n} \ge 2\sqrt{5} > 1$. Now, we have:

$$n+1 \ge \sqrt{n} + \sqrt{n+1} + 1 = \sqrt{n+1} + (\sqrt{n}+1) > \sqrt{n+1} + \sqrt{n+2}.$$

Title: Show that $n \ge \sqrt{n+1}+\sqrt{n}$


问:
Getting the sequence $\{1, 0, -1, 0, 1, 0, -1, 0, \ldots\}$ without trig?
What's the simplest way to write a function that outputs the sequence:

```<code>{1, 0, -1, 0, 1, 0, -1, 0, ...}
</code>```

... without using any trig functions?

I was able to come up with a very complex sequence involving -1 to some complicated formula, but I was hoping there is a more simple solution.

$n$ should start at 0 and go to infinity.

Update:

All the solutions you guys provided are great! I wasn't aware there were so many of them. I should have mentioned that I prefer a solution which doesn't use recursion; imaginary numbers; matrices; functions with if statements; or functions such as ceil, floor, or mod. I'm looking for something using basic algebra: addition/subtraction, multiplication/division, exponents, etc. However, I will accept anything since I didn't  include this clause originally.

This is what I came up with:

$$a_n=\frac{\left(-1\right)^n+1}{2}\cdot \left(-1\right)^{\left(\frac{n}{2}-\frac{\left(-1\right)^{n+1}+1}{4}\right)}$$

Is there a less complicated way (i.e. fewer terms) to get this same sequence?

答:
In binary (twos compliment), $a_n$ is the following two-bit number:
$$(n_1 \cdot\bar{n}_0)  \ \ \ \  (\bar{n}_0)$$

where $n_1$ and $n_0$ are the two least significant bits of $n$, $\bar{x}$ denotes "NOT $x$" and $\cdot$ denotes the boolean "AND" operator.

That's how it would be done in some picoseconds using a handful of transistors. This would be written as a function in a hardware description language, which is then synthesized into digital hardware.


问:
Induction problem
I need to prove using induction that
$$
\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3} + \cdots +\frac{1}{n(n+1)} >\frac{9n-1}{10(n+1)} \ ,\ for \ \ n \in \mathbb{N}
$$
I know how to prove it without induction, but when I try using induction I get 0>0 which is false.

答: I think your problem is that the RHS of your induction step should say only $$\frac{9(k+1) - 1}{10(k+2)}$$ because that is the substitution you get from the RHS of the original statement with $n=k+1$, whereas what you have is the RHS with $n=k$ and then also the next term of the series, but the series was from the LHS to begin with.

Title: Induction problem


问:
A subset of an infinite-dimensional simplex where no element is in the convex hull of other elements of the subset.
Let

$$S=\{x\in R_{+}^\infty \mid \sum_{n=1}^\infty x_n=1\}$$

be an infinite simplex.

Suppose $\hat{S}$ is a countable subset of $S$ such that

for each $x\in \hat{S}$ only at most finitely many coordinates are non-zero, denote the set of those coordinates $N(x)$;

for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\neq N(x’)$.

How could one prove that no $x$ is in a convex hull of other elements of $\hat{S}$, i.e., no $x\in \hat{S}$ can be represented as a convex combination of other elements of $\hat{S}$?

EDIT: Condition (2) should be read: "for any $x\neq x’$, $x, x’ \in \hat{S}$, $N(x)\triangle (\cup_{x'\in \hat{S}}N(x’)) \neq \emptyset$". Here $\triangle$ stands for a symmetric difference of sets.

Thank you! (These two infinities intimidate me...)

答: No countable set is convex [unless it is empty] (since in a convex set, an interval between two points is contained in the set, and an interval has uncountably many points. However, your statement after the i.e. has nothing to do with convexity.</p>


问:
Proving that $f(x)=\frac{\sin(\frac{1}{x})}{\frac{\pi}{2}-\arctan x}$ is not uniformly continuous in $(0,1)$
I have the following function: $f(x)=\frac{\sin(\frac{1}{x})}{\frac{\pi}{2}-\arctan x}$. I'm trying to prove that it's not uniformly continuous in $(0,1)$. As I understand, I need to show that $\lim_{x\to 0^+} f(x)$ does not exist. But I'm struggling to show it. I tried to use some sequences in order to get to some different answers but it looks to difficult to find the limit. How can I do it?

答:
$\textbf{Hint}$: No matter what you do the denominator will always approach $\frac{\pi}{2}$ so choose subsequences so that

$$\frac{1}{x} = \pm\frac{\pi}{2}+2\pi n$$

then conclude the limit cannot exist.


答: Since $\frac{9k-1}{10(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{9k+8}{10(k+2)}$, the inductive step is simple enough. Your problem was an incorrect guess as to how to "prove it backwards". We all prove things backwards: we work out what would be enough to finish the step. In your case, instead of guessing the above $=$, you tried a $>$, which doesn't work.

Title: Induction problem

Body:
I need to prove using induction that
$$
\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3} + \cdots +\frac{1}{n(n+1)} >\frac{9n-1}{10(n+1)} \ ,\ for \ \ n \in \mathbb{N}
$$
I know how to prove it without induction, but when I try using induction I get 0>0 which is false.


问:
$f(z)$ goes to $\infty$ as $z$ goes to $z_0$, which is an isolated singularity. Show that $f(x)$ has a pole at $z_0$.
I guess I need to show that $z_0$ cannot be a removable singularity or an essential singularity.

For the case of removable singularity, can I prove it by this method: Since $z_0$ is removable, when $f(x)$ is expanded into Laurent expansion, only the analytic part ($k \ge 0$ ones) remains, so $f(x)$ cannot go to infinity as $z$ goes to $z_0$.

And I have no idea on how to prove $z_0$ cannot be an essential singularity.

Thanks for the help.

答: Look at the Casorati-Weierstrass Theorem.</p>


问:
Numbers in $-A - (-B)$ form for $ 2's$ Complement
I know for $2's$ complement representation, you are supposed to perform an operation like $A - B$ into addition but perform $2's$ complement on $B$.

So $0101 - 1001 = 0101 + 0111$

Now, if I had a question in the form of $-A - (-B)$, would I combine the signs and it would become $-A + B$, or would I do $2's$ complement on $-B$ and get $-A - B$?

答:
Having $n$ digits in $2$s complement representation we define complement of number $A$, between $1$ and $2^n-1$, by formula $-A=2^n-A$.

So $-A - (-B)$ is same with $-A +B$:
$$-A - (-B) = 2^n-A +(2^n- (2^n-B)) = 2^n-A+B=-A +B$$


问:
Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$
Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.

and please don't use any $\sum_{cyc}$ because I get confused with it.

答:
Here's the inevitable 'expand and see if everything works out' solution.
$$
\frac{a^2}{a^2 + bc} + \frac{b^2}{b^2 + ac}+\frac{c^2}{c^2 + ab} < 2 \\
\iff a^2 (b^2 + ac)(c^2 + ab) + b^2 (a^2 + bc)(c^2 + ab) + c^2(a^2 + bc)(b^2 + ac) < 2 (a^2 + bc)(b^2 + ac)(c^2 + ab) \\
\iff 3a^2b^2c^2 + 2(a^3b^3 + b^3c^3 + c^3a^3) + a^4 bc + b^4 ac + c^4ab < 2(2a^2b^2c^2 + a^3 b^3 + a^3c^3 + b^3 c^3 + a^4 bc + b^4ac + c^4 ab)
$$
It's important when doing these sort of calculations to make sure you didn't miss a term. Here the number of terms on the left (before summing them up) is $12 = 3 \times 2^2$ and the number of terms on the right is $8 = 2^3$ as expected.

This inequality simplifies to
$$
a^2b^2c^2 +a^4 bc + b^4ac + c^4 ab > 0
$$
Which is evidently true.


问:
Determination of Fourier coefficient (Euler's Formulae)
Is the integral from -π to π the same with integral from 0 to 2π?</p>

答:
We generally use Fourier series decompositions for functions that are periodic, or that are defined on a bounded interval such that you can extend them into a periodic function by making copies of that interval over and over again.

The formula for finding Fourier coefficients always assumes you've scaled the sine and cosine functions to match the period of the function (or the width of the interval), but otherwise it doesn't matter how you take the integral because it's periodic - as long as the integral covers a whole period, you can chop it up and shift it in every direction.

For the case where the period is $2\pi$, then our function $f$ will always follow $f(x) = f(x + 2k\pi)$ for any integer $k$. Thus, for example,

$\begin{eqnarray}a_1 & = & \int_{-\pi}^\pi f(x) \sin x\ dx \\
& = & \int_{-\pi}^0 f(x) \sin x\ dx + \int_0^\pi f(x) \sin x\ dx \\
& = & \int_\pi^{2\pi} f(u - 2\pi) \sin (u - 2\pi)\ du + \int_0^\pi f(x) \sin x\ dx & \mbox{(where } u = x + 2\pi \mbox{)} \\
& = & \int_\pi^{2\pi} f(u) \sin u \ du + \int_0^\pi f(x) \sin x \ dx \\
& = & \int_\pi^{2\pi} f(x) \sin x \ dx + \int_0^\pi f(x) \sin x \ dx \\
& = & \int_0^\pi f(x) \sin x \ dx + \int_\pi^{2\pi} f(x) \sin x \ dx \\
& = & \int_0^{2\pi} f(x) \sin x \ dx
\end{eqnarray}$

And similarly for the other coefficients, so it doesn't actually matter where the interval is located. That means that the choice of whether to work on $[-\pi, \pi]$ or $[0, 2\pi]$ is generally just a matter of convention - sometimes it's nice to have the symmetry, sometimes it's easier to work with things that start at 0 (especially if you're using it for differential equations with boundary conditions, since it's quite common for 0 to be one of those boundaries).


问:
How do I estimate log10 of log10 of 8 billionth element of A000670?
How do I estimate log10 of log10 of the 8 billionth element of A000670
as an actual number? I believe it's between 10 and 100, but am not
sure why I believe this.

I realize http://oeis.org/A000670 offers some approximations of
A000670 (in terms of n), but nothing that quickly leads to an actual
value of log10(log10(a(n)))

Goal: I'm trying to calculate the number of possible "prejudices" in
the world, and it's the nth element of A000670, where n is the current
world population. I don't need an exact number, just something like
10^(10^r) [to the nearest integer value for r], and I'm estimating n at about 8 billion right now.

Ideally, I'd like a good general estimate of log10(log10(a(n))) but
that may be asking too much.

EDIT: It's easy to show A000670 is bounded by 2^n*n! (aka http://oeis.org/A000165), but this bound is loose and unhelpful.

http://oeis.org/A000670 notes "Unreduced denominators in convergent to log(2) = lim[n->inf, na(n-1)/a(n)].". Does this mean that: n!*log(2)^n would be an estimate, at least in terms of order of magnitude? It seems a bit low. Since log(2) < 1 this can't be right. Maybe

```<code>n!/(log(2)^n)
</code>```

答:
The OEIS comment

Unreduced denominators in convergent to log(2) = lim_{n->infinity} n*a(n-1)/a(n).

suggests that there could be an asymptotic approximation proportional to $\frac{n!}{\log(2)^n}$ - you seem to have accidentally put the $\log(2)$ in the numerator. This is indeed correct and you also say OEIS A000670 offers some approximations.  One is particularly good, and consistent with that earlier comment:

a(n) is asymptotic to (1/2)*n!*log_2(e)^(n+1), where log_2(e) = 1.442695... [Barthelemy80, Wilf90].

i.e. $\dfrac{ n! }{2\, \log_e(2)^{n+1}}$

For example, while the actual term for $n=10$ is $102247563$, the approximation gives $102247563.005271$, almost exactly the same, and the relative error gets smaller for larger $n$.

If you apply the asymptotic formula to $n=8 \times 10^9$ you get about $3.6\times 10^{77023760357}$ which has a base-$10$ logarithm of about $77023760357.557 \approx 7.7\times 10^{10}$ and the base-$10$ logarithm of that base-$10$ logarithm is then about $10.8866247$

If you apply your upper bound of $2^n\, n!$ to $n=8 \times 10^9$ you get about $2.4\times 10^{78158604011}$ which has a base-$10$ logarithm of about $78158604011.372 \approx 7.8\times 10^{10}$ and the base-$10$ logarithm of that base-$10$ logarithm is then about $10.8929768$

There is an obvious lower bound of $n!$ (strict orderings) and applying that to $n=8 \times 10^9$ you get about $1.1\times 10^{75750364046}$ which has a base-$10$ logarithm of about $75750364046.060 \approx 7.6\times 10^{10}$ and the base-$10$ logarithm of that base-$10$ logarithm is then about $10.8793847$

So the answer is between $10^{10^{10}}$ and $10^{10^{11}}$, though you can be much more precise than that.


问:
Integral substitution to find constant $\lambda$
Whilst using the substitution $x = 4\sin^2\theta$, then show that this integral:

$$ \int^{3}_{0}\sqrt{\frac{x}{4-x}}dx $$

Is equivalent to:

$$ \int^{\frac{\pi}{3}}_{0}\sin^2\theta d\theta $$

What I have tried:

$$ \int^{3}_{0}\sqrt{\frac{x}{4-x}}dx=\left(\frac{x}{4-x}\right)^{\frac{1}{2}}=\left((4-x)^{\frac{1}{2}}\right)^{-x}=\int(4-x)^{-\frac{x}{2}}dx $$

Then substituting in $x = 4\sin^2\theta$

$$ \int(4-4\sin^2\theta)^{-2\sin^2\theta}=\int \frac{2\sin^2\theta}{4-4\sin^2\theta}d\theta=\int\frac{1}{2}-\frac{\sin^2\theta}{-2\sin^2\theta}d\theta $$

Have I got the right idea?

Second approach:

$\int \sqrt{\frac{sin^2\theta}{1-sin^2\theta}}\cdot8\sin\theta\cos\theta d\theta$

identity:
$1-sin^2\theta = cos^2\theta$

$\int \sqrt{\frac{sin^2\theta}{cos^2\theta}}=\sqrt{tan^2\theta}=\tan\theta$

Then using the this identity:
$\tan x=\frac{sinx}{cosx}$

$8\int\frac{\sin\theta\sin\theta\cos\theta}{\cos \theta}d\theta=8\int sin^2\theta d\theta$

However, how do I get the upper integral to go from $3$ to $\frac{\pi}{3}$ likeso:
$\int^{3}_{0} \implies\int^{\frac{\pi}{3}}_{0}$?

答:
Hint: It's much easier to make the substitution before integrating - and, since you want to show two integrals are equivalent, you may not even need to integrate at all. Try plugging in $x = 4\sin^2\theta$ first:

$$ \sqrt{\frac{x}{4-x}} = \sqrt{\frac{4\sin^2\theta}{4-4\sin^2\theta}} , $$

then simplify that expression algebraically. Don't also forget to make the appropriate substitutions for $dx$ and for the limits of the integral too!


问:
Integrating $\frac1{2\sigma\sqrt{2\pi}}\int_{-\infty}^\infty(1+erf(\frac{z-\mu_i}{\sigma_i\sqrt2}))\exp( -\frac12(\frac{z-\mu}\sigma)^2)dz$
I am interested in the integral of
$$\frac{1}{2\sigma \sqrt{2\pi}}\int_{-\infty}^\infty \left( 1+\operatorname{erf}\left(\frac{z-\mu_i}{\sigma_i\sqrt{2}}\right)\right)\exp\left( -\frac{1}{2}\left( \frac{z-\mu}{\sigma}\right)^2\right)dz$$

where $\operatorname{erf}$ is the error function defined as the CDF of the standard normal (up to a constant).

In this case $\sigma, \mu, \mu_i, \sigma_i$ are constants.

I truly have no clue how to even start something like this. Is Mathematica even capable of this type of integral?

For those curious, this is evaluated when considering $Pr(f_i \leq f \leq f_j)$ where $f_j, f_i, f$ follow a normal distribution with different parameters. The full integral is

$$\frac{1}{2\sigma \sqrt{2\pi}}\int_{-\infty}^\infty \left( 1+erf\left(\frac{z-\mu_i}{\sigma_i\sqrt{2}}\right)\right)\exp\left( -\frac{1}{2}\left( \frac{z-\mu}{\sigma}\right)^2\right)dz \\
  - \frac{1}{4 \sigma \sqrt{2 \pi}} \int_{-\infty}^\infty \left( 1+erf \left(\frac{z-\mu_i}{\sigma_i \sqrt{2}}\right)\right)\left(1+erf\left(\frac{z-\mu_j}{\sigma_j \sqrt{2}}\right)\right)\exp\left(-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^2\right)dz$$

But I thought that might be inappropriate ;)

答:
Integral of one error function and Gaussian

Mathematica does not evaluate the integral. Surprisingly, there is an exact result. By changing variables, $z-\mu=x$, your integral of $\operatorname{erf} \times \text{Gaussian}$ may be written as

$$
I(b)=\int\limits_{-\infty}^\infty dx \ \operatorname{erf}(ax+b) \exp \left(-cx^2 \right)
$$

Since $\partial_t \operatorname{erf}(t)= \frac{2}{\sqrt{\pi}}e^{-t^2}$, differentiating with respect to $b$ produces

$$
\partial_b I(b)=\int\limits_{-\infty}^\infty dx \ \frac{2}{\sqrt{\pi}} \exp\left( -(ax+b)^2-cx^2 \right)
$$

This is just a Gaussian integral over $x$, the result is

$$
\partial_b I(b)= \frac{2}{\sqrt{a^2+c}} \exp \left( -\frac{b^2 c}{a^2 +c} \right)
$$

Now we integrate to recover $I(b)$

$$
I(b)-I(0)=\int\limits_0^b \ dt \frac{2}{\sqrt{a^2+c}} \exp \left( - t^2\frac{ c}{a^2 +c} \right)
$$

$I(0)=\int dx \ ( \text{odd function})=0$, and the integral on the right is simply the definition of $\operatorname{erf}$, so we have

$$
I(b)=\sqrt{\frac{\pi}{c}} \operatorname{erf} \left( b \sqrt{\frac{c}{a^2+c}} \right)
$$

$$
\tag*{$\blacksquare$}
$$

Integral of two error functions and Gaussian

We may use a similar method to find that

$$
\int\limits_{-\infty}^\infty dx \ \operatorname{erf}(a_1 x) \operatorname{erf}(a_2 x) e^{-cx^2} = \frac{2}{\sqrt{c\pi}} \tan^{-1}\left( \frac{a_1 a_2}{ \sqrt{c(a_1^2+a_2^2+c)}} \right)
$$

However, once we introduce $b\neq 0$, we meet the more complicated integral in an intermediate step

$$
\int\limits_0^b dt \ \operatorname{erf}(\alpha t + \beta) e^{-\gamma t^2}
$$

Update! Using the result from this post, we can make progress. Let

$$
J(\beta)=\int\limits_{-\infty}^\infty dx \ \operatorname{erfc}(\alpha x+\beta) \operatorname{erf}(a x+b) \exp \left(-cx^2 \right)
$$

Then $J(\infty)=0$. Differentiating

$$
\partial_{\beta} J(\beta)=-\frac{2}{\sqrt{\pi}}  \int\limits_{-\infty}^\infty dx \  \operatorname{erf}(a x+b) \exp \left(-cx^2 - (\alpha x +\beta)^2 \right)
$$

Which we can evaluate using our first result

$$
\partial_{\beta} J(\beta)=-\frac{2}{\sqrt{\pi}} \sqrt{\frac{\pi}{c+\alpha^2}} \operatorname{erf}\left( \left(b-\frac{a \alpha \beta}{c+\alpha^2} \right) \sqrt{\frac{c+\alpha^2}{c+\alpha^2+a^2}} \right) \exp \left( -\beta^2 \left( 1+\frac{\alpha^2}{c+\alpha^2}\right) \right)
$$

With $\Lambda=\frac{2}{\sqrt{\pi}} \sqrt{\frac{\pi}{c+\alpha^2}}$, $A=-\frac{a \alpha }{c+\alpha^2}  \sqrt{\frac{c+\alpha^2}{c+\alpha^2+a^2}}$, $B=b\sqrt{\frac{c+\alpha^2}{c+\alpha^2+a^2}}$, and $C=\left( 1-\frac{\alpha^2}{c+\alpha^2}\right)$. Now we integrate

$$
J(\infty)-J(\beta)=-\int\limits_\beta^\infty \ dt \Lambda \operatorname{erf}(At+B)e^{-Ct^2}
$$

$$
J(\beta)=\int\limits_\beta^\infty dt \ \Lambda \operatorname{erf}(At+B)e^{-Ct^2}= \frac{\Lambda}{\sqrt{2C}}  \int\limits_{\sqrt{2C}\beta}^\infty du \ \operatorname{erf}(Au/\sqrt{2C}+B)e^{-u^2/2}
$$

Which may be written in terms of the 'Generalized Owen-T' function

$$
J(\beta)=\frac{2 \sqrt{\pi} \Lambda}{\sqrt{C}}T\left( \sqrt{2C}\beta, A/\sqrt{C},\sqrt{2}B\right)
$$

Using identity ii we have

$$
J(\beta)=\frac{2 \sqrt{\pi} \Lambda}{\sqrt{C}} \left[ \frac{1}{2\pi} \left( \tan^{-1}(A/\sqrt{C})  -\tan^{-1}(A/\sqrt{C} + B/\beta\sqrt{C}) \\ -\tan^{-1}(B^{-1}(\beta \sqrt{C} +AB/\sqrt{C} + A^2 \beta /\sqrt{C} )) \right)  +\frac{1}{4} \operatorname{erf}(B (1+A^2/C)^{-1/2}) \\ +T(\beta \sqrt{2C},(A\beta^{-1}+B)/\beta \sqrt{C})+T(\sqrt{2}B (1+A^2/C)^{-1/2}, B^{-1}(\beta \sqrt{C} +AB/\sqrt{C} + A^2 \beta /\sqrt{C}  ) \right]
$$

Where $T$ is the Owen T function. I've checked it numerically, I hope there are no typos here.

Approximate integral of two or more error functions with Gaussian

We can approximate the integral using Laplace's method. I'll demonstrate with two error functions, but in principle you could have more. Let

$$
I(c)=\int\limits_{-\infty}^\infty dx \ \operatorname{erf}(a_1 x+ b_1) \operatorname{erf}(a_2 x+ b_2) e^{-cx^2}
$$

Expand the product of $\operatorname{erf}$s around the Gaussian peak: $x=0$

$$
\operatorname{erf}(a_1 x+ b_1) \operatorname{erf}(a_2 x+ b_2) = \operatorname{erf}( b_1) \operatorname{erf}( b_2) + x \left( \text{junk} \right)  + x^2 \frac{2 e^{-b_1^2-b_2^2}}{\pi} \left( 2 a_1 a_2 -a_2^2 b_2 e^{b_1^2} \sqrt{\pi} \operatorname{erf}(b_1) - a_1^2b_1 e^{b_2^2} \sqrt{\pi} \operatorname{erf}(b_2) \right) + \mathcal{O}(x^3)
$$

Term by term integration produces the asymptotic series. The $(\text{junk})$ and all $x^{\text{odd}}$ terms integrate to zero. Noting that

$$
\int\limits_{-\infty}^\infty dx \ x^{2n} e^{-cx^2} = c^{-n-1/2} \Gamma(n+1/2)
$$

For fixed $a_i, \ b_i$ we have

$$
I(c) \sim \sqrt{\frac{\pi}{c}}  \operatorname{erf}( b_1) \operatorname{erf}( b_2) \\ +\frac{1}{c^{3/2}}  \frac{ e^{-b_1^2-b_2^2}}{\sqrt{\pi}} \left( 2 a_1 a_2 -a_2^2 b_2 e^{b_1^2} \sqrt{\pi} \operatorname{erf}(b_1) - a_1^2b_1 e^{b_2^2} \sqrt{\pi} \operatorname{erf}(b_2) \right) \ \ , \ \ c \to \infty
$$

Which is the two term approximation.


问:
Equivalence of definitions of $H^1_0(\Omega)$
I would like to prove that the following definitions of $H^1_0(\Omega)$ are equivalent. In what follows $\Omega$ is supposed to be a bounded domain with smooth boundary.</p>

<p>$$1) \qquad H^1_0(\Omega) = \overline{C_0^{\infty}}^{\|\cdot\|_{H^1}}$$
$$2) \qquad H_0^1(\Omega) = \{u \in H^1(\Omega)\ :\ u_{|_{\partial \Omega}} = 0 \}$$
In the second definition $u_{|_{\partial \Omega}} = 0$ means that $Tu = 0$, where $T$ is the trace operator.
Can anyone help me? </p>

答:
Let
$$ A = \overline{C_0^{\infty}}^{\|\cdot\|_{H^1}}$$
and
$$B = \{u \in H^1(\Omega)\ :\ u_{|_{\partial \Omega}} = 0 \}.$$
We also know $H^1(\Omega) = \overline{C^\infty}^{\|\cdot\|_{H^1}}$. Then, $A \subset B$ is apparent, since $\varphi|_{\partial\Omega} = 0$ for all $\varphi \in C_0^\infty(\Omega)$ and since the trace operator is continuous.

For $B \subset A$ let $v \in B$ be given. Then, there is a $\varphi \in C^\infty(\Omega)$ with $\|v - \varphi\|_{H^1} \le \varepsilon$. Finally, you can take a smooth cut-off function $\psi \in C_0^\infty(\Omega)$, such that $\| v - \varphi\,\psi\|_{H^1} \le 2 \, \varepsilon$.


答:
First of all calculate the differential: $\mathrm{d}x = 8\sin\theta\cos\theta\,\mathrm{d}\theta$. Your boundaries are right.
The integral is
$$\int\limits_{0}^{\pi/3}\sqrt{\frac{4\sin^2\theta}{4(1-\sin^2\theta)}}{8\sin\theta\cos\theta}\,{\mathrm{d}\theta} = {8}\int\limits_{0}^{\pi/3}{\sin^2\theta}\,{\mathrm{d}\theta} = 8{\sqrt3},$$
since $\cos\theta= \sqrt{1-\sin^2\theta}$.

Edit 1: In your second approach you asked how to get to the boundaries. You want to substitute $x=4\sin^2\theta$, which means that $\theta=\arcsin\sqrt{x/4}$. Putting in your earlier values for $x$ in this formula gives you your new boundaries.

Edit 2: Solve for $\theta$:
$$\begin{align}
x&=4\sin^2\theta\quad\mid  :4 \\\\
\frac{x}{4}&=\sin^2\theta\quad\mid  \sqrt{} \\\\
\sqrt{\frac{x}{4}}&=\sin\theta\quad\mid  \sqrt{} \\\\
\sqrt{\frac{x}{4}}&=\sin\theta\quad\mid  \arcsin{} \\\\
\arcsin\sqrt{\frac{x}{4}}&=\theta
\end{align}$$
Now you can set $x=0$ and $x=3$. That should give you $\theta=0$ and $\theta=\pi/3$.

Title: Integral substitution to find constant $\lambda$

Body:
Whilst using the substitution $x = 4\sin^2\theta$, then show that this integral:

$$ \int^{3}_{0}\sqrt{\frac{x}{4-x}}dx $$

Is equivalent to:

$$ \int^{\frac{\pi}{3}}_{0}\sin^2\theta d\theta $$

What I have tried:

$$ \int^{3}_{0}\sqrt{\frac{x}{4-x}}dx=\left(\frac{x}{4-x}\right)^{\frac{1}{2}}=\left((4-x)^{\frac{1}{2}}\right)^{-x}=\int(4-x)^{-\frac{x}{2}}dx $$

Then substituting in $x = 4\sin^2\theta$

$$ \int(4-4\sin^2\theta)^{-2\sin^2\theta}=\int \frac{2\sin^2\theta}{4-4\sin^2\theta}d\theta=\int\frac{1}{2}-\frac{\sin^2\theta}{-2\sin^2\theta}d\theta $$

Have I got the right idea?

Second approach:

$\int \sqrt{\frac{sin^2\theta}{1-sin^2\theta}}\cdot8\sin\theta\cos\theta d\theta$

identity:
$1-sin^2\theta = cos^2\theta$

$\int \sqrt{\frac{sin^2\theta}{cos^2\theta}}=\sqrt{tan^2\theta}=\tan\theta$

Then using the this identity:
$\tan x=\frac{sinx}{cosx}$

$8\int\frac{\sin\theta\sin\theta\cos\theta}{\cos \theta}d\theta=8\int sin^2\theta d\theta$

However, how do I get the upper integral to go from $3$ to $\frac{\pi}{3}$ likeso:
$\int^{3}_{0} \implies\int^{\frac{\pi}{3}}_{0}$?


问:
how many shuffles are there in a deck of cards?
I need to solve this question:

How many shuffles are there of a deck of cards (52 cards), such that A$\heartsuit$ is not directly on top of K$\heartsuit$, and A$\spadesuit$ is not directly on top of K$\spadesuit$?

My solution:

For the shuffles' quantity, I'll get the total of shuffles of a deck and subtract from the number of shuffles that A$\heartsuit$ is on the top of K$\heartsuit$ and A$\spadesuit$ is on the top of K$\spadesuit$.

With this, we have:

$$52! - 48!$$

is my solution correct?

答:
There are $52!$ shuffles in total.

There are $50!$ shuffles where the top hearts are on top of each other and the top spades too (merge them to create $50$ "combo/solo cards")

There are $51!$ of the type where only one of these pairs is on guaranteed to be on top of each other, once for spades, once for hearts.

So there are $51! + 51! - 50!$ (for the double counting where both are on top of each other) "forbidden" shuffles, and subtract that number from $52!$ to get the one you want.

So $$52! - 2\cdot 51! + 50!$$ as the final result.


问:
If $R$ is a Noetherian ring and $M$ is a maximal ideal in $R[X]$, is $M \cap R$ a maximal ideal of $R$?
Theorem B on the first page of this paper states the following : If $R$ is a Noetherian ring and $M$ is a maximal ideal in $R[X_1,X_2,...,X_n]$, then $M \cap R$ is a prime ideal of $R$.

With this, as every prime ideal of a ring is not maximal, I am wondering if there is a counterexample to the following : If $R$ is a Noetherian ring and $M$ is a maximal ideal of $R[X]$, then $M \cap R$ is a maximal ideal of $R$. This claim has been speculated to be true by the instructor in my algebraic geometry course.

Thank you !

答:
For a counterexample, take $R=\mathbb{Z}_{(p)}$ and $M=(pX-1)$.  Then $R[X]/M\cong R[1/p]=\mathbb{Q}$ is a field so $M$ is maximal, but $M\cap R=0$ is not maximal in $R$.

A commutative ring $R$ with the property that every maximal ideal in $R[X]$ has maximal intersection with $R$ is called a Jacobson ring.  This property has many equivalent formulations; for instance, it is equivalent to saying that every prime ideal in $R$ is an intersection of maximal ideals.  A general version of the Nullstellensatz says that any finitely generated algebra over a Jacobson ring is Jacobson.  It follows that if $R$ is Jacobson, then any maximal ideal in any finitely generated $R$-algebra pulls back to a maximal ideal in $R$.  (In the case that $R=k[X]$ for an algebraically closed field $k$, this implies the classical Nullstellensatz over $k$.)


问:
Proof of $\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}=\frac{17\pi^4}{3240}$
Recently, I was able to prove that

$$\sum_{n=1}^\infty \frac{1}{n \binom{2n}{n}}= \frac{\pi}{3\sqrt{3}}$$

$$\sum_{n=1}^\infty \frac{1}{n^2 \binom{2n}{n}}= \frac{\pi^2}{18}$$

But does anybody know how to prove that $\displaystyle\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}= \frac{17}{36}\zeta(4)$?

I obtained the formula

$$\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3 \binom{2n}{n}}= 4\int_0^x \frac{(\arcsin t)^2}{t}dt$$

I am trying to divide by $x$ and integrate both sides, but so far I have been unsuccessful.

答:
Here is my solution:

Step 1. Reduction to an integral representation

Let $S$ denote the summation in question. Then by the successive application of integration by parts, we obtain

\begin{align*}
S &= 8 \int_{0}^{1} \frac{1}{x} \int_{0}^{x/2} \frac{\arcsin^{2} t}{t} \, dt dx
 = -8\int_{0}^{1} \frac{\arcsin^{2} (x/2)}{x} \log x \, dx \\
&= 8 \int_{0}^{1} \frac{\arcsin (x/2)}{\sqrt{1 - (x/2)^{2}}} \log^{2} x \, \frac{dx}{2}.
\end{align*}

Thus with the substitution $x = 2 \sin\theta$, we have

$$ S = 8 \int_{0}^{\frac{\pi}{6}} \theta \log^{2} (2 \sin\theta) \, d\theta. $$

To evaluate this integral, note that for $0 < \theta < \frac{\pi}{6}$ we have

$$ e^{i\theta} \cdot 2 \sin \theta = i \cdot (1 - e^{2i\theta}). $$

Taking logarithm (with the branch cut $(-\infty, 0]$ as usual) to both sides, it follows that

$$ i\theta + \log (2\sin\theta) = \frac{i\pi}{2} + \log(1 - e^{2i\theta}). $$

Cubing both sides and integrating on $\left( 0, \frac{\pi}{6} \right)$ and taking imaginary parts only,

$$ S = \frac{2}{3} \left( \frac{\pi}{6} \right)^{4} + \frac{8}{3} \Im \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta. \tag{1} $$

Step 2. Some complex-analysis techniques

Now we focus on the integral in the imaginary part:

$$ I := \int_{0}^{\frac{\pi}{6}} \left( \frac{i\pi}{2} + \log(1 - e^{2i\theta}) \right)^{3} \, d\theta \tag{2}. $$

Once we evaluate the imaginary part of $I$, the identity $(1)$ immediately gives us the answer. We first make the substitution $z = 1 - e^{2i\theta}$ and $\omega = e^{-i\pi/3}$ to obtain

$$ I = \int_{0}^{\omega} \left( \frac{i\pi}{2} + \log z \right)^{3} \frac{dz}{2i(z-1)}. $$

Here, the path of integration is a circular arc joining from $0$ to $\omega$ centered at $1$ (green-colored path in the figure below).

But since the integrand is analytic for $0 < \Re z < 1$, we may change the path of integration as $z = \omega t$ for $0 \leq t \leq 1$ (blue-colored path in the figure above). This gives

$$ I = \frac{1}{2i} \int_{0}^{1} \left( \frac{i\pi}{6} + \log t \right)^{3} \frac{\omega \, dt}{\omega t - 1}. $$

Plugging $t = e^{-x}$, $I$ reduces to

\begin{align*}
I
&= \frac{1}{2} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} \frac{\omega e^{-x}}{1 - \omega e^{-x}} \, dx
 = \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \int_{0}^{\infty} \left( \frac{\pi}{6} + ix \right)^{3} e^{-nx} \, dx \\
&= \frac{1}{2} \sum_{n=1}^{\infty} \omega^{n} \left( -\frac{6 i}{n^4}-\frac{\pi }{n^3}+\frac{i \pi ^2}{12 n^2}+\frac{\pi ^3}{216 n} \right).
\end{align*}

Taking the imaginary part,

\begin{align*}
\Im I
&= -3 \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^4} + \frac{\pi}{2} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n^3} + \frac{\pi^2}{24} \sum_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^2} - \frac{\pi^3}{432} \sum_{n=1}^{\infty} \frac{\sin (n\pi/3)}{n}. \tag{3}
\end{align*}

Step 3. Evaluation of a series

Note that for $0 < \theta < \pi$, we have

$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n} = \Im \sum_{n=1}^{\infty} \frac{e^{in\theta}}{n} = - \Im \log(1 - e^{i\theta}) = \frac{\pi-\theta}{2}. $$

Integrating both sides, we obtain

$$ \sum_{n=1}^{\infty} \frac{1-\cos n\theta}{n^{2}} = \frac{\theta (2 \pi -\theta )}{4}
\quad \Longrightarrow \quad
\sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{2}} = \frac{\theta ^2}{4}-\frac{\pi  \theta }{2}+\frac{\pi ^2}{6}.$$

Repeating this procedure, we obtain

$$ \sum_{n=1}^{\infty} \frac{\sin n\theta}{n^{3}} = \frac{\theta ^3}{12}-\frac{\pi  \theta ^2}{4}+\frac{\pi ^2 \theta }{6}$$

and

$$ \sum_{n=1}^{\infty} \frac{\cos n\theta}{n^{4}} = -\frac{\theta ^4}{48}+\frac{\pi  \theta ^3}{12}-\frac{\pi ^2 \theta ^2}{12}+\frac{\pi^4}{90}.$$

Plugging $\theta = \frac{\pi}{3}$, we have

\begin{align*}
\sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{2}} &= \frac{\pi}{3} \\
\sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{2}} &= \frac{\pi^3}{36} \\
\sum_{n=1}^{\infty} \frac{\sin (n \pi / 3)}{n^{3}} &= \frac{5 \pi^3}{162} \\
\sum_{n=1}^{\infty} \frac{\cos (n \pi / 3)}{n^{4}} &= \frac{91 \pi ^4}{19440}
\end{align*}

Plugging these to $(3)$, we have

$$ \Im I = \frac{23 \pi^4}{12960} \quad \Longrightarrow \quad S = \frac{17 \pi^4}{3240} = \frac{17}{36}\zeta(4) $$

as desired.


问:
Problem involving the imaginary number i
You are given the equation c = (a + bi)^3 - 107i. You need to find the only values for a and b (if you are unfamilar with i it is the square root of -1). My orignal approach was to remove "i" and then just reached a point where I had an equation which isn't exactly intutive to find one solution to find:
2 = log(a^2 + b^2) ^3 + 107 / log c
derived because c^2 = (a^2 + b^2)^3 + 107  so log c^2 = log(a^2 + b^2)^3 + 107 then 2logc = log(a^2 + b^2)^3 + 107 and finally 2 = log(a^2 + b^2) ^3 + 107 / log c
please can someone helo as I haven't been taught this. The problem is from Mathcup and has a solution but no explanation as the users make the solutions(not the answers) https://www.mathcup.com/solutions?globalId=10455</p>

答:
Just finished editing my answer.

$c = a^3 - 3ab^2 + (3a^2b - b^3 - 107)i.$

Equating coefficients of imaginary parts:

$b(3a^2-b^2) = 107.$

$107\ $ is prime and $a,b>0,\ $ so either

$b = 1\ $ and $3a^2 - b^2 = 107 \implies 3a^2 = 108 \implies a = 6,\ $ or

$3a^2 - b^2 = 1\ $ and $\ b = 107 \implies a^2 = \frac{11450}{3} \implies a \notin \mathbb{N} \Rightarrow\Leftarrow. $

So it must be the first case: $b = 1\ $ and $\ a = 6,\ $ and so $\ c = 198.$


问:
Calculating $\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx$ without using complex numbers
One can find the antiderivate with help of the partial fraction method introducing complex numbers:
$$\frac{x\ln(x+1)}{x^2+1}=\frac{1}{2}\left(\frac{\ln(x+1)}{x-i}+\frac{\ln(x+1)}{x+i}\right).$$
The result is a complicated function containing dilogarithms and logarithms with complex arguments. The value of the definite integral turns out to be
$$\int_0^1 \frac{x\ln(x+1)}{x^2+1} dx = \frac{1}{96}(\pi^2+12\ln^2(2)).$$

Is there a way to find that value more directly - without introducing complex numbers?

答:
Rewrite the integral as

$$I=\int_0^1 \frac{x\log (x+1)}{x^2+1}dx =\int_0^1\int_0^1\frac{x^2\:dydx}{(yx+1)(x^2+1)}$$

A partial fraction and integration order reverse later we get

$$\int_0^1\int_0^1 \frac{xy}{(y^2+1)(x^2+1)}-\frac{1}{(y^2+1)(x^2+1)}   +\frac{1}{xy+1}-\frac{y^2}{(y^2+1)(xy+1)}\:dxdy = \int_0^1\frac{y}{2(y^2+1)}\log(2) -\frac{\pi}{4}\frac{1}{y^2+1}+ \frac{\log(1+y)}{y}-\frac{y\log(y+1)}{y^2+1}\:dy$$

The first two terms simplify easily and the last term is the integral we started with. The third term is given by Taylor series

$$\sum_{n=1}^\infty \frac{(-1)^n}{n}\int_0^1x^{n-1}dx = \sum_{n=1}^\infty \frac{(-1)^n}{n^2} =\frac{\pi^2}{12}$$

thus the equation simplifies to
$$\implies I = \frac{\log^22}{4}-\frac{\pi^2}{16}+\frac{\pi^2}{12}- I$$

therefore $I = \frac{1}{96}(\pi^2 +12\log^22)$


问:
Trigonometry - worm thread related word problem

The Question

The machine tool diagram shows a symmetric worm
thread, in which a circular roller of diameter 1.5 inches sits.
Find the amount d that the top of the roller rises above the
top of the thread, given the information in the diagram. (Hint:
Extend the slanted sides of the thread until they meet at a point.)

[

My Understanding

I found w and h using basic trig

For w

$$\sin15°= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{0.75}{w}$$   $$w = \frac{0.75}{\sin15°}$$ $$\sin15°= \frac{\sqrt{6} - \sqrt{2}}{4}$$
$$\Longrightarrow w = 2.897$$

For h $$\tan15° =  \frac{0.85}{h}$$ $$h = 3.172$$

$$h-w = 0.2744$$

I've been trying to find angles I could use, but nothing comes to mind, so I'll be grateful if you give me some hints. English is not my first language, so sorry for possible mistakes.

答:
HINT

From the diagram, we have $h+d=r+w$. Rearrange to find $d$:
$$d=...?$$

Can you finish it from there? If you have any questions please don't hesitate to ask.


问:
Show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic for $\theta$.
Let $X_{1}, X_{2}, ..., X_{n}$ be a random sample from $\text{Uniform}(\theta,\theta+1)$ population with $-\infty<\theta<\theta+1< \infty$ show that $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$. Also, show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic.

For the first part I did the following
$$f(x|\theta,\theta+1)=I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})$$
then
$$\frac{f(x|\theta,\theta+1)}{f(y|\theta,\theta+1)}=\frac{I_{(\theta,\theta+1)}(x_{(1)},x_{(n)})}{I_{(\theta,\theta+1)}(y_{(1)},y_{(n)})}$$

This is a constant function in $\theta$ iff $x_{(1)}=y_{(1)}$ and $x_{(n)}=y_{(n)}$ s.t. $T(X)=(X_{(1)},X_{(n)})$ is a minimal sufficient statistic for $\theta$.

However, I am not sure how to proceed to show that $T(X)=(R,V)=\left( X_{(n)}-X_{(1)},\frac{X_{(n)}+X_{(1)}}{2} \right)$ is a minimal sufficient statistic. Can some help me with this?

答: Any invertible function of a minimal sufficient statistic is also minimal sufficient.  Such a function is given by $$M(x,y) = (y-x, (x+y)/2)$$ which has the inverse $$M^{-1}(x,y) = (y - x/2, y + x/2).$$  This can also be conceptualized as a $2 \times 2$ matrix:  $$M = \begin{bmatrix} -1 & 1 \\ 1/2 & 1/2 \end{bmatrix}.$$


问:
How do I prove that the sine function on the domain $[-1/2, 1/2]$ is injective?
How do I prove that the sine function on the domain $[-1/2, 1/2]$ is injective?

I completely understand the concept but I'm having trouble writing a proof for this. Thanks in advance.

答:
If ZettaSuro's assumption may not be used (or you can't be bothered to prove it) try using this:
Generally, when proving injectivity, show that for $f(a_1) = f(a_2)$ we must have $a_1 = a_2$ . So choose any such such values in the image, and show that they may only be equal if  $a_1 = a_2$. I like to think of this as proving no horizontal line crosses the graph of the function $f:[\frac{-1}{2},\frac{1}{2}] \rightarrow \mathbb{R} \ \ \ \ f(x)=\sin(x)$ twice. You must of course decide the level of rigour required yourself.</p>


问:
Close-form for triple integral $ \int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$
I am able to work out the double integral

$$\int_0^b \int_0^a \sqrt{x^2+y^2} dx dy $$
with brute-force (i.e. integrating $x$, then $y$) to arrive at the close-form result

$$\frac13ab\sqrt{a^2+b^2} +\frac16a^3\sinh^{-1}\frac ba +\frac16 b^3 \sinh^{-1}\frac ab$$

which has the expected parity between $a$ and $b$. However, it gets unwieldy to tackle the triple-integral extension $$\int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$$
this way and I am unable to slug it out. Does anyone know the corresponding close-form expression for the triple version?

答:
A start

Spherical coordinates. $(\rho,\theta,\phi)$.  They are related to rectangular coordinates $(x,y,z)$ by:
\begin{align}
x &= \rho\sin\theta\cos\phi\\
y &= \rho\sin\theta\sin\phi\\
z &= \rho\cos\theta
\end{align}
and in reverse by
\begin{align}
\rho &= \sqrt{x^2+y^2+z^2}\\
\theta &=\arccos\frac{z}{\sqrt{x^2+y^2+z^2}}\\
\phi &= \arctan\frac{y}{x}
\end{align}
The box we want is
$$
0 \le x \le a,\\
0 \le y \le b,\\
0 \le z \le c.
$$
In spherical coorcinates:
$$
0 \le \rho\sin\theta\cos\phi \le a,\\
0 \le \rho\sin\theta\sin\phi \le b,\\
0 \le \rho\cos\theta \le c.
$$
If we think we will express our integrals using $\rho$ a function of $\theta,\phi$, do this as
$$
0 \le \rho \le a\csc\theta\sec\phi,\\
0 \le \rho \le b\csc\theta\csc\phi,\\
0 \le \rho \le c\sec\phi
$$
Our triple integral will have three terms, depending on which of the three
$a\csc\theta\sec\phi,b\csc\theta\csc\phi,c\sec\phi$ is smallest.  That is,
dependingon which of the three faces the ray from the origin with angles $\theta,\phi$ intersects.


问:
Number of 4-digit numbers that do not contain the digits $1, 2, 7$ and are divisible by $7$.
This is an elementary combinatorics problem.

I have discovered via computer generation that the answer is 294. This happens to be $(6 \cdot 7 \cdot 7 \cdot 7)/7$ which is the answer you'd naively expect to get if you simply counted all the numbers without those digits and divided the result by 7.

My question now is: why does this work? I thought the answer would be much more complicated than this and  that the remainders modulo 7 wouldn't cancel out exactly.

答:
Set the first three digits however you like to get $ABC*$  There are $6$ choices for the first, and $7$ each for the next two.   There is then exactly one choice for the units digit  $*$ which makes the expression divisible by $7$.  This is because the available digits $\{0,3,4,5,6,8,9\}$ are congruent to $\{0,3,4,5,6,1,2\}\pmod 7$ so we see that each residue class is represented exactly once.

Example:  Suppose you pick $333$ for the first three digits.  Then we remark that $3330\equiv 5\pmod 7$.  Now, the additive inverse of $5\pmod 7$ is $2$ and in our collection $2$ is represented by $9$.  Thus we must choose $9$ for the units place and we note that $3339$ is divisible by $7$ as desired.


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:
It appears that the troublesome aspect of the proof is the negative contradictory form of induction that is employed, i.e. the contrapositive of complete induction (Fermat's infinite descent). This is easily eliminated by rewriting the proof to use complete induction, as is done below.

Suppose as inductive hypothesis that the prime factorization of every natural $< m\,$ is unique up to order. We prove that the same holds true for $\,m.\,$ Consider any two prime factorizations of $\,m$

$$\tag{1} m\, =\, p_1 p_2\cdots p_r =\, q_1 q_2\cdots q_s$$

where the $p$'s and $q$'s are primes. By reordering if need be, we may assume that

$$ p_1 \le p_2 \le \cdots \le p_r,\quad q_1 \le q_2 \le \cdots \le q_s$$

We prove $\,\color{#0a0}{p_1 = \ q_1}.\,$ We may assume $\,p_1 \le q_1$ (else change notation: swap $p$ and $q).\,$
Let $$\tag{2} m' = m -  p_1 q_2 q_3\cdots q_s$$

By substituting for $\,m\,$ its two factorizations in equation $(1)$ we obtain
$$\begin{eqnarray}
 \tag{3} m' &=\,& p_1 p_2\cdots p_r -\, p_1 q_2\cdots q_s &=\,& p_1 (p_2\cdots p_r -\, q_2 q_3 \cdots q_s) \\
 \tag{4} m' &=\,& q_1 q_2\cdots q_s -\, p_1 q_2\cdots q_s &=\,& (q_1-p_1)(q_2 q_3\cdots q_s) \end{eqnarray}$$

Suppose $\,\color{#c00}{p_1 \ne q_1}\,$ Then $\,p_1 < q_1\,$ so it follows by $(4)$ that $\,m'>0$, while $\,m'<m\,$ by $(2).\,$
So, by induction, the prime factorization of $\,m'\,$ is unique, up to order.
From  $(3)$ it follows that $\,p_1\,$ is a factor of $\, m'.\,$ Thus, by uniqueness,
$\,p_1\,$ must also appear as a factor in $(4)$, i.e. as a factor of either $\,q_1 - p_1$ or $\,q_2\cdots q_s.\,$ The latter is impossible,
since all the $\,q_i\,$ are primes larger than $\,p_1.\,$ Hence $\,p_1\,$ is a factor of $\,q_1 - p_1,\,$
i.e. there is an integer $\,k\,$ such that $\, q_1 - p_1 = k p_1,\,$ so $\,q_1 = (k\!+\!1) p_1.\,$ Thus  $\,p_1\,$ is a factor of $\,q_1,\: $ contra prime $\:q_1 > p_1.\, $  This contradiction proves that our assumption $\,\color{#c00}{p_1\ne q_1}\,$ is false, so $\,\color{#0a0}{p_1 = q_1}\,$ as claimed.

Cancelling $\,\color{#0a0}{p_1\! = q_1}\,$  from both sides of equation $(1)$ and applying the inductive hypothesis,
we infer that the smaller number $\, m/p_1 = p_2\cdots p_r =\, q_2\cdots q_s\,$ has unique factorization,
so $\, r = s\,$ and $\, p_2\! = q_2,\ p_3\! = q_3,\ldots, p_s\! = q_s.\,$ Hence the factorizations
in $(1)$ are the same up to order. $ $ QED

Title: Mathematical Induction prime help


答: The smallest number you can consider is 3000; the largest is 9999. If we consider numbers modulo 7, 3000 and 10000 are the same (we can consider the number following 9999 to be 3000). Therefore, we simply need to divide the total number of options by 7.</p>

Title: Number of 4-digit numbers that do not contain the digits $1, 2, 7$ and are divisible by $7$.

Body:
This is an elementary combinatorics problem.

I have discovered via computer generation that the answer is 294. This happens to be $(6 \cdot 7 \cdot 7 \cdot 7)/7$ which is the answer you'd naively expect to get if you simply counted all the numbers without those digits and divided the result by 7.

My question now is: why does this work? I thought the answer would be much more complicated than this and  that the remainders modulo 7 wouldn't cancel out exactly.


问:
Quadratic Forms on a (finite dimensional real) vector space with same zero set are scalar multiples?
Let $V$ be a finite-dimensional vector space over $\Bbb{R}$, let $g,h:V \times V \to \Bbb{R}$ be bilinear symmetric functionals, and let $Q_g,Q_h:V \to \Bbb{R}$ be the associated quadratic forms. Suppose that $Q_g, Q_h$ have the same zero set; $Q_g^{-1}(\{0\}) = Q_h^{-1}(\{0\})$.

Questions:

Is it then true that $Q_g$ and $Q_h$ are scalar multiples of each other?

If (1) isn't true as stated, could it perhaps be made true by making certain restrictions on the quadratic forms, such as restricting them to have Lorentz signature (i.e with the notation below, $p=1, n$ arbitrary and $k=0$). This is the case I'm mainly interested in, but I'd of course like to know it in the more general case if it's true.

Can this result (if true) be generalized to arbitrary multilinear, symmetric functionals on $V$ and their associated homogeneous polynomials (as opposed to bilinear functionals and their quadratic forms). If so could you outline such a proof/ provide a reference.

My Attempt

This reminds me of a similar result in linear algebra, namely that if $\phi,\psi \in V^*$ have the same kernel then they are scalar multiples of each other. So, my attempt at "proving" (1) was to mimic that proof as much as possible. I know that every quadratic form over $\Bbb{R}$ can be "diagonalized" (Sylvester's Law?), in the sense that we can find a basis $\beta$ for $V$ such that the matrix representation of $g$ is of the type
\begin{align}
[g]_{\beta} &=
\begin{pmatrix}
I_p &  & \\
 & -I_n & \\
 & & 0_k
\end{pmatrix}
\end{align}
such that $p+n+k = \dim V$. But from here, I'm not sure how to proceed. Any help is appreciated.

答:
(1) isn't true as stated because it is certainly possible to have inner products which are not scalar multiples; we need to make certain restrictions on the signature. In particular, everything holds for Lorentzian signature. Finally, I'm still not sure about (3).

Theorem ($2$):

Let $n,p\geq 1$ be integers, $V$ a vector space over $\Bbb{R}$ of dimension
$d=n+p$, and let $h$ be a pseudo-inner product of signature $(n,p)$ ($n$ negatives ,
$p$ positive). If $g$ is any bilinear, symmetric functional on $V$ such that
$Q_h^{-1}(\{0\})\subset Q_g^{-1}(\{0\})$,
then there is a constant $C\in\Bbb{R}$ such that $g=Ch$. Furthermore, if we
assume $n\neq p$, and that $g$ also has signature $(n,p)$, then $C>0$.

Before proving this let's establish some conventions: I shall use greek indices to run through $\{1,\dots, n\}$ (i.e the "time-like" indices) and latin ones for $\{n+1,\dots, n+p\}$ (the "spacelike indices").

Now, fix a basis $\{v_1,\dots v_d\}$ of $V$ such that $h$ has the matrix representation $\begin{pmatrix} -I_n & 0 \\0 & I_p\end{pmatrix}$ (this is always possible by a type of Grahm-Schmidt argument, I believe it's called Sylvester's law of inertia?). For each $x\in\Bbb{R}^n,y\in \Bbb{R}^p$ having unit norms, consider the vector $w=x^{\alpha}v_{\alpha} + y^iv_i\in V$ (summation convention implied, and for convenience I shall write $y=(y^{n+1},\dots, y^{n+p})$ just to make the indices match up). Then, $h(w,w)=-\lVert x\rVert^2+ \lVert y\rVert^2=0$, which means by assumption, $g(w,w)=0$ as well. If we write this out more explicitly, we get
\begin{align*}
 g_{\alpha\beta}x^{\alpha}x^{\beta} + 2g_{\alpha i}x^{\alpha}y^i
 + g_{ij}y^iy^j &= 0. \tag{$*$}
\end{align*}
First replace $y$ with $-y$ and subtract the equations to get
\begin{align}
g_{\alpha i}x^{\alpha}y^i &= 0.
\end{align}
Since this is true for all $x,y$ with unit norms, it follows each $g_{\alpha i}=0$. With this, $(*)$ reduces to
\begin{align*}
 g_{\alpha\beta}x^{\alpha}x^{\beta} &= -g_{ij}y^iy^j. \tag{$**$}
\end{align*}
Since we have "separated" the $x$ and $y$'s and they are arbitrary, it follows all the "off-diagonal" terms vanish. More explicitly, for each $\alpha\neq \beta$, take $x_{\pm}=\frac{1}{\sqrt{2}}(e_{\alpha}\pm e_{\beta})$. Then, since the RHS doesn't depend on $x$ we find that $g_{\alpha\beta}=-g_{\alpha\beta}$, and hence it is $0$. Similarly, if $i\neq j$, then $g_{ij}=0$. Finally, letting $x$ range over $e_1,\dots, e_n$ then letting $y$ range over $e_{n+1},\dots, e_{n+p}$, we see that

\begin{align*}
  [g]&=-g_{11}
  \begin{pmatrix}
   -I_p & 0\\
   0& I_q
  \end{pmatrix},
 \end{align*}
or equivalently, $g=-g_{11}\cdot h$. Therefore, we have shown that $g$ and $h$ are related by a scale factor. Finally, if we assume $n\neq p$ and that both $g,h$ have signature $(n,p)$ then it follows that $-g_{11}>0$ (otherwise the signatures wouldn't match up).
This completes the proof.

Note in particular that for Lorentzian signature $(-,+,+,+)$ (or $(+,-,-,-)$, it doesn't matter) that if $g,h$ have the same light cones then $g=Ch$ for some $C>0$; and this is exactly the statement made in Landau and Lifshitz when they say something along the lines of "since $ds^2$ and $ds'^2$ are infinitesimals of the same order they are proportional".