问:
Inequality with the Lambert's function $x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq 1$
Let $0<x<1$ then we have :

$$x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq 1$$

The equality case is $x=0.5$.

To show it I have tried to follow the lemma 7.1 and 7.2 of this paper by Vasile Cirtoaje.The problem is that the resulting expression is awful !

I have tried also Bernoulli's inequality with any effect because it's not sharp enough.

Update 18/12/2020 :

It's a another tried .We can build an approximation like this :

Let $0<\beta<x\leq 0.5$ then we have to determine the constants such that:

$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)$$

We have numerically speaking $\frac{115}{100}<\alpha<\frac{125}{100}$

To reduces the gap I have tried to introduce a linear function :
$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)+ax+b$$

But again it's not enough to work so we may consider a general polynomial like :

$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)+\sum_{k=0}^{n}a_nx^n$$

Well it's a first step and in the future I shall tried to find the coefficients of this general polynomial .

Update 20/12/2020 :

We can reformulate the problem as :

Let $x,y>0$ such that $ye^y+xe^x=2e$ then we have :

$$\left(\frac{xe^x}{2e}\right)^{(x)^{\frac{xe^x}{e}}}+\left(\frac{ye^y}{2e}\right)^{(y)^{\frac{ye^y}{e}}}\leq 1$$

Where I use the inverse function of the Lambert's function .

Well using the form $f(x)=\left(\frac{xe^x}{2e}\right)^{(x)^{\frac{xe^x}{e}}}=g(x)^{h(x)}$ I can show that the function $f(x)$ is convex on $(0,W(2e))$ so (I have tried) we can use Slater's inequality to find an upper bound .Like this  it doesn't work . On the other hand we can use Karamata's inequality but I have not tried !

Well If we use Karamata's I have a strategy :

We have by Karamata's inequality and $0\leq\varepsilon_n'\leq\varepsilon_n<y<x$:

$$f(x)+f(y)\leq f(x+\varepsilon_n)+f(y-\varepsilon_n')$$

With $(y-\varepsilon_n')e^{y-\varepsilon_n'}+(x+\varepsilon_n)e^{x+\varepsilon_n}\geq 2e$

Now we want to repeat the process to get a series of inequalities of the kind :

$$f(x)+f(y)\leq f(x+\varepsilon_n)+f(y-\varepsilon_n')\leq f(x+\varepsilon_{n-1})+f(y-\varepsilon_{n-1}')< 1$$

But it's very complicated.

It doesn't work for all the value but I think we have the inequality $y> 0.5 \geq x$ :

$$p(x)=(1-x^{xe^{x-1}})^2+x^{xe^{x-1}} \frac{xe^{x-1}}{2} (2-x^{xe^{x-1}})-x^{xe^{x-1}} \frac{xe^{x-1}}{2} (1-x^{xe^{x-1}}) \ln\left(\frac{xe^{x-1}}{2}\right)$$
We have :
$$f(x)+f(y)\leq p(y)+2^{-\varepsilon}p^{1+\varepsilon}(x)< 1$$

With $0\leq \varepsilon \leq\frac{1}{10}$

Where I use the Lemma 7.2 of the paper above .

The last idea :

Using the majorization theorem :

Let $a\geq b>0$ and $c\geq d >0$ and $n$ a natural number large enough such that :

$$a\geq c$$

And :

$$\left(a\frac{n}{n+1}+c\frac{1}{n+1}\right)\left(b\frac{n}{n+1}+d\frac{1}{n+1}\right)\geq cd$$

Then we have :

$$a+b\geq c+d$$

Proof:It's a direct consequence of the Karamata's inequality.

We have another theorem :

Let $2>x,y>0$ ,$n$ a natural number large enough and $\varepsilon>0 $

If we have :

$$xy<1-\varepsilon $$
$$x+y<2-\varepsilon$$
then we have :

$$\ln\left(\frac{n}{n+1}+x\frac{1}{n+1}\right)+\ln\left(\frac{n}{n+1}+y\frac{1}{n+1}\right)\leq 0$$

Example :

Using the theorem of majorization we have ($x=0.4$):

$$(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}< 1-\operatorname{W}^{1.25}(2ex)0.5$$

And :

$$\left(\frac{1}{4000}x^{{\operatorname{W}(2ex)}^{2x}}+\frac{3999}{4000}\operatorname{W}^{1.25}(2ex)0.5\right)\left(\frac{1}{4000}(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}+\frac{3999}{4000}(1-\operatorname{W}^{1.25}(2ex)0.5)\right)< (1-\operatorname{W}^{1.25}(2ex)0.5)\operatorname{W}^{1.25}(2ex)0.5$$

Dividing both side by the RHS and using the second theorem remarking that :

$$\frac{x^{{\operatorname{W}(2ex)}^{2x}}(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}}{\operatorname{W}^{1.25}(2ex)0.5(1-\operatorname{W}^{1.25}(2ex)0.5)}<1-\varepsilon$$

And :

$$\frac{x^{{\operatorname{W}(2ex)}^{2x}}}{\operatorname{W}^{1.25}(2ex)0.5}+\frac{(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}}{(1-\operatorname{W}^{1.25}(2ex)0.5)}<2-\varepsilon\quad (I)$$

Now I think it's easier because we can take the logarithm and study the behavior of the function .

To prove the $(I)$ we can use the bound :

Let $0<x<\frac{1}{100}$ :

$$e^x<(1+x)^2-x$$

Obviously if we study separately the differents elements of the LHS .

Then to study $(I)$ we have a quite good approximation :

Let $0< x \leq \frac{1}{2}$ then we have :

$${\operatorname{W}(2ex)}^{2x}\geq (2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}$$

In fact we have the following refinement on $(0,0.5]$ :

$$x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}+ (1-x)^{(2(1-x))^{\frac{915}{1000}\left((1-x)\right)^{\left(\frac{87}{100}\right)}}}\leq 1$$

Remarks : The method using the majorization theorem have two advantages . We need to choose two values of the same order with respect to the values in the LHS . One can be inferior (and the other necessary superior).On the other hand the bound with the exponential ,his accuracy depends of the initial approximation in $(I)$ . Finally if we split in two the LHS in $(I)$ and if for a one we prove a stronger result then the other element is slighlty easier to show .

I build an approximation on $(0,1)$ wich have the form :

$$x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\simeq \left(\left(2^{(2x)^{x^{1.25}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.25}}}\right)^{0.5}\quad (S)$$

You can play with the coefficients $-0.25$ and $1.25$ wich are not the best (make me a comment if you have better please:-))

We can slightly improve $(S)$ in using the logarithm we have on $[0.5,1)$:

$$x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\simeq \left(\left(2^{(2x)^{x^{3}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.2}}}\right)^{0.5}-0.5\ln\left(\left(\left(2^{(2x)^{x^{3}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.2}}}\right)^{0.5}\right)+0.5\ln\left(x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\right)\quad (S')$$

We can replace the coefficient $\frac{915}{1000}$ by $\frac{912}{1000}$,and $3$ by $3.5$ and finally $-0.2$ by $-0.19$ and I think it's the same order so we can apply the majorization theorem .Ouf!

Any idea to solve it ?

Thanks

Body:
To make the problem more symmetric, let $x=t+\frac 12$ and expand the function as Taylor series around $t=0$.

You will have
$$f(t)=1+\sum_{n=1}^p a_n t^{2n}$$ where the $a_n$'s are polynomials of degree $2n$ in $k=\log(2)$
$$a_1=\left\{2,-\frac{13}{4},\frac{1}{2}\right\}$$
$$a_2=\left\{\frac{15}{4},-\frac{1607}{192},\frac{439}{96},-\frac{23}{24},\frac{1}{24}\right\}$$
$$a_3=\left\{\frac{14453}{2880},-\frac{331189}{23040},\frac{142597}{11520},-\frac{7
   9}{16},\frac{541}{576},-\frac{11}{160},\frac{1}{720}\right\}$$
$$a_4=\left\{\frac{294983}{53760},-\frac{10787687}{573440},\frac{19112773}{860160},
   -\frac{1149103}{92160},\frac{368011}{92160},-\frac{5243}{7680},\frac{15}{2
   56},-\frac{43}{20160},\frac{1}{40320}\right\}$$ All these coefficients are negative (this is not the case for $n \geq 5$).

Making the coefficients rational
$$g(t)=1-\frac{64 t^2}{5119}-\frac{121 t^4}{738}-\frac{261 t^6}{598}-\frac{182
   t^8}{865}+\frac{2309 t^{10}}{1084}+\frac{16024 t^{12}}{1381}+\frac{26942
   t^{14}}{613}+O\left(t^{16}\right)$$

Using the above terms, the match is almost perfect for $0\leq t\leq 0.4$ .

Between these bounds,
$$\int_0^{0.4}\Big[f(t)-g(t)\big]^2\,dt=1.91\times 10^{-10}$$

What would be amazing is to prove that the minimum value of the function is a bit greater than $0.99$.


问:
Question about geometric series when $a = 1$ and $r=\frac{1}{2}$
Let $A$ and $B$ be subsets of $\mathbb N$ such that $A\cap B\ne\emptyset$ and $A\ne B$. Is it possible that $$\sum_{i\in A} \frac{1}{2^i}=\sum_{j\in B} \frac{1}{2^j}$$

答:
Take $A = \{1,2\}$, $B = \mathbb N \setminus \{0,2\}$, and note

$$\frac12 + \frac14 = \frac 12 + \left(\frac18 + \frac1{16} + \dots\right)$$

ContentLicense: CC BY-SA 4.0


问:
Irreducibility of $X^{p-1} + \cdots + X+1$

Can someone give me a hint how to the irreducibility of $X^{p-1} + \cdots + X+1$, where $p$ is a prime, in $\mathbb{Z}[X]$ ?

Our professor gave us already one, namely to substitute $X$ with $X+1$, but I couldn't make much of that.

答:
In the solution for this problem I will use the well-know result as $\color{blue}{\text{Eisenstein's Irreducibility Criterion}}$.

Eisenstein's Irreducibility Criterion: Let $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots a_{0}$$ be a polynomial with coefficients $a_{n},a_{n-1},\ldots,a_{0}\in \mathbb{Z}$. Suppose that there exists a prime $p$, such that:

$p\not|a_{n}$. In words: $p$ does not divide $a_{n}$.

$p| a_{n-1},a_{n-2},\ldots,a_{1},a_{0}$. In words: $p$ divide each $a_{i}$ for $0\leq i <n$.

$p^{2}\not| a_{0}$. In words: $p^{2}$ does not divide $a_{0}$.

Then $f(x)$ is is irreducible over the integers $\mathbb{Z}[x]$.

It is a corollary of $\color{blue}{\text{Gauss's lemma}}$ that $f(x)$ is also irreducible over the rational numbers .

Gauss's lemma: A non-constant polynomial in $\mathbb{Z}[x]$ is irreducible in $\mathbb{Z}[x]$ if and only if it is both irreducible in $\mathbb{Q}[x]$ and primitive in $\mathbb{Z}[x]$.

Now, your question is well-know, as that $\color{blue}{\text{Cyclotomic polynomials}}$ are irreducibles.

Problem: Let $p$ un prime numbe, so $$\Phi_{p}(x)=x^{p-1}+x^{p-2}+\cdots+1=\frac{x^{p}-1}{x-1}$$ is irreducible.

Proof: Note that $$\Phi_{p}(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+1$$
does not satisfy the conditions of Eisenstein's criterion, but we can see that $$\Phi_{p}(x+1)=\frac{(x+1)^{p}-1}{x}=x^{p-1}+\binom{p}{1}x^{p-2}+\binom{p}{2}x^{p-3}+\cdots+\binom{p}{p-2}x+\binom{p}{p-1}$$ does. Indeed, note that $\boxed{1}$ we can see that $p\not| 1$, in $\boxed{2}$ we can see that $p| \binom{p}{1}, \binom{p}{2},\ldots,\binom{p}{p-2},\binom{p}{p-1}$ and finally in $\boxed{3}$ we have that $p^{2}\not| \binom{p}{p-1}=p$. So, $\Phi_{p}(x+1)$ is irreducible. But, if $\Phi_{p}(x)$ factored as $f(x)g(x)$ so then $\Phi_{p}(x+1)=f(x+1)g(x+1)$. So, $\Phi_{p}(x)$ is irreducible.  $\boxed{}$

ContentLicense: CC BY-SA 3.0


问:
Clarification for the proof of uncountability of real numbers
The following is from the Understanding Analysis 2nd ed., Stephen Abbot, page 28

If we let $x_1 = f(1), x_2 = f(2)$ and so on, then our assumption that $f$ is onto means that we can write $\mathbb{R} = \{x_1, x_2, \dots\}$ and be confident that every real number appears somewhere on the list. We will now use the Nested Interval Property to produce a real number that is not there.
Let $I_1$ be a closed interval that does not contain $x_1$. Next, let $I_2$ be a closed interval, contained in $I_1$ that does not contain $x_2$. The existence of such an $I_2$ is easy to verify.

This is a pedantic question, but in this scenario, where we have first assumed that $\mathbb{R}$ is countable, how would we verify the existence of $I_2$? Just bluntly exclude $x_2$ from $I_1$? Given that most if not all things in the book so far have been proven, why can we just state this? Is it due to the countability assumption on $\mathbb{R}$?

答: The interval $I_1$ is of the form $[a,b]$ for some real numbers $a$ and $b$ such that $a<b$. If $x_2\notin[a,b]$, you can just take $I_2=I_1$. Now, suppose that $x_2\in[a,b]$. If $x_2\leqslant\frac{a+b}2$, you can take $I_2=\left[\frac{a+2b}3,b\right]$. And, if $x_2\geqslant\frac{a+b}2$, you can take $I_2=\left[a,\frac{2a+b}3\right]$.


问:
How to compute $y = z\left(\frac {-2\ln r^2} {r^2}\right) $ in Box–Muller transform?
In PRML chapter 11, the book using Box-Muller method to generate Gaussian distribution.

According to wiki,

$P(y_1,y_2) = \int\int\frac{1}{2\pi}e^{-\frac{r^2}{2}}dy_1dy_2 = 1$

after some polar tansform, you can get cdf of R:

$P(R<=r)= \int^r_0 \int^{2\pi}_0 \frac{e^{-\frac{r^2}{2}}}{2\pi}rd\theta dr
= \int^r_0 e^{-\frac{r^2}{2}} r dr
= 1 - e^{-\frac{r^2}{2}}$

adding uniform distribute variable $z\in (0,1)$ into the equation we have:

$P(R<=r)=1 - e^{-\frac{r^2}{2}} $

But I cannot compute the equation of $y = z(\frac {-2\ln(r^2)} {r^2})^{1/2}$

The paragraph discribe as follow:

答: PRML actually describes an alternative, the Marsaglia polar method. You've computed the distribution of the standard Box-Muller method's radial variable, with support $[0,\,\infty)$. But in Marsaglia, a different radial variablre satisfies $R^2\sim U(0,\,1)$. The angle $\Theta$ in $Z_1=R\cos\Theta,\,Z_2=R\sin\Theta$ is $U(0,\,2\pi)$. We have $Y_1=\color{red}{\sqrt{-2\ln Z_1}}\color{blue}{Z_1/R}$; the red factor is an inverse CDF converting $Z_1$ to a Box-Muller radius, while the blue factor is the cosine. Doing the same with $Z_2$ uses a sine, obtaining $Y_2$.


问:
Exponential decay of Heat equation solution
I'm refereeing a paper and the authors go to great lengths to prove the following fact.

Let $W(t,x)$ be the solution to the linear heat equation on the half-line: $\partial_t W = D \partial_{xx} W $, $t>0$, $x>0$ with initial condition $W(0,x) = F(x)$. $D$ is a positive constant. And $F \in C^\infty$ is positive with $\int_0^\infty F(x) dx >0$.

Let $u(t,x) = e^{a t} e^{bx} W(t,x)$ where $a,b$ are constants, $b>0$. Their "new" result is:

Theorem: If $a >0$ then $u(t,x) \to \infty$ as $t \to \infty$ on compact subsets of $(0,\infty)$.

This seems standard. However, I'm not a PDE expert and haven't been able to pinpoint a reference I can include in my referee report. Can anyone help?

答:
The factor $e^{bx}$ seems to be irrelevant for the theorem.

I haven't a reference, but it's straightforward to get an estimate, using an explicit
formula for solution via Green's function:
$$
W(t,x)=\int_0^\infty G(x,y,t)F(y)\,dy.
$$
It can be assumed that $D=1$. For Dirichlet condition, say,
$$
G(x,y,t)=\Gamma(x-y,t)-\Gamma(x+y,t),
$$
where
$$
\Gamma(x,y,t)=\frac{e^{-\frac{x^2}{4 t}}}{\sqrt{4 \pi t}}, \quad t>0,
$$
is a fundamental solution for the heat equation.

Let $\alpha,\beta$ and $\varepsilon>0$ be s.t. $F(x)\ge \varepsilon$ on $[\alpha,\beta]$.
Since $G(x,y,t)$ is positive for $x,y,t>0$,
$$
W(t,x)\ge \int_\alpha^\beta G(x,y,t)F(y)\,dy\ge
\varepsilon\int_\alpha^\beta (\Gamma(x-y,t)-\Gamma(x+y,t))\,dx=
$$
$$
\frac\varepsilon2 \left(\text{erf}\left(\frac{x-\alpha }{2 \sqrt{t}}\right)+\text{erf}\left(\frac{\alpha +x}{2
   \sqrt{t}}\right)-\text{erf}\left(\frac{x-\beta }{2 \sqrt{t}}\right)-\text{erf}\left(\frac{\beta +x}{2
   \sqrt{t}}\right)\right)=
$$
$$
\frac{\varepsilon   x \left(\beta^2-\alpha^2\right)}{4 \sqrt{\pi }t^{3/2}}+O\left(\left(\frac{1}{t}\right)^{5/2}\right),\quad t\to+\infty,
$$
where $O$ is uniform on compact subsets of $(0,\infty)$. So the solution decreases no faster than $t^{-3/2}$ and it's enough to multiply it on power function $t^a$, $a>3/2$, to obtain growth when $t\to+\infty$.


问:
Show that if a matrix $A$ satisfies $A=-A^T$, then all the diagonal elements of the matrix are $0$.

Show that if a matrix $A$ satisfies $A=-A^T$, then all the diagonal elements of the matrix are $0$.

I can show that $A+A^T=0$ and from there I can see that all the elements would cancel eachother out. But how does this imply that the tr$(A)$ would be $0$ as well?

答: Since transpose operation leaves diagonal of $A$ intact. If diagonal of $A$ is non-zero then the sum $A + A^T$ will have non-zero diagonal as well. Therefore diagonal must be zero.


问:
Prove that $\int_0^\infty\frac1{x^x}\, dx<2$

Prove that $$\int_0^\infty\frac1{x^x}\, dx<2.$$

 Note: This inequality is rather tight. The integral approximates to $1.9955$.

Integration by parts is out of the question. If we let $f(x)=\dfrac1{x^x}$ and $g'(x)=1$ then $f'(x)=-x^{-x}(\ln x + 1)$ by implicit differentiation and $g(x)=x$. The integral $\int f'(x)g(x)\, dx$ looks even harder to evaluate.

Expressing the left-hand side as a Frullani integral $$\int_0^\infty\frac{f(ax)-f(bx)}x\, dx=(f(0)-f(\infty))\ln\frac ba$$ means that $f(ax)-f(bx)=x^{1-x}$. However, I can't seem to find a continuous function $f$ that satisfies the functional equation. Is there such a function?

 (For context, user371838's post below proves sophomore's dream which I also asked about originally.)

答:
This comment is not the proof, just wanted to share a quick idea. Lots of assumptions, because no proofs are provided.

Assume we know that $a(x)=\int_0^\infty\frac1{x^x}\, dx = a< \infty$. Then $a(x)/a$ is the valid probability density of some random variable, say $B$. Due to the Markov inequality we have
$$
P(B>1)={1 \over a} \int_1^\infty\frac{1}{x^x}dx < {1 \over a} \int_0^\infty\frac{x}{x^x}dx,
$$
wherefrom we get
$$
\int_1^\infty\frac{1}{x^x}dx < \int_0^\infty\frac{1}{x^{x-1}}dx.
$$

Assume the integral $\int_0^\infty\frac{1}{x^{x-1}}dx=
\int_0^\infty a_1(x) dx=a_1<\infty$. Then $a_1(x)/a_1$
is the valid probability density of some random variable, say $B_1$. Again due to the Markov inequality we have
$$
P(B_1>1)={1 \over a_1} \int_1^\infty\frac{1}{x^{x-1}}dx < {1 \over a_1} \int_0^\infty\frac{1}{x^{x-2}}dx.
$$
Proceeding in the same way we get
$$
\int_0^\infty\frac{1}{x^x}dx <  \int_0^1\frac{1}{x^{x}}dx +  \int_0^1\frac{1}{x^{x-1}}dx
+
\int_0^1\frac{1}{x^{x-2}}dx
+ \int_0^1\frac{1}{x^{x-3}}dx
+\dots
=
\sum_{n=0}^\infty
\int_0^1 {x^{n-x}}dx
$$

Since Markov bound is very crude, not sure if this converges.


答: If the matrix is given over a field with $\mathrm{char}(K)\neq 2$, then due to there being no zero divisors, entries on the diagonal must be $0$.

Body:
Show that if a matrix $A$ satisfies $A=-A^T$, then all the diagonal elements of the matrix are $0$.

I can show that $A+A^T=0$ and from there I can see that all the elements would cancel eachother out. But how does this imply that the tr$(A)$ would be $0$ as well?


答:
If $A$ and $B$ are finite sets, then it is clearly impossible.

Let $d = \max(A,B)$ and multiply both summations by $2^d$.

Edit

It was suggested to first subtract the elements common to both $A$ and $B$.  That is unnecessary.  In the analysis below, which depends on constructed numbers base 2, for the constructed numbers to be distinct, it is sufficient that there is at least one difference in the binary digits.

Then you have two distinct base 2 representations, so they must represent distinct integers.

Therefore, since each summation is $\frac{\text{distinct integer}}{2^d}$,

the summations must be different.

Allowing one of the sets to be infinite is similar to recognizing that

$0.999\cdots = 1.0.$

Title: Question about geometric series when $a = 1$ and $r=\frac{1}{2}$

Body: Let $A$ and $B$ be subsets of $\mathbb N$ such that $A\cap B\ne\emptyset$ and $A\ne B$. Is it possible that $$\sum_{i\in A} \frac{1}{2^i}=\sum_{j\in B} \frac{1}{2^j}$$


问:
How sum of this series is equal to 1?
Here is a problem in my hand-

Show that
$$
\sum_{n=1}^{\infty} \frac{e^n}{(1+e)(1+e^2) \cdots (1 + e^n)} = 1
$$

Some facts about this series can be easily observed. Let $a_n = \frac{e^n}{(1+e)(1+e^2) \cdots (1 + e^n)}$. Then

$\lim_{ n \to \infty} a_n = 0$

The series is convergent by ratio test.

How can I show the sum is $1$?

答: Write the $n$th term as $\frac{1+e^n}{(1+e)\cdots(1+e^n)}-\frac{1}{(1+e)\cdots(1+e^n)}=\frac{1}{(1+e)\cdots(1+e^{n-1})}-\frac{1}{(1+e)\cdots(1+e^n)}$, then telescope.

Title: How sum of this series is equal to 1?


问:
general Complex solution of irreducible quartic polynomial
I encountered a question asking to find the general solution of
$x^4+(x-1)^4=0$, surely it has no real solution.
I tried $x=(±\sqrt{i}(x-1))$ and $±(\sqrt{-i}(x-1))$, but I was able to find some particular solution only in the form $$\frac{1}{2}+\frac{i\cot(θ/8)}{2}$$
where I got $π/8,7π/8,-π/8,3π/8$ but unable to get the general solution.
In the book the solution
$$\frac{1}{2}+\frac{i\cot(2k+1)π/8}{2}$$
where $k$ runs from $0$ to $8$.
Can anyone please help me in finding the correct form.
I am getting $3-4k$ instead of $2k+1$ which is wrong. Also why the book writes $0$ to $8$ for $k$ although its quartic.

答:
Let's go through this slower: clearly $0$ is not a solution, so we can divide by $x^4$:
$$1+\left(\frac{x-1}x\right)^4=0$$
$$\left(\frac{x-1}x\right)^4=-1$$
$$1-\frac1x=e^{ik\pi/4}$$
where $k\in\{\pm1,\pm3\}$. Finally
$$x=\frac1{1-e^{ik\pi/4}}$$
which, as you have noted, is equal to
$$x=\frac12\left(1+i\cot\frac{k\pi}8\right)$$


Title: How sum of this series is equal to 1?

Body:
Here is a problem in my hand-

Show that
$$
\sum_{n=1}^{\infty} \frac{e^n}{(1+e)(1+e^2) \cdots (1 + e^n)} = 1
$$

Some facts about this series can be easily observed. Let $a_n = \frac{e^n}{(1+e)(1+e^2) \cdots (1 + e^n)}$. Then

$\lim_{ n \to \infty} a_n = 0$

The series is convergent by ratio test.

How can I show the sum is $1$?

Body:
Hint:
\begin{eqnarray*}
\frac{e^n}{(1+e)(1+e^2) \cdots (1 + e^n)} = \frac{-1}{(1+e)(1+e^2) \cdots (1 + e^n)} + \frac{1}{(1+e)(1+e^2) \cdots (1 + e^{n-1})}.
\end{eqnarray*}


问:
Limit of $(1+3x)^\frac{-2}{3x^3}$
I would appreciate your help solving the following, here is what I did so far:

$$\lim_{x \to 0^+}(1+3x)^\frac{-2}{3x^3}=\lim_{x \to 0^+}[(1+3x)^\frac{1}{3x}]^\frac{-2}{x^2}$$

And,

$$\lim_{x \to 0^+}(1+3x)^\frac{1}{3x}=\lim_{t \to 0^+}(1+t)^\frac{1}{t}= e$$

So, there is a $\delta>0$ such that for every $0<x<\delta$ : $|(1+3x)^\frac{1}{3x}|< e-2$

So $(1+3x)^\frac{1}{3x}>2$

And because $0<x<\delta$ I can say that $\frac{-2}{x^2}<0$ So $[(1+3x)^\frac{1}{3x}]^\frac{-2}{x^2} \leq 2^\frac{-2}{x^2}$

From here I don't know what to do...

Thanks!

Title: Limit of $(1+3x)^\frac{-2}{3x^3}$

Body:
We need to calculate $$\lim_{x\to 0^{+}}(1+3x)^{-\frac{2}{3x^{3}}}$$

The limit has the form $\lim (1+f(x))^{g(x)}$ so the trick is to use $\lim e^{\ln(1+f(x))^{g(x)}}$.

Note that
\begin{eqnarray*}
\lim_{x\to 0^{+}}(1+3x)^{-\frac{2}{3x^{3}}}&=&\lim_{x\to 0^{+}}e^{\ln(1+3x)^{-\frac{2}{3x^{3}}}}\\
&=&e^{\lim_{x\to 0^{+}}\left(-\frac{2}{3x^{3}}\ln(1+3x)\right)}\\
&=&e^{-\frac{-2}{3}\lim_{x\to 0^{+}}\frac{\ln(1+3x)}{x^{3}}}\\
&=&e^{-\frac{2}{3}\lim_{x\to 0^{+}}\frac{1}{x^{2}(3x+1)}}, \quad (\text{L'Hopital's theorem.})\\
&=&e^{-\frac{2}{3}e^{-2\lim_{x\to 0^{+}}\ln(x)}}, \quad \lim_{x \to 0^{+}}\ln(x)="-\infty"\\
&=&0
\end{eqnarray*}
Therefore, $$\boxed{\lim_{x\to 0^{+}}(1+3x)^{-\frac{2}{3x^{3}}}=0}$$


问:
Two relations defined on same sets be equal?
We say two relations $R_1$ and $R_2$ are equal if:

$$R_1\subseteq \prod_{i=1}^n A_i$$
$$R_2\subseteq \prod_{i=1}^m B_i$$
then,
$$n = m ,$$
$$A_i=B_i\quad \forall i,\,1\le i \le n,$$
$$let,s_j\,\,and\,\,t_j\,\,be\,\,tuples\,\,of\,\,R_1\,\,and\,\,R_2\,\,respectively$$
$$s.t.,\,\,s_j=<a_1,a_2,a_3,\ldots,a_j>\epsilon\,R_1\land$$
$$t_j=<b_1,b_2,b_3,\ldots,b_j>\epsilon\,R_2$$
$$\Rightarrow\,(a_1=b_1\,\land a_2=b_2\,\land\ldots\land a_n=b_n)$$
$$\forall \,j\,\epsilon \mathbb N)$$

where $A_i$ and $B_i$ are sets.

So, will the below relations be equal?
$$Name = \{x,y\}\,\,and\,\,City = \{a,b\}$$
let, $$R_1=\{
|\quad p\,\epsilon\,Name\land q\,\epsilon\,City,\quad p\,\,was\,\,born\,\,in\,\,city\,\,q\}$$
$$R_2=\{
|\quad p\,\epsilon\,Name\land q\,\epsilon\,City,\quad p\,\,lives\,\,in\,\,city\,\,q\}$$
suppose we get,
$$R_1=\{<x,a>,<y,b>\}$$
$$R_2=\{<x,a>,<y,b>\}$$
That is, x was born in the same city which he/she lives in.

Can we say,$$R_1=R_2$$

答:
The relations $R_1$ and $R_2$ are equal if and only if both $x$ and $y$ live in the same city as they were born in. That is, for all $p\in\{x,y\}$ and all $q\in\{a,b\}$ we have
$$
\text{$p$ was born in city $q$} \quad\Longleftrightarrow\quad \text{$p$ lives in city $q$.}
$$

You wrote down the same set of pairs to define $R_1$ and $R_2$, so indeed they are equal.

Equality of relations is a special case of equality of sets, and the sets you wrote down are equal.


Title: Limit of $(1+3x)^\frac{-2}{3x^3}$

Body:
I would appreciate your help solving the following, here is what I did so far:

$$\lim_{x \to 0^+}(1+3x)^\frac{-2}{3x^3}=\lim_{x \to 0^+}[(1+3x)^\frac{1}{3x}]^\frac{-2}{x^2}$$

And,

$$\lim_{x \to 0^+}(1+3x)^\frac{1}{3x}=\lim_{t \to 0^+}(1+t)^\frac{1}{t}= e$$

So, there is a $\delta>0$ such that for every $0<x<\delta$ : $|(1+3x)^\frac{1}{3x}|< e-2$

So $(1+3x)^\frac{1}{3x}>2$

And because $0<x<\delta$ I can say that $\frac{-2}{x^2}<0$ So $[(1+3x)^\frac{1}{3x}]^\frac{-2}{x^2} \leq 2^\frac{-2}{x^2}$

From here I don't know what to do...

Thanks!

Body:
If
$$L=\lim_{x\to a} f(x)^{g(x)} \to 1^{\infty}$$, then
$$L=\exp[\lim_{x \to a} g(x)[f(x)-1]$$
So here
$$L=\exp[\lim_{ x\to 0} \frac{-2}{3x^3} 3x]= \exp[\lim_{x\to 0} \frac{-2}{3x^2}]$$
So $$L=\exp[\lim_{x \to 0}\frac{-2}{3x^2}]=0$$


问:
Number of ways in which $8$ persons can be seated in three different taxis.
Number of ways in which 8 person can be seated in three different taxies each having 3 seats for passenger and duly numbered. internal arrangement matters.

My approach:
Method (1):- Since seats are numbered, we can consider all seats to be different. I want to seat eight persons on $9$ seats, which can be done by $C(9,8) \cdot 8! = 9!$.

Method (2):- First, I'll make $3$ unnumbered group of sizes $3,3,2$, which can be done in $\frac {8!}{3! \cdot 3! \cdot 2! \cdot 2!}$ ways. Now I will distribute these group to $3$ taxis, which can be done in $3!$ ways. And internal arrangement of seats in the taxis can be done in $3!$, $3!$, $2!$. So final answer will be $$\frac {8!}{3! \cdot 3! \cdot 2! \cdot 2!} \cdot {3! \cdot 3! \cdot 3! \cdot 2!}$$  That is equal to $3 \cdot 8!$.

What is going wrong in second method?
I want to know mistake in second problem.

答:
There are three seats available in the taxi which holds only two people.  The two people in that taxi can be seated in $3! = 3 \cdot 2$ ways since there are three ways to seat the younger person and two ways to seat the older person.  Replacing the factor of $2!$ by $3!$ in your second method yields
$$\frac{8!}{3!3!2!2!} \cdot 3! \cdot 3! \cdot 3! \cdot 3! = 8! \cdot 3 \cdot 3 = 9!$$
in agreement with the correct answer you obtained using your first (simpler) method.


答:
Let us consider generalization of this problem.

Let $a_n$ be sequence of positive numbers such that $ \prod_{n=1}^{\infty} a_n $ diverges.

Let $S_m = \sum_{n=1}^{m} \frac{a_n}{b_n} $ where $b_n = (1 + a_1) \cdots (1+ a_n)$.
Then observe that

$
\frac{a_n}{b_n} = \frac{1}{b_{n-1}} - \frac{1}{b_n}
$

So $S_m = 1 - \frac{1}{b_m}$. Clearly as $m \to \infty$ we have sum equal $1$ since the product of $a_i + 1$ diverges to $ \infty$

Title: How sum of this series is equal to 1?


答:
If you consider the empty seat to be person number $9$, then you quickly get $9!$ as your answer.

The other answer treats the cab with 2 people as a 2-seater, and therefore counts:

$(A,B,*), (A,*,B), (*,A,B)$

as the same arrangement.

Title: Number of ways in which $8$ persons can be seated in three different taxis.

Body:
Number of ways in which 8 person can be seated in three different taxies each having 3 seats for passenger and duly numbered. internal arrangement matters.

My approach:
Method (1):- Since seats are numbered, we can consider all seats to be different. I want to seat eight persons on $9$ seats, which can be done by $C(9,8) \cdot 8! = 9!$.

Method (2):- First, I'll make $3$ unnumbered group of sizes $3,3,2$, which can be done in $\frac {8!}{3! \cdot 3! \cdot 2! \cdot 2!}$ ways. Now I will distribute these group to $3$ taxis, which can be done in $3!$ ways. And internal arrangement of seats in the taxis can be done in $3!$, $3!$, $2!$. So final answer will be $$\frac {8!}{3! \cdot 3! \cdot 2! \cdot 2!} \cdot {3! \cdot 3! \cdot 3! \cdot 2!}$$  That is equal to $3 \cdot 8!$.

What is going wrong in second method?
I want to know mistake in second problem.


问:
Figures sum of a number
Let $S(n)$ be the digits sum of a number.

Example: $S(4) = 4$, $S(12) = 3$

Then there exist a number such that:

$S(n) = 10$

$S(n^2) = 100$

The first thing I realized is that if such number exist, it is a even number.
Not only, a realized that $n^2$ need to have at least 12 digits long.
So I think $n$ need to be a big number.
In fact, $n$ need to be greater than $10^5 \sqrt{2}$

But, anyway, I have no idea how to answer it.

答:
Observe that:

\begin{align}
11^2 &=121\\
111^2 &= 12321\\
1111^2 &= 1234321\\
&:\\
111111111^2 &= 12345678987654321
\end{align}

The last number has digit sum of $81$. However, for $1, 111, 111, 111$ the pattern breaks as there is a carry.

To fix that, we consider a number:

$$11110000\dots 0000111111 = 1111\times 10^k + 111111$$

where $k$ is sufficiently large. This number has digit sum of $10$. We also have

\begin{align}(1111\times 10^k + 111111)^2 &= 1111^2 \times 10^{2k} + 2\times1111\times 111111\times 10^k+ 111111^2\\
&=1234321000\dots000246888642000\dots 000012345654321
\end{align}

which has a digit sum of $16+48+36 = 100$.

Considering our construction, $k=12$ should be sufficient, as $111111^2$ has $11$ digits (while in reality, we just need one zero to avoid the carry).

Title: Figures sum of a number


答:
By playing around a bit, I found for $n=20110002100111$, we have $n^2 = 404412184466468830466212321$.

Here $S(n) = 10$ and $S(n^2) = 100$.

Some heuristics of how I found these. If each individual digits are large, then the number of digits in $n$ and $n^2$ will be small. So I started experimenting with 1,2. And using 0's to pad.

Title: Figures sum of a number

Body:
Let $S(n)$ be the digits sum of a number.

Example: $S(4) = 4$, $S(12) = 3$

Then there exist a number such that:

$S(n) = 10$

$S(n^2) = 100$

The first thing I realized is that if such number exist, it is a even number.
Not only, a realized that $n^2$ need to have at least 12 digits long.
So I think $n$ need to be a big number.
In fact, $n$ need to be greater than $10^5 \sqrt{2}$

But, anyway, I have no idea how to answer it.


问:
A limit on binomial coefficients
Let $$x_n=\frac{1}{n^2}\sum_{k=0}^n \ln\left(n\atop k\right).$$ Find the limit of $x_n$.

What I can do is just use Stolz formula. But I could not proceed.

答: For a two terms estimate of the behaviour of $\displaystyle\prod_{k=0}^{n}\binom{n}{k}$ one can check out March 2013 volume of Asymmetry problem V2-4 here (http://akotronismaths.blogspot.gr/p/asymmetry-electronic-mathematical.html)</p>

Title: A limit on binomial coefficients


问:
Estimation of probability of the distribution of a discrete random variable
I am trying to find the probability of the distribution of a discrete random variable $X$ which is defined $P(X=x)=y(5−)$ where $X\in\{1,2,3,4\}$.

Can anybody help me to find $y$?

Body:
$$y+2y+3y+4y=1$$

$$y=\frac{1}{10}$$


问:
Clarify about the geometric interpretation of the group $H_2(S^2)$
In class we saw that $H_2(S^2)$ is isomorphic ton $H_1(S^1)$, and the isomorphism is obtained in this way:

$H_1(S^1)\simeq H_2(D,\partial D)$ by the connecting homomorphism $\delta_2$ of the pair $(D,\partial D)$, where obviously $D$ is the closed two dimensional disk;

$H_2(D,\partial D)\simeq H_2(D^+, \partial D^+)$, where $D^+$ is the upper (closed) hemisphere of $S^2$ (and so $\partial D^+$ is the equator);

$H_2(D^+, \partial D^+)\simeq H_2(S^2,D^-)$ because the inclusion $(D^+, \partial D^+) \hookrightarrow (S^2,D^-)$ induces an isomorphism $\alpha$ in homology by the excision theorem.

Until this point is all clear: we have simply said that if we take the generator of $H_1(S^1)$, this is sent by $\delta_2^{-1}$ in the equivalence class of all the disk $D$ in $H_2(D,\partial D)$, and this class is sent by $\alpha$ in the class of $H_2(S^2,D^-)$ with the $2$-chains (of $S^2$) that contain $D^+$ (Call this class $x$). Now comes my problem: consider the projection $S_2(S^2) \twoheadrightarrow S_2(S^2,D^-)$, and call $\pi$ the morphism induced in homology. My teacher said that finally $H_2(S^2,D^-)$ is isomorphic to $H_2(S^2)$ because $\pi^{-1}(x)$ is the whole sphere considered as a class of $H_2(S^2)$. However this doesn't convince me for two reasons:

heuristically it's clear that $\pi^{-1}(x)$ is the whole sphere, because the sphere itself is the only chain of $S_2(S^2)$ that contain $D^+$ and has no boundary (remember that $\pi$ is defined from $H_2(S^2)$ to $H_2(S^2,D^-)$). However I don't think that we can just assume this evidence as true;

I'm not convinced also because if we make the same reasoning not with the sphere but with $H_2(D,\partial D)$, which is still isomorphic to $H_2(S^2)$, we can't find an element of the disk that contains the whole disk and has no boundary.

I don't know, it seems to me that this last passage is a forcing, but probably I am misunderstanding something. I know that (always for the excision) one could contract $D^-$ to a point and obtain again $S^2$ (and the same for $\partial D$), but I'd like to uderstand how is reasoning my teacher. Thanks in advance for any help

答:
I guess you work with singular homology. It is not true that "the sphere itself is the only chain of $S_2(S^2)$ that contains $D^+$ and has no boundary". The sphere is no chain at all.

A chain is a formal sum of singular simplices, and of course you can define various chains triangulating the whole sphere, but (1) this is not unique and (2) also all multiples of such chains have the property mentioned in your question. And it is not at all clear why there shouldn't be also completely different chains with this property. I think heuristics will not help, you need very formal arguments.

The final step is to show that $H_2(S^2,D^-) \approx H_2(S^2)$. This follows from the exact sequence of $(S^2,D^-)$ :

$$ 0 = H_2(D^-) \to H_2(S^2) \to H_2(S^2,D^-) \to H_1(D^-) = 0$$


问:
Normalization of $y^3=(x^3-1)(x^2+1)^2$, blow up or something else?
I need to normalize the curve $\mathcal{C}:y^3=(x^2+1)^2(x^3-1)$ which is singular at $(\pm i,0)$. I have moved the x-positive one to the origin getting $y^3=x^2(x-2i)^2((x-i)^3-1)$ and I have blown up using the quadratic transformation $(x=x_1y_1;y=y_1)$. Now I have the strict transform $y_1=x_1^2(x_1y_1-2i)^2((x_1y_1-i)^3-1)$ and I need to now the preimages of the singular point in the normalization. $(0,0)$ is given by $y_1=0$ and is still the origin, what about $(2i,0)$? I can't find its preimage. Have I done something wrong?

Please help! Any correction or hint will be appreciated!

答:
We consider a point $p=(p_0,p_1)$ of a curve $\mathcal{C}$ in $\mathbb{A}^2$. The blow up of $\mathcal{C}$ in the point $p$ is the strict transform of the blow up of $\mathbb{A}^2$ in the same point:

$Bl_p(\mathcal{C})=cl(\pi^{-1}(\mathcal{C}\setminus p))$

where $\pi: Bl_p(\mathbb{A}^2)\to \mathbb{A}^2$ and

$Bl_p(\mathbb{A}^2)=\{(q,r)\in \mathbb{A}^2\times \mathbb{P}^1: q-p\in r\}$

The condition $q\in r$ corresponds to the equation
$rk\begin{pmatrix}
x-p_0 && y-p_1 \\
r_0 && r_1
\end{pmatrix}=1
$, i.e. $r_1(x-p_0)=r_0(y-p_1)$ but it could be more complicated in higher dimension.

There is another useful description of the blow up  $Bl_p(\mathbb{A}^2)$ in terms of the trivializations maps:

$\psi_0: \{r_0\neq 0\}\to \mathbb{C}^2$ sending $((x,y),r)\to ( x,\frac{r_1}{r_0})$

and the inverse is

$\psi_0^{-1}:\mathbb{C}^2 \to  \{r_0\neq 0\} $ sending $z=(z_{00},z_{01})\to (( z_{00},z_{01}(z_{00}-p_0)+p_1),|1:z_{01}|)$

As regards the other chart, one get

$\psi_1: \{r_1\neq 0\}\to \mathbb{C}^2$ sending $((x,y),r)\to (\frac{r_0}{r_1},y)$

and the inverse is

$\psi_1^{-1}:\mathbb{C}^2 \to  \{r_1\neq 0\} $ sending $z=(z_{10},z_{11})\to (( z_{10}(z_{11}-p_{1})+p_0,z_{11}),|z_{10}:1|)$

Thus the trivialization map corresponds to

$\psi_{10}: \mathbb{C}^2\setminus \{z_{01}\neq 0\}\to \mathbb{C}^2\setminus \{z_{10}\neq 0\}$ which sends $(z_{00},z_{01})\to (1/z_{01}, z_{01}(z_{00}-p_0)+p_1)$

This means the blow up can be seen as $\mathbb{C}^2\sqcup_{\psi_{10}}\mathbb{C}^2$.

If you want to blow up more than one point, then

$Bl_{p,q}(\mathbb{A}^2)=\{ ((x,y),r_p,r_q)\in \mathbb{A}^2\times \mathbb{P}^1\times \mathbb{P}^1 : (x,y)-p\in r_p, (x,y)-q\in r_q\}$

Thus the blow up in two points can be viewed as gluing of open sets of $\mathbb{C}^2$ in the following way:

$Bl_{p,q}(\mathbb{A}^2)=((\mathbb{C}^2\setminus (\pi\circ \psi_0)^{-1}(q))_{(z_{00}^p,z_{01}^p)}\sqcup_{\psi_{10}}(\mathbb{C}^2\setminus (\pi\circ \psi_1)^{-1}(q))_{(z_{10}^p,z_{11}^p)})
\sqcup
((\mathbb{C}^2\setminus (\pi\circ \psi_0)^{-1}(p))_{(z_{00}^q,z_{01}^q)}\sqcup_{\psi_{10}} (\mathbb{C}^2\setminus (\pi\circ \psi_1)^{-1}(p))_{(z_{10}^q,z_{11}^q)})$

Now we are in the position to resolve your singularities:

$Bl_{0,q}(\mathcal{C})=cl(\pi^{-1}(\mathcal{C}\setminus 0))=\{(z_{00}^p,z_{01}^p): z_{00}^3z_{01}^3=z_{00}^2(z_{00}-2i)^2((z_{00}-i)^3-1)\}\sqcup \{(0,0)\}= \{(z_{00}^p,z_{01}^p): z_{00}z_{01}^3=(z_{00}-2i)^2((z_{00}-i)^3-1)\}\sqcup\{(0,0)\}$

in the first chart $(\mathbb{C}^2\setminus (\pi\circ \psi_0)^{-1}(q))_{(z_{00}^p,z_{01}^p)}\sqcup \{point \  at  \ infinity\}$, while in the third chart $(\mathbb{C}^2\setminus (\pi\circ \psi_0)^{-1}(p))_{(z_{00}^q,z_{01}^q)}\sqcup \{point \ at \ infinity\}$, $q=(2i,0)$,  we get

$Bl_{0,q}(\mathcal{C})=cl(\pi^{-1}(\mathcal{C}\setminus q))=\{(z_{00}^q,z_{01}^q): (z_{00}-2i)^3z_{01}^3=z_{00}^2(z_{00}-2i)^2((z_{00}-i)^3-1)\}\sqcup\{(0,0)\}= \{(z_{00}^q,z_{01}^q): (z_{00}-2i)z_{01}^3=z_{00}^2((z_{00}-i)^3-1)\}\sqcup\{(0,0)\}$

Thus the inverse image of the exceptional point $0$ of the blow up of $\mathcal{C}$ will be

$\pi^{-1}(0)=\{(0,0)\}$ where $(0,0)$ is the origin of the second chart $(\mathbb{C}^2\setminus (\pi\circ \psi_1)^{-1}(q))_{(z_{10}^p,z_{11}^p)})$

and

$\pi^{-1}((2i,0))=\{(0,0)\}$ where $(0,0)$ is the origin of the fourth chart $(\mathbb{C}^2\setminus (\pi\circ \psi_1)^{-1}(p))_{(z_{10}^q,z_{11}^q)})$

These two points corresponds to the transversal intersection between the two exceptional locus of the blow up of $\mathbb{A}^2$ and the strict transform of the curve.

In general there is the possibility this intersection is not transversal, but if you do certain number of blow ups in the intersection points, then you will get a smooth curve that will intersect the exceptional locus in a transversal way. That curve is the resolution of your curve.

If you have some questions, I'll help you.


问:
If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdots(x-a_n)-1$ is irreducible over integers.

If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdots(x-a_n)-1$ is irreducible over integers.

This question is from Pathfinder for Olympiad Mathematics by Vikash Tiwari and V. Seshan. I tried using Eisenstein's Irreducibility Criterion theorem or using contradiction method. I tried to equate it with $f(x)\cdot g(x)$ and prove that either $f(x)$ or $g(x)$ is equal to $1$ but I'm unable to prove that. Please help me with this.

答: Hint: Write $PQ=-1+\prod\limits_{i=1}^n{(x-a_i)}$ with $P,Q$ monic with integer coefficients of degree less than $n$. Show that each $a_i$ is a root of $P+Q$.


问:
Function with zero Wirtinger derivative (with respect to conjugate)
Given $f(z)=(a_0+a_1z+a_2z^2)+(b_0+b_1z+b_2z^2)\bar{z}+(c_0+c_1z+c_2z^2)\bar{z}^2$ where the $a,b,c$'s are all constant, such that its (Wirtinger) derivative w.r.t. $\bar{z}$ is zero everywhere on the plane, I want to show that all $b_i$'s and $c_i$'s must be zero, or give a counterexample otherwise.

Here's my attempt. Since the derivative can be easily determine (e.g., this derivative for $C\bar{z}^k$ is just $kC\bar{z}^{k-1}$, and $z$ is considered constant w.r.t. $\bar{z}$), I get $b_0=0$ as desired by putting $z=0$, but I'm not sure if this is the correct method to get all the values of $b$'s and $c$'s.

Or, does the hypothesis mean that $f(z)$ is entire?

Any hint or suggestion is appreciated. Thanks!

答:
Answering my question based on the hint by @WimC's comment:

By writing the $\bar{z}$-derivative of $f$ being zero everywhere, it is readily seen that $2(c_0+c_1z+c_2z^2)\bar{z}$ is a polynomial, hence entire. This means that its $\bar{z}$-derivative is also zero everywhere, so that $c_0+c_1z+c_2z^2=0$ everywhere, forcing all $c_0,c_1,c_2$ to be zero as desired. Hence, $f(z)$ now only involves the first and second terms. Again by writing the $\bar{z}$-derivative of this $f$ which is zero everywhere, this gives $b_0+b_1z+b_2z^2=0$ everywhere. Thus, $b_0,b_1,b_2$ are zero as desired.


问:
Evaluate $\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$

Evaluate $$\int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) \ dx \ dy$$

My attempt. Taking $\sin x \sin y=t$, $\iint\sin x  \ dx \ dy$.
I am not sure but I think the limits would change to $x=0$ to $\pi/2$ and $t=0$ to $1$.

I do not know how to change the elemental area i.e $dx\, dy$. Can I apply Jacobian where I am only changing one variable (i.e $y$ to $t$)?

Is this the correct way. Is there any easier way to go about it?

答:
I understand I am a bit late to the party, but there is a more rudimentary method to solve this.
$$I = \int_0^{\pi/2}\int_0^{\pi/2} \sin x \sin^{-1}(\sin x \sin y) dx dy$$
Substituting $\sin x\sin y=\sin\theta$, we get
$$I = \int_0^{\pi/2}\int_0^{x} \sin x \frac{\theta \cos \theta}{\sin x\sqrt{1-\frac{\sin^2\theta}{\sin^2x}}} d\theta dx$$
We change order of integration to get
$$I = \int_0^{\pi/2}\int_{\theta}^{\pi /2} \frac{\theta \cos \theta}{\sqrt{1-\frac{\sin^2\theta}{\sin^2x}}} dx d\theta = \int_0^{\pi/2}\int_{\theta}^{\pi /2} \frac{\theta \cos \theta \sin x}{\sqrt{\cos^2 \theta- \cos^2 x}} dx d\theta$$
This is now integrable:
$$I = \int_0^{\pi/2} \theta \cos \theta [-\sin^{-1}(\frac{\cos x}{\cos \theta})]_\theta ^{\pi /2} d\theta = \frac \pi 2 \int_0^{\pi/2}\theta \cos \theta d\theta = \frac \pi 2 (\frac \pi 2 -1)$$


问:
Limit cycles of the system $ \dot{r}=\sin\pi r, \dot{\theta}=\cos\pi r. $

In polar coordinates, consider the system
$$
\dot{r}=\sin\pi r,\quad \dot{\theta}=\cos\pi r.
$$
Prove that the system has two limit cycles with one lying interior to the other and with no equilibrium points between them. Moreover, the two limit cycles are oriented in opposite sense.

I just learned the Poincaré-Bendixson theorem, but do not know how to use it here.

Actually, we can solve the ODE directly and obtain
$$
\dfrac{1}{2}\ln\dfrac{1-\cos\pi r}{1+\cos\pi r}=\pi t+C,
$$
but it also seems of little use.

Appreciate any help, and if you may, give a clear clafication of the theorem you use.

答:
When does $\dot{r}=0$? For example, when $r=1,2$ (and for infinitely many other values actually). In these cases, $\dot{\theta}$ is constant and with opposite sign for each cycle (which gives the opposite orientation).

So your flows are two concentric circles, one flowing clockwise and one counterclockwise. These are indeed both limit cycles, and to show that there are no equilibrium points between them, note that $\dot{r}$ does not vanish in between them.


问:
Planar networks
Let $(b_1,...,b_n)$ be a sequence of positive integers. Define a $(b_1,...,b_n)$-network as the following graph $G$:

$V(G)=V_1\cup...\cup V_n$

$V_i\cap V_j=\emptyset$ ($1\leq i<j\leq n$)

$|V_i|=b_i$ ($1\leq i\leq n$)

$E(G)=\{\{u,v\}:u\in V_i,v\in V_{i+1} \text{ for some }i<n\}$

Example of the $(3,5,3)$-network:

It is straightforward that the condition $b_i\leq 2$ or $b_{i+1}\leq 2$ ($\text{for all }i<n$) is necessary for a $(b_1,...,b_n)$-network to be planar. Is it true that this is also the sufficient condition? If not, what the sufficient condition could be?

答:
Just as a reply to another answer

The explicit way to show the planarity of graphs from the @bonsoon's answer:

Title: Planar networks


问:
Question about convergence of functions of an operator
Let $A$ be a bounded self-adjoint operator on a separable Hilbert space. Assume that $f_n:\mathbb{R}\to\mathbb{C}$ is a sequence of continuous bounded functions such that $f_n \to f$ point wise. Then $f_n(A)\to f(A)$ strongly (Reed & Simon Thm. VII.2 (d)). If we further assume that $f_n \to f$ uniformly, then $f_n(A)\to f(A)$ in operator norm, right?

Are there weaker conditions on $f_n \to f$ so that $f_n(A)\to f(A)$ in operator norm?

答: Functional calculus $f\mapsto f(A)$ is an isometry from $C(\sigma(A))$ to $B(H)$. Thus $f_n\to f$ uniformly on $\sigma(A)$ if and only if $f_n(A)\to f(A)$ in operator norm.


问:
Question on set-builder notation
Wikipedia's article on set-builder notation explains that the set $\{f(x) : P(x)\}$ is equal to $\{y : \exists x [y=f(x) \land P(x)]\}$. As far as I understood, $S = \{f(x) : P(x)\}$ can be expressed in predicate logic as $\forall x[f(x) \in S \leftrightarrow P(x)]$, while $S = \{y : \exists x [y=f(x) \land P(x)]\}$ can be written as $\forall y [y \in S \leftrightarrow \exists x[y=f(x) \land P(x)]]$. However, these two formulas are logically equivalent, only if the function $f$ is invertible. Is that correct?

答:
It is true that, in general (that is, for a generic function $f$), the sentences
\begin{align}\tag{1}
\forall x \, [f(x) \in S &\leftrightarrow P(x)]
\\
\tag{2}
\forall y \, [y \in S &\leftrightarrow \exists x [y = f(x) \land P(x)]]
\end{align}
are not logically equivalent. More precisely:

$(1)$ does not imply $(2)$, but the implication holds if the function $f$ is surjective in $S$ (i.e. $S$ is included in the image of $f$).

$(2)$ does not imply $(1)$, but the implication holds if the function $f$ is injective in $S$ (i.e. for all $x$ and $x'$ in the domain of $f$, if $f(x) = f(x') \in S$ then $x = x'$);

The sentences $(1)$ and $(2)$ are logically equivalent if the function $f$ is invertible in $S$ (i.e. injective in $S$ and surjective in $S$). But it is still possible, a priori, that for some function $f$ non-invertible in $S$, $(1)$ and $(2)$ are equivalent.

However, the point is that the definition $S = \{f(x) : P(x)\}$ is not expressed in predicate logic by  means of $(1)$, but only by means of $(2)$.
Indeed, setting $S = \{f(x) : P(x)\}$ (or equivalently, $S = \{y : \exists x \, [y = f(x) \land P(x)]\}$) means that the elements of $S$ are all and only the images via $f$ of the points with the property $P$.

Why does the sentence $(1)$ not mean that $S = \{f(x) : P(x)\}$?

First, the sentence $(1)$ does not say anything about the elements of $S$ that are not of the form $f(x)$, hence $(1)$ does not exclude the possibility that $S$ can have some elements that are not in the image of the function $f$. But the definition $S = \{f(x) : P(x)\}$ and the sentence $(2)$ do exclude this possibility.
Note that that possibility is excluded whenever $f$ is surjective in $S$, that is, when $S$ is a subset of the image of $f$: this is the reason why $(1)$ implies $(2)$ when $f$ is surjective in $S$.

Moreover, the sentence $(2)$ and the definition $S = \{f(x) : P(x)\}$ do not exclude the possibility that $P(x')$ holds but $P(x)$ does not, for some $x \neq x'$ with $f(x) = f(x')$; hence, in that case $f(x) \in S$ (because $f(x) = f(x')$ and $P(x')$ holds) but $P(x)$ does not hold. But the sentence $(1)$ do exclude this possibility.
Note that that possibility is excluded whenever $f$ is injective in $S$, that is, when every element in $S$ is the image via $f$ of at most one point in the domain of $f$: this is the reason why $(2)$ implies $(1)$ when $f$ is injective in $S$.

Summing up, defining $S = \{y : \exists x \, [y = f(x) \land P(x)]\}$ (or equivalently $S = \{f(x) : P(x)\}$, which is just a shorter notation of the same definition) means what is expressed by $(2)$.

Title: Question on set-builder notation


问:
Prove that $\overline{A}$ is the union of $A$ with the set of its accumulation points.
Let $X$ be a topological space and $A\subseteq X$. Prove that $\overline{A}$ is the union of $A$ with the set of its accumulation points. Give an example to show that this union may not be disjoint.

Definition 1. A point $x \in X$ is the accumulation point of $A$ if for all $V \in \tau$ such that $x \in V$ has $(V - \{x\}) \cap A \neq \emptyset$.

Definition 2. The closing of $A$ is the intersection of all closed ones that contain $A$ and will be denoted by $\overline{A}$.

We need to prove that $\overline{A}$ is the union of $A$ with the set of its accumulation points.

If $x \in A$ then it is clear that $x \in A$, because $A \subset \overline{A}$, and therefore $x$ is in  union of $A$ with the set of its accumulation points. Now take $x$ in union of $A$ with the set of its accumulation points. This implies that $x \in A$ or is set of its accumulation points. If $x \in A$, it follows that $x \in \overline{A}$, because $A \subset \overline{A}$. And if $x$ is set of its accumulation points so for all $V \in \tau$ such that $x \in V$ has $(V - \{x\}) \cap A \neq \emptyset$ (How to finish?).

My question is if every accumulation point an adherent point? if so, how to prove it?

答:
If $x$ is in $A^{\prime},$ every neighborhood of $x$ intersects $A$ (in a point different trom $x$ ). Therefore, $x$ belongs to $\bar{A}$. Hence $A^{\prime} \subset \bar{A} .$ Since by definition $A \subset \bar{A},$ it follows that $A \cup A^{\prime} \subset \bar{A}$

To demonstrate the reverse inclusion, we let $x$ be a point of $\bar{A}$ and shuw that $x \in A \cup A^{\prime} .$ If $x$ happens to lie in $A,$ it is trivial that $x \in A \cup A^{\prime} ;$ suppose that $x$ does not lie in $A .$ Since $x \in \bar{A},$ we know that every neighborhood $U$ of $x$ intersects $A$ because $x \notin A,$ the set $U$ must intersect $A$ in a point different from $x .$ Then $x \in A^{\prime}$ so that $x \in A \cup A^{\prime},$ as desired.


问:
Best algorithm (Time Complexity) to find Minimum spanning tree of an complete, positive weighted, undirected, graph
Suppose that we have a complete undirected positive weighted graph G={V,E} and I want to find one MST of it. My problem is what is the best (in time complexity) algorithm for it?
The best prime complexity is O(V^2)
The best Kruskal complexity is O(E.Log(E))=~O(V^2. Log(V))
any idea for lower time complexity?</p>

答: I posted my question in computer science part and this is a resonable answer that I found:</p>


问:
Applying simple boundary conditions to a multi-variable function.
If I have the following general differential equation solution (to Laplace equation):
$$
\phi(r,\theta)= A+B\ln(r)+\sum_{n=1}^{\infty}\left(C_nr^n+\frac{D_n}{r^n}\right)\left( E_n \cos(n\theta) + F_n \sin(n\theta) \right) \quad \ \quad  (*)
$$
subject to $\phi(r=0,\theta=0)=0$, why is the particular solution
$$
\phi(r,\theta)= \sum_{n=1}^{\infty}C_nF_n r^n \sin(n\theta) \quad ?
$$
A priori, why $A=B=D_n=0 \ \forall \ n$ is clear. With these conditions $(*)$ becomes
$$
\phi(r,\theta)=\sum_{n=1}^{\infty}C_nr^n\left( E_n \cos(n\theta) + F_n \sin(n\theta) \right)
$$
However, I dont see why $E_n =0$, the above already satisfies the boundary conditions. Moreover couldn’t one keep $D_n\neq0$ provided $E_n=0$?

To me it seems as if there is not enough information to uniquely determine $\phi(r,\theta)$.

答:
I agree with you. There is not enough information to get your particular solution from your general solution given the only boundary condition provided.

From your general solution, setting $\phi(r = 0, \theta) =0$, leads to $A = B = 0$ and $D_n = 0$. You are then left with
$$ \phi(r,\theta)=\sum_{n=1}^{\infty}C_nr^n\left( E_n \cos(n\theta) + F_n \sin(n\theta) \right),$$
or, with a redefinition of coefficients,
$$ \phi(r,\theta)=\sum_{n=1}^{\infty}r^n\left( \alpha_n \cos(n\theta) + \beta_n \sin(n\theta) \right).$$
The values of $\alpha_n$ and $\beta_n$ cannot be uniquely determined from the only condition you provide $\phi(r=0, \theta) =0$. Instead, we would need some additional condition like $F(\theta) = \phi(r = R, \theta)$, which would allow us to compute $\alpha_n$ and $\beta_n$ according to
$$\alpha_n = \frac{1}{\pi R^n}\int^{2\pi}_{0} F(\theta) \cos(n \theta), \quad \beta_n = \frac{1}{\pi R ^n}\int^{2\pi}_{0} F(\theta) \sin(n \theta).$$
Sometimes the integration domain is $[-\pi, \pi]$ in which case we could use symmetry to eliminate one of the integrals. In either case, I think you're missing a boundary condition.


问:
sequential continuous and filter convergence ,continuous
Let $f:X\to Y$ be the map between two topological space, sequential continuous does not always imply continuous,but only in sequential space.

Here is another version of continuous that is continuous in the filter of neighborhood.I was wondering does convergence in filter implies $f$ be continuous

I found it holds but need more concept the theorem stated as below:

a function $f:X\to Y$ is continuous iff for every filter $\mathcal{F}$ on $X$ if $\mathcal{F} \to x$ then $f_{\star}\mathcal{F}\to f(x)$

Where $f_*\mathcal{F} = \{B\subset Y\mid \text{exist} A \in \mathcal{F} \text{with } f(A) \subset B\}$

答:
Pick an open $U\subseteq Y$, we want to show that $f^{-1}(U)$ is open in $X$.

Pick some $x\in f^{-1}(U)$. Clearly $F_x\to x$, where $F_x$ is the filter of nbhds of $x$, so $f_\ast F_x\to f(x)$ by assumption, which means that for some $V\in F_x$, we have $f(V)\subseteq U$. But then $V\subseteq f^{-1}(U)$, and since $x$ was arbitrary in $f^{-1}(U)$ we have that $f^{-1}(U)$ contains a neighbourhood of each of its points, that is $f^{-1}(U)$ is open and $f$ is continuous.


问:
Minimum polynomial and matrix multiplication
May you help me with the following proving?

Let $A,B$ be square matrices over $\mathbb C$ and suppose that there exist rectangular matrices $P,Q$ over $\mathbb C$ such that $A=PQ$ and $B=QP$.

If $h(X)$ is any polynomial with complex coefficients, prove that $Ah(A)=Ph(B)Q$.

Hence show that $Am_B(A)=0=Bm_B(B)$. Deduce that one of the following holds: $$m_A(X)=m_B(X),~~ m_A(X)=Xm_B(X),~~ m_B(X)=Xm_A(X).$$

At the notes on how to answer are mentioned that $(PQ)^{i+1}=P(QP)^iQ$, how is that possible?
$$$$ Added: May you help me with the deduction part? $m_A(X)$ stands for minimum polynomial

Deduce that one of the following holds: $$m_A(X)=m_B(X),~~ m_A(X)=Xm_B(X),~~ m_B(X)=Xm_A(X).$$

Thank you in advance.

答:
The equality $$(PQ)^{i+1}=P(QP)^iQ\tag{eq}$$ stems from the fact that on both sides we have the product of $i+1$ copies of $PQ$.

So if $h(X) = c_0 + c_1 X + \dots + c_n X^n$, we have, using (eq)
$$
P h(B) Q = P (c_0 I + c_1 B + \dots + c_n B^n) Q
=
P h(B) Q = P (c_0 I + c_1 QP + \dots + c_n (QP)^n) Q.
= a_0 PQ + a_1 (PQ)^2 + \dots +  c_n (QP)^{n+1}
= a_0 A + a_1 A^2 + \dots +  c_n A^{n+1}
= A h(A).
$$


问:
floor number sum
For $a$ and $n$, there is formula to calculate

$$a + 2a + 3a + \cdots + na = \frac{n(n+1)}{2} a.$$

Is there formula:

$$\lfloor a\rfloor + \lfloor 2a\rfloor + \lfloor 3a\rfloor + \cdots + \lfloor na\rfloor $$

答:
There is also a way to evaluate this sum using Pick´s theorem.

$$\sum_{i=1}^n\left\lfloor \frac{p}{n} i\right\rfloor = \frac{pn+p-n + gcd(p,n)}{2}$$

Title: floor number sum


问:
Find all polynomials $f(x)$ satisfying $f(x^2 ) + f(x) \cdot f(x + 1) = 0$ ∀ $x ∈ R$
Find all polynomials $f(x)$ satisfying $f(x^2 ) + f(x) \cdot f(x + 1) = 0$  ∀  $x ∈ R$

Obvious polynomials are $f(x)=0$ and $f(x) = -1$

There are no linear polynomials satisfying the equation.

I took $f(x) = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...a_nx^n$

For $f(x)$ to satisfy the condition

$a_0(1+a_0+a_1+a_2+...a_n) = 0$

$a_n + {a_n}^2=0$

Taking simple values of $x$ like $x=1,x=0$ did not help and I am unable to proceed further.

Can anybody help?

答:
I would not consider this as a complete or rigourous answer since I've given some statements which I've not given the proof to. Consider this more like a hint.

You've noticed that there were no possible linear solutions, in fact, we can proof that any polynomial with an odd degree cannot be solution to that equation.

Let $ n\in 2\mathbb{N} $. Denoting $ f\left(x\right)=\sum\limits_{k=0}^{n}{a_{k}x^{k}} $. We have :

\begin{aligned} f\left(x\right)f\left(x+1\right)&=\left(\sum_{i=0}^{n}{a_{i}x^{i}}\right)\left(\sum_{i=0}^{n}{\left(\sum_{j=i}^{n}{a_{j}\binom{j}{i}}\right)x^{i}}\right)\\&=\sum_{k=0}^{2n}{\left(\sum_{p=\max\left(0,k-n\right)}^{\min\left(k,n\right)}{a_{k-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}\right)x^{k}} \end{aligned}

Suppose $ f\left(x\right)\neq 0 $. Let's try to find the first coefficients $ a_{n-k} $, $ k\in\mathbb{N} $ of our polynomial.

\begin{aligned}a_{n}+\sum_{p=n}^{n}{a_{2n-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0 &\iff a_{n}+a_{n}^{2}=0 \\&\ \ \Longrightarrow \ \ \ \ \ \ \ \ \ \ \ a_{n}=-1\end{aligned}

\begin{aligned} \sum_{p=n-1}^{n}{a_{2n-1-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0&\iff a_{n-1}a_{n}+a_{n}\left(a_{n-1}+n a_{n}\right)=0\\ &\ \ \Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ a_{n-1}=\frac{n}{2} \end{aligned}

\begin{aligned} a_{n-1}+\sum_{p=n-2}^{n}{a_{2n-2-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0 \Longrightarrow a_{n-2}=-\frac{\frac{n}{2}\left(\frac{n}{2}-1\right)}{2} \end{aligned}

$$ \sum_{p=n-3}^{n}{a_{2n-3-p}\sum_{j=p}^{n}{a_{j}\binom{j}{p}}}=0\Longrightarrow a_{n-3}=\frac{\frac{n}{2}\left(\frac{n}{2}-1\right)\left(\frac{n}{2}-2\right)}{6}$$

So on and so forth.

We can proof by strong induction that $ \left(\forall k\in\mathbb{N}\right),\ a_{n-k}=\left(-1\right)^{k+1}\binom{\frac{n}{2}}{k} $.

Then we can conclude that : $$ \fbox{$\begin{array}{rcl}\displaystyle f\left(x\right)=\sum_{k=0}^{n}{\left(-1\right)^{k+1}\binom{\frac{n}{2}}{k}x^{n-k}}\end{array}$} $$

Since $ n $ must be even, then all the polynomials satisfying to the equation are $ 0 $ and $ \sum\limits_{k=0}^{n}{\left(-1\right)^{k+1}\binom{n}{k}x^{2n-k}}=-x^{n}\left(1-x\right)^{n} $, $ n\in\mathbb{N} $.

The set of solutions would, thus, be : $$\fbox{$\begin{array}{rcl} \displaystyle\mathbb{S}=\left\lbrace 0,-x^{n}\left(1-x\right)^{n}\mid n\in\mathbb{N}\right\rbrace\end{array}$} $$


问:
Show that if we have a matrix $A$, which can be written as $A=DD^T$ for some matrix $D$, then...
we have $\bar{x}^TA\bar{x}\geq 0$ for any column vector $\bar{x}$.

I do not really know where to start, but I do know that I can somehow show the scalar $\bar{x}^TA\bar{x}$ is equal to $||D^T\bar{x}||^2$ and that would be enough. Unfortunately, I  cannot see how.

答:
\begin{align*}
||D^T\bar{x}||^2 &=\langle D^T\bar{x} , D^T\bar{x}\rangle\\
&= (D^T\bar{x})^TD^T\bar{x}\\
&=\bar{x}^TDD^T\bar{x}\\
&=\bar{x}^TA\bar{x}
\end{align*}


答:
Thanks to @CalvinLin for pinpointing that we cannot "assume WLOG" that the five variables are sorted as $a \le b \le c \le d \le e$.

We can still assume that the maximum element among them is $e$ $^{(*)}$. Then, by comparing $\color{red}{\text{red}}$ variables, we observe

$$3d = (\color{red}e + a + b)^3 \ge (a + b + \color{red}c)^3 = 3e \\ \implies d \ge e \overset*\implies \boxed{d=e} \tag{1}$$

$$3a = (b + c + \color{red}d)^3 \ge (\color{red}a + b + c)^3 = 3e \\\implies a \ge e \overset*\implies \boxed{a = e} \tag{2}$$

$$3a = (\color{red}b + c + d)^3  \le  (c + d + \color{red}e)^3 = 3b \\\implies b \ge a = e \overset*\implies \boxed{b = e} \tag{3}$$

$$3c = (d + e + \color{red}a)^3 \ge (\color{red}c + d + e)^3 = 3b \\ \implies c \ge b \implies \boxed{c = b} \tag{4}$$

Finally, $(1)\land(2)\land(3)\land(4) \implies \boxed{a = b = c = d = e}$.

Also, @AlbusDumbledore provided an equivalent solution in AoPS

Title: Solving the system $3a=(b+c+d)^3$, $3b=(c+d+e)^3$, ..., $3e=(a+b+c)^3$ for real $a$, $b$, $c$, $d$, $e$

Body:
$$\begin{align}
3a&=(b+c+d)^3 \\
3b&=(c+d+e)^3 \\
3c&=(d+e+a)^3 \\
3d&=(e+a+b)^3 \\
3e&=(a+b+c)^3
\end{align}$$

I tried to use inequality for
$((\sum \alpha)/n)^r\leq \sum \alpha^r/n$ by taking $\alpha=a_1+a_2+a_3$ where $a_1,a_2,a_3\in \{a,b,c,d,e\}$.


答:
Let's say $x \in \mathbb{R}^n$. Since $A = D D^T$, it's easy to see that $A$ must be square matrix so $A \in \mathbb{R}^{n \times n}$. Let $D \in \mathbb{R}^{n \times m}$.

\begin{align}
x^T A x &= x^T D D^T x\\
&\stackrel{(1)}{=} (D^T x)^T (D^T x)\\
&\stackrel{(2)}{=} ||D^T x||^2\\
&\geq 0
\end{align}

Equality $(1)$ follows by the standard result $(AB)^T = B^T A^T$ and $(B^T)^T = B$ for any matrices $A$ and $B$.

Equality $(2)$ follows because $D^T x \in \mathbb{R}^m$, i.e. $D^T x$ is a column vector. For any column vector $v$, we have $||v||^2 = v^T v$

Title: Show that if we have a matrix $A$, which can be written as $A=DD^T$ for some matrix $D$, then...

Body:
we have $\bar{x}^TA\bar{x}\geq 0$ for any column vector $\bar{x}$.

I do not really know where to start, but I do know that I can somehow show the scalar $\bar{x}^TA\bar{x}$ is equal to $||D^T\bar{x}||^2$ and that would be enough. Unfortunately, I  cannot see how.


问:
Central limit theorem for weighted random variable
Let $(X_n)_n$ be a sequence of i.i.d random variable, $(x_n)_n$ be sequence of $\mathbb{R}^*.$ Let $y^2_n=\sum_{k=1}^nx^2_k.$

Suppose that $E[X_1]=0,E[X_1^2]=1,x_n=o(y_n),y_n \to +\infty.$ Prove that the sequence of the weighted random variable $(x_nX_n)_n$ fulfills the central limit theorem: $$\frac{1}{y_n}\sum_{k=1}^nx_kX_k \implies N(0;1).$$

Remark (optional): more generally, the following holds: $E[X_1^2]<+\infty$ if and only if there exist a sequence of real numbers $(w_n)_n$ such that $\frac{1}{y_n}\sum_{k=1}^nx_kX_k-w_n$ converges in distribution to an arbitrary random variable $Y.$ In this case, $Y$ is normal distributed.

A way to prove that the CLT is fulfilled, is to prove that Lindeberg condition holds. So let $\epsilon>0.$
$$\frac{1}{y_n^2}\sum_{k=1}^nx_k^2E[X_k^2 1_{|x_kX_k|>\epsilon y_n}]=\frac{1}{y_n^2}\sum_{k=1}^nx_k^2E[X_1^2 1_{|x_kX_1|>\epsilon y_n}],$$ can't see how to continue from here, especially how to remove $x_k$ from $1_{|x_kX_1|>\epsilon y_n}?$

Any suggestions are welcomed.

答:
I think that we have to assume that $y_n\to\infty$, otherwise the convergence
would reduce to the one of a series of independent random variables and the limit may not be normal.

Let$c_{n,k}:=E[X_1^2 1_{|x_kX_1|>\epsilon y_n}]$. For $n>k_0$, using  non-decreasingness of $(y_n)$ and that for $1\leqslant k\leqslant n$, $0\leqslant c_{n,k}\leqslant c_{k,k}\leqslant 1$, we derive that
$$
\frac{1}{y_n^2}\sum_{k=1}^nx_k^2c_{n,k}= \frac{1}{y_n^2}\sum_{k=1}^{k_0}x_k^2c_{n,k}+\frac{1}{y_n^2}\sum_{k=k_0+1}^{n}x_k^2c_{n,k}\leqslant \sum_{k=1}^{k_0}\frac{x_k^2}{y_k^2}+\sup_{k\geqslant k_0}c_{k,k}
$$
hence for each $k_0$, $$\limsup_{n\to\infty}
\frac{1}{y_n^2}\sum_{k=1}^nx_k^2c_{n,k} \leqslant  \sup_{k\geqslant k_0}c_{k,k}.
$$
Using the assumption $x_k/y_k\to 0$ finishes the proof.


问:
How to integrate $\frac{1}{x\sqrt{1+x^2}}$ using substitution?
How you integrate

$$\frac{1}{x\sqrt{1+x^2}}$$

using following substitution? $u=\sqrt{1+x^2} \implies du=\dfrac{x}{\sqrt{1+x^2}}\, dx$

And now I don't know how to proceed using substitution rule.

答:
It is quickest to substitute only in part of the integral. The substitution $u=\sqrt{1+x^2}$ satisfies

$$
\frac{1}{xu}=\frac{1}{x}-\frac{1}{1+u}\frac{x}{u} \quad \text{and} \quad \mathrm{d}u=\frac{x}{u}\,\mathrm{d}x.
$$

So it follows that

$$
\int \frac{\mathrm{d}x}{x\sqrt{1+x^2}}=\int \frac{\mathrm{d}x}{x}-\int \frac{\mathrm{d}u}{1+u}=\log \frac{|x|}{1+u}+\mathrm{const}=-\operatorname{arsinh}\frac{1}{|x|}+\mathrm{const}.
$$

Here we have used the following identity for the inverse hyperbolic sine:

$$
\operatorname{arsinh}t=\log\,(t+\sqrt{t^2+1}).
$$


问:
Generated $\sigma$-algebra question
Reading through my lecture notes, and I'm stuck a bit on this concept.

Let $A$ be a set of subsets of $E$. Define $$ \sigma(A) = \{ A \subseteq E \ : \  A \in F \text{ for all } \sigma\text{-algebras } F \text{ containing }A \} .$$ Then $\sigma(A)$ is a $\sigma$-algebra, which is called the $\sigma$-algebra generated by $A$. It is the smallest $\sigma$-algebra containing $A$.

So let's say: $E = \{1,2,3\}$. All the possible subsets of $E$ will be $ = \{ \{ \emptyset \}, \{1,2,3\}, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\} \}$. So if we choose $A= \{1 \}$, then what would $\sigma(\{1\})$ equals to?

$$ \sigma(\{1\}) = \{ \{1 \} \subseteq E \ \colon \ \{1 \} \in F \text{ for all } \sigma\text{-algebras } F \text{ containing } \{1\} \} .$$

How would I find all the $\sigma$-algebras $F$ containing $\{1\}$? Thanks.

答:
Actually $A=\{\{1\}\}$ since it must be a subset of the Power set. Now to find $\sigma(A)$, I would do the following:

First $\emptyset, E, \{1\}$ must be in $\sigma(A)$ and then the complement of $\{1\}$ which is $\{2,3\}$. So that
$$
\sigma(A)=\{\emptyset, E, \{1\}, \{2,3\} \}
$$

Title: Generated $\sigma$-algebra question


问:
Find Maclaurin series of $ y(t)=(t^2+t-2)e^{t^2+1} $

Find Maclaurin series of
$$
y(t)=(t^2+t-2)e^{t^2+1}
$$

I acted as usual:
$$
y(t)=e(t^2+t-2)\sum_{k=0}^\infty\frac{t^{2k}}{k!}=\sum_{k=0}^\infty\frac{et^{2k+2}}{k!}+\sum_{k=0}^\infty\frac{et^{2k+1}}{k!}-\sum_{k=0}^\infty\frac{2et^{2k}}{k!}
$$

And then I got stuck, since I don't understand how to combine even and odd powers together.

答: Now you do\begin{align}\sum_{k=0}^\infty\frac{et^{2k+2}}{k!}+\sum_{k=0}^\infty\frac{et^{2k+1}}{k!}-\sum_{k=0}^\infty\frac{2et^{2k}}{k!}&=\sum_{k=1}^\infty\frac{et^{2k}}{(k-1)!}+\sum_{k=0}^\infty\frac{et^{2k+1}}{k!}-\sum_{k=0}^\infty\frac{2et^{2k}}{k!}\\&=-2e+\sum_{k=1}^\infty\left(\frac e{(k-1)!}-\frac e{k!}\right)t^{2k}+\\&\phantom=+\sum_{k=0}^\infty\frac{et^{2k+1}}{k!}.\end{align}In other words, the MacLaurin series is $\sum_{k=0}^\infty a_kt^k$ with$$a_k=\begin{cases}-2e&\text{ if }k=0\\\frac e{\left(\frac{k-1}2\right)!}&\text{ if $k$ is odd}\\\frac e{\left(\frac k2-1\right)!}-\frac e{\left(\frac k2\right)!}&\text{ if $k$ is even and }k>0.\end{cases}$$


问:
Proving that $\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$
I have to prove this
$$\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$$

My proof is the following:

By using $$n!>n^{\frac{n}{2}}$$
$$\frac{1}{n!}<\frac{1}{n^{\frac{n}{2}}}$$
$$\left(\frac{1}{n!}\right)^{\frac{1}{n}}<\left( \frac{1}{n}\right)^{\frac{n}{2}\cdot\frac{1}{n}}=\frac{1}{\sqrt{n}}\underset{n\to \infty}{\to}0$$
Using the Squeeze theorem
$$0<\frac{1}{\sqrt[n]{n!}}<\frac{1}{\sqrt{n}} \implies \frac{1}{\sqrt[n]{n}}\underset{n\to \infty}{\to} 0$$

I think I have done something wrong. Can I even use the first inequality here? and if yes-do I have to prove that too? and is it Ok that I write too little words in my proofs?. Could you please help me?

答:
To start of, the first inequality is not correct for all possibilities of n, say for example $n = 2$, then $2! = 2$ and $2^{\frac{2}{2}} = 2$, the strict inequality also doesn't work for $n=1$. So either use a non strict inequality or define it only for $\forall n \in \mathbb{N}: n > 2 $. If you propagate the non strict inequality in the rest of your proof then the steps you take are as far as I know correct.

Secondly, if you want to use the first inequality you will have to prove it, by proving it you would have noticed that it doesn't work for these smaller values. You can try proving the first inequality by thinking about what both sides of the inequality look like when written in full. I won't spoil the proof for that, but if you have trouble finding it you can always ask follow up questions in the comments.

To answer the second question it depends, if you are just writing the proof for yourself it would be sufficient to just write this to check if you're proof works. But if you're writing this for an assignment or an exam or a paper or anything else than a scribble on paper more text with clarification of what you're doing would always be better to write a clear proof. If a proof is correct in all steps that doesn't make it a good proof, a good proof should be as clear as possible for the reader. Also the third expression could also be split for a clearer proof in the form of first showing the inequality then showing the limit and reffering to those in your Squeeze theorem. E.g. using the found inequality we apply the Squeeze theorem wich results in...

Good luck!


问:
Example of a set in $\mathcal{B}(\mathbb R^{I})$ but not in $\mathcal{B}(\mathbb R)^{I}$ for $I$ uncountable
Let $I$ be an uncountable set and consider the Borel sigma algebra on $\mathbb R^{I}$, i.e. $\mathcal{B}(\mathbb R^{I})=\sigma(\tau_{I})$ where $\tau_{I}$ is the product topology on $\mathbb R^{I}$. By definition, the projections $p_{i}$ are continuous for all $i\in I$ and thus the projections are $\sigma(\tau_{I})$-measurable. Since the product sigma algebra $\mathcal{B}(\mathbb R)^{I}$ is the smallest sigma algebra such that the projections are measurable with respect to $\mathcal{B}(\mathbb R)^{I}$. Therefore we have $\mathcal{B}(\mathbb R)^{I} \subseteq \mathcal{B}(\mathbb R^{I})$.

Now I am interested in finding a set $A$ such that is in $\mathcal{B}(\mathbb R^{I})$ but not in $\mathcal{B}(\mathbb R)^{I}$ to show that the inclusion is strict. Any ideas?

答: Any set in $\mathbb B(\mathbb R)^{I}$ depends only on a countable number of coordinates. This means if $E \in \mathbb B(\mathbb R)^{I}$ then there exists a countable subset $(i_n)$ of $I$ such that $E=\{f: (f(i_1),f(i_2),...) \in F$ for some $F$ in $\mathbb R^{\infty}$. [To prove this show that  the collection of all sets with this property is a sigma algebra]. Hence any set which does not depend only on a countable number of coordinates will be an example. You can take $\{f: |f(i)| \leq 1 \forall i \in I\}$.


问:
Planar networks
Let $(b_1,...,b_n)$ be a sequence of positive integers. Define a $(b_1,...,b_n)$-network as the following graph $G$:

$V(G)=V_1\cup...\cup V_n$

$V_i\cap V_j=\emptyset$ ($1\leq i<j\leq n$)

$|V_i|=b_i$ ($1\leq i\leq n$)

$E(G)=\{\{u,v\}:u\in V_i,v\in V_{i+1} \text{ for some }i<n\}$

Example of the $(3,5,3)$-network:

It is straightforward that the condition $b_i\leq 2$ or $b_{i+1}\leq 2$ ($\text{for all }i<n$) is necessary for a $(b_1,...,b_n)$-network to be planar. Is it true that this is also the sufficient condition? If not, what the sufficient condition could be?

答:
The answer to the first question is no due to the following theorem which gives the necessary and sufficient condition.

Theorem. Let $G$ be a $(b_1,...,b_n)$-network, then:

If $n=1$, then $G$ is planar

If $n=2$, then $G$ is planar iff $b_1\leq2\vee b_2\leq2$

If $n\geq3$, then $G$ is planar iff $\forall(1<i<n):b_i>2\rightarrow b_{i-1}=b_{i+1}=1$

Proof. The first case is obvious. In the second case, if the condition doesn't hold, then $G$ contains $K_{3,3}$ which is not planar. Otherwise, $G$ is a subgraph of a planar graph $K_{2,x}$ $(x\geq1)$. Consider the case $n\geq3$. Suppose that the condition doesn't hold. Then for some $1<i<n$ we have $b_i>2$ and at least one of $b_{i-1}$ or $b_{i+1}$ is greater than $2$. Again, this implies that $G$ contains $K_{3,3}$. Finally, let's prove by induction that the last condition is sufficient for $G$ to be planar.

Base case. Suppose $n=3$ and the condition holds. The cases when $b_2=1$ or $b_1=b_3=1$ are straightforward. Now, if $b_2=2$, then $G$ is just $K_{2,b_1+b_3}$ which is planar.

Induction step. Suppose $G$ is a $(b_1,...,b_{n+1})$-network and the condition holds for $n$ ($n\geq3$). Then the condition $b_i>2\text{ }(1<i<n+1)\rightarrow b_{i-1}=b_{i+1}=1$ implies the $(b_1,...,b_n)$-network $G'$ is planar. There are two cases:

$b_{n+1}\geq2$. We have $b_n\leq2$. If $b_n=1$, then it's obvious how to use $G'$ to draw $G$ on a plane. Assume $b_n=2$. Then we have $b_{n-1}\leq2$. If $b_{n-1}=2$, consider the sets $V_{n-1}=\{u_1,u_2\}$ and $V_{n}=\{u_3,u_4\}$ from the definition. When drawn on a plane, $G'$ contains a face $F$ that is the cycle $u_1-u_3-u_2-u_4-u_1$. We can draw $G$ planarly putting $b_{n+1}$ vertices inside of $F$. If $b_{n-1}=1$, then both vertices from $V_n$ in $G'$ have the degree $1$, and it's straightforward to draw $G$ on a plane.

$b_{n+1}=1$. If $b_n>2$, then $b_{n-1}=1$ and all the vertices from $V_n$ in $G'$ have the degree $1$, so $G$ can be drawn on a plane. The case $b_n\leq2$ is analogous to the previous step.

$\blacksquare$

The following picture demonstrates why $(2,3,2)$-network can not be planar:

Title: Planar networks


问:
Let $x\in (0,1)$ then check for convergence or divergence of series $\sum_{n=1}^\infty x^{\sqrt{n}}$
Let $x\in(0,1)$ then check for convergence or divergence of series $$\sum_{n=1}^\infty x^{\sqrt{n}}$$

It seems to be convergent but I am having problem in showing that. Every test is seems to fail. Any hint is careful.

答:
Divide the summation into blocks from $m^2$ to $(m+1)^2$. Each block has $2m + 1$ items and the largest is the first one, so the sum of elements in the block can be estimated by
$(2m + 1)x^m$. But this series converges by ratio (d'Alambert) test.


问:
Is every countable space first countable?
All of the examples of non-first countable spaces I have seen are uncountable (for instance any uncountable set with the cofinite topology). I would like to know if every countably infinite $T_1$ space $X$ is first countable. Since $\{A\subseteq X|x\in A\}$ for a given $x\in X$ is uncountable, there doesn't seem to be a 1-line proof; a proof should require use of the axioms of a topology.

Perhaps I haven't seen a proof or counterexample anywhere because I haven't looked in the right place or am simply missing an "obvious" proof or counterexample. Maybe someone can point me to a reference or exercise in a textbook where this shows up.

答:
Other interesting counterexamples:

Appert Space

Maximal Compact Topology

One Point Compactification fo the Rationals

Single Ultrafilter Topology


问:
If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdots(x-a_n)-1$ is irreducible over integers.

If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdots(x-a_n)-1$ is irreducible over integers.

This question is from Pathfinder for Olympiad Mathematics by Vikash Tiwari and V. Seshan. I tried using Eisenstein's Irreducibility Criterion theorem or using contradiction method. I tried to equate it with $f(x)\cdot g(x)$ and prove that either $f(x)$ or $g(x)$ is equal to $1$ but I'm unable to prove that. Please help me with this.

答:
If $a_1, a_2, \ldots, a_n$ are distinct integers, then prove that $(x-a_1)(x-a_2)\cdot\ldots\cdot(x-a_n)-1$ is irreducible over integers.

If that polynomial were reducible over integers, then there would exist two monic polynomial $\;P(x)\;$ and $\;Q(x)\;$ with integer coefficients of degree less than $\;n\;$ such that

$P(x)Q(x)=(x-a_1)(x-a_2)\cdot\ldots\cdot(x-a_n)-1\;.$

Consequently, it results that

$P(a_i)Q(a_i)=-1\quad\forall i\in\{1,2,\ldots,n\}\;,$

but $\;P(a_i)\;$ and $\;Q(a_i)\;$ are integers, so there are only two possibilities:

$\;P(a_i)=1\;$ and $\;Q(a_i)=-1\;,$

$\;P(a_i)=-1\;$ and $\;Q(a_i)=1\;.$

In any case we get that

$P(a_i)+Q(a_i)=0\quad\forall i\in\{1,2,\ldots,n\}\;,$

but it is impossible because $\;P(x)+Q(x)\not\equiv0\;$ (sum of monic polynomials) has degree less than $\;n\;,\;$ so according to the fundamental theorem of algebra, it cannot have $\;n\;$ different roots $\;a_1, a_2, \ldots, a_n\;.$

Hence, there do not exist such polynomials $\;P(x)\;$ and $\;Q(x)\;,\;$ consequently

$(x-a_1)(x-a_2)\cdot\ldots\cdot(x-a_n)-1\;$ is irreducible.


问:
For what $x_{0}$ this sequence converges?
$$X_{n+1} = X_{n}^2 - X_{n}/2, \quad n \geq 0$$

For what values of $x_{0}$ this sequence converges? For what value?

I think to solve this problem separating it in three steps:

The series is constant. Ok

The series is crescent:

$X_{n+1} > X_{n}$

And for this case we need to require this:
$X_{n} - X_{n-1} > X_{n+1} - X_{n} $

The series is decrescent:

$X_{n} > X_{n+1}$

And for this case we need to require this:
$X_{n} - X_{n+1} < X_{n-1} - X_{n}$

Now, i think this is not the best approach, it is long and extremely easy to be made an algebraic mistake. There is another way to do that?

答: Hint: Use induction to show that $X_{n+1} \geq X_n$ for all $n$ if $x_0 \geq \frac 3 2 $ and $X_{n+1} \leq X_n$ for all $n$ if $x_0 \leq \frac 3 2 $.


答:
The problem I see in your proof is the first inequality $n! >n^{n/2}$, I don't think it's correct even for $n$ sufficiently large because $\frac{n!}{n^n} \rightarrow 0$ (although this is a bit different from what you say).

If I had to give a simple and basic justification for that I'd do de following.
Consider any fixed $m \in \mathbb{N}$, then for $n > m , n \in \mathbb{N}$ we have
$$n!=m! \cdot (m+1) \cdots (n-1)n > m! \cdot m \cdot m \cdots m = m! \cdot m^{n-m}$$
Takin n-roots you get $\sqrt[n]{n!} > \sqrt{m!} \cdot m^{1-\frac{m}{n}} > m^{1-\frac{m}{n}}$. Now for $n$ big enough you have $\frac{m}{n} <\frac{1}{2}$ and thus $1-\frac{m}{n} > \frac{1}{2}$. So for $n$ big enough you deduce $\sqrt[n]{n!} > \sqrt{m}$

And since you can make $\sqrt{m}$ as large as you want, you deduce $\sqrt[n]{n!}\rightarrow +\infty$

Keep in mind that these approximations are really really suboptimal, but still, they get the job done.

Title: Proving that $\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$

Body:
I have to prove this
$$\lim_{n \to \infty} \frac{1}{\sqrt[n]{n!}}=0$$

My proof is the following:

By using $$n!>n^{\frac{n}{2}}$$
$$\frac{1}{n!}<\frac{1}{n^{\frac{n}{2}}}$$
$$\left(\frac{1}{n!}\right)^{\frac{1}{n}}<\left( \frac{1}{n}\right)^{\frac{n}{2}\cdot\frac{1}{n}}=\frac{1}{\sqrt{n}}\underset{n\to \infty}{\to}0$$
Using the Squeeze theorem
$$0<\frac{1}{\sqrt[n]{n!}}<\frac{1}{\sqrt{n}} \implies \frac{1}{\sqrt[n]{n}}\underset{n\to \infty}{\to} 0$$

I think I have done something wrong. Can I even use the first inequality here? and if yes-do I have to prove that too? and is it Ok that I write too little words in my proofs?. Could you please help me?


问:
Proof for a Statement on Series
In my textbook after this theorem comes another statement(I wasn't sure to call it a theorem). The proof for the theorem is given but I don't know why the statement is true. Can you help me prove it?

Theorem If $\Sigma^{\infty}_{n=1}a_n$ converges, then $\lim_{n\to\infty}a_n = 0$. Therefore, if $\lim_{n\to\infty}a_n$ does not exist, or exists but is not zero, then the series $\Sigma^{\infty}_{n=1}a_n$ is divergent.

Statement If the sequence of terms $\{a_n\}$ tends to a non-zero limit $L$, then $\Sigma^{\infty}_{n = 1}a_n$ diverges to infinity if $L > 0$ and diverges to negative infinity if $L < 0$.

答: If $\lim_{n\to\infty}a_n=L>0$, then there is some $N\in\Bbb N$ such that$$n\geqslant N\implies|a_n-L|<\frac L2\implies a_n>\frac L2.$$Therefore, if $n>N$,$$\sum_{k=1}^na_k=\sum_{k=1}^Na_k+\sum_{k=N+1}^na_k\geqslant\sum_{k=1}^Na_k+(n-N)\frac L2$$and so, since$$\lim_{n\to\infty}\sum_{k=1}^Na_k+(n-N)\frac L2=\infty,$$your series diverges to $\infty$. The argument is similar if $L<0$.

Title: Proof for a Statement on Series


问:
When does $Ax=0$ have non zero solutions?
Let $k > n$. Let $A$ be an $n\times k$ real matrix. I am interested in the non-zero solutions of the system of $Ax=0$. More precisely, I am looking for the most general conditions that $A$ must satisfy such that the system $Ax=0$ admits non zero solutions. Any idea ?

答:
let $A_1,A_2,...,A_k$ be column vectors of $A$ and $x=\begin{pmatrix} x_1\\ x_2 \\ \vdots \\x_k \end{pmatrix}$

$Ax=A\begin{pmatrix} x_1\\ x_2 \\ \vdots \\x_k \end{pmatrix}=x_1A_1+x_2A_2+...+x_kA_k=0$

$k>n \Rightarrow RankA \leq n$

$RankA \leq n <k \Rightarrow $ $\exists A_i:A_i=c_1A_1+...+c_{k-1}A_k$

$A_i=c_1A_1+...+c_{k-1}A_k \Rightarrow x_1A_1+...+x_iA_i+x_kA_k=(x_1+c_1x_i)A_1+...+(x_k+c_{k-1}x_i)A_k=0$

note that $x_i=1$ and $x_a=-c_a$ for all $i \neq a$ is a non-zero solution if $A_i\neq 0$

if $ A_i=0 $ then $x_1=...=x_k=0\neq x_i $ and $x_i=1$ is a non zero solution for equation

so $k > n$ is most general condition that $A$ has a non zero solution


问:
Is it possible to write any bounded continuous function as a uniform limit of smooth functions

Is $C^\infty(\mathbb{R})\subset C_b(\mathbb{R})$ dense? I.e. is any continuous bounded function $f:\mathbb{R}\to\mathbb{R}$ the uniform limit of smooth functions?

On any bounded interval this is true since by Stone-Weierstrass polynomials are smooth and dense in the continuous functions. I guess the result is still true for $\mathbb{R}$ but I don't know how to prove it. Since I don't have any integrability conditions on $f$ the technique of convoluting with mollifiers is probably not applicable.

答:
Given continuous $f$ and $\epsilon>0$, find for each $n\in\mathbb Z$ a smooth approximation $g_n\colon [n-1,n+1]\to\mathbb R$ with $\left\lVert f|_{[n-1,n+1]}-g_n\right\rVert_\infty<\epsilon$.
Glue these together with a smooth partition of one.

For example: Start with a smooth function $\phi\colon[0,1]\to [0,1]$ with $\phi(x)=1$ for $x\le\frac13$ and $\phi(x)=0$ for $x\ge \frac23$.
Then $$g(x):=\phi(x-\lfloor x\rfloor)g_{\lfloor x\rfloor}(x)+(1-\phi(x-\lfloor x\rfloor))g_{\lceil x\rceil}(x)$$
is smooth (the trick is that $g(x)=g_n(x)$ if $|x-n|<\frac13$) and at each $x$ we have $|f(x)-g(x)|<\epsilon$ because it is a convex combination. Therefore, $\lVert f-g\rVert_\infty<\epsilon$.


问:
Which automorphism groups are well-understood?
Obviously the automorphism groups of finite dimensional vector spaces are studied in great detail. By changing presentations of the vector space (by which I mean isomorphisms to some $K^{\oplus n}$) we can characterize automorphisms as matrices in $GL(n,K)$ and bring them in suitable shapes (thinking of Smith normal form, Jordan normal form etc).

Categorically this might be due to the very special fact that every vector space is free (in particular there are no relations between generators) and that finite products and finite coproducts coincide, making computing automorphisms rather easy.

But other than linear algebra I could not find other fields in mathematics, where automorphism groups are studied in that detail, which might be due to the presence of said relations. In fact it is common to reduce the study of some structure to linear algebra (e.g. via representation theory) to exploit matrices.

So I was wondering, whether there are other categories, in which automorphism groups are of similar use, while being somewhat accessible/computable. Let me give some examples, to clarify what I mean.

In Galois theory automorphisms of a field extension $L\mid K$ play a central role. I don’t know how easily they can be computed in specific instances though...

Do the classification results of linear algebra carry over to automorphisms of finitely generated abelian groups? It seems like there we can exploit the structure theorem to get at least a block diagonal matrix form. I was unable to find references...

Automorphism groups of simple graphs seem to be rather difficult to compute, so it doesn’t seem like there is a general theory of automorphism groups of simple graphs. I would guess there are certain subcategories of good graphs, where automorphisms are  better behaved though...

答:
Let's say we are talking about groups; though rings and fields, among others, are also interesting.

Due to Cayley's theorem, every group can be embedded in a symmetric group.

For $\mathfrak S_n$ (finite symmetric groups), other than $n=6$, all the automorphisms are inner.

If $n\ne2,6$, then $\rm{Aut}\mathfrak S_n\cong\mathfrak S_n$.

Also, due to the fact that there is only one way to extend an automorphism from $\mathfrak A_n$, the $n$-th  alternating group, to $\mathfrak S_n$, we get that $\rm{Aut}\mathfrak S_n\cong\rm{Aut}\mathfrak A_n$.

Next, I would like to mention the automorphisms of finite cyclic groups, $\Bbb Z_n$.  They are fairly easy to understand, since an automorphism has to take a generator to a generator.  We easily get that $\rm{Aut}\Bbb Z_n\cong\Bbb Z_n^\times$.

The infinite cyclic group, $\Bbb Z$,  on the other hand, has just two automorphisms, one the trivial one and one sending $1$ to $-1$.


问:
Is it possible to do the following approximation $\Pr({\frac{f}{{XY}}<Z<\frac{f}{a}})\approx\Pr({Z<\frac{f}{a}})-\Pr({\frac{f}{{XY}}<z})$?
Is it possible to do the following approximation ?

$\Pr \left( {\frac{f}{{XY}} < Z < \frac{f}{a}} \right) \approx \Pr \left( {Z < \frac{f}{a}} \right) - \Pr \left( {\frac{f}{{XY}} < z} \right)$

Where $X,Y,Z$ are exponential random variable and $f,a>0$.

Thank you very much !

答:
These are two different values. Is it true that $0.42 \approx 0.55$? In order to speak about approximation, we must have limit.

We have  $X \sim \exp(\lambda_x)$, $Y \sim  \exp(\lambda_y)$, $Z \sim  \exp(\lambda_z)$. Hence $\lambda_x X \sim \exp(1)$ and $\lambda_y Y \sim \exp(1)$.

Suppose that $\lambda_x \lambda_y \to 0$, $f$, $a$ and $\lambda_Z$ i are fixed. Then $\frac{f}{XY} \to 0$ in distribution.

As $P \left( {\frac{f}{{XY}} < Z < \frac{f}{a}} \right)
 \approx \Pr \left( 0 < Z < \frac{f}{a} \right) =\Pr \left( Z \le \frac{f}{a} \right) $

$$ \Pr \left( {Z < \frac{f}{a}} \right) - \Pr \left( Z <  {\frac{f}{{XY}} } \right)  \approx
\Pr \left( {Z < \frac{f}{a}} \right) -  \Pr \left( {Z < 0} \right)  = \Pr \left( Z \le \frac{f}{a} \right). $$
Thus
$$P \left( {\frac{f}{{XY}} < Z < \frac{f}{a}} \right) \approx \Pr \left( {Z < \frac{f}{a}} \right) - \Pr \left( Z <  {\frac{f}{{XY}} } \right)  $$
$$ \approx \Pr \left( Z \le \frac{f}{a} \right) = 1 - e^{-\lambda_z\frac{f}a}$$


答:
We can test for monotonicity like this:

$x_{n+1}-x_n={x_n}^2-\frac{x_n}{2}-x_n=x_n(x_n-3/2)$

This means that for $x_0<0$ and $x_0>3/2$ the sequence is monotonically increasing, and for $0<x_0<3/2$ the sequence is monotonically decreasing. The sequence is constant (and therefore converges) for $x_0=0$ and for $x_0=3/2$.

If we take the case where $0<x_0<3/2$, we just need to show (using induction) that $0<x_n$, because we know that for these values of $x_0$ the sequence will be monotonically decreasing. Therefore, by finding a lower bound we can prove that the sequence converges. Then we can prove that the limit of the sequence when $n$ goes to infinity will be $0$.

For $x_0<0$ we can do the same but this time finding an upper bound (that upper bound will be zero), as the sequence is monotonically increasing for those values of $x_0$.
Lastly, for $x_0>3/2$ it is easy to show that the sequence will diverge, because it is monotonically increasing and there are no possible upper bounds.

Hope this helps

Title: For what $x_{0}$ this sequence converges?

Body:
$$X_{n+1} = X_{n}^2 - X_{n}/2, \quad n \geq 0$$

For what values of $x_{0}$ this sequence converges? For what value?

I think to solve this problem separating it in three steps:

The series is constant. Ok

The series is crescent:

$X_{n+1} > X_{n}$

And for this case we need to require this:
$X_{n} - X_{n-1} > X_{n+1} - X_{n} $

The series is decrescent:

$X_{n} > X_{n+1}$

And for this case we need to require this:
$X_{n} - X_{n+1} < X_{n-1} - X_{n}$

Now, i think this is not the best approach, it is long and extremely easy to be made an algebraic mistake. There is another way to do that?


问:
Differential equation of Geodesics of a Cone
I am working through Tom Kibbles' Classical Mechanics and am struggling to derive an equation in the excercises on calculus of variations.

The question is to find the geodesics of a cone with semi-vertical angle $\alpha$. Let the cone by specified by the distance r from the vertex and azimuth angle $\phi$ about the axis. You are told to prove:

$$ r\frac{d^2r}{d\phi^2} - 2\left ( \frac{dr}{d\phi } \right )^2 -r^2sin^2\alpha = 0  $$

My Attempt:

I started with $ ds = \sqrt{dr^2 + r^2sin^2(\alpha) d\phi^2} $

Then my function to maximize was $ F = \sqrt{\left ( \frac{dr}{d\phi } \right )^2 + r^2sin^2(\alpha)}$

However the problem I am having is when I calculate $\frac{\partial F}{\partial r} $ and $ \frac{\partial F}{\partial r'}$ I get the answer above without the factor of 2 infront of $r'^2$.

My equations for $\frac{\partial F}{\partial r} $ and $ \frac{\partial F}{\partial r'}$ where:

$$ \frac{\partial F}{\partial r} = \frac{rsin^2(\alpha)}{\sqrt{\left ( \frac{dr}{d\phi} \right )^2+r^2sin^2(\alpha)}}$$

$$ \frac{\partial F}{\partial r} = \frac{\frac{dr}{d\phi}}{\sqrt{\left ( \frac{dr}{d\phi} \right )^2+r^2sin^2(\alpha)}}$$

Any advice on where I may have made a mistake would be appreciated thank you.

答:
The partial derivatives $\frac{\partial F}{\partial r},\frac{\partial F}{\partial r'}$ are well developed. The problem perhaps appears on $\frac{d}{d\phi}\left(\frac{\partial F}{\partial r'}\right)$ which gives

$$
\frac{d}{d\phi}\left(\frac{\partial F}{\partial r'}\right)=\frac{\sin ^2(\alpha ) r(\phi ) \left(r(\phi ) r''(\phi )-r'(\phi )^2\right)}{\left(\sin ^2(\alpha ) r(\phi )^2+r'(\phi )^2\right)^{3/2}}
$$
so following

$$
\frac{\partial F}{\partial r}-\frac{d}{d\phi}\left(\frac{\partial F}{\partial r'}\right)=\frac{\sin ^2(\alpha ) r(\phi ) \left(\sin ^2(\alpha ) r(\phi )^2-r(\phi ) r''(\phi )+2 r'(\phi )^2\right)}{\left(\sin ^2(\alpha ) r(\phi )^2+r'(\phi
   )^2\right)^{3/2}}=0
$$


答:
Think about it this way: if your sequence $a_n$ converges to a positive number $L>0$, then the series, which is the sum of the infinite terms of the sequence, will come to a point where it will basically be summing $L$ an infinite number of times. This can only result in the series diverging to $+\infty$.

On the other hand, if $L<0$, you are adding a negative value an infinite number of times. This can only result in the series diverging to $-\infty$.

The only way that a series can converge is if it comes to a point where the infinite terms you are adding are basically equal to zero. This requires $L=0$.

Title: Proof for a Statement on Series

Body:
In my textbook after this theorem comes another statement(I wasn't sure to call it a theorem). The proof for the theorem is given but I don't know why the statement is true. Can you help me prove it?

Theorem If $\Sigma^{\infty}_{n=1}a_n$ converges, then $\lim_{n\to\infty}a_n = 0$. Therefore, if $\lim_{n\to\infty}a_n$ does not exist, or exists but is not zero, then the series $\Sigma^{\infty}_{n=1}a_n$ is divergent.

Statement If the sequence of terms $\{a_n\}$ tends to a non-zero limit $L$, then $\Sigma^{\infty}_{n = 1}a_n$ diverges to infinity if $L > 0$ and diverges to negative infinity if $L < 0$.


问:
Is $V$ a vector space in this case?
This shows up in the exercise of Halmos's book of Linear Algebra.

Let $P$ be the set of all polynomials, with complex coefficients, in a variable $t$. Let $V$ be a subset of $P$, consisting of vectors(polynomials) $x$ for which $x(t)=x(1-t)$ for all $t$. The question asks me to decide if $V$ is a vector space. I'm not sure if I understand the given equation. Can somebody explain this?

答: We can substitute $t = 1/2+y$ to get $x(1/2+y) = x(1/2 - y)$. Thus $x(t)$ is symmetric about $t=1/2$. Does this help visualise better?

Title: Is $V$ a vector space in this case?


答:
Assume $(n!)^\frac{1}{n}$ is bounded above by some $K$, i.e.

$$(n!)^\frac{1}{n}\le K$$

Then $$n!\le K^n$$

Let $n=K^2+K$. Then $$n!=1\ \cdot 2 \cdots K\cdot (K+1) \cdots K^2 \cdot (K^2+1) \cdots (K^2+K)$$

The product of the first $K$ terms with the last $K$ terms is greater than $K^{2K}$, and every remaining term is greater than $K$.

We have $K^2+K$ terms in total, each one (or a pairing of) is greater than $K$ (or $K^2$), hence $(K^2+K)!\gt K^{K^2+K}$, a contradiction.


问:
product of zeta(k)
The product $\prod_{k\geq 2}\zeta(k)$ converges. Is the limit a known constant ?

(This infinite product is involved in the estimation of the covolume of $SL_n(\mathbf Z)\backslash SL_n(\mathbf R)$...)

Thanks !

答:
It seems to be known only as $\ \prod_{k\geq 2}\zeta(k)$.

See OEIS entry A021002 for references especially Bernd Kellner's paper 'On asymptotic constants related to products of Bernoulli numbers and factorials'.

It appeared in Steven Finch's excellent 'Mathematical Constants' page 274 (the link appeared broken at OEIS) as 'the average number of non-isomorphic abelian groups of any given order' with additional references or in connection with the 'arithmetic properties of class numbers of finite simple groups' in Henry Cohen's classical book of Computational Algebraic Number Theory.

It appeared too in Simon Plouffe table of constants (returning these various expansions).

Title: product of zeta(k)


问:
How can I integrate $\int_0^1 (1-x^{\frac{1}{q}})^p\, dx$
I start with a substitution:
$$\int_0^1 (1-x^{\frac{1}{q}})^p\, dx $$
we let $u = x^{\frac{1}{q}}$ meaning that $du = \frac{x^{q-1}}{q}dx$:
$$ q\int_0^1 (1-u)^pu^{q(1-q)}du $$
Using integration by parts I get:
$$ \int_0^1 (1-u)^pu^{q(1-q)}du = q\left(\left[\frac{u^{q(1-q)} (1-u)^{p+1}}{p+1} \right]_0^1 -\frac{q(1-q)}{p+1}\int_0^1u^{q(1-q)-1}(1-u)^{p+1}du \right) $$
Simplifying:
$$ \int_0^1 (1-u)^pu^{q(1-q)}du =  -\frac{q^2(1-q)}{p+1}\int_0^1u^{q(1-q)-1}(1-u)^{p+1}du $$
I'm currently a bit lost as I feel like this is just going in circles. What should I do thanks!

答:
Edit: There is a mistake in your calculations: $u = x^{\frac{1}{q}} \implies du = \frac{x^\frac{1-q}{q}}{q}dx$

Then $I = q\int_0^1 (1-u)^pu^{q-1}du = \beta(p+1,q)$.

This is assuming $p+1>0, q>0$.

I am leaving my previous solution with your mistake in the block below

You can use beta integral right after the substitution.

$$ \require{enclose} \enclose{horizontalstrike}{\ q\int_0^1 (1-u)^pu^{q(1-q)}du = q\beta (p+1, q-q^2+1)} $$
This is assuming $\enclose{horizontalstrike}{p+1>0, q-q^2+1>0}$.

Here's a link:
https://en.m.wikipedia.org/wiki/Beta_function


问:
$\log(1-\cos (x-t)) \in \mathcal{L}^2([0,2\pi]^2)$

I'm currently studying on symmetric kernels of integral equations and here is the problem from Raisinghania textbook "Integral Equation and Boundary Value Problems". It is obvious that the kernel is symmetric since $$K(x,t)=\log(1-\cos (x-t))=\log(1-\cos (t-x))=K(t,x).$$
My question here is how can we know that $\log(1-\cos (x-t)) \in \mathcal{L}^2([0,2\pi]^2)$. It's not in the exercise questions, it's just my thoughs. We see that $\forall x,t:$ $x=t$ we have $K(x,x)=\log(0)= - \infty$. This part confuses me the most to prove that $K(x,t) \in \mathcal{L}^2([0,2\pi]^2)$. Any explanation would be helpful.

答:
From the decomposition
$$ K(x,t)=-\log(2)-2\sum_{n\geq 1}\frac{\cos(nx)\cos(nt)}{n}-2\sum_{n\geq 1}\frac{\sin(nx)\sin(nt)}{n} $$
it is straightforward that $K\in \mathcal{L}^2((0,2\pi)^2)$, since $\left\{\frac{1}{n}\right\}_{n\geq 1}\in\ell_2$.


问:
An example of two functions whose limits don't exist at a but the limit of their product does
Show by example that the $\lim \limits_{x \to a} [f(x) \cdot g(x)]$ may exist even though neither $\lim \limits_{x \to a} [f(x)]$ nor $\lim \limits_{x \to a} [g(x)]$ exist.

答:
Consider the limit at $0$ for the functions:

$$f(x) = \chi_{\mathbb R_{\ge 0}} = \begin{cases} 1 &\text{ for } x \ge 0\\0 &\text{ for } x < 0\end{cases}, \quad g(x) = 1-f(x)$$

then both $\lim_{x \to 0} f(x)$ and $\lim_{x \to 0} g(x)$ do not exist.

However $f(x)g(x)\equiv 0$ for all $x$, so $\lim_{x \to 0} f(x)g(x) = 0$.

Alternatively, if $a = \pm \infty$, the limits of $\tan x$ and $\cot x$ do not exist at $a$.

However $\tan x \cot x \equiv 1$ almost everywhere, so the limit of their product is $1$.


问:
Irreducible polynomials have distinct roots?
I know that irreducible polynomials over fields of zero characteristic have distinct roots in its splitting field.

Theorem 7.3 page 27
seems to show that irreducible polynomials over $\Bbb F_p$ have distinct roots in its splitting field (and all the roots are powers of one root). Is the proof correct? I have never seen this result anywhere else. The proof is very convincing to me.

Does the result hold for $\Bbb F_q$ where $q$ is a power of prime? I don't think it holds because I've heard there are irreducible polynomials with repeated roots?

Please help.

答:
The result holds over any finite field. One way of seeing this is that if $h(x) \in \Bbb{F}_{q}[x]$ (where $q$ is a power of the prime $p$) is irreducible over $\Bbb{F}_{q}$, and $\alpha$ is one of its roots, then $\alpha$ is also algebraic over $\Bbb{F}_{p}$. If $f(x) \in \Bbb{F}_{p}[x]$ is the minimal polynomial of $\alpha$ over $\Bbb{F}_{p}$, then $h(x)$ divides $f(x)$, and you know that the latter has distinct roots.

There are examples, though, of irreducible polynomials of degree $> 1$ over an infinite field of positive characteristic which have only one root.


问:
Does there exist a closed non-orientable $2$-manifold embedded in $\mathbb{R}^4$ such that it's complement is disconnected with $2$ components?
So to give some context, in my physics class we were introduced to Gauss's law which states that for a closed orientable surface in $\mathbb{R}^3$ the total flow through it is related to the electrical charge enclosed in it, by the formula: $$\epsilon_0 \Phi = q_{enc},$$
where $\epsilon_0$ is the permitivity constant of space.

So naturally I noticed that the orientability clause gives us access to notions such as inside and outside of the surface. But I started to wonder if such notions are meaningful for non-orientable surfaces embedded in higher dimensional spaces as well.

So I would like to know if (Q1) there exists a closed non-orientable $2$-manifold embedded in $\mathbb{R}^4$ such that it's complement is disconnected with $2$ components, one bounded and the other, unbounded.

Since I suspect that higher codimension might ruin this notion altogether, (Q2) what happens if we drop the non-orientability requirement in (Q1)?

答:
A smooth $2$-dimensional surface cannot separate $\mathbb{R}^4$ into two components. More precisely, one can apply the following result:

If $M\subset N$ is an a an embedded submanifold with $\operatorname{codim}M\ge 2$, and $N$ is connected, then $N\setminus M$ is also connected.

This result can be shown using transversality. Essentially, given any smooth path $\gamma:[0,1]\to N$ between points $a,b\in N\setminus M$, we can always deform $\gamma$ very slightly to avoid any intersections with $M$. This means that any points connected by a path in $N$ are also connected by a path in $N\setminus M$.