问:
How do you prove that the recurrence $T(n) = 2T(\lfloor{n/2}\rfloor) + n$ is $\Omega(n \lg n)$?
I'm working through CLRS (Introduction to Algorithms, 3rd edition) and I'm in the section about using the "substitution method" and induction to prove asymptotic bounds for recurrences. I'm trying to solve this problem (4.3-3) and am running into problems dealing with the floor function.

Show that $T(n) = 2T(\lfloor{n/2}\rfloor) + n$ is $\Omega(n \lg n)$

This was my original attempt for the substitution step, and I also see this published online in a non-official solution guide:

$$
\begin{align}
T(n) &\ge 2c\lfloor n/2 \rfloor \lg (\lfloor n/2 \rfloor) + n\\
&\ge cn \lg (n/2) + n\\
&= cn \lg n - c n \lg 2 + n\\
&= cn \lg n \quad \text{if} \space c \le 1
\end{align}
$$

But removing the floor functions directly between lines 1 and 2 seems like it could invalidate the inequality. I found another solution guide that gets around this by substituting $n/4$ for $\lfloor n/2 \rfloor$ but ends with this, which also seems to invalidate the inequality:

$$
\begin{align}
T(n) &\ge c(n/2)\lg n - cn + n\\
&\ge cn \lg n \quad \text{if} \space c \le 1
\end{align}
$$
This also doesn't seem to hold, for example, if $c = 1$. Are these approaches both just wrong or am I missing something? What's the best way to deal with the floor division in this problem?

答:
You want to show that there is some $c>0$ such that for all $n$ large enough $T(n) \geqslant cn\lg n$.

I present here a hopefully intuitive inductive approach.
We will first check how the inductive step goes.

Suppose that for some $k\geqslant 1$ we have $T(k)\geqslant ck\lg k$.

Can we show that this relation also holds for some other (greater) values $n$?

If we follow the recursive definition of $T$, we'll see that $T(2k)$ and $T(2k+1)$ can both be expressed in terms of $T(k)$.

First, consider the case $n = 2k$.
Then $T(n) = 2T(k) + 2k$ and we'll have $T(n) \geqslant c(2k)\lg(k) + 2k$.

It's easy to check that $c\leqslant 1$ implies $c(2k)\lg(k)+2k \geqslant c(2k)\lg(2k)$, regardless of the value of $k$.
In this case, $T(n)\geqslant cn\lg n$ as desired.

Now, consider the case $n = 2k+1$.
Then $T(n) = 2T(k) + 2k + 1$ and we'll have $T(n)\geqslant c(2k)\lg(k)+(2k+1)$.
Under what conditions on $c$ do we have

$$c(2k)\lg(k)+(2k+1) \geqslant c(2k+1)\lg(2k+1)\,\,?$$

That is equivalent to

$$c\leqslant \frac{2k+1}{(2k+1)\lg(2k+1)-2k\lg(k)}.$$

You only really need this to be $\geqslant \epsilon>0$ as $k\to\infty$, and it's easy to show that.
For instance, you could show that the RHS converges to $1$ as $k\to\infty$.

That said, you could also show that the minimum is attained at $k=1$ $($with value $\log 2/\log 3\approx 0.63 < 1)$, and this will save us some work in the end with base cases by allowing us to consider base cases right from the get go ($k=1$).

We hence conclude that $c\leqslant \log 2/\log 3$ implies $T(2k+1) \geqslant c(2k+1)\lg(2k+1)$, or $T(n)\geqslant cn\lg n$, once again regardless of the value of $k$.

Combining both cases, we conclude that for $c\leqslant \log 2/\log 3$, $T(k)\geqslant ck\lg k$ implies that both $T(2k)\geqslant c(2k)\lg(2k)$ and $T(2k + 1)\geqslant c(2k + 1)\lg(2k + 1)$.

This will be our inductive step.

To complete the induction, we must now handle the base cases.
Our inductive step shows that the relation can be propagated forward from a base case, but not in the traditional way (from $n$ to $n+1$).
Rather, our inductive step is a sort of doubling ($n$ to $2n$ and $2n+1$), which in general leaves gaps.

Indeed, for the sake of example, suppose we took a single base case of $k = 7$.
Our inductive step would propagate the relation to $n\in\{14, 15\}$, and from these we'd get $n\in\{28,29,30,31\}$, and so on and so forth.
It's easy to see that we'd miss many values.

We need a modified induction for our modified inductive step.

Lemma (Modified induction):
Let $P(n)$ be some proposition about $n$.

Suppose that if $P(n)$ is true, then $P(2n)$ and $P(2n+1)$ are also true.

Additionally, suppose that there is some $j\geqslant 0$, such that $P(n)$ is true for all $n$ with $2^j\leqslant n <2^{j+1}$.

Then $P(n)$ is true for all $n\geqslant 2^j$.

Once you've proved the lemma, we can finish the exercise.
If we take $j=0$ in the lemma, we'd need only show that $P(1)$ is true.
And indeed, it is.

We can calculate the first few values of $T$ by hand: $T(0) = 0$ and $T(1) = 1$.
The proposition $P(1)$ is then $T(1) = 1 \geqslant c\lg 1 = 0$, which is obviously true and concludes the induction.

This approach also helps show where tightness was lacking in the answers provided. Looser values of $c$ (meaning values closer to $1$) will require larger values of $k$ in order for the inductive step $k\mapsto 2k+1$ to work.

And, indeed, for $c\geqslant 1$ (including $c=1$) the inductive step does not hold for any $k$.


问:
simple and uniform convergence of $ \sum ( x+\frac{1}{n})^{n+\frac{x}{n}} $
It's my first post in this forum ! Hello ! (I have a very bad english sorry in advance...) I come with an exercice that I can not solve. I hope you can help me...
Q1 : simple and uniform convergence of $ \sum ( x+\frac{1}{n})^{n+\frac{x}{n}} $

Here my guess and what I have done the convergence is |x|< 1 (that I didn't proved just guessed)
I tried limited development it didn't worked.

Thank you for reading !

答:
For the convergence, using the ratio test, let
$$a_n=\left( x+\frac{1}{n}\right)^{n+\frac{x}{n}}\implies \log(a_n)=\left(n+\frac{x}{n}\right) \log \left(x+\frac{1}{n}\right)$$ Now, using Taylor series for large values of $n$
$$\log(a_n)=n \log (x)+\frac{1}{x}+\frac{2x^3 \log (x)-1}{2x^2n}+O\left(\frac{1}{n^2}\right)$$ Apply it twice and continue with Taylor series
$$\log(a_{n+1})-\log(a_n)=\log (x)+\frac{1-2x^3 \log
   (x)}{2x^2n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} } {a_n}=\exp\big(\log(a_{n+1})-\log(a_n) \big)=x+\frac{1-2 x^3 \log (x)}{2 n^2 x}+O\left(\frac{1}{n^3}\right)$$

ContentLicense: CC BY-SA 4.0


问:
Integration of $1/(1+\sin x)$
I solved it using $t=\tan(\frac{x}{2})$ substitution and got $-2/(1+\tan(x/2))+C$, but in my math book solution is $\tan(x/2-\pi/4)+C$. Are those the same expressions and if they are, how do I transform from one to another, or are one(or both) solutions incorrect ? </p>

答:
To some extent:
$$I=\int \frac{d x}{1+\varepsilon \sin x}=\int \frac{\sin^2 \frac{x}{2}+\cos^2 \frac{x}{2}}{\sin^2 \frac{x}{2}+\cos^2 \frac{x}{2}+2\varepsilon\sin(\frac{x}{2})\cos(\frac{x}{2})}dx\\=2 \int \frac{d t}{(1+t^2)+2\varepsilon t} \quad(t=\tan\frac{x}{2})\\=2\int\frac{dt}{(t+\varepsilon)^2-\varepsilon^2+1}=2\int\frac{d(t+\varepsilon)}{(t+\varepsilon)^2-\varepsilon^2+1}$$
Let $\varepsilon=\pm1$, then
$$I=2\int\frac{dm}{m^2}=-\frac{2}{m}+C=-\frac{2}{t\pm1}+C=-\frac{2}{\pm1+\tan\frac{x}{2}}+C$$
Moreover, for $\varepsilon\in(-1,1)$:
$$I=\int \frac{d x}{1+\varepsilon \sin x}=\frac{2}{\sqrt{1-\varepsilon^2}}\int\frac{dm}{1+m^2}\\=\frac{2}{\sqrt{1-\varepsilon^2}}\arctan m+C=\frac{2}{\sqrt{1-\varepsilon^2}}\arctan(\frac{\tan\frac{x}{2}+\varepsilon}{\sqrt{1-\varepsilon^2}})+C$$
for $\varepsilon>1$ or $\varepsilon<-1$:
$$I=\int \frac{d x}{1+\varepsilon \sin x}=-2\int\frac{d(t+\varepsilon)}{-(t+\varepsilon)^2+\varepsilon^2-1}\\=\frac{1}{\sqrt{\varepsilon^2-1}}\ln |\frac{\tan\frac{x}{2}+\varepsilon-\sqrt{\varepsilon^2-1}}{\tan\frac{x}{2}+\varepsilon+\sqrt{\varepsilon^2-1}}|+C$$

ContentLicense: CC BY-SA 3.0

Title: Integration of $1/(1+\sin x)$


问:
How to compute $\int_0^\infty \frac{\tanh\left(\pi x\right)}{x\left(1+x^2\right)} \, \mathrm{d}x$?
How do I compute the following integral?

$$\int_0^\infty \frac{\tanh\left(\pi x\right)}{x\left(1+x^2\right)} \, \mathrm{d}x$$

I tried some basic substitutions but they only make it more complicated. WolframAlpha says the answer is $2$ but I have no clue how to get there.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\tanh\pars{\pi x}
\over x\pars{1 + x^{2}}}\,\dd x}
\\ = &\
\int_{0}^{\infty}{1
\over x\pars{1 + x^{2}}}\ \overbrace{\bracks{%
{8x \over \pi}\sum_{n = 0}^{\infty}
{1 \over
\pars{x^{2} + 1}\bracks{4x^{2} + \pars{2n + 1}^{2}}}}}
^{\ds{\tanh\pars{\pi x}}}\dd x
\\[5mm] = &\
{8 \over \pi}\sum_{n = 0}^{\infty}
\int_{0}^{\infty}{\dd x \over \pars{x^{2} + 1}\bracks{4x^{2} + \pars{2n + 1}^{2}}}
\\[2mm] &\ \pars{\substack{\ds{{\large x}\mbox{-integration is straightforward with}}\\[1mm] \ds{Partial\ Fraction\ Decomposition}}}
\\[2mm] = &\
\sum_{n = 0}^{\infty}{1 \over n^{2} + 2n + 3/4} =
\sum_{n = 0}^{\infty}
{1 \over \pars{n + 3/2}\pars{n + 1/2}}
\\[5mm] = &\
\Psi\pars{3 \over 2} - \Psi\pars{1 \over 2}
\label{1}\tag{1}
\\[5mm] = &\
\bracks{\Psi\pars{1 \over 2} + {1 \over 1/2}} -
\Psi\pars{1 \over 2} = \bbx{2}\label{2}\tag{2} \\ &
\end{align}

(\ref{1}): $\ds{\Psi:\ Digamma\ Function}$.

(\ref{2}): $\ds{\Psi}$-$\ds{Recurrence}$.


问:
Characteristic polynomial of the differentiation map
Determine the characteristic and minimum polynomial of the differentiation map $D:  \mathbb{R_n}[X]\longrightarrow \mathbb{R_n}[X]$ (where $\mathbb{R_n}[X]$ is a set of polynomials of degree at most $n$ with real coefficients).
$$$$ My attempt was to find a matrix of $f$ relative to the natural ordered basis {$1,X,X^2,...,X^n$} which is $$\pmatrix{0 & 1 & 0&...&0 \\ 0 & 0 & 2&...&0 \\: & : & :&...&:  \\0 & 0 & 0&...&n \\0 & 0 & 0&...&0 \\ }$$ hence, the characteristic polynomial turns to be $(-1)^{n+1}X^{n+1}$ while the original answer shows that it suppose to be  $(-1)^{n-1}X^{n-1}$. So where did I go wrong?
$$$$Thank you in advance.</p>

答:
$$\dim_{\Bbb R}\Bbb R_n[X]=n+1\implies\;\text{ the char. polynomial must be of degree}\;n+1$$

and this means "the original answer" is wrong.


问:
Antiderivative of $\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}$
Is it possible to explicitly find the antiderivative of $\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}$? It is rather straight-forward to find it when $a=0$, or $b=0$ or $c=0$. This is also easy in the case where $a=b$, or $b=c$, or $a=c$. But I have been struggling to do this in the general case. Any help is appreciated. Thanks!

答:
Without loss of generality let $a>b>c>0$. Substitute $u=x^2$ to get
$$I=\int_p^q\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}\,dx=\frac12\int_{p^2}^{q^2}\sqrt{\frac{(u-a^2)(u-b^2)(u-c^2)}u}\,du$$
This is an elliptic integral and its solution depends on where the bounds are. As an example, let $c=p\le q\le b$ and as a notational convenience let an uppercase variable denote the corresponding lowercase variable squared. Then Byrd and Friedman 254.37 gives
$\newcommand{sn}{\operatorname{sn}}\newcommand{cn}{\operatorname{cn}}\newcommand{dn}{\operatorname{dn}}$
$$I=\frac12\int_P^Q\sqrt{\frac{(u-A)(u-B)(u-C)}u}\,du$$
$$=\sqrt{\frac{A-C}B}(B-C)^2\frac CB\int_0^{u_1}\frac{\sn^2u\cn^2u\dn^2u}{(1-n\sn^2u)^3}\,du$$
where $n=\frac{B-C}B$, the hidden parameter $m$ in the Jacobian elliptic functions is $\frac{(B-C)A}{(A-C)B}$ and $u_1=F(\varphi,m)$ where $\sin^2\varphi=\sn^2u_1=\frac{(Q-C)B}{(B-C)Q}$. In turn, the last integral is given by B&F 362.24 as
$$\frac1{n^3}\left(-mF(\varphi,m)+(3m-nm-n)\Pi(n,\varphi,m)+(2nm+2n-3m-n^2)V_2+(n-1)(n-m)V_3\right)$$
where $V_2$ and $V_3$ are in turn given by B&F 336.00 to .03 as
$$V_0=F(\varphi,m)\qquad V_1=\Pi(n,\varphi,m)$$
$$V_2=\frac1{2(n-1)(m-n)}\left(nE(\varphi,m)+(m-n)F(\varphi,m)+(2nm+2n-n^2-3m)\Pi(n,\varphi,m)-\frac{n^2\sn u_1\cn u_1\dn u_1}{1-n\sn^2u_1}\right)$$
$$V_{j+3}=\frac1{2(j+2)(1-n)(m-n)}\left((2j+1)mV_j + 2(j+1)(nm+n-3m)V_{j+1} + (2j+3)(n^2-2nm-2n+3m)V_{j+2} + \frac{n^2\sn u_1\cn u_1\dn u_1}{(1-n\sn^2u_1)^{j+2}}\right)$$
As you can see, the general solution can get extremely complicated very quickly.


问:
How do we find $P(Z > -1.40)$?
Can we simply do $P(Z \le 1.4)$ ?

I'm confused because in cases like
$P(-1.4 < Z < 1.63)  = P(Z < 1.63) + P( Z < 1.4) - 1$

Why do we subtract one in the second case?

This is the case for $Z \sim (0, 1)$

答:
Strating from @Александр Пальма's answer,
$$\mathbb{P}[Z\leq a]=\int_{-\infty}^{a}f(x)dx=\int_{-\infty}^{a}\frac{1}{\sqrt{2\pi}}\exp\left( -\frac{x^{2}}{2} \right)dx=\frac{1}{2} \left(1+\text{erf}\left(\frac{a}{\sqrt{2}}\right)\right)$$ the error function being available is most environments.

You could also use the approximation proposed by  Sergei Winitzki
$$\text{erf}(t)\sim\text{sgn}(t)\sqrt{1-\exp\left(-t^2\,\frac{\frac{4}{\pi }+\alpha t^2}{1+\alpha t^2} \right)}\quad \text{with} \quad \alpha=\frac{8 (\pi -3)}{3\pi (4-\pi )  }$$ with $a=1.4$, this would give $0.919353$.

Title: How do we find $P(Z > -1.40)$?


问:
Prove that $1+ \frac{5}{4}x ≤(1+x)^{\frac{5}{4}}≤1+ \frac{5}{4} x+\frac{5}{16} x^2$
How to prove this inequality:
$$1+ \frac54x \le (1+x)^{5/4} \le 1+ \frac54 x+\frac5{16} x^2$$

I tried applying binomial theorem to get this inequality, but I am not able to get it precisely. Please help.

答:
Binomial expansion of $$(1+x)^{5/4} = 1+\frac54 \frac{x}{1!} + \frac54 \frac14 \frac{x^2}{2!}$$

Now since $x^2$ is nonnegative we have your claim.


问:
Generator of $SL(2,R)$
Because of Iwasawa Decomposition(link), I know that $SL(2,\mathbb{R})$ is also decomposable by $K,A,$ and $N$, where $K$ is $SO(2)$, $A$ is the set of diagonal matrices with $(x,\frac{1}{x})$, and $N$ is the group of matrices $((1,x),(0,x))$.

However, when I read "Ergodic theory, with a view towards number theory," which is written by Manfred Einsiedler, it states that on the page 284,

Now the subgroup
  U = $\left\{\begin{pmatrix}1 & b \\ 0 & 1 \end{pmatrix}: b \in \mathbb{R}\right\} \leq SL_{2}{(\mathbb{R})}$
  together with

$w = \begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$
  generates $SL_{2}{(\mathbb{R})}$, since
  $wUw^{-1} = \left\{\begin{pmatrix}1 & 0 \\ -b & 1 \end{pmatrix}: b \in \mathbb{R} \right\}$.

But, I don't know how to this $w$ and $U$ generates the whole $SL_{2}(\mathbb{R})$. How can I show that they are generator of $SL_{2}(\mathbb{R})$?

答:
This is 4 years late, but here's an answer for others. All lower and upper triangular matrices mentioned below are to have 1s on the diagonal, they are taken from the given set of generators.

First, we look at the product
$$\begin{bmatrix}1&0\\\alpha&1\end{bmatrix}
\begin{bmatrix}x&0\\0&1/x\end{bmatrix}
\begin{bmatrix}1&\beta\\0&1\end{bmatrix} =
\begin{bmatrix} x&\beta x\\\alpha x& \alpha\beta x+1/x\end{bmatrix}$$

Given $\begin{bmatrix}a&b\\c&d\end{bmatrix}\in SL_2(\mathbb{R})$, when $a\neq 0$, we can solve for $\alpha, x, \beta$. Since we have lower and upper triangular matrices, it then suffices to show that the generators in question also generate the diagonal matrices. We will come to the $a = 0$ case later.

Now we look at a product of the form LULU:
$$\begin{bmatrix} 1&0\\\alpha&1\end{bmatrix}
\begin{bmatrix} 1&\beta\\0&1\end{bmatrix}
\begin{bmatrix} 1&0\\\gamma&1\end{bmatrix}
\begin{bmatrix} 1&\delta\\0&1\end{bmatrix}=
\begin{bmatrix}
1+\beta\gamma&\delta(1+\beta\gamma)+\beta\\
\alpha(1+\beta\gamma)+\gamma&\alpha\delta(1+\beta\gamma)+\alpha\beta+\gamma\delta+1
\end{bmatrix}$$

Assuming $1+\beta\gamma\neq 0$, to get the anti-diagonal terms to be $0$, we take
$$\alpha = \frac{-\gamma}{1+\beta\gamma}, \delta = \frac{-\beta}{1+\beta\gamma}$$
then the bottom right term becomes
$$\frac{\beta\gamma}{1+\beta\gamma}+\frac{-2\beta\gamma}{1+\beta\gamma}+1 = \frac{1}{1+\beta\gamma}$$
as required.

Therefore, a matrix of the form $\begin{bmatrix}x&0\\0&1/x\end{bmatrix}$ is generated by lower and upper diagonal matrices with ones on the diagonal for suitable choice of $\beta, \gamma$.

When $1+\beta\gamma = 0$, the above product becomes
$$\begin{bmatrix}
0&\beta\\
\gamma& 1+\alpha\beta+\gamma\delta
\end{bmatrix}$$

We can take $\delta = 0$ and make suitable choices so that the matrix becomes $\begin{bmatrix} 0&b\\c&d\end{bmatrix}\in SL_2(\mathbb{R})$.

So, any matrix in $SL_2(\mathbb{R})$ is of the form LUL or LULU where the matrices have 1s on the diagonal. The same argument holds good over other fields as well.

Title: Generator of $SL(2,R)$


问:
I am curious about series containing factorials.
Does anyone know any well known series containing factorials that do not have their origins in either Europe nor Asia or have a shared origin and is related to Pi.</p>

答:
$$\pi=\sum_{n=0}^{\infty}{(25n-3)n!(2n)!\over2^{n-1}(3n)!}$$ is due to the American mathematician, Bill Gosper.

$${1\over\pi}={12\over\sqrt{640320^3}}\sum_{k=0}^{\infty}(-1)^k{(6k)!\over(k!)^3(3k)!}{13591409+545140134k\over(640320^3)^k}$$ is due to the American mathematicians (born in Kiev) D. and G. Chudnovsky.


问:
Basis for linearly dependent vectors in R3
Suppose I have a subspace S where $(x_1, x_2, x_3) \in \mathbb{R}^3$ and for $a, b, c\in R$, not all equal to zero, we have
$ax_1 + bx_2 + cx_3 = 0$. What would be the basis for this subspace?

I tend to think that the basis would be less than 3 vectors, because I believe that S should have a dimension of less than 3 due to the linear dependence. But I can't think of how to find 2 vectors that would fulfill this requirement.

答:
Looking at this from a geometrical perspective, you basically have

$$(a,b,c) \cdot (x_1, x_2, x_3) = 0$$

This basically is describing a plane of vectors which is normal to the given vector $(a,b,c)$ - hence the basis would be any two non parallel vectors in this plane


问:
Why is 8 so special?
I have been reading about multi-dimensional numbers, and found out that it's been proven that the Octonions are the composition algebra of the largest dimension. I was wondering why, despite having infinitely many different dimensions of numbers, the only composition algebras are of 1, 2, 4, and 8 dimensions. What's so special about 8?</p>

答: You might be interested in Hurwitz's proof of his theorem (which is not as strong as Wikipedia's statement). Here is the original German and an English translation. The maximal $n$ turns out to be the solution of $2^{n-2} = n^2$.</p>

Title: Why is 8 so special?


问:
If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge?
Let $ \{a_n\}_{n=1}^\infty $ be a sequence such that $\displaystyle \sum a_n $ that is divergent to $+\infty$.

What can be said about the convergence of $\displaystyle \sum \frac{a_n}{1 + a_n} $?

Any hints, thoughts or leads would be greatly appreciated.

Thanks!

答:
It seems that one assumes that $a_n\geqslant0$ for every $n$. Let $b_n=\frac{a_n}{1+a_n}$. If $a_n\geqslant1$ infinitely often, $b_n\geqslant\frac12$ infinitely often hence $\sum\limits_nb_n$ diverges. Otherwise, $a_n\leqslant1$ for every $n$ large enough, say for every $n\geqslant N$, hence $b_n\geqslant\frac12a_n$ for every $n\geqslant N$ and $\sum\limits_{n\geqslant N}b_n\geqslant\frac12\sum\limits_{n\geqslant N}a_n$, which implies that $\sum\limits_nb_n$ diverges.

Finally, if  $a_n\geqslant0$ for every $n$ and if $\sum\limits_na_n$ diverges, then $\sum\limits_n\frac{a_n}{1+a_n}$ diverges.

A classical counterexample when the nonnegativity condition fails is $a_n=\frac{(-1)^{n+1}}{\sqrt{n+1}+(-1)^n}$ for every $n\geqslant1$. Then  $\sum\limits_na_n$ diverges because $a_n=\frac{(-1)^{n+1}}{\sqrt{n+1}}+\frac1{n+1}+O\left(\frac1{n^{3/2}}\right)$ but $b_n=\frac{(-1)^{n+1}}{\sqrt{n+1}}$ hence  $\sum\limits_nb_n$ converges.


问:
Prove that $ d(x,y)=\left\{\begin{matrix} 2-x-y &, \, \text{if}\ x\neq y \\ 0 &,\, \text{if}\ x=y \end{matrix}\right. $ is a metric space
prove if this is a metric space or not:

for $x,y \in [0,1]$, is defined $d:[0,1]\times [0,1]\rightarrow \mathbb{R}$ such that:

$$
d(x,y)=\left\{\begin{matrix}
2-x-y &,\, \text{if}\  x\neq y \\
0 &,\, \text{if}\  x=y
\end{matrix}\right.
$$

I have already proved for a) $d(x,y)\geq 0$ and b) $d(x,y)=d(y,x)$

but I am not sure how to do:

c) $d(x,x)=0$

d) $d(x,y)\leq d(x,z)+d(z,y)$

I am thinking that it is not a metric space but don´t know how to do c) and d) to prove it, if you can give me some insight I would apreciate it a lot

答:
The second one comes straight from the definition.

We can split the third one into two cases:

If $z=x$ or $z=y$, then the equality holds (from the definition), for example if $z = x$, then the inequality can be rewritten to $d(x, y) \leq d(x, x) + d(x, y) = 0 + d(x, y) = d(x, y)$.

Suppose $x \neq y \neq z \neq x$.

$$d(x, z) +d(z, y) = 4 - x - 2z - y = 2 - x - y + (2 - 2z) = d(x, y) + 2(1 - z) \geq d(x, y) $$

Because $1 \geq z$.

P.S.: To address Glare's comment, if $x \neq y$, then suppose $x < y$ (*). Now from the definition we know, that $x < 1$ (strong inequality because of (*)). From that $d(x, y) = 2 - x - y > 0$.


问:
How to find the limit of $f(x)$.
Let $$f(x) = {\frac{e^{e^x - 1} - \frac{1}{1 - x}}{\ln(\frac{1 + x}{1-x}) - 2\sin x}}$$

I need to find a $ \lim_{x \to 0}f(x)$. As I understand, we have to do some Taylor series decomposition and problem should be pretty easy. But I have had some troubles with it.

P.S : I can decompose $\ln(\frac{1 + x}{1-x})$ as $\ln(1 + \frac{2x}{1-x})$ to use some standard Taylor formulas. I know $\sin x$ Taylor series, but I've got some troubles with $e^{e^x-1}$ and some further transformations.

答:
The only problem with Taylor series approach is in finding the series for $e^{e^x-1}$. The series for other terms are available from memory.

Let $t=e^{e^x-1}$ so that $$\frac{dt} {dx} =te^x$$ so that if $$t=a_0+a_1x+a_2x^2+\dots$$ then $a_0=1$ and $$a_1+2a_2x+3a_3x^2+\dots=\left(1+a_1x+a_2x^2+\dots\right)\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots \right) $$ Equating coefficients we get $$a_1=1,2a_2=a_1+1,3a_3=\frac{1}{2}+a_1+a_2$$ so that $$a_0=1,a_1=1,a_2=1,a_3=\frac{5}{6}$$ and thus we get $$e^{e^x-1}=1+x+x^2+\frac{5}{6}x^3+o(x^3)$$ Now you can get the desired limit easily as $$\frac{(5/6)-1}{(2/3)-(-1/3)}=-\frac{1}{6}$$ (only the coefficients upto $x^3$ matter in each term and moreover coefficients of $x^0,x^1,x^2$ cancel).

Title: How to find the limit of $f(x)$.


问:
Solving inequality including logarithm
I am trying to solve an inequality which includes a logarithm. This is to show for what input one algorithms is faster than another. I do not know how to change the inequality so that I can solve it. The problem is as follows:

$8n\log(2n) > n^2$

Thank you for your help.

答:
Since $n>0$, divide both sides and consider the function
$$f(x)=8\log(2x)-x$$ for which
$$f'(x)=\frac{8}{x}-1 \qquad \text{and} \qquad f''(x)=-\frac{8}{x^2}<0 \quad \forall x$$ The first derivative cancels at $x=8$ and by the second derivative test, this is a maximum.

Since $f(8)=32 \log (2)-8 \sim 14.18$, then, if $n$ is a natural number, the inequality holds for $1 \leq n \leq 14$.

Title: Solving inequality including logarithm


问:
Is this textbook's combinatorial reasoning for $n \times n! / (b!(n - b)!)$ incorrect?
I recently asked this question on cs.stackexchange.

I am currently studying the textbook Artificial Intelligence: A Modern Approach, 4th Edition, by Russell and Norvig. Chapter 3 Solving Problems by Searching says the following:

Another type of grid world is the sokoban puzzle, in which the agent's goal is to push a number of boxes, scattered about the grid, to designated storage locations. There can be at most one box per cell. When an agent moves forward into a cell containing a box and there is an empty cell on the other side of the box, then both the box and the agent move forward. The agent can't push a box into another box or a wall. For a world with $n$ non-obstacle cells and $b$ boxes, there are $n \times n! / (b!(n - b)!)$ states; for example on an $8 \times 8$ grid with a dozen boxes, there are over 200 trillion states.

I understand the combinatorial reasoning for why we get $n! / (b!(n - b)!)$ (the binomial coefficient), since we are looking for the number of ways we can choose (with replacement) $b$ boxes from a world with $n$ non-obstacle cells. But I don't understand where the $n$ factor in $n \times n! / (b!(n - b)!)$ comes from. What is the combinatorial reasoning for this $n$ factor?

Tanner Swett gave the following answer:

Part of the state of the world is the location of the agent. Since there are $n$ cells, this variable has $n$ possible values, producing the factor of $n$.

Actually, there may be a mistake in Russell and Norvig's formula. After we have chosen the location of the agent, there are only $n - 1$ possible locations for the $b$ boxes, giving us $(n - 1)! / (b! (n - b - 1)!)$ choices for where to put all of the boxes. So, the number of states is $n \times (n - 1)!/(b!(n - b - 1)!)$, which simplifies to $n! / (b! (n - b - 1)!)$. (This is the trinomial coefficient $n \choose 1, b$.)

I wanted to ask here, since the users of this website are more mathematically-focused: Is Tanner Swett's analysis of this correct? Is this an error in the textbook?

答:
I agree with Tanner that the factor of $n$ is intended to be for the location of the agent and that the stated formula is not quite correct, assuming that the agent can't occupy the same cell as a box which seems like a reasonable assumption to me. An equivalent way to count it is to place the boxes first (which can be done in ${n \choose b}$ ways) and then notice that this leaves $n-b$ unoccupied cells to place the agent. So the true count is

$$(n-b) {n \choose b}$$

which is equivalent to Tanner's count; this actually gives a bijective proof that $(n-b) {n \choose b} = n {n-1 \choose b}$. But this doesn't make a big difference as far as the order-of-magnitude of the number of possible states goes.


问:
For a recursion based on $x+y+z+2\sqrt{xy+yz+zx},$ does what happens in $\mathbb Z\left[\sqrt n\right]$ stay in $\mathbb Z\left[\sqrt n\right]$?
This problem has a geometric origin which I'll outline below, but I believe the concepts and explanation are algebraic.

Given a function on triples
$$K((x,y,z))=x+y+z+2\sqrt{xy+yz+zx}$$

we build a general recursion as follows.

Start with an integer triple $X=(a,b,c)$, and a set $S=\{a,b,c\}$.

Compute $k=K(X)$ and add $k$ to $S$.  Writing $X$ as $(X_0,X_1,X_2)$, repeat step 2 for the triples $X=(k,X_1,X_2),(X_0,k,X_2),(X_0,X_1,k)$ respectively.

In general, the members of $S$ will contain nested radicals.  For example, if we start with the triple $(1,2,3)$ and use a restricted recursion  $x_{n+1}=K((1,2,x_n)),x_1=3$ we get

$$
S'=\left\{3,6+2 \sqrt{11},9+2 \sqrt{11}+2 \sqrt{2+3 \left(6+2 \sqrt{11}\right)},\\12+2
   \sqrt{11}+2 \sqrt{2+3 \left(6+2 \sqrt{11}\right)}+2 \sqrt{2+3 \left(9+2
   \sqrt{11}+2 \sqrt{2+3 \left(6+2 \sqrt{11}\right)}\right)},\dots \right\}
$$

However, Mathematica is able to denest these to

$$
S'=\left\{3,6+2 \sqrt{11},15+4 \sqrt{11},30+6 \sqrt{11},\dots\right\}
$$

suggesting the conjectures that

starting with initial triple $(1,2,3)$,  $S'\subset S\subset\mathbb Z[\sqrt {11}]$

starting with integer triple $(a,b,c)$, $S\subset \mathbb Z[\sqrt n]$, where $n=ab+bc+ca.$

Question: is the latter conjecture $S\subset \mathbb Z[\sqrt n]$ true, and if so why?

Bonus question: is there some characterization or invariant of the triples generated in the general recursion?

I've done some hand and computer calculations to understand why the radicals denest in particular cases, but I'm guessing that if the conjecture is true it will be "obvious" to someone more versed in this area.

Some background and observations:

This problem originates in playing with the construction of Apollonian gaskets and the function $K()$ comes from a theorem of Descartes concerning the curvatures of 4 mutually tangent circles.

In the case that $n=ab+bc+ca$ is a square, then all the numbers generated will be integers. And this leads to integral Apollonian circle packings.  But I have not seen any discussions of the case when curvatures are in some $\mathbb Z[\sqrt n]$.

see also this demo

答:
We simply define $x_1=3$ and $x_{n+1}=1+2+x_n+2\sqrt{1\cdot2+2x_n+1x_n}$ and thus:
$$x_{n+1}=3+x_n+\sqrt{2+3x_n}$$
We simply prove by induction that $\sqrt{2+3x_n} \in \mathbb{Z}[\sqrt{11}]$. You have already covered the base case. Taking the value $2+3x_n=(a+b\sqrt{11})^2$, we then have:
$$x_{n+1}=3+x_n+a+b\sqrt{11}$$
$$2+3x_{n+1}=11+3x_n+3(a+b\sqrt{11})$$
$$(2+3x_{n+1})=(2+3x_n)+2\cdot3\cdot(a+b\sqrt{11})+3^2$$
$$(2+3x_{n+1})=(a+b\sqrt{11}+3)^2$$

We may also form the general expression for the values in the sequence. Notice that the value $\sqrt{2+3x_{n+1}}=3+\sqrt{2+3x_n}$ and hence, we get:
$$\sqrt{2+3x_{n}}=3(n-1)+\sqrt{2+3x_1}=3(n-1)+\sqrt{11}$$
Hence, we have:
$$x_n=\frac{1}{3} \bigg( (3n-3+\sqrt{11})^2-2\bigg)$$
$$x_n=(3n^2-6n+6)+2(n-1)\sqrt{11}$$
This proves the required with characterization as well.


问:
Let $(x_n)_n$ be a sequence of positives st $\exists n_0$ st $0<\frac{x_{n+1}}{x_n}\leq c<1,\ \forall n\geq n_0$. Show that $\lim_{n\to\infty}x_n=0$
I can show that this sequence converges because it is monotonic and bounded, but I'm having trouble showing that the limit is 0.

My attempt:

Let $\lim_{n\to\infty}x_n = L$, thus $$\forall\epsilon>0\exists n_0\in\mathbb{N}\forall n\geq n_0 \ (|x_n - L|<\epsilon)$$
For each natural $k$, let $\epsilon=\frac{c}{k}$, hence there exist $n_{0,k}$ s.t. $\forall n>n_{0,k}$

$$L-\frac{c}{k} < x_n < L+\frac{c}{k}$$
and
$$L-\frac{c}{k} < x_{n+1} < L+\frac{c}{k}$$

but $$0<x_{n+1} < cx_n<x_n$$
so
$$L-\frac{c}{k} < cx_n < L+\frac{c}{k}$$
$$\frac{L}{c}-\frac{1}{k} < x_n < \frac{L}{c}+\frac{1}{k}\implies |x_n-\frac{L}{c}|<\frac{1}{k} ,\ \forall n\geq n_{0,k}$$

I thought of using this to somehow show that $L$ cannot be greater than 0 by making $k$ large enough, but don't know exactly how.

答: Here is an easy approach. You already know that $x_n$ converges and being positive the limit $L$ must be non-negative. Suppose that $L>0$ then $x_{n+1}/x_n\to L/L=1$. But this contradicts $x_{n+1}/x_n\leq c<1$. Hence we have $L=0$.


问:
Continuous curve for 2-adic valuation of x
I am trying to find a continuous curve that will go trough all the points in the graph created if you were to plot the 2-adic valuation of x which is defined as how many times you can divide a number by two before the result gives an odd number. This can be defined by the function: $$f(x)=\left\{\begin{array}
 &0\text{ when x is odd}\\
 f(\frac{x}{2})+1\text{ when x is even}\\
 \end{array}\right. $$
The problem with this is that I cant differentiate since I need a continuous curve, and right now this only gives me what the values go like: 0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,etc.
$$$$
What I am currently trying is to find a trigonometric function that represents it. For example $\cos^2{\left(\frac{\pi}{2}x\right)}$ works for some values (just the alternating 0's and 1's) but I dont know how I could modify this to fit all values.

答:
$v_2(n)$ is the 2-adic valuation of $n$. Then look at things like

$$g(z)=e^{z^2}\frac{\sin(\pi z)}{\pi}+\sum_{n\ne 0} \frac{e^{-n^2} v_2(n)}{z-n}$$ which is entire.

It gives $g(0)=0$, add $\frac{\sin(\pi z)}{z^3}$ if you want a meromorphic function such that $f(n)=v_2(n)$ for all $n\in \Bbb{Z}$


问:
Solve integral $\int^1_0 \frac{1-x^2}{{(1+x^2)}\sqrt{1+x^4}}dx$ using subsituition $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$
Hi this question has been posed in my integration book where it has been asked to solve it using the given substitution or any other substitution, I've found identical question posted here {1} with different substitutions

The problem is when using the given substitution I end up with integral

$$\int^{\pi/4}_0 \frac{\sin{\theta}}{2x}d\theta $$

Solving for $x$ : $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$

According to WolframAlpha: $$x = \frac{|\csc{\theta}||1 \pm \sqrt{\cos{2\theta}}|}{\sqrt{2}} $$

I do know the sign for x is $+ve$ so I'll take $+ve$ solutions after opening the modulus, I do not know which sign to prefer in  $ \pm \sqrt{\cos{2\theta}} $  term

Eitherway, I solved for both cases :

$$  \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 + \sqrt{\cos{2\theta}}} d\theta -(I) $$
$$  \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 - \sqrt{\cos{2\theta}}} d\theta -(II) $$

Which leads to :
$$  \frac{1}{\sqrt{2}}\int^{\pi/4}_0 1 \pm \sqrt{\cos{2\theta}} d\theta = \frac{\pi}{4\sqrt{2}}
\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}} $$

Here I cheated and used WolframAlpha as I was exhausted.

Both the cases lead to a similar answer with difference of a function unknown to me: $\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}}$

The correct answer is :

$$\frac{\pi}{4\sqrt{2}}$$

My main questions are:

What am I doing wrong?

How did author arrive at this ingenious  substitution $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ , this doesn't appear to me trial n' error kind of substitution, I have spent hours doing algebraic transformation on this and no matter how you procced and plug the variable your integrand will only have one of these terms only $2x$, $1-x^2$, $1+x^2$ and $ \sqrt{1+x^4}$ and $2x$ being the simplest.

Related question: How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$

答:
Divide top and bottom by $x^2$:

$$\int_0^1\frac{\frac{1}{x^2}-1}{\left(\frac{1}{x}+x\right)\sqrt{\frac{1}{x^2}+x^2}}dx$$

which suggests using the substitution $t = \frac{1}{x}+x$:

$$\int_2^\infty \frac{dt}{t\sqrt{t^2-2}} = \frac{1}{\sqrt{2}}\sec^{-1}\left(\frac{t}{\sqrt{2}}\right)\Biggr|_2^\infty = \frac{\pi}{4\sqrt{2}}$$


问:
Determine the maximum value of a so that $P(S_{50}≥a)≥0.999$
A toymaker makes $Xn$ specialized toy parts on day n, where $Xn$ are independent and identically distributed random variables with mean 6 and variance 9. Let $Sn$ be the total number of specialized parts produced from day 1 to day n. Using central limit theorem, determine the total number of parts, a, the said factory can guarantee to produce by day 50 with at least 99.9% certainty, i.e. determine the maximum value of a so that $P(S_{50}≥a)≥0.999$. The maximum value must be a whole number.

So I used the formula

$Zn = \frac{\sigma - \mu}{\sigma \sqrt n}  $

and from z-table,  $Zn = -3.10$

where,

$\sigma = \frac{Zn \sigma}{\sqrt n} + n \mu = \frac{(-3.10)(3)}{\sqrt 50} + (50)(6) = 298.69 \approx 299$

However my answer was incorrect. Any help would be appreciated.

答:
In order to apply the central limit theorem you need to consider

$$\bar X_{50} = \frac 1{50}S_{50}$$

Hence, the probability in question is

$$P\left(\bar X_{50}\geq\frac a{50}\right) \geq 0.999$$

The standardization gives

$$Z \approx \frac{\bar X_{50} - \mu}{\sigma / \sqrt{50}}$$

Hence, you look for

$$P\left(Z \geq \frac{\frac a{50} - 6}{3 / \sqrt{50}}\right)\geq 0.999 \text{ or } P\left(Z < \frac{\frac a{50} - 6}{3 / \sqrt{50}}\right)\leq 0.001$$

Your $z$-score is correct: $z=-3.1$. So, you only need to find the highest integer $a$ with

$$\frac{\frac a{50} - 6}{3 / \sqrt{50}} < -3.1 \Leftrightarrow a < 234.239\ldots $$$$\Rightarrow \boxed{a=234}$$


问:
Proof for $\frac{b}{h}=\frac{c(c-1)}{k(k-1)}$, a combinatorial identity
Imagine we have $b$ teachers and $c$ students in a school such that

each teacher teaches $k$ students,

every two students have $h$ similar teachers.

I'd like to prove that
$$\frac{b}{h}\:=\:\frac{c(c-1)}{k(k-1)}\,.$$
I thought about it for a little bit but weirdly enough, I have no insights on this.

A hint, an answer, or both would be appreciated.

答:
Adress to each teacher a pair of students he/she teaches (so he adresses ${k\choose 2}$ pairs) and vice versa, to each pair of students a teacher they are listening (so they adresses $h$ teachers).

Then by double counting we have: $$b\cdot {k\choose 2} = {c\choose 2}\cdot h$$

Can you finsih now?


问:
Secret-Santa: Probability of two people drawing each other.
When playing Secret Santa this year, where a group of $n$ people buy presents and these $n$ presents get randomly distributed to the other people, excluding the possibility of someone getting his or her own present (we did it with a online-distributer, so this possibility really is excluded), someone in this round asked the question about the probabilty of two people giving presents to each other. Since then I've been thinking about this problem, but I didnt get very far. The only thing that comes to my mind is working with the symmetric group $S_{n}$, and considering the possibilities $S_{n}\setminus F$, with $F$ being the set of all permuations with at least one fixpoint, as this is not possible in the game of Secret Santa. The probabilty I'm interested in now can be calculated by coming up with the cardinality of the set $\{\pi \in S_{n}\setminus F\ \mid \pi $ contains at least a two-cycle$\}$. Any futher ideas?

答:
Define an $r$-derangement to be a permutation whose smallest cycle has length strictly greater than $r$. Let $D_{n,r}$ denote the number of $r$-derangements in $S_n$. It is a standard result that for fixed $r$, we have that $\frac{D_{n,r}}{n!} \sim e^{-H_{r}}$ where $H_r$ is the $r$-th Harmonic number. This is proven, for instance, in Analytic Combinatorics by Flajolet and Sedgewick.

The probability that two people do not give presents to each other should thus be approximately $\frac{e^{-H_{2}}}{e^{-1}} = e^{-1/2}$.


答:
Here's a fun fact: you might know that the probability that a random permutation has no fixed points is approximately $e^{-1}$ (and approaches this value quite rapidly). This generalizes: the probability that a random permutation has no cycles of length $\le \ell$ is approximately

$$e^{-H_{\ell}} = e^{- \left( 1 + \frac{1}{2} + \dots + \frac{1}{\ell} \right)}$$

where $H_{\ell}$ is the $\ell^{th}$ harmonic number. So the probability that a random permutation has no $2$-cycles given that it has no fixed points is approximately $e^{-\frac{1}{2}}$, which means the probability you want is the complement of this, namely

$$\boxed{ 1 - e^{- \frac{1}{2} } \approx 0.393 \dots }.$$

This can be proven using a result called the permutation form of the exponential formula, which implies that for fixed $\ell$ the sequence $f(n, \ell)$ counting the number of permutations of $n$ elements with no cycles of length $\le \ell$ has exponential generating function

$$\sum_{n \ge 0} f(n, \ell) \frac{t^n}{n!} = \exp \left( \sum_{k \ge \ell+1} \frac{t^k}{k} \right) = \frac{\exp \left( - \sum_{k=1}^{\ell} \frac{t^k}{k} \right)}{1 - t}.$$

The result follows (with very good error terms) from looking at the pole at $t = 1$. You can plug this generating function into a computer algebra system like WolframAlpha to compute specific coefficients and it will tell you, for example, that when $n = 8$ there are exactly $14833$ permutations without fixed points and $8988$ permutations without cycles of length $\le 2$, so the exact probability for $n = 8$ is

$$\frac{14833 - 8988}{14833} = 0.394 \dots$$

so the above is quite a close approximation already for this small value of $n$. We have $14833 = \left[ \frac{8!}{e} \right]$ where $[x]$ denotes the closest-integer function (this is a well-known formula for the number of derangements); we might hope that $8988$ is equal to $\left[ \frac{8!}{e^{1 + \frac{1}{2}}} \right]$ but this is actually equal to $8987$. So close but not quite! Probably a very slight adjustment to this will produce an exact formula for all but possibly very small $n$.

Title: Secret-Santa: Probability of two people drawing each other.

Body: When playing Secret Santa this year, where a group of $n$ people buy presents and these $n$ presents get randomly distributed to the other people, excluding the possibility of someone getting his or her own present (we did it with a online-distributer, so this possibility really is excluded), someone in this round asked the question about the probabilty of two people giving presents to each other. Since then I've been thinking about this problem, but I didnt get very far. The only thing that comes to my mind is working with the symmetric group $S_{n}$, and considering the possibilities $S_{n}\setminus F$, with $F$ being the set of all permuations with at least one fixpoint, as this is not possible in the game of Secret Santa. The probabilty I'm interested in now can be calculated by coming up with the cardinality of the set $\{\pi \in S_{n}\setminus F\ \mid \pi $ contains at least a two-cycle$\}$. Any futher ideas?


问:
Global Injectivity Criterion of Meromorphic Functions
Suppose we have a locally injective meromorphic function $f$ on some domain $D$.  The three interesting cases for $D$ are the plane, the disc and an arbitrary domain.  Suppose its poles are $P = \lbrace p_1, \dots, p_n \rbrace$ (assumed finite).

Question: Is there a tree contained in $D$ with endpoints $P$ on which $f$ is injective, aside from (possibly) the poles themselves?

答: If $D$ is a disk, it is not true, see the picture ($f$ exists by the Riemann mapping theorem). Not sure about the case when $D$ is the entire plane.


问:
If $x^p P(|X|>x|)=o(1)$, then $E(|X|^{p-\epsilon})<\infty$ for $0<\epsilon
0$ and $x^p P(|X|>x|)=o(1)$ as $x\to\infty$, then $E(|X|^{p-\epsilon})<\infty$ for $0<\epsilonn)<\infty$ but I don't see what to do.

答:
A consequence of Fubini-Tonelli's theorem is that for any non-negative random variable $Y$ and $q>0$,
$$E(Y^q)=q\int_0^{+\infty}t^{q-1}P(|Y|>t)dt.$$
Using this with $q:=p-\varepsilon$ gives the result.  </p>

Title: If $x^p P(|X|>x|)=o(1)$, then $E(|X|^{p-\epsilon})<\infty$ for $0<\epsilon<p$

Body: If $p>0$ and $x^p P(|X|>x|)=o(1)$ as $x\to\infty$, then $E(|X|^{p-\epsilon})<\infty$ for $0<\epsilonn)<\infty$ but I don't see what to do.


问:
Surface area of $z=4-x^2-y^2$ over a square region

Find the surface area of the paraboloid $z=4-x^2-y^2$ over the square region $-2\le x\le 2 $ and $-2\le y\le 2$.

I can parametrize this surface with $x=u,\ y=v,\text{and}  \>z=4-u^2-v^2$,
where $-2\le u\le 2$ and $-2\le v\le 2$. Then I can parametrize the surface with a vector function as follows:
$$\mathbf r(u,v)=\langle u, v ,4-u^2-v^2\rangle $$
Then
$\mathbf r_u(u,v)=\langle 1,0,-2u\rangle,\>
\mathbf r_v(u,v)=\langle 0,1,-2v\rangle$ and
$$\mathbf r_u\times\mathbf r_v=\left|\matrix{\mathbf i&\mathbf j&\mathbf k\cr 1 & 0 & -2u\cr 0 & 1 & -2v}\right|=\langle 2u,2v,1\rangle$$
so $\|\mathbf r_u\times\mathbf r_v\|=\sqrt{4u^2+4v^2+1}$.
The surface area is then defined by
$$A=\int\int_D\|\mathbf r_u\times \mathbf r_v\|\,dA=\int_{-2}^2\int_{-2}^2\sqrt{4u^2+4v^2+1}\,dv\,du$$
But now, how do I perform this integral? Here's an image of this surface.

Update

Due to symmetry, we can calculate the surface area over the region $R=\{(u,v):\ 0\le u\le 2\text{ and }0\le v\le 2\}$, then multiply the result by 4. That is, the surface area is

$$A=4\int_{0}^2\int_{0}^2\sqrt{4u^2+4v^2+1}\,dv\,du$$

Now, it was suggested that I try polar coordinates. Here's an image of the region $R=\{(u,v):\ 0\le u\le 2\text{ and }0\le v\le 2\}$.

Note that the angle of the dashed segment from $(0,0)$ to $(2,2)$ has an angle of $\pi/4$ with the $u$-axis. Also, note that

\begin{align*}
\cos\theta&=\frac2r\\
r\cos\theta&=2\\
r&=\frac{2}{\cos\theta}\\
r&=2\sec\theta
\end{align*}

Again, because of the symmetry of our image, we can determine the surface area over the region $\{(r,\theta):\ 0\le \theta\le \pi/4\text{ and } 0\le r\le 2\sec\theta\}$ and multiply the result by 8. Thus, using
$$ u=r\cos\theta\qquad\text{and}\qquad v=r\sin\theta$$
the surface area is

\begin{align*}
A&=8\int_0^{\pi/4}\int_0^{2\sec\theta}\sqrt{4r^2\cos^2\theta+4r^2\sin^2\theta+1}\ r\,dr\,d\theta\\
A&=8\int_0^{\pi/4}\int_0^{2\sec\theta}\sqrt{4r^2+1}\ r\,dr\,d\theta\\
\end{align*}

Then I was able to integrate and manipulate a bit:
\begin{align*}
A&=8\int_0^{\pi/4}\left[\frac1{12}(4r^2+1)^{3/2}\right]_0^{2\sec\theta}\,d\theta\\
A&=\frac23\int_0^{\pi/4}\left[(16\sec^2\theta+1)^{3/2}-1\right]\,d\theta\\
A&=\frac23\int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta-\frac23\int_0^{\pi/4}d\theta\\
A&=\frac23\int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta-\frac{\pi}{6}
\end{align*}

But now I am unable to calculate this current integral.

Thanks.

Update

Click the link below to see the work I had to do to understand Quanto's nice suggestion.

Surface Area Solution

Body:
Let $\sinh t= {\frac{4}{\sqrt{17}}}\tan\theta$ to evaluate

\begin{align}
I &= \int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta\\
&=\frac{17^2}4\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{\cosh^4 t}{1+\frac{17}{16}\sinh^2t }dt
=34\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{(1+\cosh 2t)^2}{\frac{15}{17}+\cosh2t}dt\\
 &=\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\left(34 \cosh2t + 38 + \frac{8}{{15}+17\cosh2t}\right)dt\\
&= 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\ 4\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}\frac{d(\tanh t)}{16+\tanh^2t}\\
 &= 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\cot^{-1}\sqrt{33}
\end{align}

Thus, the surface area is
$$A=\frac23I-\frac{\pi}{6}=\frac23\left(
 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\cot^{-1}\sqrt{33} -\frac\pi4\right)
$$


答:
The numbers of ways to distribute $n$ gifts according to the rules are OEIS A000166. The numbers of those ways where two people give gifts to each other (i.e. there is a $2$-cycle) is OEIS A158243. The probability you seek is then an entry of the first sequence divided by that of the second.

Asymptotically, as given in the formulas section of both sequences, the numerator tends to $n!(e^{-1}-e^{-3/2})$ and the denominator to $n!(e^{-1})$, leading to a limiting probability of $1-\frac1{\sqrt e}=0.393469\dots$


问:
Find the largest prime factor of $3(3(3(3(3(3(3(3(3(3(3+1)+1)+1)+1)+1)+1)+1)+1)+1)+1)+1$.
What I have done is that I found the expression is equal to $1+3+3^2+3^3+\cdots \cdots3^{11}=\frac{3^{12}-1}{2}$.

Then is there any way to find the prime factors of a number in such expression?

Body:
Hint:

$a^6-1=(a^3-1)\times(a^3+1)=(a-1)(a^2+a+1) \times (a+1)(a^2-a+1)$
$a^6+1=(a^2)^3 + 1^3 = (a^2+1)(a^4-a^2+1)$


问:
What is the other way of finding probability that coin is gold.
The question says :

'Given three identical boxes 1,2,3 each containing two. In box1 both coins are gold, box2 both coins are silver and in box3 there is one gold and one silver. Find the probability that the other coin is also gold.

Now the answer equates the probability to be found out with probability that the coin was from box1 and uses Bayes theorem to solve it.
But can we find the answer without equating it to the another probability and directly find the probability of obtaining the second coin as gold too?

答:
To gain an appreciation for how easily we can be led astray through flawed reasoning, consider the following "intuitive" approach:

Incorrect:  Among all boxes, there are $3$ gold coins and $3$ silver coins.  Given that a gold coin was obtained in the first draw, this means that either the first or last box was chosen.  In only one of these two cases is the other coin gold, therefore the answer is $1/2$.

Now, before we proceed with the correct line of reasoning, ask yourself why the above reasoning is flawed.

Correct:  Label the coins in each box as follows:  $$\begin{array}{ccc}
\text{Box 1}& \text{Box 2} & \text{Box 3} \\
\hline \{g_1, g_2\} & \{s_1, s_2\} & \{g_3, s_3\} \end{array}$$  Then the act of selecting a box at random and drawing one of the two coins in that box, followed by drawing the second coin in that same box, yields the possible outcomes $$(g_1, g_2), (g_2, g_1), \\
(s_1, s_2), (s_2, s_1), \\
(g_3, s_3), (s_3, g_3).$$
Unlike in our previous use of notation, notice here that the order matters.  Each of the six outcomes is equally likely, and for instance, $(g_1, g_2)$ is distinct from $(g_2, g_1)$ even though both coins are gold, because you have labeled the coins.

Then we can immediately see that $3$ of the outcomes have $g$ in the first position:  $$(g_1, g_2), (g_2, g_1), (g_3, s_3).$$  And we can see that in exactly two of the three outcomes, $g$ is also in the second position.  Therefore, the probability is $2/3$.


问:
If $\sum a_n $ is a positive series that diverges, does $\sum \frac{a_n}{1+a_n}$ diverge?
Let $ \{a_n\}_{n=1}^\infty $ be a sequence such that $\displaystyle \sum a_n $ that is divergent to $+\infty$.

What can be said about the convergence of $\displaystyle \sum \frac{a_n}{1 + a_n} $?

Any hints, thoughts or leads would be greatly appreciated.

Thanks!

Body:
If $0 < a_n/(1+a_n) < 1/2$ then $a_n < 1$, and $0 < a_n < 2 (a_n/(1+a_n))$.

So if $\sum a_n/(1+a_n)$ converges, so does $\sum a_n$.


问:
Show that these estimators are consistent
I'm working on the following question

Consider $ \Theta\in\mathbb{R} $ and a random sample $ X_1,\cdots,X_N $ of uniformly distributed random variables on $ [\Theta,\Theta+1] $.
Is the estimator $ \hat{\Theta}_N(X_1,\cdots,X_N)=\text{min}_{i=1\cdots,N}X_i $ consistent for the parameter $ \Theta $?

My script says the estimator is consistent iff it converges in probability to $ \Theta $. And just looking at the definition of convergence in probability it makes sense to me that it should do that. However, I don't know how to prove this formally. Could someone give me a hint? Thanks in advance!

edit: Took out a part of the question which solved itself right there and then.

答:
A couple of hints first and then I'll put the solution underneath.

Hint 1: $\left|\hat{\Theta}_N - \Theta\right|$ (the first term in the "convergence in probability" definition) is equal to $\hat{\Theta}_N - \Theta$ since by definition, $\hat{\Theta}_N \geq \Theta$.

Hint 2: For any constant $c$, $\mathbb{P}\left(\hat{\Theta}_N > c\right)$ is the probability that the minimum of the $X_i$'s exceeds $c$. The minimum of the $X_i$'s exceeds $c$ if and only if $X_1 > c, X_2 > c, \dots, X_n > c$.

Solution:
$\hat{\Theta}_N$ converges in probability to $\Theta$ if, for any $\epsilon > 0$,
$$\lim_{N \rightarrow \infty} \mathbb{P}\left(\left|\hat{\Theta}_N - \Theta\right| > \epsilon\right) = 0.$$

Now
$$\mathbb{P}\left(\hat{\Theta}_N - \Theta > \epsilon\right) = \mathbb{P}\left(\hat{\Theta}_N > \Theta + \epsilon\right) = \mathbb{P}\left(X_1 > \Theta + \epsilon, X_2 > \Theta + \epsilon, \dots, X_N > \Theta + \epsilon \right).$$

Since the samples are independent and identically distributed,
$$\mathbb{P}\left(X_1 > \Theta + \epsilon, \dots, X_N > \Theta + \epsilon \right) = \mathbb{P}\left(X_1 > \Theta + \epsilon\right)\mathbb{P}\left(X_2 > \Theta + \epsilon\right) \cdots \mathbb{P}\left(X_N > \Theta + \epsilon \right).$$

We know that $X_i \sim \mathbb{U}[\Theta, \Theta + 1]$, so $\mathbb{P}\left(X_i > \Theta + \epsilon \right) = 1 - \epsilon$ (as long as $\epsilon < 1$, otherwise the probability is $0$).

Hence
$$\mathbb{P}\left(\hat{\Theta}_N - \Theta > \epsilon\right) = \begin{cases}
\left(1-\epsilon\right)^N & \text{if } \epsilon < 1,\\
0 & \text{if } \epsilon \geq 1.
\end{cases}.$$

Taking the limit as $N \rightarrow \infty$,
$$\lim_{N \rightarrow \infty} \mathbb{P}\left(\left|\hat{\Theta}_N - \Theta\right| > \epsilon\right) = \lim_{N \rightarrow \infty} \begin{cases}
\left(1-\epsilon\right)^N & \text{if } \epsilon < 1,\\
0 & \text{if } \epsilon \geq 1 \end{cases} = 0.$$


答:
So the motive of this equation is that the value of $n$ is added by $1$ then multiplied by $n$ repeating this for $11$ times. That value after some calculation is $797,160$.

So now let's try dividing this by all prime numbers up to the result.

The biggest value at this point is $73$, so that's my guess.

So as mentioned of the hint by @Adola, I decided to give it a go and try to prove it.

What I got was: $728$ at the first equation then $730$ in the other one where $ a = 3 $. However, they're even numbers so that means they're not the values we should look for. If its any value other than $3$ that's used, there could be a possibility that there'd be more prime numbers.

Title: Find the largest prime factor of $3(3(3(3(3(3(3(3(3(3(3+1)+1)+1)+1)+1)+1)+1)+1)+1)+1)+1$.

Body:
What I have done is that I found the expression is equal to $1+3+3^2+3^3+\cdots \cdots3^{11}=\frac{3^{12}-1}{2}$.

Then is there any way to find the prime factors of a number in such expression?


问:
Prove $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$ for all $n$.
Prove for all $n$: $\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}<\frac{1}{\sqrt{3n}}$.

Using induction, I tried the brain-dead method and went straight for $$\frac{2n+1}{2n+2}\cdot\frac{1}{\sqrt{3n}}<\frac{1}{\sqrt{3n+3}}$$ $$...$$ $$1<0.$$ After embarrassing myself, I looked around and I found this thread. Using induction, we can then easily prove $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot...\cdot\frac{2n-1}{2n}\leq\frac{1}{\sqrt{3n+1}}<\frac{1}{\sqrt{3n}}.$$ This gets me to the original problem. But in a problem solving standpoint, how do you think to use $\frac{1}{\sqrt{3n+1}}$? Is there some point in the first induction that leads to this idea? Or is there a better method than the above?

答:
Write $a_n$ for the $n$th term in your sequence. Look at the square of $a_n$, and rotate the numerators left by one position. Starting at $n=2$ you observe
$$
a_2^2=\frac12\frac12\frac34\frac34=\frac12\left(\frac32\frac34\right)\frac14\\
a_3^2=\frac12\frac12\frac34\frac34\frac56\frac56=\frac12\left(\frac32\frac34\right)\left(\frac54\frac56\right)\frac16\\
a_4^2=\frac12\frac12\frac34\frac34\frac56\frac56\frac78\frac78=\frac12\left(\frac32\frac34\right)\left(\frac54\frac56\right)\left(\frac76\frac78\right)\frac18
$$
and so on. The inequality $1+x\le e^x$ then gives
$$
a_2^2\le \frac18 \exp\left(\frac18\right)\\
a_3^2\le\frac1{12}\exp\left(\frac18+\frac1{24}\right)\\
a_4^2\le\frac1{16}\exp\left(\frac18+\frac1{24}+\frac1{48}\right)
$$
and in general for $n\ge 2$
$$a_n^2\le \frac1{4n}\exp\left[\frac14\left(\frac12+\frac16+\frac1{12}+\cdots+\frac1{n(n-1)}\right)\right].$$
The series $\frac12+\frac16+\frac1{12}+\cdots+\frac1{n(n-1)}$ telescopes to $1$, yielding
$$a_n^2\le \frac{e^{1/4}}{4n}$$
which also holds for $n=1$. Since $e^{1/4}\approx 1.284 < 4/3$, this proves $a_n^2< \frac1{3n}$.


问:
Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$.
Let $x$ and $y$ be real numbers such that $x^2 + y^2 = 1$. Find the maximum value of $2x - 5y$.

I do know how to solve this problem using trigonometry, however I need to solve it by using vectors. Except for defining vector $v=(x, y)$, such that $||v||=1$, I do not see any way in which I could apply vectors to solving this problem. Any ideas are appreciated. Thank you!

答:
The maximum value $k$ occurs when the line $2x-5y = k$ is tangent to the circle $x^2+y^2=1$.

Solving for $x$, we get $x = \frac{1}{2} (k+5y)$, so $\frac{1}{4}(k^2+10ky+25y^2) + y^2 = 1$. Now set up the quadratic and set $\Delta = 0$, as we only want there to be one value of $y$ for the intersection of the circle and line:

$$k^2+10ky + 25y^2 + 4y^2 = 4$$
$$\Rightarrow 29y^2+10ky+(k^2-4)=0$$
$$\Delta = b^2-4ac = 0: (10k)^2-4(29)(k^2-4)=0$$
$$\Rightarrow -16k^2+464=0, k = ±\sqrt{\frac{464}{16}} = ±\sqrt{29}$$

and we are looking for the maximum value of $k$. Thus $k = \boxed{\sqrt{29}}$.


答:
It is worth observing that if $$Y \sim \operatorname{Gamma}(r, \lambda)$$ where $r$ is a shape parameter and $\lambda$ is a rate parameter, then $W = 1/Y$ is called an inverse gamma random variable, which has density
$$f_{W}(w) = f_Y(1/w) \cdot \frac{1}{w^2} = \frac{\lambda^r (1/w)^{r-1} e^{-\lambda/w}}{w^2 \Gamma(r)} = \frac{\lambda^r e^{-\lambda/w}}{w^{r+1} \Gamma(r)}, \quad w > 0.$$
If we accept that $$\int_{w = 0}^\infty f_W(w) \, dw = 1,$$ that is to say, $f_W$ is a density for any choice of parameters $r, \lambda > 0$, then it is easy to compute the moments.  Let $0 \le k < r$.  Then
$$\begin{align}
\operatorname{E}[W^k]
&= \int_{w=0}^\infty w^k f_W(w) \, dw \\
&= \frac{\lambda^r}{\Gamma(r)} \int_{w=0}^\infty \frac{e^{-\lambda/w}}{w^{r-k+1}} \, dw \\
&= \frac{\lambda^r \Gamma(r-k)}{\lambda^{r-k} \Gamma(r)} \int_{w=0}^\infty \frac{\lambda^{r-k} e^{-\lambda/w}}{w^{r-k+1}\Gamma(r-k)} \, dw \\
&= \lambda^k \frac{\Gamma(r-k)}{\Gamma(r)},
\end{align}$$
where in the last step, the integrand is the density of an inverse gamma distribution with shape $r-k$ and scale $\lambda$.  Then for $k = 1$ and $k = 2$, we obtain
$$\operatorname{E}[W] = \frac{\lambda}{r-1}, \\ \operatorname{E}[W^2] = \frac{\lambda^2}{(r-1)(r-2)},$$ and $$\operatorname{Var}[W] = \frac{\lambda^2}{(r-1)^2(r-2)},$$ provided that $r > 2$.
Your solution is basically correct, but you just need to recognize that $$\frac{\Gamma(r-2)}{\Gamma(r)} = \frac{\Gamma(r-2)}{(r-1)(r-2)\Gamma(r-2)}$$ since $z \Gamma(z) = \Gamma(z+1)$.

Title: Variance of $1/Y$ where $Y$~Gamma

Body:
I need help computing for $Var(\frac{1}{Y})$ where $Y$~Gamma(r,λ)

I know that $$Var(\frac{1}{Y}) = E(\frac{1}{Y}^2) - [E(\frac{1}{Y})]^2$$

I have computed for:

$$E(\frac{1}{Y}) = \frac{λ}{r-1}$$

Here is my solution so far for $E(\frac{1}{Y^2})$:

$$ = \int_0^∞ \frac{1}{y^2} \frac{λ^r}{Γ(r)} y^{r-1} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \int_0^∞ y^{(r-1)-2} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \int_0^∞ y^{(r-2)-1} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \int_0^∞ \frac{Γ(r-2)}{λ^{r-2}}\frac{λ^{r-2}}{Γ(r-2)}y^{(r-2)-1} e^{-λy} dy$$
$$= \frac{λ^r}{Γ(r)} \frac{Γ(r-2)}{λ^{r-2}} \int_0^∞ \frac{λ^{r-2}}{Γ(r-2)}y^{(r-2)-1} e^{-λy} dy$$
The integral is the PDF of Gamma (r-2,λ) so
$$= \frac{λ^r}{Γ(r)} \frac{Γ(r-2)}{λ^{r-2}} (1)$$
$$= \frac{λ^r}{λ^{r-2}} \frac{Γ(r-2)}{Γ(r)}$$
$$= λ^2 \frac{Γ(r-2)}{Γ(r)}$$

I am not sure where to go from here.


答:
Also observe that

If $Y\sim Gamma$ then $\frac{1}{Y}\sim$ Inverse Gamma thus mean and variance are known indicators

Title: Variance of $1/Y$ where $Y$~Gamma


问:
Determining the number of times a rectangular object must be tossed.
Suppose the four faces of a regular rectangular object are marked with letters A, B, C, and D, respectively. How can we determine the number of times this object must be tossed so that the probability of the ratio of numbers of D's to the number of tosses being between 0.15 and 0.35 is at least 0.95?</p>

答:
As stated, the problem seems incomplete. Please check your statement carefully. I will try to get you started on what I suppose the problem must be.

I assume you mean, that when a tetrahedral object is tossed, each of its four sides
has an equal chance of facing downward. Thus the number $X$ of D's in $n$ independent tosses
has $X \sim \mathsf{Binom}(n, 1/4),$ so that $E(X) = n/4$ and
$Var(X) = 3n/16.$
Then you seek $n$ just large enough so that
$P(0.15 < X/n < 0.35) > 0.95$

I suppose you are expected to standardize and use the normal approximations to these binomial distributions in order to solve for $n$.

Hint: Searching in R, I found the probabilities below.

```<code>diff(pbinom(c(.15,.35)*75, 75, .25))
[1] 0.9555702
diff(pbinom(c(.15,.35)*74, 74, .25))
[1] 0.9411401
</code>```

Body:
As stated, the problem seems incomplete. Please check your statement carefully. I will try to get you started on what I suppose the problem must be.

I assume you mean, that when a tetrahedral object is tossed, each of its four sides
has an equal chance of facing downward. Thus the number $X$ of D's in $n$ independent tosses
has $X \sim \mathsf{Binom}(n, 1/4),$ so that $E(X) = n/4$ and
$Var(X) = 3n/16.$
Then you seek $n$ just large enough so that
$P(0.15 < X/n < 0.35) > 0.95$

I suppose you are expected to standardize and use the normal approximations to these binomial distributions in order to solve for $n$.

Hint: Searching in R, I found the probabilities below.

diff(pbinom(c(.15,.35)*75, 75, .25))
[1] 0.9555702
diff(pbinom(c(.15,.35)*74, 74, .25))
[1] 0.9411401


问:
Understanding a calculation in Riemannian manifold
Let $(M,\langle,\rangle)$ be a Riemannian manifold. If $f$ is a smooth function in $M$ and $X,Y$ are smooth vector fields on $M$ satisfying the following condition
$$f|X|^2=\langle \nabla_YX,X\rangle.$$
Now I have done the following calculation from the above one:
$$f|X|^2=\langle \nabla_YX,X\rangle=\frac{1}{2}Y\langle X,X\rangle=\frac{1}{2}Y|X|^2.$$
Therefore, we get $f=\frac{1}{2}Y$. But the left hand side is a scalar function whereas the right hand side is a vector. I can not find where I have done mistake. Please help.

答:
It seems here your function $f$ is just, on the set of $p\in M$ where $X(p)$ is not zero:
$$
f(p) = \frac{\langle \left(\nabla_YX\right)(p),X(p)\rangle}{\|X(p)\|^2}e
$$
and is extended in some way on $p\in M$ where $X(p)=0$. That is, it measures the covariant derivative in the direction of $Y$ of $X$, renormalized by $X$.

Your mistake is in your understanding of the notation $f\|X\|^2$, as if $f$ was a function with variable $\|X\|^2$. But it is not, and this notation just says that the product of the two functions $f$ and $\|X\|^2$ is equal to the right hand side. THe only input we could give to $f$ is $p \in M$.


问:
Surface integral over ellipsoid $ax^2+by^2+cz^2=1$
I'm not sure how to compute the integral
$$\int_{s} (a^2x^2+b^2y^2+c^2z^2)^{-1/2} d\vec{S}\cdot \vec{n}$$
over the surface of the ellipsoid
$$ax^2+by^2+cz^2=1$$, $$z>0$$

Where $\vec{n}$ is the unitary normal vector to the surface.

Any help is appreciated. Thank you.

答:
We propose to attack the problem with a blunt instrument: on the surface
$$\vec{r}=\langle x,y,z\rangle=\left\langle x,y,\pm\sqrt{\frac{1-ax^2-by^2}{c}}\right\rangle$$
so
$$\begin{align}d\vec{r}&=\left\langle1,0,\mp\frac{\frac{ax}c}{\sqrt{\frac{1-ax^2-by^2}{c}}}\right\rangle\,dx+\left\langle0,1,\mp\frac{\frac{by}c}{\sqrt{\frac{1-ax^2-by^2}{c}}}\right\rangle\,dy\\
&=\left\langle1,0,-\frac{ax}{cz}\right\rangle\,dx+\left\langle0,1,-\frac{by}{cz}\right\rangle\,dy\end{align}$$
So that the vector areal element is
$$\begin{align}d^2\vec{A}&=\pm\left\langle1,0,-\frac{ax}{cz}\right\rangle\,dx\times\left\langle0,1,-\frac{by}{cz}\right\rangle\,dy\\
&=\pm\left\langle\frac{ax}{cz},\frac{by}{cz},1\right\rangle\,dx\,dy=\frac{\langle ax,by,cz\rangle}{c|z|}dx\,dy\end{align}$$
Then
$$\hat{n}=\frac{d^2\vec{A}}{\left\lVert d^2\vec{A}\right\rVert}=\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}$$
$$\begin{align}I&={\large\bigcirc}\kern-1.55em\iint_S\frac1{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\cdot d^2\vec{A}\\
&=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\sqrt{\frac{1-ax^2}b}}^{\sqrt{\frac{1-ax^2}b}}\frac1{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\frac{\langle ax,by,cz\rangle}{\left(a^2x^2+b^2y^2+c^2z^2\right)^{1/2}}\cdot\frac{\langle ax,by,cz\rangle}{c|z|}dy\,dx\\
&=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\sqrt{\frac{1-ax^2}b}}^{\sqrt{\frac{1-ax^2}b}}\frac1{\sqrt c\sqrt{1-ax^2-by^2}}\,dy\,dx=2\int_{-\frac1{\sqrt a}}^{\frac1{\sqrt a}}\int_{-\frac{\pi}2}^{\frac{\pi}2}\frac1{\sqrt{bc}}\,d\theta\,dx=\frac{4\pi}{\sqrt{abc}}\end{align}$$
Where we have used the substitution
$$y=\sqrt{\frac{1-ax^2}b}\sin\theta$$
to solve the $y$-integral. Note that if $a=b=c$ then $a^2x^2+b^2y^2+c^2z^2=a$ on the surface and $d^2\vec{A}\cdot\hat{n}=d^2A$ and the integral is over the surface of a sphere of radius $1/\sqrt a$ so we should get
$$I=\frac1{\sqrt a}\frac{4\pi}{a}$$
so this checks.


问:
Basis for linearly dependent vectors in R3
Suppose I have a subspace S where $(x_1, x_2, x_3) \in \mathbb{R}^3$ and for $a, b, c\in R$, not all equal to zero, we have
$ax_1 + bx_2 + cx_3 = 0$. What would be the basis for this subspace?

I tend to think that the basis would be less than 3 vectors, because I believe that S should have a dimension of less than 3 due to the linear dependence. But I can't think of how to find 2 vectors that would fulfill this requirement.

答:
WLOG we can assume that $a \ne 0.$ Then we have

$x_1= - \frac{b}{a}x_2- \frac{c}{a}x_3,$

$x_2=x_2$,

$x_3=x_3.$

With $t=x_2$ and $s= x_3$ we derive

$$(x_1,x_2,x_3)=t(- \frac{b}{a},1,0)+s(- \frac{c}{a},0,1).$$

This is an equation of a plane. A basis is given by

$$ \{(-b,a,0),(-c,0,a)\}.$$


问:
How to derive the 3D equation of a torus?
I'm doing a presentation on 3D surfaces for college and one of the equations I am using is a Torus. I know that the equation is

$$z^2 = 25 - \left(10 - \sqrt{x^2 + y^2}\right)^2$$

For a torus with radius 5, with the inner circle 10 units away from the z axis but I was wondering how this is derived from a circle of the form

$$x^2 + y^2 = 15^2$$

Being the centre for the other circles in the $oxz$ and $oxy$ planes (15 being the distance from the $z$ axis + the radius.

Thanks, Any help will be appreciated

答:
First of all, to understand how the equation for a torus relates to that of a circle you must become familiar with the object: $\sqrt{x^2+y^2}$. As you can probably guess, this takes any point $p$ on the $x$-$y$ plane and generates $p$'s distance from the origin. The useful aspect of this tool is that it treats any point within the same distance of the origin as the same. Therefore, in order to create objects that are rotationally symmetric about the $z$-axis, it is essential to treat $z$ as a function of radius $r$ and then later substitute $r$ with $\sqrt{x^2+y^2}$.
Since we know that the torus is rotationally symmetric, we can simplify it by finding the equation to its cross section (by any origin intersecting plane perpendicular to the $x$-$y$ plane). We then get the graph of a circle translated horizontally.

Keep in mind that the horizontal axis is radius $r$ and the vertical axis is $z$. Now you may now see how the equation for a circle arises-- it's just the graph of $\pm\sqrt{a^2-(r-c)^2}=z$. Variable c is the length from the origin to the center of the circle while a is the radius of the circle.
Now we can expand this into three dimension by replacing the r with $\sqrt{x^2+y^2}$.

By then you have two parameters

$a$ = radius of the "tubing"

$c$ = distance from the center of the torus to the middle of the "tubing"

$z=\pm\sqrt{a^2-(\sqrt{x^2+y^2}-r)^2}$


问:
Integral $\int_0^\infty \frac{\ln(x)}{1-x^2} \; dx$ using residue theorem
I want to compute the integral
$$
I = \int_0^\infty \frac{\ln(x)}{1-x^2} \; dx.
$$
Here is my attempt: First, in order to extend the integral to the interval $(-\infty,\infty)$ I write
$$
I = \frac{1}{2}\int_0^\infty \frac{\ln(x^2)}{1-x^2} \; dx = \frac{1}{4}\int_{-\infty}^\infty \frac{\ln(x^2)}{1-x^2} \; dx.
$$
Now, let $R>2$ and $0 < \epsilon< 1/2$ and define
\begin{align*}
C_0 &= \{Re^{it} \, : \, 0\leq t\leq \pi\}\\
C_1 &= \{\epsilon e^{it} \, : \, 0\leq t\leq \pi\}\\
C_2 &= \{-1+\epsilon e^{it} \, : \, 0\leq t\leq \pi\}\\
C_3 &= \{1+\epsilon e^{it} \, : \, 0\leq t\leq \pi\}\\
C_4 &= [-R,R]\setminus(]-1-\epsilon,-1+\epsilon[\cup ]-\epsilon,\epsilon[\cup]1-\epsilon,1+\epsilon[),
\end{align*}
and write
$$
C = C_0 \cup C_1 \cup C_2\cup C_4 \cup C_4.
$$
The function $f(z) = \frac{\ln(z^2)}{1-z^2}$ is holomorphic, where $\ln$ denotes a branch of the logarithm. Then, we have that
$$
\int_C f(z)\; dz = 0,
$$
hence
$$
\int_{C_4} f(z)\; dz = -\sum_{j=0}^3 \int_{C_j} f(z)\; dz,
$$
and
$$
\int_{-\infty}^\infty \frac{\ln(x^2)}{1-x^2} \; dx = \lim_{\begin{subarray}{c} \epsilon\to 0 \\ R\to\infty  \end{subarray}} -\sum_{j=0}^3 \int_{C_j} f(z)\; dz.
$$
My problem (if the above is correct) is that I'm unable to compute the integrals over the curves $C_j$. I think that the integral over the curve $C_0$ tends to $0$ as $R\to\infty$ but I cannot have information about the remaining ones.

答: For large $R$,$$\int_{C_0}f(z)dz=\int_0^\pi f(Re^{it})iRe^{it}dt=\int_0^\pi\frac{2\ln R+2it}{1-R^2e^{2it}}iRe^{it}dt\sim-\frac{\ln R}{R}\int_0^\pi\frac{2idt}{e^{it}}\to0.$$The small arcs are clockwise, so pick up a $-$ sign. For small $\epsilon$,$$\int_{C_1}f(z)dz=-\int_0^\pi\frac{2\ln\epsilon+2it}{1-\epsilon^2e^{2it}}i\epsilon e^{it}dt\sim-\epsilon\ln\epsilon\int_0^\pi 2ie^{it}dt\to0.$$Whereas a careful branch choice for $\ln z^2=2\ln z$ is required for $z\approx-1<0$ viz.$$\int_{C_2}f(z)dz=-\int_0^\pi\frac{\ln[(\epsilon e^{it}-1)^2]}{1-(\epsilon e^{it}-1)^2}i\epsilon e^{it}dt=-\int_0^\pi\frac{2\pi i+2\epsilon e^{it}}{1-(\epsilon e^{it}-1)^2}i\epsilon e^{it}dt\sim-\int_0^\pi\pi dt=\pi^2,$$we conclude with the more straightforward$$\int_{C_3}f(z)dz=-\int_0^\pi\frac{\ln[(\epsilon e^{it}+1)^2]}{1-(\epsilon e^{it}+1)^2}i\epsilon e^{it}dt\sim\epsilon\int_0^\pi ie^{it}dt\to0$$so$$I=\frac14\int_{\Bbb R}f(x)dx=-\frac{\pi^2}{4}.$$


问:
Can we write $\sin(\frac{\pi}{14})$ as an finite expression using only basic operations?
First, let us recall some trigonometric values

$\sin(0)=0$

$\sin(\pi/6)=\frac{1}{2}$

$\sin(\pi/3)=\frac{\sqrt{3}}{2}$

$\sin(\pi/2)=1$

$\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$

$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$

Here, we can observe that for some values of $\theta$, $\sin(\theta)$ can be expressed as a sum of finite radical terms. It is also easy to see that, there exists infinite such $\theta$'s. Say, I have given $\sin(\pi/14)$, can we determine whether it can also be expressed as sum of finite radical terms? What about the general case? Are there any criteria that $\theta$ must obey in order to do that?

答:
It is not so bad !

As @Doug M commented, it is one of the roots of

$$8x^3 - 4x^2 - 4x + 1 = 0$$

Using the trigonometric method for three real roots,we have
$$\sin \left(\frac{\pi }{14}\right)=\frac{1}{6} \left(1+2 \sqrt{7} \sin \left(\frac{1}{3} \csc ^{-1}\left(2
   \sqrt{7}\right)\right)\right)$$

Otherwise, using radicals
$$\sin \left(\frac{\pi }{14}\right)=\frac{1}{6}-\frac{7^{2/3} \left(1-i \sqrt{3}\right)}{6\ 2^{2/3} \sqrt[3]{-1+3 i
   \sqrt{3}}}-\frac{1}{12} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3
   i \sqrt{3}\right)}$$

The second and third terms are two of the roots of
$$46656 x^6-1512 x^3+343=0$$ which is a quadratic in $x^3$.

The solutions of
$$46656 X^2-1512 X+343=0$$ are
$$X_{1,2}=\frac{7}{432} \left(1\pm3 i \sqrt{3}\right)$$


问:
How one can guaranty that $λ₁λ₄λ₅+λ₂λ₃λ₆$ is not a zero
Let $A,B,C$ be three pairiwise coprime positive integers, i.e., there exist six integers $λ_{1},λ_{2},λ_{3},λ_{4},λ_{5},λ_{6}$ such that

$$λ_{1}A+λ_{2}B=1$$

$$λ_{3}A+λ_{4}C=1$$

$$λ_{5}B+λ_{6}C=1$$

Solving with respect to $A,B,C$, we get

$$A=-(λ₂λ₄-λ₂λ₆-λ₄λ₅)/(λ₁λ₄λ₅+λ₂λ₃λ₆)$$

$$B= (λ₁λ₄-λ₁λ₆+λ₃λ₆)/(λ₁λ₄λ₅+λ₂λ₃λ₆)$$

$$C= (λ₂λ₃+λ₁λ₅-λ₃λ₅)/(λ₁λ₄λ₅+λ₂λ₃λ₆)$$

Then I am asking if there is any problem with representing $A,B,C$ in terms of $λ_{1},λ_{2},λ_{3},λ_{4},λ_{5},λ_{6}$. In particular, how one can guaranty that $λ₁λ₄λ₅+λ₂λ₃λ₆$ is not a zero.

答:
There's no guarantee that it will not be zero.

Example: Let $A=2\times11$, $B=3\times13$, $C=5\times7$. Then $$\begin{matrix}16A-9B=1\\8A-5C=1\\9B-10C=1\end{matrix}$$ yet $16.5.9=9.8.10$.


问:
Why does $\pi$ appear in the probability of a number being square-free?
The probability of a number being square-free (i.e., the number has no divisor that is a square, cube, etc.) is $6/\pi^2$. I have seen many appearances of $\pi$, and this is also similar to them. All of them can be explained intuitively why pi appears in them. But I don't see any connection between square-free numbers and $\pi$. So my question is:

Why $\pi$ appears here?

Note: I don't want rigorous proofs, I want intuitive explanations.

答: The most intuitive thing I can think of would be the Riemann Zeta function. You are working with squares, and you will be adding a bunch of squares. In a similar way, if you wanted cube free numbers, you would get Apéry's constant.</p>


问:
functional equation with two functions: $ f ( x + y ) = f ( x ) g ( y ) + f ( y ) $

Find all functions $ g : \mathbb R \to \mathbb R $ with the property:

There exists a strictly monotonic function $ f : \mathbb R \to \mathbb R $ such that
$$ f ( x + y ) = f ( x ) g ( y ) + f  ( y ) , \forall x , y \in \mathbb R $$

We have one official solution, which shows that $ g ( x ) $ must be of the form $ e ^ { k x } $, or simply $ a ^ x $, $ a > 0 $. I'm looking for other solutions, just trying to get some new ideas. If non is posted, I'll post the solution I have.

答:
As $ f $ is strictly monotonic, there is $ c \in \mathbb R $ with $ f ( c ) \ne 0 $. Let $ a = f ( c ) $ and $ b = g ( c ) $. Note that by swapping $ x $ and $ y $ in
$$ f ( x + y ) = f ( x ) g ( y ) + f ( y ) \text , \tag 0 \label 0 $$
you can deduce
$$ f ( x ) g ( y ) + f ( y ) = f ( y ) g ( x ) + f ( x ) \text , $$
which for $ y = c $ gives
$$ g ( x ) = \frac { b - 1 } a f ( x ) + 1 \text . \tag 1 \label 1 $$
Multiplying \eqref{0} by $ \frac { b - 1 } a $ and using \eqref{1} you get
$$ g ( x + y ) = g ( x ) g ( y ) \text . \tag 2 \label 2 $$
Putting $ x = y = 0 $ in \eqref{2}, you have $ g ( 0 ) \in \{ 0 , 1 \} $. If $ g ( 0 ) = 0 $, then setting $ y = 0 $ in \eqref{2} shows that $ g $ is constantly zero; but that can't happen, since then by \eqref{1} you can deduce that $ f $ is constant, and that contradicts with $ f $ being strictly monotonic. Thus, $ g ( 0 ) = 1 $. Letting $ y = - x $ in \eqref{2} you get $ g ( x ) g ( - x ) = 1 $, which in particular shows that $ g ( x ) \ne 0 $. As Substituting $ \frac x 2 $ for both $ x $ and $ y $ in \eqref{2} we get $ g ( x ) = g \left( \frac x 2 \right) ^ 2 \ge 0 $, we can conclude that $ g $ is strictly positive at all points. This lets us define $ h : \mathbb R \to \mathbb R $ with $ h ( x ) = \log g ( x ) $. Then \eqref{2} shows that $ h $ satisfies Cauchy's functional equation. Also, as $ f $ is strictly monotonic, it is locally bounded, and thus by \eqref{1} and the definition of $ h $, $ h $ is locally bounded. By a well-known fact, this implies that there must be a constant $ k \in \mathbb R $ such that $ h ( x ) = k x $ for all $ x \in \mathbb R $. This means that we must have $ g ( x ) = \exp ( k x ) $ for all $ x \in \mathbb R $, and thus every $ g $ for which there is a suitable $ f $ must be of this form. It remains to check that for any function $ g $ of this form there is a suitable $ f $. To see this, it's sufficient to let $ f ( x ) = \exp ( k x ) - 1 $ in case $ k \ne 1 $, and $ f ( x ) = x $ in case $ k = 1 $. In both cases, it's straightforward to verify that $ f $ is strictly monotonic and \eqref{0} is satisfied.


问:
Basis for subspace
Consider $V= \left \{ u= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$.

What is the basis of $V \subset K^3$

The format of the subspace is proving itself difficult to me. I know the span of the vectors in the basis is equal to the subset and that the vectors in the basis are linearly independent. I furthermore know the dimension is equal to the ammount og vectors in the basis of the subspace. Normally I would reduce to reduced echeleon form and be done. What approach would work here? I cant seem to make a matrix for the subspace.

答:
Well, $V$ is given by the kernel of the linear mapping $f:K^3\rightarrow K: (x_1,x_2,x_3)^T\mapsto x_1-x_2+x_3$.

The dimension formula $\dim K^3 = \dim ker(f) + \dim image(f)$ applies and shows that
$\dim V = \dim ker(f)=2$.
Now choose two linearly independent vectors of $V$ as shown by the comment above: $(1,1,0)^T$ and $(0,1,1)^T$. By the dimension of $V$, they form a basis of $V$.

Title: Basis for subspace


问:
How do I work out what percent of my customers will be girls and what percent will be boys?
I know that 33.3333% of all girls questions would buy my product and that 80% of all boys questioned would buy it.

What i don't know is how to work out is statistically what percentage of our customers will be boys and what percentage will girls.

答:
Hint: Let $G$ = the person is a girl and $C$ = the person is a customer. Assume that a customer is defined to be anyone who buys your product. Then we are given that:

$P(C|G)=0.333...=1/3$

$P\left(C|\overline{G}\right) = 0.8 = 4/5$

We are asked to find: $P(G|C)$ and $P\left(\overline{G}|C\right)$

The following identity will be useful:

$$P(A|B) = \frac{P(A \text{ and } B)}{P(B)} = \frac{P(A)P(B|A)}{P(A)P(B|A) + P\left(\overline{A}\right)P\left(B|\overline{A}\right)}$$


问:
convert price to time in minutes
I'm stuck in my head. If "5" is cost per minute.

How would the formula be to find number of minutes spent based on a given cost?

Say cost is calculated to be:

24,6 = minutes?

38,8 = minutes?

16,4 = minutes?

This is used in EXCEL, and I'm trying to figure out how much time any given cost is equal to in time/minutes.

It is to verify that the cost converts to a reasonable time for a machine run in my printing environment.

If not, I will have to adjust some other parameters to try and get to something reasonable.

Thanks in advance for any input :)

答:
I'm stuck in my head

really?

Just a multiplication

$$\$24.6\times \frac{1'}{\$5}=4.92'=4'.55''$$

That is 4 minutes, 55 seconds (..and 2 tenth of seconds)

Title: convert price to time in minutes


问:
Assuming the expression converges, determine the largest integer $n\le 9,000,000$ for which $\sqrt {n+\sqrt {n+\sqrt {n+\cdots}}}$ is rational.
What I have done is that I supposed $y=\sqrt {n+\sqrt {n+\sqrt {n+\cdots}}}$.

So, $y^2-n=y$, which is $y^2-y+\frac{1}{4}=n+\frac{1}{4}$. So, $(y-\frac{1}{2})^2=n+{1\over 4}=\frac{4n+1}{4}$.

Hence, $y={1\over 2}\pm\frac{\sqrt {4n+1}}{2}$ is a rational, which means $4n+1$ must be a perfect square.

Then I let $m^2=4n+1$. So, $(m-1)(m+1)=4n$, and that is where I got stuck.

Obviously $m$ must be an odd integer and $n$ CANNOT be a perfect square. But I have no idea what to do next because $n$ can be very big... Could anyone share some ideas?

答:
Let $m=2k+1$. Varying $k$ gives all possible values of $n$ for $y$ to be rational:
$$m^2=4k^2+4k+1=4(k^2+k)+1\implies n=k^2+k=k(k+1)\le9\cdot10^6$$
It is easily seen that the solution corresponds to $k=2999$, i.e. $n=8997000$.


问:
Suppose that r is a vector differentiable function of t, and w is a real function differentiable from t. Show that ⅆ/ⅆt [r(w(t))]=r^' (w(t)) w^' (t)
I do not know how to do demonstrations, I have really try, but I can not solve this.</p>

答:
Let $I$ be an interval in $ \mathbb R$ and $r:I \to \mathbb R^n$ a differentiable function, then $r(t)=(r_1(t),...,r_n(t))$ for $t \in I.$ We have that each $r_j:I \to \mathbb R$ is differentiable on $I$. If $J$ is a further interval in $ \mathbb R$ and $w:J \to I$ is differentiable, then we have , by the usual chain rule, that

$$\frac{d}{dt}(r_j(w(t))=r_j'(w(t)) w'(t)$$

for $j=1,...,n$ and $t \in J.$

This gives the result.


问:
A question on exact sequences and group cohomology
Question:

If $0\longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 0,$ is an exact sequence of $G$-modules, then show that $$0 \longrightarrow A^G \longrightarrow B^G \longrightarrow C^G \longrightarrow H^1(G,A) \longrightarrow  H^1(G,B) \longrightarrow H^1(G,C)$$ is exact, where if A is a G-module, then $A^G$ is the set of $G$-invariants of $A.$

Attempt:

The map $A^G \longrightarrow B^G$ is the restriction of $A \longrightarrow B$ to $A^G.$ (and similarly, the map $B^G \longrightarrow C^G$ is defined). The map $\eta:H^1(G,A) \longrightarrow H^1(G,B)$ is defined as follows. If $j:A\longrightarrow B$ and $h\in Z^1(G,A),$ then $\eta(\bar{h})=\overline{j\circ h}$ where bar denotes the quotients from $B^1(G,A)$ and $B^1(G,B).$

I have proved everyting except for the fact that ${\rm im}(H^1(G,A) \longrightarrow H^1(G,B))=\ker(H^1(G,B) \longrightarrow H^1(G,C)).$ Could someone help me with this please? Thank you!!

答:
Let $f: A \to B, g:B \to C$ be the corresponding maps. Then the map $H^1(G,A) \to H^1(G,B)$ is defined by $\phi \mapsto f \circ \phi$, and similarly for $H^1(G,B) \to H^1(G,C)$.

It follows that $\text{Im}(H^1(G,A) \to H^1(G,B)) \subseteq \text{Ker}(H^1(G,B) \to H^1(G,C))$. For the hard part let $\psi \in \text{Ker}(H^1(G,B) \to H^1(G,C))$. Then $g \circ \psi=0$ in $H^1(G,C)$, this means that there exists some $x \in C$ such that $g \circ \psi(\sigma)=\sigma x - x$. Since $g$ is surjective, $x=g(y)$ for some $y \in B$. Therefore

$$g(\psi(\sigma))=\sigma g(y)-g(y)=g(\sigma y- y).$$

Let $\psi'$ be defined by $\psi'(\sigma)=\psi(\sigma)-(\sigma y- y)$. It represents the same element as $\psi$ in $H^1(G,B)$, and $g \circ \psi'(\sigma)=0$ for all $\sigma$. Thus for each $\sigma \in G$,  $\psi'(\sigma) \in \text{Ker}(g)=\text{Im}(f)$. Since $f$ is injective there exists a unique $a_\sigma$ such that $\psi'(\sigma)=f(a_\sigma)$. Let $\phi: G \to A$ be given by $\sigma \mapsto a_\sigma$. Now just check that $\phi \in Z^1(G,A)$ i.e $\phi(\sigma \tau)=\sigma \phi(\tau)+ \phi(\sigma)$ for all $\sigma, \tau$.

Of course the question is trivial if you know about derived functors and the fact that $H^{*}(G,-)=\text{Ext}_{\mathbb{Z}[G]}(\mathbb{Z},-)$ is the right derived functor of $\text{Hom}_{\mathbb{Z}[G]}(\mathbb{Z},-)=(-)^G$. You can read about that in a homological algebra book.


问:
Laplace transform of $f(t)=e^{-Kt^n} $
Could someone help me find the Laplace transform of the time funtion $f(t)=e^{-Kt^n} $ where $K$ is real and positive and $0<n<1$ ? Thank you.

答:
We have that :
$$
\frac{\Gamma(\omega)}{m^{\omega}}=\int_{0}^{\infty}x^{\omega^{-1}}e^{-mn}\;\text{d}x
$$
Now applying the Laplace Transform of $e^{-Kt^{n}}$, we get :
$$
\mathcal{L}\left(e^{-Kt^{n}}\right)=\int_{0}^{\infty}e^{-Kt^{n}}e^{-st}\;\text{d}t
\tag{1}$$
Since :
$$
e^{-Kt^{n}}=\sum_{r=0}^{\infty}\frac{(-K)^{r}}{r!}t^{nr}
\tag{2}$$
Then by substituting $(2)$ in $(1)$ we get :
$$
\int_{0}^{\infty}\sum_{r=0}^{\infty}\frac{(-K)^{r}}{r!}t^{nr}e^{-st}\;\text{d}t
$$
Hence, we get that :
$$
\boxed{F(s)=\sum_{r=0}^{\infty}\frac{(-K)^{r}}{r!}\frac{\Gamma(1+nr)}{s^{1+nr}}}
$$

[Note] : $1+nr>0$ to avoid $\Gamma$ diverging to $\pm\infty$

Title: Laplace transform of $f(t)=e^{-Kt^n} $


问:
When can i get away with approximating the expected value of a ratio as the ratio of expected values
I'm actually an engineering student so I'm not too good with probability and was hoping someone may be able to help with the following:

So I have a ratio of discrete random variables. I want to be able to know when I can get away with approximating its expected value as a ratio of expected values. For example, for the continuous case I came across the following formulation from a Taylor series expansion

$\text{E}[x/y]=\text{E}[x]/ \text{E}[y]- \text{Cov}[x ,y]/ \text{E}[y]^2 + \text{E}[x] \text{Var}[y]/ \text{E}[y]^3$

Does this apply for discrete random variables? if not, what is its analogous expression?

thank you!
hadi

答:
A  simple, general, good approximation purely in terms of expectations does not seem achievable.  For example, let $\epsilon$ be a very small positive number, and let $X$ and $Y$ be independent, each taking on the value $\epsilon$ with probability $1/2$, and $1$ with probability $1/2$.  Then $E(X/Y)$ is very large, but the various moments are all quite reasonable.

Possibly some reasonable approximations are possible if all values of $Y$ are well away from $0$.


问:
Finding the limit of a function?
The limit is actually easy: $\displaystyle \lim \limits_{t\to\infty}\dfrac{t^{k+1}}{e^t}$

One can use hopitals rule and say that ultimately the upper function will be reduced to a constant while the lower function will remain the same. Hence, the limit is zero. I was wondering if there is another way or a more mathematical representation of the answer.

答:
Without l'Hôpital:

We start with  $$\tag1e^x\ge 1+x\quad\text{for all }x\ge 0,$$
 which follows (and in fact holds even for all $x\in\mathbb R$)

from Bernoulli's inequality applied to $\left(1+\frac xn\right)^n$ if one introduces $e^x$ as $\lim_{n\to\infty}\left(1+\frac xn\right)^n$

from $e^x\ge 1$ for $x\ge0$ and $e^x-e^0=(x-0)e^\xi\ge x$ with $0<\xi< x$ (mean value theorem) if one introduces $e^x$ as solution of $y'=y$ with $y(0)=1$

by dropping all higher powers if one introduces $e^x$ via the power series $\sum_{k=0}^\infty\frac{x^k}{k!}$

By substituting $x=\frac t{k+1}$ in $(1)$ and raising to $(k+1)$st power, this becomes
$$ e^t=\left(e^{t/(k+1)}\right)^{k+1}\ge \left(1+\frac t{k+1}\right)^{k+1}\ge \frac1{(k+1)^{k+1}}\cdot t^{k+1}\quad\text{for }t\ge 0.$$
Therefore,
$$\left|\frac{t^{k+1}}{e^t}\right|\le (k+1)^{k+1}\quad\text{for }t\ge 0$$
and
$$\left|\frac{t^{k}}{e^t}\right|\le (k+1)^{k+1}\cdot\frac1t\quad\text{for }t> 0,$$
i.e.
$$\lim_{t\to+\infty}\frac{t^{k}}{e^t}=0.$$

Title: Finding the limit of a function?


答:
$6/\pi^2$ appears in the density of square-free numbers because  $\zeta(2)=\pi^2/6$ and every integer is uniquely of the form $k n^2$ with $k$ square-free.

Then $\zeta(2)=\pi^2/6$ because $$\frac{\pi^2}{\sin^2(\pi z)} = \sum_n \frac1{(z-n)^2}$$
This latter formula is magic: the LHS minus the RHS is a bounded entire function, thus constant, which is the great achievement of complex analysis.

If you don't like it then the video mentioned in the comment is saying that $$\sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=1/4$$ and that $$\lim_{k\to \infty} \sum_{n=0}^{2^k} \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}=\sum_{n=-\infty}^\infty \lim_{k\to \infty}  \frac1{|2^k (e^{2i \pi (2n+1)/2^{k+1}}-1)|^2}$$ $$ = \sum_{n=-\infty}^\infty \frac1{ |i\pi (2n+1)|^2} = \frac1{\pi^2}2(1-1/4)\zeta(2)$$

Title: Why does $\pi$ appear in the probability of a number being square-free?

Body:
The probability of a number being square-free (i.e., the number has no divisor that is a square, cube, etc.) is $6/\pi^2$. I have seen many appearances of $\pi$, and this is also similar to them. All of them can be explained intuitively why pi appears in them. But I don't see any connection between square-free numbers and $\pi$. So my question is:

Why $\pi$ appears here?

Note: I don't want rigorous proofs, I want intuitive explanations.


问:
Invertible elements
Find the invertible elements of $Z_6$, $Z_7$, and $Z_8$.

From the book:
Solution By ‘trial and error’, the invertible elements are:
$$Z_6 : 1, 5\\
Z_7 : 1, 2, 3, 4, 5, 6\\
Z_8 : 1, 3, 5, 7$$

I don't understand what is done with 'trial and error'.

An element $r$ in $Z_m$ is invertible if there is some $x$ in $Z_m$ such that $rx=1$.

$$Z_6={\{1,2,3,4,5}\}\\
1\cdot 1=1\\
5\cdot \frac{1}{5}=1$$

But $\frac{1}{5}$ is not in $Z_6$. Why is that invertible?

EDIT:
Have i understood the math correct (based on the answer)?

If I test everything with $5$ from $Z_6$:
$$5*1=5(mod6)\\
5*2=4(mod6)\\
5*3=3(mod6)\\
5*4=2(mod6)\\
5*5=1(mod6)$$

$5*5=1(mod6)$ gives the answer $=1$. Therefor $5$ from $Z_6$ is invertible.

$$3*1=3(mod6)\\
3*2=0(mod6)\\
3*3=3(mod6)\\
3*4=0(mod6)\\
3*5=3(mod6)$$

None give the answer $=1$. Therefor $3$ from $Z_6$ is not ivnertible.

答:
by the definition of the inverse, you can notice that :

$$5\times 5=25=6\times 4+1=1[6] $$

Hence, 5 is the inverce of 5 in $\mathbb{Z}_6$

Title: Invertible elements


答:
Let $B = \{(1,1,0) , ( 0, 2,1) \}$ and we claim that this is a basis.

Observe that since $x +2z = y$ we have $(x,y,z) = x(1,1,0) + z(0,2,1)$ for any $x,y,z$ in $K$. So $B$ spans $V$. I leave it to you to verify that $B$ is linearly independent.

Title: Basis for subspace

Body:
Consider $V= \left \{ u= \begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix} : x_1-x_2+2x_3=0\right \}\subset K^3$.

What is the basis of $V \subset K^3$

The format of the subspace is proving itself difficult to me. I know the span of the vectors in the basis is equal to the subset and that the vectors in the basis are linearly independent. I furthermore know the dimension is equal to the ammount og vectors in the basis of the subspace. Normally I would reduce to reduced echeleon form and be done. What approach would work here? I cant seem to make a matrix for the subspace.


问:
Functional equation $f(x+y)=f(x)f(y)-f(xy)+1$
I'm new to solving functional equations and found the following functional equation from a collection of functional equations.

Find all functions $f: \mathbb{Q}\to \mathbb{R}$ for $$f(x+y)=f(x)f(y)-f(xy)+1$$ for all $x,y\in \mathbb{Q}$.

By substituting $x,y\to 0$, we get
$$f(0)=f(0)^2-f(0)+1$$
$$f(0)^2-2f(0)+1=0$$
$$f(0)=1$$
And then doing the substitution $x\to 1,\ y\to -1$, we get
$$f(0)=f(1)f(-1)-f(-1)+1$$
$$f(1)f(-1)-f(-1)=0$$
$$f(-1)(f(1)-1)=0$$
$$f(-1)=0\vee f(1)=1$$

So, no we have three possible cases to consider and I have no idea on how to proceed on any of them.
I would appreciate any hints before giving the full solution as in the best case that's all I need right now.

答:
$f(0)=f(0+0)=f(0)^2-f(0)+1$ so $(f(0)-1)^2=0$, $\therefore f(0)=1$.

Let $f(1)=a+1$, then noting that $f(n+1)=af(n)+1$ \begin{align}f(2)&=f(1+1)=a^2+a+1\\
f(4)&=f(3+1)=\cdots=a^4+a^3+a^2+a+1\\
&=f(2+2)=f(2)^2-f(4)+1\end{align} $$\therefore 2f(4)=f(2)^2+1$$ Solving for $a$ gives $a=-1,0,1$.

If $a=-1$, $f(1)=0$, then $f(n)=\begin{cases}1&n \textrm{ odd}\\0&n \textrm{ even}\end{cases}$. This gives the formula $f(x+1)=1-f(x)$. This function does not extend to $\mathbb{Q}$ otherwise $f(1)=f(2\times\tfrac{1}{2})=f(2)f(\tfrac{1}{2})-f(2+\tfrac{1}{2})+1$, which is the same as $0=f(\tfrac{1}{2})-f(\tfrac{1}{2})+1$.

If $a=0$, then $f(n)=1$ for all $n$. This extends to all of $\mathbb{Q}$ by noting that $f(x+1)=f(x)f(1)-f(x)+1=1$.

If $a=1$, then $f(n)=n+1$ for all $n$. This extends as in this answer to $\mathbb{Q}$.

The last two functions $f(x)=x+1$ and $f(x)=1$ satisfy the general formula for all $x\in \mathbb{Q}$.


问:
Unable to understand the notation in the chapter on manifolds
While self studying Topology from Wayne Patty I was unable to understand this notation on page 218 of the textbook. Subsection is Manifolds.

Consider this statement(which is given as theorem): If X and Y are compact surface, then $\chi (X\mathbin\#Y) = \chi(X)+\chi(Y)- 2$.

Here $\chi(X)$ is Euler characteristic of $X$.

The author then writes:

Assume $X$ and $Y$ are triangulated. Form $X\mathbin\#Y$ by removing the interior of a triangle from each and identifying the edges and vertices of the vertices of the boundaries of the triangles that have been removed.

Is there specific name (mathematical, name of the symbol is $\#$ is clear to me) for the construction $\#$ given in the theorem? How it is constructed is clear to me.

答: In topology, the operation "#" is called connected sum. Its effect is to join two given manifolds together near a chosen point.


问:
Find the largest prime factor of $3(3(3(3(3(3(3(3(3(3(3+1)+1)+1)+1)+1)+1)+1)+1)+1)+1)+1$.
What I have done is that I found the expression is equal to $1+3+3^2+3^3+\cdots \cdots3^{11}=\frac{3^{12}-1}{2}$.

Then is there any way to find the prime factors of a number in such expression?

答: It's just 111 111 111 111 base 3, which one can find the factorisation from the cunningham project.  The largest factor is in a12, which is 81-9+1 = 73.  The factors of a2, a3, a4, a6 are 4, 13, 10, 7.  So i guess it's $2^3 \times 5 \times 7 \times 13 \times 73$.


答:
We have
\begin{align*}
\left \{ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} : x_1-x_2+2x_3=0\right \}
&=
\left \{ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} : x_1=x_2-2x_3\right \}
\\&=
\left \{ \begin{bmatrix} x_2 - 2 x_3 \\ x_2 \\ x_3 \end{bmatrix} : x_2, x_3 \in K\right \}
\\&=
\left \{ x_2 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} : x_2, x_3 \in K\right \}
\\&= \operatorname{span} \left \{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} , \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \right \}
\end{align*}
and the two vectors are linearly independent.

Title: Basis for subspace


问:
How to do hypothesis test between two different distributions?
I meet a practical problem: How to do hypothesis test between two different distributions, for example:

$H_0$: distribution is a normal distribution
$p(x) = \frac{1}{\sqrt{2\pi} \sigma}\exp(\frac{x^2}{2\sigma^2}) $

$H_1$: distribution is Laplace distribution $p(x) = \frac{1}{2b}\exp(-\frac{|x|}{b})$

with $\sigma$ and $b$ fixed.

How can I construct some statistics to do hypothesis test? How to construct uniformly most powerful(UMP) test?

Right now, I only know how to do hypothesis test of same kind distribution with different parameters. In general, what's the procedure to do hypothesis test between two different kinds of distribution and construct some meaningful statistics and UMP? Can you please provide some references, thanks.

答:
If $\sigma$ and $b$ are fixed constants, we have two simple hypotheses.

The best test (formally: the most powerful) in this situation is given by Neyman–Pearson lemma.

The optimal criteria has the next form: $$S = \{ x: \frac{p_{H_1}(x)}{p_{H_0}(x)} \ge c\},$$

where $c$ is a solution of the equation $\int_{S} p_{H_0}(x)dx = \alpha$, where $\alpha$ is the significance level.

References: Testing Statistical Hypotheses, Lehmann, or any book about statistics.


答:
Let $\varphi$ be the Euler phi function such that, for any $n \in \mathbb N$, we have $$\varphi(n)=|\{m\in \mathbb N:m<n\text{, gcd(m,n)=1}\}|.$$

In $\mathbb Z_n$ the number of invertible elements is given by $\varphi(n)$. In your case, $\mathbb Z_6$ has two invertible elements, infact $\varphi(6)=|\{1,5\}|=2$.

The inverse of $5$ in $\mathbb Z_6$ is $5$ because $5\cdot5=25=1+4\cdot6\equiv_6 1.$

Rmk: let $p\in \mathbb N$ be a prime number $\implies$ the cardinality of the set of invertible elements in $\mathbb Z_p$ is $p-1$ because $|U(\mathbb Z_p)|=\varphi(p)=p-1.$

Title: Invertible elements

Body:
Find the invertible elements of $Z_6$, $Z_7$, and $Z_8$.

From the book:
Solution By ‘trial and error’, the invertible elements are:
$$Z_6 : 1, 5\\
Z_7 : 1, 2, 3, 4, 5, 6\\
Z_8 : 1, 3, 5, 7$$

I don't understand what is done with 'trial and error'.

An element $r$ in $Z_m$ is invertible if there is some $x$ in $Z_m$ such that $rx=1$.

$$Z_6={\{1,2,3,4,5}\}\\
1\cdot 1=1\\
5\cdot \frac{1}{5}=1$$

But $\frac{1}{5}$ is not in $Z_6$. Why is that invertible?

EDIT:
Have i understood the math correct (based on the answer)?

If I test everything with $5$ from $Z_6$:
$$5*1=5(mod6)\\
5*2=4(mod6)\\
5*3=3(mod6)\\
5*4=2(mod6)\\
5*5=1(mod6)$$

$5*5=1(mod6)$ gives the answer $=1$. Therefor $5$ from $Z_6$ is invertible.

$$3*1=3(mod6)\\
3*2=0(mod6)\\
3*3=3(mod6)\\
3*4=0(mod6)\\
3*5=3(mod6)$$

None give the answer $=1$. Therefor $3$ from $Z_6$ is not ivnertible.


问:
On the Cyclic Subgroups
I recently read in a book on group theory that, a (non-abelian) $2$-generator group all of whose proper subgroups are cyclic, has been constructed by Ol`sanski; these are infinite simple groups. (see Groups '93 GalwaySt Andrews vol. 2, p.389). Can anybody give structure of the group here, and certain lower dimensional matrix representation of the group? (I mean, viewing group as group of matrices over some ring/field,..).

Also, are there non-abelian groups of similar type, where all proper subgroups are infinite cyclic?

(I couldn't see this group (means I know only the existance, beliving onOl'sanski), I couldn't see also the paper in which it is given, due to accessibility of journals. I like to know what is this group; one can write here it in terms of generators and relations or as a group of matrices etc.)

答:
This group (s) is (are) also called Tarski Monster : it's a simple non-abelian group generated by two elements with the rather astonishing property that any non-trivial subgroup is finite, cyclic and of order a given, fixed prime $\;p\;$ .

For sure these groups exist for prime $\,p>10^{75}\,$ , and as far as I know this bound hasn't been improved greatly in the last 34 years.

Title: On the Cyclic Subgroups


问:
Prove that $1+ \frac{5}{4}x ≤(1+x)^{\frac{5}{4}}≤1+ \frac{5}{4} x+\frac{5}{16} x^2$
How to prove this inequality:
$$1+ \frac54x \le (1+x)^{5/4} \le 1+ \frac54 x+\frac5{16} x^2$$

I tried applying binomial theorem to get this inequality, but I am not able to get it precisely. Please help.

答:
You can make as strong inequalities as you which since, using the binomial expansion
$$(1+x)^{\frac{5}{4}}=\sum_{n=0}^\infty \binom{\frac{5}{4}}{n}x^n$$ and this is an alternating series after $x^2$
$$(1+x)^{\frac{5}{4}}=1+\frac{5 x}{4}+\frac{5 x^2}{32}-\frac{5 x^3}{128}+\frac{35 x^4}{2048}-\frac{77
   x^5}{8192}+\frac{385 x^6}{65536}+O\left(x^7\right)$$


问:
Integration of Lie derivatives
I'm now working on two exercises:

Exercise 1 Let $M^n$ be an oriented manifold without boundary, and $\alpha \in \Omega^{s}(M)$, $\beta \in \Omega^{n-s}(M)$ be differential forms on $M$. Let $X \in \mathcal{X}(M)$ be a smooth vector field on $M$ with compact support. Show that
$$
\int_M \mathcal{L}_X(\alpha) \wedge \beta = - \int_M \alpha \wedge \mathcal{L}_X(\beta).
$$

Exercise 2 Let $M^n$ be an oriented closed manifold (compact without boundary), and $g$ a Riemannian metric on $M$. If $X \in \mathcal{X}(M)$ be a smooth vector field on $M$ and $f \in C^{\infty}(M)$ be a smooth function on $M$. Show that
$$
\int_M X(f) \cdot \Omega = - \int_M f \cdot \mathcal{L}_X(\Omega).
$$
where $\Omega$ is the volume form of $(M,g)$.

I have been trying on the two exercises and meetting the same difficulty.

For 1: We know that $\mathcal{L}_X(\alpha) \wedge \beta + \alpha \wedge \mathcal{L}_X(\beta) = \mathcal{L}_X(\alpha \wedge \beta)$. So applying the integration on $M$ on both sides, it suffices to show that the integration vanishes:
$$
\int_M \mathcal{L}_X(\alpha \wedge \beta) = 0.
$$

For 2: We know that $\mathcal{L}_X(f \cdot \Omega) = X(f) \cdot \Omega + f \cdot \mathcal{L}_X \Omega$.So applying the integration on $M$ on both sides, it suffices to show that the integration vanishes:
$$
\int_M \mathcal{L}_X(f \cdot \Omega) = 0.
$$

Yet I don't know how to carry on to show that the two integrations are zero. It seems that there is some machinary that I'm not awareness of or not familiar with. So my question is:How to prove that the two integrations above is zero?

My attempts: I have tried to use Cartan's magic identity, namely
$$
\mathcal{L}_X = i_X \circ \mathrm{d} + \mathrm{d} \circ i_X,
$$
where $i_X$ is the interior multiplication. In both cases, the manifold $M$ has no boundary, so by Stokes' theorm, the integration on the summand $\mathrm{d} \circ i_X$ is zero. So it remains to show that the integration on the summand $i_X \circ \mathrm{d}$ is zero as well. I got stuck here. :(

Thank you all for commenting and answering!

答: For any top-form $\omega$, $$\int_M L_X(\omega) = \int_{M} d(i_X(\omega)) + i_X(d\omega) = 0, $$ as by Stokes' theorem, the integral of any exact top-form on a manifold without boundary is zero, and because $d\omega$ is zero, as it is an $(n+1)$-form on an $n$-dimensional manifold.

Title: Integration of Lie derivatives


问:
Prove U = $\mathbb{R}^2$\ {(0,0)} is open in $\mathbb{R}^2$
I have to prove that  $U = \mathbb{R}^2\setminus \{(0,0)\}$ is open in $\mathbb{R}^2$. I know that $U$ is open if $U = \operatorname{inw}(U)$ (collection of all interior points of $U$) (this is the Dutch notation, not sure about the English one). I know the definition of a interior point too, but I don't know how I can solve this question. I think you can use the balls to cover $U$ including the point $(0,0)$?

答: Hint: For each $(x,y)\neq(0,0)$ in $\mathbb R^2$ the open ball of radius $\sqrt{x^2+y^2}$ centered at $(x,y)$ does not contain the origin.