问:
Permutation involving different outcomes
So say I have want to find, with the lowercase alphabet, how many strings of 9 different letters that can be formed that have the letter z in front, a block of letters xy (among other random distinct letters), a vowel (a, e, i, o, u) at the end.

This is my working:

There is only one way to pick 1 letter, so that would be for the case for z.

With 7 letters remaining for the string of length 9, and since three predetermined letters are among the 9, then it's just 23 letters than remain and with the permutation:

$P(23, 7) = \frac{23!}{(23-7)!}$

The last vowel is simply $P(5,1) = \frac{5!}{(5-1)!}$

And the result is the product of those 3. Is that correct?

答:
I am not sure what three factors the query is referring to, since I only saw two factors: P(23,7) and P(5,1).

Anyway, this is how I would calculate it.

You have $P_1 = 5$ possibilities for the rightmost position.

Assume that "a" is chosen for that.

Then, the letters x,y,a,z are taken.

There are $P_2 = \binom{22}{5}$ ways of selecting the 5 other letters needed.

Here, I am going to assume that the (y,x) block is not allowed, and that instead, the letters must appear in the order (x,y).  If this assumption is false, multiply my final answer by 2.

Finally, since the (x,y) letters must appear as a block, you can construe them as a unit.  Then you have 6 units to be permuted.

This can be done in $P_3 = 6!$ ways.

Final answer is :

$$P_1 \times P_2 \times P_3 = 5 \times \binom{22}{5} \times 6!.$$


问:
Need help writing a recursive formula for the following sequence
$$a_n = \lceil n−1/n \rceil$$

Hey Everyone, I am new here made an account because this question has stomped me for a bit, for a normal sequence I have a pretty easy time writing a recursive sequence but this one has completely stomped me if anyone can help guide me with some work and steps it would really help me

ContentLicense: CC BY-SA 4.0

Body: $a_{1} = 0, a_{2} = 2$ and $a_{n+1} = 1+ a_{n}$ for $n > 1$.


问:
Journals specializing in point-set topology
Are there any scholarly journals specializing in point-set topology? I would like to know which ones. I do not want something that has mostly algebraic topology articles.</p>

答:
Topology and its Applications is one possibility; it has a higher proportion of papers in general and set-theoretic topology than most with which I’m at all familiar, and the current editors-in-chief work in those fields. From the journal’s web site:

The mathematical focus of the journal is that suggested by the title: Research in Topology. It is felt that it is inadvisable to attempt a definitive description of topology as understood for this journal. Certainly the subject includes the algebraic, general, geometric, and set-theoretic facets of topology as well as areas of interactions between topology and other mathematical disciplines, e.g. topological algebra, topological dynamics, functional analysis, category theory. Since the roles of various aspects of topology continue to change, the non-specific delineation of topics serves to reflect the current state of research in topology.


问:
Show that $I(y)=\int_0^1 \frac{x-y}{(x+y)^3}dx$ exists
Show that the Riemann Integral

$I(y)=\int_0^1 \frac{x-y}{(x+y)^3}dx$

exists for $y\in(0,1]$.

$A=\int_0^1(\int_0^1 \frac{x-y}{(x+y)^3}dx)dy$, $B=\int_0^1(\int_0^1 \frac{x-y}{(x+y)^3}dy)dx$

Show that $A$ and $B$ exist and $A\neq B$.

Showing $A\neq B$ wasn't that difficult, it was just integration by parts, I got $A=-\frac{1}{2}$ and $B=\frac{1}{2}$. But I don't know how to show that these integrals exist. I was able to calculate them and they're continuous but I don't really believe that that's enough.
I do realize that $I(y)$ wouldn't exist if $y\in[0,1]$ because you'd divide by $0$.

答:
The integrand $\displaystyle x \mapsto \frac{x-y}{(x+y)^3}$ is continuous on the closed interval $[0,1]$ for $0 < y \leqslant 1$ and, therefore, $I(y)$ exists as a Riemann integral.

We also have $\displaystyle \frac{x-y}{(x+y)^3}= \frac{\partial}{\partial x}\frac{-x}{(x+y)^2},$ which gives us for $y \in (0,1]$,

$$I(y) = \int_0^1\frac{x-y}{(x+y)^3} \, dx = \left.\frac{-x}{(x+y)^2}\right|_0^1 = \frac{-1}{(1+y)^2}$$

Technically $I(0)$ does not exist since, when $y = 0$, the integrand $x^{-2}$ is not integrable on $[0,1]$ in any sense.  However, the value of $I$ at $y=0$ does not affect the Riemann integral  since $\{0\}$ is a set of measure zero.  We can extend $I|_{(0,1]}$ as a Riemann integrable function on $[0,1]$ so that we have existence of the iterated integral

$$ \int_0^1\left(\int_0^1 \frac{x-y}{(x+y)^3}\,dx\right)\,dy = \int_0^1 \frac{-1}{(1+y)^2}\, dy =  \frac{1}{2}$$

By a similar argument, the iterated integral with the order switched exists.

Since the iterated integrals are not equal it follows that the function

$$(x,y) \mapsto \frac{x-y}{(x+y)^3},$$

is not absolutely (or Lebesgue) integrable over $[0,1]\times [0,1]$. Otherwise, the iterated integrals would have to be equal as a consequence of Fubini's theorem.


问:
How to compute $\int_0^1 \cos^2{\pi x}\ln \Gamma(x)dx$
Problem

Find the exact value of
$$\int_0^1 \cos^2{\pi x}\ln \Gamma(x)dx$$

My Attempt

\begin{align*}
I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)dx\\
2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)dx+\int_0^1 \cos^2{(\pi (1-x))}\ln \Gamma(1-x)dx \tag{Using reflection}\\
2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{(\pi (1-x))}\ln \Gamma(1-x)dx\\
2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{\pi x}\ln \Gamma(1-x)dx\\
2I&=\int_0^1 \cos^2{(\pi x)}[\ln \Gamma(x)+\ln \Gamma(1-x)]dx\\
2I&=\int_0^1 \cos^2{(\pi x)}\left[\ln\left(\frac{\pi}{\sin(\pi x)}\right)\right]dx\tag{Using Euler's reflection formula}
\end{align*}

And here is where I am stuck, I have no idea how to evaluate the integral. Any help is appreciated.

答:
Consider reduce the power of cosine term:

$$
\begin{aligned}
\int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx
&=\frac12\left[\int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx+\int_0^1\log\Gamma(x)\mathrm dx\right]
\end{aligned}
$$

For the first part, by the property that

$$
\int_a^bf(x)\mathrm dx=\frac12\int_a^b[f(x)+f(a+b-x)]\mathrm dx
$$

we have

$$
\begin{aligned}
\int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx
&=\frac12\int_0^1\cos(2\pi x)\log[\Gamma(x)\Gamma(1-x)]\mathrm dx \\
&=\frac12\int_0^1\cos(2\pi x)\log\left[\pi\over\sin(\pi x)\right]\mathrm dx \\
&={\log\pi\over2}\int_0^1\cos(2\pi x)\mathrm dx-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx \\
&=-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx
\end{aligned}
$$

In fact, by the Fourier expansion of log sine

$$
\log\sin\theta=-\log2-\sum_{k=1}^\infty{\cos(2k\theta)\over k}
$$

we can deduce

$$
\begin{aligned}
-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx
&={\log2\over2}\int_0^1\cos(2\pi x)\mathrm dx+\frac12\sum_{k=1}^\infty\frac1k\int_0^1\cos(2\pi x)\cos(2\pi kx)\mathrm dx \\
&=\frac14
\end{aligned}
$$

For the last part, by Raabe's integral we have

$$
\int_0^1\log\Gamma(x)\mathrm dx=\frac12\log(2\pi)
$$

All things combined, we obtain

$$
\int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx=\frac18+{\log(2\pi)\over4}
$$


问:
$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$
I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.

Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4}$. I noticed that
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\sqrt{n^3} \Big(\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n-1}}\Big) $$

That doesn't seem to lead to $-\frac{1}{4}$ when $n\to \infty$ . I tried another form of the original expression:
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)$$

If I multiply by the conjugate, we obtain
$$2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$

Now that doesn't seem to be of any use either. Any ideas?

答:
Dividing both the numerator and the denominator by $\sqrt{n^3}$, you have
\begin{eqnarray}
&&\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}\\
&=&\frac{2}{\big(\sqrt{1+\frac1n}+\sqrt{1-\frac1n}+2\big)\big( \sqrt{1-\frac1{n^2}}+1\big)}.
\end{eqnarray}
Now you can take the limit to get the result.


问:
Proving the given inequality
Let $a_1,...,a_n$ be positive numbers. Show that $(a_1+...+a_n)(\frac{1}{a_1}+...+\frac{1}{a_n})\ge n^2$. Just not really sure how to approach this problem, any tips/solutions would be greatly appreciated.

答:
You are going to see which Inequality suits this problem, once it is written in this form:
$$
\tag{1}\label{1}\frac{a_1 + a_2 + \cdots + a_n}{n} \geqslant \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}
$$
Equivalently,
$$
\left(a_1 + a_2 + \cdots + a_n\right)
\left(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}\right) \geqslant n^2
$$

Title: Proving the given inequality


问:
If $P(X_i > x) = e^{-x}$, how to prove $\max_{1\le m\le n}X_m/\log n \to 1 a.s$?
Here is a question from Rick Durrett's Probability. Exercise 2.3.18.

$X_1,X_2,...$ are i.i.d. with $P(X_i > x) = e^{-x}$, $M_n = \max_{1\le m\le n} X_m$. Prove that $M_n/\log n \to 1$ a.s.

My solution:

Clearly, we can show by Borel-Cantelli lemma that $\limsup_{n\to\infty} X_n/\log n = 1$ a.s.. By this it is easy to prove that $P(M_n \le (1+\varepsilon)\log n) < \infty$ and hence $\limsup_{n\to\infty} M_n / \log n \le 1$. On the other hand, we need to show that $P(M_n \le (1-\varepsilon)\log n) = 0$ as $n\to \infty$. However, I could only prove that $P(M_n \le (1-\varepsilon)\log n) = 1 - n^{-(1-\varepsilon)}$ and when $n\to\infty$ this probability goes to $1$. It is strange, because that means $M_n \le \log n$ when $n\to\infty$ and this contradicts the question.

I see the answer wrote by Durrett says:$P(M_n < (1-\varepsilon )\log n) = (1 - n^{-(1-\varepsilon)})^n \le e^{-n^\varepsilon}$, and it is summable, by B-C lemma shows $P(M_n < (1-\varepsilon )\log n ,i.o.) = 0$.

However, why is $P(X_i < (1-\varepsilon )\log n) = (1 - n^{-(1-\varepsilon)})^n$?

$P(X_i\le x) = 1 - e^{-x}$, thus $P(X_i \le(1-\varepsilon)\log n) = 1- e^{\log (n^{-(1-\varepsilon)})} = 1 - n^{-(1-\varepsilon)}$ and there is no such $(\cdot)^n$.

Is there anything wrong in my proof? Or the question itself is wrong?

答:
Since $M_n = \max_{1\le m\le n} X_m$ thus,

$P(M_n < (1-\varepsilon)\log n) = P(\max_{1\le m\le n} X_m < (1-\varepsilon)\log n) = P(X_1 < (1-\varepsilon)\log n,\cdots,X_n < (1-\varepsilon)\log n) = \prod_{i=1}^n P(X_i< (1-\varepsilon)\log n) = (1-n^{-(1-\varepsilon)})^n.$

The last equation comes from  $X_n$ are i.i.d.

We have the inequality: $\lim_{n\to\infty} (1-\frac{k}{n})^n\le e^{-k}$, thus $\lim_{n\to\infty} (1-\frac{1}{n^{1-\varepsilon}})^n = \lim_{n\to\infty} (1-\frac{n^\varepsilon}{n})^n \le  e^{-n^\varepsilon}$, and this gives the result.


问:
showing a sequence of sequences converges
Suppose we have an infinite sequence of finite $n$-tuples with:

$$f_1 = (a_{11},\space a_{12},\space ... , \space a_{1n})$$
$$f_2 = (a_{21},\space a_{22}, \space ... ,\space a_{2n})$$
$$f_3 = (a_{31}, \space a_{32}, \space ... , \space a_{3n})$$
$$\vdots \qquad \qquad  \qquad \vdots$$

Further suppose that the "$k^{th}$ vertical sequence" i.e. the sequence formed by the $k^{th}$ element of each sequence i.e.  $(a_{1k}, \space a_{2k}, \space a_{3k}, ... )$ converges to some real number $a_k$, and that this is true for all the "vertical" sequences meaning for all integers $k$ with $1 \leq k\leq n.$

Then, we can show that under the following metric $$ d((b_1, b_2,...), (c_1, c_2, ...)) = \sup_k \space\{|b_j - c_j|\}$$ the sequence of $n$-tuples $f_1, f_2, ...$ converges to $ f = (a_1, a_2, ...)$:

Since the sequence $(a_{1k}, \space a_{2k}, \space a_{3k}, ... )$ converges to $a_k$, given $\epsilon > 0 ,$ there exists a positive integer $N_k$ such that for all $m \geq N_k$
$$|a_{mk} - a_k| < \epsilon$$

Let $N = \max\{N_1, N_2, ... N_n\}$.

Then, given $\epsilon > 0$, for all $m \geq N$, for all $\textbf{k} \in \{1, 2, ..., n\}$ $$|a_{mk} - a_k| < \epsilon $$

Therefore,
$$\sup_k \space \{|a_{mk} - a_k|\} <\epsilon$$

Thus, for all $k \geq N$,
$$d(f_k, f) < \epsilon$$

So $\{f_k \}$ converges to $f$.

Question: My question is suppose instead of finite $n$-tuples, we have each $f_k$ is an infinite sequence, can this proof be modified to show that the $f_k$ converge to $f$?

i.e. suppose we have an infinite sequence of infinite sequences with:

$$f_1 = (a_{11},\space a_{12},\space ... )$$
$$f_2 = (a_{21},\space a_{22}, \space ... )$$
$$f_3 = (a_{31}, \space a_{32}, \space ... )$$
$$\vdots \qquad \qquad  \qquad \vdots$$

It seems like there is trouble because we can't say "let $N = \max \{N_1, N_2, ...\}$" since this will be an infinite set.

How can we fix this? Or is the statement not true?

答:
A counterexample: With
$$
 f_1 = (1, 1, 1, 1, \ldots) \\
 f_2 = (0, 1, 1, 1, \ldots) \\
 f_3 = (0, 0, 1, 1, \ldots) \\
 f_4 = (0, 0, 0, 1, \ldots) \\
\vdots
$$
you have “column-wise” convergence to
$$
 f_0 = (0, 0, 0, 0, \ldots) \\
$$
but
$$
d(f_j, f_0) = \sup_k |f_{jk} - f_{0k}| = 1
$$
does not converge to zero.

Remark: With a different metric such as
$$d(f, g)=\sum^\infty_{n=k}\frac{1}{2^k}\frac{|f_k-g_k|}{1+|f_k-g_k|}
$$
convergence in each component is equivalent to convergence with respect to $d$, see for example Convergence with respect to $d_1(x,y)=\sum^\infty_{i=1}\frac{1}{2^i}\frac{|x_i-y_i|}{1+|x_i-y_i|}$ is equivalent to pointwise convergence?.


答:
The last expression
$$A=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$
leads to the result.

Since $n$ is large
$$\sqrt{n+1}\sim \sqrt{n} \qquad \sqrt{n-1}\sim \sqrt{n}\qquad \sqrt{n^2-1}\sim \sqrt{n^2}=n$$

$$A \sim -\frac{2\sqrt{n^3}}{\big(\sqrt{n}+\sqrt{n}+2\sqrt{n}\big)\big( n+n\big)}=-\frac{2n\sqrt{n}}{\big(4\sqrt{n}\big)\big( 2n\big)}=-\frac 14$$

Title: $\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$

Body:
I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.

Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4}$. I noticed that
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\sqrt{n^3} \Big(\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n-1}}\Big) $$

That doesn't seem to lead to $-\frac{1}{4}$ when $n\to \infty$ . I tried another form of the original expression:
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)$$

If I multiply by the conjugate, we obtain
$$2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$

Now that doesn't seem to be of any use either. Any ideas?


答:
Multiplying $\sqrt{n+1} - \sqrt n$ by $\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}$ yields $\dfrac 1 {\sqrt{n+1}+\sqrt n}.$

Similarly $\sqrt n - \sqrt{n-1} = \dfrac 1 {\sqrt n + \sqrt{n-1}}.$

So then we have
\begin{align}
& \big(\sqrt{n+1} - \sqrt n\big) - \big(\sqrt n - \sqrt{n-1} \big) \\[8pt]
= {} & \frac 1 {\sqrt{n+1}+\sqrt n} - \dfrac 1 {\sqrt n + \sqrt{n-1}} \\[12pt]
= {} & \frac{\sqrt{n-1}- \sqrt{n+1}}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})} \\[12pt]
= {} & \frac{\sqrt{n-1}- \sqrt{n+1}}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})} \cdot \frac{\sqrt{n-1} + \sqrt{n+1}}{\sqrt{n-1} + \sqrt{n+1}} \\[12pt]
= {} & \frac{-2}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})(\sqrt{n-1} + \sqrt{n+1})}
\end{align}
If this is multiplied by $\sqrt{n^3}$ it becomes
$$
-2 \cdot \frac {\sqrt n} {\sqrt{n+1}+\sqrt n} \cdot \frac{\sqrt n}{ \sqrt n + \sqrt{n-1}} \cdot \frac {\sqrt n}{\sqrt{n-1} + \sqrt{n+1}} \longrightarrow \frac{-1} 4.
$$


问:
The Chinese remainder theorem
I'm trying to do some questions on the Chinese remainder theorem, I've being reading the Wikipedia explanation but I still don't get it. Can someone explain it to me, please?

Here is the question I'm working with. If you have a simpler example, that would be great.

Find an integer $x$ such that $x\equiv2\bmod5$, $x\equiv1\bmod13$ and $x\equiv5\bmod17$.

答:
$$\begin{align}
   x &\equiv 2 \pmod5 \\
   x &\equiv 1 \pmod{13} \\
   x &\equiv 5 \pmod{17}
\end{align}$$

This is not the fastest way; but it is the most straightforward way. This method only works when the moduli are pairwise prime:

$$\gcd(5,13) = \gcd(5,17) = \gcd(13,17) = 1$$

We start with $ x \equiv 2 \pmod5$. Then for some integer, $u$, we must have
$$x = 5u+2$$ Then $x \equiv 1 \pmod{13}$ becomes

\begin{align}
   x &\equiv 1 \pmod{13} \\
   5u+2 &\equiv 1 \pmod{13} \\
   5u &\equiv -1 \equiv12 \equiv 25\pmod{13}
      &\text{(Add $13$ until you get a multiple of $5$.)} \\
   u &\equiv 5 \pmod{13} \\
\hline
   u &= 13v + 5
\end{align}

So now we have $$x = 5u+2 = 5(13v+5)+2 = 65v + 27$$

Next we take $x \equiv 5 \pmod{17}$. (The first few multiples of $17$ are
$17, 34, 51, 68, 85$).

\begin{align}
   x &\equiv 5 \pmod{17} \\
   65v + 27 &\equiv 5 \pmod{17} \\
   -3v&\equiv -5 \pmod{17} \\
   3v &\equiv 5 \equiv 22 \equiv 39 \pmod{17} \\
   v &\equiv 13 \pmod{17} \\
\hline
   v &= 13 + 17w
\end{align}

So $x = 65v + 27 = 65(17w + 13)+27 = 1105w + 872$

Which we can write as

$$x \equiv 872 \pmod{1105}$$

ContentLicense: CC BY-SA 3.0

Title: The Chinese remainder theorem


问:
Dummit and Foote 4.3.13: Find all the finite groups which have exactly two conjugacy classes.
Task: Find all the finite groups which have exactly two conjugacy classes.

Ideas:
I have been show the class equation and the orbit stabilizer formula and I wonder if I can put them to use. My general intuition so far is such;

Given $$|G|=\Sigma_{i=1}^{r}|\mathcal{O}_a|+|Z(G)|,$$ we need $\Sigma|\mathcal{O}_a|=2$ and $|Z(G)|\geq 1$ because the identity is always in the center. It could be that either each of the orbits are size 1 or a single order is size 2.

My other idea is to use the orbit stabilizer formula: $|G:G_a|=|\mathcal{O}_a|=2.$

Maybe split the possibilities into $G$ abelian and $G$ not abelian.

I'm new to this material so please let me know if either or these ideas are on the right track! Pointers are appreciated.

答: Since $\{e\}$ is its own conjugacy class, let $\mathcal O_a$ be the other conjugacy class. The conjugacy classes of $G$ partition $G$, so $|G|=1+|\mathcal O_a|$.  But the size of a conjugacy class must divide the order of the group, $|\mathcal O_a|$ divides $|\mathcal O_a|+1$.  The only way this can happen is if $|\mathcal O_a|=1$.  Therefore, $|G|=1+1=2$, so $G\cong C_2$.


问:
Complex integration help
The integral given is

$$\int_{-\infty}^{\infty} \frac{\cos(x)-1}{x^2}\,dx $$

Ok, so, I've used the upper semi circular contour with the function

$$f(z) = \frac{e^{iz}-1}{z^2}$$

Now the residue I get is $i$ which when used with the residue theorem gives me the answer of $-2\pi$ as the answer to the integral of the $f(z)$ and through this I get my answer to the original integral as $-2\pi$. However in the answers to this question the answer states that the solution is $-\pi$. I've double checked with Wolfram Alpha and Wolfram Alpha also gives this as an answer. Can anyone please correct me where I'm going wrong? Thanks.

答:
Condensing what you did and comments, define

$$f(z):=\frac{e^{iz}-1}{z^2}\;,\;\;\gamma_r:=\{\,z\in\Bbb C\;;\;z=re^{it}\;,\;t\in(0,\pi)\;,\;r\in\Bbb R_+\,\}$$

and define the path

$$\Gamma_{R,\epsilon}:=[-R,\epsilon]\cup\left(-\gamma_\epsilon\right)\cup[\epsilon,R]\cup\gamma_R\;,\;\;R>>\epsilon>0$$

where $\,-\gamma_r\,$ means the integration path is taken in the clockwise direction (the negative one, thus).

Since the pole at $\,z=0\,$ is simple (why?), we get

$$\text{Re}_{z=0}(f)=\lim_{z\to 0}\,zf(z)\stackrel{\text{l'Hospital}}=ie^0=i$$

so by the lemma and in particular its corollary in an answer here, and by CIT we get

$$2\pi i\cdot i=\oint\limits_{\Gamma_{R,\epsilon}}\,f(z)\,dz\implies$$

$$ -2\pi=\lim_{R\to\infty\,,\,\epsilon\to 0}\left(\int\limits_{-R}^{-\epsilon}\,f(x)dx+\int\limits_{\gamma_\epsilon}f(z)\,dz+\int\limits_\epsilon^Rf(x)\,dx+\int\limits_{\gamma_R}f(z)\,dz\right)=$$

$$=\int\limits_{-\infty}^\infty\frac{\cos x-1+i\sin x}{x^2}dx-\pi $$

And comparing real and imaginary parts you get what you need.

Note: the integral on $\,\gamma_R\,$ vanishes as $\,R\to\infty\,$ either by Jordan's Lemma or directly estimating the integral's modulus.

Title: Complex integration help


问:
Expectation of $X^2$ where $X = x_1 + x_2 + x_ 3+ \dots +x_k$ with $P(x_i = 0) = 1-p$ and $P(x_i = x) = p$
We are told that $X = x_1 + x_2 + x_ 3+ \dots +x_k$ and that the $x$'s are independent. The probability of $x_i = x$ is $p$, while the probability that $x_i = 0$ is $1-p$. Use this to find $E(X^2)$.

The first thing I noticed was that if $P(X = x)$ that's just like saying what's the probability that only one of the $x_i$ is equal to $x$. So this makes me think it's binomial distribution. If that is the case $E(X)$ is just the sum of binomial distributions. However, I am not sure how you can handle $E(X^2)$ as $E(X^2) = E((x_1 + x_2 + x_ 3+ \dots +x_k)^2)$.

Any suggestion?

答:
It is possible to perform the calculation directly.

First, we need to observe that for each $x_i$, we have $x_i^2 = x_i$, because if $x_i = 1$, then $x_i^2 = 1$; and if $x_i = 0$, then $x_i^2 = 0$.  Then

$$\begin{align}
\operatorname{E}[X^2]
&= \operatorname{E}\,\left[\left(\sum_{i=1}^k x_i \right)^2\right] \\
&= \operatorname{E}\left[ \sum_{i=1}^k \sum_{j=1}^k x_i x_j \right] \\
&= \operatorname{E}\left[ \sum_{i=1}^k x_i^2 + \sum_{i \ne j} x_i x_j \right] \\
&= \sum_{i=1}^k \operatorname{E}[x_i^2] + \sum_{i \ne j} \operatorname{E}[x_i x_j] \\
&= \sum_{i=1}^k \operatorname{E}[x_i] + \sum_{i \ne j} \operatorname{E}[x_i] \operatorname{E}[x_j] \\
&= kp + k(k-1)p^2 \\
&= kp(kp - p + 1).
\end{align}$$

The first step is to write $X$ as the sum of the $x_i$.  The second simply expands the square by writing it as a double sum.  The third step separates the terms for which $i = j$, of which there are $k$ such terms, versus $i \ne j$, of which there are $k(k-1)$ such terms.  The fourth step is simply the linearity of expectation.  The fifth step uses the fact $x_i^2 = x_i$ we explained earlier, and the pairwise independence of the $x_i$.  The rest is just algebra.


问:
$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$?
According to an example in my text book:

$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2}{\sqrt{1+\frac{1}{\sqrt{x}}}+\sqrt{1-\frac{1}{\sqrt{x}}}}$$

I don't see how this works. The closest I can get is:

$$\frac{(x + \sqrt{x}) - (x-\sqrt{x})}{\sqrt{x+\sqrt{x}}+\sqrt{x-\sqrt{x}}} = \frac{2\sqrt{x}}{\sqrt{x}(\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1})} = \frac{2}{\sqrt{\frac{1}{\sqrt{x}} + 1}+\sqrt{\frac{1}{\sqrt{x}} -1}}$$

Which is slightly off. But I'm not even sure if I calculated that correctly. What am I missing, what's the way to think?

答:
$$ \frac{(x + \sqrt{x}) - (x - \sqrt{x} )}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}} = \frac{2 \sqrt{x}}{\sqrt{x + \sqrt{x}} + \sqrt{x - \sqrt{x}}}$$
Divide by $\sqrt{x}$ throughout. It simplifies to
$$\frac{2}{\dfrac{\sqrt{x + \sqrt{x}} }{\sqrt{x}} + \dfrac{\sqrt{x - \sqrt{x}}}{\sqrt{x}}}.$$
Now the denominator simplifies to
$\sqrt{\dfrac{x + \sqrt{x}}{x}} + \sqrt{\dfrac{x - \sqrt{x}}{x}} $ and the answer comes out nicely.</p>


问:
Subgroups of $\mathbb{Z}^*_{29}$
Assume the group $G = \mathbb{Z}^*_{29} = \{1,2,\dots, 28 \}$ with operation of multiplication modulo $29$.

Since $G$ is cyclic, for every $d \in \{1,2,4,7,14,28 \}$, there exists a $H \leq G$ with $|H| = d$.

$d=1$: $H_1 = \{1\}$

$d = 2$: $H_2 = \{1, 28\}$, since $28^2 \mod 29 = 1$

$d= 4$: $H_3 = \{1,12,17,28 \}$, since $12 \cdot 17 \mod 29 = 1$ and we have closure.

Is there a way to yield the subgroups of order $7$ and $14$, without having to check all the possible combinations to assert closure?

答:
It's well known that $\Bbb Z_p^*=(\Bbb Z_p)^×\cong\Bbb Z_{p-1}$ for $p$ prime.

Once you find a primitive root $\bmod{29}$, you can use it to get your subgroups.

By Lagrange's theorem, the order of $2$ divides $28$.  You can rule out each of $1,2,4,7$ and $14$, and then you have that $2$ is primitive.

Now $2^4$ has order $7$, and $2^2$ has order $14$, by basic theory of cyclic groups.

Title: Subgroups of $\mathbb{Z}^*_{29}$


问:
Solving basic inequality
I have the following steps to a solution

$1.$ $\frac{10}{1 - x} ≥ 15 + \frac{-12x}{1 - x}$

$2.$ $10 ≥ 15 - 15x+ (-12)x + 27x ≥ 5$

$3.$ $x ≥ \frac 5{27}$

I am confused about the transition between step $1$ and $2$.

Wouldn't multiplying $(1-x)$ to both sides of the inequality get me

$10 ≥ 15(1 - x) + (-12x)$

$10 ≥ 15 - 27x$

What am I doing wrong instead of getting the result in step $2$?

答:
Let's start here:

$$\frac{10}{1 - x} ≥ 15 + \frac{-12x}{1 - x}$$

You are mostly right.  Multiplying by $1-x$ results in

$$10 \ge 15(1-x) -12x, \quad \text{ whenever } x<1,\quad(1)$$ and

$$10 \le 15(1-x) -12x, \quad \text{ whenever } x>1.\quad(2)$$

Multiplying an inequality by a positive number maintains the inequality, multiplying by a negative number switches the direction of the inequality.

Both parts of (2) cannot be true, so we are left with:

$$10 \ge 15-27x \texttt{ AND } x<1$$

$$\frac{5}{27} \le x <1.$$

Title: Solving basic inequality


问:
maximizing the result of a minimizing mixed integer program
I have a problem, where I need to construct an investment proposal $P$ from a variety of investment instruments, subject to some constraints. I am able to formulate the selection as a mixed-integer problem (MIP) and solve it using Python's mip package.

Then, from this problem, I am extracting the worst possible sub-selection portfolio $W$ subject to constraints, which can also be formulated and solved as a MIP.

I can do this on a data set, but what I would really like is to find the investment proposal, which would yield the best possible worst-case portfolio. This ends up something like a max-min problem, where both max and min steps can be expressed as MIPs.

More formally, if $\mathcal{P}$ denotes the set of all possible proposals, and $\Omega(P)$ denotes the set of all possible selections from the proposal $P \in \mathcal{P}$, and $V:\Omega(P) \to \mathbb{R}$ denotes the evaluation of the selection portfolio, we would like to
$$
\max_{P \in \mathcal{P}} \min_{s \in \Omega(P)} V(s).
$$

One super-dumb way to implement this is to run minimization, which solves 2 MIPs at each iteration, which is both extremely slow and not necessarily so well-defined, since at any step, either of the MIPs may be infeasible.

Is there a way to tackle this more intelligently? I am asking for both references and advice, if possible, please.

POSSIBLY RELATED QUESTIONS

A Mixed Integer Model with Mixed Integer sub-Problems

UPDATE

Given a set of investment instruments and some grouping criteria (by longevity or by riskiness, for example), I can select a proposal which would minimize the cost of the resulting combination with some distribution in each sub-group. Then from this proposal, I select a portfolio of smaller size and other more strict rules.

E.g. if my proposal has 10 apples (at \$4 each) and 20 bananas (at \$2 each), I may create a sub-selection that will be between \$20 and \$25 and will have maximum weight or best nutritional content.

答: It sounds like you are describing a bilevel optimization problem.  One approach is to use Benders decomposition.</p>


问:
Image in freely homotopic group and conjugacy
Let $f: \pi_1 (X, x_0) \rightarrow [S^1, X]$. Then $f([g]) = f([h])$ iff [g] and[h] are conjugate in $\pi_1(X, x_0)$. Could anyone give some suggestion on how I can go about proving this? I think what is holding me back is that I'm a bit lost as to what $f$ is exactly.

答:
$f$ is a function.

The input of $f$ is an element $[g] \in \pi_1(X,x_0)$.

To determine the output $f([g])$, recall that $g : [0,1] \to X$ is a path, i.e. a continuous function, with $g(0)=g(1)=x_0$. Also, $[g]$ is the path homotopy class of $g$, meaning the homotopy class relative to the endpoints. The output $f([g])$ is an element of $[S^1,X]$, which is the set of homotopy classes of continuous functions $S^1 \mapsto X$. So, using $g$, one must write a formula for a continuous function $G : S^1 \mapsto X$. This formula will do, once one checks that it is well defined when $t=0,1$:
$$G(e^{2 \pi i t}) = g(t)
$$
Once you've done that, $f([g]) = [G] \in [S^1,X]$ is the homotopy class of the function $G$. Now you have to prove that this is a well-defined formula, by showing that if $[g]=[h]$, namely if $g,h : [0,1] \to X$ have endpoints at $x_0$ and are path homotopic, then $[G]=[H]$, namely $G,H : S^1 \to X$ are homotopic.

Of course, you still have to prove what it asks you to prove, namely: given two closed paths $g,h : [0,1] \to X$, the equation $[G]=[H]$ holds if and only $[g]$ and $[h]$ are conjugate elements of the group $\pi_1(X,x_0)$, which by definition means that there exists $[k] \in \pi_1(X,x_0)$ such that $[k] [g] [k]^{-1} = [f]$, which translates to the statement that there exists a closed path $k : [0,1] \to X$ with endpoints at $x_0$ such that the concatenation $k * g * \bar k$ is path homotopic to $f$.


问:
Show $\|\cdot\|$ is a norm of the space of $C^1[0,1]$
Let $C^1([0,1])$ be the space of all functions having continuous derivatives. For each $f \in C^1([0,1])$, set

$$\|f\| = \left(\int_{0}^{1} (|f|^2 + |f'|^2) dx \right)^{1/2}$$

Show that $\|\cdot\|$ is a norm of the space of $C^1([0,1])$.

I mainly want to know to answer this question do I simply need to show the following properties: positivity, non-degenerate, homogeneity & the triangle inequality?

答:
Positivity,non-degenerate, homogeneity are obvious.We can prove the triangle inequality by using Minkowski's inequality.
The inequality is very popular,so please look into it yourself.


问:
Convert ODE to first-order form
I am struggling to understand reducing higher order DE to first-order form. Say I have an equation of motion,

$$m \frac{d^2}{dt^2}
\left(
\begin{array}{c}
x \\
y
\end{array}
\right) =
-\frac{GmM}{{(x^2+y^2)}^\frac{3}{2}}
\left(
\begin{array}{c}
x \\
y
\end{array}
\right),$$

how to reduce to the first-order form? I want to use Euler or Runge-Kutta on the result.

I think I understand that I should create some variable, $r$, which do something along the lines of
$$\mathbf{r} =
\left(
\begin{array}{c}
x \\
y
\end{array}
\right),
\frac{d}{dt}\left(\frac{d\mathbf{r}}{dt}\right) = f(x, x', t) =
\left(
\begin{array}{c}
f_x \\
f_y
\end{array}
\right),$$ and then
$$\mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{d}{dt}\left(\begin{array}{c}x\\y\end{array}\right)
=
\left(
\begin{array}{c}
v_x \\
v_y
\end{array}
\right)
,$$
and ultimately
$$
\frac{d}{dt}
\left(
\begin{array}{c}
x \\
y \\
v_x \\
v_y
\end{array}
\right)
=
\left(
\begin{array}{c}
v_x \\
v_y \\
f_x \\
f_y
\end{array}
\right)
$$

This is from some lectures I've found in various corners of the internet; I am truly lost. I realize I've not yet made any changes to the original equation, but I just don't know what to do! Can any of you please give me some pointers to where I'm going or what I'm doing? Much appreciated. It does feel like I am wandering about not knowing which direction I'm supposed to go.

答:
You have the first-order form as
$$
\frac{d}{dt}
\left(
\begin{array}{c}
x \\
y \\
v_x \\
v_y
\end{array}
\right)
=
\left(
\begin{array}{c}
v_x \\
v_y \\
f_x \\
f_y
\end{array}
\right).
$$

So you have $\frac{d}{dt}{v_x} = f_x$ and $\frac{d}{dt}{v_y} = f_y$, where $f_x$ and $f_y$ are given in terms of $x$ and $y$ with some constants like masses and the gravitational constant. Similarly, you have first-order differential equations $\frac{dx}{dt} = v_x$ and $\frac{dy}{dt} = v_y$. You should have the time interval and initial conditions $x(t_0)$ and $y(t_0)$, and $v_x(t_0)$ and $v_y(t_0)$.

These are now ready for numerical integrations. But first, try to understand what's going on.

Imagine you are starting on some initial time, for instance, $t_0 = 0$. At that point, you know positions and velocities (you have your initial conditions). Since you know the initial velocity, you can predict where your $x$ and $y$ will be at the next time step $t_1$, approximately (fairly accurate if your $dt$ is small enough). You then basically have your $x_1$ and $y_1$ (you can draw a line from $x_0$ to $x_1$, for instance.

With the same logic as above, you have your original equation of motion, where you can plug in your initial conditions and predict where your $v_x$ and $v_y$ will be at $t_1$ (since you have a "slope" equation for them).

With your new data points (positions and velocities) at the next $t_1$, you just iterate over and over and you will be able to graph trajectories for both positions and velocities.

Numerical integration methods (Euler or R-K) more ore less carry out the above calculations for you.

Title: Convert ODE to first-order form


问:
Find $N(\varepsilon)$ for the current sequence.
Let $x_1 = a > 0$. And $x_{n + 1} = 6 \frac{1 + x_n}{7 + x_n}$.

We know, that: $\lim_{n \to +\infty}(x_n) = x$ ($x_n$ has limit in $\mathbb{R}$).

We need to find $N(\varepsilon)$, such that:

$\forall \varepsilon > 0  \ \forall n > N(\varepsilon) \  |x_n - x| < \varepsilon$.

The main trouble I have is to compare $x_n$ and $\varepsilon$, because it's hard to get the formula, which depends from $n$.

答:
If you look at this question, I described to steps to follow for this kind of first-order rational difference equations.

Your problem is a bit simpler and you should arrive at
$$x_n=\frac{-27 (a-2) 4^n-8 (a+3) 9^n}{9 (a-2) 4^n-4 (a+3) 9^n}\tag 1$$ and a said in comments, the limit is either $-3$ or $+2$.

Since $a>0$, using $(1)$ and making $n \to \infty$, the limit is $x=2$.

Simplifying, assuming $a \neq 2$ this case would be more tha easy), we have
$$|x_n-x|=\frac 5 {\frac{a+3 }{a-2}\left(\frac{9}{4}\right)^{n-1}-1 }$$
Now, you look for
$$   \frac{ 5  N(\varepsilon)}{\frac{a+3 }{a-2}\left(\frac{9}{4}\right)^{n-1}-1 }< \varepsilon$$ which does not seem to be too difficult.


问:
Creating $n$ Poisson processes from an exponential variable and a categorical variable over $n$ objects
Suppose I take an exponential random variable $T\sim\mathrm{Exp}(\lambda)$ and create a process that "fires" every $T$ seconds. As far as I know, this is a Poisson process.

Since time windows are independent in Poisson processes, I should be able to uniformly remove samples with probability $p$ (i.e., skip "firing" but leave the time interval intact) and still have a Poisson process. On the other hand, the $p$ I removed is also a Poisson process.

This implies that I can more generally label each sample randomly with one of $n$ labels using a categorical variable over $n$ objects, thereby creating $n$ Poisson processes interleaved in time.

Further, if label $l$ has probability $p_l$, then the interval between occurrences might be $T_l\sim\mathrm{Exp}(p_l\lambda)$. (This is a hand-wavy guess based on the fact that the minimum over $n$ exponential variables gives you another exponential variable plus a categorical variable. The fallacy is assuming that all Poisson processes are exponential.)

Does this sound correct? In particular, I'd like to know if the $n$ processes can be considered independent.

Edit: It appears, at least empirically, that the respective time intervals in the $n$ resulting processes have exponential distributions, but I haven't figured out how to check for independence. I am convinced that they're Poisson processes; it's just a matter of determining if that necessarily means that the intervals are exponentially distributed.

Edit 2: For background, this is useful for real-time simulations because it allows a single thread to generate $n$ exponential timers. It could also be useful for increasing the numerical resolution of the tail of the distribution, e.g., with $\mathrm{Exp}(k\lambda)$ and $\mathrm{Bernoulli}(k^{-1})$ to increase the resolution by a factor of $k$.

答:
Correct.  You are describing the thinning of a homogeneous Poisson process.  The interarrival times for events in the same category are exponential with intensity $p_i \lambda$, that is to say the rate at which they occur is equal to the product of the rate of the main process and the probability of that event being tagged in that respective category.

It is not difficult to prove it mathematically because it is simply a generalization of the Bernoulli case in which the main process is split into two disjoint sub-processes, to the categorical case in which there may be any positive integer number of categories.  We may even use an inductive argument for this generalization; e.g., we thin first with probabilities $p_1$ and $1-p_1$; then $p_2/(1-p_1)$ versus $1 - p_2/(1-p_1)$; etc.

The intervals over which successive events of the same type occur will in general overlap with those intervals for other events; e.g., even with only two types of event, you could observe event times $$\{0.82227, 3.34239, 3.49035, 5.59936, 6.47796, 7.60404, 7.98742, 8.43683, 10.7558, 11.0469\}$$ and the corresponding event types might be $$\{A, B, B, A, B, A, B, B, B, A\},$$ in which case the first $A$-to-$A$ interval completely encloses the first $B$-to-$B$ interval.

To answer your question about the independence of the processes, they are independent in the sense that for any fixed interval of observation, the number of events of type $i$ occurring in that interval does not depend on the number of events of any other type occurring in the same interval.


问:
What is the origin of the quantifier symbols ∀ and ∃?
It's pretty natural how they stand for the phrases "For ∀ll" and "There ∃xists", but I was wondering who was the first to use these symbols, and if they had any related meaning before this use.</p>

答:
I haven't verified these references myself, but according to this site...

Existence (existential quantifier), $\exists$:

Peano used the symbol for
existence in volume II, number 1, of his Formulaire de mathematiques,
which was published in 1897 (Cajori vol. 2, page 300).

Kevin C. Klement writes, "While Peano had the backwards E for a
predicate of classes, Russell was the first to use the backwards E as
a variable binding operator, and there are the wonderful manuscripts
printed in CPBR vol 4 in which Russell's makes large dots out of
Peano's backwards epsilons to change over from the Peano-notation for
existence to a more Fregean one."

For all, $\forall$:

According to M. J. Cresswell and Irving H. Anellis, the letter A
appearing upside-down originated in Gerhard Gentzen, "Untersuchungen
ueber das logische Schliessen," Math. Z., 39, (1935), p, 178. In
footnote 4 on that page, Gentzen explains how he came to use the sign.
It is the "All-Zeichen," an analogy with the symbol for existence for
the existential quantifier which Gentzen says that he borrowed from
Russell.


问:
Integrate $\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\lambda(y)$
This is question from Axler's Measure, Integration, & Real Analysis Problem 5.B.1, probability measures.

Let $\lambda$ denote Lebesgue measure on $[0,1]$. Show that
$$\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\lambda(y) = \frac{\pi}{4}$$ and
$$\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(y) d\lambda(x) = -\frac{\pi}{4}.$$

How does one calculate this integral? Since the answer is $\dfrac{\pi}{4}$ is there is a trig substitution  involved?

I also graphed this function and don't understand what the meaning of the graph is from a geometric point of view.

答:
Note that $$\frac {x^2-y^2}{(x^2+y^2)^2} = \frac{\partial}{\partial y}  \left(\frac{y}{x^2 + y^2}\right) = -\frac{\partial}{\partial x}  \left(\frac{x}{x^2 + y^2}\right),  $$

Riemann and Lebesgue integrals coincide on finite intervals when the integrand is Riemann itegrable.

We can evaluate the inner integral as a Riemann integral using FTC to get for $y \in (0,1]$,

$$F(y) = \int_{[0,1]}\frac {x^2-y^2}{(x^2+y^2)^2}\, d\lambda(x)=  \int_0^1 \frac {x^2-y^2}{(x^2+y^2)^2} \, dx = \left.\frac{-x}{x^2 + y^2}\right|_0^1= \frac{-1}{1+y^2}$$

Now $F$ is integrable over $(0,1]$ and easily evaluated as

$$\int_{[0,1]} F(y) \, d\lambda(y) = \int_0^1\frac{-1}{1+y^2} \, dy = -\arctan(1) = - \frac{\pi}{4}$$


问:
Solutions to the Laplace Equation $\Delta u =0$, where $u= \log p$
Find all real solutions to the two dimensional Laplace equation $U_{xx} + U_{yy} =0$ of the form $u=\log p(x,y)$, where $p$ is a quadratic polynomial.

Solution:

Let $p(x,y) = Ax^2 + By^2 +Cxy + D$ be a quadratic polynomial such that $A, B \not= 0$. Then

$$U_{x} = \frac{2Ax + Cy}{\ln(10)(Ax^2 + By^2 +Cxy + D},$$

$$U_{xx} = \frac{2A ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2},$$

$$U_{y} =  \frac{2By + Cx}{\ln(10)(Ax^2 + By^2 +Cxy + D)},$$

$$U_{yy} = \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2}.$$

This implies

$$U_{xx} + U_{yy} = \frac{2A \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2Ax + Cy)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} + \frac{2B \ln(10)(Ax^2 + By^2 +Cxy + D) - \ln(10)(2By + Cx)^2}{\ln(10) (Ax^2 + By^2 +Cxy + D)^2} = 0. $$

I feel like I am not doing this right. Is there a simpler way? Thanks. And also how do I find such solutions.

答:
The solution that is proposed makes use of the wolfram language for symbolic computations.

In the following link you can download the notebook to explore all the details:

https://community.wolfram.com/groups/-/m/t/2108153


问:
reduce differential equations system of first order and using Euler
Given differential equations $$\ddot x=Gm_1\frac{y-x}{|y-x|^3}\hspace{2cm}\ddot y=Gm_2\frac{x-y}{|y-x|^3}$$ with constant $G,m_1,m_2$ I want to solve them with the Euler method. I know I have to reduce the equations to a system of first order. So I did for $\ddot x$

$$\frac{d}{dt}\begin{pmatrix}v_0\\v_1\end{pmatrix}=\begin{pmatrix}v_1\\Gm_1\frac{y-v_0}{|v_0-y|^3}\end{pmatrix}$$and the same for $\ddot y$. But now my problem is that the right side depends on the other equation and the other way round. How can you get a system of first order so you can apply Euler to it?

答:
I'm assuming, these are two-body gravitational interaction equations, not that it matters from math standpoint.

Anyhow, you've started well, just missed one point. You have to combine all four equations into one, even though system gets coupled.

So, before doing it, first write them down.
$$
\begin{align}
\dot x & = u \\
\dot u & = G m_1 \frac {y-x}{|y-x|^3} \\
\dot y & = v \\
\dot v & = G m_1 \frac {x-y}{|x-y|^3}
\end{align}
$$
So, now it's well formed (from Euler method standpoint) system of ODEs of general type
$$
\dot{\mathbf y} = \mathbf{f}(\mathbf y, t)
$$
where
$$
\mathbf y = [x,u,y,v]^T
$$
and
$$
\mathbf f = \left [
\begin{array}{c}
x \\ G m_1 \frac {y-x}{|y-x|^3} \\ y \\ G m_2 \frac {x-y}{|x-y|^3}
\end{array}
\right ]
$$
Now, you can proceed to the Euler method from this point.
$$
\mathbf y_{n+1} = \mathbf y_n + h \mathbf f(\mathbf y_n)
$$
or, explicitly
$$
\left [
\begin{array}{c}
x_{n+1} \\ u_{n+1} \\ y_{n+1} \\ v_{n+1}
\end{array}
\right ] = \left [
\begin{array}{c}
x_n \\ u_n \\ y_n \\ v_n
\end{array} \right ]+ h \cdot \left [
\begin{array}{c}
u_n \\ G m_1 \frac {y_n-x_n}{|y_n-x_n|^3} \\ v_n \\ G m_2 \frac {x_n-y_n}{|x_n-y_n|^3}
\end{array}
\right ]
$$
So, if you do it iteratively with sufficiently small $h$, you'll get $1^{st}$ order approximation of solution of the ODEs given.

PS

You need some initial values $[x_0, u_0, y_0, v_0]^T$ as a starting point, and those are usually know as physical initial conditions. They are of the form
$$
\begin{align}
x(0) &= something \\
u(0) &= \dot x(0) = something \\
y(0) &= something \\
v(0) &= \dot y(0) = something
\end{align}
$$


答:
Most probably, in step $2$, it is

$10+27x\geq15-15x+(-12)x+27x \implies27x\geq5$.

If these steps are mentioned in a book, then it's more likely that there's an error in the print. Otherwise, you did the whole thing right.

Also, the above inequality holds if $x<1$.

Title: Solving basic inequality

Body:
I have the following steps to a solution

$1.$ $\frac{10}{1 - x} ≥ 15 + \frac{-12x}{1 - x}$

$2.$ $10 ≥ 15 - 15x+ (-12)x + 27x ≥ 5$

$3.$ $x ≥ \frac 5{27}$

I am confused about the transition between step $1$ and $2$.

Wouldn't multiplying $(1-x)$ to both sides of the inequality get me

$10 ≥ 15(1 - x) + (-12x)$

$10 ≥ 15 - 27x$

What am I doing wrong instead of getting the result in step $2$?


问:
Find convergence in distribution limit of sequence of random variables
I've been stuck with this problem for a while now:

Let $X$ and $Y$ be random variables, and define $Z_n := \frac{1}{n}X + (1-\frac{1}{n})Y$ for $n \in \mathbb{N}$, find its convergence in distribution limit.

My intuition tells me $Z_n$ would converge in distribution to $Y$ as $n \rightarrow \infty$, but I can't seem to write a formal argument. I've tried proving it converges to $Y$ in probability, by showing $$\lim_{n\to\infty}P(|Z_n-Y|>\varepsilon)=0 \Leftrightarrow \lim_{n\to\infty}P\left(\frac{|X-Y|}{n}>\varepsilon \right)=0$$

What could be a way of showing this?

答: Hint: for any random variable $W$, what is $$\bigcap_{n =1}^\infty \{|W| > n \epsilon\}$$ ?


问:
$x$ as the shortest alternating sum of $1 \ldots n$
If I have an positive integer $x \in \mathbb{N}$ and I have $Z = \sum_{i = 0}^{n}{i}$ such that $Z \geq x$ and $Z - x \equiv 0 \bmod 2$ and $n$ is the smallest such integer it is possible to create and alternating sum from $1 \ldots n$ such that it equals $x$.

For example:

Say $x = 11$, then $Z = 15$ and $n = 5$ and the sum is $1 - 2 + 3 + 4 + 5$.

I see why the condition is necessary I don't see why it is sufficient.

答:
Let $S(n)=\{x:x=\pm1 \pm2 \cdots \pm n\}$.
  Let $T(n)=\{x:|x|\le \frac{n(n+1)}{2}\}\cap\{x:x\equiv \frac{n(n+1)}{2}\pmod{2}\}$.

Claim: $S(n)=T(n)$ for all $n\in \mathbb{N}$.

Proof: Induction on $n$.  $n=1,2$: clear

$S(n+1)\subseteq T(n+1)$: consider all $+$ or all $-$

$T(n+1)\subseteq S(n+1)$: $$[0,\frac{(n+1)(n+2)}{2}]\subseteq[-\frac{n(n+1)}{2}+n,\frac{n(n+1)}{2}+n]$$ and $$[-\frac{(n+1)(n+2)}{2},0]\subseteq[-\frac{n(n+1)}{2}-n,\frac{n(n+1)}{2}-n]$$

Now, let $ x\in T(n+1)$ and $x\equiv \frac{(n+1)(n+2)}{2}\pmod{2}$.  By the above pair of inclusions, either $x+n$ or $x-n$ is in $T(n-1)$.  But also $\frac{(n+1)(n+2)}{2}- n =\frac{n^2+3n+2- 2n}{2}\equiv \frac{n^2+n+2}{2}\equiv \frac{n(n+1)}{2} \pmod{2}$. [and $n\equiv -n\pmod{2}$].  Hence we're done by the inductive hypothesis.


问:
Is $f_{XY}(x,y)$ the same as $f(x,y)$?
I have a slight confusion with the notation, sometimes I find the joint density function like this:
$$f_{XY}(x,y)$$
and on other occasions like this:
$$f(x,y)$$
Is the same or what is the difference?
Thanks!

答: In the case of $f_{XY}(x,y)$, you are specifying that the random variables you are taking the joint density of are $X$ and $Y$.  If you just write $f(x,y)$, you had better be stating somewhere which random variables this joint density belongs to.

Title: Is $f_{XY}(x,y)$ the same as $f(x,y)$?


问:
on dedekind's cut
In dedekind's cut the right set considered as open or closed in occasions where the left set is always open. but why this is the only way? won't it be a dedekind's cut if I keep the left set open or closed in occasions and the right set always open?</p>

答:
It's a convention.

You can define your own "cut", say "KK cut", in such a way that the right set is always open, as you described.

Then there is an obvious bijection between the set of "Dedekind cuts" and the set of "KK cuts", and everything we do with Dekekind cuts can translate to KK cuts.

Title: on dedekind's cut


答:
You should avoid multiplying terms like $1-x$ unless you are sure about its sign. Instead, combine the fractions:

$$
0 \leqslant \frac{10}{1-x}-15 - \frac{-12x}{1-x} = \frac{10+12x}{1-x} + \frac{-15+15x}{1-x}=\frac{27x-5}{1-x} \\
\iff (27x-5)(x-1)\leqslant 0 \text{ and } x \neq 1.
$$

Therefore
$$
\frac{5}{27} \leqslant x < 1. \blacksquare
$$

Title: Solving basic inequality


问:
What are the "humpty" and "dumpty" points?
Apologies if this isn't the right place to answer this. What are the HM points and their isogonal conjugates, also known as the Humpty and Dumpty points? I found some information via a google search, but there wasn't much on line that was a comprehensive overview. I ask, now, for a short list of properties of these points.

I am asking this because I came across an old USAMO problem (2008/2) and a friend told me it was very easy by noticing that a point (F) was in fact that A-dumpty point. This made me curious, so I decided to ask for some resources here.

thanks for your time.

答:
Taken from Kapil Pause's note "On Two Special Points in a Triangle" (PDF link via wordpress.com)

Humpty point.

Definition 1. In $\triangle ABC$ the $A$-Humpty point $P_A$ is defined to be a point inside triangle such that $\angle P_ABC = \angle P_AAB$ and $\angle P_ACB = \angle P_AAC$.

Facts about $P_A$:

lies on $A$ median of $\triangle ABC$

lies on $A$ appolonius circle of $\triangle ABC$; that is, $\frac{AB}{AC} =\frac{P_AB}{P_AC}$.

$B$, $P_A$, $H$, $C$ are concyclic, where $H$ is the orthocentre of $\triangle ABC$.

$HP_A \perp AP_A$

Dumpty point.

Definition 2. In $\triangle ABC$, the $A$-Dumpty point $Q_A$ is defined to be a point inside [the] triangle such that $\angle Q_ABA = \angle Q_AAC$ and $\angle Q_AAB = \angle Q_ACA$.

Facts about $Q_A$:

lies on $A$ symmedian

It is the centre of spiral symilarity [sic] sending $\triangle A_QAC$ to $\triangle CQ_AB$; ie, sending $AC$ to $BA$.

$B$, $Q_A$, $O$, $C$ are concyclic where $O$ is circumcentre of $\triangle ABC$.

$OQ_A \perp AQ_A$

Also notice that the Humpty Dumpty points are isogonal conjugates of each other.

We recommend you to try to prove these facts by yourself. This will help you in befriending the humpty-dumpty points well

Back to your question, which is also addressed in Pause's note:

Question: (USAMO 2008). $ABC$ is a triangle. $M$, $N$ are respectively the midpoints of $AB$, $AC$. The perpendicular bisector of $AB$, $AC$ meet the $A$ median at $D$, $E$ respectively. $BD$ and $CE$ meet at $F$. Prove that $A$, $M$, $F$, $N$ are concyclic.

Proof: Ah, yes $F$ is none other than the $A$−Dumpty point of $ABC$. To reach
the desired conclusion, see that $OF\perp AF$, $ON \perp AC$, $OM \perp AB$ directly implies $A$, $M$, $O$, $N$, $F$ are concyclic with $AO$ as diameter. Another way to finish would have been to see that since a spiral similiarity centered at $F$ maps $AC$ to $BA$, it also maps the midpoint of $AB$ to midpoint of $AC$ that is, $M$ to $N$. The angle of spiral
similarity is $\angle AFC$ which is $180^\circ − A$, and this must also be the $\angle MFN$ and we finish.


问:
Any non-empty finite subset $\{x_1,x_2,...,x_k\}$ of $B$ has a g.l.b and l.u.b in $B$

Prove that any non-empty finite subset $\{x_1,x_2,...,x_k\}$ of $B$ has a g.l.b and l.u.b in $B$ where $(B,\leq)$ forms a lattice, i.e.

$(B, \leq)$ is a partially ordered set

Any two elements $x, y\in B$ have a g.l.b(greatest lower bound) $x \land y$ and an l.u.b(least upper bound) $x \lor y$

I've been thinking of induction, but I'm not sure that it'd work since we only have a partial order on $B$. Nevertheless,

Base case: Only one element, which is the g.l.b and l.u.b both

Induction hypothesis: Let's say the statement holds for sets of size $n-1$ and less

Consider a set of size $n$, namely $\{x_1, x_2,...,x_n\}$. $\{x_1, x_2,...,x_{n-1}\} \subset \{x_1, x_2,...,x_n\}$ has a g.l.b (say $x_g$) and an l.u.b (say $x_l$) in $\{x_1, x_2,...,x_{n-1}\}$. All that remains to be shown is that g.l.b($x_1,...,x_n$) = g.l.b($x_g,x_n$). Similarly for l.u.b.

I'm not sure how to proceed from here!

答:
We proceed by induction on $k$.

Base case: $k=1$. $|X| = 1$, and $x_1$ is its g.l.b and l.u.b both, since $x_1 \leq x_1$.

Induction Hypothesis: Let the statement hold for $|X| \leq k-1$.

Induction Step: Consider $X = \{x_1,x_2,...,x_k\}$ = $\{x_1,...,x_{k-1}\} \cup {x_k}$. $\{x_1,...,x_{k-1}\}$ has a g.l.b, say $x_g \in B$ (by the induction hypothesis). Moreover, $x_g$ and $x_k$ have a g.l.b, say $x'_g$. It remains to show that $x'_g$ is in fact the g.l.b of $X$.
We have that $x'_g \leq x_g,x_k$, and if $z\leq x_g,x_k$ then $z\leq x'_g$. In addition, $x_g \leq x_1,...,x_{k-1}$ and if $z \leq x_1,...,x_{k-1}$ then $z\leq x_g$. Combining the above two statements, $x'_g\leq x_1,...,x_k$ and if $z\leq x_1,...,x_k$ then $z\leq x'_g$.

Similarly for l.u.b!


问:
Criteria for being a true martingale
Could you kindly list here all the criteria you know which guarantee that a continuous local martingale is in fact a true martingale? Which of these are valid for a general local martingale (non necessarily continuous)? Possible references to the listed results would be appreciated.

答:
Here you are :

From Protter's book "Stochastic Integration and Differential Equations" Second Edition (page 73 and 74)

First :
Let $M$ be a local martingale. Then $M$ is a martingale with
$E(M_t^2) < \infty, \forall t > 0$, if and only if $E([M,M]_t) < \infty, \forall t > 0$. If $E([M,M]_t) < \infty$, then $E(M_t^2) = E([M,M]_t)$.

Second :

If $M$ is a local martingale and $E([M, M]_\infty) < \infty$, then $M$ is a
square integrable martingale (i.e. $sup_{t>0} E(M_t^2) = E(M_\infty^2) < \infty$). Moreover $E(M_t^2) = E([M, M]_t), \forall t \in [0,\infty]$.

Third :

From George Lowther's Fantastic Blog, for positive Local Martingales that are (shall I say) weak-unique solution of some SDEs.

Take a look at it by yourself :
http://almostsure.wordpress.com/category/stochastic-processes/

Fourth :

For a positive continuous local martingales $Y$ that can written as Doléans-Dade exponential of a (continuous)-local martingale $M$, if $E(e^{\frac{1}{2}[M,M]_\infty})<\infty$ is true (that's Novikov's condition over $M$), then  $Y$ is a uniformely integrable martingale.(I think there are some variants around the same theme)

I think I can remember I read a paper with another criteria but i don't have it with me right now. I 'll to try find it and give this last criteria when I find it.

Regards

Title: Criteria for being a true martingale


问:
Fluid dynamics - Form of fluid surface
An ideal fluid is rotating under gravity g with a constant angular velocity about the z-axis, which is vertical, so that in the fixed Cartesian axes u(x,y,z,t)=(-wy,wx,0). Find the form of the fluid surface.</p>

答:
Essentially, the surface of your fluid is orthogonal to the resultant force acting on the fluid. Since this is the sum of gravitation and the centrifugal force (when looked at from the rotating frame) it should not be too hard to compute the shape of the surface.

The centrifugal force is easily computed by taking the derivative of the speed $u$, namely

$$-a=-\frac{du}{dt}=-(-w\frac{dy}{dt},w\frac{dx}{dt},0)=-(-w^2x,-w^2y,0)=(w^2x,w^2y,0)$$

Add to that the effect of gravity $(0,0,-g)$ and you get that the gradient of the implicit equation of the surface ($f(x,y,z)=0$) should be propotional to

$$\nabla f(x,y,z) \sim (w^2x,w^2y,-g)$$

This leads to a set of partial differential equations that can be solved to give

$$f(x,y,z)=k\left(\frac{1}{2}w^2x^2 + \frac{1}{2}w^2y^2 - gz + c\right)$$

in which $k$ is some proportionality factor that does not really matter.

Thus, the equation of the surface is that of a paraboloid:

$$z= \frac{w^2}{2g}x^2 + \frac{w^2}{2g}y^2 + \kappa$$

Title: Fluid dynamics - Form of fluid surface


问:
Proving the Kochen-Stone lemma using the Paley-Zygmund inequality
I am trying to understand a proof to a lemma by Kochen and Stone which appears here, using the Paley-Zygmund inequality.

I'll repeat the proof in a detailed manner, and explain what bothers me about it.

Lemma (Kochen-Stone). $\ $ Let $A_n$ be a sequence of events with $\sum\mathbb{P}(A_n)=\infty$ and
\begin{equation*}
  \liminf_{k\to\infty}\frac{\sum_{1\le m,n \le k}\mathbb{P}(A_m\cap
  A_n)}{\left(\sum_{n=1}^k\mathbb{P}(A_n)\right)^2}<\infty
\end{equation*}
then, there is a positive probability that $A_n$ occur infinitely often.

Proof (partial). $\ $ Fix
$\ell<k$. Let $X=\sum_{n=\ell}^{k}1_{A_n}$; it follows that
\begin{equation*}
  \mathbb{E}(X)=\sum_{n=\ell}^{k}\mathbb(A_n)
\end{equation*}
and
\begin{equation*}
  \mathbb{E}(X^2)=\sum_{\ell\le m,n \le k}\mathbb{P}(A_n\cap A_m).
\end{equation*}

Using Paley-Zygmund inequality for $\theta=0$ (it's not mentioned in the wikipedia page, but the inequality holds for $\theta=0$ as well), we obtain
\begin{eqnarray*}
  \mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right)
  &=& \mathbb{P}(X>0)\\
  &\ge& \frac{\left(\sum_{n=\ell}^{k}\mathbb{P}(A_n)\right)^2}
           {\sum_{\ell\le m,n \le k}\mathbb{P}(A_n\cap A_m)}\\
  &\ge& \frac{\left(\sum_{n=1}^{k}\mathbb{P}(A_n)
            -\sum_{n=1}^{\ell-1}\mathbb{P}(A_n)\right)^2}
           {\sum_{1\le m,n \le k}\mathbb(A_n\cap A_m)
            -\sum_{1\le m,n < \ell}\mathbb(A_n\cap A_m)}
\end{eqnarray*}

Now, it holds that $\mathbb{P}(A_n\text{ occurs i.o.}) = \lim_{\ell\to\infty}\lim_{k\to\infty}\mathbb{P}\left(\bigcup_{n=\ell}^{k}A_n\right)$; however, I can't see how I can bound that probability away from 0. Am I missing some minor detail here?

答:
Here is a proof from Rick Durrett's Probability.

Let $X_n = \sum_{k\le n} 1_{A_k}$, $Y_n = \frac{X_n}{E(X_n)}$, the Kochen-Stone lemma means $\limsup_{n\to\infty} \frac{1}{E(Y_n^2)}<\infty$.
Here is an inequality we will use: $P(Y_n > a) \ge \frac{(1-a)^2}{E(Y_n^2)}$, here $a$ is a constant. This comes from the Paley-Zygmund inequality, we can easily prove it with Cauchy-Schwarz inequality. By Borel-Cantelli lemma, we have $P(A_k, i.o.) = 1$. Since $P(\cup_{m\ge n} Y_m) \ge \max_{m\ge n} P(Y_m)$, it follows that $P(\limsup Y_n)\ge \limsup P(Y_n)$. Thus

$1 = P(A_n,i.o.) \ge P(\limsup Y_n > a) \ge \limsup P(Y_n > a) \ge (1-a)^2 \frac{1}{E(Y_n^2)}$,

Which means $\limsup_{n\to\infty} \frac{1}{E(Y_n^2)} < \infty$.


问:
Expected Value Problem with a 10 Sided Die
I have a $10$ sided die numbered $1-10$. I keep rolling it until it lands on a prime number. Let $X$ be the number of times I roll the die.

What is the expected value $E(X)$?

What I have so far is the probability of rolling a prime is $P(\text{prime}) = \frac{4}{10}$.

How do I calculate the expected value? I tried the formula: $$\sum_{s\in S} P(s) \cdot X(s) = P(1)X(1) + P(2)X(2) +P(3)X(3)+P(4)X(4)$$

Where $P(s)$ is the probability of rolling a prime and $X(s)$ is the number of rolls. I plug in $\frac{4}{10}$ for each $P$ and $X$ is incremented from $1$. I am getting $4$ and the answer is $\frac{10}4$. Can someone guide me to what is going wrong here?

答:
With all the help above I found the solution.

The probability of a success (getting a prime number from the die) is $P(success) = \frac{4}{10}$. Each trial is independent and we have no set number of trails to get a success, leading to the conclusion that this is a geometric distribution. Therefore, the solution we get is:

$$E(X) = \frac{1}{P(success)} = \frac{1}{\frac{4}{10}} = \frac{10}{4}$$.


问:
For $f_n(x)=nx e^{-nx^2}$ can operations of integration and limit be interchanged?
For $f_n(x)=nx e^{-nx^2}$  can operations of integration and limit be interchanged?

$$
  \lim_{n \to \infty} f_n(x)\, dx = 0 \Rightarrow \\
  \int_0^1\lim_{n \to \infty} f_n(x)\, dx = 0 \\
  \\
  \int_0^1 f_n(x)\, dx = \left.-\frac{1}{2} e^{-nx^2} \right|_{0}^{1} = -\frac{1}{2}(e^{-n}-1) \Rightarrow \\
  \lim_{n \to \infty} \int_0^1 f_n(x)\, dx = \frac{1}{2} \\
  \lim_{n \to \infty} \int_0^1 f_n(x)\, dx \ne \int_0^1\lim_{n \to \infty} f_n(x)\, dx
$$
So, from calculating each side can conclude that the operations of integration and limit can not be interchanged.

But the following can lead to a different conclusion:
$$
\begin{align}
 \left| \int_0^1 f_n(x)\, dx - \int_0^1\lim_{n \to \infty} f_n(x)\, dx \right| &= \left| \int_0^1(f_n(x)- \lim_{n \to \infty} f_n(x))\,dx \right| \\
    &<= \int_0^1 \left |f_n(x)- \lim_{n \to \infty} f_n(x) \right | \,dx
\end{align}
$$
For any $x\in (0,1)$ $f_n(x)$ converge as $n\to \infty$.
$$
 \lim_{n\to\infty} f_n(x) = f
$$
So, for $\forall \epsilon>0\,$ $\exists N > 0$ for $n \ge N$

$$
   \left |f_n(x)- f  \right | < \epsilon \Rightarrow  \\
   \int_0^1 \left |f_n(x)- \lim_{n \to \infty} f_n(x) \right | \,dx < \epsilon
$$
Can get that
for $\forall \epsilon>0\,$ $\exists N > 0$ for $n \ge N$
$$
\left| \int_0^1 f_n(x)\, dx - \int_0^1\lim_{n \to \infty} f_n(x)\, dx \right| < \epsilon \Rightarrow \\
\lim_{n \to \infty} \int_0^1 f_n(x)\, dx = \int_0^1\lim_{n \to \infty} f_n(x)\, dx
$$
which contradicts with the conclusion get before.

I cannot get why two leads to different conclusions, please help.

答:
Let $f_n(x)=nxe^{-nx^2}$ and $f(x)=\lim_{n\to \infty }f_n(x)=0$.

So, it is true that for each $x\in [0,1]$, given any $\varepsilon>0$, there exists a number $N(x,\varepsilon)>0$ that can depend on both $x$ and $\varepsilon$, such that whenever $n>N(x,\varepsilon)$, $|f_n(x)-f(x)|<\varepsilon$.  We call this pointwise convergence of $f_n(x)$.

But note that we cannot say that for each $\varepsilon>0$, there exists a number $N(\varepsilon)$ such that whenever $n>N(\varepsilon)$ and $x\in [0,1]$, $|f_n(x)-f(x)|<\varepsilon$.  We call this type of convergence uniform convergence.

That $f_n(x)$ fails to converge uniformly can be seen by observing that

$$\sup_{x\in[0,1]}|f_n(x)|=\sup_{x\in[0,1]}|nxe^{-nx^2}|=\sqrt{\frac n2}e^{-1/2}$$

and $\lim_{n\to\infty}\sup_{x\in[0,1]}|f_n(x)|= \infty$.

Hence given any $\varepsilon>0$, we cannot find a number $N(\varepsilon)$ that is independent of $x$ such that $|f_n(x)-f(x)|<\varepsilon$.  And we cannot write, therefore, that for all $\varepsilon>0$ there exists a number $N(\varepsilon)$ such that whenever $n>N$, $\int_0^1 |f_n(x)-f(x)|\,dx<\varepsilon$.


答:
Check that this is true for $n \le 2$ (which implies $x=0,1,3$)

Then, proceed by induction : first do the $+n$ jump. Then you are left with jumps of length $1 \ldots n-1$ and you are trying to move by $|x-n|$. Since $|x-n| \equiv x-n \equiv 1+2+ \ldots +n-n = 1+2+ \ldots + (n-1) \pmod 2$ and $|x-n| \le \max (n, x-n) \le \max (n, 1+2+\ldots+(n-1)) \le 1+2+\ldots+(n-1)$, you can apply the inductive hypothesis (or use recursion, programmatically)

Title: $x$ as the shortest alternating sum of $1 \ldots n$

Body:
If I have an positive integer $x \in \mathbb{N}$ and I have $Z = \sum_{i = 0}^{n}{i}$ such that $Z \geq x$ and $Z - x \equiv 0 \bmod 2$ and $n$ is the smallest such integer it is possible to create and alternating sum from $1 \ldots n$ such that it equals $x$.

For example:

Say $x = 11$, then $Z = 15$ and $n = 5$ and the sum is $1 - 2 + 3 + 4 + 5$.

I see why the condition is necessary I don't see why it is sufficient.


问:
Stratonovich integral $\int_0^T W_t \circ dW_t$
I am trying to teach myself some stochastic calculus and am struggling to derive a Stratonovich integral. I am trying to understand how, considering the integral
\begin{equation}
\int_0^T W_t \circ dW_t
\end{equation}

we get the Riemann sum
\begin{align}
\Rightarrow& \sum_{j=0}^{n-1} W(\frac{t_{j+1} - t_j}{2}) ( W(t_{j+1}) - W(t_{j}))\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots\\
&=\frac{1}{2} (W^2(T) -W^2(0)) + \sum_{j=0}^{n-1}( W(t_{j+1}) - W(t_{j})).
\end{align}

Can anyone provide the identity used to get the last step above (simplifying the Riemann sum) or show how to get to the final step from the initial definition?

答:
Your expression for Stratanovich integral is wrong - the correct sum should be

$$\sum\frac{W(t_{j+1})+W(t_j)}{2}(W(t_{j+1})-W(t_j))=\frac12\sum W^2(t_{j+1})-W^2(t_j)=\frac12 W^2(T)-\frac12 W^2(0).$$


问:
Find this limit, When n to infinity
I don´t understand this problem, I need help with this

Find the next limit. Justify your procedure
$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} $$

I Tried do this, but I think it's wrong

$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} = \displaystyle \lim_{n \to \infty} {n + c}^n $$

After that, I tried use a stolz criterium $$\displaystyle \lim_{n \to \infty} \sqrt[b_{n}]{a_{n}} = \displaystyle \lim_{n \to \infty} \sqrt[b_{n+1}-b_{n}]{\dfrac{a_{n+1}}{a_{n}}}$$
where $$a_{n} = n+c , b_{n} =n , a_{n+1} = n+c+1 , b_{n+1} = n+1 , \displaystyle \lim_{n \to \infty} b_{n} = \displaystyle \lim_{n \to \infty} n = \infty$$
so $$\displaystyle \lim_{n \to \infty} \sqrt[n]{n+c} = \displaystyle \lim_{n \to \infty} \sqrt[n+1-n]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \sqrt[1]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \dfrac{n+c+1}{n+c} = \dfrac{\displaystyle \lim_{n \to \infty} n+c+1 }{\displaystyle \lim_{n \to \infty} n+c }$$

I got this but I don't know if this is right or not,

答:
Another way to do it.
$$a_n= \sqrt[n]{n + c}\implies\log(a_n)=\frac 1 n \log(n+c)=\frac 1 n\left( \log \left(1+\frac{c}{n}\right)+\log (n)\right)$$ Since $n$ is large

$$\log \left(1+\frac{c}{n}\right)\sim \frac{c}{n}\implies
\log(a_n)\sim \frac {\log(n)}n+\frac{c}{n^2}\to 0\implies a_n \to 1$$

Title: Find this limit, When n to infinity


问:
For every natural number $n$, prove that $n^n > 1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n-1)$.

For every natural number $n$, prove that $n^n > 1 \cdot 3 \cdot 5 \cdots (2n-1)$.

I tried solving this by assuming that the above equation is true for $n$ and tried proving it is also true for $n+1$, but did not reach anywhere.

答: Hint: Pair off factors starting from the beginning and the end of the second product, and use the identity $n^2>n^2-k^2=(n-k)(n+k)$.


问:
Find CDF of constant derived random variable
I have a problem where I'm asked to find the CDF, PDF, and expected value of a function $f_Y(y)$ where:

$$F_X(x) = \cases{
 0, & x < -1 \cr
 x/3 + 1/3, & -1 $\le$ x < 0 \cr
 x/3 + 2/3, & 0 $\le$ x < 1 \cr
 1, & 1 $\le$ x
}$$

and $Y = g(X)$ where:

$$g(X) = \cases{
 0,& X < 0\cr
 100,& X $\ge$ 0
}$$

I have found $f_X(x) = 1/3 + 1/3\cdot\delta(x), -1 \le x \le 1$, but I'm not sure how to combine that with a constant Y, where Y doesn't seem to be a valid PDF in the first place.

答:
Clearly $\mathsf P(Y{=}y)=\begin{cases}\mathsf P(X{<}0)&:& y=0\\\mathsf P(X{\geq}0)&:& y=100\\0&:& \textsf{elsewhere}\end{cases}$

Well, indeed, this will not have a proper probability density function, as its support consists of two massive points.

However, that did not prevent you from denoting a probability density function for $X$ which had one massive point at $x=0$ , denoted using the delta function times the point's probability mass of $1/3$.

So ... use two delta functions and the appropriate probability masses.


问:
Solve $2^{2x} + 9e^{-2x} = 6$ for x using substitution.
This is the equation I have:

$$2^{2x} + 9e^{-2x} = 6$$

I want to solve for x using the substitution method.

I've turned it into

$$4^x+\frac{9}{e^{2x}} - 6 = 0$$

But I do not know what to substitute and how to solve it.

答:
Hints:

$$2^{2x}=e^{2x\log2}\implies 2^{2x} + 9e^{-2x} = 6\implies e^{2x(1+\log2)}-6e^{2x}+9=0\implies\ldots$$

Edit: Perhaps, as suggested in the comments, there's a typo in the OP, which misled me to commit a typo myself and a mistake: if the equation were

$$e^{2x}+9e^{-2x}=6\implies\left(e^{2x}\right)^2-6e^{2x}+9=0$$

and all is nice and dandy with the quadratic $\,t^2-6t+9=(t-3)^2\,$ ... As it stands now it looks like a mean, evil exponential equation of the form $\,t^{1+\log 2}-6t+9=0\,$ ...


问:
Show that if $r,s \in E$, then $[r,s] \subset E$
Let $f:X \to C$ be measurable, $||f||_{\infty} > 0$, and let $\phi(p)= \int_X|f|^pd\mu$, and let
$$E=\{p \in (0,\infty): \phi(p) < \infty\}$$
I would like to show that if $r,s \in E$, then $[r,s] \subset E$. I am lost on what to prove for this. Thanks in advance!

答:
Let $F=\{x\in X:|f|\leq 1\}$. Let $r,s\in E$. We first prove the case when $r\geq 1$. Since $r,s\in E$, then
\begin{align*}
\phi(r)&=\int_{X}|f|^{r}\,d\mu<\infty
\\
\phi(s)&=\int_{X}|f|^{s}\,d\mu<\infty
\end{align*}
and assume also $1\leq r<s$ (since if $r=s$, the result is true trivially). Then for any $t\in[r,s]$,
\begin{align*}
\phi(t)&=\int_{X}|f|^{t}\,d\mu
\\
&=\int_{F}|f|^{t}\,d\mu+\int_{X\setminus F}|f|^{t}\,d\mu
\\
&\leq\int_{F}|f|^{r}\,d\mu+\int_{X\setminus F}|f|^{s}\,d\mu
\\
&\leq\int_{X}|f|^{r}\,d\mu+\int_{X}|f|^{s}\,d\mu
\\
&<\infty
\end{align*}
which shows that $[r,s]\in E$.

For the case when $r<1$ and $s>1$, you can use a similar approach by considering $t\in[r,1]$ and $t\in[1,s]$, using the above argument when $t\in[1,s]$.

Any comments to improve or shorten the proof is welcome.


答:
If
$n^n > 1 \cdot 3 \cdot 5 \cdots (2n-1)
=\prod_{k=1}^n (2k-1)
$
then
we want to show that
$(n+1)^{n+1}
\gt\prod_{k=1}^{n+1} (2k-1)
$.

For $n=2$,
this is
$2^2 > 1\cdot3$
which is true.

By the induction hypothesis,
$\prod_{k=1}^{n+1} (2k-1)
=(2n+1)\prod_{k=1}^{n} (2k-1)
\lt (2n+1)n^n
$

so we want
$(2n+1)n^n
\le (n+1)^{n+1}
$

or
$2n+1
\le (n+1)(1+1/n)^n
$

or
$(1+1/n)^n
\ge \dfrac{2n+1}{n+1}
=2-\dfrac1{n+1}
$.

By Bernoulli's inequality,
$(1+1/n)^n > 2$
and we are done.

Title: For every natural number $n$, prove that $n^n > 1 \cdot 3 \cdot 5 \cdot 7 \cdots (2n-1)$.

Body:
For every natural number $n$, prove that $n^n > 1 \cdot 3 \cdot 5 \cdots (2n-1)$.

I tried solving this by assuming that the above equation is true for $n$ and tried proving it is also true for $n+1$, but did not reach anywhere.


问:
Network Source Node
The following is a question from Ahujas book Network Flows:

Show that if we add a constant to the length of every arc emanating from the source node, the shortest path tree remains the same.

I do not know how to prove this. Any recommendations?

答: If the constant is $k$, the adjusted objective function is $$\sum_{i,j} (c_{i,j}+k[i=s])x_{i,j}= \sum_{i,j} c_{i,j} x_{i,j}+k\sum_j x_{s,j}= \sum_{i,j} c_{i,j} x_{i,j}+k,$$ which differs from the original objective function  by the constant $k$ and hence has the same minimizer.

Title: Network Source Node


问:
Factorize $x^{2n}+1$ to evaluate $\int\tan^{1/n}(x)dx$
I've been attempting to find a formula for $\int\tan(x)^{1/n}dx, n\in\Bbb N$.

I started out by performing the substitution $$u=\tan(x)^{1/n}$$

$$dx=\frac {nu^{n-1}}{1+u^{2n}}du$$ to transform the integral into $$n\int \frac{u^n}{1+u^{2n}}du$$
This is a relatively simple rational function, so it should be possible to solve via a partial fraction expansion. For the simple case where $n=2$, I was able to rewrite the integral as $$\frac{\sqrt2}{2}\left(\int\frac{u}{u^2-\sqrt2u+1}du - \int\frac{u}{u^2+\sqrt2u+1}du\right)$$

Which can then be solved by completing the square to yield $$\frac{\sqrt2}{4}\ln\left(\left(\sqrt{\tan{x}}-\frac{\sqrt2}{2}\right)^2+\frac{1}{2}\over\left(\sqrt{\tan{x}}+\frac{\sqrt2}{2}\right)^2+\frac{1}{2}\right)+\frac{\sqrt2}{2}\left(\arctan\left({\sqrt{2\tan{x}}-1}\right)+\arctan\left({\sqrt{2\tan{x}}+1}\right)\right)+C$$

However, I can't figure out how to factor the denominator for the partial fraction expansion of the general case. It will always be a product of $n$ irreducible quadratic factors, clearly, but I don't know how to locate them. I have managed to find the complex linear factors, using Euler's identity to find that they're of the form $$x\pm e^{im\pi\over{2n}}$$ where m is allowed to vary from 1 to $2n$. How can I combine these linear factors into real, irreducible quadratics for use in my partial fraction expansion? Alternatively, is there a simpler method of evaluating this integral? Any help would be vastly appreciated.

答:
We are going to the fact that the natural logarithm is a periodic function to find the $2n$ distinct roots of $x^{2n}+1$. Observe that  $$\forall z,w\in \mathbb{C},z\neq 0,\ln(z^w)\equiv w\ln(z)\pmod{2\pi i}$$
which is equivalent to
$$\forall z,w\in \mathbb{C},z\neq 0,\exists k\in \mathbb{Z},\ln(z^w)=w\ln(z)+2\pi ik$$
Then
\begin{align}
x^{2n}+1&=0\\
\Leftrightarrow x^{2n}&=-1\\
\end{align}
Since $x=0$ is not a root, it follows that
\begin{align}
x^{2n}&=-1\\
\Leftrightarrow \ln(x^{2n})&=\ln(-1)=\pi i\\
\Leftrightarrow \exists k\in \mathbb{Z},2n\ln(x)&=\pi i+2\pi i k\\
\Leftrightarrow \exists k\in \mathbb{Z},\ln(x)&=\frac{\pi i(1+2k)}{2n}\\
\Leftrightarrow \exists k\in \mathbb{Z},x&=\exp(\frac{\pi i(1+2k)}{2n}) \tag{1}
\end{align}
Since $\forall m\in\mathbb{Z},\exp(z)=\exp(z+2\pi im)$, we only need the first $2n$ integers in (1), that is $k\in\{1,2,\cdots,2n\}$. Hence, the $2n$ roots of $x^{2n}+1$ are given by
\begin{align}
r_1&=\exp(\frac{\pi i(1+2(1))}{2n})\\
r_2&=\exp(\frac{\pi i(1+2(2))}{2n})\\
&\vdots\\
r_{2n}&=\exp(\frac{\pi i(1+2(2n))}{2n})
\end{align}
An element $\gamma$ in a field K is a root of a polynomial with coefficients in K, iff $x-\gamma$ divides the polynomial, then
\begin{align}
x^{2n}+1&=\prod_{k=1}^{2n}x-r_k=\prod_{k=1}^{2n}x-\exp(\frac{\pi i(1+2k)}{2n})\\
\end{align}
Using Euler's formula $e^{i\theta}=\cos(\theta)+i\sin(\theta)$
\begin{align}
x^{2n}+1&=\prod_{k=1}^{2n}x-\Big(\cos(\frac{\pi (1+2k)}{2n})+i\sin(\frac{\pi (1+2k)}{2n})\Big)
\end{align}
You can finish the integration using partial fraction decomposition. A useful theorem will be the following:

Theorem 1.0 [1] Let $P(x)$ & $Q(x)$ be polynomials of real coefficients where $\deg(P)<\deg(Q)$ & $Q(x)=(x-r_1)(x-r_2)\cdots(x-r_n)$, then
$$\frac{P(x)}{Q(x)}=\sum_{i=1}^n\frac{P(r_i)}{Q'(r_i)}\frac{1}{x-r_i}$$
Theorem 1.0 can be used if $n<2n$, that is, if $0<n$. Let $0<n$, observe that
$$\int \tan^{\frac{1}{n}}(x)dx=\int \frac{nu^n}{u^{2n}+1}du$$
Let $P(u)=u^n$ & $Q(u)=u^{2n}+1=\prod_{k=1}^{2n}x-\big(\cos(\frac{\pi (1+2k)}{2n})+i\sin(\frac{\pi (1+2k)}{2n})\big)$, then $\deg(P)=n<2n=\deg(Q)$ & $Q'(u)=2nu^{2n-1}$, it follows by Theorem 1.0 that
\begin{align}
\int \frac{nu^n}{u^{2n}+1}du&=n\int \sum_{k=1}^{2n}\frac{P(r_k)}{Q'(r_k)}\frac{1}{u-r_k}du=n\int \sum_{k=1}^{2n}\frac{(r_k)^{n}}{2n(r_k)^{2n-1}}\frac{1}{u-r_k}du\\
&=\frac{1}{2}\int \sum_{k=1}^{2n}\frac{(r_k)^{n}}{(r_k)^{2n-1}}\frac{1}{u-r_k}du
=\frac{1}{2}\sum_{k=1}^{2n}\frac{(r_k)^{n}}{(r_k)^{2n-1}}\int\frac{1}{u-r_k}du\\
&=\frac{1}{2}\sum_{k=1}^{2n}\frac{(r_k)^{n}}{(r_k)^{2n-1}}\ln|u-r_k|+c
\end{align}
where $r_k$ is the kth root of $u^{2n}+1$. Substituting $r_k=\cos(\frac{\pi (1+2k)}{2n})+i\sin(\frac{\pi (1+2k)}{2n})$ and $u=\tan^{\frac{1}{n}}(x)$ we obtain
$$\frac{1}{2}\sum_{k=1}^{2n}\frac{\Big(\cos(\frac{\pi (1+2k)}{2n})+i\sin(\frac{\pi (1+2k)}{2n})\Big)^{n}}{\Big(\cos(\frac{\pi (1+2k)}{2n})+i\sin(\frac{\pi (1+2k)}{2n})\Big)^{2n-1}}\ln|\tan^{\frac{1}{n}}(x)-\cos(\frac{\pi (1+2k)}{2n})-i\sin(\frac{\pi (1+2k)}{2n})|+c$$

[1] https://en.wikipedia.org/wiki/Partial_fraction_decomposition#Procedure


问:
Finding integration of $\exp(2\overline z)$
Let $C$ denote the unit circle in the complex plane. Find the line integral of $\exp(2\overline z)$ over $C$. I tried to find this integral by using direct definition of contour integral but couldn't work out the whole thing. My attempt:

We denote $C$ in parametric form:$\exp(it)$ $0\leq t\leq 2\pi$

By the definition of contour integral

$\oint _C$ $\exp(2\overline z) dz$ $=$ $\int_{t=0}^{2\pi} \exp(2\exp(-it)) i\exp(it) dt$
=$\int_{t=0}^{2\pi} \exp(2(\cos t -i \sin t) i\exp(it) dt$
=$\int_{t=0}^{2\pi} i \exp(2(\cos t))  \exp i(t-2\sin t)dt$
=$\int_{t=0}^{2\pi} \exp(2(\cos t))(\cos (t-2\sin t)+i\sin (t-2\sin t) )dt$

How to proceed further?
(I am not pretty good at using latex . Apologies for messy writing)

答:
$$e^{2\overline{z}}= \sum_{n=0}^\infty \frac{(2\overline{z})^n}{n!}$$

$$J=\oint_C e^{2\overline{z}}\,dz = \oint_C \sum_{n=0}^\infty \frac{(2\overline{z})^n}{n!} \, dz =  \sum_{n=0}^\infty \frac{2^n}{n!} \oint_C{\overline{z}^n} \, dz$$

$$\oint_C \overline{z}^n \, dz = \int_0^{2\pi}e^{-i n \theta}\, i\,e^{i\theta}\,d\theta = i \int_0^{2\pi} e^{i(1-n)\theta} \, d\theta=  \left\{ \begin{aligned}2\pi i, & \quad n=1 \\0, & \quad \text{otherwise} \end{aligned} \right.$$

$$J= {2\cdot 2\pi i}=4\pi i.$$


答:
To elaborate on the idea of using binomial theorem, for $n\ge 2$:

$$(n+1)^{n+1}=\sum_{k=0}^{n+1}\binom {n+1}k1^kn^{n+1-k}\ge\binom {n+1}0n^{n+1}+\binom {n+1}1n^n=(2n+1)n^n$$


答:
Here's two ways to prove that
$$n^n\geq \frac{(2n)!}{2^n n!}$$

First way, by induction:

For $n=1$, $1^1=1$ so the posited inequality holds true. For $n=2$, $2^2>3$, which is also true. For an arbitrary natural number $n=m$ assume that the inequality holds true.
Then for $n=m+1$ it holds that
$$(m+1)^{m+1}=m\left(1+\frac{1}{m}\right)^{m+1}m^m>m\left(1+\frac{1}{m}\right)^{m+1}1\cdot3...(2m-1)$$
so it suffices to show that $m\left(1+\frac{1}{m}\right)^{m+1}>2m+1$. Indeed if we use the inequality $(1+x)^a\geq 1+ax$ on the quantity in parentheses, we obtain
$$m\left(1+\frac{1}{m}\right)^{m+1}>m\left(1+\frac{m+1}{m}\right)=2m+1$$
and we are done!

Second way:

Group each term starting from the end with a term starting from the beginning. By virtue of this we can write

$$\frac{(2n)!}{2^n n!}=\prod_{i=1}^{n/2}i(2n-i)~,~ n=2k$$
$$\frac{(2n)!}{2^n n!}=n\prod_{i=1}^{(n-1)/2}i(2n-i)~,~ n=2k+1$$
or more succinctly
$$\frac{(2n)!}{2^n n!}=n^{2\{n/2\}}\prod_{i=1}^{\lfloor n/2\rfloor}i(2n-i)$$
One can easily show however that
$$(x-n)^2\geq 0\iff x(2n-x)\leq n^2, x\in \mathbb{R}$$
Therefore by applying this formula to each term we get
$$\frac{(2n)!}{2^n n!}\leq n^{2\lfloor n/2 \rfloor+2\{n/2\}}=n^{2\frac{n}{2}}=n^n$$
and this completes the proof.


问:
Complex integration help
The integral given is

$$\int_{-\infty}^{\infty} \frac{\cos(x)-1}{x^2}\,dx $$

Ok, so, I've used the upper semi circular contour with the function

$$f(z) = \frac{e^{iz}-1}{z^2}$$

Now the residue I get is $i$ which when used with the residue theorem gives me the answer of $-2\pi$ as the answer to the integral of the $f(z)$ and through this I get my answer to the original integral as $-2\pi$. However in the answers to this question the answer states that the solution is $-\pi$. I've double checked with Wolfram Alpha and Wolfram Alpha also gives this as an answer. Can anyone please correct me where I'm going wrong? Thanks.

答:
DonAntonio gave a fine answer (+1) but let's finish too...

Let's start with :
\begin{align}
\int_{-\infty}^{\infty} \frac{\cos(x)-1}{x^2}\,dx&=2\int_0^{\infty} \frac{\cos(x)-1}{x^2}\,dx\\
&=\int_0^{\infty} \frac{e^{ix}+e^{-ix}-2}{x^2}\,dx\\
&=\lim_{\epsilon\to 0^+}\left[\int_{\epsilon}^{\infty} \frac{e^{ix}-1}{x^2}\,dx+\int_{\epsilon}^{\infty} \frac{e^{-ix}-1}{x^2}\,dx\right]\\
&=PV\int_{-\infty}^{\infty} \frac{e^{ix}-1}{x^2}\,dx\\
\end{align}
(we had to replace $0$ by $\epsilon>0$ because of the singular part $\frac 1x$ of the integral at $0$)

Now let's integrate $\displaystyle \frac{e^{iz}-1}{z^2}$ over this contour (with a small circle of radius $\epsilon$ excluding $0$) :

direct link on upper half-disk http://math.fullerton.edu/mathews/c2003/integralsindentedcontour/IntegralsIndentedContourModHome/Images/IntegralsIndentedContourModHome_gr_545.gif

The integral over $C_R$ will go to $0$ as $R\to\;+\infty$ (because $e^{iz}=e^{ix-y}$ with $y\ge 0$) and there is no residue at all in the contour so that we get (supposing $z=\epsilon\;e^{i\theta}$):
$$PV\int_{-\infty}^{\infty} \frac{e^{ix}-1}{x^2}\,dx+\lim_{\epsilon\to 0}\int_{\pi}^0 \frac{e^{i\epsilon e^{i\theta}}-1}{\epsilon^2\;e^{2i\theta}}\,\epsilon\;i\;e^{i\theta}\;d\theta=0$$

Use $e^{a\epsilon}=1+a\epsilon+O(\epsilon^2)$ to get $\pi$ in the second integral and conclude !

Title: Complex integration help


问:
How to check if a ray intersects a circle
Given a ray (Origin(x,y) and Direction(x,y)) and a circle (Center(x,y) and Radius) determine if the ray intersects the circle (touches one or two points in the circle).

I've followed this solution (both geometic and analityc) but I can't get it to work on my python code. Anyone can see my mistake?

The code is the following:

```<code>def dot(x1, y1, x2, y2):
    return (x1*x2) + (y1*y2)

def intersects(x, y, vx, vy, cx, cy, r):
# r is radius, x and y are the ray origin, vx and vy are the ray direction and cx and cy are the circle center

    dirx = vx-x
    diry = vy-y

    Lx = cx - x
    Ly = cy - y

    tca = dot(Lx,Ly,dirx,diry)

    if tca < 0:
        return False

    Lsq = dot(Lx,Ly,Lx,Ly)
    tcasq = tca*tca

    dsq = Lsq - tcasq

    rsq = r*r

    if dsq>rsq:
        return False

    thc = math.sqrt(rsq-dsq)

    t0 = tca - thc

    t1 = tca + thc

    if t0 < 0:
        t0 = t1
        if t0 < 0:
            return False

    return True

</code>```

答:
I would suggest using this more simple approach rather than the one you're using. You can do this by trying to analytically find the minimum distance between the points on the line and the centre of the circle.

Define the variable $\lambda$ by:

$$ \lambda = - \dfrac{(x-c_x)v_x + (y-c_y)v_y}{v_x^2 + v_y^2} $$

Now use lambda to calculate the variable below:

$$ r^* = \sqrt{(x + \lambda v_x - c_x)^2 + (y + \lambda v_y - c_y)^2} $$

$r^{*}$ is the minimum distance between the "ray" and the centre of the circle. Clearly if $r^* > r$, the line and the circle do not intersect.(You can derive the relation I provided for $\lambda$ by differentiating this equation and setting it to $0$.) Otherwise they intersect. If $r^* < r$ they intersect at two points. Since you're using Python the chances of $r^* = r$ is really low because of rounding errors, so the downside to this method is you likely won't be able to tell if the line is exactly tangent to the circle or not.


问:
Is cos(t+1) in this vector space?
The solution set of the differential equation
\begin{gather*}
\frac{d^2 y}{d t^2} = -y(t)
\end{gather*}
is equal to the span of two linearly independent solutions,
for example $\cos(t)$ and $\sin(t)$.

$\cos(t+1)$ satisfies the equation, so $\cos(t+1)$ is a linear combination of $\cos(t)$ and $\sin(t)$.

But is it? How can I write $\cos(t + 1)$ as a linear combination of $\cos(t)$ and $\sin(t)$?

答:
How can I write $\cos(t + 1)$ as a linear combination of $\cos(t)$ and $\sin(t)$?

By the angle sum formula, $\cos(t+1)=\cos(1)\cos(t)-\sin(1)\sin(t)$.

Title: Is cos(t+1) in this vector space?


问:
determining whether $H$ is a nonplanar graph

Determine, with justification, whether the graph $H$ below is nonplanar.

I think it's nonplanar as I can find a $K_{3,3}$ subdivision for it. I can describe the subdivision as follows. Join vertex $8$ to vertex $9$ (as they are connected by the path $812349$). Then add an edge from vertex $4$ to vertex $9$ (as they are connected by the path $439$). Add an edge from $4$ to $7$ as they're connected by the path $4397$. Now delete the edges $18, 12, 23, 34, 14, 39, 37, 79,29$ (since we added edges $49$ and $89$, there are $9$ edges remaining). Delete the vertices of the resulting graph with degree zero, which are $1, 2,3.$ The result is the $K_{3,3}$ graph with vertices $5,7,9$ on one side of the graph and vertices $8,6,4$ on the other, which shows that $H$ has a $K_{3,3}$ subdivision, and hence is nonplanar.

I think my reasoning may be incorrect as I obtained the subdivision incorrectly. If so, is there a way to obtain a correct $K_{3,3}$ subdivision? I know the $5$ vertices of degree $4$ are all joined by a path to each other, so that might be useful.

答: If you contract edges $81$, $12$, $65$, $43$, you will get $K_5$.


问:
Scheme representing locally continuous maps
Let $\Lambda$ be a discrete finite group. Set $X_\Lambda:= \bigsqcup_{a \in \Lambda} X_a$, i.e. $\Lambda$-copies of a scheme $X$.

Then it is claimed in page 46 that this scheme reprenets in the cat. of $Schemes_{/X}$ the functor
$$ Schemes_{/X} \rightarrow Set$$
$$U \mapsto \Lambda ^{\pi_0 U}$$,

where $\pi_0 Y$ is set of components of the underlying topological space of $Y$.  I would appreciate if someone explain how this works.

I have written my thoughts below which one may safely ignroe.

So I took the basic case when $X=Spec \Bbb Z$. I see that
$$ U \rightarrow \bigsqcup Spec \Bbb Z$$
has

i) as underlying of spaces  locally constant maps $U \rightarrow \Bbb Z$, which is equivalent to  $\pi_0U \rightarrow \Lambda$

ii) But as scheme, we have to take account of the structure sheaf - which seems to be complicated for me.

On the other hand we also have the alternate expression

$$Hom_{Sch}(U, \bigsqcup Spec \Bbb Z) \simeq Hom_{Rng}(\Bbb Z \times \cdots \times \Bbb Z, O_U(U))  $$

In both attempts, I am able too see the why $X_\Lambda$ represents the functor.

答:
In the absolute case you consider:

There is no problem with either of the two formulations. $\mathrm{Spec}\,\mathbb{Z}$ is a final object in the category of schemes, and so you don't need to argue too much about what the maps $U \rightarrow \mathbb{Z}$ are - there is just one such map. Once you decompose $U$ as a disjoint union of its components, you will see the bijection rather quickly (in a map $U' \rightarrow \coprod \mathrm{Spec}\,\mathbb{Z}$ where $U'$ is connected, one just needs to pick one of the copies of $\mathrm{Spec}\,\mathbb{Z}$ where the image lands, and the induced map needs to be the unique one).

In the formulation $\mathrm{Ring}(\mathbb{Z} \times \dots \times \mathbb{Z}, \mathcal{O}_U(U)),$ one can observe that any such ring hom. is determined by the images $e_\lambda$ of the "canonical vectors" $(0, 0, \dots, 1, 0, \dots, 0).$ As these canonical vectors multiply pairwise to $0$ and sum up altogehter to $1$, the relations that these $e_\lambda$'s need to satisfy are $e_\lambda e_{\lambda'}=0$ when $\lambda \neq \lambda', e_{\lambda}^2=e_{\lambda}$ and $\sum_\lambda e_\lambda=1$. That is, the homs correspond precisely to complete families $(e_\lambda)_\lambda$ of orthogonal idempotents in $\mathcal{O}_U(U)$ indexed by $\Lambda$. It is not hard to see from this description that these are in bijection with $\Lambda^{\pi_{0}(U)}$.

In the general case: The situation is quite similar due to the fact that the morphisms one considers are over $X$.

In more detail: the constant (group) scheme $\coprod_{\lambda \in \Lambda}X_{\lambda}$ is implicitly endowed with a structure map $\coprod_{\lambda \in \Lambda}X_{\lambda}\stackrel{\nabla}\rightarrow X$ given by the identity map $X_\lambda \rightarrow X$ on each component. The represented functor then takes a scheme $U$ over $X$, $U \stackrel{f}{\rightarrow}X$, to the set of all morphisms $U \stackrel{g}\rightarrow \coprod_{\lambda \in \Lambda}X_{\lambda}$ over $X$, i.e. such that $\nabla \circ g = f$.

Here is the full argument if you are interested. You can see that it uses only very little information about schemes per se (namely the underlying topology + the fact that connected components give decomposition as coproduct of schemes):

First suppose that $U$ is connected. Then, just by topology, $U$ is necessarily mapped into one of the copies $X_{\lambda_0}$ by such map $g$. But then, as the restriction of $\nabla$ to $X_{\lambda_0}$ is just the identity on $X$, it follows that the "corestricted" map $U \rightarrow X_{\lambda_0}$ is just $f$. Conversely, any choice of $\lambda_0$ gives a unique such $g$ given as $U \stackrel{f}\rightarrow X_{\lambda_0} \hookrightarrow \coprod_{\lambda} X_{\lambda}.$ Therefor in this case the functor gives what it should, i.e. $\Lambda^1$.

In general, consider the decomposition $U=\coprod_{i \in \pi_0(U)} U_i$ of $U$ into its connected components and let $f_i$ be the restriction of $f$ to $U_i$. Then it is easy to see that $U$ is the coproduct of such $U_i$'s even in the category of schemes over $X$ (where $U_i$ has the structure of an $X$-scheme given by $f_i$), and we have
$$\mathrm{Sch}_X(U, \coprod_{\lambda}X_\lambda)=\mathrm{Sch}_X(\coprod_{i \in \pi_0(U)} U_i, \coprod_{\lambda}X_\lambda)=\prod_{i \in \pi_0(U)}\mathrm{Sch}_X(U_i, \coprod_{\lambda}X_\lambda)=\prod_{i \in \pi_0(U)}\Lambda=\Lambda^{\times \pi_0(U)},$$
using the universal property of coproduct in the middle step. Thus, the claim is true in the general case as well.


问:
Shortest Path Equality
The following is a proposition found in Ahuja et al Network Flows Book:

Let the vector $d$ represent the shortest path distances. Then a directed path $P$ from the source node to node $j$ is a shortest path if and only if $d(j) = d(i) + c_{ij}$ for every arc $(i,j)$
Where $c_{ij}$ is the length of the arc from $i$ to $j$

Intuitively I understand how it works, however the proposition is left without proof. How should I approach it?

答: Write the shortest path problem as a linear program with flow variables $x_{i,j}$, construct the dual LP with dual variables $d(i)$, and apply complementary slackness.

Title: Shortest Path Equality


问:
A property of a prime divisor of a number consisting of 1s

For $n>0$ let $A(n) = \underbrace{111 \ldots 11}_{n}$. Prove that if $A(n)$ is divisible by a prime number $p>3$, then $\gcd(n, p-1) > 1$.

It is no huge discovery that if $n$ is even, then $2$ is a common divisor of $n$ and $p-1$, thus the implication holds. I don't know how to justify the general case though, so I would appreciate some hints.

Body:
If $10^n\equiv 1 \pmod p,(n,p-1)=1,(p,10)=1$,then there exist integer $x,y$, satisfies $xn+y(p-1)=1,$ hence
$$10^1=10^{xn+y(p-1)}=(10^n)^x\cdot(10^{p-1})^y\equiv1^x\cdot  1^y=1 \pmod p$$
We get $p=3.$


答:
Here is yet another solution  taking $\lim_{n\rightarrow\infty}\sqrt[n]{n}=1$ as a known fact.
For all $n> |c|$ we have that

$$ 1+|c|+c\leq n+c\leq 2n$$

Hence
$$ (1+|c|+ c)^{1/n}\leq (n+c)^{1/n}\leq 2^{1/n}n^{1/n}$$

the conclusion follows from the squeeze lemma and the fact that for any $a>0$, $\lim_{n\rightarrow\infty}\sqrt[n]{a}=1$.

About your attempt:

There is a well known result that says that for any sequence $\{a_n:n\in\mathbb{N}\}\subset\mathbb{C}$

$$
\liminf_{n\rightarrow\infty}\sqrt[n]{|a_n|}\leq\liminf_{n\rightarrow\infty}\frac{|a_{n+1}|}{|a_n|}\leq \limsup_{n\rightarrow\infty}\frac{|a_{n+1}|}{|a_n|}\leq \limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}
$$

That is closer to the argument you try to use.

Title: Find this limit, When n to infinity

Body:
I don´t understand this problem, I need help with this

Find the next limit. Justify your procedure
$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} $$

I Tried do this, but I think it's wrong

$$\displaystyle \lim_{n \to \infty} \sqrt[n]{n + c} = \displaystyle \lim_{n \to \infty} {n + c}^n $$

After that, I tried use a stolz criterium $$\displaystyle \lim_{n \to \infty} \sqrt[b_{n}]{a_{n}} = \displaystyle \lim_{n \to \infty} \sqrt[b_{n+1}-b_{n}]{\dfrac{a_{n+1}}{a_{n}}}$$
where $$a_{n} = n+c , b_{n} =n , a_{n+1} = n+c+1 , b_{n+1} = n+1 , \displaystyle \lim_{n \to \infty} b_{n} = \displaystyle \lim_{n \to \infty} n = \infty$$
so $$\displaystyle \lim_{n \to \infty} \sqrt[n]{n+c} = \displaystyle \lim_{n \to \infty} \sqrt[n+1-n]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \sqrt[1]{\dfrac{n+c+1}{n+c}} = \displaystyle \lim_{n \to \infty} \dfrac{n+c+1}{n+c} = \dfrac{\displaystyle \lim_{n \to \infty} n+c+1 }{\displaystyle \lim_{n \to \infty} n+c }$$

I got this but I don't know if this is right or not,


问:
Equation of line through two tangents of the circle
I have a circle with equation $(x-a)^2+(y-b)^2=r^2$ and a point $P=(x_1,y_1)$ outside the circle. I am asked to find the equation of the line connecting the tangents from $P$ to the circle (polar line).

I set up a system of equations, but it was too long and very tedious. Any other way? I think the answer is $(x_1-a)(x-a)+(y_1-b)(y-b)=r^2$.

答:
You are being asked to find yhe equation of polar of the circle about thr pole $P(h,k)$.
The circle is $f(x,y)=x^2+y^2-2ax-2by+a^2+b^2-r^2=0$ its ploar about $P(h,k)$ will be given by
$$hx+ky-a(x+h)-b(y+k)+a^2+b^2-r^2=0. f(h,k) \ne 0 $$

The polar becomes the tangent at $(h,k)$ if $f(h,k)=0$.


问:
Proof by induction for $n\in\mathbb N$
I need some help with the following proof by induction problem:

Prove that for all natural numbers $n\in\Bbb N$:

$$\dfrac{4^{n}}{n + 1} < \dfrac{(2n)!}{(n!)^{2}}$$

first I proof this first for $n=2$, and we got this
$\dfrac{4^{2}}{2 + 1} < \dfrac{(2 \times 2)!}{(2!)^{2}}$ and this meets inequality.

now for $n=k$  $$\dfrac{4^{k}}{k + 1} < \dfrac{(2k)!}{(k!)^{2}}$$ and if this is true so $$\dfrac{4^{k}}{k + 1} +(k+1) < \dfrac{(2k)!}{(k!)^{2}} +(k+1)$$

After that I don't know how follow this problem. Any help would be sincerely appreciated. Thanks!

答:
Let $P(n)$ be the given statement.
$$P(n):\ \frac{4n}{n+1}<\frac{2n!}{n!^2}$$
For the base case $n=2$,
$$P(1):\ \frac 8 3 < 6$$
which is true.

Assume $P(k)$ is true for some $\mathbb{N}\ni k\geq 2$.
$$\therefore \frac{4k}{k+1}<\frac{2k!}{k!^2}\quad (*)$$
Now, we wish to prove that $P(k+1)$ is also true.
Note that
$$\frac{2(k+1)!}{(k+1)!^2} = \frac{(2k+2)(2k+1)}{(k+1)^2}\cdot \frac{2k!}{k!^2}$$
By using $(*)$, we get
$$\frac{2(k+1)!}{(k+1)!^2}>\frac{2(2k+1)}{k+1}\cdot \frac{4k}{k+1}$$
$$\implies\frac{2(k+1)!}{(k+1)!^2}>\frac{16k^2+8k}{k^2+2k+1}\quad (**)$$
Now, note that $16k^2\geq 4$ and $8k>4k$ for all $\mathbb{N}\ni k\geq 2$.
$$16k^2+8k>4k+4\quad (1)$$
Similarly, we have
$$k^2+2k+1>4k+4\quad (2)$$
Using $(1)$ and $(2)$ in $(**)$, we get
$$\frac{2(k+1)!}{(k+1)!^2}>\frac{4(k+1)}{(k+1)+1}$$
Thus, we have established that $P(k+1)$ is true whenever $P(k)$ is true.

Hence, by the principle of mathematical induction, we conclude that $P(n)$ is true for all $\mathbb{N}\ni n\geq 2$.

Hope this helps :)

Title: Proof by induction for $n\in\mathbb N$


问:
Finding positive integer(s) $n$ for which $3^{n-1} + 5^{n-1} | 3^{n} + 5^{n}$: Is my solution correct?
The problem is the following

Find all positive integers $n$ such that $3^{n-1} + 5^{n-1}$ divides $3^{n} + 5^{n}$.

And I attempted as the following:

As  $3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n}$,

$$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 5(3^{n-1} + 5^{n-1}) = -2\cdot 3 ^ {n-1}$$
And
$$3^{n-1} + 5^{n-1}\ |\ 3^{n} + 5^{n} - 3(3^{n-1} + 5^{n-1}) = 2\cdot 5 ^ {n-1}$$
These expressions say that
$$2\cdot 3 ^ {n-1} = p(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \  p \in \mathbb{Z}^{+}$$
And
$$2\cdot 5 ^ {n-1} = q(3^{n-1} + 5^{n-1}) \quad \textrm{for some odd} \  q \in \mathbb{Z}^{+}$$
By addition, it becomes clear that $q = 2 - p$.

Hence, the last expression is equivalent to $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} + 5^{n-1})$.

Solving for $p$, it is equal to $\frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$.

Here, $p$ is odd only for $n = 1$, hence it is the only solution.

Is this attempt a solution for the problem? And for the effort, Thanks!

答:
What you did is mostly correct, but your argument that $p = \frac{2\cdot3^{n-1}}{3^{n-1} + 5^{n-1}}$ is only odd for $n = 1$ is not true since for all odd $n$, the result is either not an integer or it's an odd integer. Instead, $p$ is odd, as you stated, but with $2 - p \gt 0$ from $2\cdot 5 ^ {n-1} = (2 - p)(3^{n-1} + 5^{n-1})$ since the left side is positive, the right side must be as well, means only $p = 1$ works. Thus, $2\cdot 3 ^ {n-1} = p(3^{n-1} + 5^{n-1})$ gives $3^{n-1} = 5^{n-1} \implies n = 1$, so your solution is correct.

A somewhat easier way to see this is that, from what you wrote, $3^{n-1} + 5^{n-1} \mid -2\cdot 3^{n-1} \implies 3^{n-1} + 5^{n-1} \le \left|-2\cdot 3^{n-1}\right| = 2\cdot 3^{n-1}$. For $n \gt 1$, we have $5^{n-1} \gt 3^{n-1} \implies 3^{n-1} + 5^{n-1} \gt 2\cdot 3^{n-1}$, so only $n = 1$ is a possible solution. Since $n = 1$ gives $2 \mid 8$, this means only $n = 1$ works.


问:
Constant rank mappings and submersions
Just to be clear: the objects we're talking about here are smooth ($C^\infty$) manifolds (without a boundary) and submersion is defined as a map between manifolds, which has constant rank that is equal to the dimension of the codomain.

While trying to do an exercise I kept stumbling upon the idea that "submersion is locally a projection and therefore an open map". This idea is not the problem. The "problem" is the theorem that states the following:

Let $M$ and $N$ be smooth manifolds, $dim M=m$, $dim N=n$, and let $f:M\to N$ be a smooth mapping of a constant rank $r$. For every $p\in M$ there is a (smooth) chart $(U, \varphi)$ at $p$ and chart $(V, \psi)$ at $f(p)$, such that $f(U)\subset V$ and such that $f$ has coordinate representation $$\psi \circ f \circ \varphi ^{-1} (x_1,\ldots ,x_r,x_{r+1}, \ldots ,x_m)=(x_1,\ldots ,x_r,0,\ldots ,0)$$

Doesn't this mean that every constant rank mapping, not only submersions, is locally a projection? It seems to me that I've completely misunderstood the idea of  "locally being a projection", because that should be something that's very characteristic of submersions. Also, if every constant rank mapping is locally a projection, that would mean there are no constant rank mappings from compact manifolds to euclidean space.

So, my question is: what do people mean when they say "submersion is locally a projection"?

答:
This local form $(x_1,\ldots, x_r,x_{r+1},\ldots, x_m)\to (x_1,\ldots, x_r,0,\ldots,0)$ is written slightly misleadingly. If we examine a few special cases, we can see that these maps are not all "projections" in the sense which you intend. If $m\ge n=r$, we have that the map is of the form
$$ (x_1,\ldots, x_m)\mapsto(x_1,\ldots, x_r)$$
and is indeed locally a bona fide projection. If $m\ge n>r$, then the map looks like
$$ (x_1,\ldots, x_m)\mapsto (x_1,\ldots,x_r)\mapsto (x_1,\ldots, x_r,0,\ldots,0)$$
and hence a composition of a projection and then an inclusion. In the case where $r=m\le n$, this map becomes
$$ (x_1,\ldots, x_m)\mapsto(x_1,\ldots, x_m,0,\ldots, 0)$$
which is an inclusion. If $r<m\le n$, we get
$$ (x_1,\ldots, x_m)\mapsto (x_1,\ldots, x_m,0,\ldots,0)\mapsto (x_1,\ldots, x_r,0,\ldots, 0)$$
which is  composition of an inclusion and a projection.

The moral of the story is that there are a few varied behaviors. For an example, take the inclusion of $S^2\hookrightarrow \Bbb{R}^3$. This is a constant rank $2$ map, so the theorem tells us that locally it looks like $(x_1,x_2)\mapsto (x_1,x_2,0)$. I.e. locally it is a standard inclusion. It is an example of an inclusion of a compact manifold into Euclidean space, and does not contradict the theorem.

Title: Constant rank mappings and submersions