问:
To qualify for the Labor Day 400 auto race, each driver must complete two laps of the track at an average speed of 100 mi/h
I'm going through my old College Algebra book to shake off some of the dust, and I came across this example:

"To qualify for the Labor Day 400 auto race, each driver must complete two laps of the track at an average speed of 100 mi/h. Due to some problems at the start, Naomi averages only 50 mi/h on her first lap. How fast must she go on her second lap to qualify for the race?"

I first tried to solve it on my own, and I saw "average" so did this:

$(50 + s)/2 = 100$

$s=150$,

which is, as it turns out, incorrect. The book arrived at the conclusion that there is no speed she could hit that would guarantee her a qualification. I understand how they arrived at the conclusion, but I'll provide their process for reference:

$50 mi/h = \frac{L mi}{t_1 h},$ where $L$ is the length of the track in miles and $t_1$ is the time it took her to complete it in hours.

$(S)mi/h = \frac{L mi}{t_2 h},$ $S$ being the speed she needs to average at for the second lap and $L$ being the length again.

$100mi/h = \frac{2L mi}{(t_1 + t_2) h}.$

This comes out to $t_2=0$, so there is no way for her to qualify.

I understand why this works, using units and all, so I'm good there. My question is why my original method isn't sufficient? There's obviously some conceptual understanding that I'm missing, and I'm hoping to close that gap. What about this problem should tip me off to the unit method and not the formula for average? Any help would be appreciated!

答:
Your first "arithmetic mean" of speed is when the object moves at each speed for the same time $t$. Then the average speed is

$$\frac{50t + st}{2t} = \frac{50+s}{2}$$

But for the average speed when the object moves at each speed for the same distance $L$, the average would be a harmonic mean:

$$\frac{2L}{\dfrac L{50} + \dfrac L{s}} = \frac{2}{\dfrac 1{50} + \dfrac1{s}}$$

For your question, the two speeds happen each for a lap, so their harmonic mean should be used.


问:
Asymptotic behaviour of of $\sum_{n=0}^k a_n b_{k-n}$
Let $(a_n)_n$ and $(b_n)_n$ be two sequences of positive numbers such that:

-$a_n\sim a/n^{3/2}$ for some $a>0$;

-$b_n\sim b/n^{3/2}$ for some $b>0$;

Is it true that as $k\to \infty$,

$\sum_{n=0}^k a_n b_{k-n}\sim \frac{1}{k^{3/2}}(a\sum_{n=0}^{\infty}b_n+b\sum_{n=0}^{\infty}a_n) ?$

If not true, under what additional conditions the statement is true?

答:
It's true that:  $\ \displaystyle \sum_{k=0}^n a_k b_{n-k}\sim \frac{1}{n^{3/2}}(a\sum_{k=0}^{\infty}b_k+b\sum_{k=0}^{\infty}a_k) $

Here is a demonstration.

$\displaystyle \sum_{k=0}^na_kb_{n-k} =t_n+u_n+v_n+w_n \quad $ with:

$t_n=\displaystyle \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor}a_kb_{n-k}   \quad  , \quad u_n= \displaystyle \sum_{k=\lfloor \root{3}\of{n}\rfloor+1}^{\lfloor n/2\rfloor }a_kb_{n-k} \quad ,\quad v_n = \sum_{k=\lfloor n/2\rfloor +1}^{n-\lfloor \root{3}\of{n}\rfloor} a_kb_{n-k} \quad ,\quad w_n=\sum_{k=n-\lfloor \root{3}\of{n}\rfloor +1}^n a_kb_{n-k}$

We know that there exists B such that $\ \forall n\geqslant 1 \ , \ b_n\leqslant \dfrac{B}{n^{3/2}}   $. Then $ \ \displaystyle u_n\leqslant \sum_{k= \lfloor \root{3}\of{n}\rfloor+1}^{\lfloor n/2\rfloor }  a_k \dfrac{B}{(n-\lfloor n/2\rfloor)^{3/2}}   $.
$\displaystyle 0\leqslant u_n\leqslant 2^{3/2}B \sum_{k=\lfloor \root{3}\of{n}\rfloor+1}^{+\infty}a_k \dfrac{1}{n^{3/2}} \quad $. So $\ u_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$

By the same idea  $\ v_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$

$t_n-\displaystyle \dfrac{b}{n^{3/2}}\sum_{k=0}^{+\infty}a_k = x_n+y_n+z_n\quad   $ with

$x_n=\displaystyle \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} a_kb \left(\dfrac{1}{(n-k)^{3/2}}-\dfrac{1}{n^{3/2}} \right)  , \quad y_n= \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} a_k\left(b_{n-k}- \dfrac{b}{(n-k)^{3/2}}\right), \quad   z_n=-\dfrac{b}{n^{3/2}}\sum_{k=\lfloor \root{3}\of{n} \rfloor +1}^{+\infty} a_k $

It's clear that $\ z_n={\rm o}\left(\dfrac{1}{n^{3/2}}\right)$

$\displaystyle 0\leqslant x_n\leqslant \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} a_kb \left(\dfrac{1}{(n-\lfloor \root{3}\of{n}\rfloor)^{3/2}}-\dfrac{1}{n^{3/2}} \right) \leqslant \dfrac{1}{n^{3/2}} \sum_{k=0}^{+\infty}a_kb \left( \dfrac{1}{\left( 1- \frac{\lfloor \root{3}\of{n}\rfloor }{n}\right)^{3/2}}-1\right)\ $. So  $\ x_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$

$|y_n| \leqslant \displaystyle \sum_{k=0}^{\lfloor \root{3}\of{n}\rfloor} \dfrac{a_k}{(n-k)^{3/2}} \left| (n-k)^{3/2}b_{n-k}-b\right| \leqslant \dfrac{1}{(n-\root{3}\of{n})^{3/2}}\sum_{k=0}^{+\infty}a_k \sup_{p\geqslant n-\lfloor \root{3}\of{n}\rfloor} |p^{3/2}b_p-b|\ $ .
And  $\ y_n={\rm o}\left( \dfrac{1}{n^{3/2}}\right)$

So, we have  $\ t_n \sim \dfrac{1}{n^{3/2}} b\displaystyle\sum_{k=0}^{+\infty} a_k$

With the same ideas $\ w_n \sim \dfrac{1}{n^{3/2}} a\displaystyle\sum_{k=0}^{+\infty} b_k$

And we can conclude (we may add the equivalents because they are positive).

ContentLicense: CC BY-SA 4.0


问:
Set Notation (Axiom of Infinity)
I'm having trouble understanding the notation used in describing the axiom of infinity (which is number 6 in the Wolfram MathWorld page). I understand what the axiom is saying, but I'm trying to understand how the following sentence:

$$\exists  S \space [ \varnothing \in S \wedge (\forall x \in S) [x \cup \{x\} \in S]]$$

describes the existence of an infinity set. If someone could just break this apart for me and describe its pieces and how it all works that would be greatly appreciated.

答:
Let's begin by understanding the letters and their meanings, then we'll give context to everything.

$\exists$ is a quantifier. It means that the following symbol is a variable (a set, in the case of set theory) and we assert there is an object which the properties which we require that symbol to have.

$S$ is that symbol. It is a placeholder that will be used to refer to some object in the universe, and describe some properties of that object.

$\varnothing$ is the empty set.

$\in$ is the membership relation, so when we say $\varnothing\in S$ we say that the empty set is a member of $S$.

$\land$ is the symbol for conjunction, it means that we want both the part in the left of the symbol to be true, and the part on its right.

$(\forall x\in S)$ is a bounded quantification. $\forall$ is the quantifier for "for all", so it says that we want that all the members of $S$ will have a certain property.

$x\cup\{x\}$ is the union of the set $x$ with the set $\{x\}$. Remember that in set theory all the variables refer to sets.

So all in all what do we have? The axiom of infinity says the following thing:

There exists a set $S$, such that the empty set is a member of $S$, and whenever $x$ is a member of $S$, so is $x\cup\{x\}$.

Then what do we have? $\varnothing\in S$, and therefore $\varnothing\cup\{\varnothing\}=\{\varnothing\}$ is a member of $S$. Therefore $\{\varnothing\}\cup\{\{\varnothing\}\}$ is a member of $S$. Therefore $\{\varnothing,\{\varnothing\}\}\cup\{\{\varnothing,\{\varnothing\}\}\}$ is a member of $S$. And so on ad infinitum.

If we think about $\varnothing=0$, and $x\cup\{x\}$ as $x+1$ we have that $0\in S$, $1\in S$, $2\in S$, and so on. So $S$ corresponds to a set which contains the natural numbers, and so it is infinite.

Of course $S$ may include other objects, but we can conclude with the other axioms that there is some $S$ which includes only the natural numbers in the way we represent them with sets. This set is commonly known as $\omega$.

ContentLicense: CC BY-SA 3.0

Title: Set Notation (Axiom of Infinity)


问:
How to prove elegantly this inequality?
Let $a$, $b$ and $c$ three reals greater than or equal to 1. Put $a+b+c=3u$, $ab+bc+ca=3v^2$ and $abc=w^3$. Show that
$$w-\frac{1}{w}\ge\frac{w^3}{v^2}-\frac{1}{u}$$

I am able to prove it by assuming without loss of generality that $c\ge b\ge a$ and posing $b=a+u$ and $c=a+u+v$ for $u,v\ge0$. The problem here is that when plug  this two expression and develop, I get a huge polynomial of $a$ whose coefficients are all positive. I am wondering if there a possibility to obtain a more elegant solution.

答:
Write the question as
$$
\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \ge\!\!? \; \sqrt[3]{abc}  \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} - \frac{1}{\sqrt[3]{abc}} \frac{3 \sqrt[3]{abc}}{a+b+c}
$$
By AM-GM, both following coefficients are at most $1$:
$$
A =  \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} \le  1\qquad
B = \frac{3 \sqrt[3]{abc}}{a+b+c}\le 1$$
Again, rewrite the question as
$$
\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \ge\!\!? \; \frac{B+A}{2}\Big(\sqrt[3]{abc}  - \frac{1}{\sqrt[3]{abc}} \Big) - \frac{B-A}{2}\Big(\sqrt[3]{abc}  + \frac{1}{\sqrt[3]{abc}} \Big)
$$
So the inequality is immediately proved for the case that
$$
 B =   \frac{3 \sqrt[3]{abc}}{a+b+c} \ge \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} = A\\
\leftrightarrow ab + bc + ca \ge (a+b+c)\sqrt[3]{abc}
$$
Now this is not generally true. By symmetry, let w.l.o.g. $c \ge b \ge a$ and realize this by writing $b=a x$ and $c = a x^2 q^3$. Since, by the original question, $a,b,c \ge 1$, we have  $a \ge 1$ and $ x\ge 1$ and $ q^3 x\ge 1$ .  Then we get the equivalent
$$
1 +  x q^3 + x^2 q^3- (1 + x +  x^2 q^3)q \ge 0 \\
\leftrightarrow q^3(1-q)(x-1/q)(x-1/q^2) \ge 0
$$
which is true (since $ x\ge 1$) for
$q \le 1$ and either $x<1/q$ or $x >1/q^2$.

So we need to consider the remaining cases 1.) $q > 1$, and  2.) $q < 1$ and $1/q<x <1/q^2$.

Taking a closer look at case 2.) we can write the condition as $q^2< q^3 x < q$ but,  since also $q <1$ in this case, this never meets the required  $ q^3 x\ge 1$. So case 2.) needs not be considered.

We have case 1.) $q > 1$: Using our substitutions, the original inequality gets

$$
a^2x^2q^2\Big( 1 -  \frac{3xq^2}{1 +  x q^3 + x^2 q^3}  \Big) \ge\!\!? \; 1 -  \frac{3xq}{1 +  x + x^2 q^3}
$$
Again with $a \ge 1$ and $ x\ge 1$, we can prove the stronger inequality
$$
q^2 -  \frac{3xq^4}{1 +  x q^3 + x^2 q^3}   \ge\!\!? \; 1 -  \frac{3xq}{1 +  x + x^2 q^3}
$$
Consider first $q$ fixed and $x$ variable. If $x$ increases, the inequality gets less tight. Inspecting  the derivatives w.r.t. $x$, the difference between the LHS and the RHS increases for $x > q^{-3/2}$ and as we have $q>1$ and $x\ge1$, this is always the case. So we only need to consider the tightest case for $x=1$. This leaves us to show
$$
q^2\Big( 1 -  \frac{3q^2}{1 +  2 q^3 }  \Big)  - \Big( 1 -  \frac{3q}{2 +   q^3} \Big)\ge\!\!? \; 0
$$
or
$$
\frac{(q-1)^3(2q^5 + 3q^4 + q^3 + q^2 + 3q + 2)}{(1 +  2 q^3) (2 +   q^3)} \ge\!\!? \; 0
$$
which is true, since we consider $q>1$.
$\qquad \Box$

Title: How to prove elegantly this inequality?


问:
Are strong convergence of measure and almost sure convergence of a random variable related?
I am studying measure theory and probability I am wondering if and how these two concepts are related.

A sequence of random variables $X_n$ defined on some probability space $(\Omega,\mathscr{A},\mu)$ converges almost surely to $X$ if $\mu\left(\{\lim_{n\rightarrow\infty}X_n=X\}\right)=1.$

A sequence of measures $\mu_n$ converges strongly to $\mu$ if $\lim_{n\rightarrow\infty}\mu_n(A)=\mu(A)$ for all measurable $A$. So $\mu$ and $\mu_n$ are defined on the same measure space $(\Omega,\mathscr{A})$.

Now, is there anyway that these two types of convergence are related?

For example, if we have $X_1,\ldots X_n,\ldots$ are $iid$ standard Gaussian, then the strong law of large numbers says that $\mu(\{\lim_{n\rightarrow\infty}\bar{X}_n=0\})=1$, where I believe $\mu$ can be written as a measure on the space $(\mathbb{R}^\infty,\mathscr{B}(\mathbb{R}^\infty))$. The space on which $\mu$ is defined is one point of confusion.

The sample mean for fixed $n$ also has a Gaussian law with variance $1/n$. We can call this measure $\mu_n$, and it can be defined on the space $(\mathbb{R},\mathscr{B}(\mathbb{R}))$.
Is $\mu_n$ a valid measure on the space of the whole sequence $(\mathbb{R}^\infty,\mathscr{B}(\mathbb{R}^\infty))$, even though it only concerns the first $n$ random variables? Can I set this problem up in a way that says $\mu_n$ converges strongly to $\mu$ ? where $\mu$ is degenerate at 0?

答:
There was such a question: "Is there anyway that these two types of convergence are related?"

Suppose we have $X_n \to X$ almost surely.

Put $P_n (A) = P(X_n \in A)$  and $P(A) = P(X \in A)$ for all Borel set $A$. Then $P_n$, $n \ge 1$ and $P$ are probability measures on real line.

$X_n$ converges to $X$ almost surely, hence $X_n$ converges to $X$ in distribution, and a sequence of measures $P_n$ converges weakly  to the measure $P$. According to Portmanteau Theorem (see Billingsley, Convergence of probability measures, 1999, 2nd edition, pp.15-16) it follows that $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$. The next example shows that condition  $P_n(A) \to P(A)$ holds not for all Borel sets $A$.

Consider a probability space $(\Omega, \mathcal{F}, \mu)$ where $\Omega = [0,1]$, $\mathcal{F}$ is sigma-algebra of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$. Put $\xi_n(\omega) = n \omega $ for $\omega \in [0, \frac{1}n ]$ and $\xi_n(\omega) = 0$ otherwise. In this case $P_n \sim U[0, \frac{1}{n}]$, $\xi_n \to X \equiv 0$, $P$ is Dirac measure at zero.
For $A = [ - 1, 0] $ we have $P_n(A) = 0 \nrightarrow P(A) = 1$.

As a result we get the next statement. If $X_n$ converges to $X$ almost surely then for corresponding measures $P_n$ and $P$ we have $P_n(A) \to P(A)$ for all Borel sets $A$ such that $P( \partial A) = 0$ but not for all $A$.

Now suppose we have convergence $P_n(A) \to P(A)$ for all $A$ from some sigma-algebra. If these measures are not probability measures of if these measures are not measures on $\mathbb{R}$ with Borel sigma-algebra, then there is no direct relationship between them and any random variables. Suppose that $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel sets $A$. Hence $P_n$ converges to $P$ weakly (Portmanteau Theorem). According to the Skorokhod's representation theorem we can construct random variables $X_n$, $X$ such that $X_n$ has distribution $P_n$, $X$ has distribution $P$, $X_n \to X$ almost surely and moreover $X_n$ and $X$ are defined of probability space $(\Omega, \mathcal{F}, \mu)$, where $\Omega = [0,1]$, $\mathcal{F}$ consists of Borel sets of $\Omega$ and $\mu$ is a standard Lebesgue measure, that is $\mu([a,b]) = b-a$.

As a result we get the next statement. If $P_n(A)$, $P(A)$ are probability measures on $\mathbb{R}$ with Borel sigma-algebra and $P_n(A) \to P(A)$ for all Borel set $A$ then there are random variables $X_n, X$ on a "simple" probability space such that $X_n$ has distribution $P_n$, $X$ has distribution $P$ and $X_n \to X$ almost surely.


问:
Show y is odd in the equation $y^3 =x^2 +2$
Found a question which asks you to find all integer solutions $y^3 =x^2 +2$ .    In order to do so, it first asks you to show that y is odd.  How would one go about showing this?

Many Thanks

答: Suppose $\,y\,$ is even, then also $\,x\,$ must be even, but then $\;x^2+2 \equiv 2\pmod 4\;$ , and since $\,y^3 \not \equiv 2 \pmod 4\;\;\forall\,y\in\Bbb Z\;$ we're done. </p>


问:
Fourier series of a piecewise continuous linear $2π-periodic$
Problem: Show that the Fourier series of a piecewise continuous linear $2π-periodic$ function converges to this function uniformly. Here the function $f (x)$ continues on the segment [a, b] is called piecewise linear, if there is a partition of the segment [a, b] in a finite number of segments $[x_{i-1}, x_i]$,
$$a = x_0 < x_1 < · · · < x_i < · · · < x_{n−1} < x_n = b$$

such that the function $f (x)$ is linear over each of the segments $[x_{i-1}, x_i], i = 1, 2,. . . , n$

I don't understand how to test it.
Can you help me?

答:
I'll firstly remind us of some common theorem.

Lemma
 Let $f$ be a $2\pi$-periodic measurable function on $\mathbb{R}$ and $f \in L^1[ -\pi,\pi]$.

Assume that $$\sum_{n \in \mathbb{Z}} |\hat{f}(n)| <+\infty$$
then there is a continuous function $g$ such that $$ \sum_{|n| \le N} \hat{f}(n)e^{inx} \xrightarrow[N \rightarrow +\infty]{\text{uniformly}} g $$
and $g = f \quad \text{a.e}$

$\square$

In light of the above theorem, if we can prove that :$$\sum_{n \in \mathbb{Z}} |\hat{f}(n)| <+\infty$$
Our work is done.

Solution

Let $g$ be the piecewise constant function which is deduced by the derivation of $f$, that is $f'(x)=g(x)$ for all $x$ except the points at which $f$ changes slope.

We have:

$$2\hat{f}(n) \underbrace{=}_{\text{periordicity}}\frac{1}{\sqrt{2\pi}}\int_{[-\pi,\pi]}\left[ f(x)-f\left(x-\frac{\pi}{n}\right) \right] e^{-inx}dx \\=\frac{1}{\sqrt{2\pi}}\int_{[-\pi,\pi]} \int_{-\pi/n}^{0} g(x+s)ds e^{-inx}dx\\=\frac{1}{\sqrt{2\pi}}\int_{-\pi/n}^{0} \int_{[-\pi,\pi]}  g(x+s)dx e^{-inx}ds\\\underbrace{=}_{\text{periordicity}}\frac{1}{\sqrt{2\pi}}\int_{-\pi/n}^{0} \int_{[-\pi,\pi]}  g(x)dx e^{-inx+ins}ds\\ = \frac{-2i}{n}\frac{1}{\sqrt{2\pi}}\int_{[-\pi,\pi]}  g(x)e^{-inx}dx $$
Repeat what we've done  at the very first equality again, we have:
$$4 \hat{f}(n) =\frac{-2i}{n}\frac{1}{\sqrt{2\pi}}\int_{[-\pi,\pi]} \left[ g(x) -g(x-\frac{\pi}{n})\right]e^{-inx}dx $$
Thus,
$$ |\hat{f}(n)| \le \frac{1}{2n} \frac{1}{\sqrt{2\pi}} \int_{[-\pi,\pi]} \underbrace{\left |g(x)-g(x-\frac{\pi}{n})\right|}_{=:h_n(x)}dx$$

Let's remind us that $g$ is piecewise constant, thus if $x$ and $x-\frac{\pi}{n}$ lie in the same constant piece, $h_n(x)=0$.

More precisely, let:

$S:= \{ a \in [-\pi,\pi] : g \text{ changes value at } a\}$

$S_n:= S+[0,\pi/n] $

If $x \not\in S_n$ ($x \in [-\pi,\pi]$) then $g_n(x)=0$ and $|S_n| \le \frac{\#S\pi}{n}$.

So let $C:= \max |g|$, from our previous result, we imply that:
$$|\hat{f}(n)| \le \frac{1}{2n}\frac{1}{\sqrt{2\pi}} \frac{\#S\pi}{n} 2C
$$
Thus $$|\hat{f}(n)| \le \frac{D}{n^2}$$.

for all $n$, for some constant $D>0$

Hence forth the conclusion.

$\square$


Body:
Let $D_{1}$ and $D_{2}$ be distances in $X \subseteq \mathbb{R^n}$ such that $D_{1}(x,y)= \sum_{r=1}^{k}|x_{r} - y_{r}|$ and $D_{2}(x,y)= \sqrt{\sum_{r=1}^{k}|x_{r} - y_{r}|^{2}}$. Prove that $\frac{1}{k}D_{1}(x,y) \leq D_{2}(x,y) \leq k D_{1}(x,y)$.

What I carry is the following

We know by definition that:

$D_{1}(x,y)= \sum_{r=1}^{k}|x_{r} - y_{r}|=|x_{1}-y_{1}|+...+|x_{k}-y_{k}|$

$D_{2}(x,y)= \sqrt{\sum_{r=1}^{k}|x_{r} - y_{r}|^{2}}= \sqrt{|x_{1}-y_{1}|^{2}+...+|x_{k}-y_{k}|^{2}}$

Then $D_{2} \leq D_{1}$ but I don't know how to proceed

Body:
1st Part
$$kD_{2}(x,y)\geq k*max{|x_{r}-y_{r}|}\geq \sum_{r=1}^{k} |x_{r}-y_{r}|$$
2nd Part
$$\frac{1}{k}D_{2}(x,y)\leq\frac{1}{k}*k*max{|x_{r}-y_{r}|}=max{|x_{r}-y_{r}|}\leq \sum_{r=1}^{k} |x_{r}-y_{r}|$$

Equality if $r=1$.


问:
Partial order with infimum and supremum
could you help me with this please

Let's define the relationship $\preceq$ in $\mathbb{N}^\mathbb{N}$ as $f\preceq g$ if $f(n)\leq g(n)$ for every $n\in \mathbb{N}$ Prove that it is a partial order and also show that any finite set in this order has supremum and infimum.

I have already managed to prove that it is a partial order, but the part of the supremum and infimun has cost me, what I have tried is to proceed by induction on the cardinality of the finite set, but I cannot see how to conclude with respect to step n + 1

答:
I don't think induction is really necessary here.

Notice that $h$ is infimum of $S\subset\mathbb N^{\mathbb N}$ if for each $n\in\mathbb N$, $h(n)=\inf S_n$ where $S_n=\{f(n)|f\in S\}$.

Any finite set in $\mathbb N$ has an infimum, which we call $\min$. Using this I think it is easy to construct the infimum sequence. Supremum follows similarly.

Title: Partial order with infimum and supremum


问:
Showing $E(X^n) = n!a$ using induction.
Let $X$ be a random variable with $a = E(X)$, and suppose $X$ has the
same distribution as $U(Y + Z)$, where

$U$ is uniformly distributed over $(0, 1)$, $U, Y, Z$ are independent and $X, Y, Z$ have the same distribution.
How can we prove using induction that $E(X^n) = n!a^n$ for $n = 1, 2,...$.

Any hint would be appreciated.

Title: Showing $E(X^n) = n!a$ using induction.

Body:
Use Strong induction (that means, suppose it holds for every $k<n+1$). Notice that
$$E(U^{n+1}(Y+Z)^{n+1})=E(U^{n+1})E\left(\sum _{k=0}^{n+1}\binom{n+1}{k}Y^kZ^{n+1-k}\right )=\frac{1}{n+2}\sum _{k=0}^{n+1}\binom{n+1}{k}E(Y^k)E(Z^{n-k+1})=\frac{1}{n+2}\left (E(Z^{n+1})+E(Y^{n+1})+\underbrace{\sum_{k=1}^{n}\binom{n+1}{k}E(Y^k)E(Z^{n-k+1})}_{\text{induction here}}\right )$$
$$E(X^{n+1})=\frac{1}{n+2}\left (2E(X^{n+1})\sum_{k=1}^{n}\binom{n+1}{k}k!a^k\cdot (n-k+1)!a^{n-k+1}\right )=\frac{2E(X^{n+1})+a^{n+1}(n+1)!\cdot n}{n+2},$$
solving for $E(X^{n+1})$ you get $E(X^{n+1})=(n+1)!a^{n+1}.$


问:
Is $E_n:=\left\{f\in L_1([0,1],m):\int_{[0,1]} |f|^3 dm\leq n \right\}$ absorbing?
I have to solve an exercise, which I am a little confused about, if I have done correctly.

Let $L_1([0,1],m)$ and $L_3([0,1],m)$ be the Lebesgue space on [0,1]. Recall that $L_3([0,1],m)\subsetneq L_1([0,1],m) $. For $n\geq 1$, define:

$E_n:=\left\{f\in L_1([0,1],m):\int_{[0,1]} |f|^3 dm\leq n \right\}$

Given $n\geq 1$, is the set $E_n\subset L_1([0,1],m) absorbing justify?

To find out if $E_n$ I absorbing, I've started by showing that it is convex, and then I want to show that for all $0\neq f\in L_1([0,1],m)$, there exists $t>0$ such that $t^{-1}f\in E_n$. I've done this by trying to show $\int_{[0,1]}|t^{-1}f|^3 dm\leq n$, and I have done that by:

$\left(\int_{[0,1]}|t^{-1}f|^3 dm\right)^{\frac{1}{3}}=t^{-1} \left(\int_{[0,1]}|f|^3 dm\right)^{\frac{1}{3}}\leq t^{-1}n\leq  n$

And then we have that $E_n$ is absorbing.

But I am not sure that it can be done this way

Body:
Take $f(t)=t^{-1/3}$. Then $f\in L^1[0,1]$, since
$$
\|f\|_1=\int_0^1x^{-1/3}\,dx=\frac32.
$$
For any $t>0$,
$$
\int_0^1 |tf|^3\,dm=t^3 \int_0^1\frac 1x\,dx=\infty.
$$
So $E_n$ is not absorbing.


问:
Reducing $2018^{2018}\pmod6$ via $\mod2$ and $\mod3$
I've been tasked to reduce $$2018^{2018}\pmod6$$using the results of reducing$$2018^{2018}\pmod2$$and $$2018^{2018}\pmod3$$

I have deduced that $$2018^{2018}\equiv0\pmod2$$and $$2018^{2018}\equiv1\pmod3$$but I don't see a way to combine these two congruences into one $\mod6$ congruence. Could someone tell me how this is achieved? I have a feeling the answer is really simple, but I just don't see it. Is there also a more generalized strategy of solving modular congruences via the divisors of the modulo number($2$ and $3$ are divisors of $6$)?

The way I have currently solved this problem is to first reduce $$2018^{2018}\pmod6$$ to $$2^{2018}\pmod6$$Then we look for patterns:$$\begin{align}2^1&\equiv2\pmod6\\2^2&\equiv4\pmod6\\2^3&\equiv2\pmod6\\2^4&\equiv4\pmod6\\&\ \ .\\&\ \ .\\&\ \ .\\2^n&\equiv\begin{cases}4&\text{if }n\equiv0\pmod2\\2&\text{if }n\equiv1\pmod2\end{cases}\pmod6\end{align}$$for $n\in\Bbb N$. In the case of $2018^{2018}$, $2018\equiv0\pmod2$, so $$2018^{2018}\equiv2^{2018}\equiv4\pmod6$$

答:
Take $x=2018^{2018}$. The Chinese Remainder theorem tells us that given that 2 and 3 are coprime. Then the system $$\begin{cases}
x = 0\ (mod\ 2)\\
x = 1\ (mod\ 3)\\
\end{cases}$$
has an unique solution $mod(2*3) = mod(6)$. Note that you can easily check that 4 is the only number that is congruent to $0\ (mod\ 2)$ and $1\ (mod\ 3)$ (between 0 and 5). Therefore, that is the unique solution.


答:
The Chinese remainder theorem states that, for these congruence equations:

$$
x\equiv 0 \pmod 2\\
x\equiv1\pmod 3
$$

Then different solutions of $x$ are congruent mod $6$, that is, $x_1 \equiv x_2 \pmod 6$.

You have calculated $2018^{2018} \bmod 2$ and $2018^{2018} \bmod 3$, and know that $x_1 = 2018^{2018}$ is a solution.

By trial and error for all the integers $0,1, \cdots, 6-1$, or by direct construction, $x_2 = 4$ is another solution. So

$$2018^{2018} \equiv 4 \pmod 6$$

Title: Reducing $2018^{2018}\pmod6$ via $\mod2$ and $\mod3$

Body:
I've been tasked to reduce $$2018^{2018}\pmod6$$using the results of reducing$$2018^{2018}\pmod2$$and $$2018^{2018}\pmod3$$

I have deduced that $$2018^{2018}\equiv0\pmod2$$and $$2018^{2018}\equiv1\pmod3$$but I don't see a way to combine these two congruences into one $\mod6$ congruence. Could someone tell me how this is achieved? I have a feeling the answer is really simple, but I just don't see it. Is there also a more generalized strategy of solving modular congruences via the divisors of the modulo number($2$ and $3$ are divisors of $6$)?

The way I have currently solved this problem is to first reduce $$2018^{2018}\pmod6$$ to $$2^{2018}\pmod6$$Then we look for patterns:$$\begin{align}2^1&\equiv2\pmod6\\2^2&\equiv4\pmod6\\2^3&\equiv2\pmod6\\2^4&\equiv4\pmod6\\&\ \ .\\&\ \ .\\&\ \ .\\2^n&\equiv\begin{cases}4&\text{if }n\equiv0\pmod2\\2&\text{if }n\equiv1\pmod2\end{cases}\pmod6\end{align}$$for $n\in\Bbb N$. In the case of $2018^{2018}$, $2018\equiv0\pmod2$, so $$2018^{2018}\equiv2^{2018}\equiv4\pmod6$$


问:
Hatcher Theorem 1.38 Example
Hatcher's theorem 1.38 states that there is a correspondence between isomorphism classes of path-connected covering spaces $p: \tilde{X} \rightarrow X$ and conjugacy classes of subgroups of $\pi_1(X,x_0)$.

The proof claims that given a covering space the basepoint $\tilde{x_0}$ within $p^{-1}(x_0)$ corresponds to changing $p_*(\pi_1(\tilde{X},\tilde{x_0}))$ to a conjugate subgroup of $\pi_1(X,x_0)$.

What's an example of a case when the conjugate subgroup that it changes to is not equal to the subgroup itself? I'm looking for a specific example of an $X$ and $\tilde{X}$.

答: There is an example to your question in the Hatcher's book itself. Earlier in section 1.3 there is a table of covering spaces of the wedge sum $S^1\vee S^1$. With $X=S^1\vee S^1$ and $\tilde{X}$ the covering space of example $(3)$ and $(4)$ you have your example.

Title: Hatcher Theorem 1.38 Example


答:
One should do a strong induction:

The base case ($n=1$) is essentially the definition of $a$.

Take $n \geq 1$ and assume inductively that, for all $k = 1,\ldots,n$, we have

$$\mathbb E (X^k) = k!a^k.$$

Then

$$
\newcommand{\EE}{\mathbb E}
\begin{alignat}4
\EE (X^{n+1}) &= \EE\big(U^{n+1}(Y+Z)^{n+1}\big) &\qquad&
\\ &= \EE (U^{n+1})\EE\big((Y+Z)^{n+1}\big) &&\text{independence}
\\ &=  \frac1{n+2}\ \ \ \ \EE\big((Y+Z)^{n+1}\big) &&\text{hint in comments}
\\ &= \frac1{n+2}\sum_{k=0}^{n+1}\binom{n+1}k\EE(Y^kZ^{n+1-k}) &&\text{linearity of expect.}
\\ &= \frac1{n+2}\sum_{k=0}^{n+1}\binom{n+1}k\EE(Y^k)\EE(Z^{n+1-k}) &&\text{independence again}
\\ &= \frac1{n+2}\sum_{k=0}^{n+1}\binom{n+1}k\EE(X^k)\EE(X^{n+1-k}) &&\text{$X,Y,Z$ ident. distributed}
\\ &= \frac2{n+2}\EE(X^{n+1})+\frac1{n+2}\sum_{k=1}^{n}\binom{n+1}k\EE(X^k)\EE(X^{n+1-k})
\\ &= \frac2{n+2}\EE(X^{n+1})+\frac1{n+2}\sum_{k=1}^{n}\frac{(n+1)!}{k!(n+1-k)!}(k!a^k)\big((n+1-k)!a^{n-1+k}\big) && \text{induct. hypothesis}
\\ &= \frac2{n+2}\EE(X^{n+1}) + n!\cdot \frac n{n+2} a^{n+1} && \text{mass cancellation}
\end{alignat}
$$

and this last equality rearranges to

$$\frac{n}{n+2} \EE(X^{n+q})= \frac n {n+2}(n+1)!a^{n+1},$$

i.e. $$\EE(X^{n+1}) = (n+1)!a^{n+1},$$

completing the inductive step, and the proof.

Token remark: this implies the moment generating function no end!

$$ M_X:\lambda ↦ \EE (e^{\lambda X}) = \sum_{k=0}^∞ \frac{k!a^k\lambda^k}{k!}=\cdots=\frac 1{1-a\lambda}.$$

Up to choosing $a$, this fixes the distribution of $X$. Does anyone recognise if it corresponds to something named?

Title: Showing $E(X^n) = n!a$ using induction.

Body:
Let $X$ be a random variable with $a = E(X)$, and suppose $X$ has the
same distribution as $U(Y + Z)$, where

$U$ is uniformly distributed over $(0, 1)$, $U, Y, Z$ are independent and $X, Y, Z$ have the same distribution.
How can we prove using induction that $E(X^n) = n!a^n$ for $n = 1, 2,...$.

Any hint would be appreciated.


问:
Inequality between 2p-norm and p-norm for random variables
Recently I was studying something about random matrix theory, and class of sub-guassian / sub-exponential random variables is of interest. In the literature it gave an inequality as following:

$\sup_{p\geq 1} \frac{\|X^2\|_p}{p} \leq 2\sup_{p\geq 1} (\frac{\|X\|_p}{\sqrt{p}})^2$

which gives a sufficient condition such that:

$\|X\|_{2p} \leq \sqrt{2}\|X\|_p$

The proof of this inequality is not provided in the literature. Suppose the random variable $X\in\mathbb{L}^\infty(\mathbb{R})$ therefore the previous defined sub-gaussian norm and sub-exponential norm both exist, then how do we proof the sufficient condition?

Any comment is greatly appreciated :)

答:
This is a direct consequence of the general fact that $\|X^2\|_p=E[|X|^{2p}]^{1/p}=\left(\|X\|_{2p}\right)^2$ for every $p\gt0$ and every  random variable $X$ in $L^{2p}$. Hence,
$$
\sup_{p\geqslant 1} \frac{\|X^2\|_p}{p}=2\,\sup_{p\geqslant 2} \left(\frac{\|X\|_p}{\sqrt{p}}\right)^2\leqslant 2\,\sup_{p\geqslant 1} \left(\frac{\|X\|_p}{\sqrt{p}}\right)^2.
$$
The rôle of the (not always true) inequality $\|X\|_{2p} \leqslant \sqrt{2}\,\|X\|_p$ is not clear.</p>


问:
bounded linear transformations
I am trying to show that a linear map from a field  to a vector space $W$ is bounded.

Here is is what I know:

Since $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, it is finite dimensional because in both cases, we have a basis consisting of one element.

Suppose $f_{1}$ is a basis for our field. Then for $f=a_{1}f_{1}$ we have:

\begin{align}
\|Tf\|_{W}&=\|Ta_{1}f_{1}\|_{W}\\
&=\|a_{1}Tf_{1}\|_{W}\\
&=|a_{1}|\|Tf_{1}\|_{W}\\
&=|a_{1}||f_{1}|\|T(1)\|_{W}\\
\end{align}

And here I am stuck. I need to show that $\|Tf\|_{V}<\infty$, but I am confused because the the a's are real numbers and hence unbounded. Any insight would be appreciated..thank you.

答:
You are done. You have that $|a_1f_1|=\|a_1f_1\|=\|f\|$. So you have
$$
\|Tf\|_V=\|T(1)\|_V\,\|f\|,
$$
showing that $T$ is bounded and $\|T\|\leq\|T(1)\|_V$.

Title: bounded linear transformations


问:
A variant of Mean Value Theorem
I was having trouble with one of the problems in Pugh's analysis text and was wondering if I could get some pointers.

Let $X \subset \mathbb{R}^m$ be compact, and $f:X \to \mathbb{R}$ be continuous. Show that given $\epsilon > 0$, there exists a constant $M \in \mathbb{R}$ such that for any $x,y \in X$, $|f(x) - f(y)| \leq M|x-y| + \epsilon$.

There's a similar(?) statement to this one: if $f$ is uniformly continuous and $X$ is bounded, then $f$ is bounded also. So I tried to use a similar idea to prove this, namely using compactness to partition $X$, but I couldn't figure out how to show anything for any pair $x,y$.

I'd appreciate some help with this problem. Thank you! (Also this is not a HW problem, I am doing practice problems for exam preparation.)

答:
Given $\varepsilon$ choose $\delta$ such that
$$\|x-y\| < \delta \Rightarrow \|f(x)-f(y)\|< \varepsilon.
$$

Then let
$$
M= \sup \left\{ \frac{ \|f(x)-f(y)\|}{\|x-y\|} : \|x-y\| \geq \delta\right\}
$$

Your inequality is now easy to prove by considering two cases depending on whether $\|x-y\|<\delta$ or not.

Title: A variant of Mean Value Theorem


问:
Nonlinear Implicit Finite Difference
How do you produce an implicit finite difference system with a nonlinear term in the pde?

For example, if you have the reaction-diffusion equation:
$$\frac{\partial{u}}{\partial{t}} = \Delta u + f(u)$$

with $f$ being non-linear, e.g. $f(u) = u^2$, how do you form a finite difference system to solve?

I understand how to derive if $f(u)=0$. For example in a 1D setting, you have the form:
$$w_{i}^{n-1} = (1+2\lambda)w_{i}^n - \lambda(w_{i+1}^n + w_{i-1}^n)$$
producing a system in the form:
$$Aw^n=w^{n-1}$$

答:
If you plan to use the so-called "Method of Lines", i.e. first you discretize in space (in this case you obtain the classical $[1, -2, 1]$ stencil) and then in time, then you obtain a (large) stiff system of ODEs of the form $$u_h' = Au_h + f(u_h)$$ where $u_h$ is the unknown vector, still continuous in time: $(u_h(t))_i=u(x_i,t)=u(x_0 + ih, t)$

Depending on how the time integration is performed, you may end up with an implicit scheme: the simplest is Backward Euler
$$u_h^{n+1} = u_h^{n} + kA u_h^{n+1} + f(u_h^{n+1})$$ which requires you to solve at each time step a non linear system of equations and it's generally expensive. Usually is $f(u)$ is non-linear, like your case, one may prefer to use an IMEX approach (IMplicit-EXplicit) and make explicit the nonlinearity $f(u)$ like @PierreCarre wrote in his answer. The advantage is that all you need to do now is to solve a linear system for each time step, which is fairly cheaper: $$u_h^{n+1} = (I-kA)^{-1}(u_h^n+f(u_h^n))$$

Title: Nonlinear Implicit Finite Difference


答: $$\|Tf\|_V = \|T(f1)\|_V = \|fT(1)\|_V = |f|\|T(1)\|_V = c|f|$$ where $c := \|T(1)\|_V$. Thus $\|Tf\|_V \leq c|f|$ for any $f \in \mathbb F$, which means that $T$ is bounded.

Title: bounded linear transformations

Body:
I am trying to show that a linear map from a field  to a vector space $W$ is bounded.

Here is is what I know:

Since $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, it is finite dimensional because in both cases, we have a basis consisting of one element.

Suppose $f_{1}$ is a basis for our field. Then for $f=a_{1}f_{1}$ we have:

\begin{align}
\|Tf\|_{W}&=\|Ta_{1}f_{1}\|_{W}\\
&=\|a_{1}Tf_{1}\|_{W}\\
&=|a_{1}|\|Tf_{1}\|_{W}\\
&=|a_{1}||f_{1}|\|T(1)\|_{W}\\
\end{align}

And here I am stuck. I need to show that $\|Tf\|_{V}<\infty$, but I am confused because the the a's are real numbers and hence unbounded. Any insight would be appreciated..thank you.


问:
Inequality $||DG(t)||\leq c||G(t)||$ implies $G\equiv 0$?
Continuous and differentiable function $G:[0,\infty)\to \mathbb{R}^n$ satisfies $G(0) = 0\in\mathbb{R}^n$ and there exists $c>0$ such that for all $t>0$ we have
$$||DG(t)||\leq c||G(t)||$$
Does it mean that $G\equiv 0$?

Remark:
$G(t) = (g_1(t),g_2(t),...,g_n(t))$, $DG(t)$ is the derivative of $G$ and $||\circ||$ is euclidean norm.

If $n=1$ then this statement is true, but every proof I've seen relied on Mean Value Theorem. As of yet I was unable to generalize it to higher dimensions or provide counter examples.

答:
Let $M= \max_{x\in [0,1]} \|G(x\|$. Then  $\|G'(x)\| \le c M$ for all $x\in [0,1]$. Now, for all $x\in [0,1]$ we have
$$G(x) = G(x) - G(0) =\int_{0}^x G'(t) dt$$
and so
$$\|G(x) \|\le c M x$$
for all $x\in [0,1]$.  Let's show that in general
$$\|G(x) \le \frac{c^n M x^n}{n!}$$
for all $x\in [0,1]$. By induction on $n$. For $n=1$, checked. Assume true for $n$. Then we have $\|G'(x)\|\le c\cdot \frac{c^n M x^n}{n!}$, and so
$$\|G(x) \| \le \int_{0}^x \|G'(t)\| dt \le c^{n+1} M \int_{0}^x \frac{t^n}{n!} = \frac{c^{n+1} M x^{n+1}}{(n+1)!}$$

Now, observe that $ \frac{c^n M x^n}{n!}\to 0$ as $n\to \infty$.

Note: We assumed that we can use the Newton-Leibniz formula.


问:
Prove that the prime factors $p$ of $n=x^2+y^2$ of the form $p=4m+3$ can only have even exponents.
I'm looking for an elementary proof of the statement in the title.

Actually I have already proved that if $-\bar{1}$ is a square then $p\equiv 1 \pmod 4$ or $p=2$.

I try to deduce from this statement that if $p=4m+3$ is a prime factor of $n$ then $p$ can only have an even exponent.

I know that I can prove it using Gaussian integers, but I'm really trying to do it using only the previous result.

答:
Let $q$ be a prime such that $q|n$ and q doesnot divides $y$ and $x$. If it divides both $x$ and $y$ then exponent of the prime would be even in $n$ which follows else,  let $v_{q}(n)=a\geq 1$ then by Euler generalization of FLT portray it by Euler totient function as $x^2+y^2\equiv 0 (\mod q^{a})$
Let $z$ be the multiplicative inverse of $y$ modulo $q^{a}$.Note that if $q$ doesnot divides $y$ this exist as $gcd(y,q^a)=1$ implies there exist c and d such that $yc+q^{a}d=1$ plugging $c=z$ we have $yz\equiv 1(\mod q^{a})$ thus $(xz)^{2}\equiv -1(\mod q^{a})$ then $ord_{q^{a}}(xz)=4$ thus $4|q^{a-1}(q-1)$ by phi function deduction. But  $q-1$ is $2$ mod $4$ if $a=1$ then $4|q-1$ then contradiction. Thus any prime dividing $n$ must be of form $4k+1$ for some $k$ in integers positive if $q$ doesnot divides both $x$ and $y$ so only possibility a prime $q$ of form $4k+3$ dividing $n$ is if it's divide both $x$ and $y$ then obviously $q^{2}|n$
Example: $3^{2}|3^{2}+6^{2}$


问:
How many axioms does ZFC have?
ZFC doesn't have finitely many axioms because it has some axiom schemas. Let's look at Axiom Schema of Specification:

$$
\forall X \exists Y \forall a \quad (a \in Y \iff a \in X \land P(a))
$$

This specifies a different axiom for each predicate $P$. Since $P$ is essentially a function from $X$ to boolean values, taken literally as $X$ is an arbitrary set, this axiom schema specifies uncontainably many axioms. That is, the class of axioms this axiom schema specifies is not a set.

Still, I think this axiom schema actually specifies countably infinite axioms, because we can construct an $X$ and a $P$ only in finitely long "text" with finitely many "characters". Is this the case?

答: You already have two answers correctly explaining that the specification schema of ZFC uses only those $P$ that can be expressed by first-order formulas in the language of set theory. But it might be of historical interest to mention that Zermelo's original axiom system (of 1908) did not have this restriction on $P$. He required only that $P$ must express a "definite property", and he gave no mathematically precise definition of what that means.  The suggestion that "definite" should mean "first-order expressible" was made by Skolem about a decade later. Zermelo didn't like it at all; if I remember correctly, he denounced it in  paper of 1930. Nevertheless, Skolem's suggestion was incorporated when setting up the theory ZFC in first-order logic.

Title: How many axioms does ZFC have?


问:
For which $x \in \mathbb{R}$ does this power series converge?
Conclude for which $x \in \mathbb{R}$, the following power series converges:
$$x\ +\ \sum_{k\ \geq \ 2} (-1)^{k-1} \frac{2k\ x^{2k-1}}{(2k-1)!}$$

I see a similiarity with the Maclaurin series for $sin(x)$, but there seems to be a factor $2k$ to much in each term.

Would mostly appreciate a hint, but a solution is also acceptable.

答: I think I might have solved finally solved it. Big thanks to @Raymond Manzoni for great hints and guiding. If you find any errors, please let me know. Here goes:$$x\ +\ \sum_{k\ \geq \ 2} (-1)^{k-1} \frac{2k\ x^{2k-1}}{(2k-1)!}=\\ x+ \sum_{k\ \geq\ 1} (-1)^k\ (1+(2k+1)) \frac{x^{2k+1}}{(2k+1)!}=\\ -x + \sum_{k \ \geq \ 0} (-1)^k (1+(2k+1)) \frac{x^{2k+1}}{(2k+1)!}=\\ -x+\sum_{k \ \geq \ 0} (-1)^k \frac{x^{2k+1}}{(2k+1)!} +x\sum_{k \ \geq \ 0} (-1)^k \ \frac{x^{2k}}{(2k)!}=\\ -x+\sin(x)\ + \ x \ cos(x)=\sin(x)+x \ (\cos(x)-1) \leq 1+2x \text{ which is convergent} \ \forall x $$


问:
Powers of Hermitian Matrix commute implies that Matrix Commutes
This could very well be a duplicate, so please close if similar. Working through old preliminary exams.

Edit: The original question is stated as follows.

Let $A$ be Hermitian $n\times n$ matrix. Suppose $B$ is an arbitrary $n\times n$ matrix. Prove that if

$A^3B=BA^3$

Then in fact

$AB=BA$.

I'm trying to prove a generalization.

Suppose $A$ is Hermitian, and let suppose there exists a $k\in\mathbb{N}$ such that

$A^kB=BA^k$

Prove that $AB=BA$.

It's easy to show from the assumption that if $p(x)$ is a polynomial then in fact

$p(A^k)B=Bp(A^k)$

So that it's sufficient to find some polynomial $p$ that $p(A^k)=A$. To that end, $A$ Hermitian implies that $\sigma(A)\subset\mathbb{R}$. The spectral mapping theorem implies that the eigenvalues of $A^k$ are $\lambda^k$ for $\lambda$ an eigenvalue of $A$. If $\{\lambda_i\}_{i=1}^{n}$ are the eigenvales of $A$, let $q$ be the interpolation polynomial of degree $n-1$ satisfying

$q(\lambda_i^k)=\lambda_i$.

Here's where I'm stuck. I want to conclude that $q(A^k)=A$. Is there a way to do that? Or is the interpolation polynomial the completely wrong approach.

Hints are appreciated if possible instead of answers!

答:
The statement is false: take a self-adjoint projection $P$ and any matrix $B$ which does not commute with $P$.  Then $B$ does not commute with
$$
A= 2P-1
$$
either.  However $A^2=1$, so clearly $A^2$ and $B$ commute.

On the other hand the statement becomes true if $A$ is assumed positive.


问:
What is the name for set of rationals including the "number" infinity?
How would you write the set of rationals $\mathbb{Q}$ including the number $\infty$? i.e. so it is closed under division? The set would correspond to points on a number circle.

i.e. given $(a,b)\in \mathbb{Z}^2$ then the pair $(a,b)$ always corresponds to a rational number $a/b$, likewise so would $(2a,2b)$ and $(-a,-b)$.

So that if two rational numbers $(a,b)$ and $(c,d)$ are equal then $ad=bc$. (This would hold only if $\infty=-\infty$ in this set.)

Is it written as $\mathbb{Q} \cup  \infty$ ?

The rationals would seem to naturally include the point $\infty$ from the pairs $(1,0)$, $(2,0)$ etc. unless one deliberately excluded it. Whereas the set of integers doesn't include infinity.

Is there a special name for the "rational infinity?"

答:
You can call it the projective line $\mathbb{P}^1(\mathbb{Q})$ over $\mathbb{Q}$, as stated in the comments. It famously appears in the theory of modular forms as the set of "cusps"; see, for example, this math.SE question.

It is not quite closed under division, but it is closed under the action of the modular group

$$PSL_2(\mathbb{Z}) = \left\{ z \mapsto \frac{az + b}{cz + d} : a, b, c, d \in \mathbb{Z}, ad - bc = 1 \right\}$$

(which acts transitively), which is related to its appearance in the theory of modular forms. Of course it is also closed under the action of the slightly larger group $PGL_2(\mathbb{Q})$.


问:
Prove that if $\sum a_n$, $\sum b_n$ and $\sum c_n$ converge to A,B and C then $C=AB$
Define $\displaystyle c_n=\sum_{k=1}^n a_kb_{n+1-k}$. Prove that if $\sum a_n=A$, $\sum b_n=B$ and $\sum c_n=C$ then $C=AB$.

My teacher did not teach us Merten's Theorem or Abel's theorem. Is there a way to prove this without those theorems?

I don't know if this is the right approach, but I tried multiplying $\sum a_n$ and $\sum b_n$ and got

$$\sum a_n \sum b_n=a_0b_0+(a_0b_1+b_0a_1)+...+(a_0b_n+a_1b_{n-1}+...+a_{n-1}b_1+a_nb_0)$$

I think this is ultimately my goal:
$$(\sum_{n=1}^N a_n)(\sum_{n=1}^N b_n)=\sum_{n=1}^N\sum_{k=1}^Na_n\cdot b_k=\sum_{n=1}^N\sum_{k=1}^na_k\cdot b_{n-k+1}$$

but I dont know how to show that $$\sum_{n=1}^N\sum_{k=1}^Na_n\cdot b_k=\sum_{n=1}^N\sum_{k=1}^na_k\cdot b_{n-k+1}$$

答: You can show that $\sum_n c_n$ is Cesaro-summable to $AB$, provided that $\sum_n a_n = A$, and $\sum_n b_n = B$ (Cesàro's theorem). If $\sum_n c_n$ is also summable, it must have sum $AB$.


问:
Show that $X$ is compact if and only if $X_{r}$ is compact for all $r \in $ {$1, ..., n$}.
Let $X_{1}, X_{2}, ..., X_{n}$ be subsets of $\mathbb{R^m}$ and $X = X_{1} \times X_{2} \times ... \times X_{n}$. Show that $X$ is compact if and only if $X_{r}$ is compact for all $r \in $ {$1, ..., n$}.

What I've done:

Let $X$ be compact, then $X$ is closed and bounded. Now because of how $X$ is defined then $X_{1}, X_{2}, ..., X_{n}$ are closed and bounded subsets, so $X_{r}$ will be compact for all $r \in$ {$1,..., n$}

That would be the departure, can someone help me with the return and tell me if the departure is okay?

答:
I am just giving you hints
If part :$X_{i}$ is compact for all $i=1,2,..,n$ then trivially Metric $X$ is compact.
For only if part: Assume some $X_{j}$ is not compact for some $j$ where $j$ is in naturals. Then show that $X$ is not compact.
Cheers!!!


问:
Higher Order Diophantine Equation
I am trying to obtain the general solution of the following Diophantine equation.
$$x^4-8x^2y^2+8y^4=z^2$$
My approach was to rewrite the equation.
\begin{align}
\begin{split}
& x^4+z^2=2(2y^2-x^2)^2\\
\implies & p^2+q^2=2r^2
\end{split}
\end{align}
Let $a^2+b^2=r^2$. After some manipulation,
\begin{align}
\begin{split}
& (a^2+b^2)+(a^2+b^2)=2r^2 \\
& (a^2+b^2+2ab)+(a^2+b^2-2ab)=2r^2 \\
& (a+b)^2+(a-b)^2=2r^2
\end{split}
\end{align}
$(a,b,c)$ forms a Pythagorean triple. So we get
\begin{align}
\begin{split}
&(a,b,c)=(k(m^2-n^2),2kmn,k(m^2+n^2))\\
&(p,q,r)=(x^2,y^2,z)=(k(m^2-n^2+2mn),k(m^2+mn),k(m^2+n^2))
\end{split}
\end{align}
I am stuck here. Any help will be appreciated.

答:
Althogh this method is different from your method, the solution is easy to derive.

Let $U=\frac{\large{x}}{\large{y}}, V=\frac{\large{z}}{\large{y^2}}$ then we get $V^2 = U^4-8U^2+8$.

This quartic equation is birationally equivalent to the elliptic curve below.

$Y^2 = X^3 + X^2 -3X + 1$

$U = \frac{Y}{(X-1)}, V = \frac{X^3-3X^2+X+1}{(X-1)^2}$

$X = \frac{1}{2}V+\frac{1}{2}U^2-1, Y = \frac{1}{2}U(V+U^2-4)$

This elliptic curve has rank $1$ with generator $(0,-1)$.

The rank and generator are obtainded using mwrank(Cremona).

Hence this curve has infinitely many rational points.

Thus, we get infinitely many integer solutions.

Example:

First, we get a rational solution $(X,Y)=(\frac{-56}{25}, \frac{153}{125})$ using mwrank or Magma.

From $U = \frac{Y}{(X-1)}, V = \frac{X^3-3X^2+X+1}{(X-1)^2}$, we get $(U,V)=(\frac{-17}{45}, \frac{-5311}{2025})$

Hence we get $(x,y,z)=(17,45,5311).$

More solutions using group law:

Let $P(X,Y)=(0,-1)$, then we get

$6P(X,Y)=(\frac{30547621}{2340900}, \frac{-173783774969}{3581577000})$

$7P(X,Y)=(\frac{-6037269840}{7661325841}, \frac{-1253783210541679}{670588189536889}).$

Hence we get the corresponding solutions

$(x,y,z)=(32721479, 8125830, 784659483951841)$

$(x,y,z)=(10712341919, 10244481689, 70259696347563139199).$

Some solutions with $x<1000000$ using PARI-GP as follows.

"hyperellratpoints($X^4-8X^2+8,1000000$)"

```<code>           [ x  y  z ]

           [1, 1, 1]
           [7, 2, 31]
           [17, 45, 5311]
           [2273, 868, 305791]
           [50639, 98069, 23436127681]
</code>```

Title: Higher Order Diophantine Equation

Body:
Althogh this method is different from your method, the solution is easy to derive.

Let $U=\frac{\large{x}}{\large{y}}, V=\frac{\large{z}}{\large{y^2}}$ then we get $V^2 = U^4-8U^2+8$.

This quartic equation is birationally equivalent to the elliptic curve below.

$Y^2 = X^3 + X^2 -3X + 1$

$U = \frac{Y}{(X-1)}, V = \frac{X^3-3X^2+X+1}{(X-1)^2}$

$X = \frac{1}{2}V+\frac{1}{2}U^2-1, Y = \frac{1}{2}U(V+U^2-4)$

This elliptic curve has rank $1$ with generator $(0,-1)$.

The rank and generator are obtainded using mwrank(Cremona).

Hence this curve has infinitely many rational points.

Thus, we get infinitely many integer solutions.

Example:

First, we get a rational solution $(X,Y)=(\frac{-56}{25}, \frac{153}{125})$ using mwrank or Magma.

From $U = \frac{Y}{(X-1)}, V = \frac{X^3-3X^2+X+1}{(X-1)^2}$, we get $(U,V)=(\frac{-17}{45}, \frac{-5311}{2025})$

Hence we get $(x,y,z)=(17,45,5311).$

More solutions using group law:

Let $P(X,Y)=(0,-1)$, then we get

$6P(X,Y)=(\frac{30547621}{2340900}, \frac{-173783774969}{3581577000})$

$7P(X,Y)=(\frac{-6037269840}{7661325841}, \frac{-1253783210541679}{670588189536889}).$

Hence we get the corresponding solutions

$(x,y,z)=(32721479, 8125830, 784659483951841)$

$(x,y,z)=(10712341919, 10244481689, 70259696347563139199).$

Some solutions with $x<1000000$ using PARI-GP as follows.

"hyperellratpoints($X^4-8X^2+8,1000000$)"

           [ x  y  z ]

           [1, 1, 1]
           [7, 2, 31]
           [17, 45, 5311]
           [2273, 868, 305791]
           [50639, 98069, 23436127681]


问:
Inverse image of sum of ideals
Lately, I've been looking for nice identities involving ring homomorphisms and ideals. For example, I've been able to prove that for an epimorphism $\varphi:R\to S$ and two ideals $I,J\subset R$ we have $\varphi(I+J)=\varphi(I)+\varphi(J)$; or that for a principal ideal $(a)$, $a\in R$, we have $\varphi( (a) )=(\varphi(a))$.

Now I'm wondering what would happen to two ideals $I',J'\subset S$, using $\varphi^{-1}$. When does it hold that $\varphi^{-1}(I'+J')=\varphi^{-1}(I')+\varphi^{-1}(J')$? I got one contention:

If $r\in\varphi^{-1}(I')+\varphi^{-1}(J')$ then $\exists\, i\in\varphi^{-1}(I'),j\in\varphi^{-1}(J'):r=i+j\Rightarrow\varphi(r)=\varphi(i)+\varphi(j)\in I'+J'\Rightarrow r\in\varphi^{-1}(I'+J')$

For the other contention I first got stuck at

If $r\in\varphi^{-1}(I'+J')$ then $\varphi(r)=i'+j',i'\in I', j'\in J'$

where I have to assume $\varphi$ is surjective to say $\exists\, i,j: \varphi(i)=i',\varphi(j)=j'$.

But then I got

$\varphi(r)=\varphi(i+j)$

and I'd say I need to conclude $r=i+j$, but for that I'd need $\varphi$ to be injective (hence an isomorphism).

Is it really necessary to assume $\varphi$ is an isomorphism? I found a counterexample where $\varphi$ is injective, but nor surjective, so I'd say we need surjectivity; but I'm not able to find a counterexample where $\varphi$ is surjective, but not injective.

Any help would be appreciated. Thanks in advance.

答:
For a counter example in which $\phi$ is injective but not surjective, consider:

$S=R\oplus R$,

$\phi(x) =(x,x)$,

$I= R\oplus 0$, and

$J=0\oplus R$.

When $\phi$ is surjective the missing inclusion may be proved as follows: write $\phi(r)=i'+j'$ and pick $s$ in $R$ such that $\phi(s)=i'$.  Then
$$
\phi(r-s) = j'$$
whence $r-s\in \phi^{-1}(J')$ and clearly $s\in \phi^{-1}(I').$

We then have
$$
r= s +(r-s) \in \phi^{-1}(I')+\phi^{-1}(J').
$$

Title: Inverse image of sum of ideals


问:
Help with proof regarding the interior of a Set
Let A, B $\subset$ R and let $x_{0}$ $\in$ R such that $x_{0}$ $\in$ int(A) and $x_{0}$ $\in$ int(B).
Show that $x_{0}$ $\in$ int(A$\cap$B)

Any insight would be helpful.
P.D i know conceptually is easy to grasp but i dont know how to formalize it

答:
$
\newcommand{\Int}{\operatorname{int}}
$
Hints:

Recall that $x \in \Int(S)$ if and only if there exists some $T \subseteq S$ which is an open neighborhood of $x$.

Since $x_0$ is in both $\Int(A)$ and $\Int(B)$, we can find such sets $A_0,B_0$ for each of $A,B$, respectively.

Observe that the intersection $A_0 \cap B_0$ contains $x_0$, and is an open neighborhood of $x_0$ contained within $A \cap B$. Reapply the definition of interior in this light.

I'll leave the justifications up to you.


问:
Variance of a Random Forest
I'm reading through Introduction to Statistical Learning (ISL) right now and I'm having trouble with understanding the variance of a random forest. Does anyone know how this variance is derived? What is $\rho$?

答:
The quote says $\rho$ is the pairwise correlation, i.e. the correlation between any two of the random variables. In other words $\mathbb E[(X_i-\mu)(X_j-\mu)]=\rho\sigma^2$ when $i \not=j$ and $\sigma^2$ when $i=j$

So if we have $B$ such random variables, the variance of their average is $$\mathbb E\left[\left(\frac1B\sum X_i-\mu\right)^2\right] \\= \frac1{B^2}\mathbb E\left[\sum_i (X_i-\mu)\sum_j (X_j-\mu)\right] \\ =\frac1{B^2} \left((B^2-B)\rho \sigma^2+B\sigma^2\right)\\= \rho \sigma^2 +\frac{1-\rho}{B}\sigma^2$$

Title: Variance of a Random Forest


问:
Determine the solutions to the following linear system of equations
This is an example to illustrate the following theorem:

Let $a,b$ be integers and let $m,n$ be positive integers. The system
\begin{eqnarray*}
x&\equiv & a\pmod{m}\\
x&\equiv & b\pmod{n}\\
\end{eqnarray*} has solution if and only if $(m,n)\mid (a-b).$ In this case, the solution can be uniquely expressed modulo $[m,n]$

Determine the solutions to the following linear system of equations
\begin{eqnarray*}
x&\equiv & 5\pmod{6}\\
x&\equiv & 8\pmod{15}\\
\end{eqnarray*}

The system will have a solution if $(15,6)\mid (5-8)$, wich is true, since $3\mid-3$. I have found that the solution of this system is $x\equiv 23\pmod{30}$, Which they argue as follows:

Since there is a solution, we express $-3$ as a linear combination of $6$ and $15$: $5-8= -3 = (-3)\cdot 6 + 1\cdot 15$, thus $x=8+15=23$.

In this step I am stuck, Is there another way to get to that 23?, if someone could explain in detail what was done in this step I would be very grateful.

答:
The idea of the given step is as follows: First, subtract $8$ to get the following equations for $y=x-8$: $$y\equiv-3\pmod6\\ y\equiv0\pmod{15}$$

By the second equation, $y$ is of the form $15a, a\in\mathbb Z$. By the first one $y$ has the form $-3-6b$. Thus $-3-6b=y=15a$ means $-3=15a+6b$, which explains why we want to write $-3$ as a linear combination.


问:
NFA to DFA: new initial state
Question 4 asks to convert this NFA to a DFA:

The solution is as follows:

Whereas I found this as a solution (excuse the messy layout):

My DFA is identical to the solutions' except the initial state (underlined).

The solutions omitted the 0 state (NFA initial state) and made 1,3 the DFA initial state (mine stays 0).

Can anybody please explain why they removed the 0 state and made 1,3 initial instead?

Edit: the original Q has been answered, but the following NFA to DFA    doesn't seem to follow:

The DFA -

I've changed the labels for comprehension:

But shouldn't the new initial state be 1,4 (not in DFA)?

答:
Okay I stumbled on another thread that explains it well:
https://math.stackexchange.com/questions/1809453/nfa-to-dfa-removing-unnecessary-states-what-is-new-start-state#:~:text=1%20Answer&text=The%20starting%20state%20of%20the,after%20reading%20the%20word%20%CF%B5.

Basically, the new DFA's initial state is the set of all states the NFA's initial state reached using only epsilon transitions.

Title: NFA to DFA: new initial state


问:
Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$
My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\theta&=4\sin\theta-3\sin2\theta-4\sin^3\theta
\end{align*}
I have faced several symmetric trigonometry problems most of them need to use product to sum identities, but this one I can't continue.

答: It is a false identity. Click here to see the graphs.</p>


问:
Measurability of an Indexed Product-Measure
If for any fixed $\omega_1$, $P_{\omega_1}$ is a probability measure and $Q_{\omega_1}$ is a stochastic kernel and both are measurable in $\omega_1$, is the indexed product measure $K_{\omega_1}:=P_{\omega_1}\otimes Q_{\omega_1}$ measurable in $\omega_1$?

To rephrase the question more precisely, let $\Omega_1$, $\Omega_2$ and $\Omega_3$ be non-empty sets, let $\mathcal{A}_1$, $\mathcal{A}_2$ and $\mathcal{A}_3$ be $\sigma$-algebras on $\Omega_1$, $\Omega_2$ and $\Omega_3$, respectively, let $P$ be a stochastic kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$ and let $Q$ be a stochastic kernel from $\mathcal{A}_1\otimes\mathcal{A}_2$ to $\mathcal{A}_3$. For each $\omega_1\in\Omega_1$, denote by $P_{\omega_1}$ the function

$$
P_{\omega_1}:\mathcal{A}_2\rightarrow\left[0,1\right],\space\space P_{\omega_1}\left(B_2\right):=P\left(\omega_1,B_2\right)
$$

and by $Q_{\omega_1}$ the function

$$
Q_{\omega_1}:\Omega_2\times\mathcal{A}_3\rightarrow\left[0,1\right],\space\space Q_{\omega_1}\left(\omega_2,B_3\right):=Q\left(\left(\omega_1,\omega_2\right),B_3\right)
$$

It is readily seen that $P_{\omega_1}$ is a probability measure on $\mathcal{A}_2$ and that $Q_{\omega_1}$ is a stochastic kernel from $\mathcal{A}_2$ to $\mathcal{A}_3$. So $P_{\omega_1}\otimes Q_{\omega_2}$ is a product measure on $\mathcal{A}_2\otimes\mathcal{A}_3$ and we can define

$$
K:\Omega_1\times\left(\mathcal{A}_2\otimes\mathcal{A}_3\right)\rightarrow\left[0,1\right],\space\space K\left(\omega_1,B_{2,3}\right):=\left(P_{\omega_1}\otimes Q_{\omega_1}\right)\left(B_{2,3}\right)
$$

Is $K$ a stochastic kernel? For $K$ to be one, it needs to satisfy two conditions:

For every $\omega_1\in\Omega_1$, $K\left(\omega_1,B_{2,3}\right)$ is a probability measure on $\mathcal{A}_2\otimes\mathcal{A}_3$,

For every $B_{2,3}\in\mathcal{A}_2\otimes\mathcal{A}_3$, $K\left(\omega_1,B_{2,3}\right)$ is $\mathcal{A}_1/\mathfrak{B}$-measurable, with $\mathfrak{B}$ being the Borel field on the real line.

It is easy to see that condition #$1$ is satisfied, but what about condition #$2$?

答:
Let $B_{2,3}\in\mathcal{A}_2\otimes\mathcal{A}_2$. We need to show that the map

$$
\omega_1\mapsto K\left(\omega_1,B_{2,3}\right)
$$

is $\mathcal{A}_1/\mathfrak{B}$-measurable.

Let $\omega_1^*\in\Omega_1$. Then

$$
K\left(\omega_1^*,B_{2,3}\right)=\int_{\Omega_2}P_{\omega_1^*}\left(d\omega_2\right)\int_{\Omega_3}Q_{\omega_1^*}\left(\omega_2,d\omega_3\right)\mathbb{1}_{B_{2,3}}\left(\omega_2,\omega_3\right)
$$

Define

$$
g:\left(\Omega_1\times\Omega_2\right)\times\Omega_3\rightarrow\left[0,1\right],\space\space g\left(\left(\omega_1,\omega_2\right),\omega_3\right):=\mathbb{1}_{\Omega_1\times B_{2,3}}\left(\omega_1,\left(\omega_2,\omega_3\right)\right)
$$

It is easy to see that $g$ is non-negative, $\left(\mathcal{A}_1\otimes\mathcal{A}_2\right)\otimes\mathcal{A}_3/\mathfrak{B}$-measurable and that

$$
K\left(\omega_1^*,B_{2,3}\right)=\int_{\Omega_2}P\left(\omega_1^*,d\omega_2\right)\space I_g\left(\omega_1^*,\omega_2\right)
$$

with

$$
I_g:\Omega_1\times\Omega_2\rightarrow\left[0,\infty\right],\space\space I_g\left(\omega_1,\omega_2\right):=\int_{\Omega_3}Q\left(\left(\omega_1,\omega_2\right),d\omega_3\right)\space g\left(\left(\omega_1,\omega_2\right),\omega_3\right)
$$

Now, consider Lemma 14.20 in Klenke's Probability Theory (2006), which i reproduce here with some cosmetic changes.

Let $\kappa$ be a finite transition kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$ and let $f:\Omega_1\times\Omega_2\rightarrow\left[0,\infty\right]$ be $\mathcal{A}_1\otimes\mathcal{A}_2$/$\overline{\mathfrak{B}}$- measurable ($\overline{\mathfrak{B}}$ being the Borel field on the extended real line). Then
  the map
  $$
I_f:\Omega_1\rightarrow\left[0,\infty\right],\space\space I_f\left(\omega_1\right):=\int_{\Omega_2} \kappa\left(\omega_1,d\omega_2\right)\space f\left(\omega_1,\omega_2\right)
$$
  is well-defined and $\mathcal{A}_1$/$\overline{\mathfrak{B}}$-measurable.

Applying this lemma with $\kappa:=Q$, $f:=g$, we see that $I_g$ is $\mathcal{A}_1\otimes\mathcal{A}_2/\overline{\mathfrak{B}}$-measurable. Since $I_g$ is clearly non-negative, we can apply the lemma a second time with $\kappa:=P$, $f:=I_g$ to arrive at the desired result.


问:
$(\sin\theta+\cos\theta)^2=1+\sin2\theta$
49)  $(\sin\theta+\cos\theta)^2=1+\sin2\theta$

Left Side:

\begin{align*}
 (\sin\theta+\cos\theta)^2=\sin^2\theta+2c\cos\theta\sin\theta+cos^2\theta=1+2\cos\theta\sin\theta
\end{align*}
This can either be $1$ or I can power reduce it. I don't know.

Right Side:

\begin{align*}
 1+\sin2\theta=1+2\sin\theta\cos\theta
\end{align*}

Thank you!

答:
Here is a detailed answer.
$$
\begin{align}(\sin\theta+\cos\theta)^2&=\sin^2\theta+2\sin\theta\cos\theta+\cos^2\theta\\
&=\color{red}{\sin^2\theta+\cos^2\theta}+2\sin\theta\cos\theta\\
&=\color{red}{1}+\color{blue}{2\sin\theta\cos\theta}\\
&=1+\color{blue}{\sin2\theta}\\
\end{align}
$$
I hope this helps.


问:
Infimum of $2^x + 2^{1-x}, x \in \mathbb{R}$. Inequality of means? Interesting points on quadratic function.
So I checked some exercises in my book and found 2 interesting ones. It is required to determine infimum of sets given by:

$$\frac{x}{x^2+1}, x \in \mathbb{R}$$
$$2^x + 2^{1-x}, x \in \mathbb{R}$$

The interesting thing about those was that usually the numbers are naturals and here I got $x \in \mathbb{R}$. Because of that I have no boundary on left side of Xes. I wasn't able to transform those in any meaningful way, aside from:
$$2^x + 2^{1-x} = 2^x + \frac{2}{2^x} = \frac{2^x*2^x +2}{2^x} \implies 2^x = t \implies \frac{t^2 +2}{t}$$

Then I saw that both functions describing both sets are alike: quadratic function in numerator, linear function in denominator. I started to play with those functions in desmos plotter and got an interesting information. There is a point in quadratic function where quadratic function in no longer growing "slower" than linear function and starts to grow faster. That is obviously the point of infimum. By that time I already got the answer for the first one - inequality of means. And here are my questions:

if that point exists for every quadratic function, how it's called (so I can read some more about it)?

how is the growth of quadratic function past that point compared to "traditional" exponential - they seem similar on picture?

is it possible to say that the inf of second set is where the infimum of expression with variable $t$ is? It stands for ever-positive variable so it would look like it can be done by inequality of means too

答:
Both of those expressions have a more important link, and that's symmetry. Consider the following argument for the second one: suppose a point $x_0$ was an extrema for the function. However notice that

$$2^{x_0}+2^{1-x_0} = 2^{1-x_0}+2^{x_0}$$

which is the point $1-x_0$. Since two distinct points evaluate to the same value on a continuous function, this implies the function has some sort of a bowl shape, and these points cannot be extrema, unless they weren't distinct. This leads the $x_0 = \frac{1}{2}$ and taking limits $x\to\pm\infty$ establishes it as a minimum.

The same argument applies for the first one, only the symmetry is $x_0 \leftrightarrow \frac{1}{x_0}$, which leads to the point $x_0 = 1$ being a maximum.


问:
If $p$ is prime, then $x^p − 2 \equiv 0 \pmod{p^k}$ has a solution for $k \ge 2$

If $p$ is prime, then $x^p − 2 \equiv 0 \pmod{p^k}$ has a solution for $k \ge 2$.

I'm supposed to either prove or disprove the statement above using Hensel's Lemma.

So far what I have is assuming $r$ is a solution to $x^p − 2 \equiv 0 \pmod{p^{k-1}}$. I can show that $f'(r) \equiv 0 \pmod p$ since $f'(r) = p(r^{p-1})$.

I think the next step should be showing that $f(r)\not\equiv0\pmod{p^k}$ which would show that there are no solutions but I don't know how to go about doing this without any way to properly represent $r$.

答:
The claim is true very, very rarely (at least for "small" $p$).  Even for $k=2$.  In fact, the challenge here is to find a single prime $p$ for which the claim holds.  Indeed, only two are known.

To see this, note that the unique solution to $x^p\equiv 2\pmod p$ is $x\equiv 2 \pmod p$.

If we try to lift this to a solution $\pmod {p^2}$  we write $x\equiv 2 +tp\pmod {p^2}$ and try to solve for $t$.  We get $$x^{p}\equiv (2+tp)^{p}\equiv 2^{p}\pmod {p^2}$$

Note that this does not depend on $t$.

Thus we want a prime $p$ such that $$2^{p}\equiv 2 \pmod {p^2}$$

Such primes are known as Weiferich primes and they are staggeringly rare for small numbers.  Indeed, only two examples are known, namely $1093$ and $3511$.  And people have searched rather a lot.  Up to about $5\times 10^{17}$ according to OEIS.


问:
How to work out a Mahalanobis distance
Currently going through my Lecturer's notes and came across the Mahalanobis distance but nothing in the notes on how to work this out and googling hasn't really turned up anything either. My maths isn't great but I've been at it for 20 minutes now trying to come up with the same answer and can't.

The Mahalanobis norm is defined by $D_M^2(x) = (x-\mu)^T C^{-1}(x-\mu)$, and for my calculation, I have $(x-\mu)^T = [-90, -100]$ with covariance matrix

$C=\begin{pmatrix}
0.00025  &  -0.00015 \\
-0.00015 &  0.00025
\end{pmatrix}$

Thanks in advance for your help.

答:
I found myself factorising things to make the computation easier.

We have $x-\mu=-10[9\ 10]^T$ and $C=0.00005\begin{bmatrix}5&-3\\-3&5\end{bmatrix}$. So $C^{-1}=0.00005^{-1}\times \frac1{16}\begin{bmatrix}5&3\\3&5\end{bmatrix}$ and so $$(x-\mu)^TC^{-1}(x-\mu)= 0.00005^{-1} \times 1/16 \times (-10) \times (-10) \times [9\ 10]\begin{bmatrix}5&3\\3&5\end{bmatrix}\begin{bmatrix}9\\10\end{bmatrix}=125000 \times [9\ 10]\begin{bmatrix}75\\77\end{bmatrix}=125000 \times 1445=180625000.$$ So $D=\sqrt{180625000}\approx 13440$.

[Maybe you should look up matrix multiplication and inversion of $2\times 2$ matrices, since that's all that's going on here].

Title: How to work out a Mahalanobis distance


问:
Monty Hall problem vs. roulette systems - how are they different?
So I got interested in the Monty Hall problem - I understand what it's about, but somehow I can't wrap my head around the idea of the final choice not being 50/50. More precisely: we all know (or maybe most of us do) the roulette progression systems which were popular a while ago - they told you that you should wait for a streak of some three-or-more consecutive reds or blacks and then bet on the contrary, always staying with this colour and eventually it has to be drawn as with every other spin, the chances become higher for the streak to be broken.

Of course the system is dumb and was shortly afterwards popularly neglected as a fallacy (though surprisingly many people still believe it works) as every spin is independent and even though you had a million reds in a row, in the next spin the odds for red, black and green are still the same as they had been these million spins ago.

And here I see an analogy to Monty Hall - sure, if we were proposed "Do you choose doors k and l (both at the same time!) or m?", it would be a straightforward 2/3 to 1/3. But why don't we cast aside the fact of some door having a goat and treat the two remaining doors as a simple 50/50 just like we do in roulette? Why aren't we interested in the past in roulette while it does matter in Monty Hall? Isn't the two remaining doors - one with a car - just "another spin"?

答:
Think of it like this.

Imagine the croupier has 2 roulette wheels; one with 12 red and 24 black slots and one with 24 red and 12 black (you can have 1 or 2 green if you like).

BEFORE you place your bet, you do not know which wheel the croupier will use (neither does he).

You place your bet and if you choose red, the croupier will tell you he is going to use wheel 1; vice-versa if you choose black.

THEN you are offered the opportunity to change your bet.

This is EXACTLY the same situation as the Monty Haul problem.


问:
The convergence of the series $\sum\limits_{n=3}^\infty \frac {1} {n^2\ln(\ln(n))}$
Is the series $$\sum\limits_{n=3}^\infty \frac {1} {n^2\ln(\ln(n))}$$ convergent?

答: It is convergent as it is positive and bounded above by $\sum 1/n^2$ which  is already convergent as can be proven using the integral test.


问:
Trivial integration...
Solve,

$$
\int\frac{1}{1+x\ln^{2}x}dx
$$

I have tried using $ \phi = e^{x} $ as a substituent... didn't quite work out for me. I don't need an answer, I just need hints. Yes, I've tried searching up online but to no avail.

答:
What makes it seem "trivial" to you? In fact it is not trivial at all, for that it doesn't have elementary close forms.

Using substitution $t=\ln x$ and the original integral(denoted as $I$) becomes
$$
I = \int \frac{1}{e^t t^2+1} \, d(e^t) = \int \frac{e^t}{e^t t^2+1} \, dt = \int \frac{1}{t^2+e^{-t}} \, dt
$$

Substitute $p=-t$ and we get
$$
I = -\int \frac{1}{p^2+e^p} \, dp
$$
Which has no elementary closed forms according to answers in this question.

Title: Trivial integration...


问:
Showing properties of the composition of a Brownian motion and a continuous function.
Let $f$ be a continuous function, and $B_t$ be a Brownian motion starting at $0$ at time $t=0$ on the filtered probability space $(\Omega, \mathcal{F}, P)$.

I am trying to establish (or find a contradiction to the claims) that $f(B_t)$ is adapted to the filtration $\mathcal{F}_t$ generated by $B_t$ and that $E[\int_{0}^{T}f(B_s)^2ds]<\infty$ P-a.s.

To show that $f(B_t)$ is adapted, i need to show that $f(B_t)$ is $\mathcal{F}_t$-measurable for all $t$. This is to show that for each borel set $B \in \mathbb{B}(\mathbb{R})$ we have
\begin{align}
\{\omega \in \Omega \ | \ f(B_t) \in B\} \in \mathcal{F_t}.
\end{align}

The argument I have in mind for this part, is to use the fact that $(f\circ B_t)^{-1} = B_t^{-1} \circ f^{-1}$. Since we have that for each Borel set $B$, $f^{-1}(B)$ is an open set and hence itself a Borel set, the measurability follows by the measurability of $B_t$. This is to say,

\begin{align}
\forall B \in \mathbb{B}(\mathbb{R}), \ \{\omega \in \Omega \ | \ f(B_t) \in B\} \in \mathcal{F_t}.
\end{align}

However, I am not finding a way with the square integrability part. I had hoped to use the fact that the brownian motion has continuous paths P-a.s to bound $|\sup_{s \in [0, t]}(B_s)(\omega') -\inf_{s \in [0, t]}(B_s)(\omega')|$ for fixed omega. Then, if this holds for almost all $\omega'$, perhaps I can show that $E[\int_{0}^{T}f(B_s)^2ds]<\infty$?

Perhaps the arguemnt is not sound or helpful?

答:
Unfortunately it's false.

Consider $f: \mathbb{R}\to \mathbb{R}$ defined as $f(x) = e^{\frac{ax^2}{2}}$ for a certain constant $a>0$.

Then, $\mathbb{E}[f(B_s)^2] = \mathbb{E}[e^{B_s^2}] = \int_\mathbb{R} \frac{1}{\sqrt{2\pi}}e^{asx^2-\frac{x^2}{2}} dx$, this is $+\infty$ if $s\geq \frac{1}{2a}$.

So, $\mathbb{E}[\int_0^Tf(B_s)^2ds] = \int_0^T \mathbb{E}[f(B_s)^2] ds = +\infty$ whenever $T>\frac{1}{2a}$.

However, as you mentioned, you can fairly easily (using the continuity of the Brownian Motion) show that $\mathbb{P}(\int_0^Tf(B_s)^2ds = +\infty) = 0$.


问:
Reducing $2018^{2018}\pmod6$ via $\mod2$ and $\mod3$
I've been tasked to reduce $$2018^{2018}\pmod6$$using the results of reducing$$2018^{2018}\pmod2$$and $$2018^{2018}\pmod3$$

I have deduced that $$2018^{2018}\equiv0\pmod2$$and $$2018^{2018}\equiv1\pmod3$$but I don't see a way to combine these two congruences into one $\mod6$ congruence. Could someone tell me how this is achieved? I have a feeling the answer is really simple, but I just don't see it. Is there also a more generalized strategy of solving modular congruences via the divisors of the modulo number($2$ and $3$ are divisors of $6$)?

The way I have currently solved this problem is to first reduce $$2018^{2018}\pmod6$$ to $$2^{2018}\pmod6$$Then we look for patterns:$$\begin{align}2^1&\equiv2\pmod6\\2^2&\equiv4\pmod6\\2^3&\equiv2\pmod6\\2^4&\equiv4\pmod6\\&\ \ .\\&\ \ .\\&\ \ .\\2^n&\equiv\begin{cases}4&\text{if }n\equiv0\pmod2\\2&\text{if }n\equiv1\pmod2\end{cases}\pmod6\end{align}$$for $n\in\Bbb N$. In the case of $2018^{2018}$, $2018\equiv0\pmod2$, so $$2018^{2018}\equiv2^{2018}\equiv4\pmod6$$

答:
I have deduced that $2018^{2018}\equiv0\pmod2$ and $2018^{2018}\equiv1\pmod3$, but I don't see a way to combine these two congruences into one $\bmod6$ congruence

One way that works in this situation

(besides trial and error and construction, which others have mentioned)

is to note that this is the constant case of the Chinese remainder theorem:

$x\equiv0\bmod2$ and $x\equiv1\bmod3 \iff x\equiv-2\bmod2$ and $x\equiv-2\bmod3$

$\iff  x\equiv-2\bmod6 \iff x\equiv4\bmod6$.


问:
Question on basic functions?
I was having trouble with this question:

"If $f(x+3) = (x-1)^2 + 4$, find $f(a-1)$

I think this is simple, but I've completely forgotten what to do. :P
Thanks!

Title: Question on basic functions?

Body: $$f(a-1)=f((a-4)+3):=((a-4)-1)^2+4\;\ldots\ldots$$


问:
Show that a portion of a contour integral is zero
I am trying to integrate the function $g(z):=\frac{\cos(\pi z)}{z^2\sin(\pi z)}$ over the square with vertices $(\pm 1\pm i)(N+1/2)$ as $N\to \infty$.

For the top and bottom sides, plugging in $z=(N+1/2)x\pm (N+1/2)i$, $\cos(\pi z)\approx 1/2e^{-i(N+1/2)x\pi}e^{(N+1/2)\pi)}$, $\sin(\pi z)\approx -1/(2i)e^{-i(N+1/2)x\pi}e^{(N+1/2)\pi)}$, so the integral become $\int_{-(N+1/2)}^{N+1/2} \frac{1}{((N+1/2)(x+i))^2}dx$ which goes to zero.

For the left and right sides, I am doing a similar thing, but it is even more sketchy because I have to argue that $e^{(N+1/2)\pi x}$ terms dominate $\cos$ and $\sin$ and then cancel in the fraction so we end up with a $1/x^2$ kind of thing again. Thus the whole integral is zero.

This all seems very messy and unappealing, and I'm not even completely convinced it is correct. Is there a better way of doing this problem? The whole point of the problem is to prove the $\sum 1/k^2=\pi^2/6$ identity, so I believe I do actually need to directly prove the integral is zero, but this just seems messy.

答:
Let us consider again the argument. To make things "convincing", it is a good idea to let $\exp$ (alone) enter the stage.

The problem is related to the estimation of the function
$$
z\mapsto
h(z)=\cot (\pi z)
=\frac{\cos (\pi z)}{\sin (\pi z)}
=2i\cdot\frac{e^{\pi iz}+e^{-\pi iz}}{e^{\pi iz}-e^{i\pi iz}}
=2i\cdot\frac{e^{2\pi i z}+1}{e^{2\pi i z}-1}\ .
$$

Let us estimate $h$ at a point of the shape $w=\left(N+\frac 12\right)+iy$, $N\in\Bbb Z$, $y\in \Bbb R$. Since $\cot$ is odd, it is enough to estimate for an $y\ge 0$.
Then $\exp(2\pi i w)= \exp(\pi i-2\pi y)=-\exp(2\pi y)$. This gives
$$
|h(w)|=2
\left|\frac{e^{2\pi i w}+1}{e^{2\pi i w}-1}\right|
=
2\frac{1-e^{-2\pi y}}{1+e^{2\pi y}}
\le 2\ .
$$

Let us estimate $h$ at a point of the shape $w=x-Mi$, $M\in\frac 12+\Bbb Z$, $x\in \Bbb R$. Since $\cot$ is odd, it is enough to estimate for an $M> 0$.
Then $\exp(2\pi i w)= \exp(2\pi i x+2\pi M)=u\exp(2\pi M)$, where $u$ is a number of modulus one. This gives
$$
|h(w)|= 2
\left|\frac{e^{2\pi i w}+1}{e^{2\pi i w}-1}\right|
\le
2
\frac{|e^{2\pi i w}|+|1|}{|e^{2\pi i w}|-|1|}
=
2\frac{e^{2\pi M}+1}{e^{2\pi M}-1}
=
2 + \frac{4}{e^{2\pi M}-1}
\ .
$$
In OP we have $M=N+\frac 12$.

So $h$ is uniformly bounded. The factor $\frac 1{z^2}\in O(N^{-2})$ is stronger than the contour length, which is $O(N)$, so the integral converges to $0$ for $N\to\infty$.


问:
how can I show that for projection matrix $P^2 = P$ without applying trigonometry
I am following MIT 18.06 lectures on linear algebra and would like to understand how one can show that $P^2 = P$ for a projection matrix.

I understand that for 2-d case:

$P = \frac{a a^t}{a^t a}$

$P^2 = \frac{a a^t}{a^t a} \frac{a a^t}{a^t a}$

I assume I should easily show it by using some properties of transpose matrices/vectors, but I couldn't quite figure out how to derive it.
(I am familiar with the proof that relies on trigonometric formulas)

答:
A projection matrix in dimension $n$ is similar to an always vanishing matrix except that it has $p \le n$ $1$ for the first $p$ diagonal entries.

And for such a matrix it is clear that $P^2=P$.


问:
Necessary and sufficient condition for weak convergence of gamma distribution
Let $(X_n)_n$ be a sequence of random variable such that $f_{X_n}(x)=\frac{1}{\Gamma(\alpha_n)}\lambda_n^{\alpha_n}x^{\alpha_n-1}e^{-\lambda_nx}1_{]0,+\infty[}(x),$ where $\alpha_n>0,\lambda_n>0.$

Suppose that $\alpha_n=1,\forall n \in \mathbb{N}.$ Find a necessary and sufficient condition on $(\lambda_n)_n$ such that $(X_n)_n$ converges in distribution.

More generally, find a necessary and sufficient condition on $\alpha_n,\lambda_n$ so that $(X_n)_n$ converges in distribution.

The first part is easy, it converges in distribution if and only if $0<\liminf_n\lambda_n=\limsup_n\lambda_n$.

Concerning part 2), is it true that a condition of weak convergence is the convergence of $(\alpha_n)$ and $(\lambda_n)$?

答:
This comment is the second part of my previous comment,
I write it separately for the convenience of typing.

What have we proved already? In case, when $\alpha_n \to 0$ or $\lambda_n \to 0$ or they both tend to zero, the answer is given by the previous theorem. Suppose now that $\alpha_n$ and  $\lambda_n$ are separated from zero. A case when both of them are separated from $\infty$  was considered also --- in this case we proved that $ \Bigl( X_n \overset{w}{\to} \Bigr) \Longleftrightarrow \exists \lim_n \lambda_n$ and  $\exists \lim_n \alpha_n$. Now we will deal with the last case, when $\alpha_n$ and  $\lambda_n$ are separated from zero and at least one of them is not separated from $\infty$.

Theorem 2.

Suppose that $\alpha_n\ge c_1 > 0$ and $\lambda_n \ge c_1 > 0$.

If $\alpha_n \le c_2$ and $\lambda_n \to \infty$ then $X_n  \overset{w}{\to} 0$.

If $\alpha_n \to \infty$ and $\lambda_n  \le c_2$ then $X_n$ diverges.

If  $\alpha_n \to \infty$ and $\lambda_n \to \infty$ then
$$X_n  \overset{w}{\to} \Longleftrightarrow \exists \lim_n \frac{\alpha_n}{\lambda_n},$$
and if $X_n$ is convergent then $X_n  \overset{w}{\to} \lim_n \frac{\alpha_n}{\lambda_n}$.

Let us prove the theorem 2.

The first and the second statements may be proved similarly to the proof of Theorem 1.

Consider the most interesting case: $\alpha_n \to \infty$ and $\lambda_n \to \infty$. Here and everywhere below we will assume that this condition holds.

As above, we have
$$Ee^{itX_n} = {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2} \cdot e^{i \alpha_n arctg \frac{t}{\lambda_n} } = e^{- \frac{\alpha_n}2 \cdot \frac{t^2}{\lambda_n^2}(1+\bar{o}(1))}  \cdot e^{i  t \frac{\alpha_n}{\lambda_n} (1+\bar{o}(1))}. $$

Suppose that $X_n$ is convergent. Hence $|Ee^{itX_n}|$ converges to a continuous function. But $|Ee^{itX_n}| = e^{- \frac{\alpha_n}2 \cdot \frac{t^2}{\lambda_n^2}(1+\bar{o}(1))}  $ thus $ \frac{\alpha_n}{\lambda_n^2} \to A \in [0, \infty)$. Moreover, $Ee^{itX_n}$ converges to a continuous function. Hence $e^{i  t \frac{\alpha_n}{\lambda_n} (1+\bar{o}(1))}$ converges to a continuous function. It means that $\xi_n = \frac{\alpha_n}{\lambda_n}$ converges weakly to some $\xi$. Put $F_{\xi_n}(x) = P(\xi_n \le x)$, $F_{\xi}(x) = P(\xi_n \le x)$. We know that $F_{\xi_n}(x) \to F_{\xi}(x) $ for all $x$ except at most countable set. Hence, $F_{\xi}(x)$ is a distribution function of some constant. It's easy to see that $\exists \lim_n \frac{\alpha_n}{\lambda_n} < \infty$ and this limit is equal to $\xi$. So we proved that

$$ \Bigl( X_n  \overset{w}{\to} \Bigr) \Longrightarrow \Bigl(  \exists \lim_n \frac{\alpha_n}{\lambda_n} \Bigr).   $$

Now suppose that $\exists \lim_n \frac{\alpha_n}{\lambda_n} = B < \infty$. Thus $\exists \lim_n \frac{\alpha_n}{\lambda_n^2} = 0$.
We know that $Ee^{itX_n} = e^{- \frac{\alpha_n}2 \cdot \frac{t^2}{\lambda_n^2}(1+\bar{o}(1))}  \cdot e^{i  t \frac{\alpha_n}{\lambda_n} (1+\bar{o}(1))}$ hence $Ee^{itX_n} = e^{i t B (1+\bar{o}(1))}$.

Thus
$$\Bigl(\exists \lim_n \frac{\alpha_n}{\lambda_n} = B \Bigr)  \Longrightarrow \Bigl(X_n  \overset{w}{\to} B\Bigr),$$
q.e.d.

Finally, let us present an informal proof of the biggest part of the theorem 2: the case when $\alpha_n \to \infty$.
It's well known that $\Gamma(\lambda, m)$ is a sum of $m$ independent exponential random variables $\exp{\lambda}$ with parameter $\lambda$. The following symbolic equality is fulfilled: $\exp{\lambda} = \frac{\exp{1}}{\lambda}$, it means that $\lambda$ is a scale parameter. Hence in case  $\alpha_n \to \infty$. we have
\begin{gather}
X_n  \overset{d}{=}  \Gamma(\lambda_n, \alpha_n) \approx \sum_{i=1}^{[\alpha_n]} \exp{\lambda_n} = \sum_{i=1}^{[\alpha_n]} \frac{\exp{1}}{\lambda_n} \nonumber =\\
=\frac{\sum_{i=1}^{[\alpha_n]} \exp{1} }{[\alpha_n]} \cdot \frac{[\alpha_n]}{\lambda_n} = \text{S.L.L.N.} = (1+\bar{o}(1)) \cdot \frac{[\alpha_n]}{\lambda_n} \approx \frac{\alpha_n}{\lambda_n}. \nonumber
\end{gather}
Now it's easy to see that if $\lambda_n \le c_2$ then $X_n$ diverges and if $\lambda_n \to \infty$ then $\Bigl(X_n  \overset{w}{\to}\Bigr)$ iff  $\exists \lim_n \frac{\alpha_n}{\lambda_n}$. Moreover, if $X_n$ is convergent then $X_n  \overset{w}{\to} \lim_n \frac{\alpha_n}{\lambda_n}$.

Now it's easy to see construct "bad" sequences $\alpha_n$ and $\lambda_n$ such that $X_n$ converges. For example, put

\begin{gather}
\alpha_{3n} = \frac1{n}, \lambda_{3n} = 2+(-1)^n,\\
\alpha_{3n+1} = 2+(-1)^n, \lambda_{3n+1} = n,\\
\alpha_{3n+2} = \frac1{n}, \lambda_{3n+2} = \frac1{n}.
\end{gather}

It follows from Theorem 1 and Theorem 2 that $X_n \overset{d}{\to} 0$.


问:
Random probability for internet connection
An internet service provider has two connection lines for its customers.

$80\%$ of customers are connected through Line $I$, the rest through Line $II$.

Line $I$ has a $\operatorname{Gamma}(3, 1/2)$ connection time (in minutes), while Line $II$ has a $U(20, 50)$ connection time (in seconds).

Find the probability that it takes a randomly selected customer more than $30$ seconds to connect to the internet.

Can somebody give me some ideas, please?

答:
Using @tommik's (+1) notation:

Assuming the second gamma parameter is rate $1/2$  (the parameterization used in R, where pgamma is a gamma CDF) we have $p_1 = 0.9978.$

```<code>p.1 = 1 - pgamma(.5, 3, .5);  p.1
[1] 0.9978385
</code>```

Also, $p_2 = 0,6667.$

```<code>p.2 = 1 - punif(30, 20, 50);  p.2
[1] 0.6666667
</code>```

So the answer is $0.9316.$

```<code>.8*p.1 + .2*p.2
[1] 0.9316041
</code>```

Note: If the second gamma parameter is scale $1/2,$ then
$p.1s$ and the answer are shown below:

```<code>p.1s = 1 - pgamma(.5, 3, 2);  p.1s
[1] 0.9196986
.8*p.1s + .2*p.2
[1] 0.8690922
</code>```

Simulation of a million randomly chosen customers, using the scale=0.5 and rate=2 for gamma is shown below.
Answers agree with analytic answers above to about 2 or 3
significant digits.

```<code>set.seed(1211)
w1 = rgamma(10^6, 3, 2)
summary(w1)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.006194 0.862998 1.336863 1.499468 1.960694 9.960564
mean(w1 > .5)
[1] 0.919711
w2 = runif(10^6, 2/6, 5/6)
summary(w2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.3333  0.4586  0.5837  0.5835  0.7085  0.8333
mean(x2 > .5)
[1] 0.666537
b = rbinom(10^6, 1, .8)
mean(b)
[1] 0.799963
w.c = b*x1 + (1-b)*x2
mean(w.c > .5)
[1] 0.869234
summary(w.c)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.005825 0.661970 1.097923 1.317951 1.775212 9.690274
</code>```

Histograms for simulated waiting times are shown below. Vertical
red lines are at 0.5 min.

R code for figure:

```<code>par(mfrow=c(3,1))
 hist(w1, prob=T, br=60, xlim=c(0,10))
  abline(v=.5, col="red")
 hist(w2, prob=T, br=seq(2/6,5/6,len=10), xlim=c(0,10))
  abline(v=.5, col="red")
 hist(w.c, prob=T, br=60, xlim=c(0,10))
  abline(v=.5, col="red")
par(mfrow=c(1,1))
</code>```

Body:
Using @tommik's (+1) notation:

Assuming the second gamma parameter is rate $1/2$  (the parameterization used in R, where pgamma is a gamma CDF) we have $p_1 = 0.9978.$

p.1 = 1 - pgamma(.5, 3, .5);  p.1
[1] 0.9978385

Also, $p_2 = 0,6667.$

p.2 = 1 - punif(30, 20, 50);  p.2
[1] 0.6666667

So the answer is $0.9316.$

.8*p.1 + .2*p.2
[1] 0.9316041

Note: If the second gamma parameter is scale $1/2,$ then
$p.1s$ and the answer are shown below:

p.1s = 1 - pgamma(.5, 3, 2);  p.1s
[1] 0.9196986
.8*p.1s + .2*p.2
[1] 0.8690922

Simulation of a million randomly chosen customers, using the scale=0.5 and rate=2 for gamma is shown below.
Answers agree with analytic answers above to about 2 or 3
significant digits.

set.seed(1211)
w1 = rgamma(10^6, 3, 2)
summary(w1)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.006194 0.862998 1.336863 1.499468 1.960694 9.960564
mean(w1 > .5)
[1] 0.919711
w2 = runif(10^6, 2/6, 5/6)
summary(w2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
 0.3333  0.4586  0.5837  0.5835  0.7085  0.8333
mean(x2 > .5)
[1] 0.666537
b = rbinom(10^6, 1, .8)
mean(b)
[1] 0.799963
w.c = b*x1 + (1-b)*x2
mean(w.c > .5)
[1] 0.869234
summary(w.c)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max.
0.005825 0.661970 1.097923 1.317951 1.775212 9.690274

Histograms for simulated waiting times are shown below. Vertical
red lines are at 0.5 min.

R code for figure:

par(mfrow=c(3,1))
 hist(w1, prob=T, br=60, xlim=c(0,10))
  abline(v=.5, col="red")
 hist(w2, prob=T, br=seq(2/6,5/6,len=10), xlim=c(0,10))
  abline(v=.5, col="red")
 hist(w.c, prob=T, br=60, xlim=c(0,10))
  abline(v=.5, col="red")
par(mfrow=c(1,1))


问:
Find $\lim\limits_{n\rightarrow \infty} \left| \ln \frac{1}{\left| x_n-A \right|} \right|^{1/n}$
The sequence $\{x_n\}$ satisfies $x_{n+1}=x_n+\sin x_n,n=1,2,\cdots $.

Prove $\{x_n\}$ is convergent and find its limit $A$.

Let $x_1\neq A$. Find the limit $$\lim_{n\rightarrow \infty} \left| \ln \dfrac{1}{\left| x_n-A \right|} \right|^{1/n}$$

My Attempt: For the first question, let $k\in \mathbb Z$ and $f(x)=x+\sin x$. When $x=k\pi$, it is trivial that $A=k\pi$. When $x\in (k\pi,(k+1)\pi)$, since $f(x)\in (k\pi,(k+1)\pi)$ and $f(x)$ monotonously increases on $\mathbb R$, obviously $\{x_n\}$ monotonously increases and is bounded from above. Take the limits of both sides of the recurrence, and we get $A=A+\sin A$, so $A=(k+1)\pi$.

My Question is, did I correctly do the first question? And how to solve the second one? I don't have any ideas yet.

答:
The sequence $\;\{x_n\}_{n\in\mathbb{N}}\;$ satisfies

$x_{n+1}=x_n+\sin x_n\quad$ for all $\;n\in\mathbb{N}\;.$

There are four possibilities:

1) $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\quad$ where $\;k\in\mathbb{Z}\;,$

2) $\;x_1\in\big]\pi+2k\pi,2\pi+2k\pi\big[\quad$ where $\;k\in\mathbb{Z}\;,$

3) $\;x_1=2k\pi\quad$ where $\;k\in\mathbb{Z}\;,$

4) $\;x_1=\pi+2k\pi\quad$ where $\;k\in\mathbb{Z}\;.$

First case

If $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;$ where $\;k\in\mathbb{Z}\;,\;$ then $\;2k\pi<x_n<x_{n+1}<\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$

I am going to prove it by induction.

Since $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that

$\begin{align}
2k\pi&<x_1<x_1+\sin x_1=x_2=f(x_1)<f\left(\pi+2k\pi\right)=\\
&=\pi+2k\pi\;.
\end{align}$

Now I suppose that $\;2k\pi<x_{n-1}<x_n<\pi+2k\pi\;$ and prove that $\;2k\pi<x_n<x_{n+1}<\pi+2k\pi\;.$

Since $\;x_n\in\big]2k\pi,\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that

$\begin{align}
2k\pi&<x_n<x_n+\sin x_n=x_{n+1}=f(x_n)<f\left(\pi+2k\pi\right)=\\
&=\pi+2k\pi\;.
\end{align}$

So I have proved by induction that

$2k\pi<x_n<x_{n+1}<\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$

Consequently,

$\exists\lim\limits_{n\to\infty}x_n=A_1=\sup\limits_{n\in\mathbb{N}}\{x_n\}\in\big]2k\pi,\pi+2k\pi\big]\;.$

Moreover,

$\lim\limits_{n\to\infty}x_{n+1}=\lim\limits_{n\to\infty}x_n+\lim\limits_{n\to\infty}\sin x_n\;,$

$A_1=A_1+\sin A_1\;,$

$A_1=\pi+2k\pi\;.$

So far I have proved that

if $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;,$ then $\;\exists\lim\limits_{n\to\infty}x_n=A_1=\pi+2k\pi\;.$

Second case

If $\;x_1\in\big]\pi+2k\pi,2\pi+2k\pi\big[\;$ where $\;k\in\mathbb{Z}\;,\;$ then $\;\pi+2k\pi<x_{n+1}<x_n<2\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$

I am going to prove it by induction.

Since $\;x_1\in\big]\pi+2k\pi,2\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that

$\begin{align}
\pi+2k\pi&=f\left(\pi+2k\pi\right)<f(x_1)=x_1+\sin x_1=\\
&=x_2<x_1<2\pi+2k\pi\;.
\end{align}$

Now I suppose that $\;\pi+2k\pi<x_n<x_{n-1}<2\pi+2k\pi\;$ and prove that $\;\pi+2k\pi<x_{n+1}<x_n<2\pi+2k\pi\;.$

Since $\;x_n\in\big]\pi+2k\pi,2\pi+2k\pi\big[\;$ and the function $\;f(x)=x+\sin x\;$ is monotonically increasing in $\;\left]-\infty,+\infty\right[\;,\;$ it follows that

$\begin{align}
\pi+2k\pi&=f\left(\pi+2k\pi\right)<f(x_n)=x_n+\sin x_n=\\
&=x_{n+1}<x_n<2\pi+2k\pi\;.
\end{align}$

So I have proved by induction that

$\pi+2k\pi<x_{n+1}<x_n<2\pi+2k\pi\quad$ for any $\;n\in\mathbb{N}\;.$

Consequently,

$\exists\lim\limits_{n\to\infty}x_n=A_2=\inf\limits_{n\in\mathbb{N}}\{x_n\}\in\big[\pi+2k\pi,2\pi+2k\pi\big[\;.$

Moreover,

$\lim\limits_{n\to\infty}x_{n+1}=\lim\limits_{n\to\infty}x_n+\lim\limits_{n\to\infty}\sin x_n\;,$

$A_2=A_2+\sin A_2\;,$

$A_2=\pi+2k\pi\;.$

So far I have proved that

if $\;x_1\in\big]2k\pi,\pi+2k\pi\big[\;,$ then $\;\exists\lim\limits_{n\to\infty}x_n=A_2=\pi+2k\pi\;.$

Third case

If $\;x_1=2k\pi\;,$ then $\;x_n=2k\pi\;$ for any $\;n\in\mathbb{N}\;.$

Consequently,

$\exists\lim\limits_{n\to\infty}x_n=A_3=2k\pi\;.$

Fourth case

If $\;x_1=\pi+2k\pi\;,$ then $\;x_n=\pi+2k\pi\;$ for any $\;n\in\mathbb{N}\;.$

Consequently,

$\exists\lim\limits_{n\to\infty}x_n=A_4=\pi+2k\pi\;.$

Hence,

if $\color{blue}{\;x_1\in\big]2k\pi,2\pi+2k\pi\big[\;}$ where $\color{blue}{\;k\in\mathbb{Z}\;}$, then $\color{blue}{\;\exists\lim\limits_{n\to\infty}x_n=A=\pi+2k\pi\;}$.

if $\color{blue}{\;x_1=2k\pi\;}$ where $\color{blue}{\;k\in\mathbb{Z}\;}$, then $\color{blue}{\;\exists\lim\limits_{n\to\infty}x_n=A^*=2k\pi\;}$.


问:
Surjective homomorphism between finite groups
I have to prove that, if $G$ and $H$ are finite groups and $f: G \rightarrow H$  an epimorphism (an surjective homomorphism)  then $|Ker(f)| = n \Rightarrow |f^{-1}[\{h\}]|=n$, $\forall h \in H$

Given an $ h \in H$ as f is surjective then $\exists g \in G $ such that $f(g)=h$ and my idea is to show that $gKer(f) = f^{-1}[\{h\}]$, is clear that if $a \in gKer(f) \Rightarrow a = gb $ for some $b \in ker(f)$ hence $f(a) = f(gb) = f(g)f(b) = h \Rightarrow a \in f^{-1}[\{h\}]$ since $b \in Ker(f) \Rightarrow f(b) = id$

$\therefore gKer(f) \subseteq f^{-1}[\{h\}]$ I need a hint to prove $f^{-1}[\{h\}] \subseteq gKer(f)$

Body: Take $a \in f^{-1}[\{h\}]$. Then $f(a) = h = f(g)$ and so $f(g^{-1}a) = f(g^{-1})f(a) = f(g)^{-1}f(a) = 1$, that is, $g^{-1}a \in \ker f$, which means $a \in g \ker f$.


问:
Moment Generating Function of $f(x) = \begin{cases} -2e^{-2x}+2e^{-x} &: 0 <x\\ 0 &: \text{else} \end{cases}$
I want to find the moment generating function of the following pdf to verify that $\mu=\frac{3}{2}$.

$f(x) = \begin{cases}
-2e^{-2x}+2e^{-x} &: 0 <x\\
0 &: \text{else}
\end{cases}$

So far this is what I have done:

$M_{X}(t)=E\left(e^{tX}\right)\int^{\infty}_{0} e^{tx}f(x) dx=\int^{\infty}_{0}e^{tx}\left(−2e^{-2x}+2e^{-x}\right)dx=\int^{\infty}_{0}-2e^{tx-2x}+2e^{tx-x}dx$

but after this is I don't know where to take the integral. Any help would be much appreciated.

Body:
Fix $t<1$. Then compute

\begin{align*}
M_{X}(t) := \mathbb{E} \left [e^{tX} \right] &= \int_{0}^{\infty} e^{tx}(2e^{-x}-2e^{-2x})\, dx \\
&= \frac{2}{t-1}e^{x(t-1)} - \frac{2}{t-2}e^{x(t-2)} \bigg |_{x=0}^{x=\infty} \\
&= \lim\limits_{x \rightarrow \infty} \left (\frac{2}{t-1}e^{x(t-1)} - \frac{2}{t-2}e^{x(t-2)} \right ) \\
&\quad - \left (\frac{2}{t-1}e^{x(t-1)} - \frac{2}{t-2}e^{x(t-2)} \right ) \bigg |_{x=0} \\
&= \frac{2}{t-2} - \frac{2}{t-1}
\end{align*}

Then the mean is given by $\mu = M_{X}'(0) = -\frac{2}{(t-2)^{2}} + \frac{2}{(t-1)^{2}} \bigg |_{t=0} = -\frac{1}{2} + 2 = \frac{3}{2}$ as desired.


问:
Show tilting measure is martingale
I am doing an exercise regarding the Girsanov theorem, $B_t$ is standard brownian motion. I already find that
$$M_t=\exp(Y_t)\quad Y_t=\int_{0}^{t}-2mB_sdB_s-\frac{1}{2} \int_{0}^{t}4m^2B_s^2ds   $$
$dM_t=(-2mB_t)M_tdB_t$, $M_t$ is already a local martingale. I need to further show that $M_t$ is actually a martingale, there are two ways(based on my knowledge) to show martingale. If we can show any of following:

$\mathbb{E}\left[M_{t}\right]=1$

$\mathbb{E}\left[\exp \left\{\frac{\langle Y\rangle_{t}}{2}\right\}\right]<\infty$

I want to try second way, we need to show
$$\mathbb{E}\left[\exp \left\{\frac{1}{2}\int_{0}^{t}4m^2B_s^2ds\right\}\right]<\infty$$

But I don't know how to proceed, can anybody help?

答:
I'd do it using Kazamaki's criterion.

We want to show that for any $t>0$, $\mathbb{E}[e^{\frac{1}{2}\int_0^t-2mB_sdB_s}]<+\infty$.

This is equivalent to $\mathbb{E}[e^{-m\int_0^tB_sdB_s}] < +\infty$.

It should be easy to conclude for $m>0$ using that $\int_0^t B_sdB_s = \frac{B_t^2-t}{2}$.

Title: Show tilting measure is martingale


问:
Christmas Tree Problem: String around Cone
This problem came up when I was trying to figure out how long of a string of lights I need for my Christmas tree. It was easy to estimate and obviously this isn't for use in the real world, but I found it a lot harder than I thought it would be at first.

A cone has a base radius $3$ ft and height of $8$ ft. A line ("string") is wrapped around the cone so that if you go straight up from any point on the string along the surface of the cone, the vertical distance (not distance along the cone's surface) to the next intersection of the string is $1$ ft.

The string is wrapped counterclockwise, looking from the top. As a more stringent restriction, a section of the string that encompasses some angle $\theta$ around the surface of the cone, the change in height from the bottom point to the top point is given by:
$$h = \frac{\theta}{2\pi}$$

What is the length of the string?

My first try at this problem assumed that the angle of the string relative to the horizontal would always be a constant. I quickly realized the string would have to get steeper as it goes up the cone ("tree"), in order to keep up with the shrinking circumferennce. I pretty much banged my head into a wall from there because I have no clue how to factor that in.

答:
To go from the bottom of the tree to the top $h$ increases from $0$ to $8$ feet, so  $\theta$ increases from $0$ to $16\pi$.  We can define $r(\theta)$ as the radius at a given $\theta$.  It decreases linearly from $3$ to $0$, so $r(\theta)=3-\frac 3{16\pi}\theta$.  The derivative of arc length with respect to $\theta$ can be found by Pythagoras.  In $\Delta \theta$ we go up $\frac {\Delta \theta}{2 \pi}$ and we go around $r \Delta \theta=\left(3-\frac 3{16\pi}\theta\right)\Delta \theta$  Now we can integrate
$$s=\int_0^{16\pi}\sqrt{\frac 1{4\pi^2}+\left(3-\frac 3{16\pi}\theta\right)^2}d\theta$$
Alpha tells me the integral is about $76.2746$.  I hit the time limit before it would give me the symbolic answer.  This seems reasonable as $8$ turns at the average radius of $1.5$ feet would give $75.4$ feet and the rise will add a bit.


问:
Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$
Let $a_k>0,k=1,2,\cdots, n$. Prove that
$$
\frac{n}{\sum_\limits{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_\limits{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}
$$

My attempt: multiply both sides by the denominators, and it is equivalent to prove:
$$
n\sum_{k=1}^n{\frac{1}{a_k}}-n\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n+1}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right) \\
\sum_{k=1}^n{\frac{1}{a_k}}-\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n\left( n+1 \right)}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right)\\ \sum_{k=1}^n{\left( \frac{1}{a_k}-\frac{1}{\frac{1}{k}+a_k} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\frac{1}{a_k}}\sum_{p=1}^n{\frac{1}{\frac{1}{p}+a_p}}
\\ \sum_{k=1}^n{\left( \frac{\frac{1}{k}}{a_k\left( \frac{1}{k}+a_k \right)} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\sum_{p=1}^n{\frac{1}{a_k\left( \frac{1}{p}+a_p \right)}}}
$$

But how to proceed further?

答:
I gave a solution years ago.

Denote $A = \sum_{k=1}^n \frac{1}{\frac{1}{k} + a_k}$,
$B = \sum_{k=1}^n \frac{1}{a_k}$.

The function $f(x) = \frac{1}{1 + \frac{1}{x}}, \ x > 0$ is concave. By Jensen's inequality, we have
\begin{align}
A &= f(\tfrac{1}{a_1}) + 2f(\tfrac{1}{2a_2})
+ 3(\tfrac{1}{3a_3}) + \cdots + n f(\tfrac{1}{na_n})\\
&\le \frac{n(n+1)}{2} f\Big( \frac{B}{\frac{n(n+1)}{2}}\Big)\\
&= \frac{n(n+1)}{2}\frac{B}{B + \frac{n(n+1)}{2}}.
\end{align}
Then we have
$$\frac{n}{A} - \frac{n}{B} \ge \frac{n}{\frac{n(n+1)}{2}\frac{B}{B + \frac{n(n+1)}{2}}} - \frac{n}{B}
= \frac{2}{n+1}\frac{B + \frac{n(n+1)}{2}}{B} - \frac{n}{B} = \frac{2}{n+1}.$$
We are done.


问:
all $A_i$ are connected sets such that $\bigcap\limits_{i\in E} A_i \neq \emptyset$ then $\bigcup\limits_{i \in E} A_i$ is connected
This is my proof

Suppose not.
Then, $\cup A_i$ has an open partition $\{U,V\}$

$U \subseteq \cup A_i$ so we only need to show two cases:

$U \subseteq \cup A_j$ with $U \neq \cup A_j$ for some $J \subseteq E$. Then there exists some $A_k$ such that $U \neq A_k$ with $U \cap A_k \neq \emptyset$. Thus $\{ U \cap A_k,V \cap A_k \}$ is an open partition of $A_k$. By assumption, $A_k$ is connected. It's a contradiction with [$\cup A_i$ is disconnected]

$U= \cup A_t$ for some $T \subseteq E$. Since $V \neq \emptyset$, there exists some $A_k$ such that $(A_k-U) \neq \emptyset$. Let $J=T \cup \{k\}$. Then by Case 1, It's a contradiction with [$\cup A_i$ is disconnected]

Is it okay??

I'm not sure about this...

答:
There are several things that I don’t understand in your proof. In particular:

$U \neq \bigcup_j A_j$ : on what set is the union performed?

Same things for case 2. with $T$.

I would just say as $$\bigcap_{j \in J} A_j$$ is supposed to be non empty, let’s take $x \in \bigcap_{j \in J} A_j$.

As by hypothesis $$\bigcup_{j \in J} A_j \subseteq U \cap V,$$ we can suppose without loss of generality than $x \in U$ (we can swap the role of $U,V$ in the other case).

Now for any $j \in J$, $A_j$ is supposed to be connected and $x \in A_j$. Therefore $A_ j \subseteq U$ and finally $$\bigcup_{j \in J} A_j \subseteq U$$ proving that the union is connected.


问:
Problem with Discrete Parseval's Theorem
I think I must be missing something obvious, but I can't for the life of me see what it is. The discrete version of Parseval's theorem can be written like this:

$$\sum_{n=0}^{N-1} |x[n]|^2 = \frac{1}{N}\sum_{n=0}^{N-1} |X[k]|^2 $$

Now, say you've got some function in time, like $x = \sin(\omega t)$. Depending on how large your value of $N$, the LHS could be arbitrarily large.

The FT of this function is a delta function at $\omega$. Everywhere else is zero, so your RHS is given by $1/N$. So how on earth does one side equal the other?! It seems to me like you've got a summation one one side and an average value on the other...

I must be missing something obvious!

答:
It seems that you're used to a normalization convention that's different from the one used by the Wikipedia article from which you quoted the theorem. With the convention used in the article,

$$
X_k = \sum_{n=0}^{N-1} x_n \cdot\mathrm e^{-\mathrm i 2 \pi k n / N}\;,
$$

the problem you describe doesn't occur. In your example, exactly one of the $X_k$ is non-zero, and it's proportional to $N$, so the left-hand side and the right-hand side are both proportional to $N$.

OwnerDisplayName: Ned Yoxall


问:
Combinatorics Summation
Let n be an even natural number.

It is required to derive/prove that:

nC1-nC2+nC3-nC4.......+nC(n/2)-........+nC(n-1)=2

(I stumbled upon this while trying to derive the formula for no. of derangements given in my textbook on my own. I checked it on the calculator  for many numberes and it is true all the time.

I have no clue how to proceed to prove this so please help)

答:
The lefthand side is the number of odd-sized subsets of $[n]=\{1,\ldots,n\}$ minus the number of even-sized non-empty, proper subsets of $[n]$. Provided that $n\ge 2$, $[n]$ has the same number of odd-sized subsets as it has even-sized subsets. If you know this, you can simply observe that the lefthand side excludes the even-sized subsets $\varnothing$ and $[n]$, and the result follows immediately.

You can prove the result about even- and odd-sized subsets of $[n]$ by induction on $n$ or by applying the binomial theorem to $(-1+1)^n$.

Title: Combinatorics Summation