问:
Is a sample space really necessary for this question?
A game involving rolling two six-sided dice followed by rolling a third die is played - to win, the number rolled on the third die must lie between the two numbers rolled previously.

(i) What is the probability that a player has no chance of winning before rolling the third die?

(ii) What is the probability that a player wins the game?

I did (i) like this. $P($no chance$) = P($same numbers$) + P($consecutive numbers$)=\frac{1}{6} + \frac{5}{6}\frac{1}{6}$ . $\frac{5}{6}$ because none of the numbers can be 6 for next number to be "larger". This is wrong, but perhaps someone can correct my thinking?

(ii) is trickier and I cannot just do $P = 1 - P($no chance$)$ so I was wondering if there was a way around that.

答:
(i) Your answer is a bit off. The first two dice are distinguishable — imagine one is red and one is green. The term of $\frac{1}{6}$ for the same value is fine, but the term of $\frac{5}{6}\cdot\frac{1}{6}$ only counts the outcomes when the first (red) die has the smaller number and the second (green) die has the next integer. But it could be the other way around, so you need to double this term. In other words: you only counted the outcomes such as $(1,2)$, $(2,3)$, and so on, but you missed outcomes such as $(2,1)$.

(ii) At first, I also thought that brute force counting was the only way to go. And it would be a little tedious, but still doable. After all, there are only six value on all dice. But then, I figured that there's a way around it.

Let's use the classical definition of probability:
$$\mathbf{Pr}\{\text{event}\}=\frac{\text{number of favorable outcomes}}{\text{number of all outcomes}}.$$

For the denominator, we're rolling three standard dice, which gives us … possible outcomes.

For the numerator, we need to count the number of ways of rolling three dice such that the value on the third is between the values on the first two. This requirement implies that the three values are different, and there are $6\choose3$ ways of picking three distinct values. The middle value is what we want to see on the third die. The other two values can appear on the first two dice in either way, so we double this number.


问:
Using definition of limit of Sequence show that, If x>0 then limit of Sequence (Arctan nx) is π/2

Question: I need to show by "using definition of limit of Sequence" that Sequence $(\arctan nx)$ converges to $π/2$ if $x>0$.

My attempt: i need to show for any $\epsilon >0$, and for each $x>0$, we have $|\arctan nx -π/2|<\epsilon$ for sufficiently large $n$. How do i proceed further, unable to solve it. Please help.

答:
Hint: following up on K.defaoite's reply, consider any integer $$N > \frac{\tan(\pi/2 - \varepsilon)}{x}$$ and use that as the lower bound for sufficiently large $n$.

Edit: I've put a full solution in the spoiler box below.

 Let $\varepsilon > 0$. We wish to show that there exists a sufficiently large integer $N$ for which $\lvert\arctan{nx} - \pi/2\rvert < \varepsilon$ for all $n \geq N$.
 Define $\varepsilon_0 = \min(\varepsilon, \pi/4)$, and take $N$ to be any integer such that $$N > \frac{\tan(\pi/2 - \varepsilon_0)}{x}.$$ Then for all $n \geq N$, \begin{align*}\lvert\arctan{nx} - \pi/2\rvert &= \pi/2 - \arctan(nx)\\ &\leq \pi/2 - \arctan(Nx) \\ &< \pi/2 - \arctan\bigg(\frac{\tan(\pi/2 - \varepsilon_0)}{x} \cdot x \bigg) \\ &= \pi/2 - (\pi/2 - \varepsilon_0) \\ &\leq \varepsilon, \end{align*} as desired. Note that the first equality holds because $\pi/2 > \arctan{x}$ for all $x \in \mathbb{R}$, and the inequalities in lines 2 and 3 hold because $\arctan$ is increasing on $\mathbb{R}$.

ContentLicense: CC BY-SA 4.0


问:
Fourier transform of $L^1$ function whose derivative is in $L^1$ and vanishes at infinity is in $L^1$
$f \in L^1(\mathbb{R})$ is a differentiable function such that $f' \in L^1(\mathbb{R}) \cap C_0(\mathbb{R})$, prove that the fourier transform of $f$ noted $\hat{f}$ is in $L^1 (\mathbb{R})$

I know if $f,f'\in L^1(\mathbb{R})$, then $\widehat{f'}(t)=it\hat{f}(t)$ but I have no ideas how to use the condition that the derivative vanish at infinity. Any ideas will be helpful.

答:
Two hints:

Use the fact that $f'$ is bounded to show that $f' \in L^2$ and
the use Plancherel.

 Note that $f'$ is bounded and since$|f'|^2 \le \sup |f'| |f'|$ we see that $f' \in L^2$. Then Plancherel shows that $\hat{f'} \in L^2$. Note that $\hat{f'}(\omega) = i\omega \hat{f(\omega)$.

Use Cauchy Schwartz and note that for $\omega \neq 0$ we have $\hat{f}(\omega) = \omega \hat{f}(\omega) \cdot {1 \over \omega}$.

 For $\omega \neq 0$ we have $|\hat{f}(\omega)| = |\omega||\hat{f}(\omega)| \cdot {1 \over |\omega|}$ and Cauchy Schwartz gives $\int|\hat{f}| = \int |\omega||\hat{f}(\omega)| \cdot {1 \over |\omega|} d \omega = \int |\hat{f'}(\omega)| \cdot {1 \over |\omega|} d \omega \le \| \hat{f'}\| \| \omega \mapsto {1 \over |\omega|} \|$.


问:
Quirky Primes Test
Wrote a program to list all primes $n$ below $1,000,000$ and noticed they never satisfied (except $n=5$)

$\frac {n-1}2 \equiv 2 \pmod{10}$

or

$\frac {n-1}2 \equiv 7 \pmod{10}$

Is that true for all primes? Thanks.

```<code>#include <stdio.h>
#include <stdint.h>

int isPrimeNumber(int number)
{
    int iLoop = 0;
    int iPrimeFlag = 1;

    if (number <= 1)
    {
        iPrimeFlag = 0;
    }
    else
    {
        for (iLoop = 2; iLoop < number; iLoop++)
        {
            if ((number % iLoop) == 0)
            {
                iPrimeFlag = 0;
                break;
            }
        }
    }
    return iPrimeFlag;
}

int main(int argc, char* argv)
{
    uint64_t n, i, m;

    for (n = 2; n < 1000000; n++)
    {
        if (isPrimeNumber(n) == 1)
        {
            i = (n - 1) / 2;
            m = i % 10;

            if (m == 2 || m == 7)
            {
                printf("n=%llu i=%llu\n", n, i);
            }
        }
    }

    return 0;
}
</code>```

答:
Just do it.  Equivalence $\pmod n$ distributes over multiplication, addition and subtraction (but not division) so

$\frac {n-1}2 \equiv 2 \pmod{10}\implies$

$2(\frac {n-1}2) + \equiv 2\cdot 2 + 1 \pmod {10}\implies$

$n \equiv 5 \pmod {10}$

And the definition of $a\equiv b\pmod n$ is $n\mid a-b$ so

$10 \mid n-5$.

Now $10 = 2\cdot 5$ so that means $2|n-5$ and $5|n-5$.

And if $a|b$ then $a|b \pm ka$ for any multiple $ka$ of $a$.

so $5|(n-5) \implies 5|(n-5)+ 5\implies 5|n$.

But if $n$ is prime that's only possible if $n = 5$.

Can you do the other one.

Title: Quirky Primes Test


问:
Proof by induction that $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$
I am a bit new to logical induction, so I apologize if this question is a bit basic.

I tried proving this by induction:

$$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$$

Starting with the base case $n=1$:

$$1^2\ge1^2$$

Then to prove that $P(n)\to P(n+1)$:

$$\left(\sum_{k=1}^{n+1}k\right)^2\ge\sum_{k=1}^{n+1}k^2$$
$$\left((n+1)\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$
$$(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$

Since they're both multiplied by $(n+1)^2$, it's easy to see that if $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$, then $(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$. But if the $\ge$ were replaced with a $\le$, the proof would still be valid, even though it's demonstrably not true.

I can see that it would take strong induction would fix this problem, but if I didn't know by observation that $\left(\sum_{k=1}^nk\right)^2\le\sum_{k=1}^nk^2$ isn't true by observation, how would I know to use strong induction?

答:
You're not quite on the right track. Your base case is just fine. For your induction step, suppose that $$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2.$$ Note then that $$\left(\sum_{k=1}^{n+1}k\right)^2=\left((n+1)+\sum_{k=0}^nk\right)^2=(n+1)^2+2(n+1)\sum_{k=1}^nk+\left(\sum_{k=1}^nk\right)^2$$ and that $$\sum_{k=1}^{n+1}k^2=(n+1)^2+\sum_{k=1}^nk^2.$$ Can you get the rest of the way from there?

Alternately, you can proceed directly, noting that $$\begin{align}\left(\sum_{k=1}^nk\right)^2 &= \left(\sum_{k=1}^nk\right)\left(\sum_{j=1}^nj\right)\\ &= \sum_{k=1}^nk\sum_{j=1}^nj\\ &= \sum_{k=1}^n\sum_{j=1}^njk\\ &= \sum_{k=1}^n\left(k^2+\underset{j\ne k}{\sum_{j=1,}^n}jk\right)\\ &= \sum_{k=1}^nk^2+\sum_{k=1}^n\underset{j\ne k}{\sum_{j=1,}^n}jk\\ &\ge \sum_{k=1}^nk^2.\end{align}$$

ContentLicense: CC BY-SA 3.0


问:
Complex integral with poles on the curve

Evaluate $\oint_{|z|=2}\frac{\cos(z)}{z^5 -32} dz$

Honestly, I have no idea how to proceed. My professor talked about integral with poles in the interior of the curves so that case was simple. I have never seen this type of integral. Can someone show me how can I proceed? Thanks before!

答:
$$\int_{|z|=2}\frac{\cos(z)}{z^5 -32} dz$$ diverges.

What is well-defined, because $2 e^{2i\pi n/5} $ are simple poles, is $$PV(\int_{|z|=2}\frac{\cos(z)}{z^5 -32} dz)$$
(principal value)

Then the theorem is that it is $$=\frac12(\int_{|z|=2+\epsilon}\frac{\cos(z)}{z^5 -32} dz+\int_{|z|=2-\epsilon}\frac{\cos(z)}{z^5 -32} dz)$$
to which we can apply the residue theorem.


问:
A step in deriving the solution to heat equation
I'm trying to understand how to solve the heat equation. I'm stuck on this particular step:

$$Q(x,t) = \dfrac{1}{2} + \dfrac{1}{\sqrt\pi}\int_{0}^{\dfrac{x}{\sqrt{4kt}}}e^{-s^2}ds, t > 0.$$

I want to compute $\frac{\partial Q}{\partial x}.$ This should turn out to be
$$ \dfrac{1}{\sqrt {4k\pi}t}e^{-\dfrac{x^2}{4kt}}.$$ I think we have to use Leibniz integral rule but not sure how to do it. Any help appreciated.

答:
Never forget the fundamental theorem of calculus!
$$\mathrm{\partial}_x\left[\int_{a(x)}^{b(x)}f(s)\mathrm{d}s\right]=f(b(x))b'(x)-f(a(x))a'(x)$$
So in your example,
$$\partial_x Q(x,t)=\partial_x\left[\frac{1}{\sqrt{\pi}}\int\limits_0^{x/\sqrt{4 kt}}\exp(-s^2)\mathrm{d}s\right]=\frac{1}{\sqrt{4\pi kt}}\exp\left(-\frac{x^2}{4kt}\right)$$


问:
If $Y = X^2$, how do I find $P\{Y = 1\}$ to compute $E[X^2]$?
$X$ is a random variable taking on any of the values $-1,0,1,$ with probabilities $P\{X= -1\} = 0.2,$ $P\{X= 0\} = 0.5,$ $P\{X= 1\} = 0.3.$ Compute $E[X^2]$.

I start off letting $Y = X^2$, then finding pmf of $Y.$ While I know $P\{Y = 0\} = P\{X= 0\} = 0.5,$ how do I find $P\{Y = 1\}$?

答:
If you know $X^2 = 1$, then you also know $X = \pm 1$. So, you have

$$ P[Y = 1] = P[X^2 = 1] $$
$$ = P[X = 1 \cup X = -1] $$
$$ = P[X = 1] + P[X = -1] - P[X = 1 \cap X = -1] . $$

Assuming $X = 1$ and $X = -1 $ are mutually exclusive, you have

$$ P[Y = 1] = P[X = 1] + P[X = -1] . $$


问:
Trigonometric substitution integral
Trying to work around this with trig substitution, but end up with messy powers on sines and cosines... It should be simple use of trigonometric properties, but I seem to be tripping somewhere.

$$\int x^5\sqrt{x^2+4}dx $$

Thanks.

答: Put $x^2=t$. You will recieve $\frac{1}{2}\int t^2\sqrt{t+1}\mbox{d}t$. Now put $t+1=s$ and you will recieve $$\frac{1}{2}\int (s-1)^2\sqrt{s}\mbox{d}s.$$ That should be easy now.</p>

Title: Trigonometric substitution integral


问:
Real-life, every day applications of number theory?
Are there any day to day situations in which number theory could be applied to? For example, I was trying to make a mosaic and I realized that diophantine equations could be really useful if I needed to find how many pieces of certain dimensions needed to be used. Are there any other real life situations in which number theory would be useful?</p>

答:
The RSA crypto-algorithm, which critically depends on number theory. That's billions of dollars in global revenue every day, which should hopefully qualify as a real life application.

https://en.wikipedia.org/wiki/RSA_(cryptosystem)


问:
Is every bounded function in $L^p$ for any $p \ge 1$?
Disclaimer: This might be very trivial, but I haven't learned any of this formally. (I didn't know what a Lebesgue integral was before looking it up today.) It is possible I am not understanding the definitions correctly, which is part of the reason I'm asking.

Context: I have a finite sample space $\Omega$ and a measure $\pi : \Omega \to \mathbb{R}$. (My event set, per the formal definition of a measure space, is trivial in the sense that it is just the power set of $\Omega$. Thus, I'm just keeping it implicit.)

I have a bounded function $f : \Omega \to \mathbb{R}$, and I want to check whether it is in $L^2(\Omega, \pi)$. I have theorized that since $f$ is bounded, it immediately follows that $f \in L^p(\Omega, \pi)$ for all $p \ge 1$. (So in particular, $f \in L^2(\Omega, \pi)$.)

My thought process: I am working in the context of a finite probability space, but since I think my hypothesis holds even if this isn't true, let's work in a more general setting. Let $(X, \mathcal{F}, \mu)$ denote a $\sigma$-finite measure space and let $1 \le p < \infty$.  According to my digging, $f \in L^p(X, \mu)$ if

$$
\int |f|^p d \mu < \infty.
$$

Per the definition of Lebesgue integration, this is equivalent to checking whether

$$
\int_{0}^\infty f^*(t) dt < \infty,
$$

where $f^*(t) = \mu(\{ x | x \in X, |f(x)|^p > t\})$. (We have converted the Lebesgue integral to an ordinary improper Riemann integral.)

Since $f$ is bounded, it follows that $|f|^p$ is bounded. This means there is some $C \ge 0$ such that $f^*(t) = 0$ for all $t \ge C$. So

$$
\int_{0}^\infty f^*(t) dt  = \int_{0}^C f^*(t) dt.
$$

We know that $f^*(t)$ is bounded since the measure space is $\sigma$-finite. (I guess if it wasn't $\sigma$-finite than my hypothesis wouldn't hold, although I don't know why you would care about such a scenario.) It follows that

$$
\int_{0}^C f^*(t) dt
$$

is finite, and thus $f \in L^p(X, \mu)$.

(I saw that formally $L^p$ spaces are equivalence classes of functions, but I'm ignoring this formalism since it doesn't seem to affect my argument.)

Results from searching: This says that any bounded function on $[0, 1]$ is automatically $L^p$ for every value of $p$. But I don't see why we need to restrict ourselves to $[0, 1]$.

Thanks for any help!

Edit: I realize now that when I wrote this question I had the definition of "$\sigma$-finite measure" confused with "finite measure." (See the accepted answer.) I also realized that you have to be more careful when taking the Lebesgue integral of signed functions, so let's retroactively say that $f$ is nonnegative. (My bounds on the improper Riemann integral incorrectly went from $-\infty$ to $\infty$ before; now they correctly go from $0$ to $\infty$.)

答:
This is true if and only if ($f$ is measurable and) the total measure is finite (this is where the restriction to $[0, 1]$ is from; it's false for $\mathbb{R}$ with the Lebesgue measure, for example). Otherwise not even the constant function $1$ is in $L^p$ for $p \in [1, \infty)$. Your argument implicitly assumes this when you say that $f^{\ast}$ is bounded (which implies finite) even though you only say you assume the measure is $\sigma$-finite.

If the total measure $\mu(X)$ is finite and $|f|$ is bounded by a constant $C$ then

$$\int_X |f|^p \, d \mu \le C^p \mu(X).$$

If the sample space is finite then $f$ is automatically measurable and the total measure is automatically finite so there's no issue there; in this very special case every function is (measurable and) bounded and so every function is in $L^p$ for every $p$.


问:
${[y] }$ denotes the greatest integer less than or equal to $ y$ , then choose the correct option.
Define $f :\mathbb{R^2} \rightarrow \mathbb{R}$ by $f(x,y) = x^{[y]}$. where ${[y] }$ denotes the greatest integer less than or equal to $ y$  ..then

choose the correct option.

$1)$ $f$ is continuous on $\mathbb{ R^2} $

$2)$for every $ y \in \mathbb{ R}$ , $x \rightarrow f(x,y)$ is continuous  on $\mathbb{R}$ \ $\{0\}$

$3)$ For every $x \in \mathbb{R}$ , $y \rightarrow f(x,y)$ is continuous on $\mathbb{R}$

$4)$ $f$  is continuous at no point of $\mathbb{R^2}$

My answer :  as  For option 1) is not correct  because  I can claim that   $f$ is  not continuous  on $\mathbb{R^2}$    becuse    if  I take  $f(x) = \frac {1}{x}$  which is not defined at $x =0$

For option $4)$ is also  False if i take $f(x) = x$

I am in confusion about option $2)$ and option $3)$

Please help me,,,

Thanks  in adavnce.

Any hints/solution will be appreciated...

答:
Given that $f :\mathbb{R^2} \rightarrow \mathbb{R}$ is defined by $f(x,y) = x^{[y]}$.

For option $\bf{(1)}$, clearly $~f~$ is not defined at $(0,0)$ and therefore it is not continuous at $(0,0)$.
Hence $~f~$ is not continuous on $ \mathbb{R}^2$ and therefore option $(1)$ is not true.

For option $\bf{(2)}$, for $~y~$ fixed in $ \mathbb{R}$, $~~f(x,y)=\phi(x) = x^{[y]}~$ is a polynomial on $ \mathbb{R}$ \ $\{0\}$.
Hence $~\phi~$ is continuous on $ \mathbb{R}$ \ $\{0\}$ and therefore option $(2)$ is true.

For option $\bf{(3)}$, for $~x~$ fixed in $ \mathbb{R}$, $$f(x,y)=g(y) = x^{[y]}=\begin{cases}
      x^0=1~~~ ;~~~ 0\le y\lt 1 \\
      x^1~~~~~~~~~~~;~~~ 1\leq y\lt 2 \\
      x^2~~~~~~~~~~~;~~~ 2\leq y\lt3\\
      \vdots~~~~~~~~~~~~~~~~~~~~~~\vdots
   \end{cases}$$ Clearly $~g~$ is not continuous at $ ~y\in\mathbb Z~$ and hence option $(3)$ is not true.

For option$\bf{(4)}$, for $~x\ne0~$ and $~y\notin\mathbb Z~$, from option $(3)$, $~f(x,y)~$ is continuous.
Therefore option $(4)$ is not true.


问:
Function that sets a maximum value of a variable and reduces its value to the maximum if it is over.
I need to know how to write some code of mine mathematically in order to make it look more elegant.
I am unfamiliar of the mathematical notation to write such a function and need some guidance.

The code I wish to translate takes the max(or min) value of a variable and reduces(or increases) its value to the maximum(or miminum) if the value is outside the bounds.
Here is the code I wish to translate into math-speak.

```<code>If x < 0    then     x = 0
x + 150
if x > 500  then     x = 500
</code>```

Many thanks from a misplaced economics student!

答:
You can project $x$ to the interval $[0,500]$ via the assignment
$$x=\min(\max(x,0),500)$$
See also Notation for forcing a real value between bounds


问:
Combinatorial proof of ${2n\choose n}n!\le (2n)^{n}$
Initially I was supposed to prove the proposition $n^n\ge \prod_{r=1}^{r=n}(2r-1)\forall n\in\mathbb{N}$. So I have basically simplified this question down to proving ${2n\choose n}n! \le (2n)^{n}$ by carrying out the following operations.

$$\begin{aligned}n^{n}\prod_{r=1}^{n}2r&\ge \prod_{r=1}^{n}(2r-1)\prod_{r=1}^{n}2r\\ 2^{n}n^nn!&\ge(2n)!\\ (2n)^{n}&\ge {2n\choose n}n!\end{aligned}$$

Any ideas on how to prove this inequality. The right side of the inequality represents the arrangement of $n$ objects chosen out of $2n$ object. And the left side represents the arrangement of $n$ objects into $2n$ vacancies. Any hints are appreciated. Thanks.

答: The lefthand side of the inequality in the title is the number of ways to put $n$ balls numbered $1$ through $n$ into $2n$ slots numbered $1$ through $2n$ if no slot is allowed to contain more than one ball. The righthand side is the number of ways to put the same $n$ balls into the same $2n$ slots with no restrictions on how many balls can occupy any slot.


问:
Expectation of random variable with differently weighted realizations
Let $Y=wX$, where $X$ is a random variable with mean $\mu$ and variance $\sigma^2$, and $w$ is the weight which could change for every realization of $Y$. In other words, $N$ realizations of $Y$ are given as $y_n$ = $w_n x_n,~n=1,...,N$. The weights $w_n,~\forall n$, are assumed to be $N$ uniformly spaced samples of a positive continuous function $W(t),~0<t<T$.

I wonder how can we express the population mean $E[Y]$?

My guess is that $E[Y]$ is defined as $E[Y]=\bar{w}E[X]=\bar{w}\mu$, where $\bar{w}=\frac{1}{T}\int^T_0 W(t) dt$.

Does that reasoning make sense?

答: Essentially $Y=wX$ is a product of two independent random variables so $E(Y)=E(w)E(X)$ where $E(w)=\overline{w}$.


问:
$\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...
Let $\{a_n\}$ be a sequence of positive real numbers such that

$a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.

Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...

(A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.

Solution attempt:

Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n>0$, the other root is rejected).

We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$, then $\lim a_n \to \infty$.

Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$. (Again, the other root is rejected due to the same reason).

Hence, $(a_n)$ increases monotonically.

Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$

Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}< 1.3$).

So, option $(A)$ is the correct answer.

Is the procedure correct?

I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.

Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)

Hints:

After calculating the first few terms, the ratio of the terms is very similar. If we estimate $ a_n \approx a_1 r^{n-1}$, this suggests we should solve for $ \frac{ 1/3} { 1 - (r_1/3) } = 1$ and $ \frac{1/3} { 1 - (r_2/3)} = 2$ to give us an idea of how to bound the sequence. This gives us $ r_1  = 2, r_2 = 5/2$, so we want to show that $ 2 a_n < a_{n+1} < \frac{5}{2} a_n$ (with some flexibility if this doesn't immediately work out).

Show that $ a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n  } } { 2} $. In particular, reject the negative root.

Show that $ a_n \geq 1$.

Show that $ 2 a_n < a_{n+1} < \frac{5}{2} a_n$.

Hence, show that $1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2 $

Note:

The LHS is true by calculating the first 5 terms.

Of course, we can't prove the RHS just by calculating enough terms.

In fact, the bounding inequality $ 2a_n < a_{n+1} < 2a_n + \frac{1}{2}$, so the ratio $ a_{n+1} / a_n$ is very close to 2, esp at (slightly) larger values of $n$.

Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25.


问:
Variance of $W=e^{\frac{X}{4}}$
If $W=e^{\frac{X}{4}}$, and we have a gamma distribution with $\alpha=4$, $\beta=2$, $\mu=8$, and $\sigma=4$, how would one find the variance of $W$ using the moment generating function for gamma distributions?

So far I have started with $M_{X}\left(t\right)=\left(1-\beta t\right)^{-\alpha}$, and I know that taking the first and second derivative of this will allow me to find the mean and variance, but I don't know how to apply this to $W$. All feedback is greatly appreciated.

答:
Mgf of Gamma distribution is
$M_X{t}=E(e^{tx})=\bigg(1-\dfrac{t}{\alpha}\bigg)^{-\beta}$

we have

$t=\frac{1}{4}$

$E(e^{\frac{x}{4}})=E(W)=\bigg(1-\dfrac{1}{4\alpha}\bigg)^{-\beta}$

$t=\frac{1}{4}$

$E((e^{\frac{x}{4}})^2)=E((e^{\frac{x}{2}}))=\bigg(1-\dfrac{1}{2\alpha}\bigg)^{-\beta}$

Substitute values  in $V(X)=E(X^2)-E(X)^2$to get the desired result.

Title: Variance of $W=e^{\frac{X}{4}}$


问:
Rudimentary topology question: interior of a subset of a space is the union of all open sets contained in it.
I'm reviewing topology and just wanted to confirm that I am doing this correctly. Professor Janich has urged the readers of his classic text on topology to prove the following:

Suppose $(X, \mathcal{O})$ is a topological space and $B \subset X$. Define the interior of the set $B$ as $\hat B$.

Claim: The interior of $B,$ namely $\hat B$, is the union of all open sets contained in $B$.

Proof of $\Longrightarrow$: Let $ x\in \hat B$. Then by definition, $B$ is a neighborhood of $x$. Thus $\exists A \subset B $ such that $A$ is open and $x \in A$. Since $x \in A$ and $A \subset B$, it must be the case that  $x \in \bigcup\{\text{all open sets contained in $B$}\}$.

Proof of $\Longleftarrow$: Suppose $x \in \bigcup\{\text{all open sets contained in $B$}\}$. Just pick any one of them and call it, say, $A$. (Question: did I just use the axiom of choice?) $x \in A \subset B$ and $A$ is open, so $B$ is a neighborhood of $x$ and thus, $x$ is an interior point of $B$.

Is all of that correct? Please feel free to offer any corrections, thank you.

答:
Recall that if $x$ and $y$ are sets, $$x \in \textstyle\bigcup y \iff (\exists z \in y) \ x \in z.$$

So, if $x$ is in the union of all the open sets contained in $B$, by definition of  union there exists an open set $A$ contained in $B$ such that $x \in A$. There is no need of choice here. By the way, write \Longleftarrow to obtain $\Longleftarrow$.


问:
$n_1, n_2,...,n_s$ such $(n_1+k)(n_2+k)\cdots(n_s+k)$ integral multiple of $n_1n_2\cdots n_s$. Prove $|n_i| = 1$ for some $i$
Let $n_1, n_2,...,n_s$ be distinct integers such that
\begin{equation}
    (n_1+k)(n_2+k)\cdots(n_s+k)
\end{equation}
is an integral multiple of $n_1n_2\cdots n_s$ for every integer $k$. Give a proof or a counterexample of the following statement: $|n_i| = 1$ for some $i$

I think this statement is always true. The way decided to approach was to arrange $n_1,...,n_s$ such that $|n_1|\geq |n_2|\geq \cdots \geq|n_s|$. I was hoping to plug in $k = \pm 1$ based on the sign of $n_2$ but I didn't get very far past that.

Some hints would be greatly appreciated!

答:
(Fill in the gaps as needed, esp when it says "Show that". If you're stuck, show your work and explain what you've tried.)

Hint: Show that $n_i \neq 0$, so henceforth $n_i \neq 0$.

Let $f(k) = \prod ( n_i + k)$.

Proof by contradiction. Suppose $ |n_i | \neq 1$.

Show that $f(0), f(1), f(-1) \neq 0$.

Hint: Show that $ 0 < f(1) f(-1) < f(0)^2$.

Hence, it is not possible for both $ f(0) \mid f(1) $ and $f(0) \mid f(-1)$, contradicting the condition that $f(0) \mid f(k)$.

Notes for how I came up with this solution:

It's initially not clear how to use the condition that $ f(0) \mid f(k)$.

The condition that $|n_i| \neq 1$ suggests that we look at $ f(1)$, $f(-1) \neq 0 $.

This suggests that we have either $ f(0) \not \mid f(1)$ or $f(0) \not \mid f(-1)$.

We then have to figure out how this can be proven. We would normally have very little control over these terms. For example, it is not generally true that $ f(1) f(-1) < f(0) ^2$.

This happens only if $ n_ i \neq 0$, so let's investigate what happens if some $n_i = 0$.

Thankfully, $n_i = 0$ isn't allowed by the condition, so now we have $f(1) f(-1) < f(0)^2$.

If so, one simple way to show that $ f(0) \not \mid f(1), f(-1)$ is that $ 0 < |f(1) f(-1)| < f(0)^2$, and we can verify that is true.


问:
How to calculate the median of a continuous random variable
$X$ is a continuous random variable with
probability density function $f(x)= \dfrac{2x}{15}$ where $1≤x≤4$.

What is the median of X?

答:
Hint: To find the median, you want to find $c$ such that $\mathbb{P}(1 \le X \le c) = \mathbb{P}(c \le X \le 4)$. That's just the definition of the median: it's the number $c$ for which the probabilities on both of its sides are the same.

The integral you get here shouldn't be hard to carry out.


问:
How to solve $(2k+1)(2k^2+1) = y^2$ over the positive integers
How could I find a solution to $(2k+1)(2k^2+1) = y^2$ over positive integers?

WolframAlpha returns some solutions, but I was wondering if there was a way to find all of them, or prove that there are infinitely many. Here is the link to the Wolfram Alpha answer:

https://www.wolframalpha.com/input/?i=solve+%282k%2B1%29%282k%5E2%2B1%29+%3D+y%5E2+over+the+integers

This is a personal question inspired by the equation $(k+2)(4k+1) = y^2$ which is doable using elementary methods.

答:
Here is another computer solution by using SAGEMATH (though we need to put it in standard form). We are indeed looking for integral points on an elliptic curve. First we transform our Diophantine equation $E_1:(2k+1)(2k^2+1) = y^2$ into the form $Y^2 = X^3 +a X^2 + b X +c$, where $a,b,c$ are integers (which SAGEMATH will compute integral points for us, assuming its algorithm is correct).

With the substitutions $k=X/64$ and $y=Y/256$, we obtain an elliptic curve in standard form with integral coefficients $E_2:Y^2 = X^3 + 32X^2+2048X+65536$. Note any integral solutions to our original problem $E_1$ will be an integral solution to this new equation $E_2$.

Using the following in SAGEMATH (which you can execute it here https://sagecell.sagemath.org/)

```
x,y = polygens(QQ,'x,y');
E = EllipticCurve(-y^2+x^3+32*x^2+2048*x+65536);
E.integral_points(both_signs=True)
```

(One can also check this curve has rank 1 by using the command E.rank())

It shows the only integral solutions to $E_2$ are (in homogeneous coordinates $(X:Y:1)$)

```
[(-32 : 0 : 1),
 (-16 : -192 : 1),
 (-16 : 192 : 1),
 (0 : -256 : 1),
 (0 : 256 : 1),
 (64 : -768 : 1),
 (64 : 768 : 1),
 (160 : -2304 : 1),
 (160 : 2304 : 1),
 (768 : -21760 : 1),
 (768 : 21760 : 1)]
```

Note, by Siegel's Theorem, there can only be finitely many integral solutions (https://mathworld.wolfram.com/SiegelsTheorem.html).

Here we see there are only six solutions such that $X$ is a multiple of 64 and $Y$ is a multiple of 256, namely

```
 (0 : -256 : 1),
 (0 : 256 : 1),
 (64 : -768 : 1),
 (64 : 768 : 1),
 (768 : -21760 : 1),
 (768 : 21760 : 1)
```

Recovering for $k$ and $y$ we have the only six integral solutions offered in Wolfram Alpha: $(k,y) = (0,\pm 1), (1,\pm 3),(12,\pm 85)$.

I only have a superficial understanding of this, but I will direct you here for more information:

https://en.wikipedia.org/wiki/Elliptic_curve

Rational Points on Elliptic Curves, by Silverman.

Body:
Here is another computer solution by using SAGEMATH (though we need to put it in standard form). We are indeed looking for integral points on an elliptic curve. First we transform our Diophantine equation $E_1:(2k+1)(2k^2+1) = y^2$ into the form $Y^2 = X^3 +a X^2 + b X +c$, where $a,b,c$ are integers (which SAGEMATH will compute integral points for us, assuming its algorithm is correct).

With the substitutions $k=X/64$ and $y=Y/256$, we obtain an elliptic curve in standard form with integral coefficients $E_2:Y^2 = X^3 + 32X^2+2048X+65536$. Note any integral solutions to our original problem $E_1$ will be an integral solution to this new equation $E_2$.

Using the following in SAGEMATH (which you can execute it here https://sagecell.sagemath.org/)

x,y = polygens(QQ,'x,y');
E = EllipticCurve(-y^2+x^3+32*x^2+2048*x+65536);
E.integral_points(both_signs=True)

(One can also check this curve has rank 1 by using the command E.rank())

It shows the only integral solutions to $E_2$ are (in homogeneous coordinates $(X:Y:1)$)

[(-32 : 0 : 1),
 (-16 : -192 : 1),
 (-16 : 192 : 1),
 (0 : -256 : 1),
 (0 : 256 : 1),
 (64 : -768 : 1),
 (64 : 768 : 1),
 (160 : -2304 : 1),
 (160 : 2304 : 1),
 (768 : -21760 : 1),
 (768 : 21760 : 1)]

Note, by Siegel's Theorem, there can only be finitely many integral solutions (https://mathworld.wolfram.com/SiegelsTheorem.html).

Here we see there are only six solutions such that $X$ is a multiple of 64 and $Y$ is a multiple of 256, namely

 (0 : -256 : 1),
 (0 : 256 : 1),
 (64 : -768 : 1),
 (64 : 768 : 1),
 (768 : -21760 : 1),
 (768 : 21760 : 1)

Recovering for $k$ and $y$ we have the only six integral solutions offered in Wolfram Alpha: $(k,y) = (0,\pm 1), (1,\pm 3),(12,\pm 85)$.

I only have a superficial understanding of this, but I will direct you here for more information:

https://en.wikipedia.org/wiki/Elliptic_curve

Rational Points on Elliptic Curves, by Silverman.


答:
Cryptography, hash functions, and random number generators. The general number field sieve obviously also uses Number Theory heavily. The same applies to Shor's Algorithm. Although at a superficial glance number theory may look to have no 'real life' applications, its principles and ideas are widely used in computer science and especially coding.

“To a mathematician, real life is a special case.”

Title: Real-life, every day applications of number theory?

Body: Are there any day to day situations in which number theory could be applied to? For example, I was trying to make a mosaic and I realized that diophantine equations could be really useful if I needed to find how many pieces of certain dimensions needed to be used. Are there any other real life situations in which number theory would be useful?</p>


问:
Chi Squared Clarification
Suppose you have $Z_1, Z_2, Z_3$ which are all independent standard Gaussian variables.

Suppose you have
$$A=\frac{(Z_1-2Z_2+Z_3)^2}{12}+\frac{(Z_1-Z_3)^2}{4}+\frac{(Z_1-Z_2)^2}{4}+\frac{(Z_1+Z_2-2Z_3)^2}{12}.$$

$A$ is $\chi_2^2$, but I don't see how this is the case.
Particularly since $\operatorname{Var}\left(\frac{Z_1-2Z_2+Z_3}{\sqrt{12}}\right)$${}=\frac{1}{2}$, which is the same issue with the other terms, i.e since none of the terms, by themselves, are standard normal before you square them.
However, I do see how:
$$B=\frac{(Z_1-2Z_2+Z_3)^2}{6}+\frac{(Z_1-Z_3)^2}{2}.$$ follows the $\chi_2^2$ distribution. I can also see that the first two terms of $A$ multiplied by $2$ is equal to $B.$

Any feedback would be appreciated.

答:
$$
\left[ \begin{array}{c} \frac{Z_1-2Z_2+Z_3}{\sqrt{12}} \\[4pt] \frac{Z_1-Z_3} 2 \\[6pt] \frac{Z_1-Z_2} 2 \\[6pt]
\frac{Z_1+Z_2-2Z_3}{\sqrt{12}}
\end{array} \right]
= \left[ \begin{array}{rrr}
1/\sqrt{12} & -2/\sqrt{12} & 1/\sqrt{12} \\
1/2 & 0 & -1/2 \\
1/2 & -1/2 & 0 \\
1/\sqrt{12} & 1/\sqrt{12} & -2/\sqrt{12}
\end{array} \right]
\left[
\begin{array}{c}
Z_1 \\ Z_2 \\ Z_3
\end{array}
\right]
\tag1
$$
Observe that the rank of this $4\times3$ matrix is only $2,$ i.e. you have only two linearly independent columns, since the second column is $-1$ times the sum of the first and third columns.

Let us find two unit vectors orthogonal to each other that span the column space of that matrix:
$$
\left[ \begin{array}{c} -\sqrt3/\sqrt8 \\ +1/\sqrt8 \\ -1/\sqrt8 \\ +\sqrt3/\sqrt8 \end{array} \right], \qquad \left[ \begin{array}{c} +1/\sqrt8 \\ +\sqrt3/\sqrt8 \\ +\sqrt3/\sqrt8 \\ +1/\sqrt8 \end{array} \right]
$$
Write the column vector on the left side of line $(1)$ above as a linear combination of these two vectors, each of the coefficients $U_1,U_2$ being a linear combination of $Z_1,Z_2,Z_3.$

Then show that $\operatorname{var}(U_1)= \operatorname{var}(U_2)=1$ and $\operatorname{cov}(U_1,U_2)=0$ and that $U_1,U_2\sim\operatorname N(0,1),$ and that $U_1^2 + U_2^2$ is equal to the sum of squares that you say should have a $\chi^2_2$ distribution.

Title: Chi Squared Clarification


问:
A very simple inequality problem
I think the following claim is clearly correct, but I cannot prove it.

Let $A$ and $B$ be sets. If $f:A \times B \to \mathbb{R}$ satisfies $f(a, b) \leq C_a$ and $f(a, b) \leq C_b$ for all $(a, b) \in A \times B$ , then $f$ is bounded above. Remark: $C_x$ is a constant that depends only on $x$.

Background of this question:
I asked the following question, and this can be solved immediately if this original question is correct. (But it was false.)
Boundedness of a bilinear operator

This original question is interesting in the following sense. If $f:A \times B \to \mathbb{R}$ satisfies $f(a, b) \leq C_a$, $f$ has a upper bound that is independent of $b \in B$. Similarly, if $f(a, b) \leq C_b$, $f$ has a upper bound that is independent of $a \in A$. Therefore, it seems to be that $f$ has a upper bound that is independent of $(a, b) \in A \times B$, namely, $f$ is bounded above. However, it was false actually.

答:
The claim is not correct, so find a counter example.

Hint:

 Let $f(x, x ) = x $

Title: A very simple inequality problem


答: Even an upper bound need not exist. Take $f:\mathbb R \times \mathbb R \to \mathbb R$ defined by $f(x,y)=x$ if $x=y$ and $0$ if $x \neq y$.

Title: A very simple inequality problem

Body:
I think the following claim is clearly correct, but I cannot prove it.

Let $A$ and $B$ be sets. If $f:A \times B \to \mathbb{R}$ satisfies $f(a, b) \leq C_a$ and $f(a, b) \leq C_b$ for all $(a, b) \in A \times B$ , then $f$ is bounded above. Remark: $C_x$ is a constant that depends only on $x$.

Background of this question:
I asked the following question, and this can be solved immediately if this original question is correct. (But it was false.)
Boundedness of a bilinear operator

This original question is interesting in the following sense. If $f:A \times B \to \mathbb{R}$ satisfies $f(a, b) \leq C_a$, $f$ has a upper bound that is independent of $b \in B$. Similarly, if $f(a, b) \leq C_b$, $f$ has a upper bound that is independent of $a \in A$. Therefore, it seems to be that $f$ has a upper bound that is independent of $(a, b) \in A \times B$, namely, $f$ is bounded above. However, it was false actually.


答:
You don't even need a trigonometric substitution. Let $u^2=x^2+4$. Then $2u\,du=2x\,dx$, so $x\,dx=u\,du$.

Rewrite $x^5\sqrt{x^2+4}\,dx$ as $x^4\sqrt{x^2+4}\,x\,dx$, and substitute. Note that $x^2=u^2-4$, so $x^4=(u^2-4)^2$. Finishing the substitution, we get $(u^2-4)^2 u^2\,du$. Expand and integrate. No trig!

Title: Trigonometric substitution integral

Body:
Trying to work around this with trig substitution, but end up with messy powers on sines and cosines... It should be simple use of trigonometric properties, but I seem to be tripping somewhere.

$$\int x^5\sqrt{x^2+4}dx $$

Thanks.


问:
The Chain Rule for Differetial of a Map
I am reading "An introduction to manifolds" by Loring Tu. And there in the proof of the chain rule for differential of a map i couldn't understand the last step. The following is the step that i am unable to understand:
$$(G_*(F_*X_p))f=(F_*X_p)(f\circ G)$$
I am attaching the screen shot of proof where this  is given and i have highlighted the step which i can't understand. Also note that they've defined the differential of a map in the previous section of which i am also attaching the screenshot.The only step that i can't understand is that is highlighted(in the screenshot) and written above.

答:
By definition, $(G_*(Y_q))f=Y_q(f\circ G)$.
Now take $Y_q=F_*(X_p)$

Title: The Chain Rule for Differetial of a Map


问:
How to get quartic polynomial function given graph, 4 points, and 1 root
I'm doing a math assignment right now, which is due in more or less four and a half hours. What I need to find is a polynomial function given this graph this graph and the points on it. Here are the points:

0,15

7,-1

22,-7

26,0

28,14

The other two points marked on the graph were just marked for another question; I'm not exactly sure if they are x intercepts because I can see that they are a few pixels above or below. Thanks in advance :)

Edit: So far, using the point 7,-1 and the x- and y-intercepts, I tried substituting and transposing the terms in the polynomial, so from y=ax^4+bx^3+cx^2+dx+e, I got 2401a + 343b + 49c + 7d + 16 = 0, and I'm stumped from there. I've also tried doing it in factored form, so y = a(x-26)(x-b)(x-c)(x-d), and I don't really know what to do next.

答:
We can set up a general quartic equation like this:

$$ y = ax^4 + bx^3 + cx^2 + dx + e . $$

We can then set up a system of equations by plugging in each of the points individually:

$$ \begin{cases}
15 = 0a + 0b + 0c + 0d + 1e \\
-1 = 7^4a + 7^3b + 7^2c + 7d + e \\
-7 = 22^4a + 22^3b + 22^2c + 22d + e \\
0 = 26^4a + 26^3b + 26^2c + 26d + e \\
14 = 28^4a + 28^3b + 28^2c + 28d + e \\
\end{cases} .$$

This is a system of linear equations with five equations and five unknowns: $a, b, c, d, e$. It's a little different from the systems of equations you might be used to, where you know the equation and not the points - in that case, you have the coefficients but not $x$ and $y$. In this case, you know the points but not the function, so it's the reverse: coefficients are unknown while $x$ and $y$ are known.

If you're familiar with matrices, this is becomes very easy to solve. The system can be rewritten as

$$\begin{bmatrix}
15 \\
-1 \\
-7 \\
0 \\
14 \\
\end{bmatrix} =
\begin{bmatrix}
0 & 0 & 0 & 0 & 1 \\
7^4 & 7^3 & 7^2 & 7 & 1 \\
22^4 & 22^3 & 22^2 & 22 & 1 \\
26^4 & 26^3 & 26^2 & 26 & 1 \\
28^4 & 28^3 & 28^2 & 28 & 1 \\
\end{bmatrix}
\begin{bmatrix}
a \\
b \\
c \\
d \\
e \\
\end{bmatrix} $$

$$\Rightarrow
\begin{bmatrix}
a \\
b \\
c \\
d \\
e \\
\end{bmatrix} =
\begin{bmatrix}
0 & 0 & 0 & 0 & 1 \\
7^4 & 7^3 & 7^2 & 7 & 1 \\
22^4 & 22^3 & 22^2 & 22 & 1 \\
26^4 & 26^3 & 26^2 & 26 & 1 \\
28^4 & 28^3 & 28^2 & 28 & 1 \\
\end{bmatrix} ^{-1}
\begin{bmatrix}
15 \\
-1 \\
-7 \\
0 \\
14 \\
\end{bmatrix} $$

This gives us the following solution:

$$ y = \frac{87}{69160}x^4 - \frac{31}{455}x^3 + \frac{82807}{69160}x^2 - \frac{268341}{34580}x + 15 $$

Here's the matrix multiplication using WolframAlpha.

If you're not familiar with matrices, you'll need to use substitution and elimination, and it'll take forever. But at least the first equation gets you $e=15$, so that's something you can start your substitution with.


答: Let $f(a,b)=min(a,b).$ Then $f$ will be bounded by $min(C_a, C_b)$, but $f(a,a)$ will diverge to $\infty$.

Title: A very simple inequality problem


问:
If $Y = X^2$, how do I find $P\{Y = 1\}$ to compute $E[X^2]$?
$X$ is a random variable taking on any of the values $-1,0,1,$ with probabilities $P\{X= -1\} = 0.2,$ $P\{X= 0\} = 0.5,$ $P\{X= 1\} = 0.3.$ Compute $E[X^2]$.

I start off letting $Y = X^2$, then finding pmf of $Y.$ While I know $P\{Y = 0\} = P\{X= 0\} = 0.5,$ how do I find $P\{Y = 1\}$?

答:
The fastest way to get $E(X^2)$ is to use the definition of expectation

$$E(X^2)=\Sigma_i x_i^2p(x_i)=(-1)^2\cdot 0.2+0^2\cdot 0.5+1^2\cdot 0.3=0.5$$

Edit: answering to comment of @Prasus Biswas here is a definiton of Expectation taken from one of the reference text (Mood Graybill Boes)


答:
The best i could come up with from Rob Pratt's link was

$x = (0∨x) + 150)∧ 500$

or

$x = min(max(0,x)+150,500)$

Title: Function that sets a maximum value of a variable and reduces its value to the maximum if it is over.

Body:
I need to know how to write some code of mine mathematically in order to make it look more elegant.
I am unfamiliar of the mathematical notation to write such a function and need some guidance.

The code I wish to translate takes the max(or min) value of a variable and reduces(or increases) its value to the maximum(or miminum) if the value is outside the bounds.
Here is the code I wish to translate into math-speak.

```<code>If x < 0    then     x = 0
x + 150
if x > 500  then     x = 500
</code>```

Many thanks from a misplaced economics student!


问:
Formula for Green function of Sturm-Liouville problem.
I'm trying to remember exact formula for Green function of Sturm-Liouville problem for operator $L\::y\longmapsto (p(t)y')' + q(t)y$, left condition $D_0$ and right condition $D_1$ (that looks like $a_0y'(0) + b_0y(0)$ and $a_1y'(1) + b_1y(1)$), when we have $h_0$ and $h_1$ such that
$$Lh_0 = 0, \;\; D_0h_0 = 0$$
$$Lh_1 = 0, \;\; D_1h_1 = 0$$
My notes tell me that Green function in that case is
$$ h_0(x)h_1(y) ;\; x < y$$
$$ h_1(x)h_0(y) ;\; x < y$$
multiplied on $$(h_0(x)h_1'(x) - h_1(x)h_0'(x)) \frac{1}{p(x)}$$
I'm in doubt if my notes are correct and would like to ask someone to verify them, because I can't find any information about this formula at all.

答:
You can apply this classical theorem to get the Green function that you want:

Theorem [Variation of Parameters]: Let $a(x),b(x),c(x)$ be continuous real
or complex functions on an open interval $(r,s)$, and suppose $a(x)$ does not
vanish on $(r,s)$. Let $f, g$ be solutions of
$$
                   ay''+by'+cy=0
$$
with a non-vanishing Wronskian $W(f,g)$. Let $h$ be a complex continuous
function on $[a,b]$. Then $y$ is solution of
$$
                    ay''+by'+cy=h
$$
iff there are complex constants $A$, $B$ such that
$$
     y = Af+Bg +g\int f\frac{1}{aW(f,g)} hdx - f\int g\frac{1}{aW(f,g)}hdx.
$$


答:
Consider the ratio:

$$\frac{2n \times 2n \times 2n \times \cdots \times 2n}
{(2n) \times (2n-1) \times (2n-2) \times \cdots \times (2n + 1 - n)}.$$

Clearly, the numerator is bigger than the denominator.

Title: Combinatorial proof of ${2n\choose n}n!\le (2n)^{n}$

Body:
Initially I was supposed to prove the proposition $n^n\ge \prod_{r=1}^{r=n}(2r-1)\forall n\in\mathbb{N}$. So I have basically simplified this question down to proving ${2n\choose n}n! \le (2n)^{n}$ by carrying out the following operations.

$$\begin{aligned}n^{n}\prod_{r=1}^{n}2r&\ge \prod_{r=1}^{n}(2r-1)\prod_{r=1}^{n}2r\\ 2^{n}n^nn!&\ge(2n)!\\ (2n)^{n}&\ge {2n\choose n}n!\end{aligned}$$

Any ideas on how to prove this inequality. The right side of the inequality represents the arrangement of $n$ objects chosen out of $2n$ object. And the left side represents the arrangement of $n$ objects into $2n$ vacancies. Any hints are appreciated. Thanks.


问:
Interpolation using homeomorphism
I have a question which is inspired from this question.

Given pairwise distinct $x_1,x_2,\dots,x_n$ in $\mathbb{R}^p$ and pairwise distinct $y_1,y_2,\dots,y_n$  in $\mathbb{R}^p$ when can we say that there is a homeomorphism $\phi $ on $\mathbb{R}^p$ such that $\phi(x_i) = y_i$?

The answer is simple when $p=1$ because a homeomorphism is a strictly increasing (or strictly decreasing) continuous function. So a condition is $(x_i - x_j)(y_i - y_j)$ has the same sign for all distinct $i,j$

But what about general p?

答:
Suppose that $p\ge2$ and prove the required claim by the induction with respect to $n$. When $n=1$ put $\phi(x)=x+y_1-x_1$ for each $x\in\Bbb R^p$. Suppose that the claim if proved for $n-1$ pairs points. Let $x_1,x_2,\dots,x_n$ be pairwise distinct points of
$\mathbb{R}^p$ and $y_1,y_2,\dots,y_n$ be pairwise distinct points of $\mathbb{R}^p$. By the induction hypothesis, there exists a homeomorphism $\phi’$ of $\Bbb R^p$ such that $\phi’(x_i)=y_i$ for each $i<n$.

Given two points $x,y\in\Bbb R^n$ let $xy$ be the point $x$, if $x=y$, and the straight line passing through $x$ and $y$, otherwise. Pick an arbitrary point $z\in\Bbb R^p$ which does not belong to any of sets $x_ny_1,\dots, x_ny_{n-1}$ and
$y_nx_1,\dots, y_nxy_{n-1}$. Then the union $C$ of segments $[x_n,z]$ and $[z,y_n]$ contains no $y_i$ with $i<n$. Here we proved that for each open cube $U$ of $\Bbb R^p$ and any two points $p$ and $q$ of  $U$ there exists an homeomorphism $h$ of $\Bbb R^p$ such that $h(p) = q$ and such that $h(x) = x$ for every $x\in \Bbb R^p\setminus U$.

For each point $x\in C$ pick an open cube $U_x$ centered at $x$ and containing no $y_i$ with $i<n$. Since the set $C$ is compact, it contains a finite subset $F$ such that $C\subset\bigcup_{x\in F} U_x$. The set $C$ is homeomorphic to a segment and this easily follows that there exists a natural number $m>1$ and a sequence $x_n=z_1,\dots, z_m=y_n$ of points of $C$ such that for each $1\le i<n$ bot points $z_{i}$ and $z_{i+1}$ are contained in a set $U_{x_i}$ for some $x_i\in F$. Pick a homeomorphism $h_i$ of $\Bbb R^p$ such that $h(z_{i}) = z_{i+1}$ and such that $h(x) = x$ for every $x\in \Bbb R^p\setminus U_{x_i}$. It remains to put $\phi=h_{m-1}\dots h_1\phi'$.

Title: Interpolation using homeomorphism


问:
Standard Bivariate/ Multivariate Gaussians and Independence
Suppose you have $Z=(Z_1,Z_2,\ldots,Z_n)^T$, where $Z$ has the standard Gaussian distribution on $\Bbb R^n$. Is it true that the $Z_i$, $i=1,2,\ldots,n$ are all independent, since they have the Standard Gaussian distribution on $\Bbb R^n$?

答:
Another approach: If the random vector $\mathbf Z\in{\mathbb R}^n$ has standard Gaussian distribution, then the joint density of $\mathbf Z:=(Z_1,\ldots,Z_n)^\top$ is
$$
f(x_1,\ldots,x_n)=\frac1{(2\pi)^{n/2}}\exp[-\frac12 {\mathbf x}^\top{\mathbf x}]=\frac1{(2\pi)^{n/2}}\exp[-\frac12(x_1^2+\cdots+x_n^2)]
$$
which can be factored into the form
$$
\prod_{i=1}^n\left(\frac1{\sqrt{2\pi}}\exp(-\frac12 x_i^2)\right)=\prod_{i=1}^n \varphi(x_i)
$$
where $\varphi$ is the standard Gaussian density. Thus we've decomposed the joint density into the product of the marginal densities, implying that $Z_1,\ldots,Z_n$ are independent and identically distributed standard Gaussian random variables.


问:
hypothesis test using a confidence interval
i need to test the hypothesis that a proportion is greater than 0.93 using a confidence interval. i cannot figure out how to do this. the basic information that i think i need to use is below. the underlying variable follows a binomial distribution.

$$\hat p = 0.95, p = 0.93, n = 100, \sigma = 0.4$$

i have already tried constructing a confidence interval for a proportion using the Z distribution, but the upper bound is larger than 1, which seems wrong.

any advice for what i should do?

答:
The problem of your exercise is that the data are NOT coherent.

Substitute the given $\sigma=0.4$ with its estimation $\hat{\sigma}=\sqrt{0.93\cdot0.07}\approx0.255$ and you will reach a nice solution.

You need also to fix a Type I error ($\alpha$)

Example: fixing $\alpha=2.5\%$ you get that the one-tail CI is

$$[0.93;0.93+\frac{1.96}{10}0.255]=[0.93;0.98]$$

You cannot reject your null hypothesis as $0.95$ is included in the interval


问:
Can you prove a set is not open if its complement is open?
Can we prove the set of natural numbers as a subset of the reals is not open by saying that its complement is a union of open intervals so its open which means the natural numbers are not open?

I know the straightforward proof is just to take open balls of distance 1/2, but I was wondering if we can prove this by taking complements.

答:
If a set and its complement are both open, then it's open and closed, or clopen.

A connected space has by definition only two clopen subsets, itself and the empty set.

So in answer to your second question,  yes, since the naturals are a proper subset of the reals, and the reals are connected, it suffices to prove the complement is open.

However this won't always work: a disconnected space has proper subsets which are clopen.  Thus the answer to the question in the title, which is more general, is no.


问:
Odd factors of $a^2+3b^2$ have this same form, when $\gcd(a,b)=1$
edit: I originally made this post to check my understanding of this proof. Now I understand this is not a good type of question to ask. Despite regretting making this post originally, I would rather fix this proof than delete it if that's ok.

My attempt at a proof of the following statement: odd factors of $a^2+3b^2$ have this same form, when $\gcd(a,b)=1$.

(i) First let us establish that the set $a^2+3b^2$ is closed under multiplication.

$\hspace{2.5cm}$$(a^2+3b^2)(c^2+3d^2)$

$\hspace{2cm}$$=(ac)^2+3(ad)^2+3(bc)^2+9(bd)^2$

$\hspace{2cm}$$=(ac)^2\pm 6abcd+9(bd)^2+3(ad)^2\mp 6abcd+3(bc)^2$

$\hspace{2cm}$$=(ac\pm 3bd)^2+3(ad\mp bc)^2=e^2+3f^2$

(ii) We also need to establish the following:

If $N=a^2+3b^2$ and $p=c^2+3d^2$, $p$ is prime and $p$ divides $N$, then $\frac{N}{p}=u^2+3v^2$ ($N,p\in Z$)

Since $p$ divides $N$, it must divide $Nd^2-pb^2$ and:

$\hspace{2cm}$$Nd^2-pb^2=(a^2+3b^2)d^2-(c^2+3d^2)b^2$

$\hspace{4.5cm}$ $=a^2d^2+3b^2d^2-c^2b^2-3b^2d^2$

$\hspace{4.5cm}$ $=(ad+cb)(ad-cb)$

Therefore $p$ divides either $ad+cb$ or $ad-cb$. Being a prime number, it cannot equal the product. Also if $p=1$ then the case is already proved as $\frac{N}{1}=N$. Now consider the expression $Np$:

$\hspace{2cm}$ $Np = (ac\pm3bd)^2+3(ad\mp bc)^2$ [using the identity from (i)]

Given that $p$ divides one of $(ad\mp bc)$, it must also divide $(ac\pm3bd)$, in other words it must divide both terms of $Np$.

Now we can divide $Np$ by $p^2$ so we have:

$\hspace{2cm}$ $\frac{Np}{p^2}=\frac{(ac\pm3bd)^2}{p^2}+\frac{3(ad\mp bc)^2}{p^2}$

$\hspace{2cm}$ $\frac{N}{p}=\left(\frac{ac\pm3bd}{p}\right)^2+3\left(\frac{ad\mp bc}{p}\right)^2$

This tells us (given the above conditions):

$Np=e^2+3f^2 \implies p$ divides both $e$ and $f$ to give $\frac{N}{p}=g^2+3h^2$

Another way to state this is that if $xp=a^2+3b^2$ then $\frac{xp}{p}=c^2+3d^2$.

Theorem: Odd factors of $a^2+3b^2$ have this same form, when $\gcd(a,b)=1$

It is enough to prove the statement for odd primes, due to (i).

Assume that we have an integer in the form $a^2+3b^2$ that has some prime factor $p$, where $p>2$, that cannot be written in this same form.

$\therefore pq=a^2+3b^2$, where $q$ is an integer and $q>1$ (otherwise $p=a^2+3b^2$). $q$ can be odd or even. These will be considered as separate cases.

Case 1: q is even.

If $q$ is even, $a^2+3b^2$ is even, so $a$ and $b$ are either both even or both odd. As we are assuming the minimal case, we can assume $a$ and $b$ are both odd.

let $a=x+y$ and $b=x-y$ where $x$ is even, $y$ is odd. $\hspace{1cm}$($a=x-y$ and $b=x+y$ also works)

$\therefore pq=(x+y)^2+3(x-y)^2=x^2+2xy+y^2+3x^2-6xy+3y^2=4x^2-4xy+4y^2=4(x^2-xy+y^2)$

Using the fact that $x^2-xy+y^2=\left(\frac{(x-2y)}{2}\right)^2+3\left(\frac{x}{2}\right)^2$ and that $x$ is even, we can say that $pq=4(c^2+3d^2)$, where $c$ and $d$ are integers and $c^2+3d^2$ is odd.

So we can produce a smaller odd factor $z$ such that $pz=c^2+3d^2$, and we only need to consider the case where $q$ is odd.

Case 2: q is odd

As $q$ is odd, $a^2+3b^2$ is odd.

Let $a=xp+r$ and $b=yp+s$ where $|r|,|s|<\frac{p}{2}$

($r$ and $s$ being some remainder after dividing $a$ and $b$ by $p$)

$\therefore pq = (xp+r)^2+3(yp+s)^2$

$\hspace{0.9cm}$$= p(x^2p+2xr+3y^2p+6ys)+r^2+3s^2$

$\therefore p\mid r^2+3s^2$ where $r^2+3s^2$ is less than $p^2$, due to the following:

As $r,s<\frac{p}{2}\implies r^2+3s^2<\left(\frac{p}{2}\right)^2+3\left(\frac{p}{2}\right)^2$

$\hspace{4.9cm}$$<\frac{p^2}{4}+\frac{3p^2}{4}$

$\hspace{4.9cm}$$<\frac{4p^2}{4} $

$\hspace{4.9cm}$$< p^2$

Hence, we have $pn=r^2+3s^2$, where $n$ is an integer, $n< p$, and $n$ can be either odd or even. However, due to case 1, we can assume $n$ is odd (or produce a smaller odd factor), and as $n$ is odd, we can repeat the above process for $n$, to produce a smaller odd factor, and so on until we produce a prime $i$ such that $i=k^2+3l^2$ (otherwise we have a contradiction by infinite descent).

We can conclude then that due to (ii) and this process of descent, $n, p$  can be shown to have the form $a^2 +3b^2$.

For example, suppose we had shown that $ni=y^2+3z^2$ and so $n=w^2+3v^2$ due to (ii). Then either $n$ is prime or has odd prime factors $n_1,n_2...$ which we would be able to show have the same form. Then $n$ must be of the same form, and therefore also $p$.

This produces a contradiction as we assumed $p$ was not of the form $a^2+3b^2$.

Therefore all odd factors of $a^2+3b^2$ have this same form.

References:

13 lectures on Fermat's Last Theorem, by Ribenboim.

Primes of the form $x^2+ny^2$ by Cox

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-n-3-a2-3b2.html

www.mathpages.com/home/kmath009/kmath009.htm

答:
I think that your argument has an error.

You have written

(i) First let us establish that the set $a^2+3b^2$ is closed under multiplication.

$\hspace{2.5cm}$$(a^2+3b^2)(c^2+3d^2)$

$\hspace{2cm}$$=(ac)^2+3(ad)^2+3(bc)^2+9(bd)^2$

$\hspace{2cm}$$=(ac)^2-6abcd+9(bd)^2+3(ad)^2+6abcd+3(bc)^2$

$\hspace{2cm}$$=(ac-3bd)^2+3(ad+bc)^2=e^2+3f^2$

$\hspace{2cm}$This also implies that $\frac{e^2+3f^2}{c^2+3d^2}=a^2+3b^2$

Then, you have written

q cannot be of the form $a^2+3b^2$

You are saying in the comments that the above argument proves the following claim :

Claim A : If $q$ is of the form $n^2+3m^2$, then $\dfrac{a^2+3b^2}{q}$ is of the form $n^2+3m^2$.

I think that you have not proven Claim A.

So, I think that you are claiming, without any proof, that $q$ cannot be of the form $a^2+3b^2$.

To see why your argument does not prove Claim A, let us consider the following claim :

Claim B : If $q$ is of the form $n^2+\color{red}7m^2$, then $\dfrac{a^2+\color{red}7b^2}{q}$ is of the form $n^2+\color{red}7m^2$.

The claim B is false. A counterexample is $\dfrac{3^2+7\times 7^2}{1^2+7\times 5^2}=2$ where $2$ is not of the form $n^2+7m^2$.

However, using your argument, I can "prove" that Claim B is true as follows :

First let us establish that the set $a^2+7b^2$ is closed under multiplication.

$\hspace{2.5cm}$$(a^2+7b^2)(c^2+7d^2)$

$\hspace{2cm}$$=(ac)^2+7(ad)^2+7(bc)^2+49(bd)^2$

$\hspace{2cm}$$=(ac)^2-14abcd+49(bd)^2+7(ad)^2+14abcd+7(bc)^2$

$\hspace{2cm}$$=(ac-7bd)^2+7(ad+bc)^2=e^2+7f^2$

$\hspace{2cm}$This also implies that $\frac{e^2+7f^2}{c^2+7d^2}=a^2+7b^2$

This shows that your argument does not prove Claim A.

If you want to use that "$q$ is not of the form $n^2+3m^2$", then, for example, you need to prove the following claim :

Claim C : If $a$ is of the form $n^2+3m^2$, and $b$ is not of the form $n^2+3m^2$, then $ab$ is not of the form $n^2+3m^2$.

If you can prove that Claim C is true, then you can say that $q$ cannot be of the form $n^2+3m^2$.


问:
Proof by induction that $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$
I am a bit new to logical induction, so I apologize if this question is a bit basic.

I tried proving this by induction:

$$\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$$

Starting with the base case $n=1$:

$$1^2\ge1^2$$

Then to prove that $P(n)\to P(n+1)$:

$$\left(\sum_{k=1}^{n+1}k\right)^2\ge\sum_{k=1}^{n+1}k^2$$
$$\left((n+1)\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$
$$(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$$

Since they're both multiplied by $(n+1)^2$, it's easy to see that if $\left(\sum_{k=1}^nk\right)^2\ge\sum_{k=1}^nk^2$, then $(n+1)^2\left(\sum_{k=1}^nk\right)^2\ge(n+1)^2\sum_{k=1}^nk^2$. But if the $\ge$ were replaced with a $\le$, the proof would still be valid, even though it's demonstrably not true.

I can see that it would take strong induction would fix this problem, but if I didn't know by observation that $\left(\sum_{k=1}^nk\right)^2\le\sum_{k=1}^nk^2$ isn't true by observation, how would I know to use strong induction?

Body:
$$\sum^n_1 k=1+2+ \dots n$$

$$(\sum^n_1 k)^2=(1+2+ \dots n)^2 \neq1^2+2^2 + \dots n^2$$


问:
Connections between Mersenne Primes and Codes
I recently watched the video for the talk The Biggest Known Prime Number (slides located here) by Keith Conrad, which is on Mersenne primes $M_n = 2^n-1$.
It is well-known that such numbers can be prime only if $n$ is prime, but that this condition is not sufficient for primality of $M_n$.
The smallest counterexample is $M_{11} = 2^{11} -1 = 23\cdot 87$.
Conrad's talk includes the remark:

That $2^{11}−1$ has factor 23 is related to an important construction in coding theory called the perfect binary Golay code(a special subset of 23-dimensional space over integers mod 2)

My questions are:

What is the relation precisely?

Are there further relations between divisors of "Mersenne composites" and coding-theoretic constructions?

答:
The connection is much simpler than I was imagining. It is well-known that the binary Golay code is perfect.
A code $\mathcal{C}\subseteq \mathbb{F}_2^n$ is called perfect if Hamming balls of radius $(d-1)/2$ centered at each codeword $c\in\mathcal{C}$ partition $\mathbb{F}_2^n$.

Note that the size of a Hamming ball of radius $r$ within $\mathbb{F}_2^n$ is $1 + \binom{n}{1}+\dots+\binom{n}{r}$.
For this to partition the whole space, this quantity must be equal to exactly $|\mathbb{F}_2^n|/|\mathcal{C}| = 2^{n-k}$, where $k$ is the rank of the code.
So we get the condition:

$$1 + \binom{n}{1}+\dots + \binom{n}{r} = 2^{n-k}\iff \binom{n}{1}+\dots + \binom{n}{r} = 2^{n-k}-1$$

As $n$ obviously divides $\sum_{i = 1}^r\binom{n}{i}$, a necessary condition for this equality (and therefore for $\mathcal{C}$ to be perfect) is for $n$ to divide $2^{n-k}-1$, hence $2^{11}-1$ being composite (and in particular being divisible by 23) is necessary for there to exist a rank 12 perfect code within $\mathbb{F}_2^{23}$.

There may be a deeper connection still, as perfect codes are much rarer than the above condition might imply. I'll probably leave this question open a few more days in case someone knows a deeper connection.


问:
For what $a$ and $b$ are there explicit expressions for $I(a, b) =\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b} $?
For what $a$ and $b$
are there explicit expressions for
$I(a, b)
=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}
$?

This is inspired by
the answer to
https://www.quora.com/What-is-displaystyle-int_-0-1-int_-0-1-frac-1-1-xy-3-mathrm-d-x-mathrm-d-y
where it is shown that
$I(1, 3)
=\dfrac34\left(\ln(3)+\dfrac{\pi\sqrt{3}}{9}\right)
$.

Here's what I've done
so far.

$\begin{array}\\
I(a, b)
&=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}\\
&=\int_0^1 \int_0^1 dx\,dy\sum_{n=0}^{\infty} (x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 \int_0^1 dx\,dy(x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 x^{an}dx \int_0^1 y^{bn}dy\\
&=\sum_{n=0}^{\infty} \dfrac{x^{an+1}}{an+1}\big|_0^1 \dfrac{y^{bn+1}}{bn+1}\big|_0^1\\
&=\sum_{n=0}^{\infty} \dfrac{1}{an+1} \dfrac{1}{bn+1}\\
&=ab\sum_{n=0}^{\infty} \dfrac{1}{abn+b} \dfrac{1}{abn+a}\\
\\
I(a.a)
&=\sum_{n=0}^{\infty} \dfrac{1}{(an+1)^2}\\
&=\dfrac1{a^2}\sum_{n=0}^{\infty} \dfrac{1}{(n+1/a)^2}\\
&=\dfrac1{a^2}\psi^{(1)}(1/a)\\
\text{If } a \ne b\\
I(a, b)
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \left(\dfrac{1}{abn+b} -\dfrac{1}{abn+a}\right)\\
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \int_0^1 \left(x^{abn+b-1} -x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\sum_{n=0}^{\infty}x^{abn+b-1} -\sum_{n=0}^{\infty}x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\sum_{n=0}^{\infty}x^{abn} -x^{a-1}\sum_{n=0}^{\infty}x^{abn}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\dfrac1{1-x^{ab}} -x^{a-1}\dfrac1{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\dfrac{x^{b-1} -x^{a-1}}{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 x^{b-1}\left(\dfrac{1 -x^{a-b}}{1-x^{ab}}\right)dx\\
\end{array}
$

But,
in general,
I can't go further.

Special cases can be handled.
For example,
if $b=1$
and $a$ is a positive integer,
we can show that
$I(a, 1)
=\dfrac{a}{(a-1)^2} \left(\ln(a)-\int_0^1 \left(\dfrac{\sum_{k=0}^{a-3}(a-2-k)x^k}{\sum_{k=0}^{a-1}x^k}\right)dx\right)
$.

This gives,
as above,
$I(3, 1)
=\dfrac34\left(\ln(3)-\int_0^1 \left(\dfrac{1}{1+x+x^2}\right)dx\right)
$
and completing the square
gives the answer above.

Also

$I(2, 1)
=2\ln(2)
$.

Similarly,

$\begin{array}\\
I(4, 1)
&=\dfrac49\left(\ln(4)-\int_0^1 \left(\dfrac{2+x}{1+x+x^2+x^3}\right)dx\right)\\
&=\dfrac49\left(\ln(4)- \dfrac{3 π + \ln(4)}{8}\right)
\quad\text{(According to Wolfy)}\\
&=\dfrac49\left(\dfrac{7\ln(4)-3 π}{8}\right)\\
\end{array}
$

We could get
$I(5, 1)$
since Wolfy gives
$$\dfrac1{1+x+x^2+x^3+x^4}
=\dfrac{-2 x + \sqrt{5} - 1}{\sqrt{5} (2 x^2 - (\sqrt{5}-1) x + 2)} + \dfrac{2 x + \sqrt{5} + 1}{\sqrt{5} (2 x^2 + (\sqrt{5}+1) x + 2)}
$$
but I'm not going to bother
to work it out.

答: The following is trivial due to expansion of digamma (formula 14 of this page) $$\sum _{n=0}^{\infty } \frac{1}{(a+n) (b+n)}=\frac{\psi ^{(0)}(a)-\psi ^{(0)}(b)}{a-b}$$ Thus OP's integral equals to $$I(a,b)=\sum _{n=0}^{\infty } \frac{a b}{(a b n+b) (a b n+a)}=\frac{\psi ^{(0)}\left(\frac{1}{b}\right)-\psi ^{(0)}\left(\frac{1}{a}\right)}{a-b}$$ Due to Gauss's digamma theorem (formula 11 of link above) for all $a,b\in \mathbb Q$ the integral can be expressed in terms of log and trig functions (even square roots, when $a,b$ are small). For instance $$I(3,5)=\frac{1}{2} \pi  \sqrt{\frac{1}{6} \left(-\sqrt{3+\frac{6}{\sqrt{5}}}+\frac{3}{\sqrt{5}}+2\right)}+\frac{1}{8} \log \left(\frac{3125}{729}\right)+\frac{1}{4} \sqrt{5} \coth ^{-1}\left(\sqrt{5}\right)$$


问:
Relating convergence of characteristic function to almost sure convergence - Exercise 10.4.10 in Cohn's Book Measure Theory
I'm working on Exercise 10.4.10 in Donald Cohn's book Measure Theory. The Exercise is the following:

Let $(\Omega,\mathscr{A},P)$ be a probability space, let $\{X_n\}$ be a sequence of independent random variables, let $S_N = \sum_{1}^{N} X_n$ be the corresponding partial sums, with distribution $P_{S_N} = PS_N^{-1} =:\mu_N$ and finally let $\{\phi_{\mu_N}\}$ be the sequence of characteristic functions $\phi_{\mu_N}(t) = \int_{\mathbb{R}}e^{ixt}d\mu_N(x)$. Prove that

If $\lim_n\phi_{\mu_N}(t)$ exists and is non-zero for a set of positive measure

Then $S_N$ converges almost surely.

We are given two Lemmas:

Lemma 1: If $\{a_n\}$ is a sequence of real numbers such that $e^{ita_n}$ converges for all $t$ in a set of positive Lebesgue measure then $a_n$ is convergent.

Lemma 2: If for a fixed $t$ the limit $\lim E(e^{itS_n})$ exists and is non-zero then $\{e^{itS_n}\}$ converges almost surely.

Since $\phi_{\mu_N}(t) = E(e^{itS_N})$ we are in the situation of applying Lemma 2. This Lemma says that $\lim_N e^{itS_N}$ exists almost surely on a set of $t$'s with positive Lebesgue measure, denote this set with $T$. Now I would like to apply Lemma 1, however this is not immediately possible. For this to work, we need to know that there is a set $A\subseteq \Omega$ with $P(A) = 1$ such that for any $\omega \in A$ the limit $\lim_N e^{itS_N(\omega)}$ exists for all $t\in T$. I don't see how this conclusion can be drawn.

答:
By the basic hypothesis there is a set  $H\subset\Bbb R$ with $0<\lambda(H)<\infty$ ($\lambda$ is Lebesgue measure) such that $\lim_nE[e^{itS_n}]$ exists and is  positive for each $t\in H$.
By Lemma 2, if $t\in H$ then there is an event $N_t$ with $P[N_t]=0$ such that $\lim_n e^{itS_n(\omega)}$ exists for all $\omega\notin N_t$.

Consider the (product measurable) set $G:=\{(\omega,t): \lim_n e^{itS_n(\omega)}$ exists$\}$. Observe that for each $t\in H$, $N_t^c$ (the complement  of $N_t$) is a subset of the section
$$
G^{(t)}:=\{\omega: (\omega,t)\in G\}.
$$
It follows that $P[G^{(t)}]=1$ for each $t\in H$.
Consequently, by Fubini's theorem,
$$
\eqalign{
P\otimes\lambda[(\Omega\times H)\cap G]
&=\int_{H}\left[\int_\Omega 1_G(\omega,t)\,P(d\omega)\right]\,\lambda(dt)\cr
&= \int_{H}P[G^{(t)}]\,\lambda(dt)\cr
&=\lambda(H)=P\otimes\lambda[\Omega\times H].\cr
}
$$
It follows that $P\otimes\lambda[(\Omega\times H)\setminus G]=0$.

Define $K:=\{\omega: \lambda(G_{(\omega)})>0\}$. We know from Lemma 1 that if $\omega\in K$ then $\lim_n S_n(\omega)$ exists. To conclude we need to check that $P[K]=1$; or, equivalently, that $P[K^c]=0$.
Using Fubini again,
$$
\eqalign{
0
&=P\otimes\lambda[(\Omega\times H)\setminus G]\cr
&=\int_\Omega\lambda(H\setminus G_{(\omega)})\,P(d\omega)\cr
&\ge\int_{K^c}\lambda(H\setminus G_{(\omega)})\,P(d\omega)\cr
&=\int_{K^c}\lambda(H)\,P(d\omega) = \lambda(H)P(K^c).\cr
}
$$
(The next-to-last equality follows from  the fact that $\lambda(G_{(\omega)})=0$ if $\omega\in K^c$.)
Because $\lambda(H)>0$, we must have $P[K^c]=0$, as desired.


答:
May be, we could start with
$$I(a, b)=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and consider
$$J_a=\int_0^1 \, _2F_1\left(1,\frac{1}{b};\frac{1}{b}+1;x^a\right)\,dx$$ and discarding the ones which contain logarithms of complex arguments, we can obtain a few of them
$$J_1=\frac{\psi \left(\frac{1}{b}\right)+\gamma }{1-b}$$
$$J_2=\frac{\psi \left(\frac{1}{b}\right)+\gamma +\log (4)}{2-b}$$
$$J_3=\frac{6 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +6 \gamma +9 \log   (3)}{18-6 b}$$
$$J_4=\frac{2 \left(\psi \left(\frac{1}{b}\right)+\gamma \right)+\pi +\log (64)}{8-2 b}$$
$$J_6=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +2 \gamma +\log
   (432)}{12-2 b}$$
$$J_8=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{2} \pi +\pi +2 \gamma +\log
   (256)+2 \sqrt{2} \coth ^{-1}\left(\sqrt{2}\right)}{16-2 b}$$
$$J_{12}=\frac{2 \psi \left(\frac{1}{b}\right)+\sqrt{3} \pi +2 \pi +2 \gamma +2
   \sqrt{3} \log \left(2+\sqrt{3}\right)+\log (1728)}{24-2 b}$$

For the particular case where $b=a$, the result seems to be quite simpler
$$K_a=\int_0^1 \, _2F_1\left(1,\frac{1}{a};\frac{1}{a}+1;x^a\right)\,dx=\frac 1 {a^2}\psi ^{(1)}\left(\frac{1}{a}\right)$$

Body:
For what $a$ and $b$
are there explicit expressions for
$I(a, b)
=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}
$?

This is inspired by
the answer to
https://www.quora.com/What-is-displaystyle-int_-0-1-int_-0-1-frac-1-1-xy-3-mathrm-d-x-mathrm-d-y
where it is shown that
$I(1, 3)
=\dfrac34\left(\ln(3)+\dfrac{\pi\sqrt{3}}{9}\right)
$.

Here's what I've done
so far.

$\begin{array}\\
I(a, b)
&=\int_0^1 \int_0^1 \dfrac{dx\,dy}{1-x^ay^b}\\
&=\int_0^1 \int_0^1 dx\,dy\sum_{n=0}^{\infty} (x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 \int_0^1 dx\,dy(x^ay^b)^n\\
&=\sum_{n=0}^{\infty} \int_0^1 x^{an}dx \int_0^1 y^{bn}dy\\
&=\sum_{n=0}^{\infty} \dfrac{x^{an+1}}{an+1}\big|_0^1 \dfrac{y^{bn+1}}{bn+1}\big|_0^1\\
&=\sum_{n=0}^{\infty} \dfrac{1}{an+1} \dfrac{1}{bn+1}\\
&=ab\sum_{n=0}^{\infty} \dfrac{1}{abn+b} \dfrac{1}{abn+a}\\
\\
I(a.a)
&=\sum_{n=0}^{\infty} \dfrac{1}{(an+1)^2}\\
&=\dfrac1{a^2}\sum_{n=0}^{\infty} \dfrac{1}{(n+1/a)^2}\\
&=\dfrac1{a^2}\psi^{(1)}(1/a)\\
\text{If } a \ne b\\
I(a, b)
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \left(\dfrac{1}{abn+b} -\dfrac{1}{abn+a}\right)\\
&=\dfrac{ab}{a-b}\sum_{n=0}^{\infty} \int_0^1 \left(x^{abn+b-1} -x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\sum_{n=0}^{\infty}x^{abn+b-1} -\sum_{n=0}^{\infty}x^{abn+a-1}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\sum_{n=0}^{\infty}x^{abn} -x^{a-1}\sum_{n=0}^{\infty}x^{abn}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(x^{b-1}\dfrac1{1-x^{ab}} -x^{a-1}\dfrac1{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 \left(\dfrac{x^{b-1} -x^{a-1}}{1-x^{ab}}\right)dx\\
&=\dfrac{ab}{a-b} \int_0^1 x^{b-1}\left(\dfrac{1 -x^{a-b}}{1-x^{ab}}\right)dx\\
\end{array}
$

But,
in general,
I can't go further.

Special cases can be handled.
For example,
if $b=1$
and $a$ is a positive integer,
we can show that
$I(a, 1)
=\dfrac{a}{(a-1)^2} \left(\ln(a)-\int_0^1 \left(\dfrac{\sum_{k=0}^{a-3}(a-2-k)x^k}{\sum_{k=0}^{a-1}x^k}\right)dx\right)
$.

This gives,
as above,
$I(3, 1)
=\dfrac34\left(\ln(3)-\int_0^1 \left(\dfrac{1}{1+x+x^2}\right)dx\right)
$
and completing the square
gives the answer above.

Also

$I(2, 1)
=2\ln(2)
$.

Similarly,

$\begin{array}\\
I(4, 1)
&=\dfrac49\left(\ln(4)-\int_0^1 \left(\dfrac{2+x}{1+x+x^2+x^3}\right)dx\right)\\
&=\dfrac49\left(\ln(4)- \dfrac{3 π + \ln(4)}{8}\right)
\quad\text{(According to Wolfy)}\\
&=\dfrac49\left(\dfrac{7\ln(4)-3 π}{8}\right)\\
\end{array}
$

We could get
$I(5, 1)$
since Wolfy gives
$$\dfrac1{1+x+x^2+x^3+x^4}
=\dfrac{-2 x + \sqrt{5} - 1}{\sqrt{5} (2 x^2 - (\sqrt{5}-1) x + 2)} + \dfrac{2 x + \sqrt{5} + 1}{\sqrt{5} (2 x^2 + (\sqrt{5}+1) x + 2)}
$$
but I'm not going to bother
to work it out.


问:
Prove if the remainder is relatively prime to the divisor, the dividend is also relatively prime to the divisor and vice versa?
The remainder of $a÷b$ is $c$, both $a$ and $b$ are positive integers.

How to prove:

$gcd(c,b)=1 \iff gcd(a,b)=1$

$gcd(a,b)=1 \iff gcd(c,b)=1$

答:
let $a = bq + c$ for some integer $q$.

Then if say $gcd(a,b) = 1$, and $gcd(b,c) = m$,

then, from $a = bq +c$ ;

we will get $m|a$ but $m$ also divides $b$ $\implies$ $m$ is a common divisor of $a$ and $b$ , And the only common divisor of $a$ and $b$ is 1.


问:
Prove equality in triangle inequality for complex numbers
We need to show that
$$ |z_{1}+z_{2}+\cdots+z_{n}|=|z_{1}|+|z_{2}|+\cdots+|z_{n}|$$
if and only if $z_{1},z_{2},\dots,z_{n}$ have the same argument (i.e. $z_{j}=r_{j}e^{i\theta}$ for $j=1,\dots,n$).

This way [$\Longleftarrow$] is easy, but for the other way around, I get stuck proving it for the sum of two complex numbers, even though the induction itself isn't that big a problem.

Does anyone happen to know how to prove this for two complex numbers?

答:
Let $z_1=a_1+ib_1$ and $z_2=a_2+ib_2$ for $a_1,a_2,b_1,b_2\in\mathbb{R}$. By simplification the equation
$$\sqrt{(a_1+a_2)^2+(b_1+b_2)^2}=\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2},$$ you will get $$\frac{a_1}{a_2}=\frac{b_1}{b_2}.$$ What does it tell about arguments?</p>


问:
Let $X$ be a topological space. Any connected component of $X$ is union of some irreducible components of $X$. How to show this?
I was thinking of writing $C$ any connected component of $X$ as union of the irreducible component of $\{x\}$ where $x$ runs over $C$. But I have doubt that the irreducible component of $X$ containing $\{x\}$ will be contained in $C$  or not. Please help how to proceed?

答: Let $C$ be a connected component of $X$, and let $x\in C$. Let $F$ be the irreducible component containing $x$. Then $F$ is a connected set containing $x$, and $C$ is the maximal connected set containing $x$, so $F\subseteq C$. (Of course in Hausdorff spaces this is not very interesting, since in that case we have $F=\{x\}$.)


问:
Is there a direct proof for $\int_0^{2\pi}\frac{r^2+r(\cos t-\sin t)}{1+2r\cos t+r^2}dt=2\pi$
Is there a direct method to prove that, if $r>1$
$$I=\int_0^{2\pi}\frac{r^2+r(\cos(t)-\sin(t))}{1+2r\cos(t)+r^2}dt=2\pi$$

I ask this question because this integral can be thought as an integral of a 1-form along a circle:
$\omega=\frac{x-y}{x^2+y^2}dx+\frac{x+y}{x^2+y^2}dy$
and $\gamma=(1+r\cos(t),r\sin(t))$ for $t\in[0,2\pi]$ and if I'm not mistaken this curve is homotopic to the unit circle centered in $(0,0)$ if $r>1$, then the integral is equal to the integral along this curve that is $2\pi.$ Is my argument correct?

答: One can use the result $$\int_0^{\pi}\frac{dx}{a+b\cos x} =\frac{\pi} {\sqrt{a^2-b^2}}, a>|b|\tag{1}$$ which can be proved using the substitution $$(a+b\cos x) (a-b\cos u) =a^2-b^2$$ The given integral in question can be written as $$\frac{1}{2}\int_{0}^{2\pi}\left(1+\frac{r^2-1-2r\sin t} {1+2r\cos t+r^2}\right)\,dt$$ which equals $$\pi+(r^2-1)\int_{0}^{\pi}\frac{dt}{1+2r\cos t+r^2}+0$$ or $$\pi+\frac{\pi(r^2-1)}{\sqrt{(1+r^2)^2-4r^2}}=2\pi$$ as $|r|>1$. The integral is $0$ if $|r|<1$ and undefined if $|r|=1$.


问:
Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$
Evaluate $360$ ($\frac{1}{15}$+$\frac{1}{105}$+$\frac{1}{315}$+$\cdots$).

My Work :-

Well we can clearly see that $15 = 1*3*5$ ; $105 = 3*5*7$ ; $315 = 5*7*9$. So I basically know the pattern, but I want to know how to apply that like maybe we can break $\frac{1}{15}$ into $x*\frac{1}{1}\pm y*\frac{1}{3}\pm z*\frac{1}{5} $ or something like that so that we can cancel some terms in the whole sum.

Body:
$T_n = \dfrac{1}{(2n-1)(2n+1)(2n+3)}$

Let $V_n= \dfrac{1}{(2n-1)(2n+1)}=(2n+3)T_n$

$V_{n} - V_{n+1}=\dfrac{1}{(2n-1)(2n+1)}-\dfrac{1}{(2n+1)(2n+3)}=\dfrac{2}{(2n-1)(2n+1)(2n+3)}=2T_n$

Let $\sum_{n=1}^{\infty}T_n=S$

Therefore,

$\sum_{n=1}^{\infty}(V_n - V_{n+1})=(V_1-V_{\infty})=2S$

$V_1=\frac{1}{3},V_{\infty}=0$

Therefore, $S=\frac{1}{6}$ and $360S = 60$

A telescopic series!


Title: Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$

Body:
Evaluate $360$ ($\frac{1}{15}$+$\frac{1}{105}$+$\frac{1}{315}$+$\cdots$).

My Work :-

Well we can clearly see that $15 = 1*3*5$ ; $105 = 3*5*7$ ; $315 = 5*7*9$. So I basically know the pattern, but I want to know how to apply that like maybe we can break $\frac{1}{15}$ into $x*\frac{1}{1}\pm y*\frac{1}{3}\pm z*\frac{1}{5} $ or something like that so that we can cancel some terms in the whole sum.

Body:
Hint: $$\frac{1}{105}=\frac{1}{4}(\frac{1}{3\cdot 5}-\frac{1}{5\cdot 7})$$ $$\frac{1}{315}=\frac{1}{4}(\frac{1}{5\cdot 7}-\frac{1}{ 7\cdot 9})$$  do you
observe a pattern........


问:
How is the depth of a parabolic mirror calculated
How is the depth of a parabolic mirror calculated. I couldn't find a similar question/answer.

I have this specific problem to solve (not related to parabolic mirrors, but is a starting point, and not interested in a focal point):

L=6"

D=4.5"

What is d?

Thanks for your help

答:
Let's take the origin of your coordinates at the vertex of your parabola.
(The minima of the parabola corresponds to its vertex.)

Since the parabola opens up, it's of the form

\begin{gather*}
y=ax^{2}\\
\end{gather*}
for any real number a.

From our figure, it seems that the abolute value of the x-coordinates of the endpoints of the parabola are the same, so let's take it as x.

The distance between the endpoints is 4.5.
So

\begin{gather*}
2x=4.5\\
x=\frac{9}{4}\\
\end{gather*}

The length of a curve is basically the integral

\begin{equation*}
\int ^{b}_{a}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} dx
\end{equation*}
From the equation of the parabola we took

\begin{equation*}
L=\int ^{\frac{9}{4}}_{-\frac{9}{4}}\sqrt{1+4a^{2} x^{2}} dx=6
\end{equation*}

The integral can be calculated by the method of integration by parts, and the resulting integrand turns out to be

\begin{equation*}
L=\frac{1}{2a}arcsinh\left(\frac{9a}{2}\right) +\frac{9}{8}\sqrt{81a^{2} +4}
\end{equation*}
Equate the length to 6 and you get the equation

\begin{equation*}
6=\frac{1}{2a}arcsinh\left(\frac{9a}{2}\right) +\frac{9}{8}\sqrt{81a^{2} +4}
\end{equation*}
This eqn is almost impossible to solve analytically.
However, since you just want an approximate answer, plugging this eqn into WolframAlpha gives the approximate solution as

\begin{equation*}
a=0.3553
\end{equation*}

As d is just the value of the function at its extreme endpoints,

\begin{equation*}
d=ax^{2} =0.3553\times \left(\frac{9}{4}\right)^{2} \approxeq 1.8
\end{equation*}
Hope this solves your queries:)


问:
Probability of guessing the likely number of question that he was sure on a multiple choice test.
Ian completed a multiple-choice test with four options for each question. He was 90% sure of his answer on some of the questions. But on questions in which he wasn't 90% sure, he randomly guessed an answer. If there were 20 questions, and he got 11 right, what was the most likely number of questions that he was 90% sure on?

Note: choices are independent of each

P.S. - This a question written by our TA and may not be exactly logical.

My attempt:
The form mutually exclusive and exhaustive set of events. So P(knows,correct)+P(guess,correct)+P(guess,incorrect)=1 and it is given P(knows,correct)=0.9 and his probability in case he doesn't know the answer and guess out of 4 option   is 0.25 .

```<code>                        So finally-> (1 x .09) + (.25 x .1)
</code>```

Answer provided:9

答:
Easiest approach here is (in the mud) trial and error.

Let $q = $ number of questions that he was 90% sure on.

Let $e(q) = $ expected number of right answers, based on $q$.

Try different values of $q$.

$e(8) = (0.9 \times 8) + (0.25 \times 12) = 10.2.$

Note that as $q$ increases by 1, $e(q)$ increases by $(0.65)$.

Therefore, $e(9) = 10.85$ and $e(10) = 11.5$.

$q = 9$ gives $e(q)$ closest to 11.

Edit

Possible criticism of my answer.

I have made the (intuitive) assumption that the most likely value of $q$ is the one that yields $e(q)$ closest to 11.  In fact, my knowledge of probability theory is insufficient to formally prove this quickly; however the assumption does seem reasonable.


答:
Let $q$ be the number of questions that Ian was confident on. Notice that because there are 20 questions, and Ian always must have been either confident or guessing on every question, he guessed on $20 - q$ questions.

So, the expected number of correct answers is $E(q) = 0.9(q) + 0.25(20-q) = 5 + 0.65q$.

Let's set $E$ equal to 11: $0.65q + 5 = 11 \to 0.65q = 6 \to q = 9.23$, which is closest to $9$.

Title: Probability of guessing the likely number of question that he was sure on a multiple choice test.

Body:
Ian completed a multiple-choice test with four options for each question. He was 90% sure of his answer on some of the questions. But on questions in which he wasn't 90% sure, he randomly guessed an answer. If there were 20 questions, and he got 11 right, what was the most likely number of questions that he was 90% sure on?

Note: choices are independent of each

P.S. - This a question written by our TA and may not be exactly logical.

My attempt:
The form mutually exclusive and exhaustive set of events. So P(knows,correct)+P(guess,correct)+P(guess,incorrect)=1 and it is given P(knows,correct)=0.9 and his probability in case he doesn't know the answer and guess out of 4 option   is 0.25 .

```<code>                        So finally-> (1 x .09) + (.25 x .1)
</code>```

Answer provided:9


问:
What do cones have to do with quadratics? Why is $2$ special?
I've always been nagged about the two extremely non-obviously related definitions of conic sections (i.e. it seems so mysterious/magical that somehow slices of a cone are related to degree 2 equations in 2 variables). Recently I came across the following pages/videos:

This 3B1B video about ellipses, which reignited my desire to understand conics

Why are quadratic equations the same as right circular conic sections?, which offers a very computational approach to trying to resolve this question

Another 3B1B video on visualizing Pythagorean triples (i.e. finding the rational points of a circle)

and Manjul Bhargava's lecture on the Birch-Swinnerton-Dyer Conjecture, where minutes ~10-15 discuss the complete solution to the problems of rational points on conics.

While 3B1B's video makes a lot of sense and is very beautiful from a geometric standpoint, it does not talk about any of the other conics, or discuss the relationship with "degree 2". Moreover, the 2nd 3B1B video I linked and then Bhargava's lecture highlights "degree 2" as something we understand well, compared to higher degrees (reminds me a little bit of Fermat's last theorem and the non-existence of solutions for $n>2$).

So, I suppose my questions are as follows:

Why, from an intuitive standpoint, should we expect cones to be deeply related to zero-sets of degree 2 algebraic equations?

and more generally:

Is there some deep reason why "$2$" is so special? I've often heard the quip that "mathematics is about turning confusing things into linear algebra" because linear algebra is "the only subject mathematicians completely understand"; but it seems we also understand a lot of nice things about quadratics as well  -- we have the aforementioned relationship with cones, a complete understanding of rational points, and the Pythagorean theorem (oh! and I just thought of quadratic reciprocity). 2 is also special in all sorts of algebraic contexts, as well as being the only possible finite degree extension of $\mathbb R$, leading to in particular $\mathbb C$ being 2-dimensional.

Also interesting to note that many equations in physics are related to $2$ (the second derivative, or inverse square laws), though that may be a stretch. I appreciate any ideas you share!

$$\rule{5cm}{0.4pt}$$

EDIT 3/12/21: was just thinking about variances, and least squares regression. "$2$" is extremely special in these areas: Why square the difference instead of taking the absolute value in standard deviation?, Why is it so cool to square numbers (in terms of finding the standard deviation)?, and the absolutely mindblowing animation of the physical realization of PCA with Hooke's law: Making sense of principal component analysis, eigenvectors & eigenvalues.

In these links I just listed, seems like the most popular (but still not very satisfying to me) answer is that it's convenient (smooth, easy to minimize, variances sum for independent r.v.'s, etc), a fact that may be a symptom of a deeper connection with the Hilbert-space-iness of $L^2$. Also maybe something about how dealing with squares, Pythagoras gives us that minimizing reconstruction error is the same as maximizing projection variance in PCA. Honorable mentions to Qiaochu Yuan's answer about rotation invariance, and Aaron Meyerowitz's answer about the arithmetic mean being the unique minimizer of sum of squared distances from a given point. As for the incredible alignment with our intuition in the form of the animation with springs and Hooke's law that I linked, I suppose I'll chalk that one up to coincidence, or some sort of SF ;)

$$\rule{5cm}{0.4pt}$$

EDIT 2/11/22:
I was thinking about Hilbert spaces, and then wondering again why they behave so nice, namely they have the closest point lemma (leading to orthogonal decomposition $\mathcal H = \mathcal M \oplus \mathcal M^\perp$ for closed subspaces $\cal M$), or orthonormal bases (leading to Parseval's identity, convergence of a series of orthogonal elements if and only if the sum of the squared lengths converge), and I came to the conclusion that the key result each time seemed to be the Pythagorean theorem (e.g. the parallelogram law is an easy corollary of Pythag). So that begs the questions, why is the Pythagorean theorem so special? The linked article in the accepted answer of this question:  What does the Pythagorean Theorem really prove? tells us essentially the Pythagorean theorem boils down to the fact that right triangles can be subdivided into two triangles both similar to the original.

The fact that this subdivision is reached by projecting the vertex onto the hypotenuse (projection deeply related to inner products) is likely also significant... ahh, indeed by the "commutativity of projection", projecting a leg onto the hypotenuse is the same as projecting the hypotenuse onto the leg, but by orthogonality of the legs, the projection of the hypotenuse onto the leg is simply the leg itself! The square comes from the fact that projection scales proportionally to the scaling of each vector, and there are two vectors involved in the operation of projection.

I suppose this sort of "algebraic understanding" of the projection explains the importance of "2" more than the geometry, since just knowing about the "self-similarity of the subdivisions" of the right triangle, one then has to wonder why say tetrahedrons or other shapes in other dimensions don't have this "self-similarity of the subdivisions" property. However it is still not clear to me why projection seems to be so fundamentally "2-dimensional". Perhaps 1-dimensionally, there is the "objective" perception of the vector, and 2-dimensionally there is the "subjective" perception of one vector in the eyes of another, and there's just no good 3-dimensional perception for 3 vectors?

There might also be some connection between the importance of projection and the importance of the Riesz representation theorem (all linear "projections" onto a 1-dimensional subspace, i.e. linear functionals, are actually literal projections against a vector in the space).

$$\rule{5cm}{0.4pt}$$

EDIT 2/18/22: again touching on the degree 2 Diophantine equations I mentioned above, a classical example is the number of ways to write $k$ as the sum of $n$ squares $r_n(k)$. There are a number of nice results for this, the most famous being Fermat's 2-square theorem, and Jacobi's 4-square theorem. A key part of this proof was the use of the Poisson summation formula for the Euler/Jacobi theta function $\theta(\tau) := \sum_{n=-\infty}^\infty e^{i \pi n^2 \tau}$, which depends on/is heavily related to the fact that Gaussians are stable under the Fourier transform. I still don't understand intuitively why this is the case (see Intuitively, why is the Gaussian the Fourier transform of itself?), but there seems to be some relation to Holder conjugates and $L^p$ spaces (or in the Gaussian case, connections to $L^2$), since those show up in generalizations to the Hardy uncertainty principle (“completing the square”, again an algebraic nicety of squares, was used in the proof of Hardy, and the Holder conjugates may have to do with the inequality $-x^p + xu \leq u^q$ -— Problem 4.1 in Stein and Shakarchi’s Complex analysis, where the LHS basically comes from computing the Fourier transform of $e^{-x^p}$) Of course why the Gaussian itself appears everywhere is another question altogether: https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics.

This (squares leading to decent theory of $r_n(k)$, and squares leading to nice properties of the Gaussian) is probably also connected to the fact that $\int_{\mathbb R} e^{-x^2} d x$ has a nice explicit value, namely $\sqrt \pi$. I tried seeing if there was a connection between this value of $\pi$ and the value of $\pi$ one gets from calculating the area of a circle "shell-by-shell" $\frac 1{N^2} \sum_{k=0}^N r_2(k) \to \pi$, but I couldn't find anything: Gaussian integral using Euler/Jacobi theta function and $r_2(k)$ (number of representations as sum of 2 squares).

$$\rule{5cm}{0.4pt}$$

EDIT 10/13/22: I was recently learning about Riemannian geometry, and indeed the metric tensor is a bilinear form (cf. above discussion on inner products), and the Riemann curvature tensor (or curvature in general) is all about the second (covariant) derivative. Taking traces we arrive at the Ricci curvature tensor (used in no less important things than Einstein's general relativity, whose "classical approximation" Newtonian gravity follows an inverse square law; and Perelman's proof of the Poincare conjecture!) or scalar curvature, which can be interpreted geometrically as the second-order change in volume of balls/base of cones (see https://math.stackexchange.com/a/469005/405572 or Equation (10) of https://arxiv.org/pdf/2201.04923v1.pdf).

And of course like Ricci flow we also have the heat/diffusion differential equations (and the Schrodinger equation, quoting James Gleck, is "diffusion through imaginary time"), and endless equations involving the Laplacian, all second-order differential equations (how AMAZING that the two great pillars of 20th century physics, general relativity and quantum mechanics both have at their hearts second-order differential equations! And the fact that observables in quantum mechanics are modeled as operators on Hilbert spaces). Relating to the Laplacian, we have the important concept of harmonicity, with beautiful manifestations/consequences in complex analysis, or PDEs, in the form of elliptic regularity. Besides tensors of 2nd derivatives, we have the ever-present Hessian matrix of second derivatives. I'll end with some quotes from a MO answer I linked here in a comment several months ago (https://mathoverflow.net/a/171743/112504) which tries to argue that "Polynomials are useful because quadratic polynomials are useful.":

So the question becomes: why are quadratic polynomials useful? There seem to be two different but interacting reasons. The first is that quadratic functions of a real variable are always either convex or concave and therefore have a unique maximum or minimum. The second is that quadratic functions are intimately related to bilinear forms and therefore can be accessed using linear algebra. The combination of these two reasons seems to explain the success of quadratic algebra in analysis and geometry (e.g. Hilbert spaces, Riemannian manifolds)

答:
The index number 2 is special in connection with the way that angles can be defined from distances.

There are many possible distance functions (norms) which can be defined, but most of them do not allow angles to be defined in a consistent way. Angles are defined from an inner product (dot product) and this is only defined if the norm obeys the quadratic expression $$||u+v||^2+||u-v||^2=2||u||^2+2||v||^2$$ for any vectors $u$ and $v$.

In a space with a different norm there are fewer rotations. There may be only a finite number of possible rotations of a circle or a sphere. A "cone" in 3d $(x,y,z)$ defined by $||x+y||=||z||$ can still be intersected by planes and a family of (nonquadratic) curves found.

In the usual geometry angles are defined, so there is a quadratic expression which must be satisfied by lengths.


问:
Find a minimal sufficient statistic for $U(\theta,\theta+c)$, where $(\theta,c)$ unknown.

Suppose $X_1,\cdots,X_n$ are $i.i.d$ from a distribution with p.d.f
$$\delta_{(\theta,c)}(x)=\frac{1}{c}\mathbb{1}_{(x\in[\theta,\theta+c])},$$
where $\theta\in\mathbb{R}$ and $c\in\mathbb{R}^+$ unknown.

Find a minimal sufficient statistic for $(\theta,c)$.

From the range of $x_i$, i.e. $x_i\in[\theta,\theta+c]$, we can determine that $\theta\leq x_i$ and $\theta+c\geq x_i$, $\forall i\in\{1,\cdots,n\}$. This implies $$\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)},$$ where $x_{(1)}=\underset{i\in\{1,\cdots,n\}}{\min}x_i$ and $x_{(n)}=\underset{i\in\{1,\cdots,n\}}{\max}x_i$.

The plots shows the area of $\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)}.$ It look like that $(x_{(1)},x_{(n)})$ is a minimal sufficient statistic for $(\theta,c)$ because $(x_{(1)},x_{(n)})$ uniquely determines the shape of the log-likelihood function. Is this correct?

答:
(Nobody answers so I post this answer which I am not sure is correct)

We show $(Y_{(1)},Y_{(n)})$ is a sufficient complete statistic, which implies $(Y_{(1)},Y_{(n)})$ is minimal sufficient. We first give the p.d.f of $(Y_{(1)},Y_{(n)})$
$$f(y_1, y_n,\theta,c)=\frac{n(n-1)}{c^n}(y_n-y_1)^{n-2},\quad\forall \theta\leq y_1\leq y_n\leq \theta+c\text{ and } (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$

Then, for any function $g(x_1,x_n)$ so that $\mathbb{E}_{\theta,c}\left[g(y_1,y_n)\right]=0,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$, we have $$0=\frac{n(n-1)}{c^n}\int_{\theta}^{\theta+c}\int_{\theta}^{y_n}g(y_1,y_n)(y_n-y_1)^{n-2} dy_1dy_n,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$
The integral area of $(y_1, y_n)$ is a triangle with vertices $(\theta,\theta)$, $(\theta,\theta+c)$ and $(\theta+c,\theta+c)$. With varying $\theta\in\mathbb{R}$ $c\in\mathbb{R}^+$, these triangles generate the Beral $\sigma$-algebra of $\mathcal{B}=\{(x,z)\in\mathbb{R}^2:x\leq z\}$. Thus
$$0=\frac{n(n-1)}{c^n}\int_{A}g(y_1,y_n)(y_n-y_1)^{n-2} d(y_1,y_n),\text{ for any Borel set }A\subset\mathcal{B}.$$
This means $$g(y_1,y_n)(y_n-y_1)^{n-2}\equiv0,a.e.\iff g\equiv 0,a.e.$$
Thus we conclude that $(Y_{(1)},Y_{(n)})$ is a complete statistic for $(\theta,c).$

Since $(Y_{(1)},Y_{(n)})$ is sufficient by factorization theorem, we conclude that $(Y_{(1)},Y_{(n)})$ is minimal sufficient.


问:
How to solve $\int_0^\infty J_0(x)\ \text{sinc}(\pi\,x)\ e^{-x}\,\mathrm dx$?
I need some help with solving this integral involving Bessel function:

$\hspace{2in}\displaystyle\int_0^\infty$$J_0(x)\ $$\text{sinc}(\pi\,x)\ $$e^{-x}\,\mathrm dx.$

答:
Prudnikov-Brychkov-Marychev (Vol. II, formula 2.12.8.5) tells us that
$$\int_0^{\infty}\frac{1-e^{-px}}{x}J_0(x)dx=\ln\left(p+\sqrt{p^2+1}\right)$$
Taking the difference of this formula with $p=1-i\pi$ and $p=1+i\pi$, one obtains
\begin{align}\int_0^{\infty}J_0(x)\mathrm{sinc}(\pi x)e^{-x}dx&=\frac{1}{2\pi i}\ln\frac{1+i\pi+\sqrt{(1+i\pi)^2+1}}{1-i\pi+\sqrt{(1-i\pi)^2+1}}=\\
&=\frac{1}{\pi}\arctan\frac{\pi\sqrt{2}+\sqrt{\pi^2-2+\sqrt{\pi^4+4}}}{
\sqrt{2}+\sqrt{2-\pi^2+\sqrt{\pi^4+4}}}.
\end{align}</p>


问:
How do you find the independent term in $(x + 1 + x^{-1})^4$?
I know you could expand it all out and take cases but is there a way to group it such that there is a general term? I've been trying to group it in order to establish a general term ie something like $((x + 1) + x^{-1})^4$ to make a binomial. Any hints?

答:
$$\left(x+\dfrac1x+1\right)^4=\sum_{r=0}^4\binom4r\left(x+\dfrac1x\right)^r$$

As the general term of $\left(x+\dfrac1x\right)^r$  is $$\binom rm x^{r-2m},$$

only the even $r$ will give $x$ independent terms

So, the required constant term $$=\binom40+\binom42\binom21+\binom44\binom42$$


答:
The contributions to the independent term are of the forms

$1,1,1,1$ ($1$ way)

$x,x^{-1},1,1$ ($4×3=12$ ways)

$x,x,x^{-1},x^{-1}$ ($\binom42=6$ ways)

Hence the independent term is $1+12+6=19$.

Title: How do you find the independent term in $(x + 1 + x^{-1})^4$?

Body: I know you could expand it all out and take cases but is there a way to group it such that there is a general term? I've been trying to group it in order to establish a general term ie something like $((x + 1) + x^{-1})^4$ to make a binomial. Any hints?


问:
Numerical method for functional equations similar to $2f(x) - f(x+2) - f(x-2) = 0$
I have some functional equations, that are of the form similar to something like
$$
2f(x) - f(x+2) - f(x-2) = 0 \text,
$$
which I need to find the solution to. I know that equations like these are generally hard to solve analytically (if even possible) so I was hoping to know if there were any numerical methods that could be used to solve functional equations of this form.

In general I'm looking to solve functional equations of the form
$$
\sum_{i=1}^n a_i(x) f\bigl(g_i(x)\bigr) = 0 \text.
$$

I know in some cases it is entirely possible to have there be multiple solutions, so that may throw a wrench in the works.

答:
Set $g(x)=f(x)-f(x-2)$ to find $g(x+2)=g(x)$. So $g$ is any $2$-periodic function, not even continuity is required. Next set $h(x)=f(x)-\frac12xg(x)$ to find
$$\begin{multline}
h(x)-h(x-2)=f(x)-f(x-2)-\frac12(xg(x)-(x-2)g(x-2))\\=g(x)-\frac12(x-(x-2))g(x)=0.
\end{multline}$$
So $h$ is another arbitrary $2$-periodic function.

In total there are many solutions $f(x)=h(x)+\frac12xg(x)$ to that functional equation. It is not clear what a numerical method should achieve.


问:
Expectation of random variable with differently weighted realizations
Let $Y=wX$, where $X$ is a random variable with mean $\mu$ and variance $\sigma^2$, and $w$ is the weight which could change for every realization of $Y$. In other words, $N$ realizations of $Y$ are given as $y_n$ = $w_n x_n,~n=1,...,N$. The weights $w_n,~\forall n$, are assumed to be $N$ uniformly spaced samples of a positive continuous function $W(t),~0<t<T$.

I wonder how can we express the population mean $E[Y]$?

My guess is that $E[Y]$ is defined as $E[Y]=\bar{w}E[X]=\bar{w}\mu$, where $\bar{w}=\frac{1}{T}\int^T_0 W(t) dt$.

Does that reasoning make sense?

答:
The Expectation of $Y$ is conditioned with the numbers of samples $n$ so

$$E(w|n) = \bar{w} = \frac{1}{n}\sum_{i=1}^{n}w_{n}$$

and so just use it in

$$E(Y|n) = E(w|n)E(X)$$.

You don't need the average of W because some values are irrelevant given a number $n$. Your calculation might only be APPROXIMATELY true if $w_{n}$  is near $W(T)$ and $n$ must be very large too.


问:
Predicting the average number of transitions given a frequency and a set size?
Say I wanted to predict the number of Software Engineers in a given year that change their job.

The total number of Software Engineers in a given market is readily available (4.4 million in the US), as are retention rates for companies (conservatively, an average of 3 years).

So, an average engineer will change jobs every 3 years or more. It feels like I should be fairly trivially able to project a number of software engineers who change jobs in a given year given this data, but, for the life of me, I don't know how.

Any help would be appreciated.

答:
Comment:

In queueing theory, Little's Formula is $L = \lambda W,$ where $L$ is average number of customers waiting, $\lambda$ is rate of service (customers/time), and $W$ is average service time.

You have $L = 4.4$ million and $W = 3\,$ yrs, so $\lambda = 4.4\; \mathrm{million}/3\, \mathrm{yrs} = 1.47$ million a year.

This is the same as in @JMoravitz's Comment, but with a "Name" put to it.
Are you getting ready to study queueing theory?