答:
Just do cases.

Either $(x-5) \ge 0$ or $(x-5) < 0$.  And either $(x+6)\ge 0$ or $(x+6) < 0$.

So you have four cases

$x-5 \ge 0; x+6 \ge 0$ and so $|x-5| = x-5$ and $|x+6| = x+6$ and you have the equation $(x+5) + (x-6) = 11$.

$x-5 \ge 0; x+6 < 0$ and so $|x+6| = -(x+6)$ and you have the equation $(x-5)-(x-6)=11$.

And so on.

But make sure your solutions agree with your initial assumptions.

And if you get an impossible answer.... that's okay; it just means your assumptions were impossible.

And you make get an answer that doesn't tell you anything about $x$.  If that answer is a true statement that just means $x$ can be anything so long as your assumptions are true.

....

That's brute force method that works.

There are some sophisticated things.

$|x-5|$ is the distance that $x$ is from $5$.  And $|x+6|$ is the distance $x$ is from $-6$.

Now the distance the distance from $5$ to $-6$ is $11$.

So if $x < -6$ then the distance $x$ is from $5$ is more than $11$.

And if $x > 5$ the distance $x$ is from $-6$ is more than $11$.

So $-6\le x < 5$.

Then the distance $x$ is from $-6$ is $x+6$ and the distance $x$ is from $5$ is $5-x$.

So $|x-5| + |x+6| = (5-x) + (6 + x) = 11$.  solving we get $11 = 11$.... which is always true.  So $x$ can be anything so that $-6 \le x \le 5$.


问:
Prove that $\frac{1}{f_n} \rightarrow \frac{1}{f}$ in measure.
Suppose that $f_n,f: E \rightarrow \mathbb{R}$ are measurable and finite almost everywhere.

If $f_n \rightarrow f$ in measure, and there is some $\delta > 0$ so that for every $n$ $f_n > \delta$ a.e., then $1/f_n \rightarrow 1/f$ in measure. (We are talking about Lebesgue measure).

Consider the set $$\left\{\left|\frac{1}{f_n}-\frac{1}{f}\right| > \epsilon \right\} = \{|f-f_n| > \epsilon \cdot|f_n|\cdot|f|\}.$$ Clearly by assumption we are given that $|f_n| > \delta$ so we can easily contain the set above by $\{|f-f_n| > \epsilon \cdot\delta\cdot|f|\}$. Here is where I am having trouble. I want to bound $f$ from below and use the fact that $f_n \rightarrow f$ in measure to complete the proof. Here is my attempt to bound it so far: Since $f_n \rightarrow f$ in measure, then $m(\{|f_n-f| \geq 1\}) < \eta$ when $n > N$. Then $|f| \geq ||f_n| - |f_n-f|| \geq |\delta-|f_n-f||$. I am stuck here since I cannot use $|f_n-f| \geq 1$ to get a lower bound, I would need the opposite inequality .. I think I am almost there but not quite.

Could someone please help me out?

ContentLicense: CC BY-SA 4.0

Body:
Hint: $$\mu (|f_n-f| >\epsilon \delta |f|)$$ $$\leq \mu (|f_n-f| >\epsilon \delta^{2}/2)+\mu (|f| \leq  \delta/2).$$
Now, $$\mu (|f| \leq  \delta /2) \leq \mu (|f_n-f| >\delta /2) +\mu (|f_n| <\delta)$$ $$=\mu (|f_n-f| >\delta /2) \to 0.$$


问:
If $m,n\in\Bbb{Z}_+$ s.t. $m<n$, is $\frac{\Gamma(n+1)\Gamma(n+2)}{\Gamma(m+1)\Gamma(m+2)\Gamma(n-m+1)\Gamma(n-m+2)}$ an integer?
Question:

If $m,n\in\Bbb{Z}_+$ s.t. $m<n$, is $$x_{n,m}=\frac{\Gamma(n+1)\Gamma(n+2)}{\Gamma(m+1)\Gamma(m+2)\Gamma(n-m+1)\Gamma(n-m+2)}$$ an integer? ($\Gamma$ is the gamma function, the extension of the factorial function)

My attempt:

We have
\begin{align}x_{n,m}=\frac{\Gamma(n+1)\Gamma(n+2)}{\Gamma(m+1)\Gamma(m+2)\Gamma(n-m+1)\Gamma(n-m+2)}\\
=\frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!}\\
=\frac{\prod_{j=1}^m(n-j+1)(n-j+2)}{m!(m+1)!}
\end{align}

I explored some basic cases:

When $m=1$, $x_{n,1}=\frac{n(n+1)}{2}$, which is clearly an integer for any $n>m$ because either $n$ or $n+1$ is even.

When $m=2$, $x_{n,2}=\frac{n^2(n+1)(n-1)}{12}$, which is still an integer for any $n>m$ because one of $n,(n+1), (n-1)$ is divisible by $3$, and two of them must be divisible by $2$, since if $2|n$, then $4|n^2$, else if $2$ is not a factor of $n$, then both $n+1$ and $n-1$ is divisible by $2$. So the whole thing is divisible by $4$ also.

When $m=3$, the denominator becomes $144$, I could also brutally prove that it's still an integer, but more importantly it seems to suggest that $x_{n,m}$ is always an integer, so I'm wondering if this could be rigorously proved by induction.

Induction Process:

The base case is already verified in the preceding section, so suppose $x_{n,m}\in\Bbb Z$ for $m=1,...,m-1$, then for $p=1,...,m$
\begin{align}
x_{n+1,p}=\frac{\prod_{j=1}^{p}(n+1-j+1)(n+1-j+2)}{p!(p+1)!}\\
=\frac{\prod_{j=1}^m(n+1-j+1)(n+1-j+2)}{m!(m+1)!}\frac{\prod_{j=m+1}^{p}(n+1-j+1)(n+1-j+2)}{\prod_{j=m+1}^pj(j+1)}\\
\end{align}

and this doesn't tell me anything except that it somehow looks like the proof of Pascal's identity... The proof of my claim could be elementary but I didn't see any plausible argument at this point...

Thanks in advance.

答:
Starting from the middle line of your attempt gives

$$\begin{equation}\begin{aligned}
& \frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!} \\
& = \left(\frac{n!}{(m+1)!(n-m)!}\right)\left(\frac{(n+1)!}{m!(n-m+1)!}\right) \\
& = \frac{1}{n+1}\left(\frac{(n+1)!}{(m+1)!(n-m)!}\right)\left(\frac{(n+1)!}{m!(n-m+1)!}\right) \\
& = \frac{1}{n+1}\binom{n+1}{m+1}\binom{n+1}{m}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

To prove \eqref{eq1A} is an integer requires

$$(n + 1) \; \left| \; \binom{n+1}{m+1}\binom{n+1}{m} \right. \tag{2}\label{eq2A}$$

Note almost the exact same thing was asked, and proven in quite different ways in  $2$ answers, in Prove using combinatorics $n \mid \binom{n}{m} \binom{n}{m-1}$, with their $n$ and $m$ being replaced by $n + 1$ and $m + 1$, respectively, in your case. For example, the answer there given by Batominovski is:

Maybe this is not a combinatorial proof, but $$\frac{1}{n}\,\binom{n}{m}\,\binom{n}{m-1}=\binom{n-1}{m-1}\,\binom{n+1}{m}-\binom{n}{m}\,\binom{n}{m-1}\,.$$
This is because
$$\frac{1}{n}\,\binom{n}{m}\,\binom{n}{m-1}=\frac{1}{m}\,\binom{n-1}{m-1}\,\binom{n}{m-1}=\frac{1}{n+1}\,\binom{n-1}{m-1}\,\binom{n+1}{m}\,.$$
The above result is due to the identity $\displaystyle\binom{k}{r}=\frac{k}{r}\,\binom{k-1}{r-1}$, which has a combinatorial proof.

Adjusting the identity in the first equation above to your use of $n$ and $m$ gives, on its right side,

$$\begin{equation}\begin{aligned}
& \binom{n}{m}\,\binom{n+2}{m+1} - \binom{n+1}{m+1}\,\binom{n+1}{m} \\
& = \left(\frac{n!}{m!(n-m)!}\right)\left(\frac{(n+2)!}{(m+1)!(n-m+1)!}\right) - \\
& \; \; \; \left(\frac{(n+1)!}{(m+1)!(n-m)!}\right)\left(\frac{(n+1)!}{m!(n-m+1)!}\right) \\
& = \left(\frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!}\right)\left((n+2) - (n+1)\right) \\
& = \frac{n!(n+1)!}{m!(m+1)!(n-m)!(n-m+1)!}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

which is the first line of \eqref{eq1A}. Since binomial coefficients are always integers, this confirms your relation is always an integer.


问:
How can you calculate $\int_{0}^{1} \exp(\sin(x))dx$ exactly?
I'm taking a computational statistics class and I have to calculate $\int_{0}^{1} \exp(\sin(x))dx$ numerically (Monte Carlo integration) and compare the numerical result to the exact result. The problem is, how can one canculate the exact value of the integral? The antiderivative doesn't seem to exist in terms of elementary functions. Mathematica gives me this:

Not only that, Mathematica also cannot calculate the definite integral:

And since Mathematica cannot do it, I don't think other softwares can, though I'm not sure. Of course, Mathematica can do the numerical integration

But that's not what I need! Thanks in advance for any help.

答:
Since $1$ is not a natural bound of integration for this problem, it is pretty clear that you would have to have an indefinite integral for this to have any chance of success. By contrast, if the upper limit were $\pi,$ you get the nice answer:

$$\int_{0}^{\pi} \exp(\sin(x))dx = \pi  (\pmb{L}_0(1)+I_0(1)),$$ while for $2\pi$ you get an even nicer answer:

$$\int_{0}^{2\pi} \exp(\sin(x))dx = 2 \pi  I_0(1).$$

(with $I$ being the Bessel function and $L$ the Struve $L$ function.


问:
What is the convex hull of such a matrix cone?
A matrix cone is in the following form

$$M := \begin{pmatrix} 1 \\ x\end{pmatrix}\begin{pmatrix}1 & x^T\end{pmatrix}$$

where $x \in F := \{x : x \in [l,u]^n \}$. How to express the convex hull of $M$?

答:
The convex hull of $C:=\left\{\begin{bmatrix}1 \\ x\end{bmatrix}\begin{bmatrix} 1 \\ x\end{bmatrix}^\top \; : \; x\in\mathbb{R}^n, \; 0\le x \le e \right\}$ is characterized using the completely positive cone. See "On the copositive representation of binaryand continuous nonconvex quadratic programs", Sam Burer, Mathematical Programming, (2009) 120:479–495. https://faculty.fuqua.duke.edu/~psun/Burer.2009.pdf

In a related paper (https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.93.3256&rep=rep1&type=pdf), Burer also characterized the convex hull of $C$ in lower dimensions $n=2,3$, which can be represented using positive semidefinite matrices and reformulation-linearization techniques. These representations are computable, while the convex hull of $C$ in higher dimensions need the completely positive cone, which is computationally hard to optimize over.


问:
Solving a Polynomial Equation by Factoring.
$n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.

Body:
Differentiate $f(n) = n^3+12n^2+48n+64$ to get $f'(n) = 3n^2 + 24n + 48 = 3(n^2 + 8n + 16) = 3(n+4)^2$.

Hence $\int f'(n) \ dn = 3 \int (n+4)^2 \ dn = 3 \cdot \left(\frac{(n+4)^3}{3} + C \right) = (n+4)^3 + C' = f(n)$, by the fundamental theorem of calculus. Since $f(-4) = C' \Rightarrow 0 = C'$, $f(n) = (n+4)^3$.


问:
Question about long proofs?
How do people write 50-page long proofs (and longer)? So they have a target in mind, but I can't get my head around them foreseeing that these 50 pages of work will actually lead them to exactly their target.</p>

答:
There are two cases:

You start by noticing something, then you prove a small claim, and slowly you add more and more claims, until you end up with a big proof of something impressive.

You set out to prove a certain thing, then you say "Ah, if only X was true", and when you think about it you realize that X is true, and you prove that. After some finitely many iterations you end up with a complete proof spanning over 50 pages or so.

Sometimes you have to develop an entirely new technology in order to prove something. Then explaining it, and proving all its basics can end up in a small book.

Of course, often when writing complicated proofs one may feel the need to add some less trivial introductory parts, which take more pages.

All in all mathematics is not something that you do from one day to the next, but rather a huge project that you never finish in your lifetime. And the proofs just accumulate.

ContentLicense: CC BY-SA 3.0

Title: Question about long proofs?


问:
Calculate if an expression is uniformly lower than another?
I have two expressions
$$
\frac{2}{n-1} \quad \text{and} \quad \frac{2n-1}{n^2}
$$
and I need to prove that
$$
\frac{2n-1}{n^2} < \frac{2}{n-1}
$$
for all $n \in \{2,3,\ldots\}$.

How can I do this?

答: Note that $n \ge 2$ to begin with. That said, then you want $\dfrac{2n-1}{n^2}- \dfrac{2}{n-1} < 0$. Rewrite this as a single fraction and you will see some thing new !


答:
Multiply away the denominators, the way you would of this were an equation and not an inequality. Note that the denominators are always positive, so this doesn't affect the inequality sign.

Take the resulting inequality and tidy it up a bit (you will, for instance, have $2n^2$ on both sides that cancel one another). The inequality should now be a bit more manageable.

Title: Calculate if an expression is uniformly lower than another?

Body:
I have two expressions
$$
\frac{2}{n-1} \quad \text{and} \quad \frac{2n-1}{n^2}
$$
and I need to prove that
$$
\frac{2n-1}{n^2} < \frac{2}{n-1}
$$
for all $n \in \{2,3,\ldots\}$.

How can I do this?


答:
You need to clear the denominators carefully. By that I mean use the following property:
$$
\frac{A}{B} < \frac{C}{D}
\quad \Longleftrightarrow \quad
AD < BC
\quad
\text{provided that }
C>0 \text{ and } D>0
$$
Since $n \geq 2$, your denominators are positive: $n^2 > 0$ and $n-1 > 0$. So the inequality is equivalent to
$$
(2n-1)(n-1) < 2n^2.
$$
After expanding, you get
$$
2n^2 - 3n + 1 < 2n^2
$$
or
$$
-3n + 1 < 0.
$$
Finally, after rearranging a bit,
$$
n > \frac{1}{3},
$$
which is certainly true as $n \geq 2$.


问:
Can anyone explain me the solutions of this $|x-5| + |x+6| = 11$?
I have an intuition of this using number line but I want to know what mathematics goes into solving these type of problems.</p>

答:
Considering,

Either $(x−5)\geq0$ or $(x−5)\leq0$. And either $(x+6)\geq0$ or $(x+6)\leq0$.

$\therefore$ We have the following four cases,

$1.$ $(x−5)\geq0$ and $(x+6)\geq0$, not possible as numbers above $5$ don't fulfill the condition.

$2.$ $(x−5)\geq0$ and $(x+6)\leq0$, not possible as numbers above $5$ don't fulfill the condition and numbers less than $-6$ don't fulfill the condition

$3.$ $(x−5)\leq0$ and $(x+6)\geq0$, absolutely possible and true.

$4.$ $(x−5)\leq0$ and $(x+6)\leq0$, not possible as numbers below $-6$ don't fulfill the condition.

Thus , we observe that $(x−5)\leq0$ and $(x+6)\geq0$ is the solution to our problem.

You could also try using graphs to find the solution. I am giving one graph here,


问:
What is the expected number of flips you need to get k changeovers?
Flip a fair coin several times. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if you flip the coin 5 times and the outcome is HHTHT, then there are 3 changeovers.

Consider n independent flips of this coin, what is the expected number of changeovers?

What is the expected number of flips you need to get k changeovers?

My Attempt

This one isn't too hard. Let $I_j$ be an indicator random variable that equals $1$ if we have a changeover and $0$ otherwise. For $I_j=1$, we need to have a heads on the $j^{th}$ flip and tails on the $(j+1)^{th}$ flip or tails first then heads for $j\in[1,n-1]$. Thus,
$$\mathbb{P}(I_j=1)=2p(1-p)$$
Our coin is fair, so $p=\frac{1}{2}$. Thus, $\mathbb{P}(I_j=1)=\frac{1}{2}$
If X is a random variable representing the number of changeovers. Then by linearity of the expectation, we have that
$$\mathbb{E}(X)=\sum_j\mathbb{P}(I_j)=(n-1)\cdot\frac{1}{2}=\frac{n-1}{2}$$

For this part, I am not so sure how to start. Would I use a binomial distribution? As in $X\sim\text{Bin}(n,\frac{1}{2})$ so that our probability is
$$\mathbb{P}(X=k)=\binom{n}{k}(\frac{1}{2})^k(1-\frac{1}{2})^{n-k}$$
I'm not too sure that this is the right approach or how to continue.

Edit

Second attempt for the second part: (I know this is wrong from the comments below, but can I get pointers on where I went wrong?)

Let X be the number of flips before k changeovers. Then $X\sim\text{Geom}(p)$. The probability is given by the formula
$$\mathbb{P}(X=k)=(1-p)^{k-1}p$$
We know that $p=\frac{1}{2}$ from the last part, so $\mathbb{P}(X=k)=\frac{1}{2^k}$ The expectation of a geometric distribution is $\mathbb{E}[X]=\frac{1}{p}$, so we have $E[X]=2^k$

答:
Your solution for the first part is correct.

For the second part, let the probability of heads be $p$ and the probability of tails be $q=1-p$.  If we've just thrown heads, then the number of tosses until the next tails is geometrically distributed with parameter $q$ and the expected number of tosses is $\frac1q$.  Similarly the expected number of tosses until the first heads after a tails is $\frac1p$.  If the first toss is heads then the expected number of tosses until the $k$th changeover is $$1+\left\lceil\frac k2\right\rceil\frac1q+\left\lfloor\frac k2\right\rfloor\frac1p,$$
since $\left\lceil\frac k2\right\rceil$ changeovers are from heads to tails, and $\left\lfloor\frac k2\right\rfloor$ changeovers are from tails to heads.  This is because the first changeover is from heads to tails, and the types of the changeovers alternate.

Similarly, if the first toss is tails then the expected number of tosses until the $k$th changeover is $$1+\left\lceil\frac k2\right\rceil\frac1p+\left\lfloor\frac k2\right\rfloor\frac1q.$$ Weighting these by the appropriate probabilities, we get that the expectation is $$1+2\left\lfloor\frac k2\right\rfloor+\left\lceil\frac k2\right\rceil\left(\frac pq+\frac qp\right).$$
With a fair coin $p=q$, and this simplifies to $2k+1$.


答:
Factor out a $64$ to get:

$$64 \left(\frac{n^3}{64} + \frac{12n^2}{64} + \frac{48n}{64} +1\right) = 64 \left(\frac{n^3}{64} + 3 \cdot\frac{n^2}{16} + 3 \cdot \frac{n}{4} +1\right)$$
$$= 64 \left( \left(\frac{n}{4}\right)^3 + 3\left(\frac{n}{4}\right)^2 + 3 \left(\frac{n}{4}\right) + 1 \right)$$
$$= 64(n/4 + 1)^3 = (4 \cdot n/4 + 4 \cdot1)^3 = (n+4)^3$$

where we have used the binomial theorem in the last line, and the fact that $4^3 = 64$.

This can be seen as the reverse of the AC method, where we instead factor out the largest perfect square of the constant term. This is $64 = 8^2$ in this question.

Title: Solving a Polynomial Equation by Factoring.

Body:
$n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.


问:
How do I know if this limit even converges?
So, there is this question which I found quite new according to my experience, and it is as follows.

Assuming that interchange of limit and integration is permissible, evaluate the following: $$\lim_{n \to \infty} {\int_0^1 \frac{nx^{n -1}}{1 + x}dx}, \quad 0 < x < 1$$

Now, I first tried integration by parts, and it doesn't give me more than an expression which is also nearly impossible to evaluate (maybe I'm doing it wrong, though) -

$$\lim_{n \to \infty} \left|\frac{1}{1+x}\int nx^{n - 1}dx  +  \int\frac{x^n}{(1 + x)^2}dx\right|_0^1$$

I'm not sure how to proceed from there, so I tried this instead:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 + x)^{-1}}dx}$$

Expanding $(1 + x)^{-1}$ using Taylor Series:

$$\lim_{n \to \infty} {\int_0^1 {nx^{n - 1}(1 - x + x^2 - x^3 + x^4 - x^5 \ldots)}dx}$$
$$ = \lim_{n \to \infty} {\int_0^1 {(nx^{n - 1} - nx^n + nx^{n + 1} - nx^{n + 2} + nx^{n + 3} - n^{x + 4} \ldots)}dx}$$
$$\lim_{n \to \infty} \left(1 - \frac{n}{n+1} + \frac{n}{n+2} - \frac{n}{n +3} + \frac{n}{n +4} - \frac{n}{n + 5}\ldots\right)$$

I think that diverges. Then, I tried putting the expression in Desmos and increased the $n$ value slowly upto $500$, $1000$ and similar big numbers - which all returned values around $0.5$.

The question was an multiple-choice based and there were both of them listed as options.

I'm not sure who is correct, the calculator, or me.

It would be a huge help if someone looks into this. Thank you in advance.

答:
For your curiosity.

$$I_n=\int n\frac{x^{n -1}}{1 + x}dx=n \left(\frac{x^n}{n}-\frac{x^{n+1}}{n+1} \, _2F_1(1,n+1;n+2;-x)\right)$$ where appears the gaussian hypergeometric function.

Integrating between the bounds, this gives the simple
$$J_n=\int_0^1 n\frac{x^{n -1}}{1 + x}dx=1+\frac{1}{2} n \left(H_{\frac{n-1}{2}}-H_{\frac{n}{2}}\right)$$ Using the asymptotics of generalized harmonic numbers
$$J_n=\frac{1}{2}+\frac{1}{4 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$


问:
Mathematically, what is the difference between averaging before or after an optimization?
Say I have an optimization problem where I want to maximize $x$ such that $f(x; p)$ is largest. Here, $p$ is a parameter, $x$ is a variable, and $f$ is a function that depends on $x$ and $p$.

However, the issue is that I don't know what $p$ is, I only know that it is somewhere in the set $P = \{p_1, ..., p_n\}$, with all entries being equally likely.

I can then try two things.

I take the average value, or the expected value $E p = \frac{p_1 + ... + p_n}{n}$, and use this to maximize $f(x;p)$ in $x$.

I solve each individual optimization problem for each $p_i$. Then I simply set the optimal $x^*$ to be the average of all the individual solutions to each subproblem.

Perhaps I can try other things too? If so, please let me know in the answers!

My question is now, considering the different ways one can solve this problem, what are the conceptual differences mathematically. When would one use over another? What is the intuition here?

答:
Your question really relates to the area of stochastic optimization.

There is no single answer, but it depends on your objective and the sequence of your decision in this decision making problem. Let me explain.

First, one situation is that you need to make the decision $x$ before you really know the true value of $p$ (this seems to be your case). This is called static decision making. However, another situation is that you can observe $p$ first, then make your decision $x$. Then, of course, you will make $x$ to be optimal for the specific $p$ you see. This is called adaptive decision making. In your question, this is a trivial case. But if imagine that you have to make a decision first before knowing $p$, then you can adjust your decision after you see $p$ and you want to minimize the overall cost. Then this becomes an interesting two-stage stochastic program.

Yes, as an example of adaptive decision making, we can formulate a two-stage stochastic program as $\min_{x_1}\{ f_1(x_1) + \mathbb{E}_p[Q(x_1,p)]\}$, where $Q(x_1,p)=\min_{x_2} f_2(x_2, x_1, p)$. Note that $Q(x_1,p)$ is the optimal objective of an optimization problem given $x_1, p$. A practical example: you need to first decide how much to invest $(x_1)$ in say stock market this month, then after you see the performance of the stocks, you can adjust your position $(x_2)$ in month 2, and you want to determine $x_1$ by considering what could happen in month 2, so you write the two-stage stochastic program.

Second, now let's talk about the objective. In your case, you can minimize the average cost as
$$\min_x \; \mathbb{E}_p[f(x,p)] \quad (1)$$
where the expectation is taken over the distribution of the random variable $p$. Why do we care about this formulation? Imagine that your decision $x$ is evaluated repeatedly over randomly drawn $p$, then the solution from (1) minimizes the long-term average cost, while the solutions from your choices 1 and 2 do not. So this stochastic program (1) gives you superior solution on average. The benefit is called the value of the stochastic solution.

Other objectives are also possible, for example, if you want to minimize certain risk measure of your solution, such as variance, value-at-risk (VaR), or conditional value at risk (CVaR).

A good reference is John Birge and Francois Louveaux's book Introduction to Stochastic Programming.


问:
Value of the following limit
I'm just a high school math student who was writing a paper on a new type of probability but I got stuck here because even the teachers who taught me were lost as to what the value of this would be. As long as the calculator could allow me, I got the answer to be 0.288788. I actually wanted to know what was the derivation of this rather than just a numeric value, for example could it be something like $\frac{1}{e^2}$ or any number which was rather peculiar.
$$
=\frac{(2^1-1)(2^2-1)(2^3-1)(2^4-1)...(2^n-1)}{2^{\frac{n(n+1)}{2}}}
$$
What will be the limit as n tends to $\infty$

In retrospect the question can also be written as
$$
\lim_{i \to \infty}\Pi_{n=1}^i \frac{2^{n}-1}{2^n}
$$
Your help would be very much appreciated, thank you.

答:
We can simplify the log of the product using a number theoretic function:
$$
\begin{align}
-\log\left(\prod_{n=1}^\infty\frac{2^n-1}{2^n}\right)
&=-\sum_{n=1}^\infty\log\left(1-2^{-n}\right)\\
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{2^{-kn}}{k}\\
&=\sum_{n=1}^\infty\sum_{d\mid n}\frac{2^{-n}}d\\
&=\sum_{n=1}^\infty\frac{\sigma_1(n)}{n}\,2^{-n}
\end{align}
$$
where $\sigma_1(n)$ is the sum of all of the divisors of $n$.

This removes the need to compute logarithms; however, it still requires computing an exponential to get the final product.

This also works for the generalization proposed by Claude Leibovici if we change all the $2$'s to $a$'s.

Title: Value of the following limit


问:
If $x^n=a_0+a_1(1+x)+\ldots+a_n(1+x)^n=b_0 + b_1(1-x)+\ldots+b_n(1-x)^n$, then for $n=101$, find $(a_{50},b_{50})$
My teacher told me add 1 on both sides of the equation, given
$$1+x^n=(a_0+1)+a_1(1+x)+\ldots+a_n(1+x)^n$$

But I don’t see how that’s useful in anyway. Am I misunderstanding the hint? What needs to be done here?

答:
Here is a slightly different hint from your teacher's.

Write $x^n$ in two different ways :

$$x^n = ((1+x)-1)^n = (1-(1-x))^n$$

Expand using binomial theorem and compare with other two appropriate polynomials.

Identify $(a_{50}, b_{50})$.


答:
1 More simply, we aren't concerned as to which face appeared on the first flip, or later on, on the $(i-1)^{th}$ flip.

With $X_i$ as the indicator variable indicating an $i_{th}$ flip changeover,

$\Bbb E[X_i] = \Bbb P[X_i] = \frac1 2$

and by linearity of expectation, summing from $i=2\,to\;  n, \,\Bbb E[X] = \frac{(n-1)}2$

2 From results of part $1, k = \frac{(n-1)}2$,

so $n = 2k+1$

Body:
Flip a fair coin several times. Say that a changeover occurs whenever an outcome differs from the one preceding it. For instance, if you flip the coin 5 times and the outcome is HHTHT, then there are 3 changeovers.

Consider n independent flips of this coin, what is the expected number of changeovers?

What is the expected number of flips you need to get k changeovers?

My Attempt

This one isn't too hard. Let $I_j$ be an indicator random variable that equals $1$ if we have a changeover and $0$ otherwise. For $I_j=1$, we need to have a heads on the $j^{th}$ flip and tails on the $(j+1)^{th}$ flip or tails first then heads for $j\in[1,n-1]$. Thus,
$$\mathbb{P}(I_j=1)=2p(1-p)$$
Our coin is fair, so $p=\frac{1}{2}$. Thus, $\mathbb{P}(I_j=1)=\frac{1}{2}$
If X is a random variable representing the number of changeovers. Then by linearity of the expectation, we have that
$$\mathbb{E}(X)=\sum_j\mathbb{P}(I_j)=(n-1)\cdot\frac{1}{2}=\frac{n-1}{2}$$

For this part, I am not so sure how to start. Would I use a binomial distribution? As in $X\sim\text{Bin}(n,\frac{1}{2})$ so that our probability is
$$\mathbb{P}(X=k)=\binom{n}{k}(\frac{1}{2})^k(1-\frac{1}{2})^{n-k}$$
I'm not too sure that this is the right approach or how to continue.

Edit

Second attempt for the second part: (I know this is wrong from the comments below, but can I get pointers on where I went wrong?)

Let X be the number of flips before k changeovers. Then $X\sim\text{Geom}(p)$. The probability is given by the formula
$$\mathbb{P}(X=k)=(1-p)^{k-1}p$$
We know that $p=\frac{1}{2}$ from the last part, so $\mathbb{P}(X=k)=\frac{1}{2^k}$ The expectation of a geometric distribution is $\mathbb{E}[X]=\frac{1}{p}$, so we have $E[X]=2^k$


问:
Conditional probability in regard to min and max of Brownian motion
Given the Brownian motion $X(t)$.

I want to ask about the conditional probability:

\begin{align*}
\Pr \left \{\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s,  X(t) < s)|X(0) = a \right \},
\end{align*}
where $s,a >0$, and $s > a$.

I tried to use the reflection principle to solve this problem but it seemed inoperative due to the intersection of both $min$ and $max$ events.

Can someone give me a hint? I really appreciate it. Thank you very much.

PS: Or else, please give me some hint about the probability
\begin{align*}
\Pr \left \{\min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right \}.
\end{align*}

Thank you all again!

答:
The law of $\left(\min_{0 \leq u \leq t} X(u), \max_{0 \leq u \leq t} X(u), X_t \right)$ is called Lévy’s triple law. According to René L. Schilling, Lothar Partzsch, Brownian Motion: An Introduction to Stochastic Processes, 6.5 Lévy’s triple law  $(6.16)$

$$P \left( m_t > a, M_t <b,  X(t) \in dx)|X(0) = 0 \right) = \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\left(e^{-\frac{(x+2n(b-a)^2)}{2t}}-e^{-\frac{(x-2a-2n(b-a)^2)}{2t}}  \right)$$
where $a<0<b$, $m_t = \min_{0 \leq u \leq t} X(u)$ and $M_t = \max_{0 \leq u \leq t} X(u)$

Hence,
\begin{align*}
P \left( m_t > a, M_t  a, M_t <b,  X(t) \in dx)|X(0) = 0 \right) \\
&= \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\int_{\mathbb{R}}\left(e^{-\frac{(x+2n(b-a)^2)}{2t}}-e^{-\frac{(x-2a-2n(b-a)^2)}{2t}}  \right)e^{-\frac{x^2}{2t}}dx\\
&= \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\frac{1}{\sqrt{2}}\left(e^{-\frac{(b-a)^2n^2}{2t}}-e^{-\frac{((b-a)n+a)^2}{2t}}  \right)\\
\end{align*}

Return back to your question, it suffices to put $(a,b) \leftarrow (-a,s-a)$

$$P\left( \min_{0 \leq u \leq t} X(u) < 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )=1-\frac{1}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\frac{1}{\sqrt{2}}\left(e^{-\frac{s^2n^2}{2t}}-e^{-\frac{(sn-a)^2}{2t}}  \right)$$

For $P\left( \min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )$, just notice that this probability is equal to
$$P\left( \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )-P\left( \min_{0 \leq u \leq t} X(u) < 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )$$

and $\left(\max_{0 \leq u \leq t} X(u)|X(0) = a\right)$ follows the same distribution as $\left(|X(t)-a|+a\right)$ according to the law of maximum of brownian motion.

Body:
The law of $\left(\min_{0 \leq u \leq t} X(u), \max_{0 \leq u \leq t} X(u), X_t \right)$ is called Lévy’s triple law. According to René L. Schilling, Lothar Partzsch, Brownian Motion: An Introduction to Stochastic Processes, 6.5 Lévy’s triple law  $(6.16)$

$$P \left( m_t > a, M_t <b,  X(t) \in dx)|X(0) = 0 \right) = \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\left(e^{-\frac{(x+2n(b-a)^2)}{2t}}-e^{-\frac{(x-2a-2n(b-a)^2)}{2t}}  \right)$$
where $a<0<b$, $m_t = \min_{0 \leq u \leq t} X(u)$ and $M_t = \max_{0 \leq u \leq t} X(u)$

Hence,
\begin{align*}
P \left( m_t > a, M_t <b\right) &= \int_{\mathbb{R}}P \left( m_t > a, M_t <b,  X(t) \in dx)|X(0) = 0 \right) \\
&= \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\int_{\mathbb{R}}\left(e^{-\frac{(x+2n(b-a)^2)}{2t}}-e^{-\frac{(x-2a-2n(b-a)^2)}{2t}}  \right)e^{-\frac{x^2}{2t}}dx\\
&= \frac{dx}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\frac{1}{\sqrt{2}}\left(e^{-\frac{(b-a)^2n^2}{2t}}-e^{-\frac{((b-a)n+a)^2}{2t}}  \right)\\
\end{align*}

Return back to your question, it suffices to put $(a,b) \leftarrow (-a,s-a)$

$$P\left( \min_{0 \leq u \leq t} X(u) < 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )=1-\frac{1}{\sqrt{2\pi t}}\sum_{n=-\infty}^{+\infty}\frac{1}{\sqrt{2}}\left(e^{-\frac{s^2n^2}{2t}}-e^{-\frac{(sn-a)^2}{2t}}  \right)$$

For $P\left( \min_{0 \leq u \leq t} X(u) > 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )$, just notice that this probability is equal to
$$P\left( \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )-P\left( \min_{0 \leq u \leq t} X(u) < 0, \max_{0 \leq u \leq t} X(u) > s|X(0) = a \right )$$

and $\left(\max_{0 \leq u \leq t} X(u)|X(0) = a\right)$ follows the same distribution as $\left(|X(t)-a|+a\right)$ according to the law of maximum of brownian motion.


问:
Showing $f$ is continuous
Let $x_1, x_2, ... , x_n$ be linearly independent vectors of a normed space $X$, so for any $ x \in X$ set $x = \alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n$ for real numbers $\alpha ... \alpha_n$. Prove that the function $f: \mathbb{R}^n \rightarrow \mathbb{R}^{\geq 0}$ defined by
$$f(\alpha_1, \alpha_2, ... \alpha_n) = || \alpha_1 x_1 + \alpha_2 x_2 + ... + \alpha_n x_n ||$$ is continuous.

Do I have to show that this is a linear operator and then show that is is bounded to imply continuity? The confusing part is this is a general normed space-- I don't have an explicit norm definition. How do I proceed?

答:
It is true that $f$ is bounded since, for $M = \max\{||x_j||; 1 \leq j \leq n\}$,
$$
     |f(\alpha_1,\ldots,\alpha_n)|
\leq \sum_{i=1}^n |\alpha_i| ||x_i||
\leq M\sum_{i=1}^n |\alpha_i|,
$$
so
$$
     \sup\{|f(\alpha_1,\ldots,\alpha_n)|: ||(\alpha_1,\ldots,\alpha_n)||_n = 1 \}
\leq M.
$$
It is not linear, however, since
$$
     f(-(\alpha_1,\ldots,\alpha_n))
\neq -f(\alpha_1,\ldots,\alpha_n).
$$
So, strictly speaking, you cannot use your approach. Nevertheless, the function
$g:\mathbb{R}^n \to X$ given by $(\alpha_1,\ldots,\alpha_n) \mapsto \sum_{i=1}^n \alpha_i x_i$ is clearly linear, bounded (as we just saw), and therefore continuous. Then just show that the function $||\cdot||: X \to \mathbb{R}$ is continuous, and your result will follow by composition.</p>

Title: Showing $f$ is continuous


问:
Changing order of limit and complex derivative
Assume $f(z,t)$ is complex differentiable at $z=z_0$, $t$ is a real parameter. Under which condition is $\lim_{t \rightarrow t_0} f(z,t)$ also complex differentiable at $z = z_0$?

I know some theorems and their proofs from real analysis, but I was curious whether I could prove the corresponding theorems in the complex domain but I am unsure what the condition of limit switching theorem might be!

EDIT

How does the result extends to interchangeability of derivatives? I.e. assume $f(z,t)$ is both $z \in \mathbb{C}$ and $t \in \mathbb{R}$ differentiable in some neighborhood of $z_0$ and $t_0$. Under which conditions we have that $\frac{\partial f}{\partial t}$ is also $z$ differentiable?

ADDENDUM
(Comment to @Bobby Ocean answer)
Consider a following: Somehow you have proven that $z^t$ is $z$ complex differentiable for each real $t$ (except branches). Then you define a function $g(z) = \lim_{t \rightarrow 0} (z^t - 1)/t$, how do you prove this is also complex differentiable? I would like to define the logaritmic function this way, and proving $g'(z)=1/z$ would require knowing whether I can interchange the derivatives.

答:
The existence of a pointwise limit at the point $t_0$ is insufficient to guarantee a differentiable function. In general, one needs uniform convergence to preserve various properties like differentiation, integration, continuity, etc.

A simple example would be $f(z,t):=z^t$ on $t>0$ and $z\in\mathbb{C}-\text{B}$ (where $\text{B}$ is some branch cut). Note, $f(z,t)$ is real-differentiable in $t\in\mathbb{R}$ and complex-differentiable in $z\in\mathbb{C}-\text{B}$. Notice, for each fixed $z$ we have, $$h(z):=\lim_{t\to 0} f(z,t) = \left\{\matrix{1 & z\neq 0 \\ 0 & z = 0}\right.$$ a function that is clearly not differentiable at $z=0$.

The the ability to interchange derivatives is not the same thing as the existence of the derivatives. Not only must you assume that the function is $C^2$ (example where $f_{xy}(0,0)\neq f_{yx}(0,0)$ because we don't have $C^2$), but you must also assume that the mixed partials already exist. You are not guaranteed the mixed partial derivatives will even exist, simply because the function is $C^2$. Let me say it another way, $C^2$ isn't good enough to allow you to take mixed partial derivatives.

Explicitly, $f(z,t)$ could be infinitely differentiable with respect to variables $t$ and $z$, and yet $\frac{\partial}{\partial z}\frac{\partial}{\partial t} f(z,t)$ could fail to exist at some point in $\mathbb{C}\times\mathbb{R}$. This is simply another way of saying that $\frac{\partial}{\partial t} f(z,t)$ could fail to be differentiable with respect to $z$ at some point $z_0$. Or as I will show by way of example, $\frac{\partial}{\partial t} f(z,t)$ could still be a discontinuous function with respect to the variable $z$.

Example.

To be clear, this is a non-obvious example that took me awhile to construct. First we define for each $\alpha\in\mathbb{C}$, $$g_\alpha(z) := \left\{\matrix{(e^{\alpha z}-1)/z & z\neq 0 \\ \alpha & z=0}\right\} = \sum_{k=1}^\infty \frac{\alpha^k}{k!}z^{k-1}$$ Note that this is a family of entire functions that are infinitely differentiable in $z$ everywhere in $\mathbb{C}$ (this is easily seen, since the power series representation has an infinite radius of convergence). Now define, $$f(z,t) := \left\{\matrix{t\cdot g_{-1/t^2}(z) & t\neq 0 \\ 0 & t=0}\right.$$ Astonishingly $f(z,t)$ is infinitely differentiable with respect to $t\in\mathbb{R}$ (including at the point $0$), likewise, $f(z,t)$ is infinitely differentiable with respect to $z\in\mathbb{C}$ (again, including the point at $0$) (literally, for each fixed $t$, $f(z,t)$ is entire). However, when we attempt to mix partial derivatives we will get a problem, take $\frac{\partial}{\partial t}$, $$\frac{\partial}{\partial t} f(z,t) = \left\{\matrix{2e^{-z/t^2}/t^2 + g_{-1/t^2}(z) & t\neq 0 \\ -1/z & t=0}\right.$$ Hence, $$\lim_{t\to 0}\frac{\partial}{\partial t} f(z,t) = -\frac{1}{z}$$ a clearly discontinuous function in the variable $z$, with no derivative at zero. Thus, even though $f(z,t)$ was infinitely differentiable at $t=0$ with respect to the variable $t\in\mathbb{R}$ and is infinitely differentiable at $z=0$ with respect to the variable $z\in\mathbb{C}$, that meant nothing in determining if $\frac{\partial}{\partial t} f(z,t)$ would be differentiable with respect $z$ at $z=0$ when $t=0$. We even point out that $\frac{\partial}{\partial t} f(z,t)$ is differentiable with respect to $z$ at $z=0$ for any $t\neq 0$.

Please note that the reason this example works is because we can utilize the "infinitely real-differentiable flat function" $e^{-1/t^2}$. Such function is not complex-differentiable at zero, even though it is real-differentiable at zero. The above example would not apply to multi-variate complex-differentiable functions, $f(z,w)$, that would be a different discussion.


问:
How can I prove that $A_1^c, A_2,A_3,A_4,\dots,A_n$ are independent events
Given $A_1,A_2,\dots,A_n$ are independent events, To prove(correct me if I am wrong) for $n$ events $A_1^c,A_2,A_3,\dots,A_n$ are mutually independent we need to check the following conditions:

$P(A_1^c\cap A_2) = P(A_1^c)\cdot P(A_2)$

$P(A_1^c\cap A_3) = P(A_1^c)\cdot P(A_3)$

and so on until:

$P(A_1^c\cap A_3) = P(A_1^c)\cdot P(A_n)$

$P(A_1^c\cap A_2\cap A_3\cap\dots\cap A_n) = P(A_1^c)\cdot P(A_2)\cdot P(A_3)\cdot \ldots \cdot P(A_n)$

$P(A_1^c\cap A_2\cap A_4)$

$P(A_1^c\cap A_2\cap A_5)$

$P(A_1^c\cap A_2\cap A_n)$

$P(A_1^c\cap A_3\cap A_8)$

$P(A_1^c\cap A_2\cap A_3\cap A_5)$

and so on until each and every intersection is included

I have made proven the 1st and 2nd condition in
My question: How does one generalize the 3rd condition so that it is not required to prove it rigorously

答:
Do induction.

For n=1, show that if $A_1$ is independent of itself, then $A_1^c$ is independent of itself. (Fun fact: The following 3 conditions are equivalent: $A$ is independent of itself. $P(A)=0$ or $1$. $A$ is independent of every other event $B$.)

For n=2, show that if $A_1,A_2$ are independent, then $A_1^c,A_2$ are independent.

Now you probably have a feel of what's going on. Otherwise continue n=3 and n=4.

Now for induction.

Assume that the independence of $A_1, A_2..., A_k$ implies independence of $A_1^c, A_2..., A_k$.

Assume the independence of $A_1, A_2..., A_k, A_{k+1}$.

Prove the independence of $A_1^c, A_2..., A_k, A_{k+1}$.


问:
What is the remainder when dividing $((1997!+11)!+29)!$ by $7$?
This question appeared in one of the national exams in Saudi Arabia which was on Saturday, November $29, 1997$.

If today is Saturday, what day of the week will it be
$((1997!+11)!+29)!$ days from now?

Actually I do not know how to start. I saw a similar problem asking $10^{{10}^{10}}$ days from now, it was a simple one.

Any help would be appreciated. THANKS!

答:
Hint :

Which day it is the $n$-th day from today, $n$ satisfying the following?
$$n \equiv 0 \pmod 7$$

Notice for $n \ge 7$, $n! \equiv 0 \pmod 7$.


问:
Intuition of joint density of min(X,Y) and max(X,Y)
The problem is to find the joint density of $U = min(X,Y)$ and $V=max(X,Y)$ when both are exponential random variables.

The solution to it is:

I can finish it after the first step but I don't understand the intuition behind

$$P(U \le u, V \le v) = P(V\le v)-P(U>u, V\le v) $$

I understand that it's because $P(A\cap B) = P(B) -  P(A^c\cap B)$ but what would make you think that it's a good idea to start off by rewriting it like this?

答:
Precision 1st and then intuition.

Precision:

$$(U \le u) \cup (U > u) = \Omega \tag{A}$$

Intersect both sides with $(V \le v)$. By distributive property of intersections and unions we get

$$(U \le u, V \le v) \cup (U > u, V \le v) = (V \le v) \tag{C}$$

Apply probability to both sides. Since the LHS of $(C)$ is a disjoint union (actually the LHS of $(A)$ is a disjoint union too), we simply replace $\cup$ with $+$:

$$P(U \le u, V \le v) + P(U > u, V \le v) = P(V \le v) \tag{B}$$

Intuition:

$(B)$ is like $(A)$ but $(V \le v)$ is like the new $\Omega$. Hopefully this answers 'what would make you think that it's a good idea to start off by rewriting it like this'


答: $((1997!+11)!+29)!$ is your expression. We know that $(1997!+11)!+29$ is a very big integer bigger than $7$, thus $((1997!+11)!+29)!$ is the product of all numbers equal to and less than it, which includes $7$, meaning that the number of days is divisible by $7$, thus the only thing that changes is the week, not the day. For example, if you add $7$ days to Sunday, you'd get Sunday. Thus, if we add a number of days divisible by $7$ to Saturday, we end on Saturday. Thus, adding $((1997!+11)!+29)!$ days, which is divisible by $7$ to Saturday will keep it a Saturday.

Title: What is the remainder when dividing $((1997!+11)!+29)!$ by $7$?

Body:
This question appeared in one of the national exams in Saudi Arabia which was on Saturday, November $29, 1997$.

If today is Saturday, what day of the week will it be
$((1997!+11)!+29)!$ days from now?

Actually I do not know how to start. I saw a similar problem asking $10^{{10}^{10}}$ days from now, it was a simple one.

Any help would be appreciated. THANKS!


Title: If $x^n=a_0+a_1(1+x)+\ldots+a_n(1+x)^n=b_0 + b_1(1-x)+\ldots+b_n(1-x)^n$, then for $n=101$, find $(a_{50},b_{50})$

Body:
My teacher told me add 1 on both sides of the equation, given
$$1+x^n=(a_0+1)+a_1(1+x)+\ldots+a_n(1+x)^n$$

But I don’t see how that’s useful in anyway. Am I misunderstanding the hint? What needs to be done here?

Body:
Here is a method using derivatives

$$x^{n}=\sum_{j=0}^{n}a_{j}(1+x)^{j}$$

Taking derivative $w.r.t$ $x$

$$nx^{n-1}=\sum_{j=1}^{n}a_{j}j(1+x)^{j-1}$$

Again derivate $w.r.t$ $x$.

$$n(n-1)x^{n-2}=\sum_{j=2}^{n}a_{j}j(j-1)(1+x)^{j-2}$$

So, $K^{th}$ derivative will look like this

$$n(n-1)...(n-k+1)x^{n-k}=\sum_{j=k}^{n}a_{j}j(j-1)...(j-k+1)(1+x)^{j-k}$$

$$\implies \frac{n!}{(n-k)!}x^{n-k}=\sum_{j=k}^{n}a_{j}\frac {j!}{(j-k)!}(1+x)^{j-k}$$

Since $n=101$ we have,

$$\frac{101!}{(101-k)!}x^{101-k}=\sum_{j=k}^{101}a_{j}\frac {j!}{(j-k)!}(1+x)^{j-k}$$

Now Choose $k$ wisely so that you will get $a_{50}$

We can find $b_{50}$ using same method.


问:
My son's Sum of Some is beautiful! But what is the proof or explanation?
My youngest son is in $6$th grade. He likes to play with numbers. Today, he showed me his latest finding. I call it his "Sum of Some" because he adds up some selected numbers from a series of numbers, and the sum equals a later number in that same series. I have translated his finding into the following equation:
$$(100\times2^n)+(10\times2^{n+1})+2^{n+3}=2^{n+7}.$$

Why is this so? What is the proof or explanation? Is it true for any $n$?

His own presentation of his finding:

Every one of these numbers is two times the number before it.
  $1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192$.

  I pick any one of them, times $100$. Then I add the next one, times $10$. Then I skip the next one. Then I add the one after that.

  If I then skip three ones and read the fourth, that one equals my sum!

答:
Here is another formula similar to your son's, using powers of $5$:

$$1000000\cdot5^n+100000\cdot5^{n+1}+10000\cdot5^{n+2}+1000\cdot5^{n+3}+100\cdot5^{n+4}+5^{n+6}=5^{n+9}$$

We can derive identities of a similar form more generally.  Consider any $m$th degree polynomial $f(x)\in\mathbb{Z}[x]$ of the following form

$$f(x)=a_0+a_1x+\ldots+a_{m-1}x^{m-1}-x^m$$

Notice that all rational roots of $f(x)$ are integers.  Suppose $f$ has a rational root $k$.  Then we have the following equality:

$$\begin{align}k^{n+m}&=k^nk^m\\&=k^n(a_0+a_1k+a_2k^2+\ldots+a_{m-1}k^{m-1})\\&=a_0k^n+a_1k^{n+1}+a_2k^{n+2}+\ldots+a_{m-1}k^{n+m-1}\end{align}$$

Your son's formula involves the $7$th degree polynomial $f(x)=100+10x+x^3-x^7$ and the root $k=2$.  Above I've used the polynomial $f(x)=10^6+10^5x+10^4x^2+10^3x^3+10^2x^4+x^6-x^9$ and the root $k=5$.

Here are other examples for $k=3$.  Consider the polynomials $f(x)=3+8x+80x^3-x^7,$ $g(x)=2130+10x+x^3-x^7$, and $h(x)=99+687x+x^3-x^7$, and notice that $f(3)=g(3)=h(3)=0$.  These give us the following identities:

$$3\cdot3^n+8\cdot 3^{n+1}+80\cdot3^{n+3}=3^{n+7}\\2130\cdot3^n+10\cdot3^{n+1}+3^{n+3}=3^{n+7}\\99\cdot3^n+687\cdot3^{n+1}+3^{n+3}=3^{n+7}$$

Try using this method to find identities for other values of $k$ (including negative integers!).


问:
Union and intersection of two transitive relations
If we have two transitive relations $R_1, R_2$ on a set $A$, then I know that if $$R_3 = R_1 \cup R_2$$ that $R_3$ is not transitive.

And if $$R_3 = R_1 \cap R_2$$ then $R_3$ is transitive. But I can't figure out why my "proof" stating the opposite of those is false.

My reasoning went as follows:
For $R_3 = R_1 \cup R_2$, Let $x,y,z\in A$ and assume $(x,y),(y,z),(x,z)\in R_1$. Since $R_1\subseteq R_1 \cup R_2 = R_3$ then $(x,y),(y,z),(x,z)\in R_3$ and similarly the same argument for $R_2$ so $R_3$ is transitive.

For $R_3 = R_1 \cap R_2$ let $A = \lbrace 1,2,3,4 \rbrace$ and have $R_1 = \lbrace (1,2),(2,3),(1,3) \rbrace$ and $R_2 = \lbrace (1,2),(2,4),(1,4) \rbrace$ hence $R_1$ and $R_2$ are transitive but $R_3 = \lbrace (1,2) \rbrace$ which is not transitive

Can someone point out my logical flaws in both arguments? I know one can show counter examples for the first one but I don't see where I'm making poor arguments in both of them as I was told we can handle relations just like any other type of set.

答:
Firstly, in order to prove transitivity of some relation $R$, you need to show that $(x,y),(y,z)\in R$ implies $(x,z)\in R$.

The problem with your "proof" for the union is that you assume that $(x,y),(y,z)$ both belong to the same relation $R_i$. In that case, you indeed have $(x,z)\in R_i$ and hence $(x,z)\in R_3$. The problem occurs when you take $(x,y)\in R_1$ and $(y,z)\in R_2$. For example, if $R_1=\{(x,y)\in\mathbb N^2\mid x-y\text{ is divisible by 2}\}$ and $R_2=\{(x,y)\in\mathbb N^2\mid x-y\text{ is divisible by 3}\}$, both $R_i$ are transitive, but $R_1\cup R_2$ isn't: Consider $(0,2)\in R_1$ and $(2,5)\in R_2$. Do you think $(0,5)\in R_1\cup R_2$?

Your "counterexample" concerning the intersection isn't actually a counterexample! $R_3$ is indeed transitive. More precisely, we need to prove
$$\forall x,y,z: (x,y),(y,z)\in R_3 \implies (x,z)\in R_3.$$

This implication is vacuously tue, that is, the antecedent (the part before $\implies$) is never satisfied, and thus the implication holds (here we use that the implications "false implies true" and "false implies false" are both correct).


问:
Showing $\varphi_i : \mathbb P^n - Z(x_i) \to \mathbb A^n$ is a homeomorphism.
From Hartshorne's Algebraic Geometry Chapter 1 section 2:

Here $U_i = \mathbb P^n - Z(x_i)$, and $\varphi_i(a_0, \dots, a_n)= (a_1/a_i, \dots, a_n/a_i)$ with $a_i/a_i$ omitted.

It doesn't seem very straightforward checking the highlighted part.

Let $\varphi_i(a_0, \dots, a_n)= (a_1/a_0, \dots, a_n/a_0) \in \varphi(Y)$.

We want to show that $(a_1/a_0, \dots, a_n/a_0) \in Z(T')$.

Let $f \in T'$. Then $f=\alpha(g(x_0, \dots, x_n))=g(1, x_1, \dots, x_n)$.

Then $g(1, a_1/a_0, \dots, a_n/a_0) = 1/a_0^{deg(g)} g(a_0^{deg(g)}, a_1, \dots, a_n)$.

Why should this be $0$?

答:
Let us show that $\varphi(Y)\subset Z(T')$. Let $p=[a_0:\ldots:a_n]$ be a point in $Y$ and let $f\in T$. Then by definition $f(a_0,\ldots,a_n)=0$. Since $f$ is homogeneous and $a_0\neq0$ it follows that

$$f(1,\frac{a_1}{a_0},\ldots,\frac{a_n}{a_0})=0\,.$$

But this is precisely saying that $\alpha(f)(\varphi(p))=0$, that is, $\varphi(p)\in Z(T')$.

The reverse inclusion $Z(T')\subset \varphi(Y)$ should also be easy and should follow by a similar analysis.


问:
Showing that the closure of a totally bounded set is totally bounded
I would like to show that if I have a subset $M$ of a metric space $(X,d)$ such that $M$ is totally bounded, then its closure $cl(M)$ is also totally bounded.

My general strategy would be to show that for every $\epsilon > 0$, there exist $x_1,...,x_n \in M$ such that $M = \bigcup B_{\epsilon/2}(x_i)$. And then I would  use the fact that this works also for $\epsilon$ since it works for $\epsilon /2$. However, should I be having two cases? I am listing the cases as follows:

$cl(M) = M;$

$cl(M) \neq M$.

答:
My answer is $6-7$ years late, but I am posting it anyway since I did not find my idea being used in any of the other solutions. Not sure what definition of totally bounded you're using, but I shall use the following characterization from Carothers' Real Analysis:

In $(M,d)$, $A\subset M$ is totally bounded iff for all $\epsilon > 0$ there are finitely many sets $A_1,A_2,\ldots,A_n\subset M$ with $\text{diam}(A_i) < \epsilon$ for $1\le i\le n$, such that $A\subset \bigcup_{i=1}^n A_i$.

In some sense, $A$ is being covered by $A_i$, $1\le i\le n$. If you're using some other definition/characterization, the equivalence should be easy to prove.

Using this, you may prove the following Lemma:

Lemma $1.1$: In $(M,d)$, $A\subset M$ is totally bounded iff for all $\epsilon > 0$ there are finitely many closed sets $A_1,A_2,\ldots,A_n\subset M$ with $\text{diam}(A_i) < \epsilon$ for $1\le i\le n$, such that $A\subset \bigcup_{i=1}^n A_i$.

This should follow easily using ideas such as $\text{diam}\ A = \text{diam}\ \overline A$, $A \subset \overline A$ and basic definitions. Once you've proved Lemma $1.1$, you're good to go.

Suppose in a metric space $(M,d)$, $A\subset M$ is totally bounded. Then, for given $\epsilon > 0$ there exist $A_1,\ldots,A_n$ such that $$A\subset\bigcup_{i=1}^n A_i$$
Using Lemma $1.1$, we may assume that all $A_i$'s are closed, and hence so is their finite union. $\overline A$ is the smallest closed set containing $A$, so it immediately follows that $$\overline A\subset\bigcup_{i=1}^n A_i$$ and we are done! We have found closed sets $A_i$ which cover $\overline A$, and Lemma $1.1$ says that we have to do no more.

At this stage, I urge you to prove the following double-implication, part of which we have already done:

In a metric space $(M,d)$, $A\subset M$ is totally bounded if and only if $\overline A$ is totally bounded.


答:
This is really two questions:

Why $R_3 = R_1 \cup R_2$ is not transitive. The property must hold for two arbitrary elements of the relation. Say $(x, y) \in R_1$ and $(y, z) \in R_2$ but not the reverse. They are both elements of the relation $R_3$ but there's no guarantee that $(x, z) \in R_3$ inherited from either $R_1$ or $R_2$. You can come up with a specific small counterexample.

Why your example doesn't contradict the fact that $R_3 = R_1 \cap R_2$ is transitive. Basically, there are no pairs of the form $(a, b), (b, c) \in R_3$, so the property holds vacuously.

Title: Union and intersection of two transitive relations

Body:
If we have two transitive relations $R_1, R_2$ on a set $A$, then I know that if $$R_3 = R_1 \cup R_2$$ that $R_3$ is not transitive.

And if $$R_3 = R_1 \cap R_2$$ then $R_3$ is transitive. But I can't figure out why my "proof" stating the opposite of those is false.

My reasoning went as follows:
For $R_3 = R_1 \cup R_2$, Let $x,y,z\in A$ and assume $(x,y),(y,z),(x,z)\in R_1$. Since $R_1\subseteq R_1 \cup R_2 = R_3$ then $(x,y),(y,z),(x,z)\in R_3$ and similarly the same argument for $R_2$ so $R_3$ is transitive.

For $R_3 = R_1 \cap R_2$ let $A = \lbrace 1,2,3,4 \rbrace$ and have $R_1 = \lbrace (1,2),(2,3),(1,3) \rbrace$ and $R_2 = \lbrace (1,2),(2,4),(1,4) \rbrace$ hence $R_1$ and $R_2$ are transitive but $R_3 = \lbrace (1,2) \rbrace$ which is not transitive

Can someone point out my logical flaws in both arguments? I know one can show counter examples for the first one but I don't see where I'm making poor arguments in both of them as I was told we can handle relations just like any other type of set.


问:
Choosing the area element of Polar double integration.
How do I divide the radius into subsections?

In one situation the region can be inside two circles and $\Delta r$ might equal for instance $\dfrac{2 - 1}{n}$. With another two radii that keep the variable in that format, like $\Delta r = \dfrac{cos(2\theta) - cos(\theta)}{n}$ how do you calculate the $\Delta r$ that's uniform everywhere? If I pick a sample of $\theta$ then I get uniform sub regions yet that seems like it should change the boundaries of integration. If the value of $\theta$ would vary then the regions don't have equal lengths of radius in the approximation. How do I always see this un evaluated expression as sub regions in the approximation walkthroughs?

答:
We first choose a differential area of $dr$  x  $ rd\theta$ (highlighted in orange color), and  separate the integrations along two different directions:

Radial direction by integrate dr from $r_1$ to $r_2$: $\int \limits_{ r_1}^{ r_2}$dr

Circumferential direction by integrating d$\theta$ from $\theta_1$ to $\theta_2$: $\int \limits_{\theta _1}^{\theta_2}d\theta$

\begin{align}\int \limits_{ r_1}^{ r_2}rdr \int \limits_{\theta _1}^{\theta_2} d\theta \end{align}

In this process, we do not include $\theta$ in expression of dr , the variation of $\theta$ is never in the picture of integration along the radial direction.

Hope this helps.


问:
How many ways to seat $n$ couples so that each husband's neighbors can only be his wife or another gentleman?
How to seat $n$ couples (husbands and wives) in a line so that each husband's neighbors can only be his wife or another gentleman?  I.e. his neighbor cannot be another woman that's not his wife.

There is a classical similar problem: In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?

But it's a bit different. In this case the husband can only be sitting with either his wife or another gentleman. Can the inclusion-exclusion principle still be applied on this problem? I can't think of a way..

[edit] through programmatical iteration, I found that the solution is http://oeis.org/A096121. But there isn't a proof to it. It's not hard to show that the rook problem is equivalent to this problem -- one column is the wives and the other column is the husbands. The sequence of the locations of the rook is how you arrange the husbands and wives.

But I wonder how do you prove the recurrence formula $a_{n+1} = n(n+1)(a_n + a_{n-1})$ ?

Another formula in OEIS was given as $a_n = 2 \cdot n! \cdot b_n$ where $b_{n+1} = n b_n + b_{n-1}$. $b_n$ is defined in http://oeis.org/A102038  This formula seems a bit more approachable. If we know the solution where the rook starts in one column, we multiply it by 2 so we get the solution where the rook in either column. $n!$ is just the permutation of all the cells in one column. We can interpret the $2 \cdot n!$ part but not sure about the rest!

答:
(Edits: Missed a factorial term, added numerical values in example.)

Denote $a_n$ to be this number of arrangements of $n$ couples.

We will show we can produce a sum formula for $a_n$, namely
$$
a_n  =  2 \sum_{k=1}^{n} {n!}{(n-k)!}{n-k+\lceil\frac{k+1}{2}\rceil-1\choose n-k}{n-k+\lceil\frac{k}{2}\rceil-1\choose n-k}.
$$
by refining it by the number of places where a husband/wife couple actually sit next to each other. With this sum expression for $a_n$ we can then algebraically demonstrate the recursion $a_{n+1}=n(n+1)(a_n+a_{n-1})$. (Note. This isn't a bijective approach as I liked, but it at least can be used to prove the recursion in principle.) Okay let's begin.

As noted by many others, a husband can only sit next to his wife but no other women, and vice versa. So we identify where this occurs. The number of such occurrence is at least 1, and at most $n$, for $n$ couples.  So a typical arrangement is something like this:
$$
(\text{some group of men})\fbox{MW}(\text{some group of women})\fbox{WM}\cdots\fbox{MW}\cdots
$$

Let us call such pair $\fbox{MW}$ or $\fbox{WM}$ a lucky couple. Denote a space to be the group of positions between lucky couples.

So let us derive an expression for $a_n$. Note that we can exchange the wife with the corresponding husband, we know $a_n = 2\cdot\sum\cdots$, and so let us look at the cases where the sequence starts with a man.

By way of example, we will do $n=5$ to illustrate $a_5$ and use it to infer $a_n$:  Assuming the arrangement starts with a man,

Case $\cdots \fbox{MW}\cdots$, exactly 1 lucky couple:

There are $(_5P_1)=5$ many ways to pick this lucky couple. Then of the remaining $4$ men, they need to be placed in one space (to the left of the lucky couple), and the remain $4$ women, they need to be placed in one space (to the right of the lucky couple). So we have
$$
(_5P_1)(4!)(4!) = 2880
$$

Case $\cdots \fbox{MW}\cdots\fbox{WM}\cdots$, exactly 2 lucky couples:

There are $(_5P_2)=5\cdot 4$ many ways to pick these two lucky couples. Then of the remaining $3$ men, they need to be arranged and placed in two possible spaces, and the remaining $3$ women need to be arranged and placed in one possible space. So we have
$$
(_5P_2)m(3,2)m(3,1) = 2880,
$$
where $m(a,b)$ denotes the number of ways to arrange $a$ people into $b$ spaces. By stars and bars this is just $m(a,b) = {a+b-1\choose a}a!$.

Case $\cdots \fbox{MW}\cdots\fbox{WM}\cdots\fbox{MW}\cdots$, exactly 3 lucky couples:

With the same reasoning as before, we see this is
$$
(_5P_3)m(2,2)m(2,2) = 2160
$$

Case $\cdots \fbox{MW}\cdots\fbox{WM}\cdots\fbox{MW}\cdots\fbox{WM}\cdots$, exactly 4 lucky couples:

With the same reasoning as before, we see this is
$$
(_5P_4)m(1,3)m(1,2) = 720
$$

Case $\cdots \fbox{MW}\cdots\fbox{WM}\cdots\fbox{MW}\cdots\fbox{WM}\cdots\fbox{MW}\cdots$, exactly 5 lucky couples:

With the same reasoning as before, we see this is
$$
(_5P_5)m(0,3)m(0,3) = 120
$$

Now observe well that if we start the arrangement with a man, having $k=1,2,3,4,5,\ldots$ lucky couples give $1,2,2,3,3,4,4,\ldots$ many spaces for the remaining men, while $1,1,2,3,3,4,4,\ldots$ many spaces for the remaining women. That is, we have $\lceil\frac{k+1}{2}\rceil$ spaces and $\lceil\frac{k}{2}\rceil$ spaces for the remaining men and women, respectively.

Putting everything together (and not forgetting a factor of 2 as we can exchange men and women), we have
$$
a_5 = 2 \sum_{k=1}^{5}(_5P_k)m(n-k,\lceil\frac{k+1}{2}\rceil)m(n-k,\lceil\frac{k}{2}\rceil)\\
=2(2880+2880+2160+720+120 )=17520 \quad(\text{checks out!})
$$

So in general we have
$$
a_n = 2 \sum_{k=1}^{n}(_nP_k)m(n-k,\lceil\frac{k+1}{2}\rceil)m(n-k,\lceil\frac{k}{2}\rceil) \\
\implies a_n  =  2 \sum_{k=1}^{n} \frac{n!}{(n-k)!}{n-k+\lceil\frac{k+1}{2}\rceil-1\choose n-k}(n-k)!{n-k+\lceil\frac{k}{2}\rceil-1\choose n-k}(n-k)!\\
\implies a_n  =  2 \sum_{k=1}^{n} {n!}{(n-k)!}{n-k+\lceil\frac{k+1}{2}\rceil-1\choose n-k}{n-k+\lceil\frac{k}{2}\rceil-1\choose n-k}.
$$
(try the formula here with SageMath .)

By being careful with even and odd cases of $k$, you may be able to simplify $a_{n}+a_{n-1}$ to obtain the recursion.


问:
Olver "Application of Lie Groups to Differential equations" Proposition 2.18.
I have trouble with proving (and even intuitively understanding) the following proposition in Olver's book "Application of Lie Groups to Differential equations".

Proposition 2.18. Let $G$ act semi-regularly on $M$ and let $\xi^1(x), ..., \xi^{m-s}(x)$ be a complete set of functionally independent invariants defined on an open subset $W \subset M$. If a subvariety $S_F = \{x: F(x) = 0\}$ is $G$-invariant, then for each solution $x_0 \in S_F$ there is a neighbourhood $U \subset W$ of $x_0$, and an equivalent $G$-invariant function $\hat{F}(x) = \hat{F}(\xi^1(x), ..., \xi^{m-s}(x))$ whose solution set coincides with that of $F$ in $U$:

$$ S_F \cap U = S_{\hat{F}}\cap U = \{ x \in U : \hat{F}(\xi^1(x), ..., \xi^{m-s}(x)) = 0 \}.$$

The solution goes (approximately) as follows. We consider a complete set of invariants near $x_0$. Then, under certain relabeling of coordinates, we can change coordinates from $(x^1, ..., x^m)$ to $(\xi^1(x), ..., \xi^{m-s}(x), x^{m-s+1}, ..., x^m)$. It has the form $\psi(x) = (\xi(x), \hat{x})$, where $\xi(x) = (\xi^1(x), ..., \xi^{m-s}(x))$, $\hat{x} = (x^{m-s+1}, ..., x^m)$. In the new coordinates, function that defines subvariety becomes $F^*(\xi(x), \hat{x}) = F(x) $. Consider arbitrary $x_0 \in S_F$ and define function (on some suitable subset around $x_0$) as $\hat{F}(\xi(x)) = F^*(\xi(x), \hat{x}_0)$, where $\hat{x}_0$ is set of last $s$ values in new coordinates of a point $x_0$.

So far, I feel comfortable and understand how $\hat{F}(\xi(x))$ is well-defined. I have problem in the following short argument where we try to show that claim is true for such choice of $\hat{F}$.

Olver's approach: since $S_F$ is $G$-invariant, and the orbits of $G$ in these coordinates are the common level sets of the invariants $(\{ \xi(x) = c \})$, we find $F^* (\xi(x), \hat{x}) = 0$ if and only if $F^* (\xi(x), \hat{x}_0) = 0$ since both points lie on the same slice.

My confusion: I think I understand comment about orbits, as I am able to prove that if $O$ is some orbit then $(O \cap U) \subset \{ \xi(x) = c \}$ for some constant $c$. Now, consider any point $x$ such that $F^*(\xi(x), \hat{x}) = 0$. This means (by definition of $F^*$) that $F(x) = 0$ and so $x \in S_F$.
We would like to argue that $F^*(\xi(x), \hat{x}_0) = 0$. If somehow one could show that $x$ and $x_0$ are on the same orbit then $x_0 = g \cdot x$ for some $g \in G$ then $\xi(x_0) = \xi(g \cdot x) = \xi(x)$ because these are invariant functions and so because $x_0 \in S_F$, $0 = F(x_0) = F^*(\xi(x_0), \hat{x}_0) = F^*(\xi(x), \hat{x}_0)$.

Question 1: Is it possible to show that $x_0$ and $x$ are on the same orbit? Intuitively, I think that is not true as my function could be $F(x) = 0$ for all $x$. Then, $S_F = M$ would be the invariant surface. But then, if I consider any two points, they do not have be on the same orbit. Am I making a mistake? If not, what alternative reasoning is used here? Should we define $x_0$ for each orbit? But then, I guess, $\hat{F}$ is not well-defined as it is not enough to know invariant function values as we also need to specify on which orbit we are.

Now, consider the other direction. Assume that  $F^*(\xi(x), \hat{x}_0) = 0$. Again, if I knew that $x$ and $x_0$ are on the same orbit, then proof is clear. But as I mentioned, I do not see why that should be true.

Question 2: How to prove this direction? What happens if $x_0$ and $x$ are not on the same orbit? Is there some conditions (maybe implied by semi-regularity) that complete set of invariants uniquely determines orbit? I think this is true only for a regular case.

答:
In general the points $x_0$ and $x$ do not lie on the same $G$-orbit.
The point here is that the 'hatted' coordinates enter in an inessential way in determining the zero set of $F$.

An example going through the elements in the above construction:  $M={\Bbb R}^3$ and $G=\{g\in \Bbb R\}$ acting by  $g.(x,y,z)=(x,y,z+g)$, translation in the $z$-direction. $\xi^1(x,y,z)=x$, $\xi^2(x,y,z)=y$ is a complete pair of invariants for $G$.
Let $S$ be the zero-set of $F(x,y,z)=x(1+y^2+z^2)$. (A somewhat silly way to describe the $G$-invariant plane: $x=0$).

Along the lines you describe, we may rewrite $F$ using the invariants plus  one remaining coordinate as $F^*(\xi^1,\xi^2,\hat{z})= \xi^1(1+(\xi^2)^2+\hat{z}^2)$. Pick a point $(0,y_0,z_0)\in S$. Then
$$\hat{F}(x,y,z)=\hat{F}(\xi^1,\xi^2) = F^*(\xi^1,\xi^2,z_0)= \xi^1(1+  (\xi^2)^2+z_0^2)=x(1+y^2+z_0^2)$$
is now $G$-invariant and will have the same zero set as $F$. But points need not be on the same $G$-orbit (because of $y$).

Regarding details in the proof of Olver: Consider $x$ in a small neighborhood of $x_0$. When $F^*(\xi(x),\hat{x})=0$ then $F^*(\xi(g.x),\widehat{g.x})=F^*(\xi(x),\widehat{g.x})=0$ for $g$ in a neighborhood of the identity. By semi-regularity of the group action (assuming that this is in the differential sense) and since $\xi$ is invariant, $g\mapsto \widehat{g.x} \in {\Bbb R}^s$ must give rise to a local diffeo from a neighborhood of ${\rm Id}\in G$ to a neighborhood of $\hat{x_0}\in {\Bbb R}^s$. So in fact, $F^*(\xi(x),\hat{y})=0$ for all $\hat{y}$ in a neighborhood of $\hat{x_0}$. In particular, $F^*(\xi(x),\hat{x})=0$ iff $F^*(\xi(x),\hat{x_0})=0$ for $x$ in a neighborhood of $x_0$. This is the key point in the proof. Note again, that $x$ and $x_0$ need not be on the same $G$-orbit.


问:
Ignoring higher order terms in computations involving real analytic functions
We have the following four analytic functions of a real variable:

\begin{align}
\lambda (\epsilon) &= 0 + \lambda_1 \epsilon + \lambda_2 \epsilon^2 + \ldots \\
\tilde{\lambda} (\epsilon) &= 0 + \tilde{\lambda}_1 \epsilon + \tilde{\lambda}_2 \epsilon^2 + \ldots \\
\kappa (\epsilon) &= \kappa_0 + \kappa_1 \epsilon + \kappa_2 \epsilon^2 + \ldots \\
\tilde{\kappa} (\epsilon) &= \tilde{\kappa}_0 + \tilde{\kappa}_1 \epsilon + \tilde{\kappa}_2 \epsilon^2 + \ldots
\end{align}

where $\kappa_0, \tilde{\kappa}_0 > 0$ and the expansions are valid for $0\leq \epsilon < R$, for some $R>0$. Now, we want to analyze a function of the form

$$ f(\epsilon) = -\lambda (\epsilon)\ln \lambda (\epsilon) +\tilde{\lambda} (\epsilon)\ln \tilde{\lambda} (\epsilon) -\kappa (\epsilon)\ln \kappa (\epsilon) +\tilde{\kappa} (\epsilon)\ln \tilde{\kappa} (\epsilon). $$

In particular, for sufficiently small $\epsilon$, we would like to "ignore" all the terms of order 2 or higher from the above expansions so as to get the following form for the derivative for $\epsilon > 0$:

$$ f'(\epsilon) = \ln (1/\epsilon) [\lambda_1 - \tilde{\lambda}_1 ] + Err(\epsilon), $$
where $Err(\epsilon)$ can be uniformly bounded as a function of $\epsilon$. Are we justified in making the above approximations. If yes, how can we give a mathematically rigorous reasoning for this justification?

答:
Consider that this is the composition of Taylor series. Changing notations, you want to compute
$$f(\epsilon)=-y_1(\epsilon)\log[y_1(\epsilon)]+y_2(\epsilon)\log[y_2(\epsilon)]-y_3(\epsilon)\log[y_3(\epsilon)]+y_4(\epsilon)\log[y_4(\epsilon)]$$
where
$$y_1(\epsilon)=a_1\epsilon+a_2\epsilon^2+O(\epsilon^3)$$
$$y_2(\epsilon)=b_1\epsilon+b_2\epsilon^2+O(\epsilon^3)$$
$$y_3(\epsilon)=c_0+c_1\epsilon+c_2\epsilon^2+O(\epsilon^3)$$
$$y_4(\epsilon)=d_0+d_1\epsilon+d_2\epsilon^2+O(\epsilon^3)$$

This will give the expansion of
$f(\epsilon)$ to $O(\epsilon^3)$ (I shall not type it).

Computing the derivative, we shall find
$$f'(\epsilon)=K+(b_1-a_1) \log (\epsilon )+O\left(\epsilon ^1\right)$$
$$K=-a_1(1+\log(a_1))+b_1(1+\log(b_1))-c_1(1+\log(c_0))+d_1(1+\log(d_0))$$

You just forgot the constant term but you have the leading order.


问:
Probability question from Brilliant
Which is more likely?

You roll two dice 5 times and, every time, one of the two comes up as 1 and the other as 6.

You roll 10 dice all at once. 5 come up as 1s and the other 5 come up as 6s.

The answer is 2 and I'm confused by it.

Here's my thinking. When rolling two dice the probability of getting 1 and 6 or 6 and 1 is $2\cdot(1/6\cdot 1/6) = 2/36 = 1/18.$ This means that probability of getting this 5 times in a row is $(2/36)^5$.

In the second case there are $2^5$ dice combinations of 1s and 6s where we get 5 of each. The probability of every combination is $(1/6)^{10}$ and so the probability of the second case is $2^5\cdot (1/6)^{10}=(2/36)^5$ which is exactly the same as the first case.

What's wrong with my thinking?

答:
the second is

$$\binom{10}{5,5}\times \left(\frac{1}{6}\right)^5\times \left(\frac{1}{6}\right)^5=\frac{10!}{5!5!}\times \left(\frac{1}{6}\right)^5\times \left(\frac{1}{6}\right)^5$$

thus it is most likely being

$$\frac{10!}{5!5!}>2^5$$

Title: Probability question from Brilliant


问:
Can anyone explain me the solutions of this $|x-5| + |x+6| = 11$?
I have an intuition of this using number line but I want to know what mathematics goes into solving these type of problems.</p>

答:
A geometric way to look at this problem is to consider the inequality version in a lifted space and the set thus defined, i.e.
$$\{(x_1, x_2)\in\mathbb{R}^2 : |x_1 - 5| + |x_2 - (-6)| \le 11,\; x_1 = x_2\}$$
This defines the intersection of $B_1 := \{(x_1, x_2)\in\mathbb{R}^2 : |x_1 - 5| + |x_2 - (-6)| \le 11\}$ with the line $x_1=x_2$.

If you have heard of $\ell_1$-norm, then $B_1$ is nothing but the ball centered at $(5,-6)$ with radius $11$ in this norm, which takes a diamond shape (i.e. a box rotated 90 degrees). It's a good exercise to try to draw it. Then intersect $B_1$ with the line $x_1 - x_2=0$ gives the solution.

Intersecting a line with an $\ell_1$-ball in two dimensions may have one, two, or infinite number of solutions. In your case, I think there are infinite number of solutions, because every point on the line segment $(-6,-6)$ to $(5,5)$ is a solution, i.e. every $x\in [-6,5]$ is a solution.


答:
1: Let $|y|=|x-5|\Rightarrow x=y+6$, we have:

$|y|+|y+5+6|=11$

This is possible only if $y=0$, hence we have:

$|x-5|=0\Rightarrow |x|=5$

which gives $0\leq x\leq 5$ and $-5\leq x\leq 0$

2: Let $|y|=|x+6|\Rightarrow  x=y-6$, we have:

$|y-6-5|+|y|=11\Rightarrow |y-11|+|y|=11$

This is possible only if $y=0\Rightarrow x=-6\Rightarrow |x|=6$ and we have:

$-6\leq x \leq 0$

So solutions in general is:

$-6\leq x \leq 0$

$0\leq x\leq 5$

Title: Can anyone explain me the solutions of this $|x-5| + |x+6| = 11$?

Body: I have an intuition of this using number line but I want to know what mathematics goes into solving these type of problems.</p>


问:
How can you calculate $\int_{0}^{1} \exp(\sin(x))dx$ exactly?
I'm taking a computational statistics class and I have to calculate $\int_{0}^{1} \exp(\sin(x))dx$ numerically (Monte Carlo integration) and compare the numerical result to the exact result. The problem is, how can one canculate the exact value of the integral? The antiderivative doesn't seem to exist in terms of elementary functions. Mathematica gives me this:

Not only that, Mathematica also cannot calculate the definite integral:

And since Mathematica cannot do it, I don't think other softwares can, though I'm not sure. Of course, Mathematica can do the numerical integration

But that's not what I need! Thanks in advance for any help.

答:
To integrate

$$\int_0^1 e^{\sin(x)}\ dx$$

take first $x = \sin^{-1}(u)$, so $dx = \frac{1}{\sqrt{1 - u^2}}\ du$, then

$$\int_{0}^{1} e^{\sin(x)}\ dx = \int_{0}^{\sin(1)} \frac{e^u}{\sqrt{1 - u^2}}\ du$$

and now note that, by using the general binomial theorem and the identity $(m)_k = (-1)^k (-m)^{(k)}$ converting the falling and rising factorial power,

$$\begin{align}\frac{1}{\sqrt{1 - u^2}} &= (1 - u^2)^{-1/2} \\
&= \sum_{k=0}^{\infty} \binom{-1/2}{k} (-1)^k u^{2k} \\
&= \sum_{k=0}^{\infty} \frac{(-1/2)_k}{k!} (-1)^k u^{2k}\\
&= \sum_{k=0}^{\infty} (1/2)^{(k)} \frac{(u^2)^k}{k!}\end{align}$$

so that, by simply multiplying the series, we have

$$\frac{e^u}{\sqrt{1 - u^2}} = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} (1/2)^{(k)} \frac{u^n}{n!} \frac{(u^2)^k}{k!}$$

And integrating, we get, noting that we should combine $u^n$ and $(u^2)^k = u^{2k}$ to $u^{(2k + n)}$,

$$\int_{0}^{\sin(1)} \frac{e^u}{\sqrt{1 - u^2}}\ du = \int_{0}^{1} e^{\sin(x)} dx = \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(1/2)^{(k)}}{(2k + n + 1)} \frac{[\sin(1)]^{2k + n + 1}}{n! k!}$$

More generally,

$$\int e^{\sin(x)}\ dx = C + \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(1/2)^{(k)}}{(2k + n + 1)} \frac{[\sin(x)]^{2k + n + 1}}{n! k!}$$

. Note that this is a hypergeometric series in two variables. To see that, you should first note the identity $\frac{2^{(n)}}{1^{(n)}} = n + 1$, thus

$$\sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(1/2)^{(k)}}{(2k + n + 1)} \frac{u^{2k + n + 1}}{n! k!} = u \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{(1/2)^{(k)} 1^{(2k+n)}}{2^{(2k+n)}} \frac{(u^2)^k u^n}{k!n!}$$

so it looks like some sort of

$$F_\mathrm{hyp}(a; b; c; x, y) := \sum_{n=0}^{\infty} \sum_{k=0}^{\infty} \frac{a^{(n)} b^{(2n+k)}}{c^{(2n+k)}} \frac{x^n y^k}{n! k!}$$

double hypergeometric function in $u = \sin(x)$ and $[\sin(x)]^2$. But I don't think this particular form has any official name, though with some work maybe it can be reduced to known ones. The Horn functions 1 come closest, with the doubled index; but none have a $2n+k$ index in both numerator and denominator of the coefficient part. Two-variable hypergeo are obscure, though, at least from Wolfram's point of view, because it does not implement them.

Welcome to this new and exotic world.


答: Here's some easy intuition as to why $2$ has to be more probable than $1$.  Suppose you roll the $10$ dice two at a time.  If the first pair comes up as a pair of $1$s, you're out of the running for scenario $1$, but you're still alive for scenario $2$.  But there is no sequence of rolls you can make that knocks you out of the running for scenario $2$ but leaves it possible to achieve scenario $1$.

Title: Probability question from Brilliant

Body:
Which is more likely?

You roll two dice 5 times and, every time, one of the two comes up as 1 and the other as 6.

You roll 10 dice all at once. 5 come up as 1s and the other 5 come up as 6s.

The answer is 2 and I'm confused by it.

Here's my thinking. When rolling two dice the probability of getting 1 and 6 or 6 and 1 is $2\cdot(1/6\cdot 1/6) = 2/36 = 1/18.$ This means that probability of getting this 5 times in a row is $(2/36)^5$.

In the second case there are $2^5$ dice combinations of 1s and 6s where we get 5 of each. The probability of every combination is $(1/6)^{10}$ and so the probability of the second case is $2^5\cdot (1/6)^{10}=(2/36)^5$ which is exactly the same as the first case.

What's wrong with my thinking?


问:
Expressing a problem generally through cartesian product
It's quite an easy problem, but can't seem to wrap my head around it.

{ { 1 } U { 2 } U { 3,  4 } U { 1 , 2 } U { 1 , 3, 4 } U { 2 , 3 , 4 } }

Questions:

This information will be in terms of a set of sets so I am thinking I can represent it with a function like the following:

```<code>Let F be the set of farms,
W the set of farmers,
w works f to mean w attends to f
so, F(w) = { f in F : w works f }.
</code>```

答:
If $F$ is your set of farms, then the power set $\mathscr{P}(F)$ is the set of all possible collections of farms. For example, if there are three farms in $F$, then $\mathscr{P}(F)$ contains all the sets like $\{1\}$, $\{1,2\}$, $\{2,3\}$, $\{1,2,3\}$, $\{3\}$, $\{\}$, etc.

If there are some farms that must be worked together, you can define them as a subset $E\subseteq \mathscr{P}(F)$. For example, in your case, $E = \{\{1\},\{2\}, \{3,4\} \} \subseteq \mathscr{P}(F)$. You can think of this as a set of jobs. (A job is a set of farms that must be worked together.)

Now what about workers?  You could use the cartesian product if you like. But I think you may have better luck with function spaces than cartesian products. If $W$ is your set of workers and $E\subseteq \mathscr{P}(F)$ is a set of possible jobs then an assignment from workers to jobs is a function $\alpha:W\rightarrow E$.

The number of possible assignments is $|E|^{|W|}$. For example, in your case, there are three different jobs ({1}, {2}, or {3,4}). When there are two workers, there are $3^2 = 9$ possible assignments.

Suppose you want to know what farms are being worked on. The image $f(W)$, which is defined as $f(W) \equiv \{ f(w) : w\in W\} \subseteq E$, is the set of all jobs that are being worked on. You can take their union to get the set of all farms being worked on, regardless of who is working on them:

$$\left[\bigcup_{A\in f(W)} A \right] \subseteq F$$

This tells you which farms are being worked on for the assignment $f$.


问:
Why is the power series of $\tan(x)$ not convergent everywhere $\cos(x)$ is non-zero?
The Taylor series expansion of $\tan(x)$ centered at $0$ has a radius of convergence of $\pi/2$, which means the power series will not converge for $|x|>\pi/2$.

How can this be when you consider $\tan(x)=\sin(x)/\cos(x)$, so Taylor series of $\tan(x)$ is just the Taylor series of $\sin(x)$ divided by the power series of $\cos(x)$, both of which converge everywhere.

At some point $|x|>\pi/2$ where $\cos(x)$ is not equal to $0$, how can the power series of $\tan(x)$ diverge? It is simply the quotient of two convergent series at that point which seems to me like it shouldn't have any problems.

答:
The representation of a function may or may not provide all the values of where a function is definable. There are whole books dedicated to writing functions in multiple different ways in order to understand the values of the function on different subsets of the complex plane (integrals, dirichlet series, power series, etc.). Take for example, the Riemann-Zeta function; this function has many many different representations, most of which only exist on some half plane of $\mathbb{C}$. Yet it is known that the Riemann-Zeta function is analytic everywhere in the complex plane and has only one simple pole at 1; a non-obvious fact if you were to look at any of the representations of the Riemann-Zeta function. The main points are:

A representation may or may not provide all the values of where a function is analytic.

A function can have many representations.

Representations have their own properties apart from the functions they might represent.

Power series are one kind of representation for analytic functions. Power series enjoy the property of uniqueness. Some functions are only really known by their power series expansion and have no "closed form". Power series expansions have what we call a radius of convergence, this radius is what defines where a power series is useful and tells us values of the underlying function. The radius of convergence is always as large as possible for the underlying function represented. If you stray outside that radius of convergence then the power series is unhelpful and does not tell you the proper values of the underlying function that is being represented. Take for example, $$f(z) = \frac{1}{1-z}$$ this function is clearly defined and analytic in all of $\mathbb{C}-\{1\}$. However, "a" representation of $f(z)$ is the power series centered at zero, namely for every $z\in B(0,1)$, we have, $$f(z) = \sum_{k=0}^\infty z^k$$ This power series is only useful on $B(0,1)$ and is not useful outside that set. The power series agrees with $f(z)$ wherever it converges, but the power series is not guaranteed to converge wherever $f(z)$ exists. This is true of any representation of a function. We could have picked another power series instead, say one centered at -1, $$f(z)=\sum_{k=0}^\infty \frac{1}{2^{k+1}}(z+1)^k$$ This power series converges with radius 2 instead of radius 1. Secretly, I know $f(z)$ has a pole at 1, and since power series always maximize their radius then I knew the radius had to be 2 for this power series because -1 is 2 units away from 1, even before I calculated the coefficients. Likewise, if I found the power series at $-i$, I already know the radius would be $\sqrt{2}$, since that is the distance from $-i$ to the pole at 1 for $f(z)$.

The power series of $\tan(z)$ at zero has radius $\frac{\pi}{2}$ because $\tan(z)$ has at least one singularity on the circle $B(0,\frac{\pi}{2})$. If it didn't have a singularity somewhere on that circle, then the radius of convergence would have been bigger. The radius of convergence always maximizes where the underlying function is analytic.


答: The mistake is subtle: When rolling the dice 5 times in 1), you are required for each roll to include a 1 and a 6. However, if you roll ten dice, you allow for the two dice to be 6, or 1, or both. So the correct probability for the first five dice in 2) is in fact higher, than the first five dice in 1). So while your calculation in 1) is correct, yours in 2) is incorrect. The probability of 2) will range between $(\frac{2}{6})^9*(\frac{1}{6})^1$ at most, and $(\frac{2}{6})^5*(\frac{1}{6})^5$ at worst, depending on what you roll. Whatever the actual answer, 2) will always have a higher probability.

Title: Probability question from Brilliant


问:
Probability of balloons with ratio

I solved this like this:

Total balloons$=1+5=6$

So the required probability is $\frac{3}{6} =0.5$

Is this correct? Can someone please say?

答:
Assuming that the selection is made without replacement, the probability changes with $k$

Your solution is correct only if $k=1$

In the bag there is a total of $6k$ balls with $k$ blue and $5k$ other colour

The probability to get at least one blue is the complement of having no blue in 3 selections, that is

$$1-\frac{\binom{5k}{3}}{\binom{6k}{3}}=1-\frac{5}{6}\cdot\frac{(5k-1)(5k-2)}{(6k-1)(6k-2)}$$

with $k\geq 1$

Title: Probability of balloons with ratio


问:
Bounded variation on [0,1] implies there is another function on [0,1] that differentiability and absolutely continuous
Problem : Let $f:[0,1]\to \Bbb R$ is of bounded variation. Prove that there exists $g:[0,1]\to \Bbb R$ which absolutely continuous and  $g'(x)=f'(x)$ for almost every $x \in [0,1]$.

I have tried to introduce $f$ as a difference of 2 monotonic functions but I'm getting stuck. I will be very happy is someone can help me with that question

答: Hint: Prove that $f'$ is Lebesgue integrable, and then consider its indefinite (Lebesgue) integral. For the former, your approach is useful.


问:
Zero Probability Events
For an uncountable number of zero-probability events, their union may not be zero-
probability.

Is there an example of such a case?
I was told that this is why we cannot have uncountable additivity in the probability axioms.

答:
Consider the uniform distribution on $[0,1]$. All singletons $\{x\}$ have probability $0$, but their (uncountable) union is the whole interval $[0,1]$ which has probability $1$.

Really, this example with singletons applies to any continuous distribution.

If you want more exotic events, you can pick the set $\mathbb{Z}$ of integers or set $\mathbb{Q}$ of rationals and all their cosets under addition - there are uncountably many cosets, each with measure $0$ (with respect to a continuous distribution), and their union is the whole real line, which has measure $1$.

Title: Zero Probability Events


问:
Find the value of this 3 digit number such that dividing by it leaves REM = 11
REM = Remainder.

Now , It is easy to think of doing it by dividing the numbers starting from 100 till whatever until you get remainder as 11. But isthere a way to solve it by formula.

Like , what I have doing is and not getting the right answer for it.

$\frac{100 + x}{13}$= y (for y it is Some value but we don’t know)and (100+x because it has to be a number either greater or or equal to 100)

Then , 100+x - 13(y) = 11.

I’m not getting how to solve further than this.

One way I tried is this but got wrong ans.

So , from here.

100 + x - 13(6) = 9 + x.

9+x = 9+x

But x gets cancel out.

答:
Firstly, the correct notation for remainder $REM(n \div p) = r$ is

$$n = pq + r \implies n\equiv r \mod p$$

Examples include $13\equiv 3\mod 5$ and $444 \equiv 3 \mod 7$

So you want to find the minimum $n\geq 100$ such that $n\equiv 11\mod 13$

$$n\equiv 11\mod 13, n\geq 100$$

$$n + 100 \equiv 11 \mod 13, n \geq 0$$

$$n+9+13 \cdot 7 \equiv 11\mod 13,n\geq 0$$

Since adding 13 won't change the remainder, we can ignore the $13\cdot 7$ term:

$$n+9\equiv 11\mod 13, n\geq 0$$

$$n\equiv 2\mod 13$$

In english terms, this means $n$ (the two-digit part) has remainder 2 when divided by $13$. The trivial example is $n=2 \implies 102\equiv 11\mod 13$, but other examples are $n=15,28,41,54,\cdots$. These give $115,128,141,154,\cdots$ which all give remainder of $11$ when divided by $13$.

Hope this helps!


问:
Circle, unit interval and unit square pairwise not homeomorphic
In my Topology exercise class I have the task to show that $S^1$, $[0,1]$ and $[0,1] \times [0,1]$ are pairwise not homeomorphic, using the fact that $S^1$ fits into the following pushout square: $\require{AMScd}$
\begin{CD} \{0,1\} @>\operatorname{id}>> [0,1] \\
@VVV @VVV \\
\{\star\} @>>> S^1
\end{CD}
I think that I understand this fact (as the universal property indeed holds for $S^1$ by commutativity requirements).
But I am not really sure how to proceed now. The exercise feels as if I should assume two of the spaces to be homeomorphic and to derive a contradiction. But is it allowed to replace, say, $S^1$ by $[0,1]$ in the diagram to get a contradiction? Does this even make sense if we do not have an explicit homeomorphism?
Any hint is really appreciated :)

答:
If $X$ is connected and $p \in X$ is such that $X\setminus \{p\}$ is disconnected, then $p$ is called a cut-point of $X$. Likewise, if $A \subseteq X$ is such that $X\setminus A$ is disconnected, $A$ is a cut-set of $X$. Easy fact: if $h: X \to Y$ is a homomorphism $A$ is a cut-set of $X$ iff $h[A]$ is a cut-set of $Y$.

So if $X \simeq Y$ they have the same numbers of cut-sets and (non-)cut-points. This is a rough but general tool to reason about certain connected spaces being homeomorphic or not:

$[0,1]$ has $2$ non-cutpoints (the endpoints), all other points are cut-points.

$\Bbb S^1$ has no cut-points but any subset of size $2$ is a cut-set.

$[0,1]^2$ has no cut-points and no subset of size $2$ is a cut-set.

So for your three spaces such simple considerations suffice.


问:
How can I show this equation is less than or equal to $\pi$?
I have that $$ \left|\int_\gamma\frac{\mathrm{d}z}{2+z^2}\right| $$
where $\gamma$ is upper half of unit circle

I know that $$\gamma(t) = e^{it}$$
and therefore $$\gamma(t) = e^{it}$$

and as $$ \left|\int_\gamma f(z)\ \mathrm{d}z\right| = \left|\int_a^b f\big(\gamma(t)\big)\gamma'(t)\ \mathrm{d}t\right| $$

I know the originally equation is equal to $$\left|\int_0^\pi\frac{1}{2 + (e^{it})^2}ie^{it}\ \mathrm{d}t\right|$$

how could I show this is $\displaystyle\left|\int_0^\pi\frac{1}{2 + (e^{it})^2}\cdot ie^{it}\ \mathrm{d}t\right|\le \pi$

答:
Why not just find an anti-derivative and use the fundamental theorem of calculus. Note, that u-substitution of $u=e^{i\theta}$ and $du=ie^{i\theta}d\theta$, would be helpful.

$$\int_0^\pi \frac{ie^{i\theta}}{2+e^{2i\theta}}d\theta=\int_{\theta=0}^{\theta=\pi}\frac{du}{2+u^2} = \left.\frac{1}{2} \tan^{-1}(u/\sqrt{2})\right|_{\theta=0}^{\theta=\pi} = \left.\frac{1}{2} \tan^{-1}(e^{i\theta}/\sqrt{2})\right|_0^\pi$$ $$= \frac{1}{2}\left(\tan^{-1}(-1/\sqrt{2})-\tan^{-1}(1/\sqrt{2})\right) = -0.615479708670387...$$


问:
Quasi-coherent sheaves on punctured affine plane
Let $A=k[x,y]$ be the polynomial ring in two variables over a field $k$,
$\mathbb{A}^2=\mathrm{Spec} A$ the affine plane ,
and $U=\mathbb{A}^2 \setminus \{0\}$ the punctured affine plane.
Then the canonical injection $j : U \to \mathbb{A}^2$ induces
a fully faithful left exact functor $j_{*}: \mathrm{QCoh} (U) \to \mathrm{QCoh} (\mathbb{A}^2)\simeq \mathrm{Mod} (A)$ between the category of quasi-coherent sheaves.
In the page 114 of this book,
the author says
\begin{equation}
\mathrm{QCoh} (U)=\{M\in \mathrm{Mod}(A) \mid \mathrm{Hom}_A(A/\mathfrak{m}, M) =\mathrm{Ext}^1 (A/\mathfrak{m}, M) =0  \}
\end{equation}
as a subcategory of $\mathrm{Mod} (k[x,y])$,
where $\mathfrak{m}=(x,y) \subset A$.

My Question:

Why can we describe $\mathrm{QCoh} (U)$ as the above?

For a scheme $X$ with a closed point $x\in X$,
consider the canonical open immersion $j:X\setminus \{x\} \to X.$
Then can we calculate $(j_* \mathcal{F})_{x}$ for a quasi-coherent sheaf on $X\setminus \{x\}$?

答:
Honestly, I can't figure out how the book wants us to explain this fact. Based on the mention of section functors and localizations I assume we are supposed to use the stuff in sections 2.13 and 2.14, so let's give this a shot to answer your question 1.

$\DeclareMathOperator{\QCoh}{QCoh}
\newcommand{\QA}{\QCoh(\mathbb A^2)}
\newcommand{\m}{\mathfrak m}
\newcommand{\QU}{\QCoh(U)}
\newcommand{\Id}{\mathrm{Id}}
\newcommand{\F}{\mathcal F}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Ext}{Ext}
\DeclareMathOperator{\Spec}{Spec}
$
Let $T$ be the full subcategory of $\QA$ consisting of sheaves set-theoretically supported at $\{0\}$. In terms of modules, these are modules such that every element is annihilated by a power of $\m$. It is clearly a Serre subcategory.

We have $j^*\colon \QA\to \QU$. It sends $T$ to $0$, so by the universal property (Theorem 13.9), it factors through $\QA\to \QA/T\to \QU$. We claim the second functor is an equivalence. The inverse is the composition $\QU\xrightarrow{j_*} \QA\to \QA/T$. Going one way, $j^*j_*\cong \Id_{\QU}$. Going the other way, we have a natural transformation on $\QA$:
$$
\Id_{\QA}\to j_*j^*.
$$
I believe that all we need to see is that it becomes an isomorphism after modding out by $T$.

Let's just identify $\QA/T$ with $\QU$ from now on.

$T$ is localizing, since $j^*\colon \QA\to \QU$ has a right adjoint, $j_*$. In terms of Section 2.14, $j_*$ is the "section functor". Proposition 14.7(6) tells us already that $j_*$ is fully faithful. Proposition 14.7(3) tells us that if $\F\in \QU$ and $M\in T$,
$$ \Hom_{\QA}(M,j_*\F)\cong \Hom_{\QU}(j^*M,j^*j_*\F) \cong \Hom_{\QU}(0,\F)=0.
$$
Proposition 14.7(3) tells us that $$\mathrm{Ext}^1_{\QA}(M,\F)=0.$$So this gives us one inclusion.

It remains to prove that if $M\in \QA$ is such that $\Hom(A/\m,M)=\Ext^1(A/\m,M)=0$, then $G$ is in the essential image of $j_*$. For this we use Theorem 14.8. It gives us an exact sequence:
$$
0\to \tau M\to M\xrightarrow{\eta} j_*j^*M\xrightarrow{\pi}M'\to 0.
$$
It further tells us that the first and last term are in $T$ and that $\eta$ is an essential map. We are going to heavily use that $A/\m$ generates $T$, in order to show that $\eta$ is an isomorphism.

First, being torsion, if $\tau M\neq 0$ has some submodule isomorphic to $A/\m$. This would mean that $M\supseteq \tau M$ does as well, which contradicts the assumption that $\Hom(A/\m,M)=0$. So $\tau M=0$.

Finally, if $M'\neq 0$ it also has a submodule isomorphic to $A/\m$. We then have a short exact sequence
$$
0 \to M\to \pi^{-1}(A/\m)\to A/\m\to 0
$$
Since $\Ext^1(A/\m,M)=0$ by assumption, this sequence splits, so $A/\m$ embeds into $\pi^{-1}(A/\m)\subseteq j_*j^*M$, into a submodule disjoint from $M$. This contradicts the fact that $\eta$ is essential. So $M'=0$, and $\eta$ is an isomorphism as desired.  This answers question 1, modulo some details.

For question 2, I'm not sure what constitutes a satisfactory answer. Since you are looking at the stalk, you can assume that $X$ is affine, say $\Spec R$, and $x = V(f_1,\ldots ,f_n)$ for some $f_i\in R$. If we let $U_i = X\setminus V(f_i) = \Spec R[f_i^{-1}]$
$$
X\setminus x = \bigcup_i U_i
$$
So by the property of being a sheaf, for any sheaf $M$ on $X\setminus x$, we have that $\Gamma(X,j_*M)=\Gamma(X\setminus x,M)$ is the kernel of
$$
\bigoplus \Gamma(U_i,M) \longrightarrow \bigoplus \Gamma(U_i\cap U_j,M)
$$
And I believe that by the exactness of localization, tensoring the above map with the local ring at $x$ will give you the stalk as the kernel. In practice, how hard is this kernel to find? I have no clue. I can show that in the case of $\mathbb A^2$, $j_*\mathcal O_{\mathbb A^2\setminus 0} \cong \mathcal O_{\mathbb A^2}$.


问:
Does compactness imply measurability?
Let $(\mathbb{R}^n,\mathcal{L})$ be a measurable space, where $\mathcal{L}$ is the Lebesgue $\sigma$-algebra. Then,

$X\subset\mathbb{R}^n$ compact $\Rightarrow X\in\mathcal{L}$

Let $(\mathbb{R}^n,\mathcal{B})$ be a measurable space, where $\mathcal{B}$ is the Borel $\sigma$-algebra. Then,

$X\subset\mathbb{R}^n$ compact $\Rightarrow X\in\mathcal{B}$

Are (1) & (2) true statements? If we replace compact for closed, are (1) & (2) true?

Bonus question: When we say that a set $X\subset\mathbb{R}^n$ is Lebesgue (or Borel) measurable, what we mean is that $X\in\mathcal{L}$ (or $X\in\mathcal{B}$), correct?

答:
Yes to both, trivially, because closed sets (being the complement of open sets which are by definition in the Borel $\sigma$-algebra) are Borel and hence Lebesgue measurable.

And yes to the "bonus question", just as $O$ open is the same as $O \in \mathcal{T}$ for a topological space $(X,\mathcal{T})$.

Title: Does compactness imply measurability?


问:
Golden Exponent? Tetration
I read somebody say “golden exponent $ x^x=x+1 $” now I didn’t understand what he meant but it really fascinated me thinking about some type of tetration version of the golden number. A number with a special or awesome property (not necessarily linked to the golden number). Now that I think about the equation mentioned $ x^x=x+1 $ that does look like an interesting equation for x to satisfy. Anyways, what does everybody think about whether such an analog for tetration exists or not?

Title: Golden Exponent? Tetration

Body: Given that $0^0=1$ let $x=0$, then $x^x=x+1$ reduces to $0^0=0+1=1$, so $x=0$.


问:
Robot moves from $(x,y)$ to $(x+y, y)$ or $(x,x+y)$
I was working on some coding related to this topic I found on Stack Overflow. This lead me to a math problem I thought would be interesting. I was wondering if one was given a starting point, what points could the robot reach. For instance, if the robot started at $(10,15)$, which coordinates would be reachable. To restate the problem,

A robot moves in the following way. If it is at the point $(x,y)$, it can move to either $(x+y,y)$ or $(x,x+y)$. If the robot starts at the point $(10,15)$, what points are reachable in a finite number of moves?

I noticed that each move preserves the greatest common divisor of the two coordinates, so all the coordinates that are reachable must have gcd equal to 5. Moreover, for relatively prime integers $a$ and $b$, if the robot can move from $(a,b)$ to $(c,d)$, then it can move from $(na, nb)$ to $(nc, nd)$. Therefore, we just have to consider the coordinates that are reachable from $(2,3)$ and just multiply each coordinate by 5. However, I'm not sure what coordinates are reachable. There are coordinates like $(35, 75)$ which are not reachable, even though their greatest common divisor is 5. Any help on the question would be great. Thanks!

答:
Premise

This answer aims to give a intuition of the solution, without going too mush into detail and leaving some demonstrations to the reader.

Hypotesis

I assume the robot can only move in points with positive coordinates.

Solution

History of a position

Given a starting point $(x, y)$:

if $x > y$ then the previous position of the robot must be $(x - y, y)$, otherwise the $y$ coordinate would become negative;

if $x < y$ then the previous position of the robot must be $(x, y - x)$, otherwise the $x$ coordinate would become negative;

if $x = y$ then there cannot be previous positions, since $(x, x)$ cannot be reached neither from $(0, x)$ nor from $(x, 0)$.

Considered that, every position on the first quadrant (excluding the diagonal) has a unique previous position. Let us define:

$
\begin{equation}
P^{-1}(x,y) =
\begin{cases}
(x - y, y) & \mbox{if } x > y,\\
(x, y - x) & \mbox{if } x < y,\\
(x, y) & \mbox{if } x = y.\\
\end{cases}
\end{equation}
$

By repeatedly computing the previous position, we can determine the whole history $H(x, y)$ of the moves made by the robot to reach the current position.

$
\begin{equation}
H(x_0,y_0) = \{(x, y) \colon\ (x, y) = P^{-n}(x_0, y_0) \text{ for some } n \in \mathbb{N} \},
\end{equation}
$

where
$P^{-n} = P^{-1} \circ P^{-1} \circ \cdots \circ P^{-1}$.

So, every point in this space encodes the whole history of moves needed for reaching it. In fact, by repeatedly subtracting $|x-y|$ from the maximum of the two coordinates, we can go backwards until reaching some element on the diagonal.

Future of a position

The previous considerations lead us to the following statement: one position $(x_1, y_1)$ can be reached from another position $(x_0, y_0)$ if and only if $H(x_1, y_1) \supseteq H(x_0, y_0)$. As a consequence, the set of all positions reachable from position $(x_0, y_0)$ with a finite number of moves is the set

$
\begin{equation}
F(x_0,y_0) = \{(x, y) \colon\ H(x, y) \supseteq H(x_0, y_0) \}.
\end{equation}
$

Visual representation of results

I don't know if this set can be expressed in some more explicit form. We can have an idea of how this sets are made, by visualizing them. The maximal sets of reachable points are of the form $F(x, x)$.

By observing that

$
\begin{equation}
F(n, n) = \{(nx, ny) \colon \ (x, y) \in F(1,1)\},
\end{equation}
$

we deduce that $F(x, x)$ are scaled versions of $F(1, 1)$, that is represented in white in the following image.

Note further that since all points on the first quadrant are reachable from a starting point on the diagonal and since no point on the diagonal is reachable from another point on the diagonal, the sets $F(x, x), x \in \mathbb{N}\setminus\{0\}$, do not intersect and their union is the whole set of positions. This property can be visualized in the following image where $F(1,1)$ is depicted in red, $F(2,2)$ in green and $F(3,3)$ in blue.


问:
Inclusion-Exclusion Principle; certain intersection has to be empty

Let $S$ be a multi-set with k distinct different elements with given repetition numbers $n_1,n_2,\ldots,n_k$ respectively. Let $r$ be a positive integer such that there exists at least one $r$-combination of the set $S$. Show that in applying the inclusion-exclusion principle to determine the number of $r$-combinations of the set $S$, one has $A_1\cap A_2\cap\ldots\cap A_k=\varnothing$.

What I've done so far is noticed that since there exists one integer $r$ such that there is at least one $r$-combination of the set $S$, then $\sum_{i=1}^k n_i\ge k$. Also notice that each $n_i\ge1$. I can see that if one of those $n_i$ is $2$ or more, then I can see that the proposition is true. However what I'm having trouble with is the case when $n_1+n_2+\ldots+n_k=r$, because in this case $A_1\cap A_2\ldots\cap A_k=1$, correct? I'm unsure of how to proceed with this this proof any help is appreciated!

答: From "there is at least one r-combination of S", we can see that r is smaller or equal to n1 + n2 + ... + nk. When trying to figure out the cases of all the intersections, we need to assume there're n1+1, n2+1, ..., nk+1 elements respectively, which is clearly bigger than r. Hence, the intersection is the empty set.</p>