问:
Understanding how the induced representation of $S_3$ under $S_2$ with the standard representation works
I am attempting to understand the representation of the induced module $\mathrm{Ind}_{S_2}^{S_3}V = \Bbbk S_3 \otimes_{\Bbbk S_2} V$, where $V = \{v \in \Bbbk^2: v_1+v_2=0\} = \Bbbk(e_1 - e_2)$, and what the actual image of a given element in this representation.  I know that $\Bbbk S_3$ is a free $\Bbbk S_2$-module with basis given by $\{(1),(13),(23)\}$, so computing the image of an element should just boil down to figuring out what the action of said element is on each member of the basis for $S_3 \otimes_{\Bbbk S_3}V$. The issue I'm having is I can't figure out how this action works.

For example, suppose I want to compute the image of $(123)$ under this representation.  Then I must figure out the action of $(123)$ on $1 \otimes (e_1 - e_2)$, $(13) \otimes (e_1 - e_2)$, and $(23) \otimes (e_1 - e_2)$.  I know in the end, I get a representation $S_3 \rightarrow GL_3(\Bbbk)$ since $\mathrm{dim}(S_3 \otimes_{\Bbbk S_3} V) = [S_3:S_2] \cdot \dim(V) = 3$, but I just don't see how to get this matrix, let alone compute the group action.  I have a misunderstanding somewhere, I'm just not sure where it is.

答:
To apply $(123)$ to e.g. $1\otimes(e_2-e_3)$, you just apply $(123)$ to the first factor $1$, to get $(123)\otimes(e_2-e_3)$. This may not appear to be one of your basis elements for $\mathrm{Ind}_{S_2}^{S_3}V$, but it actually is. You just need to slide the element of $(12)\in S_2$ across the $\otimes$ symbol:

$$ \begin{array}{ll}
(123)\otimes(e_1-e_2) & =(123)(12)^{-1}(12)\otimes (e_1-e_2) \\
& = (123)(12)^{-1}\otimes(12)(e_1-e_2) \\
& = (123)(12)\otimes(e_2-e_1) \\
& = (13)\otimes(-(e_1-e_2)).
\end{array} $$

Thus, you get $-(13)\otimes(e_1-e_2)$. I'll let you try your hand at computing other group elements applied to other basis elements to write down the matrices.


问:
Compute the integral $\int_{|z|=\rho}|z-a|^{-4}|dz|$ with $|a|\neq \rho$
I need help in computing the integral indicated above. What I've tried so far:

Parametrize the curve indicated by $|z|=\rho$ with $\gamma = z(t) = \rho \cos t + i\sin t$. Then by definition
$$
\int_\gamma f(z)|dz|=\int_\gamma f(z(t))|z'(t)| dt
$$
gives the following
\begin{align}
\int_\gamma |z-a|^{-4} |dz| & = \int_0^{2\pi} |\rho \cos t+i\rho \sin t-a_1-ia_2|^{-4}\rho dt\\
& = \rho\int_0^{2\pi}\frac{1}{(\rho^2-2a_1\cos t-2a_2\sin t + a_1^2+a_2^2)^2} dt
\end{align}
Where $a=a_1+ia_2$. It's not hard to see how this becomes complicated very easily. I want to know if there is some sort of 'trick' I'm not aware of or something I might be missing.

答:
I think there's not much way to avoid some tedious calculations. I won't provide the full calculations, but I think this roadmap should be sufficient.

The cases of $a=0$ and $\rho = 0$ are pretty straightforward. For the other cases, I think this identity is a useful hint
\begin{align*}
\left| z-a \right|^{-4} = (z-a)^{-2} \overline{(z-a)^{-2}}
\end{align*}
From here, after replacing $z$ with $\rho e^{2\pi i t}$, you have to consider the cases of $|a| > \rho$ and $|a| < \rho$ and use Cauchy's Integration formula.

ContentLicense: CC BY-SA 4.0


问:
Prove that two random variables are equal almost surely
This is a problem from Durret's probability textbook. Show that if $X,Y$ are random variables with $E(Y|\mathcal{G})=X$ and $EY^2=EX^2\leq\infty$, then $X=Y$ a.s.

My question is, by the definition of conditional expectation, $\forall A\in \mathcal{F}$,
$$\int_{A} YdP = \int_{A} E(Y|\mathcal{G})dP=\int_{A} XdP$$
Doesn't this already imply that $X=Y$ a.s? Why do we need the condition that $EY^2=EX^2\leq\infty$?

答:
The equation $\int_A XdP=\int_AYDP$ holds only for $A \in \mathcal G$ and not for all  $A \in \mathcal F$. So we cannot conlude that $X=Y$ a.s. using just the definition.

We have $EY^{2}=EX^{2}=E(E(Y|\mathcal G))^{2}\leq E E(Y^{2}|\mathcal G)) =EY^{2}$. Hence, we must have equality throughout. And the fact that $EE(Y|\mathcal G)^{2}=EE(Y^{2}|\mathcal G)$ a.s implies that $E(Y-E(Y\mathcal G))^{2}=0$ as seen by expanding the square. It follows now that $Y$ is (a.s equal to a r.v. which is)  $\mathcal G$ measurable and hence $X=E(Y|\mathcal G)=Y$ a.s.


问:
Posterior from Gamma
Suppose the prior for $μ$ is $Gamma(k,λ)$, so
$fμ(m)=Cm^{k−1}exp(−λm)I(m≥0)$.

Further, suppose the data $X_1,…,X_n$ given $μ$ is iid Poisson with parameter $μ$, so for each $[Xj∣μ=m]∼[X∣μ=m]$,
$f_X∣μ=m(i)=exp(−m)m^i/i!I(i∈{0,1,…})$,

How can I show that the posterior for μ is also Gamma? Would the parameters of the Gamma be a function of $k$ and $\lambda$?

答:
The prior is

$$\pi(\mu)\propto \mu^{k-1}e^{-\lambda\mu}$$

The likelihood is

$$p(\mathbf{x}|\mu)\propto e^{-n\mu}\mu^{\Sigma_i x_i}  $$

Thus the posterior is

$$\pi(\mu|\mathbf{x})\propto \pi(\mu)\cdot p(\mathbf{x}|\mu)  =\mu^{(k+\Sigma_i x_i)-1}e^{-(\lambda+n)\mu}$$

Thus the posterior is still gamma with parameters

$$Gamma(k+\Sigma_i x_i;\lambda+n)$$

Hint: to simplify the calculations it is suggested to waste any quantity that doesn't depend on $\mu$

Title: Posterior from Gamma


问:
Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$
Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$
I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.

答:
Some thoughts:

WLOG, assume $a\ge b \ge c$.

By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\sqrt{a+b} + \sqrt{b + c} + \sqrt{c + a}
\le \sqrt{ \left( \frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}}\right)(2 + 1 + \sqrt{3})}.$$
It suffices to prove that
$$\frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}} \le 3 + \sqrt{3}.$$
This is true. Is there nice solutions?


问:
On the inequality $\frac{\sigma(n^2)}{q^k} < \frac{n^2 - q^k}{C}$ where $C>1$ and $q^k n^2$ is an odd perfect number - Part II
(Note:  This is a continuation of this earlier question.)

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.  (In particular, we know that $q \geq 5$.)

Denote the sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of aliquot/proper divisors of $x$ by $s(x)=\sigma(x)-x$.  Finally, denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

In this accepted answer to a closely related question, it was proved that
$$\frac{\sigma(n^2)}{q^k} \leq \frac{n^2 - q^k}{\frac{188}{63}}.$$

Since we also have, in general, the following equation and inequalities
$$\frac{\sigma(n^2)}{q^k} = \frac{D(n^2)}{s(q^k)} \leq \frac{\sigma(n^2)}{q} \leq D(n^2),$$
here is my:

QUESTION: If $N=q^k n^2$ is an odd perfect number with special prime $q$, is the following inequality true in general?
$$D(n^2) \leq \frac{n^2 - q^k}{\frac{188}{63}}$$

MY ATTEMPT

The inequality under consideration in the question is equivalent to
$$376n^2 - 188\sigma(n^2) \leq 63n^2 - 63q^k.$$
Suppose to the contrary that it is false.  This implies that
$$I(n^2) < \frac{313}{188} + \frac{63q^k}{188n^2}.$$
But we have the estimate
$$\frac{q^k}{n^2} < \frac{2}{315}$$
by Broughan, Delbourgo, and Zhou (2013), which implies that
$$\frac{2(q-1)}{q} < I(n^2) < \frac{313}{188} + \frac{63q^k}{188n^2} < \frac{313}{188} + \frac{63}{188}\cdot\frac{2}{315} = \frac{1567}{940},$$
from which it follows that
$$q < \frac{1880}{313} \approx 6.00639.$$
Since $q$ is a prime which satisfies $q \equiv 1 \pmod 4$, then $q=5$.  Alas, this is where I get stuck.

答:
No, it is not true in general.

Using $$D(n^2)=2n^2-\sigma(n^2)=2n^2-\frac{2q^kn^2}{\sigma(q^k)}=2n^2-\frac{2q^kn^2(q-1)}{q^{k+1}-1}$$
we see that
$$D(n^2) \leq \dfrac{n^2 - q^k}{188/63}$$ is equivalent to $$(63q^{k+1}-376q^k+313)n^2\ge 63q^k(q^{k+1}-1)$$
which does not hold when $q=5$ and $k\gt 1$ since then LHS is negative.


问:
poisson process limit, SLLN
The number of cars that arrive at a parking lot follow a Poisson process with $\lambda > 0$.
Be $T_1,T_2 ...$ The time of arrive between each car, so that $T_1 + T_2 + T_2  ... + T_n$ is the arrival time of the nth car

to which converges in probability

$\frac{1}{n} \sum_{i=1}^{n}T_{i}^2$ when $n \rightarrow \infty  $

I am not sure what to do with the summation, i think that the solution has something to do with the exponential distribution

the mean squared + variance $\rightarrow  \frac{1}{\lambda^2} + \frac{1}{\lambda^2} = \frac{2}{\lambda^2} $

because n tends to infinity  so the the dispersion measure will tend to an exponenential distribution (SLLN)

i will appreciate if someone can clarify me more on the solution,thanks

答: $(T_i)$ is i.i.d and so is $(T_i^{2})$. Hence, SLLN shows that $\frac  1 n \sum\limits_{i=1}^{n} T_i^{2} \to ET_1^{2}$. This is just the second moment of the $exp(\lambda)$ distribution.

Title: poisson process limit, SLLN


问:
Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$
How can I solve this question with out find a,b,c,d

$$a+b+c+d=2$$

$$a^2+b^2+c^2+d^2=30$$

$$a^3+b^3+c^3+d^3=44$$

$$a^4+b^4+c^4+d^4=354$$

so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$

If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d

Is there any help?

答: I do not know how to solve it, but $a=-3,b=4,c=2,d=-1$</p>

ContentLicense: CC BY-SA 3.0


问:
$y^{\prime} = (-x-y+2)^2, y(0) = 1$
$y^{\prime} = (-x-y+2)^2, y(0) = 1$

My ideas:

$ u = -x -y +2$

$u^{\prime} = -1 -y^{\prime}$

$u^{\prime}  = -u^2 -1$

I dont know how to solve this differential equation because of the $-1$

答:
Hint:
$$u'  = -u^2 -1$$
It's separable:
$$\int \dfrac {du} {u^2 +1}=-\int dx$$
And you have that:
$$\int \dfrac {du}{u^2+1}=\arctan u+C$$

Title: $y^{\prime} = (-x-y+2)^2, y(0) = 1$


问:
Confusion about Fisher information and Cramer-Rao lower bound.
I recently learned about Fisher information and the Cramér-Rao lower bound, but there is something that is bothering me. Take for example a poission distribution, the likelihood function is defined as,

$$\mathcal{L}(\theta;\pmb{Y}=y)=\exp\{-n\theta\}\dfrac{\theta^{\sum_{i=1}^{n}y_i}}{\prod_{i=1}^{n}y_i!}$$
The log-likelihood $\ell(\theta)$ is
$$\ell(\theta)=-n\theta+\left(\sum_{i=1}^{n}y_i\right)\log\theta+\log\left(\prod_{i=1}^{n}y_i!\right)$$
Now,
$$\dfrac{\partial\ell}{\partial\theta}=-n+\dfrac{\sum_{i=1}^{n}y_i}{\theta}\Longrightarrow\dfrac{\partial\ell}{\partial\theta}=0\Longrightarrow \hat{\theta}=\dfrac{\sum_{i=1}^{n}y_i}{n}\text{( $\hat{\theta}$ is our MLE)}$$
Furthermore,
$$\dfrac{\partial^2\ell}{\partial\theta^2}=\dfrac{-\sum_{i=1}^{n}y_i}{\theta^2}$$
On the Fisher information wikipedia page, Fisher information is defined as
$$I(\theta)=-\mathbb{E}\left[\dfrac{\partial^2}{\partial\theta^2}\ell(\theta)\right]$$
But on the Cramér-Rao lower bound page, we have the fisher information defined as
$$I(\theta)=-n\mathbb{E}\left[\dfrac{\partial^2}{\partial\theta^2}\ell(\theta)\right]$$

Now with respect to the Poisson example up above, In the first case we have that the Fisher information is
$$-\mathbb{E}\left[\frac{-\sum_{i}^{n}y_i}{\theta^2}\right]=-\dfrac{(-n\theta)}{\theta^2}=\dfrac{n}{\theta}$$

So what I believed to be the Cramér-Rao lower bound is $\mathrm{var}(\hat{\theta})\geq\dfrac{1}{I(\theta)}\Rightarrow\mathrm{var}(\hat{\theta})\geq\dfrac{1}{n/\theta}=\dfrac{\theta}{n}$, which seems to be the correct CRLB, but in the second case I would get $\dfrac{\theta}{n^2}$? So I am not sure which method is correct, or if I am misunderstanding the meaning of the $n$ in the Fisher information.

答:
The difference between the formulas is that the page on Fisher information is defining the information for a single outcome of $X$ while the page on CRLB is reporting the information for a sample of $n$ observations. In both cases $\ell$ represents the likelihood for a single outcome, so the final formula for $I(\theta)$ will turn out the same.

In your context the Fisher information for a sample of $n$ Poisson observations will be $n/\theta$, not $n^2/\theta$.


问:
How to compute $\int^{\infty}_{0} t^{(\frac1n-1)}\cos t \,\mathrm{d}t$?
How to calculate the below integral?
$$
\int^{\infty}_{0} \frac{\cos t}{t^{1-\frac{1}{n}}} \textrm{d}t = \frac{\pi}{2\sin(\frac{\pi}{2n})\Gamma(1-\frac{1}{n})}
$$
where $n\in \mathbb{N}$.</p>

答:
A short cut is the Mellin transform technique (using the tables)

$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x) dx .$$

Now, the Mellin transform of $\cos(t)$ is $$ \Gamma  \left( s \right) \cos \left( \frac{\pi}{2} \,s \right) .$$

Then subs $s=\frac{1}{n}$, since $s-1=\frac{1}{n}-1$, gives

$$ I =  \Gamma  \left( \frac{1}{n} \right) \cos \left( \frac{\pi}{2n}  \right) . $$


问:
Number of possible inequalities for $n$ variables.
Let's say there are two variables $x$ and $y$. Here are the possible inequalities:

(i) $x<y$

(ii) $x>y$

(iii) $x=y$

So, there are $3$ possible inequalities. But let's say there are $n$ variables $x_1,x_2,x_3,x_4,\cdots,x_n$. How many possible distinct inequalities are there?

I assume this problem requires combinatorics , but I am not sure how to apply them in this problem.

Edit: Maybe this problem can be related with coin arranging (point me if I am wrong)

First of all , we can arrange coins (where order matters).But ,we will also allow for the coins to stack on top of each other.In the stacks (There can be multiple stacks) , order doesn't matter.Stacks will be represented with parenthesis.

For $1$ coin , there is $1$ way to arrange them.

$1$ $[x_1]$

For $2$ coins , there are $3$ ways to arrange them.

$12$ $[x_1 < x_2]$

$21$ $[x_1 > x_2]$

$(12)$ $[x_1 = x_2]$

For $3$ coins , there are $13$ ways to arrange them.

$123$ $[x_1 < x_2,x_2 < x_3,x_3 > x_1]$

$132$ $[x_1 < x_2,x_2 > x_3,x_3 > x_1]$

$213$ $[x_1 > x_2,x_2 < x_3,x_3 > x_1]$

$231$ $[x_1 > x_2,x_2 < x_3,x_3 < x_1]$

$312$ $[x_1 < x_2,x_2 > x_3,x_3 < x_1]$

$321$ $[x_1 > x_2,x_2 > x_3,x_3 < x_1]$

$(12)3$ $[x_1 = x_2,x_2 < x_3,x_3 > x_1]$

$3(12)$ $[x_1 = x_2,x_2 > x_3,x_3 < x_1]$

$1(23)$ $[x_1 < x_2,x_2 = x_3,x_3 > x_1]$

$(23)1$ $[x_1 > x_2,x_2 = x_3,x_3 < x_1]$

$2(13)$ $[x_1 > x_2,x_2 < x_3,x_3 = x_1]$

$(13)2$ $[x_1 < x_2,x_2 > x_3,x_3 = x_1]$

$(123)$ $[x_1 = x_2,x_2 = x_3,x_3 = x_1]$

I thought of thinking of stacks as variables that are equal.And a variable being left/right to another variable corresponding to a varialbe less than/ greater than another variable.So my thought is..

number of ways to arrange $n$ coins = number of distinct inequalities for $n$ variables.(Hope this helps)

答: To make an inequality, you need two of the variables and one of three operators, so $n (n-1) 3.$ Notice that the order of the variables matters, so you do not divide by two.


问:
Request for help: wrt a proof I've written. No rational $r$ satisfying $2^r = 3$
I'll start off by requesting no one gives me a solution to this problem. I would just like some direction. Please let me know if I've made a mistake where I've gone wrong and if possible let me know 'why' what I have done is wrong.

Show that there is no rational number $r$ satisfying $2^r = 3$.

If $r$ is rational then we can write $r = a/b : a,b \in$ integers, and $b\neq0$

so, $$ 2^{a/b} = 3,$$
$$log_2(2)^{a/b} = log_2(3)$$
$$a/b=log_2(3)$$
$$a=b*log_2(3)$$

This means $a$ is a multiple of $log_2(3)$ so we can write, $a = m*log_2(3)$ where $m$ is an integer and $m \neq b $

$$m*log_2(3) = b*log_2(3)$$

dividing both sides by $log_2(3)$ gives $m = b$ which is a contradiction. (I think).
The logic behind $m \neq b$ is that m can be zero.

This entire thing I've done looks almost absurd though.

Anyways, thanks in advance!

答: No, there is no contradiction in $m=b.$ The correct proof involves no log, and instead finds a contradiction in $2^a = 3^b.$


问:
Variance of Poison Random Variable
For a certain insurance company, 10% of its policies are Type, A, 50% are Type B, and 40% are Type C.

The annual number of claims for an individual Type A, Type B, and Type C policy follow Poisson distributions with respective means 1, 2, and 10.
Let X represent the annual number of claims of a randomly selected policy.
Calculate the variance of X.

So one way of arriving at the answer is using the conditional variance formula.

Var(X) = E[Var(X | type] + Var(E[X |type])
$$=E[1,2,10] + Var(1,2,10)$$
$$=5.1 + Var(1,2,10)$$

From there you calculate the second moment

$$E[{1,2,10}]^2 = 42.1$$
$$42.1 - 5.1^2=16.09$$
$$Var(X)= 5.1 +16.09 $$

And I understand this method.  But when I did it by myself, I just calculated first and second moments and used the formula $$Var(X) = E(X^2) - E(X)^2$$
getting 16.09.  Why is this not the answer?  I need help conceptualizing why 5.1 must be added to the variance here.

答:
First remember that in a poisson distribution $Po(\theta)$ the second moment is $\theta+\theta^2$

Then observe that $p(X=x)$ is a mixture of 3 poisson's pmf thus

$$\mathbb{E}[X^2]=0.1\times (1+1^2)+0.5\times (2+2^2)+0.4\times (10+10^2)=47.2$$

$$\mathbb{E}[X]=0.1\times 1+0.5\times 2+0.4\times 10=5.1$$

Thus

$$\mathbb{V}[X]=47.2-5.1^2=21.19$$

Title: Variance of Poison Random Variable


问:
Notation and proof of statement about convex functions
I am reading my course notes on convex optimisation, but the notes provide a notation that I am unfamiliar with. The notes makes the claim that if $f:\mathbb{R}^n\mapsto\mathbb{R}$ is convex, ${\bf x}$, ${\bf y}\in\mathcal{D}(f)$, ${\bf a}\in\partial f({\bf x})$ and ${\bf b}\in\partial f({\bf y})$, then $({\bf a}-{\bf b})^\mathrm{T}({\bf x}-{\bf y})\geq{\bf 0}$. I therefore would like to check the following.

$\mathcal{D}$ either represents domain or preimage, which is it? I suspect it is domain because of the it starts with 'D', but nevertheless it is worthwhile to check.

What exactly are $\partial f({\bf x})$ and $\partial f({\bf y})$? In physics, I know that $\partial V$ represents the surface bounding a volume $V$, are they related? If it is referring to a "surface", what is the proper mathematical way to define $\partial f({\bf x})$ and $\partial f({\bf y})$?

Assuming I have gotten the interpretations of these right, I wanted to prove it as follows: using without proof a claim that further appears in the notes, I have $f({\bf x}+{\bf d})-f({\bf x})\geq{\bf a}^\mathrm{T}{\bf d}$ and $f({\bf y}+{\bf d})-f({\bf y})\geq{\bf b}^\mathrm{T}{\bf d}$. If I could "subtract" these two inequalities intuitively I know I will be close, but I just can't seem to figure out how. What have I done wrong here?

答:
Here $\mathcal{D}(f)$ should be the domain of $f$:
$$
\mathcal{D}(f) := \{ x \in \mathbb{R}^n \mid f(x) < \infty \}.
$$

Moreover, $\partial f(x)$ is the subdifferential of $f$ at $x$. You can think of it as a generalization of the gradient. In particular, you always have the inequality

$$
f(y) \geq f(x) + \zeta^{\mathsf{T}}(y - x), \; \forall x, y \in \mathcal{D}(f), \; \forall \zeta \in \partial f(x).
$$

Your idea for proving the inequality is correct. Applying the above inequality twice, with the order of $y$ and $x$ reversed, you have

$$
f(y) - f(x) \geq \zeta^{\mathsf{T}}(y - x), \; \forall \zeta \in \partial f(x). \\
f(x) - f(y) \geq w^{\mathsf{T}}(x - y), \; \forall w \in \partial f(y).
$$

Add them together and the LHS cancels; rearranging will give you

$$
(\zeta - w)^{\mathsf{T}}(x - y) \geq 0,
$$

and since the choice of $x$ and $y$ was arbitrary the inequality follows.
(Note that this is exactly the proof you suggested, with $d = (y - x)$ in the first invocation and $d = (x - y)$ in the second one.)


问:
Certain Sums of Conjugacy Class Sizes of Symmetric Groups
Suppose $n$ is a natural number and $\lambda$ is an unordered integer partition of $n$ such that $\lambda$ has $a_{\lambda,j}$ parts of size $j$ for each $j$... Let $c_\lambda$ be the conjugacy class in the symmetric group of degree $n$ comprising the elements whose cycle type is $\lambda$... Then:
$$
\! |c_\lambda| = \frac{n!}{\prod_j (j)^{a_{\lambda,j}}(a_{\lambda,j}!)}
$$
(see here for more information: Conjugacy class size formula in symmetric group)

Let $\Lambda$ be the set of all partitions $\lambda$ and $M\subset\Lambda $.
For $n\ge5$ there seem to be more than one solution $M$ for the following:
$$
\sum_{\mu\in M} |c_\mu|=\frac{n!}2 \tag{1}
$$

There is at least one solution, since the splitting into odd and even permutations divides the set of all permutations properly.

For $n=5$ the even/odd splitting looks like
$$
\begin{eqnarray}
|c_{(1,1,1,1,1)}|+|c_{(3,1,1)}|+|c_{(2,2,1)}|+|c_{(5)}|&=&1+20+15+24\\
|c_{(2,1,1,1)}|+|c_{(4,1)}|+|c_{(3,2)}|&=&10+30+20,
\end{eqnarray}
$$ but there are at least $3$ more solutions:

Since $|c_{3,2}|=\frac{5!}{(3^1\cdot 1!)(2^1\cdot 1!)}=|c_{3,1,1}|=\frac{5!}{(3^1\cdot 1!)(1^2\dot 2!)}$ you can interchange the odd conjugacy class of the partition $3+2$ with the even one for the partition $3+1+1$.

and 3.  $|c_{(1,1,1,1,1)}|+|c_{(2,2,1)}|+|c_{(5)}|=1+15+24=40$, so we may again add $|c_{(3,2)}|$ or $|c_{(3,1,1)}|$

My question:

Is it possible to give an asymptotic?
  How many solutions $M$ exist for $(1)$ in general?

Body:
Ok, it was easier than I thought to find a concise way to calculate the number of solutions:

We've got to calculate
$$
\prod_{\lambda\in\Lambda} \left(1+x^{|c_\lambda|}\right)
$$

and then find the coefficient of  $\displaystyle x^\frac{n!}{2}$. Some results:

$n=5$ (supported by W|A):

$$(1+x)(1+x^{10})(1+x^{20})^2 (1+x^{15})(1+x^{30})(1+x^{24})=x^{120}+x^{119}+x^{110}+x^{109}+x^{105}+x^{104}+2 x^{100}+2 x^{99}+x^{96}+2 x^{95}+x^{94}+3 x^{90}+3 x^{89}+x^{86}+3 x^{85}+2 x^{84}+x^{81}+3 x^{80}+2 x^{79}+2 x^{76}+5 x^{75}+3 x^{74}+x^{71}+4 x^{70}+3 x^{69}+3 x^{66}+5 x^{65}+2 x^{64}+2 x^{61}+\color{red}{4 x^{60}}+2 x^{59}+2 x^{56}+5 x^{55}+3 x^{54}+3 x^{51}+4 x^{50}+x^{49}+3 x^{46}+5 x^{45}+2 x^{44}+2 x^{41}+3 x^{40}+x^{39}+2 x^{36}+3 x^{35}+x^{34}+3 x^{31}+3 x^{30}+x^{26}+2 x^{25}+x^{24}+2 x^{21}+2 x^{20}+x^{16}+x^{15}+x^{11}+x^{10}+x+1$$

$n=6$:
$$
(1+x^{1})(1+x^{15})(1+x^{40})^2(1+x^{90})^2(1+x^{45})(1+x^{144})(1+x^{120})^2(1+x^{15})\\=\dots+ \color{red}{12 x^{360}}+\dots
$$

$n=7$: $(1+x) \dots (1+x^{840}) = \dots+ \color{red}{58 x^{2520}}+\dots$

$n=8$: $(1+x) \dots (1+x^{5760}) = \dots+ \color{red}{1016 x^{20160}}+\dots$

So summarizing this we get: $0,2,2,2,4,12,58,1016$ (thanks to Jack) for the first eight symmetric groups.
The question about the asymptotics remains open...


问:
Derivative of Complex Log on Different Branches
Define for $z\in \mathbb{C}$, $\alpha\in \mathbb{R}$,

$arg_{\alpha}(z):= Arg(z) + 2n\pi$ where $n\in \mathbb{Z}$ is chosen s.t. $arg_{\alpha}(z)$ lies in $(\alpha, \alpha + 2\pi]$

and $log_{\alpha}(z):= ln|z| + iarg_{\alpha}(z)$

Now fix $\alpha\in \mathbb{R}$. Let $g(z):= log_{\alpha}(z)$ defined on $W:= \mathbb{C}\setminus\{z\in \mathbb{C} \mid z = 0 \text{ or } \arg(z) = \alpha + 2\pi\}$ (so we are taking the plane and removing the branch cut)

and $f(z):= e^z$, defined on the whole plane. I wish to prove that $g'(z)$ exists on $W$ and I would like to compute it's value. Since f is analytic on an open set containing $W$ (namely the whole plane), and $f\circ g = z$ is entire (and I think that $f'(g(z))$ is just $z$ (*), which is nonzero on $W$), and g is continuous on W, I believe this means that g is analytic on $W$ and that (by the so-called "reverse chain rule"):

$g'(z) = \frac{(f\circ g)'(z)}{f'(g(z))} = \frac{1}{z}$. (**)

So far I don't know what my mistake is. I think (**) is wrong, if I recall correctly, this derivative depends on $\alpha$, i.e., it depends on the branch chosen. If I made I mistake, I was thinking it was in (*), but $(e^z)'$ is just $e^z$ on the whole plane so that we should have:

$f'(g(z)) = e^{log_{\alpha}(z) = e^{ln|z| + iarg_{\alpha}(z)}} = |z|e^{arg_{\alpha}(z)} = z$

on W.
Right?

Does anyone know where I am going wrong? F.Y.I., I have done some googling and searching otherwise for material on specifically this matter, when we let $\alpha$ vary through $\mathbb{R}$ (not just principal "$Log(z)$"), to no avail, so an answer to this would be really appreciated.

答:
You did not make a mistake. If $g: D \to \Bbb R$ is any branch of the logarithm on any domain $D \subset \Bbb C\setminus \{ 0 \}$, then $g'(z) = 1/z$.

A “branch of the logarithm” is defined as an inverse function to the exponential function, so this does follow from the chain rule and the fact that $\exp' = \exp$:
$$
 z = \exp(g(z)) \implies 1 = \exp(g(z)) g'(z) = z g'(z) \implies g'(z) = \frac 1z \, .
$$

You can also argue that two branches of the logarithm on the same domain differ by a constant (a multiple of $2 \pi i$), and therefore have the same derivative.


问:
Calculating the partial derivative with respect to a matrix
Say, $y = W^T b$ where, $ W^T \in R^{2 \times 2} $ and $ b \in R^{2 \times 1} $. Now, we want to calculate: $\frac{\partial y}{\partial W}$.

First, if we look at the dimensions: $y \in R^{2 \times 1}$. So, we know that $\frac{\partial y}{\partial W} \in 2 \times 2 \times 2$ - so it will be a third order tensor?

Second, is the following correct: $\frac{\partial y}{\partial W} = b^T \bigotimes I_{2 \times 2}$?

Are there any good resources to read up on when it comes to matrix derivatives? especially, understanding the dimensions, etc.?

答:
According to my calculations,

$$\begin{matrix}\frac{dy_1}{dW_{11}}=b_1&\frac{dy_1}{dW_{12}}=b_2&\frac{dy_1}{dW_{21}}=0&\frac{dy_1}{dW_{22}}=0\\
\frac{dy_2}{dW_{11}}=0&\frac{dy_2}{dW_{12}}=0&\frac{dy_1}{dW_{21}}=b_1&\frac{dy_2}{dW_{22}}=b_2\end{matrix}$$

So it looks like $b^T\bigotimes I_{2\times 2}$ would give you $$\begin{pmatrix}b_1&b_2\end{pmatrix}\bigotimes\begin{pmatrix}1&0\\0&1\end{pmatrix}\\
=\begin{pmatrix}b_1\begin{pmatrix}1&0\\0&1\end{pmatrix}&b_2\begin{pmatrix}1&0\\0&1\end{pmatrix}\end{pmatrix}\\
=\begin{pmatrix}b_1&0&b_2&0\\0&b_1&0&b_2\end{pmatrix}$$

which matches

$$\begin{pmatrix}\frac{dy_1}{dW_{11}}&\frac{dy_1}{dW_{21}}&\frac{dy_1}{dW_{12}}&\frac{dy_1}{dW_{22}}\\
\frac{dy_2}{dW_{11}}&\frac{dy_2}{dW_{21}}&\frac{dy_2}{dW_{12}}&\frac{dy_2}{dW_{22}}\end{pmatrix}$$

if you write it a certain way.


问:
"The p-power bounding" right multiplication in a finite group G under some special conditions
[Roughly speaking, the following question considers a special setting in which we want to prove a property in the form of $ord(g \sigma)\ |\ p^k$.]

The Problem in Detail:

Let $G$ be a finite group, and $H \leq G$ any subgroup with order $|H|=p^k$ (where $p$ is a prime, and $k \in \mathbb{N}$. Note that $H$ may not be a Sylow $p$-subgroup). Let $g \notin H$ be an element in $G$ s.t. $ord(g)\ |\ p^k $; and $\sigma\in G$ be s.t. for every integer $n$ we have $g^n \sigma g^{-n} \in H$. We claim that $ord(g \sigma)\ |\ p^k$.

(Edit: The previous claim is $ord(g) = ord(g \sigma)$, which is wrong due to the counterexample provided by @jpvee.)

My Attempts:

I can virtually do nothing other than expanding $(g \sigma)^l$, which does not contribute much.

I am quite confused on how to even start out the first step properly.

Can anyone please give me some hints on how I should proceed?

Thank you very much for your help!
:)

答:
Sorry to disappoint you, but your conjecture is false. One counterexample I could come up with is the following:

Let $G:=S_4$ be the symmetric group on four points, let $H:=\langle\,(1,2,3,4),(1,3)\,\rangle$ be a Sylow $2$-subgroup. Now choose $g:=(1,4)$ and $\sigma:=(1,2)(3,4)$.

Then $g\not\in H$, and $g^n\sigma g^{-n}\in H$ for all $n\in\mathbb{Z}$ (since $\sigma$ is an element of the Klein subgroup which is a normal subgroup of $G$ contained in $H$).

On the other hand, we have $g\sigma=(1,3,4,2)$ and thus $ord(g)=2\neq4=ord(g\sigma)$


问:
Optimal control of a certain Poisson process
I am trying to understand this paper in which the optimal expected value of a certain Poisson process is computed.
Following the notation of op. cit, let $N_s$ ($0 \le s \le t$) be the associated counting process.
The intensity of the process is given by $\lambda_s = \lambda (p_s)$, where $\lambda$ is a known decreasing function and $p_s$ is a real-valued stochastic process depending on the control $u$.
We wish to find a control $u$ such that
$$J_u (n, t) = \mathbf{E}_u \left[ \int_0^t p_s \mathrm{d} N_s \right]$$
is maximised, where $\mathbf{E}_u$ denotes the expectation operator under control $u$.

Let $J^* (n, t) = \sup \{ J_u (n, t) : u \in \mathscr{U} \}$, where $\mathscr{U}$ is the set of controls such that $N_t \le n$ a.s.
In op. cit., there is a heuristic dynamic programming argument showing that
$$\frac{\partial J^* (n, t)}{\partial t} = \lambda (p^* (n, t)) \left( p^* (n, t) - J^* (n, t) + J^* (n - 1, t) \right)$$
where $p^* (n, t)$ is the $p$ maximising $\lambda (p) \left( p - J^* (n, t) + J^* (n - 1, t) \right)$.
This argument seems plausible enough to me, so I have no issue here.

Now suppose $\lambda (p) = a \exp (- \alpha p)$.
The following solution is given in op. cit.:
$$J^* (n, t) = p^* (n, t) \log \left( \sum_{k = 0}^n \frac{(\lambda (1 / \alpha) t)^k}{k !} \right)$$
$$p^* (n, t) = \frac{1}{\alpha} + J^* (n, t) - J^* (n - 1, t)$$
How do I verify that this is indeed a solution, or better yet, how might I have found this solution in the first place?

答:
The solution can be obtained by making some clever substitutions.

First, define $q^* (n, t) = p^* (n, t) - J^* (n, t) + J^* (n - 1, t)$.
So $q^* (n, t)$ is the $q$ that maximises $\lambda (q + J^* (n, t) - J^* (n - 1, t)) q$.
But $\lambda (p) = a \exp (- \alpha p)$, so $q^* (n, t)$ is the $q$ that maximises $a \exp (-\alpha q) q$, namely $1 / \alpha$.
Therefore,
$$p^* (n, t) = \frac{1}{\alpha} + J^* (n, t) - J^* (n - 1, t)$$
as claimed.

Now, the differential equation becomes:
$$\frac{\partial J^* (n, t)}{\partial t} = \frac{a}{\alpha e} \frac{\exp (\alpha J^* (n - 1, t))}{\exp (\alpha J^* (n, t))}$$
Define $F^* (n, t) = \exp (\alpha J^* (n, t))$ and rewrite to get:
$$\frac{1}{\alpha F^* (n, t)} \frac{\partial F^* (n, t)}{\partial t} = \frac{a}{\alpha e} \frac{F^* (n - 1, t)}{F^* (n, t)}$$
This simplifies to:
$$\frac{\partial F^* (n, t)}{\partial t} = \frac{a}{e} F^* (n - 1, t)$$
The boundary conditions are $J^* (0, t) = 0$ and $J^* (n, 0) = 0$, hence:
$$F^* (n, t) = \sum_{k = 0}^n \frac{(a t / e)^k}{k !}$$
Therefore,
$$J^* (n, t) = \frac{1}{\alpha} \log \left( \sum_{k = 0}^n \frac{(a t / e)^k}{k !} \right)$$
as claimed.


问:
What would a Tutte Polynomial =0 represent?
So I'm working on proving (via contradiction) that the flow number $\phi(G)$ of a bridgeless graph $G$ is always defined. I'm using the flow polynomial, and I got to a point where I have $0=T(0,1-u)$.

So, my question:
If $T(x,y)=0$ where $T(x,y)$ is the Tutte polynomial, what does this mean about the graph? Does it mean it has no edges at all?

Thanks!

答:
There are no graphs for which the Tutte polynomial is $0$.  One thing that would go wrong if there were such a graph:

The chromatic polynomial is contained within the Tutte polynomial; if the Tutte polynomial were $0$, then the graph would not be $k$-colourable for all $k \geq 0$.  But this is impossible since e.g. we can colour each vertex a different colour.


问:
Matrix row/column swapping problem
Given an $n \times n$ square matrix $M$ in which every row and column contains exactly one $-1$ and one $1$, and is $0$ otherwise, I'm trying to calculate whether it's always possible to reach $-M$ through successive swapping of rows/columns.

I've worked out that it's at least possible sometimes, simply by trial-and-erroring arbitrary matrices by hand, but of course this is a long way from a proof, and it only proves that it's possible for those specific matrices. I'm not too sure where to start with this.

One thing I noticed is that if $\det(M) \neq 0$ then the number of swaps must match the evenness of $n$, because every swap will flip the sign of the determinant, and then we also have

$$\det(-M)=(-1)^n \det(M)$$

However I'm not sure if this is relevant at all, it's just the only thing I've managed to deduce so far.

Any pointers?

答:
It is always possible to get $-M$ by the procedure below. But first, the determinant is always $0$, since a vector with all its coordinates equal will be in the kernel of $M$. So that argument will not take you far.

Now use row/column swaps to bring $M$ into a standard form. Note that the first row can always be transformed into $(1, -1, 0, \ldots)$. From here you can continue and get one of two options for the first two rows. Namely either $$\begin{pmatrix}1&-1&0&\ldots \\ -1&1&0&\ldots \end{pmatrix}$$ or $$\begin{pmatrix}1&-1&0&0&\ldots \\ 0&1&-1&0&\ldots \end{pmatrix}.$$ In the first case the new matrix $M’$ is in block diagonal form with a $2\times 2$ block in the upper left.

If the dimension of $M$ is greater than $2$ then use induction to argue that $M$ can indeed be transformed into $-M$.

So, using induction on the size $n$, the only case left is to consider an $m \times m$ matrix $M’$ for any $m\geq 2$ of the form $$M’ = \begin{pmatrix}
1 & -1 & 0 & \cdots & & 0 \\
0 & 1 & -1 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \ddots & & \vdots \\
0 & \cdots & & & 1 & -1 \\
-1 & 0 & \cdots & & 0 & 1
\end{pmatrix}.$$

For this case, reversing the first $m-1$ rows (swap rows $1$ and $m-1$, rows $2$ and $m-2$, etc.) and then reversing all columns results in $-M’$.

Finally retrace the steps that transformed $M$ into $M’$ to get $-M$.

Title: Matrix row/column swapping problem


问:
Why does my answer to the integral of $\frac{x^2}{x^2+9}$ need to be multiplied by $3$
This is my work below, the answer to the problem was $3$ times what my solution was. Why?

The integral was the indefinite integral of $\displaystyle \int\frac{x^2}{x^2+9}$ and my answer was $\frac{x}{3}-\arctan(\frac{x}{3})+C$ and the correct answer was $x-3\arctan(\frac{x}{2})+C$.

答:
Another way to integrate, then you will see easily why it is multiplied by $3$:

Given $\int \frac{x^2}{x^2+9}dx$, now add and subtract $9$ in the numerator, it becomes:

$$\int \frac{(x^2+9)-9}{x^2+9}dx=\int(1-\frac{9}{x^2+9})dx=\int dx-9 \int \frac{1}{x^2+9}dx$$

Recall that $\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\tan^{-1}(\frac{x}{a})+C$

Therefore,

$$\int dx-9 \int \frac{1}{x^2+9}dx=x-9(\frac{1}{3}) \tan^{-1}(\frac{x}{3})+C=x-3 \tan^{-1}(\frac{x}{3})+C$$

See the last line, the $9$ and the $3$ got simplified.


问:
What numerical lower bound on the index of an odd perfect number is implied by the results in F.-J. Chen and Y.-G. Chen's 2014 paper?
In what follows, we let $\sigma=\sigma_1$ denote the classical sum-of-divisors function.

Suppose that $N$ is an odd perfect number and $q^\alpha$ is a prime power with $q^\alpha \parallel N$.

(Euler proved that $N$ can be written in the Eulerian form $p^k n^2$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,n)=1$.)

Define the index $m = \sigma(N/q^\alpha)/q^\alpha$.  Chen and Chen (2014) proved that, if $q = p$, then $m$ cannot take any of the $30$ forms

$$q_1, {q_1}^2, {q_1}^3, {q_1}^4, {q_1}^5, {q_1}^6, {q_1}^7, {q_1}^8, q_1 q_2, {q_1}^2 q_2, {q_1}^3 q_2, {q_1}^4 q_2, {q_1}^5 q_2, {q_1}^2 {q_2}^2, {q_1}^3 {q_2}^2, {q_1}^4 {q_2}^2, q_1 q_2 q_3, {q_1}^2 q_2 q_3, {q_1}^3 q_2 q_3,$$
$${q_1}^4 q_2 q_3, {q_1}^2 {q_2}^2 q_3, {q_1}^2 {q_2}^2 {q_3}^2, q_1 q_2 q_3 q_4, {q_1}^2 q_2 q_3 q_4, {q_1}^3 q_2 q_3 q_4, {q_1}^2 {q_2}^2 q_3 q_4, q_1 q_2 q_3 q_4 q_5, {q_1}^2 q_2 q_3 q_4 q_5, q_1 q_2 q_3 q_4 q_5 q_6, q_1 q_2 q_3 q_4 q_5 q_6 q_7$$

where $q_1, q_2, q_3, q_4, q_5, q_6, q_7$ are distinct odd primes.

INITIAL QUESTION

What numerical lower bound on the index $m$ of an odd perfect number is implied by the results in F.-J. Chen and Y.-G. Chen's 2014 paper, if $q=p$?

MOTIVATION

Broughan, Delbourgo, and Zhou (2013) proved the following result:

If $q=p$, then $m=\sigma(N/q^\alpha)/q^\alpha$ cannot take any of the $11$ forms
$$p_1, {p_1}^2, {p_1}^3, {p_1}^4, {p_1}^5, {p_1}^6, p_1 p_2, {p_1}^2 p_2, {p_1}^3 p_2, {p_1}^2 {p_2}^2, p_1 p_2 p_3$$
where $p_1, p_2, p_3$ are any distinct odd primes.

Therefore, the smallest possible value of the index $m$ in this case is $3^2 \times 5 \times 7 = 315$.

MY ATTEMPT

Following Broughan, Delbourgo, and Zhou's lead, I obtain (from the results of Chen and Chen) that the smallest possible value of the index $m=\sigma(N/q^\alpha)/q^\alpha$ when $q=p$ is
$$3^2 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 = 14549535.$$

FOLLOW-UP QUESTION

Is my attempted answer correct?  If not, how can it be mended?

答:
The smallest possible value of the index $m=\sigma(N/q^\alpha)/q^\alpha$ when $q=p$ is $$3^3\times 5^3=3375$$

Proof :

Let $M$ be the number of prime divisors counted with multiplicity.

Using $M$, we can classify the $30$ forms as follows :

$$\small\begin{align}M=1&:q_1
\\\\M=2&:{q_1}^2, q_1 q_2
\\\\M=3&:{q_1}^3, {q_1}^2 q_2,q_1 q_2 q_3
\\\\M=4&:{q_1}^4, {q_1}^3 q_2,{q_1}^2 {q_2}^2,{q_1}^2 q_2 q_3,q_1 q_2 q_3 q_4
\\\\M=5&:{q_1}^5, {q_1}^4 q_2, {q_1}^3 {q_2}^2,{q_1}^3 q_2 q_3,{q_1}^2 {q_2}^2 q_3,{q_1}^2 q_2 q_3 q_4,q_1 q_2 q_3 q_4 q_5
\\\\M=6&:{q_1}^6,{q_1}^5 q_2, {q_1}^4 {q_2}^2,{q_1}^4 q_2 q_3,{q_1}^2 {q_2}^2 {q_3}^2,{q_1}^3 q_2 q_3 q_4,{q_1}^2 {q_2}^2 q_3 q_4,{q_1}^2 q_2 q_3 q_4 q_5,q_1 q_2 q_3 q_4 q_5 q_6
\\\\M=7&:{q_1}^7, q_1 q_2 q_3 q_4 q_5 q_6 q_7
\\\\M=8&:{q_1}^8\end{align}$$

For $M\le 5$, the $18$ forms are all the possible forms, so there is no form that $m$ can take.

For $M=6$, we see that ${q_1}^3{q_2}^3,{q_1}^3{q_2}^2 q_3$ are not included, so the smallest possible value of $m$ is $3^3\times 5^3=3375$.

For $M=7$, the smallest possible value of $m$ is $3^6\times 5=3645$.

For $M=8$, the smallest possible value of $m$ is $3^7\times 5=10935$.

For $M\ge 9$, the smallest possible value of $m$ is $3^9=19683$.

Therefore, it follows that the smallest possible value of the index $m=\sigma(N/q^\alpha)/q^\alpha$ when $q=p$ is $$3^3\times 5^3=3375$$


问:
Martingales and expectation on subset
Given an $\mathcal{F}_n$ martingale $X_n$, it is well known that $E[X_n]=E[X_m]$ for all $n,m$. But why is it true that if $m\geq n$, for all $A \in \mathcal{F}_n$

$$E[X_m\mathbb{1}_A]=E[X_n\mathbb{1}_A]$$

答:
It is pretty straightforward from the hints in the comments, but here is the answer for the sake of completeness.

The tower rule states that, for a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and a random variable $X$, such that $\min(\mathbb{E}[X
_+],\mathbb{E}[X
_-])<\infty$, if $\mathcal{H}\subset\mathcal{G},$ where $\mathcal{H}$ and $\mathcal{G}$ are sub $\sigma$-algebras of $\mathcal{A},$ then
$$ \mathbb {E} [\mathbb{E} [X\mid {\mathcal {G}}]\mid {\mathcal {H}}]=\mathbb{E} [X\mid {\mathcal {H}}].$$
Thus, setting $\mathcal{G}=\mathcal{F_n}$, $\mathcal{H}=\sigma(\Omega)$ and $X=X_m \mathbb{1}_A$ above, we have $$ \mathbb{E}\left[X_m\mathbb{1}_A\right]=\mathbb{E}\left[\mathbb{E}\left[X_m\mathbb{1}_A\mid \mathcal{F}_n\right]\right].$$ Moreover, since $\mathbb{1}_A$ is $\mathcal{F}_n$-measurable and $\{X_n\}$ is a martingale (hence $\mathbb{E}\left[X_m\mid \mathcal{F}_n\right]=X_n$, for $m\ge n)$,$$ \mathbb{E}\left[X_m\mathbb{1}_A\right]=\mathbb{E}\left[\mathbb{E}\left[X_m\mathbb{1}_A\mid \mathcal{F}_n\right]\right]=\mathbb{E}\left[\mathbb{1}_A\mathbb{E}\left[X_m\mid \mathcal{F}_n\right]\right]=\mathbb{E}\left[\mathbb{1}_AX_n\right],$$
for all $m\ge n$.

Title: Martingales and expectation on subset


问:
What does it mean for a sequence $x_k \in E$ such that if $x_k \to x$, then $f(x_k) \to f(x)$?
This is from a homework problem, but this is not itself the homework problem.  I'm trying to understand what everything means.  Please be patient as I'm an on and off again student and I just keep feeling hopelessly lost and too embarrassed to talk to anyone else.

What does it mean for a sequence $x_k$ $\subseteq$ E such that if $x_k \to x$, then $f(x_k) \to f(x)$?

Edit:
I mean I know what it means, that the sequence is converging onto x and so the function is converging onto x. I'm trying to see what is happening in this statement. To visualize it.

Thanks for the editing help!

答:
This is an equivalent formulation of the idea of a function being continuous (under nice enough conditions; for example, functions $f:\mathbb{R}\to\mathbb{R}$ are continuous iff they satisfy the given property).

First, how to think about this statement? Suppose we have a sequence of points $(x_n)_{n\in\mathbb{N}}$ which converges to some point $x$; this means that if we fix a given tolerance $\varepsilon$, we can ensure that the distance between $x_n$ and $x$ is always smaller than $\varepsilon$ if we look at points "far enough out" in the sequence. Formally, for every tolerance $\varepsilon>0$, there exists an index $N\in\mathbb{N}$ such that for all $n\geq N$ (i.e. all points far enough out in the sequence), the distance between $x_n$ and $x$, i.e. $|x_n-x|$, is less than $\varepsilon$.

Now, from this sequence $(x_n)_{n\in\mathbb{N}}$, we can construct a new sequence $(y_n)_{n\in\mathbb{N}}$, where $y_n=f(x_n)$ for each $n\in\mathbb{N}$. This is just giving a new name to the sequence $\big(f(x_n)\big)_{n\in\mathbb{N}}$, but it is to emphasize that the objects $f(x_n)$ are just elements of a sequence $\big(f(x_n)\big)_{n\in\mathbb{N}}=(y_n)_{n\in\mathbb{N}}$. The statement that $f(x_n)\to f(x)$ is now equivalent to saying that $y_n\to f(x)$, i.e. for every tolerance $\varepsilon$ there is an $N'\in\mathbb{N}$ such that for $n\geq N'$, we have $|y_n-f(x)|<\varepsilon$.

The implication $x_n\to x\implies f(x_n)\to f(x)$ means that, given any sequence $(x_n)_{n\in\mathbb{N}}$ which converges to some point $x$, if we look at the sequence $(y_n)_{n\in\mathbb{N}}=\big(f(x_n)\big)_{n\in\mathbb{N}}$, we have that $y_n\to f(x)$. That is, the sequence formed by applying $f$ to each element $x_n$ of the original sequence converges to $f(x)$, the value we get when we apply $f$ to the original limit $x$.

Again, this is just a reformulation of the idea of continuity. It's a good exercise to show that this is equivalent to the usual $\varepsilon-\delta$ formulation of continuity. It's also good to play around with a few examples to see why this property characterizes continuous functions. For example, if we take $f:[0,1]\to\mathbb{R}$ where $f(x)=0$ if $x\neq0$ and $f(x)=1$ if $x=0$, this function would clearly not be continuous at $0$ from the $\varepsilon-\delta$ characterization. Instead, if we look at the characterization of continuity via sequences, to prove that $f$ is not continuous at $0$, we'd need to exhibit a sequence $(x_n)_{n\in\mathbb{N}}$ converging to $0$ such that $f(x_n)\not\to f(0)$. It we take $x_n=1/n$, for example, then $x_n\to0$ and $f(1/n)=0$ for each $n\in\mathbb{N}$, but $f(0)=1$, so $f(x_n)\not\to f(0)$. Hopefully this example helps you to visualize what is going on in the definition.


问:
Strategy while solving quadratics by splitting the middle term
While solving a quadratic equation in one variable by "splitting the middle term", sometimes I just can't arrive at the two numbers I need and it takes a lot of time.

For eg: $x^2 -124x+1344 = 0$

For solving the above equation by splitting the middle term, I first multiplied the coefficient of $x$ which is $1$ in this case by $1340$ (the constant term). I prime factorised the product, and now I have to pick a group of two numbers such that their sum is $-124$ (coefficient of $x$).

Do we always have to pick random numbers? If only hit and trial works here, what must be our thinking process for picking the other pair of number if the former pair of number doesn't work?

Eg: $x^2 -124x+1344 = 0$

Two numbers from the prime factorization of ($1 \cdot 1344$) is $-64$ and $-21$ but they do not add up to $-124$. What should be my strategy to pick a different pair of numbers?

(I know about the quadratic formula, but it becomes too lengthy sometimes)

答:
In fact, we can do even better as $1344 = 2^6 \times 3 \times 7$. Dividing $x^2 - 124x + 1344$ by $2^6 = 64$ gives:

$$\frac{x^2}{64} - \frac{124x}{64} + \frac{1344}{64} = 0$$
$$\Rightarrow \left(\frac{x}{8} \right)^2 - \frac{124}{8}\frac{x}{8} + 21 =0 $$
$$\Rightarrow u = x/8: u^2 - \frac{31}{2}u + 21 = 0$$
$$\Rightarrow 2u^2 - 31u + 42 = 0$$

Factor this as $(2u+a)(u+b)$ since $2$ is prime, and the possibilities are limited as $42$ does not have that many divisors. By inspection, this factors as $(2u-3)(u-14) =0 \Rightarrow u = \frac{3}{2}, 14$, and converting back to $x$ with $u = x/8$, $x = 12, 112$.

This method is effectively the AC method in reverse, where we divide the coefficients by a convenient number and use a substitution to obtain a non-monic polynomial.


问:
Find a and b such that the following piecewise function is differentiable at x = 0
Find a and b such that the following piecewise function is differentiable at x = 0:

\begin{equation*}
f(x) = \left\{
        \begin{array}{ll}
            ax - b & \quad x \leq 0 \\
            sin(2x) - x^2cos(2/x) & \quad x > 0
        \end{array}
    \right.
\end{equation*}

Using the continuity criteria at $ x = 0$ , we can easily find that $ b = 0 $. But when we set the derivatives at $x = 0$ equal to each other, we get this:

a = $ 2cos(2x) $ - $ 2xcos(2/x)$ - $ 2sin(2/x) $

How do we find a I wonder?

答:
$a = \lim \limits_{h \to 0^+} \dfrac {f(0+h) - f(0)}{h}$

$ = \lim \limits_{h \to 0^+} \dfrac { \sin (2h) - h^2 \cos (2/h) - [f(0)=0]} {h}$

Evaluating this limit, you get $a = 2$


问:
How to establish an isomorphism between these two tensor products?
Let $V_1, ..., V_k$ be finite-dimensional vector spaces. A tensor product is defined to be the quotient space $U/I$, where $U$ denotes the free vector space generated by elements of $V_1 \times ... \times V_k$, and $I$ the subspace of $U$ generated by elements of the form

$a(v_1, ..., v_k) - (v_1, ..., av_i, ..., v_k)$

$(v_1, ..., v_i + w_i, ..., v_k) - (v_1, ..., v_i, ..., v_k)- (v_1, ..., w_i, ..., v_k)$

and the space $U/I$ is denoted $V_1 \otimes...\otimes V_k$.

Question:

How to establish an isomorphism between $(V_1 \otimes ... \otimes V_l)\otimes(V_{l+1} \otimes ... \otimes V_k)$ and $V_1 \otimes...\otimes V_k$?

This question might look silly, but I find it troublesome when dealing with the definitions above...

答:
The definition you mention is one possible realization of the tensor product, but the “universal property” is what makes it possible to manage tensor products.

The tensor product of the vector spaces $V_1$, $V_2$, $\dots$, $V_k$ can be defined as a pair $(V,\alpha)$, where $V$ is a vector space and

$$\alpha\colon V_1\times V_2\times\dots\times V_k\to V$$

is a multilinear map such that, for any multilinear map

$$\beta\colon V_1\times V_2\times\dots\times V_k\to W$$

there exists a unique linear map $f\colon V\to W$ such that $f\circ\alpha=\beta$.

It's quite easy to show that your $U/I$, together with the map

$$\gamma\colon V_1\times V_2\times\dots\times V_k\to U/I$$

defined by

$$\gamma(v_1,v_2,\dots,v_k)=v_1+v_2+\dots+v_k+I$$

has the above property (where $v_1+v_2+\dots+v_k$ denotes the element $(v_1, v_2, \ldots, v_k)$ of $U$).

If $(V,\alpha)$ is another “realization" of the tensor product, then, by the universal property, there are $f\colon V\to U/I$ and $g\colon U/I\to V$ such that

$$f\circ\alpha=\gamma,\qquad g\circ\gamma=\alpha$$

and it's easy to prove that $f\circ g$ and $g\circ f$ are both identity maps. Namely,

$$(g\circ f)\circ\alpha=g\circ(f\circ\alpha)=g\circ\gamma=
\alpha=1_V\circ\alpha$$

so, by the uniqueness, $g\circ f=1_V$. Similarly for $f\circ g$.

In this way you can denote any realization of the tensor product by $V_1\otimes V_2\otimes\dots\otimes V_k$. The multilinear map is not expressed, but it is implicit in the choice of the realization; conventionally the map is denoted by $(v_1,v_2,\dots,v_k)\mapsto v_1\otimes v_2\otimes\dots\otimes v_k$.

Any element of $V_1\otimes\dots\otimes V_k$ is a sum of elements of the form $v_1\otimes v_2\otimes\dots\otimes v_k$. For this you can use abstract reasoning or the explicit realization.

Now your problem boils down to showing that
$$
V_1\otimes V_2\otimes\dots\otimes V_k\otimes
V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l
$$
is a realization of the tensor product
$$
(V_1\otimes V_2\otimes\dots\otimes V_k)\otimes
(V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l)
$$
via a suitable bilinear map. The bilinear map
$$
(V_1\otimes V_2\otimes\dots\otimes V_k)\times
(V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l)\to
V_1\otimes V_2\otimes\dots\otimes V_k\otimes
V_{k+1}\otimes V_{k+2}\otimes\dots\otimes V_l
$$
is just
$$(v_1\otimes\dots\otimes v_k,v_{k+1}\otimes\dots \otimes v_l)\mapsto
v_1\otimes\dots\otimes v_k\otimes v_{k+1}\otimes\dots\otimes v_l
$$
extended by bilinearity.

You can have some doubts about the last map being well defined. But remember you have a given multilinear map
$$
\alpha\colon V_1\times\dots\times V_k\times V_{k+1}\times\dots\times V_l
\to
V_1\otimes\dots\otimes V_k\otimes V_{k+1}\otimes\dots\otimes V_l = T
$$

Define, for any $y\in V_{k+1}\times\dots\times V_l$, the map
$\beta_y\colon V_1\times\dots\times V_k\to T$ by $\beta_y(x)=\alpha(x,y)$. This map is multilinear, so it induces a unique linear map
$$f_y\colon V_1\otimes\dots\otimes V_k\to T$$
such that $f_y\circ\alpha_k=\beta_y$, where $\alpha_k : V_1 \times \cdots \times V_k \to V_1 \otimes \cdots \otimes V_k$ is again the canonical multilinear map.

Thus you have a multilinear map $y\mapsto f_y$
$$
V_{k+1}\times\dots\times V_l\to
\mathrm{Hom}(V_1\otimes\dots\otimes V_k,T)
$$
that induces a unique linear map
$$
g\colon V_{k+1}\otimes\dots\otimes V_l\to
\mathrm{Hom}(V_1\otimes\dots\otimes V_k,T)
$$
The bilinear map you're looking for is just
$$h(w_1,w_2)=g(w_2)(w_1)\qquad
(w_1\in V_1\otimes\dots\otimes V_k,
w_2\in V_{k+1}\otimes\dots\otimes V_l).$$

Indeed, if $w_1=v_1\otimes\dots\otimes v_k$ and $w_2=v_{k+1}\otimes\dots\otimes v_l$, showing that
$$
h(w_1,w_2)=\alpha(v_1,\dots,v_k,v_{k+1},\dots,v_l)
$$
is just a tedious verification.

Final comments

Proving the existence of the isomorphism with the “explicit” realization you start with is time consuming and not worth the trouble. The explicit realization $U/I$ is only used to show that a tensor product exists. Then you can forget about it and use only the universal property.

This is similar to how one does when constructing the real numbers from the rationals. The existence of a field with the required properties can be proved with Dedekind cuts or Cauchy sequences or whatever; then one proves that a field with those properties is unique up to isomorphisms and the actual construction is not used any more: one just employs the properties of the field.


问:
Do the critical points $p'(z)=0$ of a complex polynomial $p(z)$ all lie inside the region $L=\{z\mid |p(z)|<1\}$ if $\partial L$ is a Jordan curve?
From Gauss-Lucas theorem, I know that all the critical points lie inside the convex hull of all the roots of $p(z)$, which implies that all the critical points are inside the convex hull of all the roots of $p(z)-\exp(i t) $, for $t\in [0,2\pi)$. However the region $L$ in itself is usually not a convex set. I could not generalize the Gauss-Lucas theorem to this general case.

I have experimentally verified a lot of examples and they all satisfy the claim. I wonder why?

答:
Yes. It suffices to require that $L$ is connected. Due to the maximum modulus principle, $L$ is simply connected, and $p$ is a proper map from $L$ to the unit disk.

One can now apply the Riemann-Hurwitz formula in the following form:

Let $V$ and $W$ be domains on the Riemann sphere of finite connectivity $m$ and $n$, respectively, and let $f : V \to W$ be a $k$-sheeted (ramified) proper map having $r$ critical points (counted by multiplicity). Then $$ m − 2 = k(n − 2) + r \, .$$

(Source: Steinmetz, Norbert. (1993). The formula of riemann-hurwitz and iteration of rational functions. Complex Variables and Elliptic Equations. 22. 203-206. 10.1080/17476939308814660.)

In our case is $m=n=1$ and $k = \deg p$. It follows that $r = \deg p - 1$, i.e. all zeros of the derivative (counted with multiplicity) are in $L$.


问:
Exercise 9.8.5 in Analysis 1 by Tao (continuous function)

(a) For given $x, y \in \mathbb{R}$ with $x < y$, we know from proposition 5.4.14 that there exists $r' \in \mathbb{Q}$ such that $x < r' < y$. Since $q$ is bijection, there exists $n'$ such that $q(n') = r'$. Therefore, $f(x) < f(x) + g(q(n')) = f(x) + 2^{-n'} \le f(y)$.

(b) Note that $x > q(n') = r'$. Therefore, $f(x) = \sum_{r \in Q; r<x} g(r) \ge \sum_{r \in Q; r<r'} + 2^{-n'} = f(r')+2^{-n'}$. Consider the sequence $x_m = r' + \frac1n$. $(x_n)_{m=1}^\infty$ converges to $r'$ and every $x_m > r'$. Thus,$\lim_{m\to\infty}f(x_m) \ge f(r')+2^{-n'} >f(r')$. Thus, $f(x)$ is not continuous at rational numbers.

I am struggling with (c). I appreciate if you give some help.

答:
I'm also self-studying Tao's Analysis.
Here is my attemp (sorry about the grammar I'm not a native English speaker):
Let $n \in \mathbf{N}$, let $E_n$ be a set where
$$
    E_n = \{r \in \mathbf{Q} : g(r) \geq 2^{-n}\},
$$
and let $f_n : \mathbf{R} \to \mathbf{R}$ be a function
$$
    f_n(x) = \sum_{r \in E_n : r < x} g(r) = \sum_{r \in \mathbf{Q} : r < x \land g(r) \geq 2^{-n}} g(r).
$$
Since $q$ is bijective, there are at most $n + 1$ rationals satisfying $r \in \mathbf{Q} \land g(r) \geq 2^{-n}$, thus $E_n$ is finite and non-empty.
Since $\mathbf{R} \setminus \mathbf{Q} \subseteq \mathbf{R}$, by Lemma 9.1.11 $\forall\ x_0 \in \mathbf{R} \setminus \mathbf{Q}$, $x_0$ is an adherent point of $\mathbf{R} \setminus \mathbf{Q}$.
Let $\varepsilon \in \mathbf{R}^+$ and let $\delta = \min\{|r - x_0| : r \in E_n\}$.
Since $x_0 \notin \mathbf{Q}$, we have $\delta > 0$.
Then we have
\begin{align*}
             & \forall\ x \in \mathbf{R} \setminus \mathbf{Q}, |x - x_0| < \delta  \\
    \implies & \big(\forall\ r \in E_n, |x - x_0| < \delta \leq |r - x_0|\big) \\
    \implies & \big(\forall\ r \in E_n, |x - x_0| < |r - x_0|\big)             \\
    \implies & \big(\forall\ r \in E_n, -|r - x_0| < x - x_0 < |r - x_0|\big)  \\
    \implies & \big(\forall\ r \in E_n \land r < x_0, r - x_0 < x - x_0 < x_0 - r\big) \\
    \implies & \big(\forall\ r \in E_n, r < x_0 \land r < x\big)                       \\
    \implies & \{r \in E_n : r < x\} = \{r \in E_n : r < x_0\}                         \\
    \implies & f_n(x) = f_n(x_0)                                                       \\
    \implies & 0 = |f_n(x) - f_n(x_0)| < \varepsilon.
\end{align*}
Since $\varepsilon$ is arbitrary, by Definition 9.3.6 we have $\lim_{x \to x_0 ; x \in \mathbf{R} \setminus \mathbf{Q}} f_n(x) = f_n(x_0)$, and by Definition 9.4.1 $f_n$ is continuous at $x_0$.
Now we show that $f$ is continuous at $x_0$.
We have
\begin{align*}
     & \forall\ x \in \mathbf{R} \setminus \mathbf{Q}, |f(x) - f_n(x)|                                                                           \\
     & = |\sum_{r \in \mathbf{Q} : r < x} g(r) - \sum_{r \in E_n : r < x} g(r)|                                                                  \\
     & = |\sum_{r \in \mathbf{Q} : r < x} g(r) - \sum_{r \in \mathbf{Q} : r < x \land g(r) \geq 2^{-n}} g(r)|                                    \\
     & = |\sum_{r \in \mathbf{Q} : r < x \land g(r) < 2^{-n}} g(r)|                                           & \text{(by Proposition 8.2.6(c))} \\
     & = \sum_{r \in \mathbf{Q} : r < x \land g(r) < 2^{-n}} g(r)                                                                                \\
     & = \sum_{r \in \mathbf{Q} : r < x \land g(r) \leq 2^{-(n + 1)}} g(r)                                                                           \\
     & \leq \sum_{r \in \mathbf{Q} : g(r) \leq 2^{-(n + 1)}} g(r)                                                                                \\
     & \leq \sum_{k = n + 1}^\infty 2^{-k}                                                                                \\
     & = 2^{-(n + 1)} \bigg(\sum_{k = 0}^\infty 2^{-k}\bigg)                                                                                \\
     & = 2^{-n}.                                                                                & \text{(by Lemma 7.3.3)}
\end{align*}
By Proposition 5.4.14, $\forall\ \varepsilon \in \mathbf{R}^+$, $\exists\ n \in \mathbf{N}$ such that $2^{-n} < \varepsilon / 2$.
From the proof above we also have
$$
    \forall\ x \in \mathbf{R} \setminus \mathbf{Q}, |x - x_0| < \delta \implies f_n(x) = f_n(x_0) \\
$$
Combine the results above we have
\begin{align*}
     & |f(x) - f(x_0)|                              \\
     & = |f(x) - f_n(x) + f_n(x) - f(x_0)|          \\
     & \leq |f(x) - f_n(x)| + |f_n(x) - f(x_0)| \\
     & = |f(x) - f_n(x)| + |f_n(x_0) - f(x_0)|  \\
     & \leq 2^{-n} + 2^{-n}                             \\
     & < \varepsilon / 2 + \varepsilon / 2              \\
     & = \varepsilon.
\end{align*}
Since $\varepsilon$ is arbitrary, by Definition 9.3.6 we have $\lim_{x \to x_0 ; x \in \mathbf{R} \setminus \mathbf{Q}} f(x) = f(x_0)$, and by Definition 9.4.1 $f$ is continuous at $x_0$.


问:
Solving the equation $11x^2-6000x-27500 =0$, preferably without the quadratic formula
I obtained this form while solving an aptitude question.

$$\frac{3000}{x-50} + \frac{3000}{x+50} = 11$$

I changed it into quadratic equation

$$11x^2 -6000x - 27500 =0$$

but I don't know how to solve it.

I can't find two factor for 303500 that sums to 6000 or when I use formula the numbers become huge...
Without using calculator
how to solve it? is there any other simple way to solve [other method]? [or finding factor] I'm a beginner in math. Please explain your answer for me.

Body:
Note that $27500 = 2500 \times 11$. Now divide $11x^2-6000x-27500$ by $2500$, the largest perfect square divisor of $27500$, which gives:

$$\frac{11x^2}{2500} - \frac{6000x}{2500} - 11 = 0$$
$$\Rightarrow 11 \left(\frac{x}{50}\right)^2 - \frac{6000}{50} \frac{x}{50} -11=0$$
$$\Rightarrow X = \frac{x}{50}: 11X^2 - 120X - 11 = 0$$

and since $11$ is prime, write this as $(11X + a)(X + b)$. Since $120$ is very close to $121 = 11^2$, $-120 = -11 \times 11 + 1$, so this factors as $(11X + 1)(X - 11) = 0 \Rightarrow X = -\frac{1}{11}, 11$. As $X = x/50$, $x = -\frac{50}{11}, 550$.


问:
Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$
How can I solve this question with out find a,b,c,d

$$a+b+c+d=2$$

$$a^2+b^2+c^2+d^2=30$$

$$a^3+b^3+c^3+d^3=44$$

$$a^4+b^4+c^4+d^4=354$$

so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$

If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d

Is there any help?

答:
If you inspect the system you can see that the solutions for a,b,c,d are all going to have to be small integers. This constrains the system heavily.
 I managed to solve the system guessing some small numbers, testing them and tweaking minus signs to get the desired result.
After we have found a,b,c,d the fraction becomes trivial.

Any analytical method to solving this system would have to accommodate high order solutions and would inevitably be very complicated. I would like to see analytical solution but doubt any could be as efficient as trial and error in this case.


问:
Computing the Jacobian $J_F$ with $F = h \circ f$
Let $f : \mathbb{R}^l \rightarrow{} \mathbb{R}^m$      and    $h : \mathbb{R}^m \rightarrow{} \mathbb{R}^o$

and let  $F = h \circ  f$ with  $F : \mathbb{R}^l \rightarrow{} \mathbb{R}^o$

I want to compute the Jacobian using Forward mode accumulation in one path.

I do understand the automatic differentiation way of working in the forward mode for simple cases.

So I think based on my knowledge if I want to compute $J_F$  using as many paths as I want I could do  $l*m$ paths. Since I am constrained with only one path, I start to get confused. I know I can do something with the initialization but it's very confused in my mind. Could you please help me to understand how to implement the forward mode just in one path using the Jacobian of $f$  and $h$ along the way?

Thanks

答:
It's just the matrix multiplication of the two Jacobians. In fact, this can be seen as the reason why matrix multiplication is defined like that. If you see a $m\times n$ matrix as a linear function from $\mathbb R^n \to \mathbb R^m$, then the Jacobian of the matrix is the matrix itself. The matrix multiplication is just the composition of the two linear functions, so the Jacobian of the composed function is just the matrix multiplication of the two Jacobians.

Locally (that is, in a infinitesimal neighborhood), all differentiable functions can be seen as a linear function of the derivatives of the inputs, therefore you can use matrices to describe the local behavior of all multivariable differentiable functions.

Let's write the components of the function $f$ as $f^\mu$ where $\mu$ is an index that goes from $1$ to $m$. Similarly, you can write $g$ as $g^\mu$. In this way, can write a vector as a symbol with indices but treat them as numbers, (as in, $f^1$ is the first element of the result of $f(x)$, which is in $\mathbb R$.

Then you can write the partial derivative $\frac{\partial f^\mu}{\partial x^\nu}$ as $f^\mu{}_{,\nu}$, mind the comma. It is easy to see that the $f^\mu{}_{,\nu}$ is exactly the Jacobian of $f$, as in $f^1{}_{,2} \equiv \partial f^1/\partial x_2$, is the $(1,2)$-th element of the Jacobian. These notations are called the Ricci Calculus.

The chain rule says:
$$
\left(\frac{\partial}{\partial x^\nu}g(f(x))\right)^\mu=(g\circ f)^\mu{}_{,\nu}=\sum_\gamma g^\mu{}_{,\gamma}f^\gamma{}_{,\nu}\equiv g^\mu{}_{,\gamma}f^\gamma{}_{,\nu}
$$

The last one is to simplify the notation of the summation using Einstein notation, invented by Einstein for manipulating tensors, which is a part of Ricci Calculus as well.

You can see that $\mu$ goes from $1$ to $o$, $\gamma$ goes from $1$ to $m$, and $\nu$ goes from $1$ to $l$. This immediately gives the matrix multiplication of the two Jacobians.


问:
Given the equations $y+z=a$ and $y \cdot z = b$, how do I solve for $y$ and $z$?

Background

Recently, I have been factoring quadratic equation of the form
$$ x^2 + ax + b $$
into
$$ (x+y) \cdot (x+z) $$

e.g. From $x^2 + 3528x + 63180$ to $(x+3510) \cdot (x+18)$

Normally, it is possible to obtain $y$ and $z$ by looking into every factor of $b$, and find 2 of them, where there product is $b$ and their sum is $a$.

However, I find this method too tedious, especially when dealing with numbers that are hard or impossible to get its factors. So, I decided to use equations to directly obtain $y$ and $z$ given $a$ and $b$.

Question

According to the scenario,

$$ y+z=a $$
$$ y \cdot z = b $$

and so, how do I solve this system of equations, so that I get

$$ y = \frac{-\sqrt{a^2 - 4 \cdot b} + a}{2} \, \text{and} \, z = \frac{\sqrt{a^2 - 4 \cdot b} + a}{2}$$
$$ \text{or} \, y = \frac{\sqrt{a^2 - 4 \cdot b} + a}{2} \, \text{and} \, z = \frac{-\sqrt{a^2 - 4 \cdot b} + a}{2}$$
?

答:
Welcome to MSE. there is a method that always gets the solution (small to a large number) and it is like below.
Assume $x_1,x_2$ are roots of $ax^2+bx+c$ we can always rewrite it as$$\forall a,c,b:ax^2+bx+c=a(x-x_1)(x-x_2)$$ even in complex roots.


问:
What does this hint mean and how is it useful to solve the problem?
I am doing a problem  on convergence of random variable. There was a hint given, but I am still struggling to understand the hint.

Here is the problem:

Let $Y_n$ be uniformly distributed on $\{0,1,\ldots,n\}$: $\mathbb{P}(Y_n=k)=\frac{1}{n+1},0\le k\le n$. Prove that $X_n:=Y_n/n\xrightarrow{d}X\sim U[0,1]$ as $n\to\infty$. (where $\xrightarrow{d}$ means convergence in distribution).

Here is the diagram that I tried to draw:

And here is the hint:

The limiting DF $F(x)=x1(x\in[0,1])+1(x>1)$ (where 1 is the indicator) is continuous, so have to prove: $\forall x\in\mathbb{R},F_{X_n}\to F(x)$ as $n\to\infty$.

My question are:

1. What does the hint mean by the limiting DF?

2. And what does it mean and how does it come up with $F(x)=x1(x\in[0,1])+1(x>1)$? how is it significant to the proof? how does it relate to the diagram?

Many many thanks in advance!

答:
First, find the distribution function of $Y_n$, i.e. a $F_n$ such that $$
  \mathbb{P}(Y_n \leq x) = F_n(x) \text{.}
$$
You already drew a diagram of those $F_n$, so you'll just have to formalize that somehow. You may want to use that $x \lfloor \frac{x}{n} \rfloor = x$ if and only if $x$ is a multiple of $n$.

From that, find the distribution function $\widetilde{F}_n$ of $\widetilde{Y}_n = \frac{1}{n}$. You'll just have to rescale the $x$-axis to get from $F(n)$ to $\widetilde{F}(n)$, because $$
  \widetilde{F}(x) = \mathbb{P}(\widetilde{Y}_n \leq x) = \mathbb{P}\left(\frac{1}{n}Y_n \leq x\right) = \mathbb{P}(Y_n \leq nx) = F(nx) \text{.}
$$
But you knew that already I guess, because you also drew a nice diagram of $\widetilde{F}$.

The limiting distribution function is the distribution function of the limit of the $Y_n$, i.e. the distribution function of $U[0,1]$. So again, you're looking for an $F$ such that $$
  \mathbb{P}(U[0,1] \leq x) = F(x) \text{.}
$$
Now, $\mathbb{P}(U[0,1] \leq x) = x$ if $x \in [0,1]$. You'll just have to extent that to $\mathbb{R}$ so that the result is still $\geq 0$, monotone and bounded by $1$. There's only one way to do that...

Finally, you have to show, using the $F_n$ and $F$ you found, that $F_n \to F$ pointwise, i.e. that for every $x \in \mathbb{R}$, you have $$
  \lim_{n\to\infty} F_n(x) = F(x) \text{.}
$$
Your diagrams basically already show that too - just imagine what happens if you add more and more of those steps...


问:
Simplifying Boolean algebra... What am I missing?
I'm reviewing boolean algebra, but I'm having trouble with a basic simplification:
$$\begin{equation}\begin{aligned}
 &x'z'+ xyz +xz'\\
  &= z'(x+x')+xyz\\
  &= z'+xyz\\
  &= ??? \\
  &= z'+xy
\end{aligned}\end{equation}\tag{2}\label{eq2}$$

I can't seem to find a theorem or postulate that satisfies the final step in any table. What is it that I'm not seeing?

答:
You have

\begin{eqnarray*}z'+xyz
& = & z'(1+xy) + xyz \\
& = & (z'+z)xy + z'\\
& = & z' + xy
\end{eqnarray*}


问:
Sample of Random Bernoulli Trials Delta Method
I am trying to solve the following problem

I am trying to find the estimator of the Bernoulli trials, which since we took a sample of them, they would be computed using the Binomial Variance. I am trying to use the Delta Method to find the variance and approximate distribution. So far I have the below

n = p̂, E(Tn) = p, V(Tn) = p(1-p)/n

Using the Delta Method I know (g'(µ))^2 *V(Tn). My question is where does the f(p) come in? I am very new to the Delta method, would I take the derivative of f(p), square it and multiply it by V(Tn) to get my variance?

答: By the CLT we know that $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}\sim N(0,1)$. From this we can find that $Var(\sqrt n(\hat p-p))=p(1-p)$, so that $\sqrt n(\hat p-p)\sim N(0, p(1-p))$. Then $\sqrt n \left(e^{\frac{\hat p}{1-\hat p}}-e^{\frac{p}{1-p}}\right)\sim N\left(0, p(1-p)\frac{d}{dp}e^{\frac{p}{1-p}}\right)=N\left(0,\frac{pe^{\frac{p}{1-p}}}{1-p}\right)$ by the delta method and $\frac{pe^{\frac{p}{1-p}}}{1-p}$ is the asymptotic variance of $g(\hat p)$. The f(p) came in when you took the derivative of it. Actually, you are using f(p) and g(p) to refer to the same thing. Usually say g(p) though.


问:
How to solve this coupled linear ODE
I've tried using mathematica and for some reason it's not working
I think it's a pretty standard problem but for some reason I'm having a hard time
This is essentially a variant of the Rabbi problem I'll really appreciate the help :)
I will appreciate both explanation on how to solve this using mathematica or a manual solution
\begin{align}
i\hbar \frac{d\alpha}{dt}&=\frac{\hbar\omega_0}{2}\alpha+\frac{dE}{2}e^{i\omega t}\beta\\
i\hbar \frac{d\beta}{dt}&=-\frac{\hbar\omega_0}{2}\beta+\frac{dE}{2}e^{-i\omega t}\alpha
\end{align}

答:
We can rewrite the system as
$$
\frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}\alpha\\\beta\end{pmatrix}
=\begin{pmatrix}-i\omega_0/2&-\frac{idE}{2\hbar}e^{i\omega t}\\-\frac{idE}{2\hbar}e^{-i\omega t}&i\omega_0/2\end{pmatrix}
\begin{pmatrix}\alpha\\\beta\end{pmatrix}
$$
and let us introduce $\tilde\alpha=e^{i\omega_0 t/2}\alpha$ and $\tilde\beta=e^{-i\omega_0 t/2}\beta$, giving
$$
\frac{\mathrm{d}}{\mathrm{d}t}\begin{pmatrix}\tilde\alpha\\\tilde\beta\end{pmatrix}
=\underbrace{-\frac{idE}{2\hbar}\begin{pmatrix}&-e^{i(\omega+\omega_0) t}\\e^{-i(\omega+\omega_0) t}&\end{pmatrix}}_{A(t)}\begin{pmatrix}\tilde\alpha\\\tilde\beta\end{pmatrix}
$$
the $A$ here is much nicer to handle.

So we compute
\begin{align*}
[A]\text{-terms}&:\int_0^t A(\tau)\,\mathrm{d}\tau \\
&= \left(\frac{dE}{2\hbar(\omega+\omega_0)}\right)\begin{pmatrix}&e^{i(\omega+\omega_0)t}-1\\e^{-i(\omega+\omega_0)t}-1&\end{pmatrix}\\
[A^2]\text{-terms}&:\int_0^t A(\tau_1)\int_0^{\tau_1} A(\tau_2)\,\mathrm{d}\tau_2\,\mathrm{d}\tau_1\\
&= \left(\frac{dE}{2\hbar(\omega+\omega_0)}\right)^2
\begin{pmatrix}i(\omega+\omega_0)t+1-e^{i(\omega+\omega_0)t}\\&e^{-i(\omega+\omega_0) t}-1- i(\omega+\omega_0)t\end{pmatrix}\\
&\dots
\end{align*}
You should see that the $[A^n]$-term has two entries which are $\pm(dE/(2\hbar(\omega+\omega_0)))^n$ times the tail part of the series $e^{\pm i(\omega+\omega_0)t}$.  Hence summing gives ultimately
$$
I+\int_0^t A(\tau)\,\mathrm{d}\tau+\int_0^t A(\tau_1)\int_0^{\tau_1} A(\tau_2)\,\mathrm{d}\tau_2\,\mathrm{d}\tau_1+\dots =
$$

$$
\begin{pmatrix}
\frac1{2\omega_1} ( \omega_+ e^{i\omega_- t/2} - \omega_- e^{i\omega_+ t/2})
&
\ast
\\
\ast
&
\frac1{2\omega_1}( \omega_+ e^{i\omega_1 t/2} - \omega_- e^{-i\omega_1 t/2} )
\end{pmatrix}
$$
where
$\omega_1=\sqrt{(\omega+\omega_0)^2 + 4 (\frac{dE}{2\hbar})^2}$, $\omega_\pm= \omega+\omega_0 \pm \omega_1$ and I left two entries ($\ast$) for you to calculate.

Title: How to solve this coupled linear ODE


问:
Variance of Poison Random Variable
For a certain insurance company, 10% of its policies are Type, A, 50% are Type B, and 40% are Type C.

The annual number of claims for an individual Type A, Type B, and Type C policy follow Poisson distributions with respective means 1, 2, and 10.
Let X represent the annual number of claims of a randomly selected policy.
Calculate the variance of X.

So one way of arriving at the answer is using the conditional variance formula.

Var(X) = E[Var(X | type] + Var(E[X |type])
$$=E[1,2,10] + Var(1,2,10)$$
$$=5.1 + Var(1,2,10)$$

From there you calculate the second moment

$$E[{1,2,10}]^2 = 42.1$$
$$42.1 - 5.1^2=16.09$$
$$Var(X)= 5.1 +16.09 $$

And I understand this method.  But when I did it by myself, I just calculated first and second moments and used the formula $$Var(X) = E(X^2) - E(X)^2$$
getting 16.09.  Why is this not the answer?  I need help conceptualizing why 5.1 must be added to the variance here.

答:
You could use the total law of variance to solve this problem. You just have to be a little careful.

Instead of labelling the policy types with $A,B,$ and $C$, I am going to label them with $1,2$ and $3$ instead.

Let $T$ be a random variable with pmf $p_T$ that indicates the type of insurance policy that we randomly selected. Note we have $p_T(1)=0.1, p_T(2)=0.5,$ and $p_T(3)=0.4.$ We're given the following information in this problem: $$X|\{T=1\}\sim \text{Poisson}(1)$$ $$X|\{T=2\}\sim \text{Poisson}(2)$$ $$X|\{T=3\}\sim \text{Poisson}(10)$$ Using the fact that the expected value and variance of a random variable possessing a $\text{Poisson}(\lambda)$ distribution is $\lambda$, we see that $$E(X|T=1)=V(X|T=1)=1$$ $$E(X|T=2)=V(X|T=2)=2$$ $$E(X|T=3)=V(X|T=3)=10$$ We can summarize the above relationships nicely and say $$E(X|T)=V(X|T)=\frac{7}{2}T^2-\frac{19}{2}T+7$$ So using the total law of variance, $$V(X)=E\Big(V(X|T)\Big)+V\Big(E(X|T)\Big)$$ This becomes $$V(X)=E\bigg(\frac{7}{2}T^2-\frac{19}{2}T+7\bigg)+V\bigg(\frac{7}{2}T^2-\frac{19}{2}T+7\bigg)$$ From LOTUS, $$E\bigg(\frac{7}{2}T^2-\frac{19}{2}T+7\bigg)=\sum_{t=1}^3\Big(\frac{7t^2}{2}-\frac{19t}{2}+7\Big)p_T(t)=5.1$$ $$V\bigg(\frac{7}{2}T^2-\frac{19}{2}T+7\bigg)=\sum_{t=1}^3\Big(\frac{7t^2}{2}-\frac{19t}{2}+7-\mu_T\Big)^2p_T(t)=16.09$$ Adding the two gives $V(X)=21.19$

Title: Variance of Poison Random Variable


问:
inclusion of sets -transitive?
show that if $A\subseteq B$ and $B\subseteq C$ then $A\subseteq C$

Can I do it with using injective functions?

$A\subseteq B$  means there exists an injective fcn $f:A\to B$

$B\subseteq C$  means there exists an injective fcn $g:B\to C$

then the composition  $g\circ f:A\to C$ is also an injective function then $A\subseteq C$

in each case all the functions are identity functions

答:
No, you can't do so this way.  </p>

<p>"A⊆B means there exists an injective function f:A→B" is false.  </p>

<p>"A⊆B implies there exists an injective function f:A→B" is true.  </p>

<p>"A⊆B means there exists an injective function f:A→B" would only come as true if both</p>

<p>(1) "A⊆B implies there exists an injective function f:A→B" and </p>

<p>(2) "the existence of an injective function f:A→B implies A⊆B" were true also.  But, (2) is false, and thus "A⊆B means there exists an injective function f:A→B" is false. </p>

Title: inclusion of sets -transitive?

Body:
No, you can't do so this way.

"A⊆B means there exists an injective function f:A→B" is false.

"A⊆B implies there exists an injective function f:A→B" is true.

"A⊆B means there exists an injective function f:A→B" would only come as true if both

(1) "A⊆B implies there exists an injective function f:A→B" and

(2) "the existence of an injective function f:A→B implies A⊆B" were true also.  But, (2) is false, and thus "A⊆B means there exists an injective function f:A→B" is false.


问:
Two aeroplanes $1$ and $2$ bomb a target in succession. Probability of $1$ and $2$ hitting correctly are $0.3$ and $0.2$.

Two aeroplanes $1$ and $2$ bomb a target in succession. Probability of $1$ and $2$ hitting correctly are $0.3$ and $0.2$. The second plane will bomb only if first misses the target. Find the probability that target is bombed by the second plane

I want to know why the complementary method cannot be used here i.e. $$P(B) = 1-0.3=0.7$$

Which gives the probability that the target is never hit by the first plane.

Answer is 0.318

Note: I know the correct way to do this, I just want to know why this is wrong.

答:
The probability that target is hit by plane #2  is  the probability of a composite event: plane #1 fails AND plane #2 hits, thus

$$(1-0.3)\times 0.2 = 0.14$$

EDIT : understood that the request is " they bomb the target in succession until they hit" the solution is simply the series

$$\mathbb{P}[\text{#2}]=0.7\cdot 0.2 \sum_{i=0}^{\infty}(0.8\cdot 0.7)^i=0.7\cdot 0.2\cdot\frac{1}{1-0,56}=0.3181818...$$

Your solution is wrong because the probability that #1 never hits is not simply 0,7 but it is a result of a series too and goes to 0

In fact

$$\mathbb{P}[\text{#1}]=0.3 \sum_{i=0}^{\infty}(0.8\cdot 0.7)^i=0.3\cdot\frac{1}{1-0,56}=0.6818181...$$

Thus

$$\mathbb{P}[\text{#1}]+\mathbb{P}[\text{#2}]=1$$


答:
In the law of total variance $$\operatorname{Var}[X] = \operatorname{E}[\operatorname{Var}[X \mid T]] + \operatorname{Var}[\operatorname{E}[X \mid T]]$$ where I have used $T$ to denote the type of policy, with $$T \sim \operatorname{Categorical}(\pi_1 = 0.1, \pi_2 = 0.5, \pi_3 = 0.4), \\ \Pr[T = i] = \pi_i,$$ where the coding is $A \equiv 1$, $B \equiv 2$, and $C \equiv 3$, the first component $$\operatorname{E}[\operatorname{Var}[X \mid T]]$$ is what we call the "within-group" variance; this is the mean of the variability of $X$ that is attributable to each group.  The second component $$\operatorname{Var}[\operatorname{E}[X \mid T]]$$ is what we call the "between-groups" variance; this is the variance of the conditional means of $X$ for each group; i.e., the variability of the means between groups.

How do we compute each of these?  Since $X \mid T$ is Poisson, specifically $$X \mid T \sim \operatorname{Poisson}(\lambda_T)$$ where $\lambda_1 = 1, \lambda_2 = 2, \lambda_3 = 10$, and both the mean and variance of a Poisson distribution are equal to its rate parameter, we have $$\operatorname{E}[X \mid T] = \operatorname{Var}[X \mid T] = \lambda_T.$$  Consequently $$\operatorname{E}[\operatorname{Var}[X \mid T]] = \operatorname{E}[\lambda_T] = \lambda_1 \pi_1 + \lambda_2 \pi_2 + \lambda_3 \pi_3 = 5.1,$$ and $$\operatorname{Var}[\operatorname{E}[X \mid T]] = \operatorname{Var}[\lambda_T] = \operatorname{E}[\lambda_T^2] - \operatorname{E}[\lambda_T]^2 = (\lambda_1^2 \pi_1 + \lambda_2^2 + \pi_2 + \lambda_3^2 \pi_3) - (\lambda_1 \pi_1 + \lambda_2 \pi_2 + \lambda_3 \pi_3)^2 = 16.09.$$

Failing to take into account variability due to differences between group means--i.e, the second variance component--you are just computing how much variation arises from each individual group, ignoring that the groups may be located far apart from each other, as in this case where Type $C$ policies have a mean annual claim rate of $10$, far more than the other two types.  Another way to see this is that if $\lambda_1, \lambda_2, \lambda_3$ are all clustered very "close together," i.e., there is very little variability of the mean claim rates by policy type, then the second component will be very small relative to the first component.

Title: Variance of Poison Random Variable

Body:
For a certain insurance company, 10% of its policies are Type, A, 50% are Type B, and 40% are Type C.

The annual number of claims for an individual Type A, Type B, and Type C policy follow Poisson distributions with respective means 1, 2, and 10.
Let X represent the annual number of claims of a randomly selected policy.
Calculate the variance of X.

So one way of arriving at the answer is using the conditional variance formula.

Var(X) = E[Var(X | type] + Var(E[X |type])
$$=E[1,2,10] + Var(1,2,10)$$
$$=5.1 + Var(1,2,10)$$

From there you calculate the second moment

$$E[{1,2,10}]^2 = 42.1$$
$$42.1 - 5.1^2=16.09$$
$$Var(X)= 5.1 +16.09 $$

And I understand this method.  But when I did it by myself, I just calculated first and second moments and used the formula $$Var(X) = E(X^2) - E(X)^2$$
getting 16.09.  Why is this not the answer?  I need help conceptualizing why 5.1 must be added to the variance here.


问:
How to express a cycle decomposition of $c^d$?
Let $c$ be an $r$-cycle in $S_n$, where $r$ > 1. Let $d$ be an integer not divisible by $r$. Then the cycle decomposition of $c^d$ is the product of $k$ disjoint $s$-cycles.

I know that $k$ equals the greatest common divisor of $r$ and $d$, but what is $s$?

How can I have express $s$ in terms of $r,n,d$, $gcd(r,d)$?

答:
For convenience, think of the $r$-cycle $\sigma=(0,1,2,\ldots,r-1).$

We give two modes of argument.

First:

Notice that $\sigma^d$ sends $x$ to $x+d \pmod{r}.$ Hence, $s$ is the least positive integer such that $$x+sd = x \pmod{r}$$ for all $x\in \lbrace 0,1,\ldots,r-1\rbrace$.

Simplifying, we want the least positive integer $s$ such that $$sd = 0 \pmod{r},$$
i.e., that $sd$ is a multiple of $r$.

Since $sd$ is also a multiple of $d$, we desire $sd=lcm(r,d)$. Hence, $s=\frac{lcm(r,d)}{d}$.

Second:

Notice that if $\sigma^d(x)=x$ for any $x\in \lbrace 0,1,\ldots, r-1 \rbrace$, then $\sigma: x \mapsto x + d\pmod{r}$ means that $d$ must be divisible by $r.$ This contradicts the assumption, so that each $x\in \lbrace0,1,2\,\ldots,r-1\rbrace$ belongs to one of the $k$ cycles in the cycle decomposition of $\sigma^d$. By a symmetry argument, each of the $k$ cycles have the same length $s$. Thus, we must have $$s = \frac{r}{k}=\frac{r}{gcd(r,d)}=\frac{lcm(r,d)}{d}.$$


问:
Gentzen Style Intuitionistic Sequent Proofs
I am trying to provide intuitionistic sequent proofs (Gentzen Style) for a few statements.

The rules that I have are the rule of assumption, conjunction introduction and elimination, disjunction introduction and elimination, conditional introduction and elimination, double negation introduction, reductio ad absurdum, and ex falso.

In particular, we do not have double negation elimination since this is intuitionistic logic.

The rules above are given by Sider in his book Logic for Philosophy Section 2.4.

I am trying to prove the following statements within this logic.

$1. \Rightarrow \alpha \rightarrow (\neg \alpha \rightarrow \beta)$

$2. \neg \neg (\alpha \land \beta) \Rightarrow \neg \neg \alpha \land \neg \neg \beta$

$3. \Rightarrow \neg \neg (\neg \neg \alpha \rightarrow \alpha)$

$4. \neg \neg (\alpha \rightarrow \beta), \neg \neg \alpha \Rightarrow \neg \neg \beta$

I only managed to prove the first statement $\Rightarrow \alpha \rightarrow (\neg \alpha \rightarrow \beta)$ which is somewhat intuitive.

I'm having trouble with the other three statements in the Gentzen style. And I cannot use any other metalangauge proof theorems that may simplify the derivations.

I tried looking at other resources, but almost every logic book does sequent proofs differently. Any help would be appreciated.

答:
For 2):

$¬¬(α ∧ β)$ --- premise

$¬α$ --- assumption [a]

$α ∧ β$ --- assumption [b]

$α$ --- from 3)

$\bot$ --- from 2) and 4)

$¬(α ∧ β)$ --- from 3) and 5), discharging [b]

$\bot$ --- from 1) and 6)

$¬¬α$ --- from 2) and 7) , discharging [a]

Now repeat from:

$¬β$ --- assumption [c]

$α ∧ β$ --- assumption [d]

deriving:

$¬¬β$ --- discharging [c] and [d]

$¬¬α ∧ ¬¬β$ --- from 8) and 11).

For 3:

$\lnot (\lnot \lnot p \to p)$ --- assumption [0]

$p$ --- assumption [a]

$\lnot \lnot p \to p$ --- from 2)

$\bot$

$\lnot p$ --- from 2) and 4), discharging [a]

$\lnot \lnot \lnot p$ --- assumption [b]

$\lnot \lnot p \to p$ --- from 6) and $\vdash \lnot p \to (p \to q)$

$\bot$

$\lnot \lnot \lnot \lnot p$ --- from 6) and 8) discharging [b]

$\lnot p$ --- assumption [c]

$\lnot \lnot \lnot p$ --- from 10) and $\vdash p \to \lnot \lnot p$

$\bot$

$p$ --- from 12)

$\lnot \lnot p \to p$ --- from 10) and 13), discharging [c]

$\bot$

$\lnot \lnot (\lnot \lnot p \to p)$ --- from 1) and 15), discharging [0]

For 4):

$¬¬(α → β)$ --- premise

$¬¬α$ --- premise

$¬β$ --- assumption [a]

$α → β$ --- assumption [b]

$α$ --- assumption [c]

$β$ --- from 4) and 5)

$\bot$ --- from 3) and 6)

$¬α$ --- from 5) and 7) discharging [c]

$\bot$ --- from 2) and 8)

$¬(α → β)$ --- from 4) and 9), discharging [b]

$\bot$ --- from 1) and 10)

$¬¬β$ --- from 3) and 11), discharging [a]


问:
If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.
If f: $\mathbb{R}_{l} \rightarrow S_{\Omega}$ is continuous, then f is not injective.

I've been trying to solve this problem for a few days, but I haven't been able to see how can I do it. First, $\mathbb{R}_{l}$ refers to $\mathbb{R}$ with the lower limit topology (generated by intervals of the form $[a,b)$ ) and $S_{\Omega}$ refers to the uncountable well-ordered set with the order topology (generated by (a,b) and rays). Until now, I've tried to use the fact that both of those topologic spaces are Hausdorff and suppose that f is injective to get a contradiction, but I don't see where can I get it.

答:
Let $f: \Bbb R_l \to S_\Omega$ be an injective continuous function. We strive for a contradiction:

Let $f[\Bbb R_l]$ be the image in $S_\Omega$, which is a non-empty set so $m = \min f[\Bbb R_l]$ exists, so there is a unique (by injectivity of $f$) $x_0 \in \Bbb R_l$ with $f(x_0)=m$. In the order topology on $S_\Omega$, the set $O= \{x\mid x < m+1\}$ is open so that $f^{-1}[O] = \{x_0\}$ (identity clear from minimality of $m$ and uncity of $x_0$) is open in $\Bbb R_l$, contradicting that all non-empty open subsets of $\Bbb R_l$ are uncountable (they contain a set of the form $[a,b), a < b$).


问:
Prove measure of a set is zero
Let $(X, \mathcal{M},u)$ be a measure space with $\{E_i\}_{i=1}^{\infty}$ s.t $\sum_{i=1}^{\infty} u(E_i) < \infty$. $E$ is defined to be $E=\{ x\in X: x \in E_i \text{ for infinitely many }i \}$. Prove that $u(E) = 0$.

I notice that $E = \limsup_{k \rightarrow \infty} E_k = \cap_{n=1}^{\infty} \cup_{k \geq n} E_k$ and can have that $u(\limsup E_i) \geq \limsup u(E_i)$, but it doesn't seem to be leading to anywhere.

Thank you in advance.

答: $u(E)\leq u(\bigcup_{k \geq n} E_k)$ for each $n$. This gives $u(E) \leq \sum\limits_{k=n}^{\infty} u(E_k)$ and this tends to $0$ as $ n \to \infty$ since $\sum u(E_k) <\infty$.

Title: Prove measure of a set is zero


问:
What is the definition of $\{0,1\}^I$.
Let $I=[0,1]$, and $\Omega=\{0,1\}^I$.  what is the definition of $\Omega$

答:
It means, as suggested by Kavi, the set of functions from $I$ to {0,1}. Note this is a general set-theoretical notation, when we have any two sets $X$ and $Y$, then the set of all functions from $X$ to $Y$ is denoted $Y^{X}$.

The main reason for this notation is about cardinality, indeed the cardinality of the set of functions $Y^{X}$, denoted $|Y^{X}|$ is exactly $|Y|^{|X|}$.

Title: What is the definition of $\{0,1\}^I$.


问:
Exercise of Functional analysis
For $f$ in dual space, there exists $x$ with norm $1$ and $f(x)=‖f‖$ if space is reflexive. Use this fact to show that $\ell^ 1 $ is not reflexive.

Let $\alpha = (\alpha_{k}) \in \ell^{\infty}.$ Consider $f(x)=\sum_{k=1}^{\infty} \alpha_{k} x_{k}, \quad x=\left(x_{k}\right) \in \ell^1.$ In this case, $|| f || = || \alpha ||_{\infty}.$ I want to look for $\alpha \in \ell^{\infty}$ appropriate such that $ f (x) \ne || f ||, \forall x$ with $||x||_1 =1.$ But I haven't been able to find it yet. Can you give me an idea please?

答: Take $\alpha =((1-\frac 1 n))$. There is no $(x_n)$ of norm $1$ such that $\sum (1-\frac 1n) x_n=1$. Can you show this?

Title: Exercise of Functional analysis


问:
This series is known? $\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$
This series is known?
I was solving a differential equation by power series and my solution involves that series

$$\sum _{n=1}^{\infty }\frac{2^{n-1}x^{2n-1}}{\prod _{i=1}^n\left(2i-1\right)}$$

Body:
Note that
$$\begin{align}
\frac1{\prod_{i=1}^n(2i-1)}&=\frac{\prod_{i=1}^n(2i)}{\left(\prod_{i=1}^n(2i)\right)\left(\prod_{i=1}^n(2i-1)\right)}\\
&=\frac{\prod_{i=1}^n(2i)}{\prod_{i=1}^n(2i(2i-1))}\\
&=\frac{2^nn!}{(2n)!}.
\end{align}$$
So your sum is
$$f(x)=\sum_{n\ge1}\frac{n!}{(2n)!}(2x)^{2n-1}=\sum_{n\ge0}\frac{2^{2n}n!}{(2n)!}\frac{x^{2n+1}}{2n+1}.$$
Let $t_n=\frac{2^{2n}n!}{(2n)!}\frac{x^{2n}}{2n+1}$, so that $f(x)=x\sum_{n\ge0}t_n$. Then
$$\begin{align}
\frac{t_{n+1}}{t_n}&=\frac{\frac{2^{2n+2}(n+1)!}{(2n+2)!}\frac{x^{2n+2}}{2n+3}}{\frac{2^{2n}n!}{(2n)!}\frac{x^{2n}}{2n+1}}\\
&=(2x)^2\frac{2n+1}{2n+3}\frac{(n+1)!}{n!}\frac{(2n)!}{(2n+2)!}\\
&=(2x)^2\frac{2n+1}{2n+3}\frac{n+1}{(2n+2)(2n+1)}\\
&=\frac{n+1}{n+3/2}\frac{x^2}{n+1},
\end{align}$$
so we may write
$$t_n=\frac{(1)_n}{(3/2)_n}\frac{(x^2)^n}{n!}t_0,$$
where $$(z)_n=\frac{\Gamma(z+n)}{\Gamma(z)}=z(z+1)(z+2)\cdots(z+n-1).$$
Since $t_0=1$, we have
$$\sum_{n\ge0}t_n=\,_1F_1(1;\tfrac32;x^2),$$
and thus
$$f(x)=x\,_1F_1(1;\tfrac32;x^2),$$
where
$$_pF_q(a_1,...,a_p;b_1,...,b_q;z)=\sum_{n\ge0}\frac{(a_1)_n\cdots(a_p)_n}{(b_1)_n\cdots(b_q)_n}\frac{z^n}{n!}$$
is the generalized hypergeometric function.

Anyway, for $\Re(b)>\Re(a)>0$, we have the integral representation
$$_1F_1(a;b;z)=\frac{\Gamma(b)}{\Gamma(a)\Gamma(b-a)}\int_0^1e^{zt}t^{a-1}(1-t)^{b-a-1}dt,$$
from here.
Using this,
$$f(x)=\frac{x\Gamma(\tfrac32)}{\Gamma(1)\Gamma(\tfrac12)}\int_0^1\frac{e^{x^2 t}}{\sqrt{1-t}}dt=\frac{x}{2}\int_0^1\frac{e^{x^2t}}{\sqrt{1-t}}dt.$$
Then write $z=\sqrt{1-t}$, so that $-2dz=\frac{dt}{\sqrt{1-t}}$, and
$$f(x)=x\int_0^1e^{x^2(1-z^2)}dz=xe^{x^2}\int_0^1e^{-(xz)^2}dz.$$
Finally, writing $u=xz$, we have
$$f(x)=e^{x^2}\int_0^{x}e^{-u^2}du.$$
Recalling the definition of the Error Function $\mathrm{erf}(z)$, we have
$$f(x)=\frac{\sqrt\pi}2e^{x^2}\mathrm{erf}(x).$$
I hope this helps.


问:
Analyse the asymptotic behavior of $\sum_{p\leq x} \frac{\log^2(p)}{p}$
This question showed up in a number theory test from previous semesters so I might not know all the material needed to solve this question but here is my try:

The function $f(x)=\frac{\log^2(x)}{x}$ is bounded in $(1,\infty)$ so for some upped bound $C$ on $f(x)$ we have $$\sum_{p\leq x, p\, \text{prime}} \frac{\log^2(p)}{p}\leq C\cdot \pi(x)$$
on the other hand for large enough $x$ we have $\min\left\{\frac{\log^2(p)}{p}|p\leq x, p\, \text{prime}\right\}\geq \frac{\log^2(x)}{x}$ and so:
$$\sum_{p\leq x, p\, \text{prime}} \frac{\log^2(p)}{p}\geq \frac{\log^2(x)}{x} \pi(x)\stackrel{PNT}{=}O(\log(x))$$
Both bounds dont give a tight bound and here I am pretty much stuck.

答:
For convenience, let's define

$$
F(x)=\sum_{p\le x}{\log p\over p}
$$

Then using Riemann-Stieltjes integration we have

$$
\begin{aligned}
\sum_{p\le x}{\log^2p\over p}
&=\int_{2^-}^x\log t\mathrm dF(t)
=F(x)\log x-\int_2^x{F(t)\over t}\mathrm dt \\
&=\log^2x-\int_2^x{\log t\over t}\mathrm dt+[F(x)-\log x]\log x-\int_2^x{F(t)-\log t\over t}\mathrm dt
\end{aligned}
$$

By Mertens' first theorem, we know that $F(x)=\log x+\mathcal O(1)$, so the above formula gets simplified into

$$
\sum_{p\le x}{\log^2p\over p}=\log^2x+\mathcal O(\log x)
$$


答:
For those that aim to be precise with their set-theoretical notations and terminology, there is a difference between the fundamental notion of map and the fundamental notion of family.

Given an arbitrary family $A$ of sets indexed by set $I$, the cartesian product $\displaystyle\prod_{i \in I}A_i$ is by definition the set of all families $x$ indexed by $I$ and such that at each index $i \in I$ the component $x_i \in A_i$ is subject to this condition, of being an element of $A_i$. In the particular case when all the factors $A_i, i \in I$ of such a product were equal to one and the same set $B$ one simplifies the general notation $\displaystyle\prod_{i \in I}B$ down to the more convenient $B^I$, which is by definition the set of all families (not maps) $x$ of elements of $B$ indexed by $I$.

The relation between a map from $I$ to $B$ and a family of elements of $B$ indexed by $I$ is very close, as by definition any map from $I$ to $B$ is a triplet of the form $(I, F, B)$, where $F \subseteq I \times B$ is a functional relation fully defined on the whole of $I$ (in other words $F$ is what Bourbaki calls a graphic such that $(x, y), (x, z) \in F$ entails $y=z$ and such that the first projection of $F$ is $I$ itself), object which is nothing else than a family of elements of $B$ indexed by $I$. In other words -- somewhat loosely speaking -- a map also consists of its domain of definition and codomain, structurally encoding these in a triplet as exhibited above, whereas a family certainly knows nothing about any codomain and is defined differently, as a mathematical object.

With these amendments in mind, $\{0, 1\}^{[0, 1]}$ is the set of all families $\lambda$ indexed by the interval $[0, 1]$ and capable of taking only the two values $\{0, 1\}$ (i.e. at every index $t \in [0, 1]$ it is either the case that $\lambda_t=0$ or $\lambda_t=1$).

Title: What is the definition of $\{0,1\}^I$.

Body: Let $I=[0,1]$, and $\Omega=\{0,1\}^I$.  what is the definition of $\Omega$


问:
How to calculate the number of palindromes of a given number of characters?
A palindrome is a set of characters which equal the same thing forwards and backwards, for example; abccba.

For a set of a given amount of  characters (we can use 9 as an example) how would you calculate the amount of palindromes of characters mathematically?

This doesn't mean the words that can be formed but rather characters which are the same at start characters and end characters.

What formula can be used to calculate this for simple lengths like 9 with lowercase letters?

答:
Assuming an alphabet of size N, and a string length of k:

$N^{⌈k/2⌉}$

N to the power of the ceiling of half the string length.

(Ceiling means always round up for any decimal)

This is because:

$k = 1$ will have $N$ possibilities and the string will be in the format: $N_1$

$k = 2$ will have $N$ possibilities and the string will be in the format: $N_1N_1$

$k = 3$ will have $N^2$ possibilities and the string will be in the format: $N_1N_2N_1$

$k = 4$ will have $N^2$ possibilities and the string will be in the format: $N_1N_2N_2N_1$

$k = 5$ will have $N^3$ possibilities and the string will be in the format: $N_1N_2N_3N_2N_1$

etc...

Even though every even length string has an added a character compared to the odd string of one less in length, because it is a palindrome, this added character will be a duplicate. It's position and character is decided by its mirror.


问:
Edge-coloring with constraints (electronics)
My work is in electronics modeling, the sought procedure could be helpful in efficiently formulating circuit equations by current and voltage sources only.

For a given circuit topology, assign color/type I (current source) or V (voltage source) to components, such that

no node/vertex is only connected to type I edges

no possible mesh/loop/cycle with type V only

ground is considered type V

some components could have a fixed or preferred type.

My first approach would be constructive and trying to swap types when conflicts arise.

Are there any heuristics known? I only could find problems like minimum number of types/colors, but not the constraints as described above.

Example

Minimum spanning tree (Prim's algorithm) with netlist

(node 1 node 2 weight, node numbers see example)

```<code>0 1 1
1 2 1
2 3 2
3 0 3
</code>```

chooses the spanning (V-type) tree as

```<code>0 1
1 2
2 3
</code>```

The trick is to apply the correct weights, and these depend on the nature of component (V, I, R, L, C). Then MST gives the correct V branches.

For example, if we have a preference for C-> V (weight = 1) and L->I (weight = 3) and an assumed R||C, the branch preference becomes V, so gets the lowest cost.

答: Pick a spanning tree in the graph. Color all edges in the spanning tree type V (a tree contains no cycles) and color all remaining edges type I. A spanning tree connects all vertices, so no vertex is purely type I.</p>


问:
How can we show that $\phi$ is constant?
Hello everyone I have the following statement that I want to prove but have no idea how I can proceed, if anyone could help it would be a lot appreciated :

Let $\mathbb N = \{1,2,\dots\}$ with the topology where the open sets are $\emptyset$, $\mathbb N$ and $\{1,\dots,n\}$ for all $n \geq 1$. Also, let $\mathbb R$ with the usual topology.

How can one show that a continuous function $\phi : \mathbb N \rightarrow \mathbb R$ is constant?

Here the point is to show that $(\forall n \geq 1) : \phi(n)=k$ but I don't see how can prove it. Thanks in advance for your help.

答:
In $\mathbb R$ with the euclidean topology the singletons $\{a\}$, with $a\in\mathbb R$, are closed sets. By assumption, the function $f:\mathbb N\to\mathbb R$ is continuous, hence the preimage of a closed set is a closed set.

This means that the preimage of a generic singleton $f^{-1}(\{a\})$ is a closed set in $\mathbb N$ equipped with the topology $\mathcal T=\{1,\dots,n\}_{n\ge 1}\cup\mathbb N\cup\emptyset$.

The complement of open sets are closed sets for definition, so a closed set $K\in(\mathbb N,\mathcal T)$ can be $\mathbb N$, $\emptyset$ or $\mathcal C(\{1,\dots,n\}_{n\ge 1})\implies f^{-1}(\{a\})=\begin{cases}\mathbb N\\\emptyset\\ C(\{1,\dots,n\}_{n\ge 1}) \end{cases}$.

At this point the preimage of a point in $\mathbb R$ must be $\mathbb N$ or $\emptyset$ because in the other case, taking two distinct points $a,b\in\mathbb R$, since the fact they are closed sets, their preimages would be $f^{-1}(a)=\mathbb N$\ $\{1,\dots,n_0\}$ and $f^{-1}(b)=\mathbb N$\ $\{1,\dots,n_1\}$.

Hence $f^{-1}(a)\cap f^{-1}(b)=\mathbb N$\ $\{1,\dots,\max\{n_0,n_1\}\}$, which is a contadiction $\implies f$ constant.


问:
Need help understanding the logic behind the intersection of infinite sets.
As the post topic states, I need some help understanding the logic behind the intersection of infinite sets. I will take this time to thank anyone who reaches out with a comment or answer.

The following example has been taken from a real analysis book I've purchased.

Say the following sets exist,
$$ A_1 = N = \{1, 2, 3, ...\}\\
A_2 = \{2, 3, 4, ...\}\\
A_3 = \{3, 4, 5, ... \}\\
A_n = \{n, n+1, n+2, ...\}$$

Now, i'm okay with this. $$\bigcup_{n=1}^{\infty}A_{n} = A_1$$
This is what I can't seem to grasp.
$$\bigcap_{n=1}^{\infty}A_{n} = \emptyset$$

I can't make a mental picture of this since we have literally an infinite number of sets. I can see that $A_1 \cap A_2 = A_2$ applying that logic to each of these sets, which are by definition "infinite" there will always be some set $A_{n+1}$ for every $A_{n}$ So how can this ever be the empty set?

答:
If some number $k$ is in this intersection, then it is in all of the sets $A_n$. But $k$ is definitely not in $A_{k+1}$. Therefore no such $k$ exists; the intersection is empty.

The key is that for something to be in an intersection, it must be in each and every one of the intersecting sets.


答:
Based on the hint of Kavi Rama Murthy, I think this argument might work:

Assume by contradiction that $\phi$ is non-constant. So there exist $a,b \in \Bbb{N}$ such that $\phi(a) \neq \phi(b)$. Since $\Bbb{R}$ is Hausdorff, we can find disjoint open sets $U, V$ such that $\phi(a) \in U, \phi(b) \in V$ and $U \cap V = \emptyset.$

Since $\phi$ is continuous, $\phi^{-1}(U)$ and $\phi^{-1}(V)$ are open in $\Bbb{N}$. Now we use Kavi Rama Murthy's hint: One of these sets must be empty*. Indeed, if both of them were nonempty, say $\phi^{-1}(U) = \{1,2,...,n\}$ and $\phi^{-1}(V) = \{1,2,...,m\}$ for some $ m,n \in \Bbb{N}$ then their intersection would be nonempty. But we have that

$$\phi^{-1}(U) \cap \phi^{-1}(V) = \phi^{-1}(U \cap V) = \emptyset,$$

since the intersection $U \cap V$ is empty by the Hausdorff property.

But that is a problem: If, say, $\phi^{-1}(U)$ is empty then $\phi(a) \notin U$, a contradiction. Therefore $\phi$ must be constant.

* You can already conclude from here (see Henno Brandsma's comment)

Title: How can we show that $\phi$ is constant?

Body:
Hello everyone I have the following statement that I want to prove but have no idea how I can proceed, if anyone could help it would be a lot appreciated :

Let $\mathbb N = \{1,2,\dots\}$ with the topology where the open sets are $\emptyset$, $\mathbb N$ and $\{1,\dots,n\}$ for all $n \geq 1$. Also, let $\mathbb R$ with the usual topology.

How can one show that a continuous function $\phi : \mathbb N \rightarrow \mathbb R$ is constant?

Here the point is to show that $(\forall n \geq 1) : \phi(n)=k$ but I don't see how can prove it. Thanks in advance for your help.