问:
Absolute value of a complex exponential
I want to prove that  $ \left|e^{2iRe^{it}}\right| = e ^{ -2R\sin(t)} $  for some constant R.

I know that  $  e ^{ -2R\sin(t)}= e^{-2R\times\text{Im}(e^{it})}  $   where Im denotes the imaginary part, so i just need to prove that
$$ \left|e^{2iRe^{it}}\right| =  e^{-2R\times\text{Im}(e^{it})} $$
but i can not find the relation between the complex modulus and the imaginary part.

答: Hint: $\left| e^{x+iy} \right| =  \left| e^{x} \right| \left| \cos y + i\sin y\right| = e^x$ for $x, y \in \mathbb{R}$.

Title: Absolute value of a complex exponential


问:
How far can you go with the Martingale Roulette strategy
I am currently arguing with my friends that just learned the Martingale roulette technique for the casino that consist of doubling your initial bet as much time that you lose (with a budget of course).
So if you have 5000\$ you could initially bet 1\$ on a color (with 48% aprox. chances of winning) and if you lose you bet 2\$, then 4\$, then 8\$... up to the point where you don't have the money to double on and you lose it all.

I am certain that in the end you are losing using this technique. Yes your probability of winning a percentage of your budget are great up until a limit of wins using this technique. I'm almost certain that you hit that limit before 5000 times ( winning more than what you were willing to lose ). So in general even with millions of people who would use this the casino is still winning in total and with any maximum budget (except infinity of course). My friend base a lot of their thoughts behind the fact that this is banned from a lot of casinos along with all sort of techniques or strategies but i think it's more of a collateral damage and in general you still lose with that.

So I'm asking your help to explain why I am or they are wrong about this.
Thank you !

答:
The basic issue with the martingale is that probabilistically, "anything that can happen will happen."

If I have a nonzero probability of exhausting my bankroll via the martingale strategy, then eventually it will happen.

With a larger bankroll, I can make the probability of going broke very small, but as long as my bankroll is finite, I can never make the probability zero entirely.

Sooner or later, it will happen to me.

As regards the expected value, Wikipedia has a thorough analysis. If I'm playing an unfair game (i.e. my win probability is under $0.5$) then my net expectation is always negative, because those infrequent times that I lose the amount I budgeted for the martingale, cancel out all the gains I make.

Not to mention, as Milo and Doug point out, the casino is going to have a maximum bet at some point. So if I don't exhaust my bankroll, at a certain point I'm not going to be able to cover my losses by doubling my bet, because the casino isn't going to let me bet that much.

ContentLicense: CC BY-SA 4.0


问:
How to keep track of variations in states based on controls in optimal control problem?
Suppose that $x(t)$ is the state variable showing the level of water in a tank at time $t$, and water is leaking the tank with rate $\lambda$. Control is denoted by $u(t)$ which is the amount of water added into a tank at time $t$. I am willing to know how is it possible to consider the amount of water added into the tank during time interval $[t-\tau,t]$ and also the amount of water added into the tank in this time interval for a specific value of $u(s)$ which can be denoted as $s_{\tau}(t,u)$.

For example, $s_{\tau}(t,2)$ means the amount of water added into the tank in interval $[t-\tau,t]$ when control is exactly equal to 2 for some times in this interval.

答:
If you need to know the change of the amount of water between time $[t-\tau, t]$, you just need to store the previously measured amounts somewhere they can be accessed later (using an array is a popular method here).
If you like to test an alternative input, you first need to model the tank. This can be done using Bernoulli's principle (advised to express the input as a flow instead of an amount). As your system has some kind of leakage, the resulting system will be non-linear. This mean you can go in multiple directions:

Use ODE solvers to solve your system for a given $s_\tau(t,u)$, this will compute an accurate estimation including non-linear dynamics.

Linearize and discretize the system, computationally the least intensive method. Since the solution can be computed algebraically (even without integrals), it can be easily used to control the system. yet by linearizing the system, the estimation is only accurate for some range in height.

The first one is the better option if you know the input sequence and just want to simulate it, the latter one is pretty much required if you want to build a model-predictive-controller for it.

EDIT: To answer your question. That integral would suit if the water in the tank is not also leaking. So if the water level is denoted as $x(t)$ and the tank is not leaking, it can be expressed as the following:
$$u(t) = S\dot{x}(t)$$
Where $S$ is the surface area of the tank and $u(t)$ is the flow of water entering the tank (controlled). This means that the waterlevel at time $t$ starting from time $t-\tau$ where $\tau \geq t$ is:
$$x(t) = \frac{1}{S}\int_{t-\tau}^t u(t) dt + x(t-\tau)$$
Again, this can be only this simply expressed if the tank is not leaking. Using bernoulli's principle, if we include the leakage, the equation changes to:
$$\dot{x}(t) = \frac{1}{S}u(t) - \frac{1}{S}\alpha\sqrt{2g x(t)}$$
With valve specific parameter $\alpha$ and the gravitation acceleration $g$. Now you cannot simply solve this by integrating it over time. Sadly, the only way to properly predict the future water level is to incorporate the leakage dynamics.


问:
Find sufficient and necessary condition for $f(|x|,y^2)$ to be differentiable at $(0,0)$.
Let $f(x,y):\mathbf{R}^2\to \mathbf{R}$ be differentiable at $(0,0)$.
Let $h(x,y)=f(|x|,y^2)$.
Find sufficient and necessary condition for $h(x,y)$ to be differentiable at $(0,0)$. I suspect it has to be something about the chain rule, but $|x|$ is not differentiable at $0$. Could some give a hint?

答:
Observe what has to happen for:
$$D_1 h(0,0)=\lim_{t\to 0} \dfrac{f(|t|,0)-f(0,0)}{t}$$
to exist $(D_1 f(0,0)=0)$.

Then look that if that actually happens, then $D_1 h(0,0)=0$. Calculate $D_2 h(0,0)$ with the chain rule (possible because $g(t):=h(0,t)=f(0,t^2)$ is derivable).
Then, using what you've got, and the inequalities $||(|x|,y)||\geq ||(|x|,y^2)||$, $y\in(-1,1)$ and $||(|x|,y)||\geq |y|$ bound:
$$0\leq\left|\dfrac{h(x,y)-h(0,0)-D_1 h(0,0)\cdot x-D_2 h(0,0)\cdot y}{||(x,y)||}\right|=\\=\left|\dfrac{f(|x|,y^2)-f(0,0)}{||(|x|,y)||}\right|=\left|\dfrac{f(|x|,y^2)-f(0,0)-D_y f(0,0)\cdot y^2}{||(|x|,y)||}+\dfrac{D_y f(0,0)\cdot y^2}{||(|x|,y)||}\right|\leq\\ \leq\left|\dfrac{f(|x|,y^2)-f(0,0)-D_y f(0,0)\cdot y^2}{||(|x|,y)||}\right|+\left|\dfrac{D_y f(0,0)\cdot y^2}{||(|x|,y)||}\right|\leq\\ \leq\left|\dfrac{f(|x|,y^2)-f(0,0)-D_y f(0,0)\cdot y^2}{||(|x|,y^2)||}\right|+\left|\dfrac{D_y f(0,0)\cdot y^2}{|y|}\right|$$
And taking $(h,k)=(|x|,y^2)\to(0,0)$ first limit is $0$ by $f$ differentiability. The second one is obviously $0$. So by sandwitch $h$ is diff.


问:
How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$
Prove that

$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$

I have some question. Using this, find this integral is not converge, I'm wrong?

Thank you everyone

答:
No, this integral is not convergent.

Consider $I(k):=\int_{0}^{\pi} \sin x\sin \sqrt{2k\pi +x}dx$. For $k$ large enough, $\sqrt{2k\pi +x}-\sqrt{2k\pi}, x\in [0,\pi]$ is small enough, so

$$\int_{0}^{\pi} \sin x\sin \sqrt{2k\pi +x}dx\sim \int_{0}^{\pi}\sin x \sin \sqrt{2k\pi }dx=2\sin \sqrt{2k\pi }\not\to 0.$$

ContentLicense: CC BY-SA 3.0


问:
Give an example of a proper ideal I of a ring R such that the radical of I is equal to R i.e. √I=R.
I am having trouble with finding such example. In commutative case with identity such example does not exist as 1∈R=√I implies 1∈I and so I=R. Is there any in noncommutative case or a ring without identity?</p>

答:
Let’s $G$ be any abelian group with a nontrivial subgroup, and make $G$ into a ring using the operation $ab=0$ for all $a,b\in G$.

Every subgroup is an ideal of this ring, so any nontrivial ideal has radical $G$ ( in the usual commutative sense: all squares are zero, so everything squared gets inside of all ideals.)


问:
Solve $y''-2y'+y=\frac{e^x}{x}$
I got this question earlier in a test and we never discussed how you would go about solving such an equation. I did solve the homogeneous equation by writing the characteristic equation $\Lambda^{2}-2\Lambda+1=0$ and getting the two solutions $\Lambda_{1,2}=1$ which of course implied the general solution to the homogeneous equation is of the form $y_h=C_1e^x+C_2xe^x$ and that is where I got stuck. Since the right hand side is not of a good and easy form we learned about, I was lost. Of course, I used Wolframalpha and found the particular solution that was $y_p=xe^xln(x)$ but I still have no idea how to arrive at this solution. An exam is coming up and I would really appreciate all the help I can get. Thank you!

答:
$$y''-2y'+y=\frac{e^x}{x}$$
Multiply by $e^{-x}$:
$$y''e^{-x}-2y'e^{-x}+ye^{-x}=\frac{1}{x}$$
Rewrtite the equation as this and integrate twice:
$$(ye^{-x})''=\dfrac{1}{x}$$
First integration gives us:
$$(ye^{-x})'=\ln x +C_1$$
Then:
$$(ye^{-x})={x} \ln x - x +C_1x +C_2$$
$$y(x)=e^{x}({x} \ln x  +Cx +C_2)$$

Title: Solve $y''-2y'+y=\frac{e^x}{x}$


问:
Inequality with Gamma.
Consider the function $$f(x)=e^{\frac{x^2}{2}}\int^\infty_0e^{-t^2}t^{2n}\cos(2xt)dt,\quad x\in\Bbb R$$

I find a lot difficultes to prove $$\forall (x,n)\in\Bbb R^+\times\Bbb N:|f(x)|\leq C\Gamma(n+\frac{1}{2})$$
for some const $C$.
Thanks for any help.

My attempt:

Put $$f(x)=\frac{1}{\Gamma(n+\frac{1}{2})}e^{\frac{x}{2}}\int^\infty_0 e^{-t^2}t^{2n}\cos(2\sqrt{x}t)dt.$$

Let $x>0$, by the change of variable $t=\sqrt{n}s$ we get $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_0 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds. $$
Now we apply Laplace's method wich says:\
\textbf{Theorem} Consider the integral  $\int^b_a f(t)e^{-n g(t)}dt$ where $g$ is of class $\mathcal{C}^2$ on $[a,b[$ (here $a<b\leq+\infty)$ and $f$ is a continuous function on  $[a,b[$ such that

$f(a)\not=0$

$\forall t\in[a,b[:g'(t)>0$

$g'(a)=0$

$\forall t\in[a,b[:g''(t)>0$

Then $$\int^b_a f(t)e^{-n g(t)}dt\sim\sqrt{\frac{\pi}{2g''(a)}}\frac{e^{-ng(a)}f(a)}{\sqrt{n}},\quad n\to+\infty.$$

So we can write $$f(\frac{x}{n})=\frac{1}{\Gamma(n+\frac{1}{2})}\Big(n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds+n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds\Big).$$
Put $$I_1=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^\infty_1 e^{-n\big(s^2-2\ln(s)\big)}\cos(2\sqrt{x}s)ds$$
with $g(s)=s^2-2\ln(s), \quad s\geq 1$ and $f(s)=\cos(2\sqrt{x}s)$, by the previous theorem we have
$$I_1\sim \sqrt{\frac{\pi}{8}}n^{n}\sqrt{n}\frac{e^{-n}}{\sqrt{n}}\cos(2\sqrt{x})$$
(we have $g'(s)=2(s-\frac{1}{s})>0$ and $g''(s)=2(1+\frac{1}{s^2})>0$)
Since $n!\sim(\frac{n}{e})^{n}\sqrt{2\pi n}$ we obtain $$I_1\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$.

Using the same method we obtain $$I_2=n^{n}\sqrt{n}e^{\frac{x}{2n}}\int^0_{-1} e^{-n\big(s^2-2\ln(-s)\big)}\cos(2\sqrt{x}s)ds\sim\frac{n!}{2\sqrt{n}}\cos(2\sqrt{x})$$

Now by using the well-known result  $\frac{n!}{\sqrt{n}}\sim \Gamma(n+\frac{1}{2})$

we get $$\forall x>0,\qquad f(\frac{x}{n})\sim \cos(2\sqrt{x})$$

So, $\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow  \Big|f(\frac{x}{n})\Big|\leq C$ for some constant $C>0$.

Therefore if we replace $x$ by $nx$ we obtain:\

$\forall x>0,\qquad\exists n_0\in\Bbb N,\qquad n\geq n_0\Rightarrow  \Big|f(x)\Big|\leq C$

And hence $$\sup_{(n,x)\in \{n\in\Bbb N; n\geq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$

It remains the case $(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+$ and it is easy because if $n\leq n_0$ we have

$$f(x):=\Big(\frac{\Gamma(n+1)}{\Gamma(n+\frac{1}{2})}\Big)e^{\frac{-x}{2}}L^{\frac{-1}{2}}_{n}(x)\leq C' e^{\frac{-x}{2}}P(x) $$
for some constant $C'>0$ independent of $n$ and $P$ is polynomial with $\text{deg}(P)=n\leq n_0$.

Since $\lim_{x\to+\infty}e^{\frac{-x}{2}}P(x)=0$ we obtain $$\sup_{(n,x)\in \{n\in\Bbb N; n\leq n_0\}\times\Bbb R^*_+}\Big|f(x)\Big|<+\infty. $$

Finally: $$\sup_{(n,x)\in\Bbb N\times\Bbb R_+}\Big|f(x)\Big|<+\infty. $$

The case $x=0$ is easy because $\Psi(0)=c$.

答:
Assuming $n >0$
$$f(x)=e^{\frac{x}{2}}\int^\infty_0e^{-t^2}t^{2n}\cos(2xt)\,dt$$
$$f(x)=\frac{1}{2} e^{\frac{x}{2}}\, \Gamma \left(n+\frac{1}{2}\right) \,
   _1F_1\left(n+\frac{1}{2};\frac{1}{2};-x^2\right)$$ So, it suffices to show that $\forall  x\in\Bbb R$
$$g(x)= e^{\frac{x}{2}}\,  \,
   _1F_1\left(n+\frac{1}{2};\frac{1}{2};-x^2\right) < 2$$ If you work a little the function $g(x)$ you should notice that its maximum value is  close to $1$ (we already have $g(0)=1$).

Title: Inequality with Gamma.


答:
The integral cannot converge since the function $\sin{x}\sin{\sqrt{x}} $ doesn't converge through zero as x tends to infinity

EDIT : There is a mistake, this proof if wrong. Read the comments

Title: How prove this $\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$

Body:
Prove that

$$\int_{0}^{\infty}\sin{x}\sin{\sqrt{x}}\,dx=\frac{\sqrt{\pi}}{2}\sin{\left(\frac{3\pi-1}{4}\right)}$$

I have some question. Using this, find this integral is not converge, I'm wrong?

Thank you everyone


问:
Criterion for unramified prime in number fields
Let $L/K$ be an extension of number fields, $\mathfrak{q}$ be a prime in $\mathcal{O}_L$ and $\mathfrak{p}=\mathfrak{q}\cap\mathcal{O}_K$. If $\mathfrak{p}$ generates the maximal ideal $\mathfrak{q}\mathcal{O}_{L,\mathfrak{q}}$ of the localization ring $\mathcal{O}_{L,\mathfrak{q}}$, then how to prove the ramification index $e(\mathfrak{q}/\mathfrak{p})$ is $1$?

Thanks in advance.

答:
Let $\mathfrak q = \mathfrak q_1, ... , \mathfrak q_g$ be all the primes of $\mathcal O_L$ lying over $\mathfrak p$.  The definition of the ramification index $e = e(\mathfrak q /\mathfrak p)$ is

$$\mathfrak p \mathcal O_L = \mathfrak q^e \mathfrak q_2^{e_2} \cdots \mathfrak q_g^{e_g}$$

as ideals of $\mathcal O_L$.  If we localize these ideals to $\mathcal O_{L,\mathfrak q}$, we have

$$(\mathfrak p \mathcal O_L)_{\mathfrak q} = \mathfrak p \mathcal O_{L,\mathfrak q}$$

$$( \mathfrak q^e \mathfrak q_2^{e_2} \cdots \mathfrak q_g^{e_g})_{\mathfrak q} = \mathfrak q^e \mathcal O_{L,\mathfrak q}$$

since the other prime ideals $\mathfrak q_i$ generate the unit ideal in $\mathcal O_{L,\mathfrak q}$.


问:
Are $A^T B - I$ and $I - A B^T$ equal in Frobenius norm?
Is $\text{frob}(A^T B - I) = \text{frob}(I - A B^T)$, where $\text{frob}$ is the Frobenius norm?

Can you show why?

I ask because, I am curious what is the intuitive meaning of Eq. 4, on page 3 here:https://arxiv.org/pdf/1908.07640.pdf

It seems to be just finding the $S$ matrix s.t. $S^{-1}$ is "closest" to $R$, because the Frobenius norm of the difference between two rotations is a rotation metric known as the deviation from the identity matrix: Eq 21, https://www.cs.cmu.edu/~cga/dynopt/readings/Rmetric.pdf

答:
No, they aren't equal. You may use the classical example that $XY=0\not\to YX=0$ to construct a counterexample. Let
$$
A=\pmatrix{0&1\\ 0&0},B=\pmatrix{1&0\\ 0&0}.
$$
Then
\begin{aligned}
\|A^TB-I\|_F&=\left\|A^T-I\right\|_F=\sqrt{3},\\
\|I-AB^T\|_F&=\|I\|_F=\sqrt{2}.
\end{aligned}


问:
Formal proof for detection of intersections for constrained segments
They told me it was off-topic at stackoverflow.
So I am trying my luck here. Yes, it's a homework, but I'm looking for some guidance (or related literature) instead of complete solutions.

Please see the drawing that accompanies the question:

I need to come up with a formal proof for the following statement:

Given an arbitrary count of line segments (S1 -- S3 in the drawing) constrained by two vertical lines (M & N) an answer whether there is at least one intersection of the segments (in the whole picture) can be given by checking each of the two segments that have their left endpoints adjacent on M in descending order (in respect to their position on line M).

For example, in drawing checking could be done in such order:

S1 and S2 (no intersection)

S2 and S3 (intersection detected)

While it seems very obvious, there shall be a formal proof for that (most likely, a simple one). Any ideas?

答:
Lemma: If two points on the left line S1 and S2 define segments that intersect and there is a third point S3 between S1 and S2, then either S1 and S3 intersect or S2 and S3 intersect.

Proof of lemma: Note that S1, S2, and their intersection form a triangle that is strictly between the two vertical lines. S3 is on one edge of the triangle and the segment starting at S3 enters the interior of the triangle. Therefore, the segment starting at S3 must leave the triangle, so it must intersect either S1 or S2, since they define the other edges of the triangle.

Proof of theorem: We assume that there are two segments that intersect and we need to prove that this implies that there are two segments that intersect and whose left endpoints are adjacent.  Call the left endpoints of the intersecting segments S1 and S2. By the lemma there is a point S3 between S1 and S2 that intersects S1 or S2. If S1 and S3 or S2 and S3 are adjacent, then we are done. If not, then note that there are strictly fewer points between S1 and S3 than there are between S1 and S2, since the interval between S1 and S3 does not contain S2. This means that if we repeatedly apply this process to successively smaller intervals the process will eventually terminate, since the intervals contain fewer points and there are only a finite number of points.

Note that this assumes that none of the left-hand points are coincident. However, in this case there are two points that are adjacent, since they are the same, so the theorem is still proven.


问:
Calculate the limit of the series (measure theory)
Is this solution correct?

$$\lim_{n\to\infty} { \sum_{k=1}^{\infty} { {1}\over{k^2}} \left(1- {{\min(k,n)}\over{n} }\right)}$$ We consider two cases:
$$\min(k,n)=n$$ we have: $${{1}\over{k^2}} {(1 - 1)} = 0 \to \lim_{n\to \infty}{\sum_{k=1}^{\infty}0}= 0 $$ and for $$\min(k,n)=k $$ we have: $$ 0 \le  { {1}\over{k^2}} \left(1- {{k}\over{n} }\right) \le {{1}\over{k^2}} $$ so $$\lim_{n\to\infty} {\sum_{k=1}^{\infty} {{1}\over{k^2}}} = {{\pi^{2}}\over{6}} $$ The command says to apply the bounded convergence theorem, but this statement applies to the integral and I don't know if my solution is correct.

答:
From your observation that for $k\geq n$, the terms are $0$, we get :
\begin{align*}
\lim_{n\to\infty}  \sum_{k=1}^{\infty} { {1}\over{k^2}} \left(1- {{\min(k,n)}\over{n} }\right) &= \lim_{n\to\infty}  \sum_{k=1}^{n} { {1}\over{k^2}} \left(1- {{k}\over{n} }\right)\\
&=\lim_{n\to\infty}  \sum_{k=1}^{n}  {{1}\over{k^2}} - \frac{1}{n}\sum_{k=1}^n{{1}\over{k}}\\
\end{align*}
One can argue that $\sum_{k=1}^n \frac{1}{k}$ is of order $\ln(n)$ and so the RHS of the limit goes to $0$. And so the whole limit is $\frac{\pi^2}{6}$.


答:
To solve $$Y''(x)-2Y'(x)+Y(x)=\frac{e^{x}}{x} ~~~~(1)$$
First solve the homogeneous part $$y''-2y'+y=0 \implies y_1(x)=e^x,y_2(x)=xe^{x}$$
Wronskian of these solution is$$ w(x)=e^x (e^x+xe^x)-xe^x e^x=e^{2x}$$ Then the method of variation of parameters gives $$Y(x)=C_1(x) y_1(x)+C_2(x) y_2(x),~~~~(2)$$
where $$C_1(x)=\int y_2(x)/ w(x) dx+D_1, C_2(x)=\int y_1(x)/ w(x) dx+D_2$$
$$C_1(x)=-\int dx+D_1=-x+D_1, C_2(x)=\int \frac{dx}{x}+D_2=\ln x+D_2~~~~(3)$$
From (2) and (3), we have
$$Y(x)=(-x+D_1)e^x+(\ln x+D_2)xe^x =D_1 e^x+D_2xe^x-xe^x+xe^x \ln x$$

Title: Solve $y''-2y'+y=\frac{e^x}{x}$

Body: I got this question earlier in a test and we never discussed how you would go about solving such an equation. I did solve the homogeneous equation by writing the characteristic equation $\Lambda^{2}-2\Lambda+1=0$ and getting the two solutions $\Lambda_{1,2}=1$ which of course implied the general solution to the homogeneous equation is of the form $y_h=C_1e^x+C_2xe^x$ and that is where I got stuck. Since the right hand side is not of a good and easy form we learned about, I was lost. Of course, I used Wolframalpha and found the particular solution that was $y_p=xe^xln(x)$ but I still have no idea how to arrive at this solution. An exam is coming up and I would really appreciate all the help I can get. Thank you!


问:
Showing existence of $L: \mathbb{R}^d \to \mathbb{R}^d$ s.t $\langle x,y\rangle^{'} = \langle L(x), y\rangle$

Let $\langle \,\cdot\,,\,\cdot\, \rangle'$ be some inner product in $\mathbb{R}^d$. Show that there is a positive linear operator $L: \mathbb{R}^d \to \mathbb{R}^d$ s.t $\langle x,y\rangle^{'} = \langle L(x), y\rangle$ where $\langle\,\cdot\,,\,\cdot\,\rangle$ is the standard inner product.

I'm not even sure what a positive linear operator means, I know what is a positive definite matrix but from some reading online it seems those are different definitions(?). Help appreciated.

答:
$\langle \cdot, \cdot \rangle$ (respectively $\langle \cdot, \cdot \rangle^\prime$) induces a non degenerate form $B : \mathbb  R^d \times \mathbb R^d \to \mathbb R$ (respectively $B^\prime : \mathbb  R^d \times \mathbb R^d \to \mathbb R$).

$B$ (respectively $B^\prime$) induces a linear isomorphism

$$\begin{array}{l|rcl}
\tilde{B} : & \mathbb R^d & \longrightarrow & \left(\mathbb R^d\right)^\prime \\
    & x & \longmapsto & y \mapsto B(x,y)
\end{array}$$ where $\left(\mathbb R^d\right)^\prime$ stands for the dual of $\mathbb R^d$.

Hence $L = \tilde{B}^{-1} \circ \tilde{B}^\prime$ is a linear automorphism of $\mathbb R^d$. We have $\tilde{B} \circ L = \tilde{B}^\prime$ and therefore

$$\langle x, y \rangle^\prime = B^\prime(x,y) = (\tilde{B}^\prime(x))(y)= ((\tilde{B}\circ L)(x))(y) = B(L(x),y) = \langle L(x), y \rangle$$ for all $x,y \in \mathbb R^d$ as desired.

Moreover

$$\langle L(x), x \rangle = \langle x, x \rangle^\prime >0$$ for all $0 \neq x \in \mathbb R^d$ meaning that $L$ is positive definite.


答:
No. We can use the Dominated Convergence Theorem to solve this. We work in the space of sequences with the counting measure. Notice that $0\leq \frac{\min(k,n)}{n}\leq 1$ so $0\leq 1 - \frac{\min(k,n)}{n}\leq 1$. This is enough to show that the sequence of functions $f_n(k) = \frac{1}{k^2}\Big(1 - \frac{\min(k,n)}{n}\Big)$ are dominated by the integrable function $g(k)=\frac{1}{k^2}$ in the counting measure. Now for a fixed $k$ we have that
$$\lim_{n\to\infty}f_n(k) = \lim_{n\to\infty}\frac{1-\frac{\min(n,k)}{n}}{k^2}=\frac{1}{k^2}$$
Hence by the dominated convergence theorem we have that
$$\lim_{n\to\infty}\sum_{k=1}^\infty\frac{1}{k^2}\Big(1 - \frac{\min(k,n)}{n}\Big)= \sum_{k=1}^\infty \lim_{n\to\infty}\frac{1}{k^2}\Big(1 - \frac{\min(k,n)}{n}\Big) = \sum_{k=1}^\infty\frac{1}{k^2}=\frac{\pi^2}{6}$$

Title: Calculate the limit of the series (measure theory)

Body:
Is this solution correct?

$$\lim_{n\to\infty} { \sum_{k=1}^{\infty} { {1}\over{k^2}} \left(1- {{\min(k,n)}\over{n} }\right)}$$ We consider two cases:
$$\min(k,n)=n$$ we have: $${{1}\over{k^2}} {(1 - 1)} = 0 \to \lim_{n\to \infty}{\sum_{k=1}^{\infty}0}= 0 $$ and for $$\min(k,n)=k $$ we have: $$ 0 \le  { {1}\over{k^2}} \left(1- {{k}\over{n} }\right) \le {{1}\over{k^2}} $$ so $$\lim_{n\to\infty} {\sum_{k=1}^{\infty} {{1}\over{k^2}}} = {{\pi^{2}}\over{6}} $$ The command says to apply the bounded convergence theorem, but this statement applies to the integral and I don't know if my solution is correct.


问:
Darboux Theorem Question
Regarding:

Let $f:(0, \infty) \to \mathbb{R}$ be a differentiable function for which $f'(x)f'(\frac{1}{x})<0$ for every $0<x \ne 1$.

Then $f'(1)=0$.

--

So I am trying to prove this, and I said:

We know that $f$ is differentiable at $(0, \infty)$, so there exists a $0< \delta < 1$ such that $f$ is differentiable at $[1- \delta, 1+\delta]$ .

So there exists a $x_{0} \in [1- \delta, 1+\delta]$ such that $f'(x_{0})f'(\frac{1}{x_{0}})<0$

So by Darboux Theorem, there exists a $x_{0}<c< \frac{1}{x_{0}}$ such that $f'(c)=0$.

But now, I do not know how to link $c$ with $1$...

Any hints?

Thanks! :)

答:
You can also make it more simple without using $\delta, x_0$ if you want. But your method is better practice for general cases.

We know that $f$ is differentiable at $(0, \infty)$, thus $f$ is differentiable at $\left[\frac{1}{1.1},1.1\right]$

By hypostasis: $$f'\left(\frac{1}{1.1}\right)\cdot f'\left(1.1\right)<0$$
By Darboux's Theorem, there exists a $0.91\approx\frac{1}{1.1}<c<1.1$  such that:
$$f'\left(c\right)=0$$

Assume towards a contradiction that $f\left(c\right)\ne1$

Then By hypostasis:
$$0=f'\left(c\right)\cdot f'\left(\frac{1}{c}\right)<0$$

Contradiction!

Title: Darboux Theorem Question


问:
Body bounded with three surfaces
Let B be the body bounded with surfaces:$x^2-y^2=2, x^2-z^2=2, x=2$. Using triple integrals I need to find volume of given body. After projecting this on $OYZ$ plane i get $4$ symmetric parts. And if I take one in the first quadrant, I get bounds $0\le z\le \sqrt{2}, 0\le y\le z$. And I have a problem when it comes to find boundaries for $x$, because we have $3$ surfaces involving $x$.
Should I look which one of the first two is closer to plane $x=2$?

答:
For the volume bound between the hyperbolic cylinders and the plane, please note both $y$ and $z$ can be expressed in terms of $x$. So here are your bounds.

$-\sqrt{x^2-2} \leq y \leq \sqrt{x^2-2}$

$-\sqrt{x^2-2} \leq z \leq \sqrt{x^2-2}$

$\sqrt2 \leq x \leq 2$

Title: Body bounded with three surfaces


问:
Second grade derivate in Galton Watson Process
I'm stuck in a proof inside a galton watsonn process. My goal is to extimate the variance of $Z_n$, where $Z_n$ is the population at time $n$.

I've already given the extimate of my generating function first and second derivate, but I'm stuck where the paper I'm using says:

Now if $X$ is a nonegative integer valued random variable with probability generating function $g$

35. $EX=g'(1)$

36. $VarX=g''(1)+g'(1)-(g'(1))^2$

whenever the quantities on either side of these equations are finite

Can anyone please explain me where those two equations comes our?
Thanks in advance

答:
If $X$ has PGF $g(t)$, then
$$g(t)=\sum_{k=0}^\infty \mathbb P(X=k)\cdot t^k.$$
So for $0<t<1$,
$$g'(t)=\sum_{k=1}^\infty k\cdot\mathbb P(X=k)\cdot t^{k-1}\implies\mathbb E[X]=g'(1^-)$$
and $$g''(t)=\sum_{k=2}^\infty k(k-1)\cdot\mathbb P(X=k)\cdot t^{k-2}\implies\mathbb E[X(X-1)]=g''(1^-).$$
So $\operatorname{Var}(X)=\mathbb E[X(X-1)]+\mathbb E[X]-\mathbb E[X]^2=g''(1^-)+g'(1^-)-\left[g'(1^-)\right]^2$.


问:
Proof on showing $\frac{(b-a)}{2}(f(a) + f(b)) \leq \int_a^b f \leq (b-a) f(\frac{a+b}{2})$ for class $C^2$ function $f$
The task is as follows:

Given:

(a) function $f \in C^2$

(b) $f \geq 0$ and  (c) $f'' \leq 0$ on $[a,b]$

Goal:

Show

$$\frac{(b-a)}{2}(f(a) + f(b)) \leq \int_a^b f \leq (b-a) f(\frac{a+b}{2})$$

To get an understanding of the problem, I tried specific function $f(x) = \sqrt x$ on interval $[ 1, 4 ]$

(1) For the first area (triangle with base $b-a$):

$\frac{(b-a)}{2}(f(a) + f(b))$ = $\frac{(4-1)}{2}(f(1) + f(4))$ = $\frac{9}{2}$

(2) For the second area (integral):

$\int_1^4 \sqrt x$ = $\frac{14}{3}$

I also tried adding up areas of sub-rectangles for this one, using right rectangles.

(3) For the third area = rectangle with length $b-a$ and width $f(\frac{a+b}{2})$ = $4.7$ (approximately)

So the conclusion clearly holds for this specific case.

But I have issue on how to generalize my example >_<

Well, by given information, I break function $f$ into 3 cases:

Case 1: If $f = 0$ i.e: zero function

Then there is nothing to prove, since area is always 0

Case 2: If $f = c$ i.e: constant function

Then the proof is quite easy, since all the 3 areas "shrink" down to be the area of the "big rectangle" with base $b-a$ and width $c$

Case 3: $f$ is convex or concave

This is the part that I don't know how to generalize what I found from my example.

My thoughts:

When I do the first area, I'm dealing with a triangle, thus I'm going below (or exactly on) the function $f$

When I do the second area, I'm thinking about the upper Darboux sum. Thus the sub-rectangles exceed the original curve by some little fractional area, namely the upper left of the rectangles

When I do the third area, I'm also exceed the original curve by some fractional area, but I think this extra part is a bit more than the fractional areas formed by the sub-rectangles.  Or thinking another way, if I double up this rectangle, I get an area which is way bigger than the other two areas.

But then... how should I generalize all these ideas ?

Would someone please help me on this question?

Thank you in advance ^^

答:
This is a nice application of the slogan "convex functions lie above their tangents and below their secants", or rather its mirror image "concave functions lie below their tangents and above their secants".

Specifically this slogan says that for any $x\in[a,b]$ we have

$\frac{f(a)(b-x) + f(b)(x-a)}{b-a}\leq f(x) \leq f\left(\frac{a+b}{2}\right) + f'\left(\frac{a+b}{2}\right)\left(x-\frac{a+b}{2}\right)$.

Integrating $x$ from $a$ to $b$ now gives the result.


问:
Lifting operator on rank-one $uv^T$ matrix
Suppose I have the following rank-one matrix $$uv^T=\begin{bmatrix} u_1\\u_2\\u_3 \end{bmatrix} \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix}$$

Suppose I have the following lifting operator $\mathcal{A}: \mathbb{R}^{3\times 3}\rightarrow \mathcal{S}^{4\times 4}$ $$\mathcal{A}(Y)=\left[\begin{array}{cccc}
Y_{11}-Y_{22}-Y_{33} & Y_{12}+Y_{21} & Y_{13}+Y_{31} & Y_{23}-Y_{32} \\
Y_{12}+Y_{21} & -Y_{11}+Y_{22}-Y_{33} & Y_{23}+Y_{32} & -Y_{13}+Y_{31} \\
Y_{13}+Y_{31} & Y_{23}+Y_{32} & -Y_{11}-Y_{22}+Y_{33} & Y_{12}-Y_{21} \\
Y_{23}-Y_{32} & -Y_{13}+Y_{31} & Y_{12}-Y_{21} & Y_{11}+Y_{22}+Y_{33}
\end{array}\right]$$

Can we say something about the rank and column space of $\mathcal{A}(uv^T)$?  For example, the rank of $\mathcal{A}(uv^T)$ is one.

I believe so far it is obvious that

it is symmetric

trace$(\mathcal{A}(Y))=0$

Thanks!

答:
$\def\R{{\mathbb R}}\def\m#1{ \left[\begin{array}{c}#1\end{array}\right] }$Let
$\epsilon\in\R^{3\times 3\times 3}$ denote the Levi-Civita symbol and
$I\in\R^{3\times 3}$ denote the identity matrix. Use these to define the following scalar, vector, matrix functions of the $Y$ matrix
$$\eqalign{
\alpha &= {\rm trace}(Y) \;=\; I:Y \\
a &= \epsilon:Y \\
A &= Y + Y^T - \alpha I \;=\; A^T\\
}$$
where $(:)$ denotes the double-contraction product.

The following constants will also be useful
$$\eqalign{
z &= \m{0\\0\\0} \qquad
E &= \m{I\\z^T} \qquad
e &= \m{0\\0\\0\\1} \\
}$$
Then your lifting operator can be expressed in a form which lends itself to algebraic manipulation
$$\eqalign{
{\cal A}(Y)
 &= \m{A&a\\a^T&\alpha}
 \;\doteq\; EAE^T + Eae^T + ea^TE + \alpha ee^T \\
}$$
Setting $Y=uv^T$ yields
$$\eqalign{
\alpha &= u^Tv \qquad
a &= u\times v \qquad
A &= uv^T + vu^T - (u^Tv)I \\
}$$
where $(\times)$ denotes the vector cross product.


问:
Definition of homogeneous function
What's the actual definition of homogeneous function? This link says that the scaling factor should be positive https://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/hom/t

Some other resources claim that it can be any real number. Which one is correct and why does it matter?

答:
It depends on the author. I like to call a function $f:V\to W$ between two real vector spaces homogeneous (of degree $1$) if for all $t\in\Bbb{R}$, and all $v\in V$, $f(tv)=tf(v)$.

I would rather reserve the term "positively homogeneous" (of degree $1$) for a function $f:V\to W$ between two real vector spaces such that for all $t>0$ and all $v\in V$, $f(tv)=tf(v)$. But, as I said, it depends on the author whether they want to be picky enough to make the distinction.

The difference between these two concepts can be understood by looking at the following functions $f:\Bbb{R}\to\Bbb{R}$ and $g:\Bbb{R}\to\Bbb{R}$ defined as $f(x)=x$ and $g(x)=|x|$. Then, $f$ is homogeneous of degree $1$, while $g$ is only positively homogeneous (of degree $1$).

Title: Definition of homogeneous function


问:
Right adjoint of the induction functor.
Let $P \in {}_B\text{Mod}_A$ a $B$-$A$ bimodule (for $A, B$ rings). We can consider the induction functor
$$P \otimes_A -:{}_A\text{Mod} \to {}_B\text{Mod}:M \mapsto P\otimes_A M$$
and by the hom-tensor relation we get
$${}_B\text{Hom}(P\otimes_A M, N) \cong {}_A\text{Hom}(N, {}_B\text{Hom}(P, N)).$$
Therefore,
$${}_B\text{Hom}(P, -):{}_B\text{Mod} \to {}_A\text{Mod}:N \mapsto {}_B\text{Hom}(P, N)$$
is the right adjoint of the induction functor. However, in my representation theory course, we are studying Morita context and we saw that ${}_A\text{Mod},{}_B\text{Mod}$ are equivalent additive categories if and only if there is a strict Morita context $(B, A, P, Q, \mu, \tau)$ where $P, Q$ are in ${}_B\text{Mod}_A$, ${}_A\text{Mod}_B$ respectively. In the case that we have an equivalence of additive categories, the functors of the adjunction are exactly given by
$$P \otimes_A -:{}_A\text{Mod} \to {}_B\text{Mod} \quad \text{and} \quad Q \otimes_B -:{}_B\text{Mod} \to {}_A\text{Mod}.$$
My question is the following: In the circumstance of an equivalence of additive categories, do we then have $Q \otimes_B - \cong {}_B\text{Hom}(P, -)$ in a natural way? If the answer is yes, is there any other case where the  hom-tensor can be writen as an induction functor ?

答:
If $P\otimes_A -$ and $Q\otimes_B-$ are quasi-inverse, then this equivalence can be promoted to an adjoint equivalence (this doesn't change the functors, at most the isomorphisms to the identity functor)

In particular, $Q\otimes_B-$ is right adjoint to $P\otimes_A-$, and since right adjoints are unique, it follows that $Q\otimes_B - \cong \hom_B(P-,)$, and in fact uniquely so if you fixed the adjoint equivalence mentioned above and require that the natural isomorphism preserve all the available structure (i.e. if you require that it be an isomorphism "of right adjoints").

To answer your second question, one needs to talk about dualizable objects. Indeed, let me make the context extremely simple for a second: suppose $A=B$ is commutative and $P$ is just an $A$-module (so with the bimodule structure that comes with it). Then if $P$ is a dualizable $A$-module, one has a natural isomorphism $\hom_A(P,M)\cong \hom_A(P,A)\otimes_A M$, in which case you can write everything with tensors ($\hom_A(P,A)$ plays the role of $Q$ in that context).

More generally, Morita theory builds on the following general fact, which isn't restricted to equivalences:

Suppose $F: \mathrm{Mod}_A\to \mathrm{Mod}_B$ is a colimit-preserving functor. Then there is a unique $(B,A)$-bimodule $P$ such that $F\cong P\otimes_A -$

In particular, the right adjoint $\hom_B(P,-) : \mathrm{Mod}_B\to \mathrm{Mod}_A$ is of the form $Q\otimes_B-$ if and only if it preserves all colimits. For this, a number of things are required. Since colimits in $\mathrm{Mod}_A$ are detected in abelian groups, this is equivalent to $\hom_B(P,-) : \mathrm{Mod}_B\to \mathrm{Ab}$ preserving all colimits. In particular, this only depends on the $B$-module structure of $P$; and is in fact equivalent to $P$ being finitely presented and projective as a $B$-module.

Indeed, finitely presented is equivalent to preserving filtered colimits, so it is definitely required, and projective is equivalent to preserving epimorphisms, so it is also required. Conversrly, if you're finitely presented and projective as a $B$-module, you preserve filtered colimits, and finite colimits (finite coproducts and equalizers), so you preserve all colimits.

As a conclusion :

$\hom_B(P,-) \cong Q\otimes_B-$ for some $(A,B)$-bimodule $Q$ if and only if $P$ is finitely generated and projective as a $B$-module.

[note that for projective modules, finitely generated and finitely presented are equivalent]

When $A=B$ is commutative, this is equivalent to dualizability so we recover the example I gave earlier.

Side note:  I'm not sure whether this is in fact usually called an induction functor, in my experience this name is reserved for when it's of the form $B\otimes_A -$, given a ring morphism $A\to B$.

Title: Right adjoint of the induction functor.


问:
How to prove the inequality $\prod_{k=1}^n ((1-\beta)+\beta r_k)\le\left((1-\beta)+\beta\sqrt[n]{\prod_{k=1}^n r_k}\right)^n$
Let $\beta\geq1$, $n\in\mathbb N$ and $r_1,\dots, r_n\geq1-\frac1\beta$ be real numbers.

I want to prove that

Claim. $$\prod_{k=1}^n (1+\beta(r_k-1))\le\left(1+\beta\left(\sqrt[n]{\prod_{k=1}^n r_k}-1\right)\right)^n.$$

My thoughts.

It can be rewritten as
$$\prod_{k=1}^n ((1-\beta)+\beta r_k)\le\left((1-\beta)+\beta\sqrt[n]{\prod_{k=1}^n r_k}\right)^n,$$

which looks like generalized Hölder with the wrong direction, however, since $\beta\geq1$, we have $1-\beta\le0$ and thus classical Hölder doesn't apply.

What could be done to prove this?

Note: Using induction, I can reduce the problem to showing that for all $\beta\geq1, n\in\mathbb N, R\geq\left(1-\frac1\beta\right)^n,r\geq1-\frac1\beta$,

$$((1-\beta)+\beta r) ((1-\beta)+\beta\sqrt[n]R)^n\le((1-\beta)+\beta\ \sqrt[n+1]{Rr})^{n+1},$$

which once again looks like the classical version of Hölder with wrong direction (but again, $1-\beta\le0$).

答:
Notice that the function $f(x)\overset{\text{Def.}}=\log((1-\beta)+\beta e^x)$ is concave since $f''(x)=\frac{\beta(1-\beta)e^x}{((1-\beta)+\beta e^x)^2}\le0$.

By Jensen's inequality, we have
$$\log(\text{LHS})=\sum\limits_{i=1}^n f(\log r_i)\le n f\left(\frac{\sum_{i=1}^n\log r_i}n\right)=\log(\text{RHS}).\quad \blacksquare$$


问:
How many set of integers $a,b,c,d$ are possible such that the polynomial $ax^3+bx^2+cx+d$ equals $1$ at, $x=19$ and $2$ at $x=62$
My approach was to write $f(19) , f(62)$ in the format  where $f(x)= ax^3+bx^2+cx+d$, then I tried to make some pattern by $f(-1)$ and $f(+1)$, please guide after this

答: If $f$ is a polynomial with integer coefficients, then we have $f(x+m)\equiv f(x)\pmod m$ for all integers $x,m$. In particular, $$ f(62)\equiv f(19)\pmod{43}.$$


问:
Slicing a bar in three pieces - probability
I have the following problem: If you split a bar of length 1 in three pieces by choosing two cut points randomly. What is the probability that the lenght of at least one of them is less than 1/5?.

I think that $P(a<1/5)=P(1-b<1/5)=P(b-a<1/5)=1/5$ But now I am confused how to use this in the porpouse of compute the total probability. I would bet that 3/5 is the answer (because you either have that a<1/5 or 1-b<1/5 or a-b<1/5) but it is not crystal clear to me.

答:
In your calculations you have supposed that the pieces have lengths $a$, $b-a$, and $1-b$ and that $a$ and $b$ are independent uniformly distributed between $0$ and $1$. However, this is not the case. If $a$ and $b$ are independent and uniformly distributed between $0$ and $1$, then we do not necessarily have $a<b$. Said another way, if we suppose $a<b$ so the pieces have lengths $a$, $b-a$, and $1-b$, then $a$ and $b$ are not independent and uniformly distributed - we have to condition on $a<b$.

Next, when calculating the probability that at least one event happens, it is often easier to calculate the probability that none of the events happens, and then take the complement. I.e.,
$$P(\text{at least one segment is $<1/5$})=1-P(\text{all segments are $\geq 1/5$}).$$
Let $X$ and $Y$ be the two points at which the bar is cut, with $X$ and $Y$ independent and uniformly distributed on $[0,1]$. Then in order to have all three segments $\geq 1/5$, we require:

$1/5 \leq X \leq 4/5$

$1/5 \leq Y \leq 4/5$

$|X-Y| \geq 1/5$.

Graphically, the set of $(X,Y)$ for which the above holds lies in the intersection of the red, green, and blue shaded regions in this graph. These are two right triangles with leg length $2/5$. Since $(X,Y)$ is uniformly distributed over $[0,1]^2$, the probability of this event is the area of the region where the event occurs. Hence,
$$P(\text{all segments are $\geq 1/5$})= \frac{1}{2}\left(\frac{2}{5}\right)^2+\frac{1}{2}\left(\frac{2}{5}\right)^2=\frac{4}{25}.$$
Thus, $P(\text{at least one segment is $<1/5$})=1-4/25=21/25$.


问:
Given a scalene triangle ABC and D, E, F the middle of BC, CA, AB respectively, prove that certain lines coincide
Given a scalene triangle $\triangle ABC$ and $D,E,F$ the midpoints of $BC, CA, AB$ and $A_1, B_1, C_1$ points such that $\angle AA_1C=90^{\circ}$ and $\angle BB_1C=90^{\circ}$ and $\angle CC_1A=90^{\circ}$. $A_2$ is the intersection of $BC$ and $B_1C_1$. We define $B_2$ and $C_2$ in the same manner. Prove that the lines which from $D,E,F$ which form a right angle with $AA_2, BB_2, CC_2$ respectively, coincide.

We can easily prove that they all pass through the orthocenter $H$ of the triangle, withe use of the Brocard theorem My question is, is there any other way to prove this?

答:
Brokard's theorem is the natural way, but here's a more elementary approach.

Let $(AB_1C_1)$ meet $(ABC)$ again at $G$. The radical axis theorem on $(AB_1C_1)$, $(ABC)$ and $(BCB_1C_1)$ implies that $G$ lies on $AA_2$.

It is known that the reflection $H'$ of $H$ over $D$ is diametrically opposite $A$ on $(ABC)$ (prove it!). So $\angle AGH=\angle AGH'=90^\circ$, which implies $G$, $H$, $D$, $H'$ are collinear.

Similarly, the other two lines also pass through $H$.


问:
Finding basis for vector space of polynomials with complex coefficients

Find a basis for the vector space over $\Bbb C$ of polynomials $p(z)$ of degree at most $4$ with coefficients in $\mathbb C$ satisfying $p(z) = p(−z)$ for all $z\in\mathbb C$ and $p(−2) = 0.$

So I've found a basis but I don't think I'm correct since I only used the first condition. If I set
$$p(z) = p(-z)$$
then I get
$$az^4+bz^3+cz^2+dz+e = az^4-bz^3+cz^2-dz+e$$
$$bz^3+dz=0$$
which suggests that the basis is $\{z,z^3\}$ which seems too simple and does not require the second condition at all. Am I correct or have I made a mistake somewhere? Any help would be appreciated.

答: You made a mistake. If $p(z)=p(-z)$, then $bz^3+dz$ is the null polynomial, and therefore$$p(z)=az^4+cz^2+e.$$and$$p(-2)=0\iff16a+4c+e=0\iff e=-16a-4c.$$So, your space is$$\{az^4+cz^2-16a-4c\mid a,c\in\Bbb C\},$$of which $\{z^4-16,z^2-4\}$ is a basis.


答:
Remember that two polynomials are equal if the coefficient of each $n$th degree term are equal, so the fact that
$$
az^4 + bz^3+cz^2+dz+e = az^4 -bz^3+cz^2-dz+e
$$
Implies not that $bz^3+dz = 0$ (although this also happens to be true), but that $b = 0$ and $d = 0$. So polynomials in your vector space are of the form
$$
az^4 +cz^2+e
$$
Then you can apply your second condition to get constraints on $a,c$ and $e$, but I'll leave that to you.

Title: Finding basis for vector space of polynomials with complex coefficients

Body:
Find a basis for the vector space over $\Bbb C$ of polynomials $p(z)$ of degree at most $4$ with coefficients in $\mathbb C$ satisfying $p(z) = p(−z)$ for all $z\in\mathbb C$ and $p(−2) = 0.$

So I've found a basis but I don't think I'm correct since I only used the first condition. If I set
$$p(z) = p(-z)$$
then I get
$$az^4+bz^3+cz^2+dz+e = az^4-bz^3+cz^2-dz+e$$
$$bz^3+dz=0$$
which suggests that the basis is $\{z,z^3\}$ which seems too simple and does not require the second condition at all. Am I correct or have I made a mistake somewhere? Any help would be appreciated.


问:
Homogeneous polynomials' roots
I'm trying to understand the proof of this result

Let $f(x,y) \in K[x,y]$ be an homogeneous polynomial s.t. $deg(f) = d>0$ then $\exists$ at most d $(a,b) \in K^2$ non-trivial roots of $f$

Proof:

$f(x,y)=\sum_{i=0}^{d} c_ix^iy^{d-i}$

$deg(f)>0 \implies \exists c_i \neq 0 \:$. Since f is homogeneous, $f(a,b)=0 \implies f(ta,tb)=0 \ \forall t \in K$

B.C. $(1,0)$ is not a root $\implies c_d \neq 0$. Then $f(t,1)$ is an homogeneous polynomial in one variable, with $deg=d$ $\implies$ it as at most $d$ roots

$f(t,1)=a_0\prod_{i=1}^{d}(t-a_i)$. Let's say $\ t = \frac{x}{y}$, then

$f(t,1)=f(\frac{x}{y},1)=a_0\prod_{i=1}^{d}(\frac{x-a_i}{y})$ but $f(\frac{x}{y},1)=y^df(x,y)$, which has a root in $(1,0)$ (contradiction)

So $(1,0)$ must be a root. Let $r<d$ be its multiplicity, then
$c_d = c_{d-1} = ... = c_{d-r+1}=0$

$f(x,y)=\sum_{i=0}^{d-r} c_ix^iy^{d-i}=y^{r}\sum_{i=0}^{d-r}c_ix^iy^{d-r-i}$

And that's how it ends.

I can't understand how this works:

1)I don't think it's necessary that $f(x,y)$ has a root in $(1,0)\ $ (for example $f(x,y)=x^d+y^d$ doesn't have it), so why he has followed this line? Also, he substitutes $t$ with $\frac{x}{y}$ but weren't we considering $f(t,0)$? Is this legit?

2)How does this proves the thesis?

Thank you for help

答:
Thanks to RandyMarsh, think I can close the question now, if the following should be correct. Here is the adjusted statement

Let $f(x,y)\in K[x,y]$ be an homogeneous polynomial s.t. $deg(f) = d>0$, then $\exists$ at most $d\ (a,b) \in K^2\ projective\ non-trivial \ roots\ [(a,b) \neq (0,0)]$ of $f$, counted with their multiplicity.

Proof:

$f(x,y)= \sum_{i=0}^{d} c_ix^iy^{d-i}\\deg(f)=d>0 \implies \exists c_i \neq 0\\ \text{Since } f \text{ is homogeneous, } f(a,b)=0 \implies f(ta,tb)=0\  \forall t \in K\\ \text{We have two cases} \\
\text{1) }(1,0)\text{ is not a root}\implies c_d \neq 0 \text{. Then } f(t,1) \text{ is an one-variable polynomial of } deg = d \implies \text{ it has at most d roots}\\
2)(1,0) \text{ is a root. Let } r \leq d \text{ be its multiplicity, then } c_d = c_{d-1} = \dots = c_{d-r+1} = 0 \\
\implies f(x,y) = y^r \cdot g(x,y) \text{ where g is homogeneous with } deg(g)=d-r \implies g(1,y) \text{ is an one-variable polynomial of } deg=d-r \text{, so it has at most } d-r \text{ roots.}$.

I considered two distinct cases because i didn't see the point assuming that $(1,0)$ must be a root.

Title: Homogeneous polynomials' roots


问:
How do you calculate the probability of simultaneous events?
How do you calculate the probability of simultaneous events?  As in, given four simultaneous events each with a 10% probability, what are the odds that ONE of them occurs?  Obviously it isn't 40%, because...well, if you have ten events that probability clearly isn't 100%!

(Nope, not homework!  Video games, probability of elemental effects from a given spell)

答:
I just want to emphasize that the problem has NOT asked (at least not explicitly) for the probability of the occurrence of "at least" one of the events. What the question implies is the occurrence of "exactly" one of them, and in that case, as @ncmathsadist has said, it would be a Bernoulli experiment, i.e., an experiment with a sequence of independent trials, each only having "two" possible outcomes, e.g., occured or not-occured.

To calculate the probability of k successes(e.g., occured) in n trials in a Bernoulli experiment we would use this formula famously known as the binomial distribution: ${n \choose k} * p^k * (1-p)^{n-k}$ where p is the probability of the success of each single trial. (Note that also the logic behind this formula is fairly simple, it is summing up all the probabilities for k successfull trials among all the n independent trials.)

Now in this problem, $k$ is $1$, $n$ is $4$ and $p$ is $0.1$, so the final answer would be very easily achievable, it will give: $29\%$.

Also if what the OP really needs is the $p(at \ least \ one \ occurrence)$, it would be $1-p(no \ occurrences)$, which again contains a Bernoulli experiment, and can be calculated using the binomial distribution, which would give: $1 - {4 \choose 0}*0.1^0*0.9^4 = 1 - 0.9^4 = 34\%$ as @Vasil suggested.


问:
Finding $c_i$s that maximize $\sum_{i=1}^{r}c_i^2\sigma_i^2$ where $\sigma_1^2 \ge \sigma_2^2 \ge ...$ and $\sum_{i=1}^{r}c_i^2 =1$
I can explain the solution of the above equation in plain English, but I don't know how to explain it mathematically. It is obvious that because $\sigma_1^2$ is greater then other $\sigma_i^2$s, if we set $c_1=1$ and $c_i=0$ for all $i \ne 1$, the convex function is maximized. But could you please guide me on how to express this mathematically?

Different approaches are all welcome!

This question is from the book "Foundations of Data Science" by Blum, Hopcroft, Kannan (page 71, Exercise 3.24).

答: Substitute $x_i = c_i^2$, then the objective is to maximize a linear function over the standard simplex $\{ x : x_i \geq 0, \sum_i x_i = 1\}$. Since the feasible region is bounded, there is an optimal solution at an extreme point of the feasible region. The extreme points are the standard basis vectors $e_i$ (with a $1$ at position $i$ and zeroes elsewhere). The objective value for $x=e_1$ is at least as high as for other extreme points, so $x=e_1$ is optimal.

OwnerDisplayName: Arash Ashrafzadeh


问:
Changing bounds of double integral while converting from cartesian to polar form

Evaluate the  integral $\displaystyle \int_{0}^{\sqrt{π}} \int_{y}^{\sqrt{π}} \cos(x^2) \ \mathrm{d}x \, \mathrm{d}y $

I have replaced $x^2$ with $r^2\cos^2(θ)$ but could not change the bounds by using inequalities.

答:
If it is obligatory desire to use polar coordinates, then  $x=r \cos \phi, y= r \sin \phi$ gives
$$\left\{\begin{array}{l}
0 \leqslant x \leqslant \sqrt{\pi} \\
0 \leqslant y \leqslant x
\end{array}\right\} \to \left\{\begin{array}{l}
0 \leqslant r \cos \phi \leqslant \sqrt{\pi} \\
0 \leqslant r \sin \phi \leqslant r \cos \phi
\end{array}\right\}$$

First of equations gives restriction for radius $r \leqslant \frac{\sqrt{\pi} }{\cos \phi}$ and second for angle $\phi \in [0, \frac{\pi}{4}]$, so we have

$$ \int\limits_{0}^{\sqrt{π}} dx \int\limits_{0}^{x} \cos(x^2)dy=\int\limits_{0}^{\frac{\pi}{4}} \int\limits_{0}^{\frac{\sqrt{\pi} }{\cos \phi}}r\cos (r^2 \cos^2 \phi) drd\phi$$

Though, changing order of variables, without polar coordinates, seems to be more good way, but boss is Boss.


答:
A geometrical approach

which should be quite clear

Title: Slicing a bar in three pieces - probability

Body:
I have the following problem: If you split a bar of length 1 in three pieces by choosing two cut points randomly. What is the probability that the lenght of at least one of them is less than 1/5?.

I think that $P(a<1/5)=P(1-b<1/5)=P(b-a<1/5)=1/5$ But now I am confused how to use this in the porpouse of compute the total probability. I would bet that 3/5 is the answer (because you either have that a<1/5 or 1-b<1/5 or a-b<1/5) but it is not crystal clear to me.


问:
Obtain the conditional distribution of $X$ given $X^2=t$
Consider the model $X \sim N(\theta, 1)$ for $\theta \in \mathbb R$. Obtain the conditional distribution of $X$ given $X^2=t$ for $t\ge 0$.

Hint: This is a discrete distribution.

Current work:
$f(X=x|X^2=t)=\displaystyle \frac{f(X=x, X^2=t)}{f(X^2=t)}=\frac{f(X=x, X^2=t)}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}=\frac{f(X=\sqrt t)}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}=\frac{\frac{1}{\sqrt {2\pi}}e^{-(\sqrt t-\theta)^2/2}}{\frac{1}{2\sqrt{2\pi t}}(e^{(\sqrt t-\theta)^2/2}+e^{(\sqrt t + \theta)^2/2})}$? But this can't be right because there's no $x$ in the answer. What does one do here?

答: I'm wondering if this conditional distribution is even well$-$defined. Let $\epsilon_1,\epsilon_2$ be very small positive numbers and $t>0$ be arbitrary. Define $I_1=(-\sqrt{t}-\epsilon_1,-\sqrt{t}+\epsilon_1)$ and $I_2=(\sqrt{t}-\epsilon_2,\sqrt{t}+\epsilon_2)$. Assume $\epsilon_1, \epsilon_2$ are small enough to make $I_1,I_2$ disjoint. Notice how $$P\Big(X\in I_1 \Big|X\in I_1 \cup I_2\Big)=\frac{\int_{-\sqrt{t}-\epsilon_1}^{-\sqrt{t}+\epsilon_1}f_{X}(x)dx}{\int_{-\sqrt{t}-\epsilon_1}^{-\sqrt{t}+\epsilon_1}f_{X}(x)dx+\int_{\sqrt{t}-\epsilon_2}^{\sqrt{t}+\epsilon_2}f_{X}(x)dx}\approx \frac{\epsilon_1 f_{X}\big(-\sqrt{t}\big)}{\epsilon_1 f_{X}\big(-\sqrt{t}\big)+\epsilon_2 f_{X}\big(\sqrt{t}\big)}$$ One would expect that the above expression approaches $P\Big(X=-\sqrt{t}\Big|X^2=t\Big)$ as $(\epsilon_1,\epsilon_2)\rightarrow 0$ but such a limit doesn't exist. I think the most natural conditional pmf would be $$P(X=s|X\in S)=\frac{f_{X}(s)}{\sum_{x\in S}f_{X}(x)}$$ where $X\sim f_{X}$ is any continuous random variable supported on $\mathbb{R}$, $S\subseteq \mathbb{R}$ is any finite set, and $s\in S$ is fixed, but you could repeat a similar argument above and see something weird happening. FYI: This is not intended to be an answer, but my response is too long for a comment.


问:
Can maths functions "return" multiple different mathematical objects?
In computer science, there are statically and dynamically typed languages, with statically typed languages generally being stricter about types of variables, function parameters, etc. To what extent can maths be considered "statically typed" (obviously this isn't the correct term but hopefully you understand what I'm asking) and how strict is it?

For example, do mathematical functions always have to return the same mathematical object (e.g. number, vector, etc.), or can the "type" (to use a computer science term) vary depending on the arguments it receives? I was wondering whether something like this makes any sense:
$$
    f(x)=
\begin{cases}
    x + 2& \text{if } x > 1\\
    [1, 2, 3]              & \text{otherwise}
\end{cases}
$$ i.e. $f(x)$ can return a number or a vector.

答:
can the "type" (to use a computer science term) vary depending on the arguments it receives

That is typically what a dependent type theory allows, with dependent product types and many other constructions.

That does however not mean that a dependent type cannot be verified "statically": in a constructive type theory, type checking is usually decidable.


问:
If $p(n)$ be the $n$-th prime number, is $\lim_{n\to \infty} {p(n)}^\frac{1}{n}$ exist and how we can find it if exist?
In my research, I need a counterexample of sequences, and $p(n)$, the $n$- prime number sequence is an important sequence to me. Generally for an arbitrary sequence of the real and positive numbers, such as $a(n)$,
any classification of the $\lim_{n\to \infty} {a(n)}^\frac{1}{n}$ will be useful to me.

However, according to my calculation, this limit must be greater than $1.76151$.

答: The Prime Number Theorem implies $p(n) \sim n \log n$, so $p(n)^{1/n} \to 1$.


问:
Definition of projection ( as in Serre's Linear representation of finite group)
I'm reading through Serre's "Linear Representations of Finite Groups," and I'm a bit confused by what's probably a fairly minor point.

Serre defines the direct sum (internal) $V=W\oplus W'$,
For $x\in V$,the mapping $p(x=w+w')=w$ is the projection map of $V$ onto $ W$ associated with $V=W\oplus W'$; the image of $p$ is $W$ and $p(x)=x$ for all $x\in W$. Conversely, if $p$ is any way satisfying the above two condition; Then we can show that $V$ is the direct sum of $W$ and the kernel of $p$. A bijective correspondence is thus established between the projections of $V$ onto $W$ and the complements of $W$ in $V$.

What is the definition of projection serre uses here?

Suppose if we assume
"the mapping $p(x=w+w')=w$ is the projection map of $V$ onto $ W$ associated with $V=W+W'$" as definition.

Then by definition itself we have a bijection between projection and complement, As any projection requires $V$ to be divided into direct sum of $W$ and it's complement. In that case one doesn't need the converse statement he mentioned.

So I'm wondering what is the definition of projection serre assumes here?

[Of course if one use equivalent statement of projection like "the image of $p$ is $W$ and $p(x)=x$" or $p^2=p$.  Then one need converse part given by serre.]

I think my question is very trivial but I couldn't figured out

答:
If $W$ is a subspace of a linear space $V$, a projection from $V$ onto $W$ is by definition a linear map $p: V \rightarrow W$ such that $p(x) = x$ for all $x \in W$.  A priori this has nothing to do with internal direct sums or complements.

Note that such a map is automatically surjective, and its kernel is a complement to $W$ in $V$, which is to say that $V$ is the internal direct sum of $W$ and $\operatorname{Ker} p$.

In this way we get a mapping

$$p \mapsto \operatorname{Ker} p$$

$$\textrm{projections from $V$ onto $W$} \rightarrow \textrm{complements of $W$ in $V$}$$

Conversely, if $W'$ is a complement to $W$ in $V$, then every $v \in V$ can be uniquely written as $w+w'$ for $w \in W$ and $w' \in W'$, and we can define a projection $p: V \rightarrow W$ from $V$ onto $W$ by $p(v) = w$.  In this way we can get a mapping

$$W' \mapsto p$$
$$\textrm{complements of $W$ in $V$} \rightarrow \textrm{projections from $V$ onto $W$}$$

and these two mappings are inverse to one another.


答:
The $n$th prime is roughly $n\log n$ therefore
$$\underset{n\to \infty }{\text{lim}}(n \log n)^{1/n}=1$$

Title: If $p(n)$ be the $n$-th prime number, is $\lim_{n\to \infty} {p(n)}^\frac{1}{n}$ exist and how we can find it if exist?

Body:
In my research, I need a counterexample of sequences, and $p(n)$, the $n$- prime number sequence is an important sequence to me. Generally for an arbitrary sequence of the real and positive numbers, such as $a(n)$,
any classification of the $\lim_{n\to \infty} {a(n)}^\frac{1}{n}$ will be useful to me.

However, according to my calculation, this limit must be greater than $1.76151$.


答:
A function $f: A \to B$ maps elements of a domain set $A$ to elements of a domain set $B$.

As a set can be any collection of objects, it doesn't actually make sense to talk of objects being different "types".

Consider the $B = \{2, \pi, [1,3,7], fish, Sherlock\ Holmes, invisible\ green\ thoughts\}$.

$2$ is a natural number, $\pi$ is a transcendental irrational real number, $[1,3,7]$ is a vector, $fish$ is a group of animals, $Sherlock\ Holmes$ is a fictional character, $invisible\ green\ thoughts$ is a hypothetical idea of Noam Chomsky.  Are these the same type of object?

The answer is a emphatic:  YES!  The are all objects of the type-- Belonging to set $B$.

And $f: \mathbb N \to B$ via $$f(n) = \begin{cases} 2&\text{if }n\text{ is prime}\\\pi&\text{if }n\text{ is even composite}\\ [1,3,7]&\text{if }n\text{ is an odd perfect square}\\Sherlock\ Holmes&\text{otherwise}\end{cases}$$ is a perfectly valid (non-surjective) function that does what a good function should; it maps every element of a specific set ($\mathbb N$) to a distinct element of a specific set ($B$).

Title: Can maths functions "return" multiple different mathematical objects?

Body:
In computer science, there are statically and dynamically typed languages, with statically typed languages generally being stricter about types of variables, function parameters, etc. To what extent can maths be considered "statically typed" (obviously this isn't the correct term but hopefully you understand what I'm asking) and how strict is it?

For example, do mathematical functions always have to return the same mathematical object (e.g. number, vector, etc.), or can the "type" (to use a computer science term) vary depending on the arguments it receives? I was wondering whether something like this makes any sense:
$$
    f(x)=
\begin{cases}
    x + 2& \text{if } x > 1\\
    [1, 2, 3]              & \text{otherwise}
\end{cases}
$$ i.e. $f(x)$ can return a number or a vector.


问:
compute$\iint_D xy \ dxdy$
The question is:
$$\iint_D xy \ dxdy, \quad D=1\leq x^2+y^2\leq2, \ \ \ x^2\leq y\leq x^2+1, \ x\geq0 , \ y\geq0$$
I've tried this:
$$ u=x^2+y^2 , \  v= y-x^2 , \ \  |J|=\frac{1}{2x(1+2y)}$$
$$\frac{1}{2}\iint \frac{y}{1+2y} \ dudv$$
But how should i solve for $y$ in terms of $u$ and $v$?
Any suggestion would be great, thanks

Title: compute$\iint_D xy \ dxdy$

Body: Since $y=v+x^2=u+v-y^2$ has unique positive root $y=\tfrac{\sqrt{4(u+v)+1}-1}{2}$, $\tfrac{y}{1+2y}=\tfrac12\left(1-\tfrac{1}{\sqrt{4(u+v)+1}}\right)$. Now we evaluate$$\int_1^2du\int_0^1dv\tfrac12\left(1-\tfrac{1}{\sqrt{4(u+v)+1}}\right)=\tfrac{1-A}{2},\,A:=\int_1^2du\int_0^1dv\tfrac{1}{\sqrt{4(u+v)+1}}.$$Modulo possible mistakes in my arithmetic you should check for,$$\begin{align}A&=\int_1^2du[\tfrac12\sqrt{4u+4v+1}]_0^1\\&=\tfrac12\int_1^2du[\sqrt{4u+5}-\sqrt{4u+1}]\\&=\tfrac{1}{12}[(4u+5)^{3/2}-(4u+1)^{3/2}]_1^2\\&=\tfrac{13^{3/2}-54+5^{3/2}}{12},\end{align}$$so your original integral was $\tfrac{11}{4}-\tfrac{13^{3/2}+5^{3/2}}{24}$.


答:
Since you intuitively know the answer, you could show that any other combination is always less than or equal to it. This, in turn, is equivalent to saying that your intuitive solution is a maximum.

So, your solution is obtained by setting $c_1^2=1, c_i =0,\ \forall i\neq 1$, and in that case, your result is simply $\sigma_1^2$.

Now, lets consider another arbitrary combination. Because $\sigma_1^2 \geq \sigma_2^2 \geq ... $, you have

$$c_1^2 \sigma_1^2 + c_2^2 \sigma_2^2 + ... + c_r^2 \sigma_r^2 \leq c_1^2 \sigma_1^2 + c_2^2\sigma_1^2 + ... + c_r^2\sigma_1^2 = \sigma_1^2 \left(\sum_{i=1}^{r}c_i^2 \right) = \sigma_1^2$$

Thus, $$c_1^2 \sigma_1^2 + c_2^2 \sigma_2^2 + ... + c_r^2 \sigma_r^2 \leq \sigma_1^2$$ which is what we wanted to show.

PS. A complete different story would be to analize how many ways to obtain such a maximum exist. But, I think that anlysis is beyond the scope of this question.

Title: Finding $c_i$s that maximize $\sum_{i=1}^{r}c_i^2\sigma_i^2$ where $\sigma_1^2 \ge \sigma_2^2 \ge ...$ and $\sum_{i=1}^{r}c_i^2 =1$

Body:
I can explain the solution of the above equation in plain English, but I don't know how to explain it mathematically. It is obvious that because $\sigma_1^2$ is greater then other $\sigma_i^2$s, if we set $c_1=1$ and $c_i=0$ for all $i \ne 1$, the convex function is maximized. But could you please guide me on how to express this mathematically?

Different approaches are all welcome!

This question is from the book "Foundations of Data Science" by Blum, Hopcroft, Kannan (page 71, Exercise 3.24).


问:
No of possible binary sequences with at most 2 consecutive zeroes
The problem:

How many such sequences of ten $1's$ and ten $0's$ are possible such that they don't contain three or more consecutive zeroes? (Eg. $00101001001101110011$)

I know there are $20!/10!10!$ total sequences. How do I filter out the ones which have more than two consecutive zeroes somewhere?

答:
This answer is based upon the Goulden-Jackson Cluster Method.

We consider the words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{0,1\}$$ and the set $B=\{000\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $f(s)$ with the coefficient of $s^n$ being  the number of searched words of length $n$.

According to the paper (p.7) the generating function $f(s)$  is
\begin{align*}
f(s)=\frac{1}{1-ds-\text{w}(\mathcal{C})}\tag{1}
\end{align*}
with $d=|\mathcal{V}|=2$, the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{w}(\mathcal{C})=\text{w}(\mathcal{C}[000])
\end{align*}

We calculate according to the paper
\begin{align*}
\text{w}(\mathcal{C}[000])&=-(sz)^3-sz\cdot \text{w}(\mathcal{C}[000])-(sz)^2\cdot\text{w}(\mathcal{C}[000])\tag{2}\\
\end{align*}
where we additionally mark occurrences of zeros with $z$. We obtain
\begin{align*}
\text{w}(\mathcal{C})=\text{w}(\mathcal{C}[000])=-\frac{(sz)^3(1-(sz))}{1-(sz)^3}
\end{align*}

It follows from (1) and (2)
\begin{align*}
f(s)&=\frac{1}{1-ds-\text{w}(\mathcal{C})}\\
&=\frac{1}{1-(1+z)s+\frac{(sz)^3(1-sz)}{1-(sz)^3}}\\
&=\frac{1+sz+s^2z^2}{1-s-zs^2-z^2s^3}\\
&=1 + (z+1)s + (z+1)^2 s^2 + (3z^2+3z+1)s^3 +\cdots\\
&\qquad+ \left(z^{14}+112z^{13}+\cdots+\color{blue}{24\,068}z^{10}+\cdots+20z+1\right)s^{20}+\cdots\\
\end{align*}
The last line was calculated with some help of Wolfram Alpha. The coefficient of $z^{10}s^{20}$  shows that out of $2^{20}=1\,048\,576$ binary words of length $20$ from the alphabet $\{0,1\}$ there are $\color{blue}{24\,068}$  words containing ten $0$s and ten $1$s which do not contain $000$.


问:
loring tu proposition 23.4 help
Suppose $f:U\rightarrow \mathbb{R}$ is continuous, $U$ is open in $\mathbb{R}^n$ and $suppf\subseteq U$ is compact. Then $f$ is Riemann integrable on $U$.

Put $k=suppf$.

The proof starts of by saying: "Being continuous on a compact set, $f$ is bounded."

However, wouldn't this imply that the function $f|_{K}$, is bounded, as opposed to $f$?

Obviously, by Heine Borel, $suppf$ is closed and bounded in $\mathbb{R}^n$. Now Loring attempts to show that the extension of $f$; $\hat{f}$, is continuous on $\mathbb{R}^n$. This is easy to show, but then he sais that this implies (by Lebesgue's theorem) that $f$ is integrable on $U$. My point is that, by Lebesgue's theorem, $f|_{K}$ is integrable, not necessarily $f$.

What am I missing?

答:
$K=$supp$f$ is a compact set and $f(x)=0$ for $x \in U \setminus K$ by the definition of support.  https://en.wikipedia.org/wiki/Support_(mathematics)

If the ball $B(x,R)$ contains $K$ then the integral of $f$   on $U \cap B(x,R)$ does not depend on $R$ or $x$, so you can define the integral of $f$ on $U$ as this common value.

Title: loring tu proposition 23.4 help


问:
A confusion about the remainder theorem
today, we were taught the polynomial remainder theorem and to prove it, the following identity was written in the board by our teacher.

$$f(x) = g(x)(x-A) + R ----------->(1)$$

Here, f(x) and g(x) are polynomials with f(x) being the dividend and (x-A) being the divisor. R is a constant with real value. (Please excuse my English for I have studied mathematics in my native language)

However, since I have knowledge in simplifying fractions, I thought that the same could be written as below,

$$f(x) / (x-A)  = g(x) + R / (x-A) -------->(2)$$

In this case, we can not define the identity at x=A. (Or divide the polynomial at x=A for that matter). However we get the correct results from the connection I have marked as (1) above. How is this possible?

To further elaborate my question with a count example, we can do the following,

$$(x^3 + 6x^2 + 11x + 6) / (x+4) = x^2+2x+3-6/(x+4) ------->(3)$$

We know that we can not divide the polynomial at x = -4 in the connection I have marked as (3). However, we can write it as,
$$(x^3 + 6x^2 + 11x + 6) = (x^2+2x+3-6)(x+4) - 6 -------->(4)$$

and then find the remainder by substituting x = -4

Can someone kindly explain to me how come this is possible?

答:
So you're actually getting at a slightly subtle concept here: there is a distinction between polynomials as abstract objects and polynomials as functions. I'll answer your question from both perspectives:

Polynomials as abstract objects. In this case, we're working in a "world" where polynomials are the only things that exist. In that case, we can't write $f(x)/(x - A)$, since that isn't a polynomial (unless it turns out that $f(x)$ is a multiple of $x - a$, in which case we can write it, and everything works out fine, it just turns out that $R = 0$). Since we're dealing with polynomials as abstract objects here, the "what about when $x = A$ problem simply never comes up, because $x$ isn't a variable taking values in the real numbers - it's an extra element that we're adding to the real numbers so that we can write our polynomials down. When we do polynomial division and the like, we're often implicitly taking this approach, even if we don't explicitly say so.

Polynomials as functions. In this case, we can write $f(x)/(x - A)$, unless $x = A$. Whenever $x$ is not $A$, that's just a number divided by a non-zero number, so it's an honest-to-goodness fraction. When $x = A$, however, that's division by zero, so it is not the case that these two things are equivalent. Rather, those two things are equivalent everywhere except at $x = A$, which is often good enough for what we're doing. And, indeed, in this case, we only actually need (1), not (2), so there's no issue.

There is also another approach here: we can treat polynomials as functions taking values in either the one-point compactification of the reals. That sounds fancy, but really it just means "we can just stick in either some extra thing called $\infty$, and whenever our function would be undefined, we'll just go ahead and say it takes that new value. Then everything works out fine (for what we're doing here): we do indeed have $$\frac{f(x)}{x-A} = g(x) + \frac{R(x)}{x - A}$$ everywhere, it just so happens that at one point, those two things are both $\infty$.

Note that you can also translate between the two quite easily: we can do whatever work we can/need to with abstract polynomials, then shift over to treating them as functions when we need to evaluate them, and nothing goes wrong.

Title: A confusion about the remainder theorem


问:
Necessary and sufficient conditions for a bijection to exist
Suppose we have two sets $A$ and $B$. Suppose there exists injections $i_1:A\to B$ and $i_2:B\to A$. Does there exist a bijection from $A$ to $B$? Does a bijection exist if and only if $A$ and $B$ are finite?

This is a question I have been thinking about for a while and cannot make much progress on it. What, if any, would be the necessary and sufficient conditions on the sets $A,B$ or functions $i_1,i_2$ for there to exist a bijection from $A$ to $B$? Any tips or help would be appreciated.

答: This is Schröder–Bernstein theorem.</p>

Body: This is Schröder–Bernstein theorem.


问:
Probability that three segments can form a triangle. (NOT A DUPLICATE)
I have read a problem which says:

"Consider a segment of length $10$. Points $A$ and $B$ are chosen randomly such that $A$ and $B$ divide the segment into three smaller segments. What is the probability that the three smaller segments could form the sides of a triangle?"

I have checked the solution explanation. It was TOTALLY convenient to me, (simply begin with the triangle inequality which states that the sum of the lengths of any two sides must be greater than the length of the third side). Very well understood.

Here is the solution (taken from the book):

However, I was trying to solve it by another way, there must be a mistake, but I did not find where is that mistake:

THE WRONG SOLUTION:

Let us assume that the three segments are $x,y,$ and $z$.

Now we have $8$ possibilities to have inequalities that relate $x+y$ with $z$, relate $x+z$ with $y$, and relate $y+z$ with $x$. These $8$ possibilities are:

The possibility (V) is the only valid possibility. Hence the required probability is $\frac{1}{8}$.

One may say because of duplicate counting, if that is the mistake, please clarify where is that duplicate.

Your help would be appreciated. Thanks.

答:
As noted already in a comment, for three non-negative numbers $a,b,c,$ an inequality of the form $a+b<c$ implies that $c$ is the unique greatest number among the three,
that is, $c>a$ and $c>b.$
It is impossible to have more than one unique greatest number within a list of three numbers, so all the cases with more than one $<$ relation in them are impossible.

The cases with exactly one $<$ are I ($z$ is greatest), II ($x$ is greatest), and VI ($y$ is greatest).

It is also possible to have a greatest side that is not greater than the sum of the other two; this is case V.

So we have four cases, one of which forms a triangle, so you might jump to the conclusion that the probability is $\frac14.$ But let's not be so hasty!

We cannot determine a probability from these four cases because we have not shown any reason to believe that they are equally likely.
And indeed there is no reason to believe the four cases are equally likely
until we have explicitly specified how we "randomly" choose to break the stick.
There are at least two well-known ways to randomly break a stick in three pieces, and they give different answers.

One way to break a stick in three pieces is to choose a number uniformly between $0$ and $10,$ measure that distance from the left end of the stick, and mark point $A$ at the measured distance.
Then choose another number uniformly between $0$ and $10,$ independently from the first number, and mark point $B$ at that measured distance from the left end of the stick.
This can be represented by the sample space
$\{(u,v) \in \mathbb R\mid 0\leq u\leq 10, 0\leq v\leq 10\},$
which is a square.

Another way to break a stick in three pieces is first to break the stick in two pieces as in the procedure above and then choose one of those two pieces at random and break it at a point chosen uniformly along the length of that piece.
If $u$ is the distance where the first break is made and $v$ is the distance where the second break is made then the sample space again is at least the square
$\{(u,v) \in \mathbb R\mid 0\leq u\leq 10, 0\leq v\leq 10\}.$

A third way is to choose a number $x$ uniformly between $0$ and $10,$ break the stick at that distance from the left end; then choose a number $y$ uniformly between $0$ and $10 - x,$ and from the piece of length $10 - x$ measure a distance $y$ from one end and break the stick there.
In other words, we follow the second procedure described above, but after the first break we always take the "right-hand" piece to break again instead of choosing randomly between the two pieces.
This can be represented by the sample space
$\{(x,y) \in \mathbb R\mid 0\leq x\leq 10, 0\leq y\leq 10 - x\},$
which is the same as
$\{(x,y) \in \mathbb R\mid x + y \leq 10, x \geq 0, y \geq 0\},$
which is essentially the sample space shown in the book.
(I have $\leq$ where the book has $<$, but the entire set of extra outcomes allowed by $\leq$ has probability zero, so this makes no difference in the answer.)
It is not hard to show by symmetry that this procedure results in a triangle with the same probability as the second procedure.

Now I find the answer in the book to be very fishy,
because it constructs a sample space that looks like my third example
(and not like the first example),
represents that sample space graphically as a triangle,
and then assumes that the probability distribution over the triangle is uniform
without explaining why we should believe it is uniform.
In fact, if the stick is broken randomly according to the third procedure I gave,
the distribution over the triangle is not uniform
and the probability is not $\frac14$.

The probability of a triangle is $\frac14$ if we use the first procedure to break the stick randomly. This is proved in an answer to a previous question (for a stick of length $1$).
You might notice that this proof rests on a graphical visualization that looks quite a bit like the one in the book, except that its sample space is a square rather than a triangle and the pattern of subregions within the triangle is duplicated.
We actually have a valid reason to think the distribution over the square is uniform
(because it was produced by choosing each coordinate uniformly and independently between $0$ and the length of the stick).


问:
Can I apply a Leibniz integration rule to $\frac{d}{dr} \int_{-\infty}^{\infty} \int_{h(r)}^\infty \int_{g(r)}^\infty {rf(x,y,z)\, dz\, dy\, dx}$?
Suppose I need to compute the derivative
$$
\frac{d}{dr} \int_{-\infty}^{\infty} \int_{h(r)}^\infty \int_{g(r)}^\infty {rf(x,y,z)\, dz\, dy\, dx}
$$
Can I apply a Leibniz integration rule of some form? How?

答:
Effectively, the below constitutes a derivation of the Leibniz Integral Rule as applicable to the current setting.

Assuming sufficient conditions to allow for interchanging integration and differentiation, and existence of the integrals, you can.

Perhaps the most insightful way is to introduce some auxiliary functions. Define:

$$\begin{align}
G(x,y,b,p) &:= \int_b^\infty pf(x,y,z)\,\mathrm dz \\
H(x,b,c,p) &:= \int_c^\infty G(x,y,b,p)\,\mathrm dy \\
K(b,c,p) &:= \int_{-\infty}^\infty H(x,b,c,p)\,\mathrm dz
\end{align}$$

We want to compute $\dfrac{\mathrm d}{\mathrm dr}K(g(r),h(r),r)$, and start by the general chain rule:

$$\dfrac{\mathrm d}{\mathrm dr}K(b,c,p) = \frac{\partial K(b,c,p)}{\partial b}\frac{\mathrm db}{\mathrm dr}+\frac{\partial K(b,c,p)}{\partial c}\frac{\mathrm dc}{\mathrm dr}+\frac{\partial K(b,c,p)}{\partial p}\frac{\mathrm dp}{\mathrm dr}$$

In this expression, the quantities $\dfrac{\mathrm db}{\mathrm dr},\dfrac{\mathrm dc}{\mathrm dr}$ and $\dfrac{\mathrm dp}{\mathrm dr}$ are easily computed from $b = g(r), c=h(r),p=r$.

For the partial derivatives, we compute as follows:

$$\begin{align}
\frac{\partial K(b,c,p)}{\partial b} &= \int_{-\infty}^\infty \frac{\partial}{\partial b}H(x,b,c,p)\,\mathrm dz\\
&= \int_{-\infty}^\infty \int_c^\infty \frac{\partial}{\partial b}G(x,y,b,p)\,\mathrm dy\,\mathrm dz
\end{align}$$

Now what is $\dfrac{\partial}{\partial b}G(x,y,b,p)$? It is $-pf(x,y,b)$, as we see from the following computation:

$$\begin{align}
\lim_{h\to0} \dfrac{G(x,y,b+h,p)-G(x,y,b,p)}{h} &= \lim_{h\to 0} \frac1h \left(\int_{b+h}^\infty pf(x,y,z)\,\mathrm dz-\int_b^\infty pf(x,y,z)\,\mathrm dz\right)\\
&= \lim_{h\to0} \frac1h \left(-\int_b^{b+h}pf(x,y,z)\,\mathrm dz\right)\\
&= -pf(x,y,b)
\end{align}$$

where in the last step, the Fundamental Theorem of Calculus was used (since the penultimate line comes down to the expression of the derivative of a $z$-primitive of $pf(x,y,z)$ at $z = b$).

The other two partial differentiations can be executed in a similar fashion.


问:
Finding Probability Using a Normal Distribution Curve
If $X\sim N(y,\sigma^2)$ and $\mathbb P(X<x)=k$, find $\mathbb P(X<2y-x)$ in terms of $k$.

Note that $y$ is the mean and $\sigma$ is the standard deviation.

I have tried expressing each probability using the standard normal distribution, but I just end up with two terms that don’t help me,$Z<\frac{(x-y)}\sigma$ and $Z<\frac{(y-x)}\sigma$. I’m not sure where to go from here. If you can offer any advice, thanks.

答:
Let $X = y + \sigma Y$, where $Y \sim N(0,1)$.

We know that $k = \mathbb P(X<x) = \mathbb P(Y < \frac{x-y}\sigma)$, therefore
$$\mathbb P(X < 2y-x) = \mathbb P(Y < \frac{y-x}\sigma) = 1- \mathbb P(Y \geq \frac{y-x}\sigma) = 1 - \mathbb P(Y \leq \frac{x-y}\sigma) = 1 - k$$


问:
Find $K_{3,4}$ subgraph in a graph
I'm trying to find $K_{3,4}$ subgraph in the following graph:

I'm trying for hours however I can't find it. Can you please help me?

答:
We can partition the vertices of the complete bipartite graph $K_{3,4}$ into two disjoint sets: $V_4$ that consists of the three vertices of degree $4$ and $V_3$ that consists of the four vertices of degree $3$.

Now label the eight vertices of the graph pictured in the question clockwise starting from the top: $N$, $NE$, $E$, $SE$, $S$, $SW$, $W$ and $NW$. Note that $N$, $E$, $S$ and $W$ have degree $3$ and $NE$, $SE$, $SW$ and $NW$ have degree $5$. Therefore, every vertex in $V_4$ can only be mapped to one of $NE$, $SE$, $SW$ and $NW$.

Choose any three of $NE$, $SE$, $SW$ and $NW$ and call this set $U_4$. Now, note that the vertices in $U_4$ form a triangle. This means that for each $v \in U_4$ two out of its five edges connect to another vertex in $U_4$. This leaves only three edges to connect to vertices outside of $U_4$. Therefore $U_4$ cannot correspond to $V_4$.

Consequently, $K_{3,4}$ cannot be found in the pictured graph.

Title: Find $K_{3,4}$ subgraph in a graph


问:
Show that this set is denumerable
Let $A=\mathbb{Z}[x]_{deg\leq n}$ denote the set of all polynomials in $x$ with degree less than or equal to $n$. Show that A is denumerable.

I know that set of all polynomials with integer coefficients is countable but how can I prove this problem ?

答: $A$ is in bijection with $\mathbb Z^{n+1}$ which is countable as a finite product of countable sets.

Title: Show that this set is denumerable


问:
How to simplify $(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha})\cdot \cos\alpha \cdot \cos\beta$?
There is this problem:
$$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{1}{\tan\alpha -\tan\beta}$$
I started as $$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{\sin2\alpha \cos\beta-\sin2\beta \cos\alpha}{2(\cos\beta -\cos\alpha)}$$ but I'm stuck here, because I don't see how could I use $\sin(x-y)$ but also don't see how could I use any other identity without complicating this even more.

答:
The proposed identity is not true (even though after the correction).

Indeed, take $\alpha = \pi/4$ and $\beta = 0$. Then we obtain:
\begin{align*}
\left(\frac{\sin(\pi/4) - \sin(0)}{\cos(0) - \cos(\pi/4)}\right)\times\cos(\pi/4)\cos(0) = \frac{\sqrt{2}}{2 - \sqrt{2}}\times\frac{\sqrt{2}}{2} = \frac{1}{2-\sqrt{2}}
\end{align*}

On the other hand, one has that
\begin{align*}
\frac{1}{2(\tan(\pi/4) - \tan(0))} = \frac{1}{2}
\end{align*}

EDIT

The proposed equation is equivalent to
\begin{align*}
\frac{(\sin(a) - \sin(b))\sin(a-b)}{\cos(b) - \cos(a)} = 1
\end{align*}


问:
What does the decomposition, weak union and contraction rule mean for conditional probability and what are their proofs?
I was reading Koller's book on Probabilistic Graphical Models and was wondering what the decomposition, weak union and contraction properties of conditional probability mean.

But before I ask exactly what I am confused about let me introduce some of Koller's notation so that we are all in the same page (anything else is unclear feel free to ask in the comments). Let capital non-bold stand for random variables say $X$ is a r.v. Let little non bold stand for the assignment to a random variable say $(X = x)$. Also, let me define captial bold letters as sets of random variables. For example $\textbf{X}, \textbf{Y}, \textbf{Z}$ are three sets of random variables. Let small bold letters denote assigments to these sets $\textbf{x}, \textbf{y}, \textbf{z}$ i.e. it denotes assigments of values to the variables in these sets. Let $Val(\textbf{X})$ be the values that the set of random variables can take.

Now, these are the properties/theorems I am trying to prove and understand:

Decomposition:

$$( \textbf{X} \perp \textbf{Y}, \textbf{W} \ | \ \textbf{Z}) \implies (\textbf{X} \perp \textbf{Y} \  | \ \textbf{Z})$$

Weak union:

$$(\textbf{X} \perp \textbf{Y}, \textbf{W} \ | \ \textbf{Z}) \implies (\textbf{X} \perp \textbf{Y} \ | \ \textbf{Z}, \textbf{W})$$

Contraction:

$$(\textbf{X} \perp \textbf{W} \ | \ \textbf{Z}, \textbf{Y}) \ \& \ (\textbf{X} \perp \textbf{Y} \ | \ \textbf{Z}) \implies (\textbf{X} \perp \textbf{Y}, \textbf{W} \ | \ \textbf{Z})$$

Intersection:

$$(\textbf{X} \perp \textbf{Y} | \textbf{Z}, \textbf{W}) \ \& \  (\textbf{X} \perp \textbf{W} | \textbf{Z}, \textbf{Y}) \implies (\textbf{X} \perp \textbf{Y}, \textbf{W} | \textbf{Z})$$

Let's take the first statement. I think its a notational problem (though, not sure). Does the first statement mean, "X is conditionally independent of Y and W Given Z"? i.e. is the first statement the same as $(\textbf{X} \perp (\textbf{Y},\textbf{W}) \ \perp \ \textbf{Z})$. If that is true, then would the first statement imply two things:

$$(\textbf{X} \perp \textbf{Y}, \textbf{W} \ | \ \textbf{Z}) \implies (\textbf{X} \perp \textbf{Y} \  | \ \textbf{Z})$$

and

$$(\textbf{X} \perp \textbf{Y}, \textbf{W} \ | \ \textbf{Z}) \implies (\textbf{X} \perp \textbf{W} \  | \ \textbf{Z})$$

?

I actually intended to prove them as an exercise to myself, however, not being sure if I understood the notation or not made it pretty hard to even attempt a proof (and it also made it hard to try to understand the intuition behind each of the statements, an important thing I wanted to also do, understand it intuitively). Providing one proof as an example and explaining the notation might be good enough for me as an example so that I can attempt the other too.

Bounty Section

Now that I have tried to prove them, I have had more difficulties than I expected, I wanted to see a proof for each one as I was unable to find them on the internet.

I tried proving the first one and this is what I have so far:

we want an expression for $(\textbf{X} \perp \textbf{Y} \mid \textbf{Z}, \textbf{W} )$ using property $(\textbf{X} \perp \textbf{Y}, \textbf{W} \mid \textbf{Z} )$ so lets consider:

$P(\textbf{X},\textbf{Y} | \textbf{Z}, \textbf{W}) = \frac{P(\textbf{X} ,\textbf{Y} , \textbf{Z} , \textbf{W} )}{P(\textbf{Z} , \textbf{W})} = \frac{P(\textbf{Z})P(\textbf{X}, \textbf{Y} , \textbf{W} \mid \textbf{Z})}{P(\textbf{Z} , \textbf{W})}$

Now we can use the property we need by noting that $P(\textbf{X}, \textbf{Y} ,\textbf{W} \mid \textbf{Z}) = P( \textbf{X} | \textbf{Z} )P(\textbf{Y} , \textbf{W} \mid \textbf{Z})$ (due to assumption/property $(\textbf{X} \perp \textbf{Y}, \textbf{W} \mid \textbf{Z})$), thus:

$\frac{P(\textbf{Z})P(\textbf{X},\textbf{Y},\textbf{W} \mid \textbf{Z} )}{P(\textbf{Z}, \textbf{W})} = \frac{P(\textbf{Z})P(\textbf{X}|\textbf{Z})P(\textbf{Y} , \textbf{X} | \textbf{Z} )}{P(\textbf{Z} , \textbf{W})}$

and that was as far as I got for Weak Union. I couldn't get it to be the same as:

$P(\textbf{X},\textbf{Y}|\textbf{Z},\textbf{W}) = P(\textbf{X}|\textbf{Z},\textbf{W})P(\textbf{Y}|\textbf{Z},\textbf{W})$

答:
was wondering what the decomposition, weak union and contraction properties of conditional probability mean.

For those discovering this thread in their quest for intuitive meanings of these axioms, I would recommend the opening of Chapter 3 in Pearl's Probabilistic Reasoning in Intelligent Systems: Networks of Plausible Inference, and particularly Figure 3.1:

Graphical interpretation of the axioms governing conditional independence

View set $Z$ as "informationally separating" $X$ from $Y$. Then Decomposition expresses the idea that if $Z$ separates $X$ from the compound set $S = Y \cup W$ then it separates $X$ from every subset of $S$. Or, in other words, if two combined items of information are judged irrelevant to $X$, then each item is irrelevant to $X$ separately as well.

The Weak Union axiom states that if $Y$ is irrelevant to $X$ then finding some additional item $W$ from the same space of those irrelevant to $X$ (separated by $Z$) and augmenting the separator with it will not render $Y$ relevant to $X$ as a result.

And so on: Contraction is for reducing the size of the separating set, etc.


问:
change of order in summation
I'm reading Rudin PMA Theorem 8.3, and I have one question in mind. In the process of the proof, Rudin says

How can we justify $$\sum_{i=1}^{\infty}\sum_{j=1}^{n}a_{ij} = \sum_{j=1}^{n}\sum_{i=1}^{\infty}a_{ij}$$?

Is it possible to change the order of summation if only one of the summations is infinite?

答:
You have to assume that for each $j\in\{1,\dots, n\}$, the following sum exists $\sum\limits_{i=1}^{\infty}a_{ij}$. Then, for every $k\in\Bbb{N}$, we have (if you want to be really strict this follows by induction on the associativity and commutativity of addition)
\begin{align}
\sum_{i=1}^k\sum_{j=1}^na_{ij}=\sum_{j=1}^n\sum_{i=1}^ka_{ij}.
\end{align}
On the RHS, by assumption, for each $j\in\{1,\dots, n\}$, the limit $\lim\limits_{k\to\infty}\sum_{i=1}^ka_{ij}$ exists (afterall that was my first sentence). Thus, we can now use the fact that "limit of sum is sum of limits" (again strictly speaking that theorem is valid for sum of two sequences, so you would need induction for the case of $n$ summands) to get that
\begin{align}
\lim_{k\to\infty}\sum_{i=1}^k\sum_{j=1}^na_{ij} &=\sum_{j=1}^n\lim_{k\to\infty}\sum_{i=1}^ka_{ij}.
\end{align}
Or in other words,
\begin{align}
\sum_{i=1}^{\infty}\sum_{j=1}^na_{ij}&=\sum_{j=1}^n\sum_{i=1}^{\infty}a_{ij}.
\end{align}

Title: change of order in summation


问:
Joachimsthals Notations and tangents of conic at a point
I was going through Cut-the-Knot's "Joachimsthal's Notations" page  discussing Joachimsthal's indice method of studying conics.

The conic can be represented in a following way:

$$ s_{ij} = Ax_i x_j + B(x_i y_j + x_j y_i) + C y_i y_j + F(x_i + x_j) + G(y_i + y_j) + H=0$$

Where the $i$ denotes a pair of number $(x_i,y_i)$ and $j$ also as
$(x_j, y_j)$

It is a theorem that if $(x_i,y_i)$ is a point which lies on the
conic, then ,

$$s_i = Ax_i x + B (x_i y + x y_i) + Cy_i y + F(x_i + x) + G(y_i + y) +H=0 \tag{1}$$

Where I have let the pair of $(x_j,y_j)$ to be set as free variable

Denotes the equation of tangent of the conic $$ s= Ax^2 +2Bxy + Cy^2 + 2Fx + 2Gy +H$$

At the point $(x_i,y_i)$

My question is regarding a step in the proof of this,

I don't get what exactly is the quantity benig plugged into (1) is, it seem the point has arguement $P(x_1,x_1)$ which doesn't make sense to me. I think the correct expression should be :

$$ P(x,y)= tP(x_1,y_1) + (1-t)P(x_2,y_2)$$

Hence the point being plugged in is given as:

$$ Q = (tx_1 + (1-t)x_2 , ty_1 + (1-t)y_2) $$

However, this seems to be still algebra bashing to move from (10) to (11), is there a more elegant approach to it?

答:
This isn't particularly elegant, but it helps bring some order to the chaos of the symbol-bashing.

Introduce $u:=1-t$ to add a touch of symmetry to the linear combination of points:
$$Q = t P(x_1,y_1) + u P(x_2,y_2) = \left(t x_1 + u x_2, t y_1 + u y_2\right) \tag1$$
Also, rewrite the quadratic expression by throwing-in convenient aliases of $1$:
$$s = A x^2 + 2 B x y + C y^2 + 2 F (t+u) x + 2 G (t+u)y + H(t+u)^2 \tag2$$
Substituting $Q$ into this form of $s$ gives an expression we can write as
$$\begin{array}{rll}
A&(\color{red}{t x_1}+\color{blue}{u x_2})&(\color{red}{t x_1}+\color{blue}{u x_2}) \\
+2B&(\color{red}{t x_1}+\color{blue}{u x_2})&(\color{red}{t y_1}+\color{blue}{uy_2}) \\
+\phantom{2}C&(\color{red}{t y_1}+\color{blue}{u y_2})&(\color{red}{t y_1}+\color{blue}{u y_2}) \\
+2F&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;)&(\color{red}{tx_1}+\color{blue}{u x_2}) \\
+2G&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;)&(\color{red}{ty_1}+\color{blue}{uy_2}) \\
+\;H&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;)&(\color{red}{t}\;\;\;+\color{blue}{u}\;\;\;)
\end{array} \tag{3}$$
with which it's "easy" to gather terms by $\color{red}{t^2}$, $\color{blue}{u^2}$, and $\color{red}{t}\color{blue}{u}$ (without even explicitly expanding everything) to find
$$t^2 s_{11} + u^2s_{22} + 2 tu\, s_{12} \tag4$$
Then, Cut-the-Knot's equation $(11)$ is just the above $(4)$ after recalling that $u=1-t$. $\square$

Note. One can re-skin expression $(3)$ using the matrix form of the conic equation
$$X^TMX+NX+H$$
where
$$M:=\begin{bmatrix}A& B\\ B&C\end{bmatrix}\qquad N:=\begin{bmatrix}2F& 2G\end{bmatrix}\qquad X:=\begin{bmatrix}x\\ y\end{bmatrix}$$
Substituting-in $X=tP_1+uP_2$ (where $P_i:=(x_i,y_i)$) gives
$$(\color{red}{tP_1^T}+\color{blue}{uP_2^T})M(\color{red}{tP_1}+\color{blue}{uP_2})\;+\;N(\color{red}{t}+\color{blue}{u})(\color{red}{tP_1}+\color{blue}{uP_2})\;+\;H(\color{red}{t}+\color{blue}{u})(\color{red}{t}+\color{blue}{u}) \tag{3'}$$
where, again, it's "easy" to gather terms and get to $(4)$.

This still isn't particularly elegant, but it's a little more compact.


问:
How are the left inverses of an injection, surjective?
Consider the injection $f:{{0,1,2}} \to {0,1,2,3}:x\to x $

Now consider the left inverse of this function. How exactly is this a surjection if not all the elements in the codomain are mapped back?

答:
The given function $f$ has exactly 3 left inverses.

Note that a left inverse $g$ of $f$ must be a function $\{0,1,2,3\}\to\{0,1,2\}$ so that $g(f(x))=x$ for all $x\in\{0,1,2\}$.

So, we must have $g(0)=0,\ g(1)=1,\ g(2)=2$, and also $g(3)\in\{0,1,2\}$, which actually can be arbitrary.

Because of the equation $g(f(x))=x$ we see that indeed any such $g$ must be surjective, since for arbitrary $x$ in the codomain of $g$ (=domain of $f$) we have an element, specifically e.g. $f(x)$ which $g$ carries into $x$.


问:
How can you solve this Bessel's equation?
From what I've learned, the Bessel's equation in its standard form is
$$x^2y''+xy'+(x^2+v^2)y=0\tag{1}$$
But, for example I have this equation
$$x^2y''+xy'+\frac{1}{4}xy=0\tag{2}$$
How can you solve the 2nd equation by using the Bessel's standard form?

答:
Since WolframAlpha trivializes solving this specific ODE, I'll work in the opposite direction to show how such Bessel-like ODEs arise. Suppose we start with the modified Bessel equation $$z^2y''(z)+zy'(z)+(z^2+v^2)y(z)=0$$
with linearly independent solutions $y(z)=I_0(z),K_0(z)$. We now consider the effect of the substitution $z=bx^a$. Writing $D_z:=d/dz$ and $D_x:=d/dx$, the effect of the chain rule is that

$$zD_z = z\frac{dx}{dz} D_x =z\cdot a z^{a-1} D_x=a x D_x.$$
If we apply these operators a second time, the product rule yields

\begin{align}
(z D_z)^2&=zD_z+z^2 D_z^2,\\( ax D_x)^2&=a^2 x D_x+a^2 x^2 D_x^2.
\end{align}

This is convenient, because the first two terms of the modified Bessel equation can therefore be rewritten as

\begin{align}
z^2 y''(z)+z y'(z)
&=[z D_z+z^2 D_z^2]y(z)\\
&=(z D_z)^2 y(z)\\
&=(a x D_x)^2 y(x)\\
&=a^2 x^2 y''(x)+a^2 x y'(x).
\end{align}

Carrying out the substitution in the third term as well, the ODE now takes the form

$$a^2 x^2 y''(x)+a^2 x y'(x)+(b^2 x^{2/\alpha}+v^2)+v^2)y(z)=0.$$

If we divide through by $a^2$ and introduce $A:=b^2/a^2,B:=v^2/a^2,r:=2/a$ we obtain

$$x^2y''(x)+x y'(x)+(Ax^r+B)y(x)=0.$$

Thus we obtain something that looks very similar to the modified Bessel equation, but with $x^2+v^2$ replaced by a linear function of $x^r$. In the present case, this amounts to choosing $B=0,A=1/4,r=1$ and thus $v=0,a=1/2,b=1$. This justifies the solution set $y(x)=I_0(\sqrt{x}),K_0(\sqrt{x})$ given by WolframAlpha.


答:
Unless you find a better change of variable, my suggestion is to integrate it based on the sketch. It is split into two integral and is really not that bad.

$\displaystyle \int_a^1 y \ \bigg[ \ \int_{\sqrt{1-y^2}}^{\sqrt y} x \ dx \ \bigg] \ dy $ + $\displaystyle \int_1^b y  \ \bigg [ \int_{\sqrt{y-1}}^{\sqrt {2-y^2}} x \ dx \ \bigg ] \ dy$

$a$ is the value of $y$ at intersection of $y = x^2, x^2 + y^2 = 1 \implies y^2+y-1=0$. This gives you $a = \frac{\sqrt 5 - 1}{2}$

$b$ is the value of $y$ at intersection of $y = x^2 + 1, x^2 + y^2 = 2 \implies y^2+y-3=0$. This gives you $b = \frac{\sqrt 13 - 1}{2}$.

And the correct answer is ${\frac{36+5\sqrt5 - 13 \sqrt{13}}{48}}$

Title: compute$\iint_D xy \ dxdy$

Body:
The question is:
$$\iint_D xy \ dxdy, \quad D=1\leq x^2+y^2\leq2, \ \ \ x^2\leq y\leq x^2+1, \ x\geq0 , \ y\geq0$$
I've tried this:
$$ u=x^2+y^2 , \  v= y-x^2 , \ \  |J|=\frac{1}{2x(1+2y)}$$
$$\frac{1}{2}\iint \frac{y}{1+2y} \ dudv$$
But how should i solve for $y$ in terms of $u$ and $v$?
Any suggestion would be great, thanks


问:
Universal Quantifier $\forall$ and Generalized Conjunction $\bigwedge$ in intuitionistic logic
I have a question about $\forall xA(x)$ and $\bigwedge\! A(a_i)$ (= $A(a_1)\wedge...\wedge A(a_n)$). In classical logic, the universal statement $\forall xA(x)$ can be understood as a generalized conjunction of $A(a_i)$ if the universe of discourse is finite, say, $\{a_1,...,a_n\}$. I wonder if this generalizes to intuitionistic logic?

The concrete problem that I meet is as follows. I assume that this generalizes, then we have
$$\forall xA(x)\leftrightarrow\bigwedge\! A(a_i).\quad\quad(*)$$
However, putting $\neg\neg$ before $\forall xA(x)$, $\neg\neg\forall xA(x)\rightarrow\forall x\neg\neg A(x)$ is intuitionistically valid, but the converse is not. But putting $\neg\neg$ before $\bigwedge\! A(a_i)$, $\neg\neg\bigwedge\! A(a_i)\leftrightarrow\bigwedge\! \neg\neg A(a_i)$ (note: both directions) is intuitionistically valid. Why is there such a difference? Does this signal that $(*)$ does not hold in intuitionistic logic? Thanks!

答:
In general, $(*)$ is the wrong way to think about the universal quantifier. In both classical and intuitionistic logic, the power of the universal quantifier $\forall x$ is that it allows $x$ to range over an arbitrary, unspecified domain. There is no sense in which $(*)$ is a valid sentence in either classical or intuitionistic logic.

If you want to restrict your semantics to a fixed finite domain, you need to introduce an axiom. Let $\varphi$ be the sentence: $$\forall x\, \bigvee_{i=1}^n (x = a_i).$$ Then we indeed have $$\varphi\vdash (\forall x\,\psi(x)\leftrightarrow \bigwedge_{i=1}^n \psi(a_i))$$ for all formulas $\psi(x)$, both classically and intuitionistically. But also $$\varphi\vdash \forall x\, \lnot\lnot \psi(x) \leftrightarrow \lnot \lnot \forall x\, \psi(x)$$ for all formulas $\psi(x)$, both classically and intuitionistically. So your concerns vanish in this restricted setting.


问:
How do I find the matrix of a linear transformation from $\Bbb R^3 \longrightarrow \Bbb R^3$ of an orthogonal reflection on a planoe
So the questions is, " Find the matrix of the linear transformation of n orthogonal projection on the plane $2x-y+z=0$"

I don't know how I should tackle this problem. Any advice?

答:
Let $n=(2,-1,1)$.  Then the projection of a vector $v$ on the plane is given by
$$v-\langle v,\frac{n}{\|n\|}\rangle\frac{n}{\|n\|}
=v-\frac{\langle v,n\rangle}{\|n\|^2}n
=v-\frac16\langle v,\begin{pmatrix}2\\-1\\1\end{pmatrix}
\rangle\begin{pmatrix}2\\-1\\1\end{pmatrix}.$$
To get the desired matrix replace $v$ by the canonical basis vectors.


问:
Problem with infinite series of residues
I tried solving the integral
$$ I := \int_{-\infty}^{\infty} \frac{x \exp(\mu x)}{\exp(\nu x)-1}\,dx,~~\text{where}~~\text{Re}(\nu)>\text{Re}(\mu) >0$$

using the calculus of residues. I found the  first order poles and calculated the residues
$$ z_n = \frac{2 n \pi i}{\nu},~~n \in \mathbb{N}\setminus \lbrace{ 0 \rbrace},~~\text{Res}_{z_n} f = \frac{2 n \pi i}{\nu^2} \exp\left(2 n \pi i \frac{\mu}{\nu} \right).$$
When I now sum over all residues to get the result for the integral I encounter a series
$$ I = \dots = - \frac{4 \pi^2}{\nu^2} \sum_{n=1}^{\infty} n \exp\left(2 \pi i n \frac{\mu}{\nu} \right) = \dots = \left( \frac{\pi}{\nu} \csc\left( \frac{\pi \mu}{\nu} \right) \right)^2,$$
which I can evaluate by integrating the summands w.r.t $n$, then using the formula for the geometric series and differentiating w.r.t. $n$ again. This gives me the correct result (according to Mathematica). However to use the formula for the geometric series I have to assume that
$$ \left|\exp\left( 2 \pi i  \frac{\mu}{\nu} \right)\right| =\left|\exp\left( -2 \pi \text{Im}\left(\frac{\mu}{\nu} \right)\right)\right|<1$$
and that does not hold for all $\mu,\nu$ fulfilling $\text{Re}(\nu)>\text{Re}(\mu)>0$. Why does this approach give me the correct result although the series appears to diverge for some $\mu,\nu$ and is there some other way to solve this integral by hand?

答:
You could split up the integral into two geometric series

$$\int_{-\infty}^\infty \frac{xe^{\mu x}dx}{e^{\nu x}-1} = \sum_{n=1}^\infty \int_0^\infty xe^{(\mu-n\nu)x}dx - \sum_{n=0}^\infty \int_{-\infty}^0 xe^{(\mu+n\nu)x}dx$$

$$ = \sum_{n=-\infty}^\infty \frac{1}{(\mu-n\nu)^2} = \frac{\pi^2}{\nu^2}\csc^2\left(\frac{\pi\mu}{\nu}\right)$$