问:
Conditional entropy of linear transformation of random variables
Let $X,Y,Z$ be $n-$dimensional random vectors such that $Z=AX+AY$ where $A$ is a $n\times n$ linear transformation matrix. Let $\det(A)=1$ where $\det(A)$ stands for determinant of $A$. Let $X\perp \!\!\! \perp Y$ and let $H(X)$ be the entropy of $X$. Then:

Is this true: $H(AX|Z)=H(AX|AX+AY)=H(X|X+Y)\tag{1}$?

I know that $H(AX)=H(X)+\log(\det(A))$. So I was thinking of writing conditional entropy in terms of total entropy:

\begin{align*}
H(AX,Z) &= H(Z) + H(AX|Z) \text{ By chain rule } \\
&= H(AX+AY) + H(AX|AX+AY) \\
\implies H(AX|AX+AY) &= H(AX,AX+AY)-H(AX+AY) \\
&= H(AX,AX+AY)-H(X+Y)-\log(\det(A)) \\
&= H(AX,AX+AY)-H(X+Y) \text{ Since $\det(A)=1$}
\end{align*}

But I am not sure if the joint entropy $H(AX,AX+AY)=H(X,X+Y)$? If the previous line is true, then $(1)$ follows. Any ideas?

答:
This is true quite generically. Let $(X,Z)$ be an arbitrary pair of random variables with a reasonably nice joint density $p_{XZ}$. Recall that $$ h(X|Z) = -\int p_{XZ}(x,z) \log \frac{p_{XZ}(x,z)}{p_Z(z)} \mathrm{d}x \mathrm{d}z$$

By the standard transformation of random variables, for any invertible $A$, if $(U,V) = (AX, AZ) = B(X,Z),$ where $B = \mathrm{diag}(A,A)$ is a block diagonal matrix with blocks $A$, then $$ p_{UV}(u,v) = |\det B|^{-1} p_{XZ}(A^{-1} u, A^{-1}v).$$ Further, due to block diagonality, observe that $\det B = \det^2 A$.

By an idential calculation, $p_V(v) = |\det A|^{-1} p_Z(A^{-1}v)$

Thus, \begin{align}
h(U|V) &= -\int p_{UV}(u,v) \log \frac{p_{UV}(u,v)}{p_V(v)} \mathrm{d}u \mathrm{d}v \\
&\overset{(u,v) = B(x,z)}{=} -\int p_{UV}(Ax, Az) \log \frac{p_{UV}(Ax, Az)}{p_V(Az)} |\det B| \, \mathrm{d}x \mathrm{d}z\\
&= -\int p_{XY}(x,z) \log \frac{p_{XY}(x,z) |\det B|^{-1}}{p_Z(z) | \det A|^{-1}}
\mathrm{d}x \mathrm{d}z \\ &= h(X|Z) + \log |\det A|,\end{align} where I've used that $|\det B| = |\det A|^2$.

Your case follows on setting $X = X$ and $Z = (X+Y)$ and $\det A = 1$ in the above calculation (I chose slightly wrong notation in the beginning, because I thought that you defined $Z = X+Y$ instead of $Z = A(X+Y)$. There was too much to edit when I realised the notational snafu, so we get some mismatch. My apologies).

Notice that we made no assumptions about the joint law of $(X,Z)$ (besides regularity assumptions like $p(x,z)/p(z)$ have integrable logarithm), or about $A$ besides the fact that it is invertible. All that is really happening in this situation is that you're blowing up two random vectors with $A$, which is changing the common scale of the problem by a factor of $A$.


问:
If the distances between every pair of airports is distinct, prove that more than $5$ planes can't land at any airport

The distances between any two airports of a country are distinct. From every airport one airplane takes off and lands at the closest airport. Prove that there doesn't exist an airport where more than 5 planes will land.

I proved it in the following way:

In order for a plane to take off from a certain airport and land at another, it means that the other airport is the closest to it at a distance $r$. Hence it has a circle with radius r with center the airport which it took off from and with a point on its circumference, the airport at which it landed. If there are more than $5$ points which land at a certain airport, then you get a polygon $A_1A_2A_3...A_n$ with a point inside it $O$. If $n\gt 5$ then we have that at least one $\angle OA_iA_{i+1}\lt60^o$ which is indeed a contradiction.

This is what I managed to do for this question, but I am not completely certain I am completely correct. Could you please verify my solution and explain any other nice approaches and also explain to me why my last statement that $\angle OA_iA_{i+1}\lt60^o$ indeed is a contradiction as I worked it out only intuitively?

答:
Here are airports:

We're considering the red (at the center).  Take a nearest neighbor (blue) airport a distant $R$ from the center, whose plane must land at red.  Take the "next" blue.  We want it to also land at red but not at the other blue.  Thus it a distance $R$ from red but must be as close to the first blue as possible (so we can pack as many other airports as possible).  Assume for the moment we're allowed to put it a distance $R$ from the first blue.  (We can't, but continue reading...)  We can continue like this to get the hexagonal packing in the figure.

BUT....

the problem states the airports distances must all be different.  Thus the angle from the red center to at least two "neighboring" blues must be larger than $\pi/3$ ($60^\circ$).  We thus cannot pack all six blue neighboring airports.

Of course we can pack $5$.

QED.

ContentLicense: CC BY-SA 4.0


问:
Digit sum of multiples of 999
I have checked with Maple: The digit sum of the numbers $999\cdot k$ ($k = 1,...,  100000$; the first $100000$ multiples of $999$) is always one of the numbers $27, 36, 45, 54$. Does this pattern continue? Is it true for all $k$?
It is clear that such a digit sum is divisible by 9, and I have already read that it must be at least 27, but why should it not go beyond 54?

答: Use bigger numbers. In fact, if we choose $k$ to be numbers of the form $1001001001...$ then we get numbers of the form $999999999..$ so we should be able to get arbitrarily large strings of $9$s

Title: Digit sum of multiples of 999


问:
How to solve $e^{ix} = i$
I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.

Thanks.

答:
$$e^{ix} = i$$
Euler's formula:
$$e^{ix}=\cos(x)+i\sin(x),$$
so:
$$\cos x+i\sin x = 0+1\cdot i$$
compare real and imaginary parts
$$\sin x =1$$ and $$\cos x =0$$
$$x=\dfrac{(4n+1)\pi}{2}\;\;,n \in \mathbb W$$
(W stands for set of whole number $W=${$0,1,2,3,.......,n$}).</p>

ContentLicense: CC BY-SA 3.0

Title: How to solve $e^{ix} = i$


问:
Finding the Tree for Prufer Code
This is more of a fun question that I came up with. How would you go about building the tree for the prufer code:

$P(t) = 122333444455555666666777777788888888999999999(10)(10)(10)(10)(10)(10)(10)(10)(10)(10)$

Title: Finding the Tree for Prufer Code

Body:
I would take the easy way out and use Wolfram Alpha:

ResourceFunction["PruferCodeToLabeledTree"][{1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10}]

Resulting in


问:
Estimating the error function using Monte Carlo integration
How do I perform Monte Carlo integration on the error function

$erf(x) = \frac{2}{\sqrt(\pi)}\int_{0}^{x}e^{-t^2}dt$

I have to estimate this using python but I'm not sure how the integral works at all. Can anyone help?

答:
Monte Carlo integration is a technique used to estimate integrals. The key is to write the integrand as the product of a probability density function and some other function so that the integral effectively becomes an expected value problem.  In your case:

$$
\operatorname{erf}(x) = \frac{2 x}{\sqrt\pi}\int_{0}^{x}e^{-t^2}/x\,\mathrm dt=\frac{2 x}{\sqrt\pi}\mathsf E(\exp(-X^2)),
$$

where $X\sim\operatorname{Uniform}(0,x)$. Hence, we may estimate the error function by sampling $n$ psudeo-random variates from the $\operatorname{Uniform}(0,x)$ density and then computing
$$
\widehat{\operatorname{erf}(x)}=\frac{2x}{n\sqrt\pi}\sum_{k=1}^n\exp(-x_i^2)
$$

In MATLAB we may implement this as

```<code>data=rand(1e6,1);
erfEst=@(x) 2*x/sqrt(pi)*mean(exp(-(x*data).^2));
</code>```

Body:
Monte Carlo integration is a technique used to estimate integrals. The key is to write the integrand as the product of a probability density function and some other function so that the integral effectively becomes an expected value problem.  In your case:

$$
\operatorname{erf}(x) = \frac{2 x}{\sqrt\pi}\int_{0}^{x}e^{-t^2}/x\,\mathrm dt=\frac{2 x}{\sqrt\pi}\mathsf E(\exp(-X^2)),
$$

where $X\sim\operatorname{Uniform}(0,x)$. Hence, we may estimate the error function by sampling $n$ psudeo-random variates from the $\operatorname{Uniform}(0,x)$ density and then computing
$$
\widehat{\operatorname{erf}(x)}=\frac{2x}{n\sqrt\pi}\sum_{k=1}^n\exp(-x_i^2)
$$

In MATLAB we may implement this as

data=rand(1e6,1);
erfEst=@(x) 2*x/sqrt(pi)*mean(exp(-(x*data).^2));


问:
Find all real numbers $x$ such that $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number
I want to find all values of $x\in \mathbb R$ such that the value of $\sqrt{x+2\sqrt{x}-1}+\sqrt{x-2\sqrt{x}-1}$ is a real number.

I solved it as follows:

$x+2\sqrt{x}-1\ge 0$

$(\sqrt{x}+1)^2-2\ge 0$

$(\sqrt{x}+1)^2\ge 2$

$\sqrt{x}+1\ge 2$ or $\sqrt{x}+1\le -\sqrt{2}$

The second can't hold, from the first $x\ge 3-2\sqrt{2}$.

Similarly $(\sqrt{x}-1)^2\ge 2$, hence $x\ge 3+2\sqrt{2}$.

I find my solution to be very ugly. Is my solution correct and is there a neater approach?

答: The requirement $x\ge0$ is obvious. The expression is real iff $(\sqrt{x}\pm1)^2\ge2$ for both choices of $\pm$, i.e. neither $\sqrt{x}\pm1$ is in $(-\sqrt{2},\,\sqrt{2})$, i.e .$\sqrt{x}\notin[0,\,1+\sqrt{2})\cup[0,\,\sqrt{2}-1)=[0,\,1+\sqrt{2})$, since $\sqrt{x}\ge0$. The solution set is $[3+2\sqrt{2},\,\infty)$, as in @alex.jordan's answer.


问:
Ramification points of projection to y-line
Rick Miranda's "Algebraic Curves and Riemann Surfaces" chapter 2, lemma 4.6 says that:

Let $X$ be a smooth projective plane curve defined by a homogeneous polynomial $F(x,y,z) = 0$; consider the map $G: X \rightarrow \mathbb{P}^1$ defined by $G[x:y:z] = [x:z]$. Then $G$ is ramified at $p \in X$ if and only if $(\partial F/\partial y)(p) = 0$.

I understand that for point where either $x$ or $z$ is nonzero this theorem is true by looking at the affine parts and then considering local charts.  This only leaves the point [0:1:0] which might very well be a point in $X$.  But the map $G$ is not even defined there. Is there something missing here?

More generally, we know that for a curve $X$, ratio of two homogeneous polynomials $P,Q$ such that the denominator does not identically vanish on $X$ gives a meromorphic function, and therefore gives a holomorphic map to Riemann sphere $\mathbb{P}^1$ as $[P : Q]$. But for the points where both $P$ and $Q$ vanish, this map is not even defined?  What is going on?  Can I not write the  meromorphic function defined by $P/Q$ as $[P:Q]$ globally?  It seems to me that at points where both $P$ and $Q$ vanish I can simplify the function locally so that it either has a pole or a removable singularity. For a trivial example consider $P = Q = x$ and then $P/Q$  is just the constant function 1, but it still doesn't make sense to write $[x:x]$ because it won't be defined at points $x = 0$, assuming there is such a point on $X$.

答:
The result you're looking for here is that any rational map from a smooth curve $C$ to a projective variety $P$ extends uniquely to an honest morphism $C\to P$. This is known (in some places, at least) as the curve-to-projective extension theorem, and here's an outline of the proof: it suffices to prove the fact in the case when $C$ is affine and our map is defined on $C\setminus \{x\}$. Then our rational map $C\to P$ is specified by $c\mapsto [f_0(c):f_1(c):\cdots:f_n(c)]$ where the $f_i$ are rational functions on $C$. Since $\mathcal{O}_{C,x}$ is a DVR, we can divide through by an appropriate power of the uniformizer to arrange that none of the $f_i$ have a pole at $c$ and not all of the $f_i$ vanish at $c$, giving an honest map $C\to P$ extending our rational morphism. For a full accounting, see section 16.5 in Vakil's Foundations of Algebraic Geometry.

In the comments, some clarification regarding what happens in the case that $[0:1:0]$ is on our curve was requested. First, if $[0:1:0]$ is on $V(F)$, then $\frac{\partial F}{\partial y}([0:1:0]) =0$: if $\deg F = 1$, then $F$ is the linear equation $ax+bz$ for some $a,b\in k$; while if $\deg F \geq 2$ then every term in $F$ is divisible by either $x$ or $z$ and the same is true for $\frac{\partial F}{\partial y}$, so it must evaluate to zero.

Now I claim that the image of $[0:1:0]$ under the projection is $\left[\frac{\partial F}{\partial x}([0:1:0]):\frac{\partial F}{\partial z}([0:1:0])\right]$, where it is clear that the map on tangent spaces is zero, so the projection is ramified at $[0:1:0]$. Up to a linear change of coordinates preserving $[0:1:0]$, we may assume the tangent line to $C$ at $[0:1:0]$ is just $V(x)$. Dehomogenize by setting $y=1$: up to scaling, we can write $F=x(1+p(x))-zq(x,z)$ where $q$ has vanishing constant term. Then $$[x:z]=[x(1+p(x)):z(1+p(x))]= [zq(x,z):z(1+p(x))] = [q(x,z):1+p(x)]$$ which evaluates to $[0:1]$ when we set $x=z=0$. So the projection map is ramified at $[0:1:0]$ when that point is on our curve.


问:
How piecewise smoothness of a function is related to the Dirichlet conditions?
Do all piecewise smooth functions satisfy Dirichlet conditions for Fourier series representation? In the theorem of Fourier series can we write that being piecewise smooth is the sufficient condition for obtaining its Fourier series ?</p>

答:
Yes and no. The sufficient condition for a existence of a Fourier series of a function $f:\mathbb R\to \mathbb C$ is that $f$ has to be periodic and quadratic integrable. As example let $f$ be $2\pi$ periodic, then we have there exists the Fourier series if and only if $\int_0^{2\pi}|f(t)|^2dt<\infty$. In that case we have with $e_k(t):=e^{ikt}$
$$\lim_{n\to\infty} \underbrace{\sum_{k=-n}^n\frac 1{2\pi}\int_0^{2\pi} f(t)e^{-ikt}dt\:e_k}_{S_nf:=}=f$$
the convergence in the sence of the quadratic integrable functionspace $\mathcal L^2_{2\pi}$. That means that $$\int_0^{2\pi}|f(t)-lim_{n\to\infty}S_n(t)|^2dt=0$$
In other words $\lim_{n\to\infty}S_nf(x)=f(x)$ for allmost every $x\in \mathbb R$,  but it don't have to hold for every $x$!

If $f$ is piecewise smooth in a way that the one-sided limits $f'(x+),f'(x-)$ and $f(x+),f(x-)$ exists for every $x\in \mathbb R$. Then is $f$ quadratic integrable, so the Fourier exists. But furthermore we have now for every $x\in\mathbb R$
$$\lim_{n\to\infty}S_nf(x)=\tfrac 1 2(f(x+)+f(x-)).\tag {*}$$
So yes, you can say that if a function is piecewise smooth then  the Fourier series converges pointwise  with $(*)$.


问:
There appears to be a contradiction while solving recurrence relation
Consider the equation :
$$
y[n]-\frac{1}{2}y[n-1]=x[n]
$$
where $x[n]:=\left(\frac{1}{3}\right)^{n}u[n]$. First, I was asked to find the homogenous part of the solution having a form $y_{h}:=A\frac{1}{2^{n}}$. Indeed the homogenous solution
$$
y_{h}[n]=\frac{1}{2^{n}}y[0]
$$
where I insist $A=y[0]$. Furthermore, The particular solution is given of the form $y_{p}[n]:=B\left(\frac{1}{3}\right)^{n}u[n]$ I shall determine $B$ as follow :
$$
B\left(\frac{1}{3}\right)^{n}u[n]-\frac{1}{2}B\left(\frac{1}{3}\right)^{n-1}u[n-1]=\left(\frac{1}{3}\right)^{n}u[n]
$$
\begin{align*}
\implies B&=\frac{\displaystyle\left(\frac{1}{3}\right)^{n}u[n]}{\displaystyle\left(\frac{1}{3}\right)^{n}u[n]-\frac{1}{2}\displaystyle\left(\frac{1}{3}\right)^{n-1}u[n-1]}\\ \\
&=\frac{\displaystyle\left(\frac{1}{3}\right)^{n}}{\displaystyle\left(\frac{1}{3}\right)^{n}-\displaystyle\frac{1}{2}\left(\frac{1}{3}\right)^{n-1}}\\ \\
&=-2
\end{align*}

So now that we have $B=-2$, We are given the initial rest condition (a condition that states if $x[n]=0$, $\forall n<0$ then $y[n]=0$ $\forall n<0$). so what I did was sub $n=0$ in difference equation
$$
y[0]-\frac{1}{2}y[-1]=x[0]
$$
$$
\implies y[0]=1
$$
$$
**\text{So this means $A=y[0]=1$}**
$$
However, the major issue is that :
$$
y[0]-\frac{1}{2}y[-1]=x[0]
$$
$$
\implies y[0]=1
$$
$$
\implies y[n]:= y_{h}[n]+y_{p}[n] \implies y[0]=y_{h}[0]+y_{p}[0]=1
$$
$$
\implies A\left(\frac{1}{2}\right)^{0} -2\left(\frac{1}{3}\right)^{0}=1
$$
$$
\implies A=3
$$
So I have $A=1$ and $A=3$, an obvious contradiction and I hope someone can help me.

Remark : $u[n]$ is the unit-step function where :
$$
u[n]:=
\begin{cases}
1&\text{if $n\geq0$}\\
0&\text{if $n<0$}
\end{cases}
$$

答:
The homogeneous solution gives

$$
y_h(n) = 2^{1-n}c_0
$$

now making for the particular $y_p(n) = 2^{1-n}c_0(n)$ after substituting into the complete recurrence we have

$$
c_0(n)-c_0(n-1) = \frac{2^{n-1}}{3^n}
$$

and solving this recurrence we have

$$
c_0(n) = 1-\left(\frac23\right)^n
$$

hence

$$
y(n) = y_h(n)+y_p(n) = 2^{1-n}c_0 + 2^{1-n}\left(1-\left(\frac23\right)^n\right), \ \ \ n\ge 0
$$


问:
Understanding Higher-order Differentiability Conceptually
I understand conceptually that a function $f\colon A\to\mathbf R$ is differentiable at a point $a\in A$ if it can be well approximated by a line there; more precisely, if we can find a constant $f'(a)$ such that $$f(a+h) = f(a) + f'(a)h+o(h).$$

My goal: I want to understand (intuitively) what it means when a function is twice differentiable, or thrice, etc, and likewise what it means when it isn't.

In a very loose sense, I have the intuition that a function is somehow smoother around a point if it has higher order differentiability there, and I suppose this somehow corresponds to the fact that it can be approximated well not only by a line but even better by a polynomial. (e.g. twice differentiability is  $f(a+h) = f(a) + f'(a)h + \tfrac12f''(a)h^2 + o(h^2)$, etc.). Does this mean polynomials canonically define the notion of "smoothness"?

Perhaps I can best get across what I'm asking for with an example. When I look at the graphs of
$$y=x|x| \qquad \text{and} \qquad y=x^3$$
for instance, I see that the latter grows quicker than the former, but around zero, they both basically look "smooth" to me. Yes I know that one is made of the absolute value function which is pointy, but if you just showed me these two pictures:

I wouldn't really feel that one is somehow "smoother" around zero than the other.

Is there a better way I can think about this?

答:
I think it would be difficult to deduce that $x|x|$ is not twice differentiable at $x=0$ simply by looking at its graph, but this is how I would think about it. Every function $y=f(x)$ can trivially be parameterised with respect to the $x$-coordinate, meaning that we can define its displacement function as $s(t)=(t,f(t))$, and its velocity function as $v(t)=(1,f'(t))$. The second derivative then has a slightly more concrete interpretation as the rate of change of the vertical component of the velocity function.

In the case of $y=x^3$, $s(t)=(t,t^3)$ and $v(t)=(1,3t^2)$. In the below animation, the velocity vector is represented by the red arrow, and the vertical component of that velocity vector is also shown separately in green:

Notice how the green arrow slows down smoothly before it gets to zero, and then starts speeding up again. This is in constrast to how $y=x|x|$ behaves around $0$:

Since the derivative of $x|x|$ is $|x|$, the green vector arrow is moving downwards at a constant rate, and then it instantly changes direction and goes up at the same rate when $x>0$. Looking at how the red and green vector arrows suddenly pop up gives us a clue about the differentiability of $x|x|$ at $0$. Of course, this animation is not entirely convincing, and things get much harder when you start talking about third and fourth derivatives, but this is how like to think about intuitively. @Fakemistake's car analogy also helped me in understanding this.


问:
Exercise 2.6 (General form of the Chinese remainder theorem) from Eisenbud's Commutative Algebra

Exercise 2.6 (General form of the Chinese remainder theorem): Let $R$ be a ring, and let $Q_1,\dots,Q_n$ be ideals of $R$ such that $Q_i+Q_j = R$ for all $i\neq j$. Show that $R/(\bigcap_i Q_i) \cong \prod_{i} R/Q_i$.

Consider the map of rings $\varphi:R\to\prod_i R/Q_i$ obtained from the $n$ projection maps $R\to R/Q_i$. Show that $\ker\varphi = \bigcap_iQ_i$.

Let $\mathfrak{m}$ be a maximal ideal of $R$. Show that the hypothesis that $Q_i +Q_j=R$ for all $i\neq j$ means that at most one of the $Q_i$ is contained in $\mathfrak{m}$. Now use Corollary 2.9 to show that $\varphi$ is surjective.

Corollary 2.9 is stated in a previous post that I made.

I'm still learning about localization. The hardest part about this proof for me was considering localization of some object as both the localization of a ring and the localization of a module. I understand the main ideas of the proof, but if someone can check for me that everything I'm saying is technically correct, I would appreciate it. In other words, I'm at the stage where I understand the ideas of localization, but I'm still sometimes getting lost in all of the notation. Thanks!

Part (1) is obvious. For part (2), it is clear that any maximal ideal $\mathfrak{m}$ contains at most one of the ideals $Q_i$. I will show that localizing $\varphi$ at each maximal ideal is surjective, whence it follows by Corollary 2.9 that $\varphi$ is surjective.

Let $\mathfrak{m}$ be a maximal ideal of $R$. First, we treat $R$ as an $R$-module, and $\prod_i R/Q_i$ as an $R$-module by writing it as a direct sum $\bigoplus_i R/Q_i$ with the action of $R$ induced by the map $\varphi$; in other words, the action is given by:
$$ r\cdot ([r_1],\dots,[r_n]) = \varphi(r)([r_1],\dots,[r_n]) = ([r],\dots,[r])([r_1],\dots,[r_n]) = ([rr_1],\dots,[rr_n]) $$
However, this is the standard direct sum $R$-module structure on $\bigoplus_iR/Q_i$ induced by the usual $R$-module structure on each of the summand $R$-modules $R/Q_i$ with action $r\cdot[r']=[rr']$. So localizing $\varphi$ at $\mathfrak{m}$ we obtain:
$$ \varphi_m:R_\mathfrak{m}\to \left(\bigoplus_i R/Q_i\right)_{\mathfrak{m}} \cong \bigoplus_i (R/Q_i)_\mathfrak{m} $$
Here, in localizing $(R/Q_i)_\mathfrak{m}$ we are treating $R/Q_i$ as an $R$-module. However, this is the same as the $(R/Q_i)$-module $(R/Q_i)_{\overline{\mathfrak{m}}}$ where $\overline{\mathfrak{m}}$ is the image of $\mathfrak{m}$ under the quotient $R\to R/Q_i$. It follows that $(R/Q_i)_{\overline{\mathfrak{m}}}$ has a ring structure.

Since localization commutes with quotients, we have that $(R/Q_i)_{\overline{\mathfrak{m}}} \cong R_\mathfrak{m}/{(Q_i)}_\mathfrak{m}$. Since all except for perhaps one of the $Q_i$ are not contained in $\mathfrak{m}$ we have that $(Q_i)_{\mathfrak{m}} = R_\mathfrak{m}$ for all $Q_i$ which are not contained in $\mathfrak{m}$ and hence the summand is zero. So at most one of the $Q_j\subset \mathfrak{m}$ in which case $\varphi_\mathfrak{m}: R_\mathfrak{m}\to \left(\bigoplus_i R/Q_i\right)_\mathfrak{m} \cong \bigoplus_i \left(R/Q_i\right)_\mathfrak{m} = (R/Q_j)_\mathfrak{m}$ which is clearly surjective. So we are done by Corollary 2.9.

答:
There's actually a more constructive way to show part $\phi$ is surjective (not requiring the existence of a maximal ideal, aka not requiring the axiom of choice).

Note that every element $(x_1, x_2, ..., x_n) \in \prod_i R / Q_i$ can be written as an $R$-linear combinination of basic elements $e_i$, where $e_i$ has a 1 in the $i$th position and a 0 everywhere else. So it suffices to show that each $e_i$ is in the range of $\phi$.

Indeed, consider that for all $j \neq i$, we have $1 \in R = Q_i + Q_j$. That is, we can write $1 = w_j + k_j$ for some $w_j \in Q_i$ and $k_j \in Q_j$. Then consider $x := \prod\limits_{j \neq i} (1 - w_j)$. We see that this element is in each $Q_j$ for $j \neq i$ because $1 - w_j \in Q_j$. And we see that, reducing this value modulo $Q_i$, we get $0$, since $\pi_i(x) = \pi_i(\prod\limits_{j \neq i} (1 - w_j)) = \prod\limits_{j \neq i} \pi_i (1 - w_j) = \prod\limits_{j \neq i} (1 - \pi_i(w_j)) = \prod\limits_{j \neq i} 1 = 1$. So $\phi(x) = e_i$. Here, $\pi_i : R \to R / Q_i$ is the canonical quotient map. Therefore, we see that $\phi$ is surjective.

Your proof works too, of course.


问:
Confusion about canonical rational section of the sheaf of a divisor
In Vakil's "The Rising Sea", remark 14.2.3 says that the sheaf $O_X(D)$ for divisor $D$ comes along with a canonical rational section, namely the section corresponding to function $1$ defined in the complement of support of $D$.  My question is what if the scheme under consideration has only finitely many points and hence possibly the complement of the support is empty, e.g., we can consider a DVR.

Here our schemes are taken to be integral and normal.

答: A divisor $D$ is defined to be a formal sum of codimension-one subvarieties. No codimension-one subvariety contains the generic point of an integral scheme, so it cannot be the case that the complement of the support of $D$ is empty.


问:
Problems regarding Joint probability function of two discrete random variables

Here, I have a problem in part b. and c. For b I found the marginal probability mass function of W and V and tried to look for a value where W<V. Should I consider P(V>W)? Are they the same thing?

In part c. I know how to calculate the mathematical expectation but here I do not understand how I should multiply W and V. I mean which particular value should I multiply??

答:
Yes, of course $P(V>W)$ and $P(W<V)$ are the same. For b., if you consider $P(V-W>0)$ instead of $P(V>W)$, this problem will seem much simpler to you. We just need to add the probabilities, which give $V-W>0$:
\begin{align} P(W<V)&=P(V-W>0)\\
&=P(V=1,W=0)+P(V=2,W=0)+P(V=2,W=1)+\cdots\\
&=22c
\end{align}

For c., we shall find the conditional probability mass function, i.e., $f(z\mid V=2)=P(Z=WV=z\mid V=2).$ But this is easy thanks to the condition $\{V=2\}.$ To this end, first observe that

$$P(W=0\mid V=2)=\frac{P(W=0,V=2)}{P(V=2)}=\frac{2c}{12c}= \frac{1}{6}$$
$$P(W=1\mid V=2)=\frac{P(W=1,V=2)}{P(V=2)}=\frac{4c}{12c}= \frac{1}{3}$$
$$P(W=2\mid V=2)=\frac{P(W=2,V=2)}{P(V=2)}=\frac{6c}{12c}= \frac{1}{2}.$$
Then
$$
P(Z=z\mid V=2) = \left\{
        \begin{array}{ll}
            \frac{1}{6} & \quad z=0 \\
            \frac{1}{3} & \quad z=2 \\
            \frac{1}{2} & \quad z=4 \\
        \end{array}
    \right.
$$
Then we can simply compute the conditional expectation:
\begin{align}
E[Z\mid V=2]&=\sum_z z \ f(z\mid V=2) \\
&=\frac{1}{6}\cdot 0+\frac{1}{3} \cdot 2 + \frac{1}{2}\cdot 4\\ &=\frac{8}{3}
\end{align}


问:
Let $f:\mathbb{R} \to \mathbb{R}$ continuous with $f(f(x))=e^x$, show that $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$ (Brazilian Olympiad)
Given $f:\mathbb{R} \to \mathbb{R}$ continuous with $f(f(x))=e^x$ for all $x\in\mathbb{R}$. Show that $\lim_{x\to \infty } \frac{f(x)}{x^n}=\infty$ for all $n\in\mathbb{N}$.

I appreciate any help!

Edit 1: It is easy to see that $f$ is injective. So, if $f$ increases anywhere, $f$ will be an increasing function.

Edit 2: As said in the comments, $f$ must indeed be increasing, since
$f(1)=f(e^0)=f(f(f(0)))=e^{f(0)}>f(0)$.

答:
We'll start with a weak estimate, and then use the functional equation to improve it iteratively.

Claim: There is a constant $d > 0$ such that $ f(x) \ge x+d$ for all $x \ge 0$.

Once we have that estimate (see below for the proof), we can can continue as follows: Setting $C = e^d > 1$ we get
$$
f(x) = f(e^{\ln x}) = e^{f(\ln x)} \ge e^{\ln x + d} = xe^d = Cx
$$
for $x \ge 1$, then
$$
f(x) = e^{f(\ln x)} \ge e^{C\ln x} = x^C
$$
for $x \ge e$, and finally
$$
f(x) = e^{f(\ln x)} \ge e^{(\ln x)^C}
$$
for $x \ge e^e$.

It follows that
$$
\frac{f(e^x)}{e^{nx}} \ge \frac{e^{x^C}}{e^{nx}} = e^{x^C-nx} \ge x^C-nx+1
$$
and therefore
$$
 \lim_{y \to \infty} \frac{f(y)}{y^n} = \lim_{x \to \infty} \frac{f(e^x)}{e^{nx}} = +\infty \, .
$$

Proof of the claim: $f(0) > 0$ and $f$ has no fixed points (see for example thoughts about $f(f(x))=e^x$), therefore $f(x) > x$ for all $x \in \Bbb R$. Then
$$
 d = \min \{ f(x) - x \mid 0 \le x \le 1 \}
$$
is strictly positive. Define the sequence $(E_k)$ recursively by $E_0 = 0$, $E_{k+1} = e^{E_k}$:
$$
 0, 1, e, e^e, e^{e^e},\ldots
$$
$f(x) \ge x+d$ holds for $E_0 \le x \le E_1$, and if it holds for $E_k \le x \le E_{k+1}$ then
$$
 f(x) =  e^{f(\ln x)} \ge e^{\ln x + d} = xe^d \\
 \ge x(1+d) = x + xd \ge x+d
$$
for $E_{k+1} \le x \le E_{k+2}$. This concludes the proof of the claim, since $E_k \to \infty$.


问:
Prove that the characteristic polynomial $P(x)$ from $A$ and $Q(x)$ from $A^{-1}$ are related by...
I have to prove that $A$ characteristic's polynomial $P(x)$ and $A^{-1}$ characteristic's polynomial $Q(x)$ are related by:
$$ Q(\lambda) = \frac{\lambda^{n} P(\lambda ^{-1})}{P(0)} $$

I've tried by considering the matrix $A(A^{-1} − λI)$, but I've obteined no reults...

Any help?

答:
Hint:

Suppose $\;c\neq0\;$ is a root of a polynomial with free coefficient$\,\neq0\;$ :

$$p(x)=\sum_{k=0}^na_kx^k\implies 0=p(c)=a_0+a_1c+\ldots+ a_nc^n\implies $$

$$a_0c^{-n}=-\left(a_1c^{1-n}+a_2c^{2-n}+\ldots+a_n\right)\implies c^{-n}=\frac{-\left(a_1c^{1-n}+a_2c^{2-n}+\ldots+a_n\right)}{a_0}$$

since $\;p(x)\;$ is irreducible and $\;a_0\neq0\;$ since $\;c\neq0\;$ . The above means that $\;c^{-1}\;$ is a root of

$$\frac{a_n}{a_o}+\frac{a_{n-1}}{a_0}x+\ldots+\frac{a_2}{a_0}x^{n-2}+\frac{a_1}{a_0}x^{n-1}+x^n=\frac{a_n+a_{n-2}x+\ldots+a_1x^{n-1}+x^n}{a_0}=$$

$$=\frac{x^n\,p\left(x^{-1}\right)}{p(0)}\;\ldots\ldots$$


问:
Finiteness of the volume $G(k) \backslash G(\mathbb A)^1$ from that of $G(k) Z(\mathbb A) \backslash G(\mathbb A)$
It follows from the construction of a Siegel domain that $G(k) Z(\mathbb A) \backslash G(\mathbb A)$ has finite volume, where $G$ is a connected, reductive group over a number field $k$, $Z$ is the center of $G$, and $\mathbb A$ is the ring of adeles of $k$.  I have also heard that $G(k) \backslash G(\mathbb A)^1$ has finite volume, where $G(\mathbb A)^1$ is the group of $g \in G(\mathbb A)$ for which $|\chi(g)| = 1$ for all $k$-rational characters $\chi$ of $G$.  Does the finiteness of one of these volumes imply that of the other?

答:
Yes.  This follows from the fact that $G(\mathbb A)$ is the direct product of $G(\mathbb A)^1$ and $A_G(\mathbb R)^0$, where $A_G$ is the split component of $G$ (the unique maximal split subtorus of the center $Z$ of $G$), and $A_G(\mathbb R)^0$ denotes the connected component of the identity.

Since $G(\mathbb A) = Z(\mathbb A) G(\mathbb A)^1 = G(k) Z(\mathbb A)  G(\mathbb A)^1$, we can write

$$G(k)Z(\mathbb A) \backslash G(\mathbb A) = G(k)Z(\mathbb A) \backslash \Big( G(k)Z(\mathbb A) G(\mathbb A)^1\Big) = \Big(G(k)Z(\mathbb A) \cap G(\mathbb A)^1 \Big) \backslash G(\mathbb A)^1 = G(k) \Big(Z(\mathbb A) \cap G(\mathbb A)^1\Big) \backslash G(\mathbb A)^1$$

where $G(k)Z(\mathbb A) \cap G(\mathbb A)^1 = G(k).\Big(Z(\mathbb A) \cap G(\mathbb A)^1\Big)$ follows from the fact that $G(k) \subset G(\mathbb A)^1$.

Now, let $H = Z(\mathbb A) \cap G(\mathbb A)^1$, so that $$G(k) Z(\mathbb A) \backslash G(\mathbb A) = [G(k)H ]\backslash G(\mathbb A)^1$$

We have

$$\operatorname{vol}(G(k) H \backslash G(\mathbb A)^1) = \operatorname{vol}(G(k) \backslash G(\mathbb A)^1) \operatorname{vol}(G(k) \backslash G(k)H) = $$

and writing $G(k) \backslash G(k)H = [G(k) \cap H] \backslash H$, we see that we can rewrite the above as

$$\operatorname{vol}(G(k) Z(\mathbb A) )\backslash G(\mathbb A) = \operatorname{vol}(G(k) \cap H \backslash H) \operatorname{vol}(G(k) \backslash G(\mathbb A)^1)$$
Thus the finiteness of the volume of one of the quotients is equivalent to that of the other, provided we can show that $[G(k) \cap H] \backslash H$ has finite volume.

In fact, $[G(k) \cap H] \backslash H$ is actually compact.  Indeed, $H$ is the almost direct product of the adelic points of an anisotropic torus (which is compact) and the piece $A_G(\mathbb A)^1$ of the split component $A_G(\mathbb A)$ which has norm $1$ on all $k$-rational characters.  The compactness of $A_G(k) \backslash A_G(\mathbb A)^1$, as a finite product of copies of $\mathbb A^{\ast 1}/k^{\ast}$, implies the compactness of the given quotient.


问:
why not $F(x,0)=x_0$?
Definition of  contractible space

Let $F$ be the homotopy between $\mathrm{id}_X$ and $x_0$, that is $F:X\times [0,1]\to X$ is a continuous map such that
$$ F(x,0)=x,\quad F(x,1)=x_0$$
for all $x\in X$.

My confusion : I have some  confusion  with  $ F(x,0)=x$

why not  $F(x,0)=x_0$ ?

Here $f:[0,1]\to X$ is  a loop based $x_0$, we know that  loop is a path whose initial point is equal to its  terminal point

So i think $F(x,0)=x_0$

答:
The whole idea of a homotopy between two functions $f, g : X \to X$ is that it's a continuous map $F : X \times [0,1] \to X$ such that $F(\cdot,0) = f$ and $F(\cdot,1) = g$.

If you set $f(x) = x$ and $g(x) = x_0$, then this becomes $F(x,0) = x$ and $F(x,1) = x_0$.

Title: why not $F(x,0)=x_0$?


答:
Hint:

$$e^{xi}=i=e^{\frac\pi2i+2k\pi i}\implies x=\frac\pi2+2k\pi\;,\;\;k\in\Bbb Z$$

Title: How to solve $e^{ix} = i$

Body:
I am taking an on-line course and the following homework problem was posed: $$e^{ix} = i$$ I have no idea how to solve this problem. I have never dealt with solving equations that have imaginary parts. What are the steps to solving such equations? I am familiar with Taylor series and the Euler formula if that is any help.

Thanks.


问:
Why relative velocity must be perpendicular to velocity of one of the bodies for closest approach
I came across this question that asked the same, but I couldn't understand the reasoning. I feel it wasn't answered clearly, and the OP also sought clarification.

The answer stated,

From a physics perspective -- (here we go again, answering physics questions on a math site! ;) ) -- it may make more sense if you go into the inertial reference frame where the ship being approached is stationary. This involves subtracting the (vector) velocity of the ship being approached from the velocities of both ships. [...] one ship is stationary, its velocity is zero, so it stays at the same place all the time. It's at a point. (The red point.) The other ship moves along some straight line path past that point. The minimum distance from the point to the line is a perpendicular.

But this doesn't explain why $V_{ba} \perp V_{b}$. In fact, I think OP tried to ask the same.

But the question I'm asking John is marked in red above. The two lines in your diagram are perpendicular, the solid blue line represents the velocity vector of the patrol boat B as you rightly said. But this must mean that the dashed blue line is a relative velocity, (as well as being the shortest distance). Can you explain if the dashed line represents the velocity of B relative to S?

This is what I understand so far: (1-3 are covered in the linked question's accepted answer)

Assume $V_{a}$ is velocity of first body and $V_{b}$ the velocity of the second

$V_{ba} = V_{b} - V_{a}$ can be taken as object A is stationary and object B moves with $V_{ba}$

For distance of closest approach, it's as simple as dropping a perpendicular from A to $V_{ba}$

This part, I don't get (and unexplained in the linked question). For closest approach to happen, $V_{ba} \cdot V_{b} = 0$.

Why must $V_{ba} \perp V_{b}$?

Given $V_{ba}$ and $V_{b}$ are in different reference frames, how do I derive the relation that their dot product must be zero?

答:
In my opinion, you are right to question the answer key shown in Why must closest approach occur when relative velocity is perpendicular to motion?
It gives an incomplete explanation of how the patrol boat selects the
course to minimize the closest approach to the ship.

Provided that the patrol boat's course brings it closer to the ship than it was at the start, we can show that at the instant when the distance is minimized,
the relative velocity of the boat and ship is perpendicular to the line joining their two positions at that time.
We have seen this proved in multiple ways.

But the original question is completely unconcerned with what happens at the instant of closest approach.

To answer the actual question, which is to find the  direction the patrol boat should go in order to minimize the distance at closest approach,
notice that the difference of the two velocity vectors,
$V_B - V_S,$ is the velocity of the boat relative to the ship.
The direction of $V_S$ is fixed, but the direction  of $V_B$ can be chosen to be any direction.
Once you choose the direction of $V_B,$ the direction of $V_B - V_S$
is determined.

In the first part of the question, where the boat's speed is greater than the ship's, the tip of $V_B$ can fall anywhere on the circle shown in the figure below.
Since the tip of $V_S$ is inside that circle, the difference vector
$V_B - V_S$ (shown in red) can point in any direction.

To intercept a ship at any initial relative position $S$, you need the relative velocity $V_B - V_S$ to point in exactly the direction from $B$ to $S$, as shown for one choice of $V_B$ in the figure.
The desired direction of $V_B$ is determined by the point where the relative position vector from $B$ to $S$ intersects the circle.

If the boat is slower than the ship, we have a different set of possibilities,
as illustrated in the figure below.
Again, the tip of $V_B$ can be anywhere on the circle, and so can the tip of the relative velocity vector $V_B - V_S,$
but the direction of $V_B - V_S$ from $B$ is now limited to the directions between the two rays tangent to the circle.

We now have three possible cases, depending on the initial position of the ship $S$ relative to the boat $B.$

Case $1$.
If the direction from $B$ to $S$ is between those two tangents, as is the case if the ship is initially at $S_1,$ then it is still possible for the boat to intercept the ship. In fact there are usually two possible choices for the boat's direction in this case, though only one is shown here.
(This answer
shows both choices in a slightly different visualization.)

Case $2$.
If the ship is initially at $S_2,$ or anywhere else in the region between the rays from $B$ perpendicular to the tangents, the closest point of approach is at the starting condition; no matter which way the boat is headed, the relative velocity
$V_B - V_S$ will immediately take the boat farther from the ship and will continue to increase the distance between them indefinitely.

Case $3$.
If the ship is initially at a point such as $S_3,$ on the opposite side of one of the tangents as the circle but at a direction from $B$ not more than $90$ degrees away from the tangent direction, the closest the boat can get to the ship is at the relative position indicated by the dotted line to $S_3$ in the figure,
which connects $S_3$ to the nearest point on the nearby tangent to the circle.

The dotted line is perpendicular to the relative velocity vector along that tangent line, of course; but the fact that is relevant to the solution is that the tangent line is perpendicular to the radius that meets it, and that radius is the velocity vector $V_B$ that minimizes the distance at closest approach.

Therefore, in Case $3$, we want to choose a direction for $V_B$ that will be perpendicular to $V_B - V_S,$ and therefore we set up the right triangle shown in the second part of the answer key in Why must closest approach occur when relative velocity is perpendicular to motion?

What the answer key does not explain is how we know that Case $3$ applies.
One way to determine which case applies is to construct the diagram as if
Case $3$ applied, and then compare the direction of the relative velocity
$V_B - V_S$ in that diagram to the initial direction from $B$ to $S$.
If the direction from $B$ to $S$ goes through the triangle, Case $1$ applies and the boat can actually intercept the ship; if the direction is more than $90$ degrees on the other side of the relative velocity vector, Case $2$ applies and there is no way to get closer to the ship no matter which way the boat heads;
otherwise Case $3$ applies.

Examining the angles found in the answer key, the direction of $V_B - V_S$ is
$108.6$ degrees, which is less than $90$ degrees away from the initial direction from the boat to the ship ($50$ degrees), so Case $3$ applies.


问:
"Prove that if $G$ is a group, then $\{x \in G : x^2 =e \}$ is a subgroup of $G$": is that correct?
My teacher left this exercise:

"Prove that if $G$ is a group, then $\{x \in G : x^2 =e \}$ is a subgroup of $G$"

but I don't know if it's correct, I have seen similar exercises like that

"if $x^2 = e $ for all $x \in G$ then $G$ is abelian",

and in this case if $G$ is abelian then the proof is fairly simple but my teacher didn't say $G$ was abelian, so I'm not sure if that statement is true or if he forgot to say that $G$ is abelian.

I have tried to found a counterexample (if it exists) but haven't found any.

Any help? Thanks in advance

答: A reflection of the plane has order $2$, but the composite of two reflections in intersecting lines is a rotation, which does not necessarily have order $2$ ...


问:
Equally Distributed Data Set Measurement
I will be creating my own dataset with scores ranging from 50.00 to 100.00. How will I say that the dataset I chose is equally distributed and unbiased ? Is there a formula to know this?</p>

答:
Your question leaves room for some interpretation. Here is my interpretation. If my interpretation is not what you had in mind, please revise your question to be more informative and maybe someone else will give an answer your find more
useful.

If the population consists of numbers the $5001$ numbers $50.00, 50,01, \dots, 99.99, 100.00,$ and you select a sample of size $n=20$ with replacement, then the sample should be difficult to distinguish from a random sample of size twenty from the distribution
$\mathsf{Unif}(50,100).$ [Computations and sampling in R.]

```<code>s = seq(50, 100, by=0.01)
head(s);  length(s)
[1] 50.00 50.01 50.02 50.03 50.04 50.05  # first 6 pop values
[1] 5001  # population size

set.seed(123)
x = sample(s, 20, rep=T)
</code>```

At the 5% level, a Kolmogorov-Smirnov goodness-of-fit test, does not reject
the null hypothesis that the sample of size $n=20$ is from
the distribution $\mathsf{Unif}(50,100):$ the P-value of the
test is $0.4606 > 0.05.$

```<code>ks.test(x, punif, 50, 100)

        One-sample Kolmogorov-Smirnov test

data:  x
D = 0.183, p-value = 0.4606
alternative hypothesis: two-sided
</code>```

With samples as small as $n = 20,$ it is difficult to know
what the population might be, but the K-S test sees no evidence
that this uniform distribution could not have been the sampled
population.

```<code>par(mfrow=c(1,2))
 hist(x, prob=T, col="skyblue2")
  rug(x)
 plot(ecdf(x))
  curve(punif(x,50,100),add=T, col="blue")
par(mfrow=c(1,1))
</code>```

The K-S test statistic $D = 0.183$ is the maximum vertical
distance between the CDF (blue) of $\mathsf{Unif}(50,100)$
and the ECDF (black) of the sample of 20. [Right-hand panel.] To make the empirical CDF (ECDF) of a sample: sort the sample; begin at height $0$ on the left, jump up by $1/n$ at each sample value, end end at height $1$ on the right.

Many goodness-of-fit tests are possible, but you should use only
one of them in a practical situation. Another test is to
count the frequencies  (3, 3, 6, 2, 6)
in the five histogram bins. For a uniform distribution we
would expect $E = 4$ counts on average in each bin. A
chi.squared test finds that the disagreement between the observed
and expected frequencies is not greater than would be expected
by chance.

```<code>hist(x, plot=F)<span class="math-container">$counts
[1] 3 3 6 2 6
f = hist(x, plot=F)$</span>counts;  f
[1] 3 3 6 2 6
chisq.test(f, sim=T)

        Chi-squared test for given probabilities
        with simulated p-value (based on 2000 replicates)

data:  f
X-squared = 3.5, df = NA, p-value = 0.5502
</code>```

Notes on chisq.test in R: (1) Unless otherwise stated, the 'given probabilities' are taken to be equal in each category. (2) When expected category frequencies are small (as here), the test can
simulate an accurate P-value. (If using software without this
simulation capability, it would be better to have a sample size larger than twenty.

Body:
Your question leaves room for some interpretation. Here is my interpretation. If my interpretation is not what you had in mind, please revise your question to be more informative and maybe someone else will give an answer your find more
useful.

If the population consists of numbers the $5001$ numbers $50.00, 50,01, \dots, 99.99, 100.00,$ and you select a sample of size $n=20$ with replacement, then the sample should be difficult to distinguish from a random sample of size twenty from the distribution
$\mathsf{Unif}(50,100).$ [Computations and sampling in R.]

s = seq(50, 100, by=0.01)
head(s);  length(s)
[1] 50.00 50.01 50.02 50.03 50.04 50.05  # first 6 pop values
[1] 5001  # population size

set.seed(123)
x = sample(s, 20, rep=T)

At the 5% level, a Kolmogorov-Smirnov goodness-of-fit test, does not reject
the null hypothesis that the sample of size $n=20$ is from
the distribution $\mathsf{Unif}(50,100):$ the P-value of the
test is $0.4606 > 0.05.$

ks.test(x, punif, 50, 100)

        One-sample Kolmogorov-Smirnov test

data:  x
D = 0.183, p-value = 0.4606
alternative hypothesis: two-sided

With samples as small as $n = 20,$ it is difficult to know
what the population might be, but the K-S test sees no evidence
that this uniform distribution could not have been the sampled
population.

par(mfrow=c(1,2))
 hist(x, prob=T, col="skyblue2")
  rug(x)
 plot(ecdf(x))
  curve(punif(x,50,100),add=T, col="blue")
par(mfrow=c(1,1))

The K-S test statistic $D = 0.183$ is the maximum vertical
distance between the CDF (blue) of $\mathsf{Unif}(50,100)$
and the ECDF (black) of the sample of 20. [Right-hand panel.] To make the empirical CDF (ECDF) of a sample: sort the sample; begin at height $0$ on the left, jump up by $1/n$ at each sample value, end end at height $1$ on the right.

Many goodness-of-fit tests are possible, but you should use only
one of them in a practical situation. Another test is to
count the frequencies  (3, 3, 6, 2, 6)
in the five histogram bins. For a uniform distribution we
would expect $E = 4$ counts on average in each bin. A
chi.squared test finds that the disagreement between the observed
and expected frequencies is not greater than would be expected
by chance.

hist(x, plot=F)$counts
[1] 3 3 6 2 6
f = hist(x, plot=F)$counts;  f
[1] 3 3 6 2 6
chisq.test(f, sim=T)

        Chi-squared test for given probabilities
        with simulated p-value (based on 2000 replicates)

data:  f
X-squared = 3.5, df = NA, p-value = 0.5502

Notes on chisq.test in R: (1) Unless otherwise stated, the 'given probabilities' are taken to be equal in each category. (2) When expected category frequencies are small (as here), the test can
simulate an accurate P-value. (If using software without this
simulation capability, it would be better to have a sample size larger than twenty.


答:
Here are some thoughts:

To study, as always, I use my favourite tool of Taylor's theorem(*):

$$ f(a+h) = f(a) + h f'(a) + \frac{h^2}{2} f''(a) +O(h^3)$$

Now, let's analyze $f''(a)$, if $f(a+h)$ is not differentiable, it means the left hand derivative and right hand derivative are not equal i.e(Or may be the limit themself don't exist, but let us forget the case for now):

$$ \lim_{h \to 0} \frac{f'(a+h) - f'(a)}{h} \neq \frac{ f'(a) - f'(a-h ) }{h}$$

Let's call the right hand 2nd derivative as $f_{R}$ and left hand 2nd derivative as $f_{L}$, this leads to 'two' series of the function locally. That is, for points to the right of $f$, we have

$$ f(a+h) = f(a) + hf'(a) + \frac{h^2}{2} f_{R}''(a) + O(h^3)$$

And another series for the left of $f$ as:

$$ f(a-h) = f(a) - h f'(a) + \frac{h^2}{2} f_{L}''(a) + O(h^3)$$

Now, here's the deal, the second order derivative and higher terms only become really relevant (for most nice functions) after $h>1$, this because if $h<1$ then $h^2 <h$, so it turns out that if it is only the second derivative and above which of a function which doesn't match, the Taylor series approximates well.. but outside that bound, we need to be careful and using the piecewise definition.

*: The link is to an article I've written on it, I highly suggest reading it if you want a real insight into the theorem.

**: Right hand derivative: $ \frac{f'(a+h)-f'(a)}{h} $ and the left hand derivative $ \frac{f'(a) - f'(a-h)}{h}$

***: If we restrict ourselves to the differentiability of second derivatives, then we can think of it as the convexity of the graph suddenly changing.

Title: Understanding Higher-order Differentiability Conceptually

Body:
I understand conceptually that a function $f\colon A\to\mathbf R$ is differentiable at a point $a\in A$ if it can be well approximated by a line there; more precisely, if we can find a constant $f'(a)$ such that $$f(a+h) = f(a) + f'(a)h+o(h).$$

My goal: I want to understand (intuitively) what it means when a function is twice differentiable, or thrice, etc, and likewise what it means when it isn't.

In a very loose sense, I have the intuition that a function is somehow smoother around a point if it has higher order differentiability there, and I suppose this somehow corresponds to the fact that it can be approximated well not only by a line but even better by a polynomial. (e.g. twice differentiability is  $f(a+h) = f(a) + f'(a)h + \tfrac12f''(a)h^2 + o(h^2)$, etc.). Does this mean polynomials canonically define the notion of "smoothness"?

Perhaps I can best get across what I'm asking for with an example. When I look at the graphs of
$$y=x|x| \qquad \text{and} \qquad y=x^3$$
for instance, I see that the latter grows quicker than the former, but around zero, they both basically look "smooth" to me. Yes I know that one is made of the absolute value function which is pointy, but if you just showed me these two pictures:

I wouldn't really feel that one is somehow "smoother" around zero than the other.

Is there a better way I can think about this?


问:
What does 'i-th' mean?
I have seen a problem set for the tower of hanoi algorithm that states:

Each integer in the second line is in the range 1 to K where the i-th
  integer denotes the peg to which disc of radius i is present in the
  initial configuration.

What does i-th actually mean? If i have a line conforming to the format above such as:

4 2 4 3 1 1

How exactly does this translate to the format described above?

答:
The letter $i$ is being used as a placeholder for a natural number. So for instance the $i$-th element in the following sequence $a$ is $2^i$ $$a=(2,4,8,16,32,64,\ldots).$$

For $i=1$ we have the 1st element is $2^1=2$.

For $i=2$ we have the 2nd element is $2^2=4$.

etc

Title: What does 'i-th' mean?


问:
Understanding naturality of cup product definition.
I was reading the definition of naturality from AT (on pg. 127), it says:

For example, to say that the long exact sequence of a pair is natural means that for a map $f:(X, A) \rightarrow(Y, B),$ the diagram

\begin{array}{c}
\cdots \longrightarrow H_{n}(A) \stackrel{i_{*}}{\longrightarrow} H_{n}(X) \stackrel{j *}{\longrightarrow} H_{n}(X, A) \stackrel{\partial}{\longrightarrow} H_{n-1}(A) \longrightarrow \cdots \\
\downarrow f_{*}\quad \quad \downarrow f_{*}\quad \quad \downarrow f_{*}  \quad \quad \quad \quad \downarrow f_{*} \\
\cdots \longrightarrow H_{n}(B) \stackrel{i *}{\longrightarrow} H_{n}(Y) \stackrel{j *}{\longrightarrow} H_{n}(Y, B) \stackrel{\partial}{\longrightarrow} H_{n-1}(B) \longrightarrow \cdots
\end{array}

is commutative. Commutativity of the squares involving $i_{*}$ and $j_{*}$ follows from the obvious commutativity of the corresponding squares of chain groups, with $C_{n}$ in place of $H_{n}$. For the other square, when we defined induced homomorphisms we saw that $f_{\not} \partial=\partial f_{f}$ at the chain level. Then for a class $[\alpha] \in H_{n}(X, A)$ represented by a relative cycle $\alpha$, we have $f_{*} \partial[\alpha]=f_{*}[\partial \alpha]=\left[f_{\text {ł }} \partial \alpha\right]=\left[\partial f_{\neq} \alpha\right]=\partial\left[f_{\neq} \alpha\right]=\partial f_{*}[\alpha]$.

But then on pg.210, Proposition 3.10., proved the naturality of cup product as can be seen below:

Proposition 3.10. For a map $f: X \rightarrow Y$, the induced maps $f^{*}: H^{n}(Y ; R) \rightarrow H^{n}(X ; R)$ satisfy $f^{*}(\alpha \smile \beta)=f^{*}(\alpha) \smile f^{*}(\beta),$ and similarly in the relative case.

Proof: This comes from the cochain formula $f^{\#}(\varphi) \smile f^{\#}(\psi)=f^{\#}(\varphi \smile \psi)$ :
$$
\begin{aligned}
\left(f^{\#} \varphi \smile f^{\#} \psi\right)(\sigma) &=f^{\#} \varphi\left(\sigma \mid\left[v_{0}, \cdots, v_{k}\right]\right) f^{\#} \psi\left(\sigma \mid\left[v_{k}, \cdots, v_{k+\ell}\right]\right) \\
&=\varphi\left(f \sigma \mid\left[v_{0}, \cdots, v_{k}\right]\right) \psi\left(f \sigma \mid\left[v_{k}, \cdots, v_{k+\ell}\right]\right) \\
&=(\varphi \smile \psi)(f \sigma)=f^{\#}(\varphi \smile \psi)(\sigma)
\end{aligned}
$$

Could someone tell me how this proposition is a proof of naturality of cup product according to the definition of naturality given above?

答:
The quote from page 127 of Hatcher is not a definition of naturality, just an example.

Naturality is a general concept from category theory (which amusingly, at time of writing, is the one tag that this question doesn't use among all the other less relevant tags).  Recall that a natural transformation $\eta: F \Rightarrow G$ between functors $F, G: \mathcal{C} \to \mathcal{D}$ is a collection of morphisms $\{\eta_X: FX \to GX\}$ in $\mathcal{D}$ indexed by the objects $X$ in $\mathcal{C}$.  The key compatibility condition ("naturality") that these morphisms have to satisfy is that for any morphism $f: X \to Y$ in $\mathcal{C}$, the following diagram commutes:
$$\require{AMScd}
\begin{CD}
FX @>{Ff}>> FY \\
@V{\eta_X}VV @VV{\eta_Y}V \\
GX @>>{Gf}> GY.
\end{CD}$$

For example, the connecting homomorphism $\partial$ in the long exact sequence in homology is a natural transformation between the functors $(X,A) \mapsto H_n(X,A)$ and $(X,A) \mapsto H_{n-1}(A)$ from the category of pairs of spaces to the abelian groups.  The naturality condition is that $\partial \circ f_* = f_* \circ \partial$, i.e., the square you drew involving $\partial$ commutes.

For another example, the cup product $\smile$ is natural transformation between the functors $X \mapsto H^p(X) \otimes  H^q(X)$ and $X \mapsto H^{p+q}(X)$ from the category of spaces to abelian groups.  The naturality condition is $\smile \circ (f^*, f^*) = f^* \circ \smile$, i.e., $f^* \alpha \smile f^* \beta = f^*(\alpha \smile \beta)$ for any two cohomology classes $\alpha, \beta$.


问:
To prove the following statement by induction for positive nature number n.
There are both $n$ and $k$ in this formula and I don’t know how to prove it. Because in this case I cannot split the summation into parts contains the assumption.

$$\sum_{k=1}^n \frac{1}{(2k-1)2k} = \sum_{k=1}^n \frac{1}{n+k}$$

My basic case is $n=1$ and both LHS and RHS equals to $\frac{1}{2}$.
Assume the statement is true for some m
$$ie.  \sum_{k=1}^m \frac{1}{(2k-1)2k} = \sum_{k=1}^m \frac{1}{m+k}$$
I want to prove for $m+1$
$$ie.  \sum_{k=1}^{m+1} \frac{1}{(2k-1)2k} = \sum_{k=1}^{m+1} \frac{1}{m+1+k}$$
RHS I got
$$\sum_{k=1}^{m+1} \frac{1}{m+1+k}=\sum_{k=1}^{m} \frac{1}{m+1+k}+\frac{1}{m+1+m+1}=\sum_{k=1}^{m} \frac{1}{m+1+k}+\frac{1}{2m+2}$$
LHS i got
$$\sum_{k=1}^{m+1} \frac{1}{(2k-1)2k} = \sum_{k=1}^{m} \frac{1}{(2k-1)2k} + \frac{1}{(2(m+1)-1)2(m+1)}=\sum_{k=1}^m \frac{1}{m+k}+\frac{1}{(2m+2)(2m+1)}$$
And I got stuck here.Because RHS it is $m+1+k$ on the denominator but on LHS it is $m+k$ and they are quite different I don't know how to prove them equal.

答:
You actually can split it.

HINT: $$\frac1{2k(2k-1)}=\frac1{2k-1}-\frac1{2k}\,,$$

so

$$\sum_{k=1}^{n+1}\frac1{2k(2k-1)}-\sum_{k=1}^n\frac1{2k(2k-1)}=\frac1{2n+1}-\frac1{2(n+1)}\,.$$

Edit: And

$$\begin{align*}
\sum_{k=1}^{n+1}\frac1{n+1+k}-\sum_{k=1}^n\frac1{n+k}&=\sum_{k=2}^{n+2}\frac1{n+k}-\sum_{k=1}^n\frac1{n+k}\\
&=\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}\,.
\end{align*}$$

The only terms in the first summation that do not appear in the second are the last two, with $k=n+1$ and $k=n+2$, and the only term in the second summation that does not appear in the first is the $k=1$ term: everything else cancels out. Thus, you’ll have the induction step once you show that

$$\frac1{2n+1}-\frac1{2(n+1)}=\frac1{2n+1}+\frac1{2n+2}-\frac1{n+1}\,,$$

which is just a matter of a little algebra.


问:
Kernel of the homomorphism of an affine group onto the general linear group is the translation space
Let $E$ be an affine space attached to a $K$-vector space $T$. Let $G$ be the group of all bijective affine mappings of $E$ onto itself. Write $\phi$ for the group homomorphism associating   the $K$-linear isomorphism $\phi_u$ to $u\in G$ s.t.
$$u(t+x)=\phi_u(t)+u(x)$$
for all $t\in T$ and $x\in E$. Note that this means that $\phi$ is a surjection from $G$ onto $\mathbb{GL}(T)$. For each $t\in T$, write $h_t$ for the bijection of $E$ onto itself sending $x$ to $t+x$. Clearly $\{h_t\ |\ t\in T\}$ is a subset of $\text{Ker}(\phi)$. How can I show that $\text{Ker}(\phi)\subset\{h_t\ |\ t\in T\}$?

Let $u\in G$ such that $\phi_u=1_T$. Then $u(t+x)=t+u(x)$ for all $t\in T$ and $x\in E$. How can I show that $u=h_s$ for some $s\in T$?

答:
Fixing an $x\in E$, note that $t+x$ runs through all elements of $E$ when $t\in T$.

Then $u(t+x)=t+u(x)=(t+x)+(u(x)-x)$, where $u(x)-x\in T$, so for all $y\in E$, we get
$$u(y)=y+(u(x)-x)$$
thus $u=h_{u(x)-x}$.


问:
Explanation of the proof of $\phi(n)\cdot \frac{n}{2}=\sum_{a\in R}a$
In my number theory textbook, there is a theorem about Euler's Phi Function it says:

A Theorem:

If $R$ is a reduced residue system modulo $n$: $R=\{a\in C \mid \gcd(a,n)=1\}$ Where $C$ is a complete residue system modulo $n$, so:

$$\sum_{a\in R}a=\phi(n)\cdot \frac{n}{2}$$ Where $\sum_{a\in R}a$ is the sum of all elements in $R$.

The Proof:

since R is a reduced residue system mod $n$ so $\mid R\mid = \phi(n)$ Hence we can suppose that $R=\{a_1,a_2,...,a_{\phi(n)}\}$ Thus:
$$S=\sum_{a\in R}a=\sum_{i=1}^{\phi(n)}a_i$$ But since $\gcd(a_i,n)=1 \iff \gcd(n-a_i,n)=1$ so :$$\sum_{i=1}^{\phi(n)}a_i=\sum_{i=1}^{\phi(n)}(n-a_i)$$
$$ \implies2S = \sum_{i=1}^{\phi(n)}a_i+\sum_{i=1}^{\phi(n)}(n-a_i)=\phi(n)\cdot n$$
$$\iff S= \phi(n)\cdot \frac{n}{2}$$

Ok Everything is clear but this transition: $$\gcd(a_i,n)=1 \iff \gcd(n-a_i,n)=1$$
is not clear for me, and even if it is how can this statement implies this ?$$\sum_{i=1}^{\phi(n)}a_i=\sum_{i=1}^{\phi(n)}(n-a_i)$$

答:
Assume $\gcd(a_i,n)=1$. We want to show that $\gcd(n-a_i,n)=1$. $$\gcd(n-a_i,n)\mid -(n-a_i)+n=a_i$$ Now we see that $\gcd(n-a_i,n)$ divides all three $n-a_i,n,a_i$ therefore $\gcd(n-a_i,n)\mid \gcd(a_i,n)=1$. This is only possible if $\gcd(n-a_i,a_i)=1$.

Assume that $\gcd(n-a_i,n)=1$. We now want to show that $\gcd(a_i, n)=1$. $$\gcd(a_i,n)\mid n-a_i$$ Again, we see that $\gcd(a_i,n)$ divides all three $n-a_i, a_i, n$, in particular $\gcd(a_i,n)\mid\gcd(n-a_i, n)=1$. This is only possible if $\gcd(a_i, n)=1$.


Title: "Prove that if $G$ is a group, then $\{x \in G : x^2 =e \}$ is a subgroup of $G$": is that correct?

Body:
My teacher left this exercise:

"Prove that if $G$ is a group, then $\{x \in G : x^2 =e \}$ is a subgroup of $G$"

but I don't know if it's correct, I have seen similar exercises like that

"if $x^2 = e $ for all $x \in G$ then $G$ is abelian",

and in this case if $G$ is abelian then the proof is fairly simple but my teacher didn't say $G$ was abelian, so I'm not sure if that statement is true or if he forgot to say that $G$ is abelian.

I have tried to found a counterexample (if it exists) but haven't found any.

Any help? Thanks in advance

Body: A probably simple counterexample: in $\;S_3\;$ , we have that $\;(12)^2=(13)^2=1\;$ , yet $\;(12)(13)=(132)\;$, and $\;(132)^2\neq1\;$ , so that the set $\;\{x\in G\;|\;x^2=e\}\;$ isn't even closed...


答:
When we say "the $i$-th" it means that we have some collection of objects (in your case integers) that are indexed by naturals (so that for each object one natural number is given) and so the $i$-th is the generic object of the collection whose index (the natural used to label it) is just $i$.

So for instance, suppose we have your collection of integers $4,2,4,3,1,1$ we can label then so that each of them has a "name", we can make this by setting:

$$a_1=4 \quad a_2=2 \quad a_3=4 \quad a_4=3 \quad a_5=1\quad a_6=1$$

Then the $i$-th one is the generic $a_i$ where $i$ is some number between your labels which in this case are $1,2,3,4,5,6$.

Title: What does 'i-th' mean?

Body:
I have seen a problem set for the tower of hanoi algorithm that states:

Each integer in the second line is in the range 1 to K where the i-th
  integer denotes the peg to which disc of radius i is present in the
  initial configuration.

What does i-th actually mean? If i have a line conforming to the format above such as:

4 2 4 3 1 1

How exactly does this translate to the format described above?


问:
Help on how to proceed on this trig integral
i would appreciate if you could help me with this problem.
$$I=\int_{0}^{\pi}\frac{x \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}dx$$
I am completely overwhelmed on how to proceed with this and i am stuck. so i would appreciate some tips to start

after reading some tips i have done this
$$\int_{0}^{\pi}\frac{(\pi-x) \sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}dx$$

after that i distributed it
$$\int_{0}^{\pi}\frac{\pi\sin^{2018}(\pi-x)-x\sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}=\int_{0}^{\pi}\frac{\pi\sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}-I$$
But after that i'm lost again, i know i should be look for a way to make some terms cancel but that's as far as i got

please don't give me the direct solution, i would like to solve it by myself

Thanks in advance!

答:
HINT

Note that for even $n$ (as in your case) we have
$$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx$$

As asked for, this is a hint :) If you need any more help please don't hesitate to ask.

This is how I'd continue. We know that
$$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx$$
Adding, we find that
$$2\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx+\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx=\pi$$
So $$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\frac{\pi}{2}$$
and in particular $$\int_0^{\pi} \frac{\sin^{2018} x}{\sin^{2018}{x}+\cos^{2018}x}dx=\frac{\pi}{2}$$
We now know that
$$I=\pi\int_0^{\pi} \frac{\sin^{2018} x}{\sin^{2018}{x}+\cos^{2018}x}dx-I=\pi\times\frac{\pi}{2}-I$$
Hence $2I=\pi\times\frac{\pi}{2}=\frac{\pi^2}{2}$ and $I=\frac{\pi^2}{4}$.


答:
Note that $P(0)=det(-A)=(-1)^n det(A)$. Then we compute for $\lambda\neq 0$ $$Q(\lambda)=det(\lambda I-A^{-1}) = det (A^{-1}) det(\lambda A-I)=\lambda^n det(A^{-1}) det (A-\lambda^{-1}I)= (-\lambda)^n det(A)^{-1} P(\lambda^{-1}I-A)
=\lambda^n P(0)^{-1} P(\lambda^{-1}) .$$

Title: Prove that the characteristic polynomial $P(x)$ from $A$ and $Q(x)$ from $A^{-1}$ are related by...

Body:
I have to prove that $A$ characteristic's polynomial $P(x)$ and $A^{-1}$ characteristic's polynomial $Q(x)$ are related by:
$$ Q(\lambda) = \frac{\lambda^{n} P(\lambda ^{-1})}{P(0)} $$

I've tried by considering the matrix $A(A^{-1} − λI)$, but I've obteined no reults...

Any help?


问:
Showing the existence of a right-inverse in a von Neumann algebra
In this paper, specifically Theorem 4.1 on page 10, one of the last steps in the proof involves saying that a particular right-inverse exists. I'll try to restate what I think are the relevant pieces of info here, instead of just rewriting the whole proof.

Let $\mathfrak{A}$ be a von Neuman algebra with a faithful normal state $\rho$. In the last paragraph, the author looks at a self-adjoint, positive element $y \in \mathfrak{A}$, and then defines $q = \chi_{(1/2, 1]}(y)$, where this latter expression is meant in the sense of the Borel functional calculus on a self-adjoint $y \in \mathfrak{A}$. This $q$ satisfies
$$q \leq 2y \leq q + 1 \; (\dagger).$$
He then claims that there must exist $z \in \mathfrak{A}$ such that $\| z \|\leq 2$ and
$$1 - q = (1- y) z \; (\dagger \dagger) .$$

I'm confused by how he gets the right-hand inequality in $(\dagger)$, and how he conjures up the $z$ in $(\dagger \dagger)$. If anyone can clear this up, I'd greatly appreciate it!

答:
The first inequality is most likely a typo, it should be
$$
q\leq 2yq\leq 1+q.
$$
This is simply functional calculus on the inequalities
$$
1_{\big(\tfrac12,1\big]}(t)\leq 2t\,1_{\big(\tfrac12,1\big]}(t)\leq 2\,1_{\big(\tfrac12,1\big]}(t)\leq 1+1_{\big(\tfrac12,1\big]}(t).
$$
For $z$, take
$$
h(t)=\frac1{1-t}\,1_{\big[0,\tfrac12\big]}(t)
$$
and define $z=h(y)$. Since $|h(t)|\leq2$, you get that $\|z\|\leq2$.


问:
linear map of linear in/dependent vectors
L a linear map of $\mathbb{R}^m$ in $\mathbb{R}^n$.

1)Does a linear map of two linear dependent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear independent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?

2)a)Does a linear map of two linear independent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear dependent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?

b)And if it is what are the consequences for $ker(L)$?

1)No, because if two vectors are dependent:

 $\exists k \in \mathbb{R}$,

so that:  $k\underline{x}=\underline{y}$

and a linear map to two is linear dependent vectors $\in \mathbb{R}^n$is only possible by two vectors $\in \mathbb{R}^m$.

2)a)Yes, f.ex.: $f:\mathbb{R}^3 -> \mathbb{R}^2$,

with $f(x,y,z)^T = (x, y-z)^T$ for $(0,1,0)^T$ and $(0,0,1)^T$.

b)with the rank-nullity theorem:

$ dim V - dim (im T)= dim (ker T)$

$3 - 2 = 1 = dim (ker T)$ in the example

My questions are:

Is my explication for 1) & 2) sufficient?and if not how can I improve it?

Is 2)b) I don't really know how to interprete that...

答:
Hints:

I'd explain (1) as follows, applying explicitly a linear map:

$$x,y\in\Bbb R^m\;,\;\;a\in\Bbb R\;,\;\;T\;\text{a linear map}\;,\; x=ay\implies Tx= T(ay)=aTy$$

You don't need to worry about (2)-(b) since (1)=(2)-(a) cannot be.

Edit: After the correction to (2)-(a): the zero map is a trivial example of this possibility, and in general, if

$$x,y\in\Bbb R^m\;,\;a\in\Bbb R\;\wedge\;Ty=aTx=T(ax)\implies T(y-ax)=0\implies \ker T\neq 0$$


问:
Can a finite group act transitively on an infinite set?
I had a question on my algebra exam, asking me to show that in case of a transitive action of a finite group $G$ on a set $X$, $|X|$ divides $|G|$. In case of a finite set, this is trivial, so I tried to see if I could come up with a possible counterexample with $X$ infinite (I failed to do so). Intuitively I feel like an infinite $X$ is "too big" for a finite group to act transitively on, but how would one show this formally?

答:
Given a fixed, arbitrary $x\in X$, transitivity means that the set
$$
Gx = \{gx\mid g\in G\}\subseteq X
$$
is equal to $X$. However, this set clearly cannot be larger than $G$.


答: Let $y$ be a fixed element of $X$. For any element $x\in X$, transitivity implies there exists some $g\in G$ such that $g\cdot y=x$. This induces a injective map from $X\to G$, which is a contradiction.

Title: Can a finite group act transitively on an infinite set?

Body: I had a question on my algebra exam, asking me to show that in case of a transitive action of a finite group $G$ on a set $X$, $|X|$ divides $|G|$. In case of a finite set, this is trivial, so I tried to see if I could come up with a possible counterexample with $X$ infinite (I failed to do so). Intuitively I feel like an infinite $X$ is "too big" for a finite group to act transitively on, but how would one show this formally?


答:
Actually, if $X$ is empty, any group acts transitively on $X$, but $\lvert X \rvert = 0$ cannot divide $\lvert G \rvert$, so the claim might actually be false (depending on whether or not you allow $G$-sets to be empty).

Anyway, suppose $G$ is a finite group acting transitively on a nonempty set $X$. Since $X$ is nonempty, we may pick some $x \in X$. Then define $a : G \to X$ by $a(g) = g \cdot x$. By definition of transitive action, $a$ is surjective. Thus, $\lvert G \rvert \geq \lvert X \rvert$, so $X$ is finite.


问:
Optimization approaches to solving PDEs
In modern numerical methods, a PDE is often recast into the form of a variational problem, which is sometimes equivalent to a minimization problem.
However in my courses on numerical analysis (say, finite element methods) the focus is not (apparently) on developing optimization techniques to minimize the arosen energy functional, but rather on approximating the variational problem on a smaller subspace.

Are there interesting approaches that focus on the minimization of the energy directly? Is research being done in this field, and could you maybe provide some reference?

答:
I'll try to answer to your questions following the order of their appearance.

Are there interesting approaches that focus on the minimization of the energy directly?

Yes, there are several approaches to the solution of this problem: they are all known under the collective name of "Variational Method". This "method" has its root in the direct method in the calculus of variation which was pioneered by Leonida Tonelli, and aims to find the minimum of a functional by directly evaluating its value on a properly defined (sub)sequence of functions for which it is defined, without the need to calculate its functional derivative and solve the associated Euler-Lagrange DEs. Contrary to some widespread belief, the variational method is widely applicable to general (system of) PDEs, as we'll see below.

Let's start our description by saying that, when a functional $F$ which is equivalent to the given (system of) PDEs has been constructed, the general variational method can be described by mimicking the standard approach to direct method in the calculus of variation:$\DeclareMathOperator{\DM}{\mathrm{Dom}}$

show that the functional $F$ is (sequentially) lower semicontinuous (or upper semicontinuous, if we are searching for a maximum instead of a minimum) respect to a topology chosen on its domain of definition $\DM F$.

chose a minimizing sequence i.e. a sequence of functions $\{u_n\}$ contained in the domain $\DM F$ such that
$$
\lim_{n\to +\infty} F[u_n]=\inf_{u\in\DM F}F(u)
$$

Show that the chosen sequence $\{u_n\}$ admits a subsequence $\{u_{n_k}\}$ covering to $u\in \DM F$ respect to the given topology. Then $u$ is the sought for solution.

Differently from what happens in the direct method, where the functional is customarily known from the start, step 1. for the variational method is customarily done during the construction of $F$. For example, the very general method for giving a variational formulation to any operator equation worked out by Enzo Tonti (see for example [3] and [1], chapter IV, §17, pp. 168-182) produces the a functional which has a minimum $u$ if and only if $u$ solves the initial operator equation, implicitly proving its semicontinuity.

Finally, what are the differences between the various procedures that go under this collective name? They use different methods to construct the minimizing sequence and its subsequence: for example the method of orthonormal series, the Ritz (or Rayleigh-Ritz) Method, the Galerkin Method or its variants as the Faedo- Bunbnov- Galerking Methods, the Least Squares Method, The Courant Method and The Method of Steepest Descent are all variational methods.

Is research being done in this field, and could you maybe provide some reference?

Yes, there is currently a large deal of investigation on the variational method, due to its wide applicability to standard problems in engineering and physics: try googling one of the names given above, just to see the number of the results that you'll get. As a basic reference for the standard techniques, I'd suggest [2], part III, chapters 9-17, pp. 113-178, while the problem of constructing a functional "equivalent" to a given (system of) PDEs, technically called "the inverse problem of the calculus of variation", is comprehensively addressed in references [1] and [3]: the first one gives also a very general procedure for solving very general nonlinear equations (chapter IV, §18, pp. 182-198) due to A. D. Lyashko.

Final notes

The approach used in the solution of the inverse problem of the calculus of variation is briefly addressed also in this and this Q&A.

As I briefly alluded to in the answer to the first part of the question, it is a widespread belief that only some PDEs (or more generally some operator equations) admit a variational formulation, even if the works of Tonti, Filippov and others have shown that the inverse problem of the calculus of variation has a generalized solution for large classes of operator equations. However, the belief is not entirely unjustified since it is true that there are large classes of functional formulations of PDE problems for which the functional derivative does not lead to the Euler-Lagrange equations: for example Copson [A1] proves that the ordinary heat equation (and more generally equations of parabolic type) is not the Euler-Lagrange equation of any functional. But it also clear from the works [1] and [2] that such kind of equations do admit a variational formulation.

References

[1] Filippov, Vladimir Mikhailovich, Variational principles for nonpotential operators. With an appendix by the author and V. M. Savchin, Transl. from the Russian by J. R. Schulenberger. Transl. ed. by Ben Silver, Translations of Mathematical Monographs, 77. Providence, RI: American Mathematical Society (AMS). pp. xiii+239 (1989), ISBN: 0-8218-4529-2. MR1013998, Zbl 0682.35006.

[2] Karel Rektorys, Variational methods in mathematics, science and engineering, Translated from the Czech by Michael Basch. 2nd ed. (English), Dordrecht - Boston - London: D. Reidel Publishing Company, pp. 571 (1980), ISBN: 90-277-1060-0, MR0596582,  Zbl 0481.49002.

[3] Tonti, Enzo, "Extended variational formulation", Vestnik Rossiĭskogo Universiteta Druzhby Narodov, Seriya Matematika 2, No. 2, 148-162 (1995). Zbl 0965.35036.

Addendum reference

[A1] Edward Thomas Copson, "Partial differential equations and the calculus of variations" Proceedings of the Royal Society of Edinburgh 46, 126-135 (1926), JFM 52.0509.01.

Title: Optimization approaches to solving PDEs


问:
Regarding limits and $1^\infty$

Possible Duplicate:

  Why is $1^{\infty}$ considered to be an indeterminate form

I have some questions about limits and the undefinability of $1^\infty$.

For example, is $\lim_{x\to\infty}1^x$ indefinite? Why is it not $1$? Or do mathematicians, when saying that $1^\infty$ is indefinite, actually refer to cases such as $lim_{x\to\infty} \left(1 + \frac{a}{x}\right)^x$ where even though at a first glace the result is $1$, this is actually a special case and it is equal to $e^a$?

答: When people say that $1^{\infty}$ is an indeterminate form, what they mean is that if $f(x)$ is a function such that $\lim_{x \to r} f(x) = 1$ and $g(x)$ is a function such that $\lim_{x \to r} g(x) = \infty$, the value of $\lim_{x \to r} f(x)^{g(x)}$ is not uniquely determined in general. It is determined in the special case that $f(x) = 1$, in which case the limit is obviously just $1$.

Title: Regarding limits and $1^\infty$


问:
Existence of a continuous function $ \theta $ with $ f ( x ) = | f ( x ) | \exp \big( i \theta ( x ) \big) $

Consider a continuous function $ f : \mathbb R \to \mathbb C \setminus \{ 0 \} $. Is there a continuous function $ \theta : \mathbb R \to \mathbb R $ such that $ f ( x ) = | f ( x ) | \exp \big( i \theta ( x ) \big) $ for all $ x \in \mathbb R $?

My thoughts:

I intuitively think that this is true. Suppose that we fix some value for $ \theta ( 0 ) = \arg f ( 0 ) $ from the infinite possibilities (which differ from the one we fixed by an integer multiple of $ 2 \pi $). As $ f $ only takes nonzero values and is continuous, for all $ x \in \mathbb R $ close enough to $ 0 $, the angle between the positive real axis and the line joining $ f ( x ) $ to the origin in the complex plane is as close as we want to the corresponding angle for $ f ( 0 ) $. From all the values we can take for the measure of this angle, there is only one that is that close to $ \theta ( 0 ) $, and choosing that one for $ \theta ( x ) $, $ \theta $ will be continuous around $ 0 $. This can intuitively be done over and over again for the new points other than $ 0 $ with a fixed value of $ \theta $, and thus the result we want.

I understand that there's a flaw with this reasoning, as for example after the first step, when we've found a neighborhood $ N $ of $ 0 $ with fixed value of $ \theta ( x ) $ for $ x \in N $, when we repeat the argument for a new point $ x _ 0 \in N $, the resulting neighborhood $ N _ 0 $ for $ x _ 0 $ may be completely included in $ N $, and we don't actually get any new points with fixed value of $ \theta $. This makes me think that if $ f $ is uniformly continuous or something, so that we can ensure $ N _ 0 $ is as large as $ N $, the above intuition can be made into a rigorous proof, but not necessarily for the case where the only condition on $ f $ is mere continuity.

In any case, I neither could find a counterexample nor could find a rigorous proof and go beyond the above intuition. So is the statement really true?

P.S. I'm aware of the fact that if the domain of $ f $ was say $ \mathbb C \setminus \{ 0 \} $, the statement didn't hold, as it is wrong even for a function as simple as the identity function. But it seems to be true for the case when the domain of $ f $ is $ \mathbb R $ or an interval. While my real question is the one above, I would really appreciate an answer which gives an insight about the kind of domains that make this statement true and the ones which make it false.

答:
Proposition: For any $c\in\Bbb R$, the map$$\begin{array}{rccc}\eta\colon&(c,c+2\pi)&\longrightarrow&S^1\\&t&\mapsto&e^{it}\end{array}$$is a homeomorphism onto $S^1\setminus\{e^{ic}\}$.

Proof: It is clear that $\eta\bigl((c,c+2\pi)\bigr)=S^1\setminus\{e^{ic}\}$ and that $\eta$ is continuous and injective. What remains to be proved is that $\eta^{-1}\colon S^1\setminus\{e^{ic}\}\longrightarrow(c,c+\pi)$ is continuous. Let $F\subset(c,c+\pi)$ be a closed subset. Then the inverse image of $F$ with respect to $\eta^{-1}$ is $\eta(F)$. Since $F$ is a closed subset of $(c,c+\pi)$, $F\cup\{c,c+2\pi\}$ is a closed subset of $\Bbb R$. Actually, it is compact, since it is bounded too. But then $\{e^{iz}\mid z\in F\cup\{c,c+2\pi\}\}$ is a compact subset of $S^1$, and, in particular, it is a closed set. And$$\eta(F)=\{e^{iz}\mid z\in F\cup\{c,c+2\pi\}\}\setminus\{e^{ic}\},$$which is therefore a closed subset of $S^1\setminus\{e^{ic}\}$. Since the inverse image with respect to $\eta^{-1}$ of any closed subset of $(c,c+\pi)$ is a closed subset of $S^1\setminus\{e^{ic}\}$, $\eta^{-1}$ is continuous.

Proposition: If $f\colon\Bbb R\longrightarrow\Bbb C\setminus\{0\}$ is a continuous map, then there is a continuous map $\theta\colon\Bbb R\longrightarrow\Bbb R$ such that$$(\forall t\in\Bbb R):f(t)=|f(t)|e^{i\theta(t)}.$$

Proof: Let $F(t)=\frac{f(t)}{|f(t)|}$; we want to prove that there is a continuous map $\theta\colon\Bbb R\longrightarrow\Bbb R$ such that$$(\forall t\in\Bbb R):F(t)=e^{i\theta(t)}.$$Note that $F$ is a map from $\Bbb R$ into $S^1$. If $x\in\Bbb R$, then, since $F$ is continuous, there is some $\varepsilon_x>0$ such that $|y-x|<\varepsilon_x\implies|F(y)-F(x)|<2$. In particular, $F\bigl((x-\varepsilon_x,x+\varepsilon_x)\bigr)$ is not the whole $S^1$, since $-F(x)$ does not belong to it (because $F(x)\in S^1$ and therefore the distance from $F(x)$ to $-F(x)$ is equal to $2$). If $c\in\Bbb R$ is such that $e^{ic}=-F(x)$, then, by the previous proposition, the map$$\begin{array}{rccc}\eta\colon&(x-\pi,x+\pi)&\longrightarrow&S^1\\&t&\mapsto&e^{it}\end{array}$$is a homeomorphism onto $S^1\setminus\{-F(x)\}$. If $\mu$ is its inverse, then$$(\forall t\in(x-\varepsilon_x,x+\varepsilon_x)):F(t)=\eta\bigl(\mu\bigl(F(t)\bigr)\bigr)=e^{i\mu(F(t))}.$$In particular,$$(\forall t\in(-\varepsilon_0,\varepsilon_0)):F(t)=e^{i\mu(F(t))}.$$So, if you define $T$ as the set of those $r\in(0,\infty)$ such that, for some continuous function $\theta\colon[0,r]\longrightarrow\Bbb R$, you have$$(\forall t\in[0,r]):F(t)=e^{i\theta(t)},$$then $T\ne\emptyset$, since it contains the interval $(0,\varepsilon_0)$. If $T\ne(0,\infty)$, then it has an upper bound. Let $s=\sup S$. We have seen that there is some interval $(s-\varepsilon_s,s+\varepsilon_s)$ and some continuous function $\theta\colon(s-\varepsilon_s,s+\varepsilon_s)\longrightarrow\Bbb R$ such that we have $F(t)=e^{i\theta(t)}$ on $(s-\varepsilon_s,s+\varepsilon_s)$. Fix some $u\in T\cap(s-\varepsilon_s,s)$. Then, by the definition of $T$, there is some continuous function $\theta^\ast\colon[0,u]\longrightarrow\Bbb R$ such that$$(\forall t\in[0,u]):F(t)=e^{i\theta^\ast(t)}.$$Since, on $[u,s)$ we have $e^{i\theta(t)}=e^{i\theta^\ast(t)}$, since both $\theta$ and $\theta^\ast$ are continuous, and since $[u,s)$ is an interval, there is some $n\in\Bbb Z$ such that $(\forall t\in[u,s)):\theta(t)=\theta^\ast(t)+2n\pi$. So, if you replace $\theta^\ast$ by $\theta$ you will have$$(\forall t\in[0,s+\varepsilon_s)):F(t)=e^{i\theta(t)}.$$But then $T\supset[0,s+\varepsilon_s)$, which is impossible, since $s=\sup T$. We have reached a contradiction. So, the assumption that $T\ne(0,\infty)$ is false. In other words, there is a continuous function $\theta\colon[0,\infty)\longrightarrow\Bbb R$ such that$$(\forall t\in[0,\infty)):F(t)=e^{i\theta(t)}$$and, by a similar argument, there is a continuous function $\theta\colon(-\infty,0]\longrightarrow\Bbb R$ such that$$(\forall t\in(-\infty,0]):F(t)=e^{i\theta(t)}.$$


问:
Considering the Digital Line Topology from the set of integers, prove that there exists a singleton set that is not closed.
The digital line topology is

B(n) = {n} when n is odd

B(n) = {n-1, n, n+1} when n is even.

I've figured out how to prove that the topology is not Hausdorff, but was wondering if there was a way to prove that there is a singleton set in the digital line topology that is not open.

Right now I have that if n is even then the set {n} is not closed
since it does not have a complement in T that is open. Is this reasoning sufficient, or am I missing something?

答:
Assume shown that the collection $B(n)$ as described is a basis, and further assume $n$ is odd.  We show that $\lbrace n \rbrace$ is not closed, by showing that its complement is not open.

If its complement $U$ were open, then $n-1$ would be in $U$ and there would be a basis element $B$ containing $n-1$ in $U$.  But since $n$ is odd, we have that $n-1$ is even and thus every basis element $B$ containing $n-1$ contains the elements $n-2$ and $n$.  In particular we have that $n \in B$.  But $B \subset U = \lbrace n \rbrace^c$ was assumed to be disjoint from $\lbrace n \rbrace$, a contradiction.


问:
Show that for two vector spaces there exist a surjective linear map $\varphi$

If $V = \{(x,y,z) \in \mathbb{R^3} : x+2y-z=0 \}$ and $W=\{(x,y,z) \in \mathbb{R^3} : -x+z=0 \}$ then show that there exists a surjective linear map $\varphi: V\to W$.

How can I show this when $\varphi$ is not given? $\varphi$ is surjective iff $\operatorname{im}(\varphi)=W.$ This is just to show that two sets are the same so i'm thinking of proving this by showing $\operatorname{im}(\varphi) \subset W$ and $W \subset \operatorname{im}(\varphi)$, but i cannot do it. Is this the right way?

答:
$V\cong\Bbb R^2$ since any vector in $V$ may be written $$(x,y,z)=x(1,0,1)+y(0,1,2)$$
(any vector may be written as a linear combination of two vectors, which themselves are linearly independent. Therefore $\dim_\Bbb R(V)=2$)

Also, $W\cong\Bbb R^2$ since again any vector in $W$ may be written $$(x,y,z)=x(1,0,1)+y(0,1,0)$$

Therefore $V\cong W$, so there is an isomorphism $\varphi: V\rightarrow W$, in particular $\varphi$ is surjective.

You can always find a surjective linear map $\varphi:V\rightarrow W$ if $\dim(W)\leq\dim(V)$. Just map a basis in $V$ to basis vectors in $W$. Let $\mathfrak B_V=\{v_1,v_2,\ldots,v_n\}$, $\mathfrak B_W=\{w_1,w_2,\ldots,w_m\}$ be a basis for $V$ and $W$ respectively. Then consider the map $$\varphi:V\rightarrow W\\\varphi(v_i)=w_i$$ To see that this map is surjective let $a_1w_1+\ldots+a_mw_m\in W$ be any vector, notice that $$a_1w_1+\ldots+a_mw_m=a_1\varphi(v_1)+\ldots+a_m\varphi(w_m)\\=\varphi(a_1v_1+\ldots+a_mv_m)$$ by definition of a linear map. Thus, you have the desired vector $a_1v_1+\ldots+a_mv_m$.


答:
Well, it seems you get the idea.

To improve, introduce a letter for the linear map we are talking about, say, $T$.
Then you can write like $k\,T(x)=T(kx)=T(y)$.

For 2b, if $k\,T(x)=T(y)$ then $kx-y\in\ker T$.

Title: linear map of linear in/dependent vectors

Body:
L a linear map of $\mathbb{R}^m$ in $\mathbb{R}^n$.

1)Does a linear map of two linear dependent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear independent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?

2)a)Does a linear map of two linear independent vectors $\underline{x}$,$\underline{y} \in \mathbb{R}^m$ to two linear dependent vectors $\underline{u}$,$\underline{v} \in \mathbb{R}^n$ exist/is possible?

b)And if it is what are the consequences for $ker(L)$?

1)No, because if two vectors are dependent:

 $\exists k \in \mathbb{R}$,

so that:  $k\underline{x}=\underline{y}$

and a linear map to two is linear dependent vectors $\in \mathbb{R}^n$is only possible by two vectors $\in \mathbb{R}^m$.

2)a)Yes, f.ex.: $f:\mathbb{R}^3 -> \mathbb{R}^2$,

with $f(x,y,z)^T = (x, y-z)^T$ for $(0,1,0)^T$ and $(0,0,1)^T$.

b)with the rank-nullity theorem:

$ dim V - dim (im T)= dim (ker T)$

$3 - 2 = 1 = dim (ker T)$ in the example

My questions are:

Is my explication for 1) & 2) sufficient?and if not how can I improve it?

Is 2)b) I don't really know how to interprete that...


问:
What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?
Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?

I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$

Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.

So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.

Therefore the probability of the triangle area being greater than 2 is  $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$

However the solution says the answer is 8/9, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!

答:
If the upper vertex of the triangle is $(x,y)$, the area of the triangle formed by $(0,0)$, $(3,0)$, and $(x,y)$ is ${1 \over 2}bh$ or ${3h \over 2}$.

So the probability the area is at least $2$ is the probability that ${3h \over 2} > 2$, or equivalently the probability that $h > {4 \over 3}$. Since $h$ is uniformly distributed in $[1,4]$, the probability of this event is ${4 - {4 \over 3} \over 4 - 1} = {8 \over 9}$.


问:
Why do we make the distinction between $(A_1,B_2)$ and $(B_2, A_1)$ but not $(A_1,A_2)$ and $(A_2,A_1)$
I have the following problem:

A pair of fair dice is tossed. Find the probability of getting a total
of $8$

The problem seemed fairly simple until I looked at the solution after solving it.

The author did the following:

Of the $(6)(6)=36$ elements in the sample space, only $5$ elements $(2,6), (3,5), (4,4), (5,3)$, and $(6,2)$ add to $8$. Hence the probability of obtaining a total of $8$ is $5/36$

What I don't understand is why we didn't count $(4,4)$ twice, when solving the exercise I had the same reasoning except that I counted $(4,4)$ twice since I see that $(A_1,A_1) \neq (A_2,A_1)$

Note: the solution comes from the solution manual of the book.
Any help would be appreciated.

答:
I don't know if you'll be convinced by a simulation, but here
are results from ten million rolls of two dice--one red and one blue
as @NDTaussig suggests. With 10 million replications, probabilities approximated by simulation should be accurate to about three decimal places.

```<code>set.seed(2021)
n = 10^7
r = sample(1:6, n, rep=T)
table(r)/n
r
        1         2         3         4         5         6
0.1665514 0.1666612 0.1667533 0.1665278 0.1666923 0.1668140
b = sample(1:6, n, rep=T)
table(b)/n
b
        1         2         3         4         5         6
0.1668560 0.1666602 0.1666850 0.1666860 0.1666270 0.1664858
s = r + b
table(s)/n
s
        2         3         4         5         6         7
0.0277760 0.0556208 0.0833264 0.1111920 0.1389286 0.1665705
        8         9        10        11        12
0.1388043 0.1110687 0.0833665 0.0555016 0.0278446
mean(s == 8)
[1] 0.1388043   # aprx P(S = 8) = 5/36
5/36
[1] 0.1388889

hdr = "Distribution of Sums on Two Fair Dice"
frb = rep("wheat", 11);  frb[7]="skyblue2"
ctp=(1:12)+.5
hist(s, prob=T, br=ctp, col=frb, main=hdr)
</code>```

Body:
I don't know if you'll be convinced by a simulation, but here
are results from ten million rolls of two dice--one red and one blue
as @NDTaussig suggests. With 10 million replications, probabilities approximated by simulation should be accurate to about three decimal places.

set.seed(2021)
n = 10^7
r = sample(1:6, n, rep=T)
table(r)/n
r
        1         2         3         4         5         6
0.1665514 0.1666612 0.1667533 0.1665278 0.1666923 0.1668140
b = sample(1:6, n, rep=T)
table(b)/n
b
        1         2         3         4         5         6
0.1668560 0.1666602 0.1666850 0.1666860 0.1666270 0.1664858
s = r + b
table(s)/n
s
        2         3         4         5         6         7
0.0277760 0.0556208 0.0833264 0.1111920 0.1389286 0.1665705
        8         9        10        11        12
0.1388043 0.1110687 0.0833665 0.0555016 0.0278446
mean(s == 8)
[1] 0.1388043   # aprx P(S = 8) = 5/36
5/36
[1] 0.1388889

hdr = "Distribution of Sums on Two Fair Dice"
frb = rep("wheat", 11);  frb[7]="skyblue2"
ctp=(1:12)+.5
hist(s, prob=T, br=ctp, col=frb, main=hdr)


问:
$L^{P} ( \Omega ; W^{1,P}(Y))$ can someone define this space. $L^{P}$ and $W^{1,P}$ are sobolve and Lp spaces. How norm can define?
I have this space $L^{P} ( \Omega ; W^{1,P}_{per}(Y))$ can someone define it. As far I understands it takes a function f from $L^{P}(\Omega)$ and maps it to Sobolov space with periodicity $Y$.
Or $f : W^{1,P}_{per}(Y) \rightarrow L^{P}(\Omega)$.
i am confused here???
here $L^{P}$ and $W^{1,P}$ are sobolov and Lp spaces. Y is the periodicity of domain $\Omega$.
How norm can define?

$||f||_{L^{P} ( \Omega ; W^{1,P}_{per}(Y))} = \int_{\Omega}  ||f||_{W^{1,P}_{per}(Y)}  dx$

答:
Functions in $L^P(\Omega; W_{per}^{1, P}(Y))$ are functions $f: \Omega \rightarrow W_{per}^{1, P}(Y)$ s.t. the norm on the space is well-defined at $f$. A suitable norm is
$$
\lVert f \rVert_{L^P(\Omega; W_{per}^{1, P}(Y))} := \left (\int_\Omega \lVert f(x) \rVert_{W_{per}^{1, P}(Y)}^p ~\mathrm{d}x \right)^\frac{1}{p}.
$$


问:
Determining a polynomial $p$ from its image of rationals.
Let $f$ and $g$ be polynomials with rational coefficients, and let $F$ and $G$ denote the sets of values of $f$ and $g$ at rational numbers. Prove that $F = G$ holds if and only if $f(x) = g(ax + b)$ for some suitable rational numbers $a \not= 0$ and $b$.

This is a problem I found from "Contests in Higher Mathematics" whose problems are from the Miklos Schweitzer Competitions. I have found the problems in this competition deeply satisfying to think about and I would recommend for the reader to check them out

I was hoping to get a hint for the solution. Please do not spoil the problem for me. I am only looking for a nudge.

答:
$f(x)=\sum_{n=0}^N c_n x^n$. For $v_p(1/b)=k$ large enough we have $v_p(f(b))= v_p(c_N)-Nk$.

Thus, from $f(\Bbb{Q})$ we know $N$ and $v_p(c_N)\bmod N$.

That is to say we know the image of $c_N$ in $\Bbb{Q^*/ \pm Q}^{*N}$.

If $f(\Bbb{Q})=g(\Bbb{Q})$ then let $g_2(x)=\pm g(ax)$ where $a$ and $\pm $ are chosen such that the leading coefficient of $g_2$ is equal to $c_N$.

In fact we can assume the sign is $+$: if $N$ is odd then replace $a$ by $-a$, if $N$ is even then check if $g(\Bbb{Q})$ is bounded above or below to deduce the sign of the leading coefficient of $g$.

Let $f_3(x)=f_2(x+r_f),g_3(x)=g_2(x+r_g)$ where $r_f,r_g$ are chosen such that the coefficient of $x^{N-1}$ is $0$.

There is some integer $d$ such that  both $f_4(x)=c_N^{-1} d^{N} f_3( x/d),g_4(x)=c_N^{-1} d^{N} g_3(x/d)$ are $\in \Bbb{Z}[x]_{monic}$.

$f_4(n)\in \Bbb{Z}$ iff $n\in \Bbb{Z}$.

$f_4$ is determined by $n,f_4(n),\ldots,f_4(n+N)$.

If $N$ is odd:

For $n$ large enough $f_4(n),\ldots,f_4(n+N)$ are $N+1$ consecutive integers in $f_4(\Bbb{Q})\cap \Bbb{Z}$.

From $f_4(\Bbb{Q})\cap \Bbb{Z}$ we can pick $N+1$ large enough consecutive integers to get $f_4(n),\ldots,f_4(n+N)$ for some unknown $n$. Whence $f_4(\Bbb{Q})\cap \Bbb{Z}$ tells us $f_4$ up to a shift by an integer.

If $N$ is even and $f_4(x)\ne f_4(-x)$. Wlog, replacing $f_4$ by $f_4(-x)$ if needed we can assume that $f_4(n)>f_4(-n)$ for $n$ large enough.

Then, for $n$ large enough (as the coefficient of $x^{N-1}$ is $0$) $f_4(-n),f_4(n),f_4(-n-1),f_4(n+1),\ldots,f_4(-n-N),f_4(n+N)$ are $2N+2$ consecutive integers in $f_4(\Bbb{Q})\cap \Bbb{Z}$.

Taking one of two of such consecutive integers tells us either $f_4(n),\ldots,f_4(n+N)$ or $f_4(-n),\ldots,f_4(-n-N)$ for some unknown $n$, which tells either $f_4(x)$ or $f_4(-x)$ up to a shift.

If $f_4(x)=f_4(-x)$ then (since we know $N=\deg(f_4)$) this can be seen from the asymptotic of $f_4(\Bbb{Q})\cap \Bbb{Z}$ and it is similar to the odd $N$ case.

ie. $f(\Bbb{Q})=g(\Bbb{Q})$ implies $f_4(\Bbb{Q})=g_4(\Bbb{Q})$ so that $f_4(x)=g_4(\pm x+k)$ for some $k$ and $f(x)= g(ax+b)$.


问:
Uniform convergence, differentiability theorem's converse
Suppose $\{f_n \}$ is a sequence of differentiable functions on $[a,b]$ such that $f_n \to f$ uniformly, and $f'(x) = \lim_{n \to \infty} f_n'(x)$ (pointwise). Then is it true that $f_n'$ converges uniformly?

My initial reaction is that the answer is yes, because all the differentiable $f_n$'s will be able to fit inside an "$\epsilon \text{-thick}$ tube" around $f$, but intuition for sequences of functions rarely seem to be correct. I'd appreciate some help with this; thank you!

答:
This is not technically a counterexample, since it does not satisfy the second condition, but I believe it gives the right idea of why it will not work (I am not sure).

Consider the sequence
$$ f_{n}(x) = \frac{1}{n} \sin(n^{2}x). $$
This converges uniformly to $f(x) = 0$, but clearly the derivative is not going to converge uniformly to zero. Of course, it will not converge at all; the point, however, is that even if the change over an interval is bounded, the rate of change can be unbounded if the size of the interval shrinks fast enough.

So probably a counterexample could be constructed by replacing the $n^{2}x$ with something more complicated. I think perhaps that $nx^{-1}$ might work (basically the topologist's sine function but with the scaling factor).