问:
About calculating a Green's function
For a positive integer $d>=3$ and a real number $h_0$ I have the differential equation for a function $f$ of $x$,

$$x^2 h^2 f'' + \frac{6-d}{2}xh^2 f' = \frac{d(d-2)}{2}h^2f$$

where $h = h_0 x^{1- \frac{d}{2}}$.
(..and I believe that this explicit relationship between $h$ and $x$ is not necessary to solve the question..)

Its full Green's function is claimed as

$$G(x,x') = \theta(x-x')\frac{2x'^{d-3}}{h_0^2(3d-4)}[(x/x')^{-d/2} - (x/x')^{d-2}  ]$$

Let me state here as I would have solved this question.

First let me shift to variable $a$ such that $x = e^a$ and drop the overall factor of $h$ (it can never be zero) and then I have the differential equation as,

$[\frac{d^2}{da^2} + (2 - \frac{d}{2})\frac{d}{da} - \frac{d(d-2)}{2}]f(a) = 0$

So the Green's function equation that one wants to solve is,

$[\frac{d^2}{da^2} + (2 - \frac{d}{2})\frac{d}{da} - \frac{d(d-2)}{2}]G(a-a') = \delta (a-a')$

Now we use Dirac delta function convention as $\delta (a-a') = \int \frac{dk}{\sqrt{2\pi}}e^{ik.(a-a')} $ and hence the Green's function and}( its Fourier transform is defined as $G(a-a') = \int \frac{dk}{\sqrt{2\pi}} \tilde{G}(k) e^{ik(a-a')}$

And substituting the above one is solving the algebraic equation,

$[(ik)^2 + (2 - \frac{d}{2})ik - \frac{d(d-2)}{2}]\tilde{G}(k) =  1$

hence we have, $\tilde{G}(k) = \frac{-1}{(k-\frac{id}{2})(k-i(2-d))}$

Now with the above $\tilde{G}(k)$ we want to compute the contour integral, $ \int \frac{dk}{\sqrt{2\pi}} \tilde{G}(k) e^{ik(a-a')}$

Since $d\geq3$ one of the poles is on the positive imaginary axis, $\frac{id}{2}$ and one is on the negative imaginary axis, $i(2-d)$. For $a>a'$ I need to close the contour in the upper half plane for convergence and also I need both the poles to contribute. Hence I choose to close contour as the usual semi-circle on the UHP with the flat side on the real axis but while going near the origin I take a detour around the origin to go around the $i(2-d)$ pole in the LHP. Then I take the limit of this neck width to pinch to zero.

Then doing the usual $\sqrt{2\pi i}Residue$ analysis I get,

(after changing back to "x")

$G(x,x') = \frac{-2\sqrt{2\pi}}{3d-4}[(x/x')^{-d/2} + (x/x')^{d-2}]$

But the above is not the right answer! - What can be changed in this Fourier transform based method to get the right answer?

(the questionable thing that I am doing is to kind of deform the semi-circle contour such that it actually has to deform across the LHP pole and which is kind of tricky)

答:
There is indeed some missing information and misprints but the correct question can be reconstructed from the quoted answer.

It seems that $a$ in your equation should be replaced by $x$.

It also seems that the question is to find the Green function corresponding to the Cauchy problem on $[0,\infty)$ for the inhomogeneous differential equation
$$x^2h^2f''+\frac{6-d}{2}xh^2f'-\frac{d(d-2)}{2}h^2f=g(x).\tag{1}$$

In other words, one looks for solution of (1) satisfying the initial conditions $f(0)=f'(0)=0$
in the form
$$f(x)=\int_0^{\infty} G(x,y)g(y)dy,\tag{2}$$
which means that Green function $G(x,y)$ is a solution of (1) with $g(x)=\delta(x-y)$.

The general form of the Green function for such problem is
$$G(x,y)=\begin{cases}
\frac{f_1(x)f_2(y)-f_2(x)f_1(y)}{y^2h^2_yW(f_1(y),f_2(y))},\qquad & x\geq y\\
0,\qquad & x<y,
\end{cases}\tag{3}$$
where $f_1$, $f_2$ are two linearly independent solutions of the equation (1) with $g=0$ (which can be chosen arbitrarily) and $W(f_1,f_2)=f_1'f_2-f_1f_2'$ denotes their Wronskian.
The same formula holds for any 2nd order linear ODE if we replace the expression $y^2h^2_y$ in the denominator by the appropriate coefficient of $f''$ (expressed as function of $y$!). The easiest way to prove (3) is to plug it into (2) and then verify that (1) holds and the homogeneous initial conditions are satisfied.

In your case, the linearly independent solutions can be chosen as $f_1(x)=x^{d-2}$, $f_2(x)=x^{-d/2}$, which gives $W(f_1(y),f_2(y))=\frac{3d-4}{2}y^{d/2-3}$. Now it is an easy exercise to check that (3) is equivalent to the formula for $G(x,y)$ quoted in the question.

Added:

First of all about $h^2$. It cannot be dropped - in fact, the question of finding the Green function of homogeneous
 equation with homogeneous boundary conditions is meaningless, something should be non-zero. On the other hand, the dependence
 on this prefactor is almost trivial - if you divide by it, $f_{1,2}$ will remain the same and only $g(x)$ in (1) will change. This
 is precisely the reason why $h_y^2=h^2(y)$ is present in the denominator of (3).

Fourier transform can in principle be used but I strongly discourage this. Firstly,
 it will work only for equations reducible to linear ODEs with constant coefficients, whereas the
 formula (3) is valid for any linear 2nd order ODE. Secondly, one should well understand what is going on in the course of solution:

The solution you have written for $\tilde{G}(k)$ is not the most general one. One could add to it something like
$A\delta(k-id/2)+B\delta(k-i(2-d))$ with arbitrary $A$ and $B$. The fact that these two terms are actually absent
is related to homogeneous initial conditions.

Concerning complex integration - namely, how should we choose the contour for integrating with respect to $k$. The contour
should actually be located below
both poles. For $x<x'$ we can only close this contour in the lower half-plane, there are no poles inside and the
integral is zero (by the way this how $\theta(x-x')$ appears in the answer). For $x>x'$ we can only close in the upper half-plane,
so we will have the residue contribution of both poles in the Fourier transform of the Green function.

Title: About calculating a Green's function


问:
Problem from the 2019 Brazil Math Olympiad: $ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $
The following functional equation problem:

Determine all the functions $ f : \mathbb R \to \mathbb R $ such that for any $ x , y \in \mathbb R $ it is true that:
$$ f \big( x f ( y ) + f ( x ) \big) + f \left( y ^ 2 \right) = f ( x ) + y f ( x + y ) $$

This is the fourth problem from the university level paper here: https://www.obm.org.br/content/uploads/2019/11/Prova_Nivel_Universitario_OBM_2019.pdf

I am attaching an attempt in an answer (because would lengthen the description too much and which is good practice according to guidelines)

答:
Edit: Fully Solved

This is my attempt at a solution and is partial (eventually resolves only in $\mathbb{Q}$):

Substituting $x=0$, $y=\beta$ and then $-\beta$, for $\beta$ any non zero real number, we get the equations:
$$f(f(0)) + f(\beta^2) = f(0) + \beta f(\beta)\quad \quad \quad \quad \quad \quad \quad \quad.$$
$$f(f(0)) + f(\beta^2) = f(0) + -\beta f(-\beta)\quad \text{ subtracting, we get}$$
$$\implies f(-\beta) + f(\beta) = 0\quad \forall \beta \in \mathbb{R}-\{0\}\tag{1}$$

Substituting $x=1$, $y=-1$,
$$f(f(-1)+f(1)) + f(1) = f(1) - f(0) \quad \quad \text{ and now using (1) with }\beta=1\text{, we get:}$$
$$f(0) = -f(0) \implies f(0) = 0 \tag{2}$$

Now with this knowledge if we substitute $x=0$, we have for any $y\in\mathbb{R}$:
$$f(y^2) = yf(y) \tag{3}$$

Also if we substitute $y=0$, with all this information, we obtain $\forall x\in \mathbb{R}$,
$$f(f(x)) = f(x)\tag{4}$$

Now substitute $x=-\beta, y=\beta$ and we get for all $\beta \in \mathbb{R}$:
$$f(-\beta\cdot f(\beta)+f(-\beta)) + f(\beta^2) = f(-\beta)\quad \qquad \text{using (2)}$$
$$f((\beta+1)\cdot f(\beta)) = f(\beta^2) + f(\beta) = (\beta+1)\cdot f(\beta)\quad \text{using (1), (3)}$$
$$\text{Thus, }\qquad f((x+1)\cdot f(x)) = (x+1)\cdot f(x) \quad \forall x \in \mathbb{R}\tag{5}$$

Now put $y=x$, and we get $\forall x\in\mathbb{R}-\{0\}$ (also for $0$, but that's trivial),
$$f((x+1)\cdot f(x)) + f(x^2) = f(x) + xf(2x)$$
$$\implies (x+1)\cdot f(x) +xf(x) = f(x) + xf(2x)$$
$$\implies f(2x) = 2f(x) \tag{6}$$

Earlier I went down the following route which resolves the problem in $\mathbb{Q}$. Feel free to skip ahead to the final solution section

Now you can establish by induction that this is true for all integer
factors (not just $2$): briefly, if it is true for a factor of $k$
(say $2$), then in the inductive step, using assumption $f(kx) =
> kf(x)$, we reach a relation for $k+1$. We substitute $x=ky$:
$$f(kyf(y)+f(ky)) + yf(y) = f(ky) + yf((k+1)\cdot y)$$ $$\implies
> f(k\cdot((y+1)\cdot f(y))) + yf(y) = kf(y) + yf((k+1)\cdot y)$$
$$\implies k\cdot((y+1)\cdot f(y)) + yf(y) = kf(y) + yf((k+1)\cdot y)
> \qquad \text{using assumption and (5)}$$ $$\implies kyf(y) + kf(y) +
> yf(y) = kf(y) + yf((k+1)\cdot y)$$ $$\implies f((k+1)\cdot y) =
> (k+1)f(y)\qquad \text{ completing the induction}$$

But this takes us nowhere in the real domain. We can resolve if the
domain were $\mathbb{Q}$

Thus for rational numbers (negatives are easily determined from $(1)$ so consider positives for now), we would get, for $p, q\in
> \mathbb{Z}^+$: $$f\bigg(\frac{p}{q}\bigg) = f\bigg(p\cdot
> \frac{1}{q}\bigg) = p\cdot f\bigg(\frac{1}{q}\bigg)\qquad \text{ and }
> \qquad f(1) = f\bigg(q\cdot\frac{1}{q}\bigg) = q\cdot
> f\bigg(\frac{1}{q}\bigg)$$ $$\implies f\bigg(\frac{p}{q}\bigg) =
> \frac{p}{q}\cdot f(1)$$ Thus assuming $f(1)=c$, for rationals our
solution would be $f(x) = cx\quad\forall x\in \mathbb{Q}$ and
substituting we get possible values of $c$ as $0$ and $1$, yielding
two solutions: $$f(x)=0 \quad\forall x\in \mathbb{Q}\qquad \text{ and
> } f(x)=x\quad\forall x\in \mathbb{Q}$$

Final Solution

(Ref @1-___-'s comment)
We can observe that this is a Cauchy functional equation by adding up the $x$ and $-x$ substitutions and use $(3)$ and $(6)$ to get:
$$f(x+y) + f(y-x) = f(2y)$$
Since we can independently pick any real pair of values for $x+y$ and $y-x$, we have that $$f(\alpha)+f(\beta)=f(\alpha+\beta)\quad \forall\alpha,\beta\in\mathbb{R}$$

Now we can use this to target the original relation which now simplifies to:
$$f(xf(y)) + f(f(x)) + f(y^2) = f(x) + yf(x) + yf(y)$$
Using $(3)$ and $(4)$,
$$f(xf(y)) = yf(x)$$
Now substituting $x=1$, we get for all real $y$:
$$f(f(y)) = f(y) = y\cdot f(1)$$
Thus the solution is in the space of linear functions:
$$f(x) = c\cdot x$$
for some $c\in\mathbb{R}$. Plugging this form back into the original equation forces $c=0$ or $1$ and thus yields these two solutions in $\mathbb{R}$:
$$f(x)=0 \quad\forall x\in \mathbb{R}\qquad \text{ and } f(x)=x\quad\forall x\in \mathbb{R}$$

ContentLicense: CC BY-SA 4.0


问:
Help to clarify proof of Euler's Theorem on homogenous equations

Why is the last step (setting $\lambda = 1$ allowed?

I have trouble accepting this because if I set $\lambda =1 $ at the very start, then:

$f(\lambda x , \lambda y)=\lambda^r f(x,y)$

becomes

$f(x,y) = f(x,y)$

and

so I can't prove the theorem.

Why does setting $\lambda = 1$ at the end of the proof work?

EDIT

For example, let $f(x,y)=x^2y^2$

Therefore, $f(x,y)$ is homogeneous of degree 4.

Therefore, the second to last step will be:

$4x^2y^2=4\lambda^3x^2y^2$

The above equation is only true if $\lambda=1$

Why doesn't the theorem make a qualification that $\lambda$ must be equal to 1?

It seems to me that this theorem is saying that there is a special relationship between the derivatives of a homogenous function and its degree but this relationship holds only when $\lambda=1$. Please correct me if my observation is wrong.

答:
Consider a function $f$ that is homogeneous of degree $r$, i.e.,

$f(tx, ty) = t^rf(x, y)$ ......$(1)$

Differentiate both sides of equation (1) with respect to $t$. Then by Chain Rule:

$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}x + \frac{\partial f(tx, ty)}{\partial(ty)}y = rt^{r-1}f(x, y)
\end{equation}$ ......$(2)$

Differentiate both sides of equation $(1)$ with respect to $x$. Then by chain rule:

$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}t = t^r\frac{\partial f(x, y)}{\partial(x)}
\end{equation}$

$\implies$$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)} = t^{r-1}\frac{\partial f(x, y)}{\partial x}
\end{equation}$ ......$(3)$

Similarly,

$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(ty)} = t^{r-1}\frac{\partial f(x, y)}{\partial y}
\end{equation}$ ......$(4)$

Put $(3)$ and $(4)$ in $(2)$ to get the result:

$\begin{equation}
\frac{\partial f(x, y)}{\partial x}x + \frac{\partial f(x, y)}{\partial y}y = rf(x, y)
\end{equation}$

ContentLicense: CC BY-SA 3.0


问:
Plot in Octave of Integral on unit circle $\int_{-\pi}^\pi \frac{(1-r^2)\cos(x)}{1-2r \cos(\theta-x)+r^2}dx $
i want to get a 3D Plot in Octave of the function

$$f(r\cdot e^{i\theta})=\int_{-\pi}^\pi \frac{(1-r^2)\cdot\cos(x)}{1-2\cdot r\cdot\cos(\theta-x)+r^2}dx $$

what is called the Poisson-Integral for the function $\cos(x)$ for $0\leq r<1$. I really don't know how to get it and didn't find any answers in the internet. so maybe someone can help me.

Until now i got the following code but the output is far away from what i need:

```<code>r = linspace(0,0.999,10)
theta = pi*linspace(-1,1,10)
z = r.*exp(1i*theta)
[Y1,Y2]=meshgrid(sqrt(real(z).^2+imag(z).^2),imag(z))
for i=1:10
  for j=1:10
    n=Y1(i,j)
    m=Y2(i,j)
    y=@(x) (1-n.^2).*cos(x)/(1-2.*n.*cos(m-x)-n.^2)
    Z(i,j)= quad(y,-pi,pi)
  end
end

plot3(real(z),imag(z),Z)
</code>```

I'm thankful for every answer or try to help me.

答:
Let $t= x -\theta$
\begin{align}
\int_{-\pi}^\pi \frac{(1-r^2)\cos x}{1-2r\cos(\theta-x)+r^2}dx
=&(1-r^2) \int_{-\pi-\theta}^{\pi -\theta}\frac{\cos t\cos\theta- \sin t \sin\theta}{1-2r\cos t+r^2}dt\\
 =& 2\cos\theta (1-r^2)\int_{0}^{\pi}\frac{\cos t}{1-2r\cos t+r^2}dt\\=& 2\pi r\cos\theta
\end{align}

where, with $y=\tan\frac t2$

\begin{align}
&\int_{0}^{\pi}\frac{\cos t}{1-2r\cos t+r^2}dt
 =\frac{1}{2r}\left(-\pi+  \frac{2(1+r^2)}{(1+r)^2} \int_{0}^{\infty}\frac{dy}{y^2+ \left(\frac{1-r}{1+r}\right)^2}\right)
=\frac{\pi r}{1-r^2}
\end{align}


问:
Minimize height, above a hole, needed for getting a rectangle out of the hole.
I am looking for a solution to a particular task of manipulating a rectangle out of a hole. I will be happy for any suggestions. I wasn't able to find any simplifications (e.g.: transferring the problem to a problem of manipulating a rod), which would lead to an equivalent optimum.

Problem: Minimize height, above a hole, needed for getting a rectangle out of the hole.

Illustration picture:

Having rectangle with known sides
$$
w, h
$$
and having a hole with known width and height:
$$
a, b
$$
Minimize height, above the hole, needed for pulling the rectangle out of the hole
$$
min(x)
$$

EDIT
Two answers (one of them was edited, I will consult it later) suggested placing bottom left/right corner of a rectangle to upper left/right point of the hole and rotating rectangle as much as possible to the right/left. I think this is correct for a "short" rectangle.

I suggest following for a "long" rectangle:

When I take "long" rectangle, I think the height needed (x) is what the picture below shows:

答:
Hint: Here is what the diagram should look like

Notice that the left most point of the rectangle is on the line that is with the floor?

Could you take it from there using similar triangles?

Note that there is a gap, because if $a=w$, then the height needed to pull it out would just be $h$, so $a>w$ in this case to solve the not-so-straightforward case.


问:
Union of countable sequence of countable sets
I realize this question has been asked and answered multiple times, but I'm still not understanding the reasoning. For reference, I've read this, this, and this.

Theorem 2.12. Let {}, =1,2,…, be a sequence of countable sets, and put

$$S = \bigcup\limits_{n=1}^{\infty}E_n$$

Then  is countable.

Proof. Let every set  be arranged in a sequence {},
=1,2,3,…, and consider the infinite array

in which the elements of  form the th row. The array contains
all elements of . As indicated by the arrows, these elements can be arranged in a
sequence

11;21,12;31,22,13;41,32,23,14;…(*)

If any two of the sets  have elements in common, these will appear
more than once in (∗). Hence there is a subset  of the set of all
positive integers such that ∼, which shows that  is at most
countable. Since 1⊂, and 1 is infinite,  is infinite, and thus
countable. ◼

I understand if there's an injection $F: S\rightarrow \mathbb{N}$, S is countable. I don't understand how utilizing the diagonals leads to such an injection. Since there's 1 element in the first diagonal, 2 in the second, ..., n in the n-th, (n - 1) in the (n + 1)-th, etc., how is it possible to say that since there are n elements at most in any diagonal the set is countable?

答: The argument shows that there is a surjection $g : \Bbb{N} \to S$ (that's what the bit about arranging the elements in a sequence is telling us). Given such a surjection $g$, you can define the function $h : S \to \Bbb{N}$ that maps $s \in S$ to the least $n \in \Bbb{N}$ such that $g(n) = s$. $h$ is an injection.


问:
How can I find the surface area of a normal chicken egg?
This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

答:
Let's make a problem a little more interesting by generalizing it. I have an arbitrary convex object, and I want to find its surface area. Of the answers posted so far, only the triangulation strategy of Zach L. and Cong Xu works in this case without breaking the object into little pieces. Here's another approach.

Suppose you project the object onto a randomly oriented plane, i.e. a plane whose normal is chosen uniformly from the unit sphere. Given that the object is convex, the expected value of the projected area is exactly $1/4$ times the surface area of the object, for essentially the same reasons given by Christian Blatter for the 2D case. (Short version: each differential surface element $dA$ contributes on average $dA/4$ to the projected area; its orientation doesn't matter because we're averaging over all possible directions of projection; there is no double-counting because the object is convex). This immediately suggests a Monte Carlo algorithm:

Rotate the object into a random orientation.

Shine collimated light at the object (e.g. from the sun, or from a point-light/parabolic-mirror combination) and observe its shadow on a plane perpendicular to the light direction.

Record the area of the shadow. Maybe you have graph paper pasted on the plane, or maybe you take a picture with a calibrated camera, binarize the image, and count the number of black pixels.

Repeat lots and lots and lots of times.

The average area of the shadow, times $4$, is the surface area of the object.


问:
Computing the order of a distribution
I need to prove that the order of the distribution given by $$\langle T,\varphi\rangle=\int_\mathbb{R}(\varphi(t^{-2},\sin t)-\varphi(0,\sin t)) dt,$$ where $\varphi\in C^\infty_0(\mathbb{R^2})$ is exactly 1. I have already shown that it is less than or equal to 1 but I haven't been able to show that it can't be 0. I am allowed to use the fact that the distribution $$\langle T',\varphi\rangle=\int_0^{+\infty}(\varphi(x)-\varphi(0))\frac{dx}{x^k}$$ has order 1 whenever $1<k<2$.

Could somebody give me a hint?

答:
Using the variable substitution $t=1/s$ we get
$$
\langle T,\varphi\rangle
%= \int_\mathbb{R}(\varphi(t^{-2},\sin t)-\varphi(0,\sin t)) \, dt
%= \left(\int_{-\infty}^0 + \int_0^\infty \right) (\varphi(t^{-2},\sin t)-\varphi(0,\sin t)) \, dt
%= \{ s:=1/t \}
%= \left(\int_{-\infty}^{0} + \int_0^{\infty} \right) (\varphi(s^2,\sin \frac{1}{s})-\varphi(\infty,\sin \frac{1}{s})) \, \frac{ds}{s^2}
= \int_{\mathbb{R}}  \left( \varphi(s^2,\sin \frac{1}{s}) - \varphi(0,\sin \frac{1}{s}) \right) \, \frac{ds}{s^2}
$$

Now,
$$
\varphi(s^2,\sin \frac{1}{s})
= \varphi(0,\sin \frac{1}{s}) + s^2 \varphi'_1(0,\sin \frac{1}{s}) + s^4 o(s)
$$
so close to $s=0$ the integrand is $\varphi'_1(0,\sin \frac{1}{s})$ suggesting that $T$ has order 1.

Perhaps you can make this more precise?

Title: Computing the order of a distribution


问:
How to calculate $e^x$ with a standard calculator
Is there a simple method for calculating the $e^x$ ($x\in\mathbb{R}$) with a basic add/subtract/multiply/divide calculator that converges in reasonable time, preferably without having to memorize coefficients as in the case of the power series?

I've found one for nth roots with Newton iteration that's pretty much dead simple and I'd really like to know what else can be done.

EDIT: I'm not really that excited about approximation unless the end result is completely accurate.  There are some good ideas here, but this answer is the simplest option for a full-accuracy result using only a +/-/×/÷ calculator.

I also have a strong preference for algorithms which can be directly evaluated step-by-step on such a calculator.  Otherwise pencil and paper or an excellent memory must be exercised.

答:
In the 1970's, when four-function calculators were the norm, I made up a set of tables that fit onto a single A6 notepad page, which gives, $e^{0.n}$, down to $e^{0.00000n}$, which after, one uses $e^{0.00000xxxxxx}=1.000000xxxxxx$.  </p>

<p>When this is constructed in decimal, one multiplies the trainlig $xxxxx$ by some value, before adding in the $1$.  Otherwise, it's pretty much the same.  </p>

Body:
In the 1970's, when four-function calculators were the norm, I made up a set of tables that fit onto a single A6 notepad page, which gives, $e^{0.n}$, down to $e^{0.00000n}$, which after, one uses $e^{0.00000xxxxxx}=1.000000xxxxxx$.

When this is constructed in decimal, one multiplies the trainlig $xxxxx$ by some value, before adding in the $1$.  Otherwise, it's pretty much the same.


问:
Three Altitudes of a triangle are concurrent
I have been told that this well known fact can be shown using only Euclid's propositions from books one to three, and cyclic quadrilaterals.

I can't figure out how to start, which quadrilateral should I have in consideration. Also the only proposition about cyclic quadrilaterals is 3.22, which I am not sure how to use.

Edit: I think I have made some progress. Let CE and AF be altitudes from C and A, the idea is to show that the line going through B and D is perpendicular to AC. Which could be done by 3.22 if either AGDE or GCFD were cyclic quadrilaterals. I don't know how to show this last part.

答:
Expanded answer:

Take the diagram as shown, with $AF \perp BC$ and $CE \perp AB$ but consider the point $G$ to be merely the extension of $BD$ to meet $AC$ - angle unknown.

Now we find that $E$ and $F$ are points on the circle with $AC$ as diameter, giving $AEFC$ as cyclic, so we can see that the inscribed angle from chord $AF$ means that the internal angle $\angle ACF=\angle FEB$, the external angle at $E$.

Similarly  $E$ and $F$ are points on the circle with $BD$ as diameter, giving $DEBF$ cyclic and the inscribed angle from chord $BF$ gives $\angle FEB=\angle FDB$.

Then since $\angle ACF=\angle FDB$ we have that $\angle ACF+\angle FDG = 180^\circ$ and completing the angles in quadrilateral $GDFC$ we see that $\angle CGD = 90^\circ$ and $BG$ is the third altitude.


答: Intuitively,  a set is countable precisely when it can be put into a list.  And that's exactly what the diagonal process does:  it puts all the elements of the array into a list.

Title: Union of countable sequence of countable sets

Body:
I realize this question has been asked and answered multiple times, but I'm still not understanding the reasoning. For reference, I've read this, this, and this.

Theorem 2.12. Let {}, =1,2,…, be a sequence of countable sets, and put

$$S = \bigcup\limits_{n=1}^{\infty}E_n$$

Then  is countable.

Proof. Let every set  be arranged in a sequence {},
=1,2,3,…, and consider the infinite array

in which the elements of  form the th row. The array contains
all elements of . As indicated by the arrows, these elements can be arranged in a
sequence

11;21,12;31,22,13;41,32,23,14;…(*)

If any two of the sets  have elements in common, these will appear
more than once in (∗). Hence there is a subset  of the set of all
positive integers such that ∼, which shows that  is at most
countable. Since 1⊂, and 1 is infinite,  is infinite, and thus
countable. ◼

I understand if there's an injection $F: S\rightarrow \mathbb{N}$, S is countable. I don't understand how utilizing the diagonals leads to such an injection. Since there's 1 element in the first diagonal, 2 in the second, ..., n in the n-th, (n - 1) in the (n + 1)-th, etc., how is it possible to say that since there are n elements at most in any diagonal the set is countable?


问:
Path lifting lemma - proof clarification
So i was studying the proof of the 'path lifting lemma' on my book and I got stuck.
Here $ S^1 $ is identified with complex numbers of norm 1 and $ E: \mathbb{R} \rightarrow S^1 $ is defined by $ E(x)= e^{2\pi ix}, x\in \mathbb{R}$.

Lemma: If $ \alpha $ is a path in $ S^1 $ with starting point  1, there exist a unique path $\alpha '$ in $ \mathbb{R} $ with starting point $0$ such that $ E \alpha ' = \alpha $.

Proof: Because I is compact and $ \alpha $ is continuous, we can choose subintervals of  I, $ [t_0,t_1],...[t_{n-1},t_n],\  0=t_0 < t_1$ $ <...< t_n=1$ such that $$|\alpha (t) - \alpha (t')| < 1 \; \forall t,t' \in [t_{i-1},t_i] \;\; \forall i=1...n
\; \; \; \; \; \; \;\; \; \; \; \; \; \;\; \; \; \; \; \; \;
\textbf{(1)} $$
and so such that $$ \alpha ([t_{i-1},t_i])\mbox{ is contained in } S^1 \backslash \{ - \alpha (t_{i-1}) \} \;\; \forall i=1...n
\; \; \; \; \; \; \;\; \; \; \; \
\textbf{ (2) }
$$...(the proof goes on)

I'm ok with the rest of the proof (that I didn't write) but I don't understand this beginning: why can we choose such intervals? Where the compactness of I and the continuity of $ \alpha $ are used? Why (1) implies (2) ?

Thanks to everyone.

答:
(1)

It is well-known that any continuous functon from a compact metric space to a metric space is uniformly continuous. This implies that there exists $\delta > 0$ such that $|\alpha (t) - \alpha (t')| < 1$ for all $t, t'\in I$ with $|t - t'| < \delta$. Let $n \in \mathbb N$ such that $1/n \le \delta$. Then $t_i = i/n$ ($i=0,\dots,n$) will do.

(2)

We have $|\alpha (t_{i-1}) - (-\alpha (t_{i-1}))| = 2$, thus for $t \in [t_{i-1},t_i]$ we get
$$|\alpha (t) - (-\alpha (t_{i-1}))| = |\alpha (t_{i-1}) - (-\alpha (t_{i+1})) + \alpha (t) - \alpha (t_{i-1})| \\ \ge  |\alpha (t_{i-1}) - (-\alpha (t_{i+1}))| - | \alpha (t) - \alpha (t_{i-1})| > 2 -1 = 1 ,$$
hence $\alpha (t) \ne -\alpha (t_{i-1})$.


问:
Prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $ in circular motion
This may sound like a physics problem, but I got it in my homework in differential equations course, so Im asking here.

A particle moves on the standard sphere $ \mathbb{S}^{n-1}\subset\mathbb{R}^{n} $. I have to prove that $ \overrightarrow{v}\left(t\right)\perp\overrightarrow{x}\left(t\right) $, and given that $ |\overrightarrow{v}|=\text{const} $ I have to prove that $ \overrightarrow{a}\left(t\right)||\overrightarrow{x}\left(t\right) $.

The first part wasnt hard, if we note that $ 1=\overrightarrow{x}\cdot\overrightarrow{x} $ then $ 0=\left(\overrightarrow{x}\cdot\overrightarrow{x}\right)'=\overrightarrow{v}\cdot\overrightarrow{x}+\overrightarrow{x}\cdot\overrightarrow{v} $ and thus $ \overrightarrow{x}\cdot\overrightarrow{v}=0 $.

But Im not sure about the second part. If $|v|=const $, how can I show that $ a||x $ ?

I tried to write:

$ 0=\frac{d}{dt}\left(|v|^{2}\right)=\frac{d}{dt}\left(\sum_{k=1}^{n-1}v_{k}^{2}\left(t\right)\right)=\sum_{k=1}^{n-1}2v_{k}\left(t\right)a_{k}\left(t\right)=2\overrightarrow{v}\cdot\overrightarrow{a} $

Which shows that $ \overrightarrow{v}\perp\overrightarrow{a} $, But for general $ \mathbb{S}^{n-1} $ it does not promise that $ a||x $ as far as I understand.

Another way that I tried is to write:

$ 0=\left(\overrightarrow{x}\cdot\overrightarrow{v}\right)'=\overrightarrow{x}\cdot\overrightarrow{a}+|v|^{2} $

and thus $ \overrightarrow{x}\cdot\overrightarrow{a}=-|v|^{2} $
But Im not sure how to proceed or if it gives me anything...

Also, Im trying not to make any physical assumptions because its a mathenatical course, so a mathematical solution would be very helpful. Thanks in advance.

答:
Your hypotheses are:

$$\begin{cases}
\vec{x}(t) \cdot \vec{x}(t) & = 1\\
\vec{v}(t) \cdot \vec{v}(t) & = V^2  \neq 0 \text{(constant)}
\end{cases},$$

for all $t$.

Using them, you deduced that:

$$\begin{cases}
\vec{x}(t) \cdot \vec{v}(t) & = 0\\
\vec{a}(t) \cdot \vec{v}(t) & = 0\\
%\vec{a}(t) \cdot \vec{x}(t) &= -V^2
\end{cases},$$

for all $t$.

Using the last two equations, one gets:

$$(\vec{a}(t) + \vec{x}(t)) \cdot \vec{v}(t) = 0,$$

for all $t$.

More in general, given two real constants $m$ and $k$, we get:

$$(m\vec{a}(t) + k\vec{x}(t)) \cdot \vec{v}(t) = 0,$$

for all $t$, i.e. any linear combination of $\vec{a}(t)$ and $\vec{x}(t)$ is perpendicular to $v(t).$

Since the previous holds for all $t$, and hence for any value of $\vec{v}(t)$ satisfying the requirements (arbitrariety), we can conclude that:

$$m\vec{a}(t) + k\vec{x}(t) = \vec{0} \Rightarrow \vec{a}(t) = -\frac{k}{m}\vec{x}(t) \Rightarrow \vec{x}(t) \parallel \vec{a}(t).$$


问:
If $N$ is a normal subgroup of $G$ and if $x^2 \in N$ for all $x\in G$, prove that every nonidentity element of $G/N$ has order $2$.
I've been trying to solve this proof for a while and couldn't figure out how to prove that $(gN)^2=eN$. I've tried to set up an equation where $gN \cdot g^{-1}N=eN$ but couldn't get to the end result.

答: Let $g\in G, n,n'\in N$. We have: $gngn'=gg(g^{-1}ng)n'$ since $g^{-1}ng\in N$ and $g^2\in N$, we deduce that $gngn'\in N$.


问:
Convergence in space of distribution
Let's consider $\mathbb{T}:=\mathbb{R}/2\pi \mathbb{Z}$ and let's denote $\mathcal{E}(\mathbb{T})$ to be the space of $C^\infty$ functions defined on $\mathbb{T}$, where the topology on $\mathcal{E}(\mathbb{T})$ is defined via the following: $\phi_n \rightarrow \phi$ in $\mathcal{E}(\mathbb{T})$ iff $\phi_n^{(k)} \rightarrow \phi^{(k)}$ uniformly on $\mathbb{T}$.

Now, we consider the space of distributions $\mathcal{E}'(\mathbb{T})$.

The question that I have to solve is to check, if for every sequence of complex numbers $(a_n)_{n \in \mathbb{Z}}$ the limit $\sum_{n=-N}^N a_n e^{inx}$ exists in $\mathcal{E}'(\mathbb{T})$.

So, as far as I understand, we need to show that for every $\phi \in \mathcal{E}(\mathbb{T})$ the following limit exits
$$\lim_{N \rightarrow \infty} \int_\mathbb{T} \sum_{n=-N}^N a_n e^{inx} \phi(x) dx$$

(where we work on this integral since we consider the action of a distribution $\sum_{n=-N}^N a_n e^{inx}$ on a test function $\phi$). Could anyone share some light how we can show that something like this won't converge to any distribution $u \in \mathcal{E}'(\mathbb{T})$ for some $\phi \in \mathcal{E}(\mathbb{T})$ and $a_n$? I suspect it doesn't have to in general, since this formula is a similar one like the one we work with Fourier series, but neither of the ideas that I had were successful.

I would appreciate any help here.

答: Take $\phi(x)=\sum_{n\ge 1} 2^{-n} e^{-inx}$ and $a_n=2^n 1_{n\ge 1}$ you'll have $$\int_0^{2\pi} \sum_{n=-N}^N a_n e^{inx}\phi(x)dx=2\pi N$$

Title: Convergence in space of distribution


问:
Is $ \Phi \vdash \phi $, if and only if $ \Phi \cup \{\neg \phi \} $ being not satisfiable?
In the first order logic, $ \Phi$ is a set of formulas, and $\phi$ is a formula.
Is $ \Phi \vdash \phi $, if and only if $ \Phi \cup \{\neg \phi \} $ being not satisfiable?

答:
The basis of this proof is the following natural deduction laws:

\begin{equation*} \frac{\Phi \cup \{ \phi \}\vdash \bot}{\Phi\vdash\neg\phi} \end{equation*}

and

\begin{equation*} \frac{\Phi\vdash\phi}{\Phi \cup \{\neg\phi\}\vdash \bot} \end{equation*}

$(\rightarrow)$ If $\Phi\vdash\phi$ then $\Phi\cup \{\neg \phi \}\vdash \phi$. But $\Phi\cup \{\neg \phi \}\vdash \neg\phi$ so it can't be satified since it proves a contradiction.

$(\leftarrow)$ If $\Phi\cup \{\neg \phi \}\vdash \phi$ can't be satisfied then it has no models and so $\Phi\cup \{\neg\phi\}\models\bot$. By using the semantic completeness theorem we get $\Phi\cup \{\neg \phi \}\vdash\bot$ which by using the rules of natural deduction results in $\Phi\vdash\phi$.

Observation: This is true even in the case in which $\Phi$ is incoherent (it can prove $\bot$).


问:
Probability of Event Occurring After X Previous Attempts
Suppose you have an event that as a probability $p = 1/100$ of occurring. If I make $x=50$ independent attempts, I believe the probability $P$ of the event occurring at least once in those $50$ attempts is found by calculating the probability it will not happen $1-P$:

$$(1-P) = (1-p)^x = \left(1-\frac{1}{100}\right)^{50}$$
$$(1-P) = \left(1-\frac{1}{100}\right)^{50}$$
$$(1-P) = 60.5\%$$
$$P = 39.5\%$$

Now, let's assume the event did not occur in the $50$ attempts (a $60.5$% chance given our calculation above). What is the probability of the event occurring at least once in the next $50$ independent attempts?  My intuition says that since the attempts are independent, the scenario "resets" and does not factor in my previous attempts, keeping my odds at the same $39.5$%.  Am I correct in saying this?

答:
Yes you are correct.

Many people may be confused and say the probability would change. This is called the gambler's fallacy. Here's an example. If people kept betting on red when they're gambling, but the spinner keeps showing up black, should they stop betting red and start changing to bet on black? No, each spin is independent, so the probability that the spinner will be red is the same as the probability that the spinner will be black, so there is no reason to switch colors. The reason the spinner kept coming up black is by pure chance, but each spin should not be affected by the spins before it. Likewise, the probability that the next $50$ events happen does not depend at all on the previous $50$.


问:
Similarity between dimension of a vector space and order of a group
While studying groups, I've run into the following statement:

Let $f: G \to G'$ be a surjective homomorphism between groups with $K$ being the kernel of $f$. Then,
$$ |G| = |K|*|G'|. $$
This equation seems nearly identical to the dimension formula for vector spaces: namely, given a transformation between vector spaces $T: V \to W$,
$$ \text{dim }V = \text{dim}(\text{ker }T) + \text{dim}(\text{im }T).$$
These two equations seem extremely similar intuitively, but there's one main thing that confuses me. It seems to me that if you directly translated the statement about group order then the dimension of $V$ would be the product of the kernel and the image, not the sum. Is my confusion arising out of some misunderstanding of groups and vector spaces, or are the two equations unrelated? If they are related, could someone explain why in an intuitive manner?

答: Multiplying two exponentials with a common base is the same as adding their exponents: $x^ax^b = x^{a+b}$. To make this an analogy, note that "size" is like $x^a$ whereas "dimension" is like $a$. This is super imprecise, but I imagine it gets the point across :)


问:
Bilinear form properties
Let us look at: $B:V\times W$ be a bilinear form.

$T_{B} : W \to V^{*}$

$(T_{B}(w))(v) = B(v, w)$

and

$S_{B} : V \to W^{*}$

$(S_{B}(v))(w)=B(v,w)$

I need to prove that:

$\operatorname{rank}(B) = \dim\operatorname{Im}(S_{B}) = \dim\operatorname{Im}(T_{B})$

So how can I do that ?

All I know is that $B$ can the matrix represent according to some basis is:

\begin{pmatrix}
  I_{r} & 0\\
  0 & 0\\
  \end{pmatrix}

meaning the rank of $B$ is $r$ but how can I continue from here in order to calculate $\dim\operatorname{Im}(S_{B})$ and $\dim\operatorname{Im}(T_{B})$?

答:
So let $\mathcal B_V = \{v_1, \ldots, v_m\}$, $\mathcal B_W = \{w_1, \ldots, w_n\}$ be bases of $V$ and $W$. Then we can represent $B$ by a matrix $(b_{ij})$, where $b_{ij} = B(v_i, w_j)$. Corresponding to $\mathcal B_V$ and $\mathcal B_W$ we have the dual bases $\mathcal B_V^* = \{v_1^*, \ldots, v_m^*\}$ and $\mathcal B_W^* = \{w_1^*,\ldots, w_n^*\}$ of $V^*$ and $W^*$ respectively, given by
$$ w_i^*(w_j) = \delta_{ij}, \quad v_i^*(v_j) = \delta_{ij} $$
Let us now compute the matrix $[T_B]_{\mathcal B_W, \mathcal B_V^*}$ of $T_B$ in the bases $\mathcal B_W$ and $\mathcal B_V^*$. Let $1 \le i \le n$, write $T_B w_i = \sum_k \alpha_{ik} v_k^*$. We have for $1 \le j \le m$:
\begin{align*}
  \alpha_{ij} &= \sum_{k} \alpha_{ik} v_k^*(v_j)\\
              &= (T_B w_i)(v_j)\\
              &= B(v_j, w_i)\\
              &= b_{ji}
\end{align*}
So the matrix of $T_B$ is the transpose of the matrix for $B$, now let us consider $S_B$: We have writing $(\beta_{ij}) = [S_B]_{\mathcal B_V, \mathcal B_W^*}$ for the matrix of $S_B$ with respect to $\mathcal B_V$ and $\mathcal B_W^*$, that
\begin{align*}
  \beta_{ij} &= \sum_k \beta_{ik} w_k^*(w_j)\\
             &= (S_B v_i)(w_j)\\
             &= B(v_i, w_j)\\
             &= b_{ij}
\end{align*}
So we have $[T_B]_{\mathcal B_W^*, \mathcal B_V} = [S_B]_{\mathcal B_V^*, \mathcal B_W}^t$, and as a matrix and its transpose have the same rank, so do $S_B$ and $T_B$.</p>

Title: Bilinear form properties


答:
An alternative, but perhaps more intuitive formulation of Tsemo Aristide's answer:

A non-identity element of $G/N$ is a coset $gN$ for $g\notin N$. The binary operation on $G/N$ is given by $(gN)(hN)=(gh)N$, and thus $(gN)^2=(gN)(gN)=g^2N$, but by assumption $g^2\in N$, so $(gN)^2=g^2N=N$, which is the identity element of $G/N$, so $gN$ has order 2. Since $gN$ was an arbitrary choice of non-identity element, any non-identity element has order 2.

Title: If $N$ is a normal subgroup of $G$ and if $x^2 \in N$ for all $x\in G$, prove that every nonidentity element of $G/N$ has order $2$.

Body: I've been trying to solve this proof for a while and couldn't figure out how to prove that $(gN)^2=eN$. I've tried to set up an equation where $gN \cdot g^{-1}N=eN$ but couldn't get to the end result.


答:
Here's an answer along the lines that you tried. As $N$ is a subgroup, $NN = N$. As $N$ is normal, $gNg^{-1} = N$ for every $g \in G$. As $g^2 \in N$ for every $g \in G$, $g^2N = N$ for every $g \in G$. So, for every $g \in G$:

$$
(gN)^2 = (gN)(gN) = (gNg^{-1})(g^2N) = NN = N
$$


问:
Find $\mathbb{P}(X+Y<1)$ given the joint density function of the vector $(X,Y)$ distributed uniformly over a region $R$.
Let $X$ and $Y$ be random variables with the vector $(X, Y)$ uniformly distributed on the region
$R = \{(x, y) : 0 < y < x < 1\}$. Find $\mathbb{P}(X+Y<1)$. I am pretty sure that I calculated the pdf correctly:

$f_{X,Y}(x,y)= 2$ if $(x,y) \in R$, $0$ elsewhere.

To get the probability, I know I need to double integrate $f(x,y)$. I am not sure of the bounds though. Is the correct integration $\int_0^1 \int_0^{1-x} 2\,dy\,dx$, or should the upper bound be $x$ so that it is $\int_0^1 \int_0^x 2 \, dy \, dx$. Or are neither of these correct?

答: $P(X+Y <1)= 2 \int_0^{1}\int_0^{\min \{x,1-x\}} \, dy\,dx$ since we need $y <1-x $ as well as $y <x$. Write this as $2 \int_0^{\frac 1  2}\int_0^{x} \,dy\,dx + 2 \int_{\frac 1  2}^1 \int_0^{1-x} \,dy\,dx$. I will let you finish.


问:
Beautiful monster: Catalan's constant and the Digamma function
The problem

I have been trying for a while now to show that this monster
$$\begin{align}
&\int_0^{\pi/4}\tan(x)\sum_{n=1}^{\infty}(-1)^{n-1}\left(\psi\left(\frac{n}{2}\right)-\psi\left(\frac{n+1}{2}\right)+\frac{1}{n}\right)\sin(2nx)\,\mathrm{d}x \\
+&\int_0^{\pi/4}\cot(x)\sum_{n=1}^{\infty}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac{1}{n}\right)\sin(2nx)\,\mathrm{d}x,
\end{align}
$$
where $\psi$ denotes the Digamma function,
is equal to the beautiful Catalan's constant
$$G :=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2}.$$
I think it's a great problem; going from such a big expression to a nice, simple constant. We can see that these integrals are very similar, the main difference is probably the $\tan(\cdot)$ and the $\cot(\cdot)$. I think that a crutial step would be to find a Fourier series.

Small thank you note

The user @Quanto helps to higher the quality of this site by posting now and then very interesting integrals. I want to start doing the same. Thank you for inspiring me.

答:
Using an integral representation of the digamma function we obtain
\begin{align}
\operatorname{\psi} \left(\frac{n+1}{2}\right) - \operatorname{\psi}\left(\frac{n}{2}\right) - \frac{1}{n} &= \int \limits_0^1 \left[\frac{t^{\frac{n}{2}-1} - t^{\frac{n}{2} - \frac{1}{2}}}{1-t} - \frac{1}{2} t^{\frac{n}{2} -1}\right] \mathrm{d} t = \frac{1}{2} \int \limits_0^1 \frac{1-\sqrt{t}}{1+\sqrt{t}} \, t^{\frac{n}{2} - 1} \, \mathrm{d} t \\
&= \int \limits_0^1 \frac{1-u}{1+u} \, u^{n-1} \, \mathrm{d} u
\end{align}
for $n \in \mathbb{N}$. For $\sigma \in \{-1,1\}$ and $x \in \left(0,\frac{\pi}{4}\right)$ this implies
\begin{align}
& \phantom{=~}\sum \limits_{n=1}^\infty \sigma^{n-1} \left[\operatorname{\psi} \left(\frac{n+1}{2}\right) - \operatorname{\psi}\left(\frac{n}{2}\right) - \frac{1}{n}\right] \sin(2 n x) = \int \limits_0^1 \frac{1-u}{1+u} \, \operatorname{Im} \left[\mathrm{e}^{2\mathrm{i}x} \sum \limits_{n=1}^\infty (\sigma \mathrm{e}^{2\mathrm{i}x} u)^{n-1} \right] \, \mathrm{d} u \\
&= \int \limits_0^1 \frac{1-u}{1+u} \frac{\sin(2x)}{1 + u^2 - 2 \sigma u \cos(2x)} \, \mathrm{d} u \stackrel{u = \frac{1-v}{1+v}}{=} \frac{1}{2} \sin(2x) \int \limits_0^1 \frac{v}{1 - \frac{1 + \sigma \cos(2x)}{2} (1-v^2)} \, \mathrm{d} v \\
&= \frac{- \log \left(\frac{1 - \sigma \cos(2x)}{2}\right) \sin(2x)}{2[1 + \sigma \cos(2x)]} = \begin{cases} - \tan(x) \log(\sin(x)) , & \sigma = 1 \\ - \cot(x) \log(\cos(x)) , & \sigma = -1 \end{cases} \, .
\end{align}
Therefore, your monster equals
$$\int \limits_0^{\pi/4} \left[-\tan(x) \log(\sin(x)) \cot(x) + \cot(x) \log(\cos(x))\tan(x)\right] \, \mathrm{d} x = \int \limits_{0}^{\pi/4} \log(\cot(x)) \, \mathrm{d} x \, ,$$
which is indeed an integral representation of Catalan's constant $\mathrm{G}$.


问:
Groups acting on polynomials
I am a little confused about actions right now.

Say I have a group $G$ that acts on the roots of a polynomial, say $(x - x_1)(x - x_2)(x - x_3)$.  We know that this group permutes the roots of a polynomial, that is you can get to any of the roots by having an element of the group act on another root.  Does that mean the group fixes the polynomial?

I feel like it should, but I'm not sure how this actually works.  Supposedly we would have to write $g \cdot [(x - x_1)(x - x_2)(x - x_3)] = (x - x_1)(x - x_2)(x - x_3)$.  But is this actually possible?  Because $G$ acts on the set of the roots, not the polynomial.  I'm not sure if we can say a polynomial is fixed by certain things acting on the roots.  When they say the polynomial is fixed, does that mean the coefficients are fixed?

Also, can you say something like $g \cdot (x_1x_2 + x_3)$ = $(g \cdot x_1)(g \cdot x_2) + g \cdot x_3$?

Thanks!

答:
Let's talk a bit about "group actions" as well.

Given a group $G$ and a nonempty set $X$, a (left) group action of $G$ on $X$ means an operation $\cdot\colon G\times X\to X$ such that for all $g,h\in G$ and $x\in X$, $$\begin{align*}
h\cdot(g\cdot x) &= (hg)\cdot x;\\
1\cdot x &= x,
\end{align*}$$
An easy example of a group action is given as follows: Let $X$ be a nonempty set, and let $S_X$ be the group of all bijections from $X$ to itself (the "permutations on $X$"). Then we can define an action of $S_X$ on $X$ by $\sigma\cdot x = \sigma(x)$.

Similarly, if $H\lt S_X$,  then $H$ acts on $X$ in a natural way, by letting $h\cdot x = h(x)$ for every $h\in H$.

Another example: suppose that $G$ acts on $X$ and $H$ acts on $Y$. Then we can let $G\times H$ act on $X\times Y$ in an obvious way (coordinate-wise): given $(g,h)\in G\times H$ and $(x,y)\in X\times Y$, let $(g,h)\cdot(x,y) = (g\cdot x,h\cdot y)$.

Yet another example: If $G$ acts on both $X$ and $Y$, then $G$ acts on $X\times Y$ "coordinatewise" as well: $g\cdot (x_1,x_2) = (g\cdot x_1,g\cdot x_2)$. More generally, if $G$ acts on each set of a family $\{X_i\}_{i\in I}$, then $G$ acts on the cartesian product $\prod\limits_{i\in I}X_i$ coordinate-wise.

Another example: suppose that $G$ acts on $X$, and $\iota\colon X\hookrightarrow Y$ is a one-to-one function from $X$ to another set $Y$. Then we can define an action of $G$ on $Y$ by "extension" as follows: given any $g\in G$ and any $y\in Y$, we define $g\cdot y$ by:
$$g\cdot y = \left\{
\begin{array}{ll}
\iota(g\cdot x)&\mbox{if $y=\iota(x)$;}\\
y &\mbox{if $y\notin\iota(X)$.}
\end{array}\right.$$

Another example: Let $G$ act on $X$. We say that a subset $Y\subseteq X$ is $G$-invariant if and only if $G\cdot Y = \{g\cdot y\mid y\in Y\}\subseteq Y$. Then the restriction of the $G$-action to $Y$ gives us an action of $G$ on $Y$.

Another example: suppose that $G$ acts on $X$, and $f\colon K\to G$ is a homomorphism. Then $K$ acts on $X$ "via $f$", by letting $k\cdot x = f(k)\cdot x$.

In a sense, every action can be thought of as an action this way, in the following sense:

Theorem. Let $G$ be a group, and $X$ be a set. Then the following data are equivalent:

A left action of $G$ on $X$, $\cdot\colon G\times X\to X$.

A group homomorphism $\phi\colon G\to S_X$ from $G$ to the permutation group of $X$.

Proof. Assume $\cdot\colon G\times X\to X$ is a group action. For each $g\in G$, let $\phi_g\colon X\to X$ be the map $\cdot(g,-)\colon X\to X$ (currying $\cdot$) defined by $\phi_g\colon x\mapsto g\cdot x$. This is a bijection, because $\phi_{g^{-1}}$ is its inverse:
$$\begin{align*}
\phi_{g^{-1}}\circ\phi_g(x) &= g^{-1}\cdot(g\cdot x) = (g^{-1}g)\cdot x = 1\cdot x = x,\\
\phi_g\circ\phi_{g^{-1}}(x) &= g\cdot(g^{-1}\cdot x) = (gg^{-1})\cdot x = 1\cdot x=  x.\end{align*}$$
That is, the map $g\longmapsto \phi_g$ is a function from $G$ to $S_X$. Morover, $\phi_g\circ\phi_h = \phi_{gh}$, since $g\cdot(h\cdot x) = (gh)\cdot x$, so the map is a homomorphism. Thus, an action of $G$ on $X$ yields a homomorphism $\phi\colon G\to S_X$ by $\phi(g) = \phi_g$.

Conversely, suppose that $\phi\colon G\to S_X$ is a group homomorphism. This induces an action of $G$ on $X$ by $g\cdot x = \phi(g)(x)$, as above.

Moreover, let $\psi\colon G\to S_X$ be a homomorphism, and let $\cdot_{\psi}$ be the induced action. Then $\phi_g(x) = g\cdot_{\psi}(x) = \psi(g)(x)$, so the homomorphism we get from the action induced by $\psi$ is $\psi$. Likewise, if you start with an action $\cdot$, and we let $\phi\colon G\to S_X$ be the induced homomorphism as above, then the action $\cdot_{\phi}$ induced by this homomorphism is $g\cdot_{\phi}x = \phi_g(x) = g\cdot x$, so we get back our original action. That is, the data are equivalent. QED

Another way of defining actions is by restriction: suppose $G$ acts on a set $X$; given $x\in X$, the orbit of $x$ is the set
$$G\cdot x = \{ g\cdot x\mid g\in G\}.$$
Then $G$ acts on $G\cdot x$ in a natural way: we think of this as a *restriction * of the action of $G$, from considering what it does to all of $X$ to considering just what it does to $G\cdot x$.

An action of $G$ on $X$ induces an equivalence relation on $X$ by $x\sim y$ if and only if there exists $g\in G$ such that $g\cdot x = y$; equivalently, if and only if $y$ is in the orbit of $x$. This equivalence relation partitions $X$ into "$G$-orbits."

Two important properties of an action are faithfulness and transitivity:

Definition. Let $G$ act on $X$. The action of $G$ is faithful if
$$g\cdot x = x\text{ for all }x\in X\Longleftrightarrow g=1.$$
Equivalently, if and only if the homomorphism $G\to S_X$ induced by the action is one-to-one.

Definition. Let $G$ act on $X$. The action of $G$ is transitive if for every $x,y\in G$ there exists $g\in G$ such that $g\cdot x = y$. Equivalently, if and only if $X$ consists of a single $G$-orbit.

To see an example of an action that is not transitive, let $G=S_3$ and let $G$ act on the set $\{1,2,3,4,5,6\}$ as follows: given $\sigma\in S_3$, let $\sigma\cdot i = \sigma(i)$ if $1\leq i\leq 3$, and let $\sigma\cdot j = 3+\sigma(j-3)$ if $4\leq j\leq 6$. For example, if $\sigma=(1,3)$, then
$$\sigma(1)=3,\quad \sigma(2)=2,\quad \sigma(3)=1,\quad \sigma(4)=6,\quad \sigma(5)=5,\quad \sigma(6)=4.$$
The action is not transitive because $X$ has two $G$-orbits: $\{1,2,3\}$ and $\{4,5,6\}$.

For a more interesting example, let $X=\mathbb{Z}_2\oplus \mathbb{Z}_2$ be the Klein $4$-group, and let $G=\mathrm{Aut}(C_2\times C_2)$. Then $G$ acts on $X$ in the obvious way (apply the homomorphism). If $x=(0,0)$, then for every $\phi\in G$ we have $\phi(x)=x$ (since group automorphisms must send the identity to the identity). On the other hand, we have an automorphism that exchanges $(1,0)$ and $(0,1)$ and fixes $(1,1)$; an automorphism that exchanges $(1,0)$ and $(1,1)$ and fixes $(0,1)$; and an automorphism that exchanges $(0,1)$ and $(1,1)$ and fixes $(1,0)$. So that for every $x\in X$, $x\neq (0,0)$, and every $y\in X$, $y\neq (0,0)$, there exists $g\in G$ such that $g\cdot x = y$. That is, $X$ is partitioned into two $G$-orbits, $\{(0,0)\}$ and $\{(1,0), (0,1), (1,1)\}$.

So you see why, in your question, saying the group "permutes the roots" is not the same as saying that for any two roots there is an element of the group that maps the first root to the second. The first statement is saying that there is an action of $G$ on the set of roots; the second statement is saying that the action is transitive, and not every action is transitive.

Now, suppose that $G$ acts on a field $F$; the action may have absolutely nothing to do with the field structure of $F$ (it may just be an action on the underlying set of $F$), but these kinds of actions are not very interesting when you think of $F$ as a field. Instead, if you want to think of $F$ as a field, then you want to think of the action of $G$ as being an action "by automorphisms"; that is, we identify $G$ with a subgroup of the group $\mathrm{Aut}(F)$ of all field automorphisms of $F$, which is itself a subgroup of the group $S_F$ of all permutations on the underlying set of $F$.

Now suppose that $G$ acts on the field $F$. Then the action of $G$ on $F$ naturally induces an action of $G$ on $F[x]$, the polynomials with coefficients on $F$: remember that we can think of polynomials with coefficients in $F$ as the almost-null sequences of elements of $F$, that is, the infinite tuples $(a_0,a_1,a_2,\ldots,a_n,\ldots)$ with $a_i\in F$ and $a_i=0$ for almost all $i$; the action of $G$ on $F[x]$ is defined coordinatewise, as before.

But if the action of $G$ on $F$ is by field automorphisms, then we get more: the action of $G$ commutes with the evaluation homomorphism! Remember that for ever $a\in F$ we have an "evaluate at $a$" homomorphism $\varepsilon_a\colon F[x]\to F$ by $\varepsilon_a(p(x)) = p(a)$. If the action of $G$ is by field automorphisms, then for every $$p(x) = \alpha_nx^n + \cdots +\alpha_0\in F[x]$$ and for every $a\in F$, and we denote by $\phi_g$ the automorphism corresponding to the action of $g\in G$, we have:
$$\begin{align*}
g\cdot\Bigl(\varepsilon_a\bigl(p(x)\bigr)\Bigr) &= g\cdot\Bigl(p(a)\Bigr)\\
&= g\cdot\Bigl( \alpha_na^n + \cdots + \alpha_0\Bigr)\\
&= \phi_g\Bigl( \alpha_na^n + \cdots + \alpha_0\Bigr)\\
&= \phi_g(\alpha_n)\phi_g(a)^n + \cdots + \phi_g(\alpha_0)\\
&= (g\cdot p)(g\cdot a)\\
&= \varepsilon_{g\cdot a}\circ(g\cdot p(x))
\end{align*}$$
where $g\cdot p$ is the image of $p(x)$ under the action of $g$, namely,
$$g\cdot p = (g\cdot \alpha_n)x^n + \cdots + (g\cdot \alpha_0).$$

Because any automorphism of a field must send $0$ to $0$, we have:

Theorem. Let $F$ be a field, and let $G$ act on $F$ by automorphisms. Then for every $p(x)\in F[x]$, every $a\in F$, and every $g\in G$, $a$ is a root of $p(x)$ if and only if $g\cdot a$ is a root of $g\cdot p(x)$.

Proof. $a$ is a root of $p(x)$ if and only if $p(x)$ lies in the kernel of the evaluation map $\varepsilon_a$, if and only if $\varepsilon_a(p(x))=0$, if and only if $g\cdot\varepsilon_a(p(x)) = g\cdot 0 = 0$, if and only if $\varepsilon_{g\cdot a}(g\cdot p(x)) = 0$, which holds if and only if $g\cdot a$ is a root of $g\cdot p(x)$. QED

In particular, if $g\cdot p(x) = p(x)$ (that is, if $g$ fixes the coefficients; note that even with a faithful action you can have $g\cdot a = a$ for some $g\in G$ and some $a\in X$), then $a$ is a root of $p(x)$ if and only if $g\cdot a$ is a root of $p(x)$. That is, the roots of $p(x)$ are a $G$-invariant subset of $F$, and therefore the action of $G$ on $F$ restricts to an action of $G$ on the roots of $p(x)$.

Conversely, that the action of $G$ on $F$ is by automorphisms, that $p(x)$ is a polynomial with coefficients on $F$ that splits over $F$, and that the set of roots of $p(x)$ is a $G$-invariant subset of $F$. Since the coefficients of $p(x)$ are symmetric polynomials on the roots of $p(x)$, and symmetric polynomials are invariant under any permutation of the arguments, then it follows that $G$ fixes every coefficient of $p(x)$.

However, even in this situation the action of $G$ on the set of roots need not be transitive. For example, consider the case of the complex numbers, $F=\mathbb{C}$, and let $G$ be the group consisting of the identity and complex conjugation (so $G$ is isomorphic to the cyclic group of order $2$). If $p(x)$ is a polynomial with real coefficients, then the set of roots of $p(x)$ is $G$-invaraiant (since a complex number is a root of a real polynomial if and only if its complex conjugate is a root as well), so that the action of $G$ induces an action on the roots of $p(x)$. If $p(x) = (x^2+1)(x^2+4)(x^2-1)$, for instance, then $G$ acts on the set of roots of $p(x)$, $X=\{1,-1,i,-i, 2i, -2i\}$: the identity fixes every element, and complex conjugation fixes $1$ and $-1$, exchanges $i$ and $-i$, and exchanges $2i$ and $-2i$; the $G$-orbits of $X$ are $\{1\}$, $\{-1\}$, $\{i,-i\}$, and $\{2i,-2i\}$. The action is not transitive.

However, the situation that usually arises in Galois Theory is even more restricted. There, we are usually considering not one field, but a pair of fields in a field extension, $F\subseteq K$. The group whose action we are considering is a group of automorphisms of $K$, but not any old automorphisms, but rather automorphisms that fix $F$ pointwise; that is, for every $g\in G$, $g$ acts on $K$ by automorphisms in such a way that $g\cdot f = f$ for every $f\in F$. In that situation, any polynomial $p(x)$ with coefficients in $F$ will be fixed by the (induced) action of $G$. In particular, the set of roots of $p(x)$ that are in $K$ is $G$-invariant, and so $G$ acts on the set of roots of $p(x)$ that are in $K$; the action is necessarily "by permutations", since every action works via permutations.

Conversely, if you have a finite $G$-invariant subset $S$ of $K$, then the polynomial
$$p(x) = \prod_{s\in S}(x-s)$$
is fixed by the action of $G$, so that the coefficients lie in the "fixed field of $G$", which is the set of all $k\in K$ such that $g\cdot k = k$ for all $g\in G$. When the extension is a Galois extension, and $G$ is the collection of all automorphisms of $K$ that fix $F$ pointwise, then the fixed field of $G$ is precisely $F$ (in fact, that is one possible definition of "the extension is a Galois extension"), so that the polynomial $p(x)$ above must have coefficients in $F$.

Always, if the action of $G$ on $K$ is "by automorphisms", then you necessarily have $g\cdot(a+b) = g\cdot a + g\cdot b$ and $g\cdot(ab) = (g\cdot a)(g\cdot b)$. However, if the action is not "by automorphisms", then you generally cannot say that.

Title: Groups acting on polynomials


问:
Linking matrices to linear equations
I'd love advice on connecting two understandings of matrices as linear transformations.
If we have a system like
2x - y = 1 and
x + y = 5 ,
then we can understand it as

a scalar x times the vector (2, 1) plus a scalar y times the vector (-1, 1) results in the vector (1, 5). You can see that visually. You're trying to find how many copies of (2, 1)T plus how many copies of (-1, 1) will help you to reach (1, 5). But how do I match that with the matrix as linear transformation interpretation, that sending (1, 0) --> (2, -1) and (0, 1) --> (1, 1) results in sending (x, y) to (1, 5)? Obviously, it works out, but they feel like two very different things. In one case, x and y are scalar multipliers of the two component vectors, and in the other case (x, y) is a vector itself that gets transformed to (1, 5). How can I visually connect these two?

答:
We can think of the equation $A\mathbf x = \mathbf b$ as a system of linear equations or we can think about it as a linear transformation.  The algebra of matrix multiplication should be clear enough as to why our equation is a system of linear equations, so I am not going to talk about that much.

As a linear transformation, you are correct $\begin{bmatrix} a&b\\c&d\end{bmatrix}$ maps the vector $(1,0)$ to $(a,b)$ and $(0,1)$ to $(c,d).$  I hope you don't mind me using row vectors to represent the vectors in this transformation.

If our matrix is non-singular (the row / column vectors are linearly independent, the determinant is zero, and there exists an inverse matrix) then it is a 1-1 and onto mapping of the plane to its self.  That means that under this transformation, there is a unique vector $\mathbf x$ such that when the matrix $A$ transforms this vector, it maps it to $\mathbf b.$

I hope this helps.

Title: Linking matrices to linear equations


问:
Is $x+2$ invertible in given quotient ring?

Given the quotient ring $R=\mathbb{Q}[x]/(x^2-7)$. Is $x+2$ invertible in $R$?

I guess it is, but how to find it?

答: Hint: Use division with remainder to find $q, r$ such that $x^2 - 7 = q(x)(x + 2) + r(x)$. What can you say about the degree of $r$?


问:
How to find a constant term in CDF
Suppose I'm given a CDF: $Cf(x)$ for $x<0$ and 1 for $x\ge 0$. I need to find $C$. What is my strategy? I don't know calculus very well, unfortunately. So far I only have the following solution:

Find the derivative

Take an integral with the limits from $-\infty$ to $0$ and make it equal $1$.

Find $C$

But I am not sure whether I'm right because it's weird to take a derivative and put it into an integral assuming that I already know the antiderivative. As you see, I'm totally confused.

答:
For cumulative distribution function $F$, we have

$$\lim_{x\rightarrow \infty}F(x)=1$$

So indeed, if the variable is absolutely continuous, then

$$\int_\mathbb R f(x)dx=1,$$

where $f$ is the derivative of $F$. Hence if you have $f_2=Cf$, then

$$\int_\mathbb R f_2(x)dx=C \Rightarrow f=\frac{f_2}{C}.$$

In this case we have
$$F(0)=1\Leftrightarrow CF(0)=C,$$

so all you need to do is evaluate $CF(0)$, which is possibly

$$CF(0)=C\int_\mathbb {R_{\leq 0}} f(x)dx$$.

Title: How to find a constant term in CDF


问:
Why is $T_{\sigma}(\sigma H)$ the kernel of $\pi_{*\sigma} : T_\sigma G \rightarrow T_{\sigma H} G/H$?
I'm trying to understand the proof of theorem $3.62$ page 123 from "Foundations of differentiable manifolds and Lie groups" by Frank F. Warner

Let $\eta : G \times M \rightarrow M$ be a transitive action of a Lie group on the manifold $M$ on the left. Let $m_0 \in M$ and $H = \{g \in G : \eta(g,m_0) = m_0 \}$ be the isotropy group at $m_0$. Define a mapping
$$ \tilde \beta : G/H \rightarrow M : gH \mapsto \eta(g,m_0).$$
Then $\tilde \beta $ is a diffeomorphism.

In the proof the author states that the kernel of the differential map of $\pi$ at $\sigma \in G $ is the tangent space to $\sigma H$ at $g$. That is

$$ \ker \pi_{*\sigma} = T_{\sigma}(\sigma H) \subset T_\sigma G.$$

where

$$ \pi_{*\sigma} : T_\sigma G \rightarrow T_{\sigma H} G/H $$

I don't understand why this is the case. I presume that it is obvious since the author doesn't give a detailed explanation.

My (updated) attempt (Using Thorgott's hint.)

From previous results I know that $H$ is closed which means $\sigma H$ is closed. From this I deduce that $\sigma H$ is a Lie subgroup of $G$ which is embedded in $G$.

I can therefore identify the tangent space $T_{\sigma}{\sigma H}$ with a subspace of $T_\sigma G$ by using the differential of the inclusion map.

Let $X_\sigma \in T_\sigma \sigma H$ and $X_{i(\sigma)} = i_{*\sigma} X_{\sigma} \in T_\sigma G.$ Let $f \in C^\infty (G/H)$. Then

$$ \pi_* X_{i(\sigma)} f = X_{\sigma} (f \circ \pi \circ i).$$

But $ f \circ \pi \circ i$ is constant on $\sigma H$ since $f \circ \pi \circ i(\sigma h) = f(\sigma h H) = f (\sigma H)$. Therefore

$$\pi_* X_{i(\sigma)} = 0$$ and $T_{\sigma} \sigma H \subset \ker \pi_{*\sigma}.$

Using ThorGott's hint again I want to show that $\ker \pi_{*\sigma}$ has the same dimension as $T_\sigma \sigma H = \dim H$ to show the equality.

Since $\dim T_{gH} G/H = \dim G - \dim H$ I can conclude from the rank nullity theorem that $\dim \ker \pi_* = \dim H$ only when $\pi_*$ is surjective.

How could I show surjectivity ?

答: Hint: The map $\pi$ is constant on $\sigma H$, which gives one inclusion. Equality then follows by counting dimensions.


问:
Impact of variable change on the solution of $x^3+y^3=z$
Let $x,y,z$ be coprime integers such that $$x^3+y^3=z$$ I am trying to to gain some insights into the impact of variable substitution on the solution set of the original equation.

Here is what I mean:

Let the solution set $(x,y,z)$ exist. There are integers a,b,c such that
$$ ax+ by+ cz=0$$ with $abc\neq 0$

If I substitute for instance $x$ in the first equation, I will obtain:
$$-(by+cz)^3+(ay)^3=az$$

My question is: Is this equation equivalent to $x^3+y^3=z$. In other words, do they have the same solution set when $x,y,z$ run over the complex numbers?

答:
COMMENT.- You have an elliptic curve of difficult solution.

Consider Selmer's equation $x^3+y^3=Az^3....(*)$ where $A$ is a cube-free integer. It has been established that if $aX^3+bY^3+cZ^3=0$ where $abc=A$ has a solution then a solution of $(*)$ is given by
$$x=u^3+3u^2v-6uv^2+v^3\\y=-u^3+6u^2v-3uv^2-v^3\\Az^3=9uv(u-v)3((u^2-uv+v^2)^3$$ which is satisfied taking
$$u=aX^3\\v=-bY^3\\u-v=-cZ^3 (\text{so }z=3XYZ(u^2-uv+v^2))$$ In your case you must have to take your $z$ in the form $Az^3$ (you know your $z$ cannot be a cube) if you want to apply this. I repeat your equation is difficult, very difficult really.


Title: Is $x+2$ invertible in given quotient ring?

Body:
Given the quotient ring $R=\mathbb{Q}[x]/(x^2-7)$. Is $x+2$ invertible in $R$?

I guess it is, but how to find it?

Body:
Hint:
$$\frac{1}{\sqrt{7}+2}=\frac{1}{\sqrt{7}+2}\cdot\frac{\sqrt{7}-2}{\sqrt{7}-2}=\frac{-2}{3}+\frac{\sqrt{7}}{3}$$


问:
Is there a norm in which the vector space of all sequences with the induced metric is complete?
The question of knowing whether is possible put a norm in a vector space is something new to me. I liked the Ivo ideas and I have wondered a little more about a particular case.

Is there a norm in which the vector space of all sequences with the induced metric is complete? This question sounds harder to answer. The completion of a metric space is something that is usually done to achieve this result. Is the completion of the norm given by Ivo ideas isomorphic to the vector space of all sequences? If not, is there a norm in which the vector space of all sequences with the induced metric is complete?

答:
By "Ivo ideas" I presume you mean, given a Hamel basis $B$ of vector space $V$, the supremum norm $\|v\| = \sup_{\beta \in B} |a_\beta|$ where $v = \sum_{\beta \in B} a_\beta \beta$.

An infinite-dimensional vector space $V$ with this norm is not complete.  For example, let $\beta_i$, $i = 1,2,3,\ldots$, be a sequence of distinct members of $\beta$, and
consider the sequence $v_n = \sum_{j=1}^n 2^{-j} \beta_{j}$.  This is a Cauchy sequence, but it does not have a limit in this norm.

On the other hand, given a vector space $V$, to construct a norm in which it is complete all you need is to find a Banach space $B$ (i.e a complete normed linear space) with the same dimension (i.e. a Hamel basis of the same cardinality as that of $V$).  Then the two spaces are isomorphic as vector spaces, and you just carry the norm of $B$ over to $V$.  Specifically, if
$\| \cdot \|_B$ is the norm of $B$, $(\alpha_i)_{i \in I}$ a Hamel basis of
$V$ and $(\beta_i)_{i \in I}$ a Hamel basis of $B$ with the same index set,
you define
$$ \|v\|_V = \|\phi(v)\|_B$$
where if $v = \sum_i c_i \alpha_i$ is the representation for $v$ in the Hamel basis, $\phi(v) = \sum_i c_i \beta_i$.


问:
Existence of a "densest" sequence with natural density $0$

Determine whether there exists an increasing, infinite sequence of natural numbers $S$ with natural density $0$ such that, for almost all $n$ and any increasing, infinite sequence $P\neq S$ also with natural density $0$, $$\sum_{i=1}^{n}(\mathbb{I}_i(S)-\mathbb{I}_i(P))>0$$ (Note: Here, $\mathbb{I}_i$ represents the indicator function, which equals $1$ if $i$ is in sequence $S$ or $P$ and $0$ otherwise)

I came up with this problem while idly daydreaming, not from any textbook or math course. I'm not very experienced with this type of problem (as a high school senior, it's kind of out of my historical purview), but I think I have a solution, written below. If I've made any mistakes in notation, rigorousness, or otherwise, feel free to let me know.

答:
Assume, for the sake of contradiction, that we have such a sequence $A$. By construction, no other sequence can fulfill the conditions in the question- otherwise it would be impossible for both the sum and its negative to be $>0$ almost all of the time. Because $A$ has natural density $0$, there must exist a point at which the gap between consecutive elements of the sequence have a difference $>1$. At that point, we insert an element with a value strictly between that of the anterior and posterior elements to create a sequence $A'$. We now want to show, for the sake of contradiction, that $A'$ fulfills all of the conditions given in the question.

The addition of one element will not change an infinite sequence with natural density $0$ to one with a natural density $>0$, so this condition on $A'$ is fulfilled. Furthermore, there are only finitely many elements before the point at which we inserted the extra value, and infinitely many after that point. Therefore almost all of the possible $n$ on which to test whether $$\sum_{i=1}^{n}(\mathbb{I}_i(A')-\mathbb{I}_i(A))>0$$ do in fact give $$\sum_{i=1}^{n}(\mathbb{I}_i(A')-\mathbb{I}_i(A))=1>0$$ and so $A'$ fulfills the conditions of the question. But the original $A$ was just such a sequence that there could be no other sequence that fulfilled those conditions! This is a contradiction, and so we have shown that no such sequence can exist. QED. $\blacksquare$


问:
Conditional expectation on partition with proof
Consider a finite probability space with $\sigma-$algebra $A$ and its generating partition $P$.
Let $X,Y \in L^2(A,P)$, then it holds: $$E[X| A]= \sum_{p \in P} 1_p E[X|p] $$.

Proof: $$E\left[Y  \sum_{p \in P} 1_p E[X|p]\right]=  \sum_{p \in P}E[Y 1_p E[X|p]] =  \sum_{p \in P}E[ Y1_p X] = E[YX]$$.

I do not know why I need that extra random variable $Y$. Would it not still work without it?

答:
Without the $Y$ you are only proving that the two sides of the equation
$$E[X| A]= \sum_{p \in P} 1_p E[X|p] $$ have the same expectation. To show that $U=V$ is is not enough to show that $EU=EV$. If we show that $EYU=EYV$ for all bounded randoem variables $Y$ (or just indicator r.v.'s $Y$) we can conclude that $U=V$ a.s..

[Proof: Take $Y=1_{U>V}$ to see that $E(U-V)1_{U>V}=0$. A  non-negative  r.v.  cannot have zero expecation unless it is $0$ a.s.,  so we must have $P(U>V)=0$. Similarly, $P(U<V)=0$, so $U=V$ a.s.].


问:
Compute the following limit of the integral
Compute

$$\lim_{n \to \infty} \int_{0}^{1} \frac{n^4u^2e^{-nu}}{1 + n^2u}du$$

I'm in a class where we are discussing Lebesgue measure, and we have not discussed the dominated convergence theorem. We have learned the BND and Monotone convergence theorems though. I'm trying to do the integral in general with elementary techniques, but I'm not sure that one can. I'm also not exactly sure which theorem I can apply to move the limit inside, nor what the limit of the middle term would be. Can anyone lend a hand and get me started?

答:
Hint: Let $u=v/n.$ The integral becomes

$$\int_0^n \frac{nv}{1+nv}ve^{-v}\,dv= \int_0^\infty \left(\frac{nv}{1+nv}\chi_{(0,n)}(v)\right )ve^{-v}\,dv. $$


问:
Showing $\text{range}(A)$ dense: $A: \ell^1 \to \ell^1$, $Ax = \left(\frac{x_n}{n}\right)_{n \in \mathbb{N}}.$
Let $A : \ell^1 \to \ell^1$ be an operator where for any $x \in \ell^1$, $Ax = \left(\frac{x_n}{n}\right)_{n \in \mathbb{N}}$.

Question. I'm trying to show that $\text{range}(A)$ is dense subspace of $\ell^1$.

I know that essentially we want to show that the closure of the range is equivalent to $\ell^1$, i.e.

$$\overline{\text{range}}(A) = \{x \in \ell_1 : \exists x_n \in \text{range}(A)~\text{such that}~x_n \to x\} = \ell^1.$$

But I don't really have an idea of how to formally go about showing the above using this definition?

答: Hint: consider sequences with only finitely many nonzero terms.</p>


问:
min max optimization over compact sets
Suppose I have an infinite set $\mathcal{A}$. All I know about $\mathcal{A}$ is that elements of $\mathcal{A}$ have bounded $l_2$ norms. Suppose I have a constant $b$.
Can I solve the following problem. (Even approximate solutions are okay)

$$\min_{x \in \mathrm{R}^d} \sup_{a \in \mathcal{A}}\left|a^Tx-b\right|^2$$

EDIT: The objective of this problem is not to find an exact solution but tell something about $||x^*||_2$. Providing an upper bound for it or similar.

EDIT: So this is a follow up. Let's assume $\mathcal{A}$ is compact. Then we can say the inner supremum is always attained at some $a^*$, so now we are left with $$\min_x (a^{*^T}x - b)^2$$,
which can be solved as $x^* = (a^*a^{*^T})^\dagger ab$ where we used the pseudo inverse.
Can we now say anything about $||x^*||_2$

答:
First of all, since you seem to be interested in where the minimum occurs instead of what the minimum is, squaring makes no difference at all. Since squaring is a strictly increasing function on positive real numbers,
$$\min_{x \in \Bbb R^d} \sup_{a \in \mathcal A}\left|a^Tx-b\right| = \sqrt{\min_{x \in \Bbb R^d} \sup_{a \in \mathcal A}\left|a^Tx-b\right|^2}$$
and occurs at the same $a, x$.

But $\left|a^Tx-b\right|= \pm a^Tx \pm b$ for the appropriate sign choices. If $a^Tx > b$, then increasing $a^Tx$ will increase $\left|a^Tx-b\right|$. If $a^Tx < b$, then decreasing $a^Tx$ will increase $\left|a^Tx-b\right|$. Thus the suprema of $\left|a^Tx-b\right|$ occur at extrema of $a^Tx$. But extrema of $a^Tx$ are also extrema of $a^T(mx)$ for $0 \ne m \in \Bbb R$. Letting $$a^*(x) = \arg\max_{a \in \overline{\mathcal A}}\left|a^Tx-b\right|,$$ we see that $a^*(mx) = a^*(x)$. However, for any $x$, we can choose $m = \frac b{[a^*(x)]^Tx}$ unless $[a^*(x)]^Tx = 0$. And for that choice of $m$, $$[a^*(mx)]^Tmx - b = 0$$

If $0\ne x \in \mathcal A$, then you can choose a constant $c$ such that $|x^T(cx) - b| > b$, so $|a^*(x)^Tx\ne 0$ for such $x$. This means that there are guaranteed to be $x$ with $a^*(x)^Tx - b = 0$.

If $M = \sup \{\|a\| : a \in \mathcal A\}$, then for any $x \notin \mathcal A^\perp, x_1 = \frac b{[a^*(x)]^Tx}x$ gives the minimum possible value of $0$ for $\sup_{a \in \mathcal A}\left|a^Tx_1-b\right|$. Thus $x^*$ can be $x_1$. But note that if we limit $\|x\| = 1$ while moving it towards $\mathcal A^\perp, [a^*(x)]^Tx$ approaches $0$. Thus $\|x_1\|$ can be arbitrarily large.

Thus if $\operatorname{span}\mathcal A \ne \Bbb R^d$, then $\mathcal A^\perp \ne \{0\}$, and there is no bound on the size $\|x^*\|$. I don't think it is bounded even when $\operatorname{span}\mathcal A = \Bbb R^d$, but I haven't worked that out yet.

Title: min max optimization over compact sets


答:
You show it by demonstrating there is $x_n \in \ell^1$ (and more precisely $Ax_n$ is Cauchy), such that $Ax_n\rightarrow y,$ for any $y\in \ell^1$.

So let $y\in \ell^1$, then

$$\sum |y_i|<\infty$$

We can now have a sequence $x_n:=y_i i1(i<n)\in\ell^1$. Now $Ax_n$ converges to $y$, and the statement is proved.

Title: Showing $\text{range}(A)$ dense: $A: \ell^1 \to \ell^1$, $Ax = \left(\frac{x_n}{n}\right)_{n \in \mathbb{N}}.$

Body:
Let $A : \ell^1 \to \ell^1$ be an operator where for any $x \in \ell^1$, $Ax = \left(\frac{x_n}{n}\right)_{n \in \mathbb{N}}$.

Question. I'm trying to show that $\text{range}(A)$ is dense subspace of $\ell^1$.

I know that essentially we want to show that the closure of the range is equivalent to $\ell^1$, i.e.

$$\overline{\text{range}}(A) = \{x \in \ell_1 : \exists x_n \in \text{range}(A)~\text{such that}~x_n \to x\} = \ell^1.$$

But I don't really have an idea of how to formally go about showing the above using this definition?


问:
Computing the Riemann curvature tensor using Cartan method of moving frames
I'm trying to compute the curvature tensor from the following metric:

$$g = -(\frac{r}{M})^2 dt^2 + (\frac{M}{r})^2 dr^2$$

To compute the curvature I'm employing the first Cartan structure equation which reads:

$$d\theta^a + \omega^a_b \wedge \theta^a = 0$$

where I assumed that the manifold I'm working with is torsion free and $\theta^a$ define an orthonormal local frame given by:
$$\theta^0 = \frac{r}{M}dt, \theta^1 = \frac{M}{r}dr$$

Then from the first Cartan equation I obtain the two equations:

$$d\theta^0 + w^0_1 \wedge \theta^1 = 0$$
$$d\theta^1 + w^1_0 \wedge \theta^0 = 0$$

The first one gives:
$$\frac{1}{M} dr \wedge dt + w^0_1 \wedge \frac{r}{M}dr = 0$$
$$0 + w^1_0 \wedge \frac{r}{M}dt = 0$$

From the first equation I can see that $w^0_1$ should be given by $\frac{r}{M^2}dt$ but the second equation seems give $w^1_0=0$ which also would imply that $w^0_1$ by antisymmetry. Is there a fault in my reasoning?

答:
First of all, because you're working with a metric of signature $(1,1)$, you will have $\omega^0_1 = \omega^1_0$ rather than skew-symmetry. Second, your first line tells you that
$$\omega^0_1 = \frac{dt}r + A(r,t)dr$$
and your second line tells you that $\omega_0^1=\omega_1^0$ is a multiple of $dt$ (not necessarily $0$). Thus, we conclude that $A(r,t)=0$ and there's no problem.


答:
While routinely studying the distribution of the random variable $X+Y$ leads us to an answer (as mentioned in an answer above), for this problem you don't need to go through all those evaluation of integrals.

There is another more visually appealing solution to this problem as well as several other problems of similar kind. Assuming that you have not come across such an technique till date I am going to introduce to you this new method via this solution, leaving the remaining (non-probabilistic) computations for you.

Notice that $R = \{ ( x,y ) : 0 < y < x < 1 \}$ is basically the region strictly bounded by the line $x = y,$ the $X$ axis and the points $x=0$ and $x=1.$ Notice that $R$ is precisely the triangle whose vertices are $(0,0), (1,0) , (1,1).$

Here by ''strictly bounded'' I mean that none of the above mentioned boundaries of $R$ can be a part of $R;$ however this wouldn't matter since the boundaries anyway have probability measure zero.

Now, notice that for two real numbers $a,b$ the condition $a+b \ge 1$ is same as saying $a < 1-b$ and this in turn is same as saying that the distance between the point $(a,b)$ and the origin is at least as big as the distance between the point $(a,b)$ and the line $x = 1-y.$

Putting all this together we see that $(X,Y)$ is uniformly distributed in the region strictly bounded by the line $x=y,$ the $X$ axis and the points $x= 0,$ and $x=1,$ and the  probability $\mathbf{P} \{ X+Y \ge 1 \}$ is precisely the (conditional) probability that the point $(X,Y)$ lies in a location inside $R$ which is more closer to the line $x = 1-y$ than to the origin, thus $(X,Y)$ lies in the triangle formed by vertices $(1,0), (1/2 , 1/2) , (1,1).$

Now just compare the areas of the triangle $R,$ and the triangle formed by the vertices $(1,0) , (1/2 , 1/2) , (1,1),$ and their ratio would give us $\mathbf{P} \{ X+Y \ge 1 \}$, and thus subtracting it from $1$ (the total probability) would give us the value for $\mathbf{P} \{ X+Y < 1 \};$ thus for instance the same thing won't be true if $(X,Y)$ has the gaussian distribution or something else.

Footnote:the fact that the probability can be derived as the ratio of triangles relies heavily on the given condition that $(X,Y)$ has a uniform distribution; for a much better understanding of the importance of uniform distribution in this solution observe that from the very density (or definition) of uniform distribution it follows that if $\mathcal{X}$ is a bounded connected region in the plane with the boundary curve of $\mathcal{X}$ being smooth (although this is not strictly necessary) and having a finite non-zero area, and if $\mathcal{X}_1, \mathcal{X}_2$ are two connected disjoint (that is, $\mathcal{X}_1 \cap \mathcal{X}_2 = \phi$) subsets of $\mathcal{X}$ with $\mathcal{X} = \mathcal{X}_1 \cup \mathcal{X}_2$ and if the boundary curves of $\mathcal{X}_1$ and $\mathcal{X}_2$ are smooth, then for a random variable $Z$ distributed uniformly over $\mathcal{X}$ one has $\mathbf{P} \{ Z \in \mathcal{X}_i \} = \dfrac{ \text{ area of  } \mathcal{X}_i }{ \text{ area of } \mathcal{X}}.$

Title: Find $\mathbb{P}(X+Y<1)$ given the joint density function of the vector $(X,Y)$ distributed uniformly over a region $R$.

Body:
Let $X$ and $Y$ be random variables with the vector $(X, Y)$ uniformly distributed on the region
$R = \{(x, y) : 0 < y < x < 1\}$. Find $\mathbb{P}(X+Y<1)$. I am pretty sure that I calculated the pdf correctly:

$f_{X,Y}(x,y)= 2$ if $(x,y) \in R$, $0$ elsewhere.

To get the probability, I know I need to double integrate $f(x,y)$. I am not sure of the bounds though. Is the correct integration $\int_0^1 \int_0^{1-x} 2\,dy\,dx$, or should the upper bound be $x$ so that it is $\int_0^1 \int_0^x 2 \, dy \, dx$. Or are neither of these correct?


问:
show $\frac{1}{n}\sum (X_i - \mu_i) \xrightarrow{L^2} 0$, $X_i$ independent with mean $\mu_i$
Let $\{X_i\}_{i=1}^n$ be a sequence of uncorrelated random variables,
$$
\text{Cov}(X_i,X_j)=0,
$$
with
$$
E(X_i)=\mu_i<\infty,
$$
and
$$
\frac{V(X_i)}{i} \xrightarrow{i\to \infty} 0. \tag{1}
$$
Show
$$
\frac{ \sum_{i=1}^n \left(X_i -\mu_i\right)}{n} \xrightarrow{L^2} 0.
$$

Using the martingale
$$
M_n = \sum_{i=1}^n \frac{X_i-\mu_i}{i},
$$
which is uniformly integral if
$$
\sum_{i=1}^{\infty} \frac{V(X_i)}{i} < \infty. \tag{2}
$$
Then, by the martingale convergence theorem, $M_n$ converges a.s. to some finite random variable $M$.

By Kronecker lemma
$$
 \sum_{i=1}^n \frac{i}{n} \frac{\left(X_i -\mu_i\right)}{i} \xrightarrow{a.s.} 0
$$

Problem is, condition (2)  is very different than (1), and I get almost sure convergence instead of $L^2$ convergence.

答: Squaring $\sum_{i=1}^n (X_i-\mu_i)$ and integrating, we get $\sum_{i=1}^n V(X_i)$ because covariances vanish. So, the goal is to show $\sum_{i=1}^n V(X_i)=o(n^2)$, given that $V(X_i)=o(i)$. This is pretty standard: given $\epsilon>0$, find $k$ such that $V(X_i)<\epsilon \, i$ for $i>k$. Then $\sum_{i=1}^k V(X_i)=O(1)$ and $\sum_{i=k+1}^n V(X_i)<\epsilon \sum_{i=k+1}^n i< \epsilon\, n^2$</p>


问:
Powers of logarithm question
I have got a question about formal simple mathematics are these expressions equivalent?

$$ \log^{n} (x/a) $$

and  $$ \log^{n} (x) -\log^{n} (a) $$

I know is a dumb question and they are equal only when $ x= a$ but need some help

答: Nope. They are not equal. Raised to the power of n in both expressions make the quotient rule not working.</p>

Title: Powers of logarithm question


问:
Finding the probability of getting underweight products

When customers buy a product at a supermarket, there is a concern about the item being underweight. Suppose there are 20 "16-oz" ribeye steak on display and 3 of them are underweight. A customer buys 5 of 20 steaks at random. What is the probability that none of them is underweight?

My attempt was simply $\frac{17C5}{20C5}$ why is this wrong?

答: Possibly it is right. 20C5 expresses the number of possible size-5 samples out of 20 steaks. Then 17C5 gives the number of possible size-5 samples from the good ones.</p>


问:
Security value of log contract and stochastic calculus
I am studying the geometric Brownian motion of stock S with an expected return and volatility. And we can write the standard GBM as the form $dS(t) = \mu S(t)dt + \sigma S(t)dZ(t)$ where Z follows a standard normal distribution. And we can use it to calculate some derivatives based on stock S. And there is a special derivative called log contract $L(S,t)$ that pays off $\ln S(T)$, where S(T) is the value of the stock price at time T.
Based on previous GBM and Ito process, we can get
$$d\ln S(t) = (\mu - \frac{\sigma^2}{2})dt + \sigma dZ(t)$$
Could someone explain how to calculate the value of the security $L(S,t)$ in terms of stock price $S(t)$ at time t? And will this solution also satisfy the Black-Scholes PDE? I think the previous one satisfies.

答:
The value $L(S(t),t)$ at time $t < T$ is the discounted expectation conditioned on information known at $t$ of the contract payoff at time $T$ under the risk-neutral probability measure $Q$, that is

$$L(S(t),t)) = e^{-r(T-t)}\mathbb{E}^Q[\ln S(T) \, |\, \mathcal{F}_t],$$

where $r$ is the risk-free rate.

In the risk-neutral world, the security follows the process

$$d\ln S(t) = (r - \frac{\sigma^2}{2})\,dt + \sigma \,dZ(t)$$

Integrating over $[t,T]$, we get, in terms of a standard normal random variate $\xi \sim \mathcal{N}(0,1)$,

$$\ln S(T) = \ln S(t) + (r - \frac{\sigma^2}{2})(T-t) + \sigma\sqrt{T-t}\,\xi,$$

and, taking the expected value,

$$\tag{*}L(S(t),t))  = e^{-r(T-t)} \left[\ln S(t) + (r - \frac{\sigma^2}{2})(T-t) \right]$$

This could also be obtained by solving the Black-Scholes PDE for $L(S,t)$,

$$\frac{\partial L}{\partial t} + r S \frac{\partial L}{\partial S} + \frac{1}{2} \sigma^2 S^2 \frac{\partial L^2}{\partial S^2} - r L = 0,$$

with terminal condition $L(S,T) = \ln S$. Assuming a solution of the form $L(S,t) = \alpha(t) + \beta(t) \ln S$ and substituting into the PDE, leads to the same solution as in (*).


问:
Base Case Strong Induction Terrence Tao question
Terrence Tao formulates strong induction as follows: "(Strong principle of induction). Let $m_o$ be a natural number, and let $P(m)$ be a property pertaining to an arbitrary natural number . Suppose that for each $m \geq m_0$, we have the following implication: if $P(m')$ is true for all natural numbers $m_0 \leq m' \lt m$, then $P(m)$ is also true. (In particular, this means that $P(m_0)$ is true since in this case, the hypothesis is vacuous.) Then we can conclude that $P(m)$ is true for all natural numbers $m \geq m_0$. "

Since we are assuming that $P(m')$ is true for all $m_0 \leq m' \lt m$. Where in this statement does it suggest to prove the base case first? To me all its suggesting is to simply suppose $P(m')$ is true for all $m_0 \leq m' \lt m$ without proving a base case?

答: Well, it seems to be a bit weird. But he is saying that the base case comes in when you prove the inductive condition when $m=m_0$. Which is a bit strange because in this case the hypothesis is vacuous, so essentially you have to prove the base case. So the author covers this point in the part " (In particular, this means that $P(m_0)$ is true since in this case, the hypothesis is vacuous)".


问:
An operator is positive definite if, and only if, it is non-negative and invertable
The question is the following:

Show that an operator $T$ is positive-definite if, and only if, $T \geq 0$ and $T$ is invertable.

My initial idea was to use the fact that an operator is positive-definite if, and only if, all its eigenvalues are positive, but I'm not sure how I could relate it to being invertable.

Also, does the $T \geq 0$ part come from the definition of a positive-definite operator that $\langle Tx, x \rangle > 0$, for $x \neq 0$?

答: $T \ge 0$ (positive semi-definite) implies the eigenvalues are nonnegative. Invertibility of $T$ means the eigenvalues are nonzero. Together, they imply the eigenvalues of $T$ are positive.


问:
Chain Rule of Logarithm?
The students are taught the well known  change base rule of logarithm:

\begin{align}\log_a b = \frac{\log_c b}{\log_c a}\end{align} Most text books proves it by invoking  $(a^x)^y=a^{xy}$ to show:
\begin{align}\log_c a\times\log_a b=\log_c b\end{align}

Question:

Why don't we call it  Chain Rule of Logarithm (at least as a second name)?
Is it because the way we use it is almost alway in the change of base format or something else?
\begin{align}
\log_a b\times\log_b c\times\log_c d\times\log_d e\times\cdots\times\log_y z=\log_a z\end{align}

It is easy to remember and the students  can have fun to continue chaining  it (similar to the change rule of derivative).

答:
Well, finally I found another user or person who concurs with my idea. I discovered this relation personally.

\begin{align}
\log_a b\times\log_b c\times\log_c d\times\log_d e\times\cdots\times\log_y z=\log_a z\end{align}

but this is just a special case of

\begin{align}
\log_{b_1}{a_1} \times \log_{b_2}{a_2} \times \log_{b_3}{a_3} \times \log_{b_4}{a_4} \times \cdots \times \log_{b_n}{a_n} = \log_{b_1}{a_{\pi_1}} \times \log_{b_2}{a_{\pi_2}} \times \log_{b_3}{a_{\pi_3}} \times \log_{b_4}{a_4} \times \cdots \times \log_{b_n}{a_{\pi_n}} \end{align}

where ${\pi}$ is any permutation of the subscripts 1, ..., n.

Title: Chain Rule of Logarithm?


问:
Introducing terms in the conclusion (natural deduction, propositional logic)
I am just wanting to wrap my head around the following question;

$$A\to B \vDash ((A \land B) \to C) \to (A\to C)$$

So far all I am getting from the conclusion is that $A\to C$, if I introduce the conclusion as an assumption, but I am not sure how to tie it all together to go from the premise to the conclusion.

to clarify - I understand it is a tautology and I can translate it into English, I am struggling to show this via a fitch-style proof.

答:
$$A\to B \vDash ((A \land B) \to C) \to (A\to C)$$

So far all I am getting from the conclusion is that $A\to C$, if I introduce the conclusion as an assumption,

Never introduce the consequent as an assumption.  It is what you seek to derive.  Assuming it will just have you running in circles.

but I am not sure how to tie it all together to go from the premise to the conclusion.

What you assume is the antecedent, $(A\land B)\to C$, with the aim to derive the consequent $A\to C$; which in turn would be deduced by assuming its antecedent, $A$, and deriving its consequent, $C$.

Thus you need to nest Conditional Proofs: open with Assumptions and close with the relevant Conditional Introductions. $$\def\fitch#1#2{~~~~\begin{array}{|l} #1\\\hline #2\end{array}}\fitch{A\to B}{\fitch{(A\land B)\to C}{\fitch{A}{~\vdots\\C}\\A\to C}\\((A\land B)\to C)\to(A\to C)}$$

Well it should be clear how to derive $C$ from $A$, $A\to B$, and $(A\land B)\to C$ ...


问:
conditional expectation of gaussian random variable
Let $(X_1, \ldots , X_n)$ be a gaussian random vector with zero mean and covariance matrix $B$.  Find the distribution of the random vector $E(X_1 \mid X_2, \ldots, X_n)$.  I am not sure how to proceed.  Thanks in advance.

答:
You can find formulas for the conditional distributions from a Gaussian vector.

In particular, if $B=\begin{bmatrix} B_{11} & B_{12} \\ B_{12}^\top & B_{22} \end{bmatrix}$ where $B_{11}$ is $1 \times 1$ and $B_{22}$ is $(n-1) \times (n-1)$, then the conditional expectation is
$$E[X_1 \mid X_2 ,\ldots, X_n] = B_{12} B_{22}^{-1} \begin{bmatrix}X_2 \\ X_2 \\ \vdots \\ X_n \end{bmatrix}.$$

This is a linear transformation of the Gaussian vector $(X_2, \ldots, X_n)$ to a univariate random variable, so the distribution is univariate Gaussian. I'll leave it to you to compute the mean and variance of this univariate Gaussian.


问:
Sum of every element's product in a matrix whose entries are either $\pm1$

There is a matrix $A$ which has  $19$ rows and $19$ columns. Every element has  a value either $1$ or $-1$. Defined that $R_k$ is the product of every elements in the $k^{th}$ row and $C_k$ is the product of every element in the $k^{th}$ column. Prove that $$\sum_{i=1}^{19}(R_i+C_i)\neq0$$

help me please, this is quite hard

答: Hint: Let's first consider a matrix $A$ consisting only of $1$s. Then $C_i = R_i = 1$ for each $i$, so your sum equals 38. Whenever you "flip" a 1 to replace it with a $-1$, exactly one $C_i$ and one $R_i$ flip, that is the sum is changed by $4$, $0$ or $-4$. But $38$ is no multiple of $4$.


问:
Union of countable sequence of countable sets
I realize this question has been asked and answered multiple times, but I'm still not understanding the reasoning. For reference, I've read this, this, and this.

Theorem 2.12. Let {}, =1,2,…, be a sequence of countable sets, and put

$$S = \bigcup\limits_{n=1}^{\infty}E_n$$

Then  is countable.

Proof. Let every set  be arranged in a sequence {},
=1,2,3,…, and consider the infinite array

in which the elements of  form the th row. The array contains
all elements of . As indicated by the arrows, these elements can be arranged in a
sequence

11;21,12;31,22,13;41,32,23,14;…(*)

If any two of the sets  have elements in common, these will appear
more than once in (∗). Hence there is a subset  of the set of all
positive integers such that ∼, which shows that  is at most
countable. Since 1⊂, and 1 is infinite,  is infinite, and thus
countable. ◼

I understand if there's an injection $F: S\rightarrow \mathbb{N}$, S is countable. I don't understand how utilizing the diagonals leads to such an injection. Since there's 1 element in the first diagonal, 2 in the second, ..., n in the n-th, (n - 1) in the (n + 1)-th, etc., how is it possible to say that since there are n elements at most in any diagonal the set is countable?

答:
All you're doing is showing that you can break $\mathbb{N}$ into sequential sets of size 1, 2, 3, 4 and so on as:

$\left\{\{0\}, \{1, 2\}, \{3, 4, 5\}, \{6, 7, 8, 9\}, \ldots \right\}$

and we can match those up with equally-sized sets of ordered pairs from $\mathbb{N}^2$ as:

$\left\{\{(0, 0)\}, \{(0, 1), (1, 0)\}, \{(0, 2), (1, 1), (2, 0)\}, \{(0, 3), (1, 2), (2, 1), (3, 0)\} \ldots \right\}$

If you want to make it more formal, you can outright define the bijection between each natural number and one of the pairs by noticing that (1) in each of those sets of pairs the sum of the pair is constant and equal to the position of the set in the list (if you start numbering from 0), (2) the first value in each pair denotes the index within the set (again counting from 0), (3) in the sets of single numbers the first number in the set is a triangular number.

Let's denote $t_k = \frac{k(k+1)}{2}$ as the $k$-th triangular number. Then in one direction, we have the function:

$F: \mathbb{N}^2 \rightarrow \mathbb{N}$, $F(i, j) = t_{i+j} + i$

You can also define $F^{-1}: \mathbb{N} \rightarrow \mathbb{N}^2$ explicitly, but it's a little messier since you have to use a floor function on a quadratic formula. Instead, I'm just going to outright claim that $F$ is a bijection since:

$t_{i+j}$ is always the largest triangular number less than or equal to $F(i, j)$ because $t_{i+1} \leq F(i, j) < t_{i+j+1}$

Therefore, if $F(i, j) = F(i', j')$ then $t_{i + j} = t_{i' + j'} \implies i + j = i' + j'$ and $F(i, j) - t_{i + j} = i = i' \implies i' = j'$ so $F$ is injective

There are $k$ natural numbers in the interval $[t_k, t_{k+1} - 1]$ and $k$ ways to find natural numbers $i, j$ such that $i + j = k$, so $F$ must in some sense "fill" all of the gaps between consecutive triangular numbers.


问:
Henon maps and strange attractors
What I understand of chaos is that, it cannot occur in 2D space. A strange attractor, as seen in a Lorenze system, is a feature of chaos. Now, I am reading about the Henon map which is a 2d map with a strange attractor. My question, then, is: Does a Henon map represent a chaotic system in 2D? Or is can a strange attractor occur in 2D space without the system itself being chaotic? Said another way, does chaos imply a strange attractor but a strange attractor does not imply chaos?</p>

答:
The actual result is about continuous systems:

A two-dimensional continuous dynamical system cannot give rise to a strange attractor

For discrete systems, such as the Henon map, you can have chaos even in one dimension, e.g. the logistic map is a textbook example of chaotic behavior

$$
x_{n+1} = rx_n(1 - x_n)
$$

Title: Henon maps and strange attractors


问:
Question on limiting variance in example 10.1.8 (Casella, Statistical Inference, 2nd edition)
This is from George Casella textbook statistical inference (page 470) Example 10.1.8. Limiting variances. It says for the mean $\bar{X_n}$ of n iid normal observations with $EX=\mu$ and $VarX=\sigma^2$, if we take $T_n=\bar{X_n}$, then $lim\sqrt{n}Var\bar{X_n}=\sigma^2$ is the limiting variance of $T_n$.

I feel very confused. I think $\bar{X_n}$ is $n(\mu, \sigma^2/n)$. Then $lim\sqrt{n}*Var\bar{X_n}=lim\sqrt{n}*\sigma^2/n=\sigma^2lim\frac{\sqrt{n}}{n}=0$. The limit is 0, not $\sigma^2$.

Please point my mistake.

答:
This is a typo in the book. I have the same edition as you and upon opening my copy I see that I have already made this correction. We already know that $\mathsf{Var}\bar X_n=\sigma^2/n$. Using Casella's definition of limiting variance in definition 10.1.7 we can infer the sentence in Example 10.1.8 should read:

...if we take $T_n=\bar X_n$, then $\lim n\mathsf{Var}\bar X_n=\sigma^2$ is the limiting variance of $T_n$.

Indeed, using the same notation in the definition we have $k_n=n$, $T_n=\bar X_n$, and $\tau^2=\sigma^2$.


问:
Let $f: ~\mathbb R\rightarrow \mathbb R$ be a continuous function such that $\int_{-1}^{x}f(t)dt=0$ for all $x \in [-1,1]$
I was thinking about the problem that says:

Let $f: ~\mathbb R\rightarrow \mathbb R$ be a continuous function such that $\int_{-1}^{x}f(t)dt=0$ for all $x \in [-1,1]$. Then which of the following option(s) is/are correct?

(A) $f$ is identically $0$,

(B) $f$ is a non-zero odd function,

(C) $f$ is a non-zero even function,

(D) $f$ is a non-zero periodic function.

Please help. Thank you in advance for your time.

答: None of the $B, C, D$ is true since $A$ is a possibility. To show $A$ is true take $F(x)=0.$ Then $\forall~x\in[-1,1], F'(x)=0\implies f(x)=0$</p>